Renewal theorems and mixing for non Markov flows with infinite measure

0.3 MB·

by  Ian Melbourne

Renewal theorems and mixing for non Markov flows
7
1 with infinite measure
0
2

n
a Ian Melbourne ∗ Dalia Terhesiu †
J

9
29 January 2017
2

]
S
D
. Abstract
h
t
a We obtain results on mixing for a large class of (not necessarily Markov)
m
infinite measure semiflows and flows. Erickson proved, amongst other things,
[ a strong renewal theorem in the corresponding i.i.d. setting. Using operator

1 renewal theory, we extend Erickson’s methods to the deterministic (i.e. non-
v i.i.d.) continuous time setting and obtain results on mixing as a consequence.
0
4
4
8 1 Introduction
0
.
1 Recently, there has been increasing interest in the investigation of mixing properties
0
for infinite measure-preserving dynamical systems [2, 13, 22, 23, 24, 27, 28, 29, 30,
7
1 31, 32, 35, 36, 38]. Most of these results are for discrete time noninvertible systems.
:
v For results on semiflows preserving an infinite measure, we refer to [32] (the
i
X Markov case) and [13] (which does not assume a Markov structure). The setting
r is that F : Y → Y is a mixing uniformly expanding map defined on a probability
a
space (Y,µ) and τ : Y → R+ is a nonintegrable roof function with regularly varying
tails:
µ(y ∈ Y : τ(y) > t) = ℓ(t)t−β for various ranges of β ∈ [0,1]. (1.1)
Here, ℓ : [0,∞) → [0,∞) is a measurable slowly varying function (so
lim ℓ(λt)/ℓ(t) = 1 for all λ > 0). Consider the suspension (Yτ,µτ) and sus-
t→∞
pension semiflow F : Yτ → Yτ (the standard definitions are recalled in Section 3).
t
The aim is to prove a mixing result of the form
lim a vw◦F dµτ = vdµτ wdµτ,
t t
t→∞ ZYτ ZYτ ZYτ
∗Mathematics Institute, University of Warwick, Coventry, CV4 7AL, UK
†Department of Mathematics, University of Exeter, Exeter EX4 4QF, UK
1
forasuitablenormalisationa → ∞andsuitableclassesofobservablesv,w : Yτ → R.
t
Under certain hypotheses, [13, 32] obtained results on mixing and rates of mixing
for such semiflows. The hypotheses were of two types: (i) assumptions on “renewal
operators” associated to the transfer operator of F and the roof function τ, and (ii)
Dolgopyat-typeassumptions ofthetype used to obtainmixing ratesfor finitemeasure
(semi)flows [16].
As pointed out to us by Dima Dolgopyat, P´eter N´andori and Doma Sza´sz, mixing
for indicator functions can be regarded as a local limit theorem and hence hypotheses
of type (ii) should not be necessary.
In this paper, we show that operator renewal-theoretic assumptions (i) are indeed
sufficient for obtaining the mixing results in [13, 32]. The abstract framework in [13]
turns out again to be flexible enough to cover nonMarkov situations. Moreover, our
main results extend to flows and we are able to treat large classes of observables v,w.
(The methods in this paper are not appropriate for obtaining rates of mixing; the
best results currently are those in [13].)
The analogous probabilistic results go back to Erickson [19] who obtained strong
renewal theorems in an i.i.d. continuous time framework under the assumption
β ∈ (1,1]. (In the discrete time setting, see [21] for the i.i.d. case and [31] for
2
the deterministic case.) Our results on mixing when β ∈ (1,1] for semiflows (Corol-
2
lary 3.1 and the extensions in Section 7) and for flows (Theorem 8.5), are proved by
adapting Erickson’s methods to the deterministic setting.
