proof of inverse function theorem

proof of inverse function theorem

Since

det

⁡

D

⁢

f

⁢

(

a

)

≠

0

the

Jacobian matrix

D

⁢

f

⁢

(

a

)

is

invertible

:
let

A

=

(

D

⁢

f

⁢

(

a

)

)

-

1

be its

inverse

.
Choose

r

>

0

and

ρ

>

0

such that

B

=

B

ρ

⁢

(

a

)

¯

⊂

E

,

∥

D

⁢

f

⁢

(

x

)

-

D

⁢

f

⁢

(

a

)

∥

≤

1

2

⁢

n

⁢

∥

A

∥

∀

x

∈

B

,

r

≤

ρ

2

⁢

∥

A

∥

.

Let

y

∈

B

r

⁢

(

f

⁢

(

a

)

)

and consider the

mapping

T

y

:

B

→

ℝ

n

T

y

⁢

(

x

)

=

x

+

A

⋅

(

y

-

f

⁢

(

x

)

)

.

If

x

∈

B

we have

∥

D

⁢

T

y

⁢

(

x

)

∥

=

∥

1

-

A

⋅

D

⁢

f

⁢

(

x

)

∥

≤

∥

A

∥

⋅

∥

D

⁢

f

⁢

(

a

)

-

D

⁢

f

⁢

(

x

)

∥

≤

1

2

⁢

n

.

Let us verify that

T

y

is a

contraction mapping

. Given

x

1

,

x

2

∈

B

, by the

Mean-value Theorem

on

ℝ

n

we have

|

T

y

⁢

(

x

1

)

-

T

y

⁢

(

x

2

)

|

≤

sup

x

∈

[

x

1

,

x

2

]

⁡

n

⁢

∥

D

⁢

T

y

⁢

(

x

)

∥

⋅

|

x

1

-

x

2

|

≤

1

2

⁢

|

x

1

-

x

2

|

.

Also notice that

T

y

⁢

(

B

)

⊂

B

. In fact, given

x

∈

B

,

|

T

y

⁢

(

x

)

-

a

|

≤

|

T

y

⁢

(

x

)

-

T

y

⁢

(

a

)

|

+

|

T

y

⁢

(

a

)

-

a

|

≤

1

2

⁢

|

x

-

a

|

+

|

A

⋅

(

y

-

f

⁢

(

a

)

)

|

≤

ρ

2

+

∥

A

∥

⁢

r

≤

ρ

.

So

T

y

:

B

→

B

is a contraction mapping and hence by the contraction principle there exists one and only one solution to the equation

T

y

⁢

(

x

)

=

x

,

i.e.

x

is the only point in

B

such that

f

⁢

(

x

)

=

y

.

Hence given any

y

∈

B

r

⁢

(

f

⁢

(

a

)

)

we can find

x

∈

B

which solves

f

⁢

(

x

)

=

y

. Let us
call

g

:

B

r

⁢

(

f

⁢

(

a

)

)

→

B

the mapping which gives this solution, i.e.

f

⁢

(

g

⁢

(

y

)

)

=

y

.

Let

V

=

B

r

⁢

(

f

⁢

(

a

)

)

and

U

=

g

⁢

(

V

)

. Clearly

f

:

U

→

V

is one to one and the inverse of

f

is

g

. We have to prove that

U

is a

neighbourhood

of

a

.
However since

f

is

continuous

in

a

we know that there exists a ball

B

δ

⁢

(

a

)

such that

f

⁢

(

B

δ

⁢

(

a

)

)

⊂

B

r

⁢

(

y

0

)

and hence
we have

B

δ

⁢

(

a

)

⊂

U

.

We now want to study the differentiability of

g

. Let

y

∈

V

be any point, take

w

∈

ℝ

n

and

ϵ

>

0

so small that

y

+

ϵ

⁢

w

∈

V

.
Let

x

=

g

⁢

(

y

)

and define

v

⁢

(

ϵ

)

=

g

⁢

(

y

+

ϵ

⁢

w

)

-

g

⁢

(

y

)

.

First of all notice that being

|

T

y

⁢

(

x

+

v

⁢

(

ϵ

)

)

-

T

y

⁢

(

x

)

|

≤

1

2

⁢

|

v

⁢

(

ϵ

)

|

we have

1

2

|

v

(

ϵ

)

≥

|

v

(

ϵ

)

-

ϵ

A

⋅

w

|

≥

|

v

(

ϵ

)

|

-

ϵ

∥

A

∥

⋅

|

w

|

and hence

|

v

⁢

(

ϵ

)

|

≤

2

⁢

ϵ

⁢

∥

A

∥

⋅

|

w

|

.

On the other hand we know that

f

is

differentiable

in

x

that is we know that
for all

v

it holds

f

⁢

(

x

+

v

)

-

f

⁢

(

x

)

=

D

⁢

f

⁢

(

x

)

⋅

v

+

h

⁢

(

v

)

with

lim

v

→

0

⁡

h

⁢

(

v

)

/

|

v

|

=

0

. So we get

|

h

⁢

(

v

⁢

(

ϵ

)

)

|

ϵ

≤

2

⁢

∥

A

∥

⋅

|

w

|

⋅

|

h

⁢

(

v

⁢

(

ϵ

)

)

|

v

⁢

(

ϵ

)

→

0

when

⁢

ϵ

→

0

.

So

lim

ϵ

→

0

⁡

g

⁢

(

y

+

ϵ

)

-

g

⁢

(

y

)

ϵ

=

lim

ϵ

→

0

⁡

v

⁢

(

ϵ

)

ϵ

=

lim

ϵ

→

0

⁡

D

⁢

f

⁢

(

x

)

-

1

⋅

ϵ

⁢

w

-

h

⁢

(

v

⁢

(

ϵ

)

)

ϵ

=

D

⁢

f

⁢

(

x

)

-

1

⋅

w

that is

D

⁢

g

⁢

(

y

)

=

D

⁢

f

⁢

(

x

)

-

1

.

Title

proof of inverse function theorem

Canonical name

ProofOfInverseFunctionTheorem

Date of creation

2013-03-22 13:31:20

Last modified on

2013-03-22 13:31:20

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

6

Author

paolini (1187)

Entry type

Proof

Classification

msc 03E20

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