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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.

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    $\begingroup$ 3: if $X$ is separable, then there is a countable family of bounded linear functionals that separate points in $X$. $\endgroup$
    – Gerald Edgar
    Commented 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? $\endgroup$
    – Ramiro de la Vega
    Commented 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 $n$. $\endgroup$
    – Nate Eldredge
    Commented 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. $\endgroup$
    – Ramiro de la Vega
    Commented Apr 26, 2012 at 13:07

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

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  • $\begingroup$ 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? $\endgroup$
    – Ramiro de la Vega
    Commented Sep 18, 2013 at 10:18

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