Properties of Compactness

Corresponding to the strong (s) and weak (w) topologies for a Hilbert space $\mathbf{H}$, there are four possible interpretations of continuity for a transformation from $\mathbf{H}$ into $\mathbf{H}$: they are the ones suggested by the symbols $(\mathrm{s} \rightarrow \mathrm{s})$, $(\mathrm{w} \rightarrow \mathrm{w})$, $(\mathrm{s} \rightarrow \mathrm{w})$, and $(\mathrm{w} \rightarrow \mathrm{s})$. Thus, to say that $A$ is continuous $(\mathrm{s} \rightarrow \mathrm{w})$ means that the inverse image under $A$ of each w-open set is s-open; equivalently it means that the direct image under $A$ of a net s-convergent to $f$ is a net w-convergent to $Af$.

A linear transformation on a Hilbert space is called compact (also completely continuous) if its restriction to the unit ball is $(\mathrm{w} \rightarrow \mathrm{s})$ continuous (see Problem 169). Here "closed" refers to the norm topology, "self-adjoint" means that if $A \in \mathbf{K}$, then $A^{*} \in \mathbf{K}$, and "ideal" means that linear combinations of operators in $\mathbf{K}$ are in $\mathbf{K}$ and that products with at least one factor in $\mathbf{K}$ are in $\mathbf{K}$.

A less sharp but shorter formulation of the corollary is this: every compact normal operator is diagonal.

Let $X$ be a measure space with $\sigma$-finite measure $\mu$, and let $K$ be a complex-valued measurable function on $X \times X$ such that $\left|K\right|^{2}$ is integrable with respect to the product measure $\mu \times \mu$. It follows that, for almost every $x$, the function $y \mapsto K(x,y)$ is in $\mathrm{L}^{2}(\mu )$, and hence that the product function $y \mapsto K(x,y)f(y)$ is integrable whenever $f \in \mathrm{L}^{2}(\mu )$. Since, moreover,

(where $\left\| K\right\|$ is the norm of $K$ in $\mathrm{L}^{2}(\mu \times \mu )$), it follows that the equation

defines an operator (with kernel $K$) on $\mathrm{L}^{2}(\mu )$, with $\left\| A\right\| \leq \left\| K\right\|$. An operator of this form is called an integral operator and $K$ is called its kernel. Integral operators with kernels of this type (i.e., kernels in $\mathrm{L}^{2}(\mu \times \mu )$) are called Hilbert–Schmidt operators.

An ideal of operators is proper if it does not contain every operator. Similar results hold for non-separable spaces, but the formulations and proofs are fussier and much less interesting.

An orthonormal sequence converges weakly to $0$, and a compact operator is $(\mathrm{w} \rightarrow \mathrm{s})$ continuous. It follows that if an operator $A$ is compact, then $A$ maps each orthonormal sequence onto a strong null sequence. To what extent is the converse true? Could it be, for instance, that if an operator $A$ maps some orthonormal sequence onto a strong null sequence, then $A$ is compact? Certainly not --- that's absurd. Reason: the orthonormal sequence in question could be just half of an orthonormal basis on the other half of which $A$ is large. (Example: a projection with infinite rank and nullity.) To avoid the irrelevant distractions of large cardinal numbers, it is best to assume that the underlying Hilbert space is separable, and that the orthonormal bases to be studied come presented as sequences.

It is easy to construct non-compact operators whose square is compact; in fact, it is easy to construct non-compact operators that are nilpotent of index $2$. (Cf. Problem 96.)

Equivalently: if $C$ is compact, and if $\lambda$ is a non-zero complex number, then either $\lambda$ is an eigenvalue of $C$ or $C - \lambda$ is invertible. In this form the statement is frequently called the Fredholm alternative. It has a facetious but substantially accurate formulation in terms of the equation $(C-\lambda )f = g$, in which $g$ is regarded as given and $f$ as unknown; according to that formulation, if the solution is unique, then it exists.

Here $\Pi (A)$ denotes the approximate point spectrum of $A$ (the set of complex numbers $\lambda$ such that $A - \lambda$ is not bounded from below), and $\Pi_0(A) \subset \Pi (A)$ denotes the point spectrum of $A$ (the set of eigenvalues of $A$).

An operator $A$ is called a Fredholm operator if (1) $\operatorname {ran} A$ is closed and both $\ker A$ and $\operatorname {ran}^{\perp } A$ are finite-dimensional. (The last two conditions can be expressed by saying that the nullity and the co-rank of $A$ are finite.) An operator $A$ is invertible modulo the ideal of operators of finite rank if (2) there exists an operator $B$ such that both $1 - AB$ and $1 - BA$ have finite rank. An operator $A$ is invertible modulo the ideal of compact operators if (3) there exists an operator $B$ such that both $1 - AB$ and $1 - BA$ are compact.

The result is due to Atkinson [7].

The process of adding a compact operator to a given one is sometimes known as perturbation. The accepted attitude toward perturbation is that compact operators are "small"; the addition of a compact operator cannot (or should not) make for radical changes. Note that for $B = 0$ the statement follows from Problem 179. Here $\Pi_0(A)$ denotes the point spectrum of $A$ (the set of eigenvalues of $A$).

Weyl's theorem (Problem 182) implies that if $U$ is the unilateral shift and if $C$ is compact, then the spectrum of $U + C$ includes the unit disc. (Here is a small curiosity. The reason the spectrum of $U + C$ includes the unit disc is that $U$ has no eigenvalues. The adjoint $U^{*}$ has many eigenvalues, so that this reasoning does not apply to it, but the conclusion does. Reason: the spectrum of $U^{*} + C$ is obtained from the spectrum of $U + C^{*}$ by reflection through the real axis, and $C^{*}$ is just as compact as $C$.) More is true [137]: every point of the open unit disc is an eigenvalue of $(U+C)^{*}$.

It follows from the preceding paragraph that $U + C$ can never be invertible (the spectrum cannot avoid $0$), and it follows also that $U + C$ can never be quasinilpotent (the spectrum cannot consist of $0$ alone). Briefly: if invertibility and quasinilpotence are regarded as good properties, then not only is $U$ bad, but it cannot be improved by a perturbation.

An increasingly valuable part of the study of operators is a subject called non-commutative approximation theory. The following question is a small sample of it.