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(This was posted on math.SE over 5 days ago and has not been answered,
although a comment mentioned a similar question on this site.)

Wikipedia's statement of the implicit function theorem requires that the original function
be continuously differentiable. $\:$ Is it known whether or not that condition can be removed?
If it can't be completely removed, can it be replaced with
"differentiable function whose derivative is bounded"?

Community
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  • In the Wikipedia article they discuss a non differentiable version, so I am not sure what exactly you are asking... – Igor Rivin Sep 29 '14 at 04:15
  • If one removed the two instances of "continuously" from the text between "Writing all the hypotheses together gives the following statement." and "Regularity", would the resulting statement still be true? $\hspace{.37 in}$ –  Sep 29 '14 at 04:22
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    Answers in the linked MO post show that continuous differentiability can be replaced with differentiability everywhere. If this is not what you want, can you elaborate on your goal? – Joonas Ilmavirta Sep 29 '14 at 05:28
  • @JoonasIlmavirta : $:$ What I want is something which does show that, $\hspace{2.09 in}$ rather than just can be used to show that. $;;;$ –  Sep 29 '14 at 05:42
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    It seems to me that this is what you are looking for, or am I missing something?

    http://terrytao.wordpress.com/2011/09/12/the-inverse-function-theorem-for-everywhere-differentiable-maps/

     – Francesco Polizzi Sep 29 '14 at 08:09
  • @FrancescoPolizzi : $;;;$ Which part of that? $:$ (Remember that I'm asking about the implicit function theorem, not the inverse function theorem.) $;;;;;;;$ –  Sep 29 '14 at 14:44
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    @Ricky Demer: If $(x,y)\mapsto f(x,y)$ satisfies suitable implicit function theorem assumptions, then $(x,y)\mapsto (x,f(x,y))$ satisfies corresponding inverse function theorem assumptions. So from an inverse function theorem one gets an implicit function theorem. – TaQ Sep 29 '14 at 15:51

1 Answers1

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I haven't checked Terry Tao's proof of his inverse function theorem (Theorem 2 here), but if the proof (and hence the theorem) is correct, then from the theorem one gets the following implicit function theorem.

Let $\Omega\subset\mathbb R^n\times\mathbb R^m$ be an open set, and let $f:\Omega\to\mathbb R^m$ be an everywhere differentiable function, such that for all $(x_0,y_0)\in\Omega$ the partial derivative map $\partial_2 f(x_0,y_0):\mathbb R^m\to\mathbb R^m$ is invertible. Then for all $(x_0,y_0)\in\Omega$ with $f(x_0,y_0)=0$ there are an open neighbourhood $U$ of $x_0$ and an open neighbourhood $V$ of $y_0$ such that the set $U\times V\cap f^{-1}[\{0\}]$ is a continuous function $U\to V$ .

For the proof, one applies Tao's theorem to the map $f_0:\Omega\to\mathbb R^n\times\mathbb R^m$ defined by $(x,y)\mapsto(x,f(x,y))$ for which one first get the following result. Given any $(x_0,y_0)\in\Omega$ with $f(x_0,y_0)=0$ , there are an open neighbourhood $W$ of $(x_0,y_0)$ and an open set $W_1$ containing $(x_0,0)$ such that $f_0\,|\,W$ is a homeomorphism $W\to W_1$ . In particular, $f_1=(f_0\,|\,W)^{-1}$ is a continuous function $W_1\to W$ . Hence also the map $g_1:U_1=\{\,x:(x,0)\in W_1\,\}\to\mathbb R^m$ given by $x\mapsto{\rm pr}_2\circ f_1(x,0)$ is continuous. Noting that $\mathbb R^n\times\mathbb R^m$ has the product topology, one may suitably shrink the open sets $W$ to $U_2\times V$ and $U_1$ to $U$ so that by finally taking $g=g_1\,|\,U$ one by elementary set theoretic verifications shows the claim for $g=U\times V\cap f^{-1}[\{0\}]$ .

TaQ
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  • Is "it" the theorem or the proof? $:$ If "it" is the theorem, then one would need to derive a local lower bound on the size of the neighborhoods $U$. $:$ I think one would also need to show that being an open map to a set that includes $\mathbf{0}$ is an open property, although I imagine that would follow from some result in degree theory. $\hspace{.37 in}$ –  Sep 29 '14 at 20:14
  • @Ricky Demer: I modified and extended my answer to get it more clear. – TaQ Sep 30 '14 at 09:51
  • Oh. $:$ I misread your middle paragraph, and thought you were doing this without assuming $\hspace{.46 in}$ differentiability of $\hspace{.04 in}f$ with respect to its $\mathbb{R}^n$ argument. $;;;;$ –  Sep 30 '14 at 14:59
  • Does the theorem become false without differentiability with respect to x? (Differentiability with respect to y obviously cannot be removed.) The usual implicit function theorem does not require differentiability with respect to x (if you only want a continuous solution). – Lasse Rempe Feb 05 '24 at 09:43
  • Nb. The theorem is not by Terry Tao; as he writes himself, the result follows from work of Cernavskii, while the argument he gives is an arrangement of the proof by Saint Raymond. – Lasse Rempe Feb 05 '24 at 09:45
  • @LasseRempe "Does the theorem become false without differentiability with respect to x?" Of course not since the given stated premise is not necessary. However, it is not clear how one could reasonably weaken the premise. For example, it should not be difficult to give an example to show that replacing differentiability of $f$ by continuity together with differentiability with respect to the second variable is not sufficient. – TaQ Mar 03 '24 at 13:13