Bounded Function is of Exponential Order Zero
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 $\hointr 0 \to$.
Establish whether it is "finite subinterval" that is needed here, or what we have already defined as "Definition:Finite Subdivision".
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Pin down a specific page on which the relevant definition of "Continuous" can be found for this context.
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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
- 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Functions of Exponential Order