L contains the concatenations of all k-bit long strings. Why is it decided in PSPACE(loglogn)?
Try to implement it in constant space to see what goes wrong. You need to be able to count up to $k$, which takes $O(\log k)$ space.

More formally, consider the language
$$L = \{a\ast b : a,b \in \{0,1\}^k, b=a+1\}.$$
This is not a regular language, and so cannot be decided in constant space.