imported>Mechanikin: /* Counter-example */it just occurred to me that a locally constant function is a counterexample to the case in which the derivative were allowed to be zero

2026-01-01T07:42:28Z

Counter-example: it just occurred to me that a locally constant function is a counterexample to the case in which the derivative were allowed to be zero

← Previous revision		Revision as of 07:42, 1 January 2026
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	{{Use dmy dates\|date=December 2023}}		{{Use dmy dates\|date=December 2023}}
	{{Calculus}}		{{Calculus}}
	In [[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''.		In [[real analysis]], a branch of [[mathematics]], the '''inverse function theorem''' is a [[theorem]] that asserts that, if a [[real function]] ''f'' has a [[smoothness\|continuous derivative]] near a point where its derivative is nonzero, then, near this point, ''f'' has an [[inverse function]]. The inverse function is also [[smoothness\|continuously 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		The theorem applies verbatim to [[complex-valued function]]s of a [[complex number\|complex variable]]. It generalizes to functions from
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	== Counter-example ==		== Counter-example ==
	[[File:Inv-Fun-Thm-3.png\|thumb\|The function <math>f(x)=x+2 x^2\sin(\tfrac1x)</math> is bounded inside a quadratic envelope near the line <math>y=x</math>, so <math>f'(0)=1</math>. Nevertheless, it has local max/min points accumulating at <math>x=0</math>, so it is not one-to-one on any surrounding interval.]]		[[File:Inv-Fun-Thm-3.png\|thumb\|The function <math>f(x)=x+2 x^2\sin(\tfrac1x)</math> is bounded inside a quadratic envelope near the line <math>y=x</math>, so <math>f'(0)=1</math>. Nevertheless, it has local max/min points accumulating at <math>x=0</math>, so it is not one-to-one on any surrounding interval.]]
	If one drops the assumption that the derivative is continuous, the function no longer ~~need be invertible~~. For example <math>f(x) = x + 2x^2\sin(\tfrac1x)</math> and <math>f(0)= 0</math> has discontinuous derivative		If one drops the assumption that the derivative is continuous, the function is no longer necessarily locally injective. For example <math>f(x) = x + 2x^2\sin(\tfrac1x)</math> and <math>f(0)= 0</math> has discontinuous derivative
	<math>f'\!(x) = 1 -2\cos(\tfrac1x) + 4x\sin(\tfrac1x)</math> and <math>f'\!(0) = 1</math>, which vanishes arbitrarily close to <math>x=0</math>. These critical points are local max/min points of <math>f</math>, so <math>f</math> is not one-to-one (and not invertible) on any interval containing <math>x=0</math>. Intuitively, the slope <math>f'\!(0)=1</math> does not propagate to nearby points, where the slopes are governed by a weak but rapid oscillation.		<math>f'\!(x) = 1 -2\cos(\tfrac1x) + 4x\sin(\tfrac1x)</math> and <math>f'\!(0) = 1,</math> which vanishes arbitrarily close to <math>x=0</math>. These critical points are local max/min points of <math>f,</math> so <math>f</math> is not one-to-one (and not invertible) on any interval containing <math>x=0</math>. Intuitively, the slope <math>f'\!(0)=1</math> does not propagate to nearby points, where the slopes are governed by a weak but rapid oscillation.

			If the derivative is continuous but zero at a point, the function is no longer necessarily locally injective. A real function that is [[locally constant function\|locally constant]] at a point <math>x\in\mathbb{R}</math> in the interior of its domain is not locally injective at <math>x</math> but is trivially continuously differentiable at <math>x</math>.

