Recall the definition of T
11
and the fact that T
11
≥ T
4
to see that |π
X
X
t
− X | ≤ 5ǫ
1
for t ≤ T
4
, assuming that A
2
holds. It follows from the definition of T
8
that |Y
t
− Y | ≤ 4ǫ
1
for t ≤ T
4
. Recall that T
7
= sup{t ≤ T
4
: X
t
∈ ∂ D}. Note that X
T
4
− Y
T
4
= X
T
7
− Y
T
7
, Y
T
4
, X
T
7
∈ ∂ D, and T
7
≤ T
4
. This and the bounds
|π
X
X
t
− X | ≤ 5ǫ
1
and |Y
t
− Y | ≤ 4ǫ
1
for t ≤ T
4
, easily imply that dY
T
7
, ∂ D ≤
|X
T
7
− Y
T
7
|2, assuming that ǫ is small. Hence, T
9
≤ T
4
. This fact combined with 3.32 shows that if A
2
occurs then T
9
≤ T
4
≤ T
8
∧ T
10
. This completes the proof that A
2
⊂ A
3
. It is easy to see that
PA
2
p
2
, for some p
2
0. It follows that PA
3
p
2
. We may now apply the strong Markov property at the stopping time T
9
and repeat the argument given in the first part of the proof, which was devoted to the case r
≤ ǫ
1
2. It is straightforward to complete the proof of part i, adjusting the values of c
1
, c
2
, ǫ
, r and p
, if necessary. ii We will restart numbering of constants, i.e., we will use c
6
, c
7
, . . . , for constants unrelated to those with the same index in the earlier part of the proof.
Let c
1
, c
2
, ǫ
and r be as in part i of the lemma,
ǫ
2
= ǫ ∧ r
, and ǫ
1
≤ ǫ
2
. Recall that τ
+
ǫ
2
= inf
{t 0 : |X
t
− Y
t
| ≥ ǫ
2
}. Let T
5
= 0, and for k ≥ 1 let T
k 1
= inf{t ≥ T
k −1
5
: |X
T
k −1
5
− X
t
| ∨ |Y
T
k −1
5
− Y
t
| ≥ c
1
dY
T
k −1
5
, ∂ D} ∧ τ
+
ǫ
2
, 3.33
T
k 2
= inf{t ≥ T
k −1
5
: L
t
− L
T
k −1
5
≥ c
2
dY
T
k −1
5
, ∂ D} ∧ τ
+
ǫ
2
, 3.34
T
k 3
= inf{t ≥ T
k −1
5
: Y
t
∈ ∂ D} ∧ τ
+
ǫ
2
, 3.35
T
k 4
= T
k 1
∧ T
k 2
∧ T
k 3
, 3.36
T
k 5
= inf{t ≥ T
k 4
: X
t
∈ ∂ D} ∧ τ
+
ǫ
2
. 3.37
We will estimate EdY
T
k 5
, ∂ D. By Lemma 3.5 i and the definition of T
k 1
, on the event {T
k 4
τ
+
ǫ
2
}, P
sup
t ∈[T
k 4
,T
k 5
]
|X
t
− X
T
k 4
| ∈ [2
− j−1
, 2
− j
] | F
T
k 4
≤ c
6
dX
T
k 4
, ∂ D2
− j
≤ c
7
dY
T
k −1
5
, ∂ D2
− j
. 3.38
Write R = dY
T
k −1
5
, ∂ D, assume that T
k 4
τ
+
ǫ
2
, and let j be the largest integer such that sup
t ∈[T
k 4
,T
k 5
]
|X
t
− X
T
k 4
| ∨ ǫ
2
≤ 2
− j
. We will show that dY
T
k 5
, ∂ D ≤ R + c
8
ǫ
2
2
− j
, a.s. Note that between times T
k −1
5
and T
k 4
, the process Y
t
does not hit the boundary of D. Between times T
k 4
and T
k 5
, the process X
t
does not hit ∂ D. If Y
t
does not hit the boundary on the same interval, it is elementary to see that
dY
T
k 5
, ∂ D ≤ R + c
9
ǫ
2
2
− j
. Suppose that Y
t
∗
∈ ∂ D for some t
∗
∈ [T
k 4
, T
k 5
], and assume that t
∗
is the largest time with this property. If t
∗
= T
k 5
then dY
T
k 5
, ∂ D = 0. Otherwise we must have τ
+
ǫ
2
T
k 5
, X
T
k 5
∈ ∂ D, and X
T
k 5
− Y
T
k 5
= X
t
∗
− Y
t
∗
. Since both Y
t
∗
and X
T
k 5
belong to ∂ D, easy geometry shows that in this case
dY
T
k 5
, ∂ D ≤ c
10
ǫ
2
2
− j
. This completes the proof that dY
T
k 5
, ∂ D ≤ R + c
8
ǫ
2
2
− j
, a.s. Let j
be the smallest integer such that 2
− j
≥ diamD and let j
1
be the largest integer such that 2
− j
1
+1
≥ R. The estimate dY
T
k 5
, ∂ D ≤ R + c
8
ǫ
2
2
− j
and 3.38 imply that on the event {T
k 4
2198
τ
+
ǫ
2
},
EdY
T
k 5
, ∂ D | F
T
k 4
≤ X
j ≤ j≤ j
1
R + c
8
ǫ
2
2
− j
P sup
t ∈[T
k 4
,T
k 5
]
|X
t
− X
T
k 4
| ∈ [2
− j−1
, 2
− j
] | F
T
k 4
≤ R + X
j ≤ j≤ j
1
c
8
ǫ
2
2
− j
P sup
t ∈[T
k 4
,T
k 5
]
|X
t
− X
T
k 4
| ∈ [2
− j−1
, 2
− j
] | F
T
k 4
≤ R + X
j ≤ j≤ j
1
c
11
ǫ
2
2
− j
R2
− j
≤ R + c
12
ǫ
2
R | log R|
= dY
T
k −1
5
, ∂ D1 + c
12
ǫ
2
| log dY
T
k −1
5
, ∂ D|.
3.39 For R
≤ ǫ
4 2
we have R1 + c
12
ǫ
2
| log R| ≤ c
13
ǫ
3 2
, so R1 + c
12
ǫ
2
| log R| ≤ R1+ 4c
12
ǫ
2
| log ǫ
2
| + c
13
ǫ
3 2
. Thus, on the event
{T
k 4
τ
+
ǫ
2
},
EdY
T
k 5
, ∂ D | F
T
k 4
≤ 1 + c
12
ǫ
2
| log ǫ
2
|dY
T
k −1
5
, ∂ D + c
13
ǫ
3 2
. 3.40
Let S
∗ 1
= S
1
∧ τ
+
ǫ
2
. By the strong Markov property applied at T
k −1
5
and part i of the lemma, on the event
{S
∗ 1
T
k −1
5
}, PT
k −1
5
S
∗ 1
≤ T
k 5
| F
T
k −1
5
≥ PT
k −1
5
S
∗ 1
≤ T
k 4
| F
T
k −1
5
≥ p .
3.41 By the strong Markov property and induction,
PS
∗ 1
T
k −1
5
≤ c
14
p
k
. 3.42
This, 3.40 and 3.41 imply,
E dY
T
k 5
, ∂ D1
{S
∗ 1
T
k 5
}
1
{T
k −1
5
τ
+
ǫ
2
}
= E 1
{S
∗ 1
T
k 5
}
1
{T
k −1
5
τ
+
ǫ
2
}
E dY
T
k 5
, ∂ D | F
T
k 4
≤ E 1
{S
∗ 1
T
k 5
}
1
{T
k −1
5
τ
+
ǫ
2
}
1 + c
12
ǫ
2
| log ǫ
2
|dY
T
k −1
5
, ∂ D + c
13
ǫ
3 2
= E 1
{S
∗ 1
T
k −1
5
}
1
{S
∗ 1
T
k 5
}
1
{T
k −1
5
τ
+
ǫ
2
}
1 + c
12
ǫ
2
| log ǫ
2
|dY
T
k −1
5
, ∂ D + c
13
ǫ
3 2
≤ E 1
{S
∗ 1
T
k −1
5
}
1
{T
k −1
5
τ
+
ǫ
2
}
1 + c
12
ǫ
2
| log ǫ
2
|dY
T
k −1
5
, ∂ D + c
13
ǫ
3 2
× E1
{S
∗ 1
T
k 5
}
| F
T
k −1
5
≤ E 1
{S
∗ 1
T
k −1
5
}
1
{T
k −1
5
τ
+
ǫ
2
}
1 + c
12
ǫ
2
| log ǫ
2
|dY
T
k −1
5
, ∂ D + c
13
ǫ
3 2
1 − p ≤ 1 + c
12
ǫ
2
| log ǫ
2
|1 − p E
dY
T
k −1
5
, ∂ D1
{S
∗ 1
T
k −1
5
}
1
{T
k −2
5
τ
+
ǫ
2
}
+ c
13
1 − p ǫ
3 2
PS
∗ 1
T
k −1
5
≤ 1 + c
12
ǫ
2
| log ǫ
2
|1 − p EdY
T
k −1
5
, ∂ D1
{S
∗ 1
T
k −1
5
}
1
{T
k −2
5
τ
+
ǫ
2
}
+ c
15
1 − p ǫ
3 2
p
k
. 2199
We assume without loss of generality that p 0 is so small that 1 − p
p
−1
1. We obtain by induction,
EdY
T
k 5
, ∂ D1
{S
∗ 1
T
k 5
}
1
{T
k −1
5
τ
+
ǫ
2
}
3.43 ≤ 1 + c
12
ǫ
2
| log ǫ
2
|
k
1 − p
k
EdY
T
5
, ∂ D1
{S
∗ 1
0}
1
{T
5
τ
+
ǫ
2
}
+ c
15
1 − p ǫ
3 2
k −1
X
m=0
1 + c
12
ǫ
2
| log ǫ
2
|
m
1 − p
m
p
k −m
≤ 1 + c
12
ǫ
2
| log ǫ
2
|
k
1 − p
k
r + c
15
ǫ
3 2
p
k k
−1
X
m=0
1 + c
12
ǫ
2
| log ǫ
2
|
m
1 − p
m
p
−m
≤ 1 + c
12
ǫ
2
| log ǫ
2
|
k
1 − p
k
r + c
16
ǫ
3 2
p
k
1 + c
12
ǫ
2
| log ǫ
2
|
k
1 − p
k
p
−k
= 1 + c
12
ǫ
2
| log ǫ
2
|
k
1 − p
k
r + c
16
ǫ
3 2
1 + c
12
ǫ
2
| log ǫ
2
|
k
1 − p
k
≤ c
17
1 + c
12
ǫ
2
| log ǫ
2
|
k
1 − p
k
r + ǫ
3 2
. Note that, by 3.34 and 3.37,
L
T
j+1 2
− L
T
j 5
≤ c
2
dY
T
j 5
, ∂ D,
L
T
j+1 5
− L
T
j+1 2
= 0. Hence,
L
T
j+1 5
− L
T
j 5
≤ c
2
dY
T
j 5
, ∂ D.
3.44
2200
It follows from this and 3.43 that
EL
S
1
∧τ
+
ǫ
2
− L = EL
S
∗ 1
− L =
∞
X
k=0
E L
S
∗ 1
− L 1
{S
∗ 1
∈T
k 5
,T
k+1 5
]}
≤
∞
X
k=0
E
1
{S
∗ 1
∈T
k 5
,T
k+1 5
]} k
X
j=0
1
{T
j 5
τ
+
ǫ
2
}
L
T
j+1 5
− L
T
j 5
≤
∞
X
k=0
E
1
{S
∗ 1
∈T
k 5
,T
k+1 5
]} k
X
j=0
1
{T
j −1
5
τ
+
ǫ
2
}
c
2
dY
T
j 5
, ∂ D
= E
∞
X
k=0 k
X
j=0
1
{S
∗ 1
∈T
k 5
,T
k+1 5
]}
1
{T
j −1
5
τ
+
ǫ
2
}
c
2
dY
T
j 5
, ∂ D
= E
∞
X
j=0 ∞
X
k= j
1
{S
∗ 1
∈T
k 5
,T
k+1 5
]}
1
{T
j −1
5
τ
+
ǫ
2
}
c
2
dY
T
j 5
, ∂ D
= c
2 ∞
X
j=0
E 1
{S
∗ 1
T
j 5
}
1
{T
j −1
5
τ
+
ǫ
2
}
dY
T
j 5
, ∂ D
≤
∞
X
j=0
c
18
1 + c
12
ǫ
2
| log ǫ
2
|
j
1 − p
j
r + ǫ
3 2
. If we assume that
ǫ
2
0 is sufficiently small, this is bounded by c
19
r + ǫ
3 2
. iii We will restart numbering of constants, i.e., we will use c
6
, c
7
, . . . , for constants unrelated to those with the same index in the earlier part of the proof.
Recall that j
1
is the largest integer such that 2
− j
1
+1
≥ dY
T
k −1
5
, ∂ D. Let j
2
be the largest integer such that 2
− j
2
+1
≥ r. By 3.33 and 3.38 we have for j ≤ j
1
, on the event {T
k −1
5
τ
+
ǫ
2
}, P
sup
t ∈[T
k −1
5
,T
k 5
]
|X
t
− X
T
k 4
| ∈ [2
− j−1
, 2
− j
] | F
T
k −1
5
≤ P
sup
t ∈[T
k −1
5
,T
k 4
]
|X
t
− X
T
k −1
5
| + sup
t ∈[T
k 4
,T
k 5
]
|X
t
− X
T
k 4
| ∈ [2
− j−1
, 2
− j
] | F
T
k −1
5
≤ P
c
1
dY
T
k −1
5
, ∂ D +
sup
t ∈[T
k 4
,T
k 5
]
|X
t
− X
T
k 4
| ∈ [2
− j−1
, 2
− j
] | F
T
k −1
5
≤ c
6
dY
T
k −1
5
, ∂ D2
− j
.
2201
We will also use the trivial estimate
P
sup
t ∈[T
k −1
5
,T
k 5
]
|X
t
− X
T
k 4
| ≤ r | F
T
k −1
5
≤ 1. We use the last two estimates, 3.42 and 3.43 to obtain
E sup
≤t≤S
1
∧τ
+
ǫ
2
|X
t
− X |
= E sup
≤t≤S
∗ 1
|X
t
− X |
=
∞
X
k=0
E sup
≤t≤S
∗ 1
|X
t
− X |1
{S
∗ 1
∈T
k 5
,T
k+1 5
]}
≤
∞
X
k=0
E
1
{S
∗ 1
∈T
k 5
,T
k+1 5
]} k
X
j=0
1
{T
j 5
τ
+
ǫ
2
}
sup
T
j 5
≤t≤T
j+1 5
|X
t
− X |
≤
∞
X
k=0
E
1
{S
∗ 1
∈T
k 5
,T
k+1 5
]} k
X
j=0
E
1
{T
j 5
τ
+
ǫ
2
}
sup
T
j 5
≤t≤T
j+1 5
|X
t
− X | | F
T
k −1
5
≤
∞
X
k=0
E
1
{S
∗ 1
∈T
k 5
,T
k+1 5
]} k
X
j=0
r + X
j ≤i≤ j
2
2
−i
1
{T
j −1
5
τ
+
ǫ
2
}
c
6
dY
T
j −1
5
, ∂ D2
−i
≤
∞
X
k=0
E
1
{S
∗ 1
∈T
k 5
,T
k+1 5
]} k
X
j=0
r + c
7
| log r|1
{T
j −1
5
τ
+
ǫ
2
}
dY
T
j −1
5
, ∂ D
= E
∞
X
k=0 k
X
j=0
1
{S
∗ 1
∈T
k 5
,T
k+1 5
]}
r + c
7
| log r|1
{T
j −1
5
τ
+
ǫ
2
}
dY
T
j −1
5
, ∂ D
= E
∞
X
j=0 ∞
X
k= j
1
{S
∗ 1
∈T
k 5
,T
k+1 5
]}
r + c
7
| log r|1
{T
j −1
5
τ
+
ǫ
2
}
dY
T
j −1
5
, ∂ D
=
∞
X
j=0
E 1
{S
∗ 1
T
j 5
}
r + c
7
| log r|1
{T
j −1
5
τ
+
ǫ
2
}
dY
T
j −1
5
, ∂ D
= r
∞
X
j=0
PS
∗ 1
T
j 5
+ c
7
| log r|
∞
X
j=0
E 1
{S
∗ 1
T
j 5
}
1
{T
j −1
5
τ
+
ǫ
2
}
dY
T
j −1
5
, ∂ D
≤ r
∞
X
j=0
c
8
p
k
+ c
9
| log r|
∞
X
j=0
1 + c
10
ǫ
2
| log ǫ
2
|
j
1 − p
j
r + ǫ
3 2
. If we assume that
ǫ
2
0 is sufficiently small, this is bounded by c
11
| log r|r + ǫ
3 2
. iv Once again, we will restart numbering of constants, i.e., we will use c
6
, c
7
, . . . , for constants unrelated to those with the same index in the earlier part of the proof.
