Problem Sheets: for Fast-Slow Skew Product Systems and
Convergence to Stochastic Differential Equations
Ian Melbourne
16 April 2015
1
Problem Sheet 1
Problem 1.1 (a) Verify conditions (U4) and (P4) from Section 1.2.
(b) Show that P (vU w) = wP v for all v, w ∈ L2 (Λ).
Solution 1.1R (a) Note that URvU w = (v ◦f )(wR◦f ) = (vw)◦f . Since µ is f -invariant,
we have that Λ U v U w dµ = Λ (vw) ◦ f dµ = Λ vw dµ proving (U4).R
R Next, it follows
R from the definition of P together with (U4) that Λ P U v w dµ =
U v U w dµ = Λ vw dµ. Since w is arbitrary, (P4) follows.
Λ
(b) Let z ∈ L∞ (Λ). Then
Z
Z
Z
Z
Z
P (vU w) z dµ = (vU w)(U z) dµ =
v U (wz) dµ =
P v (wz) dµ =
wP v z dµ.
Λ
Λ
Λ
Λ
Λ
The result follows since z is arbitrary.
Problem 1.2 Theorem 1.7 proves decay of correlations for the doubling map for v
Lipschitz. Generalise this to α-H¨older observables v : Λ → R where α ∈ (0, 1) (cf.
Remark 1.8(c)).
Solution 1.2 Starting from Proposition 1.4, and proceeding as in the proof of Proposition 1.5,
y 1 x + 1 y + 1 1 x |(P v)(x) − (P v)(y)| ≤ v
−v
+
v
−
v
2
2
2
2
2
2
x y α 1
x + 1 y + 1 α
x y α
1
1
≤ |v|α − + |v|α −
= |v|α − = α |v|α |x − y|α .
2
2 2
2
2
2
2 2
2
It follows that |P v|α ≤ 1/2α |v|α .
1
R
Next, as in the proof of Corollary 1.6, |w − Λ w dµ|∞ ≤ |w|α diam Λα = |w|α for
all H¨older w, so using (P2),
R
R
|P n v − Λ v dµ|∞ = |P n v − Λ P n v dµ|∞ ≤ |P n v|α ≤ ( 21α )n |v|α .
Finally, as in the proof of Theorem 1.7,
Z
Z
Z
Z
1 n
n
n
v w ◦ f dµ − v dµ w dµ ≤ P v − v dµ |w|1 ≤ α |v|α |w|1 ,
2
∞
Λ
Λ
Λ
Λ
for all v H¨older, w ∈ L1 (Λ), n ≥ 1.
Problem 1.3 (
Let Λ = [0, 1]2 with Lebesgue measure µ and consider the skew product
( 2x , 2y mod 1 )
x < 21
map f (x, y) =
. Compute the transfer operator for f .
( 2x − 1 , 3y mod 1 ) x ≥ 12
Solution 1.3 Write
R
Λ
P v w dµ =
P5
j=1
R
Aj
A1 = [0, 12 ] × [0, 12 ],
A3 = [ 21 , 1] × [0, 13 ],
P v w dµ where
A2 = [0, 12 ] × [ 12 , 1],
A4 = [ 21 , 1] × [ 13 , 23 ],
Now for instance
Z
Z
P v w dµ =
A5 = [ 12 , 1] × [ 23 , 1].
1
v(x, y)w(2x − 1, 3y) dxdy =
6
[ 21 ,1]×[0, 13 ]
A3
Z
x + 1 y v
, w(x, y) dxdy
2
3
Λ
and similarly for the other four terms. Altogether we obtain that
1 x y 1 x y + 1
+ v ,
(P v)(x, y) = v ,
4 2 2
4 2
2
1 x + 1 y + 1 1 x + 1 y + 2
1 x+1 y
+ v
,
+ v
,
+ v
,
.
6
2
3
6
2
3
6
2
3
Problem 1.4 Verify the first statement in the proof of Proposition 1.10. Namely,
for v satisfying condition (1.2), prove that |P n v|∞ ≤ Cn−β .
Solution 1.4 Fix v and n. Let A = Cn−β and suppose for contradiction that
|P n v|∞ > A. Then there exists δ > 0 and a measurable set E with µ(E) > 0
such that |(P n v)(y)| ≥ A + δ for all y ∈ E. Let w = 1E sgn P n v. Then
Z
Z
Z
n
n
v w ◦ f dµ =
P v w dµ =
1E |P n v| dµ ≥ (A + δ)µ(E).
