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EXTENSION OF FRAMES AND BASES - II
K. MAHESH KRISHNA AND P. SAM JOHNSON

Department of Mathematical and Computational Sciences

National Institute of Technology Karnataka (NITK), Surathkal

Mangaluru 575 025, India

Emails: [email protected], [email protected],

Date:

Abstract: Operator-valued frame ( $G$ -frame), as a generalization of frame is introduced by Kaftal, Larson, and Zhang in Trans. Amer. Math. Soc., 361(12):6349-6385, 2009 and by Sun in J. Math. Anal. Appl., 322(1):437-452, 2006. It has been further extended in the paper arXiv:1810.01629 [math.OA] 3 October 2018, so as to have a rich theory on operator-valued frames for Hilbert spaces as well as for Banach spaces. The continuous version has been studied in this paper when the indexing set is a measure space. We study duality, similarity, orthogonality and stability of this extension. Several characterizations are given including a notable characterization when the measure space is a locally compact group. Variation formula, dimension formula and trace formula are derived when the Hilbert space is finite dimensional.

Keywords: Frames, weak integrals, continuous operator-valued frames, unitary representations, locally compact groups, perturbation.

Mathematics Subject Classification (2010): Primary 42C15, 47A13, 47B65, 46G10; Secondary 46E40, 28B05, 22D10.

1. Introduction

Let $\mathcal{H}$ , $\mathcal{H}_{0}$ be Hilbert spaces, $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ be the Banach space of all bounded linear operators from $\mathcal{H}$ to $\mathcal{H}_{0}$ and $\mathcal{B}(\mathcal{H})\coloneqq\mathcal{B}(\mathcal{H},\mathcal{H})$ . Letter $\mathbb{J}$ denotes an indexing set and $\mathbb{K}$ denotes the field of scalars ( $\mathbb{R}$ or $\mathbb{C}$ ).

Definition 1.1.

[14, 10] A collection $\{x_{j}\}_{j\in\mathbb{J}}$ in a Hilbert space $\mathcal{H}$ is said to be a (discrete)

(i)

frame for $\mathcal{H}$ if there exist $a,b>0$ such that

$a\|h\|^{2}\leq\sum_{j\in\mathbb{J}}|\langle h,x_{j}\rangle|^{2}\leq b\|h\|^{2},\quad\forall h\in\mathcal{H}.$
(ii)

Bessel sequence for $\mathcal{H}$ if there exists $b>0$ such that

$\sum_{j\in\mathbb{J}}|\langle h,x_{j}\rangle|^{2}\leq b\|h\|^{2},\quad\forall h\in\mathcal{H}.$

We refer [10, 51, 27, 8, 23, 24, 6, 44, 16, 12, 5, 19] for more details on frames (and a well studied class of frames) and Bessel sequences in Hilbert spaces. Most general version of Definition 1.1 is

Definition 1.2.

[33, 47] A collection $\{A_{j}\}_{j\in\mathbb{J}}$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ is said to be an operator-valued

(i)

frame in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ if the series $\sum_{j\in\mathbb{J}}A_{j}^{*}A_{j}$ converges in the strong-operator topology on $\mathcal{B}(\mathcal{H})$ to a bounded positive invertible operator.
(ii)

Bessel sequence in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ if the series $\sum_{j\in\mathbb{J}}A_{j}^{*}A_{j}$ converges in the strong-operator topology on $\mathcal{B}(\mathcal{H})$ to a bounded positive operator.

We refer [33, 47, 26, 40, 48] for more details on operator-valued frames and Bessel sequences in Hilbert spaces. In [36] we defined the following two definitions.

Definition 1.3.

[36] Let $\{\tau_{j}\}_{j\in\mathbb{J}}$ be a set of vectors in a Hilbert space $\mathcal{H}$ . A set of vectors $\{x_{j}\}_{j\in\mathbb{J}}$ in $\mathcal{H}$ is said to be a

(i)
frame with respect to (w.r.t.) $\{\tau_{j}\}_{j\in\mathbb{J}}$ if there are $c,d>0$ such that
1. (a)
  
  the map $\mathcal{H}\ni h\mapsto\sum_{j\in\mathbb{J}}\langle h,x_{j}\rangle\tau_{j}\in\mathcal{H}$ is a well-defined bounded positive invertible operator.
2. (b)
  
  $\sum_{j\in\mathbb{J}}|\langle h,x_{j}\rangle|^{2}\leq c\|h\|^{2},\forall h\in\mathcal{H};\sum_{j\in\mathbb{J}}|\langle h,\tau_{j}\rangle|^{2}\leq d\|h\|^{2},\forall h\in\mathcal{H}.$
(ii)
Bessel sequence w.r.t. $\{\tau_{j}\}_{j\in\mathbb{J}}$ if there are $c,d>0$ such that
1. (a)
  
  the map $\mathcal{H}\ni h\mapsto\sum_{j\in\mathbb{J}}\langle h,x_{j}\rangle\tau_{j}\in\mathcal{H}$ is a well-defined bounded positive operator.
2. (b)
  
  $\sum_{j\in\mathbb{J}}|\langle h,x_{j}\rangle|^{2}\leq c\|h\|^{2},\forall h\in\mathcal{H};\sum_{j\in\mathbb{J}}|\langle h,\tau_{j}\rangle|^{2}\leq d\|h\|^{2},\forall h\in\mathcal{H}.$

Definition 1.4.

[36] Define $L_{j}:\mathcal{H}_{0}\ni h\mapsto e_{j}\otimes h\in\ell^{2}(\mathbb{J})\otimes\mathcal{H}_{0}$ , where $\{e_{j}\}_{j\in\mathbb{J}}$ is the standard orthonormal basis for $\ell^{2}(\mathbb{J})$ , for each $j\in\mathbb{J}$ . A collection $\{A_{j}\}_{j\in\mathbb{J}}$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ is said to be an operator-valued

(i)
frame in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ with respect to a collection $\{\Psi_{j}\}_{j\in\mathbb{J}}$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ if
1. (a)
  
  the series $\sum_{j\in\mathbb{J}}\Psi_{j}^{*}A_{j}$ converges in the strong-operator topology on $\mathcal{B}(\mathcal{H})$ to a bounded positive invertible operator,
2. (b)
  
  both $\sum_{j\in\mathbb{J}}L_{j}A_{j}$ , $\sum_{j\in\mathbb{J}}L_{j}\Psi_{j}$ converge in the strong-operator topology on $\mathcal{B}(\mathcal{H},\ell^{2}(\mathbb{J})\otimes\mathcal{H}_{0})$ to bounded operators.
(ii)
Bessel sequence in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ with respect to a collection $\{\Psi_{j}\}_{j\in\mathbb{J}}$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ if
1. (a)
  
  the series $\sum_{j\in\mathbb{J}}\Psi_{j}^{*}A_{j}$ converges in the strong-operator topology on $\mathcal{B}(\mathcal{H})$ to a bounded positive operator,
2. (b)
  
  both $\sum_{j\in\mathbb{J}}L_{j}A_{j}$ , $\sum_{j\in\mathbb{J}}L_{j}\Psi_{j}$ converge in the strong-operator topology on $\mathcal{B}(\mathcal{H},\ell^{2}(\mathbb{J})\otimes\mathcal{H}_{0})$ to bounded operators.

All of our vector-valued integrals are in the weak-sense (i.e., they are Gelfand-Pettis integral and we refer [50, 42, 29, 41, 45, 38, 15, 39] for more details). $\Omega$ denotes a measure space with positive measure $\mu$ .

Continuous frame, as a generalization of frames was introduced independently by Ali, Antoine, Gazeau [2] and Kaiser [34].

Definition 1.5.

[2, 34] A set of vectors $\{x_{\alpha}\}_{\alpha\in\Omega}$ in $\mathcal{H}$ is said to be a continuous frame for $\mathcal{H}$ if

(i)

for each $h\in\mathcal{H}$ , the map $\Omega\ni\alpha\mapsto\langle h,x_{\alpha}\rangle\in\mathbb{K}$ is measurable,

(ii)

there exist $a,b>0$ such that

\displaystyle a\|h\|^{2}\leq\int_{\Omega}|\langle h,x_{\alpha}\rangle|^{2}\,d\mu(\alpha)\leq b\|h\|^{2},\quad\forall h\in\mathcal{H}.

We refer [34, 2, 21, 22, 18, 46, 32, 31] for more details on continuous frames. We also refer [18, 20, 3] for connections between continuous frames and discrete frames.

2. Extension of continuous operator-valued frames

In order to set continuous version of Definition 1.4, we want existence of certain operators, for which we use the following definition.

Definition 2.1.

[1, 25] Let $\Omega$ be a measure space with positive measure $\mu$ . A collection $\{A_{\alpha}\}_{\alpha\in\Omega}$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ is said to be continuous operator-valued Bessel if

(i)

for each $h\in\mathcal{H}$ , the map $\Omega\ni\alpha\mapsto A_{\alpha}h\in\mathcal{H}_{0}$ is measurable,
(ii)

there exists $b>0$ such that

(1) $\int_{\Omega}\|A_{\alpha}h\|^{2}\,d\mu(\alpha)\leq b\|h\|^{2},\quad\forall h\in\mathcal{H}.$

Let $\{A_{\alpha}\}_{\alpha\in\Omega}$ and $\{\Psi_{\alpha}\}_{\alpha\in\Omega}$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ be continuous operator-valued Bessel with bounds $b$ and $d$ , respectively. Continuity of norm and polarization identity reveal that the map $\Omega\ni\alpha\mapsto\langle A_{\alpha}h,\Psi_{\alpha}g\rangle\in\mathbb{K}$ is measurable, for each fixed $h,g\in\mathcal{H}$ . Cauchy-Schwarz inequality and Inequality (1) now tell that this map is in $\mathcal{L}^{2}(\Omega,\mathbb{K})$ , explicitly,

	$\displaystyle\left\|\int_{\Omega}\langle A_{\alpha}h,\Psi_{\alpha}g\rangle\,d\mu(\alpha)\right\|$	$\displaystyle\leq\int_{\Omega}\|\langle A_{\alpha}h,\Psi_{\alpha}g\rangle\|\,d\mu(\alpha)\leq\int_{\Omega}\\|A_{\alpha}h\\|\\|\Psi_{\alpha}g\\|\,d\mu(\alpha)$
		$\displaystyle\leq\left(\int_{\Omega}\\|A_{\alpha}h\\|^{2}\,d\mu(\alpha)\right)^{\frac{1}{2}}\left(\int_{\Omega}\\|\Psi_{\alpha}g\\|^{2}\,d\mu(\alpha)\right)^{\frac{1}{2}}\leq\sqrt{bd}\\|h\\|\\|g\\|.$

Previous inequalities also show that for each fixed $h\in\mathcal{H}$ , the map

\displaystyle\zeta_{h}:\mathcal{H}\ni g\mapsto\int_{\Omega}\langle A_{\alpha}h,\Psi_{\alpha}g\rangle\,d\mu(\alpha)\in\mathbb{K}

is a conjugate-linear bounded functional with $\|\zeta_{h}\|_{\text{op}}\leq\sqrt{bd}\|h\|$ (where $\|\cdot\|_{\text{op}}$ denotes the operator-norm). Let $\int_{\Omega}\Psi_{\alpha}^{*}A_{\alpha}h\,d\mu(\alpha)$ be that unique element (which comes from Riesz representation theorem) of $\mathcal{H}$ such that

	$\displaystyle\int_{\Omega}\langle A_{\alpha}h,\Psi_{\alpha}g\rangle\,d\mu(\alpha)$	$\displaystyle=\zeta_{h}g=\left\langle\int_{\Omega}\Psi_{\alpha}^{*}A_{\alpha}h\,d\mu(\alpha),g\right\rangle,~{}\forall g\in\mathcal{H}\text{ and }$
	$\displaystyle\\|\zeta_{h}\\|_{\text{op}}$	$\displaystyle=\left\\|\int_{\Omega}\Psi_{\alpha}^{*}A_{\alpha}h\,d\mu(\alpha)\right\\|.$

By varying $h\in\mathcal{H}$ , we get the map

\displaystyle S_{A,\Psi}:\mathcal{H}\ni h\mapsto S_{A,\Psi}h\coloneqq\int_{\Omega}\Psi_{\alpha}^{*}A_{\alpha}h\,d\mu(\alpha)\in\mathbb{K}.

Above map is a bounded linear operator, $S_{A,\Psi}^{*}=S_{\Psi,A}$ and $S_{A,A}$ is positive. Indeed,

	$\displaystyle\\|S_{A,\Psi}\\|$	$\displaystyle=\sup_{h\in\mathcal{H},\\|h\\|\leq 1}\\|S_{A,\Psi}h\\|=\sup_{h\in\mathcal{H},\\|h\\|\leq 1}\left\\|\int_{\Omega}\Psi_{\alpha}^{*}A_{\alpha}h\,d\mu(\alpha)\right\\|$
		$\displaystyle=\sup_{h\in\mathcal{H},\\|h\\|\leq 1}\\|\zeta_{h}\\|_{\text{op}}\leq\sup_{h\in\mathcal{H},\\|h\\|\leq 1}\sqrt{bd}\\|h\\|=\sqrt{bd}.$

and

	$\displaystyle\langle S_{A,\Psi}h,g\rangle$	$\displaystyle=\int_{\Omega}\langle A_{\alpha}h,\Psi_{\alpha}g\rangle\,d\mu(\alpha)=\int_{\Omega}\overline{\langle\Psi_{\alpha}g,A_{\alpha}h\rangle}\,d\mu(\alpha)$
		$\displaystyle=\overline{\int_{\Omega}\langle\Psi_{\alpha}g,A_{\alpha}h\rangle\,d\mu(\alpha)}=\overline{\langle S_{\Psi,A}g,h\rangle}=\langle h,S_{\Psi,A}g\rangle,\quad\forall h,g\in\mathcal{H},$
	$\displaystyle\langle S_{A,A}h,h\rangle$	$\displaystyle=\int_{\Omega}\\|A_{\alpha}h\\|^{2}\,d\mu(\alpha)\geq 0,\quad\forall h\in\mathcal{H}.$

We further note that Inequality (1) gives that

\theta_{A}:\mathcal{H}\ni h\mapsto\theta_{A}h\in\mathcal{L}^{2}(\Omega,\mathcal{H}_{0}),\quad\theta_{A}h:\Omega\ni\alpha\mapsto A_{\alpha}h\in\mathcal{H}_{0}

is a well-defined bounded linear operator whose adjoint is

\theta_{A}^{*}:\mathcal{L}^{2}(\Omega,\mathcal{H}_{0})\ni f\mapsto\int_{\Omega}A_{\alpha}^{*}f(\alpha)\,d\mu(\alpha)\in\mathcal{H},

where the integral is in the weak-sense. In fact,

\displaystyle\|\theta_{A}h\|^{2}=\int_{\Omega}\|\theta_{A}h(\alpha)\|^{2}\,d\mu(\alpha)=\int_{\Omega}\|A_{\alpha}h\|^{2}\,d\mu(\alpha)\leq b\|h\|^{2},\quad\forall h\in\mathcal{H}

and

	$\displaystyle\langle\theta_{A}h,f\rangle$	$\displaystyle=\int_{\Omega}\langle\theta_{A}h(\alpha),f(\alpha)\rangle\,d\mu(\alpha)=\int_{\Omega}\langle A_{\alpha}h,f(\alpha)\rangle\,d\mu(\alpha)$
		$\displaystyle=\int_{\Omega}\langle h,A_{\alpha}^{}f(\alpha)\rangle\,d\mu(\alpha)=\langle h,\theta_{A}^{}f\rangle,\quad\forall h\in\mathcal{H},\forall f\in\mathcal{L}^{2}(\Omega,\mathcal{H}_{0}).$

We next observe that Condition (ii) in Definition 2.1 holds if and only if the map $\theta_{A}$ is a well-defined bounded linear operator. With this knowledge we are ready to define the continuous version of Definition 1.4.

Definition 2.2.

A collection $\{A_{\alpha}\}_{\alpha\in\Omega}$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ is said to be a continuous operator-valued frame (in short, continuous (ovf)) in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ with respect to a collection $\{\Psi_{\alpha}\}_{\alpha\in\Omega}$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ if

(i)

for each $h\in\mathcal{H}$ , both maps $\Omega\ni\alpha\mapsto A_{\alpha}h\in\mathcal{H}_{0}$ and $\Omega\ni\alpha\mapsto\Psi_{\alpha}h\in\mathcal{H}_{0}$ are measurable,
(ii)

the map (we call as frame operator) $S_{A,\Psi}:\mathcal{H}\ni h\mapsto\int_{\Omega}\Psi_{\alpha}^{*}A_{\alpha}h\,d\mu(\alpha)\in\mathcal{H}$ (the integral is in the weak-sense) is a well-defined bounded positive invertible operator,
(iii)

both maps (we call as analysis operator and its adjoint as synthesis operator) $\theta_{A}:\mathcal{H}\ni h\mapsto\theta_{A}h\in\mathcal{L}^{2}(\Omega,\mathcal{H}_{0})$ , $\theta_{A}h:\Omega\ni\alpha\mapsto A_{\alpha}h\in\mathcal{H}_{0}$ and $\theta_{\Psi}:\mathcal{H}\ni h\mapsto\theta_{\Psi}h\in\mathcal{L}^{2}(\Omega,\mathcal{H}_{0})$ , $\theta_{\Psi}h:\Omega\ni\alpha\mapsto\Psi_{\alpha}h\in\mathcal{H}_{0}$ are well-defined bounded linear operators.

We note that $\theta_{A}^{*}:\mathcal{L}^{2}(\Omega,\mathcal{H}_{0})\ni f\mapsto\int_{\Omega}A_{\alpha}^{*}f(\alpha)\,d\mu(\alpha)\in\mathcal{H}$ , $\theta_{\Psi}^{*}:\mathcal{L}^{2}(\Omega,\mathcal{H}_{0})\ni f\mapsto\int_{\Omega}\Psi_{\alpha}^{*}f(\alpha)\,d\mu(\alpha)\in\mathcal{H}$ (both integrals are in the weak-sense). Notions of frame bounds, Parseval frame are similar to the same in Definition 2.1 in [36].

Whenever $\{A_{\alpha}\}_{\alpha\in\Omega}$ is a continuous operator-valued frame w.r.t. $\{\Psi_{\alpha}\}_{\alpha\in\Omega}$ we write $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ is continuous (ovf).

For fixed $\Omega$ , $\mathcal{H},\mathcal{H}_{0}$ and $\{\Psi_{\alpha}\}_{\alpha\in\Omega}$ , the set of all continuous operator-valued frames in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ with respect to collection $\{\Psi_{\alpha}\}_{\alpha\in\Omega}$ is denoted by $\mathscr{F}_{\Psi}.$

Remark 2.3.

Fundamental difference of continuous frames with discrete one is that we are not allowed to use orthonormal bases (indexed by $\Omega$ ).

If the condition (ii) in Definition 2.2 is replaced by “the map $S_{A,\Psi}:\mathcal{H}\ni h\mapsto\int_{\Omega}\Psi_{\alpha}^{*}A_{\alpha}h\,d\mu(\alpha)\in\mathcal{H}$ is a well-defined bounded positive operator (not necessarily invertible)”, then we say $\{A_{\alpha}\}_{\alpha\in\Omega}$ w.r.t. $\{\Psi_{\alpha}\}_{\alpha\in\Omega}$ is Bessel.

We note that (ii) in Definition 2.2 implies that there are real $a,b>0$ such that for all $h\in\mathcal{H}$ ,

\displaystyle a\|h\|^{2}\leq\langle S_{A,\Psi}h,h\rangle=\left\langle\int_{\Omega}\Psi^{*}_{\alpha}A_{\alpha}h\,d\mu(\alpha),h\right\rangle=\int_{\Omega}\langle\Psi^{*}_{\alpha}A_{\alpha}h,h\rangle\,d\mu(\alpha)=\int_{\Omega}\langle A_{\alpha}h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)\leq b\|h\|^{2},

and (iii) implies there exist $c,d>0$ such that for all $h\in\mathcal{H}$ ,

	$\displaystyle\\|\theta_{A}h\\|^{2}=\langle\theta_{A}h,\theta_{A}h\rangle=\int_{\Omega}\langle\theta_{A}h(\alpha),\theta_{A}h(\alpha)\rangle\,d\mu(\alpha)=\int_{\Omega}\langle A_{\alpha}h,A_{\alpha}h\rangle\,d\mu(\alpha)=\int_{\Omega}\\|A_{\alpha}h\\|^{2}\,d\mu(\alpha)\leq c\\|h\\|^{2};$
	$\displaystyle\\|\theta_{\Psi}h\\|^{2}=\langle\theta_{\Psi}h,\theta_{\Psi}h\rangle=\int_{\Omega}\langle\theta_{\Psi}h(\alpha),\theta_{\Psi}h(\alpha)\rangle\,d\mu(\alpha)=\int_{\Omega}\langle\Psi_{\alpha}h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)=\int_{\Omega}\\|\Psi_{\alpha}h\\|^{2}\,d\mu(\alpha)\leq d\\|h\\|^{2}.$

We note the following.

(i)

If $\{A_{\alpha}\}_{\alpha\in\Omega}$ is a continuous (ovf) w.r.t. $\{\Psi_{\alpha}\}_{\alpha\in\Omega}$ , then $\{\Psi_{\alpha}\}_{\alpha\in\Omega}$ is a continuous (ovf) w.r.t. $\{A_{\alpha}\}_{\alpha\in\Omega}$ .
(ii)

$\{h\in\mathcal{H}:A_{\alpha}h=0,\forall\alpha\in\Omega\}=\{0\}=\{h\in\mathcal{H}:\Psi_{\alpha}h=0,\forall\alpha\in\Omega\}$ , and $\overline{\operatorname{span}}\cup_{\alpha\in\Omega}A^{*}_{\alpha}(\mathcal{H}_{0})=\mathcal{H}=\overline{\operatorname{span}}\cup_{\alpha\in\Omega}\Psi^{*}_{\alpha}(\mathcal{H}_{0}).$
(iii)

$S_{A,\Psi}=S_{\Psi,A}$ .
(iv)

If $\{A_{\alpha}\}_{\alpha\in\Omega},$ $\{B_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}_{\Psi}$ , then $\{A_{\alpha}+B_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}_{\Psi}$ , and $\{\alpha A_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}_{\Psi},\forall\alpha>0.$
(v)

If $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ is tight continuous (ovf) with bound $a,$ then $S_{A,\Psi}=aI_{\mathcal{H}}.$

Proposition 2.4.

Let $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ be a continuous (ovf) in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ with an upper frame bound $b$ . If $\{\alpha\}$ is measurable and $\Psi_{\alpha}^{*}A_{\alpha}\geq 0,\forall\alpha\in\Omega,$ then $\mu(\{\alpha\})\|\Psi_{\alpha}^{*}A_{\alpha}\|\leq b,\forall\alpha\in\Omega.$

Proof.

For each $h\in\mathcal{H},\alpha\in\Omega$ we get $\mu(\{\alpha\})\langle\Psi_{\alpha}^{*}A_{\alpha}h,h\rangle=\int_{\{\alpha\}}\langle\Psi_{\beta}^{*}A_{\beta}h,h\rangle\,d\mu(\beta)\leq\int_{\{\alpha\}}\langle\Psi_{\beta}^{*}A_{\beta}h,h\rangle\,d\mu(\beta)+\int_{\Omega\setminus\{\alpha\}}\langle\Psi_{\beta}^{*}A_{\beta}h,h\rangle\,d\mu(\beta)=\int_{\Omega}\langle\Psi_{\beta}^{*}A_{\beta}h,h\rangle\,d\mu(\beta)\leq b\langle h,h\rangle$ and hence $\mu(\{\alpha\})\|\Psi_{\alpha}^{*}A_{\alpha}\|=\mu(\{\alpha\})\sup_{h\in\mathcal{H},\|h\|=1}$ $\langle\Psi_{\alpha}^{*}A_{\alpha}h,h\rangle\leq b,\forall\alpha\in\Omega.$ ∎

Proposition 2.5.

Let $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ be a continuous (ovf) in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ . Then the bounded left-inverses of

(i)

$\theta_{A}$ are precisely $S_{A,\Psi}^{-1}\theta_{\Psi}^{*}+U(I_{\mathcal{L}^{2}(\Omega,\mathcal{H}_{0})}-\theta_{A}S_{A,\Psi}^{-1}\theta_{\Psi}^{*})$ , where $U\in\mathcal{B}(\mathcal{L}^{2}(\Omega,\mathcal{H}_{0}),\mathcal{H})$ .
(ii)

$\theta_{\Psi}$ are precisely $S_{A,\Psi}^{-1}\theta_{A}^{*}+V(I_{\mathcal{L}^{2}(\Omega,\mathcal{H}_{0})}-\theta_{\Psi}S_{A,\Psi}^{-1}\theta_{A}^{*})$ , where $V\in\mathcal{B}(\mathcal{L}^{2}(\Omega,\mathcal{H}_{0}),\mathcal{H})$ .

Proof.

Similar to the proof of Proposition 2.29 in [36]. ∎

Proposition 2.6.

For every $\{A_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}_{\Psi}$ ,

(i)

$\theta_{A}^{*}\theta_{A}h=\int_{\Omega}A_{\alpha}^{*}A_{\alpha}h\,d\mu(\alpha)$ , $\forall h\in\mathcal{H}$ .
(ii)

$S_{A,\Psi}=\theta_{\Psi}^{*}\theta_{A}=\theta_{A}^{*}\theta_{\Psi}=S_{\Psi,A}.$
(iii)

$(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ is Parseval if and only if $\theta_{\Psi}^{*}\theta_{A}=I_{\mathcal{H}}.$
(iv)

$(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ is Parseval if and only if $\theta_{A}\theta_{\Psi}^{*}$ is idempotent.
(v)

$P_{A,\Psi}\coloneqq\theta_{A}S_{A,\Psi}^{-1}\theta_{\Psi}^{*}$ is idempotent and $P_{\Psi,A}=P_{A,\Psi}^{*}$ .
(vi)

$\theta_{A}$ and $\theta_{\Psi}$ are injective and their ranges are closed.
(vii)

$\theta_{A}^{*}$ and $\theta_{\Psi}^{*}$ are surjective.

Proof.

Let $h,g\in\mathcal{H}.$ We observe

\left\langle\theta_{A}^{*}\theta_{A}h,g\right\rangle=\left\langle\theta_{A}h,\theta_{A}g\right\rangle=\int_{\Omega}\langle\theta_{A}h(\alpha),\theta_{A}g(\alpha)\rangle\,d\mu(\alpha)=\int_{\Omega}\langle A_{\alpha}h,A_{\alpha}g\rangle\,d\mu(\alpha)=\left\langle\int_{\Omega}A_{\alpha}^{*}A_{\alpha}h\,d\mu(\alpha),g\right\rangle,

and

	$\displaystyle\langle S_{A,\Psi}h,g\rangle$	$\displaystyle=\left\langle\int_{\Omega}\Psi_{\alpha}^{}A_{\alpha}h\,d\mu(\alpha),g\right\rangle=\int_{\Omega}\langle\Psi_{\alpha}^{}A_{\alpha}h,g\rangle\,d\mu(\alpha)$
		$\displaystyle=\int_{\Omega}\langle A_{\alpha}h,\Psi_{\alpha}g\rangle\,d\mu(\alpha)=\langle\theta_{A}h,\theta_{\Psi}g\rangle=\langle\theta_{\Psi}^{*}\theta_{A}h,g\rangle,$

hence we get (i) and (ii). Arguments for other statements are similar to the proof of Proposition 2.30 in [36]. ∎

Proposition 2.7.

A collection $\{A_{\alpha}\}_{\alpha\in\Omega}$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ is a continuous (ovf) w.r.t. $\{\Psi_{\alpha}\}_{\alpha\in\Omega}$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ if and only if there exist $a,b,c,d>0$ such that

(i)

for each $h\in\mathcal{H}$ , both maps $\Omega\ni\alpha\mapsto A_{\alpha}h\in\mathcal{H}_{0}$ , $\Omega\ni\alpha\mapsto\Psi_{\alpha}h\in\mathcal{H}_{0}$ are measurable,
(ii)

$\int_{\Omega}\Psi_{\alpha}^{*}A_{\alpha}h\,d\mu(\alpha)=\int_{\Omega}A_{\alpha}^{*}\Psi_{\alpha}h\,d\mu(\alpha),\forall h\in\mathcal{H},$
(iii)

$a\|h\|^{2}\leq\int_{\Omega}\langle A_{\alpha}h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)\leq b\|h\|^{2},\forall h\in\mathcal{H},$
(iv)

$\int_{\Omega}\|A_{\alpha}h\|^{2}\,d\mu(\alpha)\leq c\|h\|^{2},\forall h\in\mathcal{H};\int_{\Omega}\|\Psi_{\alpha}h\|^{2}\,d\mu(\alpha)\leq d\|h\|^{2},\forall h\in\mathcal{H}$ .

Definition 2.8.

A continuous (ovf) $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ is said to be a Riesz continuous (ovf) if $P_{A,\Psi}=I_{\mathcal{L}^{2}(\Omega,\mathcal{H}_{0})}$ .

Proposition 2.9.

A continuous (ovf) $(\{A_{\alpha}\}_{\Omega},\{\Psi_{\alpha}\}_{\Omega})$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ is a Riesz continuous (ovf) if and only if $\theta_{A}(\mathcal{H})=\mathcal{L}^{2}(\Omega,\mathcal{H}_{0})$ if and only if $\theta_{\Psi}(\mathcal{H})=\mathcal{L}^{2}(\Omega,\mathcal{H}_{0}).$

Proof.

Similar to the proof of Proposition 2.36 in [36]. ∎

Following is the dilation result in discrete setting (for dilation results in Hilbert spaces we refer Theorem 2.38 in [36] and [35, 24, 11, 37]).

Theorem 2.10.

[36] Let $(\{A_{j}\}_{j\in\mathbb{J}},\{\Psi_{j}\}_{j\in\mathbb{J}})$ be a Parseval (ovf) in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ such that $\theta_{A}(\mathcal{H})=\theta_{\Psi}(\mathcal{H})$ and $P_{A,\Psi}$ is projection. Then there exist a Hilbert space $\mathcal{H}_{1}$ which contains $\mathcal{H}$ isometrically and bounded linear operators $B_{j},\Phi_{j}:\mathcal{H}_{1}\rightarrow\mathcal{H}_{0},\forall j\in\mathbb{J}$ such that $(\{B_{j}\}_{j\in\mathbb{J}},\{\Phi_{j}\}_{j\in\mathbb{J}})$ is an orthonormal (ovf) in $\mathcal{B}(\mathcal{H}_{1},\mathcal{H}_{0})$ and $B_{j}|_{\mathcal{H}}=A_{j},\Phi_{j}|_{\mathcal{H}}=\Psi_{j},\forall j\in\mathbb{J}$ .

We remark here that we don’t know any result corresponding to Theorem 2.10 when the indexing set is a measure space.

Definition 2.11.

A continuous (ovf) $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Phi_{\alpha}\}_{\alpha\in\Omega})$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ is said to be a dual of a continuous (ovf) $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ if $\theta_{\Phi}^{*}\theta_{A}=\theta_{B}^{*}\theta_{\Psi}=I_{\mathcal{H}}$ . The ‘continuous operator-valued frame’ $(\{\widetilde{A}_{\alpha}\coloneqq A_{\alpha}S_{A,\Psi}^{-1}\}_{\alpha\in\Omega},\{\widetilde{\Psi}_{\alpha}\coloneqq\Psi_{\alpha}S_{A,\Psi}^{-1}\}_{\alpha\in\Omega})$ , which is a ‘dual’ of $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ is called as the canonical dual of $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ .

Proposition 2.12.

Let $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ be a continuous (ovf) in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0}).$ If $h\in\mathcal{H}$ has representation $h=\int_{\Omega}A_{\alpha}^{*}f(\alpha)\,d\mu(\alpha)=\int_{\Omega}\Psi_{\alpha}^{*}g(\alpha)\,d\mu(\alpha),$ for some measurable $f,g:\Omega\rightarrow\mathcal{H}_{0}$ , then

\int_{\Omega}\langle f(\alpha),g(\alpha)\rangle\,d\mu(\alpha)=\int_{\Omega}\langle\widetilde{\Psi}_{\alpha}h,\widetilde{A}_{\alpha}h\rangle\,d\mu(\alpha)+\int_{\Omega}\langle f(\alpha)-\widetilde{\Psi}_{\alpha}h,g(\alpha)-\widetilde{A}_{\alpha}h\rangle\,d\mu(\alpha).

Proof.

	Right side	$\displaystyle=\int_{\Omega}\langle\widetilde{\Psi}_{\alpha}h,\widetilde{A}_{\alpha}h\rangle\,d\mu(\alpha)+\int_{\Omega}\langle f(\alpha),g(\alpha)\rangle\,d\mu(\alpha)-\int_{\Omega}\langle f(\alpha),\widetilde{A}_{\alpha}h\rangle\,d\mu(\alpha)$
		$\displaystyle\quad-\int_{\Omega}\langle\widetilde{\Psi}_{\alpha}h,g(\alpha)\rangle\,d\mu(\alpha)+\int_{\Omega}\langle\widetilde{\Psi}_{\alpha}h,\widetilde{A}_{\alpha}h\rangle\,d\mu(\alpha)$
		$\displaystyle=2\int_{\Omega}\langle\widetilde{\Psi}_{\alpha}h,\widetilde{A}_{\alpha}h\rangle\,d\mu(\alpha)+\int_{\Omega}\langle f(\alpha),g(\alpha)\rangle\,d\mu(\alpha)-\int_{\Omega}\langle f(\alpha),A_{\alpha}S_{A,\Psi}^{-1}h\rangle\,d\mu(\alpha)$
		$\displaystyle\quad-\int_{\Omega}\langle\Psi_{\alpha}S_{A,\Psi}^{-1}h,g(\alpha)\rangle\,d\mu(\alpha)$
		$\displaystyle=2\left\langle\int_{\Omega}S_{A,\Psi}^{-1}A_{\alpha}^{*}\Psi_{\alpha}S_{A,\Psi}^{-1}h\,d\mu(\alpha),h\right\rangle+\int_{\Omega}\langle f(\alpha),g(\alpha)\rangle\,d\mu(\alpha)$
		$\displaystyle\quad-\left\langle\int_{\Omega}A_{\alpha}^{}f(\alpha)\,d\mu(\alpha),S_{A,\Psi}^{-1}h\right\rangle-\left\langle S_{A,\Psi}^{-1}h,\int_{\Omega}\Psi_{\alpha}^{}g(\alpha)\,d\mu(\alpha)\right\rangle$
		$\displaystyle=2\langle S_{A,\Psi}^{-1}h,h\rangle+\int_{\Omega}\langle f(\alpha),g(\alpha)\rangle\,d\mu(\alpha)-\langle h,S_{A,\Psi}^{-1}h\rangle-\langle S_{A,\Psi}^{-1}h,h\rangle=\text{Left side.}$

∎

Theorem 2.13.

Let $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ be a continuous (ovf) with frame bounds $a$ and $b.$ Then the following statements are true.

(i)

The canonical dual (ovf) of the canonical dual (ovf) of $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ is itself.
(ii)

$\frac{1}{b},\frac{1}{a}$ are frame bounds for the canonical dual of $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ .
(iii)

If $a,b$ are optimal frame bounds for $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ , then $\frac{1}{b},\frac{1}{a}$ are optimal frame bounds for its canonical dual.

Proof.

