$\|\hat{f} \|_{\infty} = \lim _ {n \rightarrow \infty} (\|f^{(n)}\|_1)^{1/n}$

Let $f \in L^2 \cap L^1$ on the Real line, and define $f^{(n)}$ to be the $n$-fold convolution $f \circ f ... \circ f$.

I want to show that $||\hat{f} ||_{\infty} = \lim _ {n \rightarrow \infty} (||f^{(n)}||_1)^{1/n}$, using the tools of Fourier analysis on $L_1$ and $L_2$.

And actually I'm only stuck on the fact that the RHS $\le$ LHS. A formal proof would be something like this, but I'm stuck on technicalities:

\begin{align}\lim_n (\|f^{(n)}\|_1)^{1/n} &= \lim_n [\int f^{(n)} \overline{\exp{ (i \arg f^{(n)})}}]^{1/n} = \lim_n \left\langle f^{(n)}, \exp{ (i \arg f^{(n)})}\right\rangle ^{1/n}\\ & = \lim_n \left\langle\widehat{f^{(n)}}, \widehat{\exp{ (i \arg f^{(n)})}}\right\rangle^{1/n} = \lim_n \left\langle{\hat{f}^n}, \widehat{\exp{ (i \arg f^{(n)})}}\right\rangle^{1/n} \\ & = \lim_n \left[\int \hat{f}^n \overline{\widehat{\exp{ (i \arg f^{(n)})}}}\right]^{1/n} \le \lim_n \left[\| \hat{f} \|^n _\infty\int \overline{\widehat{\exp{ (i \arg f^{(n)})}}}\right]^{1/n} \le \|\hat{f}\|_\infty \end{align}

Trouble is, $\exp{ (i \arg f^{(n)})}$ is not integrable since its magnitude is always 1. I have tried to do an approach where I insert $g_k$ where $g_k$ is a compact smooth "hill" function which becomes wider and wider and limits to $1$, and this allows me to arrive at

\begin{align} \lim_n (||f^n||_1)^{1/n} &= \lim_n \lim_k [\int f^{(n)} \overline{g_k \exp{ (i \arg f^{(n)})}}]^{1/n} \\ &\le \lim_n \lim_k \left[\| \hat{f} \|^n _\infty\int \overline{\widehat{g_k \exp{ (i \arg f^{(n)})}}}\right]^{1/n}\\ & \le \| \hat{f} \|_\infty \lim_n \lim_k \int \overline{\widehat{g_k \exp{ (i \arg f^{(n)})}}}]^{1/n} \end{align}

But I can't actually take the limit $k$ because then the term will go to infinity. I thought of making $k$ a function of $n$ but then I couldn't show that this doesn't change the limit.

This strategy is taken from an analogous proof on the periodic circle with discrete Fourier transform, and I would like to see if it can be fixed somehow (because this was the hint given by the text).

• the fact that $\|\hat{f}\|_\infty = \lim_n (\|\hat{f}^n\|_2^2)^{1/2n} =\lim_n (\|f^{(n)}\|_2^2)^{1/2n}$ is probably useful (and hill function is called a bump function) Jun 26, 2016 at 23:28
• You may want to look at this answer: math.stackexchange.com/questions/1495646/… Jun 27, 2016 at 0:25
• It seems like the above link only work for finite measure spaces?
– Mark
Jun 27, 2016 at 0:52
• Almost duplicate of this question: math.stackexchange.com/questions/1664286/… (and see my answer there perhaps)
– user138530
Jul 3, 2016 at 20:31
• Yes if we restrict to the circle I think the proof will come out fine. Is there a way to reduce to that case from the real line?
– Mark
Jul 4, 2016 at 0:07

@Romain and @Glitch pointed errors or gaps in an answer I posted some time ago. As @David points out, the result is true, even without the assumption the $$f \in L^2(\mathbb{R})$$. I cannot fix my proof, hence I am deleting it.

If someone adds a proof later on, I will completely remove my post.

• How do you justify $\int \int |f^{(n-1)}(x)| |f(y-x)| dy dx = \int \int |f^{(n-1)}(x)| \; |f(y)|^{ } dy dx$
– Mark
Aug 2, 2016 at 1:27
• I have even seen triple integrals :)
– Mark
Aug 2, 2016 at 16:13
• Certainly it's true for $f\in L^1$. This is just the Spectral Radius Formula from Banach algebras, plus the fact that the Fourier transform gives all the complex homomorphisms of $L^1$. Aug 2, 2016 at 17:27
• I think there might be a mistake: a prior, you don't know that $\|{f}\|_{2/3}$ is finite, hence these estimates could be meaningless... Mar 29, 2021 at 23:46
• @Romain Thanks for pointing to the mistake. In addition to your objection, Young's inequality applies to $p,q,r \ge 1$, thus $q=2/3$ is non-sense\$|. I replaced the original argument with one that borrows from a possible proof of the Spectral Radious Formula. Mar 30, 2021 at 5:23