Recall that j is the smallest integer such that 2
− j
≥ diamD. Let j
3
be the smallest j with the property that 2
− j
≤ dY
T
k 5
, ∂ D. It follows from 3.38 that for any β
2
1, on the event {T
k 5
2202
τ
+
ǫ
2
}, E
sup
T
k 5
≤t≤T
k+1 5
|X
T
k 5
− X
t
| | F
T
k 5
≤ E
sup
T
k 5
≤t≤T
k+1 4
|X
T
k 5
− X
t
| | F
T
k 5
+ E
sup
T
k+1 4
≤t≤T
k+1 5
|X
T
k+1 4
− X
t
| | F
T
k 5
≤ c
1
dY
T
k 5
, ∂ D + E
sup
T
k+1 4
≤t≤T
k+1 5
|X
T
k+1 4
− X
t
| | F
T
k 5
≤ c
1
dY
T
k 5
, ∂ D +
j
3
X
j= j
c
6
2
− j
dY
T
k 5
, ∂ D2
− j
≤ c
7
dY
T
k 5
, ∂ D1 + | log dY
T
k 5
, ∂ D|
≤ c
8
dY
T
k 5
, ∂ D
β
2
≤ c
9
ǫ
β
2
2
. This and 3.43 imply that
E
dY
T
k 5
, ∂ D1
{S
∗ 1
T
k 5
}
1
{T
k −1
5
τ
+
ǫ
2
}
sup
T
k 5
≤t≤T
k+1 5
|X
T
k 5
− X
t
|
3.45
= E
dY
T
k 5
, ∂ D1
{S
∗ 1
T
k 5
}
1
{T
k −1
5
τ
+
ǫ
2
}
E
sup
T
k 5
≤t≤T
k+1 5
|X
T
k 5
− X
t
| | F
T
k 5
≤ c
9
ǫ
β
2
2
E dY
T
k 5
, ∂ D1
{S
∗ 1
T
k 5
}
1
{T
k −1
5
τ
+
ǫ
2
}
≤ c
10
ǫ
β
2
2
1 + c
11
ǫ
2
| log ǫ
2
|
k
1 − p
k
r + ǫ
3 2
. It follows from the definition of S
1
that |ΠX
S
∗ 1
− X
S
∗ 1
| ≤ c
11
ǫ
2 2
if S
1
σ
∗
∧ τ
+
ǫ
2
. In the case when S
∗ 1
= σ
∗
∧ τ
+
ǫ
2
, the distance between X and Y is increasing at this instance, so it is easy to see that the vector X
S
∗ 1
− Y
S
∗ 1
must also have a position such that |ΠX
S
∗ 1
− X
S
∗ 1
| ≤ c
11
ǫ
2 2
. 3.46
Recall that we assume that X ∈ ∂ D, |X
− Y | = ǫ
1
, dY
, ∂ D = r. Recall also that ǫ
∗
is the parameter used in the definition of
ξ
j
and x
∗ j
at the beginning of this section. It follows from 3.33-3.37 that if
ǫ
∗
≥ c
1
ǫ
2
then at most one ξ
i
may belong to any given interval T
k −1
5
, T
k 5
] and, moreover, if for some
ξ
i
we have ξ
i
∈ T
k −1
5
, T
k 5
] then ξ
i
= T
k 5
. This, 3.43, 3.44, 3.45 and
2203
3.46 imply that,
E
X
≤ξ
i
≤S
∗ 1
L
S
∗ 1
− L
ξ
i
|x
∗ i
− ΠX
S
∗ 1
|
=
∞
X
k=0
E
X
≤ξ
i
≤S
∗ 1
L
S
∗ 1
− L
ξ
i
|x
∗ i
− ΠX
S
∗ 1
|1
{S
∗ 1
∈T
k 5
,T
k+1 5
]}
≤
∞
X
k=0
E
1
{S
∗ 1
∈T
k 5
,T
k+1 5
]} k
X
j=0
1
{T
j 5
τ
+
ǫ }
1
{T
j 5
≤ξ
i
≤S
∗ 1
}
L
T
j+1 5
− L
T
j 5
|x
∗ i
− ΠX
S
∗ 1
|
≤
∞
X
k=0
E 1
{S
∗ 1
∈T
k 5
,T
k+1 5
]}
×
k
X
j=0
1
{T
j 5
τ
+
ǫ }
1
{T
j 5
≤ξ
i
≤S
∗ 1
}
L
T
j+1 5
− L
T
j 5
|X
S
∗ 1
− ΠX
S
∗ 1
| + |x
∗ i
− X
S
∗ 1
| ≤
∞
X
k=0
E 1
{S
∗ 1
∈T
k 5
,T
k+1 5
]}
×
k
X
j=0
j + 11
{T
j 5
τ
+
ǫ }
L
T
j+1 5
− L
T
j 5
c
11
ǫ
2 2
+ sup
T
j 5
≤t≤T
j+1 5
|X
T
j 5
− X
t
| ≤
∞
X
k=0
E 1
{S
∗ 1
∈T
k 5
,T
k+1 5
]}
×
k
X
j=0
j + 11
{T
j 5
τ
+
ǫ }
c
2
dY
T
j 5
, ∂ D
c
11
ǫ
2 2
+ sup
T
j 5
≤t≤T
j+1 5
|X
T
j 5
− X
t
|
= E
∞
X
k=0 k
X
j=0
1
{S
∗ 1
∈T
k 5
,T
k+1 5
]}
j + 11
{T
j 5
τ
+
ǫ }
c
2
dY
T
j 5
, ∂ D
c
11
ǫ
2 2
+ sup
T
j 5
≤t≤T
j+1 5
|X
T
j 5
− X
t
|
= E
∞
X
j=0 ∞
X
k= j
1
{S
∗ 1
∈T
k 5
,T
k+1 5
]}
j + 11
{T
j 5
τ
+
ǫ }
c
2
dY
T
j 5
, ∂ D
c
11
ǫ
2 2
+ sup
T
j 5
≤t≤T
j+1 5
|X
T
j 5
− X
t
|
=
∞
X
j=0
E
1
{S
∗ 1
T
j 5
}
j + 11
{T
j 5
τ
+
ǫ }
c
2
dY
T
j 5
, ∂ D
c
11
ǫ
2 2
+ sup
T
j 5
≤t≤T
j+1 5
|X
T
j 5
− X
t
|
≤
∞
X
j=0
c
12
j + 11 + c
12
ǫ
2
| log ǫ
2
|
j
1 − p
j
ǫ
2 2
+ ǫ
β
2
2
r + ǫ
3 2
. If we assume that
ǫ
2
0 is sufficiently small, this is bounded by c
13
ǫ
β
2
2
r + ǫ
3 2
. Recall definitions of
σ
∗
and S
1
, and Lemma 3.4. There exists c
14
such that if ǫ
1
≤ c
14
ǫ
2
then
2204
σ
∗
τ
+
ǫ
2
. Hence, if ǫ
1
≤ c
14
ǫ
2
then
E
X
≤ξ
i
≤S
1
L
S
1
− L
ξ
i
|x
∗ i
− ΠX
S
1
|
≤ c
13
ǫ
β
2
2
r + ǫ
3 2
. 3.47
Let b S
k
= inf{t ≥ S
k
: X
t
∈ ∂ D} ∧ σ
∗
. The following estimate can be proved just like 3.39,
E dY
b S
k
, ∂ D | F
S
k
≤ 1 + c
14
ǫ
2
| log ǫ
2
|dY
S
k
, ∂ D.
We use this estimate, 3.47, the strong Markov property at b S
k
, and the definition of S
k
to see that
E
X
S
k
≤ξ
j
≤S
k+1
L
S
k+1
− L
ξ
j
|x
∗ j
− ΠX
S
k+1
| | F
S
k
= E
X
b S
k
≤ξ
j
≤S
k+1
L
S
k+1
− L
ξ
j
|x
∗ j
− ΠX
S
k+1
| | F
S
k
= E
E
X
b S
k
≤ξ
j
≤S
k+1
L
S
k+1
− L
ξ
j
|x
∗ j
− ΠX
S
k+1
| | F
b S
k
| F
S
k
≤ E
c
13
|X
b S
k
− Y
b S
k
|
β
2
dY
b S
k
, ∂ D + |X
b S
k
− Y
b S
k
|
3
| F
S
k
≤ E
c
15
|X
S
k
− Y
S
k
|
β
2
dY
b S
k
, ∂ D + |X
S
k
− Y
S
k
|
3
| F
S
k
≤ c
15
|X
S
k
− Y
S
k
|
β
2
1 + c
14
ǫ
2
| log ǫ
2
|dY
S
k
, ∂ D
+ |X
S
k
− Y
S
k
|
3
≤ c
15
|X
S
k
− Y
S
k
|
β
2
1 + c
14
ǫ
2
| log ǫ
2
||X
S
k
− Y
S
k
|
2
+ |X
S
k
− Y
S
k
|
3
≤ c
16
|X
S
k
− Y
S
k
|
2+ β
2
.
Lemma 3.10. There exist c
1
and a 0 such that for a
1
, a
2
a , if
|X − Y
| = ǫ then a.s., for every k
≥ 1, on the event U
k
σ
∗
, ®
nΠX
U
k
, Y
U
k
− X
U
k
|Y
U
k
− X
U
k
| ¸
≤ c
1
ǫ. Proof. First we will show that one can choose c
1
, a 0 and ǫ
0 so that for a
1
a ,
ǫ ≤ ǫ ,
x ∈ ∂ D, y ∈ D, |x − y| ≤ ǫ, z ∈ ∂ D, |x − z| ≤ 2a
1
ǫ and | y − z| ≤ 2a
1
ǫ, we have
nz,
y − x
| y − x| ≥ −c
1
ǫ4. 3.48
First suppose that y ∈ ∂ D. The assumptions that x, y ∈ ∂ D, |x − y| ≤ ǫ, and ∂ D is C
2
imply that the angle between
nx and y − x is in the range [π2 − c
2
ǫ, π2 + c
2
ǫ] for some c
2
∞. It follows from the assumptions that x, z
∈ ∂ D, |x − z| ≤ 2a
1
ǫ, and ∂ D is C
2
that the angle between nx and
2205
nz is less than c
3
ǫ for some c
3
∞. Therefore, the angle between nz and y − x is in the range
[π2 − c
1
ǫ4, π2 + c
1
ǫ4], where c
1
= 4c
2
+ c
3
. This implies that
〈nz, y − x〉 ≤ |y − x| sinc
1
ǫ4 ≤ c
1
| y − x|ǫ4. Thus
nz,
y − x
| y − x| ≤ c
1
ǫ4, and, therefore,
nz,
y − x
| y − x| ≥ −c
1
ǫ4. In other words, we proved that 3.48 in the special case when y
∈ ∂ D. If a
1
0 and ǫ 0 are
sufficiently small then D ∩ Bz, 2a
1
ǫ lies above ∂ D ∩ Bz, 2a
1
ǫ in the coordinate system with the origin at z = 0 and the vertical axis containing
nz. This observation proves that 3.48 applies to
all y ∈ D satisfying all the remaining assumptions.
An argument based on similar ideas shows that if x, y ∈ D, w ∈ ∂ D, |w − z| ≤ 2a
1
ǫ and
nz,
y − x
| y − x| ≤ c
1
ǫ2, then
nw,
y − x
| y − x| ≤ c
1
ǫ. 3.49
If |X
− Y | = ǫ then |X
t
− Y
t
| ≤ c
4
ǫ for all t ≤ σ
∗
, by Lemma 3.4. It follows easily from 3.1 that we can adjust the values of c
1
and ǫ
and choose a
2
0 so that if |X − Y
| = ǫ ≤ ǫ then on the
event S
k
σ
∗
, ®
nΠX
S
k
, Y
S
k
− X
S
k
|Y
S
k
− X
S
k
| ¸
≤ c
1
ǫ2. Let
A = ¨
t ∈ [S
k
, U
k
] :
nΠX
S
k
, Y
t
− X
t
|Y
t
− X
t
| c
1
ǫ2 «
. We will show that A = ∅. Suppose otherwise and let T
1
= inf A. Then ®
nΠX
S
k
, Y
T
1
− X
T
1
|Y
T
1
− X
T
1
| ¸
= c
1
ǫ2. We must have either X
T
1
∈ ∂ D or Y
T
1
∈ ∂ D. It follows from 3.48 that either X
T
1
∈ ∂ D or Y
T
1
∈ ∂ D. Suppose without loss of generality that X
T
1
∈ ∂ D and Y
T
1
∈ ∂ D. Then by 3.48, ®
nΠX
S
k
, Y
T
1
− X
T
1
|Y
T
1
− X
T
1
| ¸
= c
1
ǫ2.
2206
By the definition of T
1
, for every δ 0, L
t
must increase on the interval [T
1
, T
1
+ δ]. It is easy to see that this implies that the function
t →
nΠX
S
k
, Y
t
− X
t
|Y
t
− X
t
| is decreasing on the interval [T
1
, T
1
+ δ
1
], for some δ
1
0. This contradicts the definition of T
1
. Hence, for all t
∈ [S
k
, U
k
],
nΠX
S
k
, Y
t
− X
t
|Y
t
− X
t
| ≤ c
1
ǫ2. In particular,
®
nΠX
S
k
, Y
U
k
− X
U
k
|Y
U
k
− X
U
k
| ¸
≤ c
1
ǫ2. The lemma follows from the above estimate and 3.49.
Lemma 3.11. There exists c
1
such that if |X
− Y | ≤ ǫ then for every k,
E X
U
k
≤σ
∗
∧τ
+
ǫ
L
S
k+1
− L
U
k
≤ c
1
ǫ| log ǫ|. Proof. We use the strong Markov property at the hitting time of
∂ D by X and Lemma 3.8 ii to see that
EL
S
1
∧τ
+
ǫ
− L
U
≤ c
2
ǫ. 3.50
We will estimate L
S
k+1
− L
U
k
1
{U
k
τ
+
ǫ}
for k ≥ 1. Fix some k ≥ 1 and assume that U
k
τ
+
ǫ. Note that
dX
U
k
, ∂ D ≤ c
3
|X
U
k
− Y
U
k
|. Let T
1
= inf{t ≥ U
k
: X
t
∈ ∂ D} ∧ σ
∗
∧ τ
+
ǫ. Let j be the
greatest integer such that 2
− j
is greater than the diameter of D and let j
1
be the least integer such that 2
− j
1
≤ |X
U
k
− Y
U
k
|. By Lemma 3.5, for j ≤ j ≤ j
1
, P
|X
U
k
− X
T
1
| ∈ [2
− j
, 2
− j+1
] | F
U
k
≤ c
4
2
j
|X
U
k
− Y
U
k
|. 3.51
Next we will estimate dY
T
1
, ∂ D. Between times U
k
and T
1
, the process X
t
does not hit ∂ D. If Y
t
does not hit the boundary on the same interval, it is elementary to see from Lemma 3.10 that for j
≤ j ≤ j
1
,
dY
T
1
, ∂ D ≤ c
5
|X
U
k
− Y
U
k
|
2
+ c
6
|X
U
k
− Y
U
k
|2
− j
≤ c
7
|X
U
k
− Y
U
k
|2
− j
. Suppose that for some t
∗
∈ [U
k
, T
1
] we have Y
t
∗
∈ ∂ D, and assume that t
∗
is the largest time with this property. If t
∗
= T
1
then dY
T
1
, ∂ D = 0. Otherwise we must have τ
+
ǫ t
∗
, X
T
1
∈ ∂ D, and X
T
1
− Y
T
1
= X
t
∗
− Y
t
∗
. Since both Y
t
∗
and X
T
1
belong to ∂ D, easy geometry shows that in this case
dY
T
1
, ∂ D ≤ c
8
|X
U
k
− Y
U
k
|2
− j
. We conclude that dY
T
1
, ∂ D ≤ c
9
|X
U
k
− Y
U
k
|2
− j
, a.s. By Lemma 3.8 ii and the strong Markov property applied at U
k
, E
L
S
k+1
− L
U
k
| U
k
τ
+
ǫ, F
T
1
≤ c
10
|X
U
k
− Y
U
k
|2
− j
+ |X
U
k
− Y
U
k
|
3
≤ c
11
|X
U
k
− Y
U
k
|2
− j
. 2207
Hence, using 3.51,
E
L
S
k+1
− L
U
k
| U
k
τ
+
ǫ, F
U
k
= E
E
L
S
k+1
− L
U
k
| U
k
τ
+
ǫ, F
T
1
F
U
k
≤
X
j ≤ j≤ j
1
c
4
|X
U
k
− Y
U
k
|2
j
c
11
|X
U
k
− Y
U
k
|2
− j
≤ c
12
|X
U
k
− Y
U
k
|
2
| log |X
U
k
− Y
U
k
||. It is elementary to check that
E
L
U
k
− L
S
k
| S
k
τ
+
ǫ, F
S
k
≥ c
13
|X
S
k
− Y
S
k
|, and the conditional distribution of L
U
k
− L
S
k
given {S
k
τ
+
ǫ} is stochastically bounded by an exponential random variable with mean c
14
|X
S
k
− Y
S
k
|. Note that |X
U
k
− Y
U
k
| ≤ c
15
|X
S
k
− Y
S
k
|. Thus,
E
L
S
k+1
− L
U
k
| U
k
τ
+
ǫ, F
U
k
≤ c
16
|X
U
k
− Y
U
k
| | log |X
U
k
− Y
U
k
|| E
L
U
k
− L
S
k
| S
k
τ
+
ǫ, F
S
k
≤ c
17
ǫ| log ǫ| E
L
U
k
− L
S
k
| S
k
τ
+
ǫ, F
S
k
.