We note that

	$\displaystyle\left\langle\int_{\Omega}\widetilde{\Psi}^{}_{\alpha}\widetilde{A}_{\alpha}h\,d\mu(\alpha),g\right\rangle=\int_{\Omega}\langle S_{A,\Psi}^{-1}\Psi_{\alpha}^{}A_{\alpha}S_{A,\Psi}^{-1}h,g\rangle\,d\mu(\alpha)=\int_{\Omega}\langle\Psi_{\alpha}^{*}A_{\alpha}S_{A,\Psi}^{-1}h,S_{A,\Psi}^{-1}g\rangle\,d\mu(\alpha)$
	$\displaystyle=\left\langle\int_{\Omega}\Psi_{\alpha}^{*}A_{\alpha}(S_{A,\Psi}^{-1}h)\,d\mu(\alpha),S_{A,\Psi}^{-1}g\right\rangle=\langle S_{A,\Psi}(S_{A,\Psi}^{-1}h),S_{A,\Psi}^{-1}g\rangle=\langle S_{A,\Psi}^{-1}h,g\rangle,\quad\forall h,g\in\mathcal{H}.$

Therefore the frame operator for the canonical dual $(\{A_{\alpha}S_{A,\Psi}^{-1}\}_{\alpha\in\Omega},\{\Psi_{\alpha}S_{A,\Psi}^{-1}\}_{\alpha\in\Omega})$ is $S_{A,\Psi}$ . Remainings are similar to the proof of Theorem 2.41 in [36]. ∎

Proposition 2.14.

(i)

$(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Phi_{\alpha}\}_{\alpha\in\Omega})$ is a dual of $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ .
(ii)

$\int_{\Omega}\Phi_{\alpha}^{*}A_{\alpha}h\,d\mu(\alpha)=\int_{\Omega}B_{\alpha}^{*}\Psi_{\alpha}h\,d\mu(\alpha)=h,\forall h\in\mathcal{H}$ .

Proof.

$\langle\theta_{\Phi}^{*}\theta_{A}h,g\rangle=\langle\theta_{A}h,\theta_{\Phi}g\rangle=\int_{\Omega}\langle\theta_{A}h(\alpha),\theta_{\Phi}g(\alpha)\rangle\,d\mu(\alpha)=\int_{\Omega}\langle A_{\alpha}h,\Phi_{\alpha}g\rangle\,d\mu(\alpha)=\langle\int_{\Omega}\Phi_{\alpha}^{*}A_{\alpha}h\,d\mu(\alpha),g\rangle,$ $\forall h,g\in\mathcal{H}$ . Similarly $\langle\theta_{B}^{*}\theta_{\Psi}h,g\rangle=\langle\int_{\Omega}B_{\alpha}^{*}\Psi_{\alpha}h\,d\mu(\alpha),g\rangle,$ $\forall h,g\in\mathcal{H}$ . ∎

Theorem 2.15.

If $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ is a Riesz continuous (ovf) in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ , then it has unique dual.

Proof.

Let $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Phi_{\alpha}\}_{\alpha\in\Omega})$ and $(\{C_{\alpha}\}_{\alpha\in\Omega},\{\Xi_{\alpha}\}_{\alpha\in\Omega})$ be continuous operator-valued frames such that both are duals of $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ . Then $\theta_{\Psi}^{*}\theta_{B}=I_{\mathcal{H}}=\theta_{A}^{*}\theta_{\Phi}=\theta_{\Psi}^{*}\theta_{C}=\theta_{A}^{*}\theta_{\Xi}$ $\Rightarrow$ $\theta_{\Psi}^{*}(\theta_{B}-\theta_{C})=0=\theta_{A}^{*}(\theta_{\Phi}-\theta_{\Xi})$ $\Rightarrow$ $\theta_{A}S_{A,\Psi}^{-1}\theta_{\Psi}^{*}(\theta_{B}-\theta_{C})=P_{A,\Psi}(\theta_{B}-\theta_{C})=I_{\mathcal{H}}(\theta_{B}-\theta_{C})=0=\theta_{\Psi}S_{\Psi,A}^{-1}\theta_{A}^{*}(\theta_{\Phi}-\theta_{\Xi})=P_{\Psi,A}(\theta_{\Phi}-\theta_{\Xi})=I_{\mathcal{H}}(\theta_{\Phi}-\theta_{\Xi})$ $\Rightarrow$ $\theta_{B}=\theta_{C},\theta_{\Phi}=\theta_{\Xi}$ $\Rightarrow$ $B_{\alpha}h=\theta_{B}h(\alpha)=\theta_{C}h(\alpha)=C_{\alpha}h,\Phi_{\alpha}h=\theta_{\Phi}h(\alpha)=\theta_{\Xi}h(\alpha)=\Xi_{\alpha}h$ , $\forall h\in\mathcal{H},\forall\alpha\in\Omega$ . ∎

Proposition 2.16.

Let $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ be a continuous (ovf) in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ . If $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Phi_{\alpha}\}_{\alpha\in\Omega})$ is dual of $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ , then there exist continuous Bessel $\{C_{\alpha}\}_{\alpha\in\Omega}$ and $\{\Xi_{\alpha}\}_{\alpha\in\Omega}$ (w.r.t. themselves) in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ such that $B_{\alpha}=A_{\alpha}S_{A,\Psi}^{-1}+C_{\alpha},\Phi_{\alpha}=\Psi_{\alpha}S_{A,\Psi}^{-1}+\Xi_{\alpha},\forall\alpha\in\Omega$ , and $\theta_{C}(\mathcal{H})\perp\theta_{\Psi}(\mathcal{H}),\theta_{\Xi}(\mathcal{H})\perp\theta_{A}(\mathcal{H})$ . Converse holds if $\theta_{\Xi}^{*}\theta_{C}\geq 0$ .

Proof.

Similar to the proof of Proposition 2.44 in [36]. ∎

Definition 2.17.

A continuous (ovf) $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Phi_{\alpha}\}_{\alpha\in\Omega})$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ is said to be orthogonal to a continuous (ovf) $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ if $\theta_{\Phi}^{*}\theta_{A}=\theta_{B}^{*}\theta_{\Psi}=0.$

Proposition 2.18.

(i)

$(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Phi_{\alpha}\}_{\alpha\in\Omega})$ is orthogonal to $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ .
(ii)

$\int_{\Omega}\Phi_{\alpha}^{*}A_{\alpha}h\,d\mu(\alpha)=\int_{\Omega}B_{\alpha}^{*}\Psi_{\alpha}h\,d\mu(\alpha)=0,\forall h\in\mathcal{H}$ .

Proposition 2.19.

Two orthogonal continuous operator-valued frames have common dual continuous (ovf).

Proof.

Similar to the proof of Proposition 2.48 in [36]. ∎

Proposition 2.20.

Let $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ and $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Phi_{\alpha}\}_{\alpha\in\Omega})$ be two Parseval continuous operator-valued frames in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ which are orthogonal. If $C,D,E,F\in\mathcal{B}(\mathcal{H})$ are such that $C^{*}E+D^{*}F=I_{\mathcal{H}}$ , then $(\{A_{\alpha}C+B_{\alpha}D\}_{\alpha\in\Omega},\{\Psi_{\alpha}E+\Phi_{\alpha}F\}_{\alpha\in\Omega})$ is a Parseval continuous (ovf) in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ . In particular, if scalars $c,d,e,f$ satisfy $\bar{c}e+\bar{d}f=1$ , then $(\{cA_{\alpha}+dB_{\alpha}\}_{\alpha\in\Omega},\{e\Psi_{\alpha}+f\Phi_{\alpha}\}_{\alpha\in\Omega})$ is a Parseval continuous (ovf).

Proof.

For all $h\in\mathcal{H}$ and $\alpha\in\Omega$ we see $\theta_{AC+BD}h(\alpha)=(A_{\alpha}C+B_{\alpha}D)h=A_{\alpha}(Ch)+B_{\alpha}(Dh)=\theta_{A}(Ch)(\alpha)+\theta_{B}(Dh)(\alpha)=(\theta_{A}(Ch)+\theta_{B}(Dh))(\alpha)=(\theta_{A}C+\theta_{B}D)h(\alpha)$ $\Rightarrow$ $\theta_{AC+BD}=\theta_{A}C+\theta_{B}D$ . Similarly $\theta_{\Psi E+\Phi F}=\theta_{\Psi}E+\theta_{\Phi}F$ . Other arguments are similar to that in the proof of Proposition 2.49 in [36]. ∎

Definition 2.21.

Two continuous operator-valued frames $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ and $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Phi_{\alpha}\}_{\alpha\in\Omega})$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ are called disjoint if $(\{A_{\alpha}\oplus B_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\oplus\Phi_{\alpha}\}_{\alpha\in\Omega})$ is continuous (ovf) in $\mathcal{B}(\mathcal{H}\oplus\mathcal{H},\mathcal{H}_{0}).$

Proposition 2.22.

If $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ and $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Phi_{\alpha}\}_{\alpha\in\Omega})$ are orthogonal continuous operator-valued frames in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ , then they are disjoint. Further, if both $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ and $(\{B_{\alpha}\}_{\alpha\in\Omega}$ , $\{\Phi_{\alpha}\}_{\alpha\in\Omega})$ are Parseval, then $(\{A_{\alpha}\oplus B_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\oplus\Phi_{\alpha}\}_{\alpha\in\Omega})$ is Parseval.

Proof.

For all $h\oplus g\in\mathcal{H}\oplus\mathcal{H}$ , $\theta_{A\oplus B}(h\oplus g)(\alpha)=(A_{\alpha}\oplus B_{\alpha})(h\oplus g)=A_{\alpha}h+B_{\alpha}g=\theta_{A}h(\alpha)+\theta_{B}g(\alpha)=(\theta_{A}h+\theta_{B}g)(\alpha)$ and for all $f\in\mathcal{L}^{2}(\Omega,\mathcal{H}_{0})$ , $\langle\theta_{\Psi\oplus\Phi}^{*}f,h\oplus g\rangle=\langle f,\theta_{\Psi\oplus\Phi}(h\oplus g)\rangle=\langle\theta_{\Psi}^{*}f,h\rangle+\langle\theta_{\Phi}^{*}f,g\rangle=\langle\theta_{\Psi}^{*}f\oplus\theta_{\Phi}^{*}f,h\oplus g\rangle.$ Thus $S_{A\oplus B,\Psi\oplus\Phi}(h\oplus g)=\theta^{*}_{\Psi\oplus\Phi}\theta_{A\oplus B}(h\oplus g)=\theta^{*}_{\Psi\oplus\Phi}(\theta_{A}h+\theta_{B}g)=\theta_{\Psi}^{*}(\theta_{A}h+\theta_{B}g)\oplus\theta_{\Phi}^{*}(\theta_{A}h+\theta_{B}g)=(S_{A,\Psi}+0)\oplus(0+S_{B,\Phi})=S_{A,\Psi}\oplus S_{B,\Phi}$ , which is bounded positive invertible with $S_{A\oplus B,\Psi\oplus\Phi}^{-1}=S_{A,\Psi}^{-1}\oplus S_{B,\Phi}^{-1}$ . ∎

3. Characterizations of the extension

Theorem 3.1.

Let $\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega}$ be in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ such that for each $h\in\mathcal{H}$ , both maps $\Omega\ni\alpha\mapsto A_{\alpha}h\in\mathcal{H}_{0}$ , $\Omega\ni\alpha\mapsto\Psi_{\alpha}h\in\mathcal{H}_{0}$ are measurable. Then $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ is a continuous (ovf) with bounds $a$ and $b$ (resp. continuous Bessel with bound $b$ )

(i)

if and only if

U:\mathcal{L}^{2}(\Omega,\mathcal{H}_{0})\ni f\mapsto\int_{\Omega}A_{\alpha}^{*}f(\alpha)\,d\mu(\alpha)\in\mathcal{H},~{}\text{and}~{}V:\mathcal{L}^{2}(\Omega,\mathcal{H}_{0})\ni g\mapsto\int_{\Omega}\Psi_{\alpha}^{*}g(\alpha)\,d\mu(\alpha)\in\mathcal{H}

are well-defined, $U,V\in\mathcal{B}(\mathcal{L}^{2}(\Omega,\mathcal{H}_{0}),\mathcal{H})$ such that $aI_{\mathcal{H}}\leq VU^{*}\leq bI_{\mathcal{H}}$ (resp. $0\leq VU^{*}\leq bI_{\mathcal{H}}$ ).

(ii)

if and only if

U:\mathcal{L}^{2}(\Omega,\mathcal{H}_{0})\ni f\mapsto\int_{\Omega}A_{\alpha}^{*}f(\alpha)\,d\mu(\alpha)\in\mathcal{H},~{}\text{and}~{}S:\mathcal{H}\ni x\mapsto Sx\in\mathcal{L}^{2}(\Omega,\mathcal{H}_{0}),Sx:\Omega\ni\alpha\mapsto\Psi_{\alpha}x\in\mathcal{H}_{0}

are well-defined, $U\in\mathcal{B}(\mathcal{L}^{2}(\Omega,\mathcal{H}_{0}),\mathcal{H})$ , $S\in\mathcal{B}(\mathcal{H},\mathcal{L}^{2}(\Omega,\mathcal{H}_{0}))$ such that $aI_{\mathcal{H}}\leq S^{*}U^{*}\leq bI_{\mathcal{H}}$ (resp. $0\leq S^{*}U^{*}\leq bI_{\mathcal{H}}$ ).

(iii)

if and only if

R:\mathcal{H}\ni h\mapsto Rh\in\mathcal{L}^{2}(\Omega,\mathcal{H}_{0}),Rh:\Omega\ni\alpha\mapsto A_{\alpha}h\in\mathcal{H}_{0},~{}\text{and}~{}V:\mathcal{L}^{2}(\Omega,\mathcal{H}_{0})\ni g\mapsto\int_{\Omega}\Psi_{\alpha}^{*}g(\alpha)\,d\mu(\alpha)\in\mathcal{H}

are well-defined, $R\in\mathcal{B}(\mathcal{H},\mathcal{L}^{2}(\Omega,\mathcal{H}_{0}))$ , $V\in\mathcal{B}(\mathcal{L}^{2}(\Omega,\mathcal{H}_{0}),\mathcal{H})$ such that $aI_{\mathcal{H}}\leq VR\leq bI_{\mathcal{H}}$ (resp. $0\leq VR\leq bI_{\mathcal{H}}$ ).

(iv)

if and only if

\mathcal{H}\ni h\mapsto Rh\in\mathcal{L}^{2}(\Omega,\mathcal{H}_{0}),Rh:\Omega\ni\alpha\mapsto A_{\alpha}h\in\mathcal{H}_{0},~{}\text{and}~{}S:\mathcal{H}\ni x\mapsto Sx\in\mathcal{L}^{2}(\Omega,\mathcal{H}_{0}),Sx:\Omega\ni\alpha\mapsto\Psi_{\alpha}x\in\mathcal{H}_{0}

are well-defined, $R,S\in\mathcal{B}(\mathcal{H},\mathcal{L}^{2}(\Omega,\mathcal{H}_{0}))$ such that $aI_{\mathcal{H}}\leq S^{*}R\leq bI_{\mathcal{H}}$ (resp. $0\leq S^{*}R\leq bI_{\mathcal{H}}$ ).

Proof.

We argue only for (i), in frame situation. $(\Rightarrow)$ Now $U=\theta_{A}^{*}$ , $V=\theta_{\Psi}^{*}$ and $VU^{*}=\theta_{\Psi}^{*}\theta_{A}=S_{A,\Psi}$ .

$(\Leftarrow)$ Now $\theta_{A}=U^{*}$ , $\theta_{\Psi}=V^{*}$ and $S_{A,\Psi}=\theta_{\Psi}^{*}\theta_{A}=VU^{*}$ . ∎

Let $\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega}$ be in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0}).$ For each fixed $\alpha\in\Omega$ , suppose $\{e_{\alpha,\beta}\}_{\beta\in\Omega_{\alpha}}$ is an orthonormal basis for $\mathcal{H}_{0}.$ From Riesz representation theorem, we get unique $u_{\alpha,\beta},v_{\alpha,\beta}\in\mathcal{H}$ such that $\langle A_{\alpha}h,e_{\alpha,\beta}\rangle=\langle h,u_{\alpha,\beta}\rangle,\langle\Psi_{\alpha}h,e_{\alpha,\beta}\rangle=\langle h,v_{\alpha,\beta}\rangle,\forall h\in\mathcal{H},\forall\beta\in\Omega_{\alpha},\forall\alpha\in\Omega$ . Now $A_{\alpha}h=\sum_{\beta\in\Omega_{\alpha}}\langle A_{\alpha}h,e_{\alpha,\beta}\rangle e_{\alpha,\beta}=\sum_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle e_{\alpha,\beta}$ , $\Psi_{\alpha}h=\sum_{\beta\in\Omega_{\alpha}}\langle h,v_{\alpha,\beta}\rangle e_{\alpha,\beta},\forall h\in\mathcal{H},\forall\alpha\in\Omega.$ We next find the adjoints of $A_{\alpha}$ ’s and $\Psi_{\alpha}$ ’s in terms of $\{u_{\alpha,\beta}\}_{\beta\in\Omega_{\alpha}}$ and $\{v_{\alpha,\beta}\}_{\beta\in\Omega_{\alpha}}$ . For all $h\in\mathcal{H}$ , $\langle h,A_{\alpha}^{*}y\rangle=\langle A_{\alpha}h,y\rangle=\sum_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle\langle e_{\alpha,\beta},y\rangle$ $=\langle h,\sum_{\beta\in\Omega_{\alpha}}\langle y,e_{\alpha,\beta}\rangle u_{\alpha,\beta}\rangle,\langle h,\Psi_{\alpha}^{*}y\rangle=\langle\Psi_{\alpha}h,y\rangle=\sum_{\beta\in\Omega_{\alpha}}\langle h,v_{\alpha,\beta}\rangle\langle e_{\alpha,\beta},y\rangle=\langle h,\sum_{\beta\in\Omega_{\alpha}}\langle y,e_{\alpha,\beta}\rangle v_{\alpha,\beta}\rangle,\forall y\in\mathcal{H}_{0}.$ Therefore $A_{\alpha}^{*}y=\sum_{\beta\in\Omega_{\alpha}}\langle y,e_{\alpha,\beta}\rangle u_{\alpha,\beta}$ , $\Psi_{\alpha}^{*}z=\sum_{\beta\in\Omega_{\alpha}}\langle z,e_{\alpha,\beta}\rangle v_{\alpha,\beta},\forall y,z\in\mathcal{H}_{0},\forall\alpha\in\Omega.$ Evaluation of these at $e_{\alpha,\beta_{0}}$ gives $u_{\alpha,\beta_{0}}=A_{\alpha}^{*}e_{\alpha,\beta_{0}},v_{\alpha,\beta_{0}}=\Psi_{\alpha}^{*}e_{\alpha,\beta_{0}},\forall\beta_{0}\in\Omega_{\alpha},\alpha\in\Omega.$

Theorem 3.2.

Let $\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega}$ be in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0}).$ Suppose $\{e_{\alpha,\beta}\}_{\beta\in\Omega_{\alpha}}$ is an orthonormal basis for $\mathcal{H}_{0},$ for each $\alpha\in\Omega.$ Let $u_{\alpha,\beta}=A_{\alpha}^{*}e_{\alpha,\beta},v_{\alpha,\beta}=\Psi_{\alpha}^{*}e_{\alpha,\beta},\forall\beta\in\Omega_{\alpha},\forall\alpha\in\Omega.$ Then $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ is a continuous

(i)

(ovf) in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ with bounds $a$ and $b$ if and only if for each $h\in\mathcal{H}$ , both maps $\Omega\ni\alpha\mapsto\sum_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle e_{\alpha,\beta}\in\mathcal{H}_{0}$ , $\Omega\ni\alpha\mapsto\sum_{\beta\in\Omega_{\alpha}}\langle h,v_{\alpha,\beta}\rangle e_{\alpha,\beta}\in\mathcal{H}_{0}$ are measurable and there exist $c,d>0$ such that the map

T:\mathcal{H}\ni h\mapsto\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle v_{\alpha,\beta}\,d\mu(\alpha)\in\mathcal{H}

is a well-defined bounded positive invertible operator such that $a\|h\|^{2}\leq\langle Th,h\rangle\leq b\|h\|^{2},\forall h\in\mathcal{H}$ , and

\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}|\langle h,u_{\alpha,\beta}\rangle|^{2}\,d\mu(\alpha)\leq c\|h\|^{2},~{}\forall h\in\mathcal{H};\quad\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}|\langle h,v_{\alpha,\beta}\rangle|^{2}\,d\mu(\alpha)\leq d\|h\|^{2},~{}\forall h\in\mathcal{H}.

(ii)

Bessel in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ with bound $b$ if and only if for each $h\in\mathcal{H}$ , both maps $\Omega\ni\alpha\mapsto\sum_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle e_{\alpha,\beta}$ $\in\mathcal{H}_{0}$ , $\Omega\ni\alpha\mapsto\sum_{\beta\in\Omega_{\alpha}}\langle h,v_{\alpha,\beta}\rangle e_{\alpha,\beta}\in\mathcal{H}_{0}$ are measurable and there exist $c,d>0$ such that the map

T:\mathcal{H}\ni h\mapsto\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle v_{\alpha,\beta}\,d\mu(\alpha)\in\mathcal{H}

is a well-defined bounded positive operator such that $0\leq\langle Th,h\rangle\leq b\|h\|^{2},\forall h\in\mathcal{H}$ , and

\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}|\langle h,u_{\alpha,\beta}\rangle|^{2}\,d\mu(\alpha)\leq c\|h\|^{2},~{}\forall h\in\mathcal{H};\quad\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}|\langle h,v_{\alpha,\beta}\rangle|^{2}\,d\mu(\alpha)\leq d\|h\|^{2},~{}\forall h\in\mathcal{H}.

(iii)

\left\|\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle v_{\alpha,\beta}\,d\mu(\alpha)\right\|\leq r\|h\|,~{}\forall h\in\mathcal{H};

\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle v_{\alpha,\beta}\,d\mu(\alpha)=\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}\langle h,v_{\alpha,\beta}\rangle u_{\alpha,\beta}\,d\mu(\alpha),~{}\forall h\in\mathcal{H};

a\|h\|^{2}\leq\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle\langle v_{\alpha,\beta},h\rangle\,d\mu(\alpha)\leq b\|h\|^{2},~{}\forall h\in\mathcal{H},~{}\text{and}

\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}|\langle h,u_{\alpha,\beta}\rangle|^{2}\,d\mu(\alpha)\leq c\|h\|^{2},~{}\forall h\in\mathcal{H};\quad\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}|\langle h,v_{\alpha,\beta}\rangle|^{2}\,d\mu(\alpha)\leq d\|h\|^{2},~{}\forall h\in\mathcal{H}.

(iv)

Bessel in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ with bound $b$ if and only if for each $h\in\mathcal{H}$ , both maps $\Omega\ni\alpha\mapsto\sum_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle e_{\alpha,\beta}$ $\in\mathcal{H}_{0}$ , $\Omega\ni\alpha\mapsto\sum_{\beta\in\Omega_{\alpha}}\langle h,v_{\alpha,\beta}\rangle e_{\alpha,\beta}$ $\in\mathcal{H}_{0}$ are measurable and there exist $c,d,r>0$ such that

\left\|\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle v_{\alpha,\beta}\,d\mu(\alpha)\right\|\leq r\|h\|,~{}\forall h\in\mathcal{H};

\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle v_{\alpha,\beta}\,d\mu(\alpha)=\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}\langle h,v_{\alpha,\beta}\rangle u_{\alpha,\beta}\,d\mu(\alpha),~{}\forall h\in\mathcal{H};

0\leq\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle\langle v_{\alpha,\beta},h\rangle\,d\mu(\alpha)\leq b\|h\|^{2},~{}\forall h\in\mathcal{H},~{}\text{and}

\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}|\langle h,u_{\alpha,\beta}\rangle|^{2}\,d\mu(\alpha)\leq c\|h\|^{2},~{}\forall h\in\mathcal{H};\quad\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}|\langle h,v_{\alpha,\beta}\rangle|^{2}\,d\mu(\alpha)\leq d\|h\|^{2},~{}\forall h\in\mathcal{H}.

Proof.

(i)

For all $h\in\mathcal{H},$

	$\displaystyle\int_{\Omega}\sum\limits_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle v_{\alpha,\beta}\,d\mu(\alpha)$	$\displaystyle=\int_{\Omega}\sum\limits_{\beta\in\Omega_{\alpha}}\langle A_{\alpha}h,e_{\alpha,\beta}\rangle\Psi_{\alpha}^{*}e_{\alpha,\beta}\,d\mu(\alpha)$
		$\displaystyle=\int_{\Omega}\Psi_{\alpha}^{}\left(\sum\limits_{\beta\in\Omega_{\alpha}}\langle A_{\alpha}h,e_{\alpha,\beta}\rangle e_{\alpha,\beta}\right)\,d\mu(\alpha)=\int_{\Omega}\Psi_{\alpha}^{}A_{\alpha}h\,d\mu(\alpha),$

	$\displaystyle\int_{\Omega}\langle A_{\alpha}h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)$	$\displaystyle=\int_{\Omega}\left\langle\sum\limits_{\beta\in\Omega_{\alpha}}\langle h,A_{\alpha}^{}e_{\alpha,\beta}\rangle e_{\alpha,\beta},\sum\limits_{\gamma\in\Omega_{\alpha}}\langle h,\Psi_{\alpha}^{}e_{\alpha,\gamma}\rangle e_{\alpha,\gamma}\right\rangle\,d\mu(\alpha)$
		$\displaystyle=\int_{\Omega}\left\langle\sum\limits_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle e_{\alpha,\beta},\sum\limits_{\gamma\in\Omega_{\alpha}}\langle h,v_{\alpha,\gamma}\rangle e_{\alpha,\gamma}\right\rangle\,d\mu(\alpha)$
		$\displaystyle=\int_{\Omega}\sum\limits_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle\langle v_{\alpha,\beta},h\rangle\,d\mu(\alpha),$

	$\displaystyle\left\\|\theta_{A}h\right\\|^{2}$	$\displaystyle=\int_{\Omega}\\|A_{\alpha}h\\|^{2}\,d\mu(\alpha)=\int_{\Omega}\sum\limits_{\beta\in\Omega_{\alpha}}\|\langle h,u_{\alpha,\beta}\rangle\|^{2}\,d\mu(\alpha);$
	$\displaystyle\left\\|\theta_{\Psi}h\right\\|^{2}$	$\displaystyle=\int_{\Omega}\sum\limits_{\beta\in\Omega_{\alpha}}\|\langle h,v_{\alpha,\beta}\rangle\|^{2}\,d\mu(\alpha).$

(ii)

Similar to (i).
(iii)

$\mathcal{H}\ni h\mapsto\int_{\Omega}\Psi_{\alpha}^{*}A_{\alpha}h\,d\mu(\alpha)\in\mathcal{H}$ exists and is bounded positive invertible if and only if there exist $c,d,r>0$ such that $\|\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle v_{\alpha,\beta}\,d\mu(\alpha)\|\leq r\|h\|,\forall h\in\mathcal{H}$ , $\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle v_{\alpha,\beta}\,d\mu(\alpha)=\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}\langle h,v_{\alpha,\beta}\rangle u_{\alpha,\beta}\,d\mu(\alpha),\forall h\in\mathcal{H}$ and $a\|h\|^{2}\leq\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle\langle v_{\alpha,\beta},h\rangle\,d\mu(\alpha)\leq b\|h\|^{2},\forall h\in\mathcal{H}$ . Also, $\theta_{A}:\mathcal{H}\ni h\mapsto\theta_{A}h\in\mathcal{L}^{2}(\Omega,\mathcal{H}_{0})$ , $\theta_{A}h:\Omega\ni\alpha\mapsto A_{\alpha}h\in\mathcal{H}_{0}$ , (resp. $\theta_{\Psi}:\mathcal{H}\ni h\mapsto\theta_{\Psi}h\in\mathcal{L}^{2}(\Omega,\mathcal{H}_{0})$ , $\theta_{\Psi}h:\Omega\ni\alpha\mapsto\Psi_{\alpha}h\in\mathcal{H}_{0}$ ) exists and is bounded if and only if there exists $c>0$ (resp. $d>0$ ) such that $\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}|\langle h,u_{\alpha,\beta}\rangle|^{2}\,d\mu(\alpha)\leq c\|h\|^{2},\forall h\in\mathcal{H}$ (resp. $\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}|\langle h,v_{\alpha,\beta}\rangle|^{2}\,d\mu(\alpha)\leq d\|h\|^{2},\forall h\in\mathcal{H}$ ).
(iv)

Similar to (iii).

∎

Similarity

Definition 3.3.

A continuous (ovf) $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Phi_{\alpha}\}_{\alpha\in\Omega})$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ is said to be right-similar to a continuous (ovf) $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ if there exist invertible $R_{A,B},R_{\Psi,\Phi}\in\mathcal{B}(\mathcal{H})$ such that $B_{\alpha}=A_{\alpha}R_{A,B},\Phi_{\alpha}=\Psi_{\alpha}R_{\Psi,\Phi}$ , $\forall\alpha\in\Omega$ .

Proposition 3.4.

Let $\{A_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}_{\Psi}$ with frame bounds $a,b,$ let $R_{A,B},R_{\Psi,\Phi}\in\mathcal{B}(\mathcal{H})$ be positive, invertible, commute with each other, commute with $S_{A,\Psi}$ , and let $B_{\alpha}=A_{\alpha}R_{A,B},\Phi_{\alpha}=\Psi_{\alpha}R_{\Psi,\Phi},\forall\alpha\in\Omega.$ Then

(i)

$\{B_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}_{\Phi}$ and $\frac{a}{\|R_{A,B}^{-1}\|\|R_{\Psi,\Phi}^{-1}\|}\leq S_{B,\Phi}\leq b\|R_{A,B}R_{\Psi,\Phi}\|.$ Assuming that $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ is Parseval, then $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Phi_{\alpha}\}_{\alpha\in\Omega})$ is Parseval if and only if $R_{\Psi,\Phi}R_{A,B}=I_{\mathcal{H}}.$
(ii)

$\theta_{B}=\theta_{A}R_{A,B},\theta_{\Phi}=\theta_{\Psi}R_{\Psi,\Phi},S_{B,\Phi}=R_{\Psi,\Phi}S_{A,\Psi}R_{A,B},P_{B,\Phi}=P_{A,\Psi}.$

Proof.

For all $h,g\in\mathcal{H}$ ,

	$\displaystyle\langle R_{\Psi,\Phi}S_{A,\Psi}R_{A,B}h,g\rangle=\langle S_{A,\Psi}R_{A,B}h,R_{\Psi,\Phi}^{*}g\rangle=\langle S_{A,\Psi}R_{A,B}h,R_{\Psi,\Phi}g\rangle$
	$\displaystyle=\int_{\Omega}\langle A_{\alpha}R_{A,B}h,\Psi_{\alpha}R_{\Psi,\Phi}g\rangle\,d\mu(\alpha)=\int_{\Omega}\langle(\Psi_{\alpha}R_{\Psi,\Phi})^{*}A_{\alpha}R_{A,B}h,g\rangle\,d\mu(\alpha)=\langle S_{B,\Phi}h,g\rangle.$

∎

Lemma 3.5.

Let $\{A_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}_{\Psi},$ $\{B_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}_{\Phi}$ and $B_{\alpha}=A_{\alpha}R_{A,B},\Phi_{\alpha}=\Psi_{\alpha}R_{\Psi,\Phi},\forall\alpha\in\Omega$ , for some invertible $R_{A,B},R_{\Psi,\Phi}\in\mathcal{B}(\mathcal{H}).$ Then $\theta_{B}=\theta_{A}R_{A,B},\theta_{\Phi}=\theta_{\Psi}R_{\Psi,\Phi},S_{B,\Phi}=R_{\Psi,\Phi}^{*}S_{A,\Psi}R_{A,B},P_{B,\Phi}=P_{A,\Psi}.$ Assuming that $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ is Parseval, then $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Phi_{\alpha}\}_{\alpha\in\Omega})$ is Parseval if and only if $R_{\Psi,\Phi}^{*}R_{A,B}=I_{\mathcal{H}}.$

Proof.

$\theta_{B}h(\alpha)=B_{\alpha}h=(A_{\alpha}R_{A,B})h=A_{\alpha}(R_{A,B}h)=\theta_{A}(R_{A,B}h)(\alpha)=(\theta_{A}R_{A,B})h(\alpha),\forall\alpha\in\Omega,\forall h\in\mathcal{H}$ $\Rightarrow$ $\theta_{B}=\theta_{A}R_{A,B},\forall\alpha\in\Omega$ . Similarly $\theta_{\Phi}=\theta_{\Psi}R_{\Psi,\Phi}$ . ∎

Theorem 3.6.

Let $\{A_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}_{\Psi},$ $\{B_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}_{\Phi}.$ The following are equivalent.

(i)

$B_{\alpha}=A_{\alpha}R_{A,B},\Phi_{\alpha}=\Psi_{\alpha}R_{\Psi,\Phi},\forall\alpha\in\Omega,$ for some invertible $R_{A,B},R_{\Psi,\Phi}\in\mathcal{B}(\mathcal{H}).$
(ii)

$\theta_{B}=\theta_{A}R_{A,B}^{\prime},\theta_{\Phi}=\theta_{\Psi}R_{\Psi,\Phi}^{\prime}$ for some invertible $R_{A,B}^{\prime},R_{\Psi,\Phi}^{\prime}\in\mathcal{B}(\mathcal{H}).$
(iii)

$P_{B,\Phi}=P_{A,\Psi}.$

If one of the above conditions is satisfied, then invertible operators in $\operatorname{(i)}$ and $\operatorname{(ii)}$ are unique and are given by $R_{A,B}=S_{A,\Psi}^{-1}\theta_{\Psi}^{*}\theta_{B},R_{\Psi,\Phi}=S_{A,\Psi}^{-1}\theta_{A}^{*}\theta_{\Phi}.$ In the case that $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ is Parseval, then $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Phi_{\alpha}\}_{\alpha\in\Omega})$ is Parseval if and only if $R_{\Psi,\Phi}^{*}R_{A,B}$ is the identity operator if and only if $R_{A,B}R_{\Psi,\Phi}^{*}$ is the identity operator.

Proof.

(ii) $\Rightarrow$ (i) $B_{\alpha}h=\theta_{B}h(\alpha)=\theta_{B}h(\alpha)=(\theta_{A}R_{A,B}^{\prime})h(\alpha)=\theta_{A}(R_{A,B}^{\prime}h)(\alpha)=A_{\alpha}(R_{A,B}^{\prime}h)=(A_{\alpha}R_{A,B}^{\prime})h$ , $\forall\alpha\in\Omega,\forall h\in\mathcal{H}$ $\Rightarrow$ $B_{\alpha}=A_{\alpha}R_{A,B}^{\prime},\forall\alpha\in\Omega$ . Similarly $\Phi_{\alpha}=\Psi_{\alpha}R_{\Psi,\Phi}^{\prime},\forall\alpha\in\Omega$ . Other arguments are similar to that in the proof of Theorem 4.4 in [36]. ∎

Corollary 3.7.

For any given continuous (ovf) $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ , the canonical dual of $(\{A_{\alpha}\}_{\alpha\in\Omega}$ , $\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ is the only dual continuous (ovf) that is right-similar to $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ .

Proof.

Similar to the proof of Corollary 4.5 in [36]. ∎

Corollary 3.8.

Two right-similar continuous operator-valued frames cannot be orthogonal.

Proof.

Similar to the proof of Corollary 4.6 in [36]. ∎

Remark 3.9.

