James convex-block norm-attainment criterion

We first record two elementary properties of $L$. For a bounded sequence $s=(s_n^{\ast })$, say $\mathrm{\parallel }{s}_n^{\ast }\mathrm{\parallel }\le C$, the function $p(u)={\mathrm{lim\hspace{0.17em}sup}}_n{s}_n^{\ast }(u)$ is finite, positively homogeneous, and subadditive: homogeneity follows directly from tail suprema and subadditivity from [F1]. Apply [F2] to the zero functional on $\{0\}$ dominated by $p$. The extension $w^{\ast }$ satisfies $w^{\ast }(u)\le p(u)$, while applying this inequality at $-u$ and using [F1] gives $w^{\ast }(u)\ge {\mathrm{lim\hspace{0.17em}inf}}_n{s}_n^{\ast }(u)$. Hence $\mathrm{\mid }{w}^{\ast }(u)\mathrm{\mid }\le C\mathrm{\parallel }u\mathrm{\parallel }$, so $w^{\ast }\in L(s)$ and $\mathrm{\parallel }{w}^{\ast }\mathrm{\parallel }\le C$. Thus $L(s)$ is nonempty and, when $s\subseteq B_{X^{\ast }}$, lies in $B_{X^{\ast }}$.

Let $V(s)$ be the set of sequences $v=(v_n^{\ast })$ for which $v_n^{\ast }\in \mathrm{co}\{s_j^{\ast }:j\ge n\}$ at every $n$. It is nonempty because $s\in V(s)$. For each $u$, every such convex combination satisfies $v_n^{\ast }(u)\le {\sup }_{j\ge n}{s}_j^{\ast }(u)$, so the receding-tail definition gives ${\mathrm{lim\hspace{0.17em}sup}}_n{v}_n^{\ast }(u)\le {\mathrm{lim\hspace{0.17em}sup}}_n{s}_n^{\ast }(u)$ and therefore $L(v)\subseteq L(s)$. Flattening two finite convex combinations proves $V(v)\subseteq V(s)$ when $v\in V(s)$. A subsequence of $s$ also belongs to $V(s)$: its $n$th index is at least $n$ by [F6].

Reindex the weights by positive integers, $b_m={\beta }_{m-1}$ for $m\ge 1$. Extend the corresponding reindexing of the dual sequence to a genuine zero-based library sequence by setting ${\widehat{x}}_0^{\ast }=x_0^{\ast }$ and ${\widehat{x}}_m^{\ast }=x_{m-1}^{\ast }$ for $m\ge 1$. The duplicate initial term does not change any scalar limsup, so $L(\widehat{x})=L(x)$, and $\{{\widehat{x}}_m^{\ast }:m\ge 1\}=\{x_n^{\ast }:n\in \mathbb{N}\}$. Put $T_m={\sum }_{j=m}^{\mathrm{\infty }}{b}_j$, so $T_1=1$, $T_m=b_m+T_{m+1}$, every $T_m>0$, and $T_m\to 0$. Choose explicitly ${\epsilon }_m=\min \{1\mathrm{/}(m+1),(1-\theta )2^{-m-1}{T}_m{T}_{m+1}\mathrm{/}{b}_m\}>0$. Then ${\epsilon }_m\to 0$ and [F6] gives \sum_{m=1}^\infty\frac{b_m\varepsilon_m}{T_mT_{m+1}}\le(1-\theta)\sum_{m=1}^\infty2^{-m-1}=\frac{1-\theta}{2}<1-\theta.\tag{1}

Now additionally assume the ultrafilter lemma and that $X$ is nonreflexive. Apply [F7] with the same $\theta$ to obtain $M$ and $(x_n^{\ast })\subseteq B_{X^{\ast }}$, pointwise null on $M$, with its convex hull at distance at least $\theta$ from $M^{\perp }$. If $w^{\ast }\in L(x)$, then for $m\in M$ the defining inequality at $m$ and $-m$ gives $w^{\ast }(m)\le 0$ and $-w^{\ast }(m)\le 0$. Hence $L(x)\subseteq M^{\perp }$, and the separation hypothesis of the technical criterion holds.

