Unique Ergodicity

Invariant Measures for Continuous Maps

4.3 Unique Ergodicity

Exercise 4.2.5.Let T :X →X be a continuous map on a compact metric space. Show that the measures inE^T(X) constrain all the ergodic averages in the following sense. Forf ∈C(X), define

m(f) = inf

μ∈E^T(X)

fdμ

and

M(f) = sup

μ∈E^T(X)

fdμ

.

Prove that

m(f)lim inf

N→∞A^f_N(x)lim sup

N→∞ A^f_N(x)M(f) for anyx∈X.

4.3 Unique Ergodicity

A natural distinguished class of transformations are those for which there is only one invariant Borel measure. This measure is automatically ergodic, and the uniqueness of this measure has several powerful consequences.

Definition 4.9.LetX be a compact metric space and let T :X →X be a continuous map. ThenT is said to beuniquely ergodicifM^T(X) comprises a single measure.

Theorem 4.10.For a continuous map T : X → X on a compact metric space, the following properties are equivalent.

(1)T is uniquely ergodic.

(2)|E^T(X)|= 1.

(3)For every f ∈C(X),

A^f_N = 1 N

N−1 n=0

f(Tⁿx)−→C_f, (4.4)

whereC_f is a constant independent ofx.

(4)For every f ∈C(X), the convergence(4.4)is uniform across X.

(5)The convergence (4.4)holds for everyf in a dense subset ofC(X).

Under any of these assumptions, the constantCf in(4.4)is

Xfdμ, whereμ is the unique invariant measure.

We will make use of Theorem 4.8 for the equivalence of (1) and (2); the equivalence between (1) and (3)–(5) is independent of it.

Proof of Theorem 4.10. (1) ⇐⇒ (2): If T is uniquely ergodic and μis the only T-invariant probability measure onX, then μ must be ergodic by Theorem4.4. If there is only one ergodic invariant probability measure onX, then by Theorem4.8, it is the only invariant probability measure onX.

(1) =⇒ (3): Let μ be the unique invariant measure for T, and apply Theorem4.1 to the constant sequence (δx). Since there is only one possible limit point andM(X) is compact, we must have

1 N

N−1 n=0

δTⁿx−→μ

in the weak*-topology, so for anyf ∈C(X) 1

N

N−1 n=0

f(Tⁿx)−→

X

fdμ.

(3) =⇒ (1): Let μ ∈ M^T(X). Then by the dominated convergence theorem, (4.4) implies that

X

fdμ=

X

lim

N→∞

1 N

N−1 n=0

f(Tⁿx) dμ=C_f

for allf ∈C(X). It follows thatCf is the integral of f with respect to any measure inM^T(X), soM^T(X) can only contain a single measure.

Notice that this also showsCf =

Xfdμfor the unique measureμ.

(1) =⇒ (4): Letμ∈M^T(X), and notice that we must haveCf = fdμ as above. If the convergence is not uniform, then there is a functionginC(X) and anε >0 such that for everyN0 there is anN > N0 and a pointxj ∈X

for which

1 N

N−1 n=0

g(Tⁿx_j)−C_g ε.

Letμ_N = _N¹ N−1

n=0 δ_Tnxj, so that

X

gdμN −Cg

ε. (4.5)

By weak*-compactness the sequence (μN) has a subsequence μ_N(k)

with μ_N(k)→ν

ask→ ∞. Thenν∈M^T(X) by Theorem4.1, and

X

gdν−Cg

ε

4.3 Unique Ergodicity 107

by (4.5). However, this shows thatμ=ν, which contradicts (1).

(4) =⇒ (5): This is clear.

(5) =⇒ (1): Ifμ, ν∈E^T(X) then, just as in the proof that (3) =⇒ (1),

X

fdν=C_f =

X

fdμ

for any functionf in a dense subset ofC(X), soν =μ.

The equivalence of (1) and (3) in Theorem4.10appeared first in the paper of Kryloff and Bogoliouboff [214] in the context of uniquely ergodic flows.

Example 4.11.The circle rotationRα:T→Tis uniquely ergodic if and only ifαis irrational. The unique invariant measure in this case is the Lebesgue measure m_T. This may be proved using property (5) of Theorem 4.10 (or using property (1); see Theorem 4.14). Assume first that α is irrational, so e^2πikα= 1 only ifk= 0. Iff(t) = e^2πikt for some k∈Z, then

1 N

N−1 n=0

f(Rⁿ_αt) = 1 N

N−1 n=0

e^2πik(t+nα)=

⎧⎨

⎩

1 ifk= 0;

1

Ne^2πikte^{2πiN kα}−1

e^2πikα−1 ifk= 0.

(4.6) Equation (4.6) shows that

1 N

N−1 n=0

f(Rⁿ_αt)−→

fdm_T=

1 ifk= 0;

0 ifk= 0.

By linearity, the same convergence will hold for any trigonometric polynomial, and therefore property (5) of Theorem4.10holds. For a curious application of this result, see Example 1.3.

If α is rational, then Lebesgue measure is invariant but not ergodic, so there must be other invariant measures.

Example 4.11 may be used to illustrate the ergodic decomposition of a particularly simple dynamical system.

Example 4.12.LetX ={z∈C| |z|= 1 or 2}, letαbe an irrational number, and define a continuous map T : X → X by T(z) = e^2πiαz. By unique ergodicity on each circle, any invariant measureμtakes the form

μ=sm₁+ (1−s)m₂,

wherem₁andm₂denote Lebesgue measures on the two circles comprisingX.

