Calculus of Real and Complex Variables

4.2 Estimates and Approximations

The following theorem follows easily from the above definitions and theorem.

Proof: Let 4.7 hold. From the proof of Theorem 4.1.4 on existence, when

< δ_m,

||∫ || 2 || f ⋅dγ − S (P)|| ≤-V (γ,[a,b]) γ m

and so

||∫ || 2 || f ⋅dγ || ≤ |S (P)|+ m-V (γ,[a,b]) γ

Then by the triangle inequality and Cauchy Schwarz inequality,

This proves 4.8 since m is arbitrary. To verify 4.9 use the above inequality to write

||∫ ∫ || ||∫ || || f ⋅dγ − fn ⋅dγ|| = || (f − fn)⋅dγ (t)|| γ γ γ

≤ max {|f ∘γ (t) − fn ∘γ (t)| : t ∈ [a,b]}V (γ,[a,b]).

Since the convergence is assumed to be uniform, this proves 4.9. ■

As an easy example of a curve of bounded variation, here is an easy lemma.

Proof: This follows from the following

Therefore it follows V ≤_∞. Here _∞ = max which exists by Theorem 2.2.25. ■

Proof: Extend γ to be defined on all ℝ according to γ

= γ if t < a and γ = γ if t > b. Now define for 0 ≤ h ≤ 1.

∫ -2h- 1-- t+(b−a)(t−a) γh (t) ≡ 2h −2h+t+(2b−ha)(t−a)γ (s)ds,γ0(t) ≡ γ (t),

where the integral is defined in the obvious way. That is, the j^th component of γ_h

will be

1 ∫ t+ (b2h−a)(t− a) 2h- -2h-- γj(s)ds −2h+t+ (b−a)(t−a)

Note that

→ γ_h is clearly continuous on the compact set × if γ₀ ≡ γ. This is from the fundamental theorem of calculus, Theorem 4.1.7 or its proof, and the observation that

( ) t+ -2h---(t− a)− − 2h + t+--2h--(t− a) = 2h (b − a) (b− a)

Thus this function of two variables is uniformly continuous on this set by Theorem 2.2.27. Also, the definition implies

∫ b+2h γ (b) = 1-- γ (s)ds = γ(b), h 2h b

∫ a γ (a) =-1- γ (s)ds = γ(a). h 2h a−2h

By continuity of γ, the chain rule from beginning calculus, and the fundamental theorem of calculus, Theorem 4.1.7,

{ ( )( ) γ′(t) =-1- γ t+ --2h--(t− a) 1+ -2h-- − h 2h b − a b − a

( ) ( ) } γ − 2h + t+-2h--(t − a ) 1 + -2h-- b− a b− a

and so γ_h ∈ C¹

. Next is a fundamental estimate.

Proof: Let a = t₀ < t₁ <

< t_n = b. Then using the definition of γ_h and changing the variables to make all integrals over ,

∑n |γh(tj) − γh(tj−1)| = j=1

n∑ || ∫ 2h[ ( ) ||1- γ s− 2h + tj + -2h-(tj − a) − j=1 |2h 0 b− a

( ) ]| γ s− 2h +t + -2h--(t − a) || j−1 b − a j−1 |

∫ n | ( ) ≤ -1- 2h∑ ||γ s− 2h+ t + -2h--(t − a) − 2h 0 j=1| j b − a j

| ( -2h-- )|| γ s− 2h + tj−1 + b− a (tj− 1 − a) |ds.

For a given s ∈

, the points, s− 2h + t_j + for j = 1,,n form an increasing list of points in the interval and so the integrand is bounded above by V = V . It follows

∑n |γh (tj)− γh (tj−1)| ≤ V (γ,[a,b])■ j=1

With this lemma the proof of Theorem 4.2.3 can be completed without too much trouble. Let H be an open set containing γ^∗ such that H is a compact subset of Ω. This is easily done as follows. For each x ∈ γ^∗ let B

be a ball centered at x which is contained in Ω. Since γ^∗ is the continuous image of a compact set, it is also compact and so there are finitely many of the balls whose union contains γ^∗. Denote these balls by _i=1ⁿ. Then let H be the union of the balls _i=1ⁿ Thus H is the union of the closures of these balls each of which is compact, so H  is also compact and contained in Ω. Let 0 < ε < min. Then there exists δ₁ such that if h < δ₁, then for all t,

due to the uniform continuity of γ. This proves 4.10. Note that for any γ ∈ γ^∗,γ ∈ Bfor some i and so γ_h, which is less than ε from γ, is contained in B ⊆ H.

From 4.3 and the above lemma, there exists δ₂ such that if

< δ₂, then for all z ∈ K,

| | | | ||∫ || ε ||∫ || ε | γ f (⋅,z)⋅dγ(t)− S (P )| < 3,|| γ f (⋅,z)⋅dγh(t)− Sh(P)|| < 3 (4.14) h

(4.14)

for all h. The reason this can be done is from uniform continuity considerations. As noted,

→ γ_h is uniformly continuous on the compact set × and f is uniformly continuous on the compact set H × K. Thus → f is uniformly continous on

[a,b]× [0,1]× K.

Therefore, there exists δ₂ > 0 such that for all z ∈ K and h ∈

, < if < δ₂. Therefore, from 4.2,

|∫ | || f (⋅,z) ⋅dγ (t)− S(P)||≤ 2V (γh,[a,b])≤ 2V-(γ,[a,b])- (4.15) | γ h | m m

(4.15)

so for large enough choice of m, the two inequalities in 4.14 are obtained. Here S

is a Riemann Steiltjes sum of the form

∑n f (γ(τi),z) ⋅(γ (ti)− γ (ti−1)) i=1

and S_h

is a similar Riemann Steiltjes sum taken with respect to γ_h instead of γ.

Because of 4.13, γ_h

has values in H ⊆ Ω. Therefore, fix the partition P, and choose h small enough that in addition to this, the following inequality is valid for all z ∈ K.

ε |S (P )− Sh(P )| < - 3

This is possible because of 4.13 and the uniform continuity of f on H×K. It follows from 4.14 that for h this small,

| | ||∫ ∫ || || f (⋅,z)⋅dγ (t)− f (⋅,z)⋅dγh(t)|| ≤ γ γh

|∫ | || f (⋅,z)⋅dγ(t)− S(P )||+ |S(P )− S (P)| | γ | h

| ∫ | || || + ||Sh (P)− γ f (⋅,z)⋅dγh(t)|| < ε. h

< ε + ε+ ε= ε ■ 3 3 3

Of course the same result is obtained without the explicit dependence of f on z.

This is a very useful theorem because if γ is C¹

, it is easy to calculate ∫ _γf ⋅ dγ and the above theorem allows a reduction to the case where γ is C¹. The next theorem shows how easy it is to compute these integrals in the case where γ is C¹.

Proof: Let P be a partition of

, P = and is small enough that whenever < ,

|f (γ(t)) − f (γ (s))| < ε (4.17)

(4.17)

and

| | ||∫ ∑n || || f ⋅dγ − f (γ (τj))⋅(γ (tj)− γ(tj−1))||< ε. |γ j=1 |

Now

n ∫ n ∑ f (γ(τ )) ⋅(γ (t)− γ (t )) = b∑ f (γ(τ ))X (s)⋅γ′(s) ds j=1 j j j−1 a j=1 j [tj−1,tj]