1, and almost surely, |Nt | 6 |Mt ||ξt | for any t ∈ [0, 1]. p Then L ∈ Dr for all p > 1 and r < 1/2. Based on this theorem we now prove: T HEOREM 2.23. – In the same situation as in the above theorem, for every T > 0, every N > 0, every p > 1, every 0 < r < 1/2 and every 0 < α < 1/2 − r there exists a constant C = C(T , N, p, r, α) such that:
L(t, x) − L(s, y) 6 C |t − s|α + |x − y|α (2.49) p,r for all s, t ∈ [0, T ] and 0 6 kxk, kyk 6 N . Proof. – First it is known that (cf. [31, Ch. VI, Exercise 1.32]):
L(t, x) − L(s, y) 6 C |t − s|1/2 + |x − y|1/2 . (2.50) p Secondly, it is easy to see from the proof of [3, Theorem 1] that
L(t, x) < ∞. (2.51) sup p,r |t |6T , |x|6N
By [19], for θ ∈ (0, 1), (2.52)
p p Dθ = Lp , D1 θ .
Hence by the reiteration theorem (cf., e.g., [37, Theorem 1.10.2]) we have: p p (2.53) Dr = Lp , Dr+α r . r+α
Consequently, according to [37, Theorem 1.9.3/f] we have by (2.50) and (2.51)
L(t, x) − L(s, y) p,r
(2.54)
1−r/(r+α)
L(t, x) − L(s, y) r/(r+α) 6 C L(t, x) − L(s, y) p p,r+α 6 C |t − s|α + |x − y|α
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for s, t ∈ [0, T ] and x, y such that max{|x|, |y|} 6 N , as claimed. 2 By Theorem 2.23 and [34, Thoerem 3.4], for every T > 0 and N > 0, p > 1 and r < 1/2 L admits a (p, r)-quasi continuous redefinition as a C([0, T ] × [−N, N], R)-valued function. On the other hand, M and hMi adimit (p, r)-quasi continuous redefinitions as C([0, T ], R)-valued e and hMi e functions, too. Using the standard diagonal procedure we see that there exists e L, M such that: e and hMi e are (p, r)-redefinitions of L, for every T > 0 and N > 0, p > 1 and r < 1/2, e L, M M and hMi as C([0, T ] × [−N, N], R)- and C([0, T ], R)-valued functions respectively. Henceforth we still denote these redefinitions by the original symbols themselves for the sake of simplicity. Now we are in a position to prove: T HEOREM 2.24. – There is a set A such that: (1) For every p > 1 and r < 1/2, Cp,r (A) = 0. (2) The equality Z
Zt Φ(Ms )(w) dhMis (w) =
(2.55)
Φ(x)L(t, x, w) dx R
0
holds for every w ∈ Ac , every t > 0 and every positive Borel function Φ. Proof. – Fix p and r and let {On,p,r , n > 1} be a common redefinition net for L, M and hMi. By [31, Ch. VI, Corollary 1.6], there is a P -negligible set B such that (2.55) holds for every w ∈SB, every t > 0 and every continuous function Φ. Therefore (2.55) holds almost everywhere c on ∞ n,p,r . According to the definition of quasi-continuity and redefinition net, it holds for n=1 OS c every w ∈ ∞ n=1 On,p,r , every t > 0 and every continuous function Φ. Varying p and r we see that it holds for every w ∈ A where: A :=
∞ [ ∞ ∞ [ [ l=1 m=1 n=1
Oc
n,l, 2
m−1 −1 2m
,
every t > 0 and every continuous function Φ. Since Cp,r (Ac ) = 0 for every p > 1 and r < 12 , the theorem follows by monotone class theorem. 2 As a consequence of Thoerem 2.19 and Theorem 2.24, we obtain the following result which includes the main results of [33,41,42] as particular cases: T HEOREM 2.25. – There is a set A such that: (1) For every n, C2,n (A) = 0. (2) The equality Z
Zt Φ(Ms )(w) dhMis (w) =
(2.56) 0
Φ(x)L(t, x, w) dx R
holds for every w ∈ Ac , every t > 0 and every positive Borel function Φ.
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3. Quasi continuity and ray continuity 3.1. Framework In this section we assume that E is a locally convex topological vector space over R which is Souslin, i.e., the continuous image of a Polish space (e.g., E is a separable Banach space). For k ∈ E define: τk (z) := z − k,
(3.1)
z ∈ E.