Set $x^{(0)}=\widehat{x}$. Suppose that $y_1^{\ast },\dots ,y_{m-1}^{\ast }$ and the sequences $x^{(0)},\dots ,x^{(m-1)}$ have been constructed. For $y^{\ast }\in \mathrm{co}\{x_j^{(m-1)}:j\ge m\}$ and $v\in V(x^{(m-1)})$, define $S_m(y^{\ast },v)=\{\mathrm{\parallel }{\sum }_{j<m}{b}_j{y}_j^{\ast }+T_m{y}^{\ast }-w^{\ast }\mathrm{\parallel }:w^{\ast }\in L(v)\},$ and let ${\alpha }_m$ be the infimum, over all such $(y^{\ast },v)$, of $\sup S_m(y^{\ast },v)$. Step 1.1 makes every $S_m$ nonempty. All displayed functionals have norm at most two, because all the convex blocks and all members of $L(v)$ lie in the dual unit ball; hence $0\le {\alpha }_m\le 2$ and [F3] makes the infimum legitimate.

For $m=1$, any admissible $y^{\ast }$ is in the convex hull of the original sequence and $L(v)\subseteq L(\widehat{x})=L(x)$ by steps 1.2–1.3. The separation hypothesis therefore gives $\mathrm{\parallel }{y}^{\ast }-w^{\ast }\mathrm{\parallel }\ge \theta$ for every admissible $w^{\ast }$, so ${\alpha }_1\ge \theta$.

Suppose $m\ge 2$. The induction will arrange that $x^{(m-1)}$ is a subsequence of some $z^{(m-1)}\in V(x^{(m-2)})$. Consequently $x^{(m-1)}\in V(x^{(m-2)})$ by step 1.2. For an admissible $y^{\ast }$ at stage $m$, both $y_{m-1}^{\ast }$ and $y^{\ast }$ lie in $\mathrm{co}\{x_k^{(m-2)}:k\ge m-1\}$, and so does $u^{\ast }=\frac{b_{m-1}{y}_{m-1}^{\ast }+T_m{y}^{\ast }}{T_{m-1}}\mathrm{.}$ Also $v\in V(x^{(m-2)})$ by flattening. Since ${\sum }_{j<m-1}{b}_j{y}_j^{\ast }+T_{m-1}{u}^{\ast }-w^{\ast }={\sum }_{j<m}{b}_j{y}_j^{\ast }+T_m{y}^{\ast }-w^{\ast },$ every stage-$m$ candidate supplies a stage-$(m-1)$ candidate of the same supremum. Hence ${\alpha }_{m-1}\le {\alpha }_m$. Together with step 3.1, $\theta \le {\alpha }_m\le 2$ for all $m$.

The definition of the positive number ${\alpha }_m$ supplies $y_m^{\ast }\in \mathrm{co}\{x_j^{(m-1)}:j\ge m\}$ and $z^{(m)}\in V(x^{(m-1)})$ such that \alpha_m\le\sup S_m(y_m^*,z^{(m)})<\alpha_m(1+\varepsilon_m).\tag{2} Because $0<{\epsilon }_m<1$, choose $w_m^{\ast }\in L(z^{(m)})$ for which the norm inside (2) is greater than ${\alpha }_m(1-{\epsilon }_m)$. By [F4] and the balance of $B_X$, there is $u_m\in B_X$ on which the same functional, without absolute-value signs, has value greater than that number. The bounded scalar sequence $z_j^{(m)}(u_m)$ has a subsequence converging to its liminf by [F1]; denote the corresponding dual sequence by $x^{(m)}$.

These choices depend on the whole finite history. Let the state set consist of all finite histories satisfying step 5.1, beginning with the empty history and $x^{(0)}=\widehat{x}$, and relate a history to each valid one-stage extension. Step 5.1 proves that every state has a successor. Applying DC once produces all $y_m^{\ast },z^{(m)},w_m^{\ast },u_m,x^{(m)}$ with (2) and the strict lower inequality. No simultaneous selection outside this DC application is being hidden.

