Implicit Function Theorem

Level Set

Let $f$ be a real valued function, $c \in R$ is a fixed constant

L_{c} = f^{- 1} (c) = {x : f (x) = c}

Implicit Function Theorem part 1

The existence of a 0 level set of a non-linear function.
Let $F : S \subseteq R^{n + k} \to R^{k}$ be $C^{1}$ . Write points in $R^{n + k}$ as $x, y$ , where $x$ in $R^{n}$ and $y \in R^{k}$ Define the partial derivative matrices:

D_{x} F = (\begin{matrix} \frac{\partial F_{1}}{\partial x_{1}} & \dots & \frac{\partial F_{1}}{\partial x_{n}} \\ ⋮ & ⋱ & ⋮ \\ \frac{\partial F_{k}}{\partial x_{1}} & \dots & \frac{\partial F_{k}}{\partial x_{n}} \end{matrix}), D_{y} F = (\begin{matrix} \frac{\partial F_{1}}{\partial y_{1}} & \dots & \frac{\partial F_{1}}{\partial y_{k}} \\ ⋮ & ⋱ & ⋮ \\ \frac{\partial F_{k}}{\partial y_{1}} & \dots & \frac{\partial F_{k}}{\partial y_{k}} \end{matrix})

Assume $F (a, b) = 0$ and $det D_{y} F (a, b) \neq 0$ . Then, there exist neighbourhoods around $a \in R^{n}$ and $b \in R^{k}$ such that for each $x$ near $a$ , there exists a unique $y = g (x)$ near $b$ satisfying

$F (x, y) = F (x, g (x)) = 0.$

This implies that $y$ can be expressed as a function of $x$ locally.

Implicit Function Theorem part 2

We can find the derivative or our implicit function where:

D g = \frac{- D_{x} F}{\partial F / \partial y}

pf:

\begin{aligned} Assume F (x, y) = F (x, g (x)) = 0 \\ Let G = (x, g (x)) \\ (x_{1}, \dots, x_{n}) \mapsto (x_{1}, \dots, x_{n}, g (x)) \end{aligned}

by chain rule

\begin{aligned} D (F \circ G) (x) & = D F (G (x)) D G (x) \\ = [\begin{array}{c} D_{x} F & \frac{\partial F}{\partial y} \end{array}] [\begin{array}{c} 1 & 0 & \dots & 0 \\ 0 & 1 & \dots & 0 \\ 0 & 0 & ⋱ \\ \frac{\partial G}{\partial x_{1}} & \frac{\partial G}{\partial x_{2}} & \dots & \frac{\partial G}{\partial x_{n}} \end{array}] \\ = [\begin{array}{c} D_{z} F & \frac{\partial F}{\partial y} \end{array}] [\begin{array}{c} I_{n} \\ D_{g} \end{array}] \\ 0 & \overset{set}{=} D_{x} F (x, g (x)) + \frac{\partial F}{\partial y} (x, g (x)) [D g (x)] since constant level set \end{aligned}

isolate $D g (x)$ and we're done.