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

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.
• You might want to have a look at Example 6.11(iv) in the book "Convex Analysis and Monotone Operator Theory in Hilbert Spaces" by Bauschke and Combettes. They provide an explicit example in infinite-dimensional and separable Hilbert space. – weirdo Oct 11 '18 at 22:53

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.