For every continuous (ovf) $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ , each of ‘continuous operator-valued frames’ $(\{A_{\alpha}S_{A,\Psi}^{-1}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega}),$ $(\{A_{\alpha}S_{A,\Psi}^{-1/2}\}_{\alpha\in\Omega},\{\Psi_{\alpha}S_{A,\Psi}^{-1/2}\}_{\alpha\in\Omega}),$ and $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}S_{A,\Psi}^{-1}\}_{\alpha\in\Omega})$ is a Parseval continuous (ovf) which is right-similar to $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega}).$ Thus every continuous (ovf) is right-similar to Parseval continuous operator-valued frames.

4. Continuous frames and representations of locally compact groups

Let $G$ be a locally compact group, $\mu_{G}$ be a left-invariant Haar measure on $G$ (we refer [17, 28, 13, 43, 4] for locally compact groups and Haar measures). Let $\lambda$ be the left regular representation of $G$ defined by $\lambda_{g}f(x)=f(g^{-1}x),\forall g,x\in G,\forall f\in\mathcal{B}(\mathcal{L}^{2}(G,\mathcal{H}_{0}))$ ; $\rho$ be the right regular representation of $G$ defined by $\rho_{g}f(x)=\Delta_{G}(g)^{1/2}f(xg),\forall g,x\in G,\forall f\in\mathcal{B}(\mathcal{L}^{2}(G,\mathcal{H}_{0}))$ , where $\Delta_{G}$ is the modular function associated with $G$ [49].

Definition 4.1.

Let $\pi$ be a unitary representation of a locally compact group $G$ on a Hilbert space $\mathcal{H}.$ An operator $A$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ is called a continuous operator-valued frame generator (resp. a Parseval frame generator) w.r.t. an operator $\Psi$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ if $(\{A_{g}\coloneqq A\pi_{g^{-1}}\}_{g\in G},\{\Psi_{g}\coloneqq\Psi\pi_{g^{-1}}\}_{g\in G})$ is a continuous (ovf) (resp. a Parseval continuous (ovf)) in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ (where the measure on $G$ is a left invariant Haar measure $\mu_{G}$ ). In this case, we write $(A,\Psi)$ is a continuous operator-valued frame generator for $\pi$ .

Proposition 4.2.

Let $(A,\Psi)$ and $(B,\Phi)$ be continuous operator-valued frame generators in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ for a unitary representation $\pi$ of a locally compact group $G$ on $\mathcal{H}.$ Then

(i)

$\theta_{A}\pi_{g}=\lambda_{g}\theta_{A},\theta_{\Psi}\pi_{g}=\lambda_{g}\theta_{\Psi},\forall g\in G.$
(ii)

$\theta_{A}^{*}\theta_{B},\theta_{\Psi}^{*}\theta_{\Phi},\theta_{A}^{*}\theta_{\Phi}$ are in the commutant $\pi(G)^{\prime}$ of $\pi(G)^{\prime\prime}.$ Further, $S_{A,\Psi}\in\pi(G)^{\prime}$ and $(AS_{A,\Psi}^{-{1/2}},\Psi S_{A,\Psi}^{-{1/2}})$ is a Parseval frame generator.

Proof.

(i)

For all $h\in\mathcal{H}$ and $f\in\mathcal{L}^{2}(G,\mathcal{H}_{0})$ ,

	$\displaystyle\langle\lambda_{g}\theta_{A}h,f\rangle$	$\displaystyle=\int_{G}\langle\lambda_{g}\theta_{A}h(\alpha),f(\alpha)\rangle\,d\mu_{G}(\alpha)=\int_{G}\langle\theta_{A}h(g^{-1}\alpha),f(\alpha)\rangle\,d\mu_{G}(\alpha)$
		$\displaystyle=\int_{G}\langle A_{g^{-1}\alpha}h,f(\alpha)\rangle\,d\mu_{G}(\alpha)=\int_{G}\langle A\pi_{(g^{-1}\alpha)^{-1}}h,f(\alpha)\rangle\,d\mu_{G}(\alpha)$
		$\displaystyle=\left\langle\pi_{g}h,\int_{G}(A\pi_{\alpha^{-1}})^{}f(\alpha)\,d\mu_{G}(\alpha)\right\rangle=\left\langle\pi_{g}h,\int_{G}A_{\alpha}^{}f(\alpha)\,d\mu_{G}(\alpha)\right\rangle$
		$\displaystyle=\langle\pi_{g}h,\theta_{A}^{*}f\rangle=\langle\theta_{A}\pi_{g}h,f\rangle$

$\Rightarrow\lambda_{g}\theta_{A}=\theta_{A}\pi_{g}$ . Similarly $\theta_{\Psi}\pi_{g}=\lambda_{g}\theta_{\Psi}.$

(ii)

$\theta_{A}^{*}\theta_{B}\pi_{g}=\theta_{A}^{*}\lambda_{g}\theta_{B}=(\lambda_{g^{-1}}\theta_{A})^{*}\theta_{B}=(\theta_{A}\pi_{g^{-1}})^{*}\theta_{B}=\pi_{g}\theta_{A}^{*}\theta_{B}$ and for all $x,y\in\mathcal{H}$

	$\displaystyle\left\langle\int_{G}(\Psi S_{A,\Psi}^{-\frac{1}{2}}\pi_{g^{-1}})^{*}(AS_{A,\Psi}^{-\frac{1}{2}}\pi_{g^{-1}})x\,d\mu_{G}(g),y\right\rangle$	$\displaystyle=\int_{G}\langle(\Psi S_{A,\Psi}^{-\frac{1}{2}}\pi_{g^{-1}})^{*}(AS_{A,\Psi}^{-\frac{1}{2}}\pi_{g^{-1}})x,y\rangle\,d\mu_{G}(g)$
		$\displaystyle=\int_{G}\langle\pi_{g}S_{A,\Psi}^{-\frac{1}{2}}\Psi^{*}AS_{A,\Psi}^{-\frac{1}{2}}\pi_{g^{-1}}x,y\rangle\,d\mu_{G}(g)$
		$\displaystyle=\int_{G}\langle S_{A,\Psi}^{-\frac{1}{2}}\pi_{g}\Psi^{*}A\pi_{g^{-1}}S_{A,\Psi}^{-\frac{1}{2}}x,y\rangle\,d\mu_{G}(g)$
		$\displaystyle=\int_{G}\langle(\Psi\pi_{g^{-1}})^{*}(A\pi_{g^{-1}})S_{A,\Psi}^{-\frac{1}{2}}x,S_{A,\Psi}^{-\frac{1}{2}}y\rangle\,d\mu_{G}(g)$
		$\displaystyle=\left\langle\int_{G}\Psi_{g}^{*}A_{g}(S_{A,\Psi}^{-\frac{1}{2}}x)\,d\mu_{G}(g),S_{A,\Psi}^{-\frac{1}{2}}y\right\rangle$
		$\displaystyle=\langle S_{A,\Psi}(S_{A,\Psi}^{-\frac{1}{2}}x),S_{A,\Psi}^{-\frac{1}{2}}y\rangle=\langle x,y\rangle$

$\Rightarrow$ $\int_{G}(\Psi S_{A,\Psi}^{-\frac{1}{2}}\pi_{g^{-1}})^{*}(AS_{A,\Psi}^{-\frac{1}{2}}\pi_{g^{-1}})x\,d\mu(g)=x,\forall x\in\mathcal{H}$ and hence the last part.

∎

Theorem 4.3.

Let $G$ be a locally compact group with identity $e$ and $(\{A_{g}\}_{g\in G},\{\Psi_{g}\}_{g\in G})$ be a Parseval continuous (ovf) in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0}).$ Then there is a unitary representation $\pi$ of $G$ on $\mathcal{H}$ for which

A_{g}=A_{e}\pi_{g^{-1}},~{}\Psi_{g}=\Psi_{e}\pi_{g^{-1}},\quad\forall g\in G

if and only if

A_{gp}A_{gq}^{*}=A_{p}A_{q}^{*},~{}A_{gp}\Psi_{gq}^{*}=A_{p}\Psi_{q}^{*},~{}\Psi_{gp}\Psi_{gq}^{*}=\Psi_{p}\Psi_{q}^{*},\quad\forall g,p,q\in G.

Proof.

$(\Rightarrow)$ Similar to the proof of ‘only if’ part of Theorem 5.3 in [36].

$(\Leftarrow)$ We claim the following three equalities among them we derive the second, two others are similar. For all $g\in G,$

\displaystyle\lambda_{g}\theta_{A}\theta_{A}^{*}=\theta_{A}\theta_{A}^{*}\lambda_{g},~{}\lambda_{g}\theta_{A}\theta_{\Psi}^{*}=\theta_{A}\theta_{\Psi}^{*}\lambda_{g},~{}\lambda_{g}\theta_{\Psi}\theta_{\Psi}^{*}=\theta_{\Psi}\theta_{\Psi}^{*}\lambda_{g}.

Let $u,v\in\mathcal{L}^{2}(G,\mathcal{H}_{0})$ . Then

	$\displaystyle\langle\lambda_{g}\theta_{A}\theta_{\Psi}^{}\lambda_{g}^{}u,v\rangle$	$\displaystyle=\langle\theta_{\Psi}^{}\lambda_{g}^{}u,\theta_{A}^{}\lambda_{g}^{}v\rangle$
		$\displaystyle=\left\langle\int_{G}\Psi_{\alpha}^{}\lambda_{g}^{}u(\alpha)\,d\mu_{G}(\alpha),\int_{G}A_{\beta}^{}\lambda_{g}^{}v(\beta)\,d\mu_{G}(\beta)\right\rangle$
		$\displaystyle=\left\langle\int_{G}\Psi_{\alpha}^{}\lambda_{g^{-1}}u(\alpha)\,d\mu_{G}(\alpha),\int_{G}A_{\beta}^{}\lambda_{g^{-1}}v(\beta)\,d\mu_{G}(\beta)\right\rangle$
		$\displaystyle=\left\langle\int_{G}\Psi_{\alpha}^{}u(g\alpha)\,d\mu_{G}(\alpha),\int_{G}A_{\beta}^{}v(g\beta)\,d\mu_{G}(\beta)\right\rangle$
		$\displaystyle=\left\langle\int_{G}\Psi_{g^{-1}p}^{}u(p)\,d\mu_{G}(g^{-1}p),\int_{G}A_{g^{-1}q}^{}v(q)\,d\mu_{G}(g^{-1}q)\right\rangle$
		$\displaystyle=\left\langle\int_{G}\Psi_{g^{-1}p}^{}u(p)\,d\mu_{G}(p),\int_{G}A_{g^{-1}q}^{}v(q)\,d\mu_{G}(q)\right\rangle$
		$\displaystyle=\int_{G}\int_{G}\langle A_{g^{-1}q}\Psi_{g^{-1}p}^{*}u(p),v(q)\rangle\,d\mu_{G}(q)\,d\mu_{G}(p)$
		$\displaystyle=\int_{G}\int_{G}\langle A_{q}\Psi_{p}^{*}u(p),v(q)\rangle\,d\mu_{G}(q)\,d\mu_{G}(p)$
		$\displaystyle=\left\langle\int_{G}\Psi_{p}^{}u(p)\,d\mu_{G}(p),\int_{G}A_{q}^{}v(q)\,d\mu_{G}(q)\right\rangle=\langle\theta_{\Psi}^{}u,\theta_{A}^{}v\rangle=\langle\theta_{A}\theta_{\Psi}^{*}u,v\rangle.$

Define $\pi:G\ni g\mapsto\pi_{g}\coloneqq\theta_{\Psi}^{*}\lambda_{g}\theta_{A}\in\mathcal{B}(\mathcal{H}).$ Using the fact that frame is Parseval, $\pi_{g}\pi_{h}=\theta_{\Psi}^{*}\lambda_{g}\theta_{A}\theta_{\Psi}^{*}\lambda_{h}\theta_{A}=\theta_{\Psi}^{*}\theta_{A}\theta_{\Psi}^{*}\lambda_{g}\lambda_{h}\theta_{A}=\theta_{\Psi}^{*}\lambda_{gh}\theta_{A}=\pi_{gh}$ for all $g,h\in G,$ and $\pi_{g}\pi_{g}^{*}=\theta_{\Psi}^{*}\lambda_{g}\theta_{A}\theta_{A}^{*}\lambda_{g^{-1}}\theta_{\Psi}=\theta_{\Psi}^{*}\theta_{A}\theta_{A}^{*}\lambda_{g}\lambda_{g^{-1}}\theta_{\Psi}=I_{\mathcal{H}},\pi_{g}^{*}\pi_{g}=\theta_{A}^{*}\lambda_{g^{-1}}\theta_{\Psi}\theta_{\Psi}^{*}\lambda_{g}\theta_{A}=\theta_{A}^{*}\lambda_{g^{-1}}\lambda_{g}\theta_{\Psi}\theta_{\Psi}^{*}\theta_{A}=I_{\mathcal{H}}$ for all $g\in G$ . We next prove that, for each fixed $h\in\mathcal{H}$ , the map $\phi_{h}:G\ni g\mapsto\pi_{g}h\in\mathcal{H}$ is continuous. So, let $h\in\mathcal{H}$ be fixed. Then $\theta_{A}h$ is fixed. Since $\lambda$ is a unitary representation, the map $G\ni g\mapsto\lambda_{g}(\theta_{A}h)\in\mathcal{L}^{2}(G,\mathbb{K})$ is continuous. Continuity of $\theta_{\Psi}^{*}$ now gives that the map $G\ni g\mapsto\theta_{\Psi}^{*}(\lambda_{g}(\theta_{A}h))\in\mathcal{H}$ is continuous, i.e., $\phi_{h}$ is continuous. This proves $\pi$ is a unitary representation. We now prove $A_{g}=A_{e}\pi_{g^{-1}},\Psi_{g}=\Psi_{e}\pi_{g^{-1}}$ for all $g\in G$ . For all $h\in\mathcal{H}$ and for all $f\in\mathcal{L}^{2}(G,\mathcal{H}_{0})$ ,

	$\displaystyle\langle A_{e}\pi_{g^{-1}}h,f\rangle$	$\displaystyle=\langle\pi_{g^{-1}}h,A_{e}^{}f\rangle=\langle\theta_{\Psi}^{}\lambda_{g^{-1}}\theta_{A}h,A_{e}^{*}f\rangle$
		$\displaystyle=\left\langle\int_{G}\Psi_{\alpha}^{}\lambda_{g^{-1}}\theta_{A}h(\alpha)\,d\mu_{G}(\alpha),A_{e}^{}f\right\rangle=\left\langle\int_{G}\Psi_{\alpha}^{}\theta_{A}h(g\alpha)\,d\mu_{G}(\alpha),A_{e}^{}f\right\rangle$
		$\displaystyle=\left\langle\int_{G}\Psi_{\alpha}^{}A_{g\alpha}h\,d\mu_{G}(\alpha),A_{e}^{}f\right\rangle=\left\langle\int_{G}\Psi_{g^{-1}\beta}^{}A_{\beta}h\,d\mu_{G}(g^{-1}\beta),A_{e}^{}f\right\rangle$
		$\displaystyle=\left\langle\int_{G}\Psi_{g^{-1}\beta}^{}A_{\beta}h\,d\mu_{G}(\beta),A_{e}^{}f\right\rangle=\int_{G}\langle A_{g^{-1}g}\Psi_{g^{-1}\beta}^{*}A_{\beta}h,f\rangle\,d\mu_{G}(\beta)$
		$\displaystyle=\int_{G}\langle A_{g}\Psi_{\beta}^{}A_{\beta}h,f\rangle\,d\mu_{G}(\beta)=\left\langle\int_{G}\Psi_{\beta}^{}A_{\beta}h\,d\mu_{G}(\beta),A_{g}^{*}f\right\rangle$
		$\displaystyle=\langle h,A_{g}^{*}f\rangle=\langle A_{g}h,f\rangle,$

and

	$\displaystyle\langle\Psi_{e}\pi_{g^{-1}}h,f\rangle$	$\displaystyle=\langle\pi_{g^{-1}}h,\Psi_{e}^{}f\rangle=\langle\theta_{\Psi}^{}\lambda_{g^{-1}}\theta_{A}h,\Psi_{e}^{*}f\rangle$
		$\displaystyle=\left\langle\int_{G}\Psi_{\alpha}^{}\lambda_{g^{-1}}\theta_{A}h(\alpha)\,d\mu_{G}(\alpha),\Psi_{e}^{}f\right\rangle=\left\langle\int_{G}\Psi_{\alpha}^{}\theta_{A}h(g\alpha)\,d\mu_{G}(\alpha),\Psi_{e}^{}f\right\rangle$
		$\displaystyle=\left\langle\int_{G}\Psi_{\alpha}^{}A_{g\alpha}h\,d\mu_{G}(\alpha),\Psi_{e}^{}f\right\rangle=\left\langle\int_{G}\Psi_{g^{-1}\beta}^{}A_{\beta}h\,d\mu_{G}(g^{-1}\beta),\Psi_{e}^{}f\right\rangle$
		$\displaystyle=\left\langle\int_{G}\Psi_{g^{-1}\beta}^{}A_{\beta}h\,d\mu_{G}(\beta),\Psi_{e}^{}f\right\rangle=\int_{G}\langle\Psi_{g^{-1}g}\Psi_{g^{-1}\beta}^{*}A_{\beta}h,f\rangle\,d\mu_{G}(\beta)$
		$\displaystyle=\int_{G}\langle\Psi_{g}\Psi_{\beta}^{}A_{\beta}h,f\rangle\,d\mu_{G}(\beta)=\left\langle\int_{G}\Psi_{\beta}^{}A_{\beta}h\,d\mu_{G}(\beta),\Psi_{g}^{*}f\right\rangle$
		$\displaystyle=\langle h,\Psi_{g}^{*}f\rangle=\langle\Psi_{g}h,f\rangle.$

∎

Corollary 4.4.

Let $G$ be a locally compact group with identity $e$ and $(\{A_{g}\}_{g\in G},\{\Psi_{g}\}_{g\in G})$ be a continuous (ovf) in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0}).$ Then there is a unitary representation $\pi$ of $G$ on $\mathcal{H}$ for which

(i)

$A_{g}=A_{e}S_{A,\Psi}^{-1}\pi_{g^{-1}}S_{A,\Psi},\Psi_{g}=\Psi_{e}\pi_{g^{-1}}$ for all $g\in G$ if and only if $A_{gp}S_{A,\Psi}^{-2}A_{gq}^{*}=A_{p}S_{A,\Psi}^{-2}A_{q}^{*},A_{gp}S_{A,\Psi}^{-1}\Psi_{gq}^{*}$ $=A_{p}S_{A,\Psi}^{-1}\Psi_{q}^{*},\Psi_{gp}\Psi_{gq}^{*}=\Psi_{p}\Psi_{q}^{*}$ for all $g,p,q\in G.$
(ii)

$A_{g}=A_{e}S_{A,\Psi}^{-1/2}\pi_{g^{-1}}S_{A,\Psi}^{1/2},\Psi_{g}=\Psi_{e}S_{A,\Psi}^{-1/2}\pi_{g^{-1}}S_{A,\Psi}^{1/2}$ for all $g\in G$ if and only if $A_{gp}S_{A,\Psi}^{-1}A_{gq}^{*}=A_{p}S_{A,\Psi}^{-1}A_{q}^{*},A_{gp}S_{A,\Psi}^{-1}\Psi_{gq}^{*}=A_{p}S_{A,\Psi}^{-1}\Psi_{q}^{*},\Psi_{gp}S_{A,\Psi}^{-1}\Psi_{gq}^{*}=\Psi_{p}S_{A,\Psi}^{-1}\Psi_{q}^{*}$ for all $g,p,q\in G.$
(iii)

$A_{g}=A_{e}\pi_{g^{-1}},\Psi_{g}=\Psi_{e}S_{A,\Psi}^{-1}\pi_{g^{-1}}S_{A,\Psi}$ for all $g\in G$ if and only if $A_{gp}A_{gq}^{*}=A_{p}A_{q}^{*},A_{gp}S_{A,\Psi}^{-1}\Psi_{gq}^{*}=A_{p}S_{A,\Psi}^{-1}\Psi_{q}^{*},\Psi_{gp}S_{A,\Psi}^{-2}\Psi_{gq}^{*}=\Psi_{p}S_{A,\Psi}^{-2}\Psi_{q}^{*}$ for all $g,p,q\in G.$

Proof.

We apply Theorem 4.3 to the Parseval continuous (ovf)

(i)

$(\{A_{g}S_{A,\Psi}^{-1}\}_{g\in G},\{\Psi_{g}\}_{g\in G})$ to get: there is a unitary representation $\pi$ of $G$ on $\mathcal{H}$ for which $A_{g}S_{A,\Psi}^{-1}=(A_{e}S_{A,\Psi}^{-1})\pi_{g^{-1}},\Psi_{g}=\Psi_{e}\pi_{g^{-1}}$ for all $g\in G$ if and only if $(A_{gp}S_{A,\Psi}^{-1})(A_{gq}S_{A,\Psi}^{-1})^{*}=(A_{p}S_{A,\Psi}^{-1})(A_{q}S_{A,\Psi}^{-1})^{*}$ , $(A_{gp}S_{A,\Psi}^{-1})\Psi_{gq}^{*}=(A_{p}S_{A,\Psi}^{-1})\Psi_{q}^{*}$ , $\Psi_{gp}\Psi_{gq}^{*}=\Psi_{p}\Psi_{q}^{*}$ for all $g,p,q\in G.$
(ii)

$(\{A_{g}S_{A,\Psi}^{-1/2}\}_{g\in G},\{\Psi_{g}S_{A,\Psi}^{-1/2}\}_{g\in G})$ to get: there is a unitary representation $\pi$ of $G$ on $\mathcal{H}$ for which $A_{g}S_{A,\Psi}^{-1/2}=(A_{e}S_{A,\Psi}^{-1/2})\pi_{g^{-1}},\Psi_{g}S_{A,\Psi}^{-1/2}=(\Psi_{e}S_{A,\Psi}^{-1/2})\pi_{g^{-1}}$ for all $g\in G$ if and only if $(A_{gp}S_{A,\Psi}^{-1/2})(A_{gq}S_{A,\Psi}^{-1/2})^{*}$ $=(A_{p}S_{A,\Psi}^{-1/2})(A_{q}S_{A,\Psi}^{-1/2})^{*},(A_{gp}S_{A,\Psi}^{-1/2})(\Psi_{gq}S_{A,\Psi}^{-1/2})^{*}=(A_{p}S_{A,\Psi}^{-1/2})(\Psi_{q}S_{A,\Psi}^{-1/2})^{*},$ $(\Psi_{gp}S_{A,\Psi}^{-1/2})(\Psi_{gq}S_{A,\Psi}^{-1/2})^{*}$ $=(\Psi_{p}S_{A,\Psi}^{-1/2})(\Psi_{q}S_{A,\Psi}^{-1/2})^{*}$ for all $g,p,q\in G.$
(iii)

$(\{A_{g}\}_{g\in G},\{\Psi_{g}S_{A,\Psi}^{-1}\}_{g\in G})$ to get: there is a unitary representation $\pi$ of $G$ on $\mathcal{H}$ for which $A_{g}=A_{e}\pi_{g^{-1}},\Psi_{g}S_{A,\Psi}^{-1}=(\Psi_{e}S_{A,\Psi}^{-1})\pi_{g^{-1}}$ for all $g\in G$ if and only if $A_{gp}A_{gq}^{*}=A_{p}A_{q}^{*},A_{gp}(\Psi_{gq}S_{A,\Psi}^{-1})^{*}=A_{p}(\Psi_{q}S_{A,\Psi}^{-1})^{*},(\Psi_{gp}S_{A,\Psi}^{-1})(\Psi_{gq}S_{A,\Psi}^{-1})^{*}$ $=(\Psi_{p}S_{A,\Psi}^{-1})(\Psi_{q}S_{A,\Psi}^{-1})^{*}$ for all $g,p,q\in G.$

∎

Corollary 4.5.

Let $G$ be a locally compact group with identity $e$ and $\{A_{g}\}_{g\in G}$ be a

(i)

Parseval continuous (ovf) (w.r.t. itself) in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ . Then there is a unitary representation $\pi$ of $G$ on $\mathcal{H}$ for which

$A_{g}=A_{e}\pi_{g^{-1}},\quad\forall g\in G$

if and only if

$A_{gp}A_{gq}^{*}=A_{p}A_{q}^{*},\quad\forall g,p,q\in G.$
(ii)

continuous (ovf) (w.r.t. itself) in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ . Then there is a unitary representation $\pi$ of $G$ on $\mathcal{H}$ for which

$A_{g}=A_{e}S_{A,\Psi}^{-1/2}\pi_{g^{-1}}S_{A,\Psi}^{1/2},\quad\forall g\in G$

if and only if

$A_{gp}S_{A,A}^{-1}A_{gq}^{*}=A_{p}S_{A,A}^{-1}A_{q}^{*},\quad\forall g,p,q\in G.$

5. Perturbations

Theorem 5.1.

Let $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ be a continuous (ovf) in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ . Suppose $\{B_{\alpha}\}_{\alpha\in\Omega}$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ is such that

(i)

$\Psi_{\alpha}^{*}B_{\alpha}\geq 0,\forall\alpha\in\Omega$ ,
(ii)

for each $h\in\mathcal{H}$ , the map $\Omega\ni\alpha\mapsto B_{\alpha}h\in\mathcal{H}_{0}$ is measurable,

(iii)

there exist $\alpha,\beta,\gamma\geq 0$ with $\max\{\alpha+\gamma\|\theta_{\Psi}S_{A,\Psi}^{-1}\|,\beta\}<1$ such that

(2)

\left\|\int_{\Omega}(A_{\alpha}^{*}-B_{\alpha}^{*})f(\alpha)\,d\mu(\alpha)\right\|\leq\alpha\left\|\int_{\Omega}A_{\alpha}^{*}f(\alpha)\,d\mu(\alpha)\right\|+\beta\left\|\int_{\Omega}B_{\alpha}^{*}f(\alpha)\,d\mu(\alpha)\right\|+\gamma\|f\|,\quad\forall f\in\mathcal{L}^{2}(\Omega,\mathcal{H}_{0}).

Then $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ is a continuous (ovf) with bounds $\frac{1-(\alpha+\gamma\|\theta_{\Psi}S_{A,\Psi}^{-1}\|)}{(1+\beta)\|S_{A,\Psi}^{-1}\|}$ and $\frac{\|\theta_{\Psi}\|((1+\alpha)\|\theta_{A}\|+\gamma)}{1-\beta}$ .

Proof.

Define $T:\mathcal{L}^{2}(\Omega,\mathcal{H}_{0})\ni f\mapsto\int_{\Omega}B_{\alpha}^{*}f(\alpha)\,d\mu(\alpha)\in\mathcal{H}$ . Then for all $f\in\mathcal{L}^{2}(\Omega,\mathcal{H}_{0})$ ,

	$\displaystyle\\|Tf\\|$	$\displaystyle=\left\\|\int_{\Omega}B_{\alpha}^{}f(\alpha)\,d\mu(\alpha)\right\\|\leq\left\\|\int_{\Omega}(B_{\alpha}^{}-A_{\alpha}^{})f(\alpha)\,d\mu(\alpha)\right\\|+\left\\|\int_{\Omega}A_{\alpha}^{}f(\alpha)\,d\mu(\alpha)\right\\|$
		$\displaystyle\leq(1+\alpha)\left\\|\int_{\Omega}A_{\alpha}^{}f(\alpha)\,d\mu(\alpha)\right\\|+\beta\left\\|\int_{\Omega}B_{\alpha}^{}f(\alpha)\,d\mu(\alpha)\right\\|+\gamma\\|f\\|$
		$\displaystyle=(1+\alpha)\left\\|\theta_{A}^{*}f\right\\|+\beta\left\\|Tf\right\\|+\gamma\\|f\\|.$

Hence

\displaystyle\|Tf\|\leq\frac{1+\alpha}{1-\beta}\left\|\theta_{A}^{*}f\right\|+\frac{\gamma}{1-\beta}\|f\|,\quad f\in\mathcal{L}^{2}(\Omega,\mathcal{H}_{0}).

Thus $T$ is bounded, therefore its adjoint exists, which is $\theta_{B}$ . Inequality (2) now gives

\|\theta_{A}^{*}f-\theta_{B}^{*}f\|\leq\alpha\|\theta_{A}^{*}f\|+\beta\|\theta_{B}^{*}f\|+\gamma\|f\|,\quad\forall f\in\mathcal{L}^{2}(\Omega,\mathcal{H}_{0}).

Other arguments are similar to the corresponding arguments used in the proof of Theorem 7.6 in [36]. (we note that Theorem 1 in [9] (we also refer [30, 7]) was used in the proof of Theorem 7.6 in [36]). ∎

Corollary 5.2.

(i)

$\Psi_{\alpha}^{*}B_{\alpha}\geq 0,\forall\alpha\in\Omega$ ,
(ii)

for each $h\in\mathcal{H}$ , the map $\Omega\ni\alpha\mapsto B_{\alpha}h\in\mathcal{H}_{0}$ is measurable,
(iii)

The map $\Omega\ni\alpha\mapsto\|A_{\alpha}-B_{\alpha}\|\in\mathbb{R}$ is measurable,
(iv)

$r\coloneqq\int_{\Omega}\|A_{\alpha}-B_{\alpha}\|^{2}\,d\mu(\alpha)<\frac{1}{\|\theta_{\Psi}S_{A,\Psi}^{-1}\|^{2}}.$

Then $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ is a continuous (ovf) with bounds $\frac{1-\sqrt{r}\|\theta_{\Psi}S_{A,\Psi}^{-1}\|}{\|S_{A,\Psi}^{-1}\|}$ and ${\|\theta_{\Psi}\|(\|\theta_{A}\|+\sqrt{r})}$ .

Proof.

Set $\alpha=0,\beta=0,\gamma=\sqrt{r}$ . Then $\max\{\alpha+\gamma\|\theta_{\Psi}S_{A,\Psi}^{-1}\|,\beta\}<1$ and

\left\|\int_{\Omega}(A_{\alpha}^{*}-B_{\alpha}^{*})f(\alpha)\,d\mu(\alpha)\right\|\leq\left(\int_{\Omega}\|A_{\alpha}^{*}-B_{\alpha}^{*}\|^{2}\,d\mu(\alpha)\right)^{\frac{1}{2}}\left(\int_{\Omega}\|f(\alpha)\|^{2}\,d\mu(\alpha)\right)^{\frac{1}{2}}=\gamma\|f\|,\quad\forall f\in\mathcal{L}^{2}(\Omega,\mathcal{H}_{0}).

Theorem 5.1 applies now. ∎

Theorem 5.3.

Let $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ be a continuous (ovf) in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ with bounds $a$ and $b$ . Suppose $\{B_{\alpha}\}_{\alpha\in\Omega}$ is continuous Bessel (w.r.t. itself) in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ such that $\theta_{\Psi}^{*}\theta_{B}\geq 0$ and there exist $\alpha,\beta,\gamma\geq 0$ with $\max\{\alpha+\frac{\gamma}{\sqrt{a}},\beta\}<1$ and for all $h\in\mathcal{H}$ ,

(3)

\left|\int_{\Omega}\langle(A_{\alpha}-B_{\alpha})h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)\right|^{\frac{1}{2}}\leq\alpha\left(\int_{\Omega}\langle A_{\alpha}h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)\right)^{\frac{1}{2}}+\beta\left(\int_{\Omega}\langle B_{\alpha}h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)\right)^{\frac{1}{2}}+\gamma\|h\|.

Then $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ is a continuous (ovf) with bounds $a\left(1-\frac{\alpha+\beta+\frac{\gamma}{\sqrt{a}}}{1+\beta}\right)^{2}$ and $b\left(1+\frac{\alpha+\beta+\frac{\gamma}{\sqrt{b}}}{1-\beta}\right)^{2}.$

Proof.

For all $h$ in $\mathcal{H}$ ,

	$\displaystyle\left(\int_{\Omega}\langle B_{\alpha}h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)\right)^{\frac{1}{2}}$	$\displaystyle\leq\left\|\int_{\Omega}\langle(B_{\alpha}-A_{\alpha})h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)\right\|^{\frac{1}{2}}+\left(\int_{\Omega}\langle A_{\alpha}h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)\right)^{\frac{1}{2}}$
		$\displaystyle\leq(1+\alpha)\left(\int_{\Omega}\langle A_{\alpha}h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)\right)^{\frac{1}{2}}+\beta\left(\int_{\Omega}\langle B_{\alpha}h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)\right)^{\frac{1}{2}}+\gamma\\|h\\|$

which implies, for all $h\in\mathcal{H}$ ,

	$\displaystyle(1-\beta)\left(\int_{\Omega}\langle B_{\alpha}h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)\right)^{\frac{1}{2}}$	$\displaystyle\leq(1+\alpha)\left(\int_{\Omega}\langle A_{\alpha}h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)\right)^{\frac{1}{2}}+\gamma\\|h\\|$
		$\displaystyle\leq(1+\alpha)\sqrt{b}\\|h\\|+\gamma\\|h\\|.$

From Inequality (3), for all $h\in\mathcal{H}$ ,

	$\displaystyle\left(\int_{\Omega}\langle A_{\alpha}h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)\right)^{\frac{1}{2}}$	$\displaystyle\leq\left\|\int_{\Omega}\langle(A_{\alpha}-B_{\alpha})h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)\right\|^{\frac{1}{2}}+\left(\int_{\Omega}\langle B_{\alpha}h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)\right)^{\frac{1}{2}}$
		$\displaystyle\leq\alpha\left(\int_{\Omega}\langle A_{\alpha}h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)\right)^{\frac{1}{2}}+(1+\beta)\left(\int_{\Omega}\langle B_{\alpha}h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)\right)^{\frac{1}{2}}+\gamma\\|h\\|$
		$\displaystyle\leq\left(\alpha+\frac{\gamma}{\sqrt{a}}\right)\left(\int_{\Omega}\langle A_{\alpha}h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)\right)^{\frac{1}{2}}+(1+\beta)\left(\int_{\Omega}\langle B_{\alpha}h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)\right)^{\frac{1}{2}}$

which produces

\displaystyle\left(1-\left(\alpha+\frac{\gamma}{\sqrt{a}}\right)\right)\left(\int_{\Omega}\langle A_{\alpha}h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)\right)^{\frac{1}{2}}\leq(1+\beta)\left(\int_{\Omega}\langle B_{\alpha}h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)\right)^{\frac{1}{2}},\quad\forall h\in\mathcal{H}.

But $\sqrt{a}\|h\|\leq\left(\int_{\Omega}\langle A_{\alpha}h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)\right)^{1/2},\forall h\in\mathcal{H}.$ Thus $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ is a continuous (ovf) with bounds $a\left(1-\frac{\alpha+\beta+\frac{\gamma}{\sqrt{a}}}{1+\beta}\right)^{2}$ and $b\left(1+\frac{\alpha+\beta+\frac{\gamma}{\sqrt{b}}}{1-\beta}\right)^{2}.$ ∎

Theorem 5.4.

(i)

$\Psi_{\alpha}^{*}B_{\alpha}\geq 0,\forall\alpha\in\Omega$ ,
(ii)

for each $h\in\mathcal{H}$ , the map $\Omega\ni\alpha\mapsto B_{\alpha}h\in\mathcal{H}_{0}$ is measurable,
(iii)

The map $\Omega\ni\alpha\mapsto\|A_{\alpha}-B_{\alpha}\|\in\mathbb{R}$ is measurable and $\int_{\Omega}\|A_{\alpha}-B_{\alpha}\|\,d\mu(\alpha)\in\mathbb{R}$ ,
(iv)

The map $\Omega\ni\alpha\mapsto\|\Psi_{\alpha}S_{A,\Psi}^{-1}\|\in\mathbb{R}$ is measurable and $\int_{\Omega}\|A_{\alpha}-B_{\alpha}\|\|\Psi_{\alpha}S_{A,\Psi}^{-1}\|\,d\mu(\alpha)\in\mathbb{R}$ ,
(v)

$\int_{\Omega}\|A_{\alpha}-B_{\alpha}\|\|\Psi_{\alpha}S_{A,\Psi}^{-1}\|\,d\mu(\alpha)<1.$

Then $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ is a continuous (ovf) with bounds $\frac{1-\int_{\Omega}\|A_{\alpha}-B_{\alpha}\|\|\Psi_{\alpha}S_{A,\Psi}^{-1}\|\,d\mu(\alpha)}{\|S_{A,\Psi}^{-1}\|}$ and $\|\theta_{\Psi}\|((\int_{\Omega}\|A_{\alpha}-B_{\alpha}\|^{2}\,d\mu(\alpha))^{1/2}+\|\theta_{A}\|)$ .

