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 $y_{0}$ anda 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 $D g$ .∎