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

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) – reuns Jun 26 '16 at 23:28
• You may want to look at this answer: math.stackexchange.com/questions/1495646/… – abnry Jun 27 '16 at 0:25
• It seems like the above link only work for finite measure spaces? – Mark Jun 27 '16 at 0:52
• Almost duplicate of this question: math.stackexchange.com/questions/1664286/… (and see my answer there perhaps) – user138530 Jul 3 '16 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 '16 at 0:07

## 1 Answer

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, $$\|\hat{f}\|_{\infty} \le L$$.

$$\dots\dots\dots$$

1. 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. $$\dots\dots\dots$$
• 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 '16 at 1:27
• I have even seen triple integrals :) – Mark Aug 2 '16 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 $L^1$. – David C. Ullrich Aug 2 '16 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... – Romain S Mar 29 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. – VictorZurkowski Mar 30 at 5:23