Scalar Multiple of Function of Exponential Order

Theorem

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

Let $\lambda$ be a complex constant.

Suppose $f$ is of exponential order $a$.

Then $\lambda f$ is also of exponential order $a$.

Proof

If $\lambda = 0$, the theorem holds trivially.

Let $\lambda \ne 0$.

\(\ds \size {\map f t}\)

\(<\)

\(\ds K e^{a t}\)

Definition of Exponential Order to Real Index

\(\ds \leadsto \ \ \)

\(\ds \size \lambda \size {\map f t}\)

\(<\)

\(\ds \size \lambda K e^{a t}\)

\(\ds \leadsto \ \ \)

\(\ds \size {\lambda \, \map f t}\)

\(<\)

\(\ds K' e^{a t}\)

Modulus of Product, $K' = \size \lambda K$

$\blacksquare$