My question concerns differences between the spectral radius $\rho$ and norm $\| \cdot \|$ of Markov operators in infinite-dimensional Banach spaces. This is far from my area of expertise, that is more "mixing for finite, typically reversible, Markov chains", where spectral properties are often simpler.

I have a uniformly-bounded and discretely-supported (see below for more details), irreducible Markov operator $P$ on a state space $\Omega = \mathbb R^n$ with (unique) equilibrium distribution $\pi$. Importantly, $P$ is not reversible wrt $\pi$. In other words, $P$ is not self-adjoint: $P \ne P^\star$. Let $B$ denote the subspace of $L^2(\Omega, \pi)$ orthogonal to the constant functions.

I have a 'spectral-gap' bound $\rho(P) \le 1 - \kappa$ of $P$ on $B$, for some $\kappa > 0$. I want to show something along the lines of

$\rho\bigl( \tfrac12 (P + P^\star) \bigr) \le \tfrac12 \bigl( 1 + \rho(P) \bigr) \le 1 - \kappa/2$.

Recall that $P$ is a Markov operator, so $\rho(P) \le \| P \| \le 1$. I'm not bothered about constants on $\kappa$.

Highly relevant is this MO question, but it doesn't address quite what I want here.

I know the standard facts.

• $\rho(T) = \inf_{k\ge1} \| T^k \|^{1/k} \le \| T \|$ for all bounded, linear operators $T$.

• $\rho(T) = \| T \|$ if $T$ is normal ($T T^\star = T^\star T$) of which self-adjoint ($T = T^\star$) is a special case.

One approach would be to prove that $\rho(P) = \| P \|$, even though $P$ is not normal in my case. Then,

$\rho\bigl( \tfrac12 (P + P^\star) \bigr) = \| \tfrac12 (P + P^\star) \| \le \tfrac12( \| P \| + \| P^\star \| ) = \| P \| = \rho(P)$,

since $P + P^\star$ is self-adjoint, $\| P \| = \| P^\star \|$ and $\rho(T) = \| T \|$ if $T = T^\star$. The following properties hold in my case:

• $\sup_{x \in \mathbb R^n} | \{ y \in \mathbb R^n \mid P(x,y) > 0 \} | < \infty$;
• $\inf_{x,y \in \mathbb R^n : P(x,y) > 0} P(x,y) > 0$;
• $\sup_{x,y \in \mathbb R^n : P(x,y) > 0} P(x,y) < 1$.

It seems very likely that $\rho(P) = \| P \|$, in my (poorly-informed) view; how much 'nicer' do you want $P$ to be? The only concrete way I know to prove this is normality: $P P^\star = P^\star P$. But, this does not hold.

Another approach would be to relate the two spectral radii quantities more generally, at least for Markov operators. Note that $\rho$ is neither subadditive nor submultiplicative, and $\rho(T) \ne \rho(T^\star)$ for general $T$, in general.