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I want to prove that, given a $2\pi-$periodic bounded function $f$, its convolution $f*g(x)$ with the function $g(x):=e^{-x^2}$ has Fourier coefficients that deacay exponentially fast.

I have tried many things, but the best I could find was this non rigorous argument:

Suppose $f$ is even (so that we have no coefficients with $\sin(\cdot)$). The Fourier series writes $$f*g(x)=\sum_{n=1}^\infty c_n\cos(nx)+c_0.$$

Now, making the Fourier transform on both sides, we get

$$\mathcal F[f](\omega)\mathcal F[g](\omega)=\sum_{n=1}^\infty c_n\pi(\delta_{n}(\omega)+\delta_{-n}(\omega))+c_0\delta_0(\omega).$$

Now, from the property of Fourier Transform, we know that

$$\mathcal F[g](\omega)= \frac{1}{\sqrt 2}e^{-\omega^2/4}.$$

Therefore, if $\mathcal F[f](\omega)$ is propter measure (i.e. $\|\mathcal F[f]\|_1<\infty$), this shlould guarantee that the coefficients decay exponentially.

Davide Maran
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    Write $f=\sum_{k\in\mathbb Z} c_ke^{ikx}$. Then, since the Fourier transform exchanges convolutions and products, $fg= \sum_{k\in\mathbb Z}( \frac{1}{\sqrt 2} e^{-k^2/4}c_k)e^{ikx}$. Since the coefficients $c_k$ are bounded (can you prove it?), the coefficients of $fg$ decay exponentially. – Lorenzo Pompili Apr 06 '23 at 08:27
  • Obscure: $g$ is defined on $R$ and is not $2\pi$ periodic. The function $fg$ is defined on $R$ and is not $2\pi$ periodic. What do you mean by the Fourier coefficients of $fg?$ – Gérard Letac Apr 06 '23 at 12:15
  • @GérardLetac If $f$ is periodic, the convolution with a gaussian remains periodic – Davide Maran Apr 06 '23 at 14:34
  • I think this may help, but I am not sure.

    This answer, https://math.stackexchange.com/a/3210338/441161 , uses a Dirac comb as the periodic function, and a non-specific pulse function, $x(t)$, and shows that, if the sum of the infinite number of shifted pulse tails converge in the intervals, then the Fourier coefficients of the convolution are samples of the Fourier Transform of the pulse function.

     – Andy Walls Apr 06 '23 at 16:20
  • Thank you Davide Maran, my bad. – Gérard Letac Apr 06 '23 at 16:59

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