Is there a dense convex cone which is not a subspace?

Question

In a finite-dimensional space any dense convex subset is the whole space, and (by the Stone-Weierstrass theorem) there are many examples of dense convex cones which are subspaces in infinite-dimensional spaces.

What I wish to know is whether there is a dense convex cone which is not a linear subspace in a (infinite-dimensional) topological vector space?

And if so, can we identify the spaces (sufficient or necessary conditions) for which there are no such cones? (meaning every dense convex cone is a linear subspace)

In particular, I am interested in the cases where the space is

• The space of measures (or more generally-)
• A dual space of a Banach space (or even more generally-)
• A locally convex space.

Answer 1

Here is a quite general construction: Let $X$ be any infinite-dimensional space with a dense subspace $D \subsetneq X$. Now, let $\hat x \in X \setminus D$ be arbitrary and consider $$ C := D + \operatorname{cone}(\hat x) = \{d + \lambda \, \hat x \mid d \in D, \lambda \geq 0\}.$$ It is clear that $C$ is a dense, convex cone but it is not a subspace.

Answer 2

One can even find an Archimedean proper¹ cone which is dense in a normed space.

Example. Let $E = \mathbb{R}[x]$ be the space of real polynomials in one variable. Choose $1 \leq a < b$ and equip $E$ with the cone $E_+$ obtained from the natural inclusion $E \hookrightarrow C[a,b]$. Then $E_+$ is an Archimedean proper cone. Furthermore, let $E$ be equipped with the coefficient-wise supremum norm $\lVert \:\cdot\: \rVert_{00}$, that is, the norm obtained from the natural isomorphism $\mathbb{R}[x] \cong c_{00} \subseteq c_0$.

We claim that $E_+$ is dense. To prove this, let $g \in E$ be arbitrary, and let $\lVert g\rVert_{[a,b]}$ denote the maximum of $|g|$ on the interval $[a,b]$. The series $\sum_{n=0}^\infty a^n$ diverges (since we assumed $a \geq 1$), so for any given $\varepsilon > 0$ we may choose $N_\varepsilon \in \mathbb{N}$ such that $\sum_{n=0}^{N_\varepsilon} a^n > \frac{\lVert g\rVert_{[a,b]}}{\varepsilon}$. By monotonicity we have $\sum_{n=0}^{N_\varepsilon} x^n > \frac{\lVert g\rVert_{[a,b]}}{\varepsilon}$ for all $x \in [a,b]$, so it follows that the polynomial $g + \varepsilon\sum_{n=0}^{N_\varepsilon} x^n$ is positive (on $[a,b]$). This shows that $d_{00}(g,E_+) \leq \varepsilon$, where $d_{00}(g,E_+) := \inf_{h\in E_+} \lVert g - h\rVert_{00}$ denotes the distance from $g$ to $E_+$. As this holds for any $\varepsilon > 0$, we find $d_{00}(g,E_+) = 0$, or in other words, $g \in \overline{E_+}$. We conclude that $E_+$ is dense, which proves our claim.

¹: By proper I mean that $E_+ \cap -E_+ = \{0\}$. Other authors call such cones salient or pointed.