# Show that $(A^o)^o = A$ if A is compact and convex.

I am working on exercises for a convex analysis course and I am slightly stuck on the following;

Suppose that $$A \subset \mathbb{R}^n$$ is closed, bounded (compact) and conex. Define the set

$$A^o = \left\{y \in \mathbb{R}^n : y^Tx \leq 1, \forall x \in A\right\}$$

Show that $$A^o$$ is convex and $$(A^o)^o = A$$.

Showing that $$A^o$$ is convex is easy; take $$u,v \in A^o$$ and $$\theta \in [0,1]\subset \mathbb{R}$$. Then for all $$x \in A$$ we have

$$(\theta u +(1-\theta)v)^Tx = \theta u^Tx+(1-\theta)v^Tx \leq \theta + (1-\theta) = 1$$

and so $$\theta u +(1-\theta)v \in A^o$$.

Also showing that $$A \subset (A^o)^o$$ is easy since for each $$x \in A$$ and $$y \in A^o$$ we have

$$x^Ty = y^Tx \leq 1.$$

Now showing that $$(A^o)^o \subset A$$ is where I am getting stuck. The way I've been trying to show this is by contradiction.

I suppose that there exists $$x \in (A^0)^0$$ so that $$x \notin A$$. Then let $$\xi \in A$$ so that

$$||x-\xi|| = \text{dist}(x,A).$$

Such a $$\xi$$ exists since $$A$$ is compact. Now since $$A$$ is convex as well as compact we know that the hyperplane

$$H = \left\{y \in \mathbb{R}^n : a^Ty =b \right\}$$

with $$a = x - \xi$$ and $$b = \frac{a^Tx+a^T\xi}{2}$$ is a strict separating hyperplane for the set $$A$$ and $$\{x\}$$ with

$$x \in \left\{y \in \mathbb{R}^n : a^Ty >b \right\}$$

and

$$A \subset \left\{y \in \mathbb{R}^n : a^Ty < b \right\}.$$

Now this is where I get fully stuck. I have tried every which way that I can think of to get a contradiction from here with no success.

How would you suggest going forward.

NOTE: Hints are also welcome, if I get to a proof with the help of a hint then I will post my final proof as the answer to this question.

Thank you in advance for any help.

• Yeah, that's what I am doing in the last part – Jandré Snyman Mar 21 at 7:55
• You need to add the assumption $0 \in A$, since you always have $0 \in (A^o)^o$. Otherwise, the result does not hold: E.g. for $A = \{1\} \subset \mathbb R$ you have $(A^o)^o = [0,1]$. – gerw Mar 21 at 11:42

edit: As per the comment of gerw, the claim is only true if $$0\in A$$.
The next step would be to think of $$A^\circ$$ not just of a set in $$\mathbb R^n$$ but as a set of hyperplanes/half-spaces.
You have shown that there are $$a,b$$ such that $$A\subset \{y\in\mathbb R^n : a^\top y Since $$0\in A$$, this implies that $$b>0$$. We can change $$a,b$$ by scaling such that $$b=1$$. Then we obtain $$a^\top y < 1 \qquad\forall y\in A$$ which implies $$a \in A^\circ.$$ Since $$a^\top x > 1$$, this shows that $$x\not\in (A^\circ)^\circ$$. This is a contradiction to your assumption.
• This does work only if you already know $b > 0$ and this should follow from $0 \in A$. On the other hand, if $0 \not\in A$, the result does not hold since $0 \in (A^o)^o$. – gerw Mar 21 at 11:42