# Alternative Differentiability Condition/Proof 1 It has been suggested that this article or section be renamed: The use of "Alternative" for definitions / proofs is suboptimal One may discuss this suggestion on the talk page.

## Theorem

Let $\mathbb K$ be either $\R$ or $\C$.

Let $f: D \to \mathbb K$ be a continuous mapping, where $D \subseteq \mathbb K$ is an open set.

Let $z \in \mathbb K$.

Then:

$f$ is differentiable at $z$
there exist $\alpha \in \mathbb K$ and $r \in \R_{>0}$ such that for all $h \in B_r \left({0}\right) \setminus \left\{ {0}\right\}$:
$f \left({z + h}\right) = f \left({z}\right) + h \left({\alpha + \epsilon \left({h}\right) }\right)$

where:

$B_r \left({0}\right)$ denotes an open ball of $0$
$\epsilon: B_r \left({0}\right) \setminus \left\{ {0}\right\} \to \mathbb K$ is a continuous mapping with $\displaystyle \lim_{h \to 0} \ \epsilon \left({h}\right) = 0$.

If the conditions are true, then $\alpha = f' \left({z}\right)$.

## Proof

This proof assumes that $\mathbb K = \R$.

### Necessary Condition

Assume that $f$ is differentiable in $z$.

By definition of open set, there exists $r \in \R_{>0}$ such that $\left({z - r \,.\,.\, z + r}\right) \subseteq D$.

We have the open ball $B_r \left({0}\right) = \left({-r \,.\,.\, r}\right)$.

Define $\epsilon: \left({-r \,.\,.\, r}\right) \setminus \left\{ {0}\right\} \to \R$ by:

$\epsilon \left({h}\right) = \dfrac{f \left({z + h}\right) - f \left({z}\right)} h - f' \left({z}\right)$

If $h \in \left({-r \,.\,.\, r}\right) \setminus \left\{ {0}\right\}$, then $z+h \in \left({z - r \,.\,.\, z + r}\right) \setminus \left\{ {z}\right\} \subseteq D$, so $\epsilon$ is well-defined.

From Continuity of Composite Mapping/Corollary, it follows that $\epsilon$ is continuous in $\left({-r \,.\,.\, r}\right) \setminus \left\{ {0}\right\}$.

As $f$ is differentiable at $z$, it follows that:

$\displaystyle \lim_{h \to 0} \ \epsilon \left({h}\right) = \lim_{h \to 0} \dfrac {f \left({z + h}\right) - f \left({z}\right)} h - f' \left({z}\right) = f' \left({z}\right) - f' \left({z}\right) = 0$

If we put $\alpha = f' \left({z}\right)$, it follows that for all $h \in \left({-r \,.\,.\, r}\right) \setminus \left\{ {0}\right\}$:

$f \left({z + h}\right) = f \left({z}\right) + h \left({\alpha + \epsilon \left({h}\right) }\right)$

$\Box$

### Sufficient condition

Rewrite the equation of the assumption to get:

$\dfrac{f \left({z + h}\right) - f \left({z}\right)} h = \alpha + \epsilon \left({h}\right)$

From Sum Rule for Limits of Functions, it follows that:

$\displaystyle \lim_{h \to 0} \dfrac{f \left({z + h}\right) - f \left({z}\right)} h = \lim_{h \to 0} \left({\alpha + \epsilon \left({h}\right)}\right) = \alpha$

By definition of differentiability, $f$ is differentiable at $z$ with $f' \left({z}\right) = \alpha$.

$\blacksquare$