Epsilon-Function Differentiability Condition

Theorem
Let $f: D \to \C$ be a continuous complex function, where $D \subseteq \C$ is an open set.

Let $z \in D$ be a complex number.

Then $f$ is differentiable at $z$ iff there exist $\alpha \in \C$ and $r \in \R_{>0}$ such that for all $h \in B_r \left({0}\right)$:


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

where $B_r \left({0}\right)$ denotes an open ball of $0$, and $o: B_r \left({0}\right) \ \to \C$ is a continuous function with $o\left({0}\right) = 0$.

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

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

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

Define $o: B_r \left({0}\right) \to \C$ by:


 * $o \left({h}\right) = \left\{ { \begin{array}{cc}\dfrac{ f \left({z + h}\right) - f \left({z}\right) }{ h } - f' \left({z}\right) & \quad \text{for} \ h \neq 0 \\ 0 & \quad \text{for}  \ h = 0 \end{array} }\right.$

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

It follows that $o \left({0}\right) = 0$ from its definition.

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

By definition of differentiability, it follows that


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

By definition of continuity, $o$ is continuous at $0$.

Then $o$ is continuous in its entire domain.

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


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

Sufficient condition
Rewrite the equation of the assumption to get:


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

By definition of continuouity, it follows that:


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

as $o$ is continuous at $0$ with $o \left({z}\right) = 0$.

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

Also see

 * Characterization of Differentiability