Q:

Upper bound and lower bound of a custom function T(n)

Assume a function $T(n)$:
$$T(n)=\log{\log^{*}{n}}$$
Let's define:
$$\log^{*}{n}=\min\{i\ge{0}:\log^{(i)}{n}\le1\}$$
and:
$$\log^{(i)}{n}=\log{(\log^{(i-1)}{n})}$$
Is the following $f(n)$ function an upper bound or a lower bound of $T(n)$?
$$f(n)=\log^{*}{\log{n}}$$

To see if $f(n)$ could possibly be an upper bound, I feel like the following limit needs to be gone through:
$$\lim_{n\to\infty}{T(n)}=\lim_{n\to\infty}{\log{\log^{*}{n}}}=?$$
Now I wonder if the above limit might boil down to this:
$$\lim_{n\to\infty}{T(n)}=\lim_{n\to\infty}{\log{(\min{\{\log\log\log\cdots\log^{(0)}{n}\}})}}$$
I'm not quite sure how to solve the above limit to get an idea of the upper bound of $T(n)$. Also, the lower bound is another story which I have little idea about.

A:

For $n$ satisfying $\log n > 1$, we have $\log^* \log n = \log^* n - 1$ (exercise). On the other hand, $\log \log^* n $ is much smaller than $\log^* n$. Hence $\log \log^* n = o(\log^* \log n)$.

