The difference between a normed space being reflexive and being isomorphic to its dual

Question

Quoting wikipedia "a normed vector space is reflexive if it coincides with its bidual". Another definition, more precise is that a normed vector space is reflexive if its evaluation map $J:X\rightarrow X^{\ast\ast}$ is surjective. Many proofs in my book, for instance the proof that a hilbert space is always reflexive end like "Then $ X$ can be identified with its bidual $X^{\ast\ast}$ so it is reflexive", without show that the evaluation map is surjective.

I know (because my professor blamed me in the last exam for what I said) that it's not true that $X$ is reflexive if and only if is isomorphic to its bidual.

My questions are:

1. What is, in this context, the difference between "to be isomorph to its bidual" and "to be identified with it's bidual"? Why the first it's not good and the second it is?

2. How to show the equivalence of the two definitions? That a space can be identified with it's bidual if and only if the evaluation map is surjective.

3. Can you show me an example of a space isomorph to its bidual but not reflexive?

I didn't found the answers of this questions anywhere I looked. Thank you for your help.

Answer 1

The point is, when the authors of your book write "hence $X$ can be identified with $X^{**}$" they mean that $J$ is onto (and hence an isomorphism, as $J$ is isometric by Hahn-Banach), but want to say that it is more or less straightforward to see it. Let us show this explicitly for Hilbert spaces. Let $H$ be a Hilbert space with inner product $\def\<#1>{\left<#1\right>}$$\<.,.>$, we know that $\Theta\colon H \to H^*$, $\Theta x = \<.,x>$ is a conjugate linear isomorphism. We define an inner product on $H^*$ by $$ \<x^*,y^*>' := \<\Theta^{-1}y^*,\Theta^{-1}x^*> $$ Let $\Theta^*\colon H^* \to H^{**}$ denote the corresponding Riesz isomorphism. Let $x^{**} \in X^{**}$, we have for any $x^* \in X^*$: \begin{align*} x^{**}(x^*) &= \<x^*, \Theta^*{}^{-1}x^{**}>'\\ &= \<\Theta^{-1}\Theta^*{}^{-1}x^{**}, \Theta^{-1}x^*>\\ &= x^*(\Theta^{-1}\Theta^*{}^{-1}x^{**})\\ &= J(\Theta^{-1}\Theta^*{}^{-1}x^{**})(x^*) \end{align*} As $x^*$ was arbitrary, that shows that $J(\Theta^{-1}\Theta^*{}^{-1}x^{**}) = x^{**}$ and $J$ is onto.

Answer 2

1. Being identified with is a loose term, it does not have a precise definition. It is used when we have a canonical isomorphism, i.e., an isomorphism that is naturally defined in terms of the objects themselves, without any choices made on our part. In our case, there is one canonical map from $X$ to $X^{**}$, namely $J$. If this map is an isomorphism, we can say that $X$ is identified with $X^{**}$, or canonically isomorphic to it. If $J$ is not an isomorphism, there may be some other operator that is an isomorphism. But that operator would not be the canonical one, which is $J$.

2. The definition of $J$ implies (by Hahn-Banach) that $J$ is an isometry: $\|Jx\|_{X^{**}} = \|x\|_X$ for all $x\in X$. So, only surjectivity remains a question. Once we have surjectivity, we'll know that $J$ is an isomorphism.

3. There are no simple examples of such spaces. There is the James Space which David Mitra already pointed out. There is also $\mathcal{W}_2$ interpolation space between $v_1$ and $\ell_\infty$: see Chapter 7 of Pisier's lecture notes.