# Alternative Differentiability Condition

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## 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 $z$ if and only if there exist $\alpha \in \mathbb K$ and $r \in \R_{>0}$ such that for all $h \in \map {B_r} 0 \setminus \set 0$:

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

where $\map {B_r} 0$ denotes an open ball of $0$, and $\epsilon: \map {B_r} 0 \setminus \set 0 \to \mathbb K$ is a continuous mapping with $\displaystyle \lim_{h \mathop \to 0} \map \epsilon h = 0$.

If the conditions are true, then $\alpha = \map {f'} z$.

## Proof 1

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 $\openint {z - r} {z + r} \subseteq D$.

We have the open ball $\map {B_r} 0 = \openint {-r} r$.

Define $\epsilon: \openint {-r} r \setminus \set 0 \to \R$ by:

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

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

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

As $f$ is differentiable at $z$, 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 $h \in \openint {-r} r \setminus \set 0$:

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

$\Box$

### Sufficient condition

Rewrite the equation of the assumption to get:

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

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

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

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

$\blacksquare$

## Proof 2

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 $h \in \map {B_r} 0 \setminus \set 0$:

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

$\Box$

### Sufficient condition

Rewrite the equation of the assumption as:

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

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

$\blacksquare$