A note on an ergodic theorem in weakly uniformly convex geodesic ...

arXiv:1411.1095v1 [math.DS] 4 Nov 2014

A note on an ergodic theorem in weakly uniformly convex geodesic spaces Laurent¸iu Leu¸stean1,2 , Adriana Nicolae3,4 1

Faculty of Mathematics and Computer Science, University of Bucharest, Academiei 14, P. O. Box 010014, Bucharest, Romania 2

Simion Stoilow Institute of Mathematics of the Romanian Academy, P. O. Box 1-764, RO-014700 Bucharest, Romania 3

Department of Mathematics, Babe¸s-Bolyai University, Kog˘ alniceanu 1, 400084 Cluj-Napoca, Romania

4

Simion Stoilow Institute of Mathematics of the Romanian Academy, Research group of the project PD-3-0152, P. O. Box 1-764, RO-014700 Bucharest, Romania E-mails: [email protected], [email protected]

Abstract Karlsson and Margulis [5] proved in the setting of uniformly convex geodesic spaces, which additionally satisfy a nonpositive curvature condition, an ergodic theorem that focuses on the asymptotic behavior of integrable cocycles of nonexpansive mappings over an ergodic measure-preserving transformation. In this note we show that this result holds true when assuming a weaker notion of uniform convexity on the space. MSC: 37A30, 53C23 Keywords: ergodic theorem, geodesic space, weak uniform convexity, Busemann convexity

1

Introduction

Let Y be a geodesic space and D ⊆ Y . Consider S a semigroup of nonexpansive (i.e. 1-Lipschitz) self-mappings defined on D and endow S with the Borel σalgebra induced by the compact-open topology on S. Assume that (X, µ) is a probability measure space, T : X → X is an ergodic measure-preserving transformation and w : X → S is a measurable map. Define the cocycles an (x) = w(x)w(T x) · · · w(T n−1 x). 1

Fix y ∈ D and suppose that Z

d(y, a1 (x)y)dµ(x) < ∞.

X

One can easily see that the sequence

Z

X

 d(y, an (x)y)dµ(x) is subadditive n

and so, by Fekete’s subadditive lemma [1], the following limit exists Z Z 1 1 0 ≤ A = lim d(y, an (x)y)dµ(x) = inf d(y, an (x)y)dµ(x) < ∞. n→∞ n X n n X As an immediate application of Kingman’s subadditive ergodic theorem [8] one gets that 1 lim d(y, an (x)y) = A, for almost every x ∈ X. n→∞ n Karlsson and Margulis [5] proved that if Y is a complete Busemann convex geodesic space satisfying a uniform convexity condition (see Section 2, (ii) for the precise definition), the following holds: if A > 0, then for almost every x ∈ X, there exists a unique geodesic ray γ in Y starting at y and depending on x such that 1 lim d(γ(An), an (x)y) = 0. n→∞ n Thus, in this case, instead of the convergence of averages as in classical ergodic results, one basically obtains for almost every x ∈ X the existence of a geodesic ray that issues at y such that, as n → ∞, the values of the nonexpansive mappings an (x) at y are ‘close’ to this geodesic ray, a property which is referred to in [4] as ray approximation. This result generalizes the multiplicative ergodic theorem of Oseledec [13] (see [5, 4]). A discussion on the asymptotic behavior of ergodic products of nonexpansive mappings and isometries of a (proper) metric space and applications thereof can be found in [4]. In this paper we show that the ergodic theorem given in [5] holds true for a more general class of geodesic spaces. More precisely, we prove that we can relax the uniform convexity assumption on Y used in [5] as follows: Y is said to be weakly uniformly convex if for any a ∈ Y , r > 0 and ε ∈ (0, 2],     1 1 1 δ(a, r, ε) = inf 1 − d a, x + y : d(a, x), d(a, y) ≤ r, d(x, y) ≥ εr > 0. r 2 2 The mapping δ is called the modulus of convexity of Y . This notion was used by Reich and Shafrir [14] in the setting of hyperbolic spaces. In addition, we assume that for every a ∈ Y and ε > 0 there exists s > 0 such that inf δ(a, r, ε) > 0.

r≥s

Thus, the main result of the paper is the following.

