Proof. Suppose, to derive a contradiction, that \(\lim_{n \rightarrow \infty}\|x_n\|_{\alpha_r, r \vee k} =0\) for all \(r \ge 1\); in particular, \(x_n \rightarrow 0\) in \(\ell_2(\overline{S})\). Hence, by a gliding hump argument, approximating by finitely disjointly supported vectors, and after passing to a subsequence and relabelling, we may assume that \(\operatorname{supp} x_n\) is finite and that \(\operatorname{supp} x_n \cap \operatorname{supp} x_m = \emptyset\) if \(m \ne n\).

Fix \(n \ge 1\) and \(F \in \mathcal{F}^{(k)}_\alpha\) satisfying \[|F \cap \operatorname{supp} x_n| \ge 2.\] Let \[N = \max \{ d(p,q) \colon p,q \in \operatorname{supp} x_n, p \ne q\}.\] It follows that \(F^\sharp \le N\), and hence \[F \in \cup_{r=1}^N \mathcal{F}_{\alpha_r}^{(r \vee k)}.\]

Let \[x_n = \sum a_\gamma e_\gamma\] and, for \(m > n\), \[x_m = \sum b^m_\gamma e_\gamma.\] Since \(x_n \rightarrow 0\) in \(\ell_2(\overline{S})\), \[\label{eq:32sumsofsquares} \lim_{n \rightarrow \infty} \sum a_\gamma^2 =0 .\tag{5}\] Since the supports of the \(x_m\)’s are disjoint, we may assume that \(a_\gamma\ge0\) and \(b^m_\gamma\ge0\).

By assumption, \(\|x_m\|_{\alpha_r, r \vee k} \rightarrow 0\) as \(m \rightarrow \infty\) for all \(r \ge 1.\) Hence \[\label{eq:32uniform1} \lim_{m \rightarrow \infty} \sum_{\gamma \in F} b^m_\gamma = 0.\tag{6}\] uniformly over all \(F \in \mathcal{F}^{(k)}_\alpha\) satisfying \(|F \cap \operatorname{supp} x_n| \ge 2\).

Note that if \(F_1,F_2,\dots,F_s\) are disjoint sets in \(\mathcal{F}^{(k)}_\alpha\) satisfying \(|F_i \cap \operatorname{supp} x_n| \ge 2\) (\(1 \le i \le s\)) then \(s \le |\operatorname{supp} x_n|\). Hence 6 implies that \[\label{eq:32uniform2} \sum_{\gamma \in \cup_{i=1}^s F_i} b^m_\gamma \rightarrow 0\tag{7}\] as \(m \rightarrow \infty\) uniformly over all such collections \((F_i)_{i=1}^s\). Let \(A_i = \sum_{\gamma \in F_i} a_\gamma\) and let \(B^m_i = \sum_{\gamma \in F_i} b^m_\gamma\). Then \[\begin{align} \sum_{i=1}^s (A_i + B^m_i)^2 &= \sum_{i=1}^s (A_i^2 +( B^m_i)^2 + 2A_iB^m_i) \\ &\le \sum_{i=1}^s A_i^2 + (\sum_{i=1}^s B^m_i)^2 + 2(\sum_{i=1}^s B^m_i) (\sum A_i^2)^{1/2}\\ &\le \sum_{i=1}^s A_i^2 + (\sum_{i=1}^s B^m_i)^2 + 2\|x_n\|_{\alpha,k} \sum_{i=1}^s B^m_i\\ & \le \sum_{i=1}^s A_i^2 + (\sum_{i=1}^s B^m_i)^2 + 2 \sum_{i=1}^s B^m_i. \end{align}\] Note that 7 implies that \(\sum_{i=1}^s B^m_i \rightarrow 0\) as \(m \rightarrow \infty\) uniformly over all such \((F_i)_{i=1}^s\). Let \(\varepsilon>0\). It follows that for all \(m \ge M(n, \varepsilon)\), \[\label{eq:32atleast2} \sum_{i=1}^s (A_i + B^m_i)^2 < \sum_{i=1}^s A_i^2 + \varepsilon \le \|x_n\|_{\alpha_k} + \varepsilon \le 1 + \varepsilon\tag{8}\] uniformly over all \((F_i)_{i=1}^s\). Moreover, it follows from 5 that for all \(n \ge N(\varepsilon)\) \[\sum a_\gamma^2 < \varepsilon^2.\] Let \(J\subset \operatorname{supp} x_n\). Consider disjoint sets \(G_\lambda \in \mathcal{F}^{(k)}_\alpha\) (\(\lambda \in J\)) satisfying \(G_\lambda \cap \operatorname{supp} x_n = \{\lambda\}\) (\(\lambda \in J\)). Let \(C^m_\lambda = \sum_{\gamma \in G_\lambda} b^m_\gamma\). Then for all \(m> n> N(\varepsilon)\), \[\label{eq:32exactlyone} \begin{align} \sum_{\lambda \in J} (a_\lambda + C^m_\lambda)^2 &\le \sum_{\lambda \in J} a_\gamma^2 + \sum_{\lambda \in J}(C^m_\lambda)^2 + 2( \sum_{\lambda \in J} a_\lambda^2)^{1/2} (\sum_{\lambda\in J}(C^m_\lambda)^2)^{1/2}\\ &\le \varepsilon + \sum_{\lambda \in J}(C^m_\lambda)^2 + 2\varepsilon \|x_m\|_{\alpha,k}\\ &\le \varepsilon + 2\varepsilon + \|x_m\|_{\alpha,k}^2 \\ &\le 1 + 3\varepsilon. \end{align}\tag{9}\] Hence, combining 8 and 9 , for all \(n \ge N(\varepsilon)\) and \(m> M(n,\varepsilon)\), \[\|x_n + x_m\|_{\alpha,k}^2 \le 2 + 4\varepsilon.\] Since \(\varepsilon>0\) is arbitrary, we have \[\label{eq:32sqrt2} \limsup_{n \rightarrow \infty} \limsup_{m \rightarrow \infty} \|x_n + x_m\|_{\alpha,k} \le \sqrt{2},\tag{10}\] which contradicts 4 . ◻

