# On norming weakly$^*$ sequences in the dual of the Banach space $c_0$

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 $$c_0$$.

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

• 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? – Taras Banakh Jun 4 '19 at 5:09
• A weak$^*$ closed absolutely convex norming set contains a multiple of the unit ball, so the answer will be positive. – Mikhail Ostrovskii Jun 4 '19 at 5:12
• 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? – Taras Banakh Jun 4 '19 at 5:15
• I think that this goes back to Dixmier (1948), but it is just a separation in the weak* topology. – Mikhail Ostrovskii Jun 4 '19 at 5:16