Differentiability of Function with Translation Property

Theorem

Let $f$ be a real function.

Let $f$ have the translation property.

Let $c$ be a real number.

Let $\map {f'} c$ exist.

Then:

$f'$ exists

$f'$ is a constant function

Proof

Let $x$ be a real number.

We have:

\(\ds \map {f'} x\)

\(=\)

\(\ds \lim_{y \mathop \to x} \frac {\map f y - \map f x} {y - x}\)

Definition 1 of Differentiable Mapping

\(\ds \)

\(=\)

\(\ds \lim_{y \mathop \to x} \frac {\map f {y - x + x - c + c} - \map f {x - c + c} } {y - x}\)

\(\ds \)

\(=\)

\(\ds \lim_{y \mathop \to x} \frac {\map f {y - x + c + x - c} - \map f {c + x - c} } {y - x}\)

\(\ds \)

\(=\)

\(\ds \lim_{y \mathop \to x} \frac {\map f {y - x + c + t} - \map f {c + t} } {y - x}\)

$t = x - c$

\(\ds \)

\(=\)

\(\ds \lim_{z \mathop + x \mathop \to x} \frac {\map f {z + c + t} - \map f {c + t} } z\)

$z = y - x$

\(\ds \)

\(=\)

\(\ds \lim_{z \mathop \to 0} \frac {\map f {z + c + t} - \map f {c + t} } z\)

\(\ds \)

\(=\)

\(\ds \lim_{z \mathop \to 0} \frac {\map f {z + c} - \map f c} z\)

Definition of Translation Property

\(\ds \)

\(=\)

\(\ds \map {f'} c\)

Definition 2 of Differentiable Mapping

We conclude that $\map {f'} x$ exists as $\map {f'} c$ exists.

Also, $\map {f'} x$ exists everywhere as $x$ is arbitrary.

In other words, $f'$ exists.

Also, $f'$ is a constant function as $\map {f'} x$ equals $\map {f'} c$ for every real number $x$.

$\blacksquare$