Proof. Note that \[\|x\|_{\alpha,k} \le \limsup_{n \rightarrow \infty} \|x_n\|_{\alpha,k} \le 1,\] since \(x_n \rightarrow x\) pointwise. Suppose, to derive a contradiction, that the conclusion is false. Then, after passing to a subsequence and relabelling, we may assume that \[\lim_{n \rightarrow\infty}\|x_n - x\|_{\alpha,k} = \delta > 0.\] Let \(x_n^\prime= x_n - x\). By assumption, for all \(r \ge 1\), \[\lim_{n \rightarrow \infty}\|x_n^\prime\|_{\alpha_r, r\vee k} = 0.\] Let \(\varepsilon>0\). Choose a finitely supported vector \(y\) such that \[\|x - y\|_{\alpha,k} < \frac{\varepsilon^2}{10}.\] By a gliding hump argument, passing to a further subsequence and relabelling, we may choose disjointly supported vectors \(y_n\) (\(n \ge 1\)), each with finite support disjoint from the support of \(y\), such that \(\|x_n^\prime - y_n\|_{\alpha,k} \rightarrow 0\) as \(n \rightarrow \infty\) and, for all \(m,n \ge 1\), \[\|y + y_n||_{\alpha,k} \le 1,\] and also \[\|2y + y_n + y_m \|_{\alpha,k} > 2 -\frac{\varepsilon^2}{4}.\] Hence \[\lim_{n \rightarrow \infty}\| y_n\|_{\alpha,k} = \delta,\] and, for all \(r \ge 1\), \[\label{eq:3232nullsequence} \lim_{n \rightarrow \infty}\| y_n\|_{\alpha_r, r \vee k} =\lim_{n \rightarrow \infty}\|x_n^\prime\|_{\alpha_r, r\vee k} =0.\tag{13}\] Without loss of generality, we may assume that \(y\ge0\) and \(y_n \ge 0\) for all \(n \ge 1\). Fix \(n \ge 1\) and let \(m>n\). Suppose that \(2y + y_n + y_m\) is normed by disjoint sets \(F_1,\dots,F_u\) in \(\mathcal{F}_{\alpha,k}\) (we suppress the dependence of \(F_i\) on \(n\) and \(m\) to simplify notation), i.e., \[|(2y + y_n + y_m; F_1,\dots,F_u)|_2 = \|2y + y_n + y_m \|_{\alpha,k}> 2 -\frac{\varepsilon^2}{4}.\] Since \[|(y+y_n; F_1,\dots,F_u)|_2 \le \|y+ y_n\|_{\alpha,k}\le 1\] and \[|(y+y_m; F_1,\dots,F_u)|_2 \le \|y+ y_m\|_{\alpha,k}\le 1,\] the uniform convexity of \(\ell_2\) yields \[|(y_n-y_m; F_1,\dots,F_u)|_2 < \varepsilon.\]

