5.1. EXISTENCE AND DEFINITION 119

and if k is sufficiently small, the same argument holds and you obtain for small negative h∣∣∣∣F (t +h)−F (t)h

− f (t)∣∣∣∣= ∣∣∣∣F (t)−F (t− k)

k− f (t)

∣∣∣∣≤ 1k

∫ t

t−k| f (s)− f (t)|ds≤ ε

provided |h|= |k| is small enough. Hence F ′ (t) = f (t) for all t ∈ (a,b) and this is also truefor one sided derivatives at the end points. ■

The next lemma is called the mean value inequality.

Lemma 5.1.9 Let Y be a normed vector space and suppose h : [a,b]→ Y is differen-tiable and satisfies ∥h′ (t)∥ ≤M, M ≥ 0. Then ∥h(b)−h(a)∥ ≤M (b−a) .

Proof: Let ε > 0 be given and let

S≡ {t ∈ [a,b] : for all s ∈ [a, t] ,∥h(s)−h(a)∥ ≤ (M+ ε)(s−a)}

Then a ∈ S. Let t = supS. Then by continuity of h it follows

∥h(t)−h(a)∥= (M+ ε)(t−a) (5.7)

Suppose t < b. Then there exist positive numbers, hk decreasing to 0 such that

∥h(t +hk)−h(a)∥> (M+ ε)(t +hk−a)

and now it follows from 5.7 and the triangle inequality that

∥h(t +hk)−h(t)∥+∥h(t)−h(a)∥= ∥h(t +hk)−h(t)∥+(M+ ε)(t−a)> (M+ ε)(t +hk−a)

and so∥h(t +hk)−h(t)∥> (M+ ε)hk

Now dividing by hk and letting k→∞, ∥h′ (t)∥ ≥M+ε , a contradiction. Thus t = 1. Sinceε is arbitrary, the conclusion of the lemma follows. ■

Corollary 5.1.10 Let f : [a,b]→ X be continuous. Suppose F ′ (x) = f (x) for all x ∈(a,b) where F is a continuous function on [a,b]. Then

∫ ba f (x)dx = F (b)−F (a) . Also if

f is real valued, there exists y ∈ (a,b) with∫ b

a f (x)dx = f (y)(b−a).

Proof: Let G(t)≡∫ t

a f (s)ds. Then

G(b)−G(a)− (F (b)−F (a)) = (G(b)−F (b))− (G(a)−F (a)) .

(G−F)′ (x) = 0 so by Lemma 5.1.9,

∥(G(b)−F (b))− (G(a)−F (a))∥= 0.

Thus, since G(a) = 0, G(b) =∫ b

a f (s)ds = F (b)−F (a). The last assertion is from themean value theorem. ■

Although the main interest is in continuous functions, here is an important case.

Theorem 5.1.11 Let f be decreasing and real valued on [a,b] then f is Riemannintegrable.