3
$\begingroup$

Let $X$ be a real Banach space.

For a continuous (not necessarily linear) function $g:X \to \mathbb{R}$ and a family $\mathcal{F} \subseteq X^*$, we´ll say that $\mathcal{F}$ determines $g$ if whenever $g(x) \neq g(y)$, there is an $f \in \mathcal{F}$ such that $f(x) \neq f(y)$. Since $\mathcal{F}=X^*$ determines any function (because in fact it "determines points"), it makes sense to define: $$s(g):=\min \{|\mathcal{F}|:\mathcal{F} \mbox{ determines } g\}.$$

For example if $g$ is constant then $s(g)=0$ and if $g$ is linear then $s(g)=1$.

Now let $$s(X)=\sup_{g \in C(X)} s(g).$$

For finite-dimensional $X$ we have that $s(X) = \dim(X)$ and the supremum is attained for example by the function $g(x)=\|x\|$.

I hope someone with more background in Banach spaces than me (that probably includes most of the regulars on this site!) can easily answer some/all of the following:

1) Is there a simple way to compute $s(X)$ for infinite-dimensional $X$?

2) Is there always a $g \in C(X)$ for which $s(X)=s(g)$?

3) Is it true that if $X$ is separable then $s(X) \leq \aleph_0$?

Any comment, answer or reference will be greatly appreciated.

$\endgroup$
4
  • 2
    $\begingroup$ 3: if $X$ is separable, then there is a countable family of bounded linear functionals that separate points in $X$. $\endgroup$ Apr 25, 2012 at 20:39
  • $\begingroup$ Thank you Gerald! I suppose that in general there is such a family of size at most the density character of $X$. Is that correct? $\endgroup$ Apr 25, 2012 at 21:16
  • $\begingroup$ By Hahn-Banach, for every $x \in X$ we can find $f_x \in X^*$ with $\| f_x\|=1$ and $f_x(x) = \|x\|$. Now if $E$ is dense in $X$, then $\{f_x : x \in E\}$ separates points of $X$. Choose $x_n \in E$ with $x_n \to x$, then $f_n(x_n) = \|x_n\| \to \|x\|$, and on the other hand $|f_n(x_n) - f_n(x)| \le \|x_n - x\| \to 0$. Thus $f_n(x) \to \|x\|$ and in particular $f_n(x) \ne 0$ for large enough $n$. $\endgroup$ Apr 26, 2012 at 3:37
  • $\begingroup$ @Nate: Thank you very much. That means $s(X) \leq dc(X)$. Michael´s answer below seems to imply that $s(X)=dc(X)=s(\|\cdot\|)$, but I still have to decrypt it. $\endgroup$ Apr 26, 2012 at 13:07

1 Answer 1

1
$\begingroup$

If the span of ${\cal F}$ is weak-* dense, it separates points. If not, it does not determine $\|x\|$.

$\endgroup$
1
  • $\begingroup$ Thanks Michael. So this just means that $s(X)=d(X^*,wk-^*)=s(||\cdot||)$ which answers the three questions. But is it obvious that if $\mathcal{F}$ determines the norm then its span must be weak-$^*$ dense? $\endgroup$ Sep 18, 2013 at 10:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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