Advanced Calculus (II)

Advanced Calculus (II)

W^EN-C^HINGL^IEN

Department of Mathematics National Cheng Kung University

2009

Ch11: Differentiability on R

Ch11.6: Inverse Function Theorem

Notation:

f :Rⁿ→ Rⁿ,

The Jacobian of f is

∆f(a) := det(Df (a)).

Let V be open inRⁿ, f : V →Rⁿ,a ∈ V, and r > 0 be so small that Br(a) ⊂ V . Suppose that f is continuous and 1-1 on Br(a), and its first-order partial derivatives exist at every point in Br(a). If ∆f 6= 0 on Br(a), then there is a ρ >0 such that Bρ(f (a)) ⊂ f (Br(a)).

Theorem (11.39)

Let V be open and nonempty inRⁿ, and f : V →Rⁿ be continuous. If f is 1-1 and has first-order partial

derivatives on V , and if ∆f 6= 0 on V , then f⁻¹is continuous on f (V ).

Let V be open inRⁿ and f : V →Rⁿ be C¹ on V . If

∆f(a) 6= 0 for some a ∈ V , then there is an r > 0 such that Br(a) ⊂ V , f is 1-1 on Br(a), ∆f(x) 6= 0 for all x ∈ Br(a), and

det ∂fi

∂xj

(cj)



n×n

6= 0 for allc1, . . . ,cn ∈ Br(a).

Theorem (11.41 Inverse Function Theorem) Let V be open inRⁿ and f : V →Rⁿ be C¹ on V . If

∆f(a) 6= 0 for some a ∈ V , then there exists an open set W containinga such that

(i) f is 1-1 on W ,

(ii) f⁻¹is C¹on f (W ), and (iii) for eachy ∈ f (W ),

D(f⁻¹)(y) = [Df (f⁻¹(y))]⁻¹,

where [ ]⁻¹ represents matrix inversion (see Theorem C.5).

Let V be open inRⁿ and f : V →Rⁿ be C¹ on V . If

∆f(a) 6= 0 for some a ∈ V , then there exists an open set W containinga such that

(i) f is 1-1 on W ,

(ii) f⁻¹is C¹on f (W ), and (iii) for eachy ∈ f (W ),

D(f⁻¹)(y) = [Df (f⁻¹(y))]⁻¹,

where [ ]⁻¹ represents matrix inversion (see Theorem C.5).

Theorem (11.41 Inverse Function Theorem) Let V be open inRⁿ and f : V →Rⁿ be C¹ on V . If

∆f(a) 6= 0 for some a ∈ V , then there exists an open set W containinga such that

(i) f is 1-1 on W ,

(ii) f⁻¹is C¹on f (W ), and (iii) for eachy ∈ f (W ),

D(f⁻¹)(y) = [Df (f⁻¹(y))]⁻¹,

where [ ]⁻¹ represents matrix inversion (see Theorem C.5).

The hypothesis ”∆f 6= 0 in Theorem 11.39 can be relaxed.

Proof.

If f (x ) = x³, then f :R → R and its inverse f⁻¹(x ) =√³ x are continuous onR, but ∆f(0) = f⁰(0) = 0

Remark (11.42)

The hypothesis ”∆f 6= 0 in Theorem 11.39 can be relaxed.

Proof.

If f (x ) = x³, then f :R → R and its inverse f⁻¹(x ) =√³ x are continuous onR, but ∆f(0) = f⁰(0) = 0

The hypothesis ”∆f 6= 0 in Theorem 11.41 cannot be relaxed. In fact, if f : Br(a) → Rⁿ is differentiable ata and its inverse f⁻¹exists and is differentiable at f (a), then

∆_f(a) 6= 0.

Proof.

Suppose to the contrary that f is differentiable ata but

∆f(a) = 0. By Exercise 8, p.338, and the Chain Rule, I =D(f⁻¹◦ f )(a) = D(f⁻¹)(f (a))Df (a).

Taking the determinant of this identity, we have 1 = ∆_f−1(f (a))∆f(a) = 0, a contradiction.

