HAL Id: hal-00456584
https://hal.archives-ouvertes.fr/hal-00456584
Submitted on 22 Mar 2018
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Examples of recurrent or transient stationary walks in d over a rotation of 2
Nicolas Chevallier, Jean-Pierre Conze
To cite this version:
Nicolas Chevallier, Jean-Pierre Conze. Examples of recurrent or transient stationary walks in
dover
a rotation of
2. Idris Assani. Ergodic theory: Papers from the Probability and Ergodic Theory
Workshops held at the University of North Carolina, American Mathematical Society, pp.71-84, 2009,
Contemporary Mathematics n° 485. �hal-00456584�
R d T 2
R
dT
2(E, A , µ) τ E
µ ϕ E R
dϕ (E, µ, τ ) R
d(ϕ, τ ) (ϕ
n)
n∈Nϕ
n:= !
n−1k=0
ϕ ◦ τ
kn ≥ 0 τ
ϕE × R
dτ
ϕ: (x, y) → (τ x, ϕ(x) + y).
τ
ϕλ := µ × m m
R
d(ϕ
n) R
d(E, µ, τ ) x (ϕ
n(x))
n≥00 τ
ϕλ
ϕ R
(E, τ, µ) (ϕ
n)
n∈Nµ(ϕ) = 0 ϕ
R
dd > 1
R
dT
2% %
2| | R
d(ϕ, τ ) (E, A , µ, τ )
R
d(k
n)
(δ
n> 0)
lim
nµ(x : | ϕ
kn(x) | ≥ δ
n) = 0 and δ
n= o(n
1/d), (ϕ, τ )
τ ( ˜ E, A ˜ , µ, ˜ τ ˜ )
(ϕ, τ ) ( ˜ ϕ, τ ˜ ) µ ˜
µ τ
B τ
ϕ(τ
ϕ"B, % ≥ 0)
λ(B) = 0 λ(B) > 0 B
E × K K R
dE
n= { x : | ϕ
kn(x) | ≤ δ
n} B
n= (E
n× K) "
B n
0λ(B
j) ≥
12λ(B) j ≥ n
0λ( ∪
nj=n0τ
ϕkjB
j) ≥
12(n − n
0)λ(B)
ρ
n:= sup
1≤k≤nδ
k(ρ
n) (δ
n) ρ
n=
o(n
1/d)
∪
nj=n0τ
ϕkjB
jE × (D(ρ
n) + K) D(ρ
n) ρ
nC
λ( ∪
nj=n0τ
ϕkjB
j) ≤ Cm(D(ρ
n)).
lim inf
n
m(D(ρ
n))
n ≥ 1
2C λ(B) > 0 ρ
n= o(n
1/d) !
% ϕ
n%
2= o(n
1d)
E
n:= { x : | ϕ
n(x) | ≤ δ
n} (δ
n) > 0 µ(E
nc) ≤ (δ
n−1% ϕ
n%
2)
2δ
n−1% ϕ
n%
2→ 0, n
−1/dδ
n→ 0.
δ
n:= (n
1/d% ϕ
n%
2)
1/2!
(E, A , µ, τ )
(ϕ, τ ) R
dd T
rr > 1
C R
dT
rT
2m
Z
2∗= Z
2\{ 0 }
x | x |
0:= max(1, | x | ) || x || := inf( { x } , { 1 − x } ) = inf
n∈Z| x − n | R
0: R
2→ R R
0(h
1, h
2) := | h
1|
0| h
2|
0%
1%
2R
2R
"1,"2: R
2→ R
R
"1,"2(h) := | %
1(h) |
0| %
2(h) |
0.
∆ T
21
∆∆ ϕ
∆:= 1
∆− m(∆)
C %
1, %
2, ...%
ph ∈ Z
2| ϕ #
∆(h) | ≤ C $
1≤i<j≤p
1 R
"i,"j(h) .
