Implicit Function Theorem/Real Functions

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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 $\tuple {a, b} \in \Omega$, with $a\in \R^n$ and $b\in \R^k$.

Let $\map f {a, b} = 0$.

For $\tuple {x_0, y_0} \in \Omega$, let $D_2 \map f {x_0, y_0}$ denote the total derivative of the function $y \mapsto \map f {x_0, y}$ at $y_0$.

Let the linear map $D_2 \map f {a, b}$ 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 $\map f {x, \map g x} = 0$ for all $x \in U$.

Moreover, $g$ is continuous.


Proof 1

Reduction to $\tuple {a, b} = \tuple {0, 0}$

We may assume $\tuple {a, b} = \tuple {0, 0}$.



Define:

$F : \Omega \to \R^k: \map F {x, y} = y - \map {D_2 \map f {a, b}^{-1} } {\map f {x, y} }$

By Linear Function is Continuous, $D_2 \map f {a, b}^{-1}$ 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 $\map B {0, r_1} \subseteq \Omega$.

We have, for $\tuple {x, y_1}, \tuple {x, y_2} \in \map B {0, r_1}$, by the Mean Value Inequality:

$\norm {\map F {x, y_2} - \map F {x, y_1} } \le \displaystyle \sup_{y \mathop \in \closedint {y_1} {y_2} } \norm {D_2 \map F {x, y} } \cdot \norm {y_2 - y_1}$

We have, for $\tuple {x, y} \in \Omega$:

$D_2 \map F {x, y} = I - \paren {D_2 \map F {0, 0} }^{-1} \circ D_2 \map F {x, y}$

where $I$ is the identity mapping.

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

By definition, $D_2 \map F {0, 0} = 0$.

By continuity, there exists $r_2 > 0$ such that:

$\norm {D_2 \map F {x, y} } \le \dfrac 1 2$ for $\norm {\tuple {x, y} } \le r_2$

From the above inequality, $F$ is a uniform contraction mapping on the open ball $\map B {0, r_2}$ for the Lipschitz Constant $\dfrac 1 2$.


Constructing a stable neighborhood

We have, for $x, y \in \map {\overline B} {0, r_2}$:

\(\displaystyle \norm {\map F {x, y} }\) \(\le\) \(\displaystyle \norm {\map F {x, y} - \map F {x, 0} } + \norm {\map F {x, 0} }\) Triangle Inequality
\(\displaystyle \) \(\le\) \(\displaystyle \frac 1 2 \norm y + \norm {\map F {x, 0} }\) Definition of $r_2$

Because $\map F {0, 0} = 0$ and $F$ is continuous, there exists $r_3>0$ such that $\norm {\map F {x, 0} } \le \dfrac {r_2} 2$.

Then:

$\norm {\map F {x, y} } \le r_2$ for $\norm y \le r_2$

and:

$\norm x \le r_3$

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

$F: \map B {0, r_3} \times \map {\overline B} {0, r_2} \to \map {\overline B} {0, r_2}$


Applying the Uniform Contraction Theorem

By Subspace of Complete Metric Space is Closed iff Complete, $\map {\overline B} {0, r_2}$ is complete.

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

Moreover, $g$ is continuous.

A mapping $h: \map B {0, r_3} \to \map {\overline B} {0, r_2}$ satisfies $\map F {x, \map h x} = \map h x$ if and only if it satisfies $\map {D_2 \map f {a, b}^{-1} } {\map f {x, \map h x} } = 0$ if and only if it satisfies $\map f {x, \map h x} = 0$, because $D_2 \map f {a, b}^{-1}$ is invertible.

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

$\blacksquare$


Proof Using the Inverse Function Theorem


Sources