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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"?

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  • $\begingroup$ In the Wikipedia article they discuss a non differentiable version, so I am not sure what exactly you are asking... $\endgroup$
    – Igor Rivin
    Commented Sep 29, 2014 at 4:15
  • $\begingroup$ 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? $\hspace{.37 in}$ $\endgroup$
    – user5810
    Commented Sep 29, 2014 at 4:22
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    $\begingroup$ 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? $\endgroup$
    – Joonas Ilmavirta
    Commented Sep 29, 2014 at 5:28
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    $\begingroup$ It seems to me that this is what you are looking for, or am I missing something? terrytao.wordpress.com/2011/09/12/… $\endgroup$
    – Francesco Polizzi
    Commented Sep 29, 2014 at 8:09
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    $\begingroup$ @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. $\endgroup$
    – TaQ
    Commented Sep 29, 2014 at 15:51

1 Answer 1

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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 $g=U\times V\cap f^{-1}[\{0\}]$ .

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  • $\begingroup$ 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. $\hspace{.37 in}$ $\endgroup$
    – user5810
    Commented Sep 29, 2014 at 20:14
  • $\begingroup$ @Ricky Demer: I modified and extended my answer to get it more clear. $\endgroup$
    – TaQ
    Commented Sep 30, 2014 at 9:51
  • $\begingroup$ 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. $\;\;\;\;$ $\endgroup$
    – user5810
    Commented Sep 30, 2014 at 14:59
  • $\begingroup$ 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). $\endgroup$
    – Lasse Rempe
    Commented Feb 5 at 9:43
  • $\begingroup$ 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. $\endgroup$
    – Lasse Rempe
    Commented Feb 5 at 9:45

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