When does the Weyl-von Neumann Theorem hold?

TL;DR

This paper characterizes the specific closed subsets of the real line for which the Weyl-von Neumann Theorem extends to unbounded self-adjoint operators, identifying when such operators with a given essential spectrum are unitarily equivalent modulo the compacts.

Contribution

It precisely determines the subsets of the real line for which the Weyl-von Neumann Theorem holds for unbounded self-adjoint operators.

Findings

Identifies the class of subsets M where the theorem applies

Provides a complete characterization of these subsets

Extends the classical theorem to certain unbounded operators

Abstract

A famous theorem due to Weyl and von Neumann asserts that two bounded self-adjoint operators are unitarily equivalent modulo the compacts, if and only if their essential spectrum agree. The above theorem does not hold for unbounded operators. Nevertheless, there exist closed subsets $M$ of $\mathbb{R}$ on which the Weyl--von Neumann Theorem hold: all (not necessarily bounded) self-adjoint operators with essential spectrum $M$ are unitarily equivalent modulo the compacts. In this paper, we determine exactly which $M$ satisfies this property.

Equations

$$\forall A,B\in \mathrm{SA}(H)[{\sigma }_{\mathrm{ess}}(A)={\sigma }_{\mathrm{ess}}(B)=M\Rightarrow \exists u\in \mathcal{U}(H)\exists K\in \mathbb{K}(H{)}_{\mathrm{sa}}(uA{u}^{\ast }+K=B)].$$

$$(\ast )d_M:=\underset{n\to \infty }{\lim }\min \left\{\underset{\lambda \in \mathbb{R}\setminus (M\cup [-n,n])}{\sup }\mathrm{dist}(\lambda ,M),1\right\}=0.$$

$$a_n:=\underset{t\in M}{\inf }|{\lambda }_n-t|,b_n:=\underset{t\in M}{\inf }|{\mu }_n-t|(n\in \mathbb{N}).$$

$$A$$

$$B$$

$${\sigma }_{\mathrm{ess}}(A)=\overline{\{{\mu }_n;n\in \mathbb{N}\}}=M={\sigma }_{\mathrm{ess}}(B).$$

$$K{\eta }_{⟨k,1⟩}=\sum _{k^{\prime }=1}^{\infty }({\lambda }_{k^{\prime }+1}-\frac{1}{4}{d}_M-{\lambda }_k)⟨{\xi }_{⟨k^{\prime },1⟩},{\eta }_{⟨k,1⟩}⟩{\xi }_{⟨k^{\prime },1⟩}+\sum _{k^{\prime }=1}^{\infty }\sum _{m=2}^{\infty }({\mu }_{k^{\prime }}-{\lambda }_k)⟨{\xi }_{⟨k^{\prime },m⟩},{\eta }_{⟨k,1⟩}⟩{\xi }_{⟨k^{\prime },m⟩}.$$

$${\lambda }_{k^{\prime }+1}-\frac{1}{4}{d}_M-{\lambda }_k>2{\lambda }_k-\frac{1}{4}{d}_M-{\lambda }_k={\lambda }_k-\frac{1}{4}{d}_M>\frac{1}{4}{d}_M.$$

$$|{\lambda }_{k^{\prime }+1}-\frac{1}{4}{d}_M-{\lambda }_k|>{\lambda }_{k^{\prime }+1}+\frac{1}{4}{d}_M>\frac{1}{4}{d}_M.$$

$$|{\lambda }_{k^{\prime }+1}-\frac{1}{4}{d}_M-{\lambda }_k|\ge \frac{1}{4}{d}_M(k\in \mathbb{N}).$$

$$|{\lambda }_k-{\mu }_{k^{\prime }}|\ge \text{dist}({\lambda }_k,M)>\frac{1}{2}{d}_M>\frac{1}{4}{d}_M.$$

$$\Vert K{\eta }_{⟨k,1⟩}{\Vert }^2$$

$$\ge \sum _{k^{\prime }=1}^{\infty }(\frac{1}{4}{d}_M{)}^2|⟨{\xi }_{⟨k^{\prime },1⟩},{\eta }_{⟨k,1⟩}⟩{|}^2+\sum _{k^{\prime }=1}^{\infty }\sum _{m=2}^{\infty }(\frac{1}{4}{d}_M{)}^2|⟨{\xi }_{⟨k^{\prime },m⟩},{\eta }_{⟨k,1⟩}⟩{|}^2$$

$$=(\frac{1}{4}{d}_M{)}^2.$$

Full text

Keywords. Weyl-von Neumann Theorem, Self-adjoint operators.

