Let $(E,\mathcal E)$ be a measurable space, $\kappa$ be a Markov kernel on $(E,\mathcal E)$, $\mu$ be a probability measure on $(E,\mathcal E)$ reversible with respect to $\kappa$ and $$L^2_0(\mu):=\left\{f\in L^2(\mu):\mu f=0\right\}.$$ We may treat $\kappa$ as a self-adjoint linear contraction on $L^2(\mu)$ via $$\kappa f:=\int\kappa(\;\cdot\;,{\rm d}y)f(y)\;\;\;\text{for }f\in L^2(\mu).$$ By contractivity, the spectrum $\sigma(\kappa)$ of $\kappa$ is contained in $[-1,1]$.
Let $\kappa_0:=\left.\kappa\right|_{L^2_0(\mu)}$. In the last paragraph on page 9 of this paper, it is claimed that if $\kappa$ is irreducible, then $\sigma(\kappa_0)\subseteq[-1,1)$. How can we prove that?
I could imagine that the guideline of the proof is as follows:
- Show that $\mathcal N(1-\kappa)=L^2_0(\mu)$ (compare with the result of Determine the orthogonal projection of $L^2$ onto $\left\{f\in L^2:\int f=0\right\}$)
- From If $\lambda$ is an eigenvalue of a self-adjoint operator, is $\lambda$ in the resolvent set of $\left.A\right|_{{\mathcal N(\lambda-A)}^\perp}$? we can conclude t that $1-\kappa_0$ is injective and hence $1$ is not an eigenvalue of $\kappa_0$
- We somehow need to conclude (probably using the irreducibility) by Theorem 12.29-c in Rudin's Functional Analysis:
How can we fill the gaps?