Multiplication Operators between Lipschitz-Type Spaces on a Tree

Robert F. Allen¹, Flavia Colonna², and Glenn R. Easley³ ¹Department of Mathematics, University of Wisconsin–La Crosse ²Department of Mathematical Sciences, George Mason University ³System Planning Corporation allen.rob3@uwlax.edu, fcolonna@gmu.edu, geasley@sysplan.com

Abstract.

Let $\displaystyle\mathcal{L}$ be the space of complex-valued functions $\displaystyle f$ on the set of vertices $\displaystyle T$ of an rooted infinite tree rooted at $\displaystyle o$ such that the difference of the values of $\displaystyle f$ at neighboring vertices remains bounded throughout the tree, and let $\displaystyle\mathcal{L}_{\textbf{w}}$ be the set of functions $\displaystyle f\in\mathcal{L}$ such that $\displaystyle|f(v)-f(v^{-})|=O(|v|^{-1})$ , where $\displaystyle|v|$ is the distance between $\displaystyle o$ and $\displaystyle v$ and $\displaystyle v^{-}$ is the neighbor of $\displaystyle v$ closest to $\displaystyle o$ . In this article, we characterize the bounded and the compact multiplication operators between $\displaystyle\mathcal{L}$ and $\displaystyle\mathcal{L}_{\textbf{w}}$ , and provide operator norm and essential norm estimates. Furthermore, we characterize the bounded and compact multiplication operators between $\displaystyle\mathcal{L}_{\textbf{w}}$ and the space $\displaystyle L^{\infty}$ of bounded functions on $\displaystyle T$ and determine their operator norm and their essential norm. We establish that there are no isometries among the multiplication operators between these spaces.

Key words and phrases:

Multiplication operators, Trees, Lipschitz space

2010 Mathematics Subject Classification:

primary: 47B38; secondary: 05C05

1. Introduction

Let $\displaystyle\mathcal{X}$ and $\displaystyle\mathcal{Y}$ be complex Banach spaces of functions defined on a set $\displaystyle\Omega$ . For a complex-valued function $\displaystyle\psi$ defined on $\displaystyle\Omega$ , the multiplication operator with symbol $\displaystyle\psi$ from $\displaystyle\mathcal{X}$ to $\displaystyle\mathcal{Y}$ is defined as

\displaystyle M_{\psi}f=\psi f,\hbox{ for all }f\in\mathcal{X}.

A fundamental objective in the study of the operators with symbol is to tie the properties of the operator to the function-theoretic properties of the symbol.

When $\displaystyle\Omega$ is taken to be the open unit disk $\displaystyle\mathbb{D}$ in the complex plane, an important space of functions to study is the Bloch space, defined as the set $\displaystyle\mathcal{B}$ of the analytic functions $\displaystyle f:\mathbb{D}\to\mathbb{C}$ for which

\displaystyle\beta_{f}=\sup_{z\in\mathbb{D}}(1-\left|z\right|^{2})\left|f^{\prime}(z)\right|<\infty.

The Bloch space can also be described as the set consisting of the Lipschitz functions between metric spaces from $\displaystyle\mathbb{D}$ endowed with the Poincaré distance $\displaystyle\rho$ to $\displaystyle\mathbb{C}$ endowed with the Euclidean distance, a fact that was proved by the second author in [4] (see also [8]). In fact, $\displaystyle f\in\mathcal{B}$ if and only if there exist $\displaystyle\beta>0$ such that

\displaystyle\left|f(z)-f(w)\right|\leq\beta\rho(z,w),

and

\displaystyle\beta_{f}=\sup_{z\neq w}\frac{\left|f(z)-f(w)\right|}{\rho(z,w)}.

More recently, considerable research has been carried out in the field of operator theory when the set $\displaystyle\Omega$ is taken to be a discrete structure, such as a discrete group or a graph. In this work, we consider the case when $\displaystyle\Omega$ is taken to be an infinite tree.

By a tree $\displaystyle T$ we mean a locally finite, connected, and simply-connected graph, which, as a set, we identify with the collection of its vertices. Two vertices $\displaystyle u$ and $\displaystyle v$ are called neighbors if there is an edge connecting them, and we use the notation $\displaystyle u\sim v$ . A vertex is called terminal if it has a unique neighbor. A path is a finite or infinite sequence of vertices $\displaystyle[v_{0},v_{1},\dots]$ such that $\displaystyle v_{k}\sim v_{k+1}$ and $\displaystyle v_{k-1}\neq v_{k+1}$ , for all $\displaystyle k$ .

Given a tree $\displaystyle T$ rooted at $\displaystyle o$ and a vertex $\displaystyle u\in T$ , a vertex $\displaystyle v$ is called a descendant of $\displaystyle u$ if $\displaystyle u$ lies in the unique path from $\displaystyle o$ to $\displaystyle v$ . The vertex $\displaystyle u$ is then called an ancestor of $\displaystyle v$ . Given a vertex $\displaystyle v\neq o$ , we denote by $\displaystyle v^{-}$ the unique neighbor which is an ancestor of $\displaystyle v$ . For $\displaystyle v\in T$ , The set $\displaystyle S_{v}$ consisting of $\displaystyle v$ and all its descendants is called the sector determined by $\displaystyle v$ .

Define the length of a finite path $\displaystyle[u=u_{0},u_{1},\dots,v=u_{n}]$ (with $\displaystyle u_{k}\sim u_{k+1}$ for $\displaystyle k=0,\dots,n$ ) to be the number $\displaystyle n$ of edges connecting $\displaystyle u$ to $\displaystyle v$ . The distance, $\displaystyle d(u,v)$ , between vertices $\displaystyle u$ and $\displaystyle v$ is the length of the path connecting $\displaystyle u$ to $\displaystyle v$ . The tree $\displaystyle T$ is a metric space under the distance $\displaystyle d$ . Fixing $\displaystyle o$ as the root of the tree, we define the length of a vertex $\displaystyle v$ , by $\displaystyle|v|=d(o,v)$ . By a function on a tree we mean a complex-valued function on the set of its vertices.

In this paper, the tree will be assumed to be rooted at a vertex $\displaystyle o$ and without terminal vertices (and hence infinite).

Infinite trees are discrete structures which exhibit significant geometric and potential-theoretic characteristics that are present in the Poincaré disk $\displaystyle\mathbb{D}$ . For instance, they have a boundary, which is defined as the set of equivalence classes of paths which differ by finitely many vertices. The union of the boundary with the tree yields a compact space. A useful resource for the potential theory on trees illustrating the commonalities with the disk is [2]. In [3] it was shown that, if the tree has the property that all its vertices have the same number of neighbors, then there is a natural embedding of the tree in the unit disk such that the edges of the tree are arcs of geodesics in $\displaystyle\mathbb{D}$ with the same hyperbolic length and the set of cluster points of the vertices is the entire unit circle.

In [5], the last two authors defined the Lipschitz space $\displaystyle\mathcal{L}$ on a tree $\displaystyle T$ as the set consisting of the functions $\displaystyle f:T\to\mathbb{C}$ which are Lipschitz with respect to the distance $\displaystyle d$ on $\displaystyle T$ and the Euclidean distance on $\displaystyle\mathbb{C}$ . For this reason, the Lipschitz space $\displaystyle\mathcal{L}$ can be viewed as a discrete analogue of the Bloch space $\displaystyle\mathcal{B}$ . It was also shown that the Lipschitz functions on $\displaystyle T$ are precisely the functions for which

\displaystyle\|Df\|_{\infty}=\sup_{v\in T^{*}}Df(v)<\infty,

where $\displaystyle Df(v)=|f(v)-f(v^{-})|$ and $\displaystyle T^{*}=T\setminus\{o\}$ . Under the norm

\displaystyle\|f\|_{\mathcal{L}}=|f(o)|+\|Df\|_{\infty},

$\displaystyle\mathcal{L}$ is a Banach space containing the space $\displaystyle L^{\infty}$ of the bounded functions on $\displaystyle T$ . Furthermore, for $\displaystyle f\in L^{\infty}$ , $\displaystyle\|f\|_{\mathcal{L}}\leq 2\|f\|_{\infty}$ .

The little Lipschitz space is defined as

\displaystyle\mathcal{L}_{0}=\left\{f\in\mathcal{L}:\lim_{|v|\to\infty}Df(v)=0\right\},

and was proven to be a separable closed subspace of $\displaystyle\mathcal{L}$ . We state the following results that will be useful in the present work.

Lemma 1.1 (Lemma 3.4 of [5]).

(a)

If $\displaystyle f\in\mathcal{L}$ and $\displaystyle v\in T$ , then

$\displaystyle|f(v)|\leq|f(o)|+|v|\\ \|Df\|_{\infty}.$

In particular, if $\displaystyle\|f\|_{\mathcal{L}}\leq 1$ , then $\displaystyle|f(v)|\leq|v|$ for each $\displaystyle v\in T^{*}$ .
(b)

If $\displaystyle f\in\mathcal{L}_{0}$ , then

$\displaystyle\lim_{|v|\to\infty}\frac{f(v)}{|v|}=0.$

Lemma 1.2 (Proposition 2.4 of [5]).

Let $\displaystyle\{f_{n}\}$ be a sequence of functions in $\displaystyle\mathcal{L}_{0}$ converging to $\displaystyle 0$ pointwise in $\displaystyle T$ and such that $\displaystyle\{\|f_{n}\|_{\mathcal{L}}\}$ is bounded. Then $\displaystyle f_{n}\to 0$ weakly in $\displaystyle\mathcal{L}_{0}$ .

In [1], we introduced the weighted Lipschitz space on a tree $\displaystyle T$ as the set $\displaystyle\mathcal{L}_{\textbf{w}}$ of the functions $\displaystyle f:T\to\mathbb{C}$ such that

\displaystyle\sup_{v\in T^{*}}|v|Df(v)<\infty.

The interest in this space is due to its connection to the bounded multiplication operators on $\displaystyle\mathcal{L}$ . Specifically, it was shown in [5] that the bounded multiplication operators on $\displaystyle\mathcal{L}$ are precisely those operators $\displaystyle M_{\psi}$ whose symbol $\displaystyle\psi$ is a bounded function in $\displaystyle\mathcal{L}_{\textbf{w}}$ . The space $\displaystyle\mathcal{L}_{\textbf{w}}$ was shown to be a Banach space under the norm

\displaystyle\|f\|_{\textbf{w}}=|f(o)|+\sup_{v\in T^{*}}|v|Df(v).

The little weighted Lipschitz space was defined as

\displaystyle\mathcal{L}_{\textbf{w},0}=\left\{f\in\mathcal{L}_{\textbf{w}}:\lim_{|v|\to\infty}|v|Df(v)=0\right\},

and was shown to be a closed separable subspace of $\displaystyle\mathcal{L}_{\textbf{w}}$ .

In this paper, we shall make repeated use of the following results proved in [1].

Lemma 1.3 (Propositions 2.1 and 2.6 of [1]).

(a)

If $\displaystyle f\in\mathcal{L}_{\textbf{w}}$ , and $\displaystyle v\in T^{*}$ , then

$\displaystyle|f(v)|\leq(1+\log|v|)\|f\|_{\textbf{w}}.$
(b)

If $\displaystyle f\in\mathcal{L}_{{\textbf{w}},0}$ , then

$\displaystyle\lim_{|v|\to\infty}\frac{f(v)}{\log|v|}=0.$

Lemma 1.4 (Proposition 2.7 of [1]).

Let $\displaystyle\{f_{n}\}$ be a sequence of functions in $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ converging to $\displaystyle 0$ pointwise in $\displaystyle T$ and such that $\displaystyle\{\|f_{n}\|_{\textbf{w}}\}$ is bounded. Then $\displaystyle f_{n}\to 0$ weakly in $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ .

In this paper, we consider the multiplication operators between $\displaystyle\mathcal{L}$ and $\displaystyle\mathcal{L}_{\textbf{w}}$ , as well as between $\displaystyle\mathcal{L}_{\textbf{w}}$ and $\displaystyle L^{\infty}$ . The multiplication operators between $\displaystyle\mathcal{L}$ and $\displaystyle L^{\infty}$ were studied by the last two authors in [6].

1.1. Organization of the paper

In Sections 2 and 3, we study the multiplication operators between $\displaystyle\mathcal{L}_{{\textbf{w}}}$ and $\displaystyle\mathcal{L}$ . We characterize the bounded and the compact operators, and give estimates on their operator norm and their essential norm. We also prove that no isometric multiplication operators exist between the respective spaces.

In Section 4, we characterize the bounded operators and the compact operators from $\displaystyle\mathcal{L}_{{\textbf{w}}}$ to $\displaystyle L^{\infty}$ and determine their operator norm and their essential norm. As was the case in Sections 2 and 3, we show that no isometries exist amongst such operators. In addition, we characterize the multiplication operators that are bounded from below.

Finally, in Section 5, we characterize the bounded and the compact multiplication operators from $\displaystyle L^{\infty}$ to $\displaystyle\mathcal{L}_{{\textbf{w}}}$ . We also determine their operator norm and their essential norm. As with all the other cases, we show that there are no isometries amongst such operators.

2. Multiplication operators from $\displaystyle\mathcal{L}_{\textbf{w}}$ to $\displaystyle\mathcal{L}$

We begin the section with the study the bounded multiplication operators $\displaystyle M_{\psi}:\mathcal{L}_{\textbf{w}}\to\mathcal{L}$ and $\displaystyle M_{\psi}:\mathcal{L}_{\textbf{w},0}\to\mathcal{L}_{0}$ .

2.1. Boundedness and Operator Norm Estimates

Let $\displaystyle\psi$ be a function on the tree $\displaystyle T$ . Define

	$\displaystyle\tau_{\psi}$	$\displaystyle=\sup_{v\in T^{*}}D\psi(v)\log(1+\|v\|),$
	$\displaystyle\sigma_{\psi}$	$\displaystyle=\sup_{v\in T}\frac{\|\psi(v)\|}{\|v\|+1}.$

In the following theorem, we give a boundedness criterion in terms of the quantities $\displaystyle\tau_{\psi}$ and $\displaystyle\sigma_{\psi}$ .

Theorem 2.1.

For a function $\displaystyle\psi$ on $\displaystyle T$ , the following statements are equivalent:

(a)

$\displaystyle M_{\psi}:\mathcal{L}_{\textbf{w}}\to\mathcal{L}$ is bounded.
(b)

$\displaystyle M_{\psi}:\mathcal{L}_{{\textbf{w}},0}\to\mathcal{L}_{0}$ is bounded.
(c)

$\displaystyle\tau_{\psi}$ and $\displaystyle\sigma_{\psi}$ are finite.

Furthermore, under these conditions, we have

\displaystyle\max\{\tau_{\psi},\sigma_{\psi}\}\leq\|M_{\psi}\|\leq\tau_{\psi}+\sigma_{\psi}.

Proof.

$\displaystyle(a)\Longrightarrow(c)$ Assume $\displaystyle M_{\psi}:\mathcal{L}_{\textbf{w}}\to\mathcal{L}$ is bounded. Applying $\displaystyle M_{\psi}$ to the constant function 1, we have $\displaystyle\psi\in\mathcal{L}$ , so that, by Lemma 1.1, we have $\displaystyle\sigma_{\psi}<\infty$ . Next, consider the function $\displaystyle f$ on $\displaystyle T$ defined by $\displaystyle f(v)=\log(1+|v|)$ . Then $\displaystyle f(o)=0$ ; for $\displaystyle v\neq o$ , a straightforward calculation shows that

\displaystyle|v|Df(v)=|v|(\log(1+|v|)-\log|v|)\leq 1

and $\displaystyle\lim\limits_{|v|\to\infty}|v|Df(v)=1$ . Thus, $\displaystyle\|f\|_{\textbf{w}}=1$ and so $\displaystyle\|M_{\psi}f\|_{\mathcal{L}}\leq\|M_{\psi}\|$ . Therefore, for $\displaystyle v\in T^{*}$ , noting that

\displaystyle D(\psi f)(v)=D\psi(v)f(v)+\psi(v^{-})Df(v),

we have

	$\displaystyle\displaystyle D\psi(v)\|f(v)\|$	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle D(\psi f)(v)+\|\psi(v^{-})\|Df(v)$
		$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\\|M_{\psi}f\\|_{\mathcal{L}}+\sigma_{\psi}\|v\|Df(v)\leq\\|M_{\psi}\\|+\sigma_{\psi}.$

Hence $\displaystyle\tau_{\psi}<\infty.$

$\displaystyle(c)\Longrightarrow(a)$ Assume $\displaystyle\tau_{\psi}$ and $\displaystyle\sigma_{\psi}$ are finite. Then, by Lemma 1.3, for $\displaystyle f\in\mathcal{L}_{\textbf{w}}$ and $\displaystyle v\in T^{*}$ , we have

$\displaystyle\displaystyle D(\psi f)(v)$	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle D\psi(v)\|f(v)\|+\|\psi(v^{-})\|Df(v)$
	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle D\psi(v)(1+\log\|v\|)\\|f\\|_{\textbf{w}}+\|v\|\sigma_{\psi}Df(v)$
	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\tau_{\psi}\\|f\\|_{\textbf{w}}+\sigma_{\psi}(\\|f\\|_{\textbf{w}}-\|f(o)\|).$

Thus, since $\displaystyle|\psi(o)|\leq\sigma_{\psi}$ , we obtain

$\displaystyle\displaystyle\\|M_{\psi}f\\|_{\mathcal{L}}$	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\|\psi(o)\|\|f(o)\|+\tau_{\psi}\\|f\\|_{\textbf{w}}+\sigma_{\psi}(\\|f\\|_{\textbf{w}}-\|f(o)\|)$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle(\tau_{\psi}+\sigma_{\psi})\\|f\\|_{\textbf{w}}+(\|\psi(o)\|-\sigma_{\psi})\|f(o)\|$
	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\left(\tau_{\psi}+\sigma_{\psi}\right)\\|f\\|_{\textbf{w}},$

proving the boundedness of $\displaystyle M_{\psi}:\mathcal{L}_{\textbf{w}}\to\mathcal{L}$ and the upper estimate.

