# When a set is convex, how does the polar set of its polar set equal the original?

I have read the following proposition, and haven't been able to connect the convexity of $X$ to the statement's main equality. Any guidance would be much appreciated.

Define $X^\text{o}$, the polar of $X$, as follows: $$X^\text{o} = \{y \in \mathbb{R}^n \; | \; x \cdot y \leq 1, \; \forall x \in X \}$$where $X$ is a nonempty subset of $\mathbb{R}^n$. Then, if $X$ is convex:

$$[X^\text{o}]^\text{o} = X.$$

• This is not true. Additionally, the closedness of $X$ is required.
– gerw
Feb 4, 2016 at 18:12
• What do you mean by "guidance"? Do you want a hint for a proof or a motivation why the statement is related with convexity?
– gerw
Feb 4, 2016 at 18:13
• A proof would be nicest, but a hint would also be appreciated. Closedness isn't mentioned in the proposition I read, which is why I may be having trouble proving it...
– MJRS
Feb 4, 2016 at 18:16

The answer given by @gerw is incorrect. As a counterexample, consider $$X=\{1\}\subset\mathbb{R}$$ which is closed and convex. Then $$X^{\circ}=(-\infty,1]$$ and $$(X^{\circ})^{\circ}=[0,1]\neq X$$.

The correct statement is: $$X=(X^{\circ})^{\circ}$$ if and only if $$X$$ is closed and convex with $$0\in X$$.

$$(\Rightarrow)$$ This is easy, since $$A^{\circ}$$ is closed and convex and $$0\in A^{\circ}$$ for every $$A\subset\mathbb{R}^n$$.

$$(\Leftarrow)$$ Clearly $$X\subset(X^{\circ})^{\circ}$$. Suppose, for contradiction, that there exists some $$z\in (X^{\circ})^{\circ}\backslash X$$. By the separating hyperplane theorem, there exists $$(\alpha,\beta)\in\mathbb{R}^n\times\mathbb{R}$$ such that $$\alpha^\top z>\beta>\alpha^\top x$$ for all $$x\in X$$. Since $$0\in X$$, then $$\beta>\alpha^\top 0=0$$. Let $$\nu=\alpha/\beta$$. Then $$\nu^\top z>1>\nu^\top x$$ for all $$x\in X$$. Therefore, $$\nu\in X^{\circ}$$ and $$z\not\in(X^{\circ})^{\circ}$$, which is a contradiction. Hence $$X=(X^{\circ})^{\circ}$$.

Note that the condition $$0\in X$$ is crucial in the proof, since it restricts $$\beta$$ to be positive, so that when we divide the inequality $$\alpha^\top z>\beta>\alpha^\top x$$ by $$\beta$$, it does not change direction.

• Yes, my answer is wrong. I will delete it.
– gerw
Nov 20, 2022 at 21:05