Let S be a convex set Prove that x e 5 is an extreme point o

Let S be a convex set. Prove that x e 5 is an extreme point of S if and only if S\\{x} is convex. Let S = {x R^n: ||x||infinity {1}. Show that ext(S) = {xeR^n:x^2_j = 1, i = 1, 2, ..., n}

Solution

1) Let S be a convex set and x be an extreme point.

To show S-{x} is convex.

Let y and z be in S-{x}.

Consider the line segment L joining y and z.

Claim: L lies (is completely contained) in S-{x}.

If not, x will lie on L and x will be a non-trivial convex linear combination of y and z.

This contradicts the fact that x is an extreme point of S.

Hence the claim and thus S-{x} is convex.

2) Converse: Now let S-{x} be convex.

To show that x is an extreme point.

If x were not an extreme point of S, x is a non-trivial convex linear combination of some points in S, say v[1],v[2],....v[k], .None of v[i] is x , from the hypothesis.

Then S-{x} cannot be convex.

Hence we conclude that x is an extreme point of S.

3) As the set S is the product of intervals [-1,1] , suffices to prove the statement for the interval J= [-1,1].

Clearly the extreme pointsof J are {-1,1}. Removal of any other point in J will result in a disconnected and hence a non-convex set.

Hene ext(S) = {x: |x[i]|=1} and we are done.