Kaba wrote: > Scrap that. Quasiconvexity does not imply triangle inequality or > convexity here. The following hold: > > Triangle inequality holds for f > <=> > The f is convex > > and > > The f is convex > ==> > The sublevel sets of f are convex (f is quasiconvex).
Well, turns out the equivalence does hold after all...
Let V be a real vector space, and f : V --> R such that
forall alpha in R: f(alpha x) = |alpha| f(x).
Then f is convex if and only if it is quasiconvex.
Assume f is quasiconvex. Then for t in [0, 1] subset R,
It holds that x / f(x) = 1 = y / f(y). Thus they are both in the sublevel set where f(x) <= 1. The equation above shows that their convex combination is in the same sublevel set. Multiplying both sides with beta in R, beta >= 0 shows that this holds for all sublevel sets. Therefore f is quasiconvex.