---
tag: MF77N00SK8
kind: statement
role: theorem
name: Compactness of integral operators
msc: [45P05, 47G10, 47B07]
uses: [ZM5G144PZG, CB5G212Q03, M2NEP1K5AE, 0381S8Z5P0, Z8FABSYQTS, FJKNJ14AD2]
about: [M2NEP1K5AE, CB5G212Q03]
sources:
  - url: https://en.wikipedia.org/wiki/Compact_operator
---

## Statement

Let $a < b$.

1. Let $K \colon [a, b] \times [a, b] \to \C$ be a [[continuous kernel|ZM5G144PZG]].
Then the operator $(T f)(x) = \int_a^b K(x, y) f(y) \, dy$ maps $C[a, b]$ to itself and is a [[compact operator|M2NEP1K5AE]] on $C[a, b]$ with the sup norm.
2. Let $(X, \mu)$ be a $\sigma$-finite measure space and let $K \in L^2(X \times X, \mu \otimes \mu)$.
Then the [[Hilbert–Schmidt integral operator|CB5G212Q03]] $(T f)(x) = \int_X K(x, y) f(y) \, d\mu(y)$ is a compact operator on $L^2(X, \mu)$, and $\norm{T} \le \norm{K}_{L^2(X \times X)}$.

## Remarks

For part 1, let $f$ run over the unit ball of $C[a, b]$.
Then $\abs{Tf(x)} \le (b - a) \max \abs{K}$, and $\abs{Tf(x) - Tf(x')} \le (b - a) \max_y \abs{K(x, y) - K(x', y)}$, which is small when $x$ and $x'$ are close, by uniform continuity of $K$.
So the image of the unit ball is bounded and equicontinuous, and the [[Arzelà–Ascoli theorem|0381S8Z5P0]] shows that it is relatively compact.

For part 2, the Cauchy–Schwarz inequality gives $\abs{Tf(x)}^2 \le \int_X \abs{K(x, y)}^2 \, d\mu(y) \cdot \norm{f}^2$, and integrating in $x$ gives the norm bound.
[[Approximating $K$ by degenerate kernels|FJKNJ14AD2]] in $L^2$ then writes $T$ as a norm limit of [[finite-rank operators|Z8FABSYQTS]], and such a limit is compact.

Example: the Volterra operator $(Vf)(x) = \int_0^x f(y) \, dy$ on $L^2[0, 1]$ has the square-integrable kernel equal to $1$ for $y \le x$ and $0$ otherwise, so it is compact.

The integrability of the kernel matters.
On $L^2(\R)$, the convolution $(Tf)(x) = \int_\R e^{-\abs{x - y}} f(y) \, dy$ has a kernel that is not in $L^2(\R^2)$.
This $T$ is not compact: it commutes with translations, so the images of the translates of a fixed nonzero function have no convergent subsequence.

## Explanations

### L1 human

An integral operator averages a function against a kernel.
Averaging smooths things out, so the outputs of all functions of size at most one form a tame family, without wild oscillations.
Compactness is the precise form of this: every sequence of outputs has a convergent subsequence.

### L2 human

Compactness is what makes integral equations of the second kind behave like finite linear systems.
Once the operator is known to be compact, [[Fredholm's theorem|XQ8DYKW4SH]] and the spectral theory of compact operators apply, and for symmetric kernels so does the [[Hilbert–Schmidt theorem|A1G25R99PY]].