Have you seen the following statement proven anywhere?


Let $G$ be a strongly regular graph with parameters $(n,k,\lambda,\mu)$ with $\lambda,\mu>0$. Then there is no set $A$ of at least $n/4$ vertices in $G$ such that the neighborhood of $A$ is of size less than $3n/4$.


A vertex in $A$ in this case would be included in the neighborhood of $A$ iff it is adjacent to another vertex in $A$. It seems like this should be known, if it's true.



Expansion in strongly regular graphs
Consider the complete tripartite graph $G = \langle X\cup Y\cup Z,E\rangle$ with $|X|=|Y|=|Z|=\frac{n}{3}$, and $E = X\times Y\cup X\times Z \cup Y\times Z$. This graph is strongly (n,2n/3,n/3,2n/3)-regular, and e.g. $X$ has more than $\frac{n}{4}$ vertices and less than $\frac{3}{4}n$ neighbors, so this is a counterexample to your conjecture.


On the other hand, since the spectrum of strongly regular graphs is known precisely (see <https://en.wikipedia.org/wiki/Strongly_regular_graph#Eigenvalues>), one can get bounds on the edge expansion of such a graph via Cheeger's inequality, which implies results like the one you conjecture.


Question: What is the minimal $\alpha>0$ such that in every strongly regular graph, every set $S \subseteq V$ whose size is at least $\alpha n$ has at least $(1-\alpha) n$ neighbors, and which strongly regular graph achieves this $\alpha$?


This seems like a natural question, but I wouldn't be surprised if no one had considered it. It looks like a good research question to me. My example implies $\alpha \geq 1/3$.


