Let $f$ be function from $\mathbb{R}\rightarrow [0,1]$ with $\int_{-\infty}^{\infty}dtf(t)=1$. Define
$\beta[f](\tau)=\max_t\int_{t}^{t+\tau}dt'f(t')$
i.e. $\beta[f](\tau)$ is the maximal value of the integral of $f$ over any (simply connected) interval of length $\tau$.
Let $(f\star f)(t)=\int_{-\infty}^{\infty}dt'f(t')f(t-t')$ be the self-convolution of $f$.
I am wondering if there are strong bounds on $\beta[f\star f](\tau)$ given by $\beta[f](\tau)$. It is relatively easy to show that $\beta[f\star f](\tau)\leq\beta[f](\tau)$, and some other bounds follow directly from this together with $\beta[f](\tau+\tau')\leq\beta[f](\tau)+\beta[f](\tau')$.
I am more looking for something like
$\beta[f\star f](\tau)\leq\beta[f](\tau')$
for some $\tau'<\tau$. Preferably $\tau'$ would be some function of $\tau$ and maybe some property of $f$, like the variance. In other words, if the integral over the self-convolution achieves a certain value over integrals of a given length, then I want to show that the original function achieves this value for strictly smaller intervals.
The intuition for this comes from the fact that under convolution, the variance increases, i.e. $\sigma_{f\star f}^2=2\sigma_{f}^2$. However one cannot derive bounds on integrals over some intervals just based on the variance, see my previous question. I am still hoping that such a bound exists nonetheless for the convolution.
If $f$ needs properties such as continuity or similar for this to hold, then they can be assumed for now.