PlanetMath: proof of implicit function theorem

Theorem 1 Let $\Omega$ be an open subset of $\R^n \times \R^m$ and let $f\in \CC^1(\Omega,\R^m)$ . Let $(x_0,y_0)\in \Omega \subset \R^n\times\R^m$ . 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 $U\subset \R^n$ of $x_0$ and a function $g \in \CC^1(U,\R^m)$ such that $$ f(x,g(x)) = f(x_0,y_0) \qquad \forall x \in U. $$

Moreover $$ Dg(x) = - (D_y f(x,g(x))) ^ {-1} D_x f(x,g(x)). $$

Proof. Consider the function $F\in \CC^1(\Omega, \R^n \times \R^m)$ defined by $$ F(x,y)=(x,f(x,y)). $$ One finds that

$\displaystyle DF(x,y) = \left(\begin{array}{c\vert c} I_m & 0 \\ \hline D_x f & D_y f \ \end{array}\right).$

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 $G\in C^1(U\times V,\R^{n+m})$ 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 $G_1\colon V\times W\to\R^n$ , $G_2\colon V\times W\to \R^m$ ) 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 $g(x)=G_2(x,0)$ to obtain $$ f(x,g(x)) = 0,\quad \forall x\in U. $$ 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 $Dg$ . $\qedsymbol$