In functional analysis, there is a concept of a self dual space (Hilbert) and a self double dual space (reflexive). I am curious as to whether a generalization exists or not (and if it exists, whether or not it is actually useful). Say a space is not reflexive, but if we keep on taking the dual many times, say $X$, $X^\*$, $X^{\*\*}$, $X^{\*\*\*}$, $X^{\*\*\*\*}$, etc, do we eventually get a cycle, i.e. the $n$th dual is isomorphic to the $m$th dual for some $m<n$?


Such an answer or a theory of the structure of the chain of duals would be a theory of the dual as a functor in the cateogry of Banach spaces, e.g. in the category of Hilbert space, the dual would be a contravariant automorphism.



Structure of chain of duals in functional analysis
No. The sequence $c\_0$, $c\_0^\* \cong l^1$, $c\_0^{\*\*} \cong l^\infty$, $\ldots$, doesn't stabilize.


Let $E$ and $F$ be Banach spaces. The following two lemmas are standard, and good exercises.


Lemma 1. If $E$ embeds isometrically in $F$, then $E^\*$ is a quotient of $F^\*$.


Lemma 2. If $E$ is a quotient of $F$, then $E^\*$ embeds isometrically in $F^\*$.


The proof of the next lemma is given in the top answer to [this question](https://math.stackexchange.com/questions/83526/the-set-of-ultrafilters-on-an-infinite-set).


Lemma 3. The set of ultrafilters on a set $X$ has cardinality $2^{2^{|X|}}$.


Now let $X\_0 = \mathbb{N}$ and inductively define $X\_{n+1}$ to be the Stone-Cech compactification of $X\_n$ with the discrete topology. By Lemma 3, $|X\_0| < |X\_1| < \cdots$.


Let $E\_0 = l^\infty$ and inductively define $E\_{n+1} = E\_n^\*$. Then $E\_0 = l^\infty(X\_0) \cong C(X\_1)$ and $l^1(X\_1)$ isometrically embeds in $E\_1$, so by Lemma 1, $l^1(X\_1)^\* \cong l^\infty(X\_1)$ is a quotient of $E\_2$. But $l^\infty(X\_1) \cong C(X\_2)$ and $l^1(X\_2)$ isometrically embeds in its dual, hence in $E\_3$ by Lemma 2. Then by Lemma 1, $l^\infty(X\_2)$ is a quotient of $E\_4$, and so on. For any $n$ we get that $l^1(X\_n)$ isometrically embeds in $E\_{2n-1}$ and $l^\infty(X\_n)$ is a quotient of $E\_{2n}$. So the cardinalities of the $E\_n$ never stabilize.


