Alternative Differentiability Condition/Proof 2

Proof
This proof assumes that $\mathbb K = \C$.

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

By definition of open set, there exists $r \in \R_{>0}$ such that the open ball $\map {B_r} z \subseteq D$.

Define $\epsilon: \map {B_r} 0 \setminus \set 0 \to \C$ by:


 * $\map \epsilon h = \dfrac {\map f {z + h} - \map f z} h - \map {f'} z$

If $h \in \map {B_r} 0 \setminus \set 0$, then $z + h \in \map {B_r} z \setminus \set z \subseteq D$, so $\epsilon$ is well-defined.

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


 * $\ds \lim_{h \mathop \to 0} \map \epsilon h = \lim_{h \mathop \to 0} \dfrac {\map f {z + h} - \map f z} h - \map {f'} z = \map {f'} z - \map {f'} z = 0$

If we put $\alpha = \map {f'} z$, it follows that for all $h \in \map {B_r} 0 \setminus \set 0$:


 * $\map f {z + h} = \map f z + h \paren {\alpha + \map \epsilon h}$

Sufficient condition
Rewrite the equation of the assumption as:


 * $\dfrac {\map f {z + h} - \map f z} h = \alpha + \map \epsilon h$

From Sum Rule for Limits of Complex Functions:


 * $\ds \lim_{h \mathop \to 0} \dfrac {\map f {z + h} - \map f z} h = \lim_{h \mathop \to 0} \paren { \alpha + \map \epsilon h} = \alpha$

By definition of differentiability, $f$ is differentiable at $z$ with $\map {f'} z = \alpha$.

Also see

 * Characterization of Differentiability