It is not necessary to have p > 2 in order to do the sort of thing just described. Here is a
major result which will have a much more difficult stochastic version presented later.
First is a simple version of an approximation theorem of Doob.
Lemma 42.2.1Let Y :
[0,T]
→ E, be ℬ
([0,T ])
measurable and suppose
Y ∈ Lp (0,T ;E) ≡ K, p ≥ 1
Then there exists a sequence of nested partitions, P_{k}⊆P_{k+1},
{ k k }
Pk ≡ t0,⋅⋅⋅,tmk
such that the step functions given by
r ∑mk ( k)
Yk (t) ≡ Y tj X[tkj−1,tkj)(t)
j=1
l ∑mk ( k )
Yk (t) ≡ Y tj− 1 X(tkj−1,tkj](t)
j=1
both converge to Y in K as k →∞ and
{|k k | }
lk→im∞ max |tj − tj+1| : j ∈ {0,⋅⋅⋅,mk } = 0.
Also, each Y
( k)
tj
,Y
(k )
tj−1
is in E. One can also assume that Y
(0)
= 0. The mesh points
{k}
tj
_{j=0}^{mk}can be chosen to miss a given set of measure zero. In addition to this, we canassume that
| |
|tkj − tkj−1| = 2−nk
except for the case where j = 1 or j = m_{nk}when this might not be so. In the case of thelast subinterval defined by the partition, we can assume
|k k | | k | −(nk+1)
|tm − tm −1| = |T − tm− 1| ≥ 2
Proof: For t ∈ ℝ let γ_{n}
(t)
≡ k∕2^{n},δ_{n}
(t)
≡
(k+ 1)
∕2^{n}, where t ∈ (k∕2^{n},
(k + 1)
∕2^{n}],
and 2^{−n}< T∕4. Also suppose Y is defined to equal 0 on
[0,T]
^{C}. Then t →
∥Y (t)∥
is in
L^{p}
(ℝ )
. Therefore by continuity of translation, as n →∞ it follows that for
t ∈
This converges to 0 as n →∞ as was shown above. Therefore,
∫ T∫ T
||Y (γn (t− s)+ s)− Y (t)||p dtds
0 0 E
also converges to 0 as n →∞. The only problem is that γ_{n}
(t− s)
+ s ≥ t − 2^{−n} and so
γ_{n}
(t − s)
+ s could be less than 0 for t ∈
[0,2− n]
. Since this is an interval whose measure
converges to 0 it follows
∫ ∫
T T |||| ( + ) ||||p
0 0 ||Y (γn(t− s)+ s) − Y (t)||E dtds
converges to 0 as n →∞. Let
∫
T |||| ( +) ||||p
mn (s) = 0 ||Y (γn(t− s)+ s) − Y (t)||Edt
Then
1∫ T
P ([mn (s) > λ]) ≤ λ mn (s)ds.
0
It follows there exists a subsequence n_{k} such that
([ 1])
P mnk (s) >-- < 2− k
k
Hence by the Borel Cantelli lemma, there exists a set of measure zero N such that for
s
∕∈
N,
m (s) ≤ 1∕k
nk
for all k sufficiently large. Picking s_{k}
∈∕
N,
( + )
Ykl(t) ≡ Y (γnk (t− sk)+ sk)
Then t → Y
( )
(γn (t− sk) +sk)+
k
is a step function of the sort described above. Of
course you can always simply define Y _{k}^{l}
(0)
≡ 0. This is because the interval affected has
length which converges to 0 as k →∞. The jumps in t → γ_{nk}
(t− sk)
determine the
mesh points of the partition. By picking s_{k} appropriately, you can have each of these
mesh points miss a given set of measure zero except for the first and last point. This is
because when you slide s_{k} it just moves the mesh points of P_{k} except for the first point
and last point. Let N_{1} be a set of measure zero and let
(a,b)
⊆
[0,T ]
. Now
let s move through
(a,b)
and denote by A_{j} the corresponding set of points
obtained by the j^{th} mesh point. Thus A_{j} has positive measure and so it is not
contained in N_{1}. Let S_{j} be the points of
(a,b)
which correspond to A_{j}∩ N_{1}. Thus
S_{j} has measure 0. Just pick s_{k}∈
(a,b)
∖∪_{j}S_{j}. You can also choose s_{k} such
that
T − sk − γn (T − sk) > 2−(nk+1)
k
which will cause the last condition mentioned above to hold.
To get the other sequence of step functions, just use a similar argument with δ_{n} in
place of γ_{n}. ■
Theorem 42.2.2Let V ⊆ H = H^{′}⊆ V^{′}be a Gelfand triple and supposeY ∈ L^{p′
}