Carathéodory's Theorem

Theorem

Let $f : I \subset \R \to \R$ be a real function.

Then $f$ is differentiable at $c \in I$ if and only if:

$\exists \varphi : I \to \R$ that is continuous at $c$ and satisfies:

1. $\forall x \in I: f(x) - f(c) = \varphi (x)(x-c)$;
2. $\varphi (c) = f'(c)$.

Proof

If:

Suppose $f$ is differentiable at $c \in I$.

Then by definition $f'(c)$ exists.

So we can define $\varphi$ as:

$\displaystyle \varphi (x) = \begin{cases} \frac{f(x)-f(c)}{x-c} & : x \ne c, x \in I\\ f'(c) & : x=c\end{cases}$

$\varphi$ is continuous, as $\displaystyle \lim_{x \to c} \varphi (x) = \varphi (c)$ follows directly from the definition of the derivative.

This definition also meets $(1)$ by multiplying both sides by $x-c$ and $(2)$ directly.

Only If

Suppose $\varphi$ exists and is continuous at $c$.

Then $\displaystyle \varphi (c) = \lim_{x \to c} \varphi (x) = \lim_{x \to c} \frac{f(x)-f(c)}{x-c}$ exists.

Therefore, the function $f$ exists and is differentiable at $c$, and $f'(c) = \varphi (c)$.

$\blacksquare$

Source of Name

This entry was named for Constantin Carathéodory.