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 Markov chain on a state space $\Omega := \{ x \in \mathbb R^n \mid \sum_i x_i = 0 \}$..

The support of the jumps is discrete: let $P(x,y) := \mathbb P_x(X_1 = y)$; then, $\sup_{x \in \mathbb R^n} | \{ y \in \mathbb R^n \mid P(x,y) > 0 \} | < \infty$.

Non-zero jump probabilities $P(x,y) \in [0,1]$ are uniformly bounded away form $0$ and $1$.

It is irreducible: given $x \in \Omega$ a positive-Lebesgue-volume set $A \subseteq \Omega$, there is a finite path $x = y_0, y_1, ..., y_k \in A$ such that $P(y_{\ell-1}, y_\ell) > 0$ for all $\ell$.

It has a unique invariant distribution, which I denote $\pi$.

Importantly, $P$

*is not*reversible wrt $\pi$—ie, $P$ is not self-adjoint: $P \ne P^\star$.

I also view $P$ as an operator on the set of functions: $(Pf)(x) := \int_\Omega f(y) P(x, dy)$. (If $\Omega$ were discrete, then $(Pf)(x) = \sum_{y \in \Omega} P(x,y) f(y)$ in the usual matrix–vector way.)

Constant functions are always eigenfunctions of $P$ with eigenvalue $1$. So, the spectral radius and norm of $P$ on $L^2(\Omega, \pi)$ is always $1$. I thus restrict to the subspace of $L^2(\Omega, \pi)$ orthogonal to the constant functions—as I believe is standard?—which I denote $B$.

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 $T = T^\star$.

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.

wasn't truefor $\rho$. Thanks! $\endgroup$