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Let $X\subset \Bbb R^n$ be a compact subset and $\alpha \in (0,n)$. Let $m:X\times X\to \Bbb R$ be a bounded measurable function. Consider the integral operator $$Tf(x)=\int_X \frac{m(x,t)}{|x-t|^\alpha}f(t)\operatorname{dt}, \quad \left(f\in L^2(X)\right).$$

I aim to prove that $T$ is a compact operator on $L^2(X)$.

In other words, that the set $\{Tf:\|f\|_{L^2}\leq 1\}\subset L^2(X)$ is relatively compact. A first instinct would be to apply Arzela-Ascoli but unfortunately we are dealing with $L^2$ instead of continuous functions. Is there any chance to remedy that?

Aside from that I tried proving, to no avail, that the set is totally bounded which would imply relative compactness.

I'm running out of ideas and I do not know what the standard approach is. Therefore, I will be very glad for a nudge in the right direction.

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Under additional constraints you have a Hilbert-Schmidt integral operator which is always compact (see http://www-m15.ma.tum.de/foswiki/pub/M15/Allgemeines/IllposedProblems/WebHome/board11.pdf).

Define the kernel $k(x, t) = \frac{m(x,t)}{|x-t|^\alpha}$. If $\alpha < \frac{1}{2}$ you have $k \in L^2(X \times X)$ and $T:L^2(X) \to L^2(X)$ is Hilbert-Schmidt hence compact.

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