The above theorem can be generalized to the case where the formula is of the
form
∫ t
BX (t) = BX0 + Y (s)ds
0
This involves an operator B ∈ℒ
(W, W ′)
and B satisfies
〈Bx, x〉 ≥ 0, 〈Bx, y〉 = 〈By,x〉
for
V ⊆ W, W ′ ⊆ V ′
Where V is dense in the Banach space W. Before giving the theorem, here is a technical
lemma. First is one which is not so technical.
Lemma 32.4.1Let V be a separable Banach space. Then there exists
{g }
k
_{k=1}^{∞}which are linearly independent and whose span is dense in V .
Proof:Let
{fk}
be a countable dense subset. Thus their span is dense. Delete f_{k1}
such that k_{1} is the first index such that f_{k} is in the span of the other vectors. That is, it
is the first which is a finite linear combination of the others. If no such vector exists, then
you have what is wanted. Next delete f_{k2} where k_{2} is the next for which f_{k} is a linear
combination of the others. Continue. The remaining vectors must be linearly
independent. If not, there would be a first which is a linear combination of
the others. Say f_{m}. But the process would have eliminated it at the m^{th} step.
■
Lemma 32.4.2Suppose V,W are separable Banach spaces such that V is dense in Wand B ∈ℒ
(W, W ′)
satisfies
〈Bx,x〉 ≥ 0, 〈Bx,y〉 = 〈By, x〉,B ⁄= 0.
Then there exists a countable set
{ei}
of vectors in V such that
〈Bei,ej〉 = δij
and for each x ∈ W,
∑∞ 2
〈Bx, x〉 = |〈Bx,ei〉| ,
i=1
and also
∑∞
Bx = 〈Bx,ei〉Bei,
i=1
the series converging in W^{′}. If B = B
(ω)
and B is ℱ measurable into ℒ
(W,W ′)
and ifthe e_{i} = e_{i}
(ω)
are as described above, then these e_{i}are measurable into V . If t → B
(t,ω)
is C^{1}
([0,T],ℒ (W, W ′))
and if for each w ∈ W,
〈B ′(t,ω)w, w〉 ≤ k (t)〈B(t,ω)w,w 〉
w,ω
Where k_{w,ω}∈ L^{1}
([0,T])
, then the vectors e_{i}
(t)
can be chosen to also be right continuousfunctions of t.
In the case of dependence on t, the extra condition is trivial if
〈B (t,ω) x,x 〉
≥ δ
∥w∥
_{W}^{2}
for example. This includes the usual case of evolution equations where W = H = H^{′} = W^{′}.
It also includes the case where B does not depend on t.
Proof:Let
{gk}
_{k=1}^{∞} be linearly independent vectors of V whose span is dense in
V . This is possible because V is separable. Thus, their span is also dense in W. Let n_{1} be
the first index such that
〈Bgn1,gn1〉
≠0.
Claim:If there is no such index, then B = 0.
Proof of claim:First note that if there is no such first index, then if x = ∑_{i=1}^{k}a_{i}g_{i}
|| ||
|〈Bx, x〉| = ||∑ a a 〈Bg ,g〉||≤ ∑ |a ||a ||〈Bg ,g 〉|
||i⁄=j i j i j|| i⁄=j i j i j
∑
≤ |ai||aj|〈Bgi,gi〉1∕2 〈Bgj,gj〉1∕2 = 0
i⁄=j
Therefore, if x is given, you could take x_{k} in the span of