## Advanced Calculus (II)

W^{EN}-C^{HING}L^{IEN}

Department of Mathematics National Cheng Kung University

2009

## Ch11: Differentiability on **R**

^{n}

### Ch11.6: Inverse Function Theorem

Notation:

f :**R**^{n}**→ R**^{n},

The Jacobian of f is

∆f(a) := det(Df (a)).

Let V be open in**R**^{n}, f : V →**R**^{n},**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 in**R**^{n}, and f : V →**R**^{n} be
continuous. If f is 1-1 and has first-order partial

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

Let V be open in**R**^{n} and f : V →**R**^{n} be C^{1} 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 all**c**1, . . . ,**c**n ∈ Br(a).

Theorem (11.41 Inverse Function Theorem)
Let V be open in**R**^{n} and f : V →**R**^{n} be C^{1} on V . If

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

(i) f is 1-1 on W ,

(ii) f^{−1}is C^{1}on f (W ), and
(iii) for each**y ∈ f (W ),**

D(f^{−1})(y) = [Df (f^{−1}(y))]^{−1},

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

Let V be open in**R**^{n} and f : V →**R**^{n} be C^{1} on V . If

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

(i) f is 1-1 on W ,

(ii) f^{−1}is C^{1}on f (W ), and
(iii) for each**y ∈ 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 in**R**^{n} and f : V →**R**^{n} be C^{1} on V . If

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

(i) f is 1-1 on W ,

(ii) f^{−1}is C^{1}on f (W ), and
(iii) for each**y ∈ 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 ) = x^{3}, then f :**R → R and its inverse f**^{−1}(x ) =√^{3}
x
are continuous on**R, but ∆**f(0) = f^{0}(0) = 0

Remark (11.42)

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

Proof.

If f (x ) = x^{3}, then f :**R → R and its inverse f**^{−1}(x ) =√^{3}
x
are continuous on**R, but ∆**f(0) = f^{0}(0) = 0

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

∆_{f}(a) 6= 0.

Proof.

Suppose to the contrary that f is differentiable at**a 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) → R^{n} is differentiable at**a and**
its inverse f^{−1}exists and is differentiable at f (a), then

∆_{f}(a) 6= 0.

Proof.

Suppose to the contrary that f is differentiable at**a 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) → R^{n} is differentiable at**a and**
its inverse f^{−1}exists and is differentiable at f (a), then

∆_{f}(a) 6= 0.

Proof.

Suppose to the contrary that f is differentiable at**a 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)

^{n} is differentiable at**a and**
its inverse f^{−1}exists and is differentiable at f (a), then

∆_{f}(a) 6= 0.

Proof.

Suppose to the contrary that f is differentiable at**a 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.

^{n} is differentiable at**a and**
its inverse f^{−1}exists and is differentiable at f (a), then

∆_{f}(a) 6= 0.

Proof.

Suppose to the contrary that f is differentiable at**a 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)

^{n} is differentiable at**a and**
its inverse f^{−1}exists and is differentiable at f (a), then

∆_{f}(a) 6= 0.

Proof.

Suppose to the contrary that f is differentiable at**a 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 C^{1} on V ” in Theorem 11.41 cannot be
relaxed.

Proof.

If f (x ) = x + 2x^{2}sin_{x}^{1}, x 6= 0, and f (0) = 0, then f :**R → R**
is differentiable on V := (−1, 1) and f^{0}(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^{−1}exists.

Remark (11.44)

The hypothesis ”f is C^{1} on V ” in Theorem 11.41 cannot be
relaxed.

Proof.

If f (x ) = x + 2x^{2}sin_{x}^{1}, x 6= 0, and f (0) = 0,then f :**R → R**
is differentiable on V := (−1, 1) and f^{0}(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^{−1}exists.

The hypothesis ”f is C^{1} on V ” in Theorem 11.41 cannot be
relaxed.

Proof.

If f (x ) = x + 2x^{2}sin_{x}^{1}, x 6= 0, and f (0) = 0, then f :**R → R**
is differentiable on V := (−1, 1) and f^{0}(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^{−1}exists.

Remark (11.44)

The hypothesis ”f is C^{1} on V ” in Theorem 11.41 cannot be
relaxed.

Proof.

If f (x ) = x + 2x^{2}sin_{x}^{1}, x 6= 0, and f (0) = 0, then f :**R → R**
is differentiable on V := (−1, 1) and f^{0}(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^{−1}exists.

The hypothesis ”f is C^{1} on V ” in Theorem 11.41 cannot be
relaxed.

