proof of implicit function theorem

We state the Theorem with a different notation:

Theorem 1.

Let $\mathrm{\Omega }$ be an open subset of ${\mathrm{R}}^n\mathrm{\times }{\mathrm{R}}^m$ and let $f\mathrm{\in }{\mathrm{C}}^{\mathrm{1}}\mathit{}\mathrm{(}\mathrm{\Omega }\mathrm{,}{\mathrm{R}}^m\mathrm{)}$. Let $\mathrm{(}{x}_{\mathrm{0}}\mathrm{,}{y}_{\mathrm{0}}\mathrm{)}\mathrm{\in }\mathrm{\Omega }\mathrm{\subset }{\mathrm{R}}^n\mathrm{\times }{\mathrm{R}}^m$. If the matrix $D_y\mathit{}f\mathit{}\mathrm{(}{x}_{\mathrm{0}}\mathrm{,}{y}_{\mathrm{0}}\mathrm{)}$ defined by

$$D_{y}f(x_0,y_0)={\left(\frac{\partial f_j}{\partial y_k}(x_0,y_0)\right)}_{j,k}\mathit{\hspace{1em}}j=1,\mathrm{\dots },m\mathit{\hspace{1em}}k=1,\mathrm{\dots },m$$

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

$$f(x,y)=f(x_0,y_0)\iff y=g(x)\mathit{\hspace{1em}\hspace{1em}}\forall (x,y)\in U\times V.$$

Moreover

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

Proof.

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

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

One finds that

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 $G_1:V\times W\to {ℝ}^n$, $G_2:V\times W\to {ℝ}^m$) 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 $g(x)=G_2(x,f(x_0,y_0))$ to obtain

$$f(x,g(x))=f(x_0,y_0),\forall x\in U.$$

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
Canonical name	ProofOfImplicitFunctionTheorem
Date of creation	2013-03-22 13:31:23
Last modified on	2013-03-22 13:31:23
Owner	paolini (1187)
Last modified by	paolini (1187)
Numerical id	8
Author	paolini (1187)
Entry type	Proof
Classification	msc 26B10