Criteria of Spectral Gap for Markov Operators

Introduction

Let $(E,\mathcal{F},\mu)$ be a probability space. Let $P$ be a Markov operator on $L^{2}(\mu)$ (i.e. $P$ is a linear operator on $L^{2}(\mu)$ such that $P1=1$ and $f\geq 0$ implies $Pf\geq 0$) such that $\mu P=\mu$ (i.e. $\mu$ is an invariant probability measure of $P$). Let $\hat{P}=\frac{1}{2}(P+P^{*})$ be the additive symmetrization of $P$, where $P^{*}$ is the adjoint operator of $P$ on $L^{2}(\mu)$. Assuming that $1$ is a simple eigenvalue of $P$, we aim to investigate the existence of spectral gap of $\hat{P}$ (i.e. $1$ is an isolated point in $\sigma(\hat{P})$, the spectrum of $\hat{P}$).

It is well known by the ergodic theorem that $1$ is a simple eigenvalue of $P$ if and only if $P$ is ergodic, i.e. for $f\in L^{2}(\mu)$,

Moreover, the ergodicity (i.e. $1$ is a simple eigenvalue) is also equivalent to the $\mu$-essential irreducibility (or resolvent-positive-improving property, see ) of $P$:

Indeed, if for some $f\in L^{2}(\mu)$ with $\mu(f)=0$ and $\mu(f^{2})=1$ such that $Pf=f$, then by the Jensen inequality we have $Pf^{+}\geq(Pf)^{+}=f^{+}$. But $\mu(Pf^{+})=\mu(f^{+})$, we conclude that $Pf^{+}=f^{+}$. Then for any $\varepsilon>0$ such that $\mu(f<-\varepsilon),\mu(f>\varepsilon)>0$, we have

so that (1.1) does not hold. On the other hand, if there exists $A,B\in\mathcal{F}$ with $\mu(A),\mu(B)>0$ such that $\mu(1_{B}P^{n}1_{A})=0$ for all $n\geq 1$, then the class

contains the non-trivial function $1_{A}$. Since the family is bounded in $L^{2}(\mu)$, we may take a sequence $\{f_{n}\}\subset\mathcal{C}$ which converges weakly to some $f\in\mathcal{C}$ such that $\mu(f)=\sup_{g\in\mathcal{C}}\mu(g)$. As $f\lor(Pf)$ is also in $\mathcal{C}$, we have $\mu((Pf)\lor f)=\mu(f)=\mu(Pf)$. Thus, $Pf=f$. Since $\mu(f)\geq\mu(A)>0$ and $\mu(1_{B}Pf)=0$, we conclude that $f-\mu(f)$ is a non-trivial eigenfunction of $P$ with respect to $1$, so that $1$ is not a simple eigenvalue of $P$.

When $P$ is symmetric, the spectrum $\sigma(P)$ of $P$ is contained in $.$ Then $P$ has a spectral gap if $P$ is ergodic and $\sigma(P)\subset\{1\}\cup[-1,\theta]$ for some $\theta\in[-1,1);$ or equivalently, the Poincaré inequality

holds. When $P$ is non-symmetric, (1.2) is equivalent to $\sigma(\hat{P}|_{\mathcal{H}_{0}})\subset[-1,\theta]$, where $\mathcal{H}_{0}:=\{f\in L^{2}(\mu):\ \mu(f)=0\}$. Thus, (1.2) holds for some $\theta\in[-1,1)$ if and only if $\hat{P}$ has a spectral gap.

Recall that $P$ is called hyperbounded if for some $p>2$

It was conjectured by Simon and H$\phi$egh-Krohn that if $P$ is symmetric, ergodic and hyperbounded, then it has a spectral gap. Although numerous papers aiming to solve this problem or to construct counterexamples have been published, see e.g. where some weaker notions such as the uniform integrability and a tail norm condition have been used to replace the hyperboundedness, the conjecture has been open for more than 40 years until Miclo found a complete proof in his recent paper .