Proof.

Let $\alpha=(\int_{\Omega}\|A_{\alpha}-B_{\alpha}\|^{2}\,d\mu(\alpha))^{1/2}$ and $\beta=\int_{\Omega}\|A_{\alpha}-B_{\alpha}\|\|\Psi_{\alpha}S_{A,\Psi}^{-1}\|\,d\mu(\alpha)$ . Fix $f\in\mathcal{L}^{2}(\Omega,\mathcal{H}_{0})$ . Then for all $h\in\mathcal{H}$

	$\displaystyle\left\|\int_{\Omega}\langle B_{\alpha}^{*}f(\alpha),h\rangle\,d\mu(\alpha)\right\|$	$\displaystyle\leq\left\|\int_{\Omega}\langle(B_{\alpha}^{}-A_{\alpha}^{})f(\alpha),h\rangle\,d\mu(\alpha)\right\|+\left\|\int_{\Omega}\langle A_{\alpha}^{*}f(\alpha),h\rangle\,d\mu(\alpha)\right\|$
		$\displaystyle=\left\|\int_{\Omega}\langle f(\alpha),(B_{\alpha}-A_{\alpha})h\rangle\,d\mu(\alpha)\right\|+\|\langle\theta_{A}^{*}f,h\rangle\|$
		$\displaystyle\leq\int_{\Omega}\\|f(\alpha)\\|\\|(B_{\alpha}-A_{\alpha})h\\|\,d\mu(\alpha)+\\|f\\|\\|\theta_{A}h\\|$
		$\displaystyle\leq\\|h\\|\int_{\Omega}\\|f(\alpha)\\|\\|B_{\alpha}-A_{\alpha}\\|\,d\mu(\alpha)+\\|f\\|\\|\theta_{A}\\|\\|h\\|$
		$\displaystyle\leq\\|h\\|\left(\int_{\Omega}\\|f(\alpha)\\|^{2}\,d\mu(\alpha)\right)^{\frac{1}{2}}\left(\int_{\Omega}\\|B_{\alpha}-A_{\alpha}\\|^{2}\,d\mu(\alpha)\right)^{\frac{1}{2}}+\\|f\\|\\|\theta_{A}\\|\\|h\\|$
		$\displaystyle=\\|h\\|\\|f\\|\alpha+\\|f\\|\\|\theta_{A}\\|\\|h\\|=(\\|f\\|\alpha+\\|f\\|\\|\theta_{A}\\|)\\|h\\|.$

Hence $\theta_{B}$ exists and $\|\theta_{B}\|=\|\theta_{B}^{*}\|=\sup\{|\langle\theta_{B}^{*}f,h\rangle|:f\in\mathcal{L}^{2}(\Omega,\mathcal{H}_{0}),h\in\mathcal{H},\|f\|\leq 1,\|h\|\leq 1\}\leq\sup\{\|h\|\|f\|\alpha+\|f\|\|\theta_{A}\|\|h\|:f\in\mathcal{L}^{2}(\Omega,\mathcal{H}_{0}),h\in\mathcal{H},\|f\|\leq 1,\|h\|\leq 1\}=\alpha+\|\theta_{A}\|$ . Therefore $S_{B,\Psi}=\theta_{\Psi}^{*}\theta_{B}$ exists and is positive. Now

	$\displaystyle\\|I_{\mathcal{H}}-S_{B,\Psi}S_{A,\Psi}^{-1}\\|$	$\displaystyle=\sup_{h,g\in\mathcal{H},\\|h\\|=1=\\|g\\|}\|\langle(I_{\mathcal{H}}-S_{B,\Psi}S_{A,\Psi}^{-1})h,g\rangle\|$
		$\displaystyle=\sup_{h,g\in\mathcal{H},\\|h\\|=1=\\|g\\|}\left\|\int_{\Omega}\langle A_{\alpha}^{}\Psi_{\alpha}S_{A,\Psi}^{-1}h,g\rangle\,d\mu(\alpha)-\int_{\Omega}\langle B_{\alpha}^{}\Psi_{\alpha}S_{A,\Psi}^{-1}h,g\rangle\,d\mu(\alpha)\right\|$
		$\displaystyle=\sup_{h,g\in\mathcal{H},\\|h\\|=1=\\|g\\|}\left\|\int_{\Omega}\langle(A_{\alpha}^{}-B_{\alpha}^{})\Psi_{\alpha}S_{A,\Psi}^{-1}h,g\rangle\,d\mu(\alpha)\right\|$
		$\displaystyle\leq\sup_{h,g\in\mathcal{H},\\|h\\|=1=\\|g\\|}\int_{\Omega}\|\langle(A_{\alpha}^{}-B_{\alpha}^{})\Psi_{\alpha}S_{A,\Psi}^{-1}h,g\rangle\|\,d\mu(\alpha)$
		$\displaystyle\leq\sup_{h,g\in\mathcal{H},\\|h\\|=1=\\|g\\|}\int_{\Omega}\\|A_{\alpha}^{}-B_{\alpha}^{}\\|\\|\Psi_{\alpha}S_{A,\Psi}^{-1}\\|\\|h\\|\\|g\\|\,d\mu(\alpha)$
		$\displaystyle=\int_{\Omega}\\|A_{\alpha}-B_{\alpha}\\|\\|\Psi_{\alpha}S_{A,\Psi}^{-1}\\|\,d\mu(\alpha)=\beta<1.$

Other arguments are similar to the corresponding arguments used in the proof of Theorem 5.1. ∎

6. Case $\mathcal{H}_{0}=\mathbb{K}$

Definition 6.1.

A set of vectors $\{x_{\alpha}\}_{\alpha\in\Omega}$ in a Hilbert space $\mathcal{H}$ is said to be a continuous frame w.r.t. a set $\{\tau_{\alpha}\}_{\alpha\in\Omega}$ in $\mathcal{H}$ if

(i)

for each $h\in\mathcal{H}$ , both maps $\Omega\ni\alpha\mapsto\langle h,x_{\alpha}\rangle\in\mathbb{K}$ and $\Omega\ni\alpha\mapsto\langle h,\tau_{\alpha}\rangle\in\mathbb{K}$ are measurable,
(ii)

the map (we call as frame operator) $S_{x,\tau}:\mathcal{H}\ni h\mapsto\int_{\Omega}\langle h,x_{\alpha}\rangle\tau_{\alpha}\,d\mu(\alpha)\in\mathcal{H}$ (the integral is in the weak-sense) is a well-defined bounded positive invertible operator,
(iii)

both maps (we call as analysis operator and its adjoint as synthesis operator) $\theta_{x}:\mathcal{H}\ni h\mapsto\theta_{x}h\in\mathcal{L}^{2}(\Omega,\mathbb{K})$ , $\theta_{x}h:\Omega\ni\alpha\mapsto\langle h,x_{\alpha}\rangle\in\mathbb{K}$ , $\theta_{\tau}:\mathcal{H}\ni h\mapsto\theta_{\tau}h\in\mathcal{L}^{2}(\Omega,\mathcal{H}_{0})$ , $\theta_{\tau}h:\Omega\ni\alpha\mapsto\langle h,\tau_{\alpha}\rangle\in\mathbb{K}$ are well-defined bounded linear operators.

We note that $\theta_{x}^{*}:\mathcal{L}^{2}(\Omega,\mathbb{K})\ni f\mapsto\int_{\Omega}f(\alpha)x_{\alpha}\,d\mu(\alpha)\in\mathcal{H}$ , $\theta_{\tau}^{*}:\mathcal{L}^{2}(\Omega,\mathbb{K})\ni f\mapsto\int_{\Omega}f(\alpha)\tau_{\alpha}\,d\mu(\alpha)\in\mathcal{H}$ (both integrals are in the weak-sense). Notions of frame bounds, Parseval frame are similar to the same in Definition 8.1 in [36]. If the condition (ii) is replaced by “the map $S_{x,\tau}:\mathcal{H}\ni h\mapsto\int_{\Omega}\langle h,x_{\alpha}\rangle\tau_{\alpha}\,d\mu(\alpha)\in\mathcal{H}$ is a well-defined bounded positive operator”, then we say $\{x_{\alpha}\}_{\alpha\in\Omega}$ w.r.t. $\{\tau_{\alpha}\}_{\alpha\in\Omega}$ is Bessel. If $\{x_{\alpha}\}_{\alpha\in\Omega}$ is continuous frame (resp. Bessel) w.r.t. $\{\tau_{\alpha}\}_{\alpha\in\Omega}$ , then we write $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is a continuous frame (resp. Bessel).

For fixed $\Omega,\mathcal{H},$ and $\{\tau_{\alpha}\}_{\alpha\in\Omega}$ , the set of all continuous frames for $\mathcal{H}$ w.r.t. $\{\tau_{\alpha}\}_{\alpha\in\Omega}$ is denoted by $\mathscr{F}_{\tau}.$

We note that (ii) in Definition 6.1 implies that there are real $a,b>0$ such that

\displaystyle a\|h\|^{2}\leq\langle S_{x,\tau}h,h\rangle=\left\langle\int_{\Omega}\langle h,x_{\alpha}\rangle\tau_{\alpha}\,d\mu(\alpha),h\right\rangle=\int_{\Omega}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu(\alpha)\leq b\|h\|^{2},\quad\forall h\in\mathcal{H},

and (iii) implies that there exist $c,d>0$ such that

	$\displaystyle\\|\theta_{x}h\\|^{2}=\langle\theta_{x}h,\theta_{x}h\rangle=\int_{\Omega}\langle\theta_{x}h(\alpha),\theta_{x}h(\alpha)\rangle\,d\mu(\alpha)=\int_{\Omega}\|\langle h,x_{\alpha}\rangle\|^{2}\,d\mu(\alpha)\leq c\\|h\\|^{2},\quad\forall h\in\mathcal{H};$
	$\displaystyle\\|\theta_{\tau}h\\|^{2}=\langle\theta_{\tau}h,\theta_{\tau}h\rangle=\int_{\Omega}\langle\theta_{\tau}h(\alpha),\theta_{\tau}h(\alpha)\rangle\,d\mu(\alpha)=\int_{\Omega}\|\langle h,\tau_{\alpha}\rangle\|^{2}\,d\mu(\alpha)\leq d\\|h\\|^{2},\quad\forall h\in\mathcal{H}.$

We note, whenever $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is a continuous frame for $\mathcal{H}$ , then $\overline{\operatorname{span}}\{x_{\alpha}\}_{\alpha\in\Omega}=\mathcal{H}=\overline{\operatorname{span}}\{\tau_{\alpha}\}_{\alpha\in\Omega}.$

Theorem 6.2.

Let $\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega}$ be in $\mathcal{H}$ . Define $A_{\alpha}:\mathcal{H}\ni h\mapsto\langle h,x_{\alpha}\rangle\in\mathbb{K}$ , $\Psi_{\alpha}:\mathcal{H}\ni h\mapsto\langle h,\tau_{\alpha}\rangle\in\mathbb{K},\forall\alpha\in\Omega$ . Then $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is a continuous frame for $\mathcal{H}$ if and only if $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ is a continuous operator-valued frame in $\mathcal{B}(\mathcal{H},\mathbb{K})$ .

Proof.

$\langle\int_{\Omega}\Psi^{*}_{\alpha}A_{\alpha}h\,d\mu(\alpha),g\rangle=\int_{\Omega}\langle\Psi^{*}_{\alpha}A_{\alpha}h,g\rangle\,d\mu(\alpha)=\int_{\Omega}\langle A_{\alpha}h,\Psi_{\alpha}g\rangle\,d\mu(\alpha)=\int_{\Omega}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu(\alpha)=\langle\int_{\Omega}\langle h,x_{\alpha}\rangle\tau_{\alpha}\,d\mu(\alpha),g\rangle,\forall h,g\in\mathcal{H}$ . ∎

Proposition 6.3.

Definition 6.1 holds if and only if there are $a,b,c,d>0$ such that

(i)

for each $h\in\mathcal{H}$ , both maps $\Omega\ni\alpha\mapsto\langle h,x_{\alpha}\rangle\in\mathbb{K}$ , $\Omega\ni\alpha\mapsto\langle h,\tau_{\alpha}\rangle\in\mathbb{K}$ are measurable,
(ii)

$a\|h\|^{2}\leq\int_{\Omega}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu(\alpha)\leq b\|h\|^{2},\forall h\in\mathcal{H},$
(iii)

$\int_{\Omega}|\langle h,x_{\alpha}\rangle|^{2}\,d\mu(\alpha)\leq c\|h\|^{2},\forall h\in\mathcal{H};\int_{\Omega}|\langle h,\tau_{\alpha}\rangle|^{2}\,d\mu(\alpha)\leq d\|h\|^{2},\forall h\in\mathcal{H},$
(iv)

$\int_{\Omega}\langle h,x_{\alpha}\rangle\tau_{\alpha}\,d\mu(\alpha)=\int_{\Omega}\langle h,\tau_{\alpha}\rangle x_{\alpha}\,d\mu(\alpha),\forall h\in\mathcal{H}$ .

If the space is over $\mathbb{C},$ then (iv) can be omitted.

Proposition 6.4.

Let $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ be a continuous frame for $\mathcal{H}$ with upper frame bound $b$ . If for some $\alpha\in\Omega$ we have $\{\alpha\}$ is measurable and $\langle x_{\alpha},x_{\beta}\rangle\langle\tau_{\beta},x_{\alpha}\rangle\geq 0,\forall\beta\in\Omega$ , then $\mu(\{\alpha\})\langle x_{\alpha},\tau_{\alpha}\rangle\leq b$ for that $\alpha.$

Proof.

	$\displaystyle\mu(\{\alpha\})\langle x_{\alpha},x_{\alpha}\rangle\langle\tau_{\alpha},x_{\alpha}\rangle$	$\displaystyle\leq\int_{\{\alpha\}}\langle x_{\alpha},x_{\beta}\rangle\langle\tau_{\beta},x_{\alpha}\rangle\,d\mu(\beta)+\int_{\Omega\setminus\{\alpha\}}\langle x_{\alpha},x_{\beta}\rangle\langle\tau_{\beta},x_{\alpha}\rangle\,d\mu(\beta)$
		$\displaystyle=\int_{\Omega}\langle x_{\alpha},x_{\beta}\rangle\langle\tau_{\beta},x_{\alpha}\rangle\,d\mu(\beta)\leq b\\|x_{\alpha}\\|^{2}.$

∎

Proposition 6.5.

For every $\{x_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}_{\tau}$ ,

(i)

$\theta_{x}^{*}\theta_{x}h=\int_{\Omega}\langle h,x_{\alpha}\rangle x_{\alpha}\,d\mu(\alpha),\theta_{\tau}^{*}\theta_{\tau}h=\int_{\Omega}\langle h,\tau_{\alpha}\rangle\tau_{\alpha}\,d\mu(\alpha),\forall h\in\mathcal{H}.$

(ii)

$S_{x,\tau}=\theta_{\tau}^{*}\theta_{x}=\theta_{x}^{*}\theta_{\tau}.$ In particular,

S_{x,\tau}h=\int_{\Omega}\langle h,x_{\alpha}\rangle\tau_{\alpha}\,d\mu(\alpha)=\int_{\Omega}\langle h,\tau_{\alpha}\rangle x_{\alpha}\,d\mu(\alpha),\quad\forall h\in\mathcal{H}~{}\operatorname{and}~{}

\langle S_{x,\tau}h,g\rangle=\int_{\Omega}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},g\rangle\,d\mu(\alpha)=\int_{\Omega}\langle h,\tau_{\alpha}\rangle\langle x_{\alpha},g\rangle\,d\mu(\alpha),\quad\forall h,g\in\mathcal{H}.

(iii)

Every $h\in\mathcal{H}$ can be written as

	$\displaystyle h$	$\displaystyle=\int_{\Omega}\langle h,S^{-1}_{x,\tau}\tau_{\alpha}\rangle x_{\alpha}\,d\mu(\alpha)=\int_{\Omega}\langle h,\tau_{\alpha}\rangle S^{-1}_{x,\tau}x_{\alpha}\,d\mu(\alpha)$
		$\displaystyle=\int_{\Omega}\langle h,S^{-1}_{x,\tau}x_{\alpha}\rangle\tau_{\alpha}\,d\mu(\alpha)=\int_{\Omega}\langle h,x_{\alpha}\rangle S^{-1}_{x,\tau}\tau_{\alpha}\,d\mu(\alpha).$

(iv)

$(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is Parseval if and only if $\theta_{\tau}^{*}\theta_{x}=I_{\mathcal{H}}.$
(v)

$(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is Parseval if and only if $\theta_{x}\theta_{\tau}^{*}$ is idempotent.
(vi)

$P_{x,\tau}\coloneqq\theta_{x}S_{x,\tau}^{-1}\theta_{\tau}^{*}$ is idempotent.
(vii)

$\theta_{x}$ and $\theta_{\tau}$ are injective and their ranges are closed.
(viii)

$\theta_{x}^{*}$ and $\theta_{\tau}^{*}$ are surjective.

Definition 6.6.

A continuous frame $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ for $\mathcal{H}$ is called a Riesz frame if $P_{x,\tau}=I_{\mathcal{L}^{2}(\Omega,\mathbb{K})}$ .

Proposition 6.7.

A continuous frame $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ for $\mathcal{H}$ is a Riesz continuous frame if and only if $\theta_{x}(\mathcal{H})=\mathcal{L}^{2}(\Omega,\mathbb{K})$ if and only if $\theta_{\tau}(\mathcal{H})=\mathcal{L}^{2}(\Omega,\mathbb{K}).$

Proof.

Similar to the proof of Proposition 8.20 in [36]. ∎

Definition 6.8.

A continuous frame $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ for $\mathcal{H}$ is said to be a dual of a continuous frame $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ for $\mathcal{H}$ if $\theta_{\omega}^{*}\theta_{x}=\theta_{y}^{*}\theta_{\tau}=I_{\mathcal{H}}$ . The ‘continuous frame’ $(\{\widetilde{x}_{\alpha}\coloneqq S_{x,\tau}^{-1}x_{\alpha}\}_{\alpha\in\Omega},\{\widetilde{\tau}_{\alpha}\coloneqq S_{x,\tau}^{-1}\tau_{\alpha}\}_{\alpha\in\Omega})$ , which is a ‘dual’ of $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is called the canonical dual of $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ .

Proposition 6.9.

Let $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ be a continuous frame for $\mathcal{H}.$ If $h\in\mathcal{H}$ has representation $h=\int_{\Omega}f(\alpha)x_{\alpha}\,d\mu(\alpha)=\int_{\Omega}g(\alpha)\tau_{\alpha}\,d\mu(\alpha),$ for some measurable $f,g:\Omega\rightarrow\mathbb{K}$ , then

\int_{\Omega}f(\alpha)\overline{g(\alpha)}\,d\mu(\alpha)=\int_{\Omega}\langle h,\widetilde{\tau}_{\alpha}\rangle\langle\widetilde{x}_{\alpha},h\rangle\,d\mu(\alpha)+\int_{\Omega}(f(\alpha)-\langle h,\widetilde{\tau}_{\alpha}\rangle)(\overline{g(\alpha)}-\langle\widetilde{x}_{\alpha},h\rangle)\,d\mu(\alpha).

Proof.

Right side $=$

	$\displaystyle\int_{\Omega}\langle S_{x,\tau}^{-1}h,\tau_{\alpha}\rangle\langle x_{\alpha},S_{x,\tau}^{-1}h\rangle\,d\mu(\alpha)+\int_{\Omega}f(\alpha)\overline{g(\alpha)}\,d\mu(\alpha)-\int_{\Omega}f(\alpha)\langle x_{\alpha},S_{x,\tau}^{-1}h\rangle\,d\mu(\alpha)$
	$\displaystyle~{}-\int_{\Omega}\overline{g(\alpha)}\langle S_{x,\tau}^{-1}h,\tau_{\alpha}\rangle\,d\mu(\alpha)+\int_{\Omega}\langle S_{x,\tau}^{-1}h,\tau_{\alpha}\rangle\langle x_{\alpha},S_{x,\tau}^{-1}h\rangle\,d\mu(\alpha)$
	$\displaystyle=2\langle S_{x,\tau}S_{x,\tau}^{-1}h,S_{x,\tau}^{-1}h\rangle+\int_{\Omega}f(\alpha)\overline{g(\alpha)}\,d\mu(\alpha)-\left\langle\int_{\Omega}f(\alpha)x_{\alpha}\,d\mu(\alpha),S_{x,\tau}^{-1}h\right\rangle-\left\langle S_{x,\tau}^{-1}h,\int_{\Omega}g(\alpha)\tau_{\alpha}\,d\mu(\alpha)\right\rangle$
	$\displaystyle=2\langle h,S_{x,\tau}^{-1}h\rangle+\int_{\Omega}f(\alpha)\overline{g(\alpha)}\,d\mu(\alpha)-\langle h,S_{x,\tau}^{-1}h\rangle-\langle S_{x,\tau}^{-1}h,h\rangle=\text{Left side.}$

∎

Theorem 6.10.

Let $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ be a continuous frame for $\mathcal{H}$ with frame bounds $a$ and $b.$ Then the following statements are true.

(i)

The canonical dual continuous frame of the canonical dual continuous frame of $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is itself.
(ii)

$\frac{1}{b},\frac{1}{a}$ are frame bounds for the canonical dual of $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega}).$
(iii)

If $a,b$ are optimal frame bounds for $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega}),$ then $\frac{1}{b},\frac{1}{a}$ are optimal frame bounds for its canonical dual.

Proof.

For $h,g\in\mathcal{H},$

	$\displaystyle\left\langle\int_{\Omega}\langle h,\widetilde{x}_{\alpha}\rangle\widetilde{\tau}_{\alpha}\,d\mu(\alpha),g\right\rangle$	$\displaystyle=\int_{\Omega}\left\langle h,S_{x,\tau}^{-1}x_{\alpha}\rangle\langle S_{x,\tau}^{-1}\tau_{\alpha},g\right\rangle\,d\mu(\alpha)$
		$\displaystyle=\left\langle\int_{\Omega}\langle S_{x,\tau}^{-1}h,x_{\alpha}\rangle\tau_{\alpha}\,d\mu(\alpha),S_{x,\tau}^{-1}g\right\rangle=\left\langle S_{x,\tau}S_{x,\tau}^{-1}h,S_{x,\tau}^{-1}g\right\rangle=\left\langle S_{x,\tau}^{-1}h,g\right\rangle.$

Thus the frame operator for the canonical dual $(\{\widetilde{x}_{\alpha}\}_{\alpha\in\Omega},\{\widetilde{\tau}_{\alpha}\}_{\alpha\in\Omega})$ is $S_{x,\tau}^{-1}.$ Therefore, its canonical dual is $(\{S_{x,\tau}S_{x,\tau}^{-1}x_{\alpha}\}_{\alpha\in\Omega},\{S_{x,\tau}S_{x,\tau}^{-1}\tau_{\alpha}\}_{\alpha\in\Omega}).$ Others can be proved as in the earlier consideration ‘continuous operator-valued frame’. ∎

Proposition 6.11.

(i)

$(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ is dual of $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ .
(ii)

$\int_{\Omega}\langle h,x_{\alpha}\rangle\omega_{\alpha}\,d\mu(\alpha)=\int_{\Omega}\langle h,\tau_{\alpha}\rangle y_{\alpha}\,d\mu(\alpha)=h,\forall h\in\mathcal{H}.$

Proof.

$\langle\theta_{\omega}^{*}\theta_{x}h,g\rangle=\langle\theta_{x}h,\theta_{\omega}g\rangle=\int_{\Omega}\theta_{x}h(\alpha)\overline{\theta_{\omega}g(\alpha)}\,d\mu(\alpha)=\int_{\Omega}\langle h,x_{\alpha}\rangle\langle\omega_{\alpha},g\rangle\,d\mu(\alpha)=\langle\int_{\Omega}\langle h,x_{\alpha}\rangle\omega_{\alpha}\,d\mu(\alpha),g\rangle,$ $\forall h,g\in\mathcal{H}$ . Similarly $\langle\theta_{y}^{*}\theta_{\tau}h,g\rangle=\langle\int_{\Omega}\langle h,\tau_{\alpha}\rangle y_{\alpha}\,d\mu(\alpha),g\rangle$ , $\forall h,g\in\mathcal{H}$ . ∎

Theorem 6.12.

If $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is a Riesz continuous frame for $\mathcal{H}$ , then it has unique dual.

Proof.

Let $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ and $(\{z_{\alpha}\}_{\alpha\in\Omega},\{\rho_{\alpha}\}_{\alpha\in\Omega})$ be dual continuous frames of $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ . Then $\theta_{x}^{*}\theta_{\omega}=\theta_{\tau}^{*}\theta_{y}=\theta_{x}^{*}\theta_{\rho}=\theta_{\tau}^{*}\theta_{z}=I_{\mathcal{H}}$ $\Rightarrow$ $\theta_{x}^{*}(\theta_{\omega}-\theta_{\rho})=0=\theta_{\tau}^{*}(\theta_{y}-\theta_{z})$ $\Rightarrow$ $I_{\mathcal{H}}(\theta_{\omega}-\theta_{\rho})=P_{\tau,x}(\theta_{\omega}-\theta_{\rho})=\theta_{\tau}S_{x,\tau}^{-1}\theta_{x}^{*}(\theta_{\omega}-\theta_{\rho})=0=\theta_{x}S_{x,\tau}^{-1}\theta_{\tau}^{*}(\theta_{y}-\theta_{z})=P_{x,\tau}(\theta_{y}-\theta_{z})=I_{\mathcal{H}}(\theta_{y}-\theta_{z})$ $\Rightarrow$ $\theta_{\omega}=\theta_{\rho}$ , $\theta_{y}=\theta_{z}$ $\Rightarrow$ $\langle h,\omega_{\alpha}\rangle=\theta_{\omega}h(\alpha)=\theta_{\rho}h(\alpha)=\langle h,\rho_{\alpha}\rangle$ , $\langle h,y_{\alpha}\rangle=\theta_{y}h(\alpha)=\theta_{z}h(\alpha)=\langle h,z_{\alpha}\rangle$ , $\forall h\in\mathcal{H},\forall\alpha\in\Omega$ $\Rightarrow$ $\omega_{\alpha}=\rho_{\alpha},y_{\alpha}=z_{\alpha},\forall\alpha\in\Omega$ . Hence the dual of $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is unique. ∎

Proposition 6.13.

Let $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ be a continuous frame for $\mathcal{H}$ . If $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ is a dual of $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ , then there exist continuous Bessel $\{z_{\alpha}\}_{\alpha\in\Omega}$ and $\{\rho_{\alpha}\}_{\alpha\in\Omega}$ (w.r.t. themselves) for $\mathcal{H}$ such that $y_{\alpha}=S_{x,\tau}^{-1}x_{\alpha}+z_{\alpha},\omega_{\alpha}=S_{x,\tau}^{-1}\tau_{\alpha}+\rho_{\alpha},\forall\alpha\in\Omega$ , and $\theta_{z}(\mathcal{H})\perp\theta_{\tau}(\mathcal{H}),\theta_{\rho}(\mathcal{H})\perp\theta_{x}(\mathcal{H})$ . Converse holds if $\theta_{\rho}^{*}\theta_{z}\geq 0$ .

Proof.

Similar to the proof of Proposition 8.28 in [36]. ∎

Proposition 6.14.

Let $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ be a continuous frame for $\mathcal{H}$ . Then the bounded left-inverses of

(i)

$\theta_{x}$ are precisely $S_{x,\tau}^{-1}\theta_{\tau}^{*}+U(I_{\mathcal{L}^{2}(\Omega,\mathbb{K})}-\theta_{x}S_{x,\tau}^{-1}\theta_{\tau}^{*})$ , where $U\in\mathcal{B}(\mathcal{L}^{2}(\Omega,\mathbb{K}),\mathcal{H})$ .
(ii)

$\theta_{\tau}$ are precisely $S_{x,\tau}^{-1}\theta_{x}^{*}+V(I_{\mathcal{L}^{2}(\Omega,\mathbb{K})}-\theta_{\tau}S_{x,\tau}^{-1}\theta_{x}^{*})$ , where $V\in\mathcal{B}(\mathcal{L}^{2}(\Omega,\mathbb{K}),\mathcal{H})$ .

Proof.

Similar to the proof of Lemma 8.30 in [36]. ∎

Definition 6.15.

A continuous frame $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ for $\mathcal{H}$ is said to be orthogonal to a continuous frame $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ for $\mathcal{H}$ if $\theta_{\omega}^{*}\theta_{x}=\theta_{y}^{*}\theta_{\tau}=0.$

Proposition 6.16.

Let $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ , $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ be continuous frames for $\mathcal{H}$ . Then the following are equivalent.

(i)

$(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ is orthogonal to $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ .
(ii)

$\int_{\Omega}\langle h,x_{\alpha}\rangle\omega_{\alpha}\,d\mu(\alpha)=0=\int_{\Omega}\langle g,\tau_{\alpha}\rangle y_{\alpha}\,d\mu(\alpha),\forall h\in\mathcal{H}$ .

Proposition 6.17.

Two orthogonal continuous frames have a common dual continuous frame.

Proof.

Similar to the proof of Proposition 8.34 in [36]. ∎

Proposition 6.18.

Let $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ and $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ be two Parseval continuous frames for $\mathcal{H}$ which are orthogonal. If $A,B,C,D\in\mathcal{B}(\mathcal{H})$ are such that $AC^{*}+BD^{*}=I_{\mathcal{H}}$ , then $(\{Ax_{\alpha}+By_{\alpha}\}_{\alpha\in\Omega},\{C\tau_{\alpha}+D\omega_{\alpha}\}_{\alpha\in\Omega})$ is a Parseval continuous frame for $\mathcal{H}$ . In particular, if scalars $a,b,c,d$ satisfy $a\bar{c}+b\bar{d}=1$ , then $(\{ax_{\alpha}+by_{\alpha}\}_{\alpha\in\Omega},\{c\tau_{j}+d\omega_{\alpha}\}_{\alpha\in\Omega})$ is a Parseval continuous frame for $\mathcal{H}$ .

Proof.

For all $h\in\mathcal{H}$ and $\alpha\in\Omega$ we see $\theta_{Ax+By}h(\alpha)=\langle h,Ax_{\alpha}+By_{\alpha}\rangle=\langle A^{*}h,x_{\alpha}\rangle+\langle B^{*}h,y_{\alpha}\rangle=\theta_{x}(A^{*}h)(\alpha)+\theta_{y}(B^{*}h)(\alpha)=(\theta_{x}A^{*}+\theta_{y}B^{*})h(\alpha).$ Similarly $\theta_{C\tau+D\omega}=\theta_{\tau}C^{*}+\theta_{\omega}D^{*}$ . Other arguments are similar to that in the proof of Proposition 8.35 in [36]. ∎

Definition 6.19.

Two continuous frames $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ and $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ for $\mathcal{H}$ are called disjoint if $(\{x_{\alpha}\oplus y_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\oplus\omega_{\alpha}\}_{\alpha\in\Omega})$ is a continuous frame for $\mathcal{H}\oplus\mathcal{H}$ .

Proposition 6.20.

If $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ and $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ are orthogonal continuous frames for $\mathcal{H}$ , then they are disjoint. Further, if both $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ and $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ are Parseval, then $(\{x_{\alpha}\oplus y_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\oplus\omega_{\alpha}\}_{\alpha\in\Omega})$ is Parseval.

Proof.

For all $h\oplus g\in\mathcal{H}\oplus\mathcal{H}$ , $\theta_{x\oplus y}(h\oplus g)(\alpha)=\langle h\oplus g,x_{\alpha}\oplus y_{\alpha}\rangle=\langle h,x_{\alpha}\rangle+\langle g,y_{\alpha}\rangle=\theta_{x}h(\alpha)+\theta_{y}g(\alpha)=(\theta_{x}h+\theta_{y}g)(\alpha)$ and for all $f\in\mathcal{L}^{2}(\Omega,\mathbb{K})$ , $\langle\theta_{\tau\oplus\omega}^{*}f,h\oplus g\rangle=\langle f,\theta_{\tau\oplus\omega}(h\oplus g)\rangle=\langle\theta_{\tau}^{*}f,h\rangle+\langle\theta_{\omega}^{*}f,g\rangle=\langle\theta_{\tau}^{*}f\oplus\theta_{\omega}^{*}f,h\oplus g\rangle.$ Thus $S_{x\oplus y,\tau\oplus\omega}(h\oplus g)=\theta^{*}_{\tau\oplus\omega}\theta_{x\oplus y}(h\oplus g)=\theta^{*}_{\tau\oplus\omega}(\theta_{x}h+\theta_{y}g)=\theta_{\tau}^{*}(\theta_{x}h+\theta_{y}g)\oplus\theta_{\omega}^{*}(\theta_{x}h+\theta_{y}g)=(S_{x,\tau}+0)\oplus(0+S_{y,\omega})=S_{x,\tau}\oplus S_{y,\omega}$ , which is bounded positive invertible with $S_{x\oplus y,\tau\oplus\omega}^{-1}=S_{x,\tau}^{-1}\oplus S_{y,\omega}^{-1}$ . ∎

Characterization

Theorem 6.21.

