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Let $X$ be a infinite dimensional Banach space, and let $\omega$ and $\omega^*$ be the weak and weak-* topologies. Let $B_X$ be the closed unit ball and $\mathbb{B}_X$ the open unit ball (for the strong topology).

  • Is $\mathbb{B}_{X^*}$ weak-*-open? Does it depend? Why?

  • I know that the restriction of the weak-* topology of $X$** over $X$ is the weak topology of $X$. So, does it mean that if $A \subset X$ is weak-open then it will be weak-$^*$-open in $X^{**}$? In particular, is $B_X$ weak-$^*$-closed in $X^{**}$ (as it is weak-closed in $X$)? If not, why?

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    Non-empty weak*-open nhoods are unbounded in norm. Goldstine's Theorem states $B_X$ is weak* dense in $B_{X^{**}}$. – 2017-01-03
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    I believe that in your first Q, you mean $\mathbb B_X$ , not $\mathbb B_{X^*}$. – 2017-01-04

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Every non-empty weak$^*$ set $U$ is unbounded in norm.

(1). (Background material).Let $Y$ be a Banach subspace of $X$ with $Y\ne X.$

For $v\in X$ \ $Y$ let <$v$>$+Y$ be the vector subspace generated by $\{v\}\cup Y.$ For (real or complex) scalar $r$ and for $y\in Y$ let $$h(rv+y)=rd(v,Y)=r\inf \;\{\|v-z\|:z\in Y\}.$$ Then

(i). <$v$>$+Y$ is closed in the norm topology on $X.$

(ii). $h$ is well-defined.

(iii). $\sup \{|h(z)|/\|z\|: f(z)\ne 0\}=1.$

(iv). By the Hahn-Banach Theorem there exists $g\in X^*$ with $\|g\|=1$ and $g|_{+Y}=h.$ Note that $Y\subset g^{-1}\{0\}.$

(2). Let $X$ be infinite-dimensional. A weak$^*$ nbhd base for $f\in X^*$ is the set of all $\{g\in X^*:\land_{i=1}^n|g(x_i)-f(x_i)|

For $f\in U,$ where $U$ is weak$^*$ open, let $B=\{g\in X^*: \land_{i=1}^n|g(x_i)-f(x_i)|

Let $Y$ be the vector subspace generated by $\{x_1,...,x_n\}.$ By induction on $n$ and by (1)(i), $Y$ is closed in $X$. By (1)(iv) there exists $g\in X^*$ with $g \ne 0$ and $Y\subset g^{-1}\{0\}.$ We have $$\{f+ng: n\in \mathbb N\} \subset B\subset U$$ $$\text { and } \quad \sup \;\{\|f+ng\|:n\in \mathbb N\}=\infty.$$

Remark: Let $\mathbb S$ be the set of scalars for $X$ , that is, $\mathbb R$ or $\mathbb C.$ Endow $\mathbb S^X$ with the Tychonoff product topology. For $f\in X^*$ let $\pi (f)=(f(x))_{x\in X}.$ Then $\pi$ is a homeomorphism from $X^*$ with the weak$^*$ topology to a subspace of $\mathbb S^X.$

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    All right! Even more, the same reasoning is also correct for the weak topology, isn't it? (Changing the nbhd) I mean, changing this a bit also would say that every non-empty weak-open is unbounded in norm. Is that true? – 2017-01-04
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    Yes. Almost verbatim. – 2017-01-04
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    Perfect! Thank you very much! Now everything is clear for me. – 2017-01-04