On the other hand, there are a lot of non-hyperbounded Markov operators having spectral gap. So, in the spirit of , we shall prove a stronger statement by using a tail norm condition to replace the hyperboundedness. The tail norm we will use is the following:

According to the Schwartz inequality, $\|P\|_{\tau}$ is smaller than the following one used in :

where the las step follows by replacing $Pf$ with $f$, since $\mu(f^{2})\leq 1$ implies $\mu(Pf)^{2})\leq 1$.

According to , $P$ is called uniformly integrable if $\|P\|_{tail}=0$. Thus, the uniformly integrability implies $\|P\|_{\tau}=0.$ In particular, $\|P\|_{\tau}=0$ holds for hyperbounded $P$. We will then strengthen the above conjecture by replacing the hyperboundedness with $\|P\|_{\tau}<1,$ which is also necessary for the existence of the spectral gap of $\hat{P}$ as shown in our following main result. See Theorem 2.1 below for two more equivalent statements on isoperimetric constants for the existence of spectral gap.

Let $P$ be an ergodic Markov operator on $L^{2}(\mu)$. Then the following statements are equivalent:

The Poincaré inequality $\eqref{P0}$ holds for some constant $\theta\in[-1,1)$, i.e. $\hat{P}$ has a spectral gap.

(a) Let $P$ be symmetric. It is easy to see that if $P^{2}$ has a spectral gap then $\|P\|_{tail}<1$. Indeed, letting $\sigma(P^{2})\subset[0,\varepsilon]$ for some $\varepsilon\in(0,1)$, for $R\geq 1$ and $\mu(f^{2})\leq 1$ we have

where $\hat{f}:=f-\mu(f).$ On the other hand, if $\|P\|_{tail}<1$ then by Theorem 1.1 $P^{2}$ has a spectral gap provided it is ergodic (i.e. $1$ is a simple eigenvalue). So, if $P$ has a spectral gap but $-1\in\sigma_{ess}(P)$ is not an eigenvalue, then $\|P\|_{tail}=1$. From this we would believe that in general the existence of spectral gap for $P$ does not imply $\|P\|_{tail}<1.$

The resolvent-uniform-positive-improving condition is weaker than the uniform-positive-improving condition used in , and the condition (E) introduced in . Recall that $P$ is called uniform-positive-improving if

In the symmetric setting this result is now improved by Theorem 1.1 since the resolvent-uniform-positive-improving property is strictly stronger than the ergodicity, which is equivalent to the resolvent-positivity-improving property (1.1). Examples to ensure this “strictly stronger” property can be constructed from the corresponding ones for Markov semigroups explained in item (c) below. More precisely, it is known that the resolvent-uniform-positive-improving property of $P$ is equivalent to the validity of the weak Poincaré inequality (see [22, Theorem 2.5]):

for some $\alpha:(0,\infty)\to(0,\infty).$ Taking $P=P_{1}$ for a symmetric Markov semigroup $P_{t}$ on $L^{2}(\mu)$ and noting that

where $(L,\mathcal{D}(L))$ and $\mathcal{E}$ are the associated generator and Dirichlet form, we see that an example such that $\mathcal{E}$ is irreducible but the weak Poincaré inequality (1.3) below is not available implies that $P$ is ergodic but does not possess the resolvent-uniform-positive-improving property.

(c) We would like to mention links of the uniform-positive-improving property of a symmetric Markov semigroup $P_{t}$ and the weak Poincaré inequality of the associated Dirichlet form $(\mathcal{E},\mathcal{D}(\mathcal{E}))$. It is well known that $P_{t}$ (for some/all $t>0$) is ergodic if and only if the Dirichlet form is irreducible, i.e. $\mathcal{E}(f,f)=0$ implies $f$ is constant. Next, according to (see also ), the uniform-positive-improving property of the semigroup (i.e. $P_{t}$ is uniform-positive-improving for some, equivalently all, $t>0,$ see ) implies the weak spectral gap property, which is equivalent to the validity of the weak Poincaré inequality (see ): for some $\alpha:(0,\infty)\to(0,\infty)$

Indeed, as explained in (b) that the resolvent-uniform-positive-improving property of $P_{t}$ for some $t>0$ also implies the existence of the weak Poincaré inequality. Moreover, it is shown in [13, §7] that there are conservative irreducible Dirichlet forms which do not satisfy the weak Poincaré inequality. Therefore, the resolvent-uniform-positive-improving property is strictly stronger than the ergodicity.

