(This was posted [on math.SE](https://math.stackexchange.com/q/943348/57159) over 5 days ago and has not been answered,
  

although a comment mentioned a similar [question on this site](https://mathoverflow.net/q/75049/5810).)


[Wikipedia's statement of the implicit function theorem](http://en.wikipedia.org/wiki/Implicit_function_theorem#Statement_of_the_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"?



Does the implicit function theorem hold for discontinuously differentiable functions?
I haven't checked Terry Tao's proof of his inverse function theorem (Theorem 2 [here](http://terrytao.wordpress.com/2011/09/12/the-inverse-function-theorem-for-everywhere-differentiable-maps/)), 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 $g=U\times V\cap f^{-1}[\{0\}]$ .


