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.
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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.
hdighfan
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Sorry I forgot to require that it be smooth in the title. $F''(x)$ is not continuous. For $0≤ x<1, F(x)=x$, $F'(x)=1$, $F''(x)=0$. But for 1≤x≤2, $F(x)=e^{x-1}$, $F'(x)=e^{x-1}$, $F''(x)=e^{x-1}≥1$ for $x≥1$. – blademan9999 Jun 09 '20 at 09:25
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You're right, of course. However, it's trivial to smooth this function whilst preserving the properties we want -- for instance, https://math.stackexchange.com/questions/45321/smooth-transition-between-two-lines-2d – hdighfan Jun 09 '20 at 09:37