Let $\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega}$ be in $\mathcal{H}$ such that for each $h\in\mathcal{H}$ , both maps $\Omega\ni\alpha\mapsto\langle h,x_{\alpha}\rangle\in\mathbb{K}$ , $\Omega\ni\alpha\mapsto\langle h,\tau_{\alpha}\rangle\in\mathbb{K}$ are measurable. Then $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is a continuous frame with bounds $a$ and $b$ (resp. Bessel with bound $b$ )

(i)

if and only if

U:\mathcal{L}^{2}(\Omega,\mathbb{K})\ni f\mapsto\int_{\Omega}f(\alpha)x_{\alpha}\,d\mu(\alpha)\in\mathcal{H},~{}\text{and}~{}V:\mathcal{L}^{2}(\Omega,\mathbb{K})\ni g\mapsto\int_{\Omega}g(\alpha)\tau_{\alpha}\,d\mu(\alpha)\in\mathcal{H}

are well-defined, $U,V\in\mathcal{B}(\mathcal{L}^{2}(\Omega,\mathbb{K}),\mathcal{H})$ such that $aI_{\mathcal{H}}\leq VU^{*}\leq bI_{\mathcal{H}}$ (resp. $0\leq VU^{*}\leq bI_{\mathcal{H}}).$

(ii)

if and only if

U:\mathcal{L}^{2}(\Omega,\mathbb{K})\ni f\mapsto\int_{\Omega}f(\alpha)x_{\alpha}\,d\mu(\alpha)\in\mathcal{H},~{}\text{and}~{}S:\mathcal{H}\ni x\mapsto Sx\in\mathcal{L}^{2}(\Omega,\mathbb{K}),~{}Sx:\Omega\ni\alpha\mapsto\langle x,\tau_{\alpha}\rangle\in\mathbb{K}

are well-defined, $U\in\mathcal{B}(\mathcal{L}^{2}(\Omega,\mathbb{K}),\mathcal{H})$ , $S\in\mathcal{B}(\mathcal{H},\mathcal{L}^{2}(\Omega,\mathbb{K}))$ such that $aI_{\mathcal{H}}\leq S^{*}U^{*}\leq bI_{\mathcal{H}}$ (resp. $0\leq S^{*}U^{*}\leq bI_{\mathcal{H}}).$

(iii)

if and only if

R:\mathcal{H}\ni h\mapsto Rh\in\mathcal{L}^{2}(\Omega,\mathbb{K}),Rh:\Omega\ni\alpha\mapsto\langle h,x_{\alpha}\rangle\in\mathbb{K},~{}\text{and}~{}V:\mathcal{L}^{2}(\Omega,\mathbb{K})\ni g\mapsto\int_{\Omega}g(\alpha)\tau_{\alpha}\,d\mu(\alpha)\in\mathcal{H}

are well-defined, $R\in\mathcal{B}(\mathcal{H},\mathcal{L}^{2}(\Omega,\mathbb{K}))$ , $V\in\mathcal{B}(\mathcal{L}^{2}(\Omega,\mathbb{K}),\mathcal{H})$ such that $aI_{\mathcal{H}}\leq VR\leq bI_{\mathcal{H}}$ (resp. $0\leq VR\leq bI_{\mathcal{H}}).$

(iv)

if and only if

R:\mathcal{H}\ni h\mapsto Rh\in\mathcal{L}^{2}(\Omega,\mathbb{K}),Rh:\Omega\ni\alpha\mapsto\langle h,x_{\alpha}\rangle\in\mathbb{K},~{}\text{and}~{}S:\mathcal{H}\ni x\mapsto Sx\in\mathcal{L}^{2}(\Omega,\mathbb{K}),~{}Sx:\Omega\ni\alpha\mapsto\langle x,\tau_{\alpha}\rangle\in\mathbb{K}

are well-defined, $R,S\in\mathcal{B}(\mathcal{H},\mathcal{L}^{2}(\Omega,\mathbb{K}))$ such that $aI_{\mathcal{H}}\leq S^{*}R\leq bI_{\mathcal{H}}$ (resp. $0\leq S^{*}R\leq bI_{\mathcal{H}}).$

Proof.

We prove the first one for continuous Bessel, others are similar.

$(\Rightarrow)$ $U=\theta_{x}^{*}$ , $V=\theta_{\tau}^{*}$ and $VU^{*}=\theta_{\tau}^{*}\theta_{x}=S_{x,\tau}$ . $(\Leftarrow)$ $\theta_{x}=U^{*}$ , $\theta_{\tau}=V^{*}$ and $S_{x,\tau}=\theta_{\tau}^{*}\theta_{x}=VU^{*}$ . ∎

Similarity

Definition 6.22.

A continuous frame $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ for $\mathcal{H}$ is said to be similar to a continuous frame $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ for $\mathcal{H}$ if there are invertible operators $T_{x,y},T_{\tau,\omega}\in\mathcal{B}(\mathcal{H})$ such that $y_{\alpha}=T_{x,y}x_{\alpha},\omega_{\alpha}=T_{\tau,\omega}\tau_{\alpha},\forall\alpha\in\Omega.$

Proposition 6.23.

Let $\{x_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}_{\tau}$ with frame bounds $a,b,$ let $T_{x,y},T_{\tau,\omega}\in\mathcal{B}(\mathcal{H})$ be positive, invertible, commute with each other, commute with $S_{x,\tau}$ , and let $y_{\alpha}=T_{x,y}x_{\alpha},\omega_{\alpha}=T_{\tau,\omega}\tau_{\alpha},\forall\alpha\in\Omega.$ Then

(i)

$\{y_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}_{\tau}$ and $\frac{a}{\|T_{x,y}^{-1}\|\|T_{\tau,\omega}^{-1}\|}\leq S_{y,\omega}\leq b\|T_{x,y}T_{\tau,\omega}\|.$ Assuming that $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is Parseval, then $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ is Parseval if and only if $T_{\tau,\omega}T_{x,y}=I_{\mathcal{H}}.$
(ii)

$\theta_{y}=\theta_{x}T_{x,y},\theta_{\omega}=\theta_{\tau}T_{\tau,\omega},S_{y,\omega}=T_{\tau,\omega}S_{x,\tau}T_{x,y},P_{y,\omega}=P_{x,\tau}.$

Proof.

For all $h,g\in\mathcal{H},$

	$\displaystyle\langle T_{\tau,\omega}S_{x,\tau}T_{x,y}h,g\rangle=\langle S_{x,\tau}T_{x,y}h,T_{\tau,\omega}^{}g\rangle=\int_{\Omega}\langle T_{x,y}h,x_{\alpha}\rangle\langle\tau_{\alpha},T_{\tau,\omega}^{}g\rangle\,d\mu(\alpha)$
	$\displaystyle=\int_{\Omega}\langle h,T_{x,y}x_{\alpha}\rangle\langle T_{\tau,\omega}\tau_{\alpha},g\rangle\,d\mu(\alpha)=\int_{\Omega}\langle h,y_{\alpha}\rangle\langle\omega_{\alpha},g\rangle\,d\mu(\alpha)=\langle S_{y,\omega}h,g\rangle.$

∎

Lemma 6.24.

Let $\{x_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}_{\tau},$ $\{y_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}_{\omega}$ and $y_{\alpha}=T_{x,y}x_{\alpha},\omega_{\alpha}=T_{\tau,\omega}\tau_{\alpha},\forall\alpha\in\Omega$ , for some invertible $T_{x,y},T_{\tau,\omega}\in\mathcal{B}(\mathcal{H}).$ Then $\theta_{y}=\theta_{x}T^{*}_{x,y},\theta_{\omega}=\theta_{\tau}T^{*}_{\tau,\omega},S_{y,\omega}=T_{\tau,\omega}S_{x,\tau}T_{x,y}^{*},P_{y,\omega}=P_{x,\tau}.$ Assuming that $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is Parseval, then $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ is Parseval if and only if $T_{\tau,\omega}T_{x,y}^{*}=I_{\mathcal{H}}.$

Proof.

$\theta_{y}h(\alpha)=\langle h,y_{\alpha}\rangle=\langle h,T_{x,y}x_{\alpha}\rangle=\langle T_{x,y}^{*}h,x_{\alpha}\rangle=\theta_{x}(T_{x,y}^{*}h)(\alpha)$ $\Rightarrow$ $\theta_{y}=\theta_{x}T^{*}_{x,y}$ . Similarly $\theta_{\omega}=\theta_{\tau}T^{*}_{\tau,\omega}$ . ∎

Theorem 6.25.

Let $\{x_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}_{\tau},$ $\{y_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}_{\omega}.$ The following are equivalent.

(i)

$y_{\alpha}=T_{x,y}x_{\alpha},\omega_{\alpha}=T_{\tau,\omega}\tau_{\alpha},\forall\alpha\in\Omega,$ for some invertible $T_{x,y},T_{\tau,\omega}\in\mathcal{B}(\mathcal{H}).$
(ii)

$\theta_{y}=\theta_{x}{T^{\prime}}_{x,y}^{*},\theta_{\omega}=\theta_{\tau}{T^{\prime}}_{\tau,\omega}^{*}$ for some invertible ${T^{\prime}}_{x,y},{T^{\prime}}_{\tau,\omega}\in\mathcal{B}(\mathcal{H}).$
(iii)

$P_{y,\omega}=P_{x,\tau}.$

If one of the above conditions is satisfied, then invertible operators in $\operatorname{(i)}$ and $\operatorname{(ii)}$ are unique and are given by $T_{x,y}=\theta_{y}^{*}\theta_{\tau}S_{x,\tau}^{-1},T_{\tau,\omega}=\theta_{\omega}^{*}\theta_{x}S_{x,\tau}^{-1}.$ In the case that $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is Parseval, then $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ is Parseval if and only if $T_{\tau,\omega}T_{x,y}^{*}=I_{\mathcal{H}}$ if and only if $T_{x,y}^{*}T_{\tau,\omega}=I_{\mathcal{H}}$ .

Proof.

(ii) $\Rightarrow$ (i) For all $h\in\mathcal{H}$ and $\alpha\in\Omega$ , $\langle h,y_{\alpha}\rangle=\theta_{y}h(\alpha)=\theta_{x}({T^{\prime}}_{x,y}^{*}h)(\alpha)=\langle{T^{\prime}}_{x,y}^{*}h,y_{\alpha}\rangle=\langle h,{T^{\prime}}_{x,y}y_{\alpha}\rangle$ which implies $y_{\alpha}={T^{\prime}}_{x,y}x_{\alpha},\omega_{\alpha}={T^{\prime}}_{\tau,\omega}\tau_{\alpha},\forall\alpha\in\Omega$ . Other arguments are similar to that in the proof of Theorem 8.45 in [36]. ∎

Corollary 6.26.

For any given continuous frame $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ , the canonical dual of $(\{x_{\alpha}\}_{\alpha\in\Omega}$ , $\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is the only dual continuous frame that is similar to $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ .

Proof.

Similar to the proof of Corollary 8.46 in [36]. ∎

Corollary 6.27.

Two similar continuous frames cannot be orthogonal.

Proof.

Similar to the proof of Corollary 8.47 in [36]. ∎

Remark 6.28.

For every continuous frame $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega}),$ each of ‘continuous frames’ $(\{S_{x,\tau}^{-1}x_{\alpha}\}_{\alpha\in\Omega}$ , $\{\tau_{\alpha}\}_{\alpha\in\Omega})$ , $(\{S_{x,\tau}^{-1/2}x_{\alpha}\}_{\alpha\in\Omega},\{S_{x,\tau}^{-1/2}\tau_{\alpha}\}_{\alpha\in\Omega}),$ and $(\{x_{\alpha}\}_{\alpha\in\Omega},\{S_{x,\tau}^{-1}\tau_{\alpha}\}_{\alpha\in\Omega})$ is a Parseval continuous frame which is similar to $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega}).$ Hence each continuous frame is similar to Parseval continuous frames.

Continuous frames and representations of locally compact groups

Let $G$ , $d\mu_{G}$ , $\lambda$ , $\rho$ be as in Section 4 and $\mathcal{H}_{0}=\mathbb{K}$ . We denote the von Neumann algebra generated by unitaries $\{\lambda_{g}\}_{g\in G}$ (resp. $\{\rho_{g}\}_{g\in G}$ ) in $\mathcal{B}(\mathcal{L}^{2}(G,\mathbb{K}))$ by $\mathscr{L}(G)$ (resp. $\mathscr{R}(G)$ ). Then $\mathscr{L}(G)^{\prime}=\mathscr{R}(G)$ and $\mathscr{R}(G)^{\prime}=\mathscr{L}(G)$ [49].

Definition 6.29.

Let $\pi$ be a unitary representation of a locally compact group $G$ on a Hilbert space $\mathcal{H}.$ An element $x$ in $\mathcal{H}$ is called a continuous frame generator (resp. a Parseval frame generator) w.r.t. $\tau$ in $\mathcal{H}$ if $(\{x_{g}\coloneqq\pi_{g}x\}_{g\in G},\{\tau_{g}\coloneqq\pi_{g}\tau\}_{g\in G})$ is a continuous frame (resp. Parseval frame) for $\mathcal{H}$ . In this case we write $(x,\tau)$ is a frame generator for $\pi$ .

Proposition 6.30.

Let $(x,\tau)$ and $(y,\omega)$ be frame generators in $\mathcal{H}$ for a unitary representation $\pi$ of $G$ on $\mathcal{H}.$ Then

(i)

$\theta_{x}\pi_{g}=\lambda_{g}\theta_{x},\theta_{\tau}\pi_{g}=\lambda_{g}\theta_{\tau},\forall g\in G.$
(ii)

$\theta_{x}^{*}\theta_{y},\theta_{\tau}^{*}\theta_{\omega},\theta_{x}^{*}\theta_{\omega}$ are in the commutant $\pi(G)^{\prime}$ of $\pi(G)^{\prime\prime}.$ Further, $S_{x,\tau}\in\pi(G)^{\prime}$ and $(S_{x,\tau}^{-1/2}x,S_{x,\tau}^{-1/2}\tau)$ is a Parseval frame generator.
(iii)

$\theta_{x}T\theta_{\tau}^{*},\theta_{x}T\theta_{y}^{*},\theta_{\tau}T\theta_{\omega}^{*}\in\mathscr{R}(G),\forall T\in\pi(G)^{\prime}.$ In particular, $P_{x,\tau}\in\mathscr{R}(G).$

Proof.

(i)

For all $h\in\mathcal{H}$ and $f\in\mathcal{L}^{2}(G,\mathbb{K})$ ,

	$\displaystyle\langle\lambda_{g}\theta_{x}h,f\rangle$	$\displaystyle=\int_{G}\lambda_{g}(\theta_{x}h)(\alpha)\overline{f(\alpha)}\,d\mu_{G}(\alpha)=\int_{G}(\theta_{x}h)(g^{-1}\alpha)\overline{f(\alpha)}\,d\mu_{G}(\alpha)$
		$\displaystyle=\int_{G}\langle h,x_{g^{-1}\alpha}\rangle\overline{f(\alpha)}\,d\mu_{G}(\alpha)=\int_{G}\langle h,\pi_{g^{-1}\alpha}x\rangle\overline{f(\alpha)}\,d\mu_{G}(\alpha)$
		$\displaystyle=\int_{G}\langle\pi_{g}h,\pi_{\alpha}x\rangle\overline{f(\alpha)}\,d\mu_{G}(\alpha)=\int_{G}\langle\pi_{g}h,x_{\alpha}\rangle\overline{f(\alpha)}\,d\mu_{G}(\alpha)$
		$\displaystyle=\int_{G}\theta_{x}(\pi_{g}h)(\alpha)\overline{f(\alpha)}\,d\mu_{G}(\alpha)=\langle\theta_{x}\pi_{g}h,f\rangle.$

(ii)

First part is similar to the proof of (ii) in Proposition 8.51 in [36]. For second, let $h,g\in\mathcal{H}$ . Then

	$\displaystyle\left\langle S_{S_{x,\tau}^{-\frac{1}{2}}x,S_{x,\tau}^{-\frac{1}{2}}\tau}h,g\right\rangle$	$\displaystyle=\int_{G}\langle h,\pi_{\alpha}S_{x,\tau}^{-\frac{1}{2}}x\rangle\langle\pi_{\alpha}S_{x,\tau}^{-\frac{1}{2}}\tau,g\rangle\,d\mu_{G}(\alpha)$
		$\displaystyle=\int_{G}\langle h,S_{x,\tau}^{-\frac{1}{2}}\pi_{\alpha}x\rangle\langle S_{x,\tau}^{-\frac{1}{2}}\pi_{\alpha}\tau,g\rangle\,d\mu_{G}(\alpha)$
		$\displaystyle=\int_{G}\langle S_{x,\tau}^{-\frac{1}{2}}h,\pi_{\alpha}x\rangle\langle\pi_{\alpha}\tau,S_{x,\tau}^{-\frac{1}{2}}g\rangle\,d\mu_{G}(\alpha)$
		$\displaystyle=\langle S_{x,\tau}(S_{x,\tau}^{-\frac{1}{2}}h),S_{x,\tau}^{-\frac{1}{2}}g\rangle=\langle h,g\rangle.$

(iii)

Similar to the proof of (iii) in Proposition 8.51 in [36].

∎

Theorem 6.31.

Let $G$ be a locally compact group with identity $e$ and $(\{x_{g}\}_{g\in G},\{\tau_{g}\}_{g\in G})$ be a Parseval continuous frame for $\mathcal{H}.$ Then there is a unitary representation $\pi$ of $G$ on $\mathcal{H}$ for which

x_{g}=\pi_{g}x_{e},~{}\tau_{g}=\pi_{g}\tau_{e},\quad\forall g\in G

if and only if

\langle x_{gp},x_{gq}\rangle=\langle x_{p},x_{q}\rangle,~{}\langle x_{gp},\tau_{gq}\rangle=\langle x_{p},\tau_{q}\rangle,~{}\langle\tau_{gp},\tau_{gq}\rangle=\langle\tau_{p},\tau_{q}\rangle,\quad\forall g,p,q\in G.

Proof.

Proof 1. $(\Rightarrow)$ Similar to the proof of ‘only if’ part of Theorem 8.52 in [36].

$(\Leftarrow)$ We state the following three, among them we prove third, others are similar.

\lambda_{g}\theta_{x}\theta_{x}^{*}=\theta_{x}\theta_{x}^{*}\lambda_{g},~{}\lambda_{g}\theta_{x}\theta_{\tau}^{*}=\theta_{x}\theta_{\tau}^{*}\lambda_{g},~{}\lambda_{g}\theta_{\tau}\theta_{\tau}^{*}=\theta_{\tau}\theta_{\tau}^{*}\lambda_{g},\quad\forall g\in G.

For all $u,v\in\mathcal{L}^{2}(G,\mathbb{K})$ ,

	$\displaystyle\langle\lambda_{g}\theta_{\tau}\theta_{\tau}^{}\lambda_{g}^{}u,v\rangle$	$\displaystyle=\langle\theta_{\tau}^{}\lambda_{g}^{}u,\theta_{\tau}^{}\lambda_{g}^{}v\rangle=\langle\theta_{\tau}^{}\lambda_{g^{-1}}u,\theta_{\tau}^{}\lambda_{g^{-1}}v\rangle$
		$\displaystyle=\left\langle\int_{G}\lambda_{g^{-1}}u(\alpha)\tau_{\alpha}\,d\mu_{G}(\alpha),\int_{G}\lambda_{g^{-1}}v(\beta)\tau_{\beta}\,d\mu_{G}(\beta)\right\rangle$
		$\displaystyle=\left\langle\int_{G}u(g\alpha)\tau_{\alpha}\,d\mu_{G}(\alpha),\int_{G}v(g\beta)\tau_{\beta}\,d\mu_{G}(\beta)\right\rangle$
		$\displaystyle=\left\langle\int_{G}u(p)\tau_{g^{-1}p}\,d\mu_{G}(g^{-1}p),\int_{G}v(q)\tau_{g^{-1}q}\,d\mu_{G}(g^{-1}q)\right\rangle$
		$\displaystyle=\left\langle\int_{G}u(p)\tau_{g^{-1}p}\,d\mu_{G}(p),\int_{G}v(q)\tau_{g^{-1}q}\,d\mu_{G}(q)\right\rangle$
		$\displaystyle=\int_{G}\int_{G}u(p)\overline{v(q)}\langle\tau_{g^{-1}p},\tau_{g^{-1}q}\rangle\,d\mu_{G}(q)\,d\mu_{G}(p)$
		$\displaystyle=\int_{G}\int_{G}u(p)\overline{v(q)}\langle\tau_{p},\tau_{q}\rangle\,d\mu_{G}(q)\,d\mu_{G}(p)$
		$\displaystyle=\left\langle\int_{G}u(p)\tau_{p}\,d\mu_{G}(p),\int_{G}v(q)\tau_{q}\,d\mu_{G}(q)\right\rangle=\langle\theta_{\tau}^{}u,\theta_{\tau}^{}v\rangle=\langle\theta_{\tau}\theta_{\tau}^{*}u,v\rangle.$

Define $\pi:G\ni g\mapsto\pi_{g}\coloneqq\theta_{\tau}^{*}\lambda_{g}\theta_{x}\in\mathcal{B}(\mathcal{H}).$ Using the Parsevalness of given frame, we get $\pi_{g}\pi_{h}=\theta_{\tau}^{*}\lambda_{g}\theta_{x}\theta_{\tau}^{*}\lambda_{h}\theta_{x}=\theta_{\tau}^{*}\lambda_{g}\lambda_{h}\theta_{x}\theta_{\tau}^{*}\theta_{x}=\theta_{\tau}^{*}\lambda_{g}\lambda_{h}\theta_{x}=\theta_{\tau}^{*}\lambda_{gh}\theta_{x}=\pi_{gh}$ for all $g,h\in G,$ and $\pi_{g}\pi_{g}^{*}=\theta_{\tau}^{*}\lambda_{g}\theta_{x}\theta_{x}^{*}\lambda_{g^{-1}}\theta_{\tau}$ $=\theta_{\tau}^{*}\theta_{x}\theta_{x}^{*}\lambda_{g}\lambda_{g^{-1}}\theta_{\tau}=I_{\mathcal{H}},\pi_{g}^{*}\pi_{g}=\theta_{x}^{*}\lambda_{g^{-1}}\theta_{\tau}\theta_{\tau}^{*}\lambda_{g}\theta_{x}=\theta_{x}^{*}\lambda_{g^{-1}}\lambda_{g}\theta_{\tau}\theta_{\tau}^{*}\theta_{x}=I_{\mathcal{H}}$ for all $g\in G$ . To prove $\pi$ is a unitary representation we use the same idea used in the proof of Theorem 4.3. Let $h\in\mathcal{H}$ be fixed. Then $\theta_{x}h$ is fixed. Since $\lambda$ is a unitary representation, the map $G\ni g\mapsto\lambda_{g}(\theta_{x}h)\in\mathcal{L}^{2}(G,\mathbb{K})$ is continuous. Continuity of $\theta_{\tau}^{*}$ now gives that the map $G\ni g\mapsto\theta_{\tau}^{*}(\lambda_{g}(\theta_{x}h))\in\mathcal{H}$ is continuous. We now establish $x_{g}=\pi_{g}x_{e},\tau_{g}=\pi_{g}\tau_{e}$ for all $g\in G$ . For all $h\in\mathcal{H}$ ,

	$\displaystyle\langle\pi_{g}x_{e},h\rangle$	$\displaystyle=\langle\theta_{\tau}^{*}\lambda_{g}\theta_{x}x_{e},h\rangle$
		$\displaystyle=\left\langle\int_{G}\lambda_{g}\theta_{x}x_{e}(\alpha)\tau_{\alpha}\,d\mu_{G}(\alpha),h\right\rangle=\left\langle\int_{G}\theta_{x}x_{e}(g^{-1}\alpha)\tau_{\alpha}\,d\mu_{G}(\alpha),h\right\rangle$
		$\displaystyle=\int_{G}\langle\theta_{x}x_{e}(g^{-1}\alpha)\tau_{\alpha},h\rangle\,d\mu_{G}(\alpha)=\int_{G}\langle\langle x_{e},x_{g^{-1}\alpha}\rangle\tau_{\alpha},h\rangle\,d\mu_{G}(\alpha)$
		$\displaystyle=\int_{G}\langle x_{g^{-1}g},x_{g^{-1}\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu_{G}(\alpha)=\int_{G}\langle x_{g},x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu_{G}(\alpha)$
		$\displaystyle=\left\langle\int_{G}\langle x_{g},x_{\alpha}\rangle\tau_{\alpha}\,d\mu_{G}(\alpha),h\right\rangle=\langle x_{g},h\rangle,$

and

	$\displaystyle\langle\pi_{g}\tau_{e},h\rangle$	$\displaystyle=\langle\theta_{\tau}^{*}\lambda_{g}\theta_{x}\tau_{e},h\rangle$
		$\displaystyle=\left\langle\int_{G}\lambda_{g}\theta_{x}\tau_{e}(\alpha)\tau_{\alpha}\,d\mu_{G}(\alpha),h\right\rangle=\left\langle\int_{G}\theta_{x}\tau_{e}(g^{-1}\alpha)\tau_{\alpha}\,d\mu_{G}(\alpha),h\right\rangle$
		$\displaystyle=\int_{G}\langle\theta_{x}\tau_{e}(g^{-1}\alpha)\tau_{\alpha},h\rangle\,d\mu_{G}(\alpha)=\int_{G}\langle\langle\tau_{e},x_{g^{-1}\alpha}\rangle\tau_{\alpha},h\rangle\,d\mu_{G}(\alpha)$
		$\displaystyle=\int_{G}\langle\tau_{g^{-1}g},x_{g^{-1}\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu_{G}(\alpha)=\int_{G}\langle\tau_{g},x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu_{G}(\alpha)$
		$\displaystyle=\left\langle\int_{G}\langle\tau_{g},x_{\alpha}\rangle\tau_{\alpha}\,d\mu_{G}(\alpha),h\right\rangle=\langle\tau_{g},h\rangle.$

Proof 2. Define $A_{g}:\mathcal{H}\ni h\mapsto\langle h,x_{g}\rangle\in\mathbb{K}$ , $\Psi_{g}:\mathcal{H}\ni h\mapsto\langle h,\tau_{g}\rangle\in\mathbb{K},\forall g\in G$ . Then, from Theorem 6.2, $(\{x_{g}\}_{g\in G},\{\tau_{g}\}_{g\in G})$ is a continuous frame for $\mathcal{H}$ if and only if $(\{A_{g}\}_{g\in G},\{\Psi_{g}\}_{g\in G})$ is a continuous (ovf) in $\mathcal{B}(\mathcal{H},\mathbb{K})$ . Further, from the proof of Theorem 6.2, we also see that $(\{x_{g}\}_{g\in G},\{\tau_{g}\}_{g\in G})$ is a Parseval continuous frame if and only if $(\{A_{g}\}_{g\in G},\{\Psi_{g}\}_{g\in G})$ is a Parseval continuous (ovf). Now applying Theorem 4.3 to the Parseval continuous (ovf) $(\{A_{g}\}_{g\in G},\{\Psi_{g}\}_{g\in G})$ yields - there is a unitary representation $\pi$ of $G$ on $\mathcal{H}$ for which

(4)

\displaystyle A_{g}=A_{e}\pi_{g^{-1}},~{}\Psi_{g}=\Psi_{e}\pi_{g^{-1}},\quad\forall g\in G

if and only if

(5)

\displaystyle A_{gp}A_{gq}^{*}=A_{p}A_{q}^{*},~{}A_{gp}\Psi_{gq}^{*}=A_{p}\Psi_{q}^{*},~{}\Psi_{gp}\Psi_{gq}^{*}=\Psi_{p}\Psi_{q}^{*},\quad\forall g,p,q\in G.

But Equation (4) holds if and only if

	$\displaystyle\langle h,x_{g}\rangle=A_{g}h=A_{e}\pi_{g^{-1}}h=\langle\pi_{g^{-1}}h,x_{e}\rangle=\langle h,\pi_{g}x_{e}\rangle,$
	$\displaystyle\langle h,\tau_{g}\rangle=\Psi_{g}h=\Psi_{e}\pi_{g^{-1}}h=\langle\pi_{g^{-1}}h,\tau_{g}\rangle=\langle h,\pi_{g}\tau_{e}\rangle,\quad\forall g\in G,\forall h\in\mathcal{H}$
	$\displaystyle\iff x_{g}=\pi_{g}x_{e},~{}\tau_{g}=\pi_{g}\tau_{e},\quad\forall g\in G.$

Also, Equation (5) holds if and only if

	$\displaystyle\langle\alpha x_{gq},x_{gp}\rangle=A_{gp}A_{gq}^{}\alpha=A_{p}A_{q}^{}\alpha=\langle\alpha x_{q},x_{p}\rangle,$
	$\displaystyle\langle\alpha\tau_{gq},x_{gp}\rangle=A_{gp}\Psi_{gq}^{}\alpha=A_{p}\Psi_{q}^{}\alpha=\langle\alpha\tau_{q},x_{p}\rangle,$
	$\displaystyle\langle\alpha\tau_{gq},\tau_{gp}\rangle=\Psi_{gp}\Psi_{gq}^{}\alpha=\Psi_{p}\Psi_{q}^{}\alpha=\langle\alpha\tau_{q},\tau_{p}\rangle,\quad\forall\alpha\in\mathbb{K}$
	$\displaystyle\iff\langle x_{gp},x_{gq}\rangle=\langle x_{p},x_{q}\rangle,~{}\langle x_{gp},\tau_{gq}\rangle=\langle x_{p},\tau_{q}\rangle,~{}\langle\tau_{gp},\tau_{gq}\rangle=\langle\tau_{p},\tau_{q}\rangle,\quad\forall g,p,q\in G.$

∎

Corollary 6.32.

Let $G$ be a locally compact group with identity $e$ and $(\{x_{g}\}_{g\in G},\{\tau_{g}\}_{g\in G})$ be a continuous frame for $\mathcal{H}.$ Then there is a unitary representation $\pi$ of $G$ on $\mathcal{H}$ for which

(i)

$x_{g}=S_{x,\tau}\pi_{g}S_{x,\tau}^{-1}x_{e},\tau_{g}=\pi_{g}\tau_{e}$ for all $g\in G$ if and only if $\langle S_{x,\tau}^{-2}x_{gp},x_{gq}\rangle=\langle S_{x,\tau}^{-2}x_{p},x_{q}\rangle,\langle S_{x,\tau}^{-1}x_{gp},\tau_{gq}\rangle=\langle S_{x,\tau}^{-1}x_{p},\tau_{q}\rangle,\langle\tau_{gp},\tau_{gq}\rangle=\langle\tau_{p},\tau_{q}\rangle$ for all $g,p,q\in G.$
(ii)

$x_{g}=S_{x,\tau}^{1/2}\pi_{g}S_{x,\tau}^{-1/2}x_{e},\tau_{g}=S_{x,\tau}^{1/2}\pi_{g}S_{x,\tau}^{-1/2}\tau_{e}$ for all $g\in G$ if and only if $\langle S_{x,\tau}^{-1}x_{gp},x_{gq}\rangle=\langle S_{x,\tau}^{-1}x_{p},x_{q}\rangle$ , $\langle S_{x,\tau}^{-1}x_{gp},\tau_{gq}\rangle=\langle S_{x,\tau}^{-1}x_{p},\tau_{q}\rangle,\langle S_{x,\tau}^{-1}\tau_{gp},\tau_{gq}\rangle=\langle S_{x,\tau}^{-1}\tau_{p},\tau_{q}\rangle$ for all $g,p,q\in G.$
(iii)

$x_{g}=\pi_{g}x_{e},\tau_{g}=S_{x,\tau}\pi_{g}S_{x,\tau}^{-1}\tau_{e}$ for all $g\in G$ if and only if $\langle x_{gp},x_{gq}\rangle=\langle x_{p},x_{q}\rangle,\langle x_{gp},S_{x,\tau}^{-1}\tau_{gq}\rangle=\langle x_{p},S_{x,\tau}^{-1}\tau_{q}\rangle,\langle\tau_{gp},S_{x,\tau}^{-2}\tau_{gq}\rangle=\langle\tau_{p},S_{x,\tau}^{-2}\tau_{q}\rangle$ for all $g,p,q\in G.$

Proof.

Apply Theorem 6.31 to

(i)

$(\{S^{-1}_{x,\tau}x_{g}\}_{g\in G},\{\tau_{g}\}_{g\in G})$ to get: there is a unitary representation $\pi$ of $G$ on $\mathcal{H}$ for which $S_{x,\tau}^{-1}x_{g}=\pi_{g}S_{x,\tau}^{-1}x_{e},\tau_{g}=\pi_{g}\tau_{e}$ for all $g\in G$ if and only if $\langle S_{x,\tau}^{-1}x_{gp},S_{x,\tau}^{-1}x_{gq}\rangle=\langle S_{x,\tau}^{-1}x_{p},S_{x,\tau}^{-1}x_{q}\rangle,\langle S_{x,\tau}^{-1}x_{gp},\tau_{gq}\rangle=\langle S_{x,\tau}^{-1}x_{p},\tau_{q}\rangle,\langle\tau_{gp},\tau_{gq}\rangle=\langle\tau_{p},\tau_{q}\rangle$ for all $g,p,q\in G.$
(ii)

$(\{S^{-1/2}_{x,\tau}x_{g}\}_{g\in G},\{S^{-1/2}_{x,\tau}\tau_{g}\}_{g\in G})$ to get: there is a unitary representation $\pi$ of $G$ on $\mathcal{H}$ for which $S_{x,\tau}^{-1/2}x_{g}=\pi_{g}(S_{x,\tau}^{-1/2}x_{e}),S_{x,\tau}^{-1/2}\tau_{g}=\pi_{g}(S_{x,\tau}^{-1/2}\tau_{e})$ for all $g\in G$ if and only if $\langle S_{x,\tau}^{-1/2}x_{gp},S_{x,\tau}^{-1/2}x_{gq}\rangle=\langle S_{x,\tau}^{-1/2}x_{p},S_{x,\tau}^{-1/2}x_{q}\rangle,\langle S_{x,\tau}^{-1/2}x_{gp},S_{x,\tau}^{-1/2}\tau_{gq}\rangle=\langle S_{x,\tau}^{-1/2}x_{p},S_{x,\tau}^{-1/2}\tau_{q}\rangle,\langle S_{x,\tau}^{-1/2}\tau_{gp},S_{x,\tau}^{-1/2}\tau_{gq}\rangle=\langle S_{x,\tau}^{-1/2}\tau_{p},$ $S_{x,\tau}^{-1/2}\tau_{q}\rangle$ for all $g,p,q\in G.$
(iii)

$(\{x_{g}\}_{g\in G},\{S^{-1}_{x,\tau}\tau_{g}\}_{g\in G})$ to get: there is a unitary representation $\pi$ of $G$ on $\mathcal{H}$ for which $x_{g}=\pi_{g}x_{e},S_{x,\tau}^{-1}\tau_{g}=\pi_{g}(S_{x,\tau}^{-1}\tau_{e})$ for all $g\in G$ if and only if $\langle x_{gp},x_{gq}\rangle=\langle x_{p},x_{q}\rangle,\langle x_{gp},S_{x,\tau}^{-1}\tau_{gq}\rangle=\langle x_{p},S_{x,\tau}^{-1}\tau_{q}\rangle,\langle S_{x,\tau}^{-1}\tau_{gp},S_{x,\tau}^{-1}\tau_{gq}\rangle=\langle S_{x,\tau}^{-1}\tau_{p},S_{x,\tau}^{-1}\tau_{q}\rangle$ for all $g,p,q\in G.$

∎

Corollary 6.33.

Let $G$ be a locally compact group with identity $e$ and $\{x_{g}\}_{g\in G}$ be a

(i)

Parseval continuous frame (w.r.t. itself) for $\mathcal{H}$ . Then there is a unitary representation $\pi$ of $G$ on $\mathcal{H}$ for which

$x_{g}=\pi_{g}x_{e},\quad\forall g\in G$

if and only if

$\langle x_{gp},x_{gq}\rangle=\langle x_{p},x_{q}\rangle,\quad\forall g,p,q\in G.$
(ii)

continuous frame (w.r.t. itself) for $\mathcal{H}$ . Then there is a unitary representation $\pi$ of $G$ on $\mathcal{H}$ for which

$x_{g}=S_{x,x}^{1/2}\pi_{g}S_{x,x}^{-1/2}x_{e},\quad\forall g\in G$

if and only if

$\langle S_{x,x}^{-1}x_{gp},x_{gq}\rangle=\langle S_{x,x}^{-1}x_{p},x_{q}\rangle,\quad\forall g,p,q\in G.$

Perturbations

Theorem 6.34.

Let $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ be a continuous frame for $\mathcal{H}.$ Suppose $\{y_{\alpha}\}_{\alpha\in\Omega}$ in $\mathcal{H}$ is such that

(i)

$\langle h,y_{\alpha}\rangle\tau_{\alpha}=\langle h,\tau_{\alpha}\rangle y_{\alpha},\langle h,y_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\geq 0,\forall h\in\mathcal{H},\forall\alpha\in\Omega$ ,
(ii)

for each $h\in\mathcal{H}$ , the map $\Omega\ni\alpha\mapsto\langle h,y_{\alpha}\rangle\in\mathbb{K}$ is measurable,

(iii)

there exist $\alpha,\beta,\gamma\geq 0$ with $\max\{\alpha+\gamma\|\theta_{\tau}S_{x,\tau}^{-1}\|,\beta\}<1$ such that

\displaystyle\left\|\int_{\Omega}f(\alpha)(x_{\alpha}-y_{\alpha})\,d\mu(\alpha)\right\|\leq\alpha\left\|\int_{\Omega}f(\alpha)x_{\alpha}\,d\mu(\alpha)\right\|+\beta\left\|\int_{\Omega}f(\alpha)y_{\alpha}\,d\mu(\alpha)\right\|+\gamma\|f\|,\quad\forall f\in\mathcal{L}^{2}(\Omega,\mathbb{K}).

