Function f is Big-O of g iff g is Big-Omega of f

Theorem
Let $f: \N \to \R, g: \N \to \R$ be two real sequences, expressed here as real-valued functions on the set of natural numbers $\N$.

Then:
 * $\map f n \in \map \OO {\map g n}$


 * $\map g n \in \map \Omega {\map f n}$
 * $\map g n \in \map \Omega {\map f n}$

where:
 * $\OO$ denotes big-$\OO$ notation
 * $\Omega$ denotes big-$\Omega$ notation.

Proof
Recall the definitions:


 * $\map \Omega f = \set {g: \N \to \R: \exists c_1 \in \R_{>0}: \exists n_1 \in \N: \forall n > n_1: 0 \le c_1 \cdot \size {\map f n} \le \size {\map g n} }$


 * $\map \OO g = \set {f: \N \to \R: \exists c_2 \in \R_{>0}: \exists n_2 \in \N: \forall n > n_2: 0 \le \size {\map f n} \le c_2 \cdot \size {\map g n} }$

for some $n_1, n_2 \in \N$.

Hence:
 * If $g \in \map \Omega f$, then $\forall n > n_1: \exists c_1 > 0 : \map g n \ge c_1 \cdot \map f n$


 * If $f \in \map \OO g$, then $\forall n > n_2: \exists c_2 > 0 : \map f n \le c_2 \cdot \map g n$

First let us assume that $n \ge \max \set {n_1, n_2}$ in all the below work.

Sufficient Condition
Let $g \in \map \Omega f$.

Then:

Hence we have:
 * $g \in \map \Omega f \implies f \in \map \OO g$

Necessary Condition
Let $f \in \map \OO g$.

Then:

Hence we have:
 * $f \in \map \OO g \implies g \in \map \Omega g$

Both conditions have been proven, hence:
 * $f \in \map \OO g \iff g \in \map \Omega f$