It follows that N
m
:=
m
X
k=1
c
18
ǫ| log ǫ|L
U
k
− L
S
k
1
{S
k
τ
+
ǫ}
− L
S
k+1
− L
U
k
1
{U
k
τ
+
ǫ}
is a submartingale with respect to the filtration F
∗ m
= F
X ,Y S
m+1
. If M = inf
{m :
m
X
k=1
L
U
k
− L
S
k
≥ 1} and M
i
= M ∧ i then
E
M
i
X
k=1
c
18
ǫ| log ǫ|L
U
k
− L
S
k
1
{S
k
τ
+
ǫ}
− L
S
k+1
− L
U
k
1
{U
k
τ
+
ǫ}
≥ 0,
and
E
M
i
X
k=1
L
S
k+1
− L
U
k
1
{U
k
τ
+
ǫ}
≤ E
M
i
X
k=1
c
18
ǫ| log ǫ|L
U
k
− L
S
k
1
{S
k
τ
+
ǫ}
. We let i
→ ∞ and obtain by the monotone convergence
E
M
X
k=1
L
S
k+1
− L
U
k
1
{U
k
τ
+
ǫ}
≤ E
M
X
k=1
c
18
ǫ| log ǫ|L
U
k
− L
S
k
1
{S
k
τ
+
ǫ}
≤ c
19
ǫ| log ǫ|. Hence,
E X
k ≥1,U
k
≤σ
∗
∧τ
+
ǫ
L
S
k+1
− L
U
k
≤ E
M
X
k=1
L
S
k+1
− L
U
k
1
{U
k
τ
+
ǫ}
≤ c
19
ǫ| log ǫ|. This and 3.50 imply the lemma.
2208
Recall parameters a
1
and a
2
and operator G
k
defined in 3.2.
Lemma 3.12. For any c
1
there exist a ,
ǫ 0 such that if a
1
, a
2
∈ 0, a and |X
− Y | = ǫ ≤ ǫ
then a.s., the following holds for all k
≥ 1. Let Θ =
Z
U
k
S
k
nY
t
d L
y t
− Z
U
k
S
k
nΠY
S
k
d L
y t
|X
S
k
− Y
S
k
| · |L
y U
k
− L
y S
k
|
−1
, with the convention that b
0 = 0. Then |Θ| ≤ c
1
and G
k
Y
S
k
− X
S
k
− Y
U
k
− X
U
k
+
nΠY
S
k
+ Θ|X
S
k
− Y
S
k
|
L
y U
k
− L
y S
k
− L
U
k
− L
S
k
+ π
ΠX
Sk
Y
S
k
− X
S
k
− Y
S
k
− X
S
k
≤ c
1
|L
U
k
− L
S
k
| · |Y
S
k
− X
S
k
|. Proof. By 2.2, for any c
2
, we can find ǫ
1
0 so small that for any x, y ∈ ∂ D with |x − y| ≤ 2ǫ
1
, |S xπ
x
x − y − n y − nx| ≤ c
2
2| y − x|. 3.52
By Lemma 3.4, if we choose a sufficiently small ǫ 0 then |Y
t
− X
t
| ≤ 2ǫ
1
for all t ≤ σ
∗
. Estimate 3.52 and C
2
-smoothness of ∂ D can be used to show that for any c
2
one can choose small a
1
, a
2
0 and ǫ 0 so that for every k ≥ 1 and all t ∈ [S
k
, U
k
] such that X
t
∈ ∂ D, |S ΠX
S
k
π
ΠX
Sk
X
S
k
− Y
S
k
− nΠY
S
k
− nX
t
| ≤ c
2
|Y
S
k
− X
S
k
|. 3.53
We obtain from 2.14 and the triangle inequality, Y
U
k
− X
U
k
− Y
S
k
− X
S
k
− S ΠX
S
k
π
ΠX
Sk
X
S
k
− Y
S
k
|L
U
k
− L
S
k
| −
nΠY
S
k
+ Θ|X
S
k
− Y
S
k
|
L
y U
k
− L
y S
k
− L
U
k
− L
S
k
= Z
U
k
S
k
nY
t
d L
y t
− Z
U
k
S
k
nX
t
d L
t
− S ΠX
S
k
π
ΠX
Sk
X
S
k
− Y
S
k
|L
U
k
− L
S
k
| −
nΠY
S
k
+ Θ|X
S
k
− Y
S
k
|
L
y U
k
− L
y S
k
− L
U
k
− L
S
k
≤ Z
U
k
S
k
nY
t
d L
y t
− Z
U
k
S
k
nΠY
S
k
d L
y t
− Θ|X
S
k
− Y
S
k
|L
y U
k
− L
y S
k
+ |Θ| |X
S
k
− Y
S
k
|L
U
k
− L
S
k
+ Z
U
k
S
k
nΠY
S
k
− nX
t
d L
t
− S ΠX
S
k
π
ΠX
Sk
X
S
k
− Y
S
k
|L
U
k
− L
S
k
| +
Z
U
k
S
k
nΠY
S
k
d L
y t
− Z
U
k
S
k
nΠY
S
k
d L
t
− nΠY
S
k
L
y U
k
− L
y S
k
− L
U
k
− L
S
k
.
2209
The expression on the last line is equal to zero for elementary reasons, so Y
U
k
− X
U
k
− Y
S
k
− X
S
k
− S ΠX
S
k
π
ΠX
Sk
X
S
k
− Y
S
k
|L
U
k
− L
S
k
| −
nΠY
S
k
+ Θ|X
S
k
− Y
S
k
|
L
y U
k
− L
y S
k
− L
U
k
− L
S
k
≤ Z
U
k
S
k
nY
t
d L
y t
− Z
U
k
S
k
nΠY
S
k
d L
y t
− Θ|X
S
k
− Y
S
k
|L
y U
k
− L
y S
k
+ |Θ| |X
S
k
− Y
S
k
|L
U
k
− L
S
k
+ Z
U
k
S
k
nΠY
S
k
− nX
t
d L
t
− S ΠX
S
k
π
ΠX
Sk
X
S
k
− Y
S
k
|L
U
k
− L
S
k
| .
The first term on the right hand side is equal to 0 by the definition of Θ. It is easy to see that this claim holds even if the definition of Θ involves the division by 0. We have obtained
Y
U
k
− X
U
k
− Y
S
k
− X
S
k
− S ΠX
S
k
π
ΠX
Sk
X
S
k
− Y
S
k
|L
U
k
− L
S
k
| 3.54
−
nΠY
S
k
+ Θ|X
S
k
− Y
S
k
|
L
y U
k
− L
y S
k
− L
U
k
− L
S
k
≤ |Θ| |X
S
k
− Y
S
k
|L
U
k
− L
S
k
+ Z
U
k
S
k
nΠY
S
k
− nX
t
d L
t
− S ΠX
S
k
π
ΠX
Sk
X
S
k
− Y
S
k
|L
U
k
− L
S
k
| .
It follows from the definitions of S
k
, U
k
and Π
x
that for sufficiently small a
1
and a
2
, we have for t
∈ [S
k
, U
k
], |Y
t
− ΠY
S
k
| ≤ 2a
1
|X
S
k
− Y
S
k
|, and a similar formula holds for X in place of Y on the left hand side. Hence, by 2.7, for some c
3
, Z
U
k
S
k
nY
t
d L
y t
− Z
U
k
S
k
nΠY
S
k
d L
y t
≤ Z
U
k
S
k
|nY
t
− nΠY
S
k
|d L
y t
≤ Z
U
k
S
k
c
3
|Y
t
− ΠY
S
k
|d L
y t
≤ Z
U
k
S
k
c
3
· 2a
1
|X
S
k
− Y
S
k
|d L
y t
≤ 2a
1
c
3
|X
S
k
− Y
S
k
| · |L
y U
k
− L
y S
k
|. This shows that if we take a
1
sufficiently small then |Θ| ≤ c
1
. We use 3.53 to derive the following estimate,
Z
U
k
S
k
nΠY
S
k
− nX
t
d L
t
− S ΠX
S
k
π
ΠX
Sk
X
S
k
− Y
S
k
|L
U
k
− L
S
k
| 3.55
≤ c
2
|X
S
k
− Y
S
k
| · |L
U
k
− L
S
k
|. 2210
We combine 3.54-3.55 to see that Y
U
k
− X
U
k
− Y
S
k
− X
S
k
− S ΠX
S
k
π
ΠX
Sk
X
S
k
− Y
S
k
|L
U
k
− L
S
k
| 3.56
−
nΠY
S
k
+ Θ|X
S
k
− Y
S
k
|
L
y U
k
− L
y S
k
− L
U
k
− L
S
k
≤ c
1
2 + c
2
|X
S
k
− Y
S
k
| · |L
U
k
− L
S
k
|. For any c
2
, we can choose small ǫ
so that π
ΠX
Sk
Y
S
k
− X
S
k
+ S ΠX
S
k
π
ΠX
Sk
X
S
k
− Y
S
k
|L
U
k
− L
S
k
| − expL
U
k
− L
S
k
S ΠX
S
k
π
ΠX
Sk
Y
S
k
− X
S
k
≤ c
2
|X
S
k
− Y
S
k
| · |L
U
k
− L
S
k
|. This and 3.56 imply that
Y
U
k
− X
U
k
− G
k
Y
S
k
− X
S
k
−
nΠY
S
k
+ Θ|X
S
k
− Y
S
k
|
L
y U
k
− L
y S
k
− L
U
k
− L
S
k
+ π
ΠX
Sk
Y
S
k
− X
S
k
− Y
S
k
− X
S
k
= Y
U
k
− X
U
k
− expL
U
k
− L
S
k
S ΠX
S
k
π
ΠX
Sk
Y
S
k
− X
S
k
−
nΠY
S
k
+ Θ|X
S
k
− Y
S
k
|
L
y U
k
− L
y S
k
− L
U
k
− L
S
k
+ π
ΠX
Sk
Y
S
k
− X
S
k
− Y
S
k
− X
S
k
≤ c
1
2 + 2c
2
|X
S
k
− Y
S
k
| · |L
U
k
− L
S
k
|. We obtain the lemma by choosing sufficiently small c
2
.
Lemma 3.13. If a
1
is sufficiently small then for some c
1
, ǫ
0 and all ǫ ǫ , if
|X − Y
| = ǫ then a.s., for all k
≥ 1, |L
y U
k
− L
y S
k
− L
U
k
− L
S
k
| ≤ c
1
|Y
S
k
− X
S
k
|
2
. Proof. Let
w = nΠX
S
k
. It follows from the definition of U
k
that |ΠX
S
k
− X
t
| ∨ |ΠX
S
k
− Y
t
| ≤ c
2
|Y
S
k
− X
S
k
|, for t
∈ [S
k
, U
k
]. This and 2.8 imply that for some c
3
and t ∈ [S
k
, U
k
], 1
− c
3
|Y
S
k
− X
S
k
|
2
≤ nX
t
, w ≤ 1,
for t such that X
t
∈ ∂ D, 3.57
1 − c
3
|Y
S
k
− X
S
k
|
2
≤ nY
t
, w ≤ 1,
for t such that Y
t
∈ ∂ D. 3.58
We appeal to 2.13 to see that if a
1
is sufficiently small and y ∈ ∂ D and z ∈ D are such that
max |z − X
S
k
|, | y − Y
S
k
| ≤ a
1
|X
S
k
− Y
S
k
| then for some c
4
, |
y
− z, w | ≤ c
4
|Y
S
k
− X
S
k
|
2
, 3.59
2211
and |
¬ Y
S
k
− X
S
k
, w
¶ | ≤ c
4
|Y
S
k
− X
S
k
|
2
. 3.60
Let I = {t ∈ [S
k
, U
k
] : Y
t
− X
t
, w
≥ 2c
4
|Y
S
k
− X
S
k
|
2
}. We claim that I = ∅. Suppose otherwise and let t
1
= inf I and t
2
= sup{t ∈ [S
k
, t
1
] : Y
t
∈ ∂ D}, with the convention that sup ∅ = S
k
. By 3.57, 3.59 and 3.60,
¬ Y
t
1
− X
t
1
, w
¶ =
¬ Y
t
2
− X
t
2
, w
¶ +
Z
t
1
t
2
nY
s
d L
y s
, w
+ −
Z
t
1
t
2
nX
s
d L
s
, w
+
≤ ¬
Y
t
2
− X
t
2
, w
¶ +
Z
t
1
t
2
nY
s
d L
y s
, w
+
= ¬
Y
t
2
− X
t
2
, w
¶ ≤ c
4
|Y
S
k
− X
S
k
|
2
. This contradicts the definition of t
1
, so we see that I = ∅. Similarly, one can prove that {t ∈ [S
k
, U
k
] : X
t
− Y
t
, w
≥ 2c
4
|Y
S
k
− X
S
k
|
2
} = ∅. Hence
{t ∈ [S
k
, U
k
] : | X
t
− Y
t
, w
| ≥ 2c
4
|Y
S
k
− X
S
k
|
2
} = ∅. This and 3.57-3.58 yield,
1 + c
3
|Y
S
k
− X
S
k
|
2
L
y U
k
− L
y S
k
− L
U
k
− L
S
k
≤ Z
U
k
S
k
nY
s
d L
y s
, w
+ −
Z
U
k
S
k
nX
s
d L
s
, w
+
= ¬
Y
U
k
− Y
S
k
− X
U
k
− X
S
k
, w
¶ ≤ 4c
4
|Y
S
k
− X
S
k
|
2
. By the definition of
σ
∗
, L
y U
k
− L
y S
k
≤ c
5
, so the above estimate implies L
y U
k
− L
y S
k
− L
U
k
− L
S
k
≤ 4c
4
|Y
S
k
− X
S
k
|
2
+ c
3
|Y
S
k
− X
S
k
|
2
L
y U
k
− L
y S
k
≤ c
6
|Y
S
k
− X
S
k
|
2
. An analogous argument gives
L
U
k
− L
S
k
− L
y U
k
− L
y S
k
≤ c
7
|Y
S
k
− X
S
k
|
2
. The lemma follows from the last two estimates.
Lemma 3.14. For some c
1
there exist a ,
ǫ 0 such that if a
1
, a
2
∈ 0, a , ǫ ≤ ǫ
and |X
− Y | = ǫ
then for all k ≥ 1,
E
π
ΠX
Sk+1
π
ΠX
Sk
Y
S
k
− X
S
k
− Y
S
k
− X
S
k
| F
S
k
≤ c
1
ǫ| log ǫ|
2
|Y
S
k
− X
S
k
|
2
.
