Let $(E,\mathcal E,\mu)$ be a probability space and $\kappa$ be a Markov kernel on $(E,\mathcal )$ symmetric$^1$ with respect to $\mu$. I'm interested in the quantity $$\inf_{\substack{g\in L^2(\mu)\setminus\{0\}\\g\ge0}}\frac{\left\|\psi g\right\|_{L^2(\mu\otimes\kappa)}^2}{\left\|g\right\|_{L^2(\mu)}^2}\tag1,$$ where $$(\psi g)(x,y):=g(x)-g(y)\;\;\;\text{for }x,y\in E\text{ and }g:E\to\mathbb R,$$ which is a bounded linear operator from $L^2(\mu)$ to $L^2(\mu\otimes\kappa)$.
Is the dependence of the fraction on $g$ concave/convex? It's clear to me that any norm is convex though. If we would consider the supremum instead of the infimum, $(1)$ would be equal to the supremum over only those $g$ with $\left\|g\right\|_{L^2(\mu)}=1$. Is there a similar result for the infimum?
$^1$ i.e. $$\int\mu({\rm d}x)\int\kappa_w(x,{\rm d}y)f(x,y)=\int\mu({\rm d}y)\kappa_w(y,{\rm d}x)f(x,y)$$ for all bounded $\mathcal E^{\otimes2}$-measurable $f:E^2\to\mathbb R$.