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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Back
Top