Let $\log_k x$ be the $k$-th iterate of the logarithm:
$\log_1 x = \log x , \log _k x = \log(\log_{k-1} x)$
Is there a continuous increasing function $f(x)$ such that $\displaystyle \lim_{x\to \infty} f(x) = \infty$ but $f(x) = o(\log _k x)$ for each fixed $k$? If so, what is an example of one?
I tried playing around with the problem to better understand it but got stuck. I found that $\log f(x) = o(\log g(x))$ is not necessarily true (let $f(x) = \sqrt{x}$ and $g(x) = x$), but I am not sure how to proceed/conclude.
Properties that may be relevant:
$f(x) = o(g(x))$ if $\displaystyle \lim_{x \to \infty} \frac{f(x)}{g(x)} = 0$.
$( \log x )^k = o (x^a)$ for every pair of constants $a>0$ and $k$.
$x^k = o(e^x)$ for every constant $k$.
Yes, there are such functions. For instance, define a sequence $(a_n)_n$ by $a_0=1$ and for $k>0$ we set $a_k$ to be the least integer such that $\ln_k(a_k)>\ln_{k-1}(a_{k-1})+1$.
If we take all the points $(a_k,\ln_k(a_k))$ and join then by straight lines, we have the graph of a function that increases to $\infty$ slower than any repeated logarithm.