The Fuglede Theorem and Some Intertwining Relations

TL;DR

This paper extends the Fuglede Theorem to unbounded operators and explores intertwining relations, providing new versions, counterexamples, and examples of operators with unique intertwining properties.

Contribution

It presents a new unbounded version of the Fuglede Theorem and constructs operators with unique intertwining characteristics, advancing understanding of operator relations.

Findings

A new version of the Fuglede Theorem for unbounded operators

Counterexample illustrating limitations of intertwining operators

Example of operators not intertwined except by the zero operator

Abstract

In this paper, we show a new and classic version of the celebrated Fuglede Theorem in an unbounded setting. A related counterexample is equally presented. In the second strand of the paper, we give a pair of a closed and self-adjoint (unbounded) operators which is not intertwined by any (bounded or closed) operator except the zero operator.

Equations

$$TA\subseteq AT⟺T{A}^{\ast }\subseteq A^{\ast }T.$$

$$TA\subseteq BT⟺T{A}^{\ast }\subseteq B^{\ast }T$$

$$BT\subset T{B}^{\ast }⟹B^{\ast }T\subset TB.$$

$$BA\subset Af(B)⟹g(B)A\subset A(g\circ f)(B).$$

$$D(A)=D(BA)\subset D(Af(B))$$

$$D(B^{2}A)=D[B(BA))]\subset D[B(Af(B))]=D[(BA)f(B)]\subset D[(Af(B))f(B)]=D[A(f(B){)}^2],$$

$$B^{2}Ax=B(BA)x=B(Af(B))x=(BA)f(B)x=(Af(B))f(B)x=A(f(B){)}^{2}x.$$

$$B^{m}A\subset A(f(B){)}^m$$

$$p(B)A\subset A(p\circ f)(B)$$

$$g(B)A=p(B)A\subset A(p\circ f)(B)=A(g\circ f)(B).$$

$$BA\subset A{B}^{\ast }⟹B^{\ast }A\subset AB.$$

$$BA\subset AB⟹B^{\ast }A\subset A{B}^{\ast }.$$

$$BA\subset AC⟹B^{\ast }A\subset A{C}^{\ast }.$$

$$\tilde{B}=\left(\begin{array}[]{cc}B&0\\ 0&C\\ \end{array}\right)$$

$$\tilde{B}\tilde{A}=\left(\begin{array}[]{cc}0&BA\\ 0&0\\ \end{array}\right)\subset\left(\begin{array}[]{cc}0&AC\\ 0&0\\ \end{array}\right)=\tilde{A}\tilde{B}$$

$${\tilde{B}}^{\ast }\tilde{A}\subset \tilde{A}{\tilde{B}}^{\ast }$$

$$BA\subset AC⟹C{A}^{\ast }\subset A^{\ast }B.$$

$$B=\left(\begin{array}[]{cc}U&0\\ 0&U^{*}\\ \end{array}\right)\text{ and }T=\left(\begin{array}[]{cc}0&A\\ A^{*}&0\\ \end{array}\right).$$

$$BT=\left(\begin{array}[]{cc}0&UA\\ U^{*}A^{*}&0\\ \end{array}\right)\text{ and }TB^{*}=\left(\begin{array}[]{cc}0&AU\\ A^{*}U^{*}&0\\ \end{array}\right).$$

$$B^{\ast }T\not\subset TB$$

$$D(B^2)=D\left({B^{\ast }}^2\right)=\{0\}$$

$$TA\subset BT⟹T{A}^2\subset BTA\subset B^{2}T.$$

$$D(T{A}^2)=D(A^2)\subset D(B^{2}T)=\{x\in H:Tx\in D(B^2)=\{0\}\}=\ker T.$$

$$H=\overline{D(A^2)}\subset \overline{\ker T}=\ker T\subset H.$$

$$SB\subset AS⟹S^{\ast }A\subset B^{\ast }{S}^{\ast }.$$

$$S^{\ast }{A}^2\subset {B^{\ast }}^2{S}^{\ast }.$$

$$A=\left(\begin{array}[]{cc}0&T\\ 0&0\\ \end{array}\right)$$

$$A^{2}=\left(\begin{array}[]{cc}0&0_{D(T)}\\ 0&0_{D(T)}\\ \end{array}\right)$$

$$TA\subset BT⟹T{A}^2\subset B^{2}T.$$

$$D(T{A}^2)=\{x\in D(A^2):0\in D(T)\}=D(A)\text{ and }D(B^{2}T)=\ker T.$$

Full text

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1. Introduction

Undoubtedly, the Fuglede Theorem is the second salient result in Operator Theory, at least, as far as normal operators are concerned. It has many applications. The most tremendous one is the fact that it improves the statement of the Spectral Theorem of normal operators. To cite only a little amount of applications of this powerful tool, we refer readers to [1], [5], [15], [16], [18], [21], [27], [29], [31], [38], [42], [43], [46] and [54].

