## Ball-covering property of Banach spaces that is not preserved under linear isomorphisms.(English)Zbl 1152.46010

Sci. China, Ser. A 51, No. 1, 143-147 (2008); erratum Isr. J. Math. 184, 505-507 (2011).
The authors continue research that originated in the first author’s paper [L.-X. Cheng, Isr. J. Math. 156, 111–123 (2006; Zbl 1139.46016)]. A Banach space $$X$$ has the ball-covering property if the unit sphere of $$X$$ can be covered by a countable collection of balls not containing the origin. It was shown in the above cited paper that $$\ell_\infty$$ in its standard norm $$\| \cdot\| _\infty$$ has this property.
In the paper under review, the authors demonstrate by means of a short and elegant proof that $$\ell_\infty$$ does not possess the ball-covering property in the equivalent norm $$\| \cdot\| := \| \cdot\| _\infty + p(\cdot)$$, where $$p(x):= \lim\sup_n| x(n)|$$. This means that the ball-covering property is not stable under isomorphisms. It is remarked also that $$\ell_\infty/c_0$$ does not have the ball-covering property, which means that the ball-covering property is not inherited by quotients. Finally, the authors state that the ball-covering property is not inherited by subspaces. The proof is based on the existence of an isometric embedding of $$(\ell_\infty, \| \cdot\| )$$ into $$(\ell_\infty, \| \cdot\| _\infty)$$. Unfortunately, the last statement is false, which leaves the question open whether the ball-covering property is inherited by subspaces. The mistake is that the authors think erroneously that weak-star separability of $$X^*$$ implies weak-star separability of $$B_{X^*}$$.
Let us explain why $$X=(\ell_\infty, \| \cdot\| )$$ cannot be isometrically embedded into $$(\ell_\infty, \| \cdot\| _\infty)$$ (or equivalently, there is no 1-norming sequence in $$X^*$$). Assume to the contrary that there is a sequence $$e_n^* \in B_{X^*}$$ such that $$\sup_n| e_n^*(x)| = \| x\|$$ for every $$x \in X$$. Let us write $$e_n^*$$ as $$f_n + g_n$$, where $$f_n \in \ell_1 \subset \ell_\infty^*$$ and $$g_n \in (c_0)^{\bot} \subset \ell_\infty^*$$. Noting that on $$c_0$$ the norm of $$X$$ coincides with the standard norm $$\| \cdot\| _\infty$$, consequently,
$1 \geq \| e_n^*\| \geq \sup_{x \in B_{c_0}}| e_n^*(x)| =\sup_{x \in B_{c_0}}| f_n(x)| = \| f_n\| _{\ell_1}.$
On the other hand, the standard fact that $$\ell_\infty/c_0$$ does not admit a countable total system of functionals implies that there is an element $$z \in X \setminus c_0$$ such that $$g_n(z)=0$$ for all $$n$$. For this $$z$$ we have
$\| z\| = \sup_n| e_n^*(z)| = \sup_n| f_n(z)| \leq \| z\| _\infty,$
i.e., $$p(z)=0$$ and $$z \in c_0$$, a contradiction.

### MSC:

 46B20 Geometry and structure of normed linear spaces 46B26 Nonseparable Banach spaces

### Keywords:

ball-covering; weak-star separable Banach space

Zbl 1139.46016
Full Text:

### References:

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