346 CHAPTER 14. FUNDAMENTALS

ω∫ b

a γ ′ (s)ds. Therefore,∣∣∣∣∫ b

aγ′ (s)ds

∣∣∣∣= ∫ b

aωγ′ (s)ds =

∫ b

aRe(ωγ′ (t))

ds≤∫ b

a

∣∣γ ′ (t)∣∣dt.■

As before, the total variation is denoted by V (γ, [a,b]). The reason I am going throughthe details again and the argument may seem a little complicated is that the function doesnot have values in Rp but in some Banach space.

Theorem 14.4.3 Let f : γ∗→ X be continuous and let γ : [a,b]→ C be continuousand of bounded variation. Then

∫γ

f dz exists. Also letting δ m > 0 be such that |t− s|< δ m

implies ∥ f (γ (t))− f (γ (s))∥< 1m ,∥∥∥∥∫

γ

f dz−S (P)∥∥∥∥≤ 2V (γ, [a,b])

m

whenever ∥P∥ < δ m. In addition to this, if γ has a continuous derivative on [a,b] , thederivative taken from left or right at the endpoints, then∫

γ

f dz =∫ b

af (γ (t))γ

′ (t)dt (14.7)

In general, if φ ∈ X ′, then φ

(∫γ

f dz)=∫

γφ ( f )dz.

Proof: The function, f ◦ γ , is uniformly continuous because it is defined on a compactset. Therefore, there exists a decreasing sequence of positive numbers, {δ m} such that if|s− t|< δ m, then ∥ f (γ (t))− f (γ (s))∥< 1

m . Let

Fm ≡ {S (P) : ∥P∥< δ m}.

Thus Fm is a closed set. (The symbol, S (P) in the above definition, means to include allsums corresponding to P for any choice of τ j.) It is shown that

diam(Fm)≤2V (γ, [a,b])

m(14.8)

and then it will follow there exists a unique point, I ∈ ∩∞m=1Fm. This is because X , the space

where f has its values is complete and Theorem 14.1.1. It will then follow I =∫

γf dz. To

verify 14.8, it suffices to verify that whenever P and Q are partitions satisfying ∥P∥ < δ mand ∥Q∥< δ m,

∥S (P)−S (Q)∥ ≤ 2m

V (γ, [a,b]) . (14.9)

Suppose ∥P∥ < δ m and Q ⊇ P. Then also ∥Q∥ < δ m. To begin with, suppose thatP≡

{t0, · · · , tp, · · · , tn

}and Q≡

{t0, · · · , tp−1, t∗, tp, · · · , tn

}. Thus Q contains only one more

point than P. Letting S (Q) and S (P) be Riemann Steiltjes sums,

S (Q)≡p−1

∑j=1

f (γ (σ j))(γ (t j)− γ

(t j−1

))+ f (γ (σ∗))(γ (t∗)− γ (tp−1))