Compactness of integral operators

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 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 in $L^2$ then writes $T$ as a norm limit of finite-rank operators, 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.

Going deeper

For a student

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 and the spectral theory of compact operators apply, and for symmetric kernels so does the Hilbert–Schmidt theorem.