TITLE: How is $O(\log(\log(n)))$ also $O( \log n)$?
QUESTION [7 upvotes]: How is $O(\log(\log(n)))$ also  $O( \log n)$?

I have seen this result somewhere with this but I still didn't quite understand how this is true.
This would also help me compute Big Omega of the same function.

REPLY [3 votes]: This is because, if $n$ is large enough, $\lvert f\rvert\le K\log(\log n)$ for some constant $K$ and $\log x \le x$ for all $x$.
More generally, if $f=O(g)$ and $g=O(h)$ then $f=O(h)\,$ $\,(O(O(h))=O(h)$).