∆ ∆
1, ..., ∆
nT
1, ..., T
n[0, 1]
2∆
kT
kT = { (x, y) ∈ [0, 1]
2: x ≤ y }
1 %
T(ξ
1, ξ
2) = 1 2iπξ
1( 1 − e
−2iπξ22iπξ
2− 1 − e
−2iπ(ξ1+ξ2)2iπ(ξ
1+ ξ
2) )
&
&
&% 1
T(ξ
1, ξ
2) & & & ≤ C( 1
| ξ
1|
0| ξ
2|
0+ 1
| ξ
1|
0| ξ
1+ ξ
2|
0+ 1
| ξ
2|
0| ξ
1+ ξ
2|
0). !
%
1%
2R
2C
1, C
2δ > 0
$
h∈Z2∗
1
R
"1,"2(h)
1+δ∈ [C
1δ
−2, C
2δ
−2].
L = (%
1, %
2) ! Λ Z
2L
h∈Λ 1 R0(h)1+δ
Λ x + [0, 1]
2M x ∈ R
2C
2$
h∈Λ
1
R
0(h)
1+δ≤ M $
h∈Z2
1
R
0(h)
1+δ< C
2δ
−2. C
1δ
−2!
%
1, %
2R
2α ∈ T
2δ > 0
K
δ(α) := sup
h∈Z2∗
{ R
"1,"2(h)
−(1+δ)% h.α %
−1} h ∈ Z
2ε > 0 A
h,δ,ε:= { α ∈ T
2: % h.α % ≤
R" ε1,"2(h)1+δ
}
ε > 0 α ∈ T
2c > 0
% k.α % ≥ c | k |
−(2+ε), ∀ k ∈ Z
2∗. K
δα K
δ(α) < + ∞ α
K
δR R
l1,l2δ > 0 m( '
h∈Z2∗
A
h,δ,ε) ≥ 1
3 min(1, δ 2
$
h∈Z2∗
m(A
h,δ,ε)).
N ≥ 1 Λ
dh = (h
1, h
2)
gcd(h
1, h
2) = d Λ
d= dΛ
1d
$
0<|h|≤N
m(A
h,δ,ε) = $
d≥1
$
h∈Λd,0<|h|≤N
m(A
h,δ,ε) = $
d≥1
$
h∈dΛ1,0<|h|≤N
2ε R(h)
1+δ≤ $
d≥1
$
h∈Λ1,0<|h|≤N
2ε
R(dh)
1+δ≤ $
h∈Λ1,0<|h|≤N
$
d≥1
2ε d
1+δR(h)
1+δ≤ 2 δ
$
h∈Λ1,0<|h|≤N
2ε
R(h)
1+δ= 2 δ
$
h∈Λ1,0<|h|≤N
m(A
h,δ,ε).
A
(N)= '
h∈Λ1,0<|h|≤N
A
h,δ,ε.
(
A(N)
$
h∈Λ1,0<|h|≤N
1
Ah,δ,εdm ≤ (m(A
(N)))
1/2( (
A(N)
( $
h∈Λ1,0<|h|≤N
1
Ah,δ,ε)
2dm)
1/2,
m(A
(N)) ≥ ( !
h∈Λ1,0<|h|≤N
m(A
h,δ,ε))
2D ,
D = $
h&=k∈Λ1,0<|h|,|k|≤N
m(A
h,δ,ε∩ A
k,δ,ε) + $
h∈Λ1,0<|h|≤N
m(A
h,δ,ε).
h k Λ
1h = ± k h k
Z
2a b
m( { α : % h.α % ≤ a } ∩ { α : % k.α % ≤ b } ) = m( { α : % h.α % ≤ a } ) m( { α : % k.α % ≤ b } ).
