Can Hölder's Inequality be strengthened for smooth functions? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T08:31:10Z http://mathoverflow.net/feeds/question/60087 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60087/can-holders-inequality-be-strengthened-for-smooth-functions Can Hölder's Inequality be strengthened for smooth functions? Kevin O'Bryant 2011-03-30T16:55:41Z 2011-09-21T08:30:41Z <p>Is there an $\epsilon>0$ so that for every nonnegative integrable function $f$ on the reals,</p> <p><code>$$\frac{\| f \ast f \|_\infty \| f \ast f \|_1}{\|f \ast f \|_2^2} &gt; 1+\epsilon?$$</code></p> <p>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.</p> <p>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 $[0,1]$.</p> <p>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.</p> <hr> <p>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 $3/2$. </p> <p>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 $\sqrt{2}$.</p> <p>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 $0 &lt; x &lt; 1/2$, $f(x)=0$ otherwise. Then $f\ast f(x) = \pi/2$ for $0 &lt; x &lt; 1/2$ and $f\ast f(x) = (\pi-4\arctan(\sqrt{2x-1}))/2$ for $1/2 &lt; x &lt; 1$, 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 $\pi/\log(16) \approx 1.133$.</p> <p><a href="http://www.uam.es/personal_pdi/ciencias/cvinuesa/papers/schbounds.pdf" rel="nofollow">In this paper</a>, Vinuesa and Matolcsi mention some proof-of-concept computations that show that $\pi/\log(16)$ is not extremal.</p> http://mathoverflow.net/questions/60087/can-holders-inequality-be-strengthened-for-smooth-functions/60210#60210 Answer by Terry Tao for Can Hölder's Inequality be strengthened for smooth functions? Terry Tao 2011-03-31T17:19:05Z 2011-04-07T13:46:04Z <p>Some initial thoughts:</p> <ul> <li>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). </li> <li>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.</li> <li>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 $f$). </li> <li>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 <a href="http://arxiv.org/abs/math/0611286" rel="nofollow">quantitative idempotent theorem</a> 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.</li> </ul> http://mathoverflow.net/questions/60087/can-holders-inequality-be-strengthened-for-smooth-functions/60259#60259 Answer by Ryan O'Donnell for Can Hölder's Inequality be strengthened for smooth functions? Ryan O'Donnell 2011-04-01T06:01:50Z 2011-04-01T06:01:50Z <p>Reminds me a bit of <a href="http://people.math.jussieu.fr/~talagran/prizes.pdf" rel="nofollow">Talagrand's 2nd <code>$</code>1000 conjecture</a>, a special case of which is the following: </p> <p>Let$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)$, where the probability is with respect to the Gaussian distribution.</p> <p>I'm pretty sure this special case is hard enough that Talagrand would give you a fraction of the <code>$</code>1000 for it.</p> http://mathoverflow.net/questions/60087/can-holders-inequality-be-strengthened-for-smooth-functions/76033#76033 Answer by Seva for Can Hölder's Inequality be strengthened for smooth functions? Seva 2011-09-21T08:30:41Z 2011-09-21T08:30:41Z <p>Not an answer, but rather an extended comment.</p> <p>Consider the following problem.</p> <p>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.)</p> <p>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.</p>