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I don't have any real background in functional analysis, so I was wondering if there is a nice sufficient condition or criterion for a convolution operator (say on $L^2\left([a,b] \times [a,b]\right) )$ to be compact.

More specifically, is convolution against $\delta(x_1 + x_2)$, a compact operator?

Thanks!

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Convolution with a delta function is just a shift operator. Well, if you're working on $L^2([a,b]\times[a,b])$ then I suppose it's a shift, truncated to fit in $[a,b]\times[a,b]$. In any case it's not compact.

It may help to view convolution with a function $g(x)$ as an integral operator with kernel function $K(x,u) = g(u-x)$. Any integral operator with $L^2$ kernel will be compact (standard fact; see Conway's Course in Functional Analysis, for example), so if $g$ is square integrable and the total measure of your space is finite then convolution with $g$ will be compact.

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