Fix k ∈ E satisfying the following condition: −1 ∼ (C) µ is k-quasi-invariant, i.e., µ ◦ τsk = µ for all s ∈ R, and the Radon–Nikodym derivatives µ
ask :=
−1 d(µ ◦ τsk ) , dµ
s ∈ R,
have the T following properties: µ µ (C1) ask ∈ q>1 Lq (E; µ), for all s ∈ R, and for all q ∈ [1, ∞) the function s 7→ kask kq is locally bounded on R; (C2) For all compact C ⊂ R Z 1 ds < ∞ for µ-a.e. z ∈ E. µ ask (z) C
Here ds denotes Lebesgue measure on R. We recall that choosing appropriate versions by [4, µ Prop. 2.4] we may always assume that ask (z) is jointly measurable in s and z and that (C.2) holds for all z ∈ E (rather than only µ-a.e. z ∈ E). For examples of measures µ satisfying condition (C) we refer to [30, Section 3]. As in [18] we define the measure Z µ −1 (3.2) (A) ds, A ∈ B(E). σk (A) := µ ◦ τsk R
∈ E0
(:= dual of E) with E 0 hl, kiE = 1 we have that σk ◦ l −1 = ds, that Note that obviously for l µ µ µ µ −1 = σk for all s ∈ R, and that σk is equivalent to µ, i.e., there exists ρk : E → (0, ∞) σk ◦ τsk such that: µ
µ
µ
µ = ρk · σk .
(3.3)
Again by [4, Prop. 2.4] choosing appropriate versions we may always assume to have the µ µ following relations between ask and ρk : µ
µ
(3.4)
ask (z) =
(3.5)
ρk (z) =
µ
ρk (z + sk)
for all z ∈ E, s ∈ R, µ ρk (z) !−1 Z µ ask (z) ds for all z ∈ E.
R
In particular, by (C2) and the remarks following it we have that: Z µ (3.6) ρk (z + sk)−1 ds < ∞ for all compact C ⊂ R and all z ∈ E. C
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Following [20], we now define a kind of Sobolev spaces which are different from the Bessel– Sobolev space defined in the above section. Let l ∈ E 0 such that E 0 hl, kiE = 1 and define: Pk (z) := E 0 hl, ziE · k,
(3.7)
z ∈ E,
and πk := IdE − Pk .
Clearly, πk (E) is a closed linear subspace of E and E = πk (E) ⊕ kR.
(3.8)
Then by (2.6) and an elementary calculation we obtain that for all f : E → R, B(R)-measurable, bounded, Z Z Z µ (3.9) f (z)µ(dz) = f (x + sk)ρk (x + sk) ds µk (dx), πk (E) R
E
where µk := µ ◦ πk−1 .
(3.10)
From now on we assume there exists a real separable Hilbert space (H, h, i) densely and continuously embedded into E; H should be thought of as a “tangent space” to E at each z ∈ E. Let K be a dense linear subspace of H such that Condition (C) in Subsection 2.1 holds for all k ∈ K. The following notions were introduced by Kusuoka and Stroock [20] (cf. also [35]): D EFINITION 3.1. – Let f : E 7→ R be a measurable function, k ∈ K. (i) f is said to be k-ray absolutely continuous (RAC for short) if there exists a measurable function fk : E 7→ R such that: f = fk ,
µ-a.e.
and for each z ∈ E, the function t 7→ fk (z + tk) is absolutely continuous. (ii) f is said to be stochastically Gateaux differentiable (SGD for short) if there exists a measurable function F such that, for any k ∈ K 1 f (z + tk) − f (z) = hF, ki t →0 t lim
in probability.
F , when existing, is determined µ-a.e. uniquely and will be denoted by ∇f . D EFINITION 3.2. – Let 1 < p < ∞; define: p (3.11) E1 := f ∈ Lp : f is RAC and SGD, ∇f ∈ Lp (H ) . p
For f ∈ E1 set kf kp,1 := kf kp + k∇f kp .
(3.12)
The following result is due to Kusuoka and Stroock [20]: p
T HEOREM 3.3. – (E1 , k · k) is a Banach space.