For β ≤ 1, additional hypotheses are needed on the tail of τ to obtain a strong
2
renewal theorem (and hence mixing) even for discrete time; see [15, 18, 21] for i.i.d.
results and [22] for deterministic results (see also [36] for higher order theory in
both the i.i.d. and deterministic settings). For the continuous time case, in work
in progress, Dolgopyat & N´andori [17] obtain strong renewal theorems for a class of
Markov semiflows with β ≤ 1 (again under extra hypotheses on the tail µ(τ > t)). In
2
the absence of additional tail hypotheses, [19] showed how to obtain a partial result
in the probabilistic setting with limit replaced by liminf. In Corollary 3.5, we obtain
such a liminf result for semiflows with β ∈ (0, 1].
2
The remainder of this paper is organised as follows. In Section 2, we describe the
operator renewal-theoretic hypotheses required in this paper and we state a strong
renewal theorem for β ∈ (1,1] as well as related results for β ≤ 1. In Section 3, we
2 2
show how these results lead to mixing properties for semiflows. Sections 4 and 5 are
devoted to the proof of the strong renewal theorem, and Section 6 contains the proofs
of the remaining results in Section 2.
Corollary 3.1 (mixing for semiflows) is stated for observables that are certain
indicator functions. This restriction is relaxed considerably in Section 7. The corre-
sponding result for flows is stated and proved in Section 8. Illustrative examples are
given in Section 9.
2
2 Strong renewal theorem for continuous time de-
terministic systems
Let (Y,µ) be a probability space and let F : Y → Y be an ergodic and mixing
measure-preserving transformation. Let τ : Y → R+ be a measurable nonintegrable
function bounded away from zero. For convenience, we suppose that essinf τ > 1.
Throughout we assume the regularly varying tail condition (1.1).
Let τ = n−1τ ◦ Fj. Given measurable sets A,B ⊂ Y, define the renewal
n j=0
measure
P
∞
U (I) = µ(y ∈ A∩F−nB : τ (y) ∈ I), (2.1)
A,B n
n=0
X
for any interval I ⊂ R. We write U (x) = U ([0,x]) for x > 0. Our aim is to
A,B A,B
generalise [19, Theorems 1 and 2] to this set up. That is, we want to obtain the
asymptotics of U (t+h)−U (t) for any h > 0.
A,B A,B
With the same notation as in [13], let H = {Res ≥ 0}. Given δ > 0 and L > 0,
let H = (H∩B (0))∪{ib : |b| ≤ L}. Define the family of operators for s ∈ H,
δ,L δ
Rˆ(s) : L1(Y) → L1(Y), Rˆ(s)v = R(e−sτv).
Here R : L1(Y) → L1(Y) is the transfer operator for F (so Rvwdµ = vw◦F dµ
Y Y
for all v ∈ L1(Y), w ∈ L∞(Y)).
R R
Let B = B(Y) be a Banach space containing constant functions, with norm kk .
B
We assume that for any L > 0, there exist constants δ > 0, α ∈ (0,1) and C > 0,
0
such that
(H) (i) kRˆ(s)nvk ≤ C(|v| +αnkvk ) for all s ∈ H , v ∈ B, n ≥ 1.
B 1 0 B δ,L
(ii) The operator R : B → B has a simple eigenvalue at 1 and the rest of the
spectrum is contained in a disk of radius less than 1.
(iii) B is embedded in Lp for some p > β/(2β−1), and is compactly embedded
in L1.
(iv) The spectrum of Rˆ(ib) : B → B does not contain 1 for all b ∈ R\{0}.
Hypothesis (H)(i)–(iii) is a slight strengthening of [13, hypothesis (H1)] (where
L = δ is fixed and arbitrarily small, and the first part of (iii) is in the statement of
the main results in [13]). (H)(iv) is a significant weakening of [13, hypothesis (H4)]
and the diophantine ratio assumption used in [32] (Dolgopyat-type condition). The
remaining hypotheses in [13], namely (H2) and (H3) (re-inducing) are not required.
Remark 2.1 As explained in Section 9, hypothesis (H) is satisfied for mixing Gibbs-
Markov maps (with B a symbolic H¨older space) and mixing AFU maps (with B
consisting of bounded variation functions) provided, for instance, there exist two
periodic orbits for F with periods q ,q such that q /q is irrational. In both cases,
t 1 2 1 2
we can take p = ∞.
