1416 CHAPTER 43. INTERPOLATION IN BANACH SPACE
Proof: Let φ 0 ∈C∞c (a,b) ,
∫ ba φ 0 (x)dx = 1 and define for φ ∈C∞
c (a,b)
ψφ (x)≡∫ x
a[φ (t)−
(∫ b
aφ (y)dy
)φ 0 (t)]dt
Then ψφ ∈C∞c (a,b) and ψ ′
φ= φ −
(∫ ba φ (y)dy
)φ 0. Then
∫ b
af (t)(φ (t))dt =
∫ b
af (t)
(ψ′φ (t)+
(∫ b
aφ (y)dy
)φ 0 (t)
)dt
=
=0 by assumption︷ ︸︸ ︷∫ b
af (t)ψ
′φ (t)dt +
(∫ b
aφ (y)dy
)∫ b
af (t)φ 0 (t)dt
=
(∫ b
a
(∫ b
af (t)φ 0 (t)dt
)φ (y)dy
).
It follows that for all φ ∈C∞c (a,b) ,∫ b
a
(f (y)−
(∫ b
af (t)φ 0 (t)dt
))φ (y)dy = 0
and so by Lemma 43.1.1,
f (y)−(∫ b
af (t)φ 0 (t)dt
)= 0 a.e. y
Theorem 43.1.9 Suppose f , f ′ both are in L1 (a,b;X) where the derivative is taken in thesense of X valued distributions. Then there exists a unique point of X , denoted by f (a)such that the following formula holds a.e. t.
f (t) = f (a)+∫ t
af ′ (s)ds
Proof:∫ b
a
(f (t)−
∫ t
af ′ (s)ds
)φ′ (t)dt =
∫ b
af (t)φ
′ (t)dt−∫ b
a
∫ t
af ′ (s)φ
′ (t)dsdt.
Now consider∫ b
a∫ t
a f ′ (s)φ′ (t)dsdt. Let Λ ∈ X ′. Then it is routine from approximating f ′
with simple functions to verify
Λ
(∫ b
a
∫ t
af ′ (s)φ
′ (t)dsdt)=∫ b
a
∫ t
aΛ(
f ′ (s))
φ′ (t)dsdt.
Now the ordinary Fubini theorem can be applied to obtain
=∫ b
a
∫ b
sΛ(
f ′ (s))
φ′ (t)dtds
= Λ
(∫ b
a
∫ b
sf ′ (s)φ
′ (t)dtds).