Then $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is a continuous frame with bounds $\frac{1-(\alpha+\gamma\|\theta_{\tau}S_{x,\tau}^{-1}\|)}{(1+\beta)\|S_{x,\tau}^{-1}\|}$ and $\frac{\|\theta_{\tau}\|((1+\alpha)\|\theta_{x}\|+\gamma)}{1-\beta}$ .

Proof.

Define $T:\mathcal{L}^{2}(\Omega,\mathbb{K})\ni f\mapsto\int_{\Omega}f(\alpha)y_{\alpha}\,d\mu(\alpha)\in\mathcal{H}$ . Then for all $f\in\mathcal{L}^{2}(\Omega,\mathbb{K})$ ,

	$\displaystyle\\|Tf\\|$	$\displaystyle=\left\\|\int_{\Omega}f(\alpha)y_{\alpha}\,d\mu(\alpha)\right\\|\leq\left\\|\int_{\Omega}f(\alpha)(y_{\alpha}-x_{\alpha})\,d\mu(\alpha)\right\\|+\left\\|\int_{\Omega}f(\alpha)x_{\alpha}\,d\mu(\alpha)\right\\|$
		$\displaystyle=(1+\alpha)\left\\|\int_{\Omega}f(\alpha)x_{\alpha}\,d\mu(\alpha)\right\\|+\beta\left\\|\int_{\Omega}f(\alpha)y_{\alpha}\,d\mu(\alpha)\right\\|+\gamma\\|f\\|$
		$\displaystyle=(1+\alpha)\left\\|\theta_{x}^{*}f\right\\|+\beta\left\\|Tf\right\\|+\gamma\\|f\\|,$

which implies

\displaystyle\|Tf\|\leq\frac{1+\alpha}{1-\beta}\left\|\theta_{x}^{*}f\right\|+\frac{\gamma}{1-\beta}\|f\|,\quad\forall f\in\mathcal{L}^{2}(\Omega,\mathbb{K})\text{ and }

\|\theta_{x}^{*}f-\theta_{y}^{*}f\|\leq\alpha\|\theta_{x}^{*}f\|+\beta\|\theta_{y}^{*}f\|+\gamma\|f\|,\quad\forall f\in\mathcal{L}^{2}(\Omega,\mathbb{K}).

Other arguments are similar to the corresponding arguments used in the proof of Theorem 5.1. ∎

Corollary 6.35.

Let $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ be a continuous frame for $\mathcal{H}.$ Suppose $\{y_{\alpha}\}_{\alpha\in\Omega}$ in $\mathcal{H}$ is such that

(i)

$\langle h,y_{\alpha}\rangle\tau_{\alpha}=\langle h,\tau_{\alpha}\rangle y_{\alpha},\langle h,y_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\geq 0,\forall h\in\mathcal{H},\forall\alpha\in\Omega$ ,
(ii)

for each $h\in\mathcal{H}$ , the map $\Omega\ni\alpha\mapsto\langle h,y_{\alpha}\rangle\in\mathbb{K}$ is measurable,
(iii)

The map $\Omega\ni\alpha\mapsto\|x_{\alpha}-y_{\alpha}\|\in\mathbb{R}$ is measurable,
(iv)

$r\coloneqq\int_{\Omega}\|x_{\alpha}-y_{\alpha}\|^{2}\,d\mu(\alpha)<\frac{1}{\|\theta_{\tau}S_{x,\tau}^{-1}\|^{2}}.$

Then $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is a continuous frame with bounds $\frac{1-\sqrt{r}\|\theta_{\tau}S_{x,\tau}^{-1}\|}{\|S_{x,\tau}^{-1}\|}$ and $\|\theta_{\tau}\|(\|\theta_{x}\|+\sqrt{r})$ .

Proof.

Take $\alpha=0,\beta=0,\gamma=\sqrt{r}$ . Then $\max\{\alpha+\gamma\|\theta_{\tau}S_{x,\tau}^{-1}\|,\beta\}<1$ and

\left\|\int_{\Omega}f(\alpha)(x_{\alpha}-y_{\alpha})\,d\mu(\alpha)\right\|\leq\left(\int_{\Omega}|f(\alpha)|^{2}\,d\mu(\alpha)\right)^{\frac{1}{2}}\left(\int_{\Omega}\|x_{\alpha}-y_{\alpha}\|^{2}\,d\mu(\alpha)\right)^{\frac{1}{2}}\leq\gamma\|f\|,\quad\forall f\in\mathcal{L}^{2}(\Omega,\mathbb{K}).

Now we can apply Theorem 6.34. ∎

Theorem 6.36.

Let $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ be a continuous frame for $\mathcal{H}$ with bounds $a$ and $b.$ Suppose $\{y_{\alpha}\}_{\alpha\in\Omega}$ in $\mathcal{H}$ is such that $\int_{\Omega}\langle h,y_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu(\alpha)$ exists for all $h\in\mathcal{H}$ and is nonnegative for all $h\in\mathcal{H}$ and there exist $\alpha,\beta,\gamma\geq 0$ with $\max\{\alpha+\frac{\gamma}{\sqrt{a}},\beta\}<1$ and for all $h\in\mathcal{H}$ ,

\displaystyle\left|\int_{\Omega}\langle h,x_{\alpha}-y_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu(\alpha)\right|^{\frac{1}{2}}\leq\alpha\left(\int_{\Omega}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu(\alpha)\right)^{\frac{1}{2}}+\beta\left(\int_{\Omega}\langle h,y_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu(\alpha)\right)^{\frac{1}{2}}+\gamma\|h\|.

Then $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is a continuous frame with bounds $a\left(1-\frac{\alpha+\beta+\frac{\gamma}{\sqrt{a}}}{1+\beta}\right)^{2}$ and $b\left(1+\frac{\alpha+\beta+\frac{\gamma}{\sqrt{b}}}{1-\beta}\right)^{2}.$

Proof.

Similar to the proof of Theorem 5.3. ∎

Theorem 6.37.

Let $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ be a continuous frame for $\mathcal{H}$ . Suppose $\{y_{\alpha}\}_{\alpha\in\Omega}$ in $\mathcal{H}$ is such that

(i)

$\langle h,y_{\alpha}\rangle\tau_{\alpha}=\langle h,\tau_{\alpha}\rangle y_{\alpha},\langle h,y_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\geq 0,\forall h\in\mathcal{H},\forall\alpha\in\Omega$ ,
(ii)

for each $h\in\mathcal{H}$ , the map $\Omega\ni\alpha\mapsto\langle h,y_{\alpha}\rangle\in\mathbb{K}$ is measurable,
(iii)

The map $\Omega\ni\alpha\mapsto\|x_{\alpha}-y_{\alpha}\|\in\mathbb{R}$ is measurable and $\int_{\Omega}\|x_{\alpha}-y_{\alpha}\|\,d\mu(\alpha)\in\mathbb{R}$ ,
(iv)

The map $\Omega\ni\alpha\mapsto\|S_{x,\tau}^{-1}\tau_{\alpha}\|\in\mathbb{R}$ is measurable and $\int_{\Omega}\|x_{\alpha}-y_{\alpha}\|\|S_{x,\tau}^{-1}\tau_{\alpha}\|\,d\mu(\alpha)\in\mathbb{R}$ ,
(v)

$\int_{\Omega}\|x_{\alpha}-y_{\alpha}\|\|S_{x,\tau}^{-1}\tau_{\alpha}\|\,d\mu(\alpha)<1.$

Then $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is a continuous frame with bounds $\frac{1-\int_{\Omega}\|x_{\alpha}-y_{\alpha}\|\|S_{x,\tau}^{-1}\tau_{\alpha}\|\,d\mu(\alpha)}{\|S_{x,\tau}^{-1}\|}$ and $\|\theta_{\tau}\|((\int_{\Omega}\|x_{\alpha}-y_{\alpha}\|^{2}\,d\mu(\alpha))^{1/2}+\|\theta_{x}\|)$ .

Proof.

Let $\alpha=(\int_{\Omega}\|x_{\alpha}-y_{\alpha}\|^{2}\,d\mu(\alpha))^{1/2}$ , $\beta=\int_{\Omega}\|x_{\alpha}-y_{\alpha}\|\|S_{x,\tau}^{-1}\tau_{\alpha}\|\,d\mu(\alpha)$ . Fix $f\in\mathcal{L}^{2}(\Omega,\mathbb{K})$ . Then for all $h\in\mathcal{H}$ ,

	$\displaystyle\left\|\int_{\Omega}\langle f(\alpha)y_{\alpha},h\rangle\,d\mu(\alpha)\right\|$	$\displaystyle\leq\left\|\int_{\Omega}\langle f(\alpha)(y_{\alpha}-x_{\alpha}),h\rangle\,d\mu(\alpha)\right\|+\left\|\int_{\Omega}\langle f(\alpha)x_{\alpha},h\rangle\,d\mu(\alpha)\right\|$
		$\displaystyle=\left\|\int_{\Omega}\langle f(\alpha)(y_{\alpha}-x_{\alpha}),h\rangle\,d\mu(\alpha)\right\|+\|\langle\theta_{x}^{*}f,h\rangle\|$
		$\displaystyle\leq\int_{\Omega}\|f(\alpha)\|\\|y_{\alpha}-x_{\alpha}\\|\\|h\\|\,d\mu(\alpha)+\\|f\\|\\|\theta_{x}h\\|$
		$\displaystyle\leq\\|h\\|\int_{\Omega}\|f(\alpha)\|\\|y_{\alpha}-x_{\alpha}\\|\,d\mu(\alpha)+\\|f\\|\\|\theta_{x}\\|\\|h\\|$
		$\displaystyle\leq\\|h\\|\left(\int_{\Omega}\|f(\alpha)\|^{2}\,d\mu(\alpha)\right)^{\frac{1}{2}}\left(\int_{\Omega}\\|y_{\alpha}-x_{\alpha}\\|^{2}\,d\mu(\alpha)\right)^{\frac{1}{2}}+\\|f\\|\\|\theta_{x}\\|\\|h\\|$
		$\displaystyle=\\|h\\|\\|f\\|\alpha+\\|f\\|\\|\theta_{x}\\|\\|h\\|=(\\|f\\|\alpha+\\|f\\|\\|\theta_{x}\\|)\\|h\\|$

and hence

	$\displaystyle\\|I_{\mathcal{H}}-S_{y,\tau}S_{x,\tau}^{-1}\\|$	$\displaystyle=\sup_{h,g\in\mathcal{H},\\|h\\|=1=\\|g\\|}\|\langle(I_{\mathcal{H}}-S_{y,\tau}S_{x,\tau}^{-1})h,g\rangle\|$
		$\displaystyle=\sup_{h,g\in\mathcal{H},\\|h\\|=1=\\|g\\|}\left\|\int_{\Omega}\langle S_{x,\tau}^{-1}h,\tau_{\alpha}\rangle\langle x_{\alpha},g\rangle\,d\mu(\alpha)-\int_{\Omega}\langle S_{x,\tau}^{-1}h,\tau_{\alpha}\rangle\langle y_{\alpha},g\rangle\,d\mu(\alpha)\right\|$
		$\displaystyle=\sup_{h,g\in\mathcal{H},\\|h\\|=1=\\|g\\|}\left\|\int_{\Omega}\langle S_{x,\tau}^{-1}h,\tau_{\alpha}\rangle\langle x_{\alpha}-y_{\alpha},g\rangle\,d\mu(\alpha)\right\|$
		$\displaystyle=\sup_{h,g\in\mathcal{H},\\|h\\|=1=\\|g\\|}\left\|\int_{\Omega}\langle h,S_{x,\tau}^{-1}\tau_{\alpha}\rangle\langle x_{\alpha}-y_{\alpha},g\rangle\,d\mu(\alpha)\right\|$
		$\displaystyle\leq\sup_{h,g\in\mathcal{H},\\|h\\|=1=\\|g\\|}\int_{\Omega}\\|x_{\alpha}-y_{\alpha}\\|\\|S_{x,\tau}^{-1}\tau_{\alpha}\\|\\|h\\|\\|g\\|\,d\mu(\alpha)$
		$\displaystyle=\int_{\Omega}\\|x_{\alpha}-y_{\alpha}\\|\\|S_{x,\tau}^{-1}\tau_{\alpha}\\|\,d\mu(\alpha)=\beta<1.$

Other arguments are similar to the corresponding arguments in the proof of Theorem 5.4.

∎

7. The finite dimensional case

Theorem 7.1.

Let $\mathcal{H}$ be a finite dimensional Hilbert space, $G$ be a locally compact group, $\{x_{\alpha}\}_{\alpha\in G},\{\tau_{\alpha}\}_{\alpha\in G}$ be a set of vectors in $\mathcal{H}$ such that

(i)

$\langle h,x_{\alpha}\rangle\tau_{\alpha}=\langle h,\tau_{\alpha}\rangle x_{\alpha},\forall h\in\mathcal{H},\forall\alpha\in G$ .
(ii)

$\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\geq 0,\forall h\in\mathcal{H},\forall\alpha\in G.$
(iii)

for each $h\in\mathcal{H}$ , both maps $G\ni\alpha\mapsto\langle h,x_{\alpha}\rangle\in\mathbb{K}$ , $G\ni\alpha\mapsto\langle h,\tau_{\alpha}\rangle\in\mathbb{K}$ are measurable.
(iv)

the map $G\ni\alpha\mapsto\|x_{\alpha}\|\|\tau_{\alpha}\|\in\mathbb{K}$ is in $\mathcal{L}^{2}(G,\mathbb{K})$ .
(v)

for each $h\in\mathcal{H}$ , the map $G\ni\alpha\mapsto\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\in\mathbb{K}$ is continuous.

Then $(\{x_{\alpha}\}_{\alpha\in G},\{\tau_{\alpha}\}_{\alpha\in G})$ is a continuous frame for $\mathcal{H}$ if and only if for every pair $G_{x},G_{\tau}$ of subsets of $G$ satisfying $G_{x}\cap G_{\tau}=\emptyset$ and $G_{x}\cup G_{\tau}=G$ one has $\operatorname{span}\{x_{\alpha},\tau_{\beta}:\alpha\in G_{x},\beta\in G_{\tau}\}=\mathcal{H}.$

Proof.

We can assume $\mathcal{H}\neq\{0\}$ .

$(\Leftarrow)$ There exists $\alpha\in G$ such that $x_{\alpha}\neq 0\neq\tau_{\alpha}$ (else $S_{x,\tau}$ is zero). Hence $\int_{G}\|x_{\alpha}\|\|\tau_{\alpha}\|\,d\mu_{G}(\alpha)>0$ . Clearly $S_{x,\tau}$ is self-adjoint and positive. Now

\displaystyle\int_{G}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu_{G}(\alpha)\leq\int_{G}|\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle|\,d\mu_{G}(\alpha)\leq\left(\int_{G}\|x_{\alpha}\|\|\tau_{\alpha}\|\,d\mu_{G}(\alpha)\right)\|h\|^{2}.

Hence the upper frame bound condition is satisfied. Define $\phi:\mathcal{H}\ni h\mapsto\int_{G}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu(\alpha)\in\mathbb{R}.$ We argue that $\phi$ is continuous. Let $h_{n}\to h$ in $\mathcal{H}$ as $n\to\infty.$ Then

	$\displaystyle\|\phi(h_{n})-\phi(h)\|$	$\displaystyle=\left\|\int_{G}\langle h_{n},x_{\alpha}\rangle\langle\tau_{\alpha},h_{n}\rangle\,d\mu_{G}(\alpha)-\int_{G}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu_{G}(\alpha)\right\|$
		$\displaystyle=\bigg{\|}\int_{G}\langle h_{n},x_{\alpha}\rangle\langle\tau_{\alpha},h_{n}\rangle\,d\mu_{G}(\alpha)-\int_{G}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h_{n}\rangle\,d\mu_{G}(\alpha)$
		$\displaystyle\quad+\int_{G}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h_{n}\rangle\,d\mu_{G}(\alpha)-\int_{G}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu_{G}(\alpha)\bigg{\|}$
		$\displaystyle=\left\|\int_{G}\langle h_{n}-h,x_{\alpha}\rangle\langle\tau_{\alpha},h_{n}\rangle\,d\mu_{G}(\alpha)+\int_{G}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h_{n}-h\rangle\,d\mu_{G}(\alpha)\right\|$
		$\displaystyle\leq\int_{G}\\|h_{n}-h\\|\\|x_{\alpha}\\|\\|\\|\tau_{\alpha}\\|\\|h_{n}\\|\,d\mu_{G}(\alpha)+\int_{G}\\|h\\|\\|x_{\alpha}\\|\\|\tau_{\alpha}\\|\\|h_{n}-h\\|\,d\mu_{G}(\alpha)$
		$\displaystyle\leq\left(\left(\sup_{n\in\mathbb{N}}\\|h_{n}\\|+\\|h\\|\right)\int_{G}\\|x_{\alpha}\\|\\|\tau_{\alpha}\\|\,d\mu_{G}(\alpha)\right)\\|h_{n}-h\\|\to 0\text{ as }n\to\infty.$

Compactness of the unit sphere of $\mathcal{H}$ gives the existence of $g\in\mathcal{H}$ with $\|g\|=1$ such that $a\coloneqq\int_{G}\langle g,x_{\alpha}\rangle\langle\tau_{\alpha},g\rangle\,d\mu_{G}(\alpha)=\inf\{\int_{G}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu_{G}(\alpha):h\in\mathcal{H},\|h\|=1\}.$ We claim that $a>0.$ If this fails: since $\langle g,x_{\alpha}\rangle\langle\tau_{\alpha},g\rangle\geq 0,\forall\alpha\in G$ , $G$ is a locally compact group and the map $G\ni\alpha\mapsto\langle g,x_{\alpha}\rangle\langle\tau_{\alpha},g\rangle\in\mathbb{K}$ is continuous, from [28] we must have $\langle g,x_{\alpha}\rangle\langle\tau_{\alpha},g\rangle=0,\forall\alpha\in G.$ Define $G_{x}\coloneqq\{\alpha\in G:\langle g,x_{\alpha}\rangle=0\}$ and $G_{\tau}\coloneqq\{\alpha\in G:\langle\tau_{\alpha},g\rangle=0\}\setminus G_{x}$ . Now using $\langle g,x_{\alpha}\rangle\langle\tau_{\alpha},g\rangle=0,\forall\alpha\in G$ we get $G_{x}\cup G_{\tau}=G$ . Clearly $G_{x}\cap G_{\tau}=\emptyset$ . Then $g\perp\operatorname{span}\{x_{\alpha},\tau_{\beta}:\alpha\in G_{x},\beta\in G_{\tau}\}=\mathcal{H}$ which implies $g=0$ which is forbidden. We claim that $a$ is a lower frame bound. For all nonzero $h\in\mathcal{H}$ ,

\displaystyle a\|h\|^{2}\leq\left(\int_{G}\left\langle\frac{h}{\|h\|},x_{\alpha}\right\rangle\left\langle\tau_{\alpha},\frac{h}{\|h\|}\right\rangle\,d\mu_{G}(\alpha)\right)\|h\|^{2}=\int_{G}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu_{G}(\alpha).

$(\Rightarrow)$ We prove by contrapositive. Suppose there are subsets $G_{x},G_{\tau}$ of $G$ satisfying $G_{x}\cap G_{\tau}=\emptyset$ and $G_{x}\cup G_{\tau}=G$ such that $\operatorname{span}\{x_{\alpha},\tau_{\beta}:\alpha\in G_{x},\beta\in G_{\tau}\}\subsetneq\mathcal{H}.$ Let $h\in\mathcal{H}$ be nonzero such that $h\perp\operatorname{span}\{x_{\alpha},\tau_{\beta}:\alpha\in G_{x},\beta\in G_{\tau}\}.$ Now because of $G_{x}\cap G_{\tau}=\emptyset$ and $G_{x}\cup G_{\tau}=G$ we get $\int_{G}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu_{G}(\alpha)=0$ which says that the lower frame bound condition fails. ∎

Proposition 7.2.

Let $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ be a continuous frame for $\mathcal{H}$ with a lower frame bound $a$ and $\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\geq 0,\forall h\in\mathcal{H},\forall\alpha\in\Omega.$ If $\Delta$ is any measurable subset of $\Omega$ such that $\Delta\ni\alpha\mapsto\|x_{\alpha}\|\|\tau_{\alpha}\|\in\mathbb{K}$ is measurable and $\int_{\Delta}\|x_{\alpha}\|\|\tau_{\alpha}\|\,d\mu(\alpha)<a,$ then $(\{x_{\alpha}\}_{\alpha\in\Omega\setminus\Delta},\{\tau_{\alpha}\}_{\alpha\in\Omega\setminus\Delta})$ is a continuous frame for $\mathcal{H}$ with lower frame bound $a-\int_{\Delta}\|x_{\alpha}\|\|\tau_{\alpha}\|\,d\mu(\alpha)$ .

Proof.

	$\displaystyle a\\|h\\|^{2}$	$\displaystyle\leq\int_{\Omega}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu(\alpha)=\int_{\Delta}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu(\alpha)+\int_{\Omega\setminus\Delta}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu(\alpha)$
		$\displaystyle\leq\\|h\\|^{2}\int_{\Delta}\\|x_{\alpha}\\|\\|\tau_{\alpha}\\|\,d\mu(\alpha)+\int_{\Omega\setminus\Delta}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu(\alpha),\quad\forall h\in\mathcal{H}.$

∎

Theorem 7.3.

Let $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ be a continuous frame for a finite dimensional complex Hilbert space $\mathcal{H}$ of dimension $m$ . Then we have the following.

(i)

The optimal lower frame bound (resp. optimal upper frame bound) is the smallest (resp. largest) eigenvalue for $S_{x,\tau}.$

(ii)

If $\{\lambda_{j}\}_{j=1}^{m}$ denotes the eigenvalues for $S_{x,\tau},$ each appears as many times as its algebraic multiplicity, then

\sum_{j=1}^{m}\lambda_{j}=\int_{\Omega}\langle x_{\alpha},\tau_{\alpha}\rangle\,d\mu(\alpha)=\int_{\Omega}\langle\tau_{\alpha},x_{\alpha}\rangle\,d\mu(\alpha).

(iii)

Condition number for $S_{x,\tau}$ is equal to the ratio between the optimal upper frame bound and the optimal lower frame bound.

(iv)

If the optimal upper frame bound is $b,$ then

b\leq\int_{\Omega}\langle x_{\alpha},\tau_{\alpha}\rangle\,d\mu(\alpha)=\int_{\Omega}\langle\tau_{\alpha},x_{\alpha}\rangle\,d\mu(\alpha)\leq mb.

(v)

	$\displaystyle\operatorname{Trace}(S_{x,\tau})=\int_{\Omega}\langle x_{\alpha},\tau_{\alpha}\rangle\,d\mu(\alpha)=\int_{\Omega}\langle\tau_{\alpha},x_{\alpha}\rangle\,d\mu(\alpha);$
	$\displaystyle\operatorname{Trace}(S^{2}_{x,\tau})=\int_{\Omega}\int_{\Omega}\langle\tau_{\alpha},x_{\beta}\rangle\langle\tau_{\beta},x_{\alpha}\rangle\,d\mu(\beta)\,d\mu(\alpha)=\int_{\Omega}\int_{\Omega}\langle\tau_{\alpha},\tau_{\beta}\rangle\langle x_{\beta},x_{\alpha}\rangle\,d\mu(\beta)\,d\mu(\alpha).$

(vi)

If the frame is tight, then the optimal frame bound $b=\frac{1}{m}\int_{\Omega}\langle x_{\alpha},\tau_{\alpha}\rangle\,d\mu(\alpha)=\frac{1}{m}\int_{\Omega}\langle\tau_{\alpha},x_{\alpha}\rangle\,d\mu(\alpha).$ In particular, if $\langle x_{\alpha},\tau_{\alpha}\rangle=1,\forall\alpha\in\Omega,$ then $b={\mu(\Omega)}/m.$ Further,

	$\displaystyle h$	$\displaystyle=\frac{1}{b}\int_{\Omega}\langle h,x_{\alpha}\rangle\tau_{\alpha}\,d\mu(\alpha)=\frac{1}{b}\int_{\Omega}\langle h,\tau_{\alpha}\rangle x_{\alpha}\,d\mu(\alpha),\quad\forall h\in\mathcal{H};$
	$\displaystyle\\|h\\|^{2}$	$\displaystyle=\frac{1}{b}\int_{\Omega}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu(\alpha)=\frac{1}{b}\int_{\Omega}\langle h,\tau_{\alpha}\rangle\langle x_{\alpha},h\rangle\,d\mu(\alpha),\quad\forall h\in\mathcal{H}.$

(vii)

If the frame is tight, then

	(Extended variation formula)	$\displaystyle\int_{\Omega}\int_{\Omega}\langle\tau_{\alpha},x_{\beta}\rangle\langle\tau_{\beta},x_{\alpha}\rangle\,d\mu(\beta)\,d\mu(\alpha)=\frac{1}{\dim\mathcal{H}}\left(\int_{\Omega}\langle x_{\alpha},\tau_{\alpha}\rangle\,d\mu(\alpha)\right)^{2}$
		$\displaystyle=\frac{1}{\dim\mathcal{H}}\left(\int_{\Omega}\langle\tau_{\alpha},x_{\alpha}\rangle\,d\mu(\alpha)\right)^{2}=\int_{\Omega}\int_{\Omega}\langle\tau_{\alpha},\tau_{\beta}\rangle\langle x_{\beta},x_{\alpha}\rangle\,d\mu(\beta)\,d\mu(\alpha).$

(viii)

If the frame is Parseval, then

\text{(Extended dimension formula)}\quad\quad\dim\mathcal{H}=\int_{\Omega}\langle x_{\alpha},\tau_{\alpha}\rangle\,d\mu(\alpha)=\int_{\Omega}\langle\tau_{\alpha},x_{\alpha}\rangle\,d\mu(\alpha).

(ix)

If the frame is Parseval, then for every $T\in\mathcal{B}(\mathcal{H}),$

\text{(Extended trace formula)}\quad\quad\operatorname{Trace}(T)=\int_{\Omega}\langle Tx_{\alpha},\tau_{\alpha}\rangle\,d\mu(\alpha)=\int_{\Omega}\langle T\tau_{\alpha},x_{\alpha}\rangle\,d\mu(\alpha).

Proof.

(i)

Using spectral theorem, $\mathcal{H}$ has an orthonormal basis $\{e_{j}\}_{j=1}^{m}$ consisting of eigenvectors for $S_{x,\tau}.$ Let $\{\lambda_{j}\}_{j=1}^{m}$ denote the corresponding eigenvalues. Then $S_{x,\tau}h=\sum_{j=1}^{m}\langle h,e_{j}\rangle S_{x,\tau}e_{j}=\sum_{j=1}^{m}\lambda_{j}\langle h,e_{j}\rangle e_{j},\forall h\in\mathcal{H}.$ Since $S_{x,\tau}$ is positive invertible, $\lambda_{j}>0,\forall j=1,...,m.$ Therefore

	$\displaystyle\min\{\lambda_{j}\}_{j=1}^{m}\\|h\\|^{2}$	$\displaystyle\leq\sum_{j=1}^{m}\lambda_{j}\|\langle h,e_{j}\rangle\|^{2}=\langle S_{x,\tau}h,h\rangle$
		$\displaystyle=\int_{\Omega}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu(\alpha)\leq\max\{\lambda_{j}\}_{j=1}^{m}\\|h\\|^{2},\quad\forall h\in\mathcal{H}.$

To get optimal frame bounds we take eigenvectors corresponding to $\min\{\lambda_{j}\}_{j=1}^{m}$ and $\max\{\lambda_{j}\}_{j=1}^{m}.$

(ii)

	$\displaystyle\sum_{j=1}^{m}\lambda_{j}$	$\displaystyle=\sum_{j=1}^{m}\lambda_{j}\\|e_{j}\\|^{2}=\sum_{j=1}^{m}\langle S_{x,\tau}e_{j},e_{j}\rangle=\sum_{j=1}^{m}\int_{\Omega}\langle e_{j},x_{\alpha}\rangle\langle\tau_{\alpha},e_{j}\rangle\,d\mu(\alpha)$
		$\displaystyle=\int_{\Omega}(\sum_{j=1}^{m}\langle\tau_{\alpha},e_{j}\rangle\langle e_{j},x_{\alpha}\rangle)\,d\mu(\alpha)=\int_{\Omega}\langle\tau_{\alpha},x_{\alpha}\rangle\,d\mu(\alpha).$

Since $S_{x,\tau}=S_{\tau,x}$ , we get $\sum_{j=1}^{m}\lambda_{j}=\int_{\Omega}\langle x_{\alpha},\tau_{\alpha}\rangle\,d\mu(\alpha).$

(iii)

This follows from (i).
(iv)

Let $\{e_{j}\}_{j=1}^{m}$ and $\{\lambda_{j}\}_{j=1}^{m}$ be as in (i). We may assume $\lambda_{1}\geq\cdots\geq\lambda_{m}$ . Then (i) gives $b=\lambda_{1}.$ Now use (ii): $b=\lambda_{1}\leq\sum_{j=1}^{m}\lambda_{j}=\int_{\Omega}\langle x_{\alpha},\tau_{\alpha}\rangle\,d\mu(\alpha)=\int_{\Omega}\langle\tau_{\alpha},x_{\alpha}\rangle\,d\mu(\alpha)\leq\lambda_{1}m=bm.$

(v)

Let $\{f_{j}\}_{j=1}^{m}$ be an orthonormal basis for $\mathcal{H}$ . Then

	$\displaystyle\operatorname{Trace}(S_{x,\tau})$	$\displaystyle=\sum_{j=1}^{m}\langle S_{x,\tau}f_{j},f_{j}\rangle=\sum_{j=1}^{m}\left\langle\int_{\Omega}\langle f_{j},x_{\alpha}\rangle\tau_{\alpha}\,d\mu(\alpha),f_{j}\right\rangle$
		$\displaystyle=\int_{\Omega}\left(\sum_{j=1}^{m}\langle f_{j},x_{\alpha}\rangle\langle\tau_{\alpha},f_{j}\rangle\right)\,d\mu(\alpha)=\int_{\Omega}\langle\tau_{\alpha},x_{\alpha}\rangle\,d\mu(\alpha),~{}\text{and}$

	$\displaystyle\operatorname{Trace}(S^{2}_{x,\tau})$	$\displaystyle=\sum_{j=1}^{m}\langle S_{x,\tau}f_{j},S_{x,\tau}f_{j}\rangle=\sum_{j=1}^{m}\left\langle\int_{\Omega}\langle f_{j},x_{\alpha}\rangle\tau_{\alpha}\,d\mu(\alpha),\int_{\Omega}\langle f_{j},\tau_{\beta}\rangle x_{\beta}\,d\mu(\beta)\right\rangle$
		$\displaystyle=\int_{\Omega}\int_{\Omega}\langle\tau_{\alpha},x_{\beta}\rangle\sum_{j=1}^{m}\langle f_{j},x_{\alpha}\rangle\langle\tau_{\beta},f_{j}\rangle\,d\mu(\beta)\,d\mu(\alpha)=\int_{\Omega}\int_{\Omega}\langle\tau_{\alpha},x_{\beta}\rangle\langle\tau_{\beta},x_{\alpha}\rangle\,d\mu(\beta)\,d\mu(\alpha),$

	$\displaystyle\operatorname{Trace}(S^{2}_{x,\tau})$	$\displaystyle=\sum_{j=1}^{m}\langle S_{x,\tau}f_{j},S_{x,\tau}f_{j}\rangle=\sum_{j=1}^{m}\left\langle\int_{\Omega}\langle f_{j},x_{\alpha}\rangle\tau_{\alpha}\,d\mu(\alpha),\int_{\Omega}\langle f_{j},x_{\beta}\rangle\tau_{\beta}\,d\mu(\beta)\right\rangle$
		$\displaystyle=\int_{\Omega}\int_{\Omega}\langle\tau_{\alpha},\tau_{\beta}\rangle\sum_{j=1}^{m}\langle f_{j},x_{\alpha}\rangle\langle x_{\beta},f_{j}\rangle\,d\mu(\beta)\,d\mu(\alpha)=\int_{\Omega}\int_{\Omega}\langle\tau_{\alpha},\tau_{\beta}\rangle\langle x_{\beta},x_{\alpha}\rangle\,d\mu(\beta)\,d\mu(\alpha).$

(vi)

Now $S_{x,\tau}=\lambda I_{\mathcal{H}},$ for some positive $\lambda.$ This gives $\lambda_{1}=\cdots=\lambda_{m}=\lambda=b.$ From (ii) we get the conclusions.

(vii)

Let the optimal frame bound be $b$ . From (v) and (vi),

	$\displaystyle\int_{\Omega}\int_{\Omega}\langle\tau_{\alpha},\tau_{\beta}\rangle\langle x_{\beta},x_{\alpha}\rangle\,d\mu(\beta)\,d\mu(\alpha)$	$\displaystyle=\int_{\Omega}\int_{\Omega}\langle\tau_{\alpha},x_{\beta}\rangle\langle\tau_{\beta},x_{\alpha}\rangle\,d\mu(\beta)\,d\mu(\alpha)$
		$\displaystyle=\operatorname{Trace}(S^{2}_{x,\tau})=\operatorname{Trace}(b^{2}I_{\mathcal{H}})=b^{2}m$
		$\displaystyle=\left(\frac{1}{m}\int_{\Omega}\langle x_{\alpha},\tau_{\alpha}\rangle\,d\mu(\alpha)\right)^{2}m=\frac{1}{m}\left(\int_{\Omega}\langle x_{\alpha},\tau_{\alpha}\rangle\,d\mu(\alpha)\right)^{2}.$

(viii)

Let $\{f_{j}\}_{j=1}^{m}$ be as in (v). Then

	$\displaystyle\dim\mathcal{H}$	$\displaystyle=\sum_{j=1}^{m}\\|f_{j}\\|^{2}=\sum_{j=1}^{m}\int_{\Omega}\langle f_{j},x_{\alpha}\rangle\langle\tau_{\alpha},f_{j}\rangle\,d\mu(\alpha)$
		$\displaystyle=\sum_{j=1}^{m}\int_{\Omega}\langle f_{j},\tau_{\alpha}\rangle\langle x_{\alpha},f_{j}\rangle\,d\mu(\alpha)=\int_{\Omega}\sum_{j=1}^{m}\langle f_{j},x_{\alpha}\rangle\langle\tau_{\alpha},f_{j}\rangle\,d\mu(\alpha)$
		$\displaystyle=\int_{\Omega}\sum_{j=1}^{m}\langle f_{j},\tau_{\alpha}\rangle\langle x_{\alpha},f_{j}\rangle\,d\mu(\alpha)=\int_{\Omega}\langle x_{\alpha},\tau_{\alpha}\rangle\,d\mu(\alpha)=\int_{\Omega}\langle\tau_{\alpha},x_{\alpha}\rangle\,d\mu(\alpha).$

(ix)

Let $\{f_{j}\}_{j=1}^{m}$ be as in (v). Then

	$\displaystyle\operatorname{Trace}(T)$	$\displaystyle=\sum_{j=1}^{m}\langle Tf_{j},f_{j}\rangle=\sum_{j=1}^{m}\left\langle\int_{\Omega}\langle Tf_{j},x_{\alpha}\rangle\tau_{\alpha}\,d\mu(\alpha),f_{j}\right\rangle$
		$\displaystyle=\int_{\Omega}\sum_{j=1}^{m}\langle\tau_{\alpha},f_{j}\rangle\langle Tf_{j},x_{\alpha}\rangle\,d\mu(\alpha)=\int_{\Omega}\sum_{j=1}^{m}\langle\tau_{\alpha},f_{j}\rangle\langle f_{j},T^{*}x_{\alpha}\rangle\,d\mu(\alpha)$
		$\displaystyle=\int_{\Omega}\langle\tau_{\alpha},T^{*}x_{\alpha}\rangle\,d\mu(\alpha)=\int_{\Omega}\langle T\tau_{\alpha},x_{\alpha}\rangle\,d\mu(\alpha).$

Similarly by using $Tf_{j}=\int_{\Omega}\langle Tf_{j},\tau_{\alpha}\rangle x_{\alpha}\,d\mu(\alpha)$ , we get $\operatorname{Trace}(T)=\int_{\Omega}\langle Tx_{\alpha},\tau_{\alpha}\rangle\,d\mu(\alpha).$

∎

Theorem 7.4.