2212
Proof. The vector w
k
:= π
ΠX
Sk
Y
S
k
− X
S
k
− Y
S
k
− X
S
k
is parallel to nΠX
S
k
. It is easy to check from the definition of S
k
that
|w
k
| ≤ c
2
|Y
S
k
− X
S
k
|
2
. Let T
1
= inf{t ≥ U
k
: X
t
∈ ∂ D}. It follows from Lemma 3.4 and definition of U
k
that dX
U
k
, ∂ D ≤
c
3
ǫ. Let j be the smallest integer such that
ǫ2
j
is greater than the diameter of D. Lemma 3.5 i shows that for some c
4
and all j = 1, 2, . . . , j ,
P |X
T
1
− X
U
k
| ≥ ǫ2
j
| F
U
k
≤ c
4
2
− j
. By Lemma 3.8 iii, the strong Markov property applied at T
1
, and Chebyshev’s inequality,
P |X
T
1
− X
S
k+1
| ≥ ǫ2
j
| F
T
1
≤ c
5
ǫ| log ǫ|ǫ2
j
= c
5
2
− j
| log ǫ|. The fact that
|X
S
k
− X
U
k
| ≤ c
6
ǫ and the last two estimates show that
P |X
S
k
− X
S
k+1
| ≥ ǫ2
j
| F
S
k
≤ c
6
2
− j
| log ǫ|. It is easy to see that
|π
ΠX
Sk+1
w
k
| ≤ c
7
ǫ2
j
|w
k
| if |X
S
k
− X
S
k+1
| ≤ ǫ2
j
. It follows that
E
π
ΠX
Sk+1
π
ΠX
Sk
Y
S
k
− X
S
k
− Y
S
k
− X
S
k
| F
S
k
≤ c
7
ǫ|w
k
| +
j
X
j=1
c
7
ǫ2
j+1
|w
k
| P|X
S
k
− X
S
k+1
| ∈ [ǫ2
j
, ǫ2
j+1
] | F
S
k
≤ c
7
ǫc
2
|Y
S
k
− X
S
k
|
2
+
j
X
j=1
c
7
ǫ2
j+1
c
2
|Y
S
k
− X
S
k
|
2
c
6
2
− j
| log ǫ| ≤ c
8
ǫ| log ǫ|
2
|Y
S
k
− X
S
k
|
2
.
Lemma 3.15. For some c
1
there exist a ,
ǫ 0 such that if a
1
, a
2
∈ 0, a , ǫ ≤ ǫ
and |X
− Y | = ǫ
then for all k ≥ 1,
E
π
ΠX
Sk+1
Y
U
k
− X
U
k
− Y
S
k+1
− X
S
k+1
| F
U
k
≤ c
1
|Y
U
k
− X
U
k
|
3
| log |Y
U
k
− X
U
k
||
2
. Proof. Fix some k and let
T
1
= inf{t ≥ U
k
: X
t
∈ ∂ D or Y
t
∈ ∂ D} and
ǫ
1
= |X
U
k
− Y
U
k
|. We will assume from now on that X
T
1
∈ ∂ D. The rest of the argument is similar if Y
T
1
∈ ∂ D. It follows from Lemma 3.4 and definition of U
k
that dX
U
k
, ∂ D ≤ c
2
ǫ
1
. Let j be the smallest
integer such that ǫ
1
2
j
is greater than the diameter of D. Lemma 3.5 shows that for some c
3
and all j = 1, 2, . . . , j
, P
|X
T
1
− X
U
k
| ≥ ǫ
1
2
j
≤ c
3
2
− j
. 3.61
By 2.9, we can choose c
4
so small that for x ∈ ∂ D ∩ BX
T
1
, 5c
4
ǫ
1
, |〈x − X
T
1
, nX
T
1
〉| ≤ a
2
ǫ
2 1
800. 3.62
2213
By the definition of σ
∗
, |Y
t
− X
t
| ≤ c
5
ǫ
1
for t ≤ σ
∗
. We make c
4
smaller, if necessary, so that, in view of 2.11,
|〈 y − x, nz〉| ≤ a
2
ǫ
2 1
400, 3.63
assuming that x, y, z ∈ ∂ D, | y − z| ≤ c
5
+ 5c
4
ǫ
1
and |x − y| ≤ 10c
4
ǫ
1
. The following definitions contain a parameter c
6
, the value of which will be chosen later. Let J = inf
{ j ≥ 1 : |X
T
1
− X
U
k
| ≤ ǫ
1
2
j
}, T
2
= inf{t ≥ T
1
: |B
t
− B
T
1
| ≥ c
4
ǫ
1
}, T
3
= inf{t ≥ T
1
: 〈nX
T
1
, B
t
− B
T
1
〉 ≤ −c
6
ǫ
2 1
2
J
}, A
1
= {T
3
≤ T
2
}. Note that neither X nor Y touches the boundary of D between times U
k
and T
1
, so Y
T
1
− X
T
1
= Y
U
k
− X
U
k
. Hence, by Lemma 3.10 and the strong Markov property applied at S
k
, ®
nΠX
U
k
, Y
T
1
− X
T
1
|Y
T
1
− X
T
1
| ¸
≤ c
7
ǫ
1
. 3.64
The angle between nΠX
U
k
and nX
T
1
is bounded by c
8
ǫ
1
2
J
because ∂ D is C
2
. This and 3.64 imply that
®
nX
T
1
, Y
T
1
− X
T
1
|Y
T
1
− X
T
1
| ¸
≤ c
9
ǫ
1
2
J
. 3.65
Let k
1
be such that c
9
ǫ
1
2
J
≤ 110 if J ≤ k
1
, and let F
1
= {J ≤ k
1
}. If F
1
holds then 3.65 implies that,
π
X
T1
Y
T
1
− X
T
1
|Y
T
1
− X
T
1
|
≥ 110. 3.66
Casei. This case is devoted to an estimate of the random variable in the statement of the lemma assuming that A
1
∩ F
1
holds. Since |Y
T
1
− X
T
1
| = ǫ
1
, 3.65 implies that
dY
T
1
, ∂ D ≤ c
10
ǫ
2 1
2
J
. 3.67
Let c
11
= 5c
4
and T
4
= inf{t ≥ T
1
: |X
t
− X
T
1
| ≥ c
11
ǫ
1
} ∧ T
2
∧ T
3
, T
5
= sup{t ≤ T
4
: X
t
∈ ∂ D}. We will show that T
4
= T
2
∧ T
3
, if ǫ and, therefore, ǫ
1
is sufficiently small. By 2.11,
〈x − y, nX
T
1
〉 ≤ c
12
ǫ
2 1
3.68 for all x, y
∈ BX
T
1
, c
11
ǫ
1
such that x ∈ ∂ D and y ∈ D. Since T
5
≤ T
3
, we have 〈B
T
5
− B
T
1
, nX
T
1
〉 ≥ −c
6
ǫ
2 1
2
J
. 3.69
2214
This and 3.68 imply that Z
T
5
T
1
nX
s
d L
s
, nX
T
1
+ =
¬ X
T
5
− X
T
1
− B
T
5
− B
T
1
, nX
T
1
¶ ≤ c
13
ǫ
2 1
2
J
. 3.70
For x ∈ ∂ D ∩ BX
T
1
, c
11
ǫ
1
we have by 2.8, for small ǫ
1
,
〈nx, nX
T
1
〉 ≥ 1 − c
14
ǫ
2 1
≥ 12. 3.71
This and 3.70 show that L
T
5
− L
T
1
≤ 2 Z
T
5
T
1
nX
s
d L
s
, nX
T
1
+ ≤ c
15
ǫ
2 1
2
J
. 3.72
For x ∈ ∂ D ∩ BX
T
1
, c
11
ǫ
1
, π
X
T1
nx ≤ c
16
ǫ
1
. 3.73
It follows from this and 3.72 that π
X
T1
Z
T
5
T
1
nX
s
d L
s
≤ c
17
ǫ
3 1
2
J
≤ c
18
ǫ
2 1
. 3.74
We can assume that ǫ
1
is so small that for x ∈ ∂ D ∩ BX
T
1
, c
11
ǫ
1
, |x − X
T
1
| ≤ 2|π
X
T1
x − X
T
1
|. 3.75
Since T
4
≤ T
2
∧ T
3
, we can use 3.74 and 3.75 to obtain, X
T
4
− X
T
1
≤ X
T
4
− X
T
5
+ X
T
5
− X
T
1
≤ X
T
4
− X
T
5
+ 2|π
X
T1
X
T
5
− X
T
1
| 3.76
≤ B
T
4
− B
T
5
+ 2 π
X
T1
B
T
5
− B
T
1
+ 2 π
X
T1
Z
T
5
T
1
nX
s
d L
s
≤ B
T
4
− B
T
1
+ B
T
1
− B
T
5
+ 2 π
X
T1
B
T
5
− B
T
1
+ 2 π
X
T1
Z
T
5
T
1
nX
s
d L
s
≤ 4c
4
ǫ
1
+ 2c
18
ǫ
2 1
. Recall that c
11
= 5c
4
. Hence, the last estimate and the definition of T
4
show that T
4
= T
2
∧ T
3
, if ǫ
1
is sufficiently small. Next we will estimate
dX
T
3
, ∂ D. Let R
1
= sup{t ≤ T
3
: X
t
∈ ∂ D}. By the definition of T
3
, 〈B
T
3
− B
R
1
, nX
T
1
〉 ≤ 0. This and the fact that X
T
3
− X
R
1
= B
T
3
− B
R
1
imply that, 〈X
T
3
− X
R
1
, nX
T
1
〉 ≤ 0. 3.77
2215
Since X
R
1
∈ ∂ D ∩ BX
T
1
, c
11
ǫ
1
, it follows from 3.62 and 3.77 that 〈X
T
3
− X
T
1
, nX
T
1
〉 = 〈X
T
3
− X
R
1
, nX
T
1
〉 + 〈X
R
1
− X
T
1
, nX
T
1
〉 ≤ a
2
ǫ
2 1
800. This and 3.62 imply that
dX
T
3
, ∂ D ≤ 2a
2
ǫ
2 1
800 = a
2
ǫ
2 1
400. 3.78
Our next goal is to estimate dY
T
3
, ∂ D. Recall that |Y
t
− X
t
| ≤ c
5
ǫ
1
for t ≤ σ
∗
. Since T
4
= T
2
∧ T
3
, the definition of T
4
implies that for t ∈ [T
1
, T
2
∧ T
3
], |Y
t
− X
T
1
| ≤ |Y
t
− X
t
| + |X
t
− X
T
1
| ≤ c
5
ǫ
1
+ c
11
ǫ
1
= c
19
ǫ
1
. 3.79
Let c
20
= 5c
4
and T
6
= inf{t ≥ T
1
: |Y
t
− Y
T
1
| ≥ c
20
ǫ
1
} ∧ T
2
∧ T
3
. If Y
t
∈ ∂ D for t ∈ [T
1
, T
6
] then L
y T
6
− L
y T
1
= 0. Suppose that Y
t
∈ ∂ D for some t ∈ [T
1
, T
6
] and let T
7
= sup{t ≤ T
6
: Y
t
∈ ∂ D}. We will show that T
6
= T
2
∧ T
3
, if ǫ and, therefore, ǫ
1
is sufficiently small. By 2.11,
〈x − y, nX
T
1
〉 ≤ c
21
ǫ
2 1
3.80 for all x, y
∈ BX
T
1
, c
19
ǫ
1
such that x ∈ ∂ D and y ∈ D. Since T
7
≤ T
3
, we have 〈B
T
7
− B
T
1
, nX
T
1
〉 ≥ −c
6
ǫ
2 1
2
J
. Since T
7
≤ T
2
∧ T
3
, we can use 3.80 and the last estimate to see that Z
T
7
T
1
nY
s
d L
y s
, nX
T
1
+ =
¬ Y
T
7
− Y
T
1
− B
T
7
− B
T
1
, nX
T
1
¶ ≤ c
22
ǫ
2 1
2
J
. 3.81
The above estimate is also valid in the case when Y
t
∈ ∂ D for t ∈ [T
1
, T
6
] because in this case L
y T
6
− L
y T
1
= 0. For x
∈ ∂ D ∩ BX
T
1
, c
19
ǫ
1
we have by 2.8, for small ǫ
1
,
〈nx, nX
T
1
〉 ≥ 1 − c
23
ǫ
2 1
≥ 12. This and 3.81 show that
L
y T
7
− L
y T
1
≤ 2 Z
T
7
T
1
nY
s
d L
y s
, nX
T
1
+ ≤ c
24
ǫ
2 1
2
J
. 3.82
For x ∈ ∂ D ∩ BX
T
1
, c
19
ǫ
1
, we have π
X
T1
nx ≤ c
25
ǫ
1
. It follows from this and 3.82 that π
X
T1
Z
T
7
T
1
nY
s
d L
y s
≤ c
26
ǫ
3 1
2
J
≤ c
27
ǫ
2 1
. 3.83
2216
We can assume that ǫ
1
is so small that for x ∈ ∂ D ∩ BX
T
1
, c
19
ǫ
1
, |x − X
T
1
| ≤ 2|π
X
T1
x − X
T
1
|. 3.84
Since T
6
≤ T
2
∧ T
3
, 3.83 and 3.84 imply that Y
T
6
− Y
T
1
≤ Y
T
6
− Y
T
7
+ Y
T
7
− Y
T
1
≤ Y
T
6
− Y
T
7
+ 2|π
X
T1
Y
T
7
− Y
T
1
| 3.85
≤ B
T
6
− B
T
7
+ 2 π
X
T1
B
T
7
− B
T
1
+ 2 π
X
T1
Z
T
7
T
1
nY
s
d L
y s
≤ B
T
6
− B
T
1
+ B
T
1
− B
T
7
+ 2 π
X
T1
B
T
7
− B
T
1
+ 2 π
X
T1
Z
T
7
T
1
nY
s
d L
y s
≤ 4c
4
ǫ
1
+ 2c
27
ǫ
2 1
. Recall that c
20
= 5c
4
. The last estimate and the definition of T
6
show that T
6
= T
2
∧ T
3
, if ǫ
1
is sufficiently small.
If ǫ
1
is small then, by 3.79, for t ∈ [T
1
, T
2
∧ T
3
], |ΠY
t
− X
T
1
| ≤ 2|Y
t
− X
T
1
| ≤ 2c
19
ǫ
1
. For x
∈ ∂ D ∩ BX
T
1
, 2c
19
ǫ
1
, by 2.9, |〈x − X
T
1
, nX
T
1
〉| ≤ c
28
ǫ
2 1
, 3.86
so, in particular, |〈ΠY
T
1
− X
T
1
, nX
T
1
〉| ≤ c
28
ǫ
2 1
. This and 3.67 imply that
|〈Y
T
1
− X
T
1
, nX
T
1
〉| ≤ |〈ΠY
T
1
− X
T
1
, nX
T
1
〉| + |〈ΠY
T
1
− Y
T
1
, nX
T
1
〉| 3.87
≤ c
28
ǫ
2 1
+ c
10
ǫ
2 1
2
J
≤ c
29
ǫ
2 1
2
J
. Recall that we assume that A
1
holds so that T
3
≤ T
2
. By 2.10, for x ∈ D ∩ BX
T
1
, c
19
ǫ
1
, 〈x − X
T
1
, nX
T
1
〉 ≥ −c
30
ǫ
2 1
, so, in view of 3.79,
〈Y
T
3
− X
T
1
, nX
T
1
〉 ≥ −c
30
ǫ
2 1
. 3.88
We now choose the parameter c
6
in the definition of T
3
so that −c
6
+ c
29
≤ −2c
30
. We will show that given this choice of c
6
, we must have Y
t
∈ ∂ D for t ∈ [T
1
, T
3
]. Suppose that Y
t
∈ ∂ D for t ∈ [T
1
, T
3
]. Then Y
t
− Y
T
1
= B
t
− B
T
1
for the same range of t’s. It follows from 3.87 and from the definition of T
3
that 〈Y
T
3
− X
T
1
, nX
T
1
〉 = 〈Y
T
3
− Y
T
1
, nX
T
1
〉 + 〈Y
T
1
− X
T
1
, nX
T
1
〉 = 〈B
T
3
− B
T
1
, nX
T
1
〉 + 〈Y
T
1
− X
T
1
, nX
T
1
〉 ≤ −c
6
ǫ
2 1
2
J
+ c
29
ǫ
2 1
2
J
≤ −2c
30
ǫ
2 1
. 2217
This contradicts 3.88, so we conclude that Y must cross ∂ D between times T
1
and T
3
. Hence, T
7
is well defined. Since we are assuming that A
1
holds, T
7
≤ T
3
= T
6
. Therefore, |Y
T
7
− Y
T
3
| ≤ |Y
T
7
− Y
T
1
| + |Y
T
1
− Y
T
3
| ≤ 2c
20
ǫ
1
= 10c
4
ǫ
1
. 3.89
By 3.79, |Y
T
7
− X
T
1
| ≤ c
5
+ 5c
4
ǫ
1
. This and 3.89 imply that the following can be derived as a special case of 3.63,
|〈Y
T
7
− x, nX
T
1
〉| ≤ a
2
ǫ
2 1
400, 3.90
for x ∈ ∂ D ∩ BY
T
7
, 2c
20
ǫ
1
. By the definition of T
3
, 〈B
T
3
− B
T
7
, nX
T
1
〉 ≤ 0. This and the fact that Y
T
3
− Y
T
7
= B
T
3
− B
T
7
imply that, 〈Y
T
3
− Y
T
7
, nX
T
1
〉 ≤ 0. We use this estimate and 3.90 to conclude that
dY
T
3
, ∂ D ≤ a
2
ǫ
2 1
400. 3.91
Recall that we are assuming that F
1
holds. It follows from 3.66 that π
X
T1
Y
T
1
− X
T
1
|Y
T
1
− X
T
1
|
≥ 110, and, therefore,
π
X
T1
Y
T
1
− X
T
1
≥ ǫ
1
10. By 3.74 and 3.83
π
X
T1
Y
T
3
− X
T
3
≥
π
X
T1
Y
T
1
− X
T
1
−
π
X
T1
Z
T
3
T
1
nX
s
d L
s
− π
X
T1
Z
T
3
T
1
nY
s
d L
y s
= π
X
T1
Y
T
1
− X
T
1
−
π
X
T1
Z
T
5
T
1
nX
s
d L
s
− π
X
T1
Z
T
7
T
1
nY
s
d L
y s
≥ ǫ
1
10 − c
18
ǫ
2 1
− c
27
ǫ
2 1
. For small
ǫ
1
, this is bounded below by ǫ
1
20. Hence, |Y
T
3
− X
T
3
| ≥ π
X
T1
Y
T
3
− X
T
3
≥ ǫ
1
20. This, 3.78 and 3.91 imply that S
k+1
≤ T
3
, assuming A
1
∩ F
1
holds. It follows from the definition of T
4
and the fact that S
k+1
≤ T
3
= T
4
that |X
S
k+1
− X
T
1
| ≤ c
11
ǫ
1
. This implies that
|ΠX
S
k+1
− X
T
1
| ≤ 2c
11
ǫ
1
, assuming that ǫ
1
is sufficiently small. Let T
8
= sup{t ∈ [T
1
, S
k+1
] : X
t
∈ ∂ D}. 2218
It is routine to check that 3.68-3.73 hold with X
T
1
replaced with ΠX
S
k+1
, and T
5
replaced with T
8
the values of the constants may have to be adjusted. Hence, we obtain as in 3.74 that π
ΠX
Sk+1
Z
S
k+1
T
1
nX
s
d L
s
= π
ΠX
Sk+1
Z
T
8
T
1
nX
s
d L
s
≤ c
31
ǫ
3 1
2
J
. 3.92
Similarly, an argument analogous to that in 3.80-3.83 yields π
ΠX
Sk+1
Z
S
k+1
T
1
nY
s
d L
y s
≤ c
32
ǫ
3 1
2
J
. This and 3.92 imply that
π
ΠX
Sk+1
Y
U
k
− X
U
k
− Y
S
k+1
− X
S
k+1
3.93
= π
ΠX
Sk+1
Y
T
1
− X
T
1
− Y
S
k+1
− X
S
k+1
3.94
= π
ΠX
Sk+1
Z
S
k+1
T
1
nX
s
d L
s
− Z
S
k+1
T
1
nY
s
d L
y s
≤ c
33
ǫ
3 1
2
J
. We obtain from this and 3.61,
E
π
ΠX
Sk+1
Y
U
k
− X
U
k
− Y
S
k+1
− X
S
k+1
1
A
1
∩F
1
| F
U
k
3.95
≤
j
X
j=1
c
34
ǫ
3 1
2
j
2
− j
≤ c
35
ǫ
3 1
| log ǫ
1
| = c
35
ǫ
2 1
| log ǫ| |Y
U
k
− X
U
k
|.