$
h&=k∈Λ1,0<|h|,|k|≤N
m(A
h,δ,ε∩ A
k,δ,ε) = $
h&=±k∈Λ1,0<|h|,|k|≤N
m(A
h,δ,ε) m(A
k,δ,ε)+ $
h∈Λ1,0<|h|≤N
m(A
h,δ,ε),
D ≤ ( $
h∈Λ1,0<|h|≤N
m(A
h,δ,ε))
2+ 2 $
h∈Λ1,0<|h|≤N
m(A
h,δ,ε).
v
2v
2+ 2v ≥ 1
3 min(1, v), ∀ v > 0 m(A
(N)) ≥ 1
3 min(1, $
h∈Λ1,0<|h|≤N
m(A
h,δ,ε))
≥ 1
3 min(1, δ 2
$
0<|h|≤N
m(A
h,δ,ε)).
N + ∞ !
K
δL
s[0, 1] s < 1
C C
'δ > 0 s ∈ ]0, 1[
C δ
ss 1 − s ≤
(
T2
K
δ(α)
sdα ≤ 1 + C
'δ
21 1 − s . σ
δ:= !
h∈Z2∗
R(h)
−(1+δ)0 < C
1< C
2σ
δδ
2∈ [C
1, C
2] s ∈ ]0, 1[
t > 0 A
st:= { α ∈ T
2| K
δ(α)
s> t } '
h&=0
A
h,δ,t−1sm(A
h,δ,t−1s) = min(1, 2R(h)
−(1+δ)t
−1s) (
T2
K
δ(α)
sdα = (
∞0
m(A
st) dt ≤ 1 + 2 (
+∞1
$
h∈Z2∗
R(h)
−(1+δ)t
−1sdt
< 1 + 2 1 − s
$
h∈Z2∗
R(h)
−(1+δ)≤ 1 + 1 1 − s
C
2δ
2.
(
T2
K
δ(α)
sdα = (
∞0
m(A
st) dt ≥ (
∞0
1
3 min(1, δ 2
$
h∈Z2∗
min(1, 2R(h)
−(1+δ)t
−1/s) dt
≥ δ 3
(
∞0
$
h∈Z2∗
min( 1 δσ
δR(h)
−(1+δ), 1
2 , R(h)
−(1+δ)t
−1/s)dt
≥ δ 3
$
h∈Z2∗
(
∞0
min( 1 δσ
δR(h)
−(1+δ), R(h)
−(1+δ)t
−1/s)dt
≥ δ 3
$
h∈Z2∗
R(h)
−(1+δ)(
∞(δσδ)s
t
−1/sdt ≥ 1 3
C
1sδ
ss 1 − s .
!
%
1%
2R
2γ > 0
α ∈ R
2$
h∈Z2∗
1
R
l1,l2(h)
2% h.α %
2−γ< + ∞ .
δ ∈ ]0, γ/3[ R R
l1,l2R(h)
1+δ% h.α % ≥ c, ∀ h ∈ Z
2∗,
c K
δ−1(α) α
λ := γ − δ
$
h∈Z2∗
1
R(h)
2% h.α %
2−γ≤ $
h∈Z2∗
1
R(h)
2(cR(h)
−(1+δ))
1−γ× 1
% h.α %
≤ B
1$
h∈Z2∗
R(h)
−(1+λ)× 1
% h.α % . n = (n
1, n
2) n
i≥ 1
$
h∈Z2∗,|"i(h)|≤ni
1
% h.α % .
(h, h
') h - = h
'Z
2|% h.α % − % h
'.α %| ≥ min( % (h + h
').α % , % (h − h
').α % )
≥ c min(R(h + h
')
−(1+δ), R(h − h
')
−(1+δ))
≥ c(R
0(2n))
−(1+δ).
