The Nash-Williams inequality

Question

I want to understand a step in the proof of the following statement

If $\{\Pi_k\}$ are disjoint edge-cutsets which separate nodes $a$ and $z$, then $$R(a\leftrightarrow z)\geq \displaystyle\sum_k\left(\displaystyle\sum_{e\in\Pi_k}c(e)\right)^{-1} $$

proof:

Let $\theta$ be a unit flow from $a$ to $z$. For any $k$, by the Cauchy-Schwarz inequality $$\displaystyle\sum_{e\in\Pi_k}c(e)\displaystyle\sum_{e\in\Pi_k}r(e)\theta(e)^2\geq \left(\displaystyle\sum_{e\in\Pi_k} \sqrt{c(e)}\sqrt(r(e))|\theta(e)|\right)^2=\left(\displaystyle\sum_{e\in\Pi_k}|\theta(e)|\right)$$

The right-hand side is bounded below by $||\theta||^2=1$, because $\Pi_k$ is a cutset and $||\theta||=1$. Therefore $$\displaystyle\sum_e r(e)\theta(e)^2\geq\displaystyle\sum_k\displaystyle\sum_{e\in\Pi_k}r(e)\theta(e)\geq \displaystyle\sum_k\left(\displaystyle\sum_{e\in\Pi_k} c(e)\right)^2$$

By Thompson's principle, we are done.

I think that I understand everything but the bound $$\left(\displaystyle\sum_{e\in\Pi_k}|\theta(e)|\right)^2\geq 1$$ why does $||\theta||= 1$ and the fact that $\Pi_k$ are cutset implies that inequality?

Thanks!

Answer 1

Consider the set $S$ $=\{x$: a and x are connected in the graph $G-\Pi_{k}\}$ where $G$ is the original graph.

Since $\theta$ is a flow from $a$ to $z$, we can write

$$||\theta||=\underset{x\in S}{\sum}\underset{\forall y}{\sum}\theta(\overset{\rightarrow}{xy})$$

$$=\underset{x\in S}{\sum}\underset{y\notin S}{\sum}\theta(\overset{\rightarrow}{xy}) \leq\underset{e \in \Pi_{k}}{\sum}|\theta(e)|$$

Since $\Pi_{k}$ is an edge cut set therefore every flow $\theta(\overset{\rightarrow}{xy})$ must go via an edge in $\Pi_{k}.$

Given that $||\theta||=1$, we have proved that

$$\left(\underset{e\in\Pi_{k}}{\sum}|\theta(e)|\right)^{2}\geq1$$