Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
What properties does the function being convoluted with have? Between what spaces are you considering this operator? – user12014 Jul 11 '12 at 21:53
Let's say it's in Hilbert space. – chaohuang Jul 12 '12 at 0:03
You mean $L^2(\mathbb{R}^n)$, I guess. But what is the precise definition of your convolution operator? – Giuseppe Negro Jul 12 '12 at 0:36
up vote 7 down vote accepted

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.

share|cite|improve this answer
You can also notice that the spectrum of $T_f$ contains $\{\hat{f}(t),t\in\mathbb{R}\}$, which is not countable – San-A Mar 22 '13 at 11:01

good questions but generally speaking, never. Over on the fourier transform side divide the set where $\hat f > \epsilon$ into infinitely many disjoint intervals then the inverse transforms of the intervals, properly normalized, map to infinitely many orthogonal guys that don't go to zero

share|cite|improve this answer
I downvoted. This answer means nothing, you should be more clear. – Giuseppe Negro Jul 11 '12 at 22:57
sorry, i see your clarification . answer is especially clear in $\mathbb L^2$ – mike Jul 12 '12 at 0:06
"the inverse transforms of the intervals". What does inverse transform of intervals mean?@mike – Harish Feb 6 '15 at 16:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.