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Suppose that we have two real functions $f$ and $g$ both belonging to $\mathcal{C}^\infty(\mathbb{R},\mathbb{R})$ analytic on $\mathbb{R}\setminus\{0\}$ but non-analytic at $x=0$. Is the convolution (assumed to be well defined) defined as:

$$(f*g)(x) = \int_\mathbb{R}f(x-z)g(z)dz$$

an analytic function ?

Otherwise, under which conditions the convolution is analytic ?

Any references or solutions will be highly appreciated.

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    $\begingroup$ Have a look at the properties of the Fourier transform of analytic functions at MO23679. If you pick your functions to be analytic at least in some strip around the real axis, I suspect that the answer will come out: iff one of $f$ or $g$ is analytic. $\endgroup$
    – Igor Khavkine
    Commented Jun 14 at 9:07
  • $\begingroup$ Thank you @IgorKhavkine for the post you mentinned. A quick question, when it is mentionned "the fourier transform should decay exponentially", we are talking about the module of the fourier tranform ?. It is to say that if $|\mathcal{F}[f*g](u)| < e^{-a|u|}$ for some $a>0$, $f*g$ is analytic ? $\endgroup$
    – NancyBoy
    Commented Jun 14 at 10:07
  • $\begingroup$ The precise condition quoted in the answer by Nate Eldredge uses the $L^2$-norm on the shifted real axis. Consult the cited reference for more details. More specialized literature might give analogous results with different norms. $\endgroup$
    – Igor Khavkine
    Commented Jun 14 at 10:20
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    $\begingroup$ @IgorKhavkine: $\mathbb{R}$ has two "ends", I can easily imagine a situation where $f$ has exponential decay on one end and $g$ on the other. $\endgroup$
    – Willie Wong
    Commented Jun 15 at 0:02
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    $\begingroup$ Cross-posted: math.stackexchange.com/questions/4940919/… $\endgroup$
    – RobPratt
    Commented Jul 3 at 3:35

1 Answer 1

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The answer to your question is negative. Take the smooth function $\chi$ defined by $$ \chi(t)=H(t) e^{-t^{-1}-t^2}, \quad H=\mathbf 1_{(0,+\infty)}. $$ This function is in $L^1(\mathbb R)$, $C^\infty$ everywhere, analytic outside $0$, and not analytic at $0$, in fact flat at $0$, i.e. all derivatives are vanishing at 0. Let us define $F=\chi\ast\chi$; we have $F\in L^1(\mathbb R)$, $F$ is $C^\infty$ everywhere, and for $k\in \mathbb N$, $$ F^{(k)}(t)=\int\chi^{(k)}(t-s)\chi(s) ds, $$ so that $ F^{(k)}(0)=\int\chi^{(k)}(-s)\chi(s) ds=0, $ since $s$ must be non-negative and non-positive in the integrand. So the function $F$ is flat at $0$ and is positive for $t>0$, so cannot be analytic at $0$.

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  • $\begingroup$ Thank you for this great answer @Bazin! Do you have any idea for the case $\chi_1(t) = H(t)e^{-1/t-t^2}$, $\chi_2(t) = H(-t)e^{-1/|t|-t^2}$ ($\chi_2$ being the symetric of $\chi_1$ w.r.t. $y$-axis) and $F=\chi_1*\chi_2$ ? $\endgroup$
    – NancyBoy
    Commented Jul 4 at 10:30
  • $\begingroup$ @NancyBoy I guess that this new $F$ is not analytic, but that the proof is not as simple as for $\chi\ast\chi$ in my answer. In fact the product of the Fourier transforms of your functions $\chi_j$ cannot be exponentially decreasing. Maybe my post "Fourier transform of Analytic Functions" (see the linked posts above on the right) may help understand that business further. $\endgroup$
    – Bazin
    Commented Jul 4 at 14:36

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