2

(1)

Theorem 1.1. Assume that Y is a complete Busemann convex geodesic space that is weakly uniformly convex and satisfies (1). If A > 0, then for almost every x ∈ X, there exists a unique geodesic ray γ in Y that issues at y and depends on x such that 1 d(γ(An), an (x)y) = 0. (2) n In Section 2 we recall some notions of uniform convexity used in nonlinear settings by various authors. All these concepts fit within the class of weakly uniform convex geodesic spaces defined as above and satisfy (1). Section 3 contains the proof of our result. lim

n→∞

2

Weakly uniformly convex geodesic spaces

A metric space (Y, d) admits midpoints if for every x, y ∈ Y there exists a point m(x, y) ∈ Y (called midpoint) such that d(x, m(x, y)) = d(y, m(x, y)) = 1 d(x, y). If Y is also complete, then it is a geodesic space. 2 Let Y be a geodesic space. A point z ∈ Y belonging to a geodesic segment joining x, y ∈ Y will be denoted by z = (1 − t)x + ty, where t = d(z, x)/d(x, y). If Y is uniquely geodesic, we use the notation [x, y] for the unique geodesic segment that joins x, y ∈ Y . Any weakly uniformly convex geodesic space is strictly convex (that is, for all
a, x, y ∈ Y with x 6= y, d a, 21 x + 12 y < max{d(a, x), d(a, y)}), and hence uniquely geodesic. A stronger notion that was first considered in a nonlinear setting in [3] is the one of uniform convexity which means that additionally the modulus of convexity δ does not depend on the points a ∈ Y , so δ = δ(r, ε). Uniform convexity was also defined in [10] in the following way: a geodesic space Y is uniformly convex if there exists a mapping η : (0, ∞) × (0, 2] → (0, 1] such that for any r > 0, ε ∈ (0, 2] and all a, x, y ∈ Y ,    d(a, x) ≤ r  1 1 d(a, y) ≤ r ⇒ d a, x + y ≤ (1 − η(r, ε))r. (3)  2 2 d(x, y) ≥ εr Such a mapping η is referred to in [10] as a modulus of uniform convexity. These two definitions of uniform convexity are equivalent, the difference being that the modulus of convexity δ(r, ε) gives the largest possible η(r, ε) for r > 0 and ε ∈ (0, 2]. One can, of course, define a notion of a modulus of weak uniform convexity. However, in this paper we will use the modulus of convexity δ. Note that the above definitions can be given in the same way in the setting of metric spaces that admit midpoints. We include next some more particular notions of (weak) uniform convexity: (i) In [7], a geodesic space Y with a convex bicombing that is uniformly convex in the above sense is also assumed to satisfy the condition that ∀s ≥ 0, ∀ε > 0, ∃α(s, ε) > 0 such that ∀r > s, δ(r, ε) > α(s, ε) > 0. 3

(ii) Karlsson and Margulis [5] defined uniform convexity in a metric space Y that admits midpoints in the following way: Y is said to be uniformly convex if there exists a strictly decreasing and continuous function g : [0, 1] → [0, 1] with g(0) = 1, so that for any a, x, y ∈ Y and midpoint m(x, y) of x and y,   d(a, m(x, y)) d(x, y) , where r = max{d(x, a), d(y, a)}. (4) ≤g r 2r Any such space is uniformly convex in the sense used in the present paper. Indeed, given r > 0 and ε ∈ (0, 2], let a, x, y ∈ Y satisfy d(x, a) ≤ r, d(y, a) ≤ r and d(x, y) ≥ εr. Set R = max{d(x, a), d(y, a)} ≤ r. Then,     ε 1 d(x, y) 1 R≤g r. d a, x + y ≤ g 2 2 2R 2 It follows that

