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$\mathbf {The \ Problem \ is}:$ If $f$ & $h$ are $L^1(\mathbb T)$ functions with $f\star h=0$ identically on $\mathbb T.$ Then can we say either $f\equiv 0$ or $h\equiv 0 ?$

$\mathbf {My \ approach}:$ By convolution theorem, $\widehat f(n)$.$\widehat h(n)=0$ for all $n\in \mathbb N.$

But, can we find $f,h \in L^1(\mathbb T)$ with $\widehat f$ is supported on odd integers & $\widehat h$ on even integers ?

I couldn't find anything . Thanks in adv. for a hint .

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    $\begingroup$ You may consider $f(x)=e^{2\pi ix}$ and $g(x)=1$. In general, choose $(a_n)_{n\in\mathbb{Z}}$ such that $\sum_n |a_n|<\infty$ and then set $$f(x)=\sum_{n\text{ odd}}a_n e^{2\pi i n x}\qquad\text{and}\qquad g(x)=\sum_{n\text{ even}}a_n e^{2\pi i n x}.$$ $\endgroup$ Jan 11 at 11:57
  • $\begingroup$ Easy example (special case of the above) - if we set $\mathbb T=[0,1]$, $f(x)=\sin(2\pi x)$, $h(x)=\sin (4\pi x)$, then $\int f(s)h(s-x)\ dx=0$. quick verification on Desmos: desmos.com/calculator/rqmudg7azm $\endgroup$ Jan 11 at 12:00
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    $\begingroup$ The similar result that I know, due to Titchmarsh: $C([0,\infty))$ under convolution is a division algebra. $\endgroup$
    – GEdgar
    Jan 11 at 12:08
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Note that a complex division Banach algebra is isomorphic to the complex numbers. Even if you consider the real scalars, the complexification gives the answer immediately.

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