Proving Finite Convex Sets Intersection is Convex

retspool
Messages
35
Reaction score
0

Homework Statement



Prove that the intersection of a number of finite convex sets is also a convex set

Homework Equations



I have a set is convex if there exists x, y in the convex S then

f(ax + (1-a)y< af(x) + (1-a)y

where 0<a<1

The Attempt at a Solution



i can prove that
f(ax + (1-a)y) < f(x) given that x is a global minimizer

then i guess that i could find another arbritary point close to x , x_1, x_2 and add their given function satisfying the convex condition to get

Sum f(axi + (1-a)y) < Sumf(xi) where i= 1, 2,...nany help would be appreciated
 
Physics news on Phys.org


Your definition of "convex set" is wrong; there is no function involved. A set S (in some real vector space V) is convex if, whenever x, y \in S and 0 \leq a \leq 1, then also ax + (1 - a)y \in S.

Once you correct that, if you find yourself working too hard, you're doing something wrong. Just chase the definitions.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top