Lecture 35:
Recall that we are in the process of proving Bowen, Theorem 1.22.
Recall statements of Lemmas 1,2 and 3 and defn. of Gibbs-Bowen
measure.
Lemma 4: For x, y ∈ M+
C with xi = yi for i = 0, . . . , m − 1,
|Smφ(x) − Smφ(y)| ≤ a :=
∞
X
vark (φ).
k=0
Proof:
|Smφ(x)−Smφ(y)| ≤
m−1
X
i
i
|φ(σ (x))−φ(σ (y))| ≤
i=0
m−1
X
varm−1−i ≤ a.
i=0
Lemma 5: µ := hν (from the Ruelle PF Theorem) is a GibbsBowen measure for φ.
Proof: Fix x ∈ MC+.
For any z ∈ MC+, There is at most one y ∈ σ −m(z) s.t.
y ∈ E := Ex0...xm−1 . By Lemmas 0(a) and Lemma 3,
X
m
(L (hχE ))(z) =
eSmφ(y)h(y)χE (y) ≤ eSmφ(x)ea||h||
y∈σ −m (z)
Thus, with λ as the eigenvalue from Ruelle PF,
Z
µ(E) = ν(hχE ) = (1/λm)L∗ν(hχE ) = (1/λm) (Lm(h · χE ))dν(z)
≤ (1/λm)eSmφ(x)ea||h||
Letting P := log λ and c2 := ea||h||, we obtain the upper bound
in the defn of Bowen-Gibbs.
1
Let N be s.t. C N > 0 (recall that C is primitive). For any
z ∈ MC+, there is at least one y 0 ∈ σ −m−N (z) s.t. y 0 ∈ E. Then
0
(Lm+N (hχE ))(z) ≥ eSm+N φ(y )h(y 0)
≥ e−N ||φ||−a(inf h)eSmφ(x)
So
µ(E) = ν(hχE ) = (1/λm+N )ν(Lm+N (hχE )) ≥ c1λ−meSmφ(x)
where c1 := λ−N e−N ||φ||−a(inf h); this gives the lower bound in the
defn of Bowen-Gibbs. Note: from Lemma 3 and proof of Lemma 5, we see that P (φ) =
P = log λ.
At this point, it can be shown that µ := hν is the unique GibbsBowen measure for φ (Bowen, last part of Theorem 1.16). However,
we will not need this fact and it will fall out later.
Lemma 6: Any Gibbs-Bowen measure for φ is an equilibrium state
for φ.
Proof: Let P = P (φ). By the variational principle,
Z
hµ(σ|MC ) + φdµ ≤ P
for all invariant measures µ (in this case, this half of the variational
principle is much easier than the general case).
Let µ be a Gibbs-Bowen measure for φ.
For each nonempty initial cylinder set E := Ea0...am−1 in MC
choose x = xE ∈ E that achieves min Smφ(x) on E. Then
Z
−µ(E) log µ(E) +
Smφ dµ ≥ −µ(E)(log µ(E) − Smφ(xE ))
E
2
≥ −µ(E)(log(c2e−P m+Smφ(xE )) − Smφ(xE )) = µ(E)(P m − log c2)
So,
Hµ(α[0,m−1]) +
Z
Smφdµ ≥ P m − log c2
So,
Z
hµ(σ|MC ) +
φdµ = lim (1/m)(Hµ(α[0,m−1]) +
m→∞
Z
Smφdµ)
≥ lim (1/m)(P m − log c2) = P
m→∞
Note: Nishant will prove another version of Lemma 6 this afternoon.
Now we will prove uniqueness of the equilibrium state. We will
need Lemmas 7,8,9.
Lemma 7: Given 0 < s ≤ 1 and ai ∈ R,
X
X
X
−
pi log pi +
piai ≤ s(log(
eai ) − log s)
i
i
i
P
assuming pi ≥ 0, i pi = s.
