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Let $X$ be a separable Banach space. Is it true that a weak* dense subspace of $X^{\ast}$ is also weak* sequentially dense?

vizietto
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2 Answers2

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Let $X=c_0$, and $W\subseteq c_0^*=\ell^1$ be defined by $$W=\{(c_n)_{n\in\mathbb{N}} \in\ell^1: c_0 = \sum_{k=1}^{\infty} c_k\} $$

For $a=(a_n)_{n\in\mathbb{N}} \in c_0$, if $$\sum_{n\in\mathbb{N}} c_na_n=0 \hspace{8mm}\forall c= (c_n)_{n\in\mathbb{N}} \in W$$ then $a=0$. Thus, $W$ is a separating subspace, so weak$^*$ dense in $\ell^1$.

Please see Theorem 1 in the Appendix of the book "Theory of Linear Operations" by Banach, which proves that $W$ is not weak$^*$ sequentially dense (and more).

Onur Oktay
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  • It's hard to read Banach's own text. The language seems outdated to me. But I believe I have a proof that $W$ is weak* sequentially dense: if $(c_k) \in \ell^1$, consider for all $n \geq 1$ the sequence $(c^{(n)}k)_k \in W$ defined by $c_k^{(n)} = c_k$ for all $k \neq n$ and $c_n^{(n)} = c_0 + c_n - \sum{k = 1}^\infty c_k$. Then for all $(a_k) \in c_0$, one has $\langle (a_k), (c_k^{(n)}) \rangle = \langle (a_k), (c_k) \rangle + (c_0 - \sum_{k=1}^\infty c_k) a_n \to 0$ as $n \to \infty$ since $a_n \to 0$, meaning that $c^{(n)} \to c$ (weak* convergence). What is wrong here? – vizietto Dec 22 '22 at 13:40
  • @vizietto your argument is correct. Instead of $W$ above, please consider the space defined in eq. (5) in the proof of Theorem 1 in Banach's book. In the proof, "weakly closed" means weak-star sequentially closed, "weak closure" means weak-star sequential closure, "weakly dense" means weak-star sequentially dense, etc.. which is the only 1930s-2020s translator we need. – Onur Oktay Dec 23 '22 at 00:33
  • I cannot see why the space defined in eq (5) is weak* dense in $\ell^1$. – vizietto Dec 23 '22 at 10:43
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I do not believe this to be true in general. However, for a subspace with the property that its weak-$\ast$-ly sequential "closure" is convex and weak-$\ast$-ly sequentially closed this is indeed true and follows from the Krein-Smulian theorem:

Let $\mathrm{cl}^*$ denote the closure operator in the weak-$\ast$ topology and define the weak-$\ast$-ly sequential "closure" operator \begin{equation} \mathrm{scl}^* A = \left\{ \phi \in X^* \mid \exists\, \text{sequence } (\phi_n)_{n \in \mathbb{N}} \text{ in } A: \phi_n \xrightarrow{\text{weak-$\ast$}} \phi \; (n \to \infty) \right\} \, . \end{equation} Note that $\mathrm{scl}^*$ is not idempotent in general. Furthermore, let $B_r(0)$ denote the open ball of radius $r > 0$ in $X^*$.

Lemma: Let $A \subseteq X^*$ be convex and weak-$\ast$-ly sequentially closed. Then $A$ is weak-$\ast$-ly closed.

proof: Suppose that there is some $r > 0$ such that $A \cap \overline{B_r(0)}$ is not weak-$\ast$-ly sequentially closed. Then there is a sequence $(\phi_n)_{n \in \mathbb{N}}$ in $A \cap \overline{B_r(0)}$ converging weak-$\ast$-ly to some $\phi \in X^* \setminus [A \cap \overline{B_r(0)}]$. Since $A$ is sequentially closed, $\phi \in A$ and thus $\Vert \phi \Vert > r$. However, the norm in $X^*$ is weak-$\ast$-ly lower semicontinuous, providing a contradiction. Consequently, $A \cap \overline{B_r(0)}$ is weak-$\ast$-ly sequentially closed and thus weak-$\ast$-ly closed since bounded sets in the weak-$\ast$ topology are metrisable since $X$ is separable by the sequential Banach-Alaoglu theorem. Hence Krein-Smulian applies.

Corollary: Let $A \subseteq X^*$ be weak-$\ast$-ly dense and

  • $\mathrm{scl}^* A$ is convex (e.g. $A$ is itself convex)
  • $\mathrm{scl}^* A$ is sequentially closed (this might be difficult in practice).

Then $A$ is weak-$\ast$-ly sequentially dense.

iolo
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