Chapter 43

Interpolation In Banach Space43.1 Some Standard Techniques In Evolution Equations

43.1.1 Weak Vector Valued DerivativesIn this section, several significant theorems are presented. Unless indicated otherwise, themeasure will be Lebesgue measure. First here is a lemma.

Lemma 43.1.1 Suppose g ∈ L1 ([a,b] ;X) where X is a Banach space. Then if∫ b

ag(t)φ (t)dt = 0

for all φ ∈C∞c (a,b) , then g(t) = 0 a.e.

Proof: Let E be a measurable subset of (a,b) and let K ⊆ E ⊆ V ⊆ (a,b) where Kis compact, V is open and m(V \K) < ε. Let K ≺ h ≺ V as in the proof of the Rieszrepresentation theorem for positive linear functionals. Enlarging K slightly and convolvingwith a mollifier, it can be assumed h ∈C∞

c (a,b) . Then∣∣∣∣∫ b

aXE (t)g(t)dt

∣∣∣∣ =

∣∣∣∣∫ b

a(XE (t)−h(t))g(t)dt

∣∣∣∣≤

∫ b

a|XE (t)−h(t)| ||g(t)||dt

≤∫

V\K||g(t)||dt.

Now let Kn ⊆ E ⊆Vn with m(Vn \Kn)< 2−n. Then from the above,∣∣∣∣∫ b

aXE (t)g(t)dt

∣∣∣∣≤ ∫ b

aXVn\Kn (t) ||g(t)||dt

and the integrand of the last integral converges to 0 a.e. as n→∞ because ∑n m(Vn \Kn)<∞. By the dominated convergence theorem, this last integral converges to 0. Therefore,whenever E ⊆ (a,b) , ∫ b

aXE (t)g(t)dt = 0.

Since the endpoints have measure zero, it also follows that for any measurable E, the aboveequation holds.

Now g ∈ L1 ([a,b] ;X) and so it is measurable. Therefore, g([a,b]) is separable. LetD be a countable dense subset and let E denote the set of linear combinations of the form∑i aidi where ai is a rational point of F and di ∈ D. Thus E is countable. Denote by Y theclosure of E in X . Thus Y is a separable closed subspace of X which contains all the valuesof g.

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