Proof: After dividing by s and transposing − log s, the inequality
above becomes
X
X
X
−
(pi/s) log(pi/s) +
(pi/s)ai ≤ log(
eai ),
i
i
P
i
subject to i pi/s = 1. So this Lemma follows from our earlier
lemma (page 4 of Lectures 30-32). Lemma 8 (a special case of Bowen, Lemma 1.23): Let µ1 ⊥ µ2 be
mutually singular invariant Borel probability measures on a two-sided
shift space. Then there exists a sequence Rm of unions of elements
of α[0,m−1] s.t. µ1(Rm) → 1 and µ2(Rm) → 0.
3
Proof: There is a Borel set B s.t. µ1(B) = 1 and µ2(B) = 0.
Let > 0 and let K ⊆ B be a compact set s.t. µ1(K) > 1 − .
Let Rm be the union of all elements of α[−m,m] which intersect K.
Then, each Rm contains K and so µ1(Rm) > 1 − .
Since the diameters of elements of α[−m,m] tend to zero and K is
compact, we have Rm ↓ K.
So, µ2(Rm) → µ2(K) ≤ µ2(B) = 0. So, for some m, µ2(Rm) < .
For a sequence n → 0, choose corresponding mn s.t. µ1(Rmn ) >
1 − n and µ2(Rmn ) < n.
By shifting the Rmn to the right, we get the desired sequence. Lemma 9: Let T be a cts. map of a compact metric space and
ν1, ν2 ∈ M(T ). If T is ergodic w.r.t. ν1 and ν2 << ν1, then ν2 = ν1.
Proof: By the Radon-Nikodym theorem, for some f ∈ L1(ν1),
f ≥ 0, we have ν2 = f ν1. Let T ∗ be the induced action of T on
M(T ). From the invariance of ν1 and ν2, one can show that f is
invariant, i.e., f = f ◦ T a.e. (ν1) – see Walters, pp. 152-153 for a
version of this. Since µ1 is ergodic, f is constant a.e. (ν1). Since ν1
and ν2 are both probability measures, the constant must be 1, and
so ν2 = ν1. Nishant’s alternative proof of Lemma 9: First we claim that
ν2 is also ergodic. To see this, first note that if T −1(A) = A, then
ν1(A) = 0 or 1. If ν1(A) = 0, then ν2(A) = 0; if ν1(A) = 1, then
ν1(Ac) = 0 and so ν2(Ac) = 0 and so ν2(A) = 1.
Let G be the set of ν1-generic points for a countable basis for the
topology, i.e.,
G = {x ∈ M : (1/n)
n−1
X
i=0
4
χU (T ix) → ν1(U )}
for all elements U of the basis.
By the ergodic theorem, ν1(G) = 1 and so ν2(G) = 1 and so for
a set of ν2-measure one, the frequency of visits to each basis element
U is ν1(U ); thus, by the ergodic theorem, ν2(U ) = ν1(U ). 5
Lecture 36:
Recall statement of Bowen’s Theorem 1.22, defn. of Gibbs-Bowen
measure, fact that µ := hν is Gibbs-Bowen and is an equilibrium
state and the following lemmas:
Lemma 7: Given 0 < s ≤ 1 and ai ∈ R,
X
X
X
−
pi log pi +
piai ≤ s(log(
eai ) − log s)
i
assuming pi ≥ 0,
i
P
i pi
i
= s.
Lemma 8: Let µ1 ⊥ µ2 be mutually singular invariant Borel probability measures on a two-sided shift space. Then there exists a
sequence Rm of unions of elements of α[0,m−1] s.t. µ1(Rm) → 1 and
µ2(Rm) → 0.
Lemma 9: Let T be a cts. map of a compact metric space and
ν1, ν2 ∈ M(T ). If T is ergodic w.r.t. ν1 and ν2 << ν1, then ν2 = ν1.
Proof of uniqueness of the equilibrium state: Let λ be an equilibrium state and µ, the Gibbs-Bowen measure given by Ruelle PF:
µ = hν. Will show λ = µ.
DO NOT CONFUSE λ with the Ruelle PF eigenvalue.
All we need regarding µ is that it is Gibbs-Bowen and ergodic.
Case 1: λ ⊥ µ
Fix m and let Rm be as in Lemma 8, with ν1 = λ and ν2 = µ.
For each E ∈ α[0,m−1], choose xE ∈ E that achieves maxx∈E Smφ(x).
