Kenneth Kuttler

2360 CHAPTER 69. GELFAND TRIPLES

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.

Now let Sn ≡ g−1 (B(yn, ||yn||/2)) where E = {yn}∞

n=1 . Thus, ∪nSn = g−1 (X \{0}) .This follows because if x ∈ Y and x ̸= 0, then in B

(x, ||x||4

)there is a point of E,yn. There-

fore, ||yn||> 34 ||x|| and so ||yn||

2 > 3||x||8 > ||x||

4 so x ∈ B(yn, ||yn||/2) . It follows that if eachSn has measure zero, then g(t) = 0 for a.e. t. Suppose then that for some n, the set, Sn haspositive measure. Then from what was shown above,

||yn|| =

∣∣∣∣∣∣∣∣ 1m(Sn)

∫Sn

g(t)dt− yn

∣∣∣∣∣∣∣∣= ∣∣∣∣∣∣∣∣ 1m(Sn)

∫Sn

g(t)− yndt∣∣∣∣∣∣∣∣

≤ 1m(Sn)

∫Sn

||g(t)− yn||dt ≤ 1m(Sn)

∫Sn

||yn||/2dt = ||yn||/2

and so yn = 0 which implies Sn = /0, a contradiction to m(Sn)> 0. This contradiction showseach Sn has measure zero and so as just explained, g(t) = 0 a.e.

Definition 69.2.2 For f ∈ L1 (a,b;X) , define an extension, f defined on

[2a−b,2b−a] = [a− (b−a) ,b+(b−a)]

as follows.

f (t)≡

 f (t) if t ∈ [a,b]f (2a− t) if t ∈ [2a−b,a]f (2b− t) if t ∈ [b,2b−a]

Definition 69.2.3 Also if f ∈ Lp (a,b;X) and h > 0, define for t ∈ [a,b] , fh (t) ≡ f (t−h)for all h < b−a. Thus the map f → fh is continuous and linear on Lp (a,b;X) . It is con-tinuous because∫ b

a|| fh (t)||p dt =

∫ a+h

a|| f (2a− t +h)||p dt +

∫ b−h

a|| f (t)||p dt

=∫ a+h

a|| f (t)||p dt +

∫ b−h

a|| f (t)||p dt ≤ 2 || f ||pp .

The following lemma is on continuity of translation in Lp (a,b;X) .

2360 CHAPTER 69. GELFAND TRIPLESand the integrand of the last integral converges to 0 a.e. as n + © because Ym (Vn \ Kn) <co, By the dominated convergence theorem, this last integral converges to 0. Therefore,whenever E C (a,b),[ Xe (the (tat =0.Since the endpoints have measure zero, it also follows that for any measurable E,, the aboveequation holds.Now g € L!({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 formy, aid; where a; is a rational point of F and d; € D. Thus F is countable. Denote by Y theclosure of EF in X. Thus Y is a separable closed subspace of X which contains all the valuesof g.Now let S, = 97! (B(yn,||n||/2)) where E = {y,}"_,. Thus, UnS, = g~! (X \ {0}).|x!This follows because if x € Y and x 4 0, then in B (s. Let) there is a point of E, y,. There-fore, ||yn|| > 2 ||x|| and so Ll > Sif > ail so x € B(yn,||Yn|| /2) . It follows that if eachS, has measure zero, then g(t) = 0 for a.e. t. Suppose then that for some n, the set, S,, haspositive measure. Then from what was shown above,lonll = |e [ea- ls , (0) —Yndtwisp fy lis ~ valde s es [lull 2d = lol /2and so y, = 0 which implies S,, = @, a contradiction to m(S,,) > 0. This contradiction showseach S,, has measure zero and so as just explained, g(t)=Oae. IfDefinition 69.2.2 For f € L' (a,b;X) , define an extension, f defined on[2a — b, 2b —a] = |a— (b—a),b+ (b—a)]as follows.P(t) ift € [a,b]f(th=¢ f(2a—t) ift € [2a—b,a}f (2b—-1) ift € [b,2b—alDefinition 69.2.3 Also if f € L? (a,b;X) and h > 0, define for t € [a,b], fy (t) = f (t—h)for all h <b—a. Thus the map f — fy is continuous and linear on L? (a,b;X). It is con-tinuous becauseb ath b—-h[\imcoira =f rea-remirare fo \r@llrarath 5 -b—h= [roars [irolrars2iiein.The following lemma is on continuity of translation in L? (a,b;X).