Bounded Function is of Exponential Order Zero

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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 $\hointr 0 \to$.

Let $f$ be bounded.

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


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

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

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


\(\ds \forall t \ge 1: \ \ \) \(\ds \size {\map f t}\) \(<\) \(\ds K\) 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 $a = 0$.