Bounded Function is of Exponential Order Zero

Theorem
Let $f: \left[{ 0 \,.\,.\, \to}\right) \to \mathbb F$ be a function, where $\mathbb F \in \left\{ {\R, \C} \right\}$.

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 $\left [{0 \,.\,.\, \to} \right)$.

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 $K > \max \left({\left \vert { U } \right \vert, \left \vert { L }\right \vert}\right)$.

Then:

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