Recall that this theorem states that if $T\in B(H)$ and $A$ is normal (not necessarily bounded), then

[TABLE]

The problem leading to this theorem was first raised by von Neumann in [36] who had already established it in a finite dimensional setting (since this is seemingly not well documented, readers may find it in e.g. Exercise 11.3.29 in [33]). Fuglede was the first one to answer this problem affirmatively in [11] (a quite different proof popped up shortly afterwards and it is due to Halmos [17]). Then Putnam [42] generalized the result to:

[TABLE]

where $A$ and $B$ are normal (not necessarily bounded) and $T\in B(H)$.

There are different proofs of the Fuglede-Putnam Theorem besides the first two due to Fuglede and Putnam (e.g. the one in [17]). Perhaps the most elegant proof is due to Rosenblum (see [48]). Then came Berberian [3], who noted that the Fuglede’s version is actually equivalent to the Putnam’s version. Other proofs which are not well-spread may be consulted in [7] or [45]. See also [39].

There is a particular terminology to the transformation which occurs in the Fuglede-Putnam Theorem.

Definition**.**

We say that $T\in B(H)$ intertwines two operators $A,B$ if $TA\subseteq BT$.

Accordingly, we may restate the Fuglede-Putnam theorem as follows: If an operator intertwines two normal operators, then it intertwines their adjoints.

There have been many generalizations of the Fuglede (-Putnam) Theorem since Fuglede’s paper. We note a generalization to the so-called "spectral operators" by Dunford [10] (and another proof of the latter by Radjavi-Rosenthal [44]). See also [13], [14], [23], [24], [28], [35], [47] and [50] (among others). See also [30]. For new versions of the Fuglede-Putnam Theorem involving unbounded operators only, readers may wish to consult [26], [32], [40] and [41]. An interesting and related paper is [19].

Most of these generalizations seem to go into one direction only, that is, towards relaxing the normality hypothesis whilst there are still some unexplored territories as regards the very first version. To get to one main problem of this paper, observe that if $A$ is self-adjoint (and unbounded), then obviously $BA\subset AB$ implies that $B^{*}A\subset AB^{*}$ for any $B\in B(H)$. In [20], it was asked whether the assumption of the self-adjointness of $A$ can be relaxed to requiring only the closedness of $A$ and imposing the normality of $B$? The referee of the same reference informed Jorgensen of Fuglede’s example (which we will be recalling below). In the same reference it was shown that $B^{*}A\subset AB^{*}$ if for instance the complement of $\sigma(B)$ is connected and the interior is empty. Readers might also be interested in [8].

Closely related to what has just been alluded at, the following conjecture was proposed in [25] (it has resisted solutions for about three years). See Theorem 2.1 and Proposition 2.6.

Conjecture 1.1**.**

Let $T$ be an operator (densely defined and closed if necessary) and let $B\in B(H)$ be normal. Then

[TABLE]

What is interesting about this conjecture is the fact that it holds when $T\in B(H)$ (as we recover the bounded version of the Fuglede-Putnam Theorem), and as it stands, it is covered by none of the known (unbounded) generalizations of Fuglede-Putnam Theorem (see e.g. [32], [41] and [51]).

In this paper, we show that this conjecture is true in case $B$ has a finite pure point spectrum (Theorem 2.1). It is, however, not true even if we assume that $A$ is self-adjoint and $B$ is unitary. In the second part of this paper, we provide a pair of a closed and self-adjoint (unbounded) operators which is not intertwined by any (bounded or closed) operator except the zero operator.

Finally, we refer readers to [52] for properties and results about matrices of unbounded operators which will be helpful in the sequel. For the general theory of unbounded operators, readers may consult [4] or [49] or [53].

2. Main Results

Theorem 2.1**.**

Let $B$ be a bounded normal operator with a finite pure point spectrum and let $A$ be a closed (possibly unbounded) operator on a separable complex Hilbert space $H$. Let $f,g:\mathbb{C}\to\mathbb{C}$ be two continuous functions. Then

[TABLE]

Proof.

The hypothesis $BA\subset Af(B)$ clearly gives

[TABLE]

where $D$ stands for domains. Hence

[TABLE]

and next successively, for any $x\in D(A)$,

[TABLE]

Hence $B^{2}A\subset A(f(B))^{2}$ and by iteration

[TABLE]

for any $m\in\mathbb{N}$. Therefore,

[TABLE]

for any polynomial $p\in\mathbb{C}(X)$.