$\displaystyle(b)\Longrightarrow(c)$ Suppose $\displaystyle M_{\psi}:\mathcal{L}_{{\textbf{w}},0}\to\mathcal{L}_{0}$ is bounded. The finiteness of $\displaystyle\sigma_{\psi}$ follows again from the fact that $\displaystyle\psi=M_{\psi}1\in\mathcal{L}_{0}$ and from Lemma 1.1. To prove that $\displaystyle\tau_{\psi}<\infty$ , let $\displaystyle 0<\alpha<1$ and, for $\displaystyle v\in T$ , define $\displaystyle f_{\alpha}(v)=(\log(1+|v|))^{\alpha}$ . Then $\displaystyle f_{\alpha}(o)=0$ and $\displaystyle|v|Df_{a}(v)\to 0$ as $\displaystyle|v|\to\infty$ ; so $\displaystyle f_{\alpha}\in\mathcal{L}_{{\textbf{w}},0}$ . Since for $\displaystyle 0<\alpha<1$ , the function $\displaystyle x\mapsto x-x^{\alpha}$ is increasing for $\displaystyle x\geq 1$ , the function $\displaystyle Df_{\alpha}(v)$ is increasing in $\displaystyle\alpha$ and $\displaystyle Df_{\alpha}(v)\leq Df(v)$ for $\displaystyle v\in T^{*}$ , where $\displaystyle f(v)=\log(1+|v|)$ , for $\displaystyle v\in T$ . Thus, $\displaystyle\|f_{\alpha}\|_{\textbf{w}}\leq\|f\|_{\textbf{w}}=1$ . Therefore, for $\displaystyle v\in T^{*}$ , we have

	$\displaystyle\displaystyle D\psi(v)\|f_{\alpha}(v)\|$	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle D(\psi f_{\alpha})(v)+\|\psi(v^{-})\|Df_{\alpha}(v)$
		$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\\|M_{\psi}f_{\alpha}\\|+\sigma_{\psi}\|v\|Df_{\alpha}(v)\leq\\|M_{\psi}\\|+\sigma_{\psi}.$

Letting $\displaystyle\alpha\to 1$ , we obtain

\displaystyle D\psi(v)\log(1+|v|)\leq\|M_{\psi}\|+\sigma_{\psi}.

Hence $\displaystyle\tau_{\psi}<\infty$ .

$\displaystyle(c)\Longrightarrow(b)$ Assume $\displaystyle\sigma_{\psi}$ and $\displaystyle\tau_{\psi}$ are finite, and let $\displaystyle f\in\mathcal{L}_{{\textbf{w}},0}$ . Then, by Lemma 1.3, for $\displaystyle v\in T^{*}$ , we have

$\displaystyle\displaystyle D(\psi f)(v)$	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle D\psi(v)\|f(v)\|+\|\psi(v^{-})\|Df(v)$
	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle D\psi(v)\log(1+\|v\|)\frac{\|f(v)\|}{\log(1+\|v\|)}+\frac{\|\psi(v^{-})\|}{\|v\|}\|v\|Df(v)$
	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\tau_{\psi}\frac{\|f(v)\|}{\log(1+\|v\|)}+\sigma_{\psi}\|v\|Df(v)\to 0$

as $\displaystyle|v|\to\infty$ . Thus, $\displaystyle\psi f\in\mathcal{L}_{0}.$ The boundedness of $\displaystyle M_{\psi}$ and the estimate $\displaystyle\|M_{\psi}f\|_{\mathcal{L}}\leq\tau_{\psi}+\sigma_{\psi}$ can be shown as in the proof of $\displaystyle(c)\Longrightarrow(a)$ .

Finally we show that, under boundedness assumptions on $\displaystyle M_{\psi}$ , $\displaystyle\|M_{\psi}\|\geq\max\{\tau_{\psi},\sigma_{\psi}\}$ . For $\displaystyle v\in T^{*}$ , let $\displaystyle f_{v}=\frac{1}{|v|+1}\chi_{v}$ , where $\displaystyle\chi_{v}$ denotes the characteristic function of $\displaystyle\{v\}$ . Then $\displaystyle\|f_{v}\|_{\textbf{w}}=1$ and

\displaystyle\|\psi f_{v}\|_{\mathcal{L}}=\frac{|\psi(v)|}{|v|+1}.

Furthermore, letting $\displaystyle f_{o}=\frac{1}{2}\chi_{o}$ , we see that $\displaystyle\|f_{o}\|_{\textbf{w}}=1$ and $\displaystyle\|\psi f_{o}\|_{\mathcal{L}}=|\psi(o)|$ . Therefore, we deduce that $\displaystyle\|M_{\psi}\|\geq\sigma_{\psi}.$

Next, fix $\displaystyle v\in T^{*}$ and for $\displaystyle w\in T$ , define

\displaystyle g_{v}(w)=\begin{cases}\log(1+|w|)&\hbox{ if }|w|<|v|,\\ \log(1+|v|)&\hbox{ if }|w|\geq|v|.\end{cases}

Then, $\displaystyle g_{v}\in\mathcal{L}_{\textbf{w}}$ and

\displaystyle\lim_{|v|\to\infty}\|g_{v}\|_{\textbf{w}}=\lim_{|v|\to\infty}|v|\left[\log(1+|v|)-\log|v|\right]=1.

Observe that for $\displaystyle w\in T^{*}$ , we have

\displaystyle D(\psi g_{v})(w)=\begin{cases}|\psi(w)\log(1+|w|)-\psi(w^{-})\log|w||&\hbox{ if }|w|<|v|,\\ D\psi(w)\log(1+|v|)&\hbox{ if }|w|\geq|v|.\end{cases}

Hence

\displaystyle\sup_{w\in T^{*}}D(\psi g_{v})(w)\geq\sup_{|w|\geq|v|}D\psi(w)\log(1+|v|)\geq D\psi(v)\log(1+|v|).

Define $\displaystyle f_{v}=\frac{g_{v}}{\|g_{v}\|_{\textbf{w}}}$ . Then $\displaystyle\|f_{v}\|_{\textbf{w}}=1$ and

\displaystyle\displaystyle\|M_{\psi}\|\geq\|M_{\psi}f_{v}\|_{\mathcal{L}}=\frac{\|D(\psi g_{v})\|_{\infty}}{\|g_{v}\|_{\textbf{w}}}\geq\frac{D\psi(v)\log(1+|v|)}{\|g_{v}\|_{\textbf{w}}}.

Taking the limit as $\displaystyle|v|\to\infty$ , we obtain $\displaystyle\|M_{\psi}\|\geq\tau_{\psi}.$ Therefore, $\displaystyle\|M_{\psi}\|\geq\max\{\tau_{\psi},\sigma_{\psi}\}.$ ∎

2.2. Isometries

In this section, we show there are no isometric multiplication operators $\displaystyle M_{\psi}$ from the spaces $\displaystyle\mathcal{L}_{\textbf{w}}$ or $\displaystyle\mathcal{L}_{\textbf{w},0}$ to the spaces $\displaystyle\mathcal{L}$ or $\displaystyle\mathcal{L}_{0}$ .

Assume $\displaystyle M_{\psi}:\mathcal{L}_{\textbf{w}}\to\mathcal{L}$ is an isometry. Then $\displaystyle\|\psi\|_{\mathcal{L}}=\|M_{\psi}1\|_{\mathcal{L}}=1.$ On the other hand, $\displaystyle|\psi(o)|=\frac{1}{2}\left\|M_{\psi}\chi_{o}\right\|_{\mathcal{L}}=\frac{1}{2}\left\|\chi_{o}\right\|_{\textbf{w}}=1.$ Thus $\displaystyle\sup\limits_{v\in T^{*}}D\psi(v)=\|\psi\|_{\mathcal{L}}-|\psi(o)|=0$ , which implies that $\displaystyle\psi$ is a constant of modulus 1. Yet, for $\displaystyle v\in T^{*}$ , letting $\displaystyle f_{v}=\frac{1}{|v|+1}\chi_{v}$ , we see that

\displaystyle 1=\|f_{v}\|_{\textbf{w}}=\|M_{\psi}f_{v}\|_{\mathcal{L}}=\frac{1}{|v|+1},

which yields a contradiction. Therefore, we obtain the following result.

Theorem 2.2.

There are no isometries $\displaystyle M_{\psi}$ from $\displaystyle\mathcal{L}_{\textbf{w}}$ to $\displaystyle\mathcal{L}$ or $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ to $\displaystyle\mathcal{L}_{0}$ .

2.3. Compactness and Essential Norm Estimates

In this section, we characterize the compact multiplication operators. As with many classical spaces, the characterization of the compact operators is a “little-oh” condition corresponding the the “big-oh” condition for boundedness. We first collect some useful results about compact operators from $\displaystyle\mathcal{L}_{\textbf{w}}$ or from $\displaystyle\mathcal{L}_{\textbf{w},0}$ to $\displaystyle\mathcal{L}$ .

Lemma 2.3.

A bounded multiplication operator $\displaystyle M_{\psi}$ from $\displaystyle\mathcal{L}_{\textbf{w}}$ to $\displaystyle\mathcal{L}$ is compact if and only if for every bounded sequence $\displaystyle\{f_{n}\}$ in $\displaystyle\mathcal{L}_{w}$ converging to 0 pointwise, the sequence $\displaystyle\{\|\psi f_{n}\|_{\mathcal{L}}\}\to 0$ as $\displaystyle n\to\infty$ .

Proof.

Assume $\displaystyle M_{\psi}$ is compact, and let $\displaystyle\{f_{n}\}$ be a bounded sequence in $\displaystyle\mathcal{L}_{\textbf{w}}$ converging to 0 pointwise. Without loss of generality, we may assume $\displaystyle\|f_{n}\|_{\textbf{w}}\leq 1$ for all $\displaystyle n\in\mathbb{N}$ . Then the sequence $\displaystyle\{M_{\psi}f_{n}\}=\{\psi f_{n}\}$ has a subsequence $\displaystyle\{\psi f_{n_{k}}\}$ which converges in the $\displaystyle\mathcal{L}$ -norm to some function $\displaystyle f\in\mathcal{L}$ . Clearly $\displaystyle\psi(o)f_{n_{k}}(o)\to\psi(o)f(o)$ , and by part (a) of Lemma 1.1, for $\displaystyle v\in T^{*}$ , we have

	$\displaystyle\|\psi(v)f_{n_{k}}(v)-f(v)\|$	$\displaystyle\leq\|\psi(o)f_{n_{k}}(o)-f(o)\|+\|v\|\\|D(\psi f_{n_{k}}-f)\\|_{\infty}$
		$\displaystyle\leq(1+\|v\|)\\|\psi f_{n_{k}}-f\\|_{\mathcal{L}}.$

Thus, $\displaystyle\psi f_{n_{k}}\to f$ pointwise on $\displaystyle T$ . Since $\displaystyle f_{n}\to 0$ pointwise, it follows that $\displaystyle f$ must be identically 0, which implies that $\displaystyle\|\psi f_{n_{k}}\|_{\mathcal{L}}\to 0$ . With 0 being the only limit point of $\displaystyle\{\psi f_{n}\}$ in $\displaystyle\mathcal{L}$ , it follows that $\displaystyle\|\psi f_{n}\|_{\mathcal{L}}\to 0$ as $\displaystyle n\to\infty$ .

Conversely, assume every bounded sequence $\displaystyle\{f_{n}\}$ in $\displaystyle\mathcal{L}_{\textbf{w}}$ converging to 0 pointwise has the property that $\displaystyle\|\psi f_{n}\|_{\mathcal{L}}\to 0$ as $\displaystyle n\to\infty$ . Let $\displaystyle\{g_{n}\}$ be a sequence in $\displaystyle\mathcal{L}_{\textbf{w}}$ with $\displaystyle\|g_{n}\|_{\textbf{w}}\leq 1$ for all $\displaystyle n\in\mathbb{N}$ . Then $\displaystyle|g_{n}(o)|\leq 1$ for all $\displaystyle n\in\mathbb{N}$ , and by part (a) of Lemma 1.2, for $\displaystyle v\in T^{*}$ , we obtain

\displaystyle|g_{n}(v)|\leq(1+\log|v|)\|g_{n}\|_{\textbf{w}}\leq 1+\log|v|.

Thus, $\displaystyle\{g_{n}\}$ is uniformly bounded on finite subsets of $\displaystyle T$ . So some subsequence $\displaystyle\{g_{n_{k}}\}$ converges pointwise to some function $\displaystyle g$ . Fix $\displaystyle v\in T^{*}$ and $\displaystyle\varepsilon>0$ . Then for $\displaystyle k$ sufficiently large, we have

\displaystyle|g(v)-g_{n_{k}}(v)|<\frac{\varepsilon}{2|v|},\text{ and }|g_{n_{k}}(v^{-})-g(v^{-})|<\frac{\varepsilon}{2|v|}.

We deduce

	$\displaystyle\|v\|Dg(v)$	$\displaystyle\leq\|v\|\|g(v)-g_{n_{k}}(v)+g_{n_{k}}(v^{-})-g(v^{-})\|+\|v\|Dg_{n_{k}}(v)$
		$\displaystyle\leq\|v\|\|g(v)-g_{n_{k}}(v)\|+\|v\|\|g_{n_{k}}(v^{-})-g(v^{-})\|+\|v\|Dg_{n_{k}}(v)$
		$\displaystyle<\varepsilon+\|v\|Dg_{n_{k}}(v)\leq\varepsilon+1,$

for all $\displaystyle k$ sufficiently large. So $\displaystyle g\in\mathcal{L}_{\textbf{w}}$ . The sequence defined by $\displaystyle f_{k}=g_{n_{k}}-g$ is bounded in $\displaystyle\mathcal{L}_{\textbf{w}}$ and converges to 0 pointwise. Thus by hypothesis, we obtain $\displaystyle\|\psi f_{k}\|_{\mathcal{L}}\to 0$ as $\displaystyle k\to\infty$ . It follows that $\displaystyle M_{\psi}g_{n_{k}}=\psi g_{n_{k}}\to\psi g$ in the $\displaystyle\mathcal{L}$ -norm, thus proving the compactness of $\displaystyle M_{\psi}$ . ∎

By an analogous argument, we obtain the corresponding compactness criterion for $\displaystyle M_{\psi}$ from $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ to $\displaystyle\mathcal{L}_{0}$ .

Lemma 2.4.

A bounded multiplication operator $\displaystyle M_{\psi}$ from $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ to $\displaystyle\mathcal{L}_{0}$ is compact if and only if for every bounded sequence $\displaystyle\{f_{n}\}$ in $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ converging to 0 pointwise, the sequence $\displaystyle\{\|\psi f_{n}\|_{\mathcal{L}}\}\to 0$ as $\displaystyle n\to\infty$ .

The following result is a variant of Lemma 1.3(a), which will be needed to prove a characterization of the compact multiplication operators from $\displaystyle\mathcal{L}_{\textbf{w}}$ to $\displaystyle\mathcal{L}$ and from $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ to $\displaystyle\mathcal{L}_{0}$ (Theorem 2.6).

Lemma 2.5.

For $\displaystyle f\in\mathcal{L}_{\textbf{w}}$ and $\displaystyle v\in T$

\displaystyle\displaystyle|f(v)|\leq|f(o)|+2\log(1+|v|)s_{\textbf{w}}(f),

(2.1)

where $\displaystyle s_{\textbf{w}}(f)=\sup\limits_{w\in T^{*}}|w|Df(w).$

Proof.

Fix $\displaystyle v\in T$ and argue by induction on $\displaystyle n=|v|.$ For $\displaystyle n=0$ , the inequality (2.1) is obvious. So assume $\displaystyle|v|=n>0$ and $\displaystyle|f(u)|\leq|f(o)|+2\log(1+|u|)s_{\textbf{w}}(f)$ for all vertices $\displaystyle u$ such that $\displaystyle|u|<n$ . Then

$\displaystyle\displaystyle\|f(v)\|$	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\|f(v)-f(v^{-})\|+\|f(v^{-})\|$	(2.2)
	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\frac{1}{\|v\|}s_{\textbf{w}}(f)+\|f(o)\|+2\log\|v\|s_{\textbf{w}}(f)$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\|f(o)\|+\left(\frac{1}{\|v\|}+2\log\|v\|\right)s_{\textbf{w}}(f).$

Next, observe that $\displaystyle\frac{1}{|v|+1}\leq\log\left(\frac{|v|+1}{|v|}\right)$ , so

\displaystyle\frac{1}{|v|}\leq\frac{2}{|v|+1}\leq 2\log\left(\frac{|v|+1}{|v|}\right).

Hence

\displaystyle\displaystyle\frac{1}{|v|}+2\log|v|\leq 2\log(|v|+1).

(2.3)

Inequality (2.1) now follows immediately from (2.2) and (2.3). ∎

Theorem 2.6.

Let $\displaystyle M_{\psi}$ be a bounded multiplication operator from $\displaystyle\mathcal{L}_{\textbf{w}}$ to $\displaystyle\mathcal{L}$ (or equivalently from $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ to $\displaystyle\mathcal{L}_{0}$ ). Then the following statements are equivalent:

(a)

$\displaystyle M_{\psi}:\mathcal{L}_{\textbf{w}}\to\mathcal{L}$ is compact.
(b)

$\displaystyle M_{\psi}:\mathcal{L}_{{\textbf{w}},0}\to\mathcal{L}_{0}$ is compact.
(c)

$\displaystyle\displaystyle\lim_{|v|\to\infty}\frac{|\psi(v)|}{|v|+1}=0$ and $\displaystyle\displaystyle\lim_{|v|\to\infty}D\psi(v)\log|v|=0$ .

Proof.

We first prove $\displaystyle(a)\Longrightarrow(c)$ . Assume $\displaystyle M_{\psi}:\mathcal{L}_{\textbf{w}}\to\mathcal{L}$ is compact. It suffices to show that for any sequence $\displaystyle\{v_{n}\}$ in $\displaystyle T$ such that $\displaystyle 2\leq|v_{n}|\to\infty$ , we have $\displaystyle\displaystyle\lim_{n\to\infty}\frac{|\psi(v_{n})|}{|v_{n}|+1}=0$ and $\displaystyle\displaystyle\lim_{n\to\infty}D\psi(v_{n})\log|v_{n}|=0$ . Let $\displaystyle\{v_{n}\}$ be such a sequence, and for each $\displaystyle n\in\mathbb{N}$ , define $\displaystyle f_{n}=\frac{1}{|v_{n}|+1}\chi_{v_{n}}$ . Then $\displaystyle f_{n}(o)=0$ , $\displaystyle f_{n}\to 0$ pointwise as $\displaystyle n\to\infty$ , and $\displaystyle\|f_{n}\|_{\textbf{w}}=1$ . By Lemma 2.3, it follows that $\displaystyle\|\psi f_{n}\|_{\mathcal{L}}\to 0$ as $\displaystyle n\to\infty$ . Furthermore

\displaystyle\|\psi f_{n}\|_{\mathcal{L}}=\sup_{v\in T^{*}}|\psi(v)f_{n}(v)-\psi(v^{-})f_{n}(v^{-})|=|\psi(v_{n})f_{n}(v_{n})|=\frac{|\psi(v_{n})|}{|v_{n}|+1}.

