Linearity of Function with Translation Property

Theorem
Let $f$ be a real function.

Let $c$ be a real number.

Then:
 * $f$ has the translation property


 * $\map {f'} c$ exists


 * $f$ is linear
 * $f$ is linear

Sufficient Condition
Let:
 * $f$ have the translation property


 * $\map {f'} c$ exist

We need to show that:
 * $f$ is linear

The fact that $f$ has the translation property means:
 * $\forall x_1, x_2, t \in \R: \map f {x_1 + t} - \map f {x_2 + t} = \map f {x_1} - \map f {x_2}$

$f$ being linear means:
 * $\forall x \in \R: \map f x = ax + b$

where $a$ and $b$ are real numbers.

We have that $f'$ exists by Differentiability of Function with Translation Property.

Also, $f'$ is a constant function by the same theorem.

Accordingly:
 * $\map {f'} x = \map {f'} c$ for every real number $x$

Let $x$ be a real number.

We have:

So, $f$ is linear.

Necessary Condition
Let:
 * $f$ be linear

We need to show that:
 * $f$ has the translation property


 * $\map {f'} c$ exist

The linearity of $f$ means that:
 * $\forall x \in \R: \map f x = ax + b$

where $a$ and $b$ are real numbers.

$f$ having the translation property means that:
 * $\forall x_1, x_2, t \in \R: \map f {x_1 + t} - \map f {x_2 + t} = \map f {x_1} - \map f {x_2}$

Let $x_1, x_2, t$ be real numbers.

We have:

So, $f$ has the translation property.

It remains to prove that $\map {f'} c$ exists.

We have:

This successful calculation of $\map {f'} c$ demonstrates the existence of $\map {f'} c$.