Let $D[0,1]$ be the space of all right-continuous left-limited functions $f\colon [0,1]\to \mathbb{R}$ equipped with the supremum norm $f\mapsto \|f\|_\infty = \sup_{t\in[0,1]} |f(t)|$. This is a non-separable Banach space whose dual $D[0,1]^\ast$ is known to be separable in the weak* topology; see, e.g., Chapter 41, p. 1756 of

Johnson, W. B. (ed.); Lindenstrauss, J. (ed.), Handbook of the geometry of Banach spaces. Volume 2, Amsterdam: North-Holland. xii, 1007-1866 (2003). ZBL1013.46001.

Is the unit ball in $D[0,1]^\ast$ separable in the weak* topology?

  • $\begingroup$ Could you briefly describe the countable dense set? $\endgroup$
    – Ruy
    May 19 at 15:34

1 Answer 1


$(D([0,1]),\|\cdot\|_{\infty})$ is a commutative $C^{\ast}$-algebra, so it is isometrically isomorphic to $C(\Delta)$ by the Gelfand map, where $\Delta$ is the character space of $D([0,1])$.

Let $h_{1+}\in\Delta$ be defined by $h_{1+}(f) = f(1)$ for all $f\in (D([0,1])$. Every $h\in\Delta\backslash\{h_{1+}\}$ is either of the form $$\forall f\in D([0,1]) \hspace{6mm} h_{c+}(f) = f(c+) = \lim_{x\to c+} f(x) $$ for some $c\in [0,1)$ or of the form $$\forall f\in D([0,1]) \hspace{6mm} h_{c-}(f) = f(c-) = \lim_{x\to c-} f(x) $$ for some $c\in (0,1]$. Let $K=\{(c,1):c\in[0,1]\}\cup\{(c,-1):c\in(0,1]\}$ with the weak parallel line topology . It is relatively straightforward to show that $\Delta$ is homeomorphic to $K$. $K$ is a separable, compact, Hausdorff space.

Since $\Delta$ and $K$ are homeomorphic, $C(\Delta)$ and $C(K)$ are isometrically isomorphic as Banach spaces. Lastly, see the implications in the first page of the paper https://doi.org/10.48550/arXiv.1112.5710 : since $K$ is separable, the unit ball of $(C(K))^{\ast}$ is weak$^{\ast}$ separable.


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