# Dual of a dual cone

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!

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 $(K^\circ)_\circ \subset \overline{K}$.

• I don't know if this is what you're looking for, David, but maybe this helps others find the argument you want. I can also add the complex case if it's of any interest. I think it should be in any book on topological vector spaces, such as Kelley-Namioka or Schaefer.
– t.b.
Sep 29, 2011 at 13:35
• I checked in Ewald's book, and there only polyhedral cones seem to be considered. The result that the dual cone of a dual cone is the cone itself is said to "follow from the definition of dual cones" (Ch. V, Lemma 2.2 (f), page 149f).
– t.b.
Oct 2, 2011 at 13:43
• An anon user proposed to change "weak" to "weak*". Just want to double check that is indeed what you meant. Oct 2, 2012 at 12:05
• @Willie: thanks, yes it's what I meant, although I'm not sure whether one should really call it weak*-topology rather than weak topology when the duality is explicitly indicated. But I'm rather agnostic on this.
– t.b.
Nov 29, 2012 at 14:25

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 $K$.

• if the separating hyperplane theorem returns a hyperplane $(\lambda, 0)$ such that for all $x\in C$, $\lambda^T x\geq 0$, then I agree straightforwardly that $K^{**}\subseteq C$. However, generally the separating hyperplane might be $(\lambda, b)$, for $b$ not necessarily $0$ or non-negative, correct? Not sure how to deal with this case.
– BoB
Apr 4, 2011 at 19:37
• You need $C$ to be closed as well as convex. Jun 23, 2017 at 22:44

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).

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.