triple dual space and more and more Let $X$ be a normed space. And Define $X^{(n)}$ by $X^{\overbrace{*****....}^{n\ times}}$ where $X^*$ means the dual space of $X$
My question is :
Is there some space $X$ such that a sequence with the initial space $X$ get bigger and bigger infinitely? i.e for all $n>m$, $X^{(n)}>X^{(n)}$ or, indepedent of choice of $X$, does $X^{(n)}$ become a reflexive space for some $n$?.
I'm very wondering this, but I don't know how to search about this. Please let me know some results, search-keyword or anything about this.
 A: I guess you're asking: does there exist a normed space $X$ such that
$$ X \subsetneq X^{\star\star} \subsetneq X^{\star\star\star\star} \subsetneq X^{\star\star\star\star\star\star} \subsetneq\dots $$
This makes more sense than the statement in your question. Remember, there is no canonical inclusion of $X$ into $X^\star$, but there does exist a canonical inclusion of $X$ into $X^{\star\star}$. (I should also clarify that I'm using $X^\star$ to denote the continuous dual of $X$.)
It's useful to know that if $X$ is a Banach space, then $X$ is reflexive if and only if $X^\star$ is reflexive. (See here or here.) Hence if $X$ is a non-reflexive Banach space (i.e. $X \subsetneq X^{\star\star}$), then $X^\star$ is also a non-reflexive Banach space (i.e. $X^\star \subsetneq X^{\star\star\star}$), and so is $X^{\star\star}$ (i.e. $X^{\star\star} \subsetneq X^{\star\star\star\star}$), and so on...
There are many examples of non-reflexive Banach spaces, including $c_0$, $l^1$ and $l^\infty$ (see here for their definitions).
And what if $X$ is not Banach? Even if $X$ itself is not Banach, $X^\star$ is Banach. (See here.) So by the same logic, $X^\star$ is non-reflexive iff $X^{\star\star}$ is non-reflexive iff $X^{\star\star\star}$ is non-reflexive iff $X^{\star\star\star\star}$ is non-reflexive, etc.
