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)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.