	==Methods of proof==		==Methods of proof==
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	In particular,<math display="block">f'(x) > \dfrac{f'(x_0)}{2} >0 \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>

	This shows that <math>f</math> is strictly increasing for all <math>\|x - x_0\| < r</math>. Let <math>\delta > 0</math> be such that <math>\delta < r</math>. Then <math>[x - \delta, x + \delta] \subseteq (x_0 - r, x_0 + r)</math>. By the intermediate value theorem, we find that <math>f</math> maps the interval <math>[x - \delta, x + \delta]</math> bijectively onto <math>[f(x - \delta), f(x + \delta)]</math>. Denote by <math>I = (x-\delta, x+\delta)</math> and <math>J = (f(x - \delta),f(x + \delta))</math>. Then <math>f: I \to J</math> is a bijection and the inverse <math>f^{-1}: J \to I</math> exists. The fact that <math>f^{-1}: J \to I</math> is differentiable follows from the differentiability of <math>f</math>. In particular, the result follows from the fact that if <math>f: I \to \R</math> is a strictly monotonic and continuous function that is differentiable at <math>x_0 \in I</math> with <math>f'(x_0) \ne 0</math>, then <math>f^{-1}: f(I) \to \R</math> is differentiable with <math>(f^{-1})'(y_0) = \dfrac{1}{f'(~~y_0~~)}</math>, where <math>y_0 = f(x_0)</math> (a standard result in analysis). This completes the proof.		This shows that <math>f</math> is strictly increasing for all <math>\|x - x_0\| < r</math>. Let <math>\delta > 0</math> be such that <math>\delta < r</math>. Then <math>[x - \delta, x + \delta] \subseteq (x_0 - r, x_0 + r)</math>. By the intermediate value theorem, we find that <math>f</math> maps the interval <math>[x - \delta, x + \delta]</math> bijectively onto <math>[f(x - \delta), f(x + \delta)]</math>. Denote by <math>I = (x-\delta, x+\delta)</math> and <math>J = (f(x - \delta),f(x + \delta))</math>. Then <math>f: I \to J</math> is a bijection and the inverse <math>f^{-1}: J \to I</math> exists. The fact that <math>f^{-1}: J \to I</math> is differentiable follows from the differentiability of <math>f</math>. In particular, the result follows from the fact that if <math>f: I \to \R</math> is a strictly monotonic and continuous function that is differentiable at <math>x_0 \in I</math> with <math>f'(x_0) \ne 0</math>, then <math>f^{-1}: f(I) \to \R</math> is differentiable with <math>(f^{-1})'(y_0) = \dfrac{1}{f'(x_0)}</math>, where <math>y_0 = f(x_0)</math> (a standard result in analysis). This completes the proof.

	=== A proof using successive approximation ===		=== A proof using successive approximation ===
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	=== Over a real closed field ===		=== Over a real closed field ===
	The inverse function theorem also holds over a [[real closed field]] ''k'' (or an [[O-minimal structure]]).<ref>Theorem 2.11. in {{cite book \|doi=10.1017/CBO9780511525919\|title=Tame Topology and O-minimal Structures. London Mathematical Society lecture note series, no. 248\|year=1998 \|last1=Dries \|first1=L. P. D. van den \|authorlink = Lou van den Dries\|isbn=9780521598385\|publisher=Cambridge University Press\|location=Cambridge, New York, and Oakleigh, Victoria }}</ref> Precisely, the theorem holds for a semialgebraic (or definable) map between open subsets of <math>k^n</math> that is continuously differentiable.		The inverse function theorem also holds over a [[real closed field]] ''k'' (or an [[o-minimal structure]]).<ref>Chapter 7, Theorem 2.11. in {{cite book \|doi=10.1017/CBO9780511525919\|title=Tame Topology and O-minimal Structures. London Mathematical Society lecture note series, no. 248\|year=1998 \|last1=Dries \|first1=L. P. D. van den \|authorlink = Lou van den Dries\|isbn=9780521598385\|publisher=Cambridge University Press\|location=Cambridge, New York, and Oakleigh, Victoria }}</ref> Precisely, the theorem holds for a semialgebraic (or definable) map between open subsets of <math>k^n</math> that is continuously differentiable.

	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>		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>

imported>AllCatsAreGrey: clean up, typo(s) fixed: On the other hand → On the other hand,

2025-05-27T16:02:52Z

clean up, typo(s) fixed: On the other hand → On the other hand,

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Inverse function theorem - Revision history

imported>Mechanikin: /* Counter-example */it just occurred to me that a locally constant function is a counterexample to the case in which the derivative were allowed to be zero

imported>AllCatsAreGrey: clean up, typo(s) fixed: On the other hand → On the other hand,