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The parametrized implicit function theorem with $C^{k}$ regularity

Statement

Let $k, m, n \geq 1$ and $p \in N$ , let $U \subseteq R^{m + n + p}$ be open, and let $F : U \to R^{n}$ be $C^{k}$ . Suppose $F (a, b, λ_{0}) = 0$ and $D_{y} F (a, b, λ_{0})$ is invertible. Then there are neighbourhoods and a unique $C^{k}$ map $φ$ solving $F (x, φ (x, λ), λ) = 0$ . More precisely, on suitable open neighbourhoods $P$ of $(a, λ_{0})$ and $Q$ of $b$ ,

$F (x, y, λ) = 0 ⟺ y = φ (x, λ),$

and

$D φ (x, λ) = - D_{y} F (x, φ (x, λ), λ)^{- 1} D_{(x, λ)} F (x, φ (x, λ), λ) .$

When no parameter block is present, this is the ordinary $C^{k}$ implicit function theorem.

Facts & Assumptions

Given: The dimensions and hypotheses in the Statement. The block map used below is $C^{k}$ by $C^{k}$ Euclidean maps are closed under componentwise algebra and composition, and matrix inversion has the regularity of Matrix inversion preserves $C^{k}$ regularity where the determinant is nonzero.

[L1]

With an invertible second-block derivative, the $C^{1}$ implicit theorem gives open neighbourhoods $P$ of $a$ and $Q$ of $b$ , and a unique $C^{1}$ map solving the equation, together with its derivative formula (The Euclidean implicit function theorem with derivative formula).

[L2]

If a map is $C^{k}$ for $k \geq 1$ , every local inverse supplied by the inverse function theorem is $C^{k}$ (A local inverse of a $C^{k}$ regular map is $C^{k}$ ).

[L3]

A $C^{1}$ map with invertible derivative has a local $C^{1}$ inverse (The Euclidean inverse function theorem).

Proof

technique · direct

1.1

Put $\tilde{U} = {(x, λ, y) : (x, y, λ) \in U}$ and $\tilde{F} (x, λ, y) = F (x, y, λ)$ . The coordinate permutation is linear, so $\tilde{U}$ is open and $\tilde{F}$ is $C^{k}$ . Regard $u = (x, λ)$ as the first variable block and apply [L1] to $\tilde{F}$ . It gives neighbourhoods $P, Q$ , a unique $C^{1}$ map $φ : P \to Q$ , the equivalence $F (x, y, λ) = 0$ if and only if $y = φ (x, λ)$ , and the combined-block derivative formula.

L1givenalgebra

2.1

On $\tilde{U}$ , the block map $H (x, λ, y) = (x, λ, F (x, y, λ))$ is $C^{k}$ . Its derivative at the base point is block triangular with identity on the first block and invertible block $D_{y} F$ on the second, so [L3] supplies a local inverse $G$ . By [L2], $G$ is $C^{k}$ . The identity $H (G (u, z)) = (u, z)$ forces $G (u, z) = (u, ψ (u, z))$ ; hence $u \mapsto ψ (u, 0)$ is a $C^{k}$ solution of the equation. After intersecting the neighbourhoods, uniqueness in step 1.1 identifies this solution with $φ$ .

step 1.1L2L3givenalgebra

3.1

Step 1.1 already supplies the displayed derivative formula for the unique $C^{1}$ solution, while step 2.1 upgrades that same solution to $C^{k}$ . Thus all regularity, equivalence, uniqueness, and derivative claims hold, including $p = 0$ .

step 1.1step 2.1∎

Depends on

Used by

Two equations implicitly determine two variables near the origin Example

Dependency tree · two levels

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Sources

J. Lebl, Basic Analysis II, Theorem 8.5.6 and Remark 8.5.8 (standard reference, not scraped)
University of Toronto MAT237, §3.3 (standard reference, not scraped)

The parametrized implicit function theorem with CkCk regularity