Natural candidates for sub-half-exponential which limit to half-exponential function from below

Question

There are no closed form candidates for half-exponential functions "Closed-form" functions with half-exponential growth.

However sub-half-exponentials (functions whose composition grows slower than exponential) have lot of candidates.

Eg: $n$ itself.

A little thought gives $f(0,a,n)={n^a}$, $f(1,a,n)={2^{(\log n)^a}}$, $f(2,a,n)={2^{2^{(\log\log n)^a}}}$, $\dots$, $f(k,a,n)={2^{^{\dots}{^{{2^{(\underbrace{\log\dots\log}_{\text{k }} n)^a}}}}}}$ etc. at fixed $a\in(1,\infty)$ and fixed $k\in\mathbb Z\cap[1,\infty]$.

The functions grow faster as $k$ increases. Define $g_{lower}(a,n)=\lim_{k\rightarrow\infty}f(k,a,n)$.

1. Is $g_{lower}(a,n)$ almost the half-exponential function? That is, does $g_{lower}(a,g_{lower}(a,n))=2^{\Omega(n^{1-\epsilon})}$ hold at every $a\in(0,1)$ and at every $\epsilon>0$?

I do not think inductive arguments depending on finite $k$ work when $k\rightarrow\infty$ without knowing speed of $f(k,a,n)$'s increase as $k$ increases. It is possible $g_{lower}(a,n)$ is the half-exponential function.

It is also possible $g_{upper}(a,n)$ and $g_{lower}(a,n)$ approach each other where $g_{upper}(a,n)$ is defined in Natural candidates for super-half-exponential which limit to half-exponential function from above.

That is $g_{lower}(a,g_{lower}(a,n))=2^{\Omega(n)}$ holds which means $g_{lower}(a,n)$ is the half-exponential function.

2. Are there other natural sequence of function candidates $h(k,n)$ (not of form form $f(k,a',n)$ where $a'\in(0,a)$) with $$f(k,a,n)\ll h(k,n)\ll f(k+1,a,n)$$ $$h(k,h(k,n))=2^{\Omega(n)}$$ at every fixed $k\in\mathbb Z\cap[1,\infty]$?

Answer 1

Not an answer Merely remarks.
Let me use superscript $[k]$ for $k$-fold composition. $\log^{[3]} n$ means $\log\log\log n$.

As I remarked on the other question, for fixed $a$ and $n$, the value $f(k,a,n)$ becomes complex for large $k$. Sequence $\log^{[k]}(n)$ decreases as $k$ increases, until it becomes negative, and then complex.

I have worked with so-called transseries. In that formalism, we are interested in the rate of growth, not the initial part of the function.

Edgar, G. A., Transseries for beginners, Real Anal. Exch. 35(2009-2010), No. 2, 253-310 (2010). ZBL1218.41019.

My work on fractional iteration applies only to "exponentiality zero", so does not include $\exp^{[1/2]}$.

Edgar, G. A., Fractional iteration of series and transseries, Trans. Am. Math. Soc. 365, No. 11, 5805-5832 (2013). ZBL1283.30001.

---

Some remarks on the question here. Is there some reason to use exponentiation base $2$ and not $e$? I will use exponential and logarithm base $e$ instead of base $2$. Let me shift your index $k$ by $1$. Then define \begin{align} f(0,a,n) &:= an, \\ f(1,a,n) &:= \exp f(0,a,\log n) = \exp(a\log n) = n^a \\ f(2,a,n) &:= \exp f(1,a,\log n) = \exp(\exp(a \log(\log n))) =\exp((\log n)^a) \\ &\qquad \dots \\ f(k,a,n) &:= \exp^{[k]}(a\log^{[k]}(n)) = \exp^{[k-1]}\Big(\big(\log^{[k-1]}(n)\big)^a\Big) \end{align} Write $M_a$ for the multiplication function: $M_a(n) = an$.
We are interested in a "limit" in some sense as $k \to \infty$. I doubt that it converges according to the natural topologies for transseries.

Note that \begin{align} f(k,a,n) &= \exp^{[k]}\circ M_a\circ \log^{[k]} (n) \\ f(k,a,f(k,a,n)) &=\exp^{[k]}\circ M_a\circ \log^{[k]}\circ \exp^{[k]}\circ M_a\circ \log^{[k]} (n) \\ &= \exp^{[k]}\circ M_a\circ M_a\circ \log^{[k]} (n) \\ &= \exp^{[k]}\circ M_{a^2}\circ \log^{[k]} (n) \\ &= f(k,a^2,n) \end{align} So if $g(a,n) = \lim_{k \to \infty} f(k,a,n)$ in some sense, then we might expect that $$ g(a,g(a,n)) = g(a^2,n) . $$ So: prove by induction on $k$ that $f(k,a,n) < e^n$. Then conclude that $g(a,n) < e^n$, and $g(a,g(a,n)) = g(a^2,n) < e^n$ (good for what we want). But also $$ g(a,g(a,g(a,n))) = g(a^3,n) < e^n \\ g(a,g(a,g(a,g(a,n)))) = g(a^4,n) < e^n $$ So that second iterate $g(a,g(a,n))$ is much, much slower than $e^n$.