ThusM^T(X) ={sm₁+ (1−s)m₂|s∈[0,1]}, with the two ergodic measures given by the extreme points s = 0 and s = 1. The decomposition of μ is described by the measureν =sδm₁+ (1−s)δm₂. A convenient notation for this isμ=

M^T(X)mdν(m).

Example 4.13.A more sophisticated version of Example4.12is a rotation on the disk. LetD={z∈C| |z|1}, letαbe an irrational number, and define a continuous mapT :D→ DbyT(z) = e^2πiαz.For each r ∈(0,1], let mr

denote the normalized Lebesgue measure on the circle {z ∈ C | |z| = r} and let m0 =δ0 (these are the ergodic measures). Then the decomposition ofμ∈M^T(X) is a measureν on{m_r|r∈[0,1]}, and

μ(A) =

M^T(X)

mr(A) dν(mr).

Both Proposition 2.16 and Example4.11are special cases of the following more general result about unique ergodicity for rotations on compact groups.

Theorem 4.14.Let X be a compact metrizable group and Rg(x) =gx the rotation by a fixed elementg∈X. Then the following are equivalent.

(1)Rg is uniquely ergodic (with the unique invariant measure beingmX, the Haar measure onX).

(2)Rg is ergodic with respect to mX.

(3)The subgroup{gⁿ}n∈Z generated by g is dense in X.

(4)X is abelian, andχ(g)= 1 for any non-trivial characterχ∈X. Proof.(1) =⇒ (2): This is clear.

(2) =⇒ (3): Let Y denote the closure of the subgroup generated byg.

If Y = X then there is a continuous non-constant function on X that is constant on each coset ofY: in fact ifdis a bi-invariant metric onX giving the topology, then

dY(x) = min{d(x, y)|y∈Y}

defines such a function (an invariant metric exists by Lemma C.2). Such a function is invariant underRg, showing thatRgis not ergodic.

(3) =⇒ (1): IfY =X thenX is abelian (since it contains a dense abelian subgroup), and any probability measure μ invariant under Rg is invariant under translation by a dense subgroup. This implies thatμis invariant under translation by anyy ∈X by the following argument. Let f ∈C(X) be any continuous function, and fix ε > 0. Then for every δ > 0 there is some n withd(y, gⁿ)< δ, so by an appropriate choice ofδwe have

|f(gⁿx)−f(yx)|< ε for allx∈X. Since

f(x) dμ(x) =

f(gⁿx) dμ(x), it follows that

f(yx) dμ(x)−

f(x) dμ(x) =

(f(yx)−f(gⁿx)) dμ(x) < ε

4.3 Unique Ergodicity 109

for allε >0, soRy preservesμ. Since this holds for ally∈X,μmust be the Haar measure. It follows thatRg is uniquely ergodic.

(4) =⇒ (2): Assume now that X is abelian andχ(g)= 1 for every non- trivial characterχ∈X. Iff ∈L²(X) is invariant underR_g, then the Fourier series

f =

χ∈Xb

c_χχ

satisfies

f =URgf =

χ∈Xb

cχχ(g)χ,

and sof is constant as required.

(2) =⇒ (4): By (3) it follows that X is abelian. If now χ ∈ X is a character withχ(g) = 1, then

χ(Rgx) =χ(g)χ(x) =χ(x)

is invariant, which by (2) implies thatχis itself a constant almost everywhere

and so is trivial.

Corollary 4.15.Let X = T, and let g = (α1, α2, . . . , α) ∈R. Then the toral rotationRg:T→T given byRg(x) =x+g is uniquely ergodic if and only if1, α1, . . . , α are linearly independent over Q.

Theorems 2.19 and4.14have been generalized to give characterizations of ergodicity for affine maps on compact abelian groups by Hahn and Parry [131]

and Parry [278], and on non-abelian groups by Chu [57].

Exercises for Sect. 4.3

Exercise 4.3.1.Prove that (3) =⇒ (1) in Theorem4.14using Pontryagin duality.

Exercise 4.3.2.Show that a surjective homomorphism T : X → X of a compact groupX is uniquely ergodic if and only if|X|= 1.

Exercise 4.3.3.Extend Theorem4.14by using the quotient spaceY\X of a compact groupX to classify the probability measures onX invariant under the rotationRg whenY =X.

Exercise 4.3.4.Show that for any Riemann-integrable function f :T→R andε >0 there are trigonometric polynomialsp⁻ andp⁺ such that

p⁻(t)< f(t)< p⁺(t)

for all t ∈ T, and 1

0(p⁺(t)−p⁻(t)) dt < ε. Use this to show that if α is irrational then for any Riemann-integrable functionf :T→R,

1 N

N−1 n=0

f(Rⁿ_αt)→

fdm_T for allt∈T.

Exercise 4.3.5.Prove Corollary 4.15 (a) using Theorem4.14;

(b) using Theorem4.10(5).

Exercise 4.3.6.⁽⁵²⁾ LetX be a compact metric space, and letT :X →X be a continuous map. Assume that μ ∈ E^T(X), and that for every x ∈ X there exists a constantC=C(x) such that for everyf ∈C(X),f 0,

lim sup

N→∞

1 N

N−1 n=0

f(Tⁿx)C

fdμ.

Show thatT is uniquely ergodic.