Thus $\displaystyle\lim\limits_{n\to\infty}\displaystyle\frac{|\psi(v_{n})|}{|v_{n}|+1}=0.$

Next, for each $\displaystyle n\in\mathbb{N}$ and $\displaystyle v\in T$ , define

\displaystyle g_{n}(v)=\begin{cases}0&\text{ if }|v|<\sqrt{|v_{n}|},\\ 2\log|v|-\log|v_{n}|&\text{ if }\sqrt{|v_{n}|}\leq|v|<|v_{n}|-1,\\ \log|v_{n}|&\text{ if }|v|\geq|v_{n}|-1.\end{cases}

Then $\displaystyle Dg_{n}(v)=0\,$ if $\displaystyle|v|\leq\sqrt{|v_{n}|}\,$ or $\displaystyle|v|>|v_{n}|-1$ . In addition, if $\displaystyle\sqrt{|v_{n}|}<|v|\leq|v_{n}|-1$ , then $\displaystyle|v|Dg_{n}(v)<4$ . Indeed, there are two possibilities. Either $\displaystyle\sqrt{|v_{n}|}\leq|v|-1$ , in which case

\displaystyle|v|Dg_{n}(v)=2|v||(\log|v|-\log(|v|-1))\leq\frac{2|v|}{|v|-1}\leq 3,

or $\displaystyle|v|-1<\sqrt{|v_{n}|}<|v|$ , in which case

$\displaystyle\displaystyle\|v\|Dg_{n}(v)$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\|v\|(2\log\|v\|-\log\|v_{n}\|)$
	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle(\sqrt{\|v_{n}\|}+1)\log\frac{(\sqrt{\|v_{n}\|}+1)^{2}}{\|v_{n}\|}$
	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\frac{2(\sqrt{\|v_{n}\|}+1)}{\sqrt{\|v_{n}\|}}\leq 2\left(1+\frac{1}{\sqrt{2}}\right)<4.$

Thus $\displaystyle\{\|g_{n}\|_{\textbf{w}}\}$ is bounded and $\displaystyle\{g_{n}\}$ converges to 0 pointwise. By Lemma 2.3, it follows that $\displaystyle\|\psi g_{n}\|_{\mathcal{L}}\to 0$ as $\displaystyle n\to\infty$ . Moreover

\displaystyle\|\psi g_{n}\|_{\mathcal{L}}\geq|\psi(v_{n})g_{n}(v_{n})-\psi(v_{n}^{-})g_{n}(v_{n}^{-})|=D\psi(v_{n})\log|v_{n}|.

Therefore $\displaystyle\lim\limits_{n\to\infty}D\psi(v_{n})\log|v_{n}|=0.$

To prove the implication $\displaystyle(c)\Longrightarrow(a)$ , suppose $\displaystyle\displaystyle\lim_{|v|\to\infty}\frac{|\psi(v)|}{|v|+1}=0$ and $\displaystyle\displaystyle\lim_{|v|\to\infty}D\psi(v)\log|v|=0$ . Clearly, if $\displaystyle\psi$ is identically 0, then $\displaystyle M_{\psi}$ is compact. So assume $\displaystyle M_{\psi}:\mathcal{L}_{\textbf{w}}\to\mathcal{L}$ is bounded with $\displaystyle\psi$ not identically 0. By Lemma 2.3, it suffices to show that if $\displaystyle\{f_{n}\}$ is bounded in $\displaystyle\mathcal{L}_{\textbf{w}}$ converging to 0 pointwise, then $\displaystyle\|\psi f_{n}\|_{\mathcal{L}}\to 0$ as $\displaystyle n\to\infty$ . Let $\displaystyle\{f_{n}\}$ be such a sequence, let $\displaystyle s=\displaystyle\sup_{n\in\mathbb{N}}\|f_{n}\|_{\textbf{w}}$ , and fix $\displaystyle\varepsilon>0$ . Note that

\displaystyle\lim_{|v|\to\infty}D\psi(v)\log(1+|v|)=\lim_{|v|\to\infty}D\psi(v)\log|v|\frac{\log(1+|v|)}{\log|v|}=0.

Thus there exists an $\displaystyle M\in\mathbb{N}$ such that

\displaystyle|f_{n}(o)|<\frac{\varepsilon}{3s\|\psi\|_{\mathcal{L}}},\ D\psi(v)\log(1+|v|)<\frac{\varepsilon}{6s}\text{ and }\frac{|\psi(v)|}{|v|+1}<\frac{\varepsilon}{3s},

for $\displaystyle|v|\geq M$ . Using Lemma 2.5, for $\displaystyle|v|>M$ , we have

	$\displaystyle D(\psi f_{n})(v)$	$\displaystyle\leq D\psi(v)\|f_{n}(v)\|+Df_{n}(v^{-})\|\psi(v^{-})\|$
		$\displaystyle\leq D\psi(v)\left(\|f_{n}(o)\|+2\log(\|v\|+1)\right)\\|f_{n}\\|_{\textbf{w}}+\\|f_{n}\\|_{\textbf{w}}\frac{\|\psi(v^{-})\|}{\|v\|}$
		$\displaystyle\leq\left(\\|\psi\\|_{\mathcal{L}}\|f_{n}(o)\|+2D\psi(v)\log(\|v\|+1)+\frac{\|\psi(v^{-})\|}{\|v\|}\right)\\|f_{n}\\|_{\textbf{w}}$
		$\displaystyle<\varepsilon.$

On the other hand, on the set $\displaystyle B_{M}=\{v\in T:|v|\leq M\}$ , $\displaystyle\{f_{n}\}$ converges to 0 uniformly, and thus $\displaystyle Df_{n}$ does as well. Moreover

	$\displaystyle D(\psi f_{n})(v)$	$\displaystyle\leq D\psi(v)\|f_{n}(v)\|+\|\psi(v^{-})\|Df_{n}(v)$
		$\displaystyle\leq\\|\psi\\|_{\mathcal{L}}\|f_{n}(v)\|+\max_{\|w\|\leq M}\|\psi(w)\|Df_{n}(v)\to 0,$

uniformly on $\displaystyle B_{M}$ . Therefore $\displaystyle D(\psi f_{n})\to 0$ uniformly on $\displaystyle T$ . Furthermore, the sequence $\displaystyle\{(\psi f_{n})(o)\}$ converges to 0 as $\displaystyle n\to\infty$ . Hence $\displaystyle\|\psi f_{n}\|_{\mathcal{L}}\to 0$ as $\displaystyle n\to\infty$ , proving that $\displaystyle M_{\psi}$ is compact.

Finally, note that the functions $\displaystyle f_{n}$ and $\displaystyle g_{n}$ defined in the proof of $\displaystyle(a)\Longrightarrow(c)$ are in $\displaystyle\mathcal{L}_{w,0}$ . So the equivalence of $\displaystyle(b)$ and $\displaystyle(c)$ is proved analogously. ∎

Recall the essential norm of a bounded operator $\displaystyle S$ between Banach spaces $\displaystyle\mathcal{X}$ and $\displaystyle\mathcal{Y}$ is defined as

\displaystyle\|S\|_{e}=\inf\;\{\|S-K\|:K\text{ is compact from $\displaystyle\mathcal{X}$ to $\displaystyle\mathcal{Y}$}\}.

For $\displaystyle\psi$ a function on $\displaystyle T$ , define the quantities

	$\displaystyle A(\psi)$	$\displaystyle=\lim_{n\to\infty}\sup_{\|v\|\geq n}\frac{\|\psi(v)\|}{\|v\|+1},$
	$\displaystyle B(\psi)$	$\displaystyle=\lim_{n\to\infty}\sup_{\|v\|\geq n}D\psi(v)\log\|v\|.$

Theorem 2.7.

Let $\displaystyle M_{\psi}$ be a bounded multiplication operator from $\displaystyle\mathcal{L}_{\textbf{w}}$ to $\displaystyle\mathcal{L}$ . Then

\displaystyle\|M_{\psi}\|_{e}\geq\max\left\{A(\psi),B(\psi)\right\}.

Proof.

For each $\displaystyle n\in\mathbb{N}$ , define $\displaystyle f_{n}=\frac{1}{n+1}\chi_{n}$ , where $\displaystyle\chi_{n}$ denotes the characteristic function of the set $\displaystyle\{v\in T:|v|=n\}$ . Then $\displaystyle f_{n}\in\mathcal{L}_{{\textbf{w}},0}$ , $\displaystyle\|f_{n}\|_{\textbf{w}}=1$ , and $\displaystyle f_{n}\to 0$ pointwise. Thus, by Lemma 1.4, $\displaystyle\{f_{n}\}$ converges to 0 weakly in $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ . Let $\displaystyle\mathcal{K}$ be the set of compact operators from from $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ to $\displaystyle\mathcal{L}_{0}$ , and let $\displaystyle K\in\mathcal{K}$ . Then $\displaystyle K$ is completely continuous [7], and so $\displaystyle\|Kf_{n}\|_{\mathcal{L}}\to 0$ as $\displaystyle n\to\infty$ . Thus

\displaystyle\|M_{\psi}-K\|\geq\limsup_{n\to\infty}\|(M_{\psi}-K)f_{n}\|_{\mathcal{L}}\geq\limsup_{n\to\infty}\|M_{\psi}f_{n}\|_{\mathcal{L}}.

Now note that

\displaystyle\|M_{\psi}f_{n}\|_{\mathcal{L}}=\sup_{|v|=n}\frac{|\psi(v)|}{n+1}.

Hence

	$\displaystyle\\|M_{\psi}\\|_{e}$	$\displaystyle\geq\inf\{\\|M_{\psi}-K\\|:K\in\mathcal{K}\}$
		$\displaystyle\geq\limsup_{n\to\infty}\\|M_{\psi}f_{n}\\|_{\mathcal{L}}$
		$\displaystyle=\lim_{n\to\infty}\sup_{\|v\|\geq n}\frac{\|\psi(v)\|}{\|v\|+1}$
		$\displaystyle=A(\psi).$

We will now show that $\displaystyle\|M_{\psi}\|_{e}\geq B(\psi)$ . This estimate is clearly true if $\displaystyle B(\psi)=0$ . So assume $\displaystyle\{v_{n}\}$ is a sequence in $\displaystyle T$ such that $\displaystyle 2\leq|v_{n}|\to\infty$ as $\displaystyle n\to\infty$ and

\displaystyle\lim_{n\to\infty}D\psi(v_{n})\log|v_{n}|=B(\psi).

For $\displaystyle n\in\mathbb{N}$ and $\displaystyle v\in T$ , define

\displaystyle h_{n}(v)=\begin{cases}\displaystyle\frac{\left[\log(|v|+1)\right]^{2}}{\log|v_{n}|}&\text{ if }0\leq|v|<|v_{n}|,\\ \log|v_{n}|&\text{ if }|v|\geq|v_{n}|.\end{cases}

Then $\displaystyle h_{n}(o)=0$ , $\displaystyle h_{n}(v_{n})=h_{n}(v_{n}^{-})=\log|v_{n}|$ , and

\displaystyle|v|Dh_{n}(v)=\begin{cases}\frac{|v|}{\log|v_{n}|}\log\left(\frac{|v|+1}{|v|}\right)\log\left[|v|(|v|+1)\right]&\text{ if }1\leq|v|<|v_{n}|,\\ \ \ \ 0&\text{ if }|v|\geq|v_{n}|.\end{cases}

The supremum of $\displaystyle|v|Dh_{n}(v)$ is attained at the vertices of length $\displaystyle|v_{n}|-1$ and is given by

\displaystyle s_{n}=\sup_{v\in T^{*}}|v|Dh_{n}(v)=(|v_{n}|-1)\log\left(\frac{|v_{n}|}{|v_{n}|-1}\right)\frac{\log\left[(|v_{n}|-1)|v_{n}|\right]}{\log|v_{n}|}.

Since $\displaystyle(|v_{n}|-1)\log\left(\frac{|v_{n}|}{|v_{n}|-1}\right)\leq 1$ , we have

\displaystyle\frac{(\log 2)^{2}}{\log|v_{n}|}\leq\|h_{n}\|_{\textbf{w}}=s_{n}\leq\frac{\log\left[(|v_{n}|-1)|v_{n}|\right]}{\log|v_{n}|}<2.

By letting $\displaystyle g_{n}=\frac{h_{n}}{\|h_{n}\|_{\textbf{w}}}$ , we have $\displaystyle g_{n}\in\mathcal{L}_{{\textbf{w}},0}$ , $\displaystyle\|g_{n}\|_{\textbf{w}}=1$ , and $\displaystyle g_{n}\to 0$ pointwise. By Lemma 1.4, the sequence $\displaystyle\{g_{n}\}$ converges to 0 weakly in $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ . Thus $\displaystyle\|Kg_{n}\|_{\mathcal{L}}\to 0$ as $\displaystyle n\to\infty$ . Therefore

\displaystyle\|M_{\psi}-K\|\geq\limsup_{n\to\infty}\|(M_{\psi}-K)g_{n}\|_{\mathcal{L}}\geq\limsup_{n\to\infty}\|\psi g_{n}\|_{\mathcal{L}}.

For each $\displaystyle n\in\mathbb{N}$ , we have $\displaystyle g_{n}(v_{n})=g_{n}(v_{n}^{-})=\frac{\log|v_{n}|}{s_{n}}$ . So

\displaystyle D(\psi g_{n})(v_{n})=\frac{1}{s_{n}}D\psi(v_{n})\log|v_{n}|.

Since $\displaystyle\displaystyle\lim_{n\to\infty}s_{n}=1$ , we have

	$\displaystyle\\|M_{\psi}\\|_{e}$	$\displaystyle\geq\inf\{\\|M_{\psi}-K\\|:K\in\mathcal{K}\}$
		$\displaystyle\geq\limsup_{n\to\infty}\sup_{v\in T^{*}}D(\psi g_{n})(v)$
		$\displaystyle\geq\lim_{n\to\infty}\frac{1}{s_{n}}D\psi(v_{n})\log\|v_{n}\|$
		$\displaystyle=B(\psi).$

Therefore, $\displaystyle\|M_{\psi}\|_{e}\geq\max\left\{A(\psi),B(\psi)\right\}$ . ∎

Now now derive an upper estimate on the essential norm.

Theorem 2.8.

Let $\displaystyle M_{\psi}$ be a bounded multiplication operator from $\displaystyle\mathcal{L}_{\textbf{w}}$ to $\displaystyle\mathcal{L}$ . Then

\displaystyle\|M_{\psi}\|_{e}\leq A(\psi)+B(\psi).

Proof.

For $\displaystyle n\in\mathbb{N}$ , define the operator $\displaystyle K_{n}$ on $\displaystyle\mathcal{L}_{\textbf{w}}$ by

\displaystyle(K_{n}f)(v)=\begin{cases}f(v)&\text{ if }|v|\leq n,\\ f(v_{n})&\text{ if }|v|>n,\end{cases}

where $\displaystyle f\in\mathcal{L}_{\textbf{w}}$ and $\displaystyle v_{n}$ is the ancestor of $\displaystyle v$ of length $\displaystyle n$ . For $\displaystyle f\in\mathcal{L}_{\textbf{w}}$ , $\displaystyle(K_{n}f)(o)=f(o)$ , and $\displaystyle K_{n}f\in\mathcal{L}_{{\textbf{w}},0}$ . Let $\displaystyle B_{n}=\{v\in T:|v|\leq n\}$ and note that $\displaystyle K_{n}f$ attains finitely many values, whose number does not exceed the cardinality of $\displaystyle B_{n}$ . Let $\displaystyle\{g_{k}\}$ be a sequence in $\displaystyle\mathcal{L}_{\textbf{w}}$ such that $\displaystyle\|g_{k}\|_{\textbf{w}}\leq 1$ for each $\displaystyle k\in\mathbb{N}$ . Then $\displaystyle a=\displaystyle\sup_{k\in\mathbb{N}}|g_{k}(o)|\leq 1$ and $\displaystyle|K_{n}g_{k}(o)|\leq a$ . Furthermore, by part (a) of Lemma 1.3, for each $\displaystyle v\in T^{*}$ and for each $\displaystyle k\in\mathbb{N}$ , we have $\displaystyle|K_{n}g_{k}(v)|\leq 1+\log n$ . Thus, some subsequence of $\displaystyle\{K_{n}g_{k}\}_{k\in\mathbb{N}}$ must converge to a function $\displaystyle g$ on $\displaystyle T$ attaining constant values on the sectors determined by the vertices of length $\displaystyle n$ . It follows that this subsequence converges to $\displaystyle g$ in $\displaystyle\mathcal{L}_{\textbf{w}}$ as well, proving that $\displaystyle K_{n}$ is a compact operator on $\displaystyle\mathcal{L}_{\textbf{w}}$ . Since $\displaystyle M_{\psi}$ is bounded as an operator from $\displaystyle\mathcal{L}_{\textbf{w}}$ to $\displaystyle\mathcal{L}$ , it follows that $\displaystyle M_{\psi}K_{n}:\mathcal{L}_{\textbf{w}}\to\mathcal{L}$ is compact for all $\displaystyle n\in\mathbb{N}$ .

Define the operator $\displaystyle J_{n}=I-K_{n}$ , where $\displaystyle I$ denotes the identity operator on $\displaystyle\mathcal{L}_{\textbf{w}}$ . Then $\displaystyle J_{n}f(o)=0$ and for $\displaystyle v\in T^{*}$ , we have

\displaystyle\displaystyle|v|D(J_{n}f)(v)=|v||(J_{n}f)(v)-(J_{n}f)(v^{-})|\leq|v|Df(v)\leq\|f\|_{\textbf{w}}.

(2.4)

By part (a) of Lemma 1.3, we see that

\displaystyle\displaystyle|(J_{n}f)(v)|\leq(1+\log|v|)\|f\|_{\textbf{w}}.

(2.5)

Using (2.4) and (2.5), we obtain

	$\displaystyle\\|(M_{\psi}-M_{\psi}K_{n})f\\|_{\mathcal{L}}$	$\displaystyle=\\|\psi(J_{n}f)\\|_{\mathcal{L}}$
		$\displaystyle=\sup_{\|v\|>n}\|\psi(v)(J_{n}f)(v)-\psi(v^{-})(J_{n}f)(v^{-})\|$
		$\displaystyle\leq\sup_{\|v\|>n}\left[\|(J_{n}f)(v)\|D\psi(v)+\|\psi(v^{-})\|D(J_{n}f)(v)\right]$
		$\displaystyle=\sup_{\|v\|>n}\left[\|(J_{n}f)(v)\|D\psi(v)+\frac{\|\psi(v^{-})\|}{\|v\|}\|v\|D(J_{n}f)(v)\right]$
		$\displaystyle\leq\sup_{\|v\|\geq n}\left[(1+\log\|v\|)D\psi(v)+\frac{\|\psi(v)\|}{\|v\|+1}\right]\\|f\\|_{\textbf{w}}$
		$\displaystyle\leq\sup_{\|v\|\geq n}\left[\log\|v\|D\psi(v)\frac{1+\log\|v\|}{\log\|v\|}+\frac{\|\psi(v)\|}{\|v\|+1}\right]\\|f\\|_{\textbf{w}}$
		$\displaystyle\leq\left[\sup_{\|v\|\geq n}\log\|v\|D\psi(v)\frac{1+\log n}{\log n}+\sup_{\|v\|\geq n}\frac{\|\psi(v)\|}{\|v\|+1}\right]\\|f\\|_{\textbf{w}}.$

Since

\displaystyle\|M_{\psi}\|_{e}\leq\limsup_{n\to\infty}\|M_{\psi}-M_{\psi}K_{n}\|=\limsup_{n\to\infty}\sup_{\|f\|_{\textbf{w}}=1}\|(M_{\psi}-M_{\psi}K_{n})f\|_{\mathcal{L}},

taking the limit as $\displaystyle n\to\infty$ , we obtain

\displaystyle\|M_{\psi}\|_{e}\leq B(\psi)+A(\psi).\;\qed

3. Multiplication operators from $\displaystyle\mathcal{L}$ to $\displaystyle\mathcal{L}_{\textbf{w}}$

We begin this section with a boundedness criterion for the multiplication operators from $\displaystyle M_{\psi}:\mathcal{L}\to\mathcal{L}_{\textbf{w}}$ and $\displaystyle M_{\psi}:\mathcal{L}_{0}\to\mathcal{L}_{\textbf{w},0}$ .

