1
$\begingroup$

Let $X$ be a Banach space, $p \in [1,\infty)$ and $l^p(\mathbb{Z},X)$ the usual sequence space taking values in $X$. Is it always true that $(l^p(\mathbb{Z},X))^* = l^q(\mathbb{Z},X^*)$ and $(c_0(\mathbb{Z},X))^* = l^1(\mathbb{Z},X^*)$, where $\frac{1}{p} + \frac{1}{q} = 1$? I do know that for a $\sigma$-finite measure $\mu$ it holds $(L^p(\mu,X))^* = L^q(\mu,X^*)$ if and only if $X^*$ has the Radon-Nikodym property w.r.t. $\mu$. I suppose that the Radon-Nikodym property is trivially satisfied here, but I am not entirely sure. As I am not an expert in this field, a short proof or a reference would be appreciated (or a counterexample if I am wrong, of course). Thanks!

$\endgroup$
1
  • $\begingroup$ A table of duals of common spaces is found in Dunford & Schwartz, volume 1. $\endgroup$ Commented Mar 24, 2015 at 1:39

0

You must log in to answer this question.

Browse other questions tagged .