Let $(E,\mathcal E,\mu)$ be a probability space and $\kappa$ be a Markov kernel on $(E,\mathcal E)$ such that $\mu$ is reversible with respect to $\kappa$ and hence $$\kappa f:=\int\kappa(\;\cdot\;,{\rm d}y)f(y)\;\;\;\text{for }f\in L^2(\mu)$$ is self-adjoint.
Assume there is a $\rho\in[0,1)$ with $$\int(\kappa f)^2\:{\rm d}\mu\le\rho^2\int f^2\:{\rm d}\mu\tag1$$ for all $\mathcal E$-measurable $f:E\to[0,\infty)$ with $\int f\:{\rm d}\mu=0$. Let $f\in\mathcal L^2(\mu)$ with $f\ge0$ and $$f_n:=\kappa^nf-\int f\:{\rm d}\mu\;\;\;\text{for all }n\in\mathbb N.$$ I want to show that $$\sum_{n=0}^\infty\left\|f_n\right\|_{L^2(\mu)}<\infty.\tag2$$
By $(1)$, $$\int(\kappa^nf)^2\:{\rm d}\mu=\int\left(\kappa(\kappa^{n-1}f)\right)^2\le\rho^2\int(\kappa^{n-1}f)^2\:{\rm d}\mu\le\cdots\le\rho^{2n}\int f^2\:{\rm d}\mu\tag3$$ for all $n\in\mathbb N$. The problem is now the additional summand in the definiton of $f_n$, but noting that $$\int\kappa^nf\:{\rm d}\mu=\int f\:{\rm d}\mu\tag4,$$ we may be able to fix this.