Inverse Function Theorem#

Theorem: Inverse Function Theorem

Let \(f: \mathcal{D} \subseteq \mathbb{R}^n \to \mathbb{R}^n\) be a real vector field on an and let \(\boldsymbol{p} \in U\).

If \(f\) is continuously partially differentiable on an open set \(U \subseteq \mathcal{D}\) and its Jacobian matrix at \(\boldsymbol{p} \in U\) is invertible, then there exists an open neighborhood \(V \subseteq U\) of \(\boldsymbol{p}\) and an open neighborhood \(W \subseteq \mathbb{R}^n\) of \(f(\boldsymbol{p})\) such that \(f\) is bijective between \(V\) and \(W\). Furthermore, its inverse \(f^{-1}\) on \(W\) is continuously partially differentiable on \(W\) and its Jacobian matrix is the inverse of \(f\)'s Jacobian matrix:

\[J_{f^{-1}} (\boldsymbol{y}) = J_{f} (f^{-1}(\boldsymbol{y}))^{-1} \qquad \forall \boldsymbol{y} \in W\]

Example:

Consider the coordinate transformation \(\mathcal{T}: [0, +\infty) \times [0, 2\uppi) \to \mathbb{R}^2\) from polar coordinates:

\[\mathcal{T}(\rho, \varphi) = \begin{bmatrix}\rho \cos \varphi \\ \rho \sin \varphi\end{bmatrix}\]

It is continuously partially differentiable on \((0, +\infty) \times (0, 2\uppi)\) with the following Jacobian matrix:

\[J_{\mathcal{T}}(\rho, \varphi) = \begin{bmatrix}\cos \varphi & - \rho \sin \varphi \\ \sin \varphi & \rho \cos \varphi\end{bmatrix}\]

The determinant is

\[\det J_{\mathcal{T}}(\rho, \varphi) = \rho \cos^2 \varphi + \rho \sin^2 \varphi = \rho\]

which is non-zero, since \(\rho \ne 0\). Therefore, for each \((\rho, \varphi) \in (0, +\infty) \times (0, 2\uppi)\), there exist an open neighborhood \(V \subseteq (0, +\infty) \times (0, 2\uppi)\) of \((\rho, \varphi)\) and an open neighborhood \(W \subseteq \mathbb{R}^2\) such that \(\mathcal{T}\) is bijective between \(V\) and \(W\). Furthermore, we have:

\[J_{\mathcal{T}^{-1}}(\mathcal{T}(\rho, \varphi)) = J_{\mathcal{T}} (\rho, \varphi)^{-1} = \begin{bmatrix} \cos \varphi & \sin \varphi \\ -\frac{\sin \varphi}{\rho} & \frac{\cos \varphi}{\rho} \end{bmatrix}\]

Proof

TODO