for all k ≥ M. Then if d,d^{′}∈ D_{m}for m > n ≥ M such that
|d− d′|
≤ T2^{−n},then
∑m
||X (d′)− X (d)|| ≤ 2 2− γj.
j=n+1
Also, there exists a constant C depending on M such that for all d,d^{′}∈ D,
′ ′γ
||X (d)− X (d)|| ≤ C |d − d | .
Proof:Suppose d^{′}< d. Suppose first m = n + 1. Then d =
(k+ 1)
T2^{−(n+1)
} and
d^{′} = kT2^{−(n+1)
}. Then from 61.2.3
n∑+1
||X (d′)− X (d)|| ≤ 2−γ(n+1) ≤ 2 2−γj.
j=n+1
Suppose the claim is true for some m > n and let d,d^{′}∈ D_{m+1} with
′
|d − d|
< T2^{−n}. If
there is no point of D_{m} between these, then d^{′},d are adjacent points either in D_{m} or in
D_{m+1}. Consequently,
m+1
||X (d′) − X (d)|| ≤ 2−γm < 2 ∑ 2−γj.
j=n+1
Assume therefore, there exist points of D_{m} between d^{′} and d. Let d^{′}≤ d_{1}^{′}≤ d_{1}≤ d
where d_{1},d_{1}^{′} are in D_{m} and d_{1}^{′} is the smallest element of D_{m} which is at least as
large as d^{′} and d_{1} is the largest element of D_{m} which is no larger than d. Then
′ ′
|d − d1|
≤ T2^{−(m+1)
} and
|d1 − d|
≤ T2^{−(m+1 )
} while all of these points are in D_{m+1}
which contains D_{m}. Therefore, from 61.2.3 and induction,
is Holder continuous on D with Holder
exponent γ. Note the constant is a measurable function of ω, depending on how many
measurable N_{k} which contain ω.
< T2^{−(M (ω)+1)
}. Thus the Holder continuous
version of X will satisfy
2 γ
||Y (t)(ω) − Y (s)(ω)|| ≤ Tγ(1−-2−γ)-(|t− s|)
provided
|t− s|
< T2^{−(M (ω)+1)
}. Does this translate into an inequality of the
form
2
||Y (t)(ω) − Y (s)(ω)|| ≤-γ-----−γ--(|t− s|)γ
T (1− 2 )
for any pair of points t,s ∈
[0,T]
? It seems it does not for any γ < 1 although it does
yield
||Y (t)(ω)− Y (s)(ω )|| ≤ C (|t− s|)γ
where C depends on the number of intervals having length less than T2^{−}
(M(ω)+1)
which
it takes to cover
[0,T]
. First note that if γ > 1, then the Holder continuity will imply
t → Y
(t)
(ω )
is a constant. Therefore, the only case of interest is γ < 1. Let s,t be any
pair of points and let s = x_{0}<
⋅⋅⋅
< x_{n} = t where
|x − x |
i i−1
< T2^{−(M (ω)+1)
}.
Then
∑n
||Y (t)(ω )− Y (s)(ω)|| ≤ ||Y (xi)(ω)− Y (xi− 1)(ω )||
i=1
n
≤ -----2----- ∑ (|x − x |)γ (61.2.9)
T γ(1− 2−γ) i=1 i i−1
(61.2.9)
How does this compare to
( n )γ
∑ |x − x | = |t − s|γ?
i=1 i i− 1
This last expression is smaller than the right side of 61.2.9 for any γ < 1. Thus for
γ < 1, the constant in the conclusion of the theorem depends on both T and
ω
∕∈
N.
In the case where α ≥ 1, here is another proof of this theorem. It is based on the one
in the book by Stroock [?].
Theorem 61.2.3Suppose X is a stochastic process on
[0,T]
having values in theBanach space E. Suppose also that there exists a constant, C and positive numbersα,β,α ≥ 1, such that
α 1+β
E (||X (t)− X (s)|| ) ≤ C |t− s| (61.2.10)
(61.2.10)
Then there exists a stochastic process Y such that for a.e. ω,t → Y
(t)
(ω)
is Höldercontinuous with exponent γ <
β
α
and for each t, P
([||X (t)− Y (t)|| > 0])
= 0. (Y is aversion of X.) Also
( )
E sup ∥Y-(t)−-Yγ(s)∥ ≤ C
0≤s<t≤T (t− s)
where C depends on α,β,T,γ.
Proof: The proof considers piecewise linear approximations of X which are
automatically continuous. These are shown to converge to Y in L^{α}
(Ω;C ([0,T],E ))
so it follows that Y must be continuous for a.e. ω. Finally, it is shown that
Y is a version of X and is Holder continuous. In the proof, I will use C to
denote a constant which depends on the quantities γ,α,β,T. Let
{tn}
k
_{k=0}^{2n
} be
a uniform partition of the interval
[0,T ]
so that t_{k+1}^{n}− t_{k}^{n} = T2^{−n}. Now
let
Mn ≡ max ∥∥X (tn)− X (tn )∥∥
k≤2n k k− 1
Then it follows that
n
α ∑2 ∥∥ n ( n )∥∥α
Mn ≤ X (tk)− X tk− 1
k=1
and so
n
α ∑2 ( − n)1+β n − n(1+β) −nβ
E (M n) ≤ C T2 = C2 2 = C2 (61.2.11)
k=1
(61.2.11)
Next denote by X_{n} the piecewise linear function which results from the values of X at
the points t_{k}^{n}. Consider the following picture which illustrates a part of the graphs of X_{n}
and X_{n+1}.