3
Remark 2.2 To verify the Lasota-Yorke inequality in (H)(i), it suffices to proceed
as follows. (a) Verify it for s = 0. (b) Verify that the family of operators s 7→ Rˆ(s)
on B is continuous at s = 0. (c) Show that for every L > 0, there exists C , n such
0 0
that kRˆ(ib)k ≤ C and kRˆ(ib)n0vk ≤ C |v| + 1kvk for all v ∈ B, |b| ≤ L.
B 0 B 0 1 2 B
Define d = 1 sinβπ, m(t) = ℓ(t)t1−β for β < 1, and d = 1, m(t) = tℓ(s)s−1ds
β π β 1
for β = 1. Throughout we suppose that A,B ⊂ Y are measurable and that 1 ∈ B.
A
R
Our main result generalizes [19, Theorem 1] to the present non i.i.d. set up:
Theorem 2.3 (Strong renewal theorem) Assume µ(τ > t) = ℓ(t)t−β where β ∈
(1,1]. Suppose that (H) holds. Then for any h > 0,
2
lim m(t)(U (t+h)−U (t)) = d µ(A)µ(B)h.
A,B A,B β
t→∞
As discussed in the introduction, additional hypotheses are needed to obtain a
strong renewal theorem when β ≤ 1. However, generalizing [19, Theorem 2] to the
2
present non i.i.d. set up, we still obtain a liminf result.
Theorem 2.4 Assume µ(τ > t) = ℓ(t)t−β where β ∈ (0,1). Suppose that (H) holds.
Then for any h > 0,
liminfm(t)(U (t+h)−U (t)) = d µ(A)µ(B)h.
A,B A,B β
t→∞
Remark 2.5 In the i.i.d. setting, results of this type are first due to [21] for discrete
time and β < 1. The results of [19] extended [21] to continuous time and incorporated
the case β = 1.
For the proof of Theorem 2.4, we will need the following result which gives the
asymptotics of U(t) for the whole range β ∈ [0,1]. This implies a property for the
semiflow F known as weak rational ergodicity [1, 4] (see Corollary 3.3 below) and
t
thus is of interest in its own right.
Theorem 2.6 Assume µ(τ > t) = ℓ(t)t−β where β ∈ [0,1]. Suppose that (H)(i)
holds for s ∈ R+ and that (H)(ii,iii) hold. Then
lim t−1m(t)U (t) = D µ(A)µ(B),
A,B β
t→∞
where D = [Γ(1−β)Γ(1+β)]−1, if β ∈ (0,1) and D = D = 1.
β 0 1
3 Mixing for infinite measure semiflows
In this section, we obtain various mixing results for semiflows as consequences of the
results in Section 2.
4
Let F : Y → Y and τ : Y → R+ be as in Section 2. Define the suspension
Yτ = {(y,u) ∈ Y ×R : 0 ≤ u ≤ τ(y)}/ ∼ where (y,τ(y)) ∼ (Fy,0). The suspension
semiflow F : Yτ → Yτ is given by F (y,u) = (y,u+t), computed modulo identifica-
t t
tions. The measure µτ = µ×Lebesgue is ergodic, F -invariant and σ-finite. Since τ
t
is nonintegrable, µτ is an infinite measure.
Throughout this section, we suppose that A = A×[a ,a ], B = B ×[b ,b ] are
1 1 2 1 1 2
measurable subsets of {(y,u) ∈ Y ×R : 0 ≤ u ≤ τ(u)} (so 0 ≤ a < a ≤ essinf τ,
1 2 A
0 ≤ b < b ≤ essinf τ), and that 1 ∈ B. Also, we continue to suppose that
1 2 B A
µ(τ > t) = ℓ(t)t−β for various ranges of β ∈ [0,1].
