7
$\begingroup$

Suppose that $\ell_\phi$ is a reflexive Orlicz sequence space such that its dual space $\ell_\phi^*$ is isomorphic to $\ell_\phi$.

Is $\ell_\phi$ isomorphic to $\ell_2$?

$\endgroup$
11
  • 2
    $\begingroup$ I would try to prove that if $1<p<2$ and $q$ is the conjugate index to $p$, then there is an Orlicz sequence space $X$ that is complementably universal for all Orlicz sequence spaces that are $p$ convex and $q$ concave. Such an $X$ would be isomorphic to $X^*$. $\endgroup$ Aug 9, 2021 at 17:11
  • 3
    $\begingroup$ Why the vote to close? $\endgroup$ Aug 9, 2021 at 17:13
  • 2
    $\begingroup$ Bill's suggestion indeed works. Lindenstrauss-Tzafriri notes this in the book. See the construction in Theorem 4.b.12 and the remark after. $\endgroup$ Aug 9, 2021 at 19:10
  • 1
    $\begingroup$ Is there a Banach space $X$ with a symmetric basis $(e_n, e_n^*)$, non-equivalent to an orthonormal basis for a Hilbert space, s.t. the map $e_n\mapsto e_n^*$ extends to an isomorphism from $X$ onto $X^*$? $\endgroup$ Aug 11, 2021 at 0:44
  • 1
    $\begingroup$ @Bunyamin: But the $x_n^*$ are not a priori biorthogonal to $x_n$. You don't know that $\langle x_n^*, x_n \rangle =\sum_{I\in F_n} a_n^2$ is bounded away from zero unless you assume what you are trying to prove. $\endgroup$ Aug 11, 2021 at 15:45

1 Answer 1

6
$\begingroup$

For a given $1<p$ and $\frac{1}{p}+\frac{1}{q}=1$ you can construct a universal Orlicz sequence space $\ell_M$ so that every Orlicz function $N$ in between $p$ and $q$ is equivalent to a function in $E_M$ which corresponds to the Orlicz subspaces spanned by constant block bases of $\ell_M$ thus complemented. Such a space is unique up to isomorphism (depending only to $p$), and $E_{M^*}$ has the same property so $\ell_{M^*}$ is isomorphic to $\ell_{M}$ by the uniqueness. This construction is given in [LT, Theorem 4.b.12]. See also the remark after.

This answer was first given by Bill Johnson in comments, I just added the reference.

Lindenstrauss, Joram; Tzafriri, Lior, Classical Banach spaces. 1: Sequence spaces. 2. Function spaces., Classics in Mathematics. Berlin: Springer-Verlag. xx, 432 p. (1996). ZBL0852.46015.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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