As applications of Theorem 1.1, we consider functional inequalities conservative symmetric Dirichlet forms. A simple consequence of the equivalence of $(1)$ and (3) is that the defective Poincaré inequality implies the tight one.

Let $(\mathcal{E},\mathcal{D}(\mathcal{E}))$ be a conservative, irreducible, symmetric Dirichlet form on $L^{2}(\mu)$. Then the Poincaré inequality

holds for some constant $C>0$ if and only if the defective Poincaré inequality

holds for some constants $C_{1},C_{2}>0.$

This result improves [13, Proposition 1.3] where the weak Poincaré inequality (1.3) is used to replace the irreducibility of the Dirichlet form. Basing on Corollary 1.2, we are able to prove the equivalence of the defective version and the tight version for more general functional inequalities. Here, we consider a family of functional inequalities introduced in , which interpolate the Poincaré inequality (1.4) and the Gross log-Sobolev inequality

for some constant $C>0.$ Let $\phi\in C()$ such that $\phi>0$ on $[1,2)$ and $\phi(2)=0$ with

When $\phi(p)=2-p$ the inequality (1.7) is equivalent to the log-Sobolev inequality, and when $\phi$ reduces to a positive constant it becomes the Poincaré inequality. See for detailed discussions on properties and applications of the inequality (1.7).

Let $(\mathcal{E},\mathcal{D}(\mathcal{E}))$ be a conservative, irreducible, symmetric Dirichlet form on $L^{2}(\mu)$. Then $(\ref{PH})$ holds for some constant $C>0$ if and only if $(\ref{DPH})$ holds for some constants $C_{1},C_{2}>0.$

The remainder of the paper is organized as follows. In Section 2, by using an approximation argument introduced in , we extend a known Cheeger type inequality for high order eigenvalues of finite-state Markov chains to the abstract setting, then use this estimate to characterize the existence of spectral gap with high-order isoperimetric constants. This characterization is then used in Section 3 to prove Theorem 1.1. Proofs of Corollaries 1.2 and 1.3 are also addressed in Section 3. Finally, in Section 4 we extend Theorem 1.1 and Corollary 1.2 to the sub-Markov setting.

Essential spectrum and isoperimetric constants

Let $P$ be a symmetric Markov operator on $L^{2}(\mu)$. We aim to characterize the essential spectrum of $P$ using high order isoperimetric constants investigated in .

We define the $n$-th isoperimetric constant by

Obviously, $\kappa_{1}=0$ and if $L^{2}(\mu)$ is infinite-dimensional then $D_{n}\neq\emptyset$ for all $n\geq 2.$ It is also easy to see that $\kappa_{n}$ is non-decreasing in $n$.

We will only consider the case that $L^{2}(\mu)$ is infinite-dimensional, since otherwise the spectrum of $P$ is finite so that the existence of spectral gap becomes trivial. By the Cheeger inequality we know that $P$ has a spectral gap if and only if $\kappa_{2}>0$, see e.g. by noting that in (1.2) we have

for the symmetric measure $J$ on $E\times E$ determined by $J(A\times B):=\mu(1_{A}P1_{B}),A,B\in\mathcal{F}$.

Consider $\lambda_{ess}(P):=\sup\sigma_{ess}(P),$ where $\sigma_{ess}(P)$ is the essential spectrum of $P$. Obviously, $P$ has a spectral gap if and only if it is ergodic and $\lambda_{ess}(P)<1.$

Let $P$ be a symmetric Markov operator on $L^{2}(\mu)$. Then $\lambda_{ess}(P)<1$ if and only if $\sup_{n\geq 1}\kappa_{n}>0.$ Consequently, when $P$ is ergodic then the existence of spectral gap of $P$ is equivalent to each of the following two statements:

\ \kappa_{2}\big{(}=\inf_{n\geq 2}\kappa_{n}\big{)}>0.

\lim_{n\to\infty}\kappa_{n}\big{(}=\sup_{n\geq 2}\kappa_{n}\big{)}>0.

