Carathéodory's Theorem (Analysis)

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) $$\varphi (c) = f'(c)$$.

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

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

So we can define $$\varphi$$ as:
 * $$\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 $$\lim_{x \to c} \varphi (x) = \varphi (c)$$ falls 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 $$\varphi (c) = \lim_{x \to c} \varphi (x) = \lim_{x \to c} \frac{f(x)-f(c)}{x-c}$$ exists.

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