We may assume that \(F_1,\dots,F_s\) have nonempty intersection with both \(\operatorname{supp} y\) and \(\operatorname{supp} y_n\), that \(F_{s+1},\dots,F_t\) intersect \(\operatorname{supp} y\) but not \(\operatorname{supp} y_n\), and that \(F_{t+1},\dots,F_u\) do not intersect \(\operatorname{supp} y\). Note that \(s \le |\operatorname{supp} y|\) and \(|F_i \cap \operatorname{supp} (y + y_n)| \ge 2\) for \(1 \le i \le s\). Hence, repeating the argument used to prove 7 , we deduce that \[\label{eq:32uniform3} \lim_{m\rightarrow\infty} \sum_{\gamma \in \cup_{i=1}^s F_i} b^m_\gamma = 0\tag{14}\] for \(y_m = \sum b^m_\gamma e_\gamma\). Hence \[|(y_m; F_1,\dots,F_s)|_2 < \frac{\varepsilon}{2}\] for all \(m > M_1(n,\varepsilon)\).

Note that \(y_n\) vanishes on \(F_i\) for \(s+1 \le i \le t\). Hence, for all \(m > M_1(n,\varepsilon)\), \[\begin{align} |(y_n+y_m; F_1,\dots,F_t)|_2 &=(\sum_{i=1}^s (\sum_{\gamma \in F_i} (b^n_\gamma + b^m_\gamma))^2 + \sum_{i = s+1}^t (\sum_{\gamma \in F_i} b^m_\gamma)^2)^{1/2} \\ &\le (\sum_{i=1}^s (\sum_{\gamma \in F_i} (b^n_\gamma -b^m_\gamma))^2 + \sum_{i = s+1}^t (\sum_{\gamma \in F_i} b^m_\gamma)^2)^{1/2} \\ & + 2 (\sum_{i=1}^s (\sum_{\gamma \in F_i} b^m_\gamma)^2)^{1/2}\\ \intertext{(by the triangle inequality in \ell_2)} &= |(y_n-y_m; F_1,\dots,F_t)|_2 + 2|(y_m; F_1,\dots F_s)|_2\\ &\le \varepsilon + \varepsilon = 2\varepsilon. \end{align}\] So \[\begin{align} |(2y + y_n + y_m; F_1,\dots,F_t)|_2 &\le 2|(y;F_1,\dots, F_t)|_2 + |(y_n+y_m;F_1,\dots,F_t)|_2\\ &\le 2\|y\|_{\alpha,k} + 2\varepsilon. \end{align}\] Thus, \[\label{eq:32firstest}\begin{align} (2 - \frac{\varepsilon^2}{4})^2 &< |(2y+y_n+y_m; F_1,\dots,F_u)|_2^2\\ &= |(2y+y_n+y_m; F_1,\dots,F_t)|_2^2 + |(y_n+y_m; F_{t+1},\dots,F_u)|_2^2\\ \intertext{(since y vanishes on F_i for t+1 \le i \le u)} &\le ( 2\|y\|_{\alpha,k} + 2\varepsilon)^2 + \|y_n + y_m\|_{\alpha,k}^2. \end{align}\tag{15}\] Since \(y\), \(y_n\), and \(y_m\) are disjointly supported, we have \[\label{eq:32secondest} \|y\|_{\alpha,k}^2 + \|y_n\|_{\alpha,k}^2 \le \|y+y_n\|_{\alpha,k}^2 \le 1\tag{16}\] and \[\label{eq:32thirdest} \|y\|_{\alpha,k}^2 + \|y_m\|_{\alpha,k}^2 \le \|y+y_m\|_{\alpha,k}^2 \le 1.\tag{17}\] Combining 15 , 16 , and 17 , \[\begin{align} 4\|y\|_{\alpha,k}^2 +2(\|y_n\|_{\alpha,k}^2 + \|y_m\|_{\alpha,k}^2) &\le 4\\ & = (2 - \frac{\varepsilon^2}{4})^2 + \varepsilon^2 -\frac{\varepsilon^4}{16}\\ &\le ( 2\|y\|_{\alpha,k} + 2\varepsilon)^2 + \|y_n + y_m\|_{\alpha,k}^2 + \varepsilon^2 \\ &\le 4\|y\|_{\alpha,k}^2 + \|y_n + y_m\|_{\alpha,k}^2 +(8\varepsilon + 5\varepsilon^2). \end{align}\] Hence for all \(m>M_1(n,\varepsilon)\), \[\label{eq:328epsilon} \|y_n + y_m\|_{\alpha,k}^2 +(8\varepsilon + 5\varepsilon^2) \ge 2(\|y_n\|_{\alpha,k}^2 + \|y_m\|_{\alpha,k}^2).\tag{18}\] Now suppose \(\varepsilon\) is chosen so that \(8\varepsilon + 5\varepsilon^2 < 2\delta^2\). Since \(\lim_{n \rightarrow \infty} \|y_n\|_{\alpha,k} = \delta\), it follows from 18 that \[\liminf_{n \rightarrow \infty} \liminf_{m \rightarrow\infty} \|y_n + y_m\|_{\alpha,k} > \sqrt2 \delta.\] which contradicts Remark 3 since, for all \(r \ge 1\), \[\lim_{n \rightarrow \infty} \|y_n\|_{\alpha_r, r \vee k} = 0.\] ◻