d := c(R
0(2n))
−(1+δ)h
0 % h.α % ≥ d % x % ≤ 1/2
x % h.α %
[d, 2d[, [2d, 3d[, ..., [rd, (r + 1)d[,
r = [
2d1] % h.α %
$
h∈Z2∗,|"i(h)|≤ni
1
% h.α % ≤
$
rk=1
1 kd ≤ 1
d (1 + ln r)
≤ 1
c R
0(2n)
1+δ(1 + ln R
0(2n)
1+δc ) ≤ B
2R
0(n)
1+2δ,
B
2n
$
n=(n1,n2), ni≥|"i(h)|0
(n
1n
2)
−(2+λ)≥ (1 + λ)
−2( | %
1(h) |
0| %
2(h) |
0)
1+λ= (1 + λ)
−2R(h)
−(1+λ),
$
h∈Z2∗
R(h)
−(1+λ)× 1
% h.α % ≤ (1 + λ)
2$
∞ n1=1,n2=1(n
1n
2)
−(2+λ)× $
h=(h1,h2)∈Z2∗,|"i(h)|≤ni
1
% h.α % . B
3, B
4$
h∈Z2∗
1
R(h)
2% h.α %
2−γ≤ B
3$
∞ n1=1,n2=1(n
1n
2)
−(2+λ)× $
h=(h1,h2)∈Z2∗,|"i(h)|≤ni
1
% h.α %
≤ B
4$
∞ n1=1,n2=1(n
1n
2)
−(2+λ)R
0(n
1, n
2)
1+2δ= B
4$
∞ n1=1,n2=1(n
1n
2)
−1−γ+3δ.
δ <
γ3!
α
C
γ(α) := $
h∈Z2∗
1
R
l1,l2(h)
2% h.α %
2−γ.
C
0(%
1, %
2) a > 0 ε > 0 a(2 − γ) = 1 − ε
(
T2
C
γ(α)
adα ≤ C
0εγ
2(1+a)( 1
γ
a+ 1 ε
a).
C
γ(α) ≥ K
δ(α)
2−γδ =
γ21−1γC
0'(
2T2
C
γ(α)
adα ≥ (
T2
K
δ(α)
1−εdα ≥ C
0'δ
1−ε1 − ε ε .
% % 2
∆
ϕ
∆:= 1
∆− m(∆) α = (α
1, α
2) ∈ T
2% %
2ϕ
∆α
S
Nαϕ
∆(x, y) :=
N−1
$
k=0
ϕ
∆(x + kα
1, y + kα
2)
ϕ
∆α
α R
dγ > 0 α ∈ T
2A
γ(α)
% S
Nαϕ
∆%
2≤ A
γ(α)N
γ.
% S
Nαϕ
∆%
22≤ $
h∈Z2∗
| ϕ ˆ
∆(h) |
2|
N
$
−1k=0
e
2πik(h.α)|
2≤ $
h∈Z2∗
| ϕ ˆ
∆(h) |
2inf(N
2, 1
|| h.α ||
2)
≤ N
2γ$
h∈Z2∗
| ϕ ˆ
∆(h) |
21
|| h.α ||
2(1−γ). γ > 0
α !
{ α : A
γ(α) < + ∞} ϕ
∆A
γ= C
1
γ2
A
γ(ϕ
∆i, i = 1, · · · , d) ϕ
∆iα = (α
1, α
2) R
dT
2(x, y ) → (x + α
1, y + α
2) Φ = (ϕ
∆i)
i=1,···,d1
[0,12[
−
12ϕ
∆Φ(x, y) = (ϕ
1(x), ϕ
2(y)) ϕ
1ϕ
2(α
1, α
2)
α
1f
T
1ε > 0 x !
n−1k=0
f (x + kα
1) = O(n
ε) Φ(x, y) = (2.1
[0,12[
(x) − 1), 2.1
[0,12[
(y) − 1) Z
2Z
2(α
1, α
2)
T
2Z
2τ
Φ: (x, y, n) → (x + α
1, y + α
2, n + Φ(x, y)).
(Φ, τ ) Φ
σ
α = (α
1, α
2) Φ
Z
2Φ
α (k
n)
lim sup
n
% S
kαnΦ %
2√ n > 0.
