Let $V$ be a quasi-complete Hausdorff locally convex space. (By quasi-complete, one means that every bounded closed subset of $V$ is complete.) For a bounded closed absolutely convex subset $B$, denote by $V\_B$ the subspace of $V$ spanned by vectors in $B$ and define a norm function $q\_B$ on $V\_B$, namely, $$q\_B(v)=\inf\{t\geqslant 0: v\in tB\},\quad v\in V\_B.$$ It is easy to see that $B=\{v\in V\_B: q\_B(v)\leqslant 1\}$ and that the inclusion map $(V\_B, q\_B)\rightarrow V$ is continuos.


Q1: Would $V\_B$ be a closed subspace of $V$?


Q2: It appears that $V\_B$ would be complete with respect to the norm function $q\_B$ and hence becomes a Banach space. (I once saw this assertion in a paper.) If it is indeed the case, what is the argument? Thanks in advance for any input.



Subspaces of quasi-complete locally convex spaces
The answer to the first question is NO: For example, consider the inclusion of $\ell\_1$ into $\ell\_2$, then the unit ball $B$ of $\ell\_1$ is closed but $V\_B=\ell\_1$ is dense in $\ell\_2$.


The argument for the second question is sometimes attributed to Wendy Robertson although it is due to Grothendieck: Given a Cauchy sequence $(x\_n)\_{n\in\mathbb N}$ in $V\_B$ it has a limit $x\_\infty$ in $V$. For each $\varepsilon>0$ and $n,m$ big enough one has
$x\_n-x\_m\in\varepsilon B$ (because this is $0$-neighbourhood in $V\_B$) and since $B$ is closed one gets $x\_n-x\_\infty\in\varepsilon B$ which proves $x\_n\to x\_\infty$ in $V\_B$. This is a primitive version of the following: Let $X$ and $Y$ be topological vector spaces and $i:X\to Y$ a continuous embedding such $X$ has a $0$-neighbourhood basis of sets $U$ such that $i(U)$ is closed in $Y$. If $Y$ is complete (or quasi-complete) then so is $X$. This in fact has not much to do with topological vector spaces and has versions for uniform spaces.