  ε 1 1 1 1 − d a, x + y ≥ 1 − g , r 2 2 2 ε > 0. from where δ(r, ε) ≥ 1 − g 2 (iii) Gelander, Karlsson and Margulis considered in [2] a strictly convex geodesic space Y to be uniformly convex if it is weakly uniformly convex and ∀ε > 0, ∃δ(ε) > 0 such that ∀r > 0, ∀a ∈ Y, δ(a, r, ε) ≥ δ(ε). (iv) In [6], a metric space Y that admits midpoints is called uniformly p-convex (where p ∈ [1, ∞]) if for every ε > 0 there exists ρp (ε) ∈ (0, 1) such that for all a, x, y ∈ Y with d(x, y) > εMp (d(a, x), d(a, y)) for p > 1 and d(x, y) > |d(a, x) − d(a, y)| + εM1 (d(a, x), d(a, y)) for p = 1 we have that d(a, m(x, y)) ≤ (1 − ρp (ε))Mp (d(a, x), d(a, y)), 1/p

where for α, β ≥ 0, Mp (α, β) = (αp + β p ) and M∞ (α, β) = max{α, β}. Note that, by [6, Lemma 4], any uniformly p-convex space is uniformly ∞convex which in turn is uniformly convex in the sense used in the present paper. Indeed, let r > 0 and ε ∈ (0, 2] and consider a, x, y ∈ Y with d(x, a) ≤ r, d(y, a) ≤ r and d(x, y) ≥ εr. Then max{d(x, a), d(y, a)} ≤ r ε and d(x, y) > max{d(x, a), d(y, a)}. This yields 2     ε   ε   1 1 max{d(x, a), d(y, a)} ≤ 1 − ρ∞ r. d a, x + y ≤ 1 − ρ∞ 2 2 2 2 It follows that

  ε 1 1 1 , 1 − d a, x + y ≥ ρ∞ r 2 2 2 ε and so δ(r, ε) ≥ ρ∞ > 0. 2 4

Geodesic spaces which are p-uniformly convex in the sense of Naor and Silberman [11] (see also [9, 12]) satisfy all four definitions (i)-(iv). These spaces are defined as follows: for a fixed 1 < p < ∞, a geodesic space (Y, d) is called p-uniformly convex with parameter k > 0 if for every a, x, y ∈ Y and t ∈ [0, 1], d(a, (1 − t)x + ty)p ≤ (1 − t)d(a, x)p + td(a, y)p −

k t(1 − t)d(x, y)p . 2

We see below that p-uniformly convex spaces indeed satisfy (i)-(iv). Remark first that, by [9, Proposition 2.5, (1)], k ≤ cp with  2(p − 1) if p ∈ (1, 2) cp = 8/2p if p ≥ 2. Let a ∈ Y , r > 0 and ε ∈ (0, 2]. Take x, y ∈ Y such that d(a, x) ≤ r, d(a, y) ≤ r and d(x, y) ≥ εr. Then, 1/p    1 1 k p 1 . 1 − d a, x + y ≥ 1 − 1 − ε r 2 2 8 This clearly implies (i) and (iii). To see that (ii) is satisfied, consider a, x, y ∈ Y and let r = max{d(a, x), d(a, y)}. Then,  p 1/p    1 1 1 k d(x, y) . d a, x + y ≤ 1 − r 2 2 8 r 1/p  k . Then, g([0, 1]) ⊆ [0, 1], g(0) = 1, g For t ∈ [0, 1], take g(t) = 1 − (2t)p 8 is continuous and strictly decreasing. Hence, (ii) holds. 1/p  1 k p > 0. It follows immediately that − ε Let ε > 0 and take ρp (ε) = 1 − 2 8 p for all a, x, y ∈ Y with d(x, y) > εM (d(a, x), d(a, y)),   1 1 d a, x + y ≤ (1 − ρp (ε))Mp (d(a, x), d(a, y)). 2 2 This means that (iv) is satisfied (see also [6, Example, page 3]). For the rest of this paper we assume that any weakly uniformly convex geodesic space also satisfies condition (1). From the above arguments it is easy to see that all uniformly convex spaces described in (i)-(iv) satisfy (1). Although we state the main result of this paper in the setting of weakly uniformly convex geodesic spaces that additionally satisfy a nonpositive curvature assumption, we remark that what we actually use in the proof is the following convexity condition which holds in any weakly uniformly convex geodesic space.

