Linearity of Function with Translation Property

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Theorem

Let $f$ be a real function.

Let $c$ be a real number.


Then:

$f$ has the translation property
$\map {f'} c$ exists

if and only if:

$f$ is linear


Proof

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:

\(\ds \map f x\) \(=\) \(\ds \map f x - \map f 0 + \map f 0\)
\(\ds \) \(=\) \(\ds \int_0^x \map {f'} y \rd y + \map f 0\) $f'$ is integrable by Integral of Constant
\(\ds \) \(=\) \(\ds \int_0^x \map {f'} c \rd y + \map f 0\) $\map {f'} y = \map {f'} c$
\(\ds \) \(=\) \(\ds \map {f'} c x + \map f 0\) Integral of Constant

So, $f$ is linear.

$\Box$


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:

\(\ds \map f {x_1 + t} - \map f {x_2 + t}\) \(=\) \(\ds a \paren {x_1 + t} + b - \paren {a \paren {x_2 + t} + b}\) Definition of Linear Real Function
\(\ds \) \(=\) \(\ds a x_1 + a t + b - \paren {a x_2 + a t + b}\)
\(\ds \) \(=\) \(\ds a x_1 + a t + b - \paren {a x_2 + b} - a t\)
\(\ds \) \(=\) \(\ds a x_1 + b - \paren {a x_2 + b}\)
\(\ds \) \(=\) \(\ds \map f {x_1} - \map f {x_2}\) Definition of Linear Real Function

So, $f$ has the translation property.


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

We have:

\(\ds \map {f'} c\) \(=\) \(\ds \lim_{h \mathop \to 0} \frac {\map f {c + h} - \map f c} h\) Definition 2 of Differentiable Mapping
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \frac {a (c + h) + b - (a c + b)} h\) Definition of Linear Real Function
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \frac {a c + a h + b - a c - b} h\)
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \frac {a h} h\)
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} a\)
\(\ds \) \(=\) \(\ds a\)

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

$\blacksquare$

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