Implicit function Theorem for R^3

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Discussion Overview

The discussion revolves around the application of the Implicit Function Theorem in the context of functions defined in three-dimensional space, specifically focusing on proving the existence of a function that satisfies certain conditions given a $C^1$ function in $\mathbb{R}^3$. Participants are exploring the nuances of extending known proofs from $\mathbb{R}^2$ to $\mathbb{R}^3$.

Discussion Character

  • Technical explanation
  • Exploratory

Main Points Raised

  • One participant presents the problem statement involving a $C^1$ function $f : U → \mathbb{R}$ and the conditions under which an implicit function $\phi$ can be defined.
  • Another participant suggests considering $\mathbb{R}^3$ as $\mathbb{R}^2 \times \mathbb{R}$ to facilitate the proof, proposing a specific substitution for the variables.
  • A subsequent reply questions the notation used in the substitution, suggesting a possible typo and seeking clarification on the correct representation of the variables.
  • The original poster acknowledges the typo and corrects it, confirming that the correct notation should be $z = (a, b)$.

Areas of Agreement / Disagreement

Participants appear to agree on the approach of using the structure of $\mathbb{R}^3$ to relate it to $\mathbb{R}^2$, but there is a minor disagreement regarding the notation used in the substitution, which was clarified by one participant.

Contextual Notes

The discussion does not resolve whether the proposed approach is the only or best method for proving the theorem in $\mathbb{R}^3$, and it remains open to further exploration of alternative methods or clarifications.

kalvin
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Let $U \subset \Bbb{R}^3$ be open and let $f : U → \Bbb{R}$ be a $C^1$
function. Let$ (a, b, c) \in U$
and suppose that$ f(a, b, c) = 0$ and $D_3f(a, b, c) \ne 0.$ Show there is an open ball$ V \subset \Bbb{R}^2$ containing $(a, b)$ and a $C^1$
function $\phi : V → \Bbb{R}$ such that $\phi(a, b) = c$ and
$f(x, y, \phi(x, y)) = 0$ for all$ (x, y) \in V$ . We call $\phi$ an implicit function determined by $f$ at $(a, b)$.

I know who how to do the proof in R^2 bout proving it for R^3 seems to be a lot trickier, any help
 
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Hi,

Consider $\Bbb{R}^{3}$ as $\Bbb{R}^{2}\times \Bbb{R}$, this is , take $(a,b,c)=(z,c)$, where $z=(a,c)$ and now the proof is identical to the one you know.
 
Fallen Angel said:
Hi,

Consider $\Bbb{R}^{3}$ as $\Bbb{R}^{2}\times \Bbb{R}$, this is , take $(a,b,c)=(z,c)$, where $z=(a,c)$ and now the proof is identical to the one you know.

Hi, I was just wondering if it was typo for z. Should z = (a, b) ? and not (a, c)? If it is not a typo do you think you could explain?

Thank you
 
Sorry, it was a typo, it should read $z=(a,b)$
 

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