Asymmetric Fuglede - Putnam theorem for Unbounded $M$-Hyponormal operators

We have $\mathcal{D}(T|_{\mathcal{M}})=\mathcal{D}(T)\cap\mathcal{M}$. Let $x\in\mathcal{D}(T|_{\mathcal{M}})$ and $P$ be an orthogonal projection on to $\mathcal{M}$. Then

Hence $T|_{\mathcal{M}}$ is a closed $M$-hyponormal operator. ∎

Define $K:ran(T-\lambda)\rightarrow ran(T^{*}-\overline{\lambda})$ by

Since $T$ is $M$-hyponormal, $K$ is a contraction with $K(T-\lambda)\subseteq\dfrac{1}{M}(T^{*}-\overline{\lambda}).$ Now we extend $K$ to $K^{\prime}\in\mathcal{B}\left(\overline{ran(T-\lambda)},\overline{ran(T^{*}-\overline{\lambda})}\right)$ such that
$K^{\prime}(T-\lambda)\subseteq\dfrac{1}{M}(T^{*}-\overline{\lambda})$. Then the contraction $A\in\mathcal{B}(\mathcal{H})$ defined by $Ax=0~{}~{}$ for all $x\in{\overline{ran(T-\lambda)}}^{\perp}$ is an extension of $K^{\prime}$. Hence

Therefore, we get

If we put $A^{*}=C_{\lambda},$ then $\dfrac{1}{M}(T-\lambda)\subseteq(T-\lambda)^{*}C_{\lambda}.$ ∎

Let $\mathcal{H}=\mathcal{H}_{1}\oplus\mathcal{H}_{2},$ where $\mathcal{H}_{1}=\mathcal{M}$ and $\mathcal{H}_{2}=\mathcal{M}^{\perp}.$ Then $T$ has the block matrix representation

where $T_{ij}:D(T)\cap\mathcal{H}_{j}\rightarrow\mathcal{H}_{i}$ is defined by $T_{ij}=P_{\mathcal{H}_{i}}TP_{\mathcal{H}_{j}}|_{D(T)\cap\mathcal{H}_{j}}$ for $j=1,2.$ Here, $P_{\mathcal{H}_{i}}$ denotes the orthogonal projection onto $\mathcal{H}_{i}.$ Since $\mathcal{M}$ is invariant under $T$, we have

Let $y\in D(T)\cap\mathcal{M}^{\perp}.$ By Lemma 2.2, we have

for every $\lambda\in\mathbb{C}$. Thus, $ran(T-\lambda)\subseteq ran(T-\lambda)^{*}$ for every $\lambda\in\mathbb{C}.$ Then there exist a densely defined operator $B$ such that $(T-\lambda)=(T-\lambda)^{*}B$ ( see [6]). Hence, $T_{12}(y)=(T_{11}-\lambda)^{*}u$ for some $u\in\mathcal{M}.$ We can choose $v$ such that $(T_{11}-\lambda)^{*}u=(T_{11}-\lambda)v.$ Therefore, $T_{12}(y)=(T_{11}-\lambda)v$ for every $\lambda\in\mathbb{C}.$ Hence,

