Raising Exponential Order

Theorem

Let $\map f t: \R \to \mathbb F$ a function, where $\mathbb F \in \set {\R, \C}$.

Let $f$ be continuous on the real interval $\hointr 0 \to$, except possibly for some finite number of discontinuities of the first kind in every finite subinterval of $\hointr 0 \to$.

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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:

$\forall t \ge M: \size {\map f t} < K e^{a t}$

From Exponential is Strictly Increasing, we have:

$K e^{a t} < K e^{b t}$

Therefore:

$\forall t \ge M: \size {\map f t} < K e^{b t}$

The result follows from the definition of exponential order.

$\blacksquare$