Finally, in order to define the moving frame, let O := Id and
O
m
x := 1
kxk
−x
1
−x
2
−x
2
x
1
:= O
−1 m
x, m = 1, . . . , 4.
Hence, dO
t
· O
−1 t
= 0 on [τ
ℓ
, τ
ℓ+1
if m
ℓ
= 0, and otherwise we use Itô’s formula to obtain dO
t
·O
−1 t
= X
2 t
x kX
t
xk
3
1
−1 0
d w
1 t
+ X
1 t
x kX
t
xk
3
−1
1
d w
2 t
+
X
2 t
x kX
t
xk
3
1
−1 0
b
1
X
t
x + X
1 t
x kX
t
xk
3
−1
1
b
2
X
t
x + 1
2 kX
t
xk
4
1 0
0 1
d t, from which the coefficient functions
α
1
, α
2
and β can be defined accordingly. Note that in any case
γ = 0. Furthermore, since nx = −x for all x ∈ ∂ G, Dnx = − Id and therefore Φ
2 t
= −Q. For simplicity we restrict ourselves now to the case b = 0. Then, the system in Theorem 2.5 can be
rewritten as follows: For t ∈ [τ
ℓ
, τ
ℓ+1
, writing Y
t
= y
1 t
, y
2 t
, we get in the case m
ℓ
= 0 that y
1 t
= 1l
{tinf C
ℓ
}
y
1 τ
ℓ
, y
2 t
= y
2 τ
ℓ
− Z
t τ
ℓ
y
2 s
d l
s
x, and in the case m
ℓ
6= 0 that y
1 t
=1l
{tinf C
ℓ
}
y
1 τ
ℓ
+ Z
t τ
ℓ
X
2 s
x kX
s
xk
3
y
2 s
d w
1 s
− Z
t τ
ℓ
X
1 s
x kX
s
xk
3
y
2 s
d w
2 s
+ Z
t τ
ℓ
1 2
kX
s
xk
4
y
1 s
ds + 1l
{t≥inf C
ℓ
}
Z
t rt
X
2 s
x kX
s
xk
3
y
2 s
d w
1 s
− Z
t rt
X
1 s
x kX
s
xk
3
y
2 s
d w
2 s
+ Z
t rt
1 2
kX
s
xk
3
y
1 s
ds y
2 t
= y
2 τ
ℓ
− Z
t τ
ℓ
X
2 s
x kX
s
xk
3
y
1 s
d w
1 s
+ Z
t τ
ℓ
X
1 s
x kX
s
xk
3
y
1 s
d w
2 s
+ Z
t τ
ℓ
1 2
kX
s
xk
4
y
2 s
ds −
Z
t τ
ℓ
y
2 s
d l
s
x, with initial value Y
τ
ℓ
as specified in Theorem 2.5.
3 Proof of the Main Result
3.1 Lipschitz Continuity w.r.t. the Initial Datum
Before adressing the question of differentiability we establish pathwise continuity properties of x 7→
X
t
x
t
w.r.t. the sup-norm topology. For this we will need that the mapping y 7→ l
t
y is bounded.
Lemma 3.1. For every t 0 we have sup
x ∈G
l
t
x ∞ a.s. Proof. See Lemma 3.3 in [4].
Proposition 3.2. Let T 0 be arbitrary and let X
t
x and X
t
y, t ≥ 0, be two solutions of 2.1 for any x, y
∈ G. Then, there exists a positive constant c only depending on T such that sup
t ∈[0,T ]
kX
t
x − X
t
yk ≤ kx − yk expcT + l
T
x + l
T
y for all x, y
∈ G.