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We define now the fractional Sobolev spaces Eα (1 < p < ∞, 0 < α < 1) by real interpolation method. D EFINITION 3.4. – For 0 < r < 1 we define p
p
Er = Lp , E1
(3.13)
r,p
,
where (·, ·) denotes the real interpolation space as in [37]. p
There are several equivalent norms in Er (cf. [37]). The one we shall use is given by Peetre’s K-method: " Z1 #1/p p dε −r (3.14) , ε K(ε, f ) kf kp,r := ε 0
where (3.15)
K(ε, f ) = inf kf1 kp + εkf2 kp,1 , f1 + f2 = f, f1 , f2 ∈ Lp . p
For convenience we denote sometimes E0 := Lp . We now prove: p
L EMMA 3.5. – For every r ∈ [0, 1], Er is uniformly convex. p
Proof. – First it is classical that Lp is uniformly convex. Now we prove so is E1 . Let Lp = Lp × Lp (H ),
(3.16) and equip Lp with the norm (3.17)
(u1 , u2 )
Lp
p 1/p
p
:= ku1 kp + ku2 kp
.
Since H , being a Hilbert space, is uniformly convex, Lp (H ) is uniformly convex. Thus by [1, p Theorem 1.22], Lp is uniformly convex. Define a mapping T from E1 to Lp as: T u := (u, ∇u).
(3.18) Set
R := Range(T ).
(3.19)
p
p
Then T is an isometric isomorphism from E1 onto R. Since E1 is complete, R is a closed p subspace of Lp . Consequently E1 is uniformly convex by [1, Theorem 1.21]. Finally, since p p (3.20) Er = Lp , E1 r,p , p
p
the uniform convexity of Er comes from that of Lp and E1 by [7,21]. 2 Parallel to the notion of capacity defined in Section 2, using the uniform convexity we define p capacity associated with Er . Since there is no danger of confusion, we still denote it by the same symbol. D EFINITION 3.6. – For an open set O, we define: p (3.21) Cp,r (O) := inf kukp,r : u ∈ Er , u > 0 a.e. and u > 1 a.e. on O . For an arbitrary subset A ⊂ E, we define: (3.22) Cp,r (A) := inf Cp,r (O), O is open and A ⊂ O .
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We also have the respective notions of redefinition, redefinition net and equilibrium potential p (whose unique existence is ensured by the uniform convexity of Er ), etc., which for simplicity will be also denoted by the same symbols used in Section 2. 3.2. Results and proofs Now we are prepared to state our results. The first one is a multi-dimensional analogue of Theorem 4.2 in [2]. T HEOREM 3.7. – Fix p > q > 1, 0 < r 6 1, T > 0 and h = (h1 , . . . , hn ) ∈ H n . There exists a constant C = C(p, q, r, h, T ) such that: Z q p/q ds p (3.23) 6 Ckf kp,r E f (· + s1 h1 + · · · + sn hn ) − f (·) |s|n+pr [0,T ]n
and
Z
Z
E f (· + s1 h1 + · · · + sn hn )
[0,T ]n [0,T ]n
q p/q − f (· + t1 h1 + · · · + tn hn )
(3.24)
ds dt p 6 Ckf kp,r |s − t|n+pr
p
for every f ∈ Er . p
Proof. – For s = (s1 , . . . , sn ) ∈ (0, 1)n , take f|s| ∈ D1 such that: (3.25) kf|s| kp,1 6 2|s|−1 K f, |s| and kf − f|s| kp 6 2K f, |s| .
(3.26)
Then by condition (C) and Hölder’s inequality we have:
! ! ! ! q n n n n
q
X X X X
si hi − f|s| · + si hi = E f · + si hi − f|s| · + si hi
f · +
q i=1 i=1 i=1 i=1 p q/p
q q (3.27) 6 C E f (·) − f|s| (·) = C f (·) − f|s| (·) 6 CK f, |s| . p
Similarly
q
" Z1 ! ! q # n n
X X d
f|s| · +t si hi − f|s| = E si hi dt
f|s| · + dt
i=1
(3.28)
q
i=1
0
Hence
i=1
0
! n + q # " Z1 * n X X q q si hi , t si hi dt 6 C|s|q kf|s| kp,1 6 CK f, |s| . Df|s| · +t =E
f
·+
n X i=1
! si hi
i=1
q
− f (·)
q
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= f
(3.29)
·+
n X
!
