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).