854
Note that the Lipschitz continuity in the initial condition, which is stated here, becomes effective since the Lipschitz constant can be controlled due to the uniform boundedness of l
T
x in x estab- lished in Lemma 3.1
Proof. The case x = y is clear and it suffices to consider the case T inf{t : X
t
x = X
t
y}. We shall proceed similarly to Lemma 3.8 in [6]. Since
∂ G is C
2
-smooth and G is connected and compact, there exists a positive constant c
1
∞ such that for all x ∈ ∂ G and all y ∈ G, 〈x − y, nx〉 ≤ c
1
kx − yk
2
. 3.1
Let T := 0 and for k
≥ 1, T
k
:= inf ¦
t ≥ T
k −1
: kX
t
x − X
t
yk 6∈
1 2
kX
T
k −1
x − X
T
k −1
yk, 2kX
T
k −1
x − X
T
k −1
yk ©
∧ T. Then, by Itô’s formula we obtain for any k
≥ 1 and t ∈ T
k −1
, T
k
], kX
t
x − X
t
yk − kX
T
k −1
x − X
T
k −1
yk =
Z
t T
k −1
X
r
x − X
r
y, bX
r
x − bX
r
y kX
r
x − X
r
yk d r
+ Z
t T
k −1
X
r
x − X
r
y, nX
r
x kX
r
x − X
r
yk d l
r
x + Z
t T
k −1
X
r
y − X
r
x, nX
r
y kX
r
x − X
r
yk d l
r
y ≤c
2
Z
t T
k −1
kX
r
x − X
r
yk d r + c
1
Z
t T
k −1
kX
r
x − X
r
yk dl
r
x + d l
r
y ≤c
3
kX
T
k −1
x − X
T
k −1
yk Z
T
k
T
k −1
d r + d l
r
x + d l
r
y, where we have used 3.1 and the Lipschitz continuity of b. Hence, for any t
∈ T
k −1
, T
k
], kX
t
x − X
t
yk kX
T
k −1
x − X
T
k −1
yk ≤ 1 + c
3
T
k
− T
k −1
+ l
T
k
x − l
T
k −1
x + l
T
k
y − l
T
k −1
y
≤ exp
c
3
T
k
− T
k −1
+ l
T
k
x − l
T
k −1
x + l
T
k
y − l
T
k −1
y
, and
kX
t
x − X
t
yk kx − yk
= kX
t
x − X
t
yk kX
T
k −1
x − X
T
k −1
yk
k −1
Y
j=1
kX
T
j
x − X
T
j
yk kX
T
j −1
x − X
T
j −1
yk ≤
k
Y
j=1
exp c
3
T
j
− T
j −1
+ l
T
j
x − l
T
j −1
x + l
T
j
y − l
T
j −1
y ≤ exp
c
3
T
k
+ l
T
k
x + l
T
k
y
≤ exp c
3
T + l
T
x + l
T
y ,
which proves the proposition. 855
Remark 3.3. By Proposition 3.2 there exists for every T 0 a random ∆
T
0 such that sup
t ∈[0,T ]
kX
t
x − X
t
yk δ
2 ,
∀ y ∈ B
∆
T
x ∩ G, with
δ as in Section 2.3. Then, by the definition of
τ
ℓ
we have for such y and for every ℓ that
X
t
y ∈ U
m
ℓ
for all t ∈ [τ
ℓ
, τ
ℓ+1
.
Lemma 3.4. For every t ∈ [0, T ] we have that for all x ∈ G the mapping y 7→ l
t
y is continuous at x.
Proof. Fix T 0 and x ∈ G and set
λ
t
y := Z
t
nX
r
y d l
r
y, y
∈ G, t
∈ [0, T ], which defines for each y
∈ G a process of bounded variation on [0, T ]. Then, we get immediately by Proposition 3.2, 2.1 and the Lipschitz property of b that
λ y converges uniformly on [0, T ] to λx as y tends to x.
Let now t ∈ [0, T ] and ℓ be such that t ∈ [τ
ℓ
, τ
ℓ+1
. Then, for all y ∈ B
∆
T
x ∩ G with ∆
T
as in Remark 3.3 we have that X
s
y, X
s
x ∈ U
m
ℓ
for all s ∈ [τ
ℓ
, t. For such y and s we get d l
s
y − dl
s
x = 〈∇u
1 m
ℓ
X
s
y, nX
s
y〉 dl
s
y − 〈∇u
1 m
ℓ
X
s
x, nX
s
x〉 dl
s
x = ∇u
1 m
ℓ
X
s
y dλ
s
y − ∇u
1 m
ℓ
X
s
x dλ
s
x = σ
1 m
ℓ
X
s
y dλ
s
y − σ
1 m
ℓ
X
s
x dλ
s
x. Hence,
l
t
y − l
t
x =l
τ
ℓ
y − l
τ
ℓ
x + Z
t τ
ℓ
σ
1 m
ℓ
X
s
x dλ
s
y − dλ
s
x +
Z
t τ
ℓ
σ
1 m
ℓ
X
s
y − σ
1 m
ℓ
X
s
x d
λ
s
y. Using Proposition 3.2 and the fact that the functions
σ
m
are uniformly Lipschitz the last term con- verges to zero as y tends to x. Recall that
λ y converges uniformly on [τ
ℓ
, t] to λx as y tends
to x. Hence, we have that the associated signed measures d λ y on [τ
ℓ
, t] converge weakly to d
λx as y tends to x. Since s 7→ σ
1 m
ℓ
X
s
x is bounded and continuous on [τ
ℓ
, t], the second term converges to zero as y tends to x. We apply the same argument for l
τ
ℓ
y − l
τ
ℓ
x on [τ
ℓ−1
, τ
ℓ
] and by iterating this procedure we obtain the claim.