2.1

A convexity assumption

Let (Y, d) be a geodesic space. We say that Y satisfies property (C) if ∀y ∈ Y, ∀r > 0, ∀ε ∈ (0, 2], ∃Ψ(y, r, ε) ∈ (0, 1] 5

such that ∀x, z ∈ Y with d(x, y) = r, d(y, z) ≥ r, if w belongs to a geodesic segment joining y and z such that d(y, w) = r then r + d(x, z) ≤ d(y, z) + Ψ(y, r, ε)r

⇒

d(w, x) ≤ εr.

In addition, we suppose that for each y ∈ Y and ε ∈ (0, 2], there exists s > 0, inf Ψ(y, r, ε) > 0.

r≥s

If Y is a normed space, Ψ does not depend on y and r, so one can take Ψ(y, r, ε) = Ψ(0, 1, ε). To see this, let y ∈ Y , r > 0, ε > 0 and x, z ∈ Y with kx − yk = r, ky−zk ≥ r. Let w be on a geodesic segment joining y and z such that ky−wk = r and r + kx − zk ≤ ky − zk + Ψ0 (1, ε)r. Dividing this inequality by r we obtain 1 1 1 + kx − zk ≤ ky − zk + Ψ0 (1, ε). r r Let x′ =

1 (x − y), r

z′ =

1 (z − y), r

w′ =

1 (w − y). r

kz ′ k =

1 kz − yk, r

kw′ k + kw′ − z ′ k = kz ′ k.

Then, kx′ − z ′ k =

1 kx − zk, r

Thus, 1+kx′ −z ′k ≤ kz ′ k+Ψ0(1, ε). By property (C) we have that kw′ −x′ k ≤ ε, that is, kw − xk ≤ εr. Lemma 2.1. Suppose Y is a weakly uniformly convex geodesic space. Then Y satisfies property (C) with Ψ(y, r, ε) = δ(y, r, ε) for all y ∈ Y , r > 0 and ε ∈ (0, 2]. Proof. Let y ∈ Y , r > 0, ε ∈ (0, 2]. Consider x, z ∈ Y with d(x, y) = r, d(y, z) ≥ r and take w ∈ [y, z] such that d(y, w) = r. Suppose that r + d(x, z) ≤ d(y, z) + δ(y, r, ε)r. We may assume that w 6= x (otherwise the conclusion is 1 1 immediate). Let m = w + x. Since d(m, z) < max{d(z, x), d(z, w)} and 2 2 d(x, z) ≤ d(y, z) − r + δ(y, r, ε)r = d(z, w) + δ(y, r, ε)r, it follows that d(m, z) < d(z, w) + δ(y, r, ε)r. Thus, d(y, z) − d(y, m) ≤ d(m, z) < d(z, w) + δ(y, r, ε)r, from where d(y, m) > (1 − δ(y, r, ε))r. By weak uniform convexity,    d(x, w) d(y, m) ≤ 1 − δ y, r, r. r Hence,   d(x, w) < δ(y, r, ε). δ y, r, r 6

Because δ is increasing with respect to the separation distance ε for fixed y ∈ Y and r > 0 this implies that d(x, w) ≤ εr. Moreover, by (1), for every y ∈ Y and ε > 0 there exists s > 0 such that inf δ(y, r, ε) > 0. r≥s

