2
$\begingroup$

Let $X$ be a Banach space and let $(x^{*}_{n})_{n}$ be a sequence in $X^{*}$. Suppose that $\sum_{n}|\langle x^{*}_{n},x\rangle |\leq \|x\|$ for all $x\in X$.

Question: Is there a probability measure $\mu$ on the closed unit ball $B_{X^{*}}$ of $X^{*}$ such that $$ \sum_{n}|\langle x^{*}_{n},x\rangle |\leq \int_{B_{X^{*}}}|\langle x^{*},x \rangle |d\mu(x^{*}), $$ for each $x\in X$ ?

Thank you !

Improve this question
$\endgroup$
4
  • $\begingroup$ This is hopeless. Take $X=\ell^1$ (maybe two-dimensional, for simplicity), $e_1, e_2\in X^*$ as your sequence for a counterexample. $\endgroup$
    – Christian Remling
    Commented Jul 5, 2017 at 18:05
  • $\begingroup$ @ChristianRemling Could you give a detailed proof of your above statement? I do not understand your statement well. $\endgroup$
    – Dongyang Chen
    Commented Jul 6, 2017 at 14:18
  • $\begingroup$ Let $\mu$ be a prob measure, and take any $\|x\|=1$. The LHS equals $1$. On the RHS $|\langle x^*, x\rangle |\le \|x^*\|\le 1$, so to make the integral $\ge 1$, we'd have to have $|\langle x^*, x\rangle|=1$ a.e. for every fixed $x$, which is clearly impossible. $\endgroup$
    – Christian Remling
    Commented Jul 6, 2017 at 16:39
  • $\begingroup$ @ChristianRemling Why is it impossible that $|<x^{*},x>|=1$ a.e. for every fixed $\|x\|=1$? $\endgroup$
    – Dongyang Chen
    Commented Jul 7, 2017 at 1:40

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .