A criterion for norming sets

Question

Let $F$ be a Banach space with the closed unit ball $B$. Let $E\subset F^*$ be a total subspace such that $B$ is complete with respect to the norm $|||f|||=\sup \limits_{e\in E,~e\ne 0} \frac{|\left<f,e\right>|}{\|e\|}$. Note that $|||\cdot|||$ is the norm of $F^{**}/E^{\bot}$, restricted on the image of $F$, and so my condition tells that the canonical inclusion from $F$ into $F^{**}/E^{\bot}$ is a semi-embedding.

Does it follow that $E$ is norming?

i.e. there is $r>0$ such that $|||f|||\ge r\|f\|$ for all $f\in F$.

Answer 1

I think that the answer is "Not necessarily" and can be shown as follows. Let $F=\ell_1=\left(\oplus_{n=1}^\infty \ell_1\right)_1$. In each summand $\ell_1$ we pick a unit vector $e_n$ (which can be regarded as the first vector of the unit vector basis in that space). The space $G:=\left(\oplus_{n=1}^\infty \ell_1^{**}\right)_1$ is canonically embedded into $F^{**}$

We construct $E^\perp$ as the weak$^*$ closure of the linear span of $e_n+p_n$, where $||p_n||\downarrow 0$ and $p_n$ has nonzero component only in $n$th summand of $G$ and the component is in $(\ell_1)^{**}\backslash \ell_1$.

Now consider any non-convergent sequence $\{f_k\}$ in the unit ball of $F$. We need to show that the limit of this sequence in $|||\cdot|||$, if exists, is in the unit ball of $F$. Passing to a subsequence we may assume that it is convergent coordinate-wise and subtracting its coordinate-wise limit we may assume that the coordinate-wise limit is $0$. By approximation we may assume that the sequence $\{f_k\}$ has finite disjoint supports. We need to show that the limit of the image of such sequence in $|||\cdot|||$ can be only $0$.

This can be done as follows. We split $f_k=s_k+t_k$, where $s_k$ is supported on $\{e_n\}$ (used above) and $t_k$ is supported on the rest of the unit vector basis of $\ell_1$. But since $\{f_n\}$ are uniformly bounded, it is easy to see that $|||s_k|||\downarrow 0$. Also it is not difficult to observe that if $|||t_k|||$ does not go to $0$, then the sequence $\{f_k\}$ is not convergent in $|||\cdot|||$.

Answer 2

It is rather common for a Banach space to have a weaker norm for which the unit ball is complete. A historically important example is the $L^1$ norm on $L^\infty$. This provides a counterexample to your conjecture---take for $E$ the space $L^\infty$ regarded as a subspace of its own dual in the usual way. This example was used by Saks (more precisely, the fact that the unit ball of $L^\infty$ has the Baire property under the $L^1$ norm) in his proof of what is now known as the Vitali-Hahn-Saks theorem.