The induction has produced the positive-indexed family $(y_m^{\ast }{)}_{m\ge 1}$. Make it a genuine sequence by putting ${\tilde{y}}_0^{\ast }=y_1^{\ast }$ and ${\tilde{y}}_m^{\ast }=y_m^{\ast }$ for $m\ge 1$. For fixed $n\ge 0$ and every $j>n$, repeated flattening of the relations in steps 4.1–5.1 gives $y_j^{\ast }\in \mathrm{co}\{x_k^{(n)}:k\ge j\}$. Ignoring the single duplicated initial term in $\tilde{y}$, the tail argument of step 1.2 therefore yields L(\widetilde y)\subseteq\bigcap_{n\ge0}L(x^{(n)})\subseteq\bigcap_{n\ge1}L(z^{(n)}).\tag{3} For the second inclusion, $x^{(n)}$ is a subsequence of $z^{(n)}$.

Fix $w^{\ast }\in L(\tilde{y})$. Since $w^{\ast }\in L(x^{(m)})$ by (3) and the $u_m$-evaluations of $x^{(m)}$ converge to the liminf of those of $z^{(m)}$, $w^{\ast }(u_m)\le {\mathrm{lim\hspace{0.17em}inf}}_j{z}_j^{(m)}(u_m)\le w_m^{\ast }(u_m)\mathrm{.}$ The last inequality follows by applying the definition $w_m^{\ast }\in L(z^{(m)})$ at $-u_m$ and using limsup reflection. Replacing $w_m^{\ast }$ by $w^{\ast }$ therefore preserves the strict lower evaluation chosen in step 5.1. The upper estimate follows from $w^{\ast }\in L(z^{(m)})$ and (2). Thus \alpha_m(1-\varepsilon_m)<\|\sum_{j<m}b_jy_j^*+T_my_m^*-w^*\|<\alpha_m(1+\varepsilon_m).\tag{4}

By steps 2.1 and 4.1, $({\alpha }_m)$ is nondecreasing and bounded above by $2$, so [F3] gives a limit $\alpha \in [\theta ,2]$. The series ${\sum }_{j\ge 1}{b}_j{y}_j^{\ast }$ is absolutely convergent and hence convergent in the Banach space $X^{\ast }$ by [F4]. If $q={\sum }_{j\ge 1}{b}_j{y}_j^{\ast }-w^{\ast }$ and the functional inside (4) is $q_m$, then $\mathrm{\parallel }q-q_m\mathrm{\parallel }\le {\sum }_{j\ge m}{b}_j\mathrm{\parallel }{y}_j^{\ast }-y_m^{\ast }\mathrm{\parallel }\le 2T_m\to 0$. Since ${\epsilon }_m\to 0$, (4) gives $\mathrm{\parallel }q\mathrm{\parallel }=\alpha$. Because $\sum b_j=1$, this is $\mathrm{\parallel }{\sum }_{j\ge 1}{b}_j(y_j^{\ast }-w^{\ast })\mathrm{\parallel }=\alpha$.

It remains to prove the strict prefix estimate. Put $P_n={\sum }_{j=1}^n{b}_j(y_j^{\ast }-w^{\ast })$, $P_0=0$, and $Q_n=P_{n-1}+T_n(y_n^{\ast }-w^{\ast })$. The upper half of (4) and ${\alpha }_n\le \alpha$ give $\mathrm{\parallel }{Q}_n\mathrm{\parallel }<\alpha (1+{\epsilon }_n)$. The exact identity $P_n=\frac{b_n}{T_n}{Q}_n+\frac{T_{n+1}}{T_n}{P}_{n-1}$ and induction yield $\mathrm{\parallel }{P}_n\mathrm{\parallel }<\alpha T_{n+1}{\sum }_{k=1}^n\frac{b_k(1+{\epsilon }_k)}{T_k{T}_{k+1}}\mathrm{.}$ Since $b_k=T_k-T_{k+1}$, ${\sum }_{k=1}^n{b}_k\mathrm{/}(T_k{T}_{k+1})=1\mathrm{/}{T}_{n+1}-1\mathrm{/}{T}_1$. Using $T_1=1$ and (1) therefore gives \|P_n\|<\alpha(1-T_{n+1})+\alpha(1-\theta)T_{n+1}=\alpha(1-\theta T_{n+1}).\tag{5} The calculation includes $n=1$, where $P_0=0$.