Case ii. We will now analyze the case when A
1
does not occur. The rest of the proof is an outline only. Most steps are very similar to those in Case i, so we omit details to save space.
Standard estimates show that
PA
c 1
| F
T
1
≤ c
36
ǫ
1
2
J
. 3.96
Recall that we have assumed that X
T
1
∈ ∂ D. Let T
9
= inf{t ≥ T
2
: Y
t
∈ ∂ D}. For some c
37
and c
38
, we let K = inf
{ j ≥ 1 : sup
t ∈[T
2
,T
9
]
|Y
t
− Y
T
2
| ≤ ǫ
1
2
j
}, T
8
= inf{t ≥ T
7
: |B
t
− B
T
7
| ≥ c
37
ǫ
1
}, T
9
= inf{t ≥ T
7
: 〈nY
T
7
, B
t
− B
T
7
〉 ≤ −c
38
ǫ
2 1
2
K
}, A
2
= {T
9
≤ T
8
}.
2219
Let T
10
= sup{t ≤ T
9
: X
t
∈ ∂ D} and note that X
T
9
− Y
T
9
= X
T
10
− Y
T
10
. Using the fact that X
T
1
∈ ∂ D and definitions of T
1
, T
2
and K, one can show that ®
nY
T
9
, Y
T
9
− X
T
9
|Y
T
9
− X
T
9
| ¸
= ®
nY
T
9
, Y
T
10
− X
T
10
|Y
T
10
− X
T
10
| ¸
≤ c
39
ǫ
1
2
K
. 3.97
This implies that dX
T
9
, ∂ D ≤ c
40
ǫ
2 1
2
K
. We can repeat the argument proving 3.94, with the roles of X and Y interchanged and T
1
replaced by T
9
, to see that if A
2
holds then S
k+1
≤ T
9
and π
ΠX
Sk+1
Y
T
9
− X
T
9
− Y
S
k+1
− X
S
k+1
≤ c
41
ǫ
3 1
2
K
. 3.98
The angle between nY
T
9
and nΠX
S
k+1
is less than c
42
ǫ
1
. We know from 3.67 that dY
T
1
, ∂ D ≤
c
43
ǫ
2 1
2
J
. These facts and 3.97 imply that
nΠX
S
k+1
, Z
T
9
T
2
nX
s
d L
s
+ =
¬
nΠX
S
k+1
, Y
T
9
− X
T
9
− Y
T
2
− X
T
2
¶ ≤ c
44
ǫ
2 1
2
J ∨K
. Let k
2
be the largest integer such that if K ≤ k
2
then for x ∈ ∂ D ∩ BY
T
2
, 2 ǫ
1
2
K
we have
〈nx, nΠX
S
k+1
〉 ≥ 12. Assume that F
2
:= {K ≤ k
2
} holds. It follows that L
T
9
− L
T
2
≤ 2 Z
T
9
T
2
nX
s
d L
s
, nΠX
S
k+1
+ ≤ c
45
ǫ
2 1
2
J ∨K
. We also have L
T
2
− L
T
1
≤ c
46
ǫ
2 1
2
J
by 3.72. Hence, L
T
9
− L
T
1
≤ c
47
ǫ
2 1
2
J ∨K
. For x
∈ ∂ D ∩ BY
T
2
, 2 ǫ
1
2
K
, we have |π
ΠX
Sk+1
nx| ≤ c
48
ǫ
1
2
K
, so π
ΠX
Sk+1
Z
T
9
T
1
nX
s
d L
s
≤ c
49
ǫ
3 1
2
J ∨K+K
. By 3.82, L
y T
2
− L
y T
1
≤ c
50
ǫ
2 1
2
J
, so π
ΠX
Sk+1
Z
T
9
T
1
nY
s
d L
y s
= π
ΠX
Sk+1
Z
T
2
T
1
nY
s
d L
y s
≤ c
51
ǫ
3 1
2
J +K
. Combining the last two estimates with 3.98, we obtain,
π
ΠX
Sk+1
Y
S
k+1
− X
S
k+1
− Y
T
1
− X
T
1
3.99
= π
ΠX
Sk+1
Y
S
k+1
− X
S
k+1
− Y
T
9
− X
T
9
+
π
ΠX
Sk+1
Y
T
9
− X
T
9
− Y
T
1
− X
T
1
≤ c
41
ǫ
3 1
2
K
+ π
ΠX
Sk+1
Z
T
9
T
1
nX
s
d L
s
+ π
ΠX
Sk+1
Z
T
2
T
1
nY
s
d L
y s
≤ c
52
ǫ
3 1
2
J ∨K+K
.
2220
This implies that
E
π
ΠX
Sk+1
Y
U
k
− X
U
k
− Y
S
k+1
− X
S
k+1
1
A
c 1
∩A
2
∩F
2
| F
U
k
3.100
=
j
X
j=1 j
X
k=1
E
π
ΠX
Sk+1
Y
U
k
− X
U
k
− Y
S
k+1
− X
S
k+1
1
A
c 1
∩A
2
∩F
2
| J = j, K = k, F
U
k
× P
J = j, K = k | F
U
k
.
By 3.67 and an estimate similar to that in Lemma 3.5 i,
P
K = k | F
T
1
≤ c
53
ǫ
2 1
2
J
ǫ
−1 1
2
−k
= c
53
ǫ
1
2
J −k
. This, 3.61 an the strong Markov property applied at T
1
yield,
P
J = j, K = k | F
U
k
≤ c
54
2
− j
ǫ
1
2
j −k
= c
54
ǫ
1
2
−k
. 3.101
For K ≥ J we have 2
J ∨K+K
= 2
2K
so the the right hand side of 3.99 is bounded by c
55
ǫ
3 1
2
2K
. This and 3.101 imply that the corresponding contribution to the expectation in 3.100 is bounded by
j
X
j=1 j
X
k= j
c
54
ǫ
1
2
−k
c
55
ǫ
3 1
2
2k
≤ c
56
ǫ
3 1
| log ǫ
1
|. 3.102
For K J we have 2
J ∨K+K
= 2
J +K
so the corresponding contribution to the expectation in 3.100 is bounded by
j
X
j=1 j
X
k=1
c
54
ǫ
1
2
−k
c
55
ǫ
3 1
2
j+k
≤ c
57
ǫ
3 1
| log ǫ
1
|. Combining this with 3.102 yields
E
π
ΠX
Sk+1
Y
U
k
− X
U
k
− Y
S
k+1
− X
S
k+1
1
A
c 1
∩A
2
∩F
2
| F
U
k
≤ c
58
ǫ
3 1
| log ǫ
1
|. 3.103
The probability that A
2
does not occur, conditional on J and K, is bounded above by c
59
ǫ
2 1
2
K
ǫ
1
= c
59
ǫ
1
2
K
. If A
c 1
∩ A
c 2
holds, we use the following crude estimate, π
ΠX
Sk+1
Y
U
k
− X
U
k
− Y
S
k+1
− X
S
k+1
≤ c
5
ǫ
1
. Therefore, using 3.101,
E
π
ΠX
Sk+1
Y
U
k
− X
U
k
− Y
S
k+1
− X
S
k+1
1
A
c 1
∩A
c 2
| F
U
k
3.104
≤
j
X
j=1 j
X
k=1
c
54
ǫ
1
2
−k
c
59
ǫ
1
2
k
c
5
ǫ
1
≤ c
60
ǫ
3 1
| log ǫ
1
|
2
.
2221
It remains to address the cases when F
1
or F
2
fail. The probability of F
c 1
∩ F
c 2
is bounded by c
61
ǫ
2 1
. Hence,
E
π
ΠX
Sk+1
Y
U
k
− X
U
k
− Y
S
k+1
− X
S
k+1
1
F
c 1
∩F
c 2
| F
U
k
≤ c
61
ǫ
2 1
c
5
ǫ
1
= c
62
ǫ
3 1
. 3.105
If F
1
fails but F
2
does not. we can repeat the analysis presented in Case ii. Hence, 3.103 holds with
1
A
c 1
∩A
2
∩F
2
replaced with 1
F
c 1
∩A
2
∩F
2
. The lemma follows from these remarks, 3.95, 3.103, 3.104 and 3.105.
Lemma 3.16. We have for some c
1
, E
m
′
X
k=0
|Y
S
k
− X
S
k
|
≤ c
1
. Proof. We will use modified versions of stopping times S
k
and U
k
by dropping σ
∗
from the definition 3.1. Let S
∗
= U
∗
= inf{t ≥ 0 : X
t
∈ ∂ D} and for k ≥ 1 define S
∗ k
= inf ¦
t ≥ U
∗ k
−1
: dX
t
,
∂ D ∨ dY
t
, ∂ D ≤ a
2
|X
t
− Y
t
|
2
© ,
U
∗ k
= inf n
t ≥ S
∗ k
: |X
t
− X
S
∗ k
| ∨ |Y
t
− Y
S
∗ k
| ≥ a
1
|X
S
∗ k
− Y
S
∗ k
| o
. Fix some k and let
T
1
= inf n
t ≥ S
∗ k
: D
B
t
− B
S
∗ k
, nΠX
S
∗ k
E ≤ −a
1
2|X
S
∗ k
− Y
S
∗ k
| o
, T
2
= inf n
t ≥ S
∗ k
: D
B
t
− B
S
∗ k
, nΠX
S
∗ k
E ≥ a
1
4|X
S
∗ k
− Y
S
∗ k
| o
, T
3
= inf ¨
t ≥ S
∗ k
: π
ΠX
S∗ k
B
t
− B
S
∗ k
≥ a
1
10|X
S
∗ k
− Y
S
∗ k
| «
, A =
{T
1
≤ T
2
≤ T
3
}, F
∗ k
= σ{B
t
, t ≤ S
∗ k
}. Let
ǫ = |X − Y
| and recall that |X
t
− Y
t
| c
2
ǫ for t ≤ σ
∗
. By Brownian scaling and the strong Markov property,
PA | F
∗ k
≥ p
1
on {S
∗ k
≤ σ
∗
}, for some p
1
0 that does not depend on ǫ or k. An argument similar to that in the proof of Lemma 3.8 i can be used to show that if
ǫ, a
1
and a
2
are small and A holds then T
1
U
∗ k
and L
T
1
− L
S
∗ k
a
1
4|X
S
∗ k
−Y
S
∗ k
|. Then L
U
∗ k
− L
S
∗ k
a
1
4|X
S
∗ k
−Y
S
∗ k
|, so
EL
U
∗ k
− L
S
∗ k
| F
∗ k
p
1
a
1
4|X
S
∗ k
− Y
S
∗ k
|.
2222
We use this estimate to see that
E
m
′
X
k=0
|Y
S
k
− X
S
k
|
= E
m
′
X
k=0
|Y
S
∗ k
− X
S
∗ k
|
3.106
= E
m
′
−1
X
k=0
|Y
S
∗ k
− X
S
∗ k
|
+ |Y
S
∗ m′
− X
S
∗ m′
| ≤ E
m
′
−1
X
k=0
c
3
E L
U
∗ k
− L
S
∗ k
| F
∗ k
+ |Y
S
∗ m′
− X
S
∗ m′
| ≤ c
3
E
m
′
−1
X
k=0
L
U
∗ k
− L
S
∗ k
+ |Y
S
∗ m′
− X
S
∗ m′
| ≤ c
3
E
σ
∗
+ |Y
S
∗ m′
− X
S
∗ m′
|. It is elementary to check that for all j,
PL
j+1
− L
j
1 | σ{B
t
, t ≤ j} ≥ p
2
0. Hence,
σ
∗
≤ σ
1
is stochastically majorized by a geometric random variable with mean depending only on D, so
E σ
∗
c
4
∞. 3.107
We have |X
S
∗ m′
− Y
S
∗ m′
| c
2
ǫ because S
∗ m
′
≤ σ
∗
. We combine this, 3.106 and 3.107 to complete the proof.
Lemma 3.17. For some c
1
there exists a 0 such that if a
1
, a
2
∈ 0, a and |X
− Y | = ǫ then,
E
m
′
X
k=0
|X
S
k
− Y
S
k
| L
y U
k
− L
y S
k
− L
U
k
− L
S
k
≤ c
1
ǫ
2
. Proof. We have by Lemmas 3.13 and 3.16,
E
m
′
X
k=0
|X
S
k
− Y
S
k
| L
y U
k
− L
y S
k
− L
U
k
− L
S
k
≤ c
2
ǫ
2
E
m
′
X
k=0
|X
S
k
− Y
S
k
|
≤ c
3
ǫ
2
.
