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Let $K(s,t)$ be a real-valued function of two real variables, and let $T: L^2(\mathbb{R}) \to L^2(\mathbb{R})$ be defined by $(Tf)(s) = \int_\mathbb{R} K(s,t) f(t) dt$.

If $||K||_{L^2({\mathbb{R}^2})} < \infty$, can we say that $T$ is a compact operator?

I think this is true if we are looking at a bounded domain for $K$ and $f$ (by an application of the Arzela-Ascoli theorem), but I am not sure if it is true in general.

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    Got it. Thanks for the suggestion.2011-01-03

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Hint:

1) Approximate $K(s,t)$ by functions of the form $\sum_{k=1}^n g_k(s)h_k(t)$.

2) A norm limit of finite-rank operators is compact.

For further reference, your operator is called a Hilbert-Schmidt integral operator. A more general version of this problem (where $\mathbb{R}$ with Lebesgue measure is replaced by an arbitrary $\sigma$-finite measure space) is Problem 173 in Halmos's Hilbert space problem book. The book contains a hint and (in case you give up or solve it but want more information) a solution.

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    @user1736: The precise sense of approximation I recommend for starters is in the norm of $L^2(\mathbb{R}^2)$. You can show that $\|T-T_n\|\leq \|K-K_n\|_2$, where the norm on the left is the operator norm (which you ultimately want to go to $0$) and the norm on the right is the $L^2$ norm. To see that the operators in question have finite rank, first note that $\int g(s)h(t)f(t)dt=g(s)\langle f,\overline{h}\rangle$, so that the range of the integral operator corresponding to $g(s)h(t)$ is the span of $g$ (unless $h=0$). Taking finite linear combinations of such gives finite dim'l ranges.2011-01-03