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Kaba
Posts:
289
Registered:
5/23/11


Re: Generalized CauchySchwarz
Posted:
Apr 13, 2012 9:04 AM


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...
Claim 
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.
Proof 
Assume f is quasiconvex. Then for t in [0, 1] subset R,
f((1  t)x + ty) = max(f((1  t)x) + f(ty)) <= f((1  t)x) + f(ty) = (1  t)f(x) + tf(y).
Assume f is convex. If both f(x) = 0 and f(y) = 0, then
f(x + y) = 2f(0.5x + 0.5y) <= 2(0.5 f(x) + 0.5 f(y)) = f(x) + f(y) = 0 = max(f(x), f(y)).
Assume either f(x) != 0 or f(y) != 0, and let
alpha = f(y) / (f(x) + f(y)).
Then
f((1  alpha) (x / f(x)) + alpha (y / f(y)) <= (1  alpha) f(x / f(x)) + alpha f(y / f(y)) = (1  alpha) + alpha = 1.
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.
 http://kaba.hilvi.org