Lemma 3.18. For some c
1
there exists a 0 such that if a
1
, a
2
∈ 0, a and |X
− Y | = ǫ then,
E
m
′
X
k=0
π
ΠX
Sk+1
nΠY
S
k
L
y U
k
− L
y S
k
− L
U
k
− L
S
k
≤ c
1
ǫ
2
| log ǫ|. Proof. First, we will show that
E
π
ΠX
Sk+1
nΠY
S
k
| F
U
k
≤ c
2
|Y
S
k
− X
S
k
| | log |Y
S
k
− X
S
k
||. 3.108
2223
Recall the notation from the proof of Lemma 3.15, in particular, ǫ
1
= |Y
U
k
− X
U
k
|, and note that by Lemma 3.4,
ǫ
1
≤ c
3
|Y
S
k
− X
S
k
|. If A
1
occurs then S
k+1
≤ T
3
≤ T
2
. This and definitions of S
k
, U
k
, T
2
, T
3
and T
4
imply that |Y
S
k
− X
S
k+1
| ≤ |Y
S
k
− X
S
k
| + |X
S
k
− X
U
k
| + |X
U
k
− X
T
1
| + |X
T
1
− X
S
k+1
| ≤ c
4
|Y
S
k
− X
S
k
|2
J
. Therefore, 2.12 shows that
π
ΠX
Sk+1
nΠY
S
k
≤ c
5
ǫ
1
2
J
. We calculate as in 3.95,
E
π
ΠX
Sk+1
nΠY
S
k
1
A
1
| F
U
k
≤
j
X
j=1
c
6
ǫ
1
2
j
2
− j
≤ c
7
ǫ
1
| log ǫ
1
|. 3.109
We obtain from 3.96,
E
π
ΠX
Sk+1
nΠY
S
k
1
A
c 1
| F
U
k
≤ E
1
A
c 1
| F
U
k
≤
j
X
j=1
c
8
ǫ
1
2
j
2
− j
≤ c
9
ǫ
1
| log ǫ
1
|. This and 3.109 prove 3.108. By 3.108 and Lemma 3.13,
E
nΠY
S
k
L
y U
k
− L
y S
k
− L
U
k
− L
S
k
| F
U
k
≤ c
10
|Y
S
k
− X
S
k
|
3
| log |Y
S
k
− X
S
k
||. We use this estimate and Lemma 3.16 to conclude that
E
m
′
X
k=0
nΠY
S
k
L
y U
k
− L
y S
k
− L
U
k
− L
S
k
≤ E
m
′
X
k=0
c
11
ǫ
2
| log ǫ||Y
S
k
− X
S
k
|
≤ c
12
ǫ
2
| log ǫ|.
Lemma 3.19. For some c
1
there exist a ,
ǫ 0 such that if a
1
, a
2
∈ 0, a , ǫ ≤ ǫ
and |X
− Y | = ǫ
then,
E
m
′
X
k=0
π
ΠX
Sk+1
π
ΠX
Sk
Y
S
k
− X
S
k
− Y
S
k
− X
S
k
≤ c
1
ǫ
2
| log ǫ|
2
. Proof. Lemmas 3.14 and 3.16 imply that
E
m
′
X
k=0
π
ΠX
Sk+1
π
ΠX
Sk
Y
S
k
− X
S
k
− Y
S
k
− X
S
k
≤ E
m
′
X
k=0
E
π
ΠX
Sk+1
π
ΠX
Sk
Y
S
k
− X
S
k
− Y
S
k
− X
S
k
| F
S
k
≤ E
m
′
X
k=0
c
2
ǫ| log ǫ|
2
|Y
S
k
− X
S
k
|
2
≤ E
m
′
X
k=0
c
3
ǫ
2
| log ǫ|
2
|Y
S
k
− X
S
k
|
≤ c
4
ǫ
2
| log ǫ|
2
.
2224
Lemma 3.20. For any c
1
, ǫ
0 there exist a 0, a random variable Λ and c
2
such that if ǫ ∈ 0, ǫ
, a
1
, a
2
a and
|X − Y
| = ǫ then |Λ| ≤ c
1
ǫ, a.s., and
E Y
σ
∗
− X
σ
∗
− G
m
′
◦ · · · ◦ G Y
− X − Λ
≤ c
2
ǫ
2
| log ǫ|
2
. Proof. Note that S
m
′
+1
= σ
∗
. We have G
m
′
◦ · · · ◦ G Y
− X − Y
σ
∗
− X
σ
∗
3.110 =
m
′
X
k=0
G
m
′
◦ · · · ◦ G
k+1
G
k
Y
S
k
− X
S
k
− Y
S
k+1
− X
S
k+1
=
m
′
X
k=0
G
m
′
◦ · · · ◦ G
k+1
G
k
Y
S
k
− X
S
k
− Y
U
k
− X
U
k
3.111
+
m
′
X
k=0
G
m
′
◦ · · · ◦ G
k+1
Y
U
k
− X
U
k
− Y
S
k+1
− X
S
k+1
.
Recall Θ from Lemma 3.12. By 2.3, Lemma 3.4 and the triangle inequality, we have the following estimate for the first sum in 3.111,
m
′
X
k=0
G
m
′
◦ · · · ◦ G
k+1
G
k
Y
S
k
− X
S
k
− Y
U
k
− X
U
k
≤ c
3 m
′
X
k=0
G
k+1
G
k
Y
S
k
− X
S
k
− Y
U
k
− X
U
k
≤ c
3 m
′
X
k=0
G
k+1
G
k
Y
S
k
− X
S
k
− Y
U
k
− X
U
k
+
nΠY
S
k
+ Θ|X
S
k
− Y
S
k
|
L
y U
k
− L
y S
k
− L
U
k
− L
S
k
+ π
ΠX
Sk
Y
S
k
− X
S
k
− Y
S
k
− X
S
k
+ c
3 m
′
X
k=0
G
k+1
nΠY
S
k
L
y U
k
− L
y S
k
− L
U
k
− L
S
k
+ c
3 m
′
X
k=0
G
k+1
Θ|X
S
k
− Y
S
k
| L
y U
k
− L
y S
k
− L
U
k
− L
S
k
+ c
3 m
′
X
k=0
G
k+1
π
ΠX
Sk
Y
S
k
− X
S
k
− Y
S
k
− X
S
k
.
2225
We combine this with 3.110 to obtain G
m
′
◦ · · · ◦ G Y
− X − Y
σ
∗
− X
σ
∗
3.112 ≤ c
3 m
′
X
k=0
G
k+1
G
k
Y
S
k
− X
S
k
− Y
U
k
− X
U
k
3.113 +
nΠY
S
k
+ Θ|X
S
k
− Y
S
k
|
L
y U
k
− L
y S
k
− L
U
k
− L
S
k
3.114 + π
ΠX
Sk
Y
S
k
− X
S
k
− Y
S
k
− X
S
k
3.115 + c
3 m
′
X
k=0
G
k+1
nΠY
S
k
L
y U
k
− L
y S
k
− L
U
k
− L
S
k
3.116 + c
3 m
′
X
k=0
G
k+1
Θ|X
S
k
− Y
S
k
| L
y U
k
− L
y S
k
− L
U
k
− L
S
k
3.117 + c
3 m
′
X
k=0
G
k+1
π
ΠX
Sk
Y
S
k
− X
S
k
− Y
S
k
− X
S
k
3.118 +
m
′
X
k=0
G
m
′
◦ · · · ◦ G
k+1
Y
U
k
− X
U
k
− Y
S
k+1
− X
S
k+1
.
3.119 We need the following elementary fact about any non-negative real numbers b
1
, b
2
and b
3
. Suppose that b
1
≤ b
2
+ b
3
. Let Λ = max0, b
1
− b
2
. Then |Λ| ≤ b
3
. Moreover, |b
1
− Λ| ≤ b
2
. To see this, suppose that b
1
≥ b
2
. Then Λ = b
1
− b
2
and |b
1
− Λ| = |b
1
− b
1
− b
2
| = b
2
. If b
1
b
2
then Λ = 0 and
|b
1
− Λ| = |b
1
| b
2
. We apply these observations to b
1
equal to 3.112, b
2
equal to the sum of the terms 3.116-3.119, and b
3
equal to 3.113-3.115. To finish the proof of the lemma, it will suffice to prove that
b
3
≤ c
1
ǫ, a.s.,
3.120 and
Eb
2
≤ c
2
ǫ
2
| log ǫ|
2
. 3.121
Fix an arbitrarily small c
1
0. By Lemma 3.4, |Y
S
k
− X
S
k
| ≤ c
4
ǫ, for all k, a.s. By Lemma 3.12, if a
1
and a
2
are sufficiently small then with probability 1, b
3
≤ c
1
c
4 m
′
X
k=0
|L
U
k
− L
S
k
| · |Y
S
k
− X
S
k
| ≤ c
1
ǫ
m
′
X
k=0
|L
U
k
− L
S
k
|. We have
P
m
′
k=0
|L
U
k
− L
S
k
| ≤ 1, so a.s., b
3
≤ c
1
ǫ, that is, 3.120 holds true. We estimate 3.116 using 2.3 and Lemma 3.18,
E
m
′
X
k=0
G
k+1
nΠY
S
k
L
y U
k
− L
y S
k
− L
U
k
− L
S
k
3.122 ≤ c
5
E
m
′
X
k=0
π
ΠX
Sk+1
nΠY
S
k
L
y U
k
− L
y S
k
− L
U
k
− L
S
k
≤ c
6
ǫ
2
| log ǫ|.
2226
Similarly, 2.3 and Lemma 3.19 yield the following estimate for 3.118,
E
m
′
X
k=0
G
k+1
π
ΠX
Sk
Y
S
k
− X
S
k
− Y
S
k
− X
S
k
3.123 ≤ c
5
E
m
′
X
k=0
π
ΠX
Sk+1
π
ΠX
Sk
Y
S
k
− X
S
k
− Y
S
k
− X
S
k
≤ c
7
ǫ
2
| log ǫ|
2
. Recall from Lemma 3.12 that
|Θ| ≤ c
8
. By 2.3 and Lemmas 3.13 and 3.16,
E
m
′
X
k=0
G
k+1
Θ|X
S
k
− Y
S
k
| L
y U
k
− L
y S
k
− L
U
k
− L
S
k
3.124 ≤ c
9
E
m
′
X
k=0
|X
S
k
− Y
S
k
| L
y U
k
− L
y S
k
− L
U
k
− L
S
k
≤ c
10
E
m
′
X
k=0
|X
S
k
− Y
S
k
|
3
≤ c
11
ǫ
2
E
m
′
X
k=0
|X
S
k
− Y
S
k
|
≤ c
12
ǫ
2
. By Lemma 3.15,
E
π
ΠX
Sk+1
Y
U
k
− X
U
k
− Y
S
k+1
− X
S
k+1
| F
U
k
≤ c
13
|Y
U
k
− X
U
k
|
3
| log |Y
U
k
− X
U
k
||
2
. Hence, using 2.3 and Lemmas 3.4 and 3.16,
E
m
′
X
k=0
G
m
′
◦ · · · ◦ G
k+1
Y
U
k
− X
U
k
− Y
S
k+1
− X
S
k+1
3.125
≤ c
14
E
m
′
X
k=0
π
ΠX
Sk+1
Y
U
k
− X
U
k
− Y
S
k+1
− X
S
k+1
≤ c
15
E
m
′
X
k=0
|Y
U
k
− X
U
k
|
3
| log |Y
U
k
− X
U
k
||
2
≤ c
16
ǫ
2
| log ǫ|
2
E
m
′
X
k=0
|Y
U
k
− X
U
k
|
≤ c
17
ǫ
2
| log ǫ|
2
. The inequality in 3.121 follows from 3.122-3.125. This completes the proof of the lemma.
Recall operator H
k
defined in 3.2.
Lemma 3.21. For any c
1
, ǫ
0 there exists a 0 such that if a
1
, a
2
a and
|X − Y
| = ǫ then,
E |G
m
′
◦ · · · ◦ G Y
− X − H
m
′
◦ · · · ◦ H Y
− X | ≤ c
1
ǫ
2
| log ǫ|.
2227
Proof. We have G
m
′
◦ · · · ◦ G Y
− X − H
m
′
◦ · · · ◦ H Y
− X 3.126
=
m
′
X
k=0
G
m
′
◦ · · · ◦ G
k+1
expL
U
k
− L
S
k
S ΠX
S
k
− expL
S
k+1
− L
S
k
S ΠX
S
k
◦ π
ΠX
Sk
H
k −1
◦ · · · ◦ H Y
− X .
By 2.6, k expL
U
k
− L
S
k
S ΠX
S
k
− expL
S
k+1
− L
S
k
S ΠX
S
k
k ≤ c
2
|L
U
k
− L
S
k+1
|. This, 2.3 and 3.126 imply that
|G
m
′
◦ · · · ◦ G Y
− X − H
m
′
◦ · · · ◦ H Y
− X | ≤ c
3
|Y − X
|
m
′
X
k=0
|L
U
k
− L
S
k+1
|. By Lemma 3.11,
E P
m
′
k=0
|L
U
k
− L
S
k+1
| ≤ c
4
ǫ| log ǫ|. Hence,
E |G
m
′
◦ · · · ◦ G Y
− X − H
m
′
◦ · · · ◦ H Y
− X | ≤ c
4
ǫ
2
| log ǫ|.
Recall notation from the beginning of this section.
Lemma 3.22. We have for any β
1
1 and some c and c
1
, assuming that |X
− Y | = ǫ and ǫ
∗
≥ c ǫ,
E
m
′
X
k=0
X
U
k
≤ξ
j
≤S
k+1
L
S
k+1
− L
ξ
j
|x
∗ j
− ΠX
S
k+1
|
≤ c
1
ǫ
1+ β
1
. Proof. By Lemma 3.8 iv, for every k,
E
X
S
k
≤ξ
j
≤S
k+1
L
S
k+1
− L
ξ
j
|x
∗ j
− ΠX
S
k+1
| | F
S
k
≤ c
2
|X
S
k
− Y
S
k
|
2+ β
1
. This and Lemma 3.16 imply that
E
m
′
X
k=0
X
U
k
≤ξ
j
≤S
k+1
L
S
k+1
− L
ξ
j
|x
∗ j
− ΠX
S
k+1
|
≤ E
m
′
X
k=0
E
X
U
k
≤ξ
j
≤S
k+1
L
S
k+1
− L
ξ
j
|x
∗ j
− ΠX
S
k+1
| | F
S
k
≤ E
m
′
X
k=0
c
2
|X
S
k
− Y
S
k
|
2+ β
1
≤ E
m
′
X
k=0
c
3
|X
U
k
− Y
U
k
|ǫ
1+ β
1
≤ c
4
ǫ
1+ β
1
.
2228
For the notation used in the following lemma and its proof, see the beginning of this section.
Lemma 3.23. We have for any
β 1, some c and c
1
, assuming that |X
− Y | = ǫ and ǫ
∗
≥ c ǫ,
E I
m
∗
◦ · · · ◦ I Y
− X − J
m
′′
◦ · · · ◦ J Y
− X ≤ c
1
ǫ
1+ β
. Proof. We will follow closely the proof of Lemma 2.13 in [BL]. We will write
S
i
= S x
′′ i
= S x
∗ i
, π
i
= π
x
′′ i
= π
x
∗ i
. Recall that m
′′
= m
∗
. We have J
m
′′
◦ · · · ◦ J Y
− X − I
m
∗
◦ · · · ◦ I Y
− X =
e
∆ℓ
∗ m∗
S
m∗
− e
ℓ
∗ m∗+1
−ℓ
′′ m∗
S
m∗
π
m
∗
◦ J
m
′′
−1
◦ · · · ◦ J Y
− X +
m
∗
X
i=1
e
∆ℓ
∗ m∗
S
m∗
π
m
∗
· · · e
∆ℓ
∗ i+1
S
i+1
π
i+1
◦
e
ℓ
∗ i+1
−ℓ
′′ i
S
i
π
i
e
∆ℓ
′′ i
−1
S
i −1
− e
∆ℓ
∗ i
S
i
π
i
e
ℓ
∗ i
−ℓ
′′ i
−1
S
i −1
◦
3.127 π
i −1
e
∆ℓ
′′ i
−2
S
i −2
· · · e
∆ℓ
′′ 1
S
1
π
1
e
∆ℓ
′′
S
π Y
− X + I
m
∗
◦ · · · ◦ I
1
e
ℓ
∗ 1
−ℓ
′′
S
− e
∆ℓ
′′
S
π
Y − X
. By virtue of 2.3 and 2.4, the last term is bounded by a constant multiple of
|ℓ
∗ 1
− ℓ
′′ 1
| |Y − X
|. Since
ℓ
′′ 1
≥ ℓ
∗ 1
, E
|ℓ
∗ 1
− ℓ
′′ 1
| |Y − X
| = ǫEℓ
′′ 1
− ℓ
∗ 1
. By the strong Markov property applied at ξ
1
and Lemma 3.8 ii,
E ℓ
′′ 1
− ℓ
∗ 1
≤ c
2
ǫ. Hence
E
I
m
∗
◦ · · · ◦ I
1
e
ℓ
∗ 1
−ℓ
′′
S
− e
∆ℓ
′′
S
π
Y − X
≤ c
3
E |ℓ
∗ 1
− ℓ
′′ 1
| |Y − X
| ≤ c
4
ǫ
2
. 3.128
We have ℓ
′′ m
∗
+1
= ℓ
∗ m
∗
+1
= 1, so by 2.3 and 2.4, the first term on the right hand side of 3.127 is bounded by a constant multiple of
|ℓ
∗ m
∗
−ℓ
′′ m
∗
| |Y − X
|. We have ℓ
′′ m
∗
≥ ℓ
∗ m
∗
so E
|ℓ
∗ m
∗
−ℓ
′′ m
∗
| |Y − X
| ≤
ǫE1 − ℓ
∗ m
∗
. The following estimate can be proved just like 3.10. We have for every x ∈ ∂ D and b
0, c
5
b ≤ H
x
|e0 − eζ| ≥ b ≤ c
6
b. 3.129
This and the exit system formula 2.16 imply that 1 − ℓ
∗ 1
is stochastically majorized by an exponen- tial random variable with mean c
7
ǫ, so E1 − ℓ
∗ 1
≤ c
7
ǫ. Hence
E e
∆ℓ
∗ m∗
S
m∗
− e
ℓ
∗ m∗+1
−ℓ
′′ m∗
S
m∗
π
m
∗
◦ J
m
′′
−1
◦ · · · ◦ J Y
− X 3.130
≤ c
8
E |ℓ
∗ m
∗
− ℓ
′′ m
∗
| |Y − X
| ≤ c
9
ǫ
2
. The compositions before and after the parentheses in 3.127 in the summation are uniformly
bounded in operator norm by 2.3, so we need only estimate the sum
m
∗
X
i=0
e
ℓ
∗ i+1
−ℓ
′′ i
S
i
π
i
e
∆ℓ
′′ i
−1
S
i −1
− e
∆ℓ
∗ i
S
i
π
i
e
ℓ
∗ i
−ℓ
′′ i
−1
S
i −1
.
