# Spectral Radius and Spectral Norm for Markov Operators

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.

• Could you please clarify a few points in your post? (i) What do you mean by a Markov operator on the state space $\Omega = \mathbb{R}^n$? In particular, on which space does the operator act a priori? (ii) Which definition of irreducibility do you use? (iii) What do you mean by 'discretely supported'? (iv) Why is $B$ invariant under the adjoint $P^*$? (v) You seem to refer to three different objects by $P$: the Markov operator with which you start; its restriction to $L^2(\Omega,\pi)$; and, in the three bullet points, to somekind of integral kernel of $P$. Could you clarify the notation? Mar 14 at 18:59
• By the way, one actually has $\rho(P) = \rho(P^*)$ for every bounded linear operator $P$ on a Hilbert space. Mar 14 at 19:00
• @JochenGlueck Certainly! (i) $P$ is the transition matrix of a Markov chain on $\Omega = \mathbb R^n$, so $P(x,y)$ is probability of being at $y \in \mathbb R^n$ after one step started at $x$; I thought people call these "operators" when $\Omega$ is continuous, not discrete, but maybe I was mistaken... (ii) Irreducibility here means that one can hit any positive-volume set. (iii) That is given near the end of the question; see the three bullets. (iv) I do not know what that means, sorry... 😬 (v) This is a fairly standard abuse, in Markov chains at least: $(Pf)(x) := \int_\Omega f(y) P(x,dy)$. Mar 16 at 10:28
• To add to (iv), I thought one typically looked at $P$ on $L^2$ restricted to non-constant functions when looking at spectral gaps. Indeed, $P$ always has constant functions as eigenfunctions with eigenvalue $1$, so $\rho(P) = 1$ if you allow constant functions. || I'll add the above to the body of the question, as well as make one other correction: $\Omega$ should be $\{ x \in \mathbb R^n \mid \sum_i x_i = 0 \}$. But, I guess that doesn't really change anything. Mar 16 at 10:30
• Good to know that $\rho(P) = \rho(P^\star)$, though. I knew that $\| P \| = \| P^\star \|$, but I actually thought that it wasn't true for $\rho$. Thanks! Mar 16 at 10:31

If I understood correctly, you are asking whether the spectral gap $$\gamma=1-\rho$$ of a non-reversible Markov chain $$P$$ provides any universal control on the Poincaré constant (which is the spectral gap of the additive reversibilization of $$P$$, or in your notation, $$1-\|P\|$$). The answer is no, even on finite state spaces: consider the Markov chain on $$\{0,1\}^n$$ which, at each step, replaces the current state $$x=(x_1,\ldots,x_n)$$ with either $$(x_2,\ldots,x_{n},0)$$ or $$(x_2,\ldots,x_{n},1)$$, each with probability a half. This chain has the maximum possible spectral gap, namely $$\gamma=1$$. Yet, its Poincaré constant tends to $$0$$ as $$n\to\infty$$.
• Thanks, @DRJ, for your answer. I'm a bit confused/sceptical of a couple of things. Before diving into those, perhaps you can correct a typo in your example? For you, $x \in \{0,1\}^n$, but then the two potential states are both in $\{0,1\}^{n-1}$. Were they supposed to be $(x_2, ..., x_n, 0/1) \in \mathbb R^n$? In words, "Kill first bit and append a uniform bit." Mar 16 at 10:53
• I think it is, unfortunately! I was thinking that any matrix $P$ with eigenvalues $1$ (constant vector) and $0$ (multiplicity $n-1$) must have all rows equal. But, that's only true if you require $P$ to be self-adjoint! (and so has a spectral decomposition). I guess I should just check that the spectral gap of the additive symmetrisation does indeed tend to $0$, but I guess this should be easy... Mar 16 at 17:44