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{{Use dmy dates|date=December 2023}}
{{Calculus}}
In [[real analysis]], a branch of [[mathematics]], the '''inverse function theorem''' is a [[theorem]] that asserts that, if a [[real function]] ''f'' has a [[continuously differentiable function|continuous derivative]] near a point where its derivative is nonzero, then, near this point, ''f'' has an [[inverse function]]. The inverse function is also [[differentiable function|differentiable]], and the ''[[inverse function rule]]'' expresses its derivative as the [[multiplicative inverse]] of the derivative of ''f''.
The theorem applies verbatim to [[complex-valued function]]s of a [[complex number|complex variable]]. It generalizes to functions from
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=== Proof for single-variable functions ===
We want to prove the following: ''Let <math>D \subseteq \R</math> be an open set with <math>x_0 \in D, f: D \to \R</math> a continuously differentiable function defined on <math>D</math>, and suppose that <math>f'(x_0) \ne 0</math>. Then there exists an open interval <math>I</math> with <math>x_0 \in I</math> such that <math>f</math> maps <math>I</math> bijectively onto the open interval <math>J = f(I)</math>, and such that the inverse function <math>f^{-1} : J \to I</math> is continuously differentiable, and for any <math>y \in J</math>, if <math>x \in I</math> is such that <math>f(x) = y</math>, then <math>(f^{-1})'(y) = \dfrac{1}{f'(x)}</math>.''
We may without loss of generality assume that <math>f'(x_0) > 0</math>. Given that <math>D</math> is an open set and <math>f'</math> is continuous at <math>x_0</math>, there exists <math>r > 0</math> such that <math>(x_0 - r, x_0 + r) \subseteq D</math> and<math display="block">|f'(x) - f'(x_0)| < \dfrac{f'(x_0)}{2} \qquad \text{for all } |x - x_0| < r.</math>
In particular,<math display="block">f'(x) > \dfrac{f'(x_0)}{2} >0 \qquad \text{for all } |x - x_0| < r.</math>
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To check that <math>g=f^{-1}</math> is C<sup>1</sup>, write <math>g(y+k) = x+h</math> so that
<math>f(x+h)=f(x)+k</math>. By the inequalities above, <math>\|h-k\| <\|h\|/2</math> so that <math>\|h\|/2<\|k\| < 2\|h\|</math>.
On the other hand, if <math>A=f^\prime(x)</math>, then <math>\|A-I\|<1/2</math>. Using the [[geometric series]] for <math>B=I-A</math>, it follows that <math>\|A^{-1}\| < 2</math>. But then
:<math> {\|g(y+k) -g(y) - f^\prime(g(y))^{-1}k \| \over \|k\|}
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=== Over a real closed field ===
The inverse function theorem also holds over a [[real closed field]] ''k'' (or an [[
The usual proof of the IFT uses Banach's fixed point theorem, which relies on the Cauchy completeness. That part of the argument is replaced by the use of the [[extreme value theorem]], which does not need completeness. Explicitly, in {{section link||A_proof_using_the_contraction_mapping_principle}}, the Cauchy completeness is used only to establish the inclusion <math>B(0, r/2) \subset f(B(0, r))</math>. Here, we shall directly show <math>B(0, r/4) \subset f(B(0, r))</math> instead (which is enough). Given a point <math>y</math> in <math>B(0, r/4)</math>, consider the function <math>P(x) = |f(x) - y|^2</math> defined on a neighborhood of <math>\overline{B}(0, r)</math>. If <math>P'(x) = 0</math>, then <math>0 = P'(x) = 2[f_1(x) - y_1 \cdots f_n(x) - y_n]f'(x)</math> and so <math>f(x) = y</math>, since <math>f'(x)</math> is invertible. Now, by the extreme value theorem, <math>P</math> admits a minimal at some point <math>x_0</math> on the closed ball <math>\overline{B}(0, r)</math>, which can be shown to lie in <math>B(0, r)</math> using <math>2^{-1}|x| \le |f(x)|</math>. Since <math>P'(x_0) = 0</math>, <math>f(x_0) = y</math>, which proves the claimed inclusion. <math>\square</math>
Alternatively, one can deduce the theorem from the one over real numbers by [[Tarski's principle]].{{
==See also==
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