Why is the log-Sobolev constant of a finite Markov chain attained?

Question

I am reading through Saloff-Coste's "Lecture notes on finite Markov Chains". He defines the log-Sobolev constant $\alpha$ of a finite Markov Chain $(K, \pi)$ [transition kernel $K$, stationary measure $\pi$] by

$$ \alpha(K) = \min \left\{ \frac{\mathcal E(f, f)}{\mathcal L(f)} : \mathcal L(f) \ne 0\right\}$$

where $\mathcal E(\bullet, \bullet)$ is the Dirichlet-form of $(K, \pi)$, and $\mathcal L(f) = \textrm{E}_\pi |f|^2 \log \frac{|f|^2}{\|f\|_2^2}$.

Under the assumption of the existence of a function $u$ which attains this minimum, he proves that $\alpha > 0$, but I don't see why such a function should exist (since the set we are optimising on isn't even closed).

[[I am asking, why is this a min, and not an inf in the definition.]]

Thanks!

Answer 1

A scaling argument shows that $$ \alpha(K)=\inf\{\mathcal E(f,f):\mathcal L(f)=1\}. $$ Now you are minimizing a continuous function over a compact set -- because the setting is finite dimensional.