By assumption, $B$ has a point spectrum with finitely many distinct eigenvalues $\lambda_{j},j\in\{1,\cdots,n\}$, and corresponding eigenprojectors $E_{j}$ adding up to the identity operator $I$, so $B=\sum_{j=1}^{n}\lambda_{j}E_{j}$ is the spectral representation of $B$. For the given continuous function $g:\mathbb{C}\to\mathbb{C}$, there exists a polynomial $p$ such that $p(f(\lambda_{j}))=g(f(\lambda_{j}))$. In fact, for any $k\in\{1,\cdots,n\}$, there is a polynomial $p_{k}$ with roots $f(\lambda_{j})$, $j\neq k$, and with the value $p(f(\lambda_{k}))=(g\circ f)(\lambda_{k})$ at $f(\lambda_{k})$. Then the polynomial $p:=\sum_{k=1}^{n}p_{k}$ has the asserted property. From the hypothesis $BA\subset Af(B)$, we obtain

[TABLE]

∎

Corollary 2.2**.**

With $A$ and $B$ as in Theorem 2.1, we have

[TABLE]

Proof.

Just apply Theorem 2.1 to the functions $f,g:z\mapsto\overline{z}$ (so that $g\circ f$ becomes the identity map on $\mathbb{C}$). ∎

A similar reasoning applies to establish the following consequence:

Corollary 2.3**.**

With $A$ and $B$ as in Theorem 2.1, we likewise have

[TABLE]

Using an idea by Berberian, we may generalize this result to the case of two normal operators whereby we obtain a Fuglede-Putnam style theorem.

Proposition 2.4**.**

Let $B$ and $C$ be bounded normal operators with a finite pure point spectrum and let $A$ be a closed (possibly unbounded) operator on a separable complex Hilbert space $H$. Then

[TABLE]

Proof.

Define $\tilde{B}$ on $H\oplus H$ by:

[TABLE]

and let \tilde{A}=\left(\begin{array}[]{cc}0&A\\ 0&0\\ \end{array}\right) with $D(\tilde{A})=H\oplus D(A)$. Since $BA\subset AC$, it follows that

[TABLE]

for $D(\tilde{B}\tilde{A})=H\oplus D(A)\subset H\oplus D(AC)=\tilde{A}\tilde{B}$.

Now, since $B$ and $C$ are normal, so is $\tilde{B}$. Finally, apply Corollary 2.3 to the pair $(\tilde{B},\tilde{A})$ to get

[TABLE]

which, upon examining their entries, yields the required result. ∎

Corollary 2.5**.**

Let $B$ and $C$ be bounded normal operators with a finite pure point spectrum and let $A$ be a densely defined operator on a separable complex Hilbert space $H$. Then

[TABLE]

Proof.

Merely use the foregoing result, then take adjoints. ∎

One may wonder whether $BT\subset TB^{*}$ implies $B^{*}T\subset TB$ in the events of the self-adjointness of $T$ and the normality of $B\in B(H)$? The next example says that this is untrue, thus providing a counterexample to Conjecture 1.1.

Proposition 2.6**.**

There is a unitary $B\in B(H)$ and a self-adjoint $T$ with domain $D(T)\subset H$ such that $BT\subset TB^{*}$ but $B^{*}T\not\subset TB$.

First, we recall the following example (which appeared in [12]):

Example 2.7**.**

There exists a unitary $B\in B(H)$ and a closed and symmetric $T$ with domain $D(T)\subset H$ such that $BT\subset TB$ but $B^{*}T\not\subset TB^{*}$.

Now, we prove Proposition 2.6.

Proof.

Consider a unitary $U\in B(H)$ and a closed $A$ such that $UA\subset AU$ and $U^{*}A\not\subset AU^{*}$. Consider

[TABLE]

Then $B$ is unitary and $T$ is self-adjoint on $D(A^{*})\oplus D(A)$ (thanks to the closedness of $A$). Besides,

[TABLE]

Since $UA\subset AU$, it follows by taking adjoints that $U^{*}A^{*}\subset A^{*}U^{*}$. Therefore, $BT\subset TB^{*}$. Since $U^{*}A\not\subset AU^{*}$ is equivalent to $UA^{*}\not\subset A^{*}U$, we may thereby get that

[TABLE]

as $D(B^{*}T)\not\subset D(TB)$. ∎

Now, we pass to the second topic of the paper. Fuglede found in [11] a closed operator which did not commute with any bounded operator except scalar ones (i.e. $\alpha I$ where $\alpha\in\mathbb{C}$). The next two results lie within the same scope. In addition, they allow us to establish the uniqueness of the solution of some particular equations.

