# Is the inverse of the Fourier transform $L^1(\mathbb R)\to (C_0(\mathbb R),\Vert \cdot \Vert_\infty)$ bounded?

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.

• 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. – David C. Ullrich Mar 8 at 11:03
• Thank you for your comment, I agree that there should be a counterexample. I edited the question accordingly. – B K Mar 8 at 11:11
• 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 $1$. – Kavi Rama Murthy Mar 8 at 12:30
• @ Kavi Rama Murthy: unfortunately, what you say concerns the opposite direction of my question, which would be to consider $L^1(S^1)$ and $l^\infty(\mathbb Z)$. – B K Mar 8 at 13:15
• ??? 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... – David C. Ullrich Mar 9 at 15:11

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: $$||\hat f_n||_\infty=1$$, $$||f_n||_1\ge\beta^n||f_0||_1\to\infty$$.

• 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. – B K Mar 8 at 14:02