TheoremBase

The Implicit Function Theorem

If F(a, b) = 0 for a function F of class Cᵏ of (x, y) ∈ ℝⁿ × ℝᵐ with values in ℝᵐ, and the derivative of F in y at (a, b) is invertible, then near (a, b) the equation F(x, y) = 0 has exactly one solution y = g(x), and g is of class Cᵏ with the derivative obtained by differentiating F(x, g(x)) = 0.

Statement

In the setting of Functions of Class Cᵏ: Notation, with points (x,y)(x,y) of Rn+m\mathbb{R}^{n+m}, sets A×BA\times B and blocks DxFD_{x}F, DyFD_{y}F of the Jacobian matrix as in Functions of Class Cᵏ: Notation (vi), let k∈Nk\in\mathbb{N}, let UU be an open subset of Rn+m\mathbb{R}^{n+m} and let F:U→RmF:U\to\mathbb{R}^{m} be of class CkC^{k} on UU. Let (a,b)∈U(a,b)\in U satisfy F(a,b)=0F(a,b)=0, with DyF(a,b)D_{y}F(a,b) invertible. Then there are open sets A⊆RnA\subseteq\mathbb{R}^{n} and B⊆RmB\subseteq\mathbb{R}^{m} with a∈Aa\in A, b∈Bb\in B and A×B⊆UA\times B\subseteq U, and a function g:A→Bg:A\to B, with the following properties.

Implicit function. For all x∈Ax\in A and y∈By\in B,

F(x,y)=0⟺y=g(x).F(x,y)=0\quad\Longleftrightarrow\quad y=g(x).

In particular g(a)=bg(a)=b and F(x,g(x))=0F(x,g(x))=0 for every x∈Ax\in A.

Uniqueness. gg is the only function h:A→Bh:A\to B with F(x,h(x))=0F(x,h(x))=0 for every x∈Ax\in A.

Regularity. gg, regarded as a function A→RmA\to\mathbb{R}^{m}, is of class CkC^{k} on AA, the matrix DyF(x,g(x))D_{y}F(x,g(x)) is invertible for every x∈Ax\in A, and

Dg(x)=−DyF(x,g(x))−1DxF(x,g(x))for every x∈A.Dg(x)=-D_{y}F\bigl(x,g(x)\bigr)^{-1}D_{x}F\bigl(x,g(x)\bigr)\qquad\text{for every }x\in A .

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