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Is there an $\epsilon>0$ so that for every nonnegative integrable function $f$ on the reals,

$$\frac{\| f \ast f \|_\infty \| f \ast f \|_1}{\|f \ast f \|_2^2} > 1+\epsilon?$$

Of course, we want to assume that all of the norms in use are finite and nonzero, and $f\ast f(c)$ is the usual convolved function $\int_{-\infty}^{\infty} f(x)f(c-x)dx$. The applications I have in mind have $f$ being the indicator function of a compact set.

A larger framework for considering this problem follows. Set $N_f(x):=\log(\| f \|_{1/x})$. Hölder's Inequality, usually stated as $$\| fg \|_1 \leq \|f\|_p \|g\|_q$$ for $p,q$ conjugate exponents, becomes (with $f=g$) $N_f(1/2+x)+N_f(1/2-x)\geq 2N_f(1/2)$. In other words, Hölder's Inequality implies that $N_f$ is convex at $x=1/2$. The generalized Hölder's Inequality gives convexity on $[0,1]$.

It is possible for $N_f$ to be linear, but only if $f$ is a multiple of an indicator function. What I am asking for is a quantitative expression of the properness of the convexity when $f$ is an autoconvolution.


Examples: The ratio of norms is invariant under replacing $f(x)$ with $a f(cx-d)$, provided that $a>0$ and $a,c,d$ are reals. This means that if $f$ is the interval of an indicator function, we can assume without loss of generality that it is the indicator function of $(-1/2,1/2)$. Now, $f\ast f(x)$ is the piecewise linear function with knuckles at $(-1,0),(0,1),(1,0)$. Therefore, $\|f\ast f\|_\infty=1$, $\|f \ast f\|_1 = 1$, $\|f \ast f \|_2^2 = 2/3$, and the ratio of norms is $3/2$.

Gaussian densities make another nice example because the convolution is easy to express. If $f(x)=\exp(-x^2/2)/\sqrt{2\pi}$, then $f\ast f(x) = \exp(-x^2/4)/\sqrt{4\pi}$, and so $\|f\ast f\|_\infty = 1/\sqrt{4\pi}$, $\|f\ast f\|_1=1$, and $\|f \ast f\|_2^2=1/\sqrt{8\pi}$. The ratio in question is then just $\sqrt{2}$.

This problem was considered (without result) by Greg Martin and myself in a series of papers concerning generalized Sidon sets. We found this ``nice'' example: $f(x)=1/\sqrt{2x}$ if $0 < x < 1/2$, $f(x)=0$ otherwise. Then $f\ast f(x) = \pi/2$ for $0 < x < 1/2$ and $f\ast f(x) = (\pi-4\arctan(\sqrt{2x-1}))/2$ for $1/2 < x < 1$, and $f\ast f$ is 0 for $x$ outside of $(0,1)$. We get $\|f \ast f\|_\infty = \pi/2$, $\|f \ast f\|_1 = 1$, $\|f \ast f \|_2^2 = \log 4$, so the norm ratio is $\pi/\log(16) \approx 1.133$.

In this paper, Vinuesa and Matolcsi mention some proof-of-concept computations that show that $\pi/\log(16)$ is not extremal.

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What happens for intervals? I ask this, because I expect the answer to be obtainable by an easy computation. In particular the limit "interval length to 0" should be relevant. –  Helge Mar 30 '11 at 19:15
    
If $f$ is the indicator function of an interval of length $I$, then $\|f \ast f\|_\infty = I$, $\|f\ast f\|_1 = I^2$ and $\|f \ast f\|_2^2 = 2/3 I^3$. The ratio in question, then, is always 3/2, independent of the length of the interval –  Kevin O'Bryant Mar 30 '11 at 20:28
    
sorry, am not that familiar with the notation. what happens when $f$ is the constant function? –  Suvrit Mar 30 '11 at 20:44
    
You might add gaussians as another example, with a slight better ratio ($\sqrt{2}$( –  Piero D'Ancona Mar 30 '11 at 21:54
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I was going to mention Sidon sets, but realized that this question is probably motivated by something similar to it. :) –  Willie Wong Mar 30 '11 at 22:27

3 Answers 3

Some initial thoughts:

