A chain of double duals

Let X be a Banach space. We know that X embeds into its double dual X** as a closed subspace isometrically via \iota: X \rightarrow X^{**} defined by [\iota(x)](\ell) = \ell(x) and we say that X is reflexive if this embedding is a surjection. One may continue this process to obtain a chain of Banach spaces X \subseteq X^{**} \subseteq \cdots \subseteq X^{(n)} \subseteq \cdots. Here X^{(0)} = X, X^{(n + 1)} = (X^{(n)})^{**}. We may ask whether the chain stabilizes. It turns out that only two extreme possbilities hold:

  1. If X is reflexive, then X = X^{**} = \cdots = X^{(n)} = \cdots.
  2. If X is not reflexive, then X \subsetneq X^{**} \subsetneq \cdots \subsetneq X^{(n)} \subsetneq \cdots.

The first part is easy. The second part amounts to the observation that closed subspace of a reflexive Banach space is reflexive (Check this post). If X^{(n)} = X^{(n + 1)} for some n, then X^{(n)} is reflexive. Since X is a  closed subspace of X^{(n)}, it follows that X itself is reflexive.


About Ken Leung

I'm a ordinary boy who likes Science and Mathematics.
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4 Responses to A chain of double duals

  1. Hon Leung says:

    Are there any results comparing the “size” of X and X^{**} in a precise way?
    What I mean is if X is infinite dimensional and X is non-reflexive, how “different” is X and X^{**}?

  2. Ken Leung says:

    I guess many things can happen… I don’t have an idea about this matter. Perhaps I’ll look into this issue later.

  3. Hon Leung says:

    Perhaps we can look at some examples first. Especially L^{1}, L^{\infty}

  4. Ken Leung says:

    I looked into a textbook on Banach space theory, it said that Lindestrauss proved that every separable Banach space is isometrically isomorphic to X**/X for some Banach space X.

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