The Implicit Function Theorem for Functions from Rn to Rn

The Implicit Function Theorem for Functions from Rn to Rn

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

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

We will now state another important result known as the Implicit Function theorem.

Theorem 1 (The Implicit Function Theorem): Let $A \subseteq \mathbb{R}^{n+k}$ be open and let $\mathbf{f} : S \to \mathbb{R}^n$ be a continuously differentiable function on $S$ ($\mathbf{f}$ is $C^1$ on $S$). If $(\mathbf{x}_0 ; \mathbf{t}_0) \in S$ is such that $\mathbf{f}(\mathbf{x}_0;\mathbf{t}_0) = \mathbf{0}$ and for which the $n \times n$ Jacobian determinant, $\mathrm{det} (D_j f_i (\mathbf{x}_0, \mathbf{t}_0)) \neq 0$ then there exists a $k$-dimension open set $T_0$ with $\mathbf{t}_0 \in T$ and a unique $\mathbf{g} : T_0 \to \mathbb{R}^n$ for which:
a) $\mathbf{g}$ is continuously differentiable on $T_0$ ($\mathbf{g}$ is $C^1$ on $T_0$).
b) $\mathbf{g}(\mathbf{t}_0) = \mathbf{x}_0$.
c) $\mathbf{f}(\mathbf{g}(t); \mathbf{t}) = \mathbf{0}$ for all $\mathbf{t} \in T_0$.

The condition that $\mathrm{det} (D_j f_i (\mathbf{x}_0, \mathbf{t}_0)) \neq 0$ is extremely important in the Implicit Function theorem. For example, consider the case when $n = 1$, $k = 1$, and the function $f : \mathbb{R}^{n+k} \to \mathbb{R}^{n}$ (i.e., $f : \mathbb{R}^2 \to \mathbb{R}$) defined by:

(1)
\begin{align} \quad f(x, t) = x^2 - t \end{align}

Consider the point $(x_0, t_0) = (0, 0)$. The $n \times n = 1 \times 1$ Jacobian determinant of $f$ is simply $2x$, and $\displaystyle{2x \biggr \lvert_{(x_0, t_0) = (0, 0)} = 0}$ and the conclusions of the Implicit Function theorem do not hold.

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