Finding a local peak in an array in O(log N)? So I watched a couple of videos regarding this and the divide and conquer made sense to me. But I am still not convinced as to how recursing on the side with the larger number guarantees us the solution.

Please can someone explain it Mathematically.
No, it is not possible to find the max (peak) element in an unsorted array better than $\mathcal{O}(n)$.

When you executed your algorithm $\mathcal{A}$ that has $c \log n$ compare operations, that means; $\mathcal{A}$ only compared $c \log n$ elements. The remaining are not compared by $\mathcal{A}$.

We can easily construct an adversary argument for any $\mathcal{O}(\log n)$ algorithm by just placing the peak, within the elements for $\mathcal{A}$ didn't touch.