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?

  • 1
    $\begingroup$ Non-empty weak*-open nhoods are unbounded in norm. Goldstine's Theorem states $B_X$ is weak* dense in $B_{X^{**}}$. $\endgroup$ – David Mitra Jan 3 '17 at 11:16
  • $\begingroup$ I believe that in your first Q, you mean $\mathbb B_X$ , not $\mathbb B_{X^*}$. $\endgroup$ – DanielWainfleet Jan 4 '17 at 9:00

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 $g|_{<v>+Y}=h.$ 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 $\{g\in X^*:\land_{i=1}^n|g(x_i)-f(x_i)|<e_i\}$ over all finite $\{x_1,..., x_n\}\subset X$ and $\{e_1,...,e_n\}\subset \mathbb R^+.$

For $f\in U,$ where $U$ is weak$^*$ open, let $B=\{g\in X^*: \land_{i=1}^n|g(x_i)-f(x_i)|<e_1\}$ be a basic weak$^*$ nbhd of $f$ with $B\subset U.$

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 $\mathbb S^X.$

  • $\begingroup$ 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? $\endgroup$ – Minkowski Jan 4 '17 at 11:19
  • 1
    $\begingroup$ Yes. Almost verbatim. $\endgroup$ – DanielWainfleet Jan 4 '17 at 11:41
  • $\begingroup$ Perfect! Thank you very much! Now everything is clear for me. $\endgroup$ – Minkowski Jan 4 '17 at 11:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.