Implicit Function Theorem/Real Functions/Proof 1

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 $y \in \R^k$.

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

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 $F(x, h(x)) = h(x)$ it satisfies $D_2f(a,b)^{-1}(f(x, h(x))) = 0$  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 $f(x, g(x)) = 0$.