Page 60 - Vector Analysis

56 CHAPTER 2. Differentiation of Functions of Several Variables

Suppose that for some x, y P M , Φ(x) = x and Φ(y) = y. Then

}x ´ y}Rn = ››Φ(x) ´ Φ(y)››Rn ď θ}x ´ y}Rn ˝
which suggests that }x ´ y}Rn = 0 or x = y. Therefore, the fixed-point of Φ is unique.

Now we state and prove the inverse function theorem.

Theorem 2.61 (Inverse Function Theorem). Let D Ď Rn be open, x0 P D, f : D Ñ Rn be
of class C 1, and (Df )(x0) be invertible. Then there exist an open neighborhood U of x0 and
an open neighborhood V of f (x0) such that

1. f : U Ñ V is one-to-one and onto;

2. The inverse function f ´1 : V Ñ U is of class C 1;

3. If x = f ´1(y), then (Df ´1)(y) = ( )(x))´1;
(Df

4. If f is of class C r for some r ą 1, so is f ´1.

Proof. We will omit the proof of 4 since it requires more knowledge about differentiation.
Assume that A = (Df )(x0). Then }A´1}B(Rn,Rn) ‰ 0. Choose λ ą 0 such that

2λ}A´1}B(Rn,Rn) = 1. Since f P C 1, there exists δ ą 0 such that

››(Df )(x) ´ A››B(Rn,Rn) = ››(Df )(x) ´ (Df )(x0)››B(Rn,Rn) ă λ whenever x P B(x0, δ) X D .

By choosing δ even smaller if necessary, we can assume that B(x0, δ) Ď D. Let U = B(x0, δ).

Claim: f : U Ñ Rn is one-to-one (hence f :U Ñ f (U) is one-to-one and onto).
Proof of claim: For each y P Rn, define φy (x) () we note
= x + A´1 y ´ f (x) (and that every

fixed-point of φy corresponds to a solution to f (x) = y). Then

(Dφy)(x) = Id ´ A´1(Df )(x) = A´1(A ´ ) ,
(Df )(x)

where Id is the identity map on Rn. Therefore,

››(Dφy )(x)››B(Rn ,Rn ) ď }A´1}B(Rn,Rn)››A ´ (Df )(x)››B(Rn,Rn) ă 1 @ x P B(x0, δ) .
2

By the mean value theorem (Theorem 2.55),

››φy (x1 ) ´ φy (x2 )››Rn ď 1 ´ x2}Rn @ x1, x2 P B(x0, δ), x1 ‰ x2 ; (2.9)
2 }x1