I have the following definition of the Big-O notation:
$$ \begin{align*} &f(x)=O(g(x))\text{ as }x\to a\\ &\qquad\text{iff}\quad \exists m>0, \;\exists \delta>0 \Big(0<|{x-a}|<\delta\;\to\;|{f(x)}|\le m\,g(x)\Big). \end{align*} $$
But given the previous definition the following property/expression, that I would like to prove, is not defined yet: $$ f\cdot O(g)=O(fg).\tag{*} $$
What does (*) means? (Technically, practically i think that it is clear to me) The definition does suppose that the $O$ is only on one side.
My proof (I hope it is correct. One has first to prove the equivalence of the definition with the $\limsup$ equivalent)
Let be $h$ a map such that $h=O(g)$, then \begin{gather*} \limsup_{x\to \infty}\left|\frac{h(x)}{g(x)}\right|<\infty \end{gather*} thus \begin{gather*} \limsup_{x\to \infty}\left|\frac{f(x)h(x)}{f(x)g(x)}\right|<\infty, \end{gather*} which proves that $fh=O(fg)$ and since $h$ was an arbitrary map $O(g)$ we have $fO(g)=O(fg)$.