Bounded Function is of Exponential Order Zero

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $f: \hointr 0 \to \to \mathbb F$ be a function, where $\mathbb F \in \set {\R, \C}$.

Let $f$ be continuous everywhere on its domain, except possibly for some finite number of discontinuities of the first kind in every finite subinterval of $\hointr 0 \to$.


This article, or a section of it, needs explaining.
In particular: Establish whether it is "finite subinterval" that is needed here, or what we have already defined as "Definition:Finite Subdivision".
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Explain}} from the code.



This article, or a section of it, needs explaining.
In particular: Pin down a specific page on which the relevant definition of "Continuous" can be found for this context.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Explain}} from the code.


Let $f$ be bounded.


Then $f$ is of exponential order $0$.


Proof

Let $U$ be an upper bound of $f$.

Let $L$ be a lower bound of $f$.

Let $K > \max \set {\size U, \size L}$.

Then:

\(\ds \forall t \ge 1: \, \) \(\ds \size {\map f t}\) \(<\) \(\ds K\) Definition of Bounded Mapping
\(\ds \) \(=\) \(\ds K e^{0 t}\) Exponential of Zero

The result follows from the definition of exponential order with $M = 1$, $K = K$, and $a = 0$.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Bounded_Function_is_of_Exponential_Order_Zero&oldid=540214"