Linearity of Function defined using Function with Translation Property/Lemma
Jump to navigation
Jump to search
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:
- $\forall n \in \Z: \map {f_x} {n l} = n \map {f_x} l$
Proof
Let $n$ be an integer.
We have:
\(\ds \map {f_x} {n l}\) | \(=\) | \(\ds \map f {x + n l} - \map f x\) | Definition of $f_x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {\map f x - \map f {x + n l} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {\sum_{k \mathop = 1}^n \map f {x + \paren {k - 1} l} - \map f {x + \paren {x + k l} } }\) | Telescoping Series | |||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {\sum_{k \mathop = 1}^n \map f x - \map f {x + l} }\) | Definition of Translation Property | |||||||||||
\(\ds \) | \(=\) | \(\ds -n \paren {\map f x - \map f {x + l} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n \paren {\map f {x + l} - \map f x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n \map {f_x} l\) | Definition of $f_x$ |
$\blacksquare$