Page 61 - Vector Analysis

§2.6 The Inverse Function Theorem 57

thus at most one x satisfies φy(x) = x; that is, φy has at most one fixed-point. As a
consequence, f : B(x0, δ) Ñ Rn is one-to-one.
Claim: The set V = f (U) is open.

Proof of claim: Let b P V. Then there is a P V with f (a) = b. Choose r ą 0 such that

B(a, r) Ď U. We observe that if y P B(b, λr), then

}φy(a) ´ a}Rn ď }A´1(y ´ ) ď }A´1}B(Rn,Rn)}y ´ b}Rn ă λ}A´1}B(Rn,Rn)r = r ;
f (a) }Rn 2

thus if y P B(b, λr) and x P B(a, r),

}φy(x) ´ a}Rn ď ››φy(x) ´ φy (a)››Rn + }φy(a) ´ a}Rn ă 1 ´ a}Rn + r ă r .
2 }x 2

Therefore, if y P B(b, λr), then φy : B(a, r) Ñ B(a, r). By the continuity of φy,

φy : B(a, r) Ñ B(a, r) .

On the other hand, (2.9) implies that φy is a contraction mapping if y P B(b, λr); thus by the
contraction mapping principle (Proposition 2.60) φy has a unique fixed-point x P B(a, r).
As a result, every y P B(b, λr) corresponds to a unique x P B(a, r) such that φy(x) = x or
equivalently, f (x) = y. Therefore,

()
B(b, λr) Ď f B(a, r) Ď f (U) = V .

Next we show that f ´1 : V Ñ U is differentiable. We note that if x P B(x0, δ),

}(Df )(x0) ´ (Df )(x)}B(Rn,Rn)}A´1}B(Rn,Rn) ă λ}A´1}B(Rn,Rn) = 1 ;
2

thus Theorem 1.87 implies that (Df )(x) is invertible if x P B(x0, δ).
Let b P V and k P Rn such that b + k P V. Then there exists a unique a P U and

h = h(k) P Rn such that a + h P U, b = f (a) and b + k = f (a + h). By the mean value

theorem and (2.9),

››φy(a + h) ´ φy (a)››Rn ă 1 ;
2 }h}Rn

thus the fact that f (a + h) ´ f (a) = k implies that

}h ´ A´1k}Rn ă 1
2 }h}Rn

which further suggests that

1 ď }A´1k}Rn ď }A´1}B(Rn,Rn)}k}Rn ď 1 . (2.10)
2 }h}Rn 2λ }k}Rn