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Does there exist two functions $f,g \in L^1(\mathbb R)$, with $f,g \neq 0$, such that $\operatorname{supp}(f)\subset[0,\infty[ $ and $ \operatorname{supp}(g) \subset[0,\infty[ $ and $f\ast g =0$ ?

I've done similar exercises where taking the Fourier transform gives a quick solution but I don't know how to make it work for that one.

Kieran McShane
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  • Well, one trope is that if their supports are in that half-line, then their Fourier transforms extend to holomorphic functions in a half-plane (upper, or lower, depending on normalizations). Of course, a product of holomorphic functions on any non-trivial open cannot be zero unless one of them is identically zero... But, still, there are non-trivial holomorphic functions on the upper half-plane with boundary-values on $\mathbb R$ equal to $0$... (due to the "one point missing"). Is this the sort of thing relevant to your situation? – paul garrett Feb 08 '23 at 21:40
  • @paul garrett Shouldn't there be a decreasing hypothesis for $\hat{f}$ and $\hat{g}$ to be extended as holomorphic functions ? It could be that $\hat{f}$ or $\hat{g}$ are not $\mathcal C^{\infty}$ on $\mathbb R_+$ if I am not mistaken. – Zag Feb 08 '23 at 21:47
  • @Zag, yes, of course, some further conditions are probably needed, depending what one is trying to get... Perhaps the context of the question needs further specifics... :) – paul garrett Feb 08 '23 at 22:10
  • If $f,g\in L^1[0,\infty)$ then $f\ast g\in L^1[0,\infty)$, $F(s)=\int_0^\infty f(t) e^{-st}dt,G(s)=\int_0^\infty g(t) e^{-st}dt$ are analytic for $s > 0$, and so $F(s)G(s) = 0$ implies either $F=0$ or $G=0$, which implies $f=0$ or $g=0$ in $L^1[0,\infty)$. – reuns Feb 08 '23 at 23:43
  • @reuns Thank you, that answers my question. – Kieran McShane Feb 09 '23 at 00:04

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