# Expander mixing lemma in combinatoric expanders

There is a well-known relation between combinatoric expansion and the gap between the first and the second largest eigenvalues (Dodziuk 1984) : $$h(G) \le d \sqrt{2 (1 - \alpha)}$$ where $$h(G) = \min\limits_{S\subseteq V;\; |S|\le|V|/2} \frac{|E(S,V\setminus S)|}{|S|},$$ $$\lambda_2 = \alpha \lambda_1 = \alpha d$$ and $$E(X,Y) = E \cap (X \times Y)$$.

The problem is that this connection holds only for $$d$$-regular graphs, for non-regular graphs the largest eigenvalue is harder to compute.

However the expansion $$h(G)$$ is defined for non-regular graphs. $$G$$ is $$(n/2, d, c)$$-expander if $$h(G) \ge c$$ and all degrees of the vertices of $$G$$ is at most $$d$$. I would like to somehow convert such graph into an algebraic expander. What I really need from an algebraic expander that combinatoric one lacks is the mixing lemma: $$\left|{|E(S,T)| \over |V|} - {d |S| |T| \over |V|}\right| \le \alpha d \sqrt{|S||T|}.$$

What I can do is add loops to $$G$$ such that it is $$d$$-regular. This modification does not affect $$h(G)$$ so the resulting graph $$G'$$ is $$(n,d,\alpha)$$ algebraic expander for some $$\alpha$$. Therefore the mixing lemma works for $$G'$$. But for disjoint $$S$$ and $$T$$ $$E_G(S,T) = E_{G'}(S,T)$$ so the mixing lemma works for disjoint $$S$$ and $$T$$ in $$G$$ as well.

Is there a flaw in this argument? And is there a better way to extract an algebraic properties from a combinatoric expander? Is there a way to get mixing lemma for non-disjoint sets?

Dodziuk, Jozef, Difference equations, isoperimetric inequality and transience of certain random walks, Trans. Am. Math. Soc. 284, 787-794 (1984). ZBL0512.39001.