Carathéodory's Theorem (Analysis)

Theorem
Let $I \subseteq \R$, and let $c \in I$ be an interior point of $I$.

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

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


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

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

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

So we can define $\varphi$ by:


 * $\displaystyle \varphi (x) = \begin{cases}

\frac {f(x) - f(c)} {x - c} & : x \ne c, x \in I \\ 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:


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

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


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

That is, $f$ is differentiable at $c$, and $f'(c) = \varphi (c)$.