# Bounded Function is of Exponential Order Zero

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## 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 $\hointr 0 \to$.

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