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
Definitions for reference :

If $g \in \Omega(f)$, then $\exists c_1 > 0 : g(n) \ge c_1 \cdot f(n) \quad \forall n > N_1$

If $f \in \OO(g)$, then $\exists c_2 > 0 : f(n) \le c_2 \cdot g(n) \quad \forall n > N_2$

Firstly, assume that $g \in \Omega(f)$. Hence $g(n) \ge c_1 \cdot f(n)$. Dividing both sides of the inequality by $c_1$ ($c_1 \neq 0$ by definition) we get $\frac 1 {c_1} \cdot g(n) \ge f(n)$. By setting $c_2 := \frac 1 {c_1}$ we reach the definition of $\OO$ (as shown above) and so $g \in \Omega(f) \implies f \in \OO(g)$.

Secondly, assume that $f \in \OO(g)$. Hence $f(n) \le c_2 \cdot g(n)$. Dividing both sides of the inequality by $c_2$ ($c_2 \neq 0$ by definition) we get $\frac 1 {c_2} \cdot f(n) \le f(n)$. By setting $c_2 := \frac 1 {c_1}$ we reach the definition of $\Omega$ (as shown above) and so $f \in \OO(f) \implies g \in \Omega(f)$.

Both conditions have been proven, hence $f \in \OO(g) \iff g \in \Omega(f) \quad\quad \blacksquare$