TITLE: $\|\hat{f} \|_{\infty} = \lim _ {n \rightarrow \infty} (\|f^{(n)}\|_1)^{1/n}$
QUESTION [9 upvotes]: 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).

REPLY [5 votes]: @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.