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.

  • $\begingroup$ Note that $X$ and $X^{**}$ can be isomorphic although $X$ is not reflexive. $\endgroup$ – Jochen Wengenroth Mar 17 '16 at 7:24
  • $\begingroup$ Even $X$ and $X^*$ may be isomorphic. Take $X=J\oplus J^*$ where $J$ is the James space. $\endgroup$ – Tomasz Kania Mar 17 '16 at 10:14

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.

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.

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