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.)

  • $\begingroup$ Thank you! I am thinking a while about it know. Couldn't construct a sequence yet but at least I know now in which direction I have to think. I appreciate your help very much. $\endgroup$ – Lukas Betz May 8 '15 at 19:05
  • $\begingroup$ 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? $\endgroup$ – Lukas Betz May 8 '15 at 22:04
  • 1
    $\begingroup$ @LeBtz: It works. $\endgroup$ – Giuseppe Negro May 14 '15 at 15:20
  • $\begingroup$ @GiuseppeNegro: Thanks for having a look on it! $\endgroup$ – Lukas Betz May 14 '15 at 15:41
  • $\begingroup$ @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. $\endgroup$ – Giuseppe Negro May 14 '15 at 16:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.