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There are two equivalent definitions of a locally convex topological vector space. Note that since the vector space is topological, addition by a fixed vector x and multiplication by a fixed scalar r are both homeomophisms, which means it is enough to give a neighborhood basis at the vector 0.

The first definition is that $0$ has a local base consisting of open sets that are convex (if $x$ and $y$ are in $S$, then $\lambda x+(1-\lambda)y$ is also in $S$ for every $\lambda\in[0,1]$), balanced, (if $x\in S$, then $r x\in S$ for every $r\in\mathbb{F}$ with $|r|=1$, where $\mathbb F$ is either the real or complex numbers), and absorbing (for any vector $y$ there exists a $\lambda\in (0,\infty)$ such that $\lambda S$ contains $y$, i.e. $S$ absorbs $y$).

The second definition is that $0$ has a local base given by the ''balls'' of radius $r$ for each $r$ of each semi-norm in some fixed collection of semi-norms (a semi-norm is just like a norm except that non-zero vectors can have a non-zero semi-norm). The two definitions are equivalent in one direction because "balls" of semi-norms are convex, balanced and absorbing, and in the other because we can define for any convex, balanced, absorbing set $S$ the semi-norm $f_S$ by assigning to $x$ the "least" (infimum) $\lambda$ for which $x$ is in $\lambda S$.

My question is: what are some examples of topological vector spaces for which $0$ does have a local base of convex open sets, but that no local base consists of convex sets that are also balanced and absorbing?

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2 Answers 2

up vote 4 down vote accepted

Unless I am misunderstanding the situation, there is no such example. That is, if $V$ is a topological vector space (over $\mathbb R$ or $\mathbb C$) which has a basis of neighbourhoods of the origin consisting of convex sets, then it also has a basis of neighbourhoods of the origin consisting of balanced convex sets.

Also, any neighbourhood of $0$ is automatically absorbent (because of continuity of scalar multiplication).

My reference is Robertson and Roberston, Topological Vector Spaces (Cambridge Tracts in Mathematics and Mathematical Physics, 53), which is a kind of summary of Bourbaki's Topological Vector Spaces volume.

The proof is not hard: one first uses continuity of scalar multiplication to show that any neighbourhood of $0$ contains a balanced neighbourhood. Now if $U$ is a convex n.h of $0$, and $W$ is a balanced n.h. of $0$ contained in $U$, then $V:= \cap_{|x| \geq 1} x U$ contains $W$, so is a n.h., is convex (being the intersection of convex sets), is balanced (by construction), and lies in $U$ (again by construction --- consider $x = 1$). Thus $V$ is a convex balanced n.h. of $0$ contained in $U$.

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NB: As Matt notes in a comment below, the following does not really work!

Consider a vector space $V$ of dimension at least $2$, pick a non-zero proper subspace $W\subseteq V$, and consider the unique linear topology on $V$ for which $\{W\}$ is a basis of neighborhoods of $0\in V$.

If you want an Hausdorff example, let now $V$ be an infinite dimensional vector space, and consider a decreasing sequence $(W_n)_{n\geq1}$ of non-zero subspaces of $V$ such that $\bigcap_{n\geq1}W_n=0$. Now put on $V$ the unique linear topology which has $\{V_n\}_{n\geq1}$ as a basis of neighborhoods of $0\in V$.

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In your first example, let $v \in V$ not lie in $W$. Then $0 \times v = 0 \in W$, but there is no n.h. $U$ of $0$ such that $U \times v \subset W.$ So it seems that scalar multiplication is not continuous. Am I blundering again? –  Matt E Nov 15 '10 at 4:29
Argh. You are of course right! A tvs is always connected, and my spaces aren't. –  Mariano Suárez-Alvarez Nov 15 '10 at 4:38

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