1605

Lemma 50.0.6 Let γ : [a,b]→C be in C1 ([a,b]) . Then V (γ, [a,b])< ∞ so γ is of boundedvariation.

Proof: This follows from the following

n

∑j=1

∣∣γ (t j)− γ(t j−1

)∣∣ =n

∑j=1

∣∣∣∣∫ t j

t j−1

γ′ (s)ds

∣∣∣∣≤

n

∑j=1

∫ t j

t j−1

∣∣γ ′ (s)∣∣ds

≤n

∑j=1

∫ t j

t j−1

∣∣∣∣γ ′∣∣∣∣∞

ds

=∣∣∣∣γ ′∣∣∣∣

∞(b−a) .

Therefore it follows V (γ, [a,b])≤ ||γ ′||∞(b−a) . Here ||γ||

∞= max{|γ (t)| : t ∈ [a,b]}.

Theorem 50.0.7 Let γ : [a,b]→ C be continuous and of bounded variation. Let Ω be anopen set containing γ∗ and let f : Ω×K→ X be continuous for K a compact set in C, andlet ε > 0 be given. Then there exists η : [a,b]→ C such that η (a) = γ (a) , γ (b) = η (b) ,η ∈C1 ([a,b]) , and

||γ−η ||< ε, (50.0.8)∣∣∣∣∫γ

f (·,z)dγ−∫

η

f (·,z)dη

∣∣∣∣< ε, (50.0.9)

V (η , [a,b])≤V (γ, [a,b]) , (50.0.10)

where ||γ−η || ≡max{|γ (t)−η (t)| : t ∈ [a,b]} .

Proof: Extend γ to be defined on allR according to γ (t)= γ (a) if t < a and γ (t)= γ (b)if t > b. Now define

γh (t)≡12h

∫ t+ 2h(b−a) (t−a)

−2h+t+ 2h(b−a) (t−a)

γ (s)ds.

where the integral is defined in the obvious way. That is,∫ b

aα (t)+ iβ (t)dt ≡

∫ b

aα (t)dt + i

∫ b

aβ (t)dt.

Therefore,

γh (b) =1

2h

∫ b+2h

bγ (s)ds = γ (b) ,

γh (a) =1

2h

∫ a

a−2hγ (s)ds = γ (a) .

Also, because of continuity of γ and the fundamental theorem of calculus,

γ′h (t) =

12h

{γ

(t +

2hb−a

(t−a))(

1+2h

b−a

)−