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Math Notes

University Mathematics

TheoremComplete

Implicit Function Theorem (Proof Sketch)

The Implicit Function Theorem is proved by reducing it to the Inverse Function Theorem. The key is to construct an auxiliary map G(x,y)=(x,F(x,y))G(\mathbf{x}, \mathbf{y}) = (\mathbf{x}, F(\mathbf{x}, \mathbf{y})) whose Jacobian is nonsingular when ∂F∂y\frac{\partial F}{\partial \mathbf{y}} is nonsingular. Inverting GG gives the implicit function.

Statement

Theorem10.1Implicit Function Theorem

Let F:Rn+m→RmF : \mathbb{R}^{n+m} \to \mathbb{R}^m be continuously differentiable with F(a,b)=0F(\mathbf{a}, \mathbf{b}) = \mathbf{0}. If ∂F∂y(a,b)\frac{\partial F}{\partial \mathbf{y}}(\mathbf{a}, \mathbf{b}) is invertible, then there exist neighborhoods UU of a\mathbf{a} and VV of b\mathbf{b} and a unique continuously differentiable function g:U→Vg : U \to V such that:

  1. g(a)=bg(\mathbf{a}) = \mathbf{b}.
  2. F(x,g(x))=0F(\mathbf{x}, g(\mathbf{x})) = \mathbf{0} for all x∈U\mathbf{x} \in U.
  3. Jg(x)=−[∂F∂y(x,g(x))]−1∂F∂x(x,g(x))Jg(\mathbf{x}) = -[\frac{\partial F}{\partial \mathbf{y}}(\mathbf{x}, g(\mathbf{x}))]^{-1} \frac{\partial F}{\partial \mathbf{x}}(\mathbf{x}, g(\mathbf{x})).

Proof Sketch

Proof

Define G:Rn+m→Rn+mG : \mathbb{R}^{n+m} \to \mathbb{R}^{n+m} by

G(x,y)=(x,F(x,y)).G(\mathbf{x}, \mathbf{y}) = (\mathbf{x}, F(\mathbf{x}, \mathbf{y})).

Step 1: Jacobian of GG. The Jacobian of GG at (a,b)(\mathbf{a}, \mathbf{b}) is

JG(a,b)=(In0∂F∂x∂F∂y).JG(\mathbf{a}, \mathbf{b}) = \begin{pmatrix} I_n & 0 \\ \frac{\partial F}{\partial \mathbf{x}} & \frac{\partial F}{\partial \mathbf{y}} \end{pmatrix}.

Since ∂F∂y(a,b)\frac{\partial F}{\partial \mathbf{y}}(\mathbf{a}, \mathbf{b}) is invertible, JG(a,b)JG(\mathbf{a}, \mathbf{b}) is invertible (block matrix with invertible diagonal blocks).

Step 2: Apply Inverse Function Theorem. By the Inverse Function Theorem, GG has a local inverse G−1G^{-1} near (a,0)(\mathbf{a}, \mathbf{0}) (since G(a,b)=(a,0)G(\mathbf{a}, \mathbf{b}) = (\mathbf{a}, \mathbf{0})). Write

G−1(x,w)=(x,h(x,w))G^{-1}(\mathbf{x}, \mathbf{w}) = (\mathbf{x}, h(\mathbf{x}, \mathbf{w}))

for some function hh (the first component is just the identity by the definition of GG).

Step 3: Define gg. Set g(x)=h(x,0)g(\mathbf{x}) = h(\mathbf{x}, \mathbf{0}). Then

G(x,g(x))=G(x,h(x,0))=G(G−1(x,0))=(x,0).G(\mathbf{x}, g(\mathbf{x})) = G(\mathbf{x}, h(\mathbf{x}, \mathbf{0})) = G(G^{-1}(\mathbf{x}, \mathbf{0})) = (\mathbf{x}, \mathbf{0}).

Thus F(x,g(x))=0F(\mathbf{x}, g(\mathbf{x})) = \mathbf{0} for all x\mathbf{x} near a\mathbf{a}.

Step 4: Derivative formula. Differentiating F(x,g(x))=0F(\mathbf{x}, g(\mathbf{x})) = \mathbf{0} using the chain rule,

∂F∂x+∂F∂yJg=0,\frac{\partial F}{\partial \mathbf{x}} + \frac{\partial F}{\partial \mathbf{y}} Jg = 0,

so Jg=−[∂F∂y]−1∂F∂xJg = -[\frac{\partial F}{\partial \mathbf{y}}]^{-1} \frac{\partial F}{\partial \mathbf{x}}.

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Summary

The Implicit Function Theorem:

  • Proved by reducing to Inverse Function Theorem via auxiliary map G(x,y)=(x,F(x,y))G(\mathbf{x}, \mathbf{y}) = (\mathbf{x}, F(\mathbf{x}, \mathbf{y})).
  • Guarantees local solvability of F(x,y)=0F(\mathbf{x}, \mathbf{y}) = \mathbf{0} for y=g(x)\mathbf{y} = g(\mathbf{x}).
  • Derivative formula: g′=−[Fy]−1Fxg' = -[F_{\mathbf{y}}]^{-1} F_{\mathbf{x}}.

See Implicit Function Theorem concept and Lagrange Multipliers.