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.