3.1. Boundedness and Operator Norm Estimates

Let $\displaystyle\psi$ be a function on the tree $\displaystyle T$ . Define the quantities

	$\displaystyle\theta_{\psi}$	$\displaystyle=\sup_{v\in T^{*}}\|v\|^{2}D\psi(v),$
	$\displaystyle\omega_{\psi}$	$\displaystyle=\sup_{v\in T}(\|v\|+1)\|\psi(v)\|.$

Theorem 3.1.

For a function $\displaystyle\psi$ on $\displaystyle T$ , the following statements are equivalent:

(a)

$\displaystyle M_{\psi}:\mathcal{L}\to\mathcal{L}_{\textbf{w}}$ is bounded.
(b)

$\displaystyle M_{\psi}:\mathcal{L}_{0}\to\mathcal{L}_{{\textbf{w}},0}$ is bounded.
(c)

$\displaystyle\theta_{\psi}$ and $\displaystyle\omega_{\psi}$ are finite.

Furthermore, under the above conditions, we have

\displaystyle\max\{\theta_{\psi},\omega_{\psi}\}\leq\|M_{\psi}\|\leq\theta_{\psi}+\omega_{\psi}.

Proof.

$\displaystyle(a)\Longrightarrow(c)$ Assume $\displaystyle M_{\psi}$ is bounded from $\displaystyle\mathcal{L}$ to $\displaystyle\mathcal{L}_{\textbf{w}}$ . The function $\displaystyle f_{o}=\frac{1}{2}\chi_{o}\in\mathcal{L}$ and $\displaystyle\left\|f_{o}\right\|_{\mathcal{L}}=1$ . Thus

\displaystyle\displaystyle|\psi(o)|=\|\psi f_{o}\|_{\textbf{w}}\leq\|M_{\psi}\|.

(3.1)

Next, fix $\displaystyle v\in T^{*}$ . Then $\displaystyle\chi_{v}\in\mathcal{L}$ and $\displaystyle\|\chi_{v}\|_{\mathcal{L}}=1$ ; so

\displaystyle\displaystyle(|v|+1)|\psi(v)|=\|\psi\chi_{v}\|_{\textbf{w}}\leq\|M_{\psi}\|.

(3.2)

Taking the supremum over all $\displaystyle v\in T$ , from (3.1) and (3.2) we see that $\displaystyle\omega_{\psi}$ is finite and

\displaystyle\displaystyle\omega_{\psi}\leq\|M_{\psi}\|.

(3.3)

With $\displaystyle v\in T^{*}$ , we now define

\displaystyle f_{v}(w)=\begin{cases}|w|&\hbox{ if }|w|<|v|,\\ |v|&\hbox{ if }|w|\geq|v|.\end{cases}

Then $\displaystyle f_{v}\in\mathcal{L}$ , $\displaystyle f_{v}(o)=0$ and $\displaystyle\|f_{v}\|_{\mathcal{L}}=1$ . By the boundedness of $\displaystyle M_{\psi}$ we obtain

	$\displaystyle\displaystyle\\|M_{\psi}\\|$	$\displaystyle\displaystyle\geq$	$\displaystyle\displaystyle\\|M_{\psi}f_{v}\\|_{\textbf{w}}\geq\sup_{1\leq\|w\|\leq\|v\|}\|w\|\|\psi(w)\|w\|-\psi(w^{-})(\|w\|-1)\|$
		$\displaystyle\displaystyle\geq$	$\displaystyle\displaystyle\sup_{1\leq\|w\|\leq\|v\|}\|w\|^{2}D\psi(w)-\sup_{1\leq\|w\|\leq\|v\|}\|w\|\|\psi(w^{-})\|,$

Therefore

\displaystyle|v|^{2}D\psi(v)\leq\sup_{1\leq|w|\leq|v|}|w|^{2}D\psi(w)\leq\|M_{\psi}\|+\omega_{\psi}.

Taking the supremum over all $\displaystyle v\in T^{*}$ , we obtain $\displaystyle\theta_{\psi}<\infty.$ From this and (3.3), we deduce the lower estimate

\displaystyle\|M_{\psi}\|\geq\max\{\theta_{\psi},\omega_{\psi}\}.

$\displaystyle(c)\Longrightarrow(a)$ Assume $\displaystyle\theta_{\psi}$ and $\displaystyle\omega_{\psi}$ are finite. Then, $\displaystyle\psi\in\mathcal{L}_{\textbf{w}}$ and by Lemma 1.1, for $\displaystyle f\in\mathcal{L}$ with $\displaystyle\|f\|_{\mathcal{L}}=1$ and $\displaystyle v\in T^{*}$ , we have

$\displaystyle\displaystyle\|v\|D(\psi f)(v)$	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\|v\|D\psi(v)\|f(v)\|+\|v\|\|\psi(v^{-})\|Df(v)$
	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\|v\|D\psi(v)\|f(o)\|+\|v\|^{2}D\psi(v)\\|Df\\|_{\infty}+\omega_{\psi}\\|Df\\|_{\infty}$
	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\|v\|D\psi(v)\|f(o)\|+(\theta_{\psi}+\omega_{\psi})\\|Df\\|_{\infty}.$

Thus, $\displaystyle\psi f\in\mathcal{L}_{\textbf{w}}$ . Note that $\displaystyle|f(o)+\|Df\|_{\infty}=1$ and

\displaystyle\|\psi\|_{\textbf{w}}=|\psi(o)|+\sup_{v\in T^{*}}|v|D\psi(v)\leq\omega_{\psi}+\sup_{v\in T^{*}}|v|^{2}D\psi(v)=\omega_{\psi}+\theta_{\psi}.

From this, we have

\displaystyle\|\psi f\|_{\textbf{w}}\leq\|\psi\|_{\textbf{w}}|f(o)|+(\theta_{\psi}+\omega_{\psi})\|Df\|_{\infty}\leq\theta_{\psi}+\omega_{\psi},

proving the boundedness of $\displaystyle M_{\psi}:\mathcal{L}\to\mathcal{L}_{\textbf{w}}$ and the upper estimate

\displaystyle\|M_{\psi}\|\leq\theta_{\psi}+\omega_{\psi}.

$\displaystyle(b)\Longrightarrow(c)$ The proof is the same as for $\displaystyle(a)\Longrightarrow(c)$ , since for $\displaystyle v\in T^{*}$ , the functions $\displaystyle\chi_{v}$ and $\displaystyle f_{v}$ used there belong to $\displaystyle\mathcal{L}_{0}$ .

$\displaystyle(c)\Longrightarrow(b)$ Assume $\displaystyle\theta_{\psi}$ and $\displaystyle\omega_{\psi}$ are finite and let $\displaystyle f\in\mathcal{L}_{0}$ . Then, by Lemma 1.1, for $\displaystyle v\in T^{*}$ , we have

$\displaystyle\displaystyle\|v\|D(\psi f)(v)$	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\|v\|D\psi(v)\|f(v)\|+\|v\|\|\psi(v^{-})\|Df(v)$
	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\|v\|^{2}D\psi(v)\frac{\|f(v)\|}{\|v\|}+\|v\|\|\psi(v^{-})\|Df(v)$
	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\theta_{\psi}\frac{\|f(v)\|}{\|v\|}+\omega_{\psi}Df(v)\to 0$

as $\displaystyle|v|\to\infty$ . Thus, $\displaystyle\psi f\in\mathcal{L}_{{\textbf{w}},0}.$ The proof of the boundedness of $\displaystyle M_{\psi}$ is similar to that in $\displaystyle(c)\Longrightarrow(a)$ . ∎

3.2. Isometries

In this section, we show there are no isometric multiplication operators $\displaystyle M_{\psi}$ from the space $\displaystyle\mathcal{L}$ to $\displaystyle\mathcal{L}_{\textbf{w}}$ or from $\displaystyle\mathcal{L}_{0}$ to $\displaystyle\mathcal{L}_{\textbf{w},0}$ .

Suppose $\displaystyle M_{\psi}:\mathcal{L}\to\mathcal{L}_{\textbf{w}}$ is an isometry. Then $\displaystyle\|\psi\|_{\textbf{w}}=\|M_{\psi}1\|_{\textbf{w}}=1.$ On the other hand,

\displaystyle|\psi(o)|=\frac{1}{2}\left\|\psi\chi_{o}\right\|_{\textbf{w}}=\frac{1}{2}\left\|\chi_{o}\right\|_{\mathcal{L}}=1.

Thus $\displaystyle\sup\limits_{v\in T^{*}}|v|D\psi(v)=\|\psi\|_{\textbf{w}}-|\psi(o)|=0$ , which implies that $\displaystyle\psi$ is a constant of modulus 1. Now observe that for $\displaystyle v\in T^{*}$ , we have

\displaystyle 1=\|\chi_{v}\|_{\mathcal{L}}=\|M_{\psi}\chi\|_{\textbf{w}}=(|v|+1)|\psi(v)|=|v|+1,

which is a contradiction. Since $\displaystyle\chi_{v}\in\mathcal{L}_{0}$ for all $\displaystyle v\in T$ , if $\displaystyle M_{\psi}:\mathcal{L}_{0}\to\mathcal{L}_{\textbf{w},0}$ is an isometry, then the above argument yields again a contradiction. Thus, we proved the following result.

Theorem 3.2.

There are no isometries $\displaystyle M_{\psi}$ from $\displaystyle\mathcal{L}$ to $\displaystyle\mathcal{L}_{\textbf{w}}$ or from $\displaystyle\mathcal{L}_{0}$ to $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ .

3.3. Compactness and Essential Norm

We now characterize the compact multiplication operators, but first we first give a useful compactness criterion for multiplication operators from $\displaystyle\mathcal{L}$ to $\displaystyle\mathcal{L}_{\textbf{w}}$ or from $\displaystyle\mathcal{L}_{0}$ to $\displaystyle\mathcal{L}_{\textbf{w},0}$ .

Lemma 3.3.

A bounded multiplication operator $\displaystyle M_{\psi}$ from $\displaystyle\mathcal{L}$ to $\displaystyle\mathcal{L}_{\textbf{w}}$ (or from $\displaystyle\mathcal{L}_{0}$ to $\displaystyle\mathcal{L}_{\textbf{w},0}$ ) is compact if and only if for every bounded sequence $\displaystyle\{f_{n}\}$ in $\displaystyle\mathcal{L}$ (respectively, $\displaystyle\mathcal{L}_{0}$ ) converging to 0 pointwise, the sequence $\displaystyle\|\psi f_{n}\|_{\textbf{w}}$ converges to 0 as $\displaystyle n\to\infty$ .

Proof.

Suppose $\displaystyle M_{\psi}$ is compact from $\displaystyle\mathcal{L}$ to $\displaystyle\mathcal{L}_{\textbf{w}}$ and $\displaystyle\{f_{n}\}$ is a bounded sequence in $\displaystyle\mathcal{L}$ converging to 0 pointwise. Without loss of generality, we may assume $\displaystyle\|f_{n}\|_{\mathcal{L}}\leq 1$ for all $\displaystyle n\in\mathbb{N}$ . Since $\displaystyle M_{\psi}$ is compact, the sequence $\displaystyle\{\psi f_{n}\}$ has a subsequence $\displaystyle\{\psi f_{n_{k}}\}$ that converges in the $\displaystyle\mathcal{L}_{\textbf{w}}$ -norm to some function $\displaystyle f\in\mathcal{L}_{\textbf{w}}$ .

By Lemma 1.3, for $\displaystyle v\in T^{*}$ we have

\displaystyle|(\psi(v)f_{n_{k}}(v)-f(v)|\leq(1+\log|v|)\|\psi f_{n_{k}}-f\|_{\textbf{w}}.

Thus, $\displaystyle\psi f_{n_{k}}\to f$ pointwise on $\displaystyle T^{*}$ . Furthermore, since $\displaystyle|(\psi(o)f_{n_{k}}(o)-f(o)|\leq\|\psi f_{n_{k}}-f\|_{\textbf{w}}$ , $\displaystyle\psi(0)f_{n_{k}}(0)\to f(0)$ as $\displaystyle k\to\infty$ . Thus $\displaystyle\psi f_{n_{k}}\to f$ pointwise on $\displaystyle T$ . Since by assumption, $\displaystyle f_{n}\to 0$ pointwise, it follows that $\displaystyle f$ is identically 0, and thus $\displaystyle\|\psi f_{n_{k}}\|_{\textbf{w}}\to 0$ . Since 0 is the only limit point in $\displaystyle\mathcal{L}_{\textbf{w}}$ of the sequence $\displaystyle\{\psi f_{n}\}$ , we deduce that $\displaystyle\|\psi f_{n}\|_{\textbf{w}}\to 0$ as $\displaystyle n\to\infty$ .

Conversely, suppose that every bounded sequence $\displaystyle\{f_{n}\}$ in $\displaystyle\mathcal{L}$ that converges to 0 pointwise has the property that $\displaystyle\|\psi f_{n}\|_{\textbf{w}}\to 0$ as $\displaystyle n\to\infty$ . Let $\displaystyle\{g_{n}\}$ be a sequence in $\displaystyle\mathcal{L}$ such that $\displaystyle\|g_{n}\|_{\mathcal{L}}\leq 1$ for all $\displaystyle n\in\mathbb{N}$ . Then $\displaystyle|g_{n}(o)|\leq 1$ , and by part (a) of Lemma 1.1, for $\displaystyle v\in T^{*}$ we have $\displaystyle|g_{n}(v)|\leq|v|$ . So $\displaystyle\{g_{n}\}$ is uniformly bounded on finite subsets of $\displaystyle T$ . Thus there is a subsequence $\displaystyle\{g_{n_{k}}\}$ , which converges pointwise to some function $\displaystyle g$ .

Fix $\displaystyle\varepsilon>0$ and $\displaystyle v\in T^{*}$ . Then $\displaystyle|g_{n_{k}}(v)-g(v)|<\frac{\varepsilon}{2}$ as well as $\displaystyle|g_{n_{k}}(v^{-})-g(v^{-})|<\frac{\varepsilon}{2}$ for $\displaystyle k$ sufficiently large. Therefore, for all $\displaystyle k$ sufficiently large, we have

\displaystyle Dg(v)\leq|g(v)-g_{n_{k}}(v)|+|g_{n_{k}}(v^{-})-g(v^{-})|+Dg_{n_{k}}(v)<\varepsilon+Dg_{n_{k}}(v).

Thus $\displaystyle g\in\mathcal{L}$ . The sequence $\displaystyle f_{n_{k}}=g_{n_{k}}-g$ is bounded in $\displaystyle\mathcal{L}$ and converges to 0 pointwise. So $\displaystyle\|\psi f_{n_{k}}\|_{\textbf{w}}\to 0$ as $\displaystyle k\to\infty$ . Thus $\displaystyle\psi g_{n_{k}}\to\psi g$ in the $\displaystyle\mathcal{L}_{\textbf{w}}$ -norm. Therefore, $\displaystyle M_{\psi}$ is compact.

The proof for the case of $\displaystyle M_{\psi}:\mathcal{L}_{0}\to\mathcal{L}_{\textbf{w},0}$ is similar. ∎

Theorem 3.4.

Let $\displaystyle M_{\psi}$ be a bounded multiplication operator from $\displaystyle\mathcal{L}$ to $\displaystyle\mathcal{L}_{\textbf{w}}$ (or equivalently from $\displaystyle\mathcal{L}_{0}$ to $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ ). Then the following are equivalent:

(a)

$\displaystyle M_{\psi}:\mathcal{L}\to\mathcal{L}_{\textbf{w}}$ is compact.
(b)

$\displaystyle M_{\psi}:\mathcal{L}_{0}\to\mathcal{L}_{{\textbf{w}},0}$ is compact.
(c)

$\displaystyle\displaystyle\lim_{|v|\to\infty}|v|^{2}D\psi(v)=0$ and $\displaystyle\displaystyle\lim_{|v|\to\infty}(|v|+1)|\psi(v)|=0$ .

Proof.

$\displaystyle(a)\Longrightarrow(c)$ Suppose $\displaystyle M_{\psi}:\mathcal{L}\to\mathcal{L}_{\textbf{w}}$ is compact. We need to show that if $\displaystyle\{v_{n}\}$ is a sequence in $\displaystyle T$ such that $\displaystyle 2\leq|v_{n}|$ increasing unboundedly, then $\displaystyle\displaystyle\lim_{n\to\infty}|v_{n}|^{2}D\psi(v_{n})=0$ and $\displaystyle\displaystyle\lim_{n\to\infty}(|v_{n}|+1)|\psi(v_{n})|=0$ . Let $\displaystyle\{v_{n}\}$ be such a sequence, and for $\displaystyle n\in\mathbb{N}$ define $\displaystyle f_{n}=\frac{|v_{n}|+1}{|v_{n}|}\chi_{v_{n}}$ . Clearly $\displaystyle f_{n}\to 0$ pointwise, and $\displaystyle\|f_{n}\|_{\mathcal{L}}\leq\frac{3}{2}$ . Using Lemma 3.3, we see that

\displaystyle\displaystyle\|\psi f_{n}\|_{\textbf{w}}\to 0\ \hbox{ as }n\to\infty.

(3.4)

On the other hand, since $\displaystyle f_{n}(o)=0$ for all $\displaystyle n\in\mathbb{N}$ , we have

\displaystyle\|\psi f_{n}\|_{\textbf{w}}=\sup_{v\in T^{*}}|v|D(\psi f_{n})(v)\geq|v_{n}|\left(\frac{|v_{n}|+1}{|v_{n}|}\right)|\psi(v_{n})|=(|v_{n}|+1)|\psi(v_{n})|.

Hence $\displaystyle\displaystyle\lim_{n\to\infty}(|v_{n}|+1)|\psi(v_{n})|=0$ .

Next, for $\displaystyle n\in\mathbb{N}$ , define

\displaystyle g_{n}(v)=\begin{cases}0&\text{ if }|v|<\left\lfloor\frac{|v_{n}|}{2}\right\rfloor,\\ 2|v|-|v_{n}|+2&\text{ if }\left\lfloor\frac{|v_{n}|}{2}\right\rfloor\leq|v|<|v_{n}|,\\ |v_{n}|&\text{ if }|v|\geq|v_{n}|.\end{cases}

Then $\displaystyle g_{n}\to 0$ pointwise, and $\displaystyle\|g_{n}\|_{\mathcal{L}}=2$ . Since $\displaystyle g_{n}(v_{n})=g_{n}(v_{n}^{-})=|v_{n}|$ , we have

\displaystyle\|\psi g_{n}\|_{w}\geq|v_{n}||\psi(v_{n})g_{n}(v_{n})-\psi(v_{n}^{-})g_{n}(v_{n}^{-})|=|v_{n}|^{2}D\psi(v_{n}).

