65.4. SOME HILBERT SPACE THEORY 2237

Proposition 65.3.3 Suppose f ≥ 0 is progressively measureable and Ft is a filtration.Then

ω →∫ t

af (s,ω)ds

is Ft adapted.

Proof: This follows right away from the fact f is B ([a, t])×Ft measurable. This isjust product measure and so the integral from a to t is Ft measurable. See also Proposition62.3.5.

65.4 Some Hilbert Space TheoryRecall the following definition which makes LU into a Hilbert space where L ∈L (U,H) .

Definition 65.4.1 Let L∈L (U,H), the bounded linear maps from U to H for U,H Hilbertspaces. For y ∈ L(U) , let L−1y denote the unique vector in

{x : Lx = y} ≡My

which is closest in U to 0.

{x : Lx = y}L−1(y)

Note this is a good definition because {x : Lx = y} is closed thanks to the continuity ofL and it is obviously convex. Thus Theorem 19.1.8 applies. With this definition define aninner product on L(U) as follows. For y,z ∈ L(U) ,

(y,z)L(U) ≡(L−1y,L−1z

)U

Thus it is obvious that L−1 : LU →U is continuous. The notation is abominable becauseL−1 (y) is the normal notation for My.

With this definition, here is one of the main results. It is Theorem 19.2.3 proved earlier.

Theorem 65.4.2 Let U,H be Hilbert spaces and let L ∈L (U,H) . Then Definition 65.4.1makes L(U) into a Hilbert space. Also L : U → L(U) is continuous and L−1 : L(U)→Uis continuous. Furthermore there is a constant C independent of x ∈U such that

∥L∥L (U,H) ||Lx||L(U) ≥ ||Lx||H (65.4.5)

If U is separable, so is L(U). Also(L−1 (y) ,x

)= 0 for all x∈ ker(L) , and L−1 : L(U)→U

is linear. Also, in case that L is one to one, both L and L−1 preserve norms.