From now on, E implicitly means that E ∈ α[0,m−1].
We have
Z
mP = mhλ(σ|MC ) + m
φdλ ≤ Hλ(α[0,m−1]) +
6
Z
Smφdλ
≤
X
λ(E)(Smφ(xE )−log λ(E))+
X
λ(E)(Smφ(xE )−log λ(E))
c
E⊆Rm
E⊆Rm
which by Lemma 7 is
X
X
Sm φ(xE )
c
≤ λ(Rm) log
e
+ (λ(Rm)) log
eSmφ(xE ) + 2K ∗
c
E⊆Rm
E⊆Rm
where K ∗ = maxs∈[0,1] −s log s. Thus
X
X
∗
Sm φ(xE )−mP
c
−2K ≤ λ(Rm) log
e
+(λ(Rm)) log
eSmφ(xE )−mP
c
E⊆Rm
E⊆Rm
and since µ is a Gibbs-Bowen measure, this is
−1
c
c
≤ λ(Rm) log c−1
1 µ(Rm ) + (λ(Rm )) log c1 (µ(Rm ))
c
c
≤ log c−1
1 + λ(Rm ) log µ(Rm ) + (λ(Rm )) log(µ(Rm ))
But, as m → ∞, the 2nd term tends to −∞ and the third term
tends to 0, a contradiction.
Perspective on proof in Case 1: Assume the special case that φ =
0. Then P = P (φ) = htop(T ).
Then the beginning of proof above shows: (letting |S|∗ denote the
number of cylinders sets of “span” m that comprise S)
c
c ∗
mhtop(T ) ≤ λ(Rm) log |Rm|∗ + λ(Rm
) log |Rm
| + 2K ∗
(1)
Since µ is a Gibbs-Bowen measure,
|Rm|∗ ≈ emhtop(T )µ(Rm),
c ∗
c
|Rm
| ≈ emhtop(T )µ(Rm
),
Since λ(Rm) → 1 and µ(Rm) → 0, we see that in (??), the first
term grows like (1 + o(1))(mhtop(T ) + log o(1)) and the second term
grows like (o(1))((mhtop(T )) + log(1 + o(1)).
Case 2: λ is not singular w.r.t. µ.
7
For an invariant measure ρ, write
Z
Pρ(φ) = hρ(T ) +
φdρ
By the Lebesgue Radon Nikodym Decomposition, we can write λ as
the sum of two measures, one of which is mutually singular w.r.t. µ
and the other absolutely continuous w.r.t. µ. Since λ is a probability
measure, these two measures are finite measures, whose measures
sum to 1. Thus, by normalizing we can write
λ = βλ0 + (1 − β)µ0
where λ0 ⊥ µ, µ0 << µ, β ∈ [0, 1].
If β = 0, then λ = µ by Lemma 9. So we may assume β 6= 0.
We claim that λ0 and µ0 are shift-invariant. To see this, first note
that
λ = λ ◦ T −1 = βλ0 ◦ T −1 + (1 − β)µ0 ◦ T −1
By uniqueness of the Decomposition, it is enough to show that a)
µ0 ◦ T −1 << µ and b) λ0 ◦ T −1 ⊥ µ.
a): If µ(E) = 0, then µ(T −1(E) = 0 and so µ0 ◦ T −1(E) = 0; so,
µ0 ◦ T −1 << µ.
b): There is a Borel set B s.t. λ0(B) = 1, µ(B) = 0. Thus, λ(B) =
βλ0(B) + (1 − β)µ0(B) = β; this together with µ0(T −1(B) = 0 gives:
β = λ(B) = λ(T −1(B)) = βλ0(T −1(B))+(1−β)(µ0(T −1(B)) = βλ0(T −1(B))
and so λ0(T −1(B)) = 1; so, λ0 ◦ T −1 ⊥ µ.
Since the entropy function is affine, we have
P = Pλ(φ) = βPλ0 (φ) + (1 − β)Pµ0 (φ)
Since Pµ0 (φ) ≤ P , β 6= 0 and by Case 1, Pλ0 (φ) < P , this is a
contradiction. 8