Fourier transform not surjective using oppen mapping theorem.

Question

I know that it is possible to prove that the Fourier transform $\displaystyle\mathcal{F}: (L^1(\mathbb R),\|\cdot\|_1) \to (\{f\in C(\mathbb R): \lim_{|x|\to\infty} f(x) = 0\}, \|\cdot\|_\infty)$ is not surjective using the open mapping theorem. But how is it done exactly? I know that $\mathcal{F}$ is continuous. Applying the open mapping theorem, $\mathcal{F}$ would be an open mapping if it was surjective. So we have to show that $\mathcal{F}$ is not open, right? But how do we do this?

Answer 1

The Fourier transform maps $L^1$ injectively into $C_0.$ If the map were onto, then it would be open by the open mapping theorem. Hence the inverse of the FT would be continuous on $C_0.$ Because these are linear operators on normed linear spaces, it would follow that there exists a constant $c>0$ such that $\|\mathcal {F}(f)\|_\infty \ge c \|f\|_1$ for all $f\in L^1.$ To show this fails, you need to show there is a sequence $f_n$ in $L^1$ such that $\|f_n\|_1 = 1$ for all $n,$ while $\|\mathcal {F}(f_n)\|_\infty \to 0.$ (I can't remember how to do this at the moment. In the case of the FT on the circle, you can take $f_n = D_n/\|D_n\|_1,$ where $D_n$ is the $n$th Dirichlet kernel.)