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.

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    $\begingroup$ 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? $\endgroup$ Mar 14 at 18:59
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    $\begingroup$ By the way, one actually has $\rho(P) = \rho(P^*)$ for every bounded linear operator $P$ on a Hilbert space. $\endgroup$ Mar 14 at 19:00
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    $\begingroup$ @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: $(Pf)(x) := \int_\Omega f(y) P(x,dy)$. $\endgroup$
    – Sam OT
    Mar 16 at 10:28
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    $\begingroup$ 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. $\endgroup$
    – Sam OT
    Mar 16 at 10:30
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    $\begingroup$ 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! $\endgroup$
    – Sam OT
    Mar 16 at 10:31

1 Answer 1


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 $n\to\infty$.

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    $\begingroup$ 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." $\endgroup$
    – Sam OT
    Mar 16 at 10:53
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    $\begingroup$ Indeed, thanks, this is now corrected. Is this not a counter-example to what you asked ? $\endgroup$
    – DRJ
    Mar 16 at 14:39
  • $\begingroup$ 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... $\endgroup$
    – Sam OT
    Mar 16 at 17:44
  • $\begingroup$ So, to be clear, my chain is nothing like yours. You have shown, very clearly, that a generic statement that I'd want doesn't hold. I guess I need to dig more into the details of my chain. I wanted to avoid that! 🙁 $\endgroup$
    – Sam OT
    Mar 16 at 17:49

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