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Let $X$ be a closed convex set.

$X^\circ$ is bounded $\Longleftrightarrow$ $\textbf 0 \in int(X)$

How is this true?

I know that $C^\circ$ is a closed convex set but not sure how to move forward from this.

Samantha
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3 Answers3

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Consider $0\in int(X)$. Then there exists some ball $B_r(0)\subset X$. Then any functional $y$ from the dual space with $\sup_{x\in X} |y(x)|\leq 1$ is fully given by it’s value on $B_r(0)$. Thus $||y||\leq 1/r $ and thus $X^\circ$ is bounded.

I do not think the converse hold though: If $V=\mathbb R$ then $X=\{1\}$ is a closed convex set, but $X^\circ=\{ x\mapsto ax : |a|\leq 1\}$ is bounded.

The reason for this is that the above proof does not actually require that the whole Ball is in $X$, but it only requires that some part of a sphere that spans the whole space in some bounded manner is in $X$. Or even less, we do not actually need a sphere, it suffices if it contains a sufficiently nice subsetset that is bounded towards $0$ and spans the whole unit sphere in some bounded manner.

Lazy
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  • The counter-example does not check out for me; isn't $x \mapsto 2 x$ in the polar? – Cloudscape Oct 03 '21 at 19:50
  • OK our definitions seem different. – Cloudscape Oct 03 '21 at 19:52
  • I’m under the assumption that $X^\circ = {y\in V' : |y(x)| \leq 1 \text{ on } x\in X}$. So in your example: $(x\mapsto 2x)(1)=2>1$. – Lazy Oct 03 '21 at 19:54
  • Another counterexample: If $X$ contains a ball $B_r(x_0)$. Then if $|y(x)|\leq 1$ for $x\in X$ then for $x\in B_r(0)$ we have $|y(x)| = |y(x+x_0)+y(x_0)| \leq |y(x+x_0)|+|y(x_0)| \leq 2$. So again $||y||\leq 2/r$. – Lazy Oct 03 '21 at 19:58
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I assume your set $X$ is in Euclidean space (or a real Hilbert space) $H$ and $0 \in X$. I also assume that you are referring to the real polar (see [1]) defined by $$X^\circ= \{ y \in H : \sup_{x \in X} \langle y,x \rangle\} \le 1.$$ If you are referring to the absolute polar (see [1]), then the claim is false without additional assumption, consider $X=[0,1]$, but it holds if you assume that $X$ is symmetric about the origin, i.e. $X=-X$.

If $0 \in$ int$(X)$ then there is an open ball $B(0,r) \subset X$. For all $y$ in the polar $X^\circ$ and $\rho<r$ we have that $\rho y/|y| \in X$ so $1 \ge \langle y,\rho y/|y| \rangle=\rho |y|$.

Conversely, if the polar is a subset of an open ball, i.e., $X^\circ \subset B(0,R)$, then we claim that $B(0,1/R) \subset X$. Indeed, suppose that there exists $z \in B(0,1/R) \setminus X$. Then there exists a hyperplane separating $z$ from $X$, i.e., there is a nonzero vector $y$ such that $$ (*) \quad \sup_{x \in X} \langle y,x \rangle < \langle y,z \rangle \,.$$ (You can take $y=z-x_1$, where $x_1$ is the vector in $X$ that is closest to $z$.) By Cauchy-Schwarz $\langle y,z \rangle \le |y| \cdot |z| <|y|/R$. Finally, (*) implies that $y_*:=Ry/|y|$ is in the real polar $X^\circ$, contradicting the hypothesis.

[1] https://en.wikipedia.org/wiki/Polar_set

Yuval Peres
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Assuming we are in a real Hilbert space:

If there is a ball of radius $r$ that contains $0$ and is contained within $X$, then $\forall x \in X: \langle x, x' \rangle \ge -1$ can only hold if $\|x'\| \le 1/r$, because otherwise you may take $x = - r x'/ \|x'\|$ and violate the condition.

Conversely, suppose that $0$ is not in the interior of $X$, and that the interior of $X$ is nonempty. By the separation theorem, we find a hyperplane $E$ inside the Hilbert space that does not intersect the interior of $X$. The orthogonal complement of that hyperplane is a line, and obviously if you take any $x \in \operatorname{int} X$ and move along that line for a sufficiently long time, you'll get $\langle x', x \rangle \ge -1$ because $x$ is not orthogonal to $x'$ since the hyperplane was chosen not to intersect the interior of $X$.

Cloudscape
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