# Hilbert-Schmidt integral operators are compact

This is mostly for my own reference, since this fact seems to come up a lot.

Theorem 1 Let ${\Omega\subset\mathbb R^n}$ be an open connected set and ${T:L^2(\Omega)\rightarrow L^2(\Omega)}$ an integral operator with kernel ${k\in L^2(\Omega^2)}$, that is,

$\displaystyle Tf(x) = \int_{\Omega} k(x,y)f(y)\ \mathsf dy$

where

$\displaystyle \|k\|_2 = \left(\int_\Omega\int_\Omega |k(x,y)|^2\ \mathsf dx\ \mathsf dy\right)^{\frac 12}<\infty.$

Then ${T}$ is compact.

Proof: Let ${\{e_i\}}$ be an orthonormal basis of ${L^2(\Omega)}$, then write the kernel as

$\displaystyle k(x,y) = \sum_{i,j=1}^\infty k_{ij}e_i(x)e_j(y),$

where

$\displaystyle k_{ij}=\int_\Omega\int_\Omega k(x,y)e_i(x)e_j(y)\ \mathsf dx\ \mathsf dy.$

Define the sequence of integral operators ${\{T_n\}}$ on ${L^2(\Omega)}$ by

$\displaystyle T_nf(x) = \int_\Omega k_n(x,y)f(y)\ \mathsf dy,$

where

$\displaystyle k_n(x,y) = \sum_{i=1}^n \sum_{j=1}^\infty k_{i,j} e_i(x)e_j(y).$

It is clear that ${T_n(\Omega)\subset\mathrm{Span}(e_1,\ldots,e_n)}$, so the ${T_n}$ are of finite rank and hence compact. For ${f\in L^2(\Omega)}$, we have

\displaystyle \begin{aligned} \|(T-T_n)f\|_2^2 &= \left\|\int_\Omega (k(x,y) -k_n(x,y))f(y)\ \mathsf dy \right\|_2^2 \\ &= \int_\Omega\left|\int_\Omega (k(x,y)-k_n(x,y))f(y)\ \mathsf dy \right|^2\ \mathsf dx\\ &\leqslant \int_\Omega \left(\int_\Omega (k(x,y)-k_n(x,y))^2\ \mathsf dy \right)\left(\int_\Omega f^2(y)\ \mathsf dy \right)\ \mathsf dx\\ &= \left(\int_\Omega\int_\Omega (k(x,y)-k_n(x,y))^2\ \mathsf dy\ \mathsf dx\right)\left(\int_\Omega f^2(y)\ \mathsf dy\right)\\ &= \left(\sum_{i=n+1}^\infty\sum_{j=1}^\infty |k_{i,j}|^2\right)\|f\|_2^2. \end{aligned}

As ${\sum_{i,j=1}^\infty |k_{i,j}|^2<\infty}$, it follows that

$\displaystyle \lim_{n\rightarrow\infty} \left(\sum_{i=n+1}^\infty\sum_{j=1}^\infty |k_{i,j}|^2\right) = 0.$

We conclude that ${T}$ is compact as the limit of a sequence of compact operators in the operator norm topology. $\Box$