Dual of a dual cone

Question

Any hint on how to prove the following please:

Let $K$ be a convex cone, and $K^*$ its dual cone. Prove that $K^{**}$ is the closure of $K$.

Thanks!

Answer 1

Disclaimer. If you're only interested in the case $X = \mathbb{R}^n$ then ignore all the comments concerning weak topologies, as they will coincide with the usual Euclidean topology on $\mathbb{R}^n$. [The topology on $X$ will always be the weak topology $\sigma(X,X^{\ast})$ and the topology on $X^{\ast}$ will always be the weak* topology $\sigma(X^{\ast},X)$.]

Suppose that $K$ is a convex cone in a locally convex real vector space $X$. Recall that this simply means that $K$ is non-empty and that for $k,k' \in K$ we have $k+k' \in K$ and for $k \in K$ and $\alpha \gt 0$ we have $\alpha k \in K$.

1. The dual cone of a non-empty subset $K \subset X$ is $$K^{\circ} = \{f \in X^{\ast}\,:\,f(k) \geq 0 \text{ for all }k \in K\} \subset X^{\ast}.$$ Note that $K^{\circ}$ is a convex cone as $0 \in K^{\circ}$ and that it is closed [in the weak* topology $\sigma(X^{\ast},X)$].

2. If $C \subset X^{\ast}$ is non-empty, its predual cone $C_{\circ}$ is the convex cone $$C_{\circ} = \{x \in X\,:\,f(x) \geq 0 \text{ for all } f \in C\} \subset X,$$ and it is closed [in the weak topology $\sigma(X,X^{\ast})$].

3. It is a tautology that $K \subset (K^{\circ})_{\circ}$: if $k \in K$ then $f(k) \geq 0$ for all $f \in K^{\circ}$, hence $k \in (K^{\circ})_{\circ}$.

If $K \subset X^\ast$ is a convex cone then its closure $\overline{K}$ is a closed and convex cone, hence $\overline{K} \subset (K^{\circ})_\circ$. Our goal is to prove that $\overline{K} = (K^{\circ})_\circ$.

Recall the following form of the Hahn-Banach separation theorem:

Let $X$ be a Hausdorff locally convex real vector space. Let $A,B \subset X$ be disjoint, closed and convex sets. If $A$ is compact then there exist a continuous linear functional $f \in X^\ast$ and constants $r \lt s$ such that $f(a) \lt r \lt s \lt f(b)$ for all $a \in A$ and $b \in B$.

Suppose that $x \notin \overline{K}$. We want to show that $x \notin (K^\circ)_\circ$. The separation theorem applied to $A = \{x\}$ and $B = \overline{K}$ gives us a continuous linear functional $f$ such that $f(x) \lt M = \inf{\{f(k)\,:\,k \in \overline{K}\}}$.

Since $0 \in \overline{K}$ we have $M \leq 0 = f(0)$, and in particular $f(x) \lt 0$. If we had $M \lt 0$ there would be $k \in \overline{K}$ such that $f(k) \lt 0$. But then, taking $\alpha = \frac{2f(x)}{f(k)} \gt 0$, we have $\alpha k \in \overline{K}$ and at the same time we would have $f(\alpha k) = 2f(x) \lt f(x) \lt 0$ contrary to the assumption on $f$. Therefore $M = 0$ and thus $f(k) \geq 0$ for all $k \in \overline{K}$. In particular $f \in K^{\circ}$. But as $f(x) \lt 0$ we have that $x \notin (K^{\circ})_\circ$.

Thus $x \notin \overline{K}$ implies $x \notin (K^{\circ})_{\circ}$, so $(K^\circ)_\circ \subset \overline{K}$.

Answer 2

First, it is clear that $K \subset K^{**}$. Second, clearly $K^{**}$ is a convex cone. Third, show that if $C \supseteq K$ is a closed convex cone then $C \supseteq K^{**}$ (hint: use the separating hyperplane theorem). Conclude that $K^{**}$ is the intersection of all closed convex cones containing $K$, so in your case it's the closure of $K$.