Corollary 3.1 Assume the setting of Theorem 2.3, so β ∈ (1,1] and (H) holds. Then
2
lim m(t)µτ(A ∩F−1B ) = d µτ(A )µτ(B ).
t→∞ 1 t 1 β 1 1
Proof Recall that essinf τ > 1. Let h ∈ (0,1) and note using (2.1) that
U (t+h)−U (t) = µ(y ∈ A : Fny ∈ B and τ (y) ∈ [t,t+h] for some n ≥ 0)
A,B A,B n
= µ(y ∈ A : F (y,0) ∈ B ×[0,h]).
t+h
After dividing rectangles into smaller subrectangles, we can suppose without loss
that b −b < 1. Set h = b −b . Then
2 1 2 1
µτ(A ∩F−1B ) = µτ{(y,u) ∈ A×[a ,a ] : F (y,u) ∈ B ×[b ,b ]}
1 t 1 1 2 t 1 2
= µτ{(y,u) ∈ A×[a ,a ] : F (y,0) ∈ B ×[0,h]}
1 2 t+u−b1
a2
= µ{y ∈ A : F (y,0) ∈ B ×[0,h]}du
t+u−b1
Za1
a2
= (U (t+u−b )−U (t+u−b −h))du. (3.1)
A,B 1 A,B 1
Za1
Hence
a2
m(t)µτ(A ∩F−1B ) = m(t)(U (t+u−b )−U (t+u−b −h))du
1 t 1 A,B 1 A,B 1
Za1
a2 m(t)
= χ(t+u−b −h)du,
1
m(t+u−b −h)
Za1 1
whereχ(t) = m(t)(U (t+h)−U (t))isboundedbyTheorem2.3. Alsom(t)/m(t+
A,B A,B
u−b −h)isboundedbyPotter’sbounds(seeforinstance[11]). Sincem(t)isregularly
1
varying, wehavelim m(t)/m(t+u−b −h) = 1foreachu ∈ [0,1]. ByTheorem2.3,
t→∞ 1
lim χ(t+u−b −h) = d µ(A)µ(B)h = d µ(A)µτ(B ) for each u ∈ [0,1]. Hence
t→∞ 1 β β 1
the result follows from the bounded convergence theorem.
Remark 3.2 The result also holds for all sets of the form F−1A and F−1B for
r 1 s 1
fixed r,s > 0. Indeed, by Corollary 3.1, using that m(t) ∼ m(t+s−r),
m(t)µτ(F−1A ∩F−1B ) = m(t)µτ(A ∩F−1 B )
r 1 t+s 1 1 t+s−r 1
→ µτ(A )µτ(A ) = µτ(F−1A )µτ(F−1A ).
1 2 r 1 s 2
5
Corollary 3.3 (Weak rational ergodicity) Assume the setting of Theorem 2.6,
so β ∈ [0,1], (H)(i) holds for s ∈ R+, and (H)(ii,iii) hold. Then
t
lim t−1m(t) µτ(A ∩F−1B )dx = D µτ(A )µτ(B ).
1 x 1 β 1 1
t→∞
Z0
Proof Continuing from (3.1) (with h = b −b ),
2 1
t a2 t
µτ(A ∩F−1B )dx = (U (x+u−b )−U (x+u−b −h))dxdu
1 x 1 A,B 1 A,B 1
Z0 Za1 Z0
a2 t a2 t−h
= U (x+u−b )dxdu− U (x+u−b )dxdu
A,B 1 A,B 1
Za1 Z0 Za1 Z−h
a2 t a2 0
= U (x+u−b )dxdu− U (x+u−b )dxdu = I +I .
A,B 1 A,B 1 1 2
Za1 Zt−h Za1 Z−h
Now
a2 0 U (x+t+u−b )
t−1m(t)I = t−1m(t)U (t) A,B 1 dxdu.
1 A,B
U (t)
Za1 Z−h A,B
By Theorem 2.6, U (t) is regularly varying so the integrand U (x + t + u −
A,B A,B
b )/U (t) is bounded for x,u bounded and converges pointwise to 1 as t → ∞.
1 A,B
Hence
a2 0 U (x+t+u−b )
A,B 1
lim dxdu = (a −a )h = (a −a )(b −b ).
