# Implicit Function Theorem/Real Functions/Proof 1

## Theorem

Let $n$ and $k$ be natural numbers.

Let $\Omega \subset \R^{n+k}$ be open.

Let $f : \Omega \to \R^k$ be continuous.

Let the partial derivatives of $f$ with respect to $\R^k$ be continuous.

Let $(a,b) \in \Omega$, with $a\in \R^n$ and $b\in \R^k$.

Let $f(a,b) = 0$.

For $(x_0,y_0)\in\omega$, let $D_2 f(x_0,y_0)$ denote the total derivative of the function $y\mapsto f(x_0, y)$ at $y_0$.

Let the linear map $D_2 f(a,b)$ be invertible.

Then there exist neighborhoods $U\subset\Omega$ of $a$ and $V\subset\R^k$ of $b$ such that there exists a unique function $g : U \to V$ such that $f(x, g(x)) = 0$ for all $x\in U$.

Moreover, $g$ is continuous.

## Outline of Proof

We apply the Uniform Contraction Mapping Theorem to the function:

$F : \R^n \times \R^k \to \R^k : F(x, y) = y - D_2f(a,b)^{-1}(f(x,y))$

for appropriate $x \in \R^n$ and $y \in \R^k$.

## Proof

#### Reduction to $(a,b)=(0,0)$

We may assume $(a,b)= (0,0)$.

Define $F : \Omega \to \R^k : F(x, y) = y - D_2f(a,b)^{-1}(f(x,y))$.

By Linear Function is Continuous, $D_2f(a,b)^{-1}$ is continuous on $\R^k$.

Thus $F$ is continuous on $\Omega$.

#### $F$ is locally a uniform contraction

Let $r_1>0$ such that the open ball $B(0,r_1)\subset\Omega$.

We have, for $(x,y_1),(x,y_2)\in B(0,r_1)$, by the Mean Value Inequality:

$\| F(x,y_2) - F(x, y_1)\| \leq \displaystyle\sup_{y\in[y_1, y_2]} \| D_2 F(x,y) \| \cdot \| y_2 - y_1 \|$.

We have, for $(x,y)\in\Omega$.

$D_2 F(x,y) = \operatorname{id} - (D_2 f(0,0))^{-1} \circ D_2 f(x,y)$

where $\operatorname{id}$ is the identity mapping.

Thus $D_2 F$ is continuous in $\Omega$.

By definition, $D_2 F(0,0) = 0$.

By continuity, there exists $r_2>0$ such that $\| D_2 F(x,y) \| \leq \frac 12$ for $\|(x,y)\| \leq r_2$.

From the above inequality, $F$ is a uniform contraction mapping on the open ball $B(0,r_2)$ for the Lipschitz Constant $\frac12$.

#### Constructing a stable neighborhood

We have, for $x, y \in \bar B(0,r_2)$:

 $\displaystyle \Vert F(x, y) \Vert$ $\leq$ $\displaystyle \Vert F(x, y) - F(x, 0) \Vert + \Vert F(x, 0) \Vert$ Triangle Inequality $\displaystyle$ $\leq$ $\displaystyle \frac12 \Vert y\Vert + \Vert F(x, 0) \Vert$ Definition of $r_2$

Because $F(0,0)=0$ and $F$ is continuous, there exists $r_3>0$ such that $\| F(x,0) \| \leq r_2/2$.

Then $\| F(x, y) \| \leq r_2$ for $\|y\| \leq r_2$ and $\|x\| \leq r_3$.

Thus the restriction of $F$ to $B(0,r_3)\times \overline B(0,r_2)$ is a uniform contraction:

$F : B(0,r_3) \times \overline B(0,r_2) \to \overline B(0,r_2)$

#### Applying the Uniform Contraction Theorem

By Subspace of Complete Metric Space is Closed iff Complete, $\overline B(0,r_2)$ is complete.

By the Uniform Contraction Mapping Theorem, there exists a unique mapping $g : B(0,r_3) \to \overline B(0,r_2)$ such that $F(x, g(x)) = g(x)$.

Moreover, $g$ is continuous.

A mapping $h : B(0,r_3) \to \overline B(0,r_2)$ satisfies $F(x, h(x)) = h(x)$ if and only if it satisfies $D_2f(a,b)^{-1}(f(x, h(x))) = 0$ if and only if it satisfies $f(x, h(x))=0$, because $D_2f(a,b)^{-1}$ is invertible.

Thus $g$ is the unique mapping $g : B(0,r_3) \to \overline B(0,r_2)$ such that $f(x, g(x)) = 0$.

$\blacksquare$