$$ \| T_x(f \ast) g \|_\infty = \|(T_xf -f) \ast g\|_\inftyīecause the shift is continuous in the $L^2$-norm (prove it for compactly supported continuous functions, then by density you get it for every $f \in L^2$) we find that $f \ast g$ is in fact uniformly contiuous. I understand the question in the following way:ġ) is $f \ast g$ well defined: Yes, certainly, and since an easy application of Cauchy-Schwartz yields One might argue that this is only a partial argument. In the preceding paragraph, $C_0$ can be replaced by $L^\infty$ throughout (the norm is the same), saving you the trouble of proving that $f*g$ vanishes at infinity. The inverse Fourier transform is continuous from $L^1$ to $C_0$. Also, $(f,g)\mapsto \hat f \hat g$ is continuous from $L^2\times L^2$ to $L^1$. The map $(f,g)\mapsto f*g$ is a continuous map from $L^2\times L^2$ into $C_0$.
For $f,g\in \mathcal S$ the result is known. You want to show that this function is $f*g$.Īrgue by density. Apply the inverse Fourier transform to it (essentially same as Fourier transform, but with opposite sign in the exponential, and maybe different normalization).
Taking the Fourier transform of a $C_0$ function is problematic: such transforms are generally not functions, but merely distributions (as Jose27 points out). For $f,g\in L^2$, the convolution $f*g$ belongs to $C_0(\mathbb R)$.