5.2. ESTIMATES AND APPROXIMATIONS 121

Lemma 5.2.2 Let γ : [a,b] → Rp be in C1 ([a,b]) . Then V (γ, [a,b]) < ∞ so γ is ofbounded variation.

Proof: You can use Proposition 5.1.7 in the first inequality.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]}

which exists by Theorem 2.5.26. ■

Theorem 5.2.3 Let γ : [a,b]→ Rp be continuous and of bounded variation. Let Ω

be an open set containing γ∗ and let f : Ω→ Rp be continuous, and let ε > 0 be given.Then there exists η : [a,b]→Rp such that η (a) = γ (a) , γ (b) = η (b) , η ∈C1 ([a,b]) , and

∥γ−η∥ ≤ ε, ∥γ−η∥ ≡max{|γ (t)−η (t)| : t ∈ [a,b]} . (5.11)∣∣∣∣∫γ

f ·dγ−∫

η

fdη

∣∣∣∣< ε, all z ∈ K (5.12)

V (η , [a,b])≤V (γ, [a,b]) , (5.13)

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 for 0≤ h≤ 1.

γh (t)≡12h

∫ t+ 2h(b−a) (t−a)

−2h+t+ 2h(b−a) (t−a)

γ (s)ds, γ0 (t)≡ γ (t) ,

where the integral is defined in the obvious way. That is, the jth component of γh (t)

will be 12h∫ t+ 2h

(b−a) (t−a)

−2h+t+ 2h(b−a) (t−a)

γ j (s)ds. Note that (t,h)→ γh (t) is clearly continuous on the

compact set [a,b]× [0,1] if γ0 (0)≡ γ (t). This is from the fundamental theorem of calculus,Theorem 5.1.8, and the observation that

t +2h

(b−a)(t−a)−

(−2h+ t +

2h(b−a)

(t−a))= 2h

Thus (t,h) → γh (t) is uniformly continuous on this set by Theorem 2.5.28. Also, thedefinition implies

γh (b) =12h

∫ b+2h

bγ (s)ds = γ (b) , γh (a) =

12h

∫ a

a−2hγ (s)ds = γ (a) .

By continuity of γ, the chain rule from beginning calculus, and the fundamental theoremof calculus, Theorem 5.1.8,γ ′h (t) =

12h

{γ

(t +

2hb−a

(t−a))(

1+2h

b−a

)−

γ

(−2h+ t +

2hb−a

(t−a))(

1+2h

b−a

)}