Λ
Λ
Λ
Also, |w|1 = µ(E) and it follows from condition (1.2) that
Z
(A + δ)µ(E) ≤
v w ◦ f n dµ ≤ A|w|1 = Aµ(E),
Λ
which contradicts the assumption that µ(E) > 0.
2
Problem 1.5 Verify the statement in Remark 1.12. (You may assume without proof
that if An →d A and Bn →d b where b is a constant, then An + Bn →d A + b.)
Hint: Estimate µ(|Yn | > c) using Markov’s inequality, where Yn = n−1/2 (vn − mn ).
(Recall that Markov’s inequality states that if X is a nonnegative random variable,
then P(X > λ) ≤ λ−1 EX for all λ > 0.)
Solution 1.5 Since χ ∈ L1 (Λ), we have |vn − mn |1 ≤ |χ ◦ f n − χ|1 ≤ 2|χ|1 . Hence
|Yn |1 = O(n−1/2 ) → 0. By Markov’s inequality
Z
−1
µ(|Yn | > a) ≤ a
|Yn | dµ → 0,
Λ
for all a > 0, so by definition Yn →d 0. But n−1/2 vn = n−1/2 mn + Yn so the result
follows.
3
2
Problem Sheet 2
Problem 2.1 Prove Proposition 2.1.
Hint: For R the off diagonal terms, Ruse polarization.
limn→∞ n−1 Λ (vni + vnj )2 dµ − limn→∞ n−1 Λ (vni − vnj )2 dµ.
That is, compute
Solution 2.1Z If i = j, then
Z we are in the one-dimensional case covered by Proposi1
tion 1.14:
(v i )2 dµ → (mi )2 dµ.
n Λ n
Λ
For the case i 6= j, consider the real-valued observable v i + v j : Λ → R with
decomposition v i +v j = mi +mj +(χi +χj )◦f −(χi +χj ). Note that vni +vnj = (v i +v j )n .
By Proposition 1.14,
Z
Z
1
i
j 2
(v + vn ) dµ → (mi + mj )2 dµ.
n Λ n
Λ
Similarly,
Z
Z
1
i
j 2
(v − vn ) dµ → (mi − mj )2 dµ.
n Λ n
Λ
Taking the difference yields
Z
Z
4
i j
v v dµ → 4 mi mj dµ,
n Λ n n
Λ
as required.
Problem 2.2 Prove Proposition 2.2.
Hint: In the d-dimensional L´evy continuity theorem, write t = sc where s ∈ R and
c ∈ Rd . (The lack of uniqueness does not matter.)
Solution 2.2 The d-dimensional L´evy continuity theorem (Lemma 1.16) states that
Yn →d Y if and only if limn→∞ E(eit·Yn ) = E(eit·Y ) for all t ∈ Rd . Taking t = sc
where s ∈ R and c ∈ Rd , it is equivalent that limn→∞ E(eis(c·Yn ) ) = E(eis(c·Y ) ) for all
s ∈ R, c ∈ Rd . By the one-dimensional case of Lemma 1.16, this holds if and only if
c · Yn →d c · Y for all c ∈ Rd .
Problem 2.3 Prove the first statement of Example 2.7(b). Namely, show that if
Wn →w W in C[0, 1], then χ(Wn ) →d χ(W ) in Rk for any continuous function
χ : C[0, 1] → Rk and any k ≥ 1.
Solution 2.3 Let c ∈ Rk and define χc : C[0, 1] → R by setting χc (g) = c · χ(g).
Then χc : C[0, 1] → R is continuous, so it follows from Theorem 2.6 that χc (Wn ) →d
χc (W ). In other words c · χ(Wn ) →d c · χ(W ). Since this holds for all c ∈ Rk , it
follows from Cramer-Wold that χ(Wn ) →w χ(W ).
4
Problem 2.4 Prove Proposition 2.9.
Hint: For part (a), consider the partition {En } where E0 = {Y = 0} and En = {|Y | ∈
(n − 1, n]} for n ≥ 1.
P
Solution 2.4 (a) Since {En } is a partition of a probability space, ∞
P(En ) = 1.
n=0
PL
It follows that for any > 0, there exists L ≥ 1 such that P(|Y | ≤ L) = n=0 P(En ) ≥
1 − .
(b) Let > 0. By part (a), there exists L0 > 0 such that P(|Y | ≤ L0 ) > 1 − /2.
