Prove that: $$ \mathcal{O}(\mathcal{O}(f)) = \mathcal{O}(f) $$
I've started with letting some $u \in \mathcal{O}(\mathcal{O}(f))$, then: $$ |u| \le k_1|v| $$ Where: $$ |v| \le k_2|f| $$
Which means: $$ |u||v| \le k_1k_2 |v||f| \iff \\ |uv| \le k_1k_2 |vf| \iff \\ |u| \le k_1k_2 |f| \\ \frac{|u|}{|f|} \le k_1k_2 = C $$ Now $u\in \mathcal{O}(\mathcal{O}(f))$ by assumption. So we may conclude that: $$ u = \mathcal{O}(\mathcal{O}(f)) = \mathcal{O}(f) $$
I'm very new to that notation and comparisons between functions. Could you please verify the above or point to the mistakes/misunderstandings?