Proof of Theorem 2. We will prove the result for a fixed \(\alpha\) and for all \(k \ge 1\) by transfinite induction on \(\alpha\). The result clearly holds for \(\alpha = 0\) since \(B(\mathcal{F}_0^{(k)}) = B(\mathcal{F}_0)= \ell_2(\overline{S})\) for all \(k \ge 1\). So suppose the result holds for all \(\beta<\alpha\) and for all \(k \ge 1\).

Case I: \(\alpha\) is a limit ordinal. So \(\mathcal{F}^{(k)}_\alpha = \cup_{r=1}^\infty \mathcal{F}^{(r \vee k)}_{\alpha_r},\) where \(\alpha_r \uparrow \alpha\). By inductive hypothesis, each \(B(\mathcal{F}^{(r \vee k)}_{\alpha_r})\) admits an equivalent \(2R\) norm \(|\!|\!|\cdot |\!|\!|_{\alpha_r, r \vee k}\). Note that \[|\!|\!|\cdot |\!|\!|_{\alpha_r, r \vee k} \le C_r \|\cdot\|_{\alpha, k}\] for some \(C_r < \infty\). Thus, \[|\!|\!|\cdot |\!|\!|_{\alpha,k}^2 := \|\cdot\|_{\alpha,k}^2 + \sum_{r=1}^\infty \frac{1}{2^rC_r^2} |\!|\!|\cdot |\!|\!|_{\alpha_r, r \vee k}^2\] defines an equivalent norm \(|\!|\!|\cdot |\!|\!|_{\alpha,k}\) on \(B(\mathcal{F}_\alpha^{(k)})\). Let us show that \(|\!|\!|\cdot |\!|\!|_{\alpha,k}\) is a \(2R\) norm. Suppose that \((x_n) \subset B(\mathcal{F}_\alpha^{(k)})\) satisfies \[\lim_{m,n \rightarrow \infty} |\!|\!|x_n + x_m |\!|\!|_{\alpha,k}^2 - 2( |\!|\!|x_n |\!|\!|_{\alpha,k}^2 + |\!|\!|x_m| |\!|\!|_{\alpha,k}^2) = 0.\] Note that \[\begin{align} & |\!|\!|x_n + x_m |\!|\!|_{\alpha,k}^2 - 2( |\!|\!|x_n |\!|\!|_{\alpha,k}^2 + |\!|\!|x_m |\!|\!|_{\alpha,k}^2)\\ &\le -(\|x_n\|_{\alpha,k} -\| x_m\|_{\alpha,k})^2 - \sum_{r=1}^\infty \frac{1}{2^rC_r^2} ( |\!|\!|x_n |\!|\!|_{\alpha_r,k} - |\!|\!| x_m |\!|\!|_{\alpha+r,k})^2. \end{align}\] It follows that \(\lim_{n \rightarrow \infty} \|x_n\|_{\alpha,k} = L\) for some \(L \ge 0\), that \[\label{eq:32limitalphak} \lim_{m,n \rightarrow \infty} \|x_n + x_m\|_{\alpha,k}^2 - 2(\|x_n\|_{\alpha,k}^2 + \|x_m\|_{\alpha,k}^2) = 0,\tag{20}\] and that, for all \(r\ge1\), \[\lim_{m,n \rightarrow \infty} |\!|\!|x_n + x_m |\!|\!|_{\alpha_r,k \vee r}^2 - 2( |\!|\!|x_n |\!|\!|_{\alpha_r,k \vee r}^2 + |\!|\!|x_m |\!|\!|_{\alpha_r,k \vee r}^2) = 0.\] Since each \(|\!|\!|\cdot |\!|\!|_{\alpha_r, r \vee k}\) is a \(2R\) norm, it follows from 3 that there exists \(x \in \ell_2(\overline{S})\) such that, for all \(r\ge 1\), \[\lim_{n \rightarrow \infty} |\!|\!|x_n - x |\!|\!|_{\alpha_r, r \vee k} = 0.\] Moreover, 20 implies that \[\lim_{m,n \rightarrow \infty}\| x_n + x_m \|_{\alpha,k} = 2L.\] So, by Lemma 3, \[\lim_{n \rightarrow \infty} \| x_n - x \|_{\alpha,k} = 0,\] and hence \[\lim_{n \rightarrow \infty} |\!|\!| x_n - x |\!|\!|_{\alpha,k} = 0,\] as desired.