By Lemma 3.3 we obtain $\displaystyle\displaystyle\lim_{n\to\infty}|v_{n}|^{2}D\psi(v_{n})\leq\lim_{n\to\infty}\|\psi g_{n}\|_{\text{w}}=0$ .

$\displaystyle(c)\Longrightarrow(a)$ Suppose $\displaystyle\displaystyle\lim_{|v|\to\infty}|v|^{2}D\psi(v)=0$ and $\displaystyle\displaystyle\lim_{|v|\to\infty}(|v|+1)|\psi(v)|=0$ . Assume $\displaystyle\psi$ is not identically zero, otherwise $\displaystyle M_{\psi}$ is trivially compact. By Lemma 3.3, to prove that $\displaystyle M_{\psi}$ is compact, it suffices to show that if $\displaystyle\{f_{n}\}$ is a bounded sequence in $\displaystyle\mathcal{L}$ converging to 0 pointwise, then $\displaystyle\|\psi f_{n}\|_{\textbf{w}}\to 0$ as $\displaystyle n\to\infty$ . Let $\displaystyle\{f_{n}\}$ be such a bounded sequence, let $\displaystyle s=\displaystyle\sup_{v\in T}\|f_{n}\|_{\mathcal{L}}$ , and fix $\displaystyle\varepsilon>0$ . There exists $\displaystyle M\in\mathbb{N}$ such that $\displaystyle(|v|+1)|\psi(v)|<\frac{\varepsilon}{2s}$ and $\displaystyle|v|^{2}D\psi(v)<\frac{\varepsilon}{2s}$ for $\displaystyle|v|\geq M$ . For $\displaystyle v\in T^{*}$ and by Lemma 1.1, we have

	$\displaystyle\|v\|D(\psi f_{n})(v)$	$\displaystyle\leq\|v\|\|\psi(v)\|Df_{n}(v)+\|v\|D\psi(v)\|f_{n}(v^{-})\|$
		$\displaystyle\leq\|v\|\|\psi(v)\|Df_{n}(v)+\|v\|D\psi(v)(\|f_{n}(o)\|+\|v\|\\|Df_{n}\\|_{\infty})$
		$\displaystyle\leq(\|v\|+1)\|\psi(v)\|Df_{n}(v)+\|v\|^{2}D\psi(v)(\|f_{n}(o)\|+\\|Df_{n}\\|_{\infty})$
		$\displaystyle=(\|v\|+1)\|\psi(v)\|Df_{n}(v)+\|v\|^{2}D\psi(v)\\|f_{n}\\|_{\mathcal{L}}.$

Since $\displaystyle f_{n}\to 0$ uniformly on $\displaystyle\{v\in T:|v|\leq M\}$ as $\displaystyle n\to\infty$ , so does $\displaystyle Df_{n}$ . So, on the set $\displaystyle\{v\in T:|v|\leq M\}$ , $\displaystyle|v|D(\psi f_{n})(v)\to 0$ as $\displaystyle n\to\infty$ . On the other hand, on $\displaystyle\{v\in T:|v|\geq M\}$ , we have

\displaystyle|v|D(\psi f_{n})(v)\leq(|v|+1)|\psi(v)|Df_{n}(v)+|v|^{2}D\psi(v)\|f_{n}\|_{\mathcal{L}}<\varepsilon.

So $\displaystyle|v|D(\psi f_{n})(v)\to 0$ as $\displaystyle n\to\infty$ . Since $\displaystyle f_{n}\to 0$ pointwise, $\displaystyle\psi(o)f_{n}(o)\to 0$ as $\displaystyle n\to\infty$ . Thus $\displaystyle\|\psi f_{n}\|_{\textbf{w}}\to 0$ as $\displaystyle n\to\infty$ . The compactness of $\displaystyle M_{\psi}$ follows at once from Lemma 3.3.

The proof of the equivalence of $\displaystyle(b)$ and $\displaystyle(c)$ is analogous. ∎

For $\displaystyle\psi$ a function on $\displaystyle T$ , define

	$\displaystyle\mathcal{A}(\psi)$	$\displaystyle=\lim_{n\to\infty}\sup_{\|v\|\geq n}\|v\|\|\psi(v)\|,$
	$\displaystyle\mathcal{B}(\psi)$	$\displaystyle=\lim_{n\to\infty}\sup_{\|v\|\geq n}\|v\|^{2}D\psi(v).$

Theorem 3.5.

Let $\displaystyle M_{\psi}$ be a bounded multiplication operator from $\displaystyle\mathcal{L}$ to $\displaystyle\mathcal{L}_{\textbf{w}}$ . Then

\displaystyle\|M_{\psi}\|_{e}\geq\max\left\{\mathcal{A}(\psi),\frac{1}{2}\mathcal{B}(\psi)\right\}.

Proof.

Fix $\displaystyle k\in\mathbb{N}$ and for each $\displaystyle n\in\mathbb{N}$ , consider the sets

	$\displaystyle\displaystyle E_{n,k}$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\{v\in T:n\leq\|v\|\leq kn,\|v\|\hbox{ even}\},$
	$\displaystyle\displaystyle O_{n,k}$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\{v\in T:n\leq\|v\|\leq kn,\|v\|\hbox{ odd}\}.$

Define the functions $\displaystyle f_{n,k}=\chi_{E_{n,k}}$ and $\displaystyle g_{n,k}=\chi_{O_{n,k}}$ . Then $\displaystyle f_{n,k},g_{n,k}\in\mathcal{L}_{0}$ , $\displaystyle\|f_{n,k}\|_{\mathcal{L}}=\|g_{n,k}\|_{\mathcal{L}}=1$ , and $\displaystyle f_{n}$ and $\displaystyle g_{n,k}\to 0$ pointwise as $\displaystyle n\to\infty$ . By Lemma 1.2, the sequences $\displaystyle f_{n,k}$ and $\displaystyle g_{n,k}$ approach 0 weakly in $\displaystyle\mathcal{L}_{0}$ as $\displaystyle n\to\infty$ . Let $\displaystyle\mathcal{K}_{0}$ be the set of compact operators from $\displaystyle\mathcal{L}_{0}$ to $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ , and note that every operator in $\displaystyle\mathcal{K}_{0}$ is completely continuous. Thus, if $\displaystyle K\in\mathcal{K}_{0}$ , then $\displaystyle\|Kf_{n,k}\|_{\textbf{w}}\to 0$ and $\displaystyle\|Kg_{n,k}\|_{\textbf{w}}\to 0$ , as $\displaystyle n\to\infty$ .

Therefore, if $\displaystyle K\in\mathcal{K}_{0}$ , then

$\displaystyle\displaystyle\\|M_{\psi}-K\\|$	$\displaystyle\displaystyle\geq$	$\displaystyle\displaystyle\limsup_{n\to\infty}\\|(M_{\psi}-K)f_{n,k}\\|_{\textbf{w}}$	(3.5)
	$\displaystyle\displaystyle\geq$	$\displaystyle\displaystyle\limsup_{n\to\infty}\\|M_{\psi}f_{n,k}\\|_{\textbf{w}}$
	$\displaystyle\displaystyle\geq$	$\displaystyle\displaystyle\limsup_{n\to\infty}\sup_{v\in E_{n,k}}(\|v\|+1)\|\psi(v)\|.$

Similarly,

\displaystyle\displaystyle\|M_{\psi}-K\|\geq\limsup_{n\to\infty}\sup_{v\in O_{n,k}}(|v|+1)|\psi(v)|.

(3.6)

Therefore, combining (3.5) and (3.6), we obtain

	$\displaystyle\\|M_{\psi}\\|_{e}$	$\displaystyle=\inf\{\\|M_{\psi}-K\\|:K\in\mathcal{K}_{0}\}$
		$\displaystyle\geq\limsup_{n\to\infty}\sup_{kn\geq\|v\|\geq n}(\|v\|+1)\|\psi(v)\|$
		$\displaystyle\geq\limsup_{n\to\infty}\sup_{kn\geq\|v\|\geq n}\|v\|\|\psi(v)\|.$

Letting $\displaystyle k\to\infty$ , we obtain $\displaystyle\|M_{\psi}\|_{e}\geq\mathcal{A}(\psi).$

Next, we wish to show that $\displaystyle\|M_{\psi}\|_{e}\geq\frac{1}{2}\mathcal{B}(\psi)$ . The result is clearly true if $\displaystyle\mathcal{B}(\psi)=0$ . So assume there exists a sequence $\displaystyle\{v_{n}\}$ in $\displaystyle T$ such that $\displaystyle 2<|v_{n}|\to\infty$ as $\displaystyle n\to\infty$ and

\displaystyle\lim_{n\to\infty}|v_{n}|^{2}D\psi(v_{n})=\mathcal{B}(\psi).

For $\displaystyle n\in\mathbb{N}$ , define

\displaystyle h_{n}(v)=\begin{cases}0&\text{ if }v=o,\\ \frac{(|v|+1)^{2}}{|v_{n}|}&\text{ if }1\leq|v|<|v_{n}|,\\ |v_{n}|&\text{ if }|v|\geq|v_{n}|.\end{cases}

Clearly, $\displaystyle h_{n}(o)=0,h_{n}(v_{n})=h_{n}(v_{n}^{-})=|v_{n}|$ , and

\displaystyle Dh_{n}(v)=\begin{cases}\frac{4}{|v_{n}|}&\text{ if }|v|=1,\\ \frac{2|v|+1}{|v_{n}|}&\text{ if }1<|v|<|v_{n}|,\\ 0&\text{ if }|v|\geq|v_{n}|.\end{cases}

The supremum of $\displaystyle Dh_{n}(v)$ is attained on the set $\displaystyle\{v\in T:|v|=|v_{n}|-1\}$ . Thus $\displaystyle\|h_{n}\|_{\mathcal{L}}=\frac{2|v_{n}|-1}{|v_{n}|}<2$ . Define $\displaystyle g_{n}=\frac{h_{n}}{\|h_{n}\|_{\mathcal{L}}}$ , and observe that $\displaystyle g_{n}\in\mathcal{L}_{0}$ , $\displaystyle\|g_{n}\|_{\mathcal{L}}=1$ , and $\displaystyle g_{n}\to 0$ pointwise on $\displaystyle T$ . By Lemma 1.2, $\displaystyle g_{n}\to 0$ weakly in $\displaystyle\mathcal{L}_{0}$ . Thus $\displaystyle\|Kg_{n}\|_{\textbf{w}}\to 0$ as $\displaystyle n\to\infty$ for any $\displaystyle K\in\mathcal{K}_{0}$ .

For each $\displaystyle n\in\mathbb{N}$ , $\displaystyle g_{n}(v_{n})=g_{n}(v_{n}^{-})=\frac{|v_{n}|^{2}}{2|v_{n}|-1}$ . Thus

	$\displaystyle\|v_{n}\|D(\psi g_{n})(v_{n})$	$\displaystyle=\|v_{n}\|\|\psi(v_{n})g_{n}(v_{n})-\psi(v_{n}^{-})g_{n}(v_{n}^{-})\|$
		$\displaystyle=\frac{\|v_{n}\|}{2\|v_{n}\|-1}\|v_{n}\|^{2}D\psi(v_{n}).$

We deduce that

	$\displaystyle\\|M_{\psi}\\|_{e}$	$\displaystyle=\inf\{\\|M_{\psi}-K\\|:K\in\mathcal{K}_{0}\}$
		$\displaystyle\geq\limsup_{n\to\infty}\\|(M_{\psi}-K)g_{n}\\|_{\textbf{w}}$
		$\displaystyle\geq\limsup_{n\to\infty}\\|M_{\psi}g_{n}\\|_{\textbf{w}}$
		$\displaystyle\geq\lim_{n\to\infty}\sup_{v\in T^{*}}\|v\|D(\psi g_{n})(v)$
		$\displaystyle\geq\lim_{n\to\infty}\|v_{n}\|D(\psi g_{n})(v_{n})$
		$\displaystyle=\lim_{n\to\infty}\frac{\|v_{n}\|}{2\|v_{n}\|-1}\|v_{n}\|^{2}D\psi(v_{n})$
		$\displaystyle\geq\frac{1}{2}\mathcal{B}(\psi).$

Therefore,

\displaystyle\|M_{\psi}\|_{e}\geq\max\left\{\mathcal{A}(\psi),\frac{1}{2}\mathcal{B}(\psi)\right\}.\;\qed

We next derive an upper estimate on the essential norm.

Theorem 3.6.

Let $\displaystyle M_{\psi}$ be a bounded multiplication operator from $\displaystyle\mathcal{L}$ to $\displaystyle\mathcal{L}_{\textbf{w}}$ . Then

\displaystyle\|M_{\psi}\|_{e}\leq\mathcal{A}(\psi)+\mathcal{B}(\psi).

Proof.

For each $\displaystyle n\in\mathbb{N}$ , consider the operator $\displaystyle K_{n}$ defined by

\displaystyle(K_{n}f)(v)=\begin{cases}f(v)&\text{ if }|v|\leq n,\\ f(v_{n})&\text{ if }|v|>n,\end{cases}

for $\displaystyle f\in\mathcal{L}$ , where $\displaystyle v_{n}$ is the ancestor of $\displaystyle v$ of length $\displaystyle n$ . Then $\displaystyle(K_{n}f)(o)=f(o)$ , and $\displaystyle K_{n}f\in\mathcal{L}_{\textbf{w},0}$ . Arguing as in the proof of Theorem 2.8, by the boundedness of $\displaystyle M_{\psi}$ , it follows that $\displaystyle M_{\psi}K_{n}$ is a compact operator from $\displaystyle\mathcal{L}$ to $\displaystyle\mathcal{L}_{\textbf{w}}$ .

Define the operator $\displaystyle J_{n}=I-K_{n}$ , where $\displaystyle I$ is the identity operator $\displaystyle I$ on $\displaystyle\mathcal{L}$ . Then,

\displaystyle D(J_{n}f)(v)\leq Df(v)\leq\|f\|_{\mathcal{L}}.

Since $\displaystyle(J_{n}f)(v)=0$ for $\displaystyle|v|\leq n$ , by Lemma 1.1, we obtain

\displaystyle|(J_{n}f)(v)|\leq|v|\|f\|_{\mathcal{L}}.

From these two estimates, we arrive at

$\displaystyle\displaystyle\\|M_{\psi}J_{n}f\\|_{\textbf{w}}$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\sup_{\|v\|>n}\|v\|\left\|\psi(v)(J_{n}f)(v)-\psi(v^{-})(J_{n}f)(v^{-})\right\|$	(3.7)
	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\sup_{\|v\|>n}\left[\|v\|D\psi(v)\|(J_{n}f)(v)\|+\|v\|\|\psi(v^{-})\|D(J_{n}f)(v)\right]$
	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\sup_{\|v\|>n}\|v\|^{2}D\psi(v)\frac{\|(J_{n}f)(v)\|}{\|v\|}+\sup_{\|v\|>n}\|v\|\|\psi(v^{-})\|D(J_{n}f)(v)\|$
	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\sup_{\|v\|>n}\|v\|^{2}D\psi(v)\\|f\\|_{\mathcal{L}}+\sup_{\|v\|>n}\|v\|\|\psi(v^{-})\|\\|f\\|_{\mathcal{L}}.$

Since

	$\displaystyle\\|M_{\psi}\\|_{e}$	$\displaystyle\leq\limsup_{n\to\infty}\\|M_{\psi}-M_{\psi}K_{n}\\|$
		$\displaystyle=\limsup_{n\to\infty}\sup_{\\|f\\|_{\mathcal{L}}=1}\\|(M_{\psi}-M_{\psi}K_{n})f\\|_{\textbf{w}}$
		$\displaystyle=\limsup_{n\to\infty}\sup_{\\|f\\|_{\mathcal{L}}=1}\\|M_{\psi}J_{n}f\\|_{\textbf{w}},$

from (3.7), taking the limit as $\displaystyle n\to\infty$ , we obtain

\displaystyle\|M_{\psi}\|_{e}\leq\mathcal{B}(\psi)+\mathcal{A}(\psi).\;\qed

4. Multiplication operators from $\displaystyle\mathcal{L}_{\textbf{w}}$ or $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ to $\displaystyle L^{\infty}$

In this section, we study the multiplication operators $\displaystyle M_{\psi}$ from the weighted Lipschitz space or the little weighted Lipschitz space into $\displaystyle L^{\infty}$ . We begin by characterizing the bounded operators and determining their operator norm. In addition, we characterize the bounded operators that are bounded from below and show that there are no isometries among them. Finally, we characterize the compact multiplication operators and determine the essential norm.

4.1. Boundedness and Operator Norm

For a function $\displaystyle\psi$ on $\displaystyle T$ , define

\displaystyle\gamma_{\psi}=\max\left\{|\psi(o)|,\sup_{v\in T^{*}}(1+\log|v|)|\psi(v)|\right\}.

Theorem 4.1.

For a function $\displaystyle\psi$ on $\displaystyle T$ , the following statements are equivalent:

(a)

$\displaystyle M_{\psi}:\mathcal{L}_{\textbf{w}}\to L^{\infty}$ is bounded.
(b)

$\displaystyle M_{\psi}:\mathcal{L}_{{\textbf{w}},0}\to L^{\infty}$ is bounded.
(c)

$\displaystyle\sup_{v\neq o}\log|v||\psi(v)|$ is finite.

Furthermore, under the above conditions, we have $\displaystyle\|M_{\psi}\|=\gamma_{\psi}.$

Proof.

The implication $\displaystyle(a)\Longrightarrow(b)$ is obvious.

$\displaystyle(b)\Longrightarrow(a)$ : We begin by showing that for each $\displaystyle f\in\mathcal{L}_{\textbf{w}}$ , the function $\displaystyle\psi f$ is bounded. Since $\displaystyle M_{\psi}$ is bounded on $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ , $\displaystyle\psi=M_{\psi}1\in L^{\infty}$ . Thus, if $\displaystyle f$ is constant, then $\displaystyle\psi f\in L^{\infty}$ . Fix $\displaystyle f\in\mathcal{L}_{\textbf{w}}$ , $\displaystyle f$ nonconstant, $\displaystyle v\in T$ , and set $\displaystyle n=|v|$ . For $\displaystyle w\in T$ , define

\displaystyle f_{n}(w)=\begin{cases}f(w)&\quad\hbox{ if }|w|\leq n\\ f(w_{n})&\quad\hbox{ if }|w|>n\end{cases}

where $\displaystyle w_{n}$ is the ancestor of $\displaystyle w$ of length $\displaystyle n$ . Then $\displaystyle f_{n}\in\mathcal{L}_{{\textbf{w}},0}$ and $\displaystyle\|f_{n}\|_{\textbf{w}}\leq\|f\|_{\textbf{w}}$ . Thus, $\displaystyle\psi f_{n}\in L^{\infty}$ and

\displaystyle\|\psi f_{n}\|_{\infty}\leq\|M_{\psi}\|\,\|f\|_{\textbf{w}}.

