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

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