Carathéodory's Theorem (Analysis)

Theorem
Let $I \subseteq \R$.

Let $c \in I$ be an interior point of $I$.

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

Then $f$ is differentiable at $c$ :
 * There exists a real function $\varphi : I \to \R$ that is continuous at $c$ and satisfies:


 * $(1): \quad \forall x \in I: \map f x - \map f c = \map \varphi x \paren {x - c}$
 * $(2): \quad \map \varphi c = \map {f'} c$

Necessary Condition
Suppose $f$ is differentiable at $c$.

Then by definition $\map {f'} c$ exists.

So we can define $\varphi$ by:


 * $\map \varphi x = \begin{cases}

\dfrac {\map f x - \map f c} {x - c} & : x \ne c, x \in I \\ \map {f'} c & : x = c \end{cases}$

Condition $(2)$, that $\varphi$ is continuous at $c$, is satisfied, since:

Finally, condition $(1)$ is vacuous for $x = c$.

For $x \ne c$, it follows from the definition of $\varphi$ by dividing both sides of $(1)$ by $x - c$.

Sufficient Condition
Suppose a $\varphi$ as in the theorem statement exists.

Then for $x \ne c$, we have that:


 * $\map \varphi x = \dfrac {\map f x - \map f c} {x - c}$

Since $\varphi$ is continuous at $c$:


 * $\ds \map \varphi c = \lim_{x \mathop \to c} \map \varphi x = \lim_{x \mathop \to c} \frac {\map f x - \map f c} {x - c}$

That is, $f$ is differentiable at $c$, and $\map {f'} c = \map \varphi c$.