Φ : T
2→ R
2Φ(x) = Φ(x
1, x
2) = (ϕ(x
1), ϕ(x
2)), where ϕ := 1
[0,15]
− 1 5 . T
2x → x + α α = (α
1, α
2)
(x
1, x
2) T
2|
N−1
$
k=0
ϕ(x
1+ kα
1) | + |
N−1
$
k=0
ϕ(x
2+ kα
2) |
N−→
→+∞+ ∞ . Φ
Φ = (ϕ
1, ϕ
2) = (1
R1− 1
5 , 1
R2− 1 5 ), R
1= [0,
15] × [0, 1] R
2= [0, 1] × [0,
15]
α = (α
1, α
2) R
2Z α + Z
2R
2(x
1, x
2) ∈ T
2| S
Nα1ϕ
1(x
1) | | S
Nα2ϕ
2(x
2) | N
(k(n))
n≥1k(1) = 1
k(n + 1) ≥ 10 × ln 3
ln 2 × k(n), ∀ n ≥ 1.
α α = (
$
∞ n=13
−k(2n+1),
$
∞ n=12
−k(2n)).
p
n= 2 n p
n= 3 n (β
n)
n≥0α
β
n:= ( $
m<n2
3
−k(2m+1), $
m≤n2
2
−k(2m)).
n ≥ 1 β
nQ
n= p
k(nn−1−1)p
k(n)nα = β
n+ ε
nε
n= ( !
m≥n2
3
−k(2m+1), !
m>n2
2
−k(2m)) k(%) % > n
| ε
n| ≤ $
m≥n2
3
−k(2m+1)+ $
m>n2
2
−k(2m)≤ 4 × 2
−k(n+1).
β
n(
pqnn,
pq#n#n
) q
n= 3
k(n−1)q
n'= 2
k(n)n q
n= 3
k(n)q
'n= 2
k(n−1)n Q
n= q
nq
'nr(j) r
'(j ) j q
nq
n'j /→ (r(j ), r
'(j)) [0, q
nq
n'− 1] [0, q
n− 1] × [0, q
n'− 1]
{ jβ
nmod Z
2, j ∈ Z} {
qjnmod 1, 0 ≤ j ≤ q − 1 } × {
qj#n#mod 1, 0 ≤ j
'≤ q
n'− 1 }
q ≤ Q
nd(qα, qβ
n) = q | ε
n| ≤ Q
n| ε
n| = p
k(nn−1−1)p
k(n)n× 4 × 2
−k(n+1)→ 0,
Z α + Z
2R
2p ≥ 1
p p = 2n + 1 s
j,2n+1j = 0, ..., 2
k(2n)− 1
jβ
2n+1+ [0, 1] × { 0 } p = 2n s
j,2nj = 0, ..., 3
k(2n−1)− 1 jβ
2n+ { 0 } × [0, 1]
q q = ap
k(nn−1−1)+ j a j 0 ≤ j <
p
k(n−1)n−1qβ
ns
j,nq ≤ Q
2nqα s
j,nd
T2(qα, s
j,n) = d
T2(q(β
n+ ε
n), s
j,n) ≤ 0 + q | ε
n| ≤ Q
2n| ε
n| .
δ
n:= Q
2n| ε
n|
B
j,n≤ 2δ
ns
j,nr
ns
j,ns
i,np
−n−1k(n−1)4 × δ
n≤ p
2k(nn−1−1)p
2k(n)n× 16 × 2
−k(n+1)≤ 3
4k(n)+32
−k(n+1)≤ p
−6k(n)+3n−1< 1
2 p
−k(n−1)n−1, B
j,nj = 0, ..., p
k(n−1)n− 1
R
is
j,ns
j,nR
i− x J
x= J
nJ
n+ 1 J
n= )
1/5 rn
*
t
n=
15− J
nr
nρ
n= min(t
n, r
n− t
n) ρ
n= min
j∈Z
&
&
&
&
&
1
5 − j p
k(nn−1−1)&
&
&
&
& ≥ 1 5 × p
k(nn−1−1).