2 1 2 1 2 1
t→∞Za1 Z−h UA,B(t)
By Theorem 2.6, t−1m(t)U (t) = D µ(A)µ(B)(1 + o(1)). Hence,
A,B β
lim t−1m(t)I = D µ(A)µ(B)(a −a )(b −b ) = µτ(A )µτ(B ). A simpler argu-
t→∞ 1 β 2 1 2 1 1 1
ment shows that t−1m(t)I = o(1).
2
Proposition 3.4 Let f : [0,∞) → R be bounded and integrable on compact sets, and
let K ∈ R. Suppose that β ∈ (0,1), that ℓ(t) is slowly varying, and that
(a) liminf ℓ(t)t1−βf(t) ≥ K,
t→∞
(b) lim ℓ(t)t−β tf(x)dx = β−1K.
t→∞ 0
Then there exists a seRt E ⊂ [0,∞) of density zero such that lim ℓ(t)t1−βf(t) =
t→∞,t6∈E
K.
In particular, liminf ℓ(t)t1−βf(t) = K.
t→∞
Proof This is the continuous time analogue of [31, Proposition 8.2] (which is itself
a version of [33, p. 65, Lemma 6.2]). We list the main steps which are proved exactly
as in [31].
6
Step 1. Without loss of generality, K = 0 and ℓ(t)t1−β is increasing.
Step 2. Define the nested sequence of sets E = {t > 0 : ℓ(t)t1−βf(t) > 1/q},
q
q = 1,2,... Then E has density zero for each q, i.e. lim 1 t1 (x)dx = 0.
q t→∞ t 0 Eq
Step 3. By Step 2, we can choose 0 = i < i < i < ··· suchRthat 1 t1 (x)dx <
0 1 2 t 0 Eq
1/q for t ≥ i , q ≥ 2. Define E = ∞ E ∩(i ,i ). Then E has density zero and
q−1 q=1 q q−1 q R
lim ℓ(t)t1−βf(t) = 0.
t→∞,t6∈E S
Corollary 3.5 Assume the setting of Theorem 2.4, so β ∈ (0,1) and (H) holds. Then
(i) liminf m(t)µτ(A ∩F−1B ) = d µτ(A )µτ(B ), and
t→∞ 1 t 1 β 1 1
(ii) There exists a set E ⊂ [0,∞) of density zero such that
lim m(t)µτ(A ∩F−1B ) = d µτ(A )µτ(B ).
t→∞,t6∈E 1 t 1 β 1 1
Proof We start from the conclusion of Theorem 2.4. Arguing as in the proof of
Corollary 3.1, but with lim replaced by liminf and using Fatou’s lemma instead of
the bounded convergence theorem, we obtain
liminfℓ(t)t1−βµτ(A ∩F−1B ) ≥ d µτ(A )µτ(B ).
1 t 1 β 1 1
t→∞
This is condition (a) in Proposition 3.4, and Corollary 3.3 is condition (b). Hence the
result follows from Proposition 3.4.
4 Main results used in the proof of Theorem 2.3
The first result needed in the proof of the strong renewal theorem, Theorem 2.3, is
an inversion formula for the symmetric measure
V (I) = 1(U (I)+U (−I)).
A,B 2 A,B A,B
Here, U(−I) = U({x : −x ∈ I}) (with U(−I) = 0 if I ⊂ [0,∞]). We find it
convenient to adapt the formulation in [19, Section 4], but such an inversion formula
goes back to [20] (see also [12, Chapter 10]).1
By (H), Tˆ(s) = (I −Rˆ(s))−1 is a bounded operator on B for all s ∈ H\{0}. Let
A, B ⊂ Y be measurable with 1 ∈ B.
A
Proposition 4.1 ( Analogue of [19, Inversion formula, Section 4]. ) Let g :
R → R be a continuous compactly supported function with Fourier transform γ(x) =
∞ eixbg(b)db satisfying γ(x) = O(x−2) as x → ∞. Then for all λ,t ∈ R,
−∞
∞ ∞
R
e−iλ(x−t)γ(x−t)dV (x) = e−itbg(b+λ)Re Tˆ(ib)1 dµdb.
