**Edit:** This is a question related to my other post, stated in a much more concrete way I think.


I am interested in anything (ideas, references) related to the following problem:


Suppose that $A \subset \mathbb{Z}\_p$ is a set of size $\delta p$ for $\delta > \frac{1}{2}$ (i.e. relatively big). What is known about the distribution of $\mu\_A\*\mu\_A$ compared to $\mu\_A$? In particular, what is known about 
$$\left\|\mu\_A - \mu\_A\*\mu\_A \right\|\_{\ell\_1(G)}$$


Taking $A = \left(\frac{p}{3},\frac{2p}{3}\right)$ we obtain that $A+A$ could be almost completely disjoint with $A$. This is however only possible if $\delta < \frac{1}{3}$ in view of the Cauchy-Davenport Theorem.


But even for $\delta>\frac{1}{3}$, it might still happen that $\mu\_A\*\mu\_A$ puts a lot of its mass outside of $A$ and the distance is large even if $A+A$ intersects $A$. Do you know any concrete quantitative bounds?


**Motivation:** The upper bound would give us the bound on the rate of convergence of $n$-fold convolution of every measure on $\mathbb{Z}\_p$, that has is $\delta$-close to the uniform measure in the total variational distance.



Anticoncentration of the convolution of two characteristic functions
I assume you use the counting norm to define the convolution, as otherwise the $\ell\_1$-norms of $\mu\_A$ and $\mu\_A\ast\mu\_A$ are just of different order of magnitude. Thus, $(\mu\_A\ast\mu\_A)(g)=|A|^{-2}r(g)$, where $r(g)$ is the number of representations of $g$ as a sum of two elements of $A$. We therefore have 
 $$ \|\mu\_A-\mu\_A\ast\mu\_A\|\_{\ell\_1} = \sum\_{g\in G} \big|1\_A(g)/|A|-r(g)/|A|^2\big|, $$
and an easy computation shows that this is equal to 
 $$ 2-\frac2{|A|^2}\sum\_{a\in A}r(a). $$
You thus ask how small can the sum $\sum\_{a\in A} r(A)$ be.


As you remarked, if $\delta<1/3$, then $2A$ can be disjoint with $A$, in which case the sum vanishes. If $\delta>1/3$ then taking $A$ to be the interval of length $\delta p$ centered around $p/2$ makes the intersection of $A$ and $2A$ to be of size about $(3\delta-1)p$, and the sum in question about $2(1+2+\dotsb+((3\delta-1)/2)p)\approx\frac14(3\delta-1)^2p^2$. Here is an argument showing that this is the worst-case scenario; that is, our sum is *always* at least as large as $\sim\frac14(3\delta-1)^2p^2$.


For each $i\ge 1$, let $S\_i:=\{g\in G\colon r(g)\ge i\}$, and fix an integer $k\ge 2|A|-p$. We have then
 $$ \sum\_{a\in A} r(a) = \sum\_{i\ge 1} |S\_i\cap A| 
 \ge \sum\_{i=1}^k (|S\_i|+|A|-p) = \sum\_{i=1}^k |S\_i| - k(p-|A|). $$
Now, a well-known [theorem of J. Pollard](http://jlms.oxfordjournals.org/content/s2-8/3/460.full.pdf) (incidentally, the uncle of J. Pollard serving his life sentence in a North Carolina prison for passing classified information to Israel) says that the sum in the right-hand side is at least $k(2|A|-k)$. It follows that
 $$ \sum\_{a\in A} r(a) \ge k(3|A|-k-p), $$
and to complete the proof we just choose $k\approx(3|A|-p)/2$.


