Let $K$ be a non-empty closed convex set in a real normed space $X$. Let $x_0$ be a boundary point of $K$ i.e. $x_0 \in \partial K$. Is there a convex set $Y$ disjoint from $K$ such that $x_0$ is also a boundary point of $Y$? Can you ensure $Y$ has an interior point?

Background: I was trying to prove supporting hyperplane theorem for normed spaces and my strategy was to use Hahn-Banach separation theorem on two convex sets ($K$ and $Y$) to obtain a continuous functional separating the sets. Then by the fact $x_0$ lies on the boundary of both sets and by continuity, it must be a maximum or minimum for the functional, giving us the result. Ideally $Y$ would also have an interior point, so I may actually apply the theorem.

  • $\begingroup$ I suspect this may be easily proven by assuming supporting hyperplane theorem. Guess I really am just looking for a proof of supporting hyperplane theorem. $\endgroup$ – E.Lim Jan 8 '17 at 2:02

As a simple example suppose $\mathbb{R}$ as a normed space.Let $K=[0,1] $ which is convex and closed and $1\in \partial K$. Now let $Y=(1,2)$ we have also $1\in \partial Y.$

Certainly we can consider $\mathbb{R}^2$ and other important spaces.

  • $\begingroup$ I mean $K$ as a given closed convex set in a real normed space. $\endgroup$ – E.Lim Jan 7 '17 at 15:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.