Since Yn →d Y , there exists N ≥ 1 such that |P(|Yn | ≤ L0 ) − P(|Y | ≤ L0 )| < /2
for all n > N . Hence P(|Yn | ≤ L0 ) > 1 − for all n > N . Again, by part (a),
for each n = 1, . . . , N there exists Ln such that P(|Yn | ≤ Ln ) > 1 − . Now let
L = max{L0 , L1 , . . . , LN ).
Problem 2.5 (a) Let f : Λ → Λ be an ergodic measure-preserving transformation.
Suppose that u ∈ Lp (Λ) for some p ≥ 1. Show that n−1/p u ◦ f n → 0 a.e.
Hint: Apply the ergodic theorem to up .
(b). Let an be a real sequence and suppose that n−b an → 0 where b > 0. Show that
n−b maxj=1,...,n |aj | → 0.
(c). Suppose that the martingale-coboundary decomposition (1.3) holds where m, χ ∈
L2 (Λ). Show that the WIP for v holds if and only if the WIP holds for m (with the
same limit process). Note that it suffices to show that maxt∈[0,1] |Wn (t) − Mn (t)| →
0 a.e.
Solution 2.5 (a) Since u ∈ Lp (Λ), we have that up ∈ L1 (Λ). By the ergodic theorem
Z
Z
n
n−1
X
X
−1 p
n
−1
p
j
−1
p
j
p
n u ◦f =n
u ◦f −n
u ◦f →
u dµ − up dµ = 0 a.e.
j=0
Λ
j=0
Λ
Now take p’th roots.
(b) Given > 0, choose n0 ≥ 1 such that n−b |an | < for all n ≥ n0 . Let M =
max{|a1 |, . . . , |an0 |} and choose n1 ≥ n0 such that n−b
1 M < .
−b
−b
Suppose n ≥ n1 . Then n maxj≤n0 |aj | ≤ n1 M < and n−b maxn0 ≤j≤n |aj | ≤
maxn0 ≤j≤n j −b |aj | ≤ maxj≥n0 j −b |aj | < , so n−b maxj≤n |aj | < .
(c) On the interval [0, 1], we have
max |Wn (t) − n−1/2 v[nt] | ≤ n−1/2 max |v ◦ f j |,
t
j≤n
max |Mn (t) − n
−1/2
t
−1/2
max |n
t
v[nt] − n
m[nt] | ≤ n
−1/2
−1/2
m[nt] | ≤ n
5
max |m ◦ f j |,
j≤n
−1/2
max |χ ◦ f j | + n−1/2 |χ|.
j≤n
Ignoring the n−1/2 χ term in the third of these (which clearly converges to zero a.e.)
each error term is of the form n−1/2 maxj≤n |u ◦ f j | where u ∈ L2 (Λ).
By part (a) (with p = 2), n−1/2 u ◦ f n → 0 a.e. By part (b) (with b = 21 and
an = u ◦ f n ), n−1/2 maxj≤n u ◦ f j → 0 a.e. Hence maxt |Wn (t) − Mn (t)| → 0 a.e.
6
3
Problem Sheet 3
Problem 3.1 Let B be the underlying σ-algebra on Λ.
(a) Show that 1f −1 B = U 1B for all B ∈ B.
(b) Show that U P is the conditional expectation operator given by U P v = E(v|f −1 B)
for v ∈ L1 (Λ).
(c) Prove that if m ∈ ker P , then E(m ◦ f n |f −(n+1) B) = 0 for all n ≥ 0.
Solution 3.1 (a) 1f −1 B (y) = 1 if and only if y ∈ f −1 B if and only if f y ∈ B if and
only if U 1B = 1B ◦ f = 1.
(b) Clearly U P v = (P v) ◦ f is f −1 B-measurable.
Let A = f −1 B ∈ f −1 B where B ∈ B. Then
Z
Z
Z
U P v dµ = U P v 1A dµ = U P v 1f −1 B dµ.
A
By part (a) and (U4),
Z
Z
Z
Z
Z
v dµ,
U P v dµ = U P v U 1B dµ = P v 1B dµ = v U 1B dµ =
A
A
as required.
(c) By Proposition 3.3(b), with π = f n : Λ → Λ, and using part (b),
E(m ◦ f n |f −(n+1) B) = E(m|f −1 B) ◦ f n = U n E(m|f −1 B) = U n+1 P m = 0.
Problem 3.2 Prove Proposition 3.9.