If a continuous frame $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ for $\mathbb{R}^{m}$ is such that

\int_{\Omega}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},g\rangle\,d\mu(\alpha)=\int_{\Omega}\langle g,x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu(\alpha),\quad\forall h,g\in\mathbb{R}^{n},

then it is also a continuous frame for $\mathbb{C}^{m}.$ Further, if $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is a tight (resp. Parseval) continuous frame for $\mathbb{R}^{m}$ , then it is also a tight (resp. Parseval) continuous frame for $\mathbb{C}^{m}.$

Proof.

Let $a,b$ be lower and upper frame bounds, in order. For $z\in\mathbb{C}^{m}$ we write $z=\operatorname{Re}z+i\operatorname{Im}z,\operatorname{Re}z,\operatorname{Im}z$ $\in\mathbb{R}^{m}.$ Then

	$\displaystyle a\\|z\\|^{2}$	$\displaystyle=a\\|\operatorname{Re}z\\|^{2}+a\\|\operatorname{Im}z\\|^{2}\leq\int_{\Omega}\langle\operatorname{Re}z,x_{\alpha}\rangle\langle\tau_{\alpha},\operatorname{Re}z\rangle\,d\mu(\alpha)+\int_{\Omega}\langle\operatorname{Im}z,x_{\alpha}\rangle\langle\tau_{\alpha},\operatorname{Im}z\rangle\,d\mu(\alpha)$
		$\displaystyle=\left(\int_{\Omega}\langle\operatorname{Re}z,x_{\alpha}\rangle\langle\tau_{\alpha},\operatorname{Re}z\rangle\,d\mu(\alpha)+i\int_{\Omega}\langle\operatorname{Im}z,x_{\alpha}\rangle\langle\tau_{\alpha},\operatorname{Re}z\rangle\,d\mu(\alpha)\right)$
		$\displaystyle\quad-i\left(\int_{\Omega}\langle\operatorname{Re}z,x_{\alpha}\rangle\langle\tau_{\alpha},\operatorname{Im}z\rangle\,d\mu(\alpha)+i\int_{\Omega}\langle\operatorname{Im}z,x_{\alpha}\rangle\langle\tau_{\alpha},\operatorname{Im}z\rangle\,d\mu(\alpha)\right)$
		$\displaystyle=\int_{\Omega}\langle\operatorname{Re}z+i\operatorname{Im}z,x_{\alpha}\rangle\langle\tau_{\alpha},\operatorname{Re}z\rangle\,d\mu(\alpha)-i\int_{\Omega}\langle\operatorname{Re}z+i\operatorname{Im}z,x_{\alpha}\rangle\langle\tau_{\alpha},\operatorname{Im}z\rangle\,d\mu(\alpha)$
		$\displaystyle=\int_{\Omega}\langle z,x_{\alpha}\rangle\langle\tau_{\alpha},\operatorname{Re}z\rangle\,d\mu(\alpha)+\int_{\Omega}\langle z,x_{\alpha}\rangle\langle\tau_{\alpha},i\operatorname{Im}z\rangle\,d\mu(\alpha)=\int_{\Omega}\langle z,x_{\alpha}\rangle\langle\tau_{\alpha},z\rangle\,d\mu(\alpha)$
		$\displaystyle\leq b\\|\operatorname{Re}z\\|^{2}+b\\|\operatorname{Im}z\\|^{2}=b\\|z\\|^{2}.$

∎

Theorem 7.5.

If $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is a continuous frame for $\mathbb{C}^{m}$ such that

\int_{\Omega}\langle h,\operatorname{Re}x_{\alpha}\rangle\langle\operatorname{Im}\tau_{\alpha},h\rangle\,d\mu(\alpha)=\int_{\Omega}\langle h,\operatorname{Im}x_{\alpha}\rangle\langle\operatorname{Re}\tau_{\alpha},h\rangle\,d\mu(\alpha),\quad\forall h\in\mathbb{C}^{m},

then $(\{\operatorname{Re}x_{\alpha}\}_{\alpha\in\Omega}\cup\{\operatorname{Im}x_{\alpha}\}_{\alpha\in\Omega},\{\operatorname{Re}\tau_{\alpha}\}_{\alpha\in\Omega}\cup\{\operatorname{Im}\tau_{\alpha}\}_{\alpha\in\Omega})$ is a continuous frame for $\mathbb{R}^{m}.$ Further, if $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is a tight (resp. Parseval) continuous frame for $\mathbb{C}^{m}$ , then $(\{\operatorname{Re}x_{\alpha}\}_{\alpha\in\Omega}\cup\{\operatorname{Im}x_{\alpha}\}_{\alpha\in\Omega}$ , $\{\operatorname{Re}\tau_{\alpha}\}_{\alpha\in\Omega}\cup\{\operatorname{Im}\tau_{\alpha}\}_{\alpha\in\Omega})$ is a tight (resp. Parseval) continuous frame for $\mathbb{R}^{m}.$

Proof.

Consider

	$\displaystyle\int_{\Omega}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu(\alpha)$	$\displaystyle=\int_{\Omega}\langle h,\operatorname{Re}x_{\alpha}+i\operatorname{Im}x_{\alpha}\rangle\langle\operatorname{Re}\tau_{\alpha}+i\operatorname{Im}\tau_{\alpha},h\rangle\,d\mu(\alpha)$
		$\displaystyle=\int_{\Omega}\langle h,\operatorname{Re}x_{\alpha}\rangle\langle\operatorname{Re}\tau_{\alpha},h\rangle\,d\mu(\alpha)+i\int_{\Omega}\langle h,\operatorname{Re}x_{\alpha}\rangle\langle\operatorname{Im}\tau_{\alpha},h\rangle\,d\mu(\alpha)$
		$\displaystyle\quad-i\int_{\Omega}\langle h,\operatorname{Im}x_{\alpha}\rangle\langle\operatorname{Re}\tau_{\alpha},h\rangle\,d\mu(\alpha)+\int_{\Omega}\langle h,\operatorname{Im}x_{\alpha}\rangle\langle\operatorname{Im}\tau_{\alpha},h\rangle\,d\mu(\alpha)$
		$\displaystyle=\int_{\Omega}\langle h,\operatorname{Re}x_{\alpha}\rangle\langle\operatorname{Re}\tau_{\alpha},h\rangle\,d\mu(\alpha)+\int_{\Omega}\langle h,\operatorname{Im}x_{\alpha}\rangle\langle\operatorname{Im}\tau_{\alpha},h\rangle\,d\mu(\alpha),\quad\forall h\in\mathbb{R}^{m}.$

Therefore, if $a,b$ are lower and upper frame bounds, respectively, then $a\|h\|^{2}\leq\int_{\Omega}\langle h,\operatorname{Re}x_{\alpha}\rangle\langle\operatorname{Re}\tau_{\alpha},h\rangle\,d\mu(\alpha)$ $+\int_{\Omega}\langle h,\operatorname{Im}x_{\alpha}\rangle\langle\operatorname{Im}\tau_{\alpha},h\rangle\,d\mu(\alpha)\leq b\|h\|^{2},\forall h\in\mathbb{R}^{m}.$ ∎

Proposition 7.6.

Let $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ be a continuous frame for $\mathcal{H}.$ Then $\mathcal{H}$ is finite dimensional if and only if

\int_{\Omega}\langle x_{\alpha},\tau_{\alpha}\rangle\,d\mu(\alpha)=\int_{\Omega}\langle\tau_{\alpha},x_{\alpha}\rangle\,d\mu(\alpha)<\infty.

In particular, if dimension of $\mathcal{H}$ is finite, $\langle x_{\alpha},\tau_{\alpha}\rangle>0,\forall\alpha\in\Omega$ and $\inf_{\alpha\in\Omega}\langle x_{\alpha},\tau_{\alpha}\rangle$ is positive, then $\mu(\Omega)<\infty.$

Proof.

Let $\{e_{k}\}_{k\in\mathbb{L}}$ be an orthonormal basis for $\mathcal{H}$ , and $a,b$ be frame bounds for $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ .

$(\Rightarrow)$ Now $\mathbb{L}$ is finite. Then

	$\displaystyle\int_{\Omega}\langle x_{\alpha},\tau_{\alpha}\rangle\,d\mu(\alpha)$	$\displaystyle=\int_{\Omega}\sum_{k\in\mathbb{L}}\langle x_{\alpha},e_{k}\rangle\langle e_{k},\tau_{\alpha}\rangle\,d\mu(\alpha)=\sum_{k\in\mathbb{L}}\int_{\Omega}\langle x_{\alpha},e_{k}\rangle\langle e_{k},\tau_{\alpha}\rangle\,d\mu(\alpha)$
		$\displaystyle\leq b\sum_{k\in\mathbb{L}}\\|e_{k}\\|^{2}=b\operatorname{Card}(\mathbb{L})<\infty.$

Since $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is a continuous frame, we must have

\int_{\Omega}\langle x_{\alpha},e_{k}\rangle\langle e_{k},\tau_{\alpha}\rangle\,d\mu(\alpha)=\overline{\int_{\Omega}\langle x_{\alpha},e_{k}\rangle\langle e_{k},\tau_{\alpha}\rangle}\,d\mu(\alpha).

Therefore $\int_{\Omega}\langle x_{\alpha},\tau_{\alpha}\rangle\,d\mu(\alpha)=\int_{\Omega}\langle\tau_{\alpha},x_{\alpha}\rangle\,d\mu(\alpha)<\infty.$

$(\Leftarrow)$ $\operatorname{dim}\mathcal{H}=\sum_{k\in\mathbb{L}}\|e_{k}\|^{2}\leq(1/a)\sum_{k\in\mathbb{L}}\int_{\Omega}\langle e_{k},x_{\alpha}\rangle\langle\tau_{\alpha},e_{k}\rangle\,d\mu(\alpha)=(1/a)\int_{\Omega}\sum_{k\in\mathbb{L}}\langle e_{k},x_{\alpha}\rangle\langle\tau_{\alpha},e_{k}\rangle\,d\mu(\alpha)=(1/a)\int_{\Omega}\langle x_{\alpha},\tau_{\alpha}\rangle\,d\mu(\alpha)<\infty.$

For the second, $0<\inf_{\alpha\in\Omega}\langle x_{\alpha},\tau_{\alpha}\rangle\mu(\Omega)=\int_{\Omega}\inf_{\alpha\in\Omega}\langle x_{\alpha},\tau_{\alpha}\rangle\,d\mu(\alpha)\leq\int_{\Omega}\langle x_{\alpha},\tau_{\alpha}\rangle\,d\mu(\alpha)<\infty.$ ∎

Proposition 7.7.

Let $I\subseteq\mathbb{R}$ be a bounded interval. Let $a_{\alpha},b_{\alpha}\geq 0,\theta_{\alpha},\phi_{\alpha}\in\mathbb{R},\forall\alpha\in I$ and $I\ni\alpha\mapsto\theta_{\alpha}\in\mathbb{R}$ , $I\ni\alpha\mapsto\phi_{\alpha}\in\mathbb{R}$ be continuous. Then $\left\{x_{\alpha}\coloneqq\begin{bmatrix}a_{\alpha}\cos\theta_{\alpha}\\ a_{\alpha}\sin\theta_{\alpha}\end{bmatrix}\right\}_{\alpha\in I}$ is a tight continuous frame for $\mathbb{R}^{2}$ w.r.t. $\left\{\tau_{\alpha}\coloneqq\begin{bmatrix}b_{\alpha}\cos\phi_{\alpha}\\ b_{\alpha}\sin\phi_{\alpha}\end{bmatrix}\right\}_{\alpha\in I}$ if and only if $\int_{I}\begin{bmatrix}a_{\alpha}b_{\alpha}\cos(\theta_{\alpha}+\phi_{\alpha})\\ a_{\alpha}b_{\alpha}\sin(\theta_{\alpha}+\phi_{\alpha})\\ a_{\alpha}b_{\alpha}\sin(\theta_{\alpha}-\phi_{\alpha})\end{bmatrix}\,d\alpha=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}.$

Proof.

Matrix of $S_{x,\tau}$ is

\displaystyle\begin{bmatrix}\int_{I}a_{\alpha}b_{\alpha}\cos\theta_{\alpha}\cos\phi_{\alpha}\,d\alpha&\int_{I}a_{\alpha}b_{\alpha}\sin\theta_{\alpha}\cos\phi_{\alpha}\,d\alpha\\ \int_{I}a_{\alpha}b_{\alpha}\cos\theta_{\alpha}\sin\phi_{\alpha}\,d\alpha&\int_{I}a_{\alpha}b_{\alpha}\sin\theta_{\alpha}\sin\phi_{\alpha}\,d\alpha\\ \end{bmatrix}.

We next observe that $S_{x,\tau}=aI_{\mathbb{R}^{2}}$ for some $a>0$ if and only if

\displaystyle\int_{I}\begin{bmatrix}a_{\alpha}b_{\alpha}\cos(\theta_{\alpha}+\phi_{\alpha})\\ a_{\alpha}b_{\alpha}\sin(\theta_{\alpha}+\phi_{\alpha})\\ a_{\alpha}b_{\alpha}\sin(\theta_{\alpha}-\phi_{\alpha})\end{bmatrix}\,d\alpha=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}.

∎

8. Further extension

Definition 8.1.

A collection $\{A_{\alpha}\}_{\alpha\in\Omega}$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ is said to be a weak continuous operator-valued frame (we write weak continuous (ovf)) in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ with respect to a collection $\{\Psi_{\alpha}\}_{\alpha\in\Omega}$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ if

(i)

for each $h\in\mathcal{H}$ , both maps $\Omega\ni\alpha\mapsto A_{\alpha}h\in\mathcal{H}_{0}$ and $\Omega\ni\alpha\mapsto\Psi_{\alpha}h\in\mathcal{H}_{0}$ are measurable,
(ii)

the map (we call as frame operator) $S_{A,\Psi}:\mathcal{H}\ni h\mapsto\int_{\Omega}\Psi_{\alpha}^{*}A_{\alpha}h\,d\mu(\alpha)\in\mathcal{H}$ is a well-defined bounded positive invertible operator.

Notions of frame bounds, optimal bounds, tight frame, Parseval frame, Bessel are in same fashion as in Definition 2.2.

For fixed $\Omega$ , $\mathcal{H},\mathcal{H}_{0},$ and $\{\Psi_{\alpha}\}_{\alpha\in\Omega}$ , the set of all weak continuous operator-valued frames in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ with respect to collection $\{\Psi_{\alpha}\}_{\alpha\in\Omega}$ is denoted by $\mathscr{F}^{\text{w}}_{\Psi}.$

Proposition 8.2.

A collection $\{A_{\alpha}\}_{\alpha\in\Omega}$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ is a weak continuous (ovf) w.r.t. $\{\Psi_{\alpha}\}_{\alpha\in\Omega}$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ if and only if there exist $a,b,r>0$ such that

(i)

for each $h\in\mathcal{H}$ , both maps $\Omega\ni\alpha\mapsto A_{\alpha}h\in\mathcal{H}_{0}$ , $\Omega\ni\alpha\mapsto\Psi_{\alpha}h\in\mathcal{H}_{0}$ are measurable,
(ii)

$\|\int_{\Omega}\Psi_{\alpha}^{*}A_{\alpha}h\,d\mu(\alpha)\|\leq r\|h\|,\forall h\in\mathcal{H},$
(iii)

$a\|h\|^{2}\leq\int_{\Omega}\langle A_{\alpha}h,\Psi_{\alpha}h\rangle\,d\mu(\alpha)\leq b\|h\|^{2},\forall h\in\mathcal{H},$
(iv)

$\int_{\Omega}\Psi_{\alpha}^{*}A_{\alpha}h\,d\mu(\alpha)=\int_{\Omega}A_{\alpha}^{*}\Psi_{\alpha}h\,d\mu(\alpha),\forall h\in\mathcal{H}$ .

If the Hilbert space is complex, then condition (iv) can be dropped.

Proposition 8.3.

Let $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ be a weak continuous (ovf) in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ with an upper frame bound $b$ . If $\{\alpha\}$ is measurable and $\Psi_{\alpha}^{*}A_{\alpha}\geq 0,\forall\alpha\in\Omega,$ then $\mu(\{\alpha\})\|\Psi_{\alpha}^{*}A_{\alpha}\|\leq b,\forall\alpha\in\Omega.$

Definition 8.4.

A weak continuous (ovf) $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Phi_{\alpha}\}_{\alpha\in\Omega})$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ is said to be dual of a weak continuous (ovf) $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ if $\int_{\Omega}B_{\alpha}^{*}\Psi_{\alpha}h\,d\mu(\alpha)=\int_{\Omega}\Phi^{*}_{\alpha}A_{\alpha}h\,d\mu(\alpha)=h,\forall h\in\mathcal{H}$ . The ‘weak continuous (ovf)’ $(\{\widetilde{A}_{\alpha}\coloneqq A_{\alpha}S_{A,\Psi}^{-1}\}_{\alpha\in\Omega},\{\widetilde{\Psi}_{\alpha}\coloneqq\Psi_{\alpha}S_{A,\Psi}^{-1}\}_{\alpha\in\Omega})$ , which is a ‘dual’ of $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ is called the canonical dual of $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ .

Proposition 8.5.

Let $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ be a weak continuous (ovf) in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0}).$ If $h\in\mathcal{H}$ has representation $h=\int_{\Omega}A_{\alpha}^{*}f(\alpha)\,d\mu(\alpha)=\int_{\Omega}\Psi_{\alpha}^{*}g(\alpha)\,d\mu(\alpha),$ for some measurable $f,g:\Omega\rightarrow\mathcal{H}_{0}$ , then

\int_{\Omega}\langle f(\alpha),g(\alpha)\rangle\,d\mu(\alpha)=\int_{\Omega}\langle\widetilde{\Psi}_{\alpha}h,\widetilde{A}_{\alpha}h\rangle\,d\mu(\alpha)+\int_{\Omega}\langle f(\alpha)-\widetilde{\Psi}_{\alpha}h,g(\alpha)-\widetilde{A}_{\alpha}h\rangle\,d\mu(\alpha).

Theorem 8.6.

Let $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ be a weak continuous (ovf) with frame bounds $a$ and $b.$ Then the following statements are true.

(i)

The canonical dual weak continuous (ovf) of the canonical dual weak continuous (ovf) of $(\{A_{\alpha}\}_{\alpha\in\Omega}$ , $\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ is itself.
(ii)

$\frac{1}{b},\frac{1}{a}$ are frame bounds for the canonical dual of $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega}).$
(iii)

If $a,b$ are optimal frame bounds for $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega}),$ then $\frac{1}{b},\frac{1}{a}$ are optimal frame bounds for its canonical dual.

Definition 8.7.

A weak continuous (ovf) $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Phi_{\alpha}\}_{\alpha\in\Omega})$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ is said to be orthogonal to a weak continuous (ovf) $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ if $\int_{\Omega}B_{\alpha}^{*}\Psi_{\alpha}h\,d\mu(\alpha)=\int_{\Omega}\Phi^{*}_{\alpha}A_{\alpha}h\,d\mu(\alpha)=0,\forall h\in\mathcal{H}$ .

Proposition 8.8.

Two orthogonal weak continuous operator-valued frames have common dual weak continuous (ovf).

Proof.

	$\displaystyle\langle S_{C,\Xi}h,g\rangle$	$\displaystyle=\int_{\Omega}\langle C_{\alpha}h,\Xi_{\alpha}g\rangle\,d\mu(\alpha)$
		$\displaystyle=\int_{\Omega}\langle(A_{\alpha}S_{A,\Psi}^{-1}+B_{\alpha}S_{B,\Phi}^{-1})h,(\Psi_{\alpha}S_{A,\Psi}^{-1}+\Phi_{\alpha}S_{B,\Phi}^{-1})g\rangle\,d\mu(\alpha)$
		$\displaystyle=\int_{\Omega}\langle A_{\alpha}S_{A,\Psi}^{-1}h,\Psi_{\alpha}S_{A,\Psi}^{-1}g\rangle\,d\mu(\alpha)+\int_{\Omega}\langle A_{\alpha}S_{A,\Psi}^{-1}h,\Phi_{\alpha}S_{B,\Phi}^{-1}g\rangle\,d\mu(\alpha)$
		$\displaystyle\quad+\int_{\Omega}\langle B_{\alpha}S_{B,\Phi}^{-1}h,\Psi_{\alpha}S_{A,\Psi}^{-1}g\rangle\,d\mu(\alpha)+\int_{\Omega}\langle B_{\alpha}S_{B,\Phi}^{-1}h,\Phi_{\alpha}S_{B,\Phi}^{-1}g\rangle\,d\mu(\alpha)$
		$\displaystyle=\langle S_{A,\Psi}(S_{A,\Psi}^{-1}h),S_{A,\Psi}^{-1}g\rangle+\langle 0,S_{B,\Phi}^{-1}g\rangle+\langle 0,S_{A,\Psi}^{-1}g\rangle+\langle S_{B,\Phi}(S_{B,\Phi}^{-1}h),S_{B,\Phi}^{-1}g\rangle$
		$\displaystyle=\langle(S_{A,\Psi}^{-1}+S_{B,\Phi}^{-1})h,g\rangle.$

Hence $S_{C,\Xi}=S_{A,\Psi}^{-1}+S_{B,\Phi}^{-1}$ which is positive invertible. Therefore $(\{C_{\alpha}\}_{\alpha\in\Omega},\{\Xi_{\alpha}\}_{\alpha\in\Omega})$ is a weak continuous (ovf) in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ . Further, for all $h,g\in\mathcal{H}$ ,

	$\displaystyle\int_{\Omega}\langle\Psi_{\alpha}^{*}C_{\alpha}h,g\rangle\,d\mu(\alpha)$	$\displaystyle=\int_{\Omega}\langle(A_{\alpha}S_{A,\Psi}^{-1}+B_{\alpha}S_{B,\Phi}^{-1})h,\Psi_{\alpha}g\rangle\,d\mu(\alpha)$
		$\displaystyle=\int_{\Omega}\langle A_{\alpha}S_{A,\Psi}^{-1}h,\Psi_{\alpha}g\rangle\,d\mu(\alpha)+\int_{\Omega}\langle B_{\alpha}S_{B,\Phi}^{-1}h,\Psi_{\alpha}g\rangle\,d\mu(\alpha)=\langle h,g\rangle+\langle 0,g\rangle,$

	$\displaystyle\int_{\Omega}\langle A_{\alpha}^{*}\Xi_{\alpha}h,g\rangle\,d\mu(\alpha)$	$\displaystyle=\int_{\Omega}\langle(\Psi_{\alpha}S_{A,\Psi}^{-1}+\Phi_{\alpha}S_{B,\Phi}^{-1})h,A_{\alpha}g\rangle\,d\mu(\alpha)$
		$\displaystyle=\int_{\Omega}\langle\Psi_{\alpha}S_{A,\Psi}^{-1}h,A_{\alpha}g\rangle\,d\mu(\alpha)+\int_{\Omega}\langle\Phi_{\alpha}S_{B,\Phi}^{-1}h,A_{\alpha}g\rangle\,d\mu(\alpha)=\langle h,g\rangle+\langle 0,g\rangle,$

and

	$\displaystyle\int_{\Omega}\langle\Phi_{\alpha}^{*}C_{\alpha}h,g\rangle\,d\mu(\alpha)$	$\displaystyle=\int_{\Omega}\langle(A_{\alpha}S_{A,\Psi}^{-1}+B_{\alpha}S_{B,\Phi}^{-1})h,\Phi_{\alpha}g\rangle\,d\mu(\alpha)$
		$\displaystyle=\int_{\Omega}\langle A_{\alpha}S_{A,\Psi}^{-1}h,\Phi_{\alpha}g\rangle\,d\mu(\alpha)+\int_{\Omega}\langle B_{\alpha}S_{B,\Phi}^{-1}h,\Phi_{\alpha}g\rangle\,d\mu(\alpha)=\langle 0,g\rangle+\langle h,g\rangle,$

	$\displaystyle\int_{\Omega}\langle B_{\alpha}^{*}\Xi_{\alpha}h,g\rangle\,d\mu(\alpha)$	$\displaystyle=\int_{\Omega}\langle(\Psi_{\alpha}S_{A,\Psi}^{-1}+\Phi_{\alpha}S_{B,\Phi}^{-1})h,B_{\alpha}g\rangle\,d\mu(\alpha)$
		$\displaystyle=\int_{\Omega}\langle\Psi_{\alpha}S_{A,\Psi}^{-1}h,B_{\alpha}g\rangle\,d\mu(\alpha)+\int_{\Omega}\langle\Phi_{\alpha}S_{B,\Phi}^{-1}h,B_{\alpha}g\rangle\,d\mu(\alpha)=\langle 0,g\rangle+\langle h,g\rangle.$

Thus $(\{C_{\alpha}\}_{\alpha\in\Omega},\{\Xi_{\alpha}\}_{\alpha\in\Omega})$ is a common dual of $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ and $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Phi_{\alpha}\}_{\alpha\in\Omega}).$ ∎

Proposition 8.9.

Let $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ and $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Phi_{\alpha}\}_{\alpha\in\Omega})$ be two Parseval weak continuous operator-valued frames in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ which are orthogonal. If $C,D,E,F\in\mathcal{B}(\mathcal{H})$ are such that $C^{*}E+D^{*}F=I_{\mathcal{H}}$ , then $(\{A_{\alpha}C+B_{\alpha}D\}_{\alpha\in\Omega},\{\Psi_{\alpha}E+\Phi_{\alpha}F\}_{\alpha\in\Omega})$ is a Parseval weak continuous (ovf) in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ . In particular, if scalars $c,d,e,f$ satisfy $\bar{c}e+\bar{d}f=1$ , then $(\{cA_{\alpha}+dB_{\alpha}\}_{\alpha\in\Omega},\{e\Psi_{\alpha}+f\Phi_{\alpha}\}_{\alpha\in\Omega})$ is a Parseval weak continuous (ovf).

Proof.

For all $h,g\in\mathcal{H}$ ,

	$\displaystyle\langle S_{AC+BD,\Psi E+\Phi F}h,g\rangle$	$\displaystyle=\int_{\Omega}\langle(A_{\alpha}C+B_{\alpha}D)h,(\Psi_{\alpha}E+\Phi_{\alpha}F)g\rangle\,d\mu(\alpha)$
		$\displaystyle=\int_{\Omega}\langle A_{\alpha}(Ch),\Psi_{\alpha}(Eg)\rangle\,d\mu(\alpha)+\int_{\Omega}\langle A_{\alpha}(Ch),\Phi_{\alpha}(Fg)\rangle\,d\mu(\alpha)$
		$\displaystyle\quad+\int_{\Omega}\langle B_{\alpha}(Dh),\Psi_{\alpha}(Eg)\rangle\,d\mu(\alpha)+\int_{\Omega}\langle B_{\alpha}(Dh),\Phi_{\alpha}(Fg)\rangle\,d\mu(\alpha)$
		$\displaystyle=\langle Ch,Eg\rangle+\langle 0,Fg\rangle+\langle 0,Eg\rangle+\langle Dh,Fg\rangle$
		$\displaystyle=\langle E^{}Ch,g\rangle+\langle F^{}Dh,g\rangle=\langle(E^{}C+F^{}D)h,g\rangle=\langle h,g\rangle.$

∎

Definition 8.10.

Two weak continuous operator-valued frames $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ and $(\{B_{\alpha}\}_{\alpha\in\Omega}$ , $\{\Phi_{\alpha}\}_{\alpha\in\Omega})$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ are called disjoint if $(\{A_{\alpha}\oplus B_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\oplus\Phi_{\alpha}\}_{\alpha\in\Omega})$ is a weak continuous (ovf) in $\mathcal{B}(\mathcal{H}\oplus\mathcal{H},\mathcal{H}_{0}).$

Proposition 8.11.

If $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ and $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Phi_{\alpha}\}_{\alpha\in\Omega})$ are orthogonal weak continuous operator-valued frames in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ , then they are disjoint. Further, if both $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ and $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Phi_{\alpha}\}_{\alpha\in\Omega})$ are Parseval weak, then $(\{A_{\alpha}\oplus B_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\oplus\Phi_{\alpha}\}_{\alpha\in\Omega})$ is Parseval weak.

Proof.

Let $h\oplus g,u\oplus v\in\mathcal{H}\oplus\mathcal{H}$ . Then

	$\displaystyle\langle S_{A\oplus B,\Psi\oplus\Phi}(h\oplus g),u\oplus v\rangle=\int_{\Omega}\langle(A_{\alpha}\oplus B_{\alpha})(h\oplus g),(\Psi_{\alpha}\oplus\Phi_{\alpha})(u\oplus v)\rangle\,d\mu(\alpha)$
	$\displaystyle=\int_{\Omega}\langle A_{\alpha}h+B_{\alpha}g,\Psi_{\alpha}u+\Phi_{\alpha}v\rangle\,d\mu(\alpha)=\int_{\Omega}\langle A_{\alpha}h,\Psi_{\alpha}u\rangle\,d\mu(\alpha)+\int_{\Omega}\langle A_{\alpha}h,\Phi_{\alpha}v\rangle\,d\mu(\alpha)$
	$\displaystyle\quad+\int_{\Omega}\langle B_{\alpha}g,\Psi_{\alpha}u\rangle\,d\mu(\alpha)+\int_{\Omega}\langle B_{\alpha}g,\Phi_{\alpha}v\rangle\,d\mu(\alpha)$
	$\displaystyle=\langle S_{A,\Psi}h,u\rangle+\langle 0,u\rangle+\langle 0,v\rangle+\langle S_{B,\Phi}g,v\rangle=\langle S_{A,\Psi}h\oplus S_{B,\Phi}g,u\oplus v\rangle$
	$\displaystyle=\langle(S_{A,\Psi}\oplus S_{B,\Phi})(h\oplus g),u\oplus v\rangle\quad\implies S_{A\oplus B,\Psi\oplus\Phi}=S_{A,\Psi}\oplus S_{B,\Phi}.$

∎

Characterization

Theorem 8.12.

(i)

T:\mathcal{H}\ni h\mapsto\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle v_{\alpha,\beta}\,d\mu(\alpha)\in\mathcal{H}

is a well-defined bounded positive invertible operator such that $a\|h\|^{2}\leq\langle Th,h\rangle\leq b\|h\|^{2},\forall h\in\mathcal{H}.$

(ii)

T:\mathcal{H}\ni h\mapsto\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle v_{\alpha,\beta}\,d\mu(\alpha)\in\mathcal{H}

is a well-defined bounded positive operator such that $0\leq\langle Th,h\rangle\leq b\|h\|^{2},\forall h\in\mathcal{H}.$

(iii)

\left\|\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle v_{\alpha,\beta}\,d\mu(\alpha)\right\|\leq r\|h\|,~{}\forall h\in\mathcal{H};

\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle v_{\alpha,\beta}\,d\mu(\alpha)=\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}\langle h,v_{\alpha,\beta}\rangle u_{\alpha,\beta}\,d\mu(\alpha),~{}\forall h\in\mathcal{H};

a\|h\|^{2}\leq\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle\langle v_{\alpha,\beta},h\rangle\,d\mu(\alpha)\leq b\|h\|^{2},~{}\forall h\in\mathcal{H}.

(iv)

\left\|\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle v_{\alpha,\beta}\,d\mu(\alpha)\right\|\leq r\|h\|,~{}\forall h\in\mathcal{H};

\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle v_{\alpha,\beta}\,d\mu(\alpha)=\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}\langle h,v_{\alpha,\beta}\rangle u_{\alpha,\beta}\,d\mu(\alpha),~{}\forall h\in\mathcal{H};

0\leq\int_{\Omega}\sum_{\beta\in\Omega_{\alpha}}\langle h,u_{\alpha,\beta}\rangle\langle v_{\alpha,\beta},h\rangle\,d\mu(\alpha)\leq b\|h\|^{2},~{}\forall h\in\mathcal{H}.

Similarity of weak continuous operator-valued frames

Definition 8.13.

A weak continuous (ovf) $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Phi_{\alpha}\}_{\alpha\in\Omega})$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ is said to be right-similar to a weak continuous (ovf) $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ in $\mathcal{B}(\mathcal{H},\mathcal{H}_{0})$ if there exist invertible $R_{A,B},R_{\Psi,\Phi}\in\mathcal{B}(\mathcal{H})$ such that $B_{\alpha}=A_{\alpha}R_{A,B},\Phi_{\alpha}=\Psi_{\alpha}R_{\Psi,\Phi},\forall\alpha\in\Omega.$

Proposition 8.14.

Let $\{A_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}^{\text{w}}_{\Psi}$ with frame bounds $a,b,$ let $R_{A,B},R_{\Psi,\Phi}\in\mathcal{B}(\mathcal{H})$ be positive, invertible, commute with each other, commute with $S_{A,\Psi}$ , and let $B_{\alpha}=A_{\alpha}R_{A,B},\Phi_{\alpha}=\Psi_{\alpha}R_{\Psi,\Phi},\forall\alpha\in\Omega.$ Then $\{B_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}^{\text{w}}_{\Phi},$ $S_{B,\Phi}=R_{\Psi,\Phi}S_{A,\Psi}R_{A,B}$ , and $\frac{a}{\|R_{A,B}^{-1}\|\|R_{\Psi,\Phi}^{-1}\|}\leq S_{B,\Phi}\leq b\|R_{A,B}R_{\Psi,\Phi}\|.$ Assuming that $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ is a Parseval weak continuous (ovf), then $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Phi_{\alpha}\}_{\alpha\in\Omega})$ is a Parseval weak continuous (ovf) if and only if $R_{\Psi,\Phi}R_{A,B}=I_{\mathcal{H}}.$

Proposition 8.15.

Let $\{A_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}^{\text{w}}_{\Psi},$ $\{B_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}^{\text{w}}_{\Phi}$ and $B_{\alpha}=A_{\alpha}R_{A,B},\Phi_{\alpha}=\Psi_{\alpha}R_{\Psi,\Phi},\forall\alpha\in\Omega$ , for some invertible $R_{A,B},R_{\Psi,\Phi}\in\mathcal{B}(\mathcal{H}).$ Then $S_{B,\Phi}=R_{\Psi,\Phi}^{*}S_{A,\Psi}R_{A,B}.$ Assuming that $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ is a Parseval weak continuous (ovf), then $(\{B_{\alpha}\}_{\alpha\in\Omega},\{\Phi_{\alpha}\}_{\alpha\in\Omega})$ is a Parseval weak continuous (ovf) if and only if $R_{\Psi,\Phi}^{*}R_{A,B}=I_{\mathcal{H}}.$

Proof.

