User blog comment:LittlePeng9/Long hierarchies of functions (on my other blog)/@comment-11227630-20171101003439/@comment-1605058-20171101092927

I think with some changes my method works just fine. We can construct from one function \(f\) a function eventually outgrowing it by taking \(\tilde f(n)=f(n)\cdot f(n+1)\cdot\dots\cdot f(2n)\), or the product up to \(n^2\) if you want your second definition. We can now define \(f\) "properly outgrowing" \(g\) if \(f\) outgrows \(g,\tilde g,\tilde{\tilde g},\dots\). I believe a variant of the lemma can be proven in just about the same way, and then the construction can be repeated.