3

Proof of the main theorem

Let E be the set of points x ∈ X for which taking any ε > 0 there exist M ∈ N and infinitely many n ∈ N such that for every k ∈ [M, n], d(y, an (x)y) − d(y, an−k (T k x)y) ≥ (A − ε)k. By [5, Proposition 4.2] we know that µ(E) = 1. 1 d(y, an (x)y) = A > 0. Then ∀(αi ) ⊆ n (0, 1], ∀(pi ) ⊆ N, ∃(Ki ) ⊆ N, ∃(ni ) ⊆ N satisfying

Lemma 3.1. Let x ∈ E such that lim

n→∞

(i) ∀i ≥ 1, pi ≤ Ki ; (ii) ∀i ≥ 2, Ki < ni−1 < ni and d(y, ani (x)y) ≥ max{d(y, ani−1 (x)y), Ani−1 }; (iii) ∀k ∈ [Ki , ni ], |d(y, ak (x)y) − Ak| ≤

Ak and 2i


1 − min{2−i , αi } d(y, ak (x)y) + d(ak (x)y, ani (x)y) ≤ d(y, ani (x)y). (5) Proof. For i ≥ 1, take εi = min



 A Aαi . Then , 1 + 2i+1 2 + αi

2εi ≤ min{2−i , αi }. A − εi

(6)

Since x ∈ Eεi , there exist Mi and infinitely many n such that for every k ∈ [Mi , n], d(y, an (x)y)) − d(y, an−k (T k x)y) ≥ (A − εi )k. (7) 1 Moreover, since lim d(y, an (x)y) = A it follows that there exists Ji such that n→∞ n for every k ≥ Ji , (A − εi )k ≤ d(y, ak (x)y) ≤ (A + εi )k.

(8)

Thus, there exist Ki = max{Mi , Ji , pi } and infinitely many n such that for every k ∈ [Ki , n], (7) and (8) are satisfied. Let i = 1 and n1 > K1 + K2 such that (7) and (8) hold for k ∈ [K1 , n1 ]. For each i ≥ 2, pick ni > max{ni−1 , Ki+1 } such that d(y, ani (x)y) ≥ max{d(y, ani−1 (x)y), Ani−1 } 7

and (7) and (8) hold for k ∈ [Ki , ni ]. Then for all i ≥ 1 and k ∈ [Ki , ni ], |d(y, ak (x)y) − Ak| ≤ εi k ≤

Ak 2i

and d(y, ani (x)y) − d(y, ani −k (T k x)y) + (A + εi )k ≥ (A − εi )k + d(y, ak (x)y). This implies that d(y, ak (x)y) + d(y, ani −k (T k x)y) ≤ d(y, ani (x)y) + 2εi k ≤ d(y, ani (x)y) + 2εi

d(y, ak (x)y) A − εi

by (8) ≤ d(y, ani (x)y) + min{2−i , αi }d(y, ak (x)y) by (6). But d(ak (x)y, ani (x)y) = d(ak (x)y, ak (x)ani −k (T k x)y) ≤ d(y, ani −k (T k x)y), so,
1 − min{2−i , αi } d(y, ak (x)y) + d(ak (x)y, ani (x)y) ≤ d(y, ani (x)y). Recall that a geodesic space Y is Busemann convex if given any pair of geodesic paths γ1 : [0, l1 ] → Y and γ2 : [0, l2 ] → Y with γ1 (0) = γ2 (0) one has d(γ1 (tl1 ), γ2 (tl2 )) ≤ td(γ1 (l1 ), γ2 (l2 )),

3.1

for every t ∈ [0, 1].

Proof of Theorem 1.1

Let x ∈ E such that lim

n→∞

1 d(y, an (x)y) = A. To simplify the writing, we denote n rn = d(y, an (x)y) → ∞.

