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:

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

Thanks in advance!

  • $\begingroup$ This is not true. Additionally, the closedness of $X$ is required. $\endgroup$
    – gerw
    Commented Feb 4, 2016 at 18:12
  • $\begingroup$ What do you mean by "guidance"? Do you want a hint for a proof or a motivation why the statement is related with convexity? $\endgroup$
    – gerw
    Commented Feb 4, 2016 at 18:13
  • $\begingroup$ 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... $\endgroup$
    – MJRS
    Commented Feb 4, 2016 at 18:16

1 Answer 1


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.

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

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