2010 Mathematics Subject Classification: 47B25

1 Introduction and Main Theorem

Let $H$ be a separable infinite-dimensional complex Hilbert space, and let ${\rm{SA}}(H)$ (resp. $\mathbb{B}(H)_{\rm{sa}}$) be the set of all self-adjoint (resp. bounded self-adjoint) operators on $H$. Also let $\mathcal{U}(H)$ (resp. $\mathbb{K}(H)_{\rm{sa}}$) be the group of unitaries (resp. compact self-adjoint operators) on $H$. The essential spectrum of $A\in{\rm{SA}}(H)$ is denoted by $\sigma_{\rm{ess}}(A)$. The celebrated Weyl–von Neumann Theorem [Wey09, Wey10, vN35] asserts that operators $A,B\in\mathbb{B}(H)_{\rm{sa}}$ are unitarily equivalent modulo the compacts (which we call Weyl–von Neumann equivalent), that is, $uAu^{*}+K=B$ for some $u\in\mathcal{U}(H)$ and $K\in\mathbb{K}(H)_{\rm{sa}}$ if and only if $\sigma_{\rm{ess}}(A)=\sigma_{\rm{ess}}(B)$. This theorem has continued to play significant roles in many fields of analysis. On the other hand, for unbounded operators the Weyl–von Neumann Theorem does not hold (Weyl–von Neumann equivalent operators always have the same essential spectrum, but the converse need not hold). In fact, the Weyl–von Neumann equivalence, viewed as an equivalence relation on the Polish space ${\rm{SA}}(H)$ endowed with the strong resolvent topology is unclassifiable by countable structures, despite the fact that its restriction to $\mathbb{B}(H)_{\rm{sa}}$ is smooth [AM15, Theorems 3.12 and 3.33]. Somewhat unexpectedly, it is shown that the Weyl–von Neumann Theorem holds on $\mathbb{R}$ [AM15, Theorem 3.17 (2)], i.e., any two operators $A,B\in{\rm{SA}}(H)$ with $\sigma_{\rm{ess}}(A)=\sigma_{\rm{ess}}(B)=\mathbb{R}$ are always Weyl–von Neumann equivalent, while it fails on $\emptyset$ or $\mathbb{N}$ [AM15, Examples 3.3 and 3.5]. It would be natural to ask on which closed subsets $M$ of $\mathbb{R}$ the Weyl–von Neumann Theorem holds. The answer to the question is the main result of the paper:

Theorem 1.1**.**

Let $M$ be a closed subset of $\mathbb{R}$. The following two conditions are equivalent.

• (i)

The Weyl-von Neumann Theorem holds on $M$. That is,

[TABLE]

• (ii)

$M$* has no large holes at infinity. That is, $M\neq\emptyset$ and*

[TABLE]

Here we assume $\sup\emptyset=0$.

Proof.

(ii)$\Rightarrow$(i): Assume that $M$ has no large holes at infinity, and let $A,B\in{\rm{SA}}(H)$ be such that $\sigma_{\rm{ess}}(A)=\sigma_{\rm{ess}}(B)=M$. By the Weyl’s compact perturbation Theorem, we may assume that $A$ and $B$ are both diagonal with eigenvalues $\{\lambda_{n}\}_{n=1}^{\infty},\{\mu_{n}\}_{n=1}^{\infty}\subset\mathbb{R}$, and moreover that all eigenvalues are of simple multiplicity. This implies that the sets of accumulation points of $\{\lambda_{n}\}_{n=1}^{\infty}$ and $\{\mu_{n}\}_{n=1}^{\infty}$ both equal $M$. Following von Neumann’s proof (see [AG61, $\S$94]), let

[TABLE]

Then we show that $\lim_{n\to\infty}a_{n}=\lim_{n\to\infty}b_{n}=0.$ Assume by contradiction that $a_{n}$ does not converge to 0 as $n\to\infty$. Then there exist $0<\delta<1$ and a subsequence $(a_{n_{k}})_{k=1}^{\infty}$ such that $a_{n_{k}}\geq\delta\ (k\in\mathbb{N})$ holds. We first observe that for a fixed $N\in\mathbb{N}$, there exist only finitely many $k\in\mathbb{N}$ for which $|\lambda_{n_{k}}|\leq N$. Indeed, assume by contradiction that there exists a subsequence $(n_{k}^{\prime})_{k=1}^{\infty}$ of $(n_{k})_{k=1}^{\infty}$ for which $|\lambda_{n_{k}^{\prime}}|\leq N\ (k\in\mathbb{N})$. Then $(\lambda_{n_{k}^{\prime}})_{k=1}^{\infty}$ must have an accumulation point, say $\lambda$ with $|\lambda|\leq N$. Then $\lambda\in M$ and there exists $k_{0}\in\mathbb{N}$ such that $\delta\leq a_{n_{k_{0}}^{\prime}}\leq|\lambda_{n_{k_{0}}^{\prime}}-\lambda|<\delta$,  which is a contradiction. Therefore by taking further subsequence of $(a_{n_{k}})_{k=1}^{\infty}$ if necessary, we may assume that $|\lambda_{n_{k}}|>k$ for every $k\in\mathbb{N}$. Then $\lambda_{n_{k}}\in\mathbb{R}\setminus(M\cup[-k,k])$. Let $0<\varepsilon<\delta(<1)$. By $(*)$, there exists $k_{0}\in\mathbb{N}$ such that for every $k\geq k_{0}$, and $\lambda\in\mathbb{R}\setminus(M\cup[-k,k]),\text{dist}(\lambda,M)<\varepsilon$ holds. This shows that $a_{n_{k_{0}}}=\text{dist}(\lambda_{n_{k_{0}}},M)<\delta$, a contradiction. Therefore $\lim_{n\to\infty}a_{n}=0$. Similarly, $\lim_{n\to\infty}b_{n}=0$ holds. Then as in [AG61, $\S$94], there exists a bijection $\pi\colon\mathbb{N}\to\mathbb{N}$ such that $\lim_{n\to\infty}|\lambda_{\pi(n)}-\mu_{n}|=0$. By (the proof of) [AG61, $\S 94$, Theorem 3], this shows that $A,B$ are Weyl-von Neumann equivalent.