So $\displaystyle|\psi(v)f(v)|=|\psi(v)f_{n}(v)|\leq\|M_{\psi}\|\,\|f\|_{\textbf{w}}.$ Therefore $\displaystyle\psi f\in L^{\infty}$ and

\displaystyle\|\psi f\|_{\infty}\leq\|M_{\psi}\|\,\|f\|_{\textbf{w}},

proving the boundedness of $\displaystyle M_{\psi}$ as an operator from $\displaystyle\mathcal{L}_{\textbf{w}}$ to $\displaystyle L^{\infty}$ .

$\displaystyle(a)\Longrightarrow(c)$ : Assume $\displaystyle M_{\psi}:\mathcal{L}_{\textbf{w}}\to L^{\infty}$ is bounded. Then $\displaystyle\psi=M_{\psi}1\in L^{\infty}$ and

\displaystyle\displaystyle\|M_{\psi}\|\geq\|\psi\|_{\infty}\geq|\psi(o)|.

(4.1)

For $\displaystyle v\in T$ , define $\displaystyle f(v)=\log(1+|v|)$ . Then $\displaystyle f(o)=0$ and since for $\displaystyle x\geq 1$ the function $\displaystyle x\mapsto x\log\left(\frac{x+1}{x}\right)$ is increasing and has limit 1 as $\displaystyle x\to\infty$ , $\displaystyle f\in\mathcal{L}_{\textbf{w}}$ and $\displaystyle\|f\|_{\textbf{w}}=1.$ Thus

\displaystyle\displaystyle\|M_{\psi}\|\geq\|\psi f\|_{\infty}=\sup_{v\in T^{*}}\log(1+|v|)|\psi(v)|,

(4.2)

proving (c). Furthermore, from (4.1) and (4.2), we obtain

\displaystyle\displaystyle\|M_{\psi}\|\geq\gamma_{\psi}.

(4.3)

$\displaystyle(c)\Longrightarrow(a)$ : Assume $\displaystyle\displaystyle\sup_{v\neq o}\log|v||\psi(v)|<\infty.$ Let $\displaystyle f\in\mathcal{L}_{\textbf{w}}$ such that $\displaystyle\|f\|_{\textbf{w}}=1$ . Then $\displaystyle|\psi(o)f(o)|\leq|\psi(o)|$ and by Lemma 1.3, for $\displaystyle v\in T^{*}$ , we have

\displaystyle|\psi(v)f(v)|\leq(1+\log|v|)|\psi(v)|\leq\gamma_{\psi}.

Thus, $\displaystyle\psi f\in L^{\infty}$ and

\displaystyle\displaystyle\|\psi f\|_{\infty}\leq\gamma_{\psi},

(4.4)

proving the boundedness of $\displaystyle M_{\psi}$ as an operator from $\displaystyle\mathcal{L}_{\textbf{w}}$ to $\displaystyle L^{\infty}$ . Taking the supremum over all functions $\displaystyle f\in\mathcal{L}_{\textbf{w}}$ such that $\displaystyle\|f\|_{\textbf{w}}=1$ , from (4.4) we obtain $\displaystyle\|M_{\psi}\|\leq\gamma_{\psi}.$ Therefore, from (4.3) we conclude that $\displaystyle\|M_{\psi}\|=\gamma_{\psi}.$ ∎

4.2. Boundedness From Below

Recall that an operator $\displaystyle S$ from a Banach space $\displaystyle\mathcal{X}$ to a Banach space $\displaystyle\mathcal{Y}$ is bounded below if there exists a constant $\displaystyle C>0$ such that for all $\displaystyle x\in X$

\displaystyle\|Sx\|\geq C\|x\|.

Theorem 4.2.

A bounded multiplication operator $\displaystyle M_{\psi}$ from $\displaystyle\mathcal{L}_{\textbf{w}}$ or $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ to $\displaystyle L^{\infty}$ is bounded below if and only if

\displaystyle\inf_{v\in T}\frac{|\psi(v)|}{|v|+1}>0.

Proof.

Assume $\displaystyle M_{\psi}$ is bounded below and, arguing by contradiction, assume there exists $\displaystyle v\in T$ such that $\displaystyle\psi(v)=0$ . Then $\displaystyle M_{\psi}\chi_{v}$ is identically 0. Since operators that are bounded below are necessarily injective [7], it follows that $\displaystyle M_{\psi}$ is not bounded below. Therefore, if $\displaystyle M_{\psi}$ is bounded below, then $\displaystyle\psi$ is nonvanishing.

Next assume $\displaystyle\psi$ is nonvanishing and $\displaystyle\inf\limits_{v\in T}\frac{|\psi(v)|}{|v|+1}=0.$ Then, there exists a sequence $\displaystyle\{v_{n}\}$ in $\displaystyle T$ with $\displaystyle 1\leq|v_{n}|\to\infty$ , such that $\displaystyle\frac{|\psi(v_{n})|}{|v_{n}|+1}\to 0$ as $\displaystyle n\to\infty$ . For $\displaystyle n\in\mathbb{N}$ , define $\displaystyle f_{n}=\frac{1}{|v_{n}|+1}\chi_{v_{n}}$ . Then $\displaystyle\|f_{n}\|_{\textbf{w}}=1$ , but

\displaystyle\|\psi f_{n}\|_{\infty}=\frac{|\psi(v_{n})|}{|v_{n}|+1}\to 0.

Thus, $\displaystyle M_{\psi}$ is not bounded below.

Conversely, assume $\displaystyle\inf\limits_{v\in T}\frac{|\psi(v)|}{|v|+1}=c>0$ and that $\displaystyle M_{\psi}$ is not bounded below. Then, for each $\displaystyle n\in\mathbb{N}$ , there exists $\displaystyle f_{n}\in\;$ $\displaystyle\mathcal{L}_{\textbf{w}}$ such that $\displaystyle\|f_{n}\|_{\textbf{w}}=1$ and $\displaystyle\|\psi f_{n}\|_{\infty}<\frac{1}{n}$ . Then, for each $\displaystyle v\in T$ , we have

\displaystyle c(|v|+1)|f_{n}(v)|\leq|\psi(v)f_{n}(v)|<\frac{1}{n},

so that the sequence $\displaystyle\{g_{n}\}$ defined by $\displaystyle g_{n}(v)=(|v|+1)f_{n}(v)$ converges to 0 uniformly.

On the other hand, for $\displaystyle v\in T^{*}$ , we have

	$\displaystyle\displaystyle\|v\|Df_{n}(v)$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\left\|\frac{\|v\|}{\|v\|+1}g_{n}(v)-g_{n}(v^{-})\right\|$
		$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\|g_{n}(v)\|+\|g_{n}(v^{-})\|\to 0$

uniformly as $\displaystyle n\to\infty$ . Since $\displaystyle|\psi(o)f_{n}(o)|<1/n$ , yet $\displaystyle\|f_{n}\|_{\textbf{w}}=1$ , this yields a contradiction. ∎

4.3. Isometries

In this section, we show there are no isometries among the multiplication operators from the spaces $\displaystyle\mathcal{L}_{\textbf{w}}$ or $\displaystyle\mathcal{L}_{\textbf{w},0}$ into $\displaystyle L^{\infty}$ .

Suppose $\displaystyle M_{\psi}$ is an isometry from $\displaystyle\mathcal{L}_{\textbf{w}}$ or $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ to $\displaystyle L^{\infty}$ . Then, for $\displaystyle v\in T$ the function $\displaystyle f_{v}=\frac{1}{|v|+1}\chi_{v}$ is in $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ , $\displaystyle\|f_{v}\|_{\textbf{w}}=1$ , and

\displaystyle\frac{1}{|v|+1}|\psi(v)|=\|M_{\psi}f_{v}\|_{\infty}=\|f_{v}\|_{\textbf{w}}=1.

Thus, $\displaystyle|\psi(v)|=|v|+1$ . On the other hand, since $\displaystyle M_{\psi}$ is bounded, by Theorem 4.1, we have $\displaystyle\lim\limits_{|v|\to\infty}|v|\log|v||\psi(v)=0$ ; so $\displaystyle\psi(v)\to 0$ as $\displaystyle|v|\to\infty$ , which yields a contradiction. Thus, we proved the following result.

Theorem 4.3.

The are no isometric multiplication operators $\displaystyle M_{\psi}$ from $\displaystyle\mathcal{L}_{\textbf{w}}$ or $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ to $\displaystyle L^{\infty}$ .

4.4. Compactness and Essential Norm

We begin by giving a useful compactness criterion for the bounded operators from $\displaystyle\mathcal{L}_{\textbf{w}}$ or $\displaystyle\mathcal{L}_{\textbf{w},0}$ into $\displaystyle L^{\infty}$ .

Lemma 4.4.

A bounded multiplication operator $\displaystyle M_{\psi}$ from $\displaystyle\mathcal{L}_{\textbf{w}}$ to $\displaystyle L^{\infty}$ is compact if and only if for every bounded sequence $\displaystyle\{f_{n}\}$ in $\displaystyle\mathcal{L}_{\textbf{w}}$ converging to 0 pointwise, the sequence $\displaystyle\|\psi f_{n}\|_{\infty}$ approaches 0 as $\displaystyle n\to\infty$ .

Proof.

Assume $\displaystyle M_{\psi}$ is compact on $\displaystyle\mathcal{L}_{\textbf{w}}$ and let $\displaystyle\{f_{n}\}$ be a bounded sequence in $\displaystyle\mathcal{L}_{\textbf{w}}$ converging to 0 pointwise. By rescaling the sequence, if necessary, we may assume $\displaystyle\|f_{n}\|_{\textbf{w}}\leq 1$ for all $\displaystyle n\in\mathbb{N}$ . By the compactness of $\displaystyle M_{\psi}$ , $\displaystyle\{f_{n}\}$ has a subsequence $\displaystyle\{f_{n_{k}}\}$ such that $\displaystyle\{\psi f_{n_{k}}\}$ converges in the supremum-norm to some function $\displaystyle f\in L^{\infty}$ . In particular, $\displaystyle\psi f_{n_{k}}\to f$ pointwise. Since by assumption, $\displaystyle f_{n}\to 0$ pointwise, it follows that $\displaystyle f$ must be identically 0. Thus, the only limit point of the sequence $\displaystyle\{\psi f_{n}\}$ in $\displaystyle L^{\infty}$ is 0. Hence $\displaystyle\|\psi f_{n}\|_{\infty}\to 0$ .

Conversely, assume that for every bounded sequence $\displaystyle\{f_{n}\}$ in $\displaystyle\mathcal{L}_{\textbf{w}}$ converging to 0 pointwise, the sequence $\displaystyle\|\psi f_{n}\|_{\infty}$ approaches 0 as $\displaystyle n\to\infty$ . Let $\displaystyle\{g_{n}\}$ be a sequence in $\displaystyle\mathcal{L}_{\textbf{w}}$ with $\displaystyle\|g_{n}\|_{\textbf{w}}\leq 1$ . Fix $\displaystyle w\in T$ and, by replacing $\displaystyle g_{n}$ with $\displaystyle g_{n}-g_{n}(w)$ , assume $\displaystyle g_{n}(w)=0$ for all $\displaystyle n\in\mathbb{N}$ . Then, for each $\displaystyle v\in T$ , $\displaystyle|g_{n}(v)|=|g_{n}(v)-g_{n}(w)|\leq d(v,w)$ . Therefore, $\displaystyle g_{n}$ is uniformly bounded on finite subsets of $\displaystyle T$ , and so some subsequence $\displaystyle\{g_{n_{k}}\}_{k\in\mathbb{N}}$ converges pointwise to some function $\displaystyle g$ on $\displaystyle T$ . Fix $\displaystyle\varepsilon>0$ and $\displaystyle v\in T^{*}$ . Then, $\displaystyle|g(o)-g_{n_{k}}(o)|<\displaystyle\frac{\varepsilon}{2}$ , $\displaystyle|g_{n_{k}}(v)-g(v)|<\displaystyle{\varepsilon}{2|v|}$ and $\displaystyle|g_{n_{k}}(v^{-})-g(v^{-})|<\frac{\varepsilon}{2|v|}$ for all $\displaystyle k$ sufficiently large. Thus,

	$\displaystyle\displaystyle\|v\|Dg(v)$	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\|v\|\|g(v)-g(v^{-})-(g_{n_{k}}(v)-g_{n_{k}}(v^{-}))\|+\|v\|Dg_{n_{k}}(v)$
		$\displaystyle\displaystyle<$	$\displaystyle\displaystyle\varepsilon+\|v\|Dg_{n_{k}}(v),$

for $\displaystyle k$ sufficiently large. Consequently, $\displaystyle g\in\mathcal{L}_{\textbf{w}}$ we have

$\displaystyle\displaystyle\\|g\\|_{\textbf{w}}$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\|g(o)\|+\sup_{v\in T^{*}}\|v\|Dg(v)$
	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\|g(o)-g_{n_{k}}(o)\|+\|g_{n_{k}}(o)\|+\varepsilon+\sup_{v\in T^{*}}Dg_{n_{k}}(v)$
	$\displaystyle\displaystyle<$	$\displaystyle\displaystyle 2\varepsilon+\\|g_{n_{k}}\\|_{\textbf{w}}\leq 2\varepsilon+1.$

Since $\displaystyle\varepsilon$ was arbitrary, it follows that $\displaystyle\|g\|_{\textbf{w}}\leq 1$ . Therefore, the sequence $\displaystyle\{f_{k}\}$ defined by $\displaystyle f_{k}=g_{n_{k}}-g$ is bounded in $\displaystyle\mathcal{L}_{\textbf{w}}$ and converges to 0 pointwise, hence, by the hypothesis, $\displaystyle\|\psi f_{k}\|_{\infty}\to 0$ as $\displaystyle n\to\infty$ . We conclude that $\displaystyle\psi g_{n_{k}}\to\psi g$ in $\displaystyle L^{\infty}$ , proving the compactness of $\displaystyle M_{\psi}$ . ∎

By an analogous argument, we obtain the corresponding compactness criterion for $\displaystyle M_{\psi}:\mathcal{L}_{\textbf{w},0}\to L^{\infty}$ .

Lemma 4.5.

A bounded multiplication operator $\displaystyle M_{\psi}$ from $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ to $\displaystyle L^{\infty}$ is compact if and only if for every bounded sequence $\displaystyle\{f_{n}\}$ in $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ converging to 0 pointwise, the sequence $\displaystyle\|\psi f_{n}\|_{\infty}$ approaches 0 as $\displaystyle n\to\infty$ .

Theorem 4.6.

For a bounded operator $\displaystyle M_{\psi}$ from $\displaystyle\mathcal{L}_{\textbf{w}}$ to $\displaystyle L^{\infty}$ (or equivalently from $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ to $\displaystyle L^{\infty}$ ) the following statements are equivalent:

(a)

$\displaystyle M_{\psi}:\mathcal{L}_{\textbf{w}}\to L^{\infty}$ is compact.
(b)

$\displaystyle M_{\psi}:\mathcal{L}_{{\textbf{w}},0}\to L^{\infty}$ is compact.
(c)

$\displaystyle\lim\limits_{|v|\to\infty}\log|v||\psi(v)|=0.$

Proof.

(a) $\displaystyle\Longrightarrow$ (b) is trivial.

(b) $\displaystyle\Longrightarrow$ (c): Let $\displaystyle\{v_{n}\}$ be a sequence of vertices such that $\displaystyle 1\leq|v_{n}|\to\infty$ . We need to show that

\displaystyle\lim\limits_{n\to\infty}\log|v_{n}||\psi(v_{n})|=0.

For $\displaystyle n\in\mathbb{N}$ define

\displaystyle f_{n}(v)=\begin{cases}\ \ 0&\hbox{ if }v=0,\\ \frac{(\log|v|)^{2}}{\log|v_{n}|}&\hbox{ if }1\leq|v|<|v_{n}|,\\ \log|v_{n}|&\hbox{ if }|v|\geq|v_{n}|.\end{cases}

Then $\displaystyle\{f_{n}\}$ converges to 0 pointwise. Using the fact that $\displaystyle|v|(\log|v|-\log(|v|-1))\leq 1$ for any choice of $\displaystyle v$ in $\displaystyle T^{*}$ , we have

\displaystyle|v|Df_{n}(v)=\frac{|v|\left[(\log|v|)^{2}-(\log(|v|-1))^{2}\right]}{\log|v_{n}|}\leq\frac{\log|v|+\log(|v|-1)}{\log|v_{n}|}\leq 2,

for $\displaystyle 1\leq|v|\leq|v_{n}|$ . Moreover, $\displaystyle|v|Df_{n}(v)=0$ for $\displaystyle|v|>|v_{n}|$ . Thus, $\displaystyle f_{n}\in\mathcal{L}_{{\textbf{w}},0}$ and $\displaystyle\{\|f_{n}\|_{\textbf{w}}\}$ is bounded. By the compactness of $\displaystyle M_{\psi}$ as an operator acting on $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ and by Lemma 4.5, we deduce

\displaystyle\log|v_{n}||\psi(v_{n})|\leq\|\psi f_{n}\|_{\infty}\to 0

as $\displaystyle n\to\infty$ .

(c) $\displaystyle\Longrightarrow$ (a): Assume $\displaystyle\{f_{n}\}$ is a sequence in $\displaystyle\mathcal{L}_{\textbf{w}}$ converging to 0 pointwise and such that $\displaystyle a=\sup\limits_{n\in\mathbb{N}}\|f_{n}\|_{\textbf{w}}<\infty$ . By Lemma 1.3, for all $\displaystyle v\in T^{*}$ and all $\displaystyle n\in\mathbb{N}$ , we have

\displaystyle|\psi(v)f_{n}(v)|\leq a(1+\log|v|)|\psi(v)|.

Fix $\displaystyle\varepsilon>0$ . There exists $\displaystyle N\in\mathbb{N}$ such that $\displaystyle N\geq 3$ and for $\displaystyle|v|\geq N$ , $\displaystyle\log|v||\psi(v)|<\displaystyle\frac{\varepsilon}{2a}.$ Thus, for $\displaystyle|v|\geq N$ and for all $\displaystyle n\in\mathbb{N}$ , $\displaystyle|\psi(v)f_{n}(v)|\leq 2a\log|v||\psi(v)|<\varepsilon.$ On the other hand, since $\displaystyle f_{n}\to 0$ pointwise, for each vertex $\displaystyle v$ such that $\displaystyle|v|<N$ and $\displaystyle\psi(v)\neq 0$ , we obtain $\displaystyle|f_{n}(v)|<\displaystyle\frac{\varepsilon}{|\psi(v)|}$ for all $\displaystyle n$ sufficiently large. Hence $\displaystyle|\psi(v)f_{n}(v)|<\varepsilon$ for all $\displaystyle v\in T$ and all $\displaystyle n$ sufficiently large. Therefore, $\displaystyle\|M_{\psi}f_{n}\|_{\infty}\to 0$ as $\displaystyle n\to\infty$ , which, by Lemma 4.4, proves the compactness of $\displaystyle M_{\psi}$ . ∎

Next, we determine the essential norm of the bounded multiplication operators $\displaystyle M_{\psi}$ from $\displaystyle\mathcal{L}_{\textbf{w}}$ or $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ to $\displaystyle L^{\infty}$ .

