Cheeger constant on regular trees and relation with amenability Let $G$ be a graph. The Cheeger constant (or isoperimetric number) $h(G)$ of $G$ is
$$h(G) = \inf_S\dfrac{|\partial S|}{|S|},$$
where $S$ is a finite nonempty set of vertices in $G$ and $\partial S,$ the boundary od $S,$ consists of all vertices in $V\backslash S$ that have a neighbor in $S.$
A graph $G$ is called amenable if $h(G)=0.$
QUESTION 1: I would like to show that the Cheeger constant for a k-regular tree is k-2, that is given a k-regular tree $T_k$ (all vertices of $T_k$ have a degree of $k+1)$ how can I prove that $h(T_k) = k-2?$
QUESTION 2: What is the relation between Chegeer constant and amenability? In other words, how can I use the Cheeger constant for show that a graph $T_k \times \mathbb{Z}$ is nonamenable?
 A: Here is an answer to the first question.  First of all note that to obtain the result, $T_d$ needs to be an infinite tree and consequently the definition of the Cheeger constant is adjusted so that the infimum is taken over subsets of $V$ that are finite.
The first step is to note that we need only consider subsets $W$  that induce connected subgraphs of $T_d$.  Suppose otherwise that $W$ induces a subgraph of $T_d$ that is disconnected into two components $W_1$ and $W_2$.  Since $T_d$ is a tree with no cycles there is one path exactly between $W_1$ and $W_2$ with a $v_1$ in $W_1$ at the one end of the path and a vertex $v_2$ in $W_2$ at the other end of the path.  Now consider a subset $W'$ that induces the same components $W_1$ and $W_2$ but with $v_1$~$v_2$.  This is possible because of the regular structure of $T_d$.
$|W'|=|W_1|+|W_2|=|W|$ but $|\delta  W'|\leq|\delta  W|-2$ as we have connected $v_1$ to $v_2$.  From which we see we have found a connected induced subgraph with $|\delta  W|/|W|$ less than the disconnected induced subgraph on the same number of vertices.  The general case for multiple disconnected components follows by a simple induction.
Now let $W \subset V$ be finite and induce a connected subgraph and let $v_0\in W$.  Let $W_k \in W$ be the subset of all vertices of $W$ at distance $k$ of $v_0$.  Since $W$ is finite and induces a connected subgraph there exists $N \in \mathbb N$ such that $W_k = \varnothing$ for $k\geq N$.  We have:
$|W|=1+|W_1|+...+|W_N|$
$|\delta W_1|=d-|W_1|$, $|\delta W_i|=(d-1)|W_{i-1}|-|W_i|$ for $i=2,...,N-1$ and $|\delta W_N|=(d-1)|W_{N-1}|$.  So,
$|\delta W|=| \delta W_1|+....+|\delta W_N|$=$d+(d-2)|W_1|+...+(d-2)|W_{N-1}|$.  Hence,
${|\delta W| \over |W|}$=$2+(d-2)(1+|W_k|+...+|W_{n-1}|) \over 1+|W_k|+...+|W_{n-1}|$=$(d-2)+ \frac{2}{|W|}$
Taking the limit as |W|$\rightarrow \infty$ then gives $h(T_d)=d-2.$
