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? $\endgroup$ 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. $\endgroup$ 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? $\endgroup$ 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. $\endgroup$ Jun 4, 2019 at 5:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.