We fix now an arbitrary x ∈ G, v ∈ R
d
and T 0. Then, we set x
ǫ
:= x + ǫ v for all ǫ ∈ [a
x
, b
x
] with a
x
≤ 0 and b
x
≥ 0 such that x
ǫ
∈ G for all ǫ ∈ [a
x
, b
x
]. Furthermore, we define for such ǫ and t
∈ [0, T ],
M
x, ℓ
t
ǫ :=
if t ∈ [0, τ
ℓ
, R
t τ
ℓ
b
1 m
ℓ
X
r
x
ǫ
d r + R
t τ
ℓ
σ
1 m
ℓ
X
r
x
ǫ
d w
r
if t ∈ [τ
ℓ
, τ
ℓ+1
], M
x, ℓ
τ
ℓ+1
ǫ if t
τ
ℓ+1
. 856
The index x is there to indicate that the stopping times τ
ℓ
are the same as in the definition of M
x, ℓ
that are depending on x and not on ǫ. In particular, M
x, ℓ
ǫ is a well-defined object, since the coefficient functions b
1 m
and σ
1 m
have been extended to the whole domain G. Note that M
x, ℓ
t
= M
x, ℓ
t
0, t ∈ [0, T ]. Finally, we set ∆M
x, ℓ
t
ǫ, ǫ
′
:= M
x, ℓ
t
ǫ − M
x, ℓ
t
ǫ
′
, t
∈ [0, T ], ǫ, ǫ
′
∈ [a
x
, b
x
], so that
∆M
x, ℓ
t
ǫ, 0 = Z
t τ
ℓ
b
1 m
ℓ
X
r
x
ǫ
− b
1 m
ℓ
X
r
x d r +
Z
t τ
ℓ
σ
1 m
ℓ
X
r
x
ǫ
− σ
1 m
ℓ
X
r
x d w
r
, for t
∈ [τ
ℓ
, τ
ℓ+1
and ǫ ∈ [a
x
, b
x
]. In the next lemma we show that M
x, ℓ
t
ǫ is pathwise jointly continuous in t and
ǫ.
Lemma 3.5. Let ∆
T
be as in Remark 3.3. Then, for a.e. ω ∈ Ω the following holds. For every δ
1
∈ 0, 1 and
δ
2
∈ 0,
1 2
there exists a random constant K = Kω, δ
1
, δ
2
, T such that ∆ M
x, ℓ
t
ǫ, ǫ
′
− ∆M
x, ℓ
s
ǫ, ǫ
′
≤ K |ǫ − ǫ
′
|
1 −δ
1
|t − s|
1 2
−δ
2
, ∀s, t ∈ [0, T ],
3.2 for all
ǫ, ǫ
′
such that x
ǫ
, x
ǫ
′
∈ B
∆
T
x ∩ G. In particular, for every δ ∈ 0, 1 we have for all such ǫ M
x, ℓ
t
ǫ − M
x, ℓ
t
≤ K |ǫ|
1 −δ
, ∀t ∈ [0, T ],
for some random constant K = K ω, δ, T .
Proof. In a first step we use Kolmogorov’s continuity theorem to show the existence of a modification of M
ℓ,x
ǫ
t, ǫ
satisfying the above estimate and in a second step we show the claim by a continuity argument.
