Are dual spaces to separable normed spaces first-countable?

Question

In [1] I found the following theorem (roughly translated by me)

Satz 13.10 If $X$ is a separable normed $k$-vector space ($k = \mathbb R$ or $\mathbb C$) with continuous dual space $X'$, then the closed unit ball $B' \subset X'$ is a first-countable space with respect to the weak-*-subspace topology.

They also make the observation that this holds then for any bounded subset of $X'$, but they don't elaborate any further. I would have thought that the above Satz implies actually that $X'$ itself is first-countable, because being first-countable is a local property, and every $f \in X'$ is contained in the interior of some big ball $\lambda B'$, for $\lambda > ||f||$.

But I was wondering why the authors would not claim this. Maybe it's not true that $f$ is in the interior of $\lambda B'$? Is the "open" unit ball $$B'^\circ = \{f \in X' : ||f||<1\}$$ not open with respect to the weak-*-topology? If so, an example where this fails would be helpful.

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[1] Hirzebruch, Scharlau, Einführung in die Funktionalanalysis

Answer 1

The best you can get is metrizability of the unit ball and, in general, of any bounded subset of $X'$.

The weak-star interior of the open ball $B_{X'}$ of $X'$ is empty. This is because any weakly open set $U$ in an infinite dimensional B-space is unbounded as it contains a non-trivial subspace.

EDIT: I wanted to point out that it is indeed the case that $(X',w^*)$ is never metrizable whenever $X$ is an infinite dimensional Banach space but this can fail if $X$ is not complete.

Consider $c_{00}$, the space of eventually zero sequences, endowed with the $\ell_2$ norm. Then $c_{00}^*= \ell_2$ via the isometry
$$ λ = (λ_n) \in \ell_2 \mapsto f_{ \lambda}, \\ f_{\lambda}(x) = \sum_{n=1}^\infty \lambda_n x_n, \quad x \in c_{00}$$ Then the weak-star topology on $c_{00}^* = \ell_2 $ is the topology of pointwise convergence which is metrizable (as it is generated by the countable family of seminorms $p_i( \lambda) = |\lambda_i|$.)

Answer 2

It is possible to show that if $X$ is separable, the ball $B_{X'}$ endowed with the weak star topology is metrizable with a metric given by: $$d(x^*,y^*)=\sum_n\frac{|(x^*-y^*)(x_n)|}{2^n}$$ Where $x_n$ is a countable dense set in the ball $B_X$. Clearly, any metric space is first countable, because if we take $y^*$, then $\{B(y^*,1/n)\: |\: n \in \mathbb{N}\}$ is a countable neighborhood basis for $y^*$.

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More generally, if there is a bounded set in $X'$, it is contained in a large enough ball wich is metrizable. Clearly, a subset of a metrizable space is metrizable with the restriction of $d$. Hence, it will also be first countable.

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EDIT: You asky why this argument fails for $X'$.

If we are not to take $x_n$ in the unitary ball but rather try taking all of the $x_n$ to define a global metric in $X'$, then the norm of $\lVert x_n \rVert$ may be very large and we have no guarantee that the series we defined above converges, so $d$ makes no sense...