James nonreflexivity sequence separated from an annihilator

We first derive the exact Countable Choice instance used below. For a sequence $(E_n)$ of nonempty sets, let $S$ consist of all finite histories $s$ with $s(j)\in E_j$ for $j<\mathrm{dom}s$, starting with the empty history, and relate $s$ to every one-term extension by a member of $E_{\mathrm{dom}s}$. This relation is entire. DC gives a chain of successively extended histories, whose union chooses one element of every $E_n$. Thus the assumed DC supplies every application of ${\mathrm{A}\mathrm{C}}_{\omega }$ below; we do not use the unproved bibliographic remark “DC implies ${\mathrm{A}\mathrm{C}}_{\omega }$” as a theorem.

By the contrapositive of [F1], choose a norm-bounded sequence $(z_n)$ in $X$ with no weakly convergent subsequence. If $R$ bounds all $\mathrm{\parallel }{z}_n\mathrm{\parallel }$, then $R>0$, since an $R=0$ sequence is constantly zero. Replacing $z_n$ by $z_n\mathrm{/}R$, which preserves and reflects weak convergence of subsequences, we may assume $z_n\in B_X$. Put $M=\overline{{\mathrm{span}}_{\mathbb{R}}\{z_n:n\in \mathbb{N}\}}$. By [F2], $M$ is a separable closed Banach subspace, and its intrinsic weak topology is the relative weak topology inherited from $X$.

The space $M$ is not reflexive. Otherwise [F1], applied to the bounded sequence $(z_n)$ in the Banach space $M$, would give a subsequence converging weakly in $M$. Equality of the two weak topologies in [F2] would make the same subsequence weakly convergent in $X$, contrary to step 1.2. In particular $M\mathrm{\ne }\{0\}$.

Let $J_M:M\to M^{\ast \ast }$ be the canonical map and $Y=J_M(M)$. By [F3], $J_M$ is an isometry, so $Y$ is isometric to the complete space $M$. Step 1.1 and [F5] therefore make $Y$ norm closed in $M^{\ast \ast }$. It is proper because $M$ is not reflexive.

Choose $q\in M^{\ast \ast }\setminus Y$ and put $d=\mathrm{dist}(q,Y)$. Closedness of $Y$ gives $d>0$. Since $d\mathrm{/}\theta >d$ and $d$ is the infimum of the nonempty set $\{\mathrm{\parallel }q-y\mathrm{\parallel }:y\in Y\}$, choose $y\in Y$ with $\mathrm{\parallel }q-y\mathrm{\parallel }<d\mathrm{/}\theta$. Define $F=(q-y)\mathrm{/}\mathrm{\parallel }q-y\mathrm{\parallel }$. Then $\mathrm{\parallel }F\mathrm{\parallel }=1$, translation by $y\in Y$ does not change distance to the linear subspace $Y$, and hence $\mathrm{dist}(F,Y)=\frac{d}{\mathrm{\parallel }q-y\mathrm{\parallel }}>\theta \mathrm{.}$ No simultaneous family is chosen here.

Since $M$ is nonzero and separable, take a nonempty at most countable norm-dense subset $D\subseteq M$ and, by [F6], one surjection $j\mapsto m_j$ from $\mathbb{N}$ onto $D$. Repetitions are harmless.

For $n\in \mathbb{N}$ set $K_n=\{u\in M^{\ast }:u(m_j)=0\text{ for }0\le j\le n\},E_n=\mathrm{span}\{J_M{m}_j:0\le j\le n\}\mathrm{.}$ Then $E_n=K_n^{\perp }$ inside $M^{\ast \ast }$. The inclusion $E_n\subseteq K_n^{\perp }$ follows by evaluation. Conversely, if $H\in M^{\ast \ast }$ vanishes on $K_n$, define $T:M^{\ast }\to {\mathbb{R}}^{n+1}$ by $T(u)=(u(m_0),\dots ,u(m_n))$. Since $\ker T=K_n$, the rule $\lambda (Tu)=H(u)$ is a well-defined linear functional on $\mathrm{im}T$. Finite-dimensional linear algebra extends $\lambda$ to a functional $(t_0,\dots ,t_n)\mapsto {\sum }_{j=0}^n{a}_j{t}_j$ on ${\mathbb{R}}^{n+1}$, using only finitely many choices. Thus $H={\sum }_{j=0}^n{a}_j{J}_M{m}_j\in E_n$.

