This is inspired by  (which still needs answers).
What is the tight lower bound (or optimal algorithms) for the "finding repeated elements" problem: Given a sorted integer array of size $n$, how to determine whether there exists some element which occurs more than $\lfloor n/k \rfloor$ times?
Note 1: The original problem  focuses on $k=2$, i.e., the "majority detection" problem. There is a simple $\Theta(\log n)$ algorithm for it: Only the median, called $m$, can be the majority. We can learn the length of the $m$ contiguous block using binary search.
However, is $\Omega(\log n)$ a lower bound for $k = 2$?
Note 2: For unsorted array,  gives an optimal $O(n \log k)$ algorithm. The lower bound is proved using the decision-tree technique (section 5). I failed to extend it to the sorted case.
 Lower bound of finding majority element in an ordered array. @ cs.exchange
 Finding Repeated Elements by J. Misra and David Gries, TR, 1982.