# proof of implicit function theorem

We state the Theorem with a different notation:

###### Theorem 1.

Let $\Omega$ be an open subset of $\mathbb{R}^{n}\times\mathbb{R}^{m}$ and let $f\in\mathcal{C}^{1}(\Omega,\mathbb{R}^{m})$. Let $(x_{0},y_{0})\in\Omega\subset\mathbb{R}^{n}\times\mathbb{R}^{m}$. If the matrix $D_{y}f(x_{0},y_{0})$ defined by

 $D_{y}f(x_{0},y_{0})=\left(\frac{\partial f_{j}}{\partial y_{k}}(x_{0},y_{0})% \right)_{j,k}\quad j=1,\ldots,m\quad k=1,\ldots,m$

is invertible, then there exists a neighbourhood $U\subset\mathbb{R}^{n}$ of $x_{0}$, a neighbourhood $V\subset\mathbb{R}^{m}$ of $y_{0}$ and a function $g\in\mathcal{C}^{1}(U,V)$ such that

 $f(x,y)=f(x_{0},y_{0})\Leftrightarrow y=g(x)\qquad\forall(x,y)\in U\times V.$

Moreover

 $Dg(x)=-(D_{y}f(x,g(x)))^{-1}D_{x}f(x,g(x)).$
###### Proof.

Consider the function $F\in\mathcal{C}^{1}(\Omega,\mathbb{R}^{n}\times\mathbb{R}^{m})$ defined by

 $F(x,y)=(x,f(x,y)).$

One finds that

 $DF(x,y)=\left(\begin{array}[]{c|c}I_{m}&0\\ \hline D_{x}f&D_{y}f\\ \end{array}\right).$

Being $D_{y}f(x_{0},y_{0})$ invertible, $DF(x_{0},y_{0})$ is invertible too. Applying the inverse function Theorem to $F$ we find that there exist a neighbourhood $U$ of $x_{0}$ and $V$ of $y_{0}$ and a function $G\in C^{1}(U\times V,\mathbb{R}^{n+m})$ such that $F(G(x,y))=(x,y)$ for all $(x,y)\in U\times V$. Letting $G(x,y)=(G_{1}(x,y),G_{2}(x,y))$ (so that $G_{1}\colon V\times W\to\mathbb{R}^{n}$, $G_{2}\colon V\times W\to\mathbb{R}^{m}$) we hence have

 $(x,y)=F(G_{1}(x,y),G_{2}(x,y))=(G_{1}(x,y),f(G_{1}(x,y),G_{2}(x,y)))$

and hence $x=G_{1}(x,y)$ and $y=f(G_{1}(x,y),G_{2}(x,y))=f(x,G_{2}(x,y))$. So we only have to set $g(x)=G_{2}(x,f(x_{0},y_{0}))$ to obtain

 $f(x,g(x))=f(x_{0},y_{0}),\quad\forall x\in U.$

Differentiating with respect to $x$ we obtain

 $D_{x}f(x,g(x))+D_{y}f(x,g(x))Dg(x)=0$

which gives the desired formula for the computation of $Dg$. ∎

Title proof of implicit function theorem ProofOfImplicitFunctionTheorem 2013-03-22 13:31:23 2013-03-22 13:31:23 paolini (1187) paolini (1187) 8 paolini (1187) Proof msc 26B10