I'm familiar with the classical convex hull calculation algorithms. The lower bound for computing the CH of a set of points $P$ is $n\log(n)$.
However, what if I'm given a sequence of points and told they form a CH? One check is to "walk the points" in order and check for "left turns" (CCW). However, this doesn't handle the case of self-intersections (you could imagine a counter-example that "winds around starting from the center" making only CCW turns.
I've been told this "validation" can be done in $O(n)$ time, but to check for self-intersection naively, I see $O(n^2)$ time.
My next thought is to run Graham's Scan. Since the points are presumably on a CH, it takes $O(n)$ to find the lower left point, say $p_0$, and from there, I THINK you can guarantee that all points $p_1, p_2,\dots,p_n$ are already sorted by angle between $p_0$ and $x$-axis. As you move the three-point-window forward along the hull, the points ahead are still properly sorted. The reasoning for $O(n)$ then follows per normal reasoning for Graham's scan, minus the need for $O(n\log n)$ sorting.
1) Is my assumption, in bold, correct?
2) Is there an easier method that wouldn't involve running Graham's scan?