2229
Using the fact that π
i
commutes with S
i
, we can rewrite the i-th term in this sum as e
∆ℓ
∗ i
S
i
◦ π
i
◦
e
ℓ
∗ i
−ℓ
′′ i
S
i
− e
ℓ
∗ i
−ℓ
′′ i
S
i −1
e
∆ℓ
′′ i
−1
S
i −1
≤ e
∆ℓ
∗ i
S
i
e
ℓ
∗ i
−ℓ
′′ i
S
i
− e
ℓ
∗ i
−ℓ
′′ i
S
i −1
e
∆ℓ
′′ i
−1
S
i −1
. From 2.3 and 2.5, this last expression is bounded by c
10
ℓ
∗ i
− ℓ
′′ i
x
′′ i
− x
′′ i
−1
. By Lemma 3.22, for any
β 1,
E
m
∗
X
i=1
ℓ
∗ i
− ℓ
′′ i
x
′′ i
− x
′′ i
−1
≤ c
11
ǫ
1+ β
. This combined with 3.128 and 3.130 yields the lemma.
Once again, we ask the reader to consult the beginning of this section concerning notation used in the next lemma and its proof.
Lemma 3.24. Suppose that ǫ
∗
= c ǫ, where c
is as in Lemma 3.23. For some c
1
, if we assume that |X
− Y | = ǫ then,
E H
m
′
◦ · · · ◦ H Y
− X − J
m
′′
◦ · · · ◦ J Y
− X ≤ c
1
ǫ
4 3
| log ǫ|. Proof. Note that
H
k
= exp∆ℓ
′ k
S x
′ k
π
x
′ k
. Let
{ℓ
k
, x
k
}
≤k≤m+1
be the sequence containing all the distinct elements of the union of {ℓ
′ k
, x
′ k
}
≤k≤m
′
+1
and {ℓ
′′ k
, x
′′ k
}
≤k≤m
′′
+1
. We will explain how the sequence {ℓ
k
, x
k
}
≤k≤m+1
is ordered but first we note that ℓ
′ k
’s need not be distinct, and neither do ℓ
′′ k
’s, and, moreover, some ℓ
′ k
’s may be equal to some ℓ
′′ k
’s. We order the sequence {ℓ
k
, x
k
}
≤k≤m+1
in such a way that i
ℓ
k
≤ ℓ
k+1
for all k. ii If
ℓ
k
1
= ℓ
′ j
1
, ℓ
k
2
= ℓ
′ j
2
, ℓ
′ j
1
= L
S
j1
, ℓ
′ j
2
= L
S
j2
, and S
j
1
S
j
2
then k
1
k
2
. iii If
ℓ
k
1
= ℓ
′′ j
1
, ℓ
k
2
= ℓ
′′ j
2
, ℓ
′′ j
1
= λℓ
∗ j
3
, ℓ
′′ j
2
= λℓ
∗ j
4
, and ℓ
∗ j
3
ℓ
∗ j
4
then k
1
k
2
. iv If
ℓ
k
1
, x
k
1
= ℓ
′ j
1
, x
′ j
1
, ℓ
k
2
, x
k
2
= ℓ
′′ j
2
, x
′′ j
2
and ℓ
′ j
1
= ℓ
′′ j
2
then k
1
k
2
. It is easy to check that the above conditions define one and only one ordering of
{ℓ
k
, x
k
}
≤k≤m+1
. We introduce the following shorthand notations, ∆
i
= ℓ
i+1
− ℓ
i
, x
i
= γ
′
ℓ
i
, ex
i
= γ
′′
ℓ
i
, S
i
= S x
i
, f
S
i
= S ex
i
, π
i
= π
x
i
, e
π
i
= π
ex
i
. Observing that
π e
π = π
and e
π
m+1
J
m
′′
◦ · · · ◦ J Y
− X = J
m
′′
◦ · · · ◦ J Y
− X , we have,
H
m
′
◦ · · · ◦ H Y
− X − J
m
′′
◦ · · · ◦ J Y
− X =
m
X
i=0
e
∆
m
S
m
π
m
· · · e
∆
i+1
S
i+1
π
i+1
e
∆
i
S
i
π
i
− e
π
i+1
e
∆
i
f S
i
e π
i
· · · e
∆
1
f S
1
e π
1
e
∆ f
S
e π
Y − X
.
2230
By 2.3, the compositions of operators before and after the parentheses in the summation above are uniformly bounded in operator norm by a constant. Therefore,
|H
m
′
◦ · · · ◦ H Y
− X − J
m
′′
◦ · · · ◦ J Y
− X |
3.131 ≤ c
2 m
X
i=0
π
i+1
◦ e
∆
i
S
i
◦ π
i
− e
π
i+1
◦ e
∆
i
f S
i
◦ e
π
i
|Y − X
|. Using the fact that
S
i
and π
i
commute, as do f S
i
and e
π
i
, we obtain, π
i+1
◦ e
∆
i
S
i
◦ π
i
− e
π
i+1
◦ e
∆
i
f S
i
◦ e
π
i
3.132 =
π
i+1
◦ π
i
◦ e
∆
i
S
i
− e
∆
i
f S
i
◦ e
π
i
+ π
i+1
◦ π
i
− e
π
i+1
◦ e
π
i
◦ e
∆
i
f S
i
. We will deal with each of these terms separately.
For the first term, we have by 2.5, π
i+1
◦ π
i
◦ e
∆
i
S
i
− e
∆
i
f S
i
◦ e
π
i
≤ e
∆
i
S
i
− e
∆
i
f S
i
≤ c
3
∆
i
|x
i
− ex
i
|. 3.133
For the second term on the right hand side of 3.132, Lemma 2.2 and 2.3 allow us to conclude that
π
i+1
◦ π
i
− e
π
i+1
◦ e
π
i
◦ e
∆
i
f S
i
≤ c
4
x
i+1
− x
i
x
i
− ex
i
+ x
i+1
− ex
i+1
ex
i+1
− ex
i
e
∆
i
f S
i
≤ c
5
x
i+1
− x
i
x
i
− ex
i
+ x
i+1
− ex
i+1
ex
i+1
− ex
i
.
3.134 We will now analyze 3.133. Suppose that ∆
i
0 and x
i
6= ex
i
. Let j and k be defined by x
i
= γ
′
ℓ
′ j
and ex
i
= γ
′′
ℓ
′′ k
. Suppose that
ℓ
i
= ℓ
′ j
= ℓ
′′ k+1
. Then, by our ordering of ℓ
r
’s, ℓ
i+1
= ℓ
′′ k+1
= ℓ
i
, so ∆
i
= 0. For the same reason, we have ∆
i
= 0 if any of the following conditions holds: ℓ
′′ k
= ℓ
i
= ℓ
′ j
or ℓ
i
= ℓ
′′ k
= ℓ
′ j+1
. For this reason we consider only sharp versions of the corresponding inequalities in 3.135-3.138
below. We have assumed that x
i
6= ex
i
so one of the following four events holds, F
1 i
= {ℓ
′′ k
ℓ
i
= ℓ
′ j
ℓ
′′ k+1
, ξ
k
S
j
≤ t
′′ k+1
}, 3.135
F
2 i
= {ℓ
′′ k
ℓ
i
= ℓ
′ j
ℓ
′′ k+1
, t
′′ k+1
S
j
≤ ξ
k+1
}, 3.136
F
3 i
= {ℓ
′ j
ℓ
i
= ℓ
′′ k
ℓ
′ j+1
, S
j
ξ
k
≤ U
j
≤ S
j+1
}, 3.137
F
4 i
= {ℓ
′ j
ℓ
i
= ℓ
′′ k
ℓ
′ j+1
, S
j
U
j
≤ ξ
k
≤ S
j+1
}. 3.138
If F
1 i
holds then, {ξ
k
≤ S
j
≤ t
′′ k+1
} ∩ {|x
i
− ex
i
| a} ⊂ [
1 ≤r≤m
sup
ξ
r
tt
′′ r+1
|x
′′ r
− X
t
| a .
3.139
2231
This and Lemma 3.6 yield,
E
m
X
i=0
∆
i
|x
i
− ex
i
|1
F
1 i
≤ E max
≤k≤m
∗
sup
ξ
k
tt
∗ k+1
|x
∗ k
− X
t
|
m
X
i=0
∆
i
3.140
= E max
≤k≤m
∗
sup
ξ
k
tt
∗ k+1
|x
∗ k
− X
t
| ≤ c
6
ǫ
1 3
= c
7
ǫ
1 3
. If F
2 i
holds then ∆
i
= 0, because X does not hit ∂ D in the interval t
′′ k+1
, ξ
k+1
, and, therefore, the local time L
t
does not increase on this time interval. Hence,
m
X
i=0
∆
i
|x
i
− ex
i
|1
F
2 i
= 0. 3.141
If F
3 i
holds, the definition of U
j
implies that |x
i
− ex
i
| ≤ c
8
ǫ. Thus
m
X
i=0
∆
i
|x
i
− ex
i
|1
F
3 i
≤
m
X
i=0
c
8
∆
i
ǫ = c
8
ǫ. 3.142
Suppose that F
4 i
occurred. It follows from the condition U
j
≤ ξ
k
≤ S
j+1
and the definition of ℓ
′′ k
that ℓ
′′ k
= ℓ
′ j+1
. We have already shown that in this case, ∆
i
= 0. Hence,
m
X
i=0
∆
i
|x
i
− ex
i
|1
F
4 i
= 0. 3.143
Next we will consider the right hand side of 3.134. We start our discussion with the terms of the form
x
i+1
− x
i
x
i
− ex
i
. Recall that we have defined j and k by x
i
= γ
′
ℓ
′ j
and ex
i
= γ
′′
ℓ
′′ k
. We will consider all possibilities listed in 3.135-3.138. If ∆
i
= 0 then ℓ
i
= ℓ
i+1
and x
i
= γ
′
ℓ
i
= γ
′
ℓ
i+1
= x
i+1
. It follows that in this case, x
i+1
− x
i
x
i
− ex
i
= 0. Hence, we can limit ourselves to 3.135-3.138, with sharp inequalities in the definitions.
Suppose that F
1 i
∪ F
2 i
occurred. Then ξ
k
S
j
, x
i
= X
S
j
and ex
i
= X
ξ
k
. By Lemma 3.8 iii and the strong Markov property applied at
ξ
k
, E
x
i
− ex
i
1
F
1 i
∪F
2 i
| F
ξ
k
= E
X
S
j
− X
ξ
k
1
F
1 i
∪F
2 i
| F
ξ
k
3.144
≤ c
9
| log dY
ξ
k
, D |dY
ξ
k
, D + ǫ
3
≤ c
10
ǫ| log ǫ|. We have x
i+1
= X
t
for some t ∈ S
j
, S
j+1
]. By Lemma 3.5 ii, the strong Markov property applied at the stopping time R
1
= inf{t ≥ S
j
: X
t
∈ ∂ D} and Lemma 3.8 iii,
E
x
i+1
− x
i
1
F
1 i
∪F
2 i
| F
S
j
≤ E sup
S
j
≤t≤S
j+1
X
t
− X
S
j
1
F
1 i
∪F
2 i
| F
S
j
3.145
≤ E
sup
S
j
≤t≤R
1
X
t
− X
S
j
1
F
1 i
∪F
2 i
| F
S
j
+ E
sup
R
1
≤t≤S
j+1
X
t
− X
R
1
1
F
1 i
∪F
2 i
| F
S
j
≤ c
11
ǫ| log ǫ|. 2232
It follows from this and 3.144 that
E
x
i+1
− x
i
x
i
− ex
i
1
F
1 i
∪F
2 i
| F
ξ
k
3.146
= E
x
i
− ex
i
E x
i+1
− x
i
1
F
1 i
∪F
2 i
| F
S
j
| F
ξ
k
≤ c
12
ǫ
2
| log ǫ|
2
. By 3.129 and the exit system formula 2.16, the expected value of m
∗
is bounded by c
13
ǫ. It follows from this estimate and 3.146 that
E
m
X
k=0
x
i+1
− x
i
x
i
− ex
i
1
F
1 i
∪F
2 i
≤ E
m
∗
X
k=1
E
x
i+1
− x
i
x
i
− ex
i
1
F
1 i
∪F
2 i
| F
ξ
k
≤ c
14
ǫ| log ǫ|
2
. 3.147
Next suppose that F
3 i
occurred. Then x
i
= X
S
j
and ex
i
= X
ξ
k
. Since ξ
k
≤ U
j
, we have x
i
− ex
i
≤ c
15
ǫ. As in the previous case, we have x
i+1
= X
t
for some t ∈ S
j
, S
j+1
], so we can use estimate 3.145. It follows that
E
x
i+1
− x
i
x
i
− ex
i
1
F
3 i
| F
ξ
k
≤ c
16
ǫ
2
| log ǫ|. The following estimate is analogous to 3.147,
E
m
X
k=0
x
i+1
− x
i
x
i
− ex
i
1
F
3 i
≤ E
m
∗
X
k=1
E
x
i+1
− x
i
x
i
− ex
i
1
F
3 i
| F
ξ
k
≤ c
17
ǫ| log ǫ|. 3.148
We have already shown that if F
4 i
holds then ∆
i
= 0 and, therefore, x
i+1
− x
i
x
i
− ex
i
= 0. Hence
E
m
X
k=0
x
i+1
− x
i
x
i
− ex
i
1
F
4 i
= 0. 3.149
We continue our discussion of the right hand side of 3.134. We now consider the terms of the form x
i+1
− ex
i+1
ex
i+1
− ex
i
. The overall structure of our argument is similar to that used to analyze the terms of the form
x
i+1
− x
i
x
i
− ex
i
. Suppose that x
i+1
6= ex
i+1
. Let j and k be defined by x
i+1
= γ
′
ℓ
′ j
and ex
i+1
= γ
′′
ℓ
′′ k
. We have assumed that x
i+1
6= ex
i+1
so one of the following four events holds, F
5 i
= {ℓ
′′ k
ℓ
i+1
= ℓ
′ j
ℓ
′′ k+1
, ξ
k
S
j
≤ t
′′ k+1
}, 3.150
F
6 i
= {ℓ
′′ k
ℓ
i+1
= ℓ
′ j
ℓ
′′ k+1
, t
′′ k+1
S
j
≤ ξ
k+1
}, 3.151
F
7 i
= {ℓ
′ j
ℓ
i+1
= ℓ
′′ k
ℓ
′ j+1
, S
j
ξ
k
≤ U
j
≤ S
j+1
}, 3.152
F
8 i
= {ℓ
′ j
ℓ
i+1
= ℓ
′′ k
ℓ
′ j+1
, S
j
U
j
≤ ξ
k
≤ S
j+1
}. 3.153
Suppose that ℓ
i+1
= ℓ
′ j
= ℓ
′′ k
. Then because of the way we ordered ℓ
i
, x
i
, we have ℓ
i
, x
i
= ℓ
′ j
, x
′ j
and ℓ
i+1
, x
i+1
= ℓ
′′ k
, x
′′ k
. Therefore ℓ
i
= ℓ
i+1
. It follows that ex
i
= γ
′′
ℓ
i
= γ
′′
ℓ
i+1
= ex
i+1
. In this case,
x
i+1
− ex
i+1
ex
i+1
− ex
i
= 0. We can reach the same conclusion in the same way in case 2233
we have ℓ
′′ k+1
= ℓ
i+1
= ℓ
′ j
or ℓ
i+1
= ℓ
′′ k
= ℓ
′ j+1
. Hence, we can limit ourselves to 3.150-3.153, with sharp inequalities in the definitions.
