Inverse mapping and implicit mapping theorem

Inverse mapping theorem:

$W$ is an open set of $\mathbb{R}^n$ , $f:W \rightarrow \mathbb{R}^n$ is a $C^k(k\geq 1)$ mapping.

if $det(\mathcal{J}f(x_0)) \neq 0$ , then there is a open neighborhood $U\subset W$ of $x_0$ and open neighborhood $V \subset \mathbb{R}^n$ of $y_0=f(x_0)$

such that

1. $f|_U:U \rightarrow V$ is an invertible mapping.
2. Its inverse is still a $C^k$ mapping.

Proof:

Translation do not change the topology or differential properties of the function

so we can assume $x_0=0$ , $y_0=f(x_0)=0$ and set $A=\mathcal{J}f(x_0)$ .

From $det(\mathcal{J}f(x_0)) \neq 0$ we know $A$ is invertible.

$\mathcal{J}(f \circ A^{-1}(0))$ is an identity mapping because $\mathcal{J}(f \circ A^{-1}(0))=\mathcal{J}f(A^{-1}(0))\cdot \mathcal{J}A^{-1}(0)$ (chain rule)

$A$ is a linear mapping so $A^{-1}(0)=0$ ,

$\mathcal{J}f(A^{-1}(0))\cdot \mathcal{J}(0)=\mathcal{J}f(0)\cdot \mathcal{J}A^{-1}(0)=\mathcal{J}f(0)\cdot \mathcal{J}A^{-1}(0)=\mathcal{J}f(0)\cdot A^{-1}=AA^{-1}=I$ ($A=\mathcal{J}f(x_0)$).

Linear transformation do not change the topology or differential properties of the function

so we can assume $\mathcal{J}f(x_0)=I_n$.

Let’s review a classic conclusion:

A linear mapping $A : \mathbb{R}^n \rightarrow \mathbb{R}^n$ is invertible and has $\det A \neq 0$, i.e., $A$ is injective.

Furthermore, we have the following:

for a linear mapping $B : \mathbb{R}^n \rightarrow \mathbb{R}^n$ with $\|B\| < 1$, then $I_n - B$ is invertible.

In fact, suppose $(I_n - B)v = 0$, then $\|v\| = \|Bv\| \leq \|B\| \cdot \|v\|$, which implies that $v = 0$.

This example means that a invertible mapping such as an identity map is still invertible after a small twist.

In general, any invertible linear mapping remains invertible under perturbation.

Back to inverse mapping theorem,

we need to represent the difference between $f$ and $I$ , that is $g : x \mapsto f(x) - x$ .

So $g$ is a $C^k$ mapping from $W$ to $\mathbb{R}^n$ and

$\mathcal{J}g(0)=\mathcal{J}f(0)-x'=I-I=0$ , $\exists \epsilon_0>0$ , $Jg(x) \| \leq \displaystyle{\frac{1}{2}}, \quad \forall x \in \overline{B}_{\epsilon_0}(0) \subseteq W$ .

Accroding to differential mean value theorem ,

$\displaystyle{g(x_1) - g(x_2) \| \leq \| Jg(\xi) \| \cdot \| x_1 - x_2 \| \leq \frac{1}{2} \| x_1 - x_2 \|, \quad \forall x_1, x_2 \in \overline{B}_{\epsilon_0}(0)}$

Given $x \in \overline{B}_{\frac{\epsilon_0}{2}}(0)$ , solve equation $f(x) = y, \quad x \in B_{\epsilon_0}(0)$

It is equivalent to finding the fixed point on $g_y(x)=x+y-f(x)=y-g(x)$ in $B_{\epsilon_0}(0)$ .

Firstly, $\displaystyle{\|g_y(x) \| = \| y - g(x) \| \leq \| y \| + \| g(x) \| < \frac{\epsilon_0}{2} + \frac{1}{2} \| x \| \leq \frac{\epsilon_0}{2} + \frac{\epsilon_0}{2} = \epsilon_0, \quad \forall x \in \overline{B}_{\epsilon_0}(0)}$ .

Further, $\displaystyle{\|g_y(x_1) - g_y(x_2)\| = \|g(x_2) - g(x_1)\| \leq \frac{1}{2} \|x_1 - x_2\|, \quad \forall x_1, x_2 \in \bar{B}_\epsilon(0)}$ , it is a constraction mapping.

