Does there exist a smooth function that eventually becomes n-logarithimcally convex for each n?

Question

Does there exist a smooth function s.t. for all $n \in \mathbb{N}$ $\exists x_0$ s.t. $f(x)$ is n-logarithimcally conves for $x \in (x_0, \infty )$. Here n logarithimcally convex is definded as follows. f(x) is n logarithimcally convex if log(f(x)) is n-1 logarithimcally convex. So for example, $e^{e^{x^2}}$ is 2-logarithimcally convex because $ln(e^{e^{x^2}})=e^{x^2}$ which is logarithimcally convex.

Answer 1

Yes.

Consider tetration $f(n,x,e)=e^{e^{⋰^x}}$ where there are $n$ $e$s. Then consider the function $$F(x)=f(\lfloor x\rfloor,\{x\},e).$$

Logging this function $n$ times, and considering $g(x)=(\ln^n\circ F)(x+n)$, we see that $g=f$ so it's sufficient for $f$ to be increasing, which is trivial to verify across each interval.

(if the function definition doesn't make sense, what I'm doing is

$f(x)=x$ for $0\le x<1$,

$f(x)=e^{x-1}$ for $1\le x < 2,$

$f(x)=e^{e^{x-2}}$ for $2\le x<3,$

and so on.