# Self-dual Orlicz sequence spaces

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 $$\ell_2$$?

• 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 $X^*$. Aug 9, 2021 at 17:11
• Why the vote to close? Aug 9, 2021 at 17:13
• Bill's suggestion indeed works. Lindenstrauss-Tzafriri notes this in the book. See the construction in Theorem 4.b.12 and the remark after. Aug 9, 2021 at 19:10
• 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 $X^*$? Aug 11, 2021 at 0:44
• @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. Aug 11, 2021 at 15:45

For a given $$1 and $$\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.