Let $E$ be a separable or reflexive infinite dimensional normed space, then there exists a sequence in $E'$ weak$^{*}$ convergent to $0$.

Question

The following exercise I found in a book in which I found errors, I think that in the following exercise I can remove hypothesis.

Let $E$ be a infinite dimensional normed space, suppose that $E$ is separable or reflexive. Show that there exists a senquence $(x_{n}')\subseteq E'$ such that $\left\|x_{n}'\right\|=1$, $n\in\mathbb{N}$, and $x_{n}'\overset{w^{*}}{\rightarrow} 0$ (that is, $x_{n}'(x)\rightarrow 0$ for all $x \in E$).

The problem: I tink that the hypothesis that $E$ is separable or reflexive can be removed. I want to know if my thinking is correct?

My attepmt: Assuming we only have $E$ infinite dimensional normed vector space, we denote $S=\left\{x'\in E'\: :\: \left\|x'\right\|=1\right\}$, $\overline{B'}=\left\{x'\in E'\: :\: \left\|x'\right\|\leq 1\right\}$, and $\sigma(E',E)$ the weak$^{*}$ topology in $E'$. We know that $$\overline{S}^{\sigma(E',E)}=\overline{B'} \tag{$\bigstar$}$$ where $\overline{S}^{\sigma(E',E)}$ is the closure of $S$ respect to topology $\sigma(E',E)$. (for a proof of ($\bigstar$) see this post ).

Therefore, $0\in \overline{S}^{\sigma(E',E)}$, then there exists $(x_{n}')\subseteq S$ such that $x_{n}'\rightarrow 0$ in the topology $\sigma(E',E)$, that is, $\left\|x_{n}'\right\|=1$ and $x_{n}'\overset{w^{*}}{\rightarrow} 0$.

Remark: Note that in my attempt I do not use the fact that $E$ is separable or reflexive. Am I making any mistakes?.

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Addendum

The comment of @GiuseppeNegro is correct, but we know that $\overline{B}'$ is $\sigma(E', E)$-compact. If $E$ is separable, then we know that $\overline{B}'$ is metrizable with induced topology of $\sigma(E', E)$. Therefore, my attempt is the proof for the case $E$ separable.

The problem is when $E$ is reflexive. I do not find a relation between reflexivity and metrizable. I'm thinking that when $E$ is reflexive it's not true that $\overline{B}'$ is $\sigma(E', E)$-metrisable, but I can not find a counter-example.

Answer 1

Right, $\overline{B'}$ is $\sigma(E',E)$-compact by Banach-Alaoglu, and if $E$ is separable, then $\sigma(E',E)\rvert_{\overline{B'}}$ is metrisable, so $\overline{B'}$ is also the sequential closure of $S$.

If $E$ is reflexive, we can use the result for separable spaces. Consider a linearly independent sequence $(y_n)$ in $S$, and let $F = \overline{\operatorname{span}} \{ y_n : n \in \mathbb{N}\}$. Then $F$ is a separable closed subspace of $E'$. As a closed subspace of a reflexive normed space, $F$ is itself reflexive, so $F = (F')'$ (with the usual abuse of notation). The separability of $(F')'$ implies the separability of $F'$, so by the first part we know that there is a sequence $(x_n)$ in $S \cap F$ such that $x_n \to 0$ in $\sigma(F,F')$.

It remains to see that $\sigma(F,F') = \sigma(E',E)\lvert_F$. The continuity of the inclusion $F \hookrightarrow E'$ gives $\sigma(E',E'')\lvert_F \subset \sigma(F,F')$. For the other inclusion, consider $\lambda \in F'$. By Hahn-Banach, there is a $\mu \in E''$ with $\lambda = \mu\lvert_F$. Now

$$\{ f \in F : \lvert \lambda(f)\rvert < \varepsilon\} = \{ x \in E' : \lvert \mu(x)\rvert < \varepsilon\} \cap F$$

is immediate, so every $\sigma(F,F')$-neighbourhood of $0$ is also a $\sigma(E',E'')\lvert_F$-neighbourhood of $0$. By reflexivity of $E$, $\sigma(E',E'') = \sigma(E',E)$. Thus $x_n \to 0$ in $\sigma(E',E)$.

I haven't yet come up with an example of a non-separable non-reflexive Banach space in which there is no sequence $(x_n)$ in $S$ with $x_n \to 0$ in $\sigma(E',E)$. Will think about that further after sleeping.

I still haven't found an example. For such an example, $E$ must not have an infinite-dimensional separable (or reflexive) complemented subspace, which rules out spaces like $c_0(T)$ or $\ell^1(T)$ for uncountable $T$. If $E \cong F \oplus G$ with separable (or reflexive) infinite-dimensional $F$, then the identification $F' \cong G^{\perp} = \{\lambda \in E' : G\subset \ker \lambda\}$ gives a sequence in $S \cap G^{\perp}$ that converges weakly to $0$. I suspect that $\ell^{\infty}(\mathbb{N})$ would work, but so far I cannot prove (or disprove) that.

Answer 2

Let $E$ be an infinite dimensional Banach Space.

If $E^*$ is separable, there is $x_n$ such that $\lVert x_n \rVert=1$ and $x_n\rightharpoonup0$.

We take $\{f_n\}$ a dense set of functions in $E^*$. We define $g_n:E\rightarrow \mathbb{R}^n$ by:

$$g_n(x)=(f_1(x),...,f_n(x))$$

By the rank and nullity theorem, $\dim \ker g_n =\dim \cap_{i=1}^n \ker f_i=\infty$. This means that we may take $\lVert x_n \rVert=1$ with $x_n\in \cap_{i=1}^n \ker f_i$. In this case, for every $f\in E^*$, there is a certain $j_o(\varepsilon)$ such that $\lVert f-f_{j_o}\rVert<\varepsilon$. If $j\geq j_o(\varepsilon)$, $x_j\in \ker f_{j_o}$ and it follows that:

$$\lVert f(x_j) \rVert=\lVert (f-f_{j_o})(x_j) \rVert\leq \lVert f-f_{j_o}\rVert<\varepsilon$$

Hence $f(x_n)\rightarrow 0$ and $x_n\rightharpoonup 0$.

Let $E$ be an infinite dimensional Banach Space.

If $E$ is reflexive, there is $x_n$ with $\lVert x_n \rVert=1$ and $x_n\rightharpoonup 0$.

Take $e_n$ a linearly independent countable set of vectors in the circle. $Y=\overline{\text{span}\{e_1,...,e_n,...\}}$ is a closed subspace of a reflexive space, hence reflexive. $D=\cup_n \mathbb{Q}e_1+...+\mathbb{Q} e_n$ is a dense countable set of elements, hence $Y$ is separable, $J_Y(Y)=Y^{**}$ is separable and $Y^*$ is separable so that we may find (by the first result) unitary vectors such that $f(x_n)\rightarrow 0$ for any $f\in Y^*$.

Take any $f\in E^*$. $\lVert f|_Y\rVert \leq \lVert f\rVert$ from which it follows that $f|_Y\in Y^*$ and therefore:

$$f(x_n)=f|_{Y}(x_n)\rightarrow 0\Rightarrow x_n\rightharpoonup 0$$