  • the question is basically asking whether $f*f$ can be close to a constant multiple $c1_E$ of an indicator function (these are the only non-negative functions for which Holder is sharp).
  • the hypothesis that $f$ is non-negative is going to be crucial. Note that any Schwartz function can be expressed as $f*f$ for some complex-valued f by square-rooting the Fourier transform, and so by approximating an indicator function by a Schwartz function we see that there is no gain.
  • On the other hand the hypothesis that $f$ is an indicator function (or a constant multiple thereof) is only of limited utility, because any non-negative function in $L^1$ can be expressed as the weak limit of constant multiples of indicator functions (similarly to how a grayscale image can be rendered using black and white pixels in the right proportion; the indicator is of a random union of small intervals whose intensity is proportional to $f$).
  • If $f*f$ is close to $c1_E$, and we normalise $c=1$ and $|E|=1$, then $f$ has $L^1$ norm close to $1$ and the Fourier transform has $L^4$ norm close to $1$ (i.e. the Gowers $U^2$ norm is close to $1$). Using the quantitative idempotent theorem of Green and Sanders we also see that the $L^2$ norm of $f$ (which controls the Wiener norm of $f*f$) is much larger than $1$. But it's not clear to me where to go next from here.
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Terry, instead of using the idempotent theorem I think a result of Rudin, improved by Saeki is more relevant: MR0225102 (37 #697) Saeki, Sadahiro On norms of idempotent measures. Proc. Amer. Math. Soc. 19 1968 600–602. In R, where there are no interesting compact subgroups, this will tell you that $\|\hat{1}_E\| > 1.2$. WIll have to think about the details. –  Ben Green Mar 31 '11 at 20:21
    
Hmm, it's not so clear. The problem is that it seems to be hard to say much about f. For example, f could consist of spikes of height $N$ and width $1/N^2$ about a Sidon set of size about $N$; then $f \ast f$ will look a bit like the characteristic function of a union of $N$ intervals of width $\sim 1/N$; call this set $E$. Unfortunately I can't conclude anything useful about $\Vert \hat{1}_E \Vert_1 = \Vert f \Vert_2^2$ being small. –  Ben Green Mar 31 '11 at 20:58
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Let $A$ be Sidon set with $|A|$, and take $f(x)=\sum_{a\in A} 1_{(a-1/2,a+1/2)(x)$, so that $f*f$ is piecewise linear with knuckles at $(k,r(k))$, where $r(k)$ is the number of reps of the integer $k$ as a sum of two elements of $A$. Then $\|f*f\|_{\infty}=2$ (by Sidon-ness), and $\|f*f\|_1=n$ (by $|A|=\sqrt{n}$). But $\|f*f\|_2^2$ depends on the number of intervals in $A+A$, or (equivalently) on the sum $\sum_k r(k)r(k+1)$. This is a difference between the continuous and discrete settings: in the discrete setting we always have $\|f*f\|_2^2=2|A|^2-|A|$, and norm ratio would tend to 1. –  Kevin O'Bryant Mar 31 '11 at 21:18

Reminds me a bit of Talagrand's 2nd $1000 conjecture, a special case of which is the following:

Let $f$ be a nonnegative function on the reals and let $g = U_t f$, where $U_t$ is the Ornstein-Uhlenbeck semigroup and $t$ is some fixed positive number; say, $t = 1$. Then Markov's inequality is not tight for $g$; i.e., $\Pr[g > c \mathrm{E}[g]] = o(c)$, where the probability is with respect to the Gaussian distribution.

I'm pretty sure this special case is hard enough that Talagrand would give you a fraction of the $1000 for it.

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Not an answer, but rather an extended comment.

Consider the following problem.

Given a set of integers $A\subset [1,N]$, denote by $\nu(n)$ the number of representations of $n$ as a sum of two elements of $A$. Thus, $\nu=1_A\ast1_A$ up to normalization, and, trivially, we have $$ \sum_n \nu^2(n) \le |A|^2 \max_n \nu(n). $$ Does there exist an absolute constant $\varepsilon>0$ such that if $$ \sum_n \nu^2(n) > (1-\varepsilon) |A|^2 \max_n\nu(n), $$ then $\alpha:=|A|/N\to 0$ as $N\to\infty$? (The flavor of this question to me is as follows: we want to draw a conclusion about a finite set, given that its ``additive energy'' is large -- but not as large as in Balog-Szemeredi-Gowers.)

What is the relation between this and the original problem? Although I cannot establish formally equivalence in both directions, it is my understanding that the two problems are ``essentially equivalent''; at least, if in the original problem we confine ourselves to indicator functions of open sets.

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