Define the asserted zero-based sequence by $y_n^{\mathrm{o}\mathrm{u}\mathrm{t}\ast }=y_{n+1}^{\ast }$. It and $\tilde{y}$ differ only by a one-place shift and a duplicated first term, so their scalar limsups agree and $L(y^{\mathrm{o}\mathrm{u}\mathrm{t}})=L(\tilde{y})$. Step 9.1 is therefore the asserted infinite-series equality for every $w^{\ast }\in L(y^{\mathrm{o}\mathrm{u}\mathrm{t}})$, while (5) with $n$ replaced by $n+1$ is exactly $\mathrm{\parallel }{\sum }_{j=0}^n{\beta }_j(y_j^{\mathrm{o}\mathrm{u}\mathrm{t}\ast }-w^{\ast })\mathrm{\parallel }<\alpha (1-\theta {\sum }_{j>n}{\beta }_j)\mathrm{.}$ Relabeling $y^{\mathrm{o}\mathrm{u}\mathrm{t}}$ as $y$ proves the technical criterion, including the first term $n=0$ and every positive weight sequence.

Put $\mathrm{\Delta }={\theta }^2\mathrm{/}4$ and $r=\mathrm{\Delta }\mathrm{/}2$, and take the positive zero-based weights ${\beta }_j=(1-r)r^j$. By [F6] they sum to one; their tails $R_k={\sum }_{j\ge k}{\beta }_j$ satisfy $R_{k+1}=r{R}_k<\mathrm{\Delta }{R}_k$. Apply the technical criterion to obtain $\alpha$, $y$, and choose one $w^{\ast }\in L(y)$, which is possible by step 1.1. The absolutely convergent series $z^{\ast }={\sum }_{j=0}^{\mathrm{\infty }}{\beta }_j(y_j^{\ast }-w^{\ast })$ has $\mathrm{\parallel }{z}^{\ast }\mathrm{\parallel }=\alpha \ge \theta >0$.

Fix $u\in B_X$. Step 1.1 gives $\mathrm{\parallel }{w}^{\ast }\mathrm{\parallel }\le 1$ and ${\mathrm{lim\hspace{0.17em}inf}}_j{y}_j^{\ast }(u)\le w^{\ast }(u)$. Since ${\theta }^2-2\mathrm{\Delta }={\theta }^2\mathrm{/}2>0$, there is an arbitrarily late $k$, and in particular one with $k\ge 1$, such that $(y_k^{\ast }-w^{\ast })(u)<{\theta }^2-2\mathrm{\Delta }\le \alpha \theta -2\mathrm{\Delta }$. Split $z^{\ast }(u)$ before $k$, at $k$, and after $k$. The prefix estimate at $k-1$, the displayed scalar inequality, and $\mathrm{\parallel }{y}_j^{\ast }-w^{\ast }\mathrm{\parallel }\le 2$ give $z^{\ast }(u)<\alpha (1-\theta R_k)+(\alpha \theta -2\mathrm{\Delta }){\beta }_k+2R_{k+1}<\alpha (1-\theta R_k)+(\alpha \theta -2\mathrm{\Delta }){\beta }_k+2\mathrm{\Delta }{R}_k=\alpha -(\alpha \theta -2\mathrm{\Delta })R_{k+1}<\alpha ,$ because $\alpha \theta -2\mathrm{\Delta }\ge {\theta }^2-2\mathrm{\Delta }>0$. Applying the same argument to $-u$ gives $-z^{\ast }(u)<\alpha$. Therefore $\mathrm{\mid }{z}^{\ast }(u)\mathrm{\mid }<\alpha =\mathrm{\parallel }{z}^{\ast }\mathrm{\parallel }$ for every $u\in B_X$: $z^{\ast }$ is nonzero but attains its norm nowhere on the closed unit ball.

The first part uses DC only in step 6.1 and HB only in step 1.1. The ultrafilter lemma is absent there and enters solely through [F7] in the final nonreflexive consequence. The empty space cannot meet either separation or nonreflexivity; singleton convex combinations and the first prefix occur in steps 3.1 and 11.1; all tail denominators are positive because every weight is positive; and both strict endpoints $0<\theta <1$ are used in (1) and step 13.1.