6

It seems true that, given $K \subseteq \mathbb{R}^n$ a convex set with $K^\circ \neq \emptyset$, then $\overline{K^{\circ}} = \overline{K}$ and $\left ( \overline{K} \right )^\circ = K^\circ$.

I am able to prove the first equality by making use of the "segment Lemma", which states that if $y \in K$ and $x \in K^\circ$, then $[x, y[ \subseteq K^\circ$ (here $[x, y[$ is the segment joining $x$ and $y$ without taking $y$).

However I have not found any correct proof of the second equality, and neither a counter-example has come to my mind.

Thanks all!

Hugo
  • 437
  • 2
    (Prove and) use the following. Given a convex set in $K\subset\mathbb{R}^n$, let $H_K$ be the set of closed halfspaces containing $K$. Then
    1. $\bar K=\bigcap_{h\in H_K} h$
    2. $\mathring{K}=\bigcap_{h\in H_k} \mathring{h}$
     – Andrea Bianchi Feb 25 '16 at 12:24

2 Answers2

2

Either use the suggestion by Andrea, or you can prove directly this way: $K \subseteq \bar K$, so $\mathring K \subseteq (\bar K)^\circ$. On the other hand, if $x \in (\bar K)^\circ$ then there is a neighborhood $x \in U_x \subseteq \bar K$. Now take a small simplex in $U_x$ that contains $x$ in its interior. Up to perturbing slightliy its vertices, you can make them to belong to $K$ while $x$ still belongs to the interior of the simplex. Then by convexity the simplex is in $K$, so $x \in \mathring K$.

I realized after writing that my answer is essentially the same as https://math.stackexchange.com/a/7509.

Community
  • 1
Giovanni Mascellani
  • 742
  • BTW, you do not appear to need the hypothesis $\mathring K \neq \emptyset$. – Giovanni Mascellani Feb 25 '16 at 13:13
  • 1
    Thanks! Where I looked for it they treat all as a trivial thing and miss some key-passage, but in fact it is not so obvious, since you must perturbate small simplexes, etc. The non-empty interior hypotesis is mainly for the first part, maybe it is not necessary for the second indeed! – Hugo Feb 25 '16 at 15:37
2

Here is a more functional-analytic proof, which avoids the argument with simplexes: assume by contradiction that there exists $x\in(\overline{K})^\circ\setminus K^\circ$. By Hahn-Banach theorem (applied to $K^\circ$ and $\{x\}$) and the fact that $(\mathbb{R}^n)^*=\mathbb{R}^n$, you can find $v\neq 0$ such that $\langle v,K^\circ\rangle \le \langle v,x\rangle$. So $$ \langle v,K^\circ\rangle + \epsilon|v|^2\le\langle v,x+\epsilon v\rangle. $$ But $x+\epsilon v\in\overline{K}$ if $\epsilon>0$ is small. In particular you can find $x'\in K$ such that $|(x+\epsilon v)-x'|<\frac{\epsilon}{2}|v|$, which gives $$ \langle v,K^\circ\rangle + \frac{\epsilon}{2}|v|^2\le\langle v,x'\rangle. $$ But by the "segment lemma", for any fixed $z\in K^\circ$, $[z,x')\subseteq K^\circ$, so $\langle v,[z,x')\rangle\le\langle v,x'\rangle-\frac{\epsilon}{2}|v|^2$ and by continuity $\langle v,x'\rangle\le\langle v,x'\rangle-\frac{\epsilon}{2}|v|^2$, contradiction.

We assumed $K^\circ\neq 0$, but you can always find an affine subspace containing $K$ where this happens and you can repeat the proof here.

Mizar
  • 5,911
  • 21
  • 47
  • (When I wrote $\langle v,K^\circ\rangle$, it is a shorthand to say that the inequality holds for all elements of $K^\circ$) – Mizar Feb 25 '16 at 21:36
  • I guess you should specify that $x'$ is of the form $x + \delta v$ for some $\delta < \epsilon/2$. Anyway it seems it works, even if I did not want to pass through Hahn-Banach, because this was one of the first results in a Convex Analysis course I am following. Thank you! – Hugo Feb 26 '16 at 14:45
  • A priori you do not know that you can choose $x'$ of that form. I meant $|(x+\epsilon v)-x'|<\frac{\epsilon}{2}|v|$, now I corrected the post: in fact you get $\langle v,x'\rangle=\langle v,x\rangle+\langle v,x'-x\rangle\ge\langle v,x\rangle-|v|\cdot|x-x'|\ge\langle v,x\rangle-\frac{\epsilon}{2}|v|^2$ (sorry that I skipped this computation). – Mizar Feb 26 '16 at 14:59
  • My fault, you cannot choose $x'$ as I said indeed. Your proof works perfectly! – Hugo Feb 26 '16 at 15:21