Case II: \(\alpha= \beta^+\) is a successor ordinal. The proof is very similar to the limit ordinal case. By the inductive hypothesis, \(B(\mathcal{F}_\beta^{(k)})\) admits an equivalent \(2R\) norm \(|\!|\!| \cdot |\!|\!|_{\beta,k}\). Let \[|\!|\!| \cdot |\!|\!|^2_{\alpha, k}= \|\cdot\|_{\alpha,k}^2 + |\!|\!| \cdot |\!|\!|^2_{\beta,k}.\] Using Lemma 4 instead of Lemma 3 and repeating the argument of Case I shows that \(|\!|\!| \cdot |\!|\!|_{\alpha, k}\) is a \(2R\) norm. ◻

Proof. We may assume that \(y \ge0\) and \(y_n \ge.\) Choose positive norming vectors \(x, x_n \in B(\mathcal{F})\) with \[(x,y) = \|x\| = \|y\|_* =1, \quad (x_n,y_n)=\|x_n\|= \|y_n\|_*=1,\] where \((\cdot,\cdot)\) denotes the duality pairing for \(B(\mathcal{F})\times B(\mathcal{F})^*\). Note that \(x\) and \(x_n\) have disjoint finite supports (\(n \ge 1\)). Fixing \(n \ge 1\), choose disjoint \(F_i \in \mathcal{F}\) (\(1 \le i \le N\)) such that \[\|x + \delta x_n\| = |(x+ \delta x_n; F_1,\dots,F_N)|_2.\] We may assume that only \(F_1,\dots,F_k\) have non-empty intersection with both \(\operatorname{supp} x\) and \(\operatorname{supp} x_n\). Note that \[k \le M:= |\operatorname{supp} x|.\] For each \(1 \le i \le k\), \[\|y_n \cdot 1_{F_i}\|_* \le \|y_n\|_\infty.\] Hence \[\sum_{i=1}^k \|y_n \cdot 1_{F_i}\|_* \le M \|y_n\|_\infty \rightarrow 0 \quad\text{as n \rightarrow \infty}.\] Let \(F = \cup_{i=1}^k F_i\). (To simplify notation we suppress the dependence of \(F\) on \(n\).) Then \[\begin{align} \|x_n - x_n \cdot 1_F\| &\ge (x_n -x_n\cdot 1_F, y_n)\\ &= (x_n,y_n -y_n\cdot 1_F)\\ &= 1 - (x_n, y_n\cdot 1_F)\\ &\ge 1 - \sum_{i=1}^k \|y_n \cdot 1_{F_i}\|_*\\ &\rightarrow 1 \quad \text{as n \rightarrow \infty.} \end{align}\]Hence \[\lim_{n \rightarrow \infty} (\|x_n\|^2 - \|x_n - x_n\cdot 1_F\|^2)= 1 - \lim_{n \rightarrow \infty} \|x_n - x_n\cdot 1_F\|^2=0.\] Since \((B(\mathcal{F},\|\cdot\|)\) satisfies a lower \(2\)-estimate, it follows that \(\|x_n \cdot 1_F\| \rightarrow 0\) as \(n \rightarrow \infty\). Hence \[\begin{align} 1 + \delta^2 &\le \liminf_{n \rightarrow \infty}\|x + \delta x_n\|^2\\ & \le \limsup_{n \rightarrow \infty}\|x + \delta x_n\|^2\\ & = \limsup_{n \rightarrow \infty} |(x+\delta x_n; F_1,\dots F_N)|_2^2\\ &= \limsup_{n \rightarrow \infty} |(x+ \delta x_n - \delta x_n \cdot 1_F; F_1,\dots,F_N)|_2^2\\ & \le \limsup_{n \rightarrow \infty}[ \|x\|^2+ \delta^2 \| x_n - x_n \cdot 1_F\|^2]\\ \intertext{(since no F_i intersects both \operatorname{supp} x and \operatorname{supp} (x_n - x_n \cdot 1_F))} &= 1 + \delta^2. \end{align}\] Thus, \[\lim_{n \rightarrow \infty}\|x+ \delta x_n\| ^2= 1 + \delta^2,\] and hence \[\begin{align} (1 + \delta^2)^{1/2} &\ge \limsup_{n \rightarrow \infty} \|y + \delta y_n\|_*\\ \intertext{(since \|\cdot\|_* satisfies an upper 2-estimate)} &\ge \liminf_{n \rightarrow \infty} \|y + \delta y_n\|_*\\ & \ge \liminf_{n \rightarrow \infty} \frac{(x + \delta x_n, y + \delta y_n)}{\|x + \delta x_n\|}\\ &= \frac{1 + \delta^2}{(1 + \delta^2)^{1/2}}\\ &=(1 + \delta^2)^{1/2}. \end{align}\] ◻

