For an $(n,d,\lambda )$-graph with $\lambda <d$, every nontrivial cut has many crossing edges

Let ${\mathbf{1}}_S$ be the indicator vector of $S$, and put $x:={\mathbf{1}}_S-\frac{\mathrm{\mid }S\mathrm{\mid }}{n}\mathbf{1}$. Then $x\mathrm{\ne }0$ because $S$ is nonempty and proper, and $x$ is orthogonal to $\mathbf{1}$. Since $G$ is $d$-regular by [F1], the vector $\mathbf{1}$ is an adjacency eigenvector with eigenvalue $d$, so [L1] gives $x^{\mathsf{T}}Ax\le \lambda x^{\mathsf{T}}x$.

A direct computation gives $x^{\mathsf{T}}x=\frac{\mathrm{\mid }S\mathrm{\mid }(n-\mathrm{\mid }S\mathrm{\mid })}{n}$ and $x^{\mathsf{T}}Ax=d\frac{\mathrm{\mid }S\mathrm{\mid }(n-\mathrm{\mid }S\mathrm{\mid })}{n}-e(S,V(G)\setminus S)$, because ${\mathbf{1}}_S^{\mathsf{T}}A{\mathbf{1}}_S$ counts twice the edges internal to $S$, while ${\mathbf{1}}_S^{\mathsf{T}}A\mathbf{1}=d\mathrm{\mid }S\mathrm{\mid }$. Substituting these expressions into step 1.1 yields $d\frac{\mathrm{\mid }S\mathrm{\mid }(n-\mathrm{\mid }S\mathrm{\mid })}{n}-e(S,V(G)\setminus S)\le \lambda \frac{\mathrm{\mid }S\mathrm{\mid }(n-\mathrm{\mid }S\mathrm{\mid })}{n}$, which rearranges to the claimed cut bound.

If $\lambda <d$ and $G$ were disconnected, [L2] would provide a connected component $C$ with $\mathrm{\varnothing }\mathrm{\ne }C\mathrm{\ne }V(G)$ and $e(C,V(G)\setminus C)=0$. But step 2.1 would then force $0\ge \frac{(d-\lambda )\mathrm{\mid }C\mathrm{\mid }(n-\mathrm{\mid }C\mathrm{\mid })}{n}>0$, a contradiction. So $G$ is connected.

Step 2.1 gives the edge-expansion inequality, and step 3.1 gives the connectedness consequence.