78 CHAPTER 2. BASIC TOPOLOGY AND ALGEBRA

26. Let Y be a normed vector space and suppose h : [a,b]→ Y is differentiable on (a,b)meaning

limh→0

f (t +h)− f (t)h

= f ′ (t) ,

continuous on [a,b] and h′ (t) = 0. Then ∥h(b)−h(a)∥= 0. Show this. Hint: Let

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

Then let t ≡ supS. By continuity, ∥h(t)−h(a)∥ ≤ ε (t−a) . Suppose t < b. If strictinequality holds, then this will persist for s near t and violate the definition of t.Therefore, ∥h(t)−h(a)∥ = ε (t−a). Then, still assuming t < b, there exists hk ↓ 0and

ε (t−a+hk)< ∥h(t +hk)−h(a)∥ ≤ ∥h(t +hk)−h(t)∥+=ε(t−a)

∥h(t)−h(a)∥

Now we have ε < 1hk∥h(t +hk)−h(t)∥ and passing to a limit, ε < ∥h′ (t)∥ a contra-

diction.

27. Let Y be a normed vector space and suppose h : [a,b]→ Y is differentiable on (a,b)meaning

limh→0

f (t +h)− f (t)h

= f ′ (t) ,

continuous on [a,b] and ∥h′ (t)∥ ≤M for all t ∈ (a,b) . Then

∥h(b)−h(a)∥ ≤M |b−a| .

This is called the mean value inequality. Show this. Hint: Let

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