Proof.

^{2}sin_{x}^{1}, x 6= 0, and f (0) = 0, then f :**R → R**
is differentiable on V := (−1, 1) and f^{0}(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^{−1}exists.

Remark (11.44)

The hypothesis ”f is C^{1} on V ” in Theorem 11.41 cannot be
relaxed.

Proof.

^{2}sin_{x}^{1}, x 6= 0, and f (0) = 0, then f :**R → R**
is differentiable on V := (−1, 1) and f^{0}(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^{−1}exists.

The hypothesis ”f is C^{1} on V ” in Theorem 11.41 cannot be
relaxed.

Proof.

^{2}sin_{x}^{1}, x 6= 0, and f (0) = 0, then f :**R → R**
is differentiable on V := (−1, 1) and f^{0}(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^{−1}exists.

Remark (11.44)

The hypothesis ”f is C^{1} on V ” in Theorem 11.41 cannot be
relaxed.

Proof.

^{2}sin_{x}^{1}, x 6= 0, and f (0) = 0, then f :**R → R**
is differentiable on V := (−1, 1) and f^{0}(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^{−1}exists.

Suppose that V is open in**R**^{n+p}, and F = (F1, . . . ,Fn) :
V →**R**^{n}is C^{1}on V . Suppose further that F (x0,**t**0) = **0 for**
some (x0,**t**0) ∈V , where**x**0**∈ R**^{n}and**t**0 **∈ R**^{p}. If

∂(F1, . . . ,Fn)

∂(x1, . . . ,xn)(x0,**t**0) 6=0,

then there is an open set W ⊂**R**^{p} containing**t**0and a
unique continuously differentiable function g : W →**R**^{n}
such that g(t0) = **x**0, and F (g(t), t) = 0 for all t ∈ W .

Example (11.49)

Prove that there exist function u, v :**R**^{4}**→ 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

u^{2}+v^{2}+w^{2}=29, u^{2}
x^{2} + v^{2}

y^{2} +w^{2}
z^{2} =17
both hold for all (x , y , z, w ) in B.

Set n = 2, p = 4, and

F (u, v , x , y , z, w )= (u^{2}+v^{2}+w^{2}− 29,u^{2}
x^{2}+v^{2}

y^{2}+w^{2}

z^{2} − 17).

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

∂(F1,F2)

∂(u, v ) =det2u 2v

2u
x^{2}

2v
y^{2}

=4uv 1
y^{2} − 1

x^{2}

.

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 ) = (u^{2}+v^{2}+w^{2}− 29,u^{2}
x^{2}+v^{2}

y^{2}+w^{2}

z^{2} − 17).

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

∂(F1,F2)

∂(u, v ) =det2u 2v

2u
x^{2}

2v
y^{2}

=4uv 1
y^{2} − 1

x^{2}

.

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 ) = (u^{2}+v^{2}+w^{2}− 29,u^{2}
x^{2}+v^{2}

y^{2}+w^{2}

z^{2} − 17).

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

∂(F1,F2)

∂(u, v ) =det2u 2v

2u
x^{2}

2v
y^{2}

=4uv 1
y^{2} − 1

x^{2}

. 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 ) = (u^{2}+v^{2}+w^{2}− 29,u^{2}
x^{2}+v^{2}

y^{2}+w^{2}

z^{2} − 17).

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

∂(F1,F2)

∂(u, v ) =det2u 2v

2u
x^{2}

2v
y^{2}

=4uv 1
y^{2} − 1

x^{2}

. 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 ) = (u^{2}+v^{2}+w^{2}− 29,u^{2}
x^{2}+v^{2}

y^{2}+w^{2}

z^{2} − 17).

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

∂(F1,F2)

∂(u, v ) =det2u 2v

2u
x^{2}

2v
y^{2}

=4uv 1
y^{2} − 1

x^{2}

. 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 ) = (u^{2}+v^{2}+w^{2}− 29,u^{2}
x^{2}+v^{2}

y^{2}+w^{2}

z^{2} − 17).

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

∂(F1,F2)

∂(u, v ) =det2u 2v

2u
x^{2}

2v
y^{2}

=4uv 1
y^{2} − 1

x^{2}

. 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 ) = (u^{2}+v^{2}+w^{2}− 29,u^{2}
x^{2}+v^{2}

y^{2}+w^{2}

z^{2} − 17).

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

∂(F1,F2)

∂(u, v ) =det2u 2v

2u
x^{2}

2v
y^{2}

=4uv 1
y^{2} − 1

x^{2}

. 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.