Perturbation and semicontinuity of the spectrum

Upper semicontinuity for a set-valued function such as $\operatorname {spec}$ has at least two definitions, a metric one and a sequential one. Metric definition: to each open set $\Lambda_0$ that includes $\operatorname {spec} A$ there corresponds a positive number $\varepsilon$ such that if $\left\| A-B\right\| < \varepsilon$, then $\operatorname {spec} B \subset \Lambda_0$. The sequential definition has to be preceded by an auxiliary concept: if $\left\{ \Lambda_n\right\}$ is a sequence of sets of complex numbers, then $\operatorname {limsup}_n \Lambda_n$ is the set of all limit points of those sequences $\left\{ \lambda_n\right\}$ for which $\lambda_n \in \Lambda_n$ for each $n$. Sequential definition of upper semicontinuity: whenever $A_n \to A$, it follows that

The two definitions are equivalent.

Note that the example of Problem 102 gives no information; in that case the spectral radius is equal to $1$ for each term of the sequence and also for the limit.

If $\left\{ \Lambda_n\right\}$ is a sequence of sets of complex numbers, define $\operatorname {liminf}_n \Lambda_n$ to be the set of all limits of those convergent sequences $\left\{ \lambda_n\right\}$ for which $\lambda_n \in \Lambda_n$ for each $n$. (Recall that $\operatorname {limsup}_n \Lambda_n$ was defined as the set of all limit points of not necessarily convergent sequences of this kind; see Problem 103.) The sequence $\left\{ \Lambda_n\right\}$ is called convergent $(\Lambda_n \to \Lambda )$ if and only if

in that case the common value deserves to be called $\lim_n \Lambda_n$. To say that $\operatorname {spec}$ is continuous at a particular operator $A$ means that $A_n \to A$ implies

Since upper semicontinuity is true for all $A$ (Problem 103), a necessary and sufficient condition for continuity at $A$ is that

Recall also that, for normal operators, spectral radius is equal to norm.

If small is interpreted in the sense of norm, a perturbation that produces only a small change in the spectrum does not have to be small: if, for instance, $A = \begin{pmatrix} \alpha & 0 \\ 0 & \beta \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, then $\operatorname {spec}(A+B) = \operatorname {spec} A = \left\{ \alpha , \beta \right\}$, but $B$ is not small; in fact $\operatorname {spec}(A+nB) = \operatorname {spec} A$ for all $n$. The answer is clearly yes in some degenerate cases: if, for instance, $A$ is a scalar, and if $\operatorname {spec}(A+nB) = \operatorname {spec} A$ for just one non-zero value of $n$, then $\operatorname {spec} B = 0$.

The spectrum of an operator varies upper semicontinuously with respect to the norm topology (Problem 103): if an operator is replaced by one that is near it in the norm topology, then the spectrum can increase only a little. The question here is what happens to this semicontinuity when the strong topology is used in place of the norm topology -- that is, whether $A_n \to A$ (strongly) implies that $\operatorname {spec} A$ cannot be much larger, in a suitable limiting sense, than the spectra $\operatorname {spec} A_n$.