TITLE: Nontrivial expansion in sumsets
QUESTION [5 upvotes]: Let $A \subset \mathbb{Z}/p$, let $f$ be a function on $\mathbb{Z}/p$ and let $B:=\{f(a): a \in A\}$.
Can we conclude that $|A+B|$ is large if $f$ is a sufficiently "nice" function?  For instance say that $f(a)=a^2$.  Then can we say that if $|A|=1000\sqrt{p}$ or even if $|A|=\frac{p}{\log^{*}(p)}$, then $|A+B| \ge p/100$?  
The idea should be that $A+B$ is only small if $A,B$ are related by some additive structure, and if $f$ is a sufficiently random function then it should kill any structure so that $A+B$ should be large.

REPLY [6 votes]: Here is a partial "yes", to complement Sam Zbarsky's negative answer to the question.
There are a bunch of different papers on this topic focusing on different functions $f$ and in different ranges. To simplify things, I will focus on $f(x)=x^2$, since you mentioned it.
For $|A|< p^{5/8}$, a result of Pham, Vinh and de Zeeuw gives $|A+f(A)| \gg |A|^{6/5}$.
For larger sets, I cannot find a reference from the top of my head, but I am quite confident that exponential sum techniques could prove that
\begin{equation} \label{claim}
|A+f(A)| \gg \min \left \{ \sqrt{p|A|}, \frac{|A|^2}{\sqrt{p}} \right \}.
\end{equation}
See for example Theorem 7 in this paper of Balog, Broughan and Shparlinksi, which implies this result for $f(x)=x^{-1}$. I am pretty sure that I have seen this result with $f(x)=x^2$ in the literature at some point, but cannot remember where right now. If this claim is correct then it is optimal, as you can construct a set $A$ of any given size with $|A+A^2| \ll \sqrt{|A|p}$.