## 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.

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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.