As is well-known, the Fourier-transform is an injective bounded linear operator from $L^1(\mathbb R)$ to $(C_0(\mathbb R),\Vert \cdot \Vert_\infty)$, where the latter is the space of continuous functions on $\mathbb R$, vanishing at infinity, equipped with the supremum norm.

My question: is the inverse operator (defined on the image) bounded?

I.e., is there a $C>0$ such that $$\Vert \hat f \Vert_\infty \geq C \Vert f \Vert_{L^1}\qquad \forall\; f\in L^1(\mathbb R) ?$$

Edit: Maybe I should add (in view of the first comment) that I expect the answer to be negative $-$ in fact, I've spent some time trying to construct simple counterexamples but without success. Therefore, a way to re-phrase this "provocative" question is:

How can we see that the answer is no? What is a nice counterexample?

2nd Edit: Perhaps it is easier to answer the analogous question for the discrete Fourier transform $l^1(\mathbb Z)\to (C(S^1),\Vert \cdot \Vert)$, where $S^1$ is the circle.

  • $\begingroup$ No. It seems there should be a very simple counterexample, not sure what it is right now. If you can't find a simpler example a modification of the Rudin-Shapiro polynomials should work. $\endgroup$ Mar 8, 2020 at 11:03
  • $\begingroup$ Thank you for your comment, I agree that there should be a counterexample. I edited the question accordingly. $\endgroup$
    – B K
    Mar 8, 2020 at 11:11
  • 1
    $\begingroup$ A simple counter-example in the discrete case is given by the Dirichlet kernel: $\|D_n\|_1 \to \infty$ but the Fourier coefficients of $D_n$ are bonded by $1$. $\endgroup$ Mar 8, 2020 at 12:30
  • $\begingroup$ @ Kavi Rama Murthy: unfortunately, what you say concerns the opposite direction of my question, which would be to consider $L^1(S^1)$ and $l^\infty(\mathbb Z)$. $\endgroup$
    – B K
    Mar 8, 2020 at 13:15
  • $\begingroup$ ??? One can certainly regard $D_n$ as an element of $L^2(S^1)$; I don't follow your objection. In any case, I wrote out the details of using the Dirchlet kernel here in an edit below... $\endgroup$ Mar 9, 2020 at 15:11

1 Answer 1


Here we have two counterexamples. I suspect that many readers will consider the first example "simpler". I actually feel like the second is simpler, because it starts from less; I actually see every detail of why the second example works, while the first depends on mysterious previous knowledge. Anyway

First Example

As suggested in a comment but disputed in another comment, one can use the Dirichlet kernel to give a simple counterexample.

Say $\phi\in L^1$, $\delta>0$, $$|\phi(t)|\ge \delta\quad(|t|\le\delta),$$and $\hat \phi$ is supported in $(-1/2,1/2)$. Let $$f_n(t)=\phi(t)D_n(t)=\phi(t)\sum_{j=-n}^ne^{ijt}.$$Then $$||f_n||_1\ge\delta\int_{-\delta}^\delta|D_n(t)|\,dt\to\infty$$but $$||\hat f_n||_\infty=||\hat\phi||_\infty.$$

Second Example

Lemma. If $f\in L^1(\Bbb R)$ then $\lim_{\xi\to\pm\infty}\int_{-\infty}^\infty|1-e^{i\xi t}||f(t)|\,dt=\alpha||f||_1$, where $\alpha=\frac1{2\pi}\int_0^{2\pi}|1-e^{it}|\,dt.$

(WLOG $f$ is uniformly continuous with compact support...)

Note that $\alpha>1$. Fix $\beta\in(1,\alpha)$.

Now say $f_0\in L^1$ with $||\hat f_0||_\infty=1$ and $\hat f_0$ supported in $[0,1]$. We're going to let $$f_1(t)=(1-e^{i\xi t})f(t)$$for a suitable constant $\xi$. If $\xi$ is large enough we have $$||\hat f_1||_\infty=||f_0||_\infty=1,$$ $$||f_1||_1\ge\beta||f_0||_1.$$Repeat: $||\hat f_n||_\infty=1$, $||f_n||_1\ge\beta^n||f_0||_1\to\infty$.

  • $\begingroup$ That is a nice argument. Somehow I have the feeling that one might be able to simpliy it further... I'll think about it while waiting for more answers. $\endgroup$
    – B K
    Mar 8, 2020 at 14:02

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