Remark (11.43)

The hypothesis ”∆f 6= 0 in Theorem 11.41 cannot be relaxed. In fact, if f : Br(a) → Rⁿ is differentiable ata and its inverse f⁻¹exists and is differentiable at f (a), then

∆_f(a) 6= 0.

Proof.

Suppose to the contrary that f is differentiable ata but

∆f(a) = 0.By Exercise 8, p.338, and the Chain Rule, I = D(f⁻¹◦ f )(a)=D(f⁻¹)(f (a))Df (a).

Taking the determinant of this identity, we have 1 = ∆_f−1(f (a))∆f(a) = 0, a contradiction.

The hypothesis ”∆f 6= 0 in Theorem 11.41 cannot be relaxed. In fact, if f : Br(a) → Rⁿ is differentiable ata and its inverse f⁻¹exists and is differentiable at f (a), then

∆_f(a) 6= 0.

Proof.

Suppose to the contrary that f is differentiable ata but

∆f(a) = 0. By Exercise 8, p.338, and the Chain Rule, I =D(f⁻¹◦ f )(a) = D(f⁻¹)(f (a))Df (a).

Taking the determinant of this identity, we have 1 = ∆_f−1(f (a))∆f(a) = 0, a contradiction.

Remark (11.43)

The hypothesis ”∆f 6= 0 in Theorem 11.41 cannot be relaxed. In fact, if f : Br(a) → Rⁿ is differentiable ata and its inverse f⁻¹exists and is differentiable at f (a), then

∆_f(a) 6= 0.

Proof.

Suppose to the contrary that f is differentiable ata but

∆f(a) = 0. By Exercise 8, p.338, and the Chain Rule, I = D(f⁻¹◦ f )(a)=D(f⁻¹)(f (a))Df (a).

Taking the determinant of this identity, we have 1 = ∆_f−1(f (a))∆f(a) = 0, a contradiction.

The hypothesis ”∆f 6= 0 in Theorem 11.41 cannot be relaxed. In fact, if f : Br(a) → Rⁿ is differentiable ata and its inverse f⁻¹exists and is differentiable at f (a), then

∆_f(a) 6= 0.

Proof.

Suppose to the contrary that f is differentiable ata but

∆f(a) = 0. By Exercise 8, p.338, and the Chain Rule, I = D(f⁻¹◦ f )(a) = D(f⁻¹)(f (a))Df (a).

Taking the determinant of this identity, we have 1 = ∆_f−1(f (a))∆f(a) = 0, a contradiction.

Remark (11.43)

The hypothesis ”∆f 6= 0 in Theorem 11.41 cannot be relaxed. In fact, if f : Br(a) → Rⁿ is differentiable ata and its inverse f⁻¹exists and is differentiable at f (a), then

∆_f(a) 6= 0.

Proof.

Suppose to the contrary that f is differentiable ata but

∆f(a) = 0. By Exercise 8, p.338, and the Chain Rule, I = D(f⁻¹◦ f )(a) = D(f⁻¹)(f (a))Df (a).

Taking the determinant of this identity, we have 1 = ∆_f−1(f (a))∆f(a) = 0, a contradiction.

The hypothesis ”f is C¹ on V ” in Theorem 11.41 cannot be relaxed.

Proof.

If f (x ) = x + 2x²sin_x¹, x 6= 0, and f (0) = 0, then f :R → R is differentiable on V := (−1, 1) and f⁰(0) = 1 6= 0.

However, since

f

 2

(4k − 1)π



<f

 2

(4k + 1)π



<f

 2

(4k − 3)π



for k ∈N, f is not 1-1 on any open set that contains 0.

Therefore, no open subset of f (V ) can be closen on which f⁻¹exists.

Remark (11.44)

The hypothesis ”f is C¹ on V ” in Theorem 11.41 cannot be relaxed.

Proof.

If f (x ) = x + 2x²sin_x¹, x 6= 0, and f (0) = 0,then f :R → R is differentiable on V := (−1, 1) and f⁰(0) = 1 6= 0.

However, since f

 2

(4k − 1)π



<f

 2

(4k + 1)π



<f

 2

(4k − 3)π



for k ∈N, f is not 1-1 on any open set that contains 0.

