174 CHAPTER 7. THE INTEGRAL

However, I need to verify that if f is continuous on [a,b] , then there is an antiderivativeF in order to use that Procedure. This is the following lemma. It is a major result. Recallthat for a function f defined on an interval [a,b] ,∥ f∥ ≡ sup{| f (x)| : x ∈ [a,b]}. Also recallthat if f is a continuous function defined on [a,b] , then there exists a sequence of polyno-mials {pn (x)} for which ∥ f − pn∥→ 0. This is by the Weierstrass approximation theoremof the chapter on continuous functions.

The message of the following lemma is that a continuous function on a closed intervalhas an antiderivative.

Lemma 7.1.3 Let f be a continuous, real valued function defined on [a,b] . Then thereexists F such that F ′ (x) = f (x) for all x ∈ (a,b) . At the end points, F ′ (x) will refer to aone sided derivative.

Proof: Assume that a< b in what follows. If not, simply switch a and b in the argument.Let {pn} be a sequence of polynomials for which ∥pn − f∥ → 0 and let P′

n (x) = pn (x) forall x ∈ (a,b). By the mean value theorem,

|Pn (x)−Pn (a)− (Pm (x)−Pm (a))|=

|Pn (x)−Pm (x)− (Pn (a)−Pm (a))|= |(pn (t)− pm (t))(x−a)| ≤ ∥pn − pm∥|b−a|≤ (∥pn − f∥+∥ f − pm∥) |b−a|

The right side converges to 0 as n,m → ∞ and so by completeness, there exists

F (x) = limn→∞

(Pn (x)−Pn (a)) ,

this for any choice of x. It remains to verify that F ′ (x) = f (x) . Say x ∈ [a,b) and let h > 0.Then by the mean value theorem,

Pn (x+h)−Pn (x)h

=(Pn (x+h)−Pn (a))− (Pn (x)−Pn (a))

h= pn (thn)

for some thn ∈ (x,x+h). By compactness, there is a subsequence, still denoted as thn forwhich limn→∞ thn = th ∈ [x,x+h]. Now

|pn (thn)− f (th)| ≤ |pn (thn)− f (thn)|+ | f (thn)− f (th)|≤ ∥pn − f∥+ | f (thn)− f (th)|

and so, letting n → ∞, this shows, from continuity of f that |pn (thn)− f (th)| → 0. Takinga limit, |pn (thn)− f (th)| → 0. Then

F (x+h)−F (x)h

= f (th) , th ∈ [x,x+h]

Now by continuity of f , we can take a limit of this as h → 0 and obtain F ′ (x) = f (x) ,where F ′ (x) is a right derivative at x = a. For x ∈ (a,b], the situation is exactly the samefor when h is restrained to be negative. Indeed,

F (x+h)−F (x)h

=−F (x− (−h))−F (x)−h

=F (x)−F (x− k)

k