1414 CHAPTER 43. INTERPOLATION IN BANACH SPACE

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

n=1 . Then, ∪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 mesure. 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 43.1.2 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 43.1.3 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) .

Lemma 43.1.4 Let f be as defined in Definition 69.2.2. Then for f ∈ Lp (a,b;X) for p ∈[1,∞),

limδ→0

∫ b

a

∣∣∣∣ f (t−δ )− f (t)∣∣∣∣p

X dt = 0.

Proof: Regarding the measure space as (a,b) with Lebesgue measure, by Lemma21.5.9 there exists g ∈ Cc (a,b;X) such that || f −g||p < ε. Here the norm is the norm inLp (a,b;X) . Therefore,

|| fh− f ||p ≤ || fh−gh||p + ||gh−g||p + ||g− f ||p≤

(21/p +1

)|| f −g||p + ||gh−g||p

<(

21/p +1)

ε + ε