Proof. (i) The first equality follows from the fact that the set on the right-hand side is the smallest solid subspace of  $H[E]$ which contains δ_x for every $x\in E$. The second equality is immediate from the inequalities

\begin{equation*} {\bigvee}_{j=1}^n | {\delta}_{x_j}|\leqslant \sum_{j=1}^n | {\delta}_{x_j}| \leqslant {\bigvee}_{j=1}^n | {\delta}_{nx_j}|\qquad (n\in\mathbb{N},\, x_1,\ldots,x_n\in E).\end{equation*}

(ii) This is a consequence of the fact that the lattice operations in $H[E]$ are defined pointwise.

(iii) This is clear because $\| x^*\|_{p,{{\text{weak}}}} = \| x^*\|$ for each $x^*\in E^*$.

(iv) This follows from the general result that the uniform limit of a sequence of continuous functions defined on a compact Hausdorff space is continuous.

(v) Clause (ii) implies that $H_{w^*}^p[E]$ is a sublattice of  $H^p[E]$. To see that it is closed, suppose that $(f_n)$ is a sequence in $H_{w^*}^p[E]$ which converges to $f\in H^p[E]$ in the norm $\|\,\cdot\,\|_{\text{FBL}^p\lbrack E\rbrack}$. Then $(f_n)$ converges uniformly on  $B_{E^*}$ to f by (iii), so (iv) implies that $f\in H_{w^*}[E]$.

(vi) This is proved for p = 1 in [Reference Avilés, Rodríguez and Tradacete6, Lemma 4.10], and the argument given therein carries over verbatim to p > 1. Indeed, (v) implies that $S:=\textrm{FBL}^p[E]\cap H_{w^*}[E]$ is a closed sublattice of  $(\textrm{FBL}^p[E],\|\,\cdot\,\|_{\textrm{FBL}^p[E]})$. Clearly ${\delta}_x\in S$ for every $x\in E$, so $S=\textrm{FBL}^p[E]$; that is, $\textrm{FBL}^p[E]\subseteq H_{w^*}[E]$.

(vii) We begin by showing that $I[E]\subseteq H^p[E]$. Take $f\in I[E]$. By (i), we can find $n\in\mathbb{N}$ and $x_1,\ldots,x_n\in E$ such that $| f| \leqslant\sum_{k=1}^n|{\delta}_{x_k}|$. Suppose that $m\in\mathbb{N}$ and $x_1^*,\ldots,x^*_m\in E^*$ with $\| (x_j^*)_{j=1}^m\|_{p,{{\text{weak}}}} \leqslant 1$. Then $\sum_{j=1}^m | x_j^*(x_k)|^p \leqslant \| x_k\|^p$ for each $k\in\{1,\ldots,n\}$, and consequently

\begin{align*} \bigg(\sum_{j=1}^m | f(x_j^*)|^p\bigg)^{\frac1p} \leqslant \bigg(\sum_{j=1}^m\Big(\sum_{k=1}^n | x_j^*(x_k)|\Big)^p\bigg)^{\frac1p} \leqslant \sum_{k=1}^n\bigg(\sum_{j=1}^m | x_j^*(x_k)|^p\bigg)^{\frac1p} \leqslant \sum_{k=1}^n \| x_k\|, \end{align*}

where the second inequality is simply the statement that the norm of the sum of n vectors in  ${\ell}_p^m$ is dominated by the sum of their norms. We conclude that $\| f\|_{\textrm{FBL}^p[E]} \leqslant \sum_{k=1}^n\| x_k\| \lt \infty$, and therefore $f\in H^p[E]$, as desired.

This implies that  $I[E]$ is the ideal of  $H^p[E]$ generated by δ_x for $x\in E$. These generators belong to $H_{w^*}^p[E]$ (as we already observed in (vi) above), and hence $I[E]\subseteq J_{w^*}^p[E]$.

(viii) We have just seen that $I[E]\subseteq H^p[E]$, so $I_{w^*}[E]\subseteq H_{w^*}^p[E]$. Now the conclusion follows from the fact that $H_{w^*}^p[E]$ is closed by (v).

