Let $X$ be a Banach space. It is natural for us to introduce a quantity measuring the separability of sets as follows: for a subset $A$ of $X$, we set

$\textrm{sep}(A)=\inf\{\epsilon>0: A\subseteq K+\epsilon B_{X}$ for some countable subset $K$ of $X\}$.

Clearly, $A$ is separable if and only if $\textrm{sep}(A)=0$.

It is elementary that a Banach space $X$ is separable if $X^{*}$ is separable. My question is about a quantitative version of this result.

Question. Does there exist a universal constant $C$ such that $$\textrm{sep}(B_{X})\leq C\cdot \textrm{sep}(B_{X^{*}}) \, ?$$


1 Answer 1


For the unit ball $B_X$ of the Banach space there are only two possibilities:

sep$(B_X)= 1$, if $B_X$ is not separable, and sep$(B_X)=0$ if $B_X$ is separable. Indeed, if sep$(B_X)<1$ there are $\varepsilon <1$ and a countable subset $K\subseteq B_X$ with $B_X\subseteq K + \varepsilon B_X$. But this can be iterated, i.e., $$ B_X\subseteq K +\varepsilon (K+\varepsilon B_X) \subseteq K_1+\varepsilon^2 B_X $$ where $K_1=K+\varepsilon K$ is again countable. Inductively, this implies $B_X\subseteq K_n +\varepsilon^n B_X$ for a countable set $K_n$ and hence sep$(B_X)=0$.

Therefore, $C=1$ satisfies the desired inequality sep$(B_X)\le$ sep$(B_{X^*})$.

  • 1
    $\begingroup$ That is great! Many thanks, Jochen. $\endgroup$ Feb 24, 2021 at 9:48
  • $\begingroup$ In particular, if $X$ is reflexive, then one has $\mathrm{sep}(B_X) = \mathrm{sep}(B_{X^\star})$, right ? $\endgroup$
    – dohmatob
    Apr 12, 2022 at 14:44
  • $\begingroup$ Sure, because sep$(B_{X^*})\le$ sep$(B_{X^{**}})=$ sep$(B_X)$. $\endgroup$ Apr 13, 2022 at 8:26

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