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Let $\mathcal{H}$ be a Hilbert space and $K\subset \mathcal{H}$ a subset. We define a cone $C$ in $\mathcal{H}$ to be a set which satisfies $x\in C\implies \alpha x\in C$ for all nonnegative $\alpha$. We define the polar cone of $K$ to be the set $K^\circ\{x\in \mathcal{H} : \left(\forall y\in K\right)\ \langle x,y\rangle \leq 0\}$. I want to show that if $K$ is a closed convex cone then $K^{\circ\circ}=K$. I begin by showing inclusion in one direction, let $x\in K$ and $y\in K^\circ$. Then, $\langle x,y\rangle \leq 0$ by definition of $K^\circ$. Since $y$ was arbitrary in $K^\circ$, $x\in K^{\circ\circ}$, thus $K\subset K^{\circ\circ}$.

However, the other direction is less trivial. Can someone give me a hint?

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$\DeclareMathOperator{\cv}{cv}$Let $A\subset\mathcal{H}$ be an arbitrary set. I will show that $A^{\circ\circ}=\overline{\cv(\{\lambda a:a\in A,\lambda\geq 0\})},$ where $\cv$ denotes the convex hull.

If $x\in\overline{\cv(\{\lambda a:a\in A,\lambda\geq 0\})}$ then $x$ is the limit of $x_n$ where $x_n$ are convex combinations of $0$ and a finite collection of points of the form $\lambda_{i,n}a_{i,n}$ where $a_{i,n}\in A$ and $\lambda_{i,n}\geq 0$. Now by the bilinearity of the inner product we obtain for any $y\in A^{\circ}$ we have $\langle x_n,y\rangle\leq 0,$ and so by passing to the limit and using the continuity of the inner product we obtain that $\langle x,y\rangle\leq 0$ as well. Hence, $x\in A^{\circ\circ}$.

Conversely, if $x\notin\overline{\cv(\{\lambda a:a\in A,\lambda\geq 0\})}$ then by the Hahn-Banach theorem we can strictly separate $\{x\}$ from $\overline{\cv(\{\lambda a:a\in A,\lambda\geq 0\})}$. That is to say, we can find some $h\in\mathcal{H}$, $\alpha\in\mathbb{R}$, and $\varepsilon>0$ such that $\langle h,x\rangle\geq\alpha+\varepsilon$ and for all $z\in\overline{\cv(\{\lambda a:a\in A,\lambda\geq 0\})}$ we have $\langle h,z\rangle\leq\alpha-\varepsilon$. Since $0\in\overline{\cv(\{\lambda a:a\in A,\lambda\geq 0\})}$ we have that $0\leq \alpha-\varepsilon$ and so $\alpha\geq\varepsilon > 0$. Now take any $z\in \overline{\cv(\{\lambda a:a\in A,\lambda\geq 0\})}$ and we compute $$\langle h,\lambda z\rangle \leq \alpha - \varepsilon\implies \langle h,z\rangle\leq \frac{\alpha-\varepsilon}{\lambda},$$ and since this holds for all $\lambda\geq 0$ we can take $\lambda\to+\infty$ to find that $\langle h,z\rangle\leq 0$ for all $z\in\overline{\cv(\{\lambda a:a\in A,\lambda\geq 0\})}$. This shows that $h\in A^{\circ}$. Since $\langle h,x\rangle\geq \alpha +\varepsilon\geq 2\varepsilon > 0$ we deduce that $x\notin A^{\circ\circ}$.

So we have shown the desired relationship $A^{\circ\circ}=\overline{\cv(\{\lambda a:a\in A,\lambda\geq 0\})}$. Now in your case where $K$ is a closed convex cone it is clear that $\overline{\cv(\{\lambda k:k\in K,\lambda\geq 0\})}=K$ and so $K=K^{\circ\circ}$,

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  • $\begingroup$ Quite the hint, thanks yousuf. $\endgroup$ – Tony Nov 21 '17 at 8:23

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