Proposition 2.8**.**

On some Hilbert space $H$, there is a self-adjoint operator $A$ and a densely defined closed operator $B$ such that $TA\subset BT$ (whenever $T\in B(H)$) implies $T=0$. Also (for the same pair $A$ and $B$), $SB\subset AS$ for any $S\in B(H)$ forces $S=0$.

Proof.

Let $H=L^{2}(\mathbb{R})\oplus L^{2}(\mathbb{R})$ and let $A$ be any unbounded self-adjoint operator with domain $D(A)\subset H$ and let $B$ be a closed operator such that

[TABLE]

(as in [9], cf. [34]). Let $T\in B(H)$. Then, clearly

[TABLE]

Hence

[TABLE]

Since $A^{2}$ is densely defined, it follows that

[TABLE]

Hence $\ker T=H$, that is, $T=0$, as required.

Now, we pass to the second part of the question. Plainly,

[TABLE]

As before, we obtain

[TABLE]

Similar arguments as above then yield $S^{*}=0$ or simply $S=0$, as needed. ∎

Remark*.*

In fact, the first case of the foregoing counterexample may be beefed up by even allowing $B$ to be also symmetric and semi-bounded (see e.g. [53] for the definition of semi-boundedness). This is based on the famous counterexample by Chernoff in [6].

Proposition 2.9**.**

On some Hilbert space $H$, there are two densely defined closed operators $A$ and $B$ such that $TA\subset BT$ implies $T=0$ whenever $T$ is closed.

Proof.

Let $H=L^{2}(\mathbb{R})\oplus L^{2}(\mathbb{R})$ and let $A$ be a densely defined closed operator with domain $D(A)\subset L^{2}(\mathbb{R})\oplus L^{2}(\mathbb{R})$ such that $A^{2}=0$ on $D(A^{2})=D(A)$ (cf. [37]). An explicit and adapted example to our case is to consider

[TABLE]

where $T$ is say an unbounded self-adjoint operator with domain $D(T)\subset L^{2}(\mathbb{R})$. By definition, $D(A)=L^{2}(\mathbb{R})\oplus D(T)$.

Then as may easily be seen

[TABLE]

with $D(A^{2})=D(A)=L^{2}(\mathbb{R})\oplus D(T)$. Now, let $B$ be a closed operator defined on $L^{2}(\mathbb{R})\oplus L^{2}(\mathbb{R})$ satisfying $D(B^{2})=\{0\}$ (as in [9]).

Now, clearly

[TABLE]

But

[TABLE]

Hence

[TABLE]

and so upon passing to the closure (w.r.t. $L^{2}(\mathbb{R})\oplus L^{2}(\mathbb{R})$)

[TABLE]

because $\ker T$ is closed for $T$ is closed. Therefore, $Tx=0$ for all $x\in D(T)$, i.e. $T\subset 0$. Accordingly, as $T$ is bounded on $D(T)$ and also closed, then $D(T)$ becomes closed and so $D(T)=H$, that is, $T=0$ everywhere, as coveted. ∎

3. Concluding Remarks and an Open Problem

It seems noteworthy that easy arguments allow us to show that $BT=TB^{*}$ does imply that $B^{*}T=TB$ when $B$ is unitary even if $T$ is any (unbounded) operator. In other words, the self-adjointness of $BT$ entails that of $B^{*}T$ if we further assume that $T$ is self-adjoint. One may therefore wonder what happens if one assumes that $B$ is only normal? The problem thus becomes: If $B\in B(H)$ is normal and if $T$ is (unbounded) self-adjoint, then

[TABLE]

Recall that if $T$ is bounded, then the self-adjointness of $BT$ gives the self-adjointness of $B^{*}T$ and vice versa. The analogous question in the case of normality of $BT$ has already a negative answer as a famous counterexample by Kaplansky shows (see [22]. Cf. [2]). Going back to the main question, observe that a naive counterexample is not available either. In other words, if $BT$ is closed, then $B^{*}T$ is necessarily closed (and conversely). Indeed, the normality of $B$ gives

[TABLE]

Hence, the graph norms of $B^{*}T$ and $BT$ coincide and hence the closedness of one implies the closedness of the other. With the closedness of $B^{*}T$ at hand, we may try to show that $B^{*}T$ is normal and having a real spectrum. But honestly, we just do not know whether this would lead anywhere or one has to look for counterexamples?