Proof. Suppose \((y_n) \subset B(\mathcal{F})^*\) satisfies 21 . By Lemma 5 there exists \(y \in B(\mathcal{F})^*\) such that \(y = w-\lim_{n \rightarrow \infty} y_n\) and \(\lim_{n \rightarrow \infty}\|y_n - y\|_\infty =0\). Suppose, to derive a contradiction, that \((y_n)\) does not converge strongly to \(y\). Passing to a subsequence and relabelling, we may assume that \(y_n = y+z_n\) where \[\lim_{n \rightarrow\infty} \|z_n\|_* = \delta>0, \lim_{n \rightarrow\infty} \|z_n\|_\infty = 0,\] and that the following limits exist: \[\lim_{n \rightarrow\infty} \|y+z_n\|_*, \lim_{n,m \rightarrow\infty} \|2y+z_n+z_m\|_* .\] Let \(\varepsilon>0\). By passing to a further subsequence, a gliding hump argument and the fact that \(\lim_{n \rightarrow\infty} \|z_n\|_\infty = 0\) show that there exist vectors \(y^\prime\) and \(z_n^\prime\) (\(n \ge 1\)) with disjoint finite supports such that \[\label{eq:32approxi} \|y - y^\prime\|_* < \varepsilon, \lim_{n \rightarrow \infty}\| z_n - z_n^\prime\|_* =0.\tag{22}\] Note that 21 implies that \[\label{eq:32limitnorm42} \lim_{m,n \rightarrow \infty} \|2y+ z_n + z_m \|_*^2- 2(\|y+z_n|\|_*^2 + \|y+z_m\|_*^2) = 0.\tag{23}\] Since \(\lim_{n \rightarrow \infty} \|z^\prime_n\|_\infty = 0\), Lemma 6 yields \[\lim_{n \rightarrow \infty} \|y^\prime + z_n^\prime\|_*^2 = \|y^\prime\|_*^2 + \lim_{n \rightarrow \infty}\|z_n^\prime\|_*^2 = \|y^\prime\|_*^2 + \delta^2.\] Since \((B(\mathcal{F},\|\cdot\|_*)\) satisfies an upper \(2\)-estimate, \[\lim_{n,m \rightarrow \infty} \|2y^\prime + z_n^\prime + z_m^\prime \|_*^2 \le 4\|y^\prime\|_*^2 + 2\delta^2.\] Hence \[\begin{align} &\limsup_{n,m \rightarrow \infty}[\|2y^\prime + z_n^\prime + z_m^\prime \|_*^2 - 2(\|y^\prime + z_n^\prime\|_*^2 + \|y^\prime + z_m^\prime\|_*^2)]\\ &\le 4\|y^\prime\|_*^2 + 2\delta^2 - 2(2\|y^\prime\|_*^2 + 2\delta^2)\\ &= -2\delta^2, \end{align}\] which contradicts 22 and 23 provided \(\varepsilon\) is sufficiently small. ◻

Proof. The space \(Y\) is defined as \(X \oplus X^*\), where \(X\) is defined below.