Solution 3.2 Clearly m
˜ ◦ f˜−j is Fj -measurable and hence Fn -measurable for each
j ≤ n. It follows that m
˜−
n is Fn -measurable for each n ≥ 1 so the first condition in
Definition 3.1 is satisfied. Moreover, E(m
˜−
˜−
˜ ◦ f˜−(n+1) |Fn ) so it
n+1 |Fn ) = m
n + E(m
remains to show that E(m
˜ ◦ f˜−(n+1) |Fn ) = 0.
It follows from Proposition 3.3(b) that
E(m
˜ ◦ f˜−(n+1) |Fn ) = E(m
˜ ◦ f˜−(n+1) |f˜n F0 ) = E(m|
˜ f˜−1 F0 ) ◦ f˜−(n+1) .
Moreover,
E(m|
˜ f˜−1 F0 ) = E(m ◦ π|π −1 f −1 F0 ) = E(m|f −1 F0 ) ◦ π.
Finally, E(m|f −1 F0 ) = U P m = 0.
7
fn ◦ f˜−n = χ(M
fn− ) where χ(g)(t) = g(1) − g(1 − t).
Problem 3.3 Prove that M
fn (t) ◦ f˜−n = M
f− (1) − M
f− (1 − t). Since
Solution 3.3 It is required to show that M
n
n
fn and M
fn− are defined using linear interpolation at the points j/n, j = 0, 1, . . . , n,
M
it suffices to check this equality at such values of t. In other words, it suffices to check
that
−1/2
n−1/2 m
˜ j ◦ f˜−n = n−1/2 m
˜−
m
˜ n−j ,
n −n
for each j, n. But
m
˜ j ◦ f˜−n =
j−1
X
−(n−j)−1
X
k
−n
˜
˜
m
˜ ◦f ◦f =
m
˜ ◦ f k,
k=0
k=−n
and
m
˜−
n
−m
˜ n−j =
−1
X
m
˜ ◦ f˜k −
k=−n
−1
X
−(n−j)−1
X
m
˜ ◦ f˜k =
k=−(n−j)
m
˜ ◦ f k.
k=−n
Problem 3.4 Prove Corollary 3.12.
Solution 3.4 Passing to the natural extension, we define the forward and backward
Birkhoff sums m
˜ n and m
˜−
˜−
n as before. Then m
n is a martingale with increments
−k
dk = m
˜ ◦ f and |dk |p = |m|p for all k, p. Hence it follows from Burkholder’s
1/2
inequality that max0≤j≤n−1 |m
˜−
where C = Cp |m|p .
j | p ≤ Cn
Now
max
0≤j≤n−1
|m
˜−
j |
=d max
0≤j≤n−1
|m
˜−
j |
−1
X
◦ f = max m
˜ ◦ f k ◦ f n
n
0≤j≤n−1
k=−j
n−1
X
k
= max m
˜ ◦ f = max |m
˜ n−j − m
˜ n|
0≤j≤n−1
k=n−j
0≤j≤n−1
≥ max (|m
˜ n−j | − |m
˜ n |) = max |m
˜ n−j | − |m
˜ n | = max |m
˜ j | − |m
˜ n |.
0≤j≤n−1
0≤j≤n−1
1≤j≤n
Hence
max |m
˜ j | p ≤ |m
˜ n |p + max |m
˜−
|
˜ n |p + Cn1/2 .
j p ≤ |m
1≤j≤n
0≤j≤n−1
Recall (by a simpler
argument)
that m
˜ n =d m
˜−
˜ n| ≤
n so by Burkholder’s inequality |m
Cn1/2 . Hence max1≤j≤n |m
˜ j | p ≤ 2Cn1/2 . Finally max1≤j≤n |mj |◦π = max1≤j≤n |m
˜ j|
1/2
so max1≤j≤n |mj | p ≤ 2Cn .
8
Problem 3.5 (a) Define the lap number N (t) = N (y, u, t) as in the proof of Theorem 3.16. Prove that
τN (t)
= τ¯ a.e.
lim
t→∞ N (t)
Hence prove that
N (t)
1
= a.e.
t→∞
t
τ¯
(b) Suppose that a : [0, ∞) → R is a function that is bounded on compact sets such
that limt→∞ a(t)/t = 0. Define un : [0, 1] → R, un (s) = a(ns)/n. Prove that
lim
lim sup un (s) = 0.
n→∞ s∈[0,1]
(Note: this is similar to Problem 2.5(b).)
(c) Let gn , g ∈ D[0, 1] be defined as in the proof of Theorem 3.16. Prove that
limn→∞ sup[0,1] |gn − g| = 0 a.e.
Solution 3.5 (a) We have the elementary estimates
C1 N (t) ≤ τN (t) ≤ t < τN (t)+1 ≤ C2 (N (t) + 1).
