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I want to find two functions $f_1, f_2 : \mathbb{R} \rightarrow \mathbb{R}$ which are not zero almost everywhere, but $f_1 * f_2$ is a zero function.

My first idea is to try function with disjoint support. So I choose two simple functions with disjoint compact supports $$f_1 = \chi_{[0,1]}, f_2 = \chi_{[2,3]}.$$ Then $$f_1 * f_2 (x) = \int_0^1 \chi_{[2,3]}(x-y) dy.$$ When $x$ is such that $x-y \in [0,1]$, the convolution wont be zero there.

Next I try a simple function with function that integrate to zero on a finite interval like $f(x) = x - 3$ on $[2,4]$ and zero outside. The convolution still not zero.

So I am not sure what type of functions I should looks for.

Any suggestion ?

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  • $\begingroup$ I’m not sure what the answer would be, but the Titchmarsh Lemma gives that they must have infinite support. $\endgroup$ – David M. May 13 '18 at 22:29
  • $\begingroup$ I don't have an explicit example but if $f= \hat {\phi},g= \hat {\psi}$ where $\phi$ and $\psi$ are smooth even functions with compact support whose supports are disjoint then Fourier transform of $f*g=0$ so $f*g=0$. $\endgroup$ – Kavi Rama Murthy May 14 '18 at 7:53
  • $\begingroup$ Smooth even function ? You mean $c^\infty$ with compact support with $f(-x) = f(x)$ ? Oh, I have never thought of them, and yeah I cannot think of one such functions. $\endgroup$ – user117375 May 14 '18 at 10:58
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Kavi had the right idea, but smoothness is unnecessary. The functions \begin{eqnarray} f(x) &=& \frac{\sin(x)}{x} \\ g(x) &=& \frac{\sin(x)}{x}\cos(3x) \end{eqnarray} have Fourier transforms with finite and disjoint support, so their convolution vanishes identically.

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