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Let $\ln_n(f(x))$ refer to the $n$-fold composition of the natural logarithm with some function $f$. Can you find a function such that $\ln_{\infty}(f(x))$ is convex (concave up)?

So $\ln_2(f(x)) = \ln\ln(f(x))$, and we take $\ln_0(f(x)) = f(x)$.

My efforts so far:

$$\frac{d}{dx} \ln_n(f(x)) = \frac{f'(x)}{\prod\limits_{k=0}^{n-1} \ln_k(f(x))}$$

$$g(x) = \prod\limits_{k=0}^{n-1} \ln_k(f(x))$$

$$\frac{d^2}{dx^2} \ln_n(f(x)) = \frac{g(x)f''(x) - f'(x)g'(x)}{g(x)^2}$$

It follows that:

$$g(x)f''(x) - f'(x)g'(x) > 0$$ (Must we also consider the case of equality?)

I'm sort of stuck from here. My first intuition is to find $g'(x)$, but I'm not exactly sure how to go about doing so, or if this problem has a much simpler solution than what I have done so far.

Edit: $$\ln(g(x)) = \sum\limits_{k=1}^n \ln_k(f(x))$$ $$g'(x) = \left(\prod\limits_{k=0}^{n-1} \ln_k(f(x))\right)\left(\sum\limits_{k=1}^n\frac{f'(x)}{\prod\limits_{m=0}^{k-1} \ln_m(f(x))}\right)$$

So our condition in full is: $$\left(\prod\limits_{k=0}^{n-1} \ln_k(f(x))\right)f''(x) - \left(\prod\limits_{k=0}^{n-1} \ln_k(f(x))\right)\left(\sum\limits_{k=1}^n\frac{f'(x)}{\prod\limits_{m=0}^{k-1} \ln_m(f(x))}\right)f'(x) > 0 $$

$$\left(\prod\limits_{k=0}^{n-1} \ln_k(f(x))\right)\left(f''(x) - \left(\sum\limits_{k=1}^n\frac{1}{\prod\limits_{m=0}^{k-1} \ln_m(f(x))}\right)(f'(x))^2\right) > 0$$

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    $\begingroup$ Does that even make sense to compose the logarithm many times? A function must be bigger than 1 and all its iterated logarithms must remain bigger than 1. Can you make an example of a function for which this iterated process makes sense indefinitely? Not even the Gamma function qualifies for that $\endgroup$ May 13 at 23:22
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    $\begingroup$ Does $\ln_\infty(f(x))$ mean $\lim_{n \to \infty} \ln_n(f(x))$? If so then the limit does not exist. One can show that for every $y$, there exists $n$ such that $\ln_n(y) \le 0$, and then $\ln_{n+1}(y)$ does not exist. $\endgroup$ May 14 at 6:19
  • $\begingroup$ Let $x$ be fixed, then $f(x)$ is finite, and so is less than $\exp^n(f(x)) = \exp\exp\cdots\exp(1)$ for some $n$, and $\log_{n+1}(f(x))$ is undefined. $\endgroup$
    – Gareth Ma
    May 14 at 8:33

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