A,B A
Z−∞ Z−∞ ZB
1The resultdoesnotrequireanyregularvariationassumptionsonµ(τ >t), butwe usethe extra
structure for simplicity.
7
The second result required in the proof of Theorem 2.3 comes directly from [19]
and does not require any modification in our set up. To state this result, for each
a > 0 we let γ (0) = 1 and for x 6= 0, define
a
2(1−cosax)
γ (x) = .
a a2x2
Proposition 4.2 ( [19, Lemma 8] ) Let {µ , t > 0} be a family of measures such
t
that µ (I) < ∞ for every compact set I and all t. Suppose that for some constant C,
t
∞ ∞
lim e−iλxγ (x)dµ (x) = C e−iλxγ (x)dx,
a t a
t→∞
Z−∞ Z−∞
for all a > 0, λ ∈ R. Then µ (I) → C|I| for every bounded interval I, where |I|
t
denotes the length of I.
Next, note that γ is the Fourier transform of
a
a−1(1−|b|/a), |b| ≤ a
g (b) = .
a
(0, |b| > a
The final result required in the proof of Theorem 2.3 is as follows.
Proposition 4.3 For all a > 0 and λ ∈ R,
∞
lim m(t) e−itbg (b+λ)Re Tˆ(ib)1 dµdb = πd g (λ)µ(A)µ(B).
a A β a
t→∞
Z−∞ ZB
Proof of Theorem 2.3 With the convention I +t = {x : x−t ∈ I}, let
µ (I) = 2m(t)V (I +t) = m(t)(U (I +t)+U (−I −t))
t A,B A,B A,B
and note that for I = [0,h] with h > 0,
m(t)(U (t+h)−U (t)) = µ (I).
A,B A,B t
Now,
∞ ∞
m(t) e−iλ(x−t)γ (x−t)dV (x) = m(t) e−iλxγ (x)dV (x+t)
a A,B a A,B
Z−∞ Z−∞
1 ∞
= e−iλxγ (x)dµ (x).
a t
2
Z−∞
Since γ satisfies the assumptions of Proposition 4.1,
a
∞ ∞
e−iλxγ (x)dµ (x) = 2m(t) e−itbg (b+λ)Re Tˆ(ib)1 dµdb.
a t a A
Z−∞ Z−∞ ZB
8
By Proposition 4.3 together with the Fourier inversion formula ∞ e−iλxγ (x)dx =
−∞ a
2πg (λ),
a
R
∞ ∞
lim e−iλxγ (x)dµ (x) = 2πd g (λ)µ(A)µ(B) = d e−iλxγ (x)dxµ(A)µ(B).
a t β a β a
t→∞
Z−∞ Z−∞
Hence, we have shown that the hypothesis of Proposition 4.2 holds with C =
d µ(A)µ(B). It now follows from Proposition 4.2 with I = [0,h] that
β
U (t+h)−U (t) = µ ([0,h]) → d µ(A)µ(B)h,
A,B A,B t β
as t → ∞.
TheproofofPropositions4.1and4.3aregiveninSection5. Theyrelyonstandard
arguments combined with the following estimates for Tˆ. Let ℓ˜(t) = tℓ(s)s−1ds. For
1
β ∈ (0,1) define c = i ∞e−iσσ−βdσ.
β 0 R
R
Proposition 4.4 (a) When β < 1, there exists δ ∈ (0,1) such that
ReTˆ(s) = (Rec−1)ℓ(1/|s|)−1|s|−β(P(0)+E(s)), for all s ∈ H∩B (0), s 6= 0,
β δ
where P(0)v = vdµ for v ∈ L1, and E(s) is a family of operators satisfying
Y
lims→0kE(s)kB→L1 = 0.
R
When β = 1, there is a corresponding statement of the form
π
ReTˆ(ib) = ℓ(1/|b|)ℓ˜(1/|b|)−2|b|−1(P(0)+E(b)), for all b ∈ R, 0 < |b| < δ.