Let \(S = \mathbb{N}\). Let \(\mathcal{F}_1\) be the collection of all finite subsets \(F\) of \(\overline{S}\) such that, if \(|F| \ge 2\), then for all \(p \in F\), \(p(1)=1\) and \(p(i) \in \{1,2,\dots,i-1\}\) for all \(i \ge 2\) and such that for all distinct \(p,q \in F\), there exists \(m \ge 3\) such that \(p(i) = q(i)\) for all \(1 \le i \le m-1\) and \(p(m) \ne q(m)\), which implies that \(F^\sharp = m-1\) and \(|F| \le m-1\) as required. The space \(X\) is defined to be the closed linear span of \[\{e_p \colon p(1)=1, p(i) \in \{1,2,\dots,i-1\}, i \ge 2\}.\] in \(B(\mathcal{F}_1)\). The successor case of the proof of Theorem 2 shows that \[|\!|\!| \cdot |\!|\!|^2 = \|\cdot\|_1^2 + \|\cdot\|^2_{\ell_2(\overline{S})}\] is an equivalent \(2R\) norm on \(B(\mathcal{F}_1)\). Hence \(|\!|\!| \cdot |\!|\!|\) restricts to an equivalent \(2R\) norm on \(X\). By Theorem 4, \(B(\mathcal{F}_1)^*\) admits an equivalent \(2R\) norm. Note that \(X^*\) is isomorphic to a quotient space of \(B(\mathcal{F}_1)^*\). It is easily seen that a quotient norm of a \(2R\) norm is \(2R\). Hence \(X^*\) admits an equivalent \(2R\) norm, \(|\!|\!| \cdot |\!|\!|^\prime\) say. Finally, \[\|(x,x^*)\| =\sqrt{|\!|\!|x|\!|\!|^2 + |\!|\!| x^* |\!|\!|^{\prime2}}\qquad ( (x,x^*) \in X \oplus X^*)\] is an equivalent \(2R\) norm on \(X \oplus X^* =Y\). ◻

Proof. First, we consider the case \(\Gamma = \mathbb{N}\). Let \((e_n)\) denote the unit vector basis of \(B(\mathcal{F})\).

Let \((F_i)_{i=1}^\infty \subset \mathcal{F}\) be a collection of disjoint elements of \(\mathcal{F}\) and suppose that \(\sum_{i=1}^\infty |a_i|^2 \le 1\). For \(x = \sum_{i=1}^\infty x_i e_i \in B(\mathcal{F})\), \[\label{eq:32functionalrep} | \sum_{i=1}^\infty a_i(\sum_{j \in F_i} x_j)| \le (\sum_{i=1}^\infty |a_i|^2)^{1/2} \|x\| \le \|x\|.\tag{24}\] Hence we may identify \(\sum_{i=1}^\infty a_i 1_{F_i} \in \ell_\infty\) with the element \(k\) in the unit ball of \(B(\mathcal{F})^*\) defined by 24 .

Suppose \(x \in B(\mathcal{F})\) has finite support and that \[\|x\| = (\sum_{i=1}^n (\sum_{j \in G_i} |x_j|) ^2)^{1/2}\] for disjoint \(G_i \in \mathcal{F}\). There exist nonnegative \(a_1,\dots,a_n\) with \(\sum_{i=1}^n a_i^2 = 1\) such that \[\sum_{i=1}^n a_i (\sum_{j \in G_i} |x_j|) = \|x\|,\] and there exist \(H_i\subseteq G_i\) (\(1 \le i \le n\)) such that \[\sum_{i=1}^n a_i|\sum_{j \in H_i} x_j| \ge \frac{1}{2} \sum_{i=1}^n a_i (\sum_{j \in G_i} |x_j|) = \frac{1}{2} \|x\|.\] Note that \(H_i \in\mathcal{F}\) since \(\mathcal{F}\) is hereditary. Hence the collection of linear functionals with a representation of the form \[k=\sum_{i=1}^\infty a_i 1_{F_i} \in \ell_\infty\qquad (\sum_{i=1}^\infty | a_i|^2 =1, (F_i)_{i=1}^\infty\, \text{disjoint sets in\,} \mathcal{F})\] is a \(2\)-norming set for \(B(\mathcal{F})\). It follows that the discretized collection \[K=\{\sum_{r=1}^\infty \pm 2^{-s(r)} 1_{F_r} \colon \sum_{r=1}^\infty 2^{-2s(r)} \le 1 \}.\] is a \(4\)-norming set.