(i)$\Rightarrow$(ii): We show the contrapositive. Assume that (ii) does not hold. If $M=\emptyset$, then the Weyl-von Neumann Theorem does not hold on $M$ (what is much worse, the Weyl-von Neumann equivalence relation restricted to $\{A\in{\rm{SA}}(H);\ \sigma_{\rm{ess}}(A)=\emptyset\}$ is still unclassifiable by countable structures [AM15, Theorem 3.32]). Hence we may assume that $M\neq\emptyset$ and $(1\geq)d_{M}>0$. Note that this in particular means that for each $n\in\mathbb{N}$, $\mathbb{R}\setminus(M\cup[-n,n])\neq\emptyset$ holds. Then at least one of $(0,\infty)\setminus M$ or $(-\infty,0)\setminus M$ is unbounded. We may therefore assume that $(0,\infty)\setminus M$ is unbounded, so that there exist numbers $1<\lambda_{1}<\lambda_{2}<\cdots$ in $\mathbb{R}\setminus M$ such that $\text{dist}(\lambda_{n},M)>\frac{1}{2}d_{M}$ and $\lambda_{n+1}>2\lambda_{n}$ for every $n\in\mathbb{N}$. Since $M\neq\emptyset$, let $\{\mu_{n}\}_{n=1}^{\infty}$ be a countable dense subset of $M$ (it is possible that some $\mu_{n}$ and $\mu_{m}$ are equal for different $n,m$). Fix a bijection $\langle\ \cdot\ ,\ \cdot\ \rangle\colon\mathbb{N}^{2}\to\mathbb{N}$ given by $\langle k,m\rangle=2^{k-1}(2m-1)\ (k,m\in\mathbb{N})$. Fix an orthonormal basis $\{\xi_{n}\}_{n=1}^{\infty}$ for $H$ and let $e_{n}$ be the orthogonal projection of $H$ onto $\mathbb{C}\xi_{n}\ (n\in\mathbb{N})$. Define $A,B\in{\rm{SA}}(H)$ by

[TABLE]

Then

[TABLE]

We show that $A$ and $B$ are not Weyl-von Neumann equivalent. Assume by contradiction that there exist $u\in\mathcal{U}(H)$ and $K\in\mathbb{K}(H)_{\rm{sa}}$ such that $uAu^{*}+K=B$ holds. Let $\eta_{n}:=u\xi_{n}\ (n\in\mathbb{N})$ and let $f_{n}=ue_{n}u^{*}$, the orthogonal projection of $H$ onto $\mathbb{C}\eta_{n}$. Then for each $k\in\mathbb{N}$, we have $uAu^{*}\eta_{\langle k,1\rangle}+K\eta_{\langle k,1\rangle}=B\eta_{\langle k,1\rangle}$, so that

[TABLE]

Since $\langle k,1\rangle\stackrel{{\scriptstyle k\to\infty}}{{\to}}\infty$, we have $\eta_{\langle k,1\rangle}\stackrel{{\scriptstyle k\to\infty}}{{\to}}0$ weakly in $H$. Since $K$ is compact, this shows that $\|K\eta_{\langle k,1\rangle}\|\stackrel{{\scriptstyle k\to\infty}}{{\to}}0$. If $k=k^{\prime}+1$, then $|\lambda_{k^{\prime}+1}-\frac{1}{4}d_{M}-\lambda_{k}|=\frac{1}{4}d_{M}$. If $k<k^{\prime}+1$, then

[TABLE]

If $k>k^{\prime}+1$, then

[TABLE]

This shows that in any case

[TABLE]

Also, for every $k^{\prime}\in\mathbb{N}$,

[TABLE]

Therefore for every $k\in\mathbb{N}$, we have

[TABLE]

This contradicts $\|K\eta_{\langle k,1\rangle}\|\stackrel{{\scriptstyle k\to\infty}}{{\to}}0$. Thus $A,B$ are not Weyl-von Neumann equivalent. ∎

Example 1.2**.**

Here are examples of closed sets $M\subset\mathbb{R}$ having no large halls at infinity.

• (a)

$M=\mathbb{R}$, or more generally $M=\mathbb{R}\setminus U$ with $U$ bounded and open.

• (b)

$M=\mathbb{R}\setminus\bigcup_{n\in\mathbb{N}}(n-r_{n},n+r_{n})$, where $r_{n}>0\ (n\in\mathbb{N})$ and $\displaystyle\lim_{n\to\infty}r_{n}=0$.