Thus, $T_{12}(y)=0$ ([20]) and so $T_{12}=0$. ∎

(i) Suppose $T$ is $M$-hyponormal. Then by Lemma 2.2, we have

Let $E$ be the spectral measure of $S$ and let $\Omega$ be a compact subset of $\mathbb{C}.$ Since $E$ is regular, it is sufficient to prove that $|A|E(\Omega)=E(\Omega)|A|$ for every compact set $\Omega$ of $\mathbb{C}.$ Since $S$ is normal, we have $ran(E(\Omega))$ reduces $S-\lambda$ and $ran(E(\Omega))\subset\mathcal{D}(S-\lambda)$ for every $\lambda\in\mathbb{C}$.
For $\lambda\notin\Omega,$ define the function $\psi(\lambda)=\int_{\Omega}\dfrac{1}{(z-\lambda)}E(dz)x$
Then $Ax=A(S-\lambda)\psi(\lambda),~{}\mbox{for}~{}~{}\lambda\notin\Omega.$ Since $AS\subseteq TA,$ we have $Ax=(T-\lambda)A\psi(\lambda),~{}\mbox{for}~{}~{}\lambda\notin\Omega.$ Hence by equation (2.1), $A^{*}Ax=A^{*}M(T-\lambda)^{*}C_{\lambda}A\psi(\lambda).$ Since $A(S-\lambda)\subseteq(T-\lambda)A,$ for $\lambda\in\mathbb{C},$ $A^{*}Ax=(S-\lambda)^{*}MA^{*}C_{\lambda}A\psi(\lambda),~{}\mbox{for}~{}~{}\lambda\notin\Omega$ and hence $A^{*}Ax\in\bigcap\limits_{z\in\mathbb{C}\setminus\Omega^{*}}ran(S^{*}-z),$ where $\Omega^{*}=\{z:\overline{z}\in\Omega\}.$ Then $E(\mathbb{C}\setminus\Omega)A^{*}Ax=0$ ([25, Theorem 2.2]). Therefore, $A^{*}Ax=E(\Omega)A^{*}Ax.$ Since $x\in ran(E(\Omega))$ is arbitrary,

Since $A^{*}A$ is selfadjoint, $A^{*}AE(\Omega)=E(\Omega)A^{*}A.$ This completes the proof.

(ii) Since $A\geq 0,$ from (i) we have $SAx=ASx=TAx$ for $x\in\mathcal{D}(S).$ Thus, $S|_{A\mathcal{D}(S)}\subseteq T.$ Since $\mathcal{D}$ is a core for $S,$ $A\mathcal{D}(S)$ is a core for $S$ from Proposition 2.4. Hence,

Therefore, $S\subseteq T$. Since $D(T)\subseteq D(T^{*})$, we have $D(T)\subseteq D(T^{*})\subseteq D(S^{*})=D(S).$ Hence, $S=T.$

∎

Let $\Omega$ be a Borel subset of $\mathbb{C}$ and let $E$ be the spectral measure of $S$. We have $E(\Omega)$ is an orthogonal projection. To prove $\mathcal{R}(A^{*})$ reduces $S,$ it is sufficient to prove that $\mathcal{R}(A^{*})=\overline{ran(|A|)}$ is invariant under $E(\Omega)$.

Let $y\in\overline{ran(|A|)}$. Then there exist a sequence $(y_{n})\in ran(|A|)$ such that $y_{n}$ converges to $y$. Since $E(\Omega)$ is bounded, $E(\Omega)y_{n}$ converges to $E(\Omega)y$ . Since $(y_{n})\in ran(|A|)$, there exist $x_{n}\in\mathcal{D}(|A|)$ such that $y_{n}=|A|x_{n}.$ Therefore, $E(\Omega)|A|x_{n}$ converges to $E(\Omega)y.$ From Theorem 2.5 (i), $|A|E(\Omega)=E(\Omega)|A|.$ Hence, $|A|E(\Omega)x_{n}$ converges to $E(\Omega)y.$ Thus, $\mathcal{R}(A^{*})$ reduces $S$.

Since $AS\subseteq TA$ and $\mathcal{R}(A^{*})$ reduces $S,$ we have $T|_{\mathcal{R}(A)}$ is a closed densely defined operator in $\mathcal{R}(A)$ and

by Proposition 2.6 (i). Since $T$ is $M$-hyponormal, $T|_{\mathcal{R}(A)}$ is a closed $M$-hyponormal operator in $\mathcal{R}(A)$ by Lemma 2.1. From ([25, Lemma 3.1(v)]), we have