Step 1: It follows directly from Proposition 3.2, the uniform Lipschitz continuity of b
1 m
and σ
1 m
and the Burkholder inequality that for every p
1 there exists a positive constant c
1
= c
1
p, T such that E
M
x, ℓ
t
ǫ − M
x, ℓ
t
ǫ
′ p
≤ c
1
|ǫ − ǫ
′
|
p
, ∀t ∈ [0, T ], ǫ, ǫ
′
∈ [a
x
, b
x
]. Moreover, the functions b
1 m
and σ
1 m
are uniformly bounded and again by using Burkholder’s inequal- ity we get that for every p
1 E
M
x, ℓ
t
ǫ − M
x, ℓ
s
ǫ
p
≤ c
2
|t − s|
p 2
, ∀s, t ∈ [0, T ], ǫ ∈ [a
x
, b
x
] for some constant c
2
= c
2
p, T . Next we will show that for every p 1 there exists a constant c
3
= c
3
p, T such that E
∆ M
x, ℓ
t
ǫ, ǫ
′
− ∆M
x, ℓ
s
ǫ, ǫ
′ p
≤ c
3
|ǫ − ǫ
′
|
p
|t − s|
p 2
, ∀s, t ∈ [0, T ], ǫ, ǫ
′
∈ [a
x
, b
x
]. For the rest of the proof the symbol c denotes a constant whose value may change from one oc-
curence to the other one. Let 0 ≤ s ≤ t ≤ T and ǫ, ǫ
′
∈ [a
x
, b
x
]. Recall that both M
x, ℓ
ǫ and 857
M
x, ℓ
ǫ
′
are defined to be constant on [0, T ]\[τ
ℓ
, τ
ℓ+1
]. Thus, it is enough to consider the case where [s, t] intersects [
τ
ℓ
, τ
ℓ+1
]. Setting ˆ s
i
:= s ∨ τ
ℓ
and ˆt := t ∧ τ
ℓ+1
we have |ˆt − ˆs| ≤ |t − s|. By
the definition of M
x, ℓ
ǫ and M
x, ℓ
ǫ
′
we have E
∆ M
x, ℓ
t
ǫ, ǫ
′
− ∆M
x, ℓ
s
ǫ, ǫ
′ p
≤c E
Z
ˆt ˆ
s
b
1 m
ℓ
X
r
x
ǫ
− b
1 m
ℓ
X
r
x
ǫ
′
d r
p
+ c E
Z
ˆt ˆ
s
σ
1 m
ℓ
X
r
x
ǫ
− σ
1 m
ℓ
X
r
x
ǫ
′
d w
r p
. By the uniform Lipschitz continuity of b
m
and Proposition 3.2 the first term can be estimated by c
|t − s|
p
E
sup
r ∈[ˆs,ˆt]
kX
r
x
ǫ
− X
r
x
ǫ
′
k
p
≤ c |ǫ − ǫ
′
|
p
|t − s|
p
. For the second term we get the following estimate by Burkholder’s inequality, the uniform Lipschitz
continuity of σ
m
and again by Proposition 3.2: c E
sup
r ∈[ˆs,ˆt]
Z
r ˆ
s
σ
1 m
ℓ
X
r
x
ǫ
− σ
1 m
ℓ
X
r
x
ǫ
′
d w
r p
≤c E
Z
ˆt ˆ
s
kX
r
x
ǫ
− X
r
x
ǫ
′
k
2
d r
p 2
≤c |t − s|
p 2
E
sup
r ∈[ˆs,ˆt]
kX
r
x
ǫ
− X
r
x
ǫ
′
k
p
≤c |ǫ − ǫ
′
|
p
|t − s|
p 2
and we obtain the desired estimate. We apply now Kolmogorov’s continuity theorem, in particular the version for double parameter random fields in Theorem 1.4.4 in [17], which implies that for
any given δ
1
∈ 0, 1 and δ
2
∈ 0,
1 2
there exists a modification of the random field M
x, ℓ
ǫ
t, ǫ
satisfying 3.2 for some random constant K = K ω, δ
1
, δ
2
, T . Step 2: The existence of a modification shown in Step 1 immediately implies that a.s. 3.2 holds
for all s, t ∈ [0, T ] ∩ Q and ǫ, ǫ
′
∈ [a
x
, b
x
] ∩ Q. The claim follows if ∆M
x, ℓ
t
ǫ, ǫ
′
− ∆M
x, ℓ
s
ǫ, ǫ
′
is pathwise continuous in s and t as well as in
ǫ and ǫ
′
. It is enough to show the continuity of M
x, ℓ
t
ǫ in t and
ǫ. The continuity in t is obvious and for every ǫ such that x
ǫ
∈ B
∆
T
x ∩ G we get by an application of Itô’s formula as in 2.2
M
x, ℓ
t
ǫ = u
1 m
ℓ
X
t
x
ǫ
− u
1 m
ℓ
X
τ
ℓ
x
ǫ
− L
t
x
ǫ
, where the right hand side is continuous in
ǫ by Proposition 3.2 and Lemma 3.4.
3.2 Convergence of Minimum Times