2
(b) For any L > 0, there exists C > 0 such that for 0 < |b| ≤ L,
Cℓ(1/|b|)−1|b|−β β < 1
kReTˆ(ib)kB→L1 ≤ (Cℓ(1/|b|)ℓ˜(1/|b|)−2|b|−1 β = 1 .
(c) For any L > 0, there exists γ > 1−β, C > 0, such that for all 0 < h < b ≤ L,
kTˆ(i(b+h))−Tˆ(ib)kB→L1 ≤ C ℓ˜(1/b)−2b−2βℓ˜(1/h)hβ +b−βhγ , β ≤ 1.
(cid:0) (cid:1)
Proof The arguments for s ∈ B (0) are written out in [13, Section 4]. In particular
δ
this covers part (a). The details for the remainder of parts (b) and (c) are given in
Appendix A.
5 Completion of the proof of Theorem 2.3
In this section, we give the proof of Propositions 4.1 and 4.3, thereby completing the
proof of Theorem 2.3.
9
5.1 Proof of Proposition 4.1
For n ≥ 0, the Fourier transform of the distribution G (x) = µ(τ (y) ≤ x, y ∈
n n
A∩F−nB) is given by 1 1 ◦Fneibτn dµ = Rˆ(−ib)n1 dµ. Hence
Y A B B A
∞ R ∞ R
eibxdV (x) = Re eibxdU (x)
A,B A,B
Z−∞ Z0
∞
= Re Rˆ(ib)n1 dµ = Re Tˆ(ib)1 dµ.
A A
n=0 ZB ZB
X
Let γ and g be as in the statement of Proposition 4.1. Choose L > 0 such that
ℓ(1/|b|)−1|b|−β β < 1
supph ∈ [−L,L]. Letψ(b) = . ByProposition4.4(b),
(ℓ(1/|b|)ℓ˜(1/|b|)−2|b|−1 β = 1
|Tˆ(ib)1 | ≪ ψ(b)k1 k for |b| ≤ L. Since ψ is integrable, it follows from Fubini’s
A 1 A B
theorem that
∞ ∞
γ(x)dV (x) = g(b)Re Tˆ(ib)1 dµdb.
A,B A
Z−∞ Z−∞ ZB
Let g (b) = e−ibtg(b+λ) and set γ (x) = ∞ eibxg (b)db = e−iλ(x−t)γ(x−t). Replac-
1 1 −∞ 1
ing γ,g with γ ,g , we obtain the conclusion of Proposition 4.1.
1 1
R
5.2 Proof of Proposition 4.3
We follow the proof of [19, Theorem 1] (an adaptation of the argument in [21]).
Let W(b) = Re Tˆ(ib)1 dµ. Fix ω > 1 and write ∞ e−itbg (b +
B A −∞ a
λ)Re Tˆ(ib)1 dµdb = I (t,ω)+I (t,ω) where
B A R 1 2 R
R ω/t
I (t,ω) = e−itbg (b+λ)W(b)db, I (t,ω) = e−itbg (b+λ)W(b)db.
1 a 2 a
Z−ω/t Z|b|>ω/t
Proposition 4.3 follows immediately from the estimates for I (t,ω) and I (t,ω)
1 2
below. Set c−1 = π/2 when β = 1.
β
Lemma 5.1 lim m(t)I (t,ω) = 2g (λ)Rec−1 ωb−βcosbdbµ(A)µ(B), for all
t→∞ 1 a β 0
ω > 1.
R
Proof It follows from the definition of g that |g (b )−g (b )| ≤ a−2|b −b |. Hence
a a 1 a 2 1 2
ω/t ω/t
I (t,ω)−g (λ) e−itbW(b)db ≤ |g (b+λ)−g (λ)||W(b)|db
1 a a a
(cid:12) Z−ω/t (cid:12) Z−ω/t
(cid:12) (cid:12) ω/t
(cid:12) (cid:12) ≤ 2a−2ωt−1 |W(b)|db.
Z0
10