Product of Big-O Estimates/Real Analysis

Theorem
Let $c$ be a real number.

Let $f, g : \hointr c \infty \to \R$ be real functions.

Let $R_1 : \hointr c \infty \to \R$ be a real function such that $f = \map \OO {R_1}$.

Let $R_2 : \hointr c \infty \to \R$ be a real function such that $g = \map \OO {R_2}$.

Then:


 * $f g = \map \OO {R_1 R_2}$

Proof
Since:


 * $f = \map \OO {R_1}$

there exists $x_1 \in \hointr c \infty$ and a real number $C_1$ such that:


 * $\size {\map f x} \le C_1 \size {\map {R_1} x}$

for $x \ge x_1$.

Similarly, since:


 * $g = \map \OO {R_2}$

there exists $x_2 \in \hointr c \infty$ and a real number $C_2$ such that:


 * $\size {\map g x} \le C_2 \size {\map {R_2} x}$

Set:


 * $x_0 = \max \set {x_1, x_2}$

Then for $x \ge x_0$ we have:

That is, by the definition of big-O notation, we have:


 * $f g = \map \OO {R_1 R_2}$