I know that there is no convex function $f$ on $(1,\infty)$ such that $f(x) \rightarrow \infty $ as $x \rightarrow \infty$ and at the same time $f(x)<\log x$ for all $x \in (1,\infty)$ because convex function is a supremum of some affine functions. What I am wondering is:
Can we extend this result to powers of logarithms? In particular, is there a convex function $g$ on $(1,\infty)$ such that $g(x) \rightarrow \infty$ as $x \rightarrow \infty$ and $g<\log ^2 $?
Any hint would be really appreciated! Thanks and regards.
The exact same argument applies to powers of logarithms. Any affine function of positive slope is eventually larger than $\log^n$ for any fixed $n$, and so no convex function going to $\infty$ can be bounded by $\log^n$.