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So here is my question,

I know that the operator $$T:L^2[0,1]\rightarrow L^2[0,1]$$ $$f\mapsto(Kf)(x)=\int_{[0,1]}k(x,y)f(y)\;dy$$ for a function $k$ continuous on $[0,1]^2$ is compact. Is this also true for $k\in L^2([0,1]\times[0,1])$? In the prove I know for $k$ continuous one uses explicitly the continuity to deduce that the image of the unit ball is equicontinuous and hence by Arzela-Ascoli the compactness follows. Does anybody know how this is with a "just" square integrable $k$?

Thanks a lot!

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You can approximate $k$ with a sequence $k_n\in C([0, 1]\times[0,1])$ converging in the $L^2([0, 1]\times[0,1])$ norm. The resulting sequence of operators $K_n$ will converge to $K$ in the operator norm, and so $K$ will be compact.

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