In Markov process, an extremal reversible measure is an extremal invariant measure.

arXiv:2312.14816v1 [math.GM] 14 Aug 2023 In Markov process, an extremal reversible measure is an extremal invariant measure. Hiroki Yagisita (Department of Mathematics, Kyoto Sangyo University) Abstract: We consider a discrete-time temporally-homogeneous conservative Markov process. We show that extremality of reversible measure implies extremality of invariant measure. Using analogue of Dirichlet form, we modify a proof that in stochastic Ising model (Glauber dynamics), an extreme Gibbs state is an extreme invariant measure. Keyword: Choquet simplex, ergodic decomposition, detailed balance condition, mutual singularity. 1 1 Introduction Let (X, B) be a measurable space. Let {Px}x∈Xbe a discrete-time temporallyhomogeneous Markov process whose state-space is (X, B). Let {p(x, E)}x∈X,E∈B be the transition function of {Px}x∈X. That is, p(x, E) := Px(η(1) ∈E). {Px}x∈Xis conservative, if and only if for any x∈X,p(x, X) = 1 holds. Definition 1 (extreme element): Let Cbe a subset of the set of all probability measures on (X, B). Let m∈C. Then, mis said to be an extreme element of C, if and only if [ t∈(0,1), m0, m1∈C, m = (1 −t)m0+tm1] implies m0=m1. Definition 2 (T): Let define the linear map Tfrom the set of all Rvalued bounded B-measurable functions to the set of all R-valued bounded B-measurable functions by (T f)(x) := Zy∈X f(y)p(x, dy).  Definition 3 (I,R,G,Ie,Re,Ge): (1) Let mbe a probability measure on (X, B). Then, mis said to be an invariant measure, if and only if for any R-valued bounded B-measurable function f, Zx∈X (T f)(x)m(dx) = Zx∈X f(x)m(dx) holds. mis said to be a reversible measure, if and only if for any R-valued bounded B-measurable functions fand g, Zx∈X (T f)(x)g(x)m(dx) = Zx∈X f(x)(T g)(x)m(dx) holds. mis said to be a conservative measure, if and only if for any R-valued bounded B-measurable function g, Zx∈X (T1)(x)g(x)m(dx) = Zx∈X g(x)m(dx) holds. 2 (2) Let define Ias the set of all invariant measures. Let define Ieas the set of all extreme elements of I. Let define Ras the set of all reversible measures. Let define Reas the set of all extreme elements of R. Let define Gas the set of all conservative reversible measures. Let define Geas the set of all extreme elements of G. Theorem 4 (Main Result): Suppose that {Px}x∈Xis conservative. Then, Re⊂ Ieholds.  3 2 Proof Definition 5 (joint distribution σm): For each probability measure mon (X, B), let define σmas the joint distribution with mas the initial distribution and {0,1}as the set of times. That is, σm(C) := Zx0∈XZx1∈X 1C(x0, x1)p(x0, dx1)m(dx0).  Lemma 6: Let m∈ R. Then, σmis symmetric. Proof: Because of σm(A×B) = Rx0∈X(Rx1∈X1A(x0)1B(x1)p(x0, dx1))m(dx0) =Rx0∈X1A(x0)(T1B)(x0)m(dx0) = Rx0∈X1B(x0)(T1A)(x0)m(dx0) = Rx0∈X (Rx1∈X1B(x0)1A(x1)p(x0, dx1))m(dx0) = σm(B×A), it is symmetric (by Hopf extension theorem).  The following lemma is the basis in this section. It was inspired by the formula (1.4.8) in [1]. Lemma 7 (quadratic form): Let m∈ G. Then, for any R-valued bounded B-measurable function f, Zx∈X ((1 −T)f)(x)f(x)m(dx) =1 2Z(x0,x1)∈X2 |f(x0)−f(x1)|2σm(d(x0, x1)) holds. Proof: From Lemma 6, 2 Rx∈X((1 −T)f)(x)f(x)m(dx) = 2 Rx∈Xf(x) (f(x)(T1)(x)−(T f)(x))m(dx) = 2 Rx0∈Xf(x0)(Rx1∈X(f(x0)−f(x1))p(x0, dx1)) m(dx0) = 2 R(x0,x1)∈X2f(x0)(f(x0)−f(x1))σm(d(x0, x1)) = R(x0,x1)∈X2f(x0) (f(x0)−f(x1))σm(d(x0, x1))+R(x0,x1)∈X2f(x1)(f(x1)−f(x0))σm(d(x0, x1)) = R(x0,x1)∈X2|f(x0)−f(x1)|2σm(d(x0, x1)) holds.  