The quotient-norm identity $\mathrm{\parallel }F{\mathrm{\mid }}_{K_n}\mathrm{\parallel }=\mathrm{dist}(F,E_n)$ holds. For $H\in E_n=K_n^{\perp }$, restriction gives $\mathrm{\parallel }F-H\mathrm{\parallel }\ge \mathrm{\parallel }F{\mathrm{\mid }}_{K_n}\mathrm{\parallel }$. Conversely [F7] extends $F{\mathrm{\mid }}_{K_n}$ to some $G\in M^{\ast \ast }$ with $\mathrm{\parallel }G\mathrm{\parallel }=\mathrm{\parallel }F{\mathrm{\mid }}_{K_n}\mathrm{\parallel }$; then $F-G$ vanishes on $K_n$, so step 6.1 puts $F-G$ in $E_n$ and yields the reverse inequality. Since $E_n\subseteq Y$, step 4.1 now gives $\mathrm{\parallel }F{\mathrm{\mid }}_{K_n}\mathrm{\parallel }=\mathrm{dist}(F,E_n)\ge \mathrm{dist}(F,Y)>\theta \mathrm{.}$

For each $n$, the last strict inequality and the dual-norm definition supply $v\in K_n$ with $\mathrm{\parallel }v\mathrm{\parallel }\le 1$ and $\mathrm{\mid }F(v)\mathrm{\mid }>\theta$. Replacing $v$ by $-v$ if necessary and then setting $u=\theta v\mathrm{/}F(v)$ gives $u\in K_n$, $\mathrm{\parallel }u\mathrm{\parallel }<1$, and $F(u)=\theta$. By [F7], $u$ has an extension $x^{\ast }\in X^{\ast }$ with $\mathrm{\parallel }{x}^{\ast }\mathrm{\parallel }=\mathrm{\parallel }u\mathrm{\parallel }<1$. Therefore the set $P_n$ of all such pairs $(u,x^{\ast })$ is nonempty.

Apply the ${\mathrm{A}\mathrm{C}}_{\omega }$ instance from step 1.1 to $(P_n)$, writing the selected pair as $(u_n,x_n^{\ast })$. Then $x_n^{\ast }\in B_{X^{\ast }}$, $x_n^{\ast }{\mathrm{\mid }}_M=u_n$, $F(u_n)=\theta$, and $u_n(m_j)=0$ whenever $j\le n$. For fixed $m\in M$ and $\epsilon >0$, density supplies $m_j$ with $\mathrm{\parallel }m-m_j\mathrm{\parallel }<\epsilon$; for $n\ge j$, $\mathrm{\mid }{x}_n^{\ast }(m)\mathrm{\mid }=\mathrm{\mid }{u}_n(m-m_j)\mathrm{\mid }\le \mathrm{\parallel }{u}_n\mathrm{\parallel }\mathrm{\parallel }m-m_j\mathrm{\parallel }<\epsilon \mathrm{.}$ Hence $x_n^{\ast }(m)\to 0$ for every $m\in M$.

Let $x^{\ast }={\sum }_{k=1}^r{a}_k{x}_{n_k}^{\ast }$ be any finite convex combination, where $r\ge 1$, $a_k\ge 0$, and ${\sum }_k{a}_k=1$, and let $w^{\ast }\in M^{\perp }$. Restriction to $M$ and step 9.1 give $F\text{\hspace{0.17em}⁣}((x^{\ast }-w^{\ast }){\mathrm{\mid }}_M)={\sum }_{k=1}^r{a}_{k}F(u_{n_k})=\theta \mathrm{.}$ Since $\mathrm{\parallel }F\mathrm{\parallel }=1$, restriction cannot increase norm, and therefore $\mathrm{\parallel }{x}^{\ast }-w^{\ast }\mathrm{\parallel }\ge \mathrm{\parallel }(x^{\ast }-w^{\ast }){\mathrm{\mid }}_M\mathrm{\parallel }\ge \theta \mathrm{.}$ Taking the infimum over both nonempty sets proves $\mathrm{dist}(M^{\perp },\mathrm{co}\{x_n^{\ast }:n\in \mathbb{N}\})\ge \theta$.

Steps 1.2 and 2.1 provide the required separable closed $M$, step 9.1 gives the pointwise-null dual-ball sequence, and step 10.1 gives the asserted annihilator separation. The endpoints $\theta =0,1$ are excluded exactly as stated; $X=\{0\}$ cannot satisfy the nonreflexivity hypothesis. The argument is real: the sign change and the order comparison in step 8.1 are not offered as a complex proof. The ultrafilter lemma and HB enter through [F1], HB also enters through [F2], [F3] and [F7], and DC enters exactly through the finite-history derivation in step 1.1.