Suppose that F
5 i
∪ F
6 i
occurred. Then x
i+1
= X
S
j
and ex
i+1
= X
ξ
k
. The following is a version of 3.144,
E
x
i+1
− ex
i+1
1
F
5 i
∪F
6 i
| F
ξ
k
≤ c
18
ǫ| log ǫ|. 3.154
We have ex
i
= X
t
for some t ∈ [ξ
k −1
, ξ
k
, so by Lemma 3.7 and the strong Markov property applied at
ξ
k −1
, E
ex
i+1
− ex
i
1
F
5 i
∪F
6 i
| F
ξ
k −1
≤ E
sup
ξ
k −1
≤t≤ξ
k
X
t
− X
ξ
k −1
| F
ξ
k −1
≤ c
19
ǫ
1 3
∗
= c
19
c
1 3
ǫ
1 3
. 3.155
It follows from this and 3.154 that
E
x
i+1
− ex
i+1
ex
i+1
− ex
i
1
F
5 i
∪F
6 i
| F
ξ
k −1
3.156
= E
ex
i+1
− ex
i
E x
i+1
− ex
i+1
1
F
5 i
∪F
6 i
| F
ξ
k
| F
ξ
k −1
≤ c
20
ǫ
4 3
| log ǫ|. Recall that the expected value of m
∗
is bounded by c
13
ǫ. It follows from this and 3.156 that
E
m
X
k=0
x
i+1
− ex
i+1
ex
i+1
− ex
i
1
F
5 i
∪F
6 i
≤ E
m
∗
X
k=1
E
x
i+1
− ex
i+1
ex
i+1
− ex
i
1
F
5 i
∪F
6 i
| F
ξ
k −1
≤ c
21
ǫ
1 3
| log ǫ|. 3.157
Next suppose that F
7 i
occurred. Then x
i+1
= X
S
j
and ex
i+1
= X
ξ
k
. Since ξ
k
≤ U
j
, we have x
i+1
− ex
i+1
≤ c
22
ǫ. As in the previous case, we have ex
i
= X
t
for some t ∈ [ξ
k −1
, ξ
k
], so we can use estimate 3.155. It follows that
E
x
i+1
− ex
i+1
ex
i+1
− ex
i
1
F
7 i
| F
ξ
k −1
≤ c
23
ǫ
4 3
. The following estimate is analogous to 3.157
E
m
X
k=0
x
i+1
− ex
i+1
ex
i+1
− ex
i
1
F
7 i
≤ E
m
∗
X
k=1
E
x
i+1
− ex
i+1
ex
i+1
− ex
i
1
F
7 i
| F
ξ
k −1
≤ c
24
ǫ
1 3
. 3.158
Suppose that F
8 i
occurred. It follows from the condition U
j
≤ ξ
k
≤ S
j+1
and the definition of ℓ
′′ k
that ℓ
′′ k
= ℓ
′ j+1
. We have already argued that in this case, x
i+1
− ex
i+1
ex
i+1
− ex
i
= 0. Hence,
m
X
k=0
x
i+1
− ex
i+1
ex
i+1
− ex
i
1
F
8 i
= 0. 3.159
2234
Recall that |X
− Y | = ǫ. The estimates in 3.140, 3.141, 3.142, 3.143, 3.147, 3.148,
3.149, 3.157, 3.158 and 3.159 are all less than or equal to c
25
ǫ
1 3
| log ǫ|. We combine these remarks with 3.131-3.134 to conclude that,
E |H
m
′
◦ · · · ◦ H Y
− X − J
m
′′
◦ · · · ◦ J Y
− X | ≤ c
26
ǫ
4 3
| log ǫ|.
Proof of Theorem 3.1. Suppose that |Y
− X | = ǫ and ǫ
∗
= c ǫ, where c
is as in Lemma 3.23. Consider an arbitrarily small c
1
0 let Λ be the random variable in the statement of Lemma 3.20. According to that lemma, for all sufficiently small
ǫ 0, we have a.s., |Λ| c
1
ǫ. 3.160
By the triangle inequality, |Y
σ
∗
− X
σ
∗
− I
m
∗
◦ · · · ◦ I Y
− X |
3.161 ≤ |Λ| +
|Y
σ
1
− X
σ
1
− G
m
′
◦ · · · ◦ G Y
− X | − Λ
+ |G
m
′
◦ · · · ◦ G Y
− X − H
m
′
◦ · · · ◦ H Y
− X |
+ |H
m
′
◦ · · · ◦ H Y
− X − J
m
′′
◦ · · · ◦ J Y
− X |
+ J
m
′′
◦ · · · ◦ J Y
− X − I
m
∗
◦ · · · ◦ I Y
− X :=
|Λ| + Ξ. By Lemma 3.20,
E Y
σ
∗
− X
σ
∗
− G
m
′
◦ · · · ◦ G Y
− X − Λ
≤ c
2
ǫ
2
| log ǫ|
2
. 3.162
By Lemma 3.21,
E |G
m
′
◦ · · · ◦ G Y
− X − H
m
′
◦ · · · ◦ H Y
− X | ≤ c
3
ǫ
2
| log ǫ|. 3.163
Lemma 3.24 implies that
E H
m
′
◦ · · · ◦ H Y
− X − J
m
′′
◦ · · · ◦ J Y
− X ≤ c
4
ǫ
4 3
| log ǫ|. 3.164
Lemma 3.23 yields for any β 1,
E J
m
′′
◦ · · · ◦ J Y
− X − I
m
∗
◦ · · · ◦ I Y
− X ≤ c
5
ǫ
1+ β
. 3.165
Combining 3.162-3.165, and using the definition of Ξ in 3.161, we see that
EΞ ≤ c
6
ǫ
4 3
| log ǫ|. 3.166
Fix some β
1
∈ 1, 43 and β
2
∈ 0, 43 − β
1
. By 3.166 and Chebyshev’s inequality,
PΞ c
7
ǫ
β
1
≤ c
8
ǫ
β
2
. 3.167
2235
Fix an arbitrary b 1 and v ∈ R
n
with |v| = 1. We apply the last estimate to a sequence of processes
Y = X
z
+ǫv
with ǫ = b
−k
, k ≥ k
, for some fixed large k . We obtain
PΞ
c
7
b
−kβ
1
≤ c
8
b
−kβ
2
, k
≥ k .
Since P
k ≥k
c
8
b
−kβ
2
∞, the Borel-Cantelli Lemma shows that only a finite number of events {Ξ c
7
b
−kβ
1
} occur. This is the same as saying that only a finite number of events {Ξb
−k
c
7
b
−kβ
1
−1
} occur. We combine this fact with 3.160 and 3.161 to see that for any c
1
0, a.s., lim sup
k →∞
X
z +b
−k
v
σ
∗
− X
σ
∗
b
−k
− I
m
∗
◦ · · · ◦ I v
≤ c
1
. Since c
1
is arbitrarily small, we have in fact, a.s., lim
k →∞
X
z +b
−k
v
σ
∗
− X
σ
∗
b
−k
− I
m
∗
◦ · · · ◦ I v
= 0. 3.168
It is easy to see that the last formula holds for all v
∈ R
n
, not only those with |v| = 1.
Consider an arbitrary compact set K ⊂ R
n
. Let c
9
be the same constant as c
1
in the statement of Lemma 3.4. It follows easily from 2.3 that
kI
m
∗
◦ · · · ◦ I k ≤ c
10
, a.s. Fix any c
11
0 and find
w
1
, . . . , w
j
1
∈ R
n
such that for every v
∈ K there exists j = jv such that |v − w
j
| c
11
2c
9
+ c
10
. Note that
|z + b
−k
v − z
+ b
−k
w
j v
| b
−k
c
11
2c
9
and, in view of 3.168, lim
k →∞
sup
1 ≤ j≤ j
1
X
z +b
−k
w
j
σ
∗
− X
σ
∗
b
−k
− I
m
∗
◦ · · · ◦ I w
j
= 0. 3.169
By Lemma 3.4, for v
∈ K and j = jv, a.s.,
X
z +b
−k
w
j
σ
∗
− X
σ
∗
b
−k
− X
z +b
−k
v
σ
∗
− X
σ
∗
b
−k
≤ c
9
|z + b
−k
v − z
+ b
−k
w
j
|b
−k
≤ c
11
2. 3.170
Since |v − w
j
| c
11
2c
10
, |I
m
∗
◦ · · · ◦ I w
j v
− I
m
∗
◦ · · · ◦ I v| ≤ c
11
2. 3.171
Combining 3.169-3.171 yields a.s., lim
k →∞
sup
v ∈K
X
z +b
−k
v
σ
∗
− X
σ
∗
b
−k
− I
m
∗
◦ · · · ◦ I v
≤ c
11
. Since c
11
0 is arbitrarily small, we have a.s., lim
k →∞
sup
v
∈K
X
z +b
−k
v
σ
∗
− X
σ
∗
b
−k
− I
m
∗
◦ · · · ◦ I v
= 0. 3.172
2236
Let c
12
= sup{|v| ∈ K}. For ǫ ∈ [b
−k
, b
−k+1
, we have, |z
+ b
−k
v
− z + ǫv|ǫ ≤ c
12
1 − 1b. Hence, by Lemma 3.4, a.s.,
X
z +b
−k
v
σ
∗
− X
σ
∗
b
−k
− X
z +ǫv
σ
∗
− X
σ
∗
ǫ ≤
X
z +b
−k
v
σ
∗
− X
σ
∗
b
−k
− X
z +b
−k
v
σ
∗
− X
σ
∗
ǫ +
X
z +b
−k
v
σ
∗
− X
σ
∗
ǫ −
X
z +ǫv
σ
∗
− X
σ
∗
ǫ ≤ 1 − 1b
X
z +b
−k
v
σ
∗
− X
σ
∗
b
−k
+ c
9
|z + ǫv − z
+ b
−k
v |ǫ
≤ 1 − 1b X
z +b
−k
v
σ
∗
− X
σ
∗
b
−k
+ c
9
c
12
1 − 1b. Let
ǫ
∗
= c b
−k
, where k is defined by ǫ ∈ [b
−k
, b
−k+1
. The last formula and 3.172 yield, lim
ǫ→0
sup
v ∈K
X
z +ǫv
σ
∗
− X
σ
∗
ǫ − I
m
∗
◦ · · · ◦ I v
≤ 1 − 1b lim sup
k →∞
sup
v ∈K
X
z +b
−k
v
σ
∗
− X
σ
∗
b
−k
+ c
9
c
12
1 − 1b. Let
ǫ
∗
= c ǫ. We can take b 1 arbitrarily close to 1, so, a.s.,
lim
ǫ→0
sup
v ∈K
X
z
+ǫv
σ
∗
− X
σ
∗
ǫ − I
m
∗
◦ · · · ◦ I v
= 0. Recall the definition of
σ
∗
from the beginning of this section. We let k
∗
→ ∞ to see that, a.s., lim
ǫ→0
sup
v ∈K
X
z +ǫv
σ
1
− X
σ
1
ǫ − I
m
∗
◦ · · · ◦ I v
= 0. We combine this with Theorem 2.5 to complete the proof of the theorem.
Proof of Corollary 3.2. According to Theorem 3.1, for every r 0 and compact set K ⊂ R
n
, we have lim
ǫ→0
sup
v ∈K
X
z +ǫv
σ
r
− X
z σ
r
ǫ − A
r
v = 0, a.s. By Fubini’s Theorem, with probability 1, for
almost all r 0, we have
lim
ǫ→0
sup
v ∈K
X
z +ǫv
σ
r
− X
z σ
r
ǫ − A
r
v = 0.
3.173
2237
Recall the definitions of E
r
, v
r
and A
r
from Section 2.3. Let E
r
1
,r
2
= {e
s
: σ
r
1
≤ s σ
r
2
} and define
v
r
1
,r
2
and A
r
1
,r
2
relative to E
r
1
,r
2
in the same way as A
r
was defined relative to E
r
and v
r
. Fix any t
0 and let y = X
z t
and Y
x s
= X
x t+s
for s ≥ 0. Let w be defined relative to v by
y
+ ǫw = X
z +ǫv
t
. By the Markov property applied at time t, Theorem 3.1 and 3.173 hold for the flow
{Y
· s
, s ≥ 0} in place of the flow {X
· s
, s ≥ 0}. In other words, if A
′ r
= A
L
t
,L
t
+r
and σ
′ r
= σ
L
t
+r
− t then with probability 1, for almost all r
0, we have lim
ǫ→0
sup
v ∈K
Y
y
+ǫw
σ
′ r
− Y
y σ
′ r
ǫ − A
′ r
w = 0.
3.174 Note that for any sequence
v
n
∈ K, n ≥ 1, there exists v
∗
∈ K such that a subsequence of v
n
converges to
v
∗
, by compactness of K. Suppose that it is not true that lim
ǫ→0
sup
v ∈K
X
z
+ǫv
t
− X
z t
ǫ − A
L
t
v = 0 with probability
1. It will suffice to show that this assumption leads to a contradiction. The assumption im- plies that we can find c
1
, p
1
0, ǫ
n
0 for n ≥ 1, v
∗
∈ K, and v
n
∈ K for n ≥ 1, such that lim
n →∞
ǫ
n
= 0, lim
n →∞
v
n
= v
∗
, and with probability greater than p
1
we have lim inf
n
X
z +ǫ
n
v
n
t
− X
z t
ǫ
n
− A
L
t
v
n
c
1
. Let A
1
be the event defined by the last formula. Let
π
H
denote the orthogonal projection on an n − 1-dimensional hyperplane H.
We can choose H so that for some c
2
0 and subsequence n
k
, we have on the event A
1
, π
H
X
z +ǫ
nk
v
nk
t
− X
z t
ǫ
n
k
− π
H
◦ A
L
t
v
n
k
c
2
. Let w
k
= X
z +ǫ
nk
v
nk
t
− X
z t
ǫ
n
k
so that the last formula can be written as
π
H
w
k
− π
H
◦ A
L
t
v
n
k
c
2
. Since D is a bounded domain with C
2
boundary, there exists x ∈ ∂ D such that the tangent hyperplane at x is parallel to H, so we can
assume that π
H
= π
x
. There exist r
1
0 and c
3
∈ 0, c
2
such that for y ∈ M := ∂ D ∩ Bx, r
1
, we have on the event A
1
, π
y
w
k
− π
y
◦ A
L
t
v
n
k
c
3
. Let T = inf
{s t : X
z s
∈ ∂ D} and A
2
= A
1
∩ {X
z T
∈ M}. By the support theorem for Brownian motion and the Markov property at time t, there exists p
2
0 such that PA
2
p
2
. If A
2
holds then π
X
z0 T
w
k
− π
X
z0 T
◦ A
L
t
v
n
k
c
3
. 3.175
It follows from the definition of A
r
and A
′ r
that lim
r ↓0
kA
′ r
− π
X
z0 T
k = 0 and lim
r ↓0
kA
L
t
+r
− π
X
z0 T
◦ A
L
t
k = 0. The rate of convergence to 0 may depend on the trajectory of the flow X . Let r
2
0 and p
3
0 be so small that with probability greater than p
3
the event A
2
holds and |A
′ r
w
k
− π
X
z0 T
w
k
| ≤ c
3
4, 3.176
|π
X
z0 T
◦ A
L
t
v
n
k
− A
L
t
+r
v
n
k
| ≤ c
3
4, for r
∈ 0, r
2
and k ≥ 1. Let A
3
be the event that the last inequalities in 3.176 hold and A
2
holds. Combining 3.175 and 3.176, we see that on the event A
3
we have |A
′ r
w
k
− A
L
t
+r
v
n
k
| ≥ c
3
2 for r
∈ 0, r
2
and k ≥ 1. By 3.173 and 3.174, with probability 1, there exist some r
∈ 0, r
2
such that, lim
k →∞
X
z +ǫ
nk
v
nk
σ
Lt +r
− X
z σ
Lt +r
ǫ
n
k
− A
L
t
+r
v
n
k
= 0
2238
and lim
k →∞
X
z +ǫ
nk
v
nk
σ
Lt +r
− X
z σ
Lt +r
ǫ
n
k
− A
′ r
w
n
k
= lim
k →∞
Y
y +ǫ
nk
w
nk
σ
′ r
− Y
y σ
′ r
ǫ
n
k
− A
′ r
w
n
k
= 0. Since
|A
′ r
w
k
− A
L
t
+r
v
n
k
| ≥ c
3
2 on the event A
3
of positive probability, the last two formulas form a contradiction and this completes the proof.
References
[A] H. Airault, Perturbations singulières et solutions stochastiques de problèmes de D. Neumann- Spencer. J. Math. Pures Appl.
55, 1976, no. 3, 233–267. MR0501184
