Given a convex absorbing neighborhood of 0 in a topological vector space is it always possible to construct a subset that is also convex absorbing neighborhood of 0 but balanced as well?
A locally convex space is topological vector space with has a neighborhood base of 0 of balanced absorbing convex sets.
The questions corresponds to the question if it is possible to drop the requirement of being balanced.
For topological vector spaces over $\mathbb{R}$ that should be indeed the case (take $-V \cap V$ if $V$ is the convex absorbing neighborhood) but what about spaces over $\mathbb{C}$?
(E.g. here the construction of balanced absorbing sets in arbitrary vector spaces is shown:
Topological vector space, and balanced sets.)
Best Answer
We’ll use Functional Analysis by Huiqiang Jiang for references. Let $X$ be a topological vector space and $U$ be a convex neighborhood of $0$ in $X$. Then $U$ is absorbing, see the last sentence at p.3. By Remark 1, $U$ contains a balanced convex neighborhood of $0$. This follows from the construction from the proof of Theorem 3. Since scalar multiplication is continuous, exists $\delta > 0$, and an open neighborhood $V$ of the zero such that $\alpha V \subset U$ whenever $|\alpha| <\delta$. Then $W=\bigcup_{|\alpha|<\delta} \alpha V$ is balanced and $W\subset U$. The convex hull of $W$ is a convex balanced absorbing neighborhood of $0$ contained in $U$.