Therefore, no open subset of f (V ) can be closen on which f⁻¹exists.

The hypothesis ”f is C¹ on V ” in Theorem 11.41 cannot be relaxed.

Proof.

If f (x ) = x + 2x²sin_x¹, x 6= 0, and f (0) = 0, then f :R → R is differentiable on V := (−1, 1) and f⁰(0) = 1 6= 0.

However, since f

 2

(4k − 1)π



<f

 2

(4k + 1)π



<f

 2

(4k − 3)π



for k ∈N, f is not 1-1 on any open set that contains 0.

Therefore, no open subset of f (V ) can be closen on which f⁻¹exists.

Remark (11.44)

The hypothesis ”f is C¹ on V ” in Theorem 11.41 cannot be relaxed.

Proof.

If f (x ) = x + 2x²sin_x¹, x 6= 0, and f (0) = 0, then f :R → R is differentiable on V := (−1, 1) and f⁰(0) = 1 6= 0.

However, since f

 2

(4k − 1)π



<f

 2

(4k + 1)π



<f

 2

(4k − 3)π



for k ∈N,f is not 1-1 on any open set that contains 0.

Therefore, no open subset of f (V ) can be closen on which f⁻¹exists.

The hypothesis ”f is C¹ on V ” in Theorem 11.41 cannot be relaxed.

Proof.

If f (x ) = x + 2x²sin_x¹, x 6= 0, and f (0) = 0, then f :R → R is differentiable on V := (−1, 1) and f⁰(0) = 1 6= 0.

However, since f

 2

(4k − 1)π



<f

 2

(4k + 1)π



<f

 2

(4k − 3)π



for k ∈N, f is not 1-1 on any open set that contains 0.

Therefore, no open subset of f (V ) can be closen on which f⁻¹exists.

Remark (11.44)

The hypothesis ”f is C¹ on V ” in Theorem 11.41 cannot be relaxed.

Proof.

If f (x ) = x + 2x²sin_x¹, x 6= 0, and f (0) = 0, then f :R → R is differentiable on V := (−1, 1) and f⁰(0) = 1 6= 0.

However, since f

 2

(4k − 1)π



<f

 2

(4k + 1)π



<f

 2

(4k − 3)π



for k ∈N,f is not 1-1 on any open set that contains 0.

Therefore, no open subset of f (V ) can be closen on which f⁻¹exists.

The hypothesis ”f is C¹ on V ” in Theorem 11.41 cannot be relaxed.

Proof.

If f (x ) = x + 2x²sin_x¹, x 6= 0, and f (0) = 0, then f :R → R is differentiable on V := (−1, 1) and f⁰(0) = 1 6= 0.

However, since f

 2

(4k − 1)π



<f

 2

(4k + 1)π



<f

 2

(4k − 3)π



for k ∈N, f is not 1-1 on any open set that contains 0.

Therefore, no open subset of f (V ) can be closen on which f⁻¹exists.

Remark (11.44)

The hypothesis ”f is C¹ on V ” in Theorem 11.41 cannot be relaxed.

Proof.

If f (x ) = x + 2x²sin_x¹, x 6= 0, and f (0) = 0, then f :R → R is differentiable on V := (−1, 1) and f⁰(0) = 1 6= 0.

However, since f

 2

(4k − 1)π



<f

 2

(4k + 1)π



<f

 2

(4k − 3)π



for k ∈N, f is not 1-1 on any open set that contains 0.

Therefore, no open subset of f (V ) can be closen on which f⁻¹exists.

Suppose that V is open inR^n+p, and F = (F1, . . . ,Fn) : V →Rⁿis C¹on V . Suppose further that F (x0,t0) = 0 for some (x0,t0) ∈V , wherex0∈ Rⁿandt0 ∈ R^p. If

∂(F1, . . . ,Fn)

∂(x1, . . . ,xn)(x0,t0) 6=0,

then there is an open set W ⊂R^p containingt0and a unique continuously differentiable function g : W →Rⁿ such that g(t0) = x0, and F (g(t), t) = 0 for all t ∈ W .

