Subspaces of quasi-complete locally convex spaces

Question

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.

Answer 1

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.