Raising Exponential Order

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

Let $f$ be continuous on the real interval $\left [{0 \,.\,.\, \to} \right)$, except possibly for some finite number of discontinuities of the first kind in every finite subinterval of $\left [{0 \,.\,.\, \to} \right)$.

Let $f$ be of exponential order $a$.

Let $b > a$.

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

Proof
From the definition of exponential order, there exist strictly positive real numbers $M$ and $K$ such that:
 * $\forall t \ge M: \left\vert {f \left({t}\right)} \right \vert < K e^{a t}$

From Exponential is Strictly Increasing, we have:
 * $K e^{a t} < K e^{b t}$

Therefore:
 * $\forall t \ge M: \left\vert {f \left({t}\right)} \right \vert < K e^{b t}$

The result follows from the definition of exponential order.