Do "close" operators have "close" spectra?

Question

Let's let $A$ and $B$ be bounded operators in a Hilbert space, and let $\sigma$ be the spectrum:

$ \sigma(A) = \{\lambda \in \mathbb{C} \mid \lambda I - A \text{ is not invertible}\} $

I'd like to know whether $\lVert A - B \rVert < \epsilon$ implies that $\sigma(A)$ and $\sigma(B)$ are "close". I'm most interested in whether $\sigma(B) \subset \sigma(A) + B_\epsilon(0)$, but if this is false maybe the right statement is $\sigma(B) \subset \sigma(A) + B_{f(\epsilon)}(0)$ for some $f = o(1)$, with $f$ perhaps depending on one of $A$ if we keep it fixed and vary $B$.

If this isn't clear, the kind of counterexample I'd be looking for is: an operator $A$ such that there are $B \to 0$ with $\sigma(A + B)$ at least a fixed distance (in the Hausdorff metric) from $\sigma(A)$.

Edit: the original claim (with linear dependence on $\epsilon$) is false; see this blog post by Terry Tao. Basically he constructs matrices $A$ and $B$ such that $\lVert B \rVert < \epsilon$, the spectrum of $A$ is all zeroes, but the eigenvalues of $A + B$ are the $n$th roots of $\epsilon$ (where $n$ is the dimensionality). However, the second option I listed is true in finite dimensions, at least if $f$ is allowed to depend on $A$, since the eigenvalues are continuous in the matrix entries.

Answer 1

Working up on Tao's example, I think that there is no relation.

Fix $\varepsilon>0$. For each $n$, let $$ A_n=\begin{bmatrix} 0&1&0&\cdots&0\\ 0&0&1&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&1\\ 0&0&0&\cdots&0 \end{bmatrix}, \ \ \ \ \ \ \ \ \ B_n=\begin{bmatrix} 0&1&0&\cdots&0\\ 0&0&1&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&1\\ \varepsilon&0&0&\cdots&0 \end{bmatrix}. $$ Then $\sigma(A)=\{0\}$, $\sigma(B)=\{w:\ w^n=\varepsilon\}$, and $\|A_n-B_n\|=\varepsilon\|E_{n1}\|=\varepsilon$.

Now construct, acting on the Hilbert space $H=\bigoplus_{n=1}^\infty\mathbb C^n$, the operators $$ A=\bigoplus_{n=1}^\infty A_n, \ \ \ \ \ B=\bigoplus_{n=1}^\infty B_n. $$ Then $$\|A-B\|=\sup_n\|A_n-B_n\|=\varepsilon,$$ $$\sigma(A)=\overline{\bigcup_n\sigma(A_n)}=\{0\},$$ and $$\sigma(B)=\overline{\bigcup_n\sigma(B_n)}=\overline{\bigcup_n\{w:\ w^n=\varepsilon\}}.$$ In particular $\varepsilon^{1/n}\in\sigma(B)$ for all $n$, which implies that $1\in \sigma(B)$. So, in the Hausdorff metric, $$\text{dist}\,(\sigma(A),\sigma(B))=1.$$

Answer 2

For general operators, this claim is false, as Terry Tao's example shows.

But if one, however restricts oneself to certain classes of operators, for example selfadjoint operators in Hilbert spaces, then you do have indeed a whole array of perturbation theory for the spectra of operators.

For instance, as noted in this paper, if we have a bounded selfadjoint operator $H$ (this is your $A$) and consider the operator $T = H +A$ (this is your $B$) for some self-adjoint perturbation $A$ (in your terms, this is $B-A$), then

$$\sigma(T) \subset \{\lambda: \operatorname{dist}(\lambda,\sigma(H))\leq \|A\|\},$$

which translates (in OPs notation) to $\sigma(B)\subset \sigma(A)+\overline{B}_{\|B-A\|}(0)$.

The cited paper does not give a reference (just states this as well-known result), but I am sure that this result can somehow be found in Kato's book on perturbation theory

Recent research is being done on perturbations of non-Hilbert-selfadjoint operators, such as selfadjoint operators in Krein spaces.

Answer 3

If $A $ is normal then it is true that $\sigma (A+E) $ is contained in a $\|E\|$- ball of the spectrum of $A $. In general, as the other answers indicate, there is no nice perturbation property of non-normal operators. However, this question is studied in many references on pseudospectra. For example, although the shift operator has bad perturbation properties (see other answers), the corresponding pseudospectra have good convergence properties under perturbation, finite dimensional approximations, etc, and can say something meaningful about the original operator.