Anticoncentration of the convolution of two characteristic functions 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.
 A: 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 (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 $k\approx(3|A|-p)/2$.
A: It means that in some sense, the Polard inequality is a quantiative estimate on the following behavior: 
Qualitative Statement. If $A,B\subset \mathbb{Z}_p$ are big enough, then the convolution $\mu_{A}*\mu_{B}$ is (in some sense) not heavily-tailed.
Quantitative Statement. Let $|A|=|B|=\delta$ be subset of $\mathbb{Z}_p$. Then
$$ \mathbf{E}_{x\sim\mu}r(x)\mathbf{1}_{r(x) \geqslant t}(x) \leqslant \psi(t) := \delta^2 p - t \left(2\delta-\frac{t}{p}\right) \xrightarrow{t\to \delta p} 0  \quad (\star) $$
This is equivalent to the inequality in the Pollard Theorem.
Having such an estimate, for every $C\subset \mathbb{Z}_p$ big enough should be (with good constant) $$ \mathbf{E}_{x\sim\mu_C} r(x) \approx \mathbf{E}_{x\sim\mu} r(x)  $$
And to finish, we use the following simple estimate 
$$ \mathbf{E}_{x\sim\mu} r(x)\mathbf{1}_{C}(x)  \geqslant \mathbf{E}_{x\sim\mu} r(x) - \psi(t)-t\cdot \mathbf{P}(C^c) $$
with suitably chosen $t$. In our case this is $t\left(3\delta-1-\frac{t}{p}\right) +t(1-\delta) $
