Compact Operators — Mathematics Knowledge System

Overview

A compact operator is a bounded linear operator between Banach spaces that maps bounded sets to precompact sets (sets whose closure is compact). Compact operators are in many respects the closest infinite-dimensional analogue of finite-rank (matrix) operators: their spectrum behaves like that of a finite matrix (at most countably many nonzero eigenvalues accumulating only at 0), and the Fredholm alternative holds—either $T - \lambda I$ is invertible or $\lambda$ is an eigenvalue with finite-dimensional eigenspace. Compact operators arise naturally as integral operators, as solution operators for elliptic PDEs via Sobolev embeddings, and in the spectral theory of differential operators.

Intuition

A compact operator 'almost has finite rank': it compresses any bounded input set into a set that looks essentially finite-dimensional (precompact). Concretely, any bounded sequence $(x_n)$ yields a subsequence $(Tx_{n_k})$ that converges. This is why the spectral theory of compact operators so closely mirrors finite-dimensional linear algebra—the Fredholm alternative and the countable-spectrum theorem both follow from this near-finite-dimensionality.

Formal Definition

Definition

A linear operator $T: X \to Y$ between Banach spaces is compact if the image $T(B)$ of the closed unit ball $B = \{x \in X : \|x\| \le 1\}$ is precompact in $Y$ (i.e., its closure is compact). Equivalently, every bounded sequence $(x_n) \subset X$ has a subsequence $(x_{n_k})$ such that $(Tx_{n_k})$ converges in $Y$. The space of compact operators is denoted $\mathcal{K}(X,Y)$.

T \in \mathcal{K}(X,Y) \iff \overline{T(B_X)} \text{ is compact in } Y

Compact operator definition

\mathcal{K}(X,Y) \subset \mathcal{B}(X,Y)

Compact operators are bounded

Properties

Compact operators form a closed two-sided ideal

\mathcal{K}(X) is a closed subspace of \mathcal{B}(X) and satisfies: if T \in \mathcal{K}(X) and S \in \mathcal{B}(X), then ST, TS \in \mathcal{K}(X).

Finite-rank operators are compact

Every bounded operator with finite-dimensional range is compact. The compact operators are the norm-closure of the finite-rank operators (in Hilbert spaces and many Banach spaces).

Spectrum of a compact operator

If T \in \mathcal{K}(X) and X is infinite-dimensional, then 0 \in \sigma(T). The nonzero spectrum consists entirely of eigenvalues, each with finite-dimensional eigenspace, and 0 is the only possible accumulation point.

Theorems

Theorem 8.1: Spectral Theorem for Compact Self-Adjoint Operators

Let T be a compact self-adjoint operator on a separable Hilbert space H. Then H has an orthonormal basis \{e_n\} consisting of eigenvectors of T, with corresponding real eigenvalues \lambda_n satisfying \lambda_n \to 0 as n \to \infty.

Theorem 8.2: Fredholm Alternative

Let T \in \mathcal{K}(X) and \lambda \neq 0. Either (i) T - \lambda I is bijective (with bounded inverse), or (ii) \lambda is an eigenvalue of T with finite-dimensional eigenspace, and \dim \ker(T-\lambda I) = \dim \ker(T^*-\lambda I^*) < \infty (index zero).

Theorem 8.3: Schauder's Theorem

T \in \mathcal{B}(X,Y) is compact if and only if its adjoint T^* \in \mathcal{B}(Y^*,X^*) is compact.

Theorem 8.4: Hilbert–Schmidt operators are compact

An integral operator T: L^2(\Omega) \to L^2(\Omega) with kernel K \in L^2(\Omega \times \Omega) is compact (in fact Hilbert–Schmidt), with \|T\|_{\mathcal{K}} \le \|K\|_{L^2(\Omega\times\Omega)}.

Worked Examples

Write $Tf(x) = \int_0^1 K(x,t)f(t)\,dt$ with kernel $K(x,t) = \mathbf{1}_{\{t \le x\}}$.
Check $K \in L^2([0,1]^2)$: $\int_0^1\int_0^1 |K(x,t)|^2\,dt\,dx = \int_0^1 x\,dx = 1/2 < \infty$.
$\|K\|_{L^2([0,1]^2)}^2 = \frac{1}{2} < \infty$
Since the kernel is in $L^2$, $T$ is a Hilbert–Schmidt operator. By Theorem 8.4, $T$ is compact.
Estimate $\|T\|_{\mathcal{B}} \le \|K\|_{L^2} = 1/\sqrt{2}$.
$\|T\| \le \frac{1}{\sqrt{2}}$

Answer: The Volterra operator is Hilbert–Schmidt (kernel in $L^2$), hence compact, with $\|T\| \le 1/\sqrt{2}$.

Practice Problems

Difficulty 8/10

Prove that the composition $ST$ of a compact operator $T \in \mathcal{K}(X,Y)$ with a bounded operator $S \in \mathcal{B}(Y,Z)$ is compact.

Difficulty 9/10

Prove the Fredholm alternative for compact operators: if $T \in \mathcal{K}(X)$ and $\lambda \neq 0$, then $\ker(T - \lambda I)$ is finite-dimensional.

Difficulty 8/10

Give an example of a bounded operator that is NOT compact, and verify it fails the compactness criterion.

Common Mistakes

Common Mistake

Compact operators on infinite-dimensional spaces can be surjective

A compact operator on an infinite-dimensional Banach space is never surjective (its range cannot be all of the infinite-dimensional space, since the image of the unit ball is precompact hence 'small'). Equivalently, $0$ is always in the spectrum.

Common Mistake

Every bounded operator is compact

The identity operator on any infinite-dimensional space is bounded but not compact. Compactness is a strict strengthening: the image of the unit ball must be precompact, which fails for the identity in infinite dimensions.

Quiz

Summary

A compact operator maps bounded sets to precompact sets; equivalently, every bounded sequence has a subsequence whose image converges.
Finite-rank operators are compact; compact operators are the norm-closure of finite-rank operators in Hilbert spaces.
The spectrum of a compact operator on an infinite-dimensional space has 0 as the only possible accumulation point; every nonzero spectral value is a finite-multiplicity eigenvalue.
The Fredholm alternative: for compact $T$ and $\lambda \neq 0$, either $(T-\lambda I)^{-1}$ is bounded or $\lambda$ is a finite-multiplicity eigenvalue (and the index is zero).
Compact self-adjoint operators on Hilbert spaces admit a complete orthonormal eigenbasis with eigenvalues tending to 0—the exact infinite-dimensional analogue of diagonalisation.

References

BookRudin, W. Functional Analysis. 2nd ed., McGraw-Hill, 1991. Chapter 4.
BookConway, J. B. A Course in Functional Analysis. 2nd ed., Springer, 1990. Chapter 6.
WebsiteWikipedia — Compact operator