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It is well known that the spectrum is continuous as function of operator. More precisely, let $\mathcal{H}$ be separable Hilbert space and $\mathcal{B}(\mathcal{H})$ the Banach algebra of linear operators acting on $\mathcal{H}$, then one has $$(\forall A,B\in\mathcal{B}(\mathcal{H}))(\forall \epsilon>0)(\exists \delta >0 )( \|A-B\|<\delta \Rightarrow \mbox{dist}(\sigma(A),\sigma(B))<\epsilon).$$

My first question is: Is it possible that this continuity is even uniform, i.e.,

$$(\forall \epsilon>0)(\exists \delta >0 )(\forall A,B\in\mathcal{B}(\mathcal{H}))( \|A-B\|<\delta \Rightarrow \mbox{dist}(\sigma(A),\sigma(B))<\epsilon)?$$

I do not expect the affirmative answer, however, I can not prove it. For that reason, I add a second question which, as I hope, could have the affirmative answer.

Let $\mathcal{S}\subset\mathcal{B}(\mathcal{H})$ such that $$\sup_{A\in\mathcal{S}}\|A\|<\infty,$$ is then true that $$(\forall \epsilon>0)(\exists \delta >0 )(\forall A,B\in\mathcal{S})( \|A-B\|<\delta \Rightarrow \mbox{dist}(\sigma(A),\sigma(B))<\epsilon)?$$

Thanks!

Edit: Nik Weaver pointed out that even the first statement denoted as "well-known" is not true. I also realized that I did not clarify the notion of distance here. Let me reformulate the statement and provide a verification. Where is the mistake in the following verification?

  1. Restatement:

$$(\forall A,B\in\mathcal{B}(\mathcal{H}))(\forall \epsilon>0)(\exists \delta >0 )( \|A-B\|<\delta \Rightarrow \sigma(B)\subset\mathcal{U}_{\epsilon}(\sigma(A)))$$ Here $\mathcal{U}_{\epsilon}(M)$ denotes the $\epsilon$-neighborhood of a set $M\subset\mathbb{C}$.

  1. "Verification": Let $\epsilon>0$ be given. And define $\delta>0$ by $$\delta^{-1}:=\max_{z\notin\mathcal{U}_{\epsilon}(\sigma(A))}\|(A-z)^{-1}\|.$$ Note that the above maximum exists and is finite since resolvent operator $(A-z)^{-1}$ is an analytic function on the resolvent set $\rho(A)$ and bounded in a neighborhood of $\infty$ for one has $$ (A-z)^{-1}\sim-\frac{1}{z}, \quad \mbox{ as } z\to\infty,$$ by the von Neumann series argument.

Now, if $\|A-B\|<\delta$ and $z\notin\mathcal{U}_{\epsilon}(\sigma(A))$, then $$ \|(A-B)(A-z)^{-1}\|<1$$ and hence the resolvent operator $$ (B-z)^{-1}=(A-z)^{-1}(1-(A-B)(A-z)^{-1})^{-1}$$ exists as a bounded operator. Consequently, $z\in\rho(B)$ and the statement follows.

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    $\begingroup$ Check out problems 102, 103 in Halmos' Hilbert space problem book. Usually, checking the mistakes in someone's proof is not on-topic here. $\endgroup$ – András Bátkai Feb 15 '16 at 20:02
  • $\begingroup$ The placement of quantifiers in the restatement is wrong: given $A$ and $B$, either $A=B$ (so that trivially $\sigma(B) = \sigma(A)$) or you can take $0 < \delta < \|A - B\|$ making the condition vacuous. $\endgroup$ – Robert Israel Feb 15 '16 at 20:21
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    $\begingroup$ What you are actually verifying is $\forall A \in {\mathcal B}({\mathcal H})\; \forall \epsilon > 0 \; \exists \delta > 0 \; \forall B \ldots$. Thus the approximating operator $B$ has its spectrum close to the spectrum of $A$. But note the asymmetry: it is not true that the spectrum of $A$ is close to the spectrum of $B$ (as the theorem cited by Nik Weaver shows). $\endgroup$ – Robert Israel Feb 15 '16 at 20:24
  • $\begingroup$ It is clear now, thank you Robert Israel! $\endgroup$ – Twi Feb 15 '16 at 20:53
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The "well-known" fact is, of course, false. According to a theorem of C. Apostol and B. Morrel (On uniform approximation of operators by simple models, Indiana Univ. Math. J. 26 (1977), 427–442), if $K$ is a nonempty compact subset of $\mathbb{C}$ and $A$ is a normal operator such that $\sigma(A)$ is connected and $K \subseteq \sigma(A)$, then $A$ is approximated in norm by operators whose spectrum is contained in $K$.

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  • $\begingroup$ This is pretty spectacular ($K=\{ 0\}$, $A=S+S^*$ on $\ell^2$, say), I didn't know this. $\endgroup$ – Christian Remling Feb 15 '16 at 17:45
  • $\begingroup$ If I remember correctly, Halmos also has an example from Kakutani in his Hibert space problem book... $\endgroup$ – András Bátkai Feb 15 '16 at 17:58
  • $\begingroup$ @ChristianRemling: yes, it's a surprisingly strong result. Actually the theorem of Apostol and Morrel is even more general (it's enough for every component of $\sigma(T)$ to meet $K$, etc.). $\endgroup$ – Nik Weaver Feb 15 '16 at 19:33
  • $\begingroup$ 2Nik Weaver: Please see the edit I have made in the posted question. Maybe the notion of distance led to a misunderstanding, maybe something is wrong in my verification using only simple properties of linear operators (therefore the "well-known"). $\endgroup$ – Twi Feb 15 '16 at 19:35
  • $\begingroup$ 2Christian Remling: Can you explain your example in more detail? I can't see what is the operator having $\{0\}$ as its spectrum and approximating $A=S+S^{*}$ in norm? $\endgroup$ – Twi Feb 15 '16 at 19:37

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