Upper bound and lower bound of a custom function T(n)
For $n$ satisfying $\log n &gt; 1$, we have $\log^* \log n = \log^* n - 1$ (exercise). On the other hand, $\log \log^* n $ is much smaller than $\log^* n$. Hence $\log \log^* n = o(\log^* \log n)$.