# Fourier transform not surjective using oppen mapping theorem.

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?

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.)
• I think I got it now. Let $\displaystyle\phi_{\epsilon} := \frac{1}{2\epsilon}\exp(-\left|\frac{x}{\epsilon}\right|)$. Then $\displaystyle\mathcal{F}(\phi_\epsilon)(x) = \frac{1}{1+(\epsilon x)^2}$. Hence $\mathcal{F}(\chi_{[-1,1]}\ast \phi_\epsilon) = \frac{2\sin}{id}\mathcal{F}(\phi_\epsilon)\in L^1$. For $\epsilon \to 0$ however the $L^1$-Norm of $\frac{2\sin}{id}\mathcal{F}(\phi_\epsilon)$ explodes while the Supremumnorm of $\mathcal{FF}(\chi_{[-1,1]}\ast \phi_\epsilon)$ is bounded. Should work, right? – Lukas Betz May 8 '15 at 22:04
• @LeBtz: You are welcome. The main idea, as you righfully found, is that the function $\sin x /x$ has infinite $L^1$ norm and its Fourier transform has finite $L^\infty$ norm. – Giuseppe Negro May 14 '15 at 16:24