(3.1)
In particular, N (t) ≥ (t−1)/C2 , so limt→∞ N (t) = ∞. Hence, by the ergodic theorem
τN (t)
τn
= lim
= τ¯ a.e.
t→∞ N (t)
n→∞ n
lim
Also, using (3.1),
τN (t)
τN (t)+1
t
≤
<
,
N (t)
N (t)
N (t)
and so by the sandwich rule t/N (t) → τ¯ as t → ∞.
(b) Let > 0. Choose t0 > 0 so that |a(t)/t| < for all t ≥ t0 . Next, choose n0 such
that supt∈[0,t0 ] |a(t)|/n0 < .
Then for n ≥ 1,
sup |un (s)| =
s∈[t0 /n,1]
sup |a(ns)|/n ≤
s∈[t0 /n,1]
sup |a(ns)|/(ns) = sup |a(t)|/t < ,
s∈[t0 /n,1]
t∈[t0 ,n]
and for n ≥ n0 ,
sup |un (s)| =
s∈[0,t0 /n]
sup |a(ns)|/n = sup |a(t)|/n < .
s∈[0,t0 /n]
t∈[0,t0 ]
Combining these estimates, sup[0,1] |un | < for all n ≥ n0 .
(b) Set a(t) = N (t) − t/¯
τ . Then a(t)/t → 0 a.e. by part (a). Using (3.1), |a(t)| ≤
(1/C1 + 1/¯
τ )t so a is bounded on compact sets. Also a(ns)/n = gn (s) − g(s) so the
result follows from part (b).
9
Problem 3.6 Prove Proposition 4.1.
Solution 3.6 By change of variables,
y (t) = y1 (t−2 ) = φt−2 .
Hence
Z
t
−1
−1
Z
Z
0
0
˙ (t) = −1 v(y (t)).
In particular, W
10
t−2
v ◦ φs ds = W (t).
v ◦ φs−2 ds = v(y (s)) ds = 0
t
4
Problem Sheet 4
Problem 4.1 Prove Lemma 5.2.
Solution 4.1 (a) We compute that
n Z
X
(v i v j ◦ f k dµ − vˆi vˆj ◦ f k ) dµ
Λ
k=1
=
=
n Z
X
k=1 Λ
n Z
X
(χi ◦ f − χi ) v j ◦ f k + vˆi (χj ◦ f − χj ) ◦ f k dµ
i
(χ ◦ f n−k+1 − χi ◦ f n−k ) v j ◦ f n + vˆi (χj ◦ f k+1 − χj ◦ f k ) dµ
Λ
Zk=1
= (χi v j − vˆi χj ◦ f ) dµ + Ln
Λ
R
(ˆ
v i χj ◦ f n+1 − χi v j ◦ f n ) dµ → 0 as n → ∞ by the mixing assumption.
P[nt]−1
(b) Write v = vˆ + u, u = χ ◦ f − χ, and Un (t) = n−1/2 k=0 u ◦ f k . Then
Z t
Z t
Z t
Z t
ij
ij
i
j
i
j
i
j
c dW
c =
c i dU j .
Wn (t) − c
Wn (t) =
Wn dWn −
W
Un dWn +
W
n
n
n
n
where Ln =
Λ
0
0
0
0
Now
Z
t
Uni
dWnj
=n
−1
0
[nt]−1 `−1
X X
[nt]−1
i
k
j
`
−1
u ◦f v ◦f =n
`=0 k=0
X
(χi ◦ f ` − χi ) v j ◦ f `
`=0
[nt]−1
[nt]−1
= n−1
X
X
R
i j
(χi v j ) ◦ f ` − n−1 χi
`=0
vj ◦ f `,
`=0
which converges to t Λ χ v dµ a.e. by the ergodic theorem.
A similar
for the
term, after changing order of summation
R t argument
R remaining
i
j
i j
c
yields that 0 Wn dUn → −t Λ vˆ χ ◦ f dµ a.e.
Hence we have shown that
Z
ij
ij
c
Wn (t) − Wn (t) → t (χi v j − vˆi χj ◦ f ) dµ.
Λ
Now combine this with part (a).
Problem 4.2 Prove Corollary 5.3.
Solution 4.2 Since vˆ = m ∈ ker P , it follows from Corollary 1.13 that the formula
for A in Lemma 5.2 reduces to the given one. Since the process A is deterministic,
the weak limit of (Wn , Wn ) is given by (W, M) + (0, A) (noting footnote 10).
11