Lemma 8: Let m∈ G. Let fbe a R-valued bounded B-measurable function. Then, the followings (a), (b), (c) and (d) are equivalent. (a) Let gbe a R-valued bounded B-measurable function. Then, m-a.s. x: (T(gf))(x) = (T g)(x)f(x) holds. 4 (b) m-a.s. x: (T f)(x) = f(x) holds. (c) Zx∈X ((1 −T)f)(x)f(x)m(dx) = 0 holds. (d) σm-a.s. (x0, x1) : f(x0) = f(x1) holds. Proof: (1) Suppose that (a) holds. We show (b). For any R-valued bounded B-measurable function g,Rx∈X(T f)(x)g(x)m(dx) = Rx∈X(T(1f))(x) g(x)m(dx) = Rx∈X(T1)(x)f(x)g(x)m(dx) = Rx∈Xf(x)g(x)m(dx) holds. (2) [(b)⇒(c)] is easy. (3) From Lemma 7, [(c)⇒(d)] holds. (4) Suppose that (d) holds. We show (a). Let gbe a R-valued bounded Bmeasurable function. Then, for any R-valued bounded B-measurable function h,Rx∈Xh(x)(T(gf))(x)m(dx) = R(x0,x1)∈X2h(x0)g(x1)f(x1)σm(d(x0, x1)) = R(x0,x1)∈X2h(x0)g(x1)f(x0)σm(d(x0, x1)) = Rx∈Xh(x)(T g)(x)f(x)m(dx) holds.  Lemma 9: Let m∈ G. Let ρbe a [0,+∞)-valued bounded B-measurable function. Then, ρm ∈ G holds, if and only if ρm ∈ I holds. Proof: (1) Suppose that ρm ∈ G holds. We show that ρm ∈ I holds. For any R-valued bounded B-measurable function f,Rx∈X(T f)(x)ρ(x)m(dx) = Rx∈X(T1)(x)f(x)ρ(x)m(dx) = Rx∈Xf(x)ρ(x)m(dx) holds. (2) Suppose that ρm ∈ I holds. We show that ρm ∈ G holds. For any R-valued bounded B-measurable function f,Rx∈Xf(x)(T ρ)(x)m(dx) = Rx∈X(T f)(x)ρ(x)m(dx) = Rx∈Xf(x)ρ(x)m(dx) holds. [m-a.s. x: (T ρ)(x) = ρ(x)] holds. From Lemma 8, for any R-valued bounded B-measurable function g, [m-a.s. x: (T(gρ))(x) = (T g)(x)ρ(x)] holds. For any R-valued bounded B-measurable functions fand g,Rx∈X(T f)(x)g(x)ρ(x)m(dx) = Rx∈Xf(x)(T(gρ))(x)m(dx) = Rx∈Xf(x)(T g)(x)ρ(x)m(dx) holds.  Lemma 10: Ge⊂ Ieholds. Proof: Suppose that m∈ Geholds. We show that m∈ Ieholds. (1) For any R-valued bounded B-measurable function f,Rx∈X(T f)(x)m(dx) =Rx∈X(T1)(x)f(x)m(dx) = Rx∈Xf(x)m(dx) holds. 5 (2) Suppose that t∈(0,1), m0, m1∈ I and m= (1 −t)m0+tm1hold. Then, m0≤1 1−tmand m1≤1 tmhold. There exist [0,+∞)-valued bounded B-measurable functions ρ0and ρ1such that m0=ρ0mand m1=ρ1mhold. From Lemma 9, m0, m1∈ G holds. m0=m1holds.  Proof of Main Result: R=Gholds. So, from Lemma 10, Re=Ge⊂ Ie holds.  The following remark seems to be similar to the proposition (4.3.5) in [2]. Lemma 7 seems to be similar to the lemma (4.4.3) in [2]. As stochastic Ising model seems to be a typical example of symmetric Feller process, a reversible measure and a Gibbs measure are equivalent from the theorem (4.2.14) in [2]. While the corollary (4.4.20) in [2] asserts that an extreme Gibbs measure is an extreme invariant measure, it seems that (4.3.5) and (4.4.3) were essential for the proof in [2]. Remark: Let m∈ G. Let ρbe a [0,+∞)-valued bounded B-measurable function. Suppose that Rx∈Xρ(x)m(dx) = 1 holds. Then, ρm ∈ G holds, if and only if [σm-a.s. (x0, x1) : ρ(x0) = ρ(x1) ] holds. Proof: From Lemma 8, it is shown. (We do not go into details, as it is not difficult.)  6 Reference: [1] Fukushima, M., Oshima, Y., Takeda, M., Dirichlet forms and symmetric Markov processes, Walter de Gruyter & Co., Berlin, 1994. [2] Liggett, T. M., Interacting particle systems, Springer-Verlag, New York, 1985. 7