 Solution

Enumerate both sets of peaks of a sequence in increasing order of index. If $a_n\to L$, the two selected subsequences both tend to $L$.

For the converse, suppose $a_{k_r}\to L$ and $a_{m_r}\to L$. Given $\epsilon >0$, choose $r$ large enough that both selected terms differ from $L$ by less than $\epsilon$. For every $n>\max \{k_r,m_r\}$, the two peak of a sequence conditions implyThe lower bound comes from the peak of a sequence of $(-a_n)$ at $m_r$. This is precisely the definition of $a_n\to L$. Thus convergence is equivalent to a common finite limit of a sequence for the upper and lower peak subsequences.