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The chain rule.

Fundamental to many of the constructions we want to consider in the following sections is the chain rule:

Theorem 1.1 (Chain Rule.)   Let $ U \subset
\mathbb{R}^n$ be open , $ f\colon U \to
\mathbb{R}^m$, $ V \subset
\mathbb{R}^m$ open and $ g \colon V \to
\mathbb{R}^k$ with $ f(U) \subset V$. Let $ x \in U$. Then:

$\displaystyle d(f \circ g) (x) = dg(f(x)) \circ df(x).
$

The composition on the right hand side is the composition of linear operators. In particular if we expand both sides in terms of the standard basis of $ \mathbb{R}^n$ then we have

$\displaystyle {\partial _j( g^i \circ f)} (x) = \sum_{l=1}^m {\partial _l g^i }
\circ {\partial _j f^l }
$

An important part of the chain rule is the fact that the composition of smooth functions is also smooth. A partial converse of this result will be important in the sequel.

Lemma 1.1   Let $ U$ be an open subset of $ \mathbb{R}^n$ and $ V $ an open subset of $ \mathbb{R}^m$. A function $ \phi \colon U \to V$ is smooth if and only if for every smooth function $ f \colon V \to
\mathbb{R}$ the composite $ f \circ \phi \colon U \to
\mathbb{R}$ is smooth.

Proof. If $ \phi$ is smooth then the result follows via the chain rule. If the result is true then take $ f$ to be the restriction to $ V $ of each of the co-ordinate functions $ x^i$. Then $ x^i$ is smooth so $ x^i \circ \phi = \phi^i$ is smooth. $ \qedsymbol$


next up previous contents
Next: Diffeomorphisms and the inverse Up: Co-ordinate independent calculus. Previous: Derivatives as linear operators.   Contents
Michael Murray
1998-09-16