# Iterated duals of Banach spaces

Let $B$ be a (non-reflexive) Banach space. Denote by $B^{(n)}$ the $n$-fold dual, i.e. $B^{(0)}=B$ and $B^{(n)}=(B^{(n-1)})^\prime$. Does there exist an $n\ge 1$ s.t. $B^{(n)}\cong B$ ? Please explain.

-
James' space is an example of non-reflexive Banach space which is isomorphic to it's bidual: James, A non reflexive Banach space isometric with its second conjugate space, Proc. Nat. Acad. Sc. USA, 37 (1951), p.174-177) –  Davide Giraudo Oct 11 '12 at 14:51
This thread (math.stackexchange.com/questions/113198/… this one (mathoverflow.net/questions/46138) can be helpful. –  Davide Giraudo Oct 11 '12 at 15:13
And you can change you nickname, as it's not a stupid question at all. –  Davide Giraudo Oct 11 '12 at 15:31
Thanks for the link, the second one solves it sufficiently. That's a somewhat surprising answer. I'll keep my nick, in case I ask a stupid question later on ;-). –  Your Ad Here Oct 11 '12 at 15:33
Note any separable Banach space $X$ with non-separable dual $X^*$ would furnish a counterexample (since a normed space is separable if its dual is). –  David Mitra Oct 11 '12 at 15:38

It may happen that $B^{(2)}\cong B$, even if $B$ is not reflexive, as James' space shows, see James, A non reflexive Banach space isometric with its second conjugate space, Proc. Nat. Acad. Sc. USA, 37 (1951), p.174-177) for instance.

But it's possible such a $n$ doesn't exists.

Lemma: If $E$ is a Banach space and $E'$ its topological dual space, and $E'$ is separable, so is $E$.

If $\{f_n\}$ is dense in the unit ball of $E'$, for each $n$, fix $x_n\in E$ of norm $1$ such that $|f_n(x_n)| \geq \frac 12\lVert f_n\rVert_{B'}$. We have to show that $F:=\operatorname{Span}\{x_k,k\geq 1\}$ is dense in $E$. Let $f$ a continuous linear functional with vanishes on $F$. We have to show that it vanished on $E$. Let $\{n_k\}$ such that $\lVert f-f_{n_k}\rVert\leq k^{-1}$ (we can assume that $f$ has a norm $\leq 1$). Then $$\lVert f_{n_k}\rVert\leq \lVert f_{n_k}(x_{n_k})-f(x_{n_k})\rVert\leq k^{-1},$$ so $\lVert f\rVert\leq 2k^{-1}$.

To apply this to our problem, take $B$ a separable Banach space with non-separable dual (as $\ell^1(\Bbb R)$). It's not reflexive, otherwise $B''$ would be separable and so would be $B'$. If $B^{(n)}\cong B$ for some $n\geq 2$, then $B^{(n)}$ would be separable, as this property is conserved by isomorphism. By the lemma $B^{(n-1)}$ would be separable, and iterating that $B'$^would be separable, a contradiction.

-
There's a typo in the Lemma, it should be \$E –  Your Ad Here Oct 12 '12 at 18:18
Right, fixed now. Thanks. –  Davide Giraudo Oct 12 '12 at 18:26