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 $N_f(x):=\log(\| f \|_{1/x})$. Holder'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, Holder's inequality implies that $N_f$ is convex at $x=1/2$. The generalized Holder'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.

An example: 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 $3/2$.