The Inverse Function Theorem for Functions from Rn to Rn Examples 1

Recall from The Inverse Function Theorem for Functions from Rn to Rn that if $A \subseteq \mathbb{R}^n$ is open and $\mathbf{f} : \mathbb{R}^n \to \mathbb{R}^n$ is a continuously differentiable function ($\mathbf{f}$ is $C^1$) on $A$ then if there exists a point $\mathbf{a} \in A$ for which the Jacobian determinant $\mathbf{J}_{\mathbf{f}} (\mathbf{a}) \neq 0$ then there exists open sets $X \subseteq A$ and $Y \subseteq f(A)$ and a function $\mathbf{g} : Y \to X$ for which:

• 1. $\mathbf{a} \in X$ and $\mathbf{f} (\mathbf{a}) \in Y$.

• 2. $\mathbf{f}$ is bijective on $X$.

• 3. $\mathbf{g}(Y) = X$ and $\mathbf{g}(\mathbf{f}(\mathbf{x})) = \mathbf{x}$ for all $\mathbf{x} \in X$.

• 4. $\mathbf{g}$ is continuously differentiable ($C^1$) on $Y$.

We will now look at an example of applying the Inverse Function theorem.

Example 1

Consider the function $f : \mathbb{R} \times (0, \infty) \times \mathbb{R} \to \mathbb{R}^3$ defined by $\mathbf{f}(x, y, z) = (x^3z + 4x, \sqrt{y}, xz^3 +z^2)$. Determine if $\mathbf{f}$ is locally invertible at the point $(x, y, z) = \left ( 1, \frac{1}{4}, -1 \right )$.

Set $A = \mathbb{R} \times (0, \infty) \times \mathbb{R}$. Then $A$ is clearly an open set and $\mathbf{f} : A \to \mathbb{R}^3$ is continuously differentiable on $A$ (since $A$ avoids the set of points in $\mathbb{R}^3$ with nonpositive $y$-coordinates).

We need to compute the Jacobian matrix of $\mathbf{f}$. We have that $\mathbf{f} = (f_1, f_2, f_3)$ with $f_1(x, y, z) = x^3z + 4x$, $f_(x, y, z) = \sqrt{y}$, and $f_3(x, y, z) = xz^3 + z^2$, and:

(1)

\begin{align} \mathbf{D} \mathbf{f} = \begin{bmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} & \frac{\partial f_1}{\partial z} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} & \frac{\partial f_2}{\partial z} \\ \frac{\partial f_3}{\partial x} & \frac{\partial f_3}{\partial y} & \frac{\partial f_3}{\partial z} \\ \end{bmatrix} = \begin{bmatrix} 3x^2z + 4 & 0 & x^3 \\ 0 & \frac{1}{2\sqrt{y}} & 0 \\ z^3 & 0 & 3xz^2 + 2z \\ \end{bmatrix} \end{align}

We now evaluate the determinant of the Jacobian matrix above at the point $(x, y, z) = \left ( 1, \frac{1}{4}, -1 \right )$:

(2)

\begin{align} \quad \mathbf{J}_{\mathbf{f}} \left (1, \frac{1}{4}, 1 \right ) = \mathrm{det} \left ( \begin{bmatrix} 3x^2z + 4 & 0 & x^3 \\ 0 & \frac{1}{2\sqrt{y}} & 0 \\ z^3 & 0 & 3xz^2 + 2z \\ \end{bmatrix}_{(x, y, z) = \left (1, \frac{1}{4}, -1 \right )} \right ) = \mathrm{det} \left ( \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \right ) = 2 \neq 0 \end{align}

So the conditions of the Inverse Function theorem are satisfied, i.e., $\mathbf{f}$ is locally invertible at $(x, y, z) = \left ( 1, \frac{1}{4}, -1 \right )$.