By property (C), for each i ∈ N there exists si > 0 such that   1 αi = inf Ψ y, r, i > 0. r≥si 2 Also, for each i ∈ N there exists pi ∈ N such that rk ≥ si for k ≥ pi . Apply Lemma 3.1 to obtain the sequences (Ki ) and (ni ) satisfying properties (i)-(iii) thereof. For each i, let γi be the geodesic segment joining y and ani (x)y.

8

Let k ∈ [Ki , ni ]. If rk > rni , then, by (5), 1 rk . (9) 2i   1 Otherwise, since k ≥ pi it follows that rk ≥ si and so αi ≤ Ψ y, rk , i . Then 2 (5) yields that   1 rk + d(ak (x)y, ani (x)y) ≤ d(y, ani (x)y) + Ψ y, rk , i rk . 2 d(ak (x)y, γi (rni )) = d(ak (x)y, ani (x)y) ≤

By property (C), we have that d(ak (x)y, γi (rk )) ≤

1 rk . 2i

(10)

Thus, since Ki+1 < ni < ni+1 and rni ≤ rni+1 we have that d(γi+1 (rni ), γi (rni )) = d(γi+1 (rni ), ani (x)y) ≤

1

rn . 2i+1 i

Fix R > 0. Let I(R) be the smallest natural number for which rnI(R) ≥ R. Since (rni )i is an increasing sequence, it follows that rni ≥ R for all i ≥ I(R). Let i ≥ I(R). Then, by Busemann convexity, d(γi+1 (R), γi (R)) ≤

R 1 R 1 rni = i+1 R. d(γi+1 (rni ), γi (rni )) ≤ i+1 rni rni 2 2

(11)

Now, by the triangle inequality, for all m > 0, d(γi+m (R), γi (R)) ≤

m X 1 R R ≤ i. i+j 2 2 j=1

(12)

This means that (γi (R))i≥I(R) is Cauchy. Define γ(R) = lim γi (R). Letting i→∞

m → ∞ in (12) we obtain that for all i ≥ I(R), d(γ(R), γi (R)) ≤

R . 2i

(13)

It is easy to see that γ is a geodesic ray starting at y. For each k there exists i such that ni−1 ≤ k < ni , which yields Ki < k < ni . Ani Note that Ak < Ani ≤ rni+1 . Also, if rk > rni , because |rni − Ani | ≤ i+1 and 2 Ak |rk − Ak| ≤ i , we have that 2       1 1 1 A 1 − i k ≤ A 1 − i+1 ni ≤ rni < rk ≤ A 1 + i k. 2 2 2 9

Ak Ak . Let mk = min{rk , rni }. It follows that |mk − Ak| ≤ i i 2 2 1 and, by (9) and (10), d(ak (x)y, γi (mk )) ≤ i rk . Note that since rni+1 > Ak 2 and rni ≥ mk we have that i + 1 ≥ I(Ak) and i ≥ I(mk ), respectively. Then, using (13) and (11), Thus, |rni − Ak| ≤

d(γ(Ak), ak (x)y) ≤ d(γ(Ak), γi+1 (Ak)) + d(γi+1 (Ak), γi+1 (mk )) + d(γi+1 (mk ), γi (mk ) + d(γi (mk ), ak (x)y) Ak 1 1 ≤ i+1 + |Ak − mk | + i+1 mk + i rk 2 2 2 6Ak 9A 3Ak ≤ i+1 + i+1 = i+1 k. 2 2 2 This implies that

1 d(γ(Ak), ak (x)y) = 0. k By Busemann convexity if follows easily that the obtained geodesic ray is unique. lim

k→∞

Acknowledgements: Laurent¸iu Leu¸stean was supported by a grant of the Romanian National Authority for Scientific Research, CNCS - UEFISCDI, project number PN-II-IDPCE-2011-3-0383. Adriana Nicolae was supported by a grant of the Romanian Ministry of Education, CNCS - UEFISCDI, project number PN-II-RU-PD-2012-3-0152.

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