n X
si hi − f|s| · +
i=1
! si hi
i=1
q
si hi − f|s| (·) + f|s| (·) − f (·) + f|s| · +
i=1 q " ! ! q n n
X X
6 C f · + si hi − f|s| · + si hi
i=1 i=1 q
! ! q # q n n
X X
+ f|s| · + si hi − f|s| (·) + f|s| (·) − f|s| · + si hi
i=1 i=1 q q q 6 CK f, |s| . !
n X
(3.30)
Consequently, Z "
E f
(3.31) [0,1]n
·+
n X
! si hi
i=1
q !#p/q Z ds 6C − f (·) |s|n+pr
p/q K f, |s|
[0,1]n
ds . |s|n+pr
Using spherical coordinates we easily see that the above integral is equal to: π
π
Z2
Z 2 Z1 ···
(3.32) C 0
q p/q t
K(f, t)
1 · · · dθn dt t n+pr
n−1 dθ
Z1 =C
0 0
t −r K(f, t)
p dt t
p
= Ckf kp,r .
0
This proves (3.23). Proof of (3.24): Assume that t > s; by (3.29), condition (C) and Hölder inequality, we have:
! ! q n n
X X q
(3.33) si hi − f · + ti hi 6 CK f, |t − s| .
f · +
i=1
i=1
q
With this and the elementary change of variables ξ = t − s, η = t + s we obtain: ! ! q !#p/q Z " Z n n X X ds dt si hi − f · + si hi E f · + I= |t − s|n+pr i=1
[0,1]n [0,1]n
Z
Z
6C
K f, |t − s|
p
[0,1]n [0,1]n
Z
=C (3.34)
|ξ |−r K f, |ξ |
p
[0,1]n p = Ckf kp,r .
i=1
ds dt |t − s|n+pr
dξ |ξ |n+pr
This proves (3.24). 2 Using this theorem we can prove: T HEOREM 3.8. – Fix T > 0, h = (h1 , . . . , hn ) ∈ H n , p > 1, 0 < r 6 1 such that pr > n, α ∈ (0, r − np−1 ) and p > q > n(r + np−1 − α)−1 . Then there exists a constant
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C = C(p, q, r, α, h, T ) such that for every f ∈ Er there exists a modification ξt (·) of f (· + th) satisfying |ξt − ξs | q q 6 Ckf kp,r . E sup α s6=t,s,t ∈[0,T ]n |t − s| Proof. – First it is easy to see that t 7→ f (· + th) is stochastically continuous. Obviously 1 < q < p and, therefore, by Theorem 3.6 we have: Z Z q p/q E f (· + s1 h1 + · · · + sn hn ) − f (· + t1 h1 + · · · + tn hn ) [0,T ]n [0,T ]n
× (3.35)
ds dt |s − t|n+pr p
6 Ckf kp,r ,
which implies by Hölder’s inequality Z Z q E f (· + s1 h1 + · · · + sn hn ) − f (· + t1 h1 + · · · + tn hn ) [0,T ]n [0,T ]n
× (3.36)
ds dt |s − t|nqp
−1 +qr
q
6 Ckf kp,r ,
or, in a form good for applying Lemma 2.4: Z Z q E f (· + s1 h1 + · · · + sn hn ) − f (· + t1 h1 + · · · + tn hn ) [0,T ]n [0,T ]n
× (3.37)
ds dt |s
−1 −1 − t|n+q(r+n(p −q ))
q
6 Ckf kp,r .
Now the conclusion follows by Lemma 2.4. 2 Next we state: T HEOREM 3.9. – Let everything be the same as in the previous theorem. Then there exists a constant C = C(p, r, T ) > 0 such that for any A ⊂ W we have: µ∗ (A + M(T ; h1 , . . . , hn )) 6 C · Cp,r (A), where µ∗ denotes the outer measure produced by µ and ( M(T ; h1 , . . . , hn ) :=
n X
) si hi , |si | 6 T .
i=1
P Proof. – We denote simply sh := ni=1 si hi . By changing signs we only need to prove: [ ∗ (3.38) (A + sh) 6 C · Cp,r (A). µ s∈[0,T ]n
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n Let S I denote the set of all rational points in [0, T ] . If O ⊂ W is an open set then so is s∈[0,T ]n (O + sh) and we have:
[
(O + sh) =
s∈[0,T ]n
[
(O + sh).
