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WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

2009

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## Ch11: Differentiability on R

n

### Ch11.6: Inverse Function Theorem

Notation:

f :Rn→ Rn,

The Jacobian of f is

f(a) := det(Df (a)).

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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)).

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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 ).

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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).

(6)

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).

(7)

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).

(8)

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).

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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

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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

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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.

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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.

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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.

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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.

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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.

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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.

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

Proof.

If f (x ) = x + 2x2sinx1, 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.

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Remark (11.44)

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

Proof.

If f (x ) = x + 2x2sinx1, 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.

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

Proof.

If f (x ) = x + 2x2sinx1, 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.

(20)

Remark (11.44)

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

Proof.

If f (x ) = x + 2x2sinx1, 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.

(21)

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

Proof.

If f (x ) = x + 2x2sinx1, 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.

(22)

Remark (11.44)

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

Proof.

If f (x ) = x + 2x2sinx1, 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.

(23)

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

Proof.

If f (x ) = x + 2x2sinx1, 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.

(24)

Remark (11.44)

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

Proof.

If f (x ) = x + 2x2sinx1, 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.

(25)

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

∂(F1, . . . ,Fn)

∂(x1, . . . ,xn)(x0,t0) 6=0,

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

(26)

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.

(27)

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 ) =det2u 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.

(28)

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 ) =det2u 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.

(29)

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 ) =det2u 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.

(30)

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 ) =det2u 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.

(31)

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 ) =det2u 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.

(32)

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 ) =det2u 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.

(33)

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 ) =det2u 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.

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## Thank you.

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung