A bounded subset $B$ of the dual $X^\*$ of a Banach space $X$ is called *norming* if the formula $\|x\|:=\sup\{|x^\*(x)|:x^\*\in B\}$ determines an equivalent norm on $X$.


Observe that the sequence $(e\_n^\*)\_{n\in\omega}\subset c\_0^\*=\ell\_1$ of coordinate functionals in the dual of $c\_0$ is norming and weakly$^\*$ null.



> 
> **Question 1.** Is it true that every absolutely convex bounded norming set $B\subset c\_0^\*$ in the dual of the Banach space $c\_0$ contains a norming weakly$^\*$ null sequence?
> 
> 
> 


The same question can be asked for subspaces of $c\_0$.



> 
> **Question 2.** Let $X$ be a closed subspace of the Banach space $c\_0$ and $B\subset X^\*$ is a norming bounded absolutely convex set. Is it true that $B$ contains a norming weakly$^\*$ null sequence? 
> 
> 
> 


**Remark.** If a Banach space $X$ admits a norming weakly$^\*$ null sequence of functionals $\{f\_n\}\_{n\in\omega}\subset X^\*$, then the operator $T:X\to c\_0$, $T:x\mapsto (f\_n(x))\_{n\in\omega}$, is an isomorphic embedding of $X$ into $c\_0$.
Therefore, $X$ is isomorphic to a subspace of the Banach space $c\_0$.



On norming weakly$^*$ sequences in the dual of the Banach space $c_0$
The answer to your Question 1 (and thus Question 2, for general
subspace) is "No".


**Example.** Let $B=B\_{\ell\_1}\cap \ker {\mathbf{1}}$, where
$B\_{\ell\_1}$ is the unit ball of $\ell\_1$ and
${\mathbf{1}}=(1,\dots,1,\dots)\in\ell\_\infty$ (it is well-known
and easy to check that $B$ is norming).


Suppose that there is a weak$^\*$ null norming sequence
$\{x\_i^\*\}\_{i=1}^\infty$ inside $B$. We may assume, slightly
decreasing the norming constant, that


(1) $x\_i^\*$ are finitely supported.


(2) eventually they get more and more zeros at the beginning.


Let $c$ be the norming constant of this sequence, that is
$\sup\_i|x\_i^\*(x)|\ge c||x||$ for every $x\in c\_0$. Let
$\alpha\in(0,1)$ be such that $1-\alpha\le c$. We consider the
following sequence $u$ in $c\_0$:


$$u=(1,\underbrace{\alpha,\dots,\alpha}\_{n\_1}, \underbrace{\alpha^2,\dots,\alpha^2}\_{n\_2},\alpha^3,\dots),$$
where $n\_1$ is such that all $x\_i^\*$ whose support contains $1$,
are supported in $I\_1:=\{1,\dots,n\_1+1\}$; $n\_2$ is such that all
$x\_i^\*$ whose support intersects $I\_1$ are supported in
$I\_2:=\{1,\dots,n\_1+n\_2+1\}$, so on.


We claim that $|x\_i^\*(u)|\le\frac{c}2$ for each $i$, thus getting
a contradiction.


In fact, the vector $u$ is such that the support of each $x\_i^\*$
is contained in the set, where the coordinates of $u$ are either
$\alpha^k$ or $\alpha^{k+1}$ for some $k\in\{0,1,2,3,\dots\}$,
also $||x^\*\_i||\_{\ell\_1}\le 1$ and ${\mathbf{1}}(x\_i^\*)=0$. Thus
$|x\_i^\*(u)|$ does not exceed the value which we get if the
positive support of $x\_i^\*$ corresponds to $\alpha^k$ and the
negative support of $x\_i^\*$ corresponds to $\alpha^{k+1}$, or the
other way around. Therefore $|x^\*\_i(u)|\le
\frac12(\alpha^k-\alpha^{k+1})\le\frac12(1-\alpha)\le \frac c2$.


