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Suppose we have a continuous strictly increasing function $f:\mathbb R^+\to\mathbb R^+$ such that $f(f(x))=\mathcal O(\exp(x))$ for $x\to\infty$.${^{[1]}}$ In other words, the function $f(x)$ exhibits (at most) half-exponential${^{[2]}}$$\!{^{[3]}}$$\!{^{[4]}}$ growth rate. It is known${^{[5]}}$ that there is no function that is built from arithmetical operators $(+,-,\times,/)$, $\exp$, $\log$, and real numbers that has exactly half-exponential growth rate. But, of course, there are elementary functions that are upper bounds for such functions, e.g. $\exp(x)$ is a very simple example of a (very loose) upper bound.

I'm interested in finding an elementary function $g:\mathbb R^+\to\mathbb R^+$ such that $f(x)=\mathcal O(g(x))$ that provides a much tighter upper bound (I mean an asymptotic upper bound, up to a coefficient). Ideally, I would like to find an infinite sequence of progressively tighter upper bounds that can approximate the half-exponential growth rate as close as we wish. If necessary, we can relax the requirement that the function must be elementary, and allow use of a well-known analytic special functions, such as Lambert $W$-function.

How can I approach this problem?

Vladimir Reshetnikov
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    Just mentioning that $f$ grows faster than any polynomial, compare https://math.stackexchange.com/q/4049486/42969. – Martin R Aug 13 '21 at 22:27
  • Yes, but this seems almost trivial: a composition of a polynomial with itself (a polynomial) is again a polynomial, so it grows much slower than $\exp(x)$. I am looking for a much tighter bound. – Vladimir Reshetnikov Aug 14 '21 at 19:28
  • It is obvious that $f(x) = O(x^n)$ cannot hold, but that does not immediately imply $\lim_{x\to \infty } \frac{f(x)}{x^n}=\infty$. But I am aware that this is a very weak lower bound. – Martin R Aug 14 '21 at 20:03
  • You should be a little bit more careful about what class of functions you want to consider. Otherwise the function $f(x)$ that is $0$ up to $T$, grows linearly from $0$ to $e^T$ between $T$ and $T+1$ and stays at $e^T$ after $T+1$ satisfies all your criteria and, obviously, no bound better than the trivial $e^x$ one is feasible for the whole family like that. – fedja Aug 16 '21 at 03:23
  • @fedja Sorry, I do not follow. What is $T$? The function you described seems to be $\mathcal O(1)$. – Vladimir Reshetnikov Aug 16 '21 at 05:58
  • @VladimirReshetnikov An arbitrary positive number. For any such function $f(f(x))\le e^x$ but if you want a uniform bound, then you are doomed. Alternatively, pile them up: choose some big $T_0>0$, put $T_{k+1}=e^{T_k}$, and raise $f$ from $e^{T_{k-1}}$ to $e^{T_{k}}$ between $T_k$ and $T_k+1$. – fedja Aug 16 '21 at 10:16
  • Perhaps, I did not express my idea clearly. I am looking for an asymptotic upper bound, up to a coefficient, in the sense of $\mathcal O(\cdot)$ notation. Wouldn't the function $f$ you constructed obey $f(x)=\mathcal O(\exp(x)/x)$? – Vladimir Reshetnikov Aug 16 '21 at 18:30
  • It won't: for $x=T_k+1$, we have $f(x)=e^{T_k}=e^{x-1}$. It is, probably, I who fails to write clearly :-) – fedja Aug 16 '21 at 22:49

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