I was going through the lectures of Graph Partitioning and Expanders - Stanford Online. In lecture 1, near the end of page 5, I came across this inequality for regular graphs: $$\phi(S) \leq h(S) \leq 2\cdot \phi(S)$$ where $\phi(S)$ is the sparsity of graph and $h(S)$ is the edge expansion of the graph.
However, I fail to see how this arises. I think it should be $h(S) \leq \phi(S)$. Is the result given in the lecture notes correct?
After getting no response for a long time, I have gone through the lecture once more, and I am pretty confident that the result given in the lecture is incorrect. The sparsity of a regular graph for a given partition is $\phi (S) = \frac{E(S,V-S)}{\frac{d}{|V|}\cdot |S| \cdot |V-S|}$ and the edge expansion of the graph is $h(S) = \frac{E(S,V-S)}{d \cdot min\{|S|,|V-S|\}}$. As for every partition $\{S,V-S\}$, $$|V|\cdot min\{|S|,|V-S|\} \ge |S|\cdot |V-S| $$ hence,$$\phi (S) \geq h(S)$$Also, because $$2 \cdot max\{|S|,|V-S|\} \geq |V|$$ $$\therefore |V|\cdot min\{|S|,|V-S|\} \le 2\cdot |S|\cdot |V-S|$$ $$\therefore 2\cdot h(S) \geq\phi(S)$$ And, as $h(S)\leq\phi(S)\leq2\cdot h(S)$ for every partition $\{S,V-S\}$, $$\therefore h(G) \leq \phi(G) \leq 2\cdot h(G)$$ PS: However, as the lectures belong to a respectable source, I would still like to hear someone else's opinion on this result.