Inverse function theorem: Difference between revisions

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Here is a proof based on the [[contraction mapping theorem]]. Specifically, following T. Tao,<ref>Theorem 17.7.2 in {{cite book|mr=3310023|last1=Tao|first1=Terence|title=Analysis. II|edition=Third edition of 2006 original|series=Texts and Readings in Mathematics|volume=38|publisher=Hindustan Book Agency|___location=New Delhi|year=2014|isbn=978-93-80250-65-6|zbl=1300.26003}}</ref> it uses the following consequence of the contraction mapping theorem.

{{math_theorem|name=Lemma|math_statement=Let <math>B(0, r)</math> denote an open ball of radius ''r'' in <math>\mathbb{R}^n</math> with center 0. If <math>g : B(0, r) \to \mathbb{R}^n</math> is a map such that <math>g(0) = 0</math> and there exists a constant <math>0 < c < 1</math> such that

:<math>|g(y) - g(x)| \le c|y-x|</math>

for all <math>x, y</math> in <math>B(0, r)</math>, then for <math>f = I + g</math> is injective on <math>B(0, r)</math>, andwe have
:<math>B(0, (1-c)r)|x - y| \subsetle |f(B(0, r)x) \subset- Bf(0y)|, (1+c)r)</math>.

in particular, ''f'' is injective. If, moreover, <math>g(0) = 0</math>, then <math>B(0, (1-c)r) \subset f(B(0, r)) \subset B(0, (1+c)r)</math>.

(More generally, the statement remains true if <math>\mathbb{R}^n</math> is replaced by a Banach space.) Also, the first part of the lemma is true for any normed space.}}

Basically, the lemma says that a small perturbation of the identity map by a contraction map is injective and preserves a ball in some sense. Assuming the lemma for a moment, we prove the theorem first. As in the above proof, it is enough to prove the special case when <math>a = 0, b = f(a) = 0</math> and <math>f'(0) = I</math>. Let <math>g = f - I</math>. The [[mean value inequality]] applied to <math>t \mapsto g(x + t(y - x))</math> says:

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As <math>k \to 0</math>, we have <math>h \to 0</math> and <math>|h|/|k|</math> is bounded. Hence, <math>g</math> is differentiable at <math>y</math> with the derivative <math>g'(y) = f'(g(y))^{-1}</math>. Also, <math>g'</math> is the same as the composition <math>\iota \circ f' \circ g</math> where <math>\iota : T \mapsto T^{-1}</math>; so <math>g'</math> is continuous.

It remains to show the lemma. First, thewe map <math>f</math> is injective on <math>B(0, r)</math> since if <math>f(x) = f(y)</math>, then <math>g(y) - g(x) = x - y</math> and sohave:

:<math>|g(x - y| - |f(x) - f(y)| \le |g(x) - g(y)| =\le c|yx - xy|,</math>,

which is to say

which is a contradiction unless <math>y = x</math>. (This part does not need the assumption <math>g(0) = 0</math>.) Next we show <math>f(B(0, r)) \supset B(0, (1-c)r)</math>. The idea is to note that this is equivalent to, given a point <math>y</math> in <math>B(0, (1-c) r)</math>, find a fixed point of the map

:<math>(1 - c)|x - y| \le |f(x) - f(y)|.</math>

which is a contradiction unless <math>y = x</math>. (This part does not needproves the assumptionfirst <math>g(0) = 0</math>part.) Next, we show <math>f(B(0, r)) \supset B(0, (1-c)r)</math>. The idea is to note that this is equivalent to, given a point <math>y</math> in <math>B(0, (1-c) r)</math>, find a fixed point of the map

:<math>F : \overline{B}(0, r') \to \overline{B}(0, r'), \, x \mapsto y - g(x)</math>

where <math>0 < r' < r</math> such that <math>|y| \le (1-c)r'</math> and the bar means a closed ball. To find a fixed point, we use the contraction mapping theorem and checking that <math>F</math> is a well-defined strict-contraction mapping is straightforward. Finally, we have: <math>f(B(0, r)) \subset B(0, (1+c)r)</math> since