Inverse Function Theorem (Statement and Proof Sketch)

The Inverse Function Theorem is proved using the Banach Fixed-Point Theorem. The key idea: solving $f(\mathbf{x}) = \mathbf{y}$ is equivalent to finding a fixed point of an auxiliary map $T(\mathbf{x}) = \mathbf{x} + A^{-1}(\mathbf{y} - f(\mathbf{x}))$ , where $A = Jf(\mathbf{a})$ . Near $\mathbf{a}$ , this map is a contraction, guaranteeing a unique solution.

Statement

Theorem10.1Inverse Function Theorem

Let $f : U \subseteq \mathbb{R}^n \to \mathbb{R}^n$ be continuously differentiable on an open set $U$ , and let $\mathbf{a} \in U$ with $\det Jf(\mathbf{a}) \neq 0$ . Then there exist neighborhoods $V$ of $\mathbf{a}$ and $W$ of $f(\mathbf{a})$ such that:

$f : V \to W$ is a bijection.
$g = f^{-1} : W \to V$ is continuously differentiable.
$Jg(\mathbf{y}) = [Jf(g(\mathbf{y}))]^{-1}$ for $\mathbf{y} \in W$ .

Proof Sketch

Proof

Step 1: Setup. Let $A = Jf(\mathbf{a})$ , which is invertible. For $\mathbf{y}$ near $f(\mathbf{a})$ , define

$T_{\mathbf{y}}(\mathbf{x}) = \mathbf{x} + A^{-1}(\mathbf{y} - f(\mathbf{x})).$

Note that $f(\mathbf{x}) = \mathbf{y}$ iff $T_{\mathbf{y}}(\mathbf{x}) = \mathbf{x}$ (i.e., $\mathbf{x}$ is a fixed point of $T_{\mathbf{y}}$ ).

Step 2: $T_{\mathbf{y}}$ is a contraction near $\mathbf{a}$ . The derivative of $T_{\mathbf{y}}$ is

$DT_{\mathbf{y}}(\mathbf{x}) = I - A^{-1} Jf(\mathbf{x}).$

At $\mathbf{x} = \mathbf{a}$ , $DT_{\mathbf{y}}(\mathbf{a}) = I - A^{-1} A = 0$ . By continuity of $Jf$ , for $\mathbf{x}$ near $\mathbf{a}$ ,

$\|DT_{\mathbf{y}}(\mathbf{x})\| < 1/2.$

By the Mean Value Theorem, $T_{\mathbf{y}}$ is a contraction on a ball around $\mathbf{a}$ with contraction constant $\lambda < 1$ .

Step 3: Existence and uniqueness of inverse. By the Banach Fixed-Point Theorem, $T_{\mathbf{y}}$ has a unique fixed point $\mathbf{x}$ in a neighborhood $V$ of $\mathbf{a}$ . Define $g(\mathbf{y}) = \mathbf{x}$ . Then $f(g(\mathbf{y})) = \mathbf{y}$ , so $g = f^{-1}$ .

Step 4: $g$ is continuously differentiable. Differentiating $f(g(\mathbf{y})) = \mathbf{y}$ using the chain rule,

$Jf(g(\mathbf{y})) \cdot Jg(\mathbf{y}) = I,$

so $Jg(\mathbf{y}) = [Jf(g(\mathbf{y}))]^{-1}$ . Since $Jf$ is continuous and $g$ is continuous, $Jg$ is continuous.

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Summary

The Inverse Function Theorem:

Proved via Banach Fixed-Point Theorem (contraction mapping).
Local invertibility when Jacobian is nonsingular.
Inverse is smooth with derivative $(f^{-1})' = (f')^{-1}$ .

See Inverse Function Theorem concept and Implicit Function Theorem.