Theorem 4.7.

Let $\displaystyle M_{\psi}$ be a bounded multiplication operator from $\displaystyle\mathcal{L}_{\textbf{w}}$ or $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ to $\displaystyle L^{\infty}$ . Then

\displaystyle\|M_{\psi}\|_{e}=\lim_{n\to\infty}\sup_{|v|\geq n}\log|v||\psi(v)|.

Proof.

Define $\displaystyle A(\psi)=\lim\limits_{n\to\infty}\sup\limits_{|v|\geq n}\log|v||\psi(v)|.$ If $\displaystyle A(\psi)=0$ , then by Theorem 4.6, $\displaystyle M_{\psi}$ is compact, hence its essential norm is 0. So assume $\displaystyle A(\psi)>0$ . We first show that $\displaystyle\|M_{\psi}\|_{e}\geq A(\psi)$ . Let $\displaystyle\{v_{n}\}$ be a sequence in $\displaystyle T$ such that $\displaystyle 1\leq|v_{n}|\to\infty$ and

\displaystyle A(\psi)=\lim_{n\to\infty}\log|v_{n}||\psi(v_{n})|.

Fix $\displaystyle p\in(0,1)$ and for each $\displaystyle n\in\mathbb{N}$ , define

\displaystyle f_{n,p}(v)=\begin{cases}\ \ 0&\quad\hbox{ if }v=0,\\ \frac{\ (\log|v|)^{p+1}}{(\log|v_{n}|)^{p}}&\quad\hbox{ if }1\leq|v|<|v_{n}|,\\ \ \log|v_{n}|&\quad\hbox{ if }|v|\geq|v_{n}|.\end{cases}

Then $\displaystyle\{f_{n,p}\}$ converges to 0 pointwise, $\displaystyle f_{n,p}\in\mathcal{L}_{{\textbf{w}},0}$ , $\displaystyle f_{n,p}(v_{n})=\log|v_{n}|$ , and

	$\displaystyle\displaystyle\\|f_{n,p}\\|_{\textbf{w}}$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\sup_{2\leq\|v\|\leq\|v_{n}\|}\frac{\|v\|}{(\log\|v_{n}\|)^{p}}\left[\left(\log\|v\|\right)^{p+1}-\left(\log(\|v\|-1)\right)^{p+1}\right]$
		$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\frac{\|v_{n}\|}{(\log\|v_{n}\|)^{p}}\left[(\log\|v_{n}\|)^{p+1}-(\log(\|v_{n}\|-1))^{p+1}\right]\leq p+1.$

By Lemma 1.4, $\displaystyle\{f_{n,p}\}$ converges to $\displaystyle 0$ weakly in $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ . Let $\displaystyle K$ be a compact operator from $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ (or equivalently, from $\displaystyle\mathcal{L}_{\textbf{w}}$ ) to $\displaystyle L^{\infty}$ . Since compact operators are completely continuous, it follows that $\displaystyle\|Kf_{n,p}\|_{\infty}\to 0$ as $\displaystyle n\to\infty$ . Thus,

$\displaystyle\displaystyle\\|M_{\psi}-K\\|$	$\displaystyle\displaystyle\geq$	$\displaystyle\displaystyle\limsup_{n\to\infty}\frac{\\|(M_{\psi}-K)f_{n,p}\\|_{\infty}}{\\|f_{n,p}\\|_{\textbf{w}}}$
	$\displaystyle\displaystyle\geq$	$\displaystyle\displaystyle\frac{1}{p+1}\limsup_{n\to\infty}\\|M_{\psi}f_{n,p}\\|_{\infty}$
	$\displaystyle\displaystyle\geq$	$\displaystyle\displaystyle\frac{1}{p+1}\limsup_{n\to\infty}\log\|v_{n}\|\|\psi(v_{n})\|.$

Taking the infimum over all such compact operators $\displaystyle K$ and passing to the limit as $\displaystyle p$ approaches 0, we obtain

\displaystyle\|M_{\psi}\|_{e}\geq\lim_{n\to\infty}\log|v_{n}||\psi(v_{n})|=A(\psi).

To prove the estimate $\displaystyle\|M_{\psi}\|_{e}\leq A(\psi)$ , for each $\displaystyle n\in\mathbb{N}$ and for $\displaystyle f\in\mathcal{L}_{\textbf{w}}$ , define

\displaystyle K_{n}f(v)=\begin{cases}f(v)&\quad\hbox{ if }|v|\leq n,\\ f(v_{n})&\quad\hbox{ if }|v|>n,\end{cases}

where $\displaystyle v_{n}$ is the ancestor of $\displaystyle v$ of length $\displaystyle n$ . In the proof of Theorem 2.8, it is was shown that $\displaystyle K_{n}$ is a compact operator on $\displaystyle\mathcal{L}_{\textbf{w}}$ . Since $\displaystyle M_{\psi}:\mathcal{L}_{\textbf{w}}\to L^{\infty}$ is bounded, it follows that $\displaystyle M_{\psi}K_{n}$ is also compact as an operator from $\displaystyle\mathcal{L}_{\textbf{w}}$ to $\displaystyle L^{\infty}$ .

Let $\displaystyle v\in T$ , and let $\displaystyle w$ be a vertex in the path from $\displaystyle o$ to $\displaystyle v$ of length $\displaystyle k\geq 1$ . Label the vertices from $\displaystyle w$ to $\displaystyle v$ by $\displaystyle v_{j}$ , $\displaystyle j=k,\dots,|v|$ . Then for $\displaystyle f\in\mathcal{L}_{\textbf{w}}$ with $\displaystyle\|f\|_{\textbf{w}}=1$ , we have

\displaystyle|f(v)-f(w)|\leq\sum_{j=k+1}^{|v|}|f(v_{j})-f(v_{j-1})|\leq\sum_{j=k+1}^{|v|}\frac{1}{j}\leq\log|v|.

Thus

\displaystyle\|(M_{\psi}-M_{\psi}K_{n})f\|_{\infty}=\sup_{|v|>n}|\psi(v)||f(v)-f(v_{n})|\leq\sup_{|v|>n}\log|v||\psi(v)|.

We deduce

\displaystyle\|M_{\psi}\|_{e}\leq\sup_{\|f\|_{\textbf{w}}=1}\|(M_{\psi}-M_{\psi}K_{n})f\|_{\infty}\leq\sup_{|v|>n}\log|v||\psi(v)|.

Taking the limit as $\displaystyle n\to\infty$ , we obtain $\displaystyle\|M_{\psi}\|_{e}\leq A(\psi)$ . ∎

5. Multiplication operators from $\displaystyle L^{\infty}$ to $\displaystyle\mathcal{L}_{\textbf{w}}$ or $\displaystyle\mathcal{L}_{{\textbf{w}},0}$

In this last section, we study the multiplication operators $\displaystyle M_{\psi}$ from $\displaystyle L^{\infty}$ into the weighted Lipschitz space or the little weighted Lipschitz space. We first characterize the bounded operators and determine the operator norm. We also show there are no isometries among such operators. Finally, we characterize the compact multiplication operators and determine the essential norm.

5.1. Boundedness and Operator Norm

For a function $\displaystyle\psi$ on $\displaystyle T$ , define

\displaystyle\eta_{\psi}=|\psi(o)|+\sup_{v\in T^{*}}|v|\left[|\psi(v)|+|\psi(v^{-})|\right].

Theorem 5.1.

For a function $\displaystyle\psi$ on $\displaystyle T$ , the following statements are equivalent:

(a)

$\displaystyle M_{\psi}:L^{\infty}\to\mathcal{L}_{\textbf{w}}$ is bounded.
(b)

$\displaystyle\sup\limits_{v\in T}|v||\psi(v)|<\infty$ .

Furthermore, under these conditions, we have

\displaystyle\|M_{\psi}\|=\eta_{\psi}.

Proof.

$\displaystyle(a)\Longrightarrow(b)$ : Assume $\displaystyle M_{\psi}:L^{\infty}\to\mathcal{L}_{\textbf{w}}$ is bounded. Fix $\displaystyle v\in T^{*}$ . Since $\displaystyle\chi_{v}\in L^{\infty}$ and $\displaystyle\|\chi_{v}\|_{\infty}=1$ , the function $\displaystyle\psi\chi_{v}\in\mathcal{L}_{\textbf{w}}$ , so

\displaystyle|v||\psi(v)|<(|v|+1)|\psi(v)|=\sup_{w\in T^{*}}|w|D(\psi\chi_{v})(w)\leq\|M_{\psi}\|.

Thus, $\displaystyle\displaystyle\sup_{v\in T}|v||\psi(v)|$ is finite.

$\displaystyle(b)\Longrightarrow(a)$ : Suppose $\displaystyle\displaystyle\sup_{v\in T}|v||\psi(v)|<\infty$ . Let $\displaystyle f\in L^{\infty}$ such that $\displaystyle\|f\|_{\infty}=1$ . Then

\displaystyle\|M_{\psi}f\|_{\textbf{w}}\leq|\psi(o)|+\sup_{v\in T^{*}}|v|\left[|\psi(v)|+|\psi(v^{-})|\right]<\infty.

Thus, $\displaystyle M_{\psi}$ is bounded and $\displaystyle\|M_{\psi}\|\leq\eta_{\psi}$ .

We next show that $\displaystyle\|M_{\psi}\|\geq\eta_{\psi}$ . The inequality is obvious is $\displaystyle\psi$ is identically 0. For $\displaystyle\psi$ not identically 0 and for $\displaystyle v\in T$ , define

\displaystyle f(v)=\begin{cases}\ \ \ \ 0&\quad\hbox{ if }\psi(v)=0,\\ \ \ \overline{\psi(v)}/|\psi(v)|&\quad\hbox{ if }\psi(v)\neq 0,|v|\hbox{ even},\\ -\overline{\psi(v)}/|\psi(v)|&\quad\hbox{ if }\psi(v)\neq 0,|v|\hbox{ odd}.\end{cases}

Then $\displaystyle\|f\|_{\infty}=1$ and for $\displaystyle v\in T^{*}$ , $\displaystyle D(\psi f)(v)=|\psi(v)|+|\psi(v^{-})|$ , so that

\displaystyle\|M_{\psi}f\|_{\textbf{w}}=|\psi(o)|+\sup_{v\in T^{*}}|v|\left[|\psi(v)|+|\psi(v^{-})|\right].

Thus, $\displaystyle\|M_{\psi}\|\geq\eta_{\psi},$ completing the proof. ∎

In the next result, we characterize the bounded multiplication operators from $\displaystyle L^{\infty}$ to $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ .

Theorem 5.2.

For a function $\displaystyle\psi$ on $\displaystyle T$ , the following statements are equivalent:

(a)

$\displaystyle M_{\psi}:L^{\infty}\to\mathcal{L}_{{\textbf{w}},0}$ is bounded.
(b)

$\displaystyle\lim\limits_{|v|\to\infty}|v||\psi(v)|=0$ .

Furthermore, under these conditions, we have,

\displaystyle\|M_{\psi}\|=\eta_{\psi}.

Proof.

$\displaystyle(a)\Longrightarrow(b)$ : Assume $\displaystyle M_{\psi}:L^{\infty}\to\mathcal{L}_{{\textbf{w}},0}$ is bounded. Applying $\displaystyle M_{\psi}$ to the constant function 1, we obtain $\displaystyle\psi=M_{\psi}1\in\mathcal{L}_{{\textbf{w}},0}$ . On the other hand, if $\displaystyle\mathcal{O}=\{v\in T:|v|\text{ is odd}\}$ , then $\displaystyle\psi\chi_{\mathcal{O}}\in\mathcal{L}_{{\textbf{w}},0}$ , so for $\displaystyle v\in T^{*}$ , we have

	$\displaystyle\displaystyle\|v\|\|\psi(v)\|$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\|v\|\|\psi(v)\|D\chi_{\mathcal{O}}(v)\leq\|v\|D(\psi\chi_{\mathcal{O}})(v)+\|v\|D\psi(v)\|\chi_{\mathcal{O}}(v^{-})\|$
		$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\|v\|D(\psi\chi_{\mathcal{O}})(v)+\|v\|D\psi(v)\to 0,$

as $\displaystyle|v|\to\infty$ , proving (b).

$\displaystyle(b)\Longrightarrow(a)$ : Suppose $\displaystyle|v||\psi(v)|\to 0$ as $\displaystyle|v|\to\infty$ . First observe that

	$\displaystyle\displaystyle\|v\|D\psi(v)$	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\|v\|\|\psi(v)\|+\frac{\|v\|}{\|v\|-1}(\|v\|-1)\|\psi(v^{-})\|$
		$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\|v\|\|\psi(v)\|+2(\|v\|-1)\|\psi(v^{-})\|\to 0$

as $\displaystyle|v|\to\infty$ . Then for $\displaystyle f\in L^{\infty}$ and $\displaystyle v\in T^{*}$ , we have

	$\displaystyle\displaystyle\|v\|D(\psi f)(v)$	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\|v\|\|\psi(v)\|Df(v)+\|v\|D\psi(v)\|f(v^{-})\|$
		$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle(2\|v\|\|\psi(v)\|+\|v\|D\psi(v))\\|f\\|_{\infty}\to 0,$

as $\displaystyle|v|\to\infty$ . Thus, $\displaystyle\psi f\in\mathcal{L}_{{\textbf{w}},0}$ . The proof of the boundedness of $\displaystyle M_{\psi}$ and of the formula $\displaystyle\|M_{\psi}\|=\eta_{\psi}$ is similar to the case when $\displaystyle M_{\psi}:L^{\infty}\to\mathcal{L}_{\textbf{w}}$ . ∎

5.2. Isometries

As for all other multiplication operators in this article, there are no isometries among the multiplication operators from $\displaystyle L^{\infty}$ into $\displaystyle\mathcal{L}_{\textbf{w}}$ or $\displaystyle\mathcal{L}_{\textbf{w},0}$ .

Assume $\displaystyle M_{\psi}$ is an isometry from $\displaystyle L^{\infty}$ to $\displaystyle\mathcal{L}_{\textbf{w}}$ or $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ . Then, for $\displaystyle v\in T$ the function $\displaystyle f_{v}=\frac{1}{|v|+1}\chi_{v}$ is in $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ with $\displaystyle\|M_{\psi}\chi_{v}\|_{\textbf{w}}=\|\chi_{v}\|_{\infty}=1$ . In particular, it follows that $\displaystyle|\psi(o)|=\frac{1}{2}$ and for $\displaystyle v\in T^{*}$ ,

\displaystyle(|v|+1)|\psi(v)|=1.

Thus, $\displaystyle|\psi(v)|=\frac{1}{|v|+1}$ . On the other hand, taking as a test function $\displaystyle f$ the characteristic function of the set $\displaystyle\{v\in T:|v|\leq 1\}$ , we obtain

\displaystyle 1=\|f\|_{\infty}=\|M_{\psi}f\|_{\textbf{w}}=|\psi(o)|+\max\left\{\sup_{|v|=1}|\psi(v)-\psi(o)|,\sup_{|v|=1}2|\psi(v)|\right\}\geq\frac{3}{2},

which yields a contradiction. Therefore, we obtain the following result.

Theorem 5.3.

The are no isometric multiplication operators $\displaystyle M_{\psi}$ from $\displaystyle L^{\infty}$ to $\displaystyle\mathcal{L}_{\textbf{w}}$ or $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ .

5.3. Compactness and Essential Norm

The following two results are compactness criteria for multiplication operators from $\displaystyle L^{\infty}$ into $\displaystyle\mathcal{L}_{\textbf{w}}$ or $\displaystyle\mathcal{L}_{\textbf{w},0}$ similar to those given in the previous sections.

Lemma 5.4.

A bounded multiplication operator $\displaystyle M_{\psi}$ from $\displaystyle L^{\infty}$ to $\displaystyle\mathcal{L}_{\textbf{w}}$ is compact if and only if for every bounded sequence $\displaystyle\{f_{n}\}$ in $\displaystyle L^{\infty}$ converging to 0 pointwise, the sequence $\displaystyle\|\psi f_{n}\|_{\textbf{w}}$ approaches 0 as $\displaystyle n\to\infty$ .

Proof.

Assume $\displaystyle M_{\psi}$ is compact and let $\displaystyle\{f_{n}\}$ be a bounded sequence in $\displaystyle L^{\infty}$ converging to 0 pointwise. By rescaling the sequence, if necessary, we may assume $\displaystyle\|f_{n}\|_{\infty}\leq 1$ for all $\displaystyle n\in\mathbb{N}$ . By the compactness of $\displaystyle M_{\psi}$ , $\displaystyle\{f_{n}\}$ has a subsequence $\displaystyle\{f_{n_{k}}\}$ such that $\displaystyle\{\psi f_{n_{k}}\}$ converges in the $\displaystyle\mathcal{L}_{\textbf{w}}$ -norm to some function $\displaystyle f\in\mathcal{L}_{\textbf{w}}$ . Since by Lemma 1.3, for $\displaystyle v\in T^{*}$ ,

\displaystyle|\psi(v)f_{n_{k}}(v)-f(v)|\leq(1+\log|v|)\|\psi f_{n_{k}}-f\|_{\textbf{w}},

and $\displaystyle|\psi(o)f_{n_{k}}(o)-f(o)|\leq\|\psi f_{n_{k}}-f\|_{\textbf{w}},$ it follows that $\displaystyle\psi f_{n_{k}}\to f$ pointwise. Since by assumption, $\displaystyle f_{n}\to 0$ pointwise, the function $\displaystyle f$ must be identically 0. Thus, the only limit point of the sequence $\displaystyle\{\psi f_{n}\}$ in $\displaystyle\mathcal{L}_{\textbf{w}}$ is 0. Hence $\displaystyle\|\psi f_{n}\|_{\textbf{w}}\to 0$ as $\displaystyle n\to\infty$ .

Conversely, suppose $\displaystyle\|\psi f_{n}\|_{\textbf{w}}$ approaches 0 as $\displaystyle n\to\infty$ for every bounded sequence $\displaystyle\{f_{n}\}$ in $\displaystyle L^{\infty}$ converging to 0 pointwise. Let $\displaystyle\{g_{n}\}$ be a sequence in $\displaystyle L^{\infty}$ with $\displaystyle\|g_{n}\|_{\infty}\leq 1$ . Then some subsequence $\displaystyle\{g_{n_{k}}\}$ converges to a bounded function $\displaystyle g$ . Thus, the sequence $\displaystyle f_{n_{k}}=g_{n_{k}}-g$ converges to 0 uniformly and $\displaystyle\|f_{n}\|_{\infty}$ is bounded. By the hypothesis, it follows that $\displaystyle\|\psi f_{n_{k}}\|_{\textbf{w}}\to 0$ as $\displaystyle k\to\infty$ . Thus, $\displaystyle\psi g_{n_{k}}\to\psi g$ in $\displaystyle\mathcal{L}_{\textbf{w}}$ . Therefore, $\displaystyle M_{\psi}$ is compact. ∎

By an analogous argument, we obtain the corresponding result for $\displaystyle M_{\psi}:L^{\infty}\to\mathcal{L}_{{\textbf{w},0}}$ .