So the equation has a unique solution.

Note $U = f^{-1}(\overline{B}_{\epsilon_0/2}(0)) \cap B_{\epsilon_0}(0), \quad V = \overline{B}_{\epsilon_0/2}(0)$ , thus $f|_{U}:U \rightarrow V$ is bijective. (The proof of (1) is done.)

Define $h$ is the inverse of $f$ on $V$ , from $f(x)=x+g(x)$ together with $f(h(y))=y$

we know $f(h(y))=h(y)+g(h(y))$ so $y-g(h(y))=h(y)$ .

$h(y_1)&#8722;h(y_2)=(y_1&#8722;g(h(y_1)))&#8722;(y_2&#8722;g(h(y_2)))=(y_1&#8722;y_2)&#8722;(g(h(y_1))&#8722;g(h(y_2)))$ .

Take the norm on both sides of the equation:

$\|h(y_1)&#8722;h(y_2)\| \leq \|y_1&#8722;y_2\|+\|g(h(y_1))&#8722;g(h(y_2))\|&#8804;\|y_1&#8722;y_2\|+\displaystyle{\frac{1}{2}}\|h(y_1)&#8722;h(y_2\|$ .

We can infer that $\forall y_1, y_2 \in V$ , $\|h(y_1)&#8722;h(y_2)\| &#8804;2\|y_1&#8722;y_2\|$ , $h$ is a continous mapping.

Further, $h: V \to U$ is a differentiable mapping. Let $y_0 \in V$ , then for $y \in V$,

we have:

$h(y) - h(y_0) = (y - y_0) - [g(h(y)) - g(h(y_0))]$ $= (y - y_0) - \mathcal{J}g(h(y_0)) \cdot (h(y) - h(y_0)) + o(\|h(y) - h(y_0)\|)$

This can be rewritten as: $[I_n + \mathcal{J}g(h(y_0))] \cdot (h(y) - h(y_0)) = (y - y_0) + o(\|y - y_0\|)$ .

Thus, $h(y) - h(y_0) = [I_n + \mathcal{J}g(h(y_0))]^{-1} \cdot (y - y_0) + o(\|y - y_0\|)$

$h$ is differentiable at $y_0$ , $h \in C^1$ .

From the proof of above:

$\mathcal{J}h(y) = [I_n + \mathcal{J}g(h(y))]^{-1} = [\mathcal{J}f(h(y))]^{-1}, \forall y \in V.$ Since $f \in C^k$ , we know $\mathcal{J}f \in C^{k-1}$ .

We have proved $\mathcal{J}h$ is a $C^1$ mapping , we finally obtain $h \in C^k$ by analogy.

Implicit mapping theorem:

Let $W$ be a subset of $\mathbb{R}^{n+m}$, and $(x, y)$ represents a point in $W$, where $x = (x_1, \dots, x_n)$, $y = (y_1, \dots, y_m)$ .

Let $f: W \to \mathbb{R}^m$ be a $C^k$ map, where

Assume that $(x^0, y^0) \in W$, $f(x^0, y^0) = 0$, and $\text{det} \mathcal{J}f_y(x^0, y^0) \neq 0$, then

Then, in a neighborhood $V \subset \mathbb{R}^n$ around $x^0$, there exists a unique $C^k$ map $g: V \to \mathbb{R}^m$ such that

1. $y^0 = g(x^0)$, $f(x, g(x)) = 0$, for $x \in V$,
2. $\mathcal{J}g(x) = -[\mathcal{J}f_y(x, g(x))]^{-1} \mathcal{J}f_x(x, g(x))$,

where

Proof:

Let $F: W \to \mathbb{R}^{n+m}$ be defined by

The original problem is transferred to solving equation $F(x,y)=0$ in the neighborhood of $(x_0,y_0)$ .

Now $F(x^0, y^0) = (x^0, 0)$, and

By the inverse function theorem, near $(x^0, y^0)$, $F$ is invertible.

If $x$ is near $x^0$ and $y^0$ exists near $g(x)$, then $F(x, g(x)) = (x, 0)$, which leads to the equation:

Thus,

Two-dimentional case:

The function $y = g(x)$ is called the implicit function defined by $f(x, y) = 0$.