Let us show that \(K \subset \ell_\infty\) is compact in the topology of pointwise convergence on \(\ell_\infty\). For \(n\ge1\), let \[k_n = \sum_{r=1}^\infty 2^{-r}( 1_{U^n_r} - 1_{V^n_r}),\] where \(U^n_r= \cup_{i=1}^{p(n,r)} F^n_i\) and \(V^n_r=\cup_{i=1}^{q(n,r)} G^n_i\), and for each \(n \ge 1\), \[\{F^n_{r,i}, G^{n}_{r,j} \colon r\ge1, 1 \le i \le p(n,r), 1 \le j \le q(n,r)\}\] is a collection of nonempty disjoint elements of \(\mathcal{F}\), and \[\sum_{r=1}^\infty 2^{-2r}( p(n,r) + q(n,r)) \le 1.\] In particular, \(p(n,r) + q(n,r) \le 2^{2r}\) for all \(n,r\ge 1\). By a diagonal argument, passing to a subsequence and relabelling, we may assume that \[p(n,r) = p_r, q(n,r) = q_r \quad\text{for all n \ge r}.\] By compactness of \(\mathcal{F}\), we may also assume that \[\lim_{n \rightarrow \infty} F^n_{r,i} = F_{r,i}\quad (1 \le i \le p_r), \lim_{n \rightarrow \infty} G^n_{r,j} = G_{r,j} \quad (1 \le j \le q_r),\] where \[\{F_{r,i}, G_{r,j} \colon r\ge1, 1 \le i \le p_r, 1 \le j \le q_r\}\] is a collection of disjoint (possibly empty) elements of \(\mathcal{F}\) and \[\sum_{r=1}^\infty 2^{-2r}( p_r + q_r) \le 1.\] Set \(F_r = \cup_{i=1}^{p_r} F_{r,i}\) and \(G_r = \cup_{i=1}^{q_r} G_{r,i}\). It follows that \[k = \sum_{r=1}^\infty 2^{-r}(1_{F_r} - 1_{G_r}) \in K\] and \(k = \lim_{n \rightarrow \infty} k_n\) pointwise in \(\ell_\infty\). So \(K\) is compact (and metrizable) in the topology of pointwise convergence.

For \(x \in B(\mathcal{F})\), define \(\hat{x}\colon K \rightarrow \mathbb{R}\) by \(\hat{x}(k) = k(x).\) Suppose that \((k_n) \subset K\) and \(k_n \rightarrow k\) pointwise in \(\ell_\infty\). Clearly, \(\hat{x}(k_n) \rightarrow \hat{x}(k)\) when \(x\) has finite support. Since the finitely supported vectors are norm-dense in \(B(\mathcal{F})\), it follows that \(\hat{x}(k_n) \rightarrow \hat{x}(k)\) for all \(x \in B(\mathcal{F})\), i.e., that \(\hat{x}\) is continuous on \(K\). Since \(K\) is \(4\)-norming for \(B(\mathcal{F})\), the mapping \(x \mapsto \hat{x}\) defines a linear isomorphism from \(B(\mathcal{F})\) onto a closed subspace of \(C(K)\).

Suppose that \((x_n) \subset B(\mathcal{F})\) is bounded and coordinatewise null with respect to \((e_n)\). It follows from 24 that \[\lim_{n\rightarrow \infty} \hat{x}_n(k) = 0 \qquad (k \in K).\] Hence, by the Riesz representation and bounded convergence theorems, \(\hat{x}_n \rightarrow 0\) weakly in \(C(K)\). In particular, if \(x_n =\sum_{i = p_{n-1}+1}^{p_n}a_i e_i\), where \(p_{n-1} < p_n\), is a bounded block basis of \((e_n)\), then \((\hat{x}_n)\) is weakly null in \(C(K)\). Hence \((x_n)\) is weakly null in \(B(\mathcal{F})\), which implies that \((e_n)\) is a shrinking basis. On the other hand, since \((e_n)\) satisfies a lower \(2\)-estimate, it is boundedly complete. It follows from a theorem of James [14] that \(B(\mathcal{F})\) is reflexive.

Next suppose that \(\Gamma\) is uncountable. Let \(\Gamma_0\) be a countably infinite subset of \(\Gamma\). Then \[X_0 = \{ x \in B(\mathcal{F}) \colon \operatorname{supp} x \subseteq \Gamma_0\}\] is the Baernstein space on \(\Gamma_0\) corresponding to the family \(\mathcal{F}_0 = \{F \cap \Gamma_0 \colon F \in \mathcal{F} \}\). By the first part of the proof, \(X_0\) is reflexive. But every separable subspace of \(B(\mathcal{F})\) is contained in \(X_0\) for some \(\Gamma_0\). Hence every separable subspace of \(B(\mathcal{F})\) is reflexive, which implies that \(B(\mathcal{F})\) is also reflexive since reflexivity is separably determined. ◻