I want to show that if $f$ and $g$ are functions then $$f\mathcal{O}(g) = \mathcal{O}(fg)$$ As we know, $\mathcal{O}(f)$ formally means that there exist a constant $C$ and $x_0$ such that $|f(x)|\le C|h(x)|$ for $x\ge x_0$ and some strictly positive (on $x\ge x_0$) function $h$. In my proof I consider three cases: (i) when $|f(x)|\ge 1$ for $x\ge x_0$, (ii) when $|f(x)|<1$ for $x\ge x_0$, and (iii) when $|f(x)|\equiv 0$ for $x\ge x_0$.
For case (iii), however, I have a question. Does every function $g$ necessarily have an order in the big-oh sense? In other words, is it true that for every function $g$ there exist a constant $D$ and $x_1$ such that $|g(x)|\le D|j(x)|$ with $j(x)$ strictly positive for $x\ge x_1$?
And if it is not true for all functions $g$ then does this imply that for $f(x)\equiv 0$ for $x\ge x_0$, $f\mathcal{O}(g) = \mathcal{O}(fg)$ is not necessarily true?