150 CHAPTER 7. THE DERIVATIVE

Definition 7.11.1 Let { fn} be a sequence of functions defined on D. Then { fn} issaid to converge uniformly to f if it converges pointwise to f and for every ε > 0 thereexists N such that for all n≥ N, | f (x)− fn (x)|< ε for all x ∈ D.

To save on notation, denote by ∥k∥ ≡ sup{|k (ξ )| : ξ ∈ D} . Then

∥k+ l∥ ≤ ∥k∥+∥l∥ (7.14)

because for each ξ ∈ D, |k (ξ )+ l (ξ )| ≤ ∥k∥+ ∥l∥and taking sup yields 7.14. From thedefinition of uniform convergence, you see that fn converges uniformly to f is the same assaying limn→∞ ∥ fn− f∥ = 0. Now here is the theorem. Note how the mean value theoremis one of the principal parts of the argument.

Theorem 7.11.2 Let (a,b) be an open interval and let fk : (a,b)→ R be differen-tiable and suppose there exists x0 ∈ (a,b) such that

{ fk (x0)} converges,{f ′k}

converges uniformly to g

Then there exists a function f defined on (a,b) such that

fk→ f uniformly,

and f ′ = g.

Proof: By the mean value theorem,

( fk (x)− fm (x))− ( fk (x0)− fm (x0)) =(

f ′k (tkm)− f ′m (tkm))(x− x0)

and so, if k,m is large enough,

| fk (x)− fm (x)| ≤ | fk (x0)− fm (x0)|+∥∥ f ′k− f ′m

∥∥< ε

2

provided m,k are large enough. Therefore, ∥ fk− fm∥< ε if k,m are large enough showingthat { fk} converges uniformly to a continuous function f by Theorem 6.9.7. That is

∥ fk− f∥→ 0

I want to show that f ′ exists and equals g.Let c ∈ (a,b) and define

gn (x,c)≡{ fn(x)− fn(c)

x−c if x ̸= cf ′n (c) if x = c

.

Thus x→ gn (x,c) is continuous.Claim : For each c, x→ gn (x,c) converges uniformly to a continuous function hc, on

(a,b) and hc (c) = g(c).