State the most general version of the implicit function theorem for real functions and outline how it can be proved using the inverse function theorem.
Solution
1. Implicit Function Theorem. Let \(F : \mathbb{R}^{n+m} \to \mathbb{R}^m\) be \(C^k\) (\(k \ge 1\)), and let \((a, b) \in \mathbb{R}^n \times \mathbb{R}^m\) with \(F(a, b) = 0\). If the \(m \times m\) matrix \(\frac{\partial F}{\partial y}(a, b)\) (the partial derivatives with respect to the last \(m\) variables) is invertible, then there are open neighborhoods \(U \ni a\) and \(V \ni b\) and a unique \(C^k\) function \(g : U \to V\) such that \(g(a) = b\) and \(F(x, g(x)) = 0\) for all \(x \in U\).
statement of the theorem.
2. Define \(\Phi : \mathbb{R}^{n+m} \to \mathbb{R}^{n+m}\) by \(\Phi(x, y) = (x, F(x, y))\).
augment \(F\) with the identity on the first \(n\) coordinates.
3. The Jacobian of \(\Phi\) at \((a, b)\) is\[D\Phi(a,b) = \begin{pmatrix} I_n & 0 \\ \frac{\partial F}{\partial x}(a,b) & \frac{\partial F}{\partial y}(a,b) \end{pmatrix},\]which is invertible because \(\frac{\partial F}{\partial y}(a,b)\) is invertible.
block matrix; its determinant is \(\det \frac{\partial F}{\partial y}(a,b) \neq 0\).
4. By the inverse function theorem, \(\Phi\) has a \(C^k\) local inverse \(\Psi\) near \((a, b)\).
3 and the inverse function theorem.
5. Write \(\Psi(x, z) = (x, \psi(x, z))\); then \(\Phi(x, \psi(x,z)) = (x, z)\), so \(F(x, \psi(x,z)) = z\).
4, matching the first \(n\) coordinates.
6. Define \(g(x) = \psi(x, 0)\).
set \(z = 0\).
7. Then \(F(x, g(x)) = F(x, \psi(x,0)) = 0\), and \(g(a) = \psi(a, 0) = b\) (since \(\Phi(a,b) = (a, 0)\)).
5 and 6.
8. Hence \(g\) is the desired implicit function, proving the implicit function theorem from the inverse function theorem.
7.
9. Q.E.D.
8.