Monday, January 28, 2013 CC-BY-NC
Carathéodory's Theorem and set-up for the chain rule

Maintainer: admin

1Motivation behind the chain rule

We can generalize the product rule and (by induction) show that
$$ (f^n)'(c) = n(f^{n-1})(c) \cdot f'(c) $$
$f^n$ is the composition of $f(x)$ and $g(x) = x^n$. Naturally the question arises: what is the derivative of two composed functions? Of course, we already know the answer: $(g \circ f)(x) = g'(f(x)) \cdot f'(x)$

If we try to prove this the usual way, we would take the route of writing

$$ \frac{g(f(x)) - g(f(c))}{x-c} = \frac{g(f(x))-g(f(c))}{f(x)-f(c)}\cdot\frac{f(x)-f(c)}{x-c} $$

While this seems like the right way to go, it also presents a problem: the first factor on the right is undefined when $f(x)-f(c) = 0$ for values of $x$ near $c$. To avoid this, we introduce Carathéodory's Theorem:

2Carathéodory's Theorem

This theorem must have some other name in real life since it doesn't seem to be here. Anyway:

Theorem: (Carathéodory's) Let $f$ be defined on an interval $I$ and $c \in I$. $f$ is differentiable at $c \in I$ if and only if there exists a function $\phi$ on $I$ continuous at $c$ such that $f(x)-f(c)=\phi(x)(x-c)$ for all $x \in I$. In this case, $f'(c) = \phi(c)$.

Proof: ($\Rightarrow$) if $f'(c)$ exists, define

$$ \phi(x) = \left\{ \begin{array}{lr} \frac{f(x)-f(c)}{x-c} & x \neq c \\ f'(c) & x = c \end{array} \right. $$

Then $\phi$ is defined on $I$ and since $f$ is differentiable at $c$, for a given $\epsilon > 0$ there exists some $\delta > 0$ such that $0 <|x-c|<\delta$ then

$$ \left|\frac{f(x)-f(c)}{x-c} - f'(c)\right| < \epsilon, \text{ i.e. } |\phi(x)-\phi(c)| < \epsilon $$

If $x=c$, then $|\phi(x) - \phi(c)| = 0 < \epsilon$ thus if $|x-c|<\delta$ then $|\phi(x) - \phi(c)| < \epsilon$

($\Leftarrow$) Suppose there exists a function $\phi$ continuous at $c$ such that $f(x)-f(c) = \phi(x)(x-c)$. Then for $x \neq c$ $\phi(x) = \frac{f(x) - f(c)}{x-c}$. Since $\phi$ is continuous at $c$, for a given $\epsilon > 0$ there exists some $\delta > 0$ such that if $|x-c|<\delta$ then $|\phi(x)-\phi(c)|<\epsilon$. So, in particular, if $0<|x-c|<\delta$ then $\left|\frac{f(x)-f(c)}{x-c} - \phi(c)\right| < \epsilon$, i.e. $f$ is differentiable at $c$ and $f'(c) = \phi(c)$.