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 $c_0$.


1 Answer 1


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 $|x^*_i(u)|\le \frac12(\alpha^k-\alpha^{k+1})\le\frac12(1-\alpha)\le \frac c2$.

  • $\begingroup$ Thank you very much for the answer to this question. But what will be the answer if we shall additionally require the norming set $B$ to be weakly$^*$ compact? $\endgroup$ Commented Jun 4, 2019 at 5:09
  • $\begingroup$ A weak$^*$ closed absolutely convex norming set contains a multiple of the unit ball, so the answer will be positive. $\endgroup$ Commented Jun 4, 2019 at 5:12
  • $\begingroup$ Great! This is what is need. Thanks. Is there any good reference to this fact? (that a weak$^*$ closed norming set contains a dual ball)? Or it is just a corollary of the Bipolar Theorem? $\endgroup$ Commented Jun 4, 2019 at 5:15
  • $\begingroup$ I think that this goes back to Dixmier (1948), but it is just a separation in the weak* topology. $\endgroup$ Commented Jun 4, 2019 at 5:16

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