For all $h,g\in\mathcal{H}$ ,

	$\displaystyle\langle S_{B,\Phi}h,g\rangle$	$\displaystyle=\int_{\Omega}\langle B_{\alpha}h,\Phi_{\alpha}g\rangle\,d\mu(\alpha)=\int_{\Omega}\langle A_{\alpha}(R_{A,B}h),\Psi_{\alpha}(R_{\Psi,\Phi}g)\rangle\,d\mu(\alpha)$
		$\displaystyle=\langle S_{A,\Psi}(R_{A,B}),R_{\Psi,\Phi}g\rangle=\langle R_{\Psi,\Phi}^{*}S_{A,\Psi}R_{A,B}h,g\rangle.$

∎

Remark 8.16.

For every weak continuous (ovf) $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ , each of ‘weak continuous operator-valued frames’ $(\{A_{\alpha}S_{A,\Psi}^{-1}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega}),$ $(\{A_{\alpha}S_{A,\Psi}^{-1/2}\}_{\alpha\in\Omega},\{\Psi_{\alpha}S_{A,\Psi}^{-1/2}\}_{\alpha\in\Omega}),$ and $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}S_{A,\Psi}^{-1}\}_{\alpha\in\Omega})$ is a Parseval weak continuous (ovf) which is right-similar to $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega}).$

The case $\mathcal{H}_{0}=\mathbb{K}$ of weak continuous operator-valued frames

Definition 8.17.

A set of vectors $\{x_{\alpha}\}_{\alpha\in\Omega}$ in a Hilbert space $\mathcal{H}$ is said to be a weak continuous frame w.r.t. a set $\{\tau_{\alpha}\}_{\alpha\in\Omega}$ in $\mathcal{H}$ if

(i)

for each $h\in\mathcal{H}$ , both maps $\Omega\ni\alpha\mapsto\langle h,x_{\alpha}\rangle\in\mathbb{K}$ and $\Omega\ni\alpha\mapsto\langle h,\tau_{\alpha}\rangle\in\mathbb{K}$ are measurable,
(ii)

the map (we call as frame operator) $S_{x,\tau}:\mathcal{H}\ni h\mapsto\int_{\Omega}\langle h,x_{\alpha}\rangle\tau_{\alpha}\,d\mu(\alpha)\in\mathcal{H}$ is a well-defined bounded positive invertible operator.

Notions of frame bounds, optimal bounds, tight frame, Parseval frame, Bessel are similar to the same in Definition 6.1.

For fixed $\Omega,\mathcal{H},$ and $\{\tau_{\alpha}\}_{\alpha\in\Omega}$ , the set of all weak continuous frames for $\mathcal{H}$ w.r.t. $\{\tau_{\alpha}\}_{\alpha\in\Omega}$ is denoted by $\mathscr{F}^{w}_{\tau}.$

Proposition 8.18.

A set of vectors $\{x_{\alpha}\}_{\alpha\in\Omega}$ in $\mathcal{H}$ is a weak continuous frame w.r.t. a set $\{\tau_{\alpha}\}_{\alpha\in\Omega}$ in $\mathcal{H}$ if and only if there are $a,b,r>0$ such that

(i)

for each $h\in\mathcal{H}$ , both maps $\Omega\ni\alpha\mapsto\langle h,x_{\alpha}\rangle\in\mathbb{K}$ , $\Omega\ni\alpha\mapsto\langle h,\tau_{\alpha}\rangle\in\mathbb{K}$ are measurable,
(ii)

$\|\int_{\Omega}\langle h,x_{\alpha}\rangle\tau_{\alpha}\,d\mu(\alpha)\|\leq r\|h\|,\forall h\in\mathcal{H},$
(iii)

$a\|h\|^{2}\leq\int_{\Omega}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},h\rangle\,d\mu(\alpha)\leq b\|h\|^{2},\forall h\in\mathcal{H},$
(iv)

$\int_{\Omega}\langle h,x_{\alpha}\rangle\tau_{\alpha}\,d\mu(\alpha)=\int_{\Omega}\langle h,\tau_{\alpha}\rangle x_{\alpha}\,d\mu(\alpha),\forall h\in\mathcal{H}.$

If the space is over $\mathbb{C},$ then (iv) can be omitted.

Theorem 8.19.

Let $\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega}$ be in $\mathcal{H}$ . Define $A_{\alpha}:\mathcal{H}\ni h\mapsto\langle h,x_{\alpha}\rangle\in\mathbb{K}$ , $\Psi_{\alpha}:\mathcal{H}\ni h\mapsto\langle h,\tau_{\alpha}\rangle\in\mathbb{K},\forall\alpha\in\Omega$ . Then $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is a weak continuous frame for $\mathcal{H}$ if and only if $(\{A_{\alpha}\}_{\alpha\in\Omega},\{\Psi_{\alpha}\}_{\alpha\in\Omega})$ is a weak continuous operator-valued frame in $\mathcal{B}(\mathcal{H},\mathbb{K})$ .

Proposition 8.20.

If $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is a weak continuous frame for $\mathcal{H}$ , then every $h\in\mathcal{H}$ can be written as

h=\int_{\Omega}\langle h,S^{-1}_{x,\tau}\tau_{\alpha}\rangle x_{\alpha}\,d\mu(\alpha)=\int_{\Omega}\langle h,\tau_{\alpha}\rangle S^{-1}_{x,\tau}x_{\alpha}\,d\mu(\alpha)=\int_{\Omega}\langle h,S^{-1}_{x,\tau}x_{\alpha}\rangle\tau_{\alpha}\,d\mu(\alpha)=\int_{\Omega}\langle h,x_{\alpha}\rangle S^{-1}_{x,\tau}\tau_{\alpha}\,d\mu(\alpha).

Proof.

For all $h,g\in\mathcal{H}$ , $\langle h,g\rangle=\langle S_{x,\tau}h,S^{-1}_{x,\tau}g\rangle=\int_{\Omega}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},S^{-1}_{x,\tau}g\rangle\,d\mu(\alpha)=\int_{\Omega}\langle h,x_{\alpha}\rangle\langle S^{-1}_{x,\tau}\tau_{\alpha},g\rangle\,d\mu(\alpha)=\langle\int_{\Omega}\langle h,x_{\alpha}\rangle S^{-1}_{x,\tau}\tau_{\alpha}\,d\mu(\alpha),g\rangle$ , $\langle h,g\rangle=\langle S_{x,\tau}S_{x,\tau}^{-1}h,g\rangle=\int_{\Omega}\langle S^{-1}_{x,\tau}h,\tau_{\alpha}\rangle\langle x_{\alpha},g\rangle\,d\mu(\alpha)=\int_{\Omega}\langle h,S^{-1}_{x,\tau}\tau_{\alpha}\rangle\langle x_{\alpha},g\rangle\,d\mu(\alpha)$ $=\langle\int_{\Omega}\langle h,S^{-1}_{x,\tau}\tau_{\alpha}\rangle x_{\alpha}\,d\mu(\alpha),g\rangle$ . ∎

Proposition 8.21.

Let $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ be a weak continuous frame for $\mathcal{H}$ with upper frame bound $b$ . If for some $\alpha\in\Omega$ we have $\{\alpha\}$ is measurable and $\langle x_{\alpha},x_{\beta}\rangle\langle\tau_{\beta},x_{\alpha}\rangle\geq 0,\forall\beta\in\Omega$ , then $\mu(\{\alpha\})\langle x_{\alpha},\tau_{\alpha}\rangle\leq b$ for that $\alpha.$

Definition 8.22.

A weak continuous frame $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ for $\mathcal{H}$ is said to be a dual of weak continuous frame $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ for $\mathcal{H}$ if $\int_{\Omega}\langle h,x_{\alpha}\rangle\omega_{\alpha}\,d\mu(\alpha)=\int_{\Omega}\langle h,\tau_{\alpha}\rangle y_{\alpha}\,d\mu(\alpha)=h,\forall h\in\mathcal{H}$ . The ‘weak continuous frame’ $(\{\widetilde{x}_{\alpha}\coloneqq S_{x,\tau}^{-1}x_{\alpha}\}_{\alpha\in\Omega},\{\widetilde{\tau}_{\alpha}\coloneqq S_{x,\tau}^{-1}\tau_{\alpha}\}_{\alpha\in\Omega})$ , which is a ‘dual’ of $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is called the canonical dual of $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ .

Proposition 8.23.

Let $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ be a weak continuous frame for $\mathcal{H}.$ If $h\in\mathcal{H}$ has representation $h=\int_{\Omega}f(\alpha)x_{\alpha}\,d\mu(\alpha)=\int_{\Omega}g(\alpha)\tau_{\alpha}\,d\mu(\alpha),$ for some measurable $f,g:\Omega\rightarrow\mathbb{K}$ , then

\int_{\Omega}f(\alpha)\overline{g(\alpha)}\,d\mu(\alpha)=\int_{\Omega}\langle h,\widetilde{\tau}_{\alpha}\rangle\langle\widetilde{x}_{\alpha},h\rangle\,d\mu(\alpha)+\int_{\Omega}(f(\alpha)-\langle h,\widetilde{\tau}_{\alpha}\rangle)(\overline{g(\alpha)}-\langle\widetilde{x}_{\alpha},h\rangle)\,d\mu(\alpha).

Theorem 8.24.

Let $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ be a weak continuous frame for $\mathcal{H}$ with frame bounds $a$ and $b.$ Then

(i)

The canonical dual weak continuous frame of the canonical dual weak continuous frame of $(\{x_{\alpha}\}_{\alpha\in\Omega}$ , $\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is itself.
(ii)

$\frac{1}{b},\frac{1}{a}$ are frame bounds for the canonical dual of $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega}).$
(iii)

If $a,b$ are optimal frame bounds for $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega}),$ then $\frac{1}{b},\frac{1}{a}$ are optimal frame bounds for its canonical dual.

Definition 8.25.

A weak continuous frame $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ for $\mathcal{H}$ is said to be orthogonal to a weak continuous frame $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ for $\mathcal{H}$ if $\int_{\Omega}\langle h,x_{\alpha}\rangle\omega_{\alpha}\,d\mu(\alpha)=\int_{\Omega}\langle h,\tau_{\alpha}\rangle y_{\alpha}\,d\mu(\alpha)=0,\forall h\in\mathcal{H}.$

Proposition 8.26.

Two orthogonal weak continuous frames have a common dual weak continuous frame.

Proof.

Let $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ and $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ be orthogonal weak continuous frames for $\mathcal{H}$ . Define $z_{\alpha}\coloneqq S_{x,\tau}^{-1}x_{\alpha}+S_{y,\omega}^{-1}y_{\alpha},\rho_{\alpha}\coloneqq S_{x,\tau}^{-1}\tau_{\alpha}+S_{y,\omega}^{-1}\omega_{\alpha},\forall\alpha\in\Omega$ . For $h,g\in\mathcal{H}$ ,

	$\displaystyle\langle S_{z,\rho}h,g\rangle$	$\displaystyle=\int_{\Omega}\langle h,z_{\alpha}\rangle\langle\rho_{\alpha},g\rangle\,d\mu(\alpha)$
		$\displaystyle=\int_{\Omega}\langle h,S_{x,\tau}^{-1}x_{\alpha}+S_{y,\omega}^{-1}y_{\alpha}\rangle\langle S_{x,\tau}^{-1}\tau_{\alpha}+S_{y,\omega}^{-1}\omega_{\alpha},g\rangle\,d\mu(\alpha)$
		$\displaystyle=\int_{\Omega}\langle S_{x,\tau}^{-1}h,x_{\alpha}\rangle\langle\tau_{\alpha},S_{x,\tau}^{-1}g\rangle\,d\mu(\alpha)+\int_{\Omega}\langle S_{x,\tau}^{-1}h,x_{\alpha}\rangle\langle\omega_{\alpha},S_{y,\omega}^{-1}g\rangle\,d\mu(\alpha)$
		$\displaystyle\quad+\int_{\Omega}\langle S_{y,\omega}^{-1}h,y_{\alpha}\rangle\langle\tau_{\alpha},S_{x,\tau}^{-1}g\rangle\,d\mu(\alpha)+\int_{\Omega}\langle S_{y,\omega}^{-1}h,y_{\alpha}\rangle\langle\omega_{\alpha},S_{y,\omega}^{-1}g\rangle\,d\mu(\alpha)$
		$\displaystyle=\langle S_{x,\tau}S_{x,\tau}^{-1}h,S_{x,\tau}^{-1}g\rangle+\left\langle\int_{\Omega}\langle S_{x,\tau}^{-1}h,x_{\alpha}\rangle\omega_{\alpha}\,d\mu(\alpha),S_{y,\omega}^{-1}g\right\rangle$
		$\displaystyle\quad+\left\langle\int_{\Omega}\langle S_{y,\omega}^{-1}h,y_{\alpha}\rangle\tau_{\alpha}\,d\mu(\alpha),S_{x,\tau}^{-1}g\right\rangle+\langle S_{y,\omega}S_{y,\omega}^{-1}h,S_{y,\omega}^{-1}g\rangle$
		$\displaystyle=\langle S_{x,\tau}^{-1}h,g\rangle+\langle 0,S_{y,\omega}^{-1}g\rangle+\langle 0,S_{x,\tau}^{-1}g\rangle+\langle S_{y,\omega}^{-1}h,g\rangle=\langle S_{x,\tau}^{-1}h+S_{y,\omega}^{-1}h,g\rangle.$

Therefore $S_{z,\rho}=S_{x,\tau}^{-1}+S_{y,\omega}^{-1}$ which tells that $(\{z_{\alpha}\}_{\alpha\in\Omega},\{\rho_{\alpha}\}_{\alpha\in\Omega})$ is a weak continuous frame for $\mathcal{H}$ . For duality, let $h,g\in\mathcal{H}$ . Then

	$\displaystyle\left\langle\int_{\Omega}\langle h,x_{\alpha}\rangle\rho_{\alpha}\,d\mu(\alpha),g\right\rangle$	$\displaystyle=\left\langle\int_{\Omega}\langle h,x_{\alpha}\rangle(S_{x,\tau}^{-1}\tau_{\alpha}+S_{y,\omega}^{-1}\omega_{\alpha})\,d\mu(\alpha),g\right\rangle$
		$\displaystyle=\int_{\Omega}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},S_{x,\tau}^{-1}g\rangle\,d\mu(\alpha)+\int_{\Omega}\langle h,x_{\alpha}\rangle\langle\omega_{\alpha},S_{y,\omega}^{-1}g\rangle\,d\mu(\alpha)=\langle h,g\rangle+\langle 0,S_{y,\omega}^{-1}g\rangle,$

	$\displaystyle\left\langle\int_{\Omega}\langle h,\tau_{\alpha}\rangle z_{\alpha}\,d\mu(\alpha),g\right\rangle$	$\displaystyle=\left\langle\int_{\Omega}\langle h,\tau_{\alpha}\rangle(S_{x,\tau}^{-1}x_{\alpha}+S_{y,\omega}^{-1}y_{\alpha})\,d\mu(\alpha),g\right\rangle$
		$\displaystyle=\int_{\Omega}\langle h,\tau_{\alpha}\rangle\langle x_{\alpha},S_{x,\tau}^{-1}g\rangle\,d\mu(\alpha)+\int_{\Omega}\langle h,\tau_{\alpha}\rangle\langle y_{\alpha},S_{y,\omega}^{-1}g\rangle\,d\mu(\alpha)=\langle h,g\rangle+\langle 0,S_{y,\omega}^{-1}g\rangle,$

and

	$\displaystyle\left\langle\int_{\Omega}\langle h,y_{\alpha}\rangle\rho_{\alpha}\,d\mu(\alpha),g\right\rangle$	$\displaystyle=\left\langle\int_{\Omega}\langle h,y_{\alpha}\rangle(S_{x,\tau}^{-1}\tau_{\alpha}+S_{y,\omega}^{-1}\omega_{\alpha})\,d\mu(\alpha),g\right\rangle$
		$\displaystyle=\int_{\Omega}\langle h,y_{\alpha}\rangle\langle\tau_{\alpha},S_{x,\tau}^{-1}g\rangle\,d\mu(\alpha)+\int_{\Omega}\langle h,y_{\alpha}\rangle\langle\omega_{\alpha},S_{y,\omega}^{-1}g\rangle\,d\mu(\alpha)=\langle 0,S_{y,\omega}^{-1}g\rangle+\langle h,g\rangle,$

	$\displaystyle\left\langle\int_{\Omega}\langle h,\omega_{\alpha}\rangle z_{\alpha}\,d\mu(\alpha),g\right\rangle$	$\displaystyle=\left\langle\int_{\Omega}\langle h,\omega_{\alpha}\rangle(S_{x,\tau}^{-1}x_{\alpha}+S_{y,\omega}^{-1}y_{\alpha})\,d\mu(\alpha),g\right\rangle$
		$\displaystyle=\int_{\Omega}\langle h,\omega_{\alpha}\rangle\langle x_{\alpha},S_{x,\tau}^{-1}g\rangle\,d\mu(\alpha)+\int_{\Omega}\langle h,\omega_{\alpha}\rangle\langle y_{\alpha},S_{y,\omega}^{-1}g\rangle\,d\mu(\alpha)=\langle 0,S_{x,\tau}^{-1}g\rangle+\langle h,g\rangle.$

∎

Proposition 8.27.

Let $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ and $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ be two Parseval weak continuous frames for $\mathcal{H}$ which are orthogonal. If $A,B,C,D\in\mathcal{B}(\mathcal{H})$ are such that $AC^{*}+BD^{*}=I_{\mathcal{H}}$ , then $(\{Ax_{\alpha}+By_{\alpha}\}_{\alpha\in\Omega},\{C\tau_{\alpha}+D\omega_{\alpha}\}_{\alpha\in\Omega})$ is a Parseval weak continuous frame for $\mathcal{H}$ . In particular, if scalars $a,b,c,d$ satisfy $a\bar{c}+b\bar{d}=1$ , then $(\{ax_{\alpha}+by_{\alpha}\}_{\alpha\in\Omega},\{c\tau_{\alpha}+d\omega_{\alpha}\}_{\alpha\in\Omega})$ is a Parseval weak continuous frame for $\mathcal{H}$ .

Proof.

For all $h,g\in\mathcal{H}$ ,

	$\displaystyle\langle S_{Ax+By,C\tau+D\omega}h,g\rangle=\int_{\Omega}\langle h,Ax_{\alpha}+By_{\alpha}\rangle\langle C\tau_{\alpha}+D\omega_{\alpha},g\rangle\,d\mu(\alpha)$
	$\displaystyle=\int_{\Omega}\langle A^{}h,x_{\alpha}\rangle\langle\tau_{\alpha},C^{}g\rangle\,d\mu(\alpha)+\int_{\Omega}\langle A^{}h,x_{\alpha}\rangle\langle\omega_{\alpha},D^{}g\rangle\,d\mu(\alpha)$
	$\displaystyle\quad+\int_{\Omega}\langle B^{}h,y_{\alpha}\rangle\langle\tau_{\alpha},C^{}g\rangle\,d\mu(\alpha)+\int_{\Omega}\langle B^{}h,y_{\alpha}\rangle\langle\omega_{\alpha},D^{}g\rangle\,d\mu(\alpha)$
	$\displaystyle=\langle A^{}h,C^{}g\rangle+\langle 0,D^{}g\rangle+\langle 0,C^{}g\rangle+\langle B^{}h,D^{}g\rangle$
	$\displaystyle=\langle CA^{}h,g\rangle+\langle DB^{}h,g\rangle=\langle(CA^{}+DB^{})h,g\rangle=\langle h,g\rangle.$

∎

Definition 8.28.

Two weak continuous frames $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ and $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ for $\mathcal{H}$ are called disjoint if $(\{x_{\alpha}\oplus y_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\oplus\omega_{\alpha}\}_{\alpha\in\Omega})$ is a weak continuous frame for $\mathcal{H}\oplus\mathcal{H}$ .

Proposition 8.29.

If $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ and $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ are orthogonal weak continuous frames for $\mathcal{H}$ , then they are disjoint. Further, if both $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ and $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ are Parseval weak, then $(\{x_{\alpha}\oplus y_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\oplus\omega_{\alpha}\}_{\alpha\in\Omega})$ is Parseval weak.

Proof.

Let $h\oplus g,u\oplus v\in\mathcal{H}\oplus\mathcal{H}$ . Then

	$\displaystyle\langle S_{x\oplus y,\tau\oplus\omega}(h\oplus g),u\oplus v\rangle=\int_{\Omega}\langle h\oplus g,x_{\alpha}\oplus y_{\alpha}\rangle\langle\tau_{\alpha}\oplus\omega_{\alpha},u\oplus v\rangle\,d\mu(\alpha)$
	$\displaystyle=\int_{\Omega}(\langle h,x_{\alpha}\rangle+\langle g,y_{\alpha}\rangle)(\langle\tau_{\alpha},u\rangle+\langle\omega_{\alpha},v\rangle)\,d\mu(\alpha)=\int_{\Omega}\langle h,x_{\alpha}\rangle\langle\tau_{\alpha},u\rangle\,d\mu(\alpha)$
	$\displaystyle\quad+\int_{\Omega}\langle h,x_{\alpha}\rangle\langle\omega_{\alpha},v\rangle\,d\mu(\alpha)+\int_{\Omega}\langle g,y_{\alpha}\rangle\langle\tau_{\alpha},u\rangle\,d\mu(\alpha)+\int_{\Omega}\langle g,y_{\alpha}\rangle\langle\omega_{\alpha},v\rangle\,d\mu(\alpha)$
	$\displaystyle=\langle S_{x,\tau}h,u\rangle+\langle 0,v\rangle+\langle 0,u\rangle+\langle S_{y,\omega}g,v\rangle=\langle S_{x,\tau}h\oplus S_{y,\omega}g,u\oplus v\rangle$
	$\displaystyle=\langle(S_{x,\tau}\oplus S_{y,\omega})(h\oplus g),u\oplus v\rangle\quad\implies S_{x\oplus y,\tau\oplus\omega}=S_{x,\tau}\oplus S_{y,\omega}.$

∎

Similarity

Definition 8.30.

A weak continuous frame $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ for $\mathcal{H}$ is said to be similar to a weak continuous frame $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ for $\mathcal{H}$ if there are invertible $T_{x,y},T_{\tau,\omega}\in\mathcal{B}(\mathcal{H})$ such that $y_{\alpha}=T_{x,y}x_{\alpha},\omega_{\alpha}=T_{\tau,\omega}\tau_{\alpha},\forall\alpha\in\Omega.$

Proposition 8.31.

Let $\{x_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}^{w}_{\tau}$ with frame bounds $a,b,$ let $T_{x,y},T_{\tau,\omega}\in\mathcal{B}(\mathcal{H})$ be positive, invertible, commute with each other, commute with $S_{x,\tau}$ , and let $y_{\alpha}=T_{x,y}x_{\alpha},\omega_{\alpha}=T_{\tau,\omega}\tau_{\alpha},\forall\alpha\in\Omega.$ Then $\{y_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}^{w}_{\tau}$ , $S_{y,\omega}=T_{\tau,\omega}S_{x,\tau}T_{x,y}$ , and $\frac{a}{\|T_{x,y}^{-1}\|\|T_{\tau,\omega}^{-1}\|}\leq S_{y,\omega}\leq b\|T_{x,y}T_{\tau,\omega}\|$ . Assuming that $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is Parseval weak continuous, then $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ is Parseval weak continuous if and only if $T_{\tau,\omega}T_{x,y}=I_{\mathcal{H}}.$

Proposition 8.32.

Let $\{x_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}^{w}_{\tau},$ $\{y_{\alpha}\}_{\alpha\in\Omega}\in\mathscr{F}^{w}_{\omega}$ and $y_{\alpha}=T_{x,y}x_{\alpha},\omega_{\alpha}=T_{\tau,\omega}\tau_{\alpha},\forall\alpha\in\Omega$ , for some invertible $T_{x,y},T_{\tau,\omega}\in\mathcal{B}(\mathcal{H}).$ Then $S_{y,\omega}=T_{\tau,\omega}S_{x,\tau}T_{x,y}^{*}.$ Assuming that $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ is Parseval weak continuous frame, then $(\{y_{\alpha}\}_{\alpha\in\Omega},\{\omega_{\alpha}\}_{\alpha\in\Omega})$ is Parseval weak continuous frame if and only if $T_{\tau,\omega}T_{x,y}^{*}=I_{\mathcal{H}}.$

Proof.

For all $h,g\in\mathcal{H}$ ,

	$\displaystyle\langle S_{y,\omega}h,g\rangle$	$\displaystyle=\int_{\Omega}\langle h,y_{\alpha}\rangle\langle\omega_{\alpha},g\rangle\,d\mu(\alpha)=\int_{\Omega}\langle h,T_{x,y}x_{\alpha}\rangle\langle T_{\tau,\omega}\tau_{\alpha},g\rangle\,d\mu(\alpha)$
		$\displaystyle=\int_{\Omega}\langle T_{x,y}^{}h,x_{\alpha}\rangle\langle\tau_{\alpha},T_{\tau,\omega}^{}g\rangle\,d\mu(\alpha)=\langle S_{x,\tau}(T^{}_{x,y}h),T^{}_{\tau,\omega}g\rangle=\langle T_{\tau,\omega}S_{x,\tau}T^{*}_{x,y}h,g\rangle.$

∎

Remark 8.33.

For every weak continuous frame $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega}),$ each of ‘weak continuous frames’ $(\{S_{x,\tau}^{-1}x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega})$ , $(\{S_{x,\tau}^{-1/2}x_{\alpha}\}_{\alpha\in\Omega},\{S_{x,\tau}^{-1/2}\tau_{\alpha}\}_{\alpha\in\Omega}),$ and $(\{x_{\alpha}\}_{\alpha\in\Omega},\{S_{x,\tau}^{-1}\tau_{\alpha}\}_{\alpha\in\Omega})$ is a Parseval weak continuous frame which is similar to $(\{x_{\alpha}\}_{\alpha\in\Omega},\{\tau_{\alpha}\}_{\alpha\in\Omega}).$

9. Acknowledgments

The first author thanks the National Institute of Technology Karnataka (NITK), Surathkal for giving financial support and the present work of the second author was partially supported by National Board for Higher Mathematics (NBHM), Ministry of Atomic Energy, Government of India (Reference No.2/48(16)/2012/NBHM(R.P.)/R&D 11/9133).

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	$\displaystyle\left\\|\theta_{A}h\right\\|^{2}$	$\displaystyle=\int_{\Omega}\\|A_{\alpha}h\\|^{2}\,d\mu(\alpha)=\int_{\Omega}\sum\limits_{\beta\in\Omega_{\alpha}}\|\langle h,u_{\alpha,\beta}\rangle\|^{2}\,d\mu(\alpha);$
	$\displaystyle\left\\|\theta_{\Psi}h\right\\|^{2}$	$\displaystyle=\int_{\Omega}\sum\limits_{\beta\in\Omega_{\alpha}}\|\langle h,v_{\alpha,\beta}\rangle\|^{2}\,d\mu(\alpha).$

	$\displaystyle\langle\lambda_{g}\theta_{A}\theta_{\Psi}^{}\lambda_{g}^{}u,v\rangle$	$\displaystyle=\langle\theta_{\Psi}^{}\lambda_{g}^{}u,\theta_{A}^{}\lambda_{g}^{}v\rangle$
		$\displaystyle=\left\langle\int_{G}\Psi_{\alpha}^{}\lambda_{g}^{}u(\alpha)\,d\mu_{G}(\alpha),\int_{G}A_{\beta}^{}\lambda_{g}^{}v(\beta)\,d\mu_{G}(\beta)\right\rangle$
		$\displaystyle=\left\langle\int_{G}\Psi_{\alpha}^{}\lambda_{g^{-1}}u(\alpha)\,d\mu_{G}(\alpha),\int_{G}A_{\beta}^{}\lambda_{g^{-1}}v(\beta)\,d\mu_{G}(\beta)\right\rangle$
		$\displaystyle=\left\langle\int_{G}\Psi_{\alpha}^{}u(g\alpha)\,d\mu_{G}(\alpha),\int_{G}A_{\beta}^{}v(g\beta)\,d\mu_{G}(\beta)\right\rangle$
		$\displaystyle=\left\langle\int_{G}\Psi_{g^{-1}p}^{}u(p)\,d\mu_{G}(g^{-1}p),\int_{G}A_{g^{-1}q}^{}v(q)\,d\mu_{G}(g^{-1}q)\right\rangle$
		$\displaystyle=\left\langle\int_{G}\Psi_{g^{-1}p}^{}u(p)\,d\mu_{G}(p),\int_{G}A_{g^{-1}q}^{}v(q)\,d\mu_{G}(q)\right\rangle$
		$\displaystyle=\int_{G}\int_{G}\langle A_{g^{-1}q}\Psi_{g^{-1}p}^{*}u(p),v(q)\rangle\,d\mu_{G}(q)\,d\mu_{G}(p)$
		$\displaystyle=\int_{G}\int_{G}\langle A_{q}\Psi_{p}^{*}u(p),v(q)\rangle\,d\mu_{G}(q)\,d\mu_{G}(p)$
		$\displaystyle=\left\langle\int_{G}\Psi_{p}^{}u(p)\,d\mu_{G}(p),\int_{G}A_{q}^{}v(q)\,d\mu_{G}(q)\right\rangle=\langle\theta_{\Psi}^{}u,\theta_{A}^{}v\rangle=\langle\theta_{A}\theta_{\Psi}^{*}u,v\rangle.$

	$\displaystyle\langle A_{e}\pi_{g^{-1}}h,f\rangle$	$\displaystyle=\langle\pi_{g^{-1}}h,A_{e}^{}f\rangle=\langle\theta_{\Psi}^{}\lambda_{g^{-1}}\theta_{A}h,A_{e}^{*}f\rangle$
		$\displaystyle=\left\langle\int_{G}\Psi_{\alpha}^{}\lambda_{g^{-1}}\theta_{A}h(\alpha)\,d\mu_{G}(\alpha),A_{e}^{}f\right\rangle=\left\langle\int_{G}\Psi_{\alpha}^{}\theta_{A}h(g\alpha)\,d\mu_{G}(\alpha),A_{e}^{}f\right\rangle$
		$\displaystyle=\left\langle\int_{G}\Psi_{\alpha}^{}A_{g\alpha}h\,d\mu_{G}(\alpha),A_{e}^{}f\right\rangle=\left\langle\int_{G}\Psi_{g^{-1}\beta}^{}A_{\beta}h\,d\mu_{G}(g^{-1}\beta),A_{e}^{}f\right\rangle$
		$\displaystyle=\left\langle\int_{G}\Psi_{g^{-1}\beta}^{}A_{\beta}h\,d\mu_{G}(\beta),A_{e}^{}f\right\rangle=\int_{G}\langle A_{g^{-1}g}\Psi_{g^{-1}\beta}^{*}A_{\beta}h,f\rangle\,d\mu_{G}(\beta)$
		$\displaystyle=\int_{G}\langle A_{g}\Psi_{\beta}^{}A_{\beta}h,f\rangle\,d\mu_{G}(\beta)=\left\langle\int_{G}\Psi_{\beta}^{}A_{\beta}h\,d\mu_{G}(\beta),A_{g}^{*}f\right\rangle$
		$\displaystyle=\langle h,A_{g}^{*}f\rangle=\langle A_{g}h,f\rangle,$

	$\displaystyle\langle\Psi_{e}\pi_{g^{-1}}h,f\rangle$	$\displaystyle=\langle\pi_{g^{-1}}h,\Psi_{e}^{}f\rangle=\langle\theta_{\Psi}^{}\lambda_{g^{-1}}\theta_{A}h,\Psi_{e}^{*}f\rangle$
		$\displaystyle=\left\langle\int_{G}\Psi_{\alpha}^{}\lambda_{g^{-1}}\theta_{A}h(\alpha)\,d\mu_{G}(\alpha),\Psi_{e}^{}f\right\rangle=\left\langle\int_{G}\Psi_{\alpha}^{}\theta_{A}h(g\alpha)\,d\mu_{G}(\alpha),\Psi_{e}^{}f\right\rangle$
		$\displaystyle=\left\langle\int_{G}\Psi_{\alpha}^{}A_{g\alpha}h\,d\mu_{G}(\alpha),\Psi_{e}^{}f\right\rangle=\left\langle\int_{G}\Psi_{g^{-1}\beta}^{}A_{\beta}h\,d\mu_{G}(g^{-1}\beta),\Psi_{e}^{}f\right\rangle$
		$\displaystyle=\left\langle\int_{G}\Psi_{g^{-1}\beta}^{}A_{\beta}h\,d\mu_{G}(\beta),\Psi_{e}^{}f\right\rangle=\int_{G}\langle\Psi_{g^{-1}g}\Psi_{g^{-1}\beta}^{*}A_{\beta}h,f\rangle\,d\mu_{G}(\beta)$
		$\displaystyle=\int_{G}\langle\Psi_{g}\Psi_{\beta}^{}A_{\beta}h,f\rangle\,d\mu_{G}(\beta)=\left\langle\int_{G}\Psi_{\beta}^{}A_{\beta}h\,d\mu_{G}(\beta),\Psi_{g}^{*}f\right\rangle$
		$\displaystyle=\langle h,\Psi_{g}^{*}f\rangle=\langle\Psi_{g}h,f\rangle.$

	$\displaystyle\\|Tf\\|$	$\displaystyle=\left\\|\int_{\Omega}B_{\alpha}^{}f(\alpha)\,d\mu(\alpha)\right\\|\leq\left\\|\int_{\Omega}(B_{\alpha}^{}-A_{\alpha}^{})f(\alpha)\,d\mu(\alpha)\right\\|+\left\\|\int_{\Omega}A_{\alpha}^{}f(\alpha)\,d\mu(\alpha)\right\\|$
		$\displaystyle\leq(1+\alpha)\left\\|\int_{\Omega}A_{\alpha}^{}f(\alpha)\,d\mu(\alpha)\right\\|+\beta\left\\|\int_{\Omega}B_{\alpha}^{}f(\alpha)\,d\mu(\alpha)\right\\|+\gamma\\|f\\|$
		$\displaystyle=(1+\alpha)\left\\|\theta_{A}^{*}f\right\\|+\beta\left\\|Tf\right\\|+\gamma\\|f\\|.$

This is the title

1. Introduction

Definition 1.1.

Definition 1.2.

Definition 1.3.

Definition 1.4.

Definition 1.5.

2. Extension of continuous operator-valued frames

Definition 2.1.

Definition 2.2.

Remark 2.3.

Proposition 2.4.

Proof.

Proposition 2.5.

Proof.

Proposition 2.6.

Proof.

Proposition 2.7.

Definition 2.8.

Proposition 2.9.

Proof.

Theorem 2.10.

Definition 2.11.

Proposition 2.12.

Proof.

Theorem 2.13.

Proof.

Proposition 2.14.

Proof.

Theorem 2.15.

Proof.

Proposition 2.16.

Proof.

Definition 2.17.

Proposition 2.18.

Proposition 2.19.

Proof.

Proposition 2.20.

Proof.

Definition 2.21.

Proposition 2.22.

Proof.

3. Characterizations of the extension

Theorem 3.1.

Proof.

Theorem 3.2.

Proof.

Definition 3.3.

Proposition 3.4.

Proof.

Lemma 3.5.

Proof.

Theorem 3.6.

Proof.

Corollary 3.7.

Proof.

Corollary 3.8.

Proof.

Remark 3.9.

4. Continuous frames and representations of locally compact groups

Definition 4.1.

Proposition 4.2.

Proof.

Theorem 4.3.

Proof.

Corollary 4.4.

Proof.

Corollary 4.5.

5. Perturbations

Theorem 5.1.

Proof.

Corollary 5.2.

Proof.

Theorem 5.3.

Proof.

Theorem 5.4.

Proof.

6. Case ℋ0=𝕂\mathcal{H}_{0}=\mathbb{K}

Definition 6.1.

Theorem 6.2.

6. Case $\mathcal{H}_{0}=\mathbb{K}$