Linearity of Function defined using Function with Translation Property
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Theorem
Let $f$ be a real function.
Let $f$ have the translation property.
Let $x$ and $l$ be real numbers.
Define:
- $\map {f_x} l = \map f {x + l} - \map f x$
Then:
- $\forall q \in \Q: \map {f_x} {q l} = q \map {f_x} l$
Proof
Lemma
Let $f$ be a real function.
Let $f$ have the translation property.
Let $x$ and $l$ be real numbers.
Define:
- $\map {f_x} l = \map f {x + l} - \map f x$
Then:
- $\forall n \in \Z: \map {f_x} {n l} = n \map {f_x} l$
$\Box$
Let $q$ be a rational number.
Choose integers $n$, $m$ such that:
- $\dfrac n m = q$
We need to prove that:
- $\map {f_x} {q l} = q \map {f_x} l$
We have:
\(\ds \map {f_x} {q l}\) | \(=\) | \(\ds \map {f_x} {\frac n m l}\) | Definition of $n$, $m$ | |||||||||||
\(\ds \) | \(=\) | \(\ds n \map {f_x} {\frac l m}\) | Lemma | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac n m m \map {f_x} {\frac l m}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac n m \map {f_x} {\frac {m l} m}\) | Lemma | |||||||||||
\(\ds \) | \(=\) | \(\ds q \map {f_x} l\) | Definition of $n$, $m$ |
$\blacksquare$