Spectrum of a Markov kernel acting on $L^2$

Question

Let $P$ be a Markov kernel on a measurable space $(E,\mathcal E)$ admitting an invariant probability measure $\pi$. $P$ acts on $L^2(\pi)$ via $$Pf:=\int\kappa(\;\cdot\;{\rm d}y)f(y).$$ The invariance means that $\int\kappa f\:{\rm d}\pi=\int f\:{\rm d}\pi$. Let $L^2_0(\pi):=\left\{f\in L^2(\pi):\int f\:{\rm d}\pi=0\right\}$. I've read the following:

The first equality in (22.2.3) holds since they've argued that $L^2_0(\pi)$ is a reducing subspace for $P$. But how does the second equality follow? Moreover, I've often read that $\operatorname{Spec}\left(P\mid L^2_0(\pi)\right)\subseteq[-1,1)$ (so, $1$ is excluded from the spectrum when restricting to $L^2(\pi)$. How does this follow?

Note that $$U:L^2(\mu)\to L^2(\mu)\;,\;\;\;f\mapsto\langle1,f\rangle_{L^2(\mu)}1$$ is an orthogonal projection with $\mathcal N(U)=L^2_0(\mu)$. So, $1-U$ is the orthogonal projection of $L^2(\mu)$ onto ${\mathcal R(U)}^\perp=L^2_0(\mu)$. Now, if $\lambda\in\mathbb R$, then $\lambda-\left.P\right|_{{L^2_0(\mu)}^\perp}$ is injective if and only if \begin{equation}\begin{split}\{0\}&=\mathcal N\left(\lambda-\left.P\right|_{{L^2_0(\mu)}^\perp}\right)\\&=\left\{g\in\mathcal R(U):(\lambda-P)g=0\right\}\\&=\left\{Uf:f\in L^2(\mu)\text{ and }(\lambda-P)Uf=0\right\}\\&=\left\{Uf:f\in L^2_0(\mu)\text{ and }(\lambda-P)Uf=0\right\}\\&\;\;\;\;\;\;\;\;\;\;\;\;\uplus\left\{Uf:f\in L^2(\mu)\setminus L^2_0(\mu)\text{ and }(\lambda-P)Uf=0\right\}\\&=\left\{0\right\}\uplus\left\{\langle1,f\rangle_{L^2(\mu)}1:f\in L^2(\mu)\setminus L^2_0(\mu)\text{ and }\lambda=1\right\}\\&=\left\{0\right\}\uplus\left\{c:c\in\mathbb R\setminus\{0\}\text{ and }\lambda=1\right\}\\&=\begin{cases}\mathbb R&\text{, if }\lambda=1\\\{0\}&\text{, otherwise}\end{cases},\end{split}\tag1\end{equation} where we've used that $P1=1$ (and we treat $c\in\mathbb R$ as the constant function $E\ni x\mapsto c$).

So, we can conclude that $\lambda\in\mathbb R$ is contained in the point spectrum of $\left.P\right|_{{L^2_0(\mu)}^\perp}$ if and only if $\lambda=1$. How can we conclude?

Answer 1

It seems the piece you missed, which is implicit after (22.2.2) is that $L^2_0(\pi)^\perp=\langle \boldsymbol{1}\rangle$ (indeed $L^2_0(\pi)$ is a hyperplane, so that $L^2_0(\pi)^\perp$ is a line, and since $\int f\boldsymbol{1} \,\mathrm{d}\pi = \int f \,\mathrm{d}\pi = 0$ for all $f\in L^2_0(\pi)$ we have $L^2_0(\pi)^\perp=\langle \boldsymbol{1}\rangle$).

Then the restriction of $\mathrm{P}$ to this space is the identity, and its spectrum is $\{1\}$.

Answer 2

Two remarks too long to be posted as comments.

1. To claim that the spectrum of a Markov operator with respect to a finite stationary measure is real one has to assume that it is self-adjoint (equivalently, that the chain is reversible). For such operators geometric ergodicity is indeed equivalent to the fact that 1 does not belong to the spectrum in $L^2_0$ (since the spectrum is closed, it means that there is a spectral gap separating the spectrum from 1).

2. If the state space is infinite, then it is well possible that there is no spectral gap in spite of irreducibility of the operator. There are examples like this even for countable state spaces.