Show that the set S x1 x2T x21 x22 lessthanorequalto 1 is

Show that the set S = {[x_1, x_2]^T: x^2_1 + x^2_2 lessthanorequalto 1} is convex. c. Show that the set S_c = {[x_1, x_2]^T: x^2_1 + x^2_2 > 1} is not convex. d. From and c above can we conclude that the complement of a convex set is not convex? Prove or provide a counter example.

Solution

b) if (x1,x2) is in the set and (y1,y2) is in the set

x1^2 + x2^2 <=1

y1^2 + y2^2 <=1

(lambda x1 + (1-lambda) y1)^2 + (lambda x2 + (1-lambda) y2)^2

= lambda^2 (x1^2 + x2^2) + (1-lambda)^2 (y1^2 + y2^2) + 2lambda(1-lambda) (x1y1 + x2 y2)

we have

(x1y1 + x2 y2)^2 < = x1^2 + x2^2 + y1^2 + y2^2

||u.v|| < = ||u|| .||v||

and we are done

c) [1.1, 0 ] is in the set

[0,1.1] is in the set

but [0.55, 0.55] is not in the set , so the set is not convex

d) no we cannot conclude. Complement may or may not be complex.

In R2 R X R⁺ is convex and also its complement

upper half ( including x axis) and lower half plane are convex and complement of each other