Example (11.49)

Prove that there exist function u, v :R⁴→ R, continuously differentiable on some ball B centered at the point

(x , y , z, w ) = (2, 1, −1, −2), such that

u(2, 1, −1, −2) = 4, v (2, 1, −1, −2) = 3, and the equations

u²+v²+w²=29, u² x² + v²

y² +w² z² =17 both hold for all (x , y , z, w ) in B.

Set n = 2, p = 4, and

F (u, v , x , y , z, w )= (u²+v²+w²− 29,u² x²+v²

y²+w²

z² − 17).

Then F (4, 3, 2, 1, −1, −2) = (0, 0), and

∂(F1,F2)

∂(u, v ) =det2u 2v

2u x²

2v y²



=4uv 1 y² − 1

x²

 .

This determinant is nonzero when u = 4, v = 3, x = 2, and y = 1. Therefore, such functions u, v exist by the Implicit Function Theorem.

Proof.

Set n = 2, p = 4, and

F (u, v , x , y , z, w ) = (u²+v²+w²− 29,u² x²+v²

y²+w²

z² − 17).

Then F (4, 3, 2, 1, −1, −2) = (0, 0), and

∂(F1,F2)

∂(u, v ) =det2u 2v

2u x²

2v y²



=4uv 1 y² − 1

x²

 .

This determinant is nonzero when u = 4, v = 3, x = 2, and y = 1. Therefore, such functions u, v exist by the Implicit Function Theorem.

Set n = 2, p = 4, and

F (u, v , x , y , z, w ) = (u²+v²+w²− 29,u² x²+v²

y²+w²

z² − 17).

Then F (4, 3, 2, 1, −1, −2) = (0, 0), and

∂(F1,F2)

∂(u, v ) =det2u 2v

2u x²

2v y²



=4uv 1 y² − 1

x²

 . This determinant is nonzero when u = 4, v = 3, x = 2, and y = 1. Therefore, such functions u, v exist by the Implicit Function Theorem.

Proof.

Set n = 2, p = 4, and

F (u, v , x , y , z, w ) = (u²+v²+w²− 29,u² x²+v²

y²+w²

z² − 17).

Then F (4, 3, 2, 1, −1, −2) = (0, 0), and

∂(F1,F2)

∂(u, v ) =det2u 2v

2u x²

2v y²



=4uv 1 y² − 1

x²

 . This determinant is nonzero when u = 4, v = 3, x = 2, and y = 1.Therefore, such functions u, v exist by the Implicit Function Theorem.

Set n = 2, p = 4, and

F (u, v , x , y , z, w ) = (u²+v²+w²− 29,u² x²+v²

y²+w²

z² − 17).

Then F (4, 3, 2, 1, −1, −2) = (0, 0), and

∂(F1,F2)

∂(u, v ) =det2u 2v

2u x²

2v y²



=4uv 1 y² − 1

x²

 . This determinant is nonzero when u = 4, v = 3, x = 2, and y = 1. Therefore, such functions u, v exist by the Implicit Function Theorem.

Proof.

Set n = 2, p = 4, and

F (u, v , x , y , z, w ) = (u²+v²+w²− 29,u² x²+v²

y²+w²

z² − 17).

Then F (4, 3, 2, 1, −1, −2) = (0, 0), and

∂(F1,F2)

∂(u, v ) =det2u 2v

2u x²

2v y²



=4uv 1 y² − 1

x²

 . This determinant is nonzero when u = 4, v = 3, x = 2, and y = 1.Therefore, such functions u, v exist by the Implicit Function Theorem.

Set n = 2, p = 4, and

F (u, v , x , y , z, w ) = (u²+v²+w²− 29,u² x²+v²

y²+w²

z² − 17).

Then F (4, 3, 2, 1, −1, −2) = (0, 0), and

∂(F1,F2)

∂(u, v ) =det2u 2v

2u x²

2v y²



=4uv 1 y² − 1

x²

 . This determinant is nonzero when u = 4, v = 3, x = 2, and y = 1. Therefore, such functions u, v exist by the Implicit Function Theorem.

Thank you.