Answer 3

I will first show that the closure of $K$ belongs to $K^{**}$. Assume the opposite, there is a sequence $(a_i)$ with $a_i\in K$ and $\lim\limits_{i \to \infty} a_i=a$ such that $a\notin K^{**}$. This means there is $v\in K^*$ such that $a\cdot v=z<0$. But since $\lim\limits_{i \to \infty} a_i=a$, there exists some $a_i$ sufficiently close to $a$ such that $a_i\cdot v\in[z-\epsilon,z+\epsilon]$, for any $\epsilon$. For a sufficiently small $\epsilon$, this means $a_i\cdot v<0$ which is a contradiction of $v\in K^*$ and $a_i\in K$.

Now I will prove that any $b$ outside the closure $closure(K)$ of $K$ does not belong to $K^{**}$. There is a strictly separating hyperplane between $closure(K)$ and $\{b\}$ such that $a\cdot v>b\cdot v~\forall a\in closure(K)$, based on a so-called simple separation theorem from [1] or "Separation theorem I" from [2]. Using $0\in K$, we have $0\cdot v>b\cdot v \implies b\cdot v<0$. Assume $\exists a'\in K$ such that $a'\cdot v <0$. Using the cone property, $ta'\in K$ for any arbitrarily large $t$. This means $ta'\cdot v$ can be arbitrarily low, easily less than $b \cdot v$, which is a contradiction. The assumption $\exists a'\in K$ such that $a'\cdot v<0$ is false. All $a\in K$ need to verify $a\cdot v\geq 0$. We are done: we produced $v\in K^*$ such that $b\cdot v<0$ for any $b$ outside $closure(K)$.

REFS

[1] Proposition 3, Section 6.4 of the course of Peter Norman at http://www.unc.edu/~normanp/890part4.pdf

[2] "Separation theorem I" presented at http://en.wikipedia.org/wiki/Hyperplane_separation_theorem stating there is a strictly separating hyperplane between two disjoint nonempty closed convex sets, one of which is compact (in our case, $\{b\}$ is compact).

Answer 4

Whilst the existing answers are extremely helpful, it may also be useful to have a self-contained proof, at least in the finite dimensional case:

Let $V=\mathbb{R}^n$, and let $K\subseteq V$ be a convex cone. Let $C$ denote its (topological) closure, and suppose $v \in V \backslash C$. Our goal is to show that $v\notin K^{**}$. That is we must find a vector $x\in V$, such that $\langle x,w\rangle\geq 0$ for all $w\in K$, but $\langle x,v\rangle< 0$. Here $\langle\,\,,\,\,\rangle$ is the standard inner product.

The intersection of $C$ with the closed ball of radius $\|v\|$ about $v$ is non-empty, closed and bounded, so the function $\|w-v\|$ attains its minimum at some point $w=u\in C$.

Then our solution will be $x=u-v$. Let $w\in C$. We will first show $\langle x,w\rangle\geq 0$:

We know $u+tw\in C$ for all $t\geq 0$. Thus: $$\|u-v\|^2\leq \|u+tw-v\|^2=\|x+tw\|^2=\|u-v\|^2+2t\langle x,w\rangle+t^2\|w\|^2.$$ This inequality will be violated for small positive $t$, unless $\langle x,w\rangle\geq 0$ as required.

It remains to show that $\langle x,v\rangle<0$. If $u=0$ this is trivial: $\langle -v,v\rangle=-\|v\|^2<0$. We may therefore assume $u\neq 0$.

We have a quadratic equation in $t$: $$\|tu-v\|^2=t^2\|u\|^2-2t\langle u,v\rangle+\|v\|^2, $$ which minimises at $t=1$. Differentiating the above expression and setting $t=1$, we therefore obtain:$$\|u\|^2=\langle u,v\rangle.$$

Then $$0<\|x\|^2=\|u\|^2-2\langle u,v\rangle+\|v\|^2=-\langle u,v\rangle+\|v\|^2=-\langle x,v\rangle,$$ as required.