Asymptotic ordering of functions

Question

Suppose we define a relation on functions $\mathbb{R}_{\geq 0}\rightarrow\mathbb{R}_{\geq 0}$ as follows: $f<<g$ if $f^n = o(g)$ for all $n$. Suppose we further mod out by the equivalence relation $f\cong g$ if $f<<g$ and $g<<f$. I'd like to know what the order type of the resulting order $<<$ on equivalence classes is. In particular, are there infinite descending sequences?

Answer 1

I believe there are. In particular, I think iterating $log$ will produce such a sequence. That is $$f_{n+1}= \log \circ f_n$$ and $f_0$ is the identity. You should try to prove this. (As I might not be right...)

I'm also pretty sure this is only a partial order, as oscillation can lead to incompatibilities.