As mentioned at the end of Introduction, to prove this result we will extend a known estimate on the hight order eigenvalues using $\kappa_{n}$ for finite-state Markov chains. So, below we first consider Markov operators on a finite set.

As shown in [10, Theorem 2], we may take $c(n)=\frac{c_{0}}{n^{4}}$ for a universal constant $c_{0}>0.$

Below we aim to extend this estimate to our abstract setting by using an approximation argument introduced in . For any $n\geq 1$, let

To see that this quantity can be regarded as the $n$-th eigenvalue of $L:=1-P$, let $\lambda_{ess}(L)=\inf\sigma_{ess}(L)$ be the bottom of the essential spectrum of $L$. Then, see e.g. [11, Theorem XIII.2], $\lambda_{n}$ is the $n$-th eigenvalue of $L$ if $\lambda_{n}<\lambda_{ess}(L),$ and $\lambda_{n}=\lambda_{ess}(L)$ otherwise. The following result was stated as Proposition 5 in , we include here a proof for completeness.

Let $c(n)$ be in $(\ref{B1})$ for $n\geq 1$. Then $\kappa_{n}\geq\lambda_{n}\geq c(n)^{2}\kappa_{n}^{2}.$

Let $L=1-P$. Since $\lambda_{1}=\kappa_{1}=0$, we only prove for $n\geq 2.$

(a) Upper bound estimate of $\lambda_{n}$. For any $f_{1},\cdots,f_{n-1}\in L^{2}(\mu)$ and any $(A_{1},\cdots,A_{n})\in D_{n}$, there exists a function $f:=\sum_{i=1}^{n}a_{i}1_{A_{i}}$ such that $\sum_{i=1}^{n}a_{i}^{2}=1$ and $\mu(ff_{i})=0,1\leq i\leq n.$ Let

Therefore, by the definition of $\lambda_{n}$ and $\kappa_{n}$, we have $\lambda_{n}\leq\kappa_{n}$.

(b) Lower bound estimate of $\lambda_{n}.$ Assume that $\lambda_{n}<c(n)^{2}\kappa_{n}^{2}$. Then there exist $f_{1},\cdots,f_{n}\in L^{2}(\mu)$ such that

where $\delta_{n}:=c(n)^{2}\kappa_{n}^{2}-\lambda_{n}>0.$ Indeed, for any $\varepsilon>0$ we may find an orthonormal family $\{f_{1},\cdots,f_{n}\}\subset L^{2}(\mu)$ such that $\mu(|Lf_{i}-\lambda_{i}f_{i}|^{2})\leq\varepsilon,1\leq i\leq n.$ Since $\max_{1\leq i\leq n}\lambda_{i}=\lambda_{n}=c(n)^{2}\kappa_{n}^{2}-\delta_{n}$ and $\mu(f_{i}f_{j})=\delta_{ij}$, (2.3) holds for small enough $\varepsilon>0.$

Let $\mathcal{F}_{\infty}=\sigma(f_{1},\cdots,f_{n}).$ Since $\mathcal{F}_{\infty}$ is separable, we may find an increasing sequence of $\sigma$-fields $\{\mathcal{F}_{N}\}_{N\geq 1}$ such that

Then $P_{N}$ is a symmetric Markov operator on $L^{2}(\mu_{N}).$

To identify $P_{N}$ with a Markov operator on a finite set, let

holds for $D_{N,n}:=\{(A_{1},\cdots,A_{n})\in D_{n}:\ A_{k}\in\mathcal{F}_{N},1\leq k\leq n\}.$ Since for $A_{k}\in\mathcal{F}_{N}$ we have $\mu_{N}(A_{k})=\mu(A_{k})$ and

which contradicts (2.4) for large $N$ such that $\varepsilon_{N}<\frac{\delta_{n}}{16n}$. ∎

If $\lambda_{ess}(P)<1$, then $\sigma(L)\cap[0,1-\lambda_{ess}(P))$ is discrete and each eigenvalue in this set is of finite multiplicity. So, in this case $\lambda_{n}>0$ for $n$ lager than the multiplicity of the first eigenvalue $\lambda_{1}=0,$ and hence by Lemma 2.2, $\kappa_{n}>0$ for large $n$. On the other hand, we aim to prove that if $\lambda_{ess}(P)=1$ then $\kappa_{n}=0$ for all $n\geq 1.$ Since $\lambda_{ess}(P)=1$, i.e. $0\in\sigma_{ess}(L)$, we have $\lambda_{n}=0$ for all $n$. Combining this with Lemma 2.2, we prove $\kappa_{n}=0$ for all $n\geq 1$. Thus, the proof of the first assertion is finished.

