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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