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 Hilbert space W. Before giving the theorem, here is a technical
lemma.
Lemma 42.3.1Suppose V,W are separable Banach spaces, W also a Hilbert space suchthat V is dense in W and 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^{′}.
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
〈Bgn ,gn 〉
1 1
≠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}
. Then Bx = lim_{r→∞}Bx_{r} = lim_{r→∞}By_{r} where
y_{r}∈span
(ei,⋅⋅⋅,ek)
. Say
∑k r
Bxr = aiBei
i=1
It follows easily that
〈Bxr,ej〉
= a_{j}^{r}. (Act on e_{j} by both sides and use
〈Bei,ej〉
= δ_{ij}.)
Now since x_{r} is bounded, it follows that these a_{j}^{r} are also bounded. Hence, defining
y_{r}≡∑_{i=1}^{k}a_{i}^{r}e_{i}, it follows that y_{r} is bounded in span
(ei,⋅⋅⋅,ek)
and so, there exists a
subsequence, still denoted by r such that y_{r}→ y ∈span
(ei,⋅⋅⋅,ek)
. Therefore,
Bx = lim_{r→∞}By_{r} = By. In other words, BW = B
(span(ei,⋅⋅⋅,ek))
as claimed. This
proves the claim.
If this happens, the process being described stops. You have found what is desired
which has only finitely many vectors involved.
As long as the process does not stop, let
gn − ∑k 〈Bgn ,ei〉 ei
ek+1 ≡ 〈--(-------∑----〈-k+1----i〉=1)----k+1--∑---〈---------〉-〉1∕2-
B gnk+1 − ki=1 Bgnk+1,ei ei ,gnk+1 − ki=1 Bgnk+1,ei ei
Thus, as in the usual argument for the Gram Schmidt process,
〈Bei,ej〉
= δ_{ij} for
i,j ≤ k + 1. This is already known for i,j ≤ k. Letting l ≤ k, and using the orthogonality
already shown,
〈 ( ) 〉
∑k 〈 〉
〈Bek+1,el〉 = C B gnk+1 − Bgnk+1,ei ei ,el
( i=〈1 〉)
= C 〈Bgk+1,el〉− Bgnk+1,el = 0
Either this equals 0 because p is never one of the n_{k} or eventually it equals 0 for some k
because g_{p} = g_{nk} for some n_{k} and so, from the construction, g_{nk} = g_{p}∈span
(e1,⋅⋅⋅,ek)
and therefore,
∑k
gp = ajej
j=1
which requires easily that
∑k
Bgp = 〈Bgp,ei〉Bei,
i=1
the above holding for all k large enough. It follows that for any x ∈span