Now, let $P$ be ergodic. Since (6) follows from (5), it suffices to prove that (1) (i.e. the existence of spectral gap of $P$) implies (5) while (6) implies (1). If (1) holds then $\lambda_{2}>0$, so that by Lemma 2.2 we have $\kappa_{2}>0$, i.e. (5) holds. On the other hand, if (6) holds, then by the first assertion of this theorem we have $\lambda_{ess}(P)<1$ such that $1$ is isolate in $\sigma(P)$. Since $1$ is a simple eigenvalue of $P$, this implies that $P$ has a spectral gap, i.e. (1) holds. ∎

Proofs of Theorem 1.1 and corollaries

By [3, Theorem 4.1(d)] with $\pi=\hat{P}$ and $r_{sp}(\pi)=1$, (1) implies (4). Next, it is obvious that (2) implies (3). Moreover, since $\|\hat{P}^{m}\|_{\tau}\leq\|\hat{P}^{m}\|_{tail}$ and by the Jensen inequality the later is decreasing in $m$, we see that (4) implies (3) for $\hat{P}$ in place of $P$. Therefore, it remains to prove that (1) implies (2), and (3) implies (1). Below we prove these two implications respectively.

(1) implies (2). Let $\theta\in(0,1)$ such that (1.2) holds. For any $f\geq 0$ with $\mu(f^{2})\leq 1$, we have

Replacing $f$ by $(f-\sqrt{R})^{+}$ for $R>1$, and noting that

Since by the Jensen inequality $(Pf-R)^{+}\leq P(f-R)^{+},$ combining this with (3.1) we arrive at

(3) implies (1). Let $\|P^{2m-1}\|_{\tau}<1$ for some $m\geq 1$. It suffices to prove for the case that $L^{2}(\mu)$ is infinite-dimensional since in the finite dimensional case the existence of spectral gap is trivial. We first assume that $P$ is symmetric. In this case, the existence of spectral gap for $P$ is equivalent to that for $P^{2m-1}$. Since $\|P^{2m-1}\|_{\tau}<1$, there exists $\delta\in(0,1)$ and $R>0$ such that

On the other hand, if $P^{2m-1}$ does not have spectral gap, by Theorem 2.1, for any $\varepsilon\in(0,1)$ and $n\geq 2$, there exists $(A_{1},\cdots A_{n})\in D_{n}$ such that

Letting $B_{k}=A_{k}\cap\{P^{2m-1}1_{A_{k}}\geq 1-\sqrt{\varepsilon}\}$, this implies

Since $A_{1},\cdots,A_{n}$ are disjoint with positive $\mu$-mass and $\mu$ is a probability measure, there exists $1\leq k\leq n$ such that $\mu(A_{k})\in(0,\frac{1}{n}).$ Take

Combining this with (3.2), (3.3) and $\mu(A_{k})\in(0,\frac{1}{n})$, we arrive at

Since $\varepsilon\in(0,1)$ and $n\geq 2$ are arbitrary, by letting $\varepsilon\to 0$ and $n\to\infty$ we obtain $\delta\geq 1$ which contradicts $\delta\in(0,1)$.

Next, for non-symmetric $P$, we consider the symmetrizing operator $\hat{P}$. Obviously, $P$ satisfies (1.1) implies that $\hat{P}$ satisfies (1.1), i.e. $\hat{P}$ is ergodic. By (3), there exists $m\geq 1$ such that $\varepsilon:=\|P^{2m-1}\|_{\tau}<1$. Moreover, as proved above if $\|{\hat{P}}^{2m-1}\|_{\tau}<1$ then $\hat{P}$ has a spectral gap. Therefore, it suffices to show that $\|{\hat{P}}^{2m-1}\|_{\tau}<1.$ Noting that for any $R>0$ and $f\geq 0$ with $\mu(f^{2})\leq 1$,

we obtain $\|{\hat{P}}^{2m-1}\|_{\tau}\leq 2^{1-2m}\varepsilon+1-2^{1-2m}<1$, since $\varepsilon<1.$∎

