Carathéodory’s Differentiation Criterion

# Carathéodory’s Differentiation Criterion

Theorem 1 (Carathéodory’s Differentiation Criterion): Let $f$ be defined on an interval $I$ and let $a \in I$. Then $f$ if differentiable at $a$ if and only if there exists a continuous function $\varphi$ defined on $I$ such that $f(x) - f(a) = \varphi(x)(x - a)$. |

**Proof:**$\Rightarrow$ Suppose that $f$ is differentiable at $a \in I$. Then $f'(a)$ exists and:

\begin{align} \quad f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \end{align}

- Define a function $\varphi$ on $I$ by:

\begin{align} \quad \varphi(x) = \left\{\begin{matrix} f'(a) & \mathrm{if} \: \mathrm{if} \: x = a\\ \frac{f(x) - f(a)}{x - a} & \mathrm{if} \: x \neq a \end{matrix}\right. \end{align}

- Then $\varphi$ is continuous on $I$ and is such that $f(x) - f(a) = \varphi(x)(x - a)$.

- $\Leftarrow$ Suppose that there exists a continuous function $\varphi$ on $I$ such that $f(x) - f(a) = \varphi(x)(x - a)$. Then for $x \neq a$ we have that:

\begin{align} \quad \varphi(x) = \frac{f(x) - f(a)}{x - a} \end{align}

- Since $\varphi$ is continuous - it's limit at $a$ exists and equals $\varphi(a)$, and so:

\begin{align} \quad \varphi(a) = \lim_{x \to a} \varphi(x) = \lim_{x \to a}\frac{f(x) - f(a)}{x - a} = f'(a) \end{align}

- So $f'(a)$ exists and hence $f$ is differentiable at $a$. $\blacksquare$