Let the conductance profile of a random walk, $\phi(u)$, be defined as:
$$\phi_S:=\frac{Q(S,S^c)}{\pi(S)}, \phi(u):=\inf_{S:\pi(S)\leq u} \phi_S$$
where $S\subset \Omega$, $\Omega$ is a finite state space, $Q(S,S^c)=\sum_{x\in S,y\in S^c}\pi(x)P(x,y)$ and $\pi$ the stationary distribution of a random walk.
Now consider a lazy random walk on a $n\times n$ square i.e. there is a holding probability of $1/2$ at each state state and the walk moves to nearest neighbors with equal probability. There is a claim in a paper(Page $249$, Example $1$)I'm reading that for $0\leq u\leq 1/2$, which I don't understand why? : $$\phi(u)\geq\frac{a}{n\sqrt{u}}$$ where $a$ is a constant.
My attempt: Let $u=1/2$, we cut the square into two almost equal pieces by a vertical line segment and consider one of the pieces to be $S$. $\pi$ is uniform and so I see that for this specific $S$, $Q(S,S^c)=O(1/n)$ but am not sure why it is also true for the $S$ that minimizes $\phi_S$ in the definition of $\phi(u)$ and also not sure where we are getting $\sqrt{u}$ from?
$Q(S,S^c)$ can be interpreted as the boundary of the set $S$ and $\pi(S)$ as it's volume. One needs to convince themselves that the minimum in the definition of $\phi(u)$ happens for a square. If we know that, then the volume of the square set $S$ for which minimum is attained, in the definition of $\phi(u)$, is $n^2\times u$ assuming $\pi$ is uniform ($\pi$ is in fact almost uniform) and the boundary or surface area is $4n\times \sqrt{u}$ and thus
$$\phi(u)\geq \frac{4n\times \sqrt{u}}{n^2\times u}=\frac{a}{n\sqrt{u}}$$