It suffices to prove (1.4) from (1.5). Let $P_{t}$ be the associated Markov semigroup. Then the irreducibility of the Dirichlet form implies that of $P_{t}$ for $t>0$. Next, by [16, Theorem 3.3] with $\phi\equiv 1$, (1.5) implies that $\|P_{t}\|_{\tau}\leq\text{\rm{e}}^{-t/C_{1}}<1$ for $t>0$. Then due to Theorem 1.1 we conclude that $P_{t}$ has a spectral gap, equivalently, the Poincaré inequality (1.4) holds for some constant $C>0.$∎

By [19, Proposition 2.1], for any $f\in L^{2}(\mu)$ and $\hat{f}:=f-\mu(f)$, we have

Next, since $\phi(2)=0$, from the proof of [19, Theorem 1.1(2)] we see that the $F$-Sobolev inequality (1.7) in holds for some nonnegative function $F$ with $F(r)\uparrow\infty$ as $r\uparrow\infty$. According to (see also ), this inequality is equivalent to the super Poincaré inequality

for some function $\beta:(0,\infty)\to(0,\infty)$. In particular, the defective Poincaré inequality holds. Thus, by Corollary 1.2, we have the Poincaré inequality

Extensions to the sub-Markov setting

In this section we let $(\mathcal{E},\mathcal{D}(\mathcal{E}))$ be a non-conservative Dirichlet form on $L^{2}(\mu)$, for which either $1\notin\mathcal{D}(\mathcal{E})$ or $1\in\mathcal{D}(\mathcal{E})$ but $\mathcal{E}(1,1)>0.$ In this case the Dirichlet form is irreducible if and only if for any $f\in\mathcal{D}(\mathcal{E})$ with $\mathcal{E}(f,f)=0$ one has $f=0.$ Let $\|P\|_{p}$ be the norm in $L^{p}(\mu)$ for $p\geq 1.$ Below is an extension of Corollary 1.2 to the present situation.

Let $(\mathcal{E},\mathcal{D}(\mathcal{E}))$ be a non-conservative irreducible Dirichlet form on $L^{2}(\mu)$. Then the Poincaré inequality

holds for some $C>0$ if and only if the defective Poincaré inequality

Assume that (4.2) holds. We aim to prove (4.1) for some constant $C>0$. According to [17, Proposition 3.2] we need only to prove the weak Poincaré inequality

for some function $\alpha:(0,\infty)\to(0,\infty).$ This is ensured by Lemma 4.3 below. ∎

Before introducing Lemma 4.3, we present an application of Theorem 4.1 to sub-Markov operators.

Let $P$ be a sub-Markov operator on $L^{2}(\mu)$; i.e. $P$ is a contraction linear operator on $L^{2}(\mu)$ with $P1\leq 1$ such that $f\geq 0$ implies $Pf\geq 0.$ Let $P^{*}$ be the adjoint operator of $P$. Assume that ${\rm Ker}(1-P^{*}P)=\{0\}$. Then

It suffices to prove $\|P\|_{2}<1$ from $\|P\|_{tail}<1$. To apply Theorem 4.1, let

Then $(\mathcal{E},\mathcal{D}(\mathcal{E}))$ is a symmetric Dirichlet form. Since Ker$\,(1-P^{*}P)=0$ and noting that the contraction of $P^{*}$ in $L^{2}(\mu)$ implies

this Dirichlet form is non-conservative and irreducible. By $\|P\|_{tail}<1$, there exist $R>0$ and $\varepsilon\in(0,1)$ such that

Thus, (4.2) holds for $C_{1}=\frac{2}{1-\varepsilon},C_{2}=\frac{R^{2}}{(1-\varepsilon)^{2}}.$ By Theorem 4.1, there exists $C>0$ such that

This implies that $\|P\|_{2}^{2}\leq\frac{C-1}{C}<1.$ ∎

Correspondingly to Remark 1.1(a) in the ergodic setting, we would believe that in the present situation $\|P\|_{\tau}<1$ is not enough to imply $\|P\|_{2}<1$. But it seems hard to give a proof or a counterexample.