Lemma 5.5.

A bounded multiplication operator $\displaystyle M_{\psi}$ from $\displaystyle L^{\infty}$ to $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ is compact if and only if for every bounded sequence $\displaystyle\{f_{n}\}$ in $\displaystyle L^{\infty}$ converging to 0 pointwise, the sequence $\displaystyle\|\psi f_{n}\|_{\textbf{w}}$ approaches 0 as $\displaystyle n\to\infty$ .

Theorem 5.6.

For a bounded operator $\displaystyle M_{\psi}$ from $\displaystyle L^{\infty}$ to $\displaystyle\mathcal{L}_{\textbf{w}}$ , the following statements are equivalent:

(a)

$\displaystyle M_{\psi}$ is compact.
(b)

$\displaystyle\lim\limits_{|v|\to\infty}|v||\psi(v)|=0$ .

Proof.

$\displaystyle(a)\Longrightarrow(b)$ : Assume $\displaystyle M_{\psi}$ is compact. Let $\displaystyle\{v_{n}\}$ be a sequence in $\displaystyle T$ such that $\displaystyle|v_{n}|\to\infty$ as $\displaystyle n\to\infty$ . For $\displaystyle n\in\mathbb{N}$ , let $\displaystyle f_{n}$ denote the characteristic function of the set $\displaystyle\{w\in T:|w|\geq|v_{n}|\}$ . Then $\displaystyle\|f_{n}\|_{\infty}=1$ and $\displaystyle f_{n}\to 0$ pointwise. By Lemma 5.4 and the compactness of $\displaystyle M_{\psi}$ , it follows that

\displaystyle|v_{n}||\psi(v_{n})|=|v_{n}|D(\psi f_{n})(v_{n})\leq\|M_{\psi}f_{n}\|_{\textbf{w}}\to 0

as $\displaystyle n\to\infty$ .

$\displaystyle(b)\Longrightarrow(a)$ : Assume $\displaystyle\lim\limits_{|v|\to\infty}|v||\psi(v)|=0$ and that $\displaystyle\psi$ is not identically 0. In particular, $\displaystyle\psi$ is bounded. Let $\displaystyle\{f_{n}\}$ be a sequence in $\displaystyle L^{\infty}$ converging pointwise to 0 and such that $\displaystyle\|f_{n}\|_{\infty}$ is bounded above by some positive constant $\displaystyle C$ . Then corresponding to $\displaystyle\varepsilon>0$ , there exists $\displaystyle N\in\mathbb{N}$ such that $\displaystyle|v||\psi(v)|<\displaystyle\frac{\varepsilon}{4C}$ for all vertices $\displaystyle v$ such that $\displaystyle|v|\geq N$ . Therefore, for $\displaystyle|v|>N$ and $\displaystyle n\in\mathbb{N}$ , we have

\displaystyle|v|D(\psi f_{n})(v)\leq|v||\psi(v)|Df_{n}(v)+|v|D\psi(v)|f_{n}(v^{-})|<\varepsilon.

Furthermore, the sequence $\displaystyle\{f_{n}\}$ converges to 0 uniformly on the set $\displaystyle\{v\in T:|v|\leq N\}$ so that $\displaystyle|f_{n}(v)|<\frac{\varepsilon}{4N\|\psi\|_{\infty}}$ for all $\displaystyle n$ sufficiently large. Hence $\displaystyle|v|D(\psi f_{n})(v)<\varepsilon$ for all $\displaystyle v\in T^{*}$ and all $\displaystyle n$ sufficiently large. Consequently, $\displaystyle\|\psi f_{n}\|_{\textbf{w}}\to 0$ as $\displaystyle n\to\infty$ . Using Lemma 5.4, we deduce that $\displaystyle M_{\psi}$ is compact. ∎

Since the above proof is also valid when $\displaystyle M_{\psi}$ is a bounded operator from $\displaystyle L^{\infty}$ to $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ , through the application of Lemma 5.5, from Theorems 5.2 and 5.6 we obtain the following result.

Corollary 5.7.

For a function $\displaystyle\psi$ on $\displaystyle T$ , the following statements are equivalent:

(a)

$\displaystyle M_{\psi}:L^{\infty}\to\mathcal{L}_{{\textbf{w}}}$ is compact.
(b)

$\displaystyle M_{\psi}:L^{\infty}\to\mathcal{L}_{{\textbf{w}},0}$ is bounded.
(c)

$\displaystyle M_{\psi}:L^{\infty}\to\mathcal{L}_{{\textbf{w}},0}$ is compact.
(d)

$\displaystyle\lim\limits_{|v|\to\infty}|v||\psi(v)|=0$ .

We now determine the essential norm of the bounded multiplication operators from $\displaystyle L^{\infty}$ to $\displaystyle\mathcal{L}_{\textbf{w}}$ .

Theorem 5.8.

Let $\displaystyle M_{\psi}:L^{\infty}\to\mathcal{L}_{\textbf{w}}$ be bounded. Then

\displaystyle\|M_{\psi}\|_{e}=\lim_{n\to\infty}\sup_{|v|\geq n}|v|\left[|\psi(v)|+|\psi(v^{-})|\right].

Proof.

Set $\displaystyle B(\psi)=\lim\limits_{n\to\infty}\sup\limits_{|v|\geq n}|v|\left[|\psi(v)|+|\psi(v^{-})|\right].$ In the case $\displaystyle B(\psi)=0$ , then $\displaystyle\lim\limits_{|v|\to\infty}|v|\psi(v)=0$ , so by Theorem 5.6, $\displaystyle M_{\psi}$ is compact and thus $\displaystyle\|M_{\psi}\|_{e}=0$ . So assume $\displaystyle B(\psi)>0$ . Then there exists a sequence $\displaystyle\{v_{n}\}$ in $\displaystyle T$ such that $\displaystyle 1\leq|v_{n}|\to\infty$ and

\displaystyle B(\psi)=\lim_{n\to\infty}|v_{n}|\left[|\psi(v_{n})|+|\psi(v_{n}^{-})|\right].

For each $\displaystyle n\in\mathbb{N}$ let $\displaystyle f_{n}$ be the function on $\displaystyle T$ defined by

\displaystyle f_{n}(v)=\begin{cases}\ \ \ \ 0&\quad\hbox{ if }|v|<|v_{n}|\hbox{ or }\psi(v)=0,\\ \ \ \overline{\psi(v)}/|\psi(v)|&\quad\hbox{ if }|v|\geq|v_{n}|,|v|\hbox{ is even, and }\psi(v)\neq 0,\\ -\overline{\psi(v)}/|\psi(v)|&\quad\hbox{ otherwise.}\end{cases}

Then $\displaystyle\|f_{n}\|_{\infty}=1$ and $\displaystyle\{f_{n}\}$ converges to 0 pointwise. Thus, for any compact operator $\displaystyle K:L^{\infty}\to\mathcal{L}_{\textbf{w}}$ , there exists a subsequence $\displaystyle\{f_{n_{k}}\}$ such that $\displaystyle\|Kf_{n_{k}}\|_{\textbf{w}}\to 0$ as $\displaystyle k\to\infty$ . Thus

	$\displaystyle\displaystyle\\|M_{\psi}-K\\|$	$\displaystyle\displaystyle\geq$	$\displaystyle\displaystyle\limsup_{k\to\infty}\\|(M_{\psi}-K)f_{n_{k}}\\|_{\textbf{w}}\geq\limsup_{k\to\infty}\\|\psi f_{n_{k}}\\|_{\textbf{w}}$
		$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\limsup_{k\to\infty}\sup_{\|v\|\geq\|v_{n_{k}}\|}\|v\|\left[\|\psi(v)\|+\|\psi(v^{-})\|\right]=B(\psi).$

Therefore, $\displaystyle\|M_{\psi}\|_{e}\geq B(\psi).$

We now show that $\displaystyle\|M_{\psi}\|_{e}\leq B(\psi)$ . For each $\displaystyle n\in\mathbb{N}$ , define the operator $\displaystyle K_{n}$ on $\displaystyle L^{\infty}$ by

\displaystyle K_{n}f(v)=\begin{cases}f(v)&\quad\hbox{ if }|v|\leq n\\ 0&\quad\hbox{ if }|v|>n.\end{cases}

Then, for $\displaystyle v\in T^{*}$ , we have

\displaystyle|v|D(K_{n}f)(v)=\begin{cases}|v|Df(v)&\quad\hbox{ for }1\leq|v|\leq n,\\ (n+1)|f(v^{-})|&\quad\hbox{ for }|v|=n+1,\\ 0&\quad\hbox{ for }|v|>n+1.\end{cases}

Thus, $\displaystyle K_{n}f\in\mathcal{L}_{\textbf{w}}$ with $\displaystyle\|K_{n}f\|_{\textbf{w}}\leq|f(o)|+2n\|f\|_{\infty}.$

Assume $\displaystyle\{f_{k}\}$ is a sequence in $\displaystyle L^{\infty}$ with $\displaystyle\|f_{k}\|_{\infty}\leq 1$ . Then there exists a subsequence $\displaystyle\{f_{k_{j}}\}$ converging pointwise to some function $\displaystyle f\in L^{\infty}$ . Thus,

$\displaystyle\displaystyle\\|K_{n}f_{k_{j}}-K_{n}f\\|_{\textbf{w}}$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\|f_{k_{j}}(o)-f(o)\|$
	$\displaystyle\displaystyle+$	$\displaystyle\displaystyle\max\left\{\sup_{1\leq\|v\|\leq n}\|v\|D(f_{k_{j}}-f)(v),\sup_{\|v\|=n+1}\|v\|\|f_{k_{j}}(v^{-})-f(v^{-})\|\right\}$
	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\|f_{k_{j}}(o)-f(o)\|$
	$\displaystyle\displaystyle+$	$\displaystyle\displaystyle 2n\max\left\{\sup_{1\leq\|v\|\leq n}D(f_{k_{j}}-f)(v),\sup_{\|v\|=n+1}\|f_{k_{j}}(v^{-})-f(v^{-})\|\right\}.$

So $\displaystyle\|K_{n}f_{k_{j}}-K_{n}f\|\to 0$ as $\displaystyle j\to\infty$ . Therefore, $\displaystyle K_{n}$ is compact, and thus, since $\displaystyle M_{\psi}$ is bounded, $\displaystyle M_{\psi}K_{n}$ is also compact.

For $\displaystyle f\in L^{\infty}$ , we have

$\displaystyle\displaystyle\\|(M_{\psi}-M_{\psi}K_{n})f\\|_{\textbf{w}}$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\sup_{\|v\|>n}\|v\|\|\psi(v)f(v)-\psi(v^{-})f(v^{-})+\psi(v^{-})K_{n}f(v^{-})\|$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\max\left\{\sup_{\|v\|=n+1}\|v\|\|\psi(v)\|\|f(v)\|,\sup_{\|v\|>n+1}\|v\|\|\psi(v)f(v)-\psi(v^{-})f(v^{-})\|\right\}$
	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\sup_{\|v\|>n}\|v\|\left[\|\psi(v)\|+\|\psi(v^{-})\|\right]\\|f\\|_{\infty}.$

Therefore, we obtain

$\displaystyle\displaystyle\\|M_{\psi}\\|_{e}$	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\limsup_{n\to\infty}\\|M_{\psi}-M_{\psi}K_{n}\\|$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\limsup_{n\to\infty}\sup_{\\|f\\|_{\infty}=1}\\|(M_{\psi}-M_{\psi}K_{n})f\\|_{\textbf{w}}$
	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle B(\psi),$

thus completing the proof. ∎

References

[1] Robert F. Allen, Flavia Colonna, and Glenn R. Easley, Multiplication operators on the weighted Lipschitz space of a tree, J. Operator Theory 69 (2013), no. 1, 209–231. MR 3029495
[2] Pierre Cartier, Fonctions harmoniques sur un arbre, Symposia Mathematica, Vol. IX (Convegno di Calcolo delle Probabilità, INDAM, Rome, 1971), 1972, pp. 203–270. MR 0353467
[3] Joel M. Cohen and Flavia Colonna, Embeddings of trees in the hyperbolic disk, Complex Variables Theory Appl. 24 (1994), no. 3-4, 311–335. MR 1270321
[4] Flavia Colonna, Bloch and normal functions and their relation, Rend. Circ. Mat. Palermo (2) 38 (1989), no. 2, 161–180. MR 1029707
[5] Flavia Colonna and Glenn R. Easley, Multiplication operators on the Lipschitz space of a tree, Integral Equations Operator Theory 68 (2010), no. 3, 391–411. MR 2735443
[6] by same author, Multiplication operators between the Lipschitz space and the space of bounded functions on a tree, Mediterr. J. Math. 9 (2012), no. 3, 423–438. MR 2954500
[7] John B. Conway, A course in functional analysis, second ed., Graduate Texts in Mathematics, vol. 96, Springer-Verlag, New York, 1990. MR 1070713
[8] Kehe Zhu, Operator theory in function spaces, second ed., Mathematical Surveys and Monographs, vol. 138, American Mathematical Society, Providence, RI, 2007. MR 2311536

	$\displaystyle\displaystyle D\psi(v)\|f(v)\|$	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle D(\psi f)(v)+\|\psi(v^{-})\|Df(v)$
		$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\\|M_{\psi}f\\|_{\mathcal{L}}+\sigma_{\psi}\|v\|Df(v)\leq\\|M_{\psi}\\|+\sigma_{\psi}.$

$\displaystyle\displaystyle\\|M_{\psi}f\\|_{\mathcal{L}}$	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\|\psi(o)\|\|f(o)\|+\tau_{\psi}\\|f\\|_{\textbf{w}}+\sigma_{\psi}(\\|f\\|_{\textbf{w}}-\|f(o)\|)$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle(\tau_{\psi}+\sigma_{\psi})\\|f\\|_{\textbf{w}}+(\|\psi(o)\|-\sigma_{\psi})\|f(o)\|$
	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\left(\tau_{\psi}+\sigma_{\psi}\right)\\|f\\|_{\textbf{w}},$

	$\displaystyle\displaystyle D\psi(v)\|f_{\alpha}(v)\|$	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle D(\psi f_{\alpha})(v)+\|\psi(v^{-})\|Df_{\alpha}(v)$
		$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\\|M_{\psi}f_{\alpha}\\|+\sigma_{\psi}\|v\|Df_{\alpha}(v)\leq\\|M_{\psi}\\|+\sigma_{\psi}.$

	$\displaystyle\|v\|Dg(v)$	$\displaystyle\leq\|v\|\|g(v)-g_{n_{k}}(v)+g_{n_{k}}(v^{-})-g(v^{-})\|+\|v\|Dg_{n_{k}}(v)$
		$\displaystyle\leq\|v\|\|g(v)-g_{n_{k}}(v)\|+\|v\|\|g_{n_{k}}(v^{-})-g(v^{-})\|+\|v\|Dg_{n_{k}}(v)$
		$\displaystyle<\varepsilon+\|v\|Dg_{n_{k}}(v)\leq\varepsilon+1,$

$\displaystyle\displaystyle\|v\|Dg_{n}(v)$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\|v\|(2\log\|v\|-\log\|v_{n}\|)$
	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle(\sqrt{\|v_{n}\|}+1)\log\frac{(\sqrt{\|v_{n}\|}+1)^{2}}{\|v_{n}\|}$
	$\displaystyle\displaystyle\leq$	$\displaystyle\displaystyle\frac{2(\sqrt{\|v_{n}\|}+1)}{\sqrt{\|v_{n}\|}}\leq 2\left(1+\frac{1}{\sqrt{2}}\right)<4.$

Multiplication Operators between Lipschitz-Type Spaces on a Tree

Abstract.

Key words and phrases:

2010 Mathematics Subject Classification:

1. Introduction

Lemma 1.1 (Lemma 3.4 of [5]).

Lemma 1.2 (Proposition 2.4 of [5]).

Lemma 1.3 (Propositions 2.1 and 2.6 of [1]).

Lemma 1.4 (Proposition 2.7 of [1]).

1.1. Organization of the paper

2. Multiplication operators from ℒw\displaystyle\mathcal{L}_{\textbf{w}} to ℒ\displaystyle\mathcal{L}

2.1. Boundedness and Operator Norm Estimates

Theorem 2.1.

Proof.

2.2. Isometries

Theorem 2.2.

2.3. Compactness and Essential Norm Estimates

Lemma 2.3.

Proof.

Lemma 2.4.

Lemma 2.5.

Proof.

Theorem 2.6.

Proof.

Theorem 2.7.

Proof.

Theorem 2.8.

Proof.

3. Multiplication operators from ℒ\displaystyle\mathcal{L} to ℒw\displaystyle\mathcal{L}_{\textbf{w}}

3.1. Boundedness and Operator Norm Estimates

Theorem 3.1.

Proof.

3.2. Isometries

Theorem 3.2.

3.3. Compactness and Essential Norm

Lemma 3.3.

Proof.

Theorem 3.4.

Proof.

Theorem 3.5.

Proof.

Theorem 3.6.

Proof.

4. Multiplication operators from ℒw\displaystyle\mathcal{L}_{\textbf{w}} or ℒw,0\displaystyle\mathcal{L}_{{\textbf{w}},0} to L∞\displaystyle L^{\infty}

4.1. Boundedness and Operator Norm

Theorem 4.1.

Proof.

4.2. Boundedness From Below

Theorem 4.2.

Proof.

4.3. Isometries

Theorem 4.3.

4.4. Compactness and Essential Norm

Lemma 4.4.

Proof.

Lemma 4.5.

Theorem 4.6.

Proof.

Theorem 4.7.

Proof.

5. Multiplication operators from L∞\displaystyle L^{\infty} to ℒw\displaystyle\mathcal{L}_{\textbf{w}} or ℒw,0\displaystyle\mathcal{L}_{{\textbf{w}},0}

5.1. Boundedness and Operator Norm

Theorem 5.1.

Proof.

Theorem 5.2.

Proof.

5.2. Isometries

Theorem 5.3.

5.3. Compactness and Essential Norm

Lemma 5.4.

Proof.

Lemma 5.5.

Theorem 5.6.

Proof.

Corollary 5.7.

Theorem 5.8.

Proof.

References

2. Multiplication operators from $\displaystyle\mathcal{L}_{\textbf{w}}$ to $\displaystyle\mathcal{L}$

3. Multiplication operators from $\displaystyle\mathcal{L}$ to $\displaystyle\mathcal{L}_{\textbf{w}}$

4. Multiplication operators from $\displaystyle\mathcal{L}_{\textbf{w}}$ or $\displaystyle\mathcal{L}_{{\textbf{w}},0}$ to $\displaystyle L^{\infty}$

5. Multiplication operators from $\displaystyle L^{\infty}$ to $\displaystyle\mathcal{L}_{\textbf{w}}$ or $\displaystyle\mathcal{L}_{{\textbf{w}},0}$