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I understand the general 'word' definitions of the Implicit Function Theorem and the simple examples such as the one on wikipedia but the version of the Implicit Function theorem given in our lecture notes seems pretty obscure (or perhaps I am just missing something). Here it is.

Let $D \subset \mathbb{R}^m \times \mathbb{R}^p$ be open and $f:D \to \mathbb{R}^n$ be continuously differentiable. Further assume $$f(\bar{x},\bar{\lambda}) = 0 \, \, \, \mathrm{and} \, \, \, \det f_1'(\bar{x},\bar{\lambda})\neq 0.$$ Then there is a neighbourhood $N$ of $\bar{\lambda}$ and a uniquely determined continuously differentiable mapping $$\phi : N \to \mathbb{R}^m \quad N \subset \mathbb{R}^p$$ such that for all $\lambda \in N$ $$f(\phi(\lambda), \lambda) = 0.$$

Can someone please try and explain this particular formulation of the Implicit Function Theorem in a different way? Thanks!

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Could you expand on what you don't understand? –  Antonio Vargas May 11 '12 at 16:23
    
Perhaps this might help: math.stackexchange.com/questions/128269/… –  copper.hat May 11 '12 at 16:45
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1 Answer 1

First of all, I think that in the statement of the theorem $n=m$; in the following I will assume so. $f$ is a function depending on $m+p$ variables: ($x_1,\dots,x_m,\lambda_1,\dots,\lambda_p)$. $f$ is a vector valued function taking values in $\mathbb{R}^m$; let $f_1,\dots,f_m$ be its components. We are given a system of $m$ equations in $m+p$ unknowns: $$\begin{align*} f_1(x_1,\dots,x_m,\lambda_1,\dots,\lambda_p)&=0,\\ f_2(x_1,\dots,x_m,\lambda_1,\dots,\lambda_p)&=0,\\ \dots\dots\dots\qquad&=0,\\ f_m(x_1,\dots,x_m,\lambda_1,\dots,\lambda_p)&=0,\\ \end{align*}$$ Our goal is to solve for $x_1,\dots,x_m$ in terms of $\lambda_1,\dots,\lambda_p$. The implicit function theorem gives sufficient conditions for this to be possible. What are these conditions?

  1. There exists at least one solution. In mathematical notation, this is written as: there exists $(\bar x,\bar\lambda)\in D$ such that $f(\bar x,\bar\lambda)=0$.
  2. A certain non-degeneracy condition is satisfied. In this case it is $$ \frac{\partial(f_1,\dots,f_m)}{\partial(x_1,\dots,x_m)}(\bar x,\bar\lambda)= \det\,\begin{pmatrix} \frac{\partial f_1}{\partial x_1}(\bar x,\bar\lambda)&\dots&\frac{\partial f_1}{\partial x_m}(\bar x,\bar\lambda)\\ \dots&\dots&\dots\\ \frac{\partial f_m}{\partial x_1}(\bar x,\bar\lambda)&\dots&\frac{\partial f_m}{\partial x_m}(\bar x,\bar\lambda)\\ \end{pmatrix}\ne0. $$

If they are satisfied, the theorem guarantees the existence of $C^1$ functions $x_k=\phi_k(\lambda_1,\dots,\lambda_p)$, $1\le k\le m$, solving the equation.

The theorem is called the implicit function theorem because the system of equations defines implicitely $x_1,\dots,x_m$ as functions of $\lambda_1,\dots,\lambda_p$.

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