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

Question

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).

Answer 1

After almost 5 years of my original message, in March 29, 2021, @Romain pointed to an error in my original post that invalidates the original argument. For completion, I am writing a proof, borrowing from the proof Spectral Radious Formula, avoiding use of analytic functions.

To begin with, we don't know the limit exists, so let's work with lim-inf and lim-sup.

1. Since $\hat f(\zeta)^n = \mathcal F(f^{(n)}) $, then we have $ \|\hat f\|_{\infty}^n = \|\hat f^n \|_{\infty} \le \|f^{(n)}\|_1 $, so $ \|\hat f\|_{\infty} \le \liminf_n (\|f^{(n)}\|_1)^{1/n} \le \limsup_n (\|f^{(n)}\|_1)^{1/n}.$
   $$\dots\dots\dots$$

2. Let $L = \limsup_n (\|f^{(n)}\|_1)^{1/n} $. Note that $ \|f^{(n)}\|_1 \le \|f\|_1^n $, therefore $L \le \|f\|_1 <\infty $.

Pick a small real number $\epsilon > 0$. Let $z$ such that $ |z| L < 1$, for example $ z = \frac{1-\epsilon}{\epsilon + L}$. By definition of $\limsup$, for $n$ sufficiently large, $\|f^{(n)}\|_1^{1/n} \le L+\epsilon$, therefore, with the above choice of $z$, for large $n$, $ |z|^n \|f^{(n)}\|_1 \le \left( \frac{1-\epsilon}{\epsilon + L} \right)^n (L+\epsilon)^n = (1-\epsilon)^n$.

Let $G \in L^2(\mathbb R^d)$, and consider the sequence $h_n = z^n f^{(n)}*G$ , for $n=1,2,...$. Then,
$$ \| z^n f^{(n)}*G \|_2 \le |z|^n \|f^{(n)}\|_1 \|G\|_2 $$ by Young's inequality, and so, for large $n$,
$$ \| z^n f^{(n)}*G \|_2 \le (1-\epsilon)^n \|G\|_2 \rightarrow 0 \text{ in } L^2 . $$

Thus, using properties of the Fourier transform, we conclude that for any $G \in L^2$, $z^n |\hat{f}(\xi)|^n \hat{G}(\xi) \rightarrow 0$ in $L^2$.

Now, let $R$ be a large number, and take $G$ so that $\hat{G}(\xi) = I\big( |\hat{f}| > \frac{1}{z}$ and $|\xi| \le R \big)$ (i.e.: $\hat{G}$ is $1$ on the set where $|\hat{f}(\xi)| > \frac{1}{z}$ and $|\xi| \le R$, and $0$ otherwise).

Then, since on the set where $\hat{G}$ is not zero we have $|\hat{f}(\xi)| > 1/z$, $$ \begin{aligned} z^n |\hat{f}(\xi)|^n \hat{G}(\xi) &\ge z^n \frac{1}{z^n} I\big( |\hat{f}| > \frac{1}{z} \text{ and } |\xi| \le R \big ) \\ & = I\big( |\hat{f}| > \frac{1}{z} \text{ and } |\xi| \le R \big ) \rightarrow 0 \text{ in } L^2. \end{aligned} $$ Hence, the set $\big( |\hat{f}| > \frac{1}{z}$ and $|\xi| \le R \big ) $ has measure $0$. This is true for all $R$, so the set where $\big( |\hat{f}| > \frac{1}{z} \big)$ has measure $0$, or $|\hat{f}(\xi)| \le \frac{1}{z} $ almost everywhere, or, more simply, $\|\hat{f}\|_{\infty} \le \frac{1}{z} = \frac{L+\epsilon}{1-\epsilon}$. Since $\epsilon$ is arbitrary, $\|\hat{f}\|_{\infty} \le L$.

$$\dots\dots\dots$$

3. Notes:

• This proof covers $\mathbb R^d$ of any dimension, the circle, and compact groups.
• The proof shows that the limit exists, without assuming it does.
• Note that the condition $f \in L^2$ is not used. $$\dots\dots\dots$$