A non-conservative Dirichlet form $(\mathcal{E},\mathcal{D}(\mathcal{E}))$ on $L^{2}(\mu)$ is irreducible if and only if there exists $\alpha:(0,\infty)\to(0,\infty)$ such that

Consequently, for any symmetric $($sub-$)$ Markov semigroup $P_{t}$ on $L^{2}(\mu)$, $\lim_{t\to\infty}\|P_{t}f\|_{2}=0$ for any $f\in L^{2}(\mu)$ if and only if

(a) Let $P_{t}$ be the associated semigroup of $(\mathcal{E},\mathcal{D}(\mathcal{E}))$. Then $(\mathcal{E},\mathcal{D}(\mathcal{E}))$ is irreducible if and only if $\mu((P_{t}f)^{2})\to 0$ as $t\to\infty$ for any $f\in L^{2}(\mu).$ Indeed, the sufficiency is obvious since $\mathcal{E}(f,f)=0$ implies

so that $\mu(|P_{t}f|^{2})=\mu(f^{2})$ for all $t\geq 0$, while the necessity holds since the irreducibility of the Dirichlet form implies that

where $\{E_{\lambda}\}_{\lambda\leq 0}$ is the spectral family of the generator such that $\text{\rm{d}}\|E_{\lambda}(f)\|^{2}$ is a finite measure on $(-\infty,0)$ for $f\in L^{2}(\mu)$ (by the irreducibility $\{0\}$ is a null-set of the measure).

On the other hand, by [13, Theorem 2.1] with $\Phi(f)=\|f\|_{\infty}^{2}$, (4.3) holds for some $\alpha$ if and only if (4.4) holds. So, the second assertion follows from the first one.

(c) Now, let $(\mathcal{E},\mathcal{D}(\mathcal{E}))$ be irreducible. We claim that (4.3) holds for some function $\alpha:(0,\infty)\to(0,\infty).$ Otherwise, there exist some $r>0$ and a sequence $\{f_{n}\}\subset\mathcal{D}(\mathcal{E})$ such that

Since $\mathcal{E}(|f_{n}|,|f_{n}|)\leq\mathcal{E}(f_{n},f_{n})$, we may and do assume that $f_{n}\geq 0$ for all $n\geq 1.$ Since $\{f_{n}\}$ is bounded in $L^{2}(\mu)$, there exist $f\in L^{2}(\mu)$ and a subsequence $\{f_{n_{k}}\}$ such that $f_{n_{k}}$ converges weakly to $f$ in $L^{2}(\mu)$.

Let $P_{t}$ be the (sub-) Markov semigroup and $(L,\mathcal{D}(L))$ the generator associated to $(\mathcal{E},\mathcal{D}(\mathcal{E}))$. Then $P_{t}f\in\mathcal{D}(L)$ for any $t>0.$ By the symmetry of $P_{t}$ and the weak convergence of $\{f_{n_{k}}\}$ to $f$ in $L^{2}(\mu)$, we have

Moreover, due to (4.5) and the symmetry of $\mathcal{E}$,

Combining this with (4.6) we conclude that $\mathcal{E}(P_{t}f,P_{t}f)=0$ for all $t>0$. Thus, by the irreducibility, $P_{t}f=0$ holds for all $t>0.$ This implies $f=0$ by the strong continuity of $P_{t}$ in $L^{2}(\mu)$. Since (4.5) implies $f_{n}\leq r^{-1/2},$ by the weak convergence of $\{f_{n_{k}}\}$ to $f=0$ in $L^{2}(\mu)$ and $1\in L^{2}(\mu)$, we obtain

This contradicts the assumption that $\mu(f_{n}^{2})=1$ for all $n\geq 1.$ Therefore, (4.3) holds for some function $\alpha:(0,\infty)\to(0,\infty).$∎

I would like to thank professor Laurent Miclo for sending me his exciting paper and useful discussions, as well as the referee, Professor L. Gross and Dr. Shaoqin Zhang for helpful comments and corrections.

References