A recent question asked whether, given a finite set of positive numbers $A$, it is always the case that the set $A+A-A$ has more positive than negative elements. Terry Tao showed that this is false (and can fail in a very strong sense), but is true if one counts with multiplicity. My question is whether there is a method of counting with partial multiplicity which is critical.
To make this precise, it is helpful to reframe this problem in terms of convolutions. For this, we take a finite set $A$ and consider the convolution $1_{A+A-A} = 1_A \ast 1_A \ast 1_{-A}$. In this language, the first question is equivalent to the integral inequality $$ \int_{-\infty}^0 (1_{A+A-A})^0 \, d \mu \leq \int_{0}^\infty (1_{A+A-A})^0 \, d \mu, $$ where the integrals are done with respect to the counting measure and we use the convention $0^0 =0$.
On the other hand, counting with multiplicity in the usual sense corresponds to showing that $$ \int_{-\infty}^0 1_{A+A-A} \, d \mu \leq \int_{0}^\infty 1_{A+A-A} \, d \mu. $$ And the latter inequality is straightforward to establish because $A-A$ is a symmetric set and adding $A$ shifts it toward the positive numbers).
This raises the possibility of counting the solutions with partial multiplicity. More precisely, is there a $p < 1$ so that $$ \int_{-\infty}^0 (1_{A+A-A})^p \, d \mu \leq \int_{0}^\infty (1_{A+A-A})^p \, d \mu $$ for all finite sets $A$? Furthermore, is there a critical value $p^\ast$ so that for any $q<p^\ast$, we can find finite sets so that $$ \int_{-\infty}^0 (1_{A+A-A})^q \, d \mu \geq \int_{0}^\infty (1_{A+A-A})^q \, d \mu $$ (or even $\geq C \int_{0}^\infty (1_{A+A-A})^q \, d \mu$ for arbitrarily large constant $C$?).
