4

If $f(x)$ ($\Bbb{R} \to \Bbb{R}$) is an $L^1$ function and $\phi (t)$ is the Fourier transform, then it is well known that

$${\left\Vert {\phi}\right\Vert }_\infty \le {\left\Vert {f}\right\Vert }_1 $$

I've noticed it often true that

$${\left\Vert {f}\right\Vert }_1 \le 2 {\left\Vert {\phi}\right\Vert }_\infty $$

My attempts to prove this using Hölder's inequality result in integrals that don't converge. Is this statement even true (perhaps under additional conditions)? Is it already known? Is it just outright wrong?

Edit: Given that the general statement is false, I have realized that I need to clarify what $f$ lead me to my conjecture. $f = g - h$ where

$${\left\Vert {g}\right\Vert }_1 = {\left\Vert {\hat h}\right\Vert }_\infty = 1$$

$${\left\Vert {h}\right\Vert }_1 = {\left\Vert {\hat g}\right\Vert }_\infty = 1$$

My Fourier transform is the one used in probability theory to generate characteristic functions.

skewray
  • 43
  • This may be related – skewray May 09 '25 at 22:57
  • 2
    It is nearly inherent in what you have written, but you should really be explicit about what version of the Fourier transform you are working with. Different conventions may lead to different bounds since you're interested in pointwise behavior. – Cameron L. Williams May 10 '25 at 13:26
  • Thank you. I clarified which FT. One that doesn't care about pointwise behaviour, or at least shouldn't. – skewray May 10 '25 at 15:22
  • 1
    That's not what I meant. There are several competing conventions, you need to state which you're using. But my point was that $\infty$ norm behavior for continuous functions is really a pointwise behavior (you are looking for the maximum of the modulus). The convention is important because some conventions have normalization factors out front or in the exponent which change the analysis a bit. – Cameron L. Williams May 10 '25 at 16:43
  • Then I don't understand your points. Sorry? Characteristic functions are uniformly continuous. I guess I am assuming that the difference between two uniformly continuous functions is still uniformly continuous. And the FT used for computing characteristic functions is well defined - E[e^ixt]. – skewray May 10 '25 at 18:12

1 Answers1

4

A simple sufficient condition for your claim to be true is $f\ge0$. Then, $$\Vert f \Vert_{L^1}=\phi(0)\le 2\Vert\phi\Vert_{L^\infty}.$$

In general however, I believe this is false. Consider the Hilbert transform $$H\ast \psi(x):=\lim_{\varepsilon\to0}\frac1\pi\int_{\varepsilon<| y| <1/\varepsilon} \frac{\psi(y)}{x-y}\,\mathrm d y.$$

Morally speaking $H= \frac{1}{\pi x}$ and the above is the convolution of $H$ with a function $\psi$. Its a well known fact that $\hat H(\xi)= -i\operatorname{sign}(\xi)$ and more generally, $\mathcal F(H\ast \psi)=-i\operatorname{sign}(\xi)\,\hat \psi.$ All this can be made rigorous in the sense of tempered distributions.

The idea to disproving the claim is to consider a sequence $\psi_\delta\longrightarrow\delta_0$, so that $$ \Vert H\ast \psi_\delta \Vert_{L^1}\longrightarrow\infty, \quad\text{while}\quad \Vert \mathcal{F}(H\ast \psi_\delta)\Vert_{L^\infty}\le C. $$ This rules out any possible estimate of the form $\Vert f\Vert_{L^1}\le C \Vert \phi \Vert_{L^\infty}$, for a constant $C>0$ to hold for all $f\in L^1$.

To get this more down to earth, let us explicitly choose $$\psi_{\delta}(x):= \begin{cases}\frac{1}{2\delta} &x\in[-\delta,\delta]\\ 0& \text{else}\end{cases}.$$ You can compute $$H\ast \psi_\delta(x)=\frac{1}{2\pi\delta}\,\log\left(\left\vert\frac{x+\delta}{x-\delta}\right\vert\right)$$ and I am fairly confident that this is an $L^1$ function.

In particular $H\ast \psi_\delta(x)\longrightarrow \frac{1}{\pi x}$ uniformly on every interval $[r,\frac 1 r]$, $r>0$, from which you can deduce $\Vert H\ast \psi_\delta \Vert_{L^1}\longrightarrow\infty$, since $\frac{1}{\pi x}$ is not integrable.

On the Fourier side, $\mathcal F(\psi_\delta) = \frac{\sin(2\pi \delta\xi)}{2\pi \delta\xi}$ and thus $$\Vert \mathcal{F}(H\ast \psi_\delta)\Vert_{L^\infty} = \left\Vert -i \operatorname{sign}(\xi) \frac{\sin(2\pi \delta\xi)}{2\pi \delta\xi}\right\Vert_{L^\infty}\le 1.$$ This should disprove your conjecture.

As a final remark, there are $L^p$-$L^q$-estimates for the Fourier transform. See e.g. the discussion in this post.

Espen Xylander
  • 54
  • Although most of what you wrote I will have to go study to understand, the bit at the end looks like a Dirichlet kernel. At any rate, the functions I am interested in are not always positive, and I clarified that with an edit to be more explicit about which functions motivated this question. – skewray May 10 '25 at 15:25