s∈I
Since eO > 0 and eO > 1 on O, we have: Z [ (O + th) = sup 1O (· + th) dµ µ t ∈I
t ∈[0,T ]n
Z sup eO (· + th) dµ. 6 t ∈I
np−1
and q ∈ (n(r + np−1 − α)−1 , p), by Theorem 3.7 there On the other hand, for α < r − n exists a modification {ξt (·), t ∈ [0, T ] } of {eO (· + th), t ∈ [0, T ]n } such that:
E
sup
s,t ∈[0,T ]n s6=t
|ξt − ξs )| |t − s|α
q
q
6 CkeO kp,r .
In particlular, taking α = 0 we obtain: q
E|ξt − ξs |q 6 C · keO kp,r . Hence
Z
Z sup eO (ω + th) dµ = t ∈I
6 6
sup |ξt | dµ t ∈I
Z
sup |ξt − ξ0 | + |ξ0 | dµ
t ∈[0,T ]
Z
sup |ξt − ξ0 | + |ξ0 |
t ∈[0,T ]
1/q
q dµ
6 C · keo kp,r = C · Cp,r (O). Thus (3.38) is proved for open sets. For general A, we have, for O ⊃ A [ [ ∗ ∗ (A + sh) 6 µ (O + sh) µ s∈[0,T ]n
s∈[0,T ]n
6 C · Cp,r (O).
(3.39) Consequently, µ
∗
[
(A + sh) 6 C · inf Cp,r (O), O ⊃ A
s∈[0,T ]n
(3.40)
= C · Cp,r (A),
as desired. 2 With the above preparations we can now prove:
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T HEOREM 3.10. – Suppose f ∈ Eα , with pα > n. Then for any h = (h1 , . . . , hn ) ∈ H n , there exists a redefinition f˜ such that the function [0, T ]n 3 s 7→ f˜(z + sh) ∈ R is continuous for all z ∈ E. Proof. – Let f ∗ be a (p, r)-redefinition of f and An be a (p, r)-continuity net of f . Set Bn = z: z + sh ∈ An , ∀s ∈ [0, T ]n . Then
Bnc = z: there exists s ∈ [0, T ]n such that z + sh ∈ Acn .
By Theorem 3.3 µ
∞ [
!c ! =µ
Bn
n=1
∞ \
! Bnc
n=1
6 C · Cp,r
∞ \
! Acn
n=1
= 0.
S ∗ ∗ Obviously for z ∈ ∞ n=1 Bn , s 7→ f (z + sh) is continuous since f |Bn is continuous for each n. ∗ Finally, modify f by setting: ∗ S f (z) if z ∈ ∞ n=1 Bn , (3.41) f˜(z) := 0 otherwise. Then obviously f˜ is what we were looking for. 2 3.3. Application to the ergodicity problem Now we apply Theorem 3.9 to the ergodicity problem of K. For this let A be a subset of E and (h1 , . . . , hn , . . .) ⊂ H . Then by Condition (C) µ is K-ergodic, i.e., if (h1 , . . . , hn , . . .) is a dense subset of K then µ(A + G) = 0 or 1 where G is the linear space spanned by (h1 , . . . , hn , . . .). Hence, if µ(A) > 0, then µ(A + G) = 1. But of course A + G may have full measure even if µ(A) = 0. So it is natural to ask how small A should be to guarantee µ(A) = 0. Using Theorem 4.9 and the fact that G=
(3.42)
∞ ∞ [ [
M(m; h1, . . . , hn )
m=1 n=1
we obtain immediately the following answer: T HEOREM 3.11. – If there exists pn , rn with pn rn → ∞ as n → ∞ such that Cpn ,rn (A) = 0 for all n, then µ(A + G) = 0. Finally, the next consequence is also direct.
T p T HEOREM 3.12. – If there exists r > 0 such that f ∈ p>1 Dr , then for any (h1 , . . . , hn , · · ·) ⊂ K, there exists a version f˜ of f such that for any integer m, any (n1 , . . . , nm ) ⊂ N, (t1 , . . . , tn ) 7→ f˜(x + t1 h1 + · · · + tn hn ) is continuous for every x ∈ E.
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