It is not necessary to have p > 2 in order to do the sort of thing just described. First is an
approximation theorem which says that a given functionin L^{p}
([0,T ];E )
can be
approximated by step functions.
Lemma 32.3.1Let Φ :
[0,T]
→ E, be Lebesgue measurable and suppose
Φ ∈ K ≡ Lp([0,T ];E ), 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)
Φ k(t) ≡ Φ tj X[tkj−1,tkj)(t)
j=1
l ∑mk ( k )
Φ k(t) ≡ Φ tj−1 X[tkj−1,tkj)(t)
j=1
both converge to Φ in K as k →∞ and
{||k k || }
lk→im∞ max tj − tj+1 : j ∈ {0,⋅⋅⋅,mk } = 0.
In the formulas, define Φ
(0)
= 0. The mesh points
{ }
tkj
_{j=0}^{mk}can be chosen to miss agiven set of measure zero.
Note that it would make no difference in terms of the conclusion of this lemma if you
defined
m
l ∑k ( k )
Φ k(t) ≡ Φ tj−1 X(tkj−1,tkj](t)
j=1
because the modified function equals the one given above off a countable subset of
[0,T]
,
the union of the mesh points.
Proof: For t ∈ ℝ let γ_{n}
(t)
≡ k∕2^{n},δ_{n}
(t)
≡
(k+ 1)
∕2^{n}, where
n n
t ∈ (k∕2 ,(k+ 1)∕2 ],
and 2^{−n}< T∕4. Also suppose Φ is defined to equal 0 on
[0,T]
^{C}× Ω. There
exists a set of measure zero N such that for ω
∈∕
N,t →
∥Φ (t,ω)∥
is in L^{p}
(ℝ )
.
Therefore by continuity of translation, as n →∞ it follows that for ω
∫ T|| ( ) ||p
mn (s) = ||||Φ (γn (t− s)+ s)+ − Φ(t)|||| dt
0 E
Then letting μ denote Lebesgue measure,
1∫ T
μ([mn (s) > λ]) ≤ λ mn (s)ds.
0
It follows there exists a subsequence n_{k} such that
([ 1])
μ mnk (s) > -- < 2−k
k
Hence by the Borel Cantelli lemma, there exists a set of measure zero N such that for
s
∕∈
N,
mnk (s) ≤ 1∕k
for all k sufficiently large. Pick such an s. Then consider t → Φ
( + )
(γnk (t− s)+ s)
. For
n_{k}, t →
(γnk (t− s)+ s)
^{+} has jumps at points of the form 0, s + l2^{−nk} where l is an
integer. Thus P_{nk} consists of points of
[0,T ]
which are of this form and these partitions
are nested. Define Φ_{k}^{l}
(0)
≡ 0, Φ_{k}^{l}
(t)
≡ Φ
((γ (t − s)+ s)+)
nk
. Now suppose N_{1} is a set
of measure zero. Can s be chosen such that all jumps for all partitions occur off N_{1}? Let
(a,b)
be an interval contained in
[0,T]
. Let S_{j} be the points of
(a,b)
which are
translations of the measure zero set N_{1} by t_{j}^{l} for some j. Thus S_{j} has measure 0. Now
pick s ∈
(a,b)
∖∪_{j}S_{j}. To get the other sequence of step functions, the right
step functions, just use a similar argument with δ_{n} in place of γ_{n}. Just apply
the argument to a subsequence of n_{k} so that the same s can hold for both.
■
Theorem 32.3.2Let V ⊆ H = H^{′}⊆ V^{′}be a Gelfand triple and supposeY ∈ L^{p′
}