I
n=
pk(n−1)n−1 −1
'
j=0
B
j,n, K
n= I
n± ρ
ne
n, E
n= I
n∪ K
n,
e
n= (1, 0) n (0, 1) n E
n3p
k(nn−1−1)× ( B
j,n) m
n= 12 × p
k(nn−1−1)δ
nN [Q
2n−1, Q
2n[
n n x ∈ T
2\ E
nM
x,Nq < N x + qα ∈ R
2s
j,nd
T2(x + s
j,n, { 0 } × [0, 1]) = d
T2(x + jβ
n, { 0 } × [0, 1])
= d
T2(x, − jβ
n+ { 0 } × [0, 1]) = d
T2(x, s
i,n)
i { 0, ..., p
k(nn−1−1)− 1 } x I
nd
T2(x, s
i,n) > 2δ
nx + s
j,n{ 0 } × [0, 1]
2δ
nd
T2(x + s
j,n, { 1
5 } × [0, 1]) = d
T2(x + jβ
n, { 1
5 } × [0, 1])
= d
T2(x, − jβ
n+ { 1
5 } × [0, 1])
= d
T2(x, − jβ
n+ { J
nr
n+ t
n} × [0, 1])
= d
T2(x, ( − jβ
n+ (J
np
k(nn−1−1), 0)) + { t
n} × [0, 1])
= d
T2(x, s
i,n+ t
ne
n) x K
nd
T2(x, s
i,n+ t
ne
n) > 2δ
nx + s
j,n{
15} × [0, 1]
2δ
nx + B
j,nR
2R
2j < p
k(nn−1−1)q < N qα
B
j,nA A + 1 A := [(N − 1) × p
−n−1k(n−1)]
J
xA ≤ M
x,N≤ J
x(A + 1).
| S
Nϕ
2(x) | = & & M
x,N−
15N & & J
x= J
n1
5 N − M
x,N≥ 1
5 N − J
x(A + 1) ≥ 1
5 N − J
n( N
p
k(n−1)n−1+ 1)
= ( 1
5 − J
np
k(nn−1−1))N − J
n≥ ρ
nN − J
n≥ 1
5 × p
k(n−1)n−1Q
2n−1− 1/5
p
−k(n−1)n−1= p
k(n−1)n−15 (p
2k(n−2)n−2− 1) J
x= J
n+ 1
M
x,N− 1
5 N ≥ (J
n+ 1)A − 1
5 N ≥ p
k(nn−1−1)5 (p
2k(n−2)n−2− 1) − 1.
(x -∈ E
n) = ⇒ | S
Nϕ
2(x) | ≥ p
k(n−1)n−15 (p
2k(n)n− 1) − 1.
n S
Nϕ
1(x)
$
n≥1
m(E
n) ≤ 12 $
n≥1
p
k(nn−1−1)δ
n= 12 $
n≥1
p
k(nn−1−1)Q
2n| ε
n|
≤ 12 $
n≥1
p
k(n−1)n−1p
2k(n−1)n−1p
2k(n)n× 4 × 2
−k(n+1)< + ∞ ,
x T
2E
nN→+∞
lim | S
Nϕ(x) | = + ∞ .
!
α
ε > 0
N
εN ≥ N
εmin {% qα % : 0 < q ≤ N } ≤ εN
−12. N ∈ [Q
n, Q
n+1[
min {% qα % : 0 < q ≤ N } ≤ | ε
n−1|
| ε
n| N
12≤ ( $
m≥n2
3
−k(2m+1)+ $
m>n2
2
−k(2m))Q
n+112.
n
( $
m≥n2
3
−k(2m+1)+ $
m>n2
2
−k(2m))Q
n+112≤ 2 $
m≥n2
3
−k(2m+1)Q
n+112≤ 4 × 3
−k(n+1)× 2
12k(n)3
12k(n+1)n
( $
m≥n2
3
−k(2m+1)+ $
m>n2
2
−k(2m))Q
n+112≤ 2 $
m>n2