Let $W=U|_{\mathcal{R}(A^{*})}$, and $V=W^{*}~{}~{}T|_{\mathcal{R}(A)}~{}~{}W.$ Also we have $W$ is unitary isomorphism and $T|_{\mathcal{R}(A)}$ is $M$-hyponormal. Then for $x\in\mathcal{R}(A^{*}),$

Hence $(U|_{\mathcal{R}(A^{*})})^{*}~{}~{}T|_{\mathcal{R}(A)}~{}~{}U|_{\mathcal{R}(A^{*})}$ is a closed $M$-hyponormal operator.
From Theorem 2.5 (ii), we get $S|_{\mathcal{R}(A^{*})}=(U_{A}|_{\mathcal{R}(A^{*})})^{*}~{}~{}T|_{\mathcal{R}(A)}~{}~{}U_{A}|_{\mathcal{R}(A^{*})}$ because $ker(|~{}A|_{\mathcal{R}(A^{*})}~{}|)=ker(A|_{\mathcal{R}(A^{*})})=\{0\}$. Thus, $T|_{\mathcal{R}(A)},~{}S|_{\mathcal{R}(A^{*})}$ are unitarily equivalent normal operators.

∎

First assume that $B\in\mathcal{C}(\mathcal{H})$ is subnormal and $T\in\mathcal{C}(\mathcal{K})$ is $M$-hyponormal. Since $B$ is subnormal, there exist a normal extension $S$ on the Hilbert space $\mathcal{L}\supseteq\mathcal{H}.$ Define $Y\in\mathcal{B}(\mathcal{K},\mathcal{L})$ by $Yx=A^{*}x,x\in\mathcal{K}.$ From ([25, Theorem 3.8]), we have $\mathcal{R}(Y^{*})=\mathcal{R}(A).$
Since $AB^{*}\subseteq TA,$ we have

Since from equation (2.3) and $Yx=A^{*}x,$ we have

Thus $YT^{*}\subseteq SY.$ Hence $Y^{*}S^{*}\subseteq TY^{*}.$ Since $\mathcal{R}(Y^{*})=\mathcal{R}(A)$, we have $T|_{\mathcal{R}(A)}$ and $S|_{\mathcal{R}(A^{*})}$ are unitarily equivalent normal operators by Theorem 2.7. Thus, $\mathcal{R}(A)$ reduces $T$ to the normal operator $T|_{\mathcal{R}(A)}$.
Since $A^{*}T^{*}\subseteq BA^{*}~{}\mbox{and}~{}\mathcal{R}(A)$ reduces $T$, we have $B|_{\mathcal{R}(A^{*})}$ is closed densely defined in $\mathcal{R}(A^{*})$ and

by Proposition 2.6 (i) and ([25, Lemma 3.1(iii)]). Since $B|_{\mathcal{R}(A^{*})}\subseteq B\subseteq S,$ $B|_{\mathcal{R}(A^{*})}$ is subnormal. Then $B|_{\mathcal{R}(A^{*})}$ is $M$-hyponormal. Thus, $\mathcal{R}(A|_{\mathcal{R}(A^{*})})^{*}=\mathcal{R}(A^{*})$ reduces $B|_{\mathcal{R}(A^{*})}$ to the normal operator from equation (2.4) and Theorem 2.7. Hence by Theorem 2.3, $\mathcal{R}(A^{*})$ reduces $B$ to the normal operator $B|_{\mathcal{R}(A^{*})}.$ The result (i) follows from Proposition 2.6 (ii).
Next we assume that $B\in\mathcal{C}(\mathcal{H})$ is $M$-hyponormal and $T\in\mathcal{C}(\mathcal{K})$ is subnormal. Since $A^{*}T^{*}\subseteq BA^{*},$ $\mathcal{R}(A)$ and $\mathcal{R}(A^{*})$ are reducing subspace for $T$ and $B$ respectively. Also the part $T|_{\mathcal{R}(A)}$ and $B|_{\mathcal{R}(A^{*})}$ are normal. By Proposition 2.6 (ii), we have $AB\subseteq T^{*}A.$

∎