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Topic: Generalized Cauchy-Schwarz
Replies: 7   Last Post: Apr 13, 2012 3:09 PM

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Kaba

Posts: 289
Registered: 5/23/11
Re: Generalized Cauchy-Schwarz
Posted: Apr 13, 2012 9:04 AM
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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



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