Is convolution operator compact?

Question

I know convolution is not a Hilbert–Schmidt integral operator, but it needs more to tell if convolution is compact or not.

Answer 1

Let us use $T_f$ to denote the operator which convolutes with $f$ and let $M_\phi$ to denote the operator which multiplies by $\phi$. Since Young's inequality provides sharp bounds, we necessarily have $f \in L^1$, if we want $T_f:L^2\to L^2$ to even be bounded. Consider $FT_f$, the composition of the Fourier transform with $T_f$. Since $F$ is unitary, $T_f$ is compact if and only if $FT_f$ is. By the convolution theorem we have

$$FT_fg = \hat{f}\cdot\hat{g} = M_{\hat{f}}Fg$$

so that $FT_f = M_{\hat{f}}F$ is a multiplication operator composed with the Fourier transform. Again, the operator on the right hand side is compact if and only if the multiplication operator $M_{\hat{f}}$ is. Now, $\hat{f} \in C_0$ and hence $\hat{f} \in L^\infty$. But the only compact multiplication operator on $L^2$ induced by a bounded measurable function is the operator that is identically zero. Hence $\hat{f}$ and therefore $f$ must be identically zero. It follow that the only compact convolution operator on $L^2$ is the operator which is identically zero.

To see that there are no non-trivial compact multiplication operators, suppose that $f \in L^\infty$ and $f \neq 0$. Then there exists a set $E$ of positive measure such that $|f| \geq \epsilon > 0$ on $E$. Thus $M_f$ would be compact with a bounded left inverse on a subspace isomorphic to $L^2(E)$ which is impossible.