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There are Banach spaces which are isomorphic to their second dual but not reflexive (most famously, the James space).

Now let $X$ be such a space and $X'$ be its dual space and let $\phi:X\to X''$ be an isomorphism between $X$ and $X''$. Now consider the weak* topology on $X'$, i.e. the topology generated by the neighborhood base $$\tilde B_r(y_1,...y_n) = \{x\ :\ |x(y_j)| < r,\ j=1,...,n\},$$ where the $y_j$'s are elements in $X$. This neighborhood base is in one-to-one correspondence with the neighborhood base for the weak topology on $X'$ i.e. with the sets $$B_r(f_1,…,f_n) = \{x\ :\ |f_j(x)| < r,\ j=1,…,n\},$$ where the $f_j$'s are elements in $X''$ (via the isomorphism $\phi$).

However, this seem to contradict the fact that the weak and the weak* topology coincide precisely when $X$ is reflexive!?

There should be some flaw in my reasoning but somehow I am blocked here and would be glad in somebody could help out…

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Dirk
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    There is a one-to-one correspondence between the Sorgenfrey line neighbourhoods $[x, x+\frac{1}{n})$ and the real line neighbourhoods $(x-\frac{1}{n}, x+\frac{1}{n})$ as well, but they're not homeomorphic at all.. – Henno Brandsma Nov 20 '14 at 15:36
  • As a matter of fact, each non-reflexive Banach space can be equivalently renormed so as not to be isometrically isomorphic to a dual space.

    W. Davis, W.B. Johnson, A renorming of nonreflexive Banach spaces. Proc. Amer. Math. Soc. 37 (1973), 486–488.

     – Tomasz Kania Nov 20 '14 at 16:32

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Let $\phi$ be an “unnatural” isometric isomorphism between $X$ and $X''$. If $X$ is not reflexive, then there exists some $y\in X$ such that for all $z\in X$, there is some $x_z\in X'$ for which $\phi(y)(x_z)\neq x_z(z)$. This destroys the possibility to identify $\tilde B_r(y)$ with $B_r(\phi(y))$ through $\phi$.

ADDED: Here is a somewhat more transparent proof that the two neighborhood bases cannot be identified through $\phi$. If you assume that the two neighborhood bases can be identified through $\phi$, then you have, in particular, that $\tilde B_r(w)=B_r(\phi(w))$ for all $w\in X$ and $r>0$. That is, for any $x\in X'$, $w\in X$, and $r>0$, $|x(w)|<r$ if and only if $|\phi(w)(x)|<r$. Therefore, $x(w)$ and $\phi(w)(x)$ must have the same absolute values for all $x\in X'$ and $w\in X$: $$|\phi(w)(x)|=|x(w)|=|\psi(w)(x)|\quad\forall x\in X',w\in X,\tag{$*$}$$ where $\psi$ is the canonical embedding.

Now pick any $f\in X''$. Then, there exists unique $w\in X$ such that $f=\phi(w)$ since $\phi$ is an (“unnatural”) isometric isomorphism between $X$ and $X''$. By ($*$), $\phi(w)$ and $\psi(w)$ (considered as continuous linear functionals over $X^*$) have the same kernel, so they must be proportional (for an easy proof of this, see here): there exists some $c\in\mathbb F$ (where $\mathbb F$ denotes the field over which $X$ is a vector space) such that $\phi(w)=c\psi(w)=\psi(cw)$. Consequently, $f=\psi(cw)$. Since $f$ has been an arbitrary element of $X''$, it follows that $X''=\psi(X)$. In other words, the canonical embedding $\psi$ is surjective, so $X$ must be reflexive—a contradiction.

This contradiction reveals that there must exist some $x\in X'$, $w\in X$, and $r>0$ such that either $|x(w)|<r$ and $|\phi(w)(x)|\geq r$, or $|x(w)|\geq r$ and $|\phi(w)(x)|< r$.

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triple_sec
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  • Just to be sure that I understand your argument correctly: In the second sentence you could drop the part "If $X$ is not reflexive" since you assume that $\phi$ is not the canonical embedding, right? – Dirk Nov 20 '14 at 16:30
  • I implicitly use the assumption that $X$ is not reflexive. Sorry about not having been clear enough. Let me clarify, Let $\psi$ be the canonical embedding. If $X$ is not reflexive, then there is some $f\in X''$ such that $f\neq\psi(z)$ for any $z\in X$. But since $\phi$ is an isometric isomorphism, there is a unique $y\in X$ such that $f=\phi(y)$. Therefore, $\phi(y)\neq\psi(z)$ for any $z\in X$. You can then choose, for each $z\in X$, some $x_z\in X'$ such that $\phi(y)(x_z)\neq\psi(z)(x_z)=x_z(z)$. – triple_sec Nov 20 '14 at 16:34
  • Got it, thanks a lot. – Dirk Nov 20 '14 at 21:20
  • @Dirk I added a more detailed—and perhaps more illuminating—proof. – triple_sec Nov 21 '14 at 15:25