James reflexivity theorem

Suppose $X$ is reflexive and let $f\in X^{\ast }$. If $f=0$, then $x=0\in B_X$ attains its norm. If $f\mathrm{\ne }0$, define $g$ on the one-dimensional scalar-linear subspace ${\mathrm{span}}_{\mathbb{K}}\{f\}\subseteq X^{\ast }$ by $g(cf)=c\mathrm{\parallel }f\mathrm{\parallel }$. Then $\mathrm{\mid }g(cf)\mathrm{\mid }=\mathrm{\parallel }cf\mathrm{\parallel }$, so $\mathrm{\parallel }g\mathrm{\parallel }=1$. By [F2] it extends to $G\in X^{\ast \ast }$ with $\mathrm{\parallel }G\mathrm{\parallel }=1$. Reflexivity and [F1] give $x\in X$ with $G=J_{X}x$ and $\mathrm{\parallel }x\mathrm{\parallel }=\mathrm{\parallel }G\mathrm{\parallel }=1$. Hence $f(x)=G(f)=\mathrm{\parallel }f\mathrm{\parallel }$, so $f$ attains its norm on $B_X$.

For the reverse implication, first suppose that $X$ is real and every member of $X^{\ast }$ attains its norm. If $X$ were nonreflexive, [F3] would supply $z^{\ast }\in X^{\ast }$ that attains its norm nowhere on $B_X$, a contradiction. Thus $X$ is reflexive.

Now suppose $X$ is complex and every complex-linear member of $X^{\ast }$ attains its norm. Let $X_{\mathbb{R}}$ be the same additive normed space with scalars restricted to $\mathbb{R}$. It remains a real Banach space because its norm and Cauchy sequences are unchanged. For $u\in (X_{\mathbb{R}}{)}^{\ast }$ define $f_u(x)=u(x)-iu(ix)\mathrm{.}$ Real linearity gives $f_u(ix)=i{f}_u(x)$, hence $f_u$ is complex linear, and $\mathrm{Re}{f}_u=u$. The inequalities $\mathrm{\parallel }u\mathrm{\parallel }\le \mathrm{\parallel }{f}_u\mathrm{\parallel }$ and $\mathrm{\parallel }{f}_u\mathrm{\parallel }\le \mathrm{\parallel }u\mathrm{\parallel }$ follow respectively from $u=\mathrm{Re}{f}_u$ and, for each $x$, choosing a unit scalar $a$ with $a{f}_u(x)=\mathrm{\mid }{f}_u(x)\mathrm{\mid }$ and observing $\mathrm{\mid }{f}_u(x)\mathrm{\mid }=u(ax)$. Thus $\mathrm{\parallel }{f}_u\mathrm{\parallel }=\mathrm{\parallel }u\mathrm{\parallel }$.

By hypothesis, $f_u$ attains its norm at some $x\in B_X$. Choose a unit scalar $a$ with $a{f}_u(x)=\mathrm{\mid }{f}_u(x)\mathrm{\mid }$ (take $a=1$ if the value is zero). Then $ax\in B_X$ and $u(ax)=\mathrm{Re}{f}_u(ax)=\mathrm{Re}(a{f}_u(x))=\mathrm{\parallel }{f}_u\mathrm{\parallel }=\mathrm{\parallel }u\mathrm{\parallel }\mathrm{.}$ Hence every member of $(X_{\mathbb{R}}{)}^{\ast }$ attains its norm. The real implication in step 1.2 shows that $X_{\mathbb{R}}$ is reflexive.

To pass back to the complex space without an unproved slogan, let $H\in X^{\ast \ast }$ be complex linear. For $u\in (X_{\mathbb{R}}{)}^{\ast }$ put $U(u)=\mathrm{Re}H(f_u)$. Step 1.3 makes $U$ a bounded real-linear functional on $(X_{\mathbb{R}}{)}^{\ast }$, with $\mathrm{\mid }U(u)\mathrm{\mid }\le \mathrm{\parallel }H\mathrm{\parallel }\mathrm{\parallel }u\mathrm{\parallel }$. Real reflexivity from step 2.1 supplies $x\in X_{\mathbb{R}}$ such that $U(u)=u(x)$ for every real-dual $u$. If $f\in X^{\ast }$ and $u=\mathrm{Re}f$, then the formula in step 1.3 gives $f_u=f$, so $\mathrm{Re}H(f)=\mathrm{Re}f(x)$. Apply the same equality to the complex functional $if$: complex linearity gives $\mathrm{Re}H(if)=-\mathrm{Im}H(f)$ and $\mathrm{Re}(if(x))=-\mathrm{Im}f(x)$. Thus $H(f)=f(x)=J_{X}x(f)$ for every $f\in X^{\ast }$. Therefore $J_X$ is onto and $X$ is complex-reflexive.

Steps 1.1 and 1.2 prove both implications over the reals; steps 1.3–3.1 prove the complex reverse implication, while step 1.1 already covers the complex forward implication. If $X=\{0\}$, its dual and bidual are zero and the unique functional attains norm zero at zero. The forward implication uses only HB; UL and DC enter the reverse implication exactly through [F3].