Chapter 7

Compact Operators

Compact operators behave almost like finite-rank operators, and for the self-adjoint ones among them, the spectral theory is completely explicit.

Definition

Definition $T \in B(H)$ is compact if it maps bounded sets to relatively compact sets: for every bounded sequence $(x_n)$ in $H$, the sequence $(Tx_n)$ has a convergent subsequence. The set of compact operators on $H$ is denoted $K(H)$.

Equivalently, $T$ is compact iff $T$ is the operator-norm limit of a sequence of finite-rank operators (operators with finite-dimensional range) — this characterization holds on any Hilbert space and is often the easiest way to verify compactness in practice.

Basic structural facts

Every compact operator is bounded, but not conversely — the identity operator on an infinite-dimensional $H$ is bounded but not compact, since the closed unit ball is not compact in infinite dimensions.
$K(H)$ is a closed, two-sided ideal in $B(H)$: if $T$ is compact and $S \in B(H)$, then both $ST$ and $TS$ are compact.
$K(H)$ is a closed subspace of $B(H)$ under the operator norm: a norm-limit of compact operators is compact.
If $T$ is compact, so is $T^*$.

Spectrum of a compact operator

Theorem (Riesz–Schauder) Let $T$ be a compact operator on an infinite-dimensional Hilbert space $H$. Then:

$0 \in \sigma(T)$;
every nonzero $\lambda \in \sigma(T)$ is an eigenvalue of finite multiplicity (i.e. $\ker(\lambda I - T)$ is finite-dimensional);
$\sigma(T) \setminus \{0\}$ is either finite or forms a sequence converging to $0$.

This is the qualitative heart of the Fredholm alternative: for $\lambda \ne 0$, the equation $(\lambda I - T)x = y$ has a unique solution for every $y$ exactly when the homogeneous equation $(\lambda I - T)x = 0$ has only the trivial solution — injectivity and surjectivity become equivalent, just as in the finite-dimensional case.

The spectral theorem for compact self-adjoint operators

Spectral theorem Let $T$ be a compact self-adjoint operator on $H$. Then there is an orthonormal basis $\{e_n\}$ of $H$ consisting of eigenvectors of $T$, with real eigenvalues $\lambda_n \to 0$ (if infinitely many are nonzero), such that $$ Tx = \sum_n \lambda_n \langle x, e_n \rangle e_n \qquad \text{for all } x \in H. $$

This is the direct infinite-dimensional generalization of diagonalizing a real symmetric matrix by an orthonormal basis of eigenvectors — the compactness ensures the "diagonal" $(\lambda_n)$ decays to $0$, which is exactly what is needed for the sum to converge in operator norm.

Where compact operators show up Integral operators with a sufficiently well-behaved (e.g. continuous, or square-integrable) kernel are typically compact — this is why compact operator theory is the natural functional-analytic setting for classical integral equations. See Chapter 8 for an explicit example.

Previous Spectrum Next Worked Examples