(ix) Suppose that f belongs to the closure of the ideal of  $H^p[E]$ generated by $M_{w^*,0}^p[E]$, and take a weak*-null net $(x^*_\alpha)$ in $B_{E^*}$. The set $M_{w^*,0}^p[E]$ is a sublattice of $H[E]$ because the vector lattice operations are defined pointwise. Therefore, for each ɛ > 0, we can find $g\in H^p[E]$ and $h\in M_{w^*,0}^p[E]$ such that $\| f-g\|_{\textrm{FBL}^p[E]} \leqslant\varepsilon$ and $| g| \leqslant | h|$. Since $h(0)=0$ by positive homogeneity, we can choose α ₀ such that $| h(x_\alpha^*)| \leqslant \varepsilon$ for every $\alpha\geqslant \alpha_0$. Then we have

\begin{align*} | f(x^*_\alpha)-f(0)| = | f(x^*_\alpha)|& \leqslant | f(x_\alpha^*)-g(x^*_\alpha)| + | g(x_\alpha^*)|\\ & \leqslant \| f-g\|_{\textrm{FBL}^p[E]} + | h(x_\alpha^*)| \leqslant 2\varepsilon\qquad (\alpha\geqslant \alpha_0), \end{align*}

which shows that $f\in M_{w^*,0}^p[E]$, and therefore $M_{w^*,0}^p[E]$ is a closed ideal of  $H^p[E]$.

This implies that $\overline{J_{w^*}^p[E]}\subseteq M_{w^*,0}^p[E]$ because $H_{w^*}^p[E]\subseteq M_{w^*,0}^p[E]$.

Proof. (a) $\Rightarrow$(b). We have $\textrm{lat}\delta[E] = H[E]$ for $E=\{0\}$ because $H[E] =\{0\}$ in this case.

Now suppose that $E=\mathbb{R}$, in which case $E^*=\mathbb{R}$, and the duality is given by ${\delta}_x(y)=\langle x,y\rangle = xy$ for $x,y\in\mathbb{R}$. Take $f\in H[\mathbb{R}]$, and suppose that $f(1)\geqslant -f(-1)$. Then, for each $y\in\mathbb{R}$, we have

\begin{align*} ({\delta}_{f(1)}\vee {\delta}_{-f(-1)})(y) &= {\delta}_{f(1)}(y)\vee {\delta}_{-f(-1)}(y) = f(1)y\vee (-f(-1)y)\\ &= \left\{\begin{array}{ll} f(1)y &\text{for}\ y\geqslant 0\\ -f(-1)y &\text{for}\ y \lt 0\end{array}\right\} = f(y) \end{align*}

by positive homogeneity. Hence $f={\delta}_{f(1)}\vee {\delta}_{-f(-1)}$. A similar argument shows that $f={\delta}_{f(1)}\wedge {\delta}_{-f(-1)}$ when $f(1) \leqslant -f(-1)$, and so $f\in\textrm{lat}\delta[\mathbb{R}]$ in both cases.

We have (b) $\Rightarrow$(c) because the inclusions $H^p[E]\subseteq H[E]$ and $\textrm{lat}\delta[E]\subseteq H_{w^*}[E]$ are always true.

(c) $\Rightarrow$(d). Suppose that $H^p[E]\subseteq H_{w^*}[E]$. Then we have $H_{w^*}^p[E] = H^p[E]$, from which (d) follows.

(d) $\Rightarrow$(e). Suppose that $H_{w^*}^p[E] = J_{w^*}^p[E]$. Then lemma 2.1(vii) implies that $I[E]\subseteq H_{w^*}^p[E]\subseteq H_{w^*}[E]$.

Finally, we prove that (e) $\Rightarrow$(a) by contraposition. Suppose that $\dim E\geqslant 2$, and choose two linearly independent functionals  $x^*_0,y^*_0\in E^*$, each having norm  $\frac12$. Then $(x_0^* + \frac1n y_0^*)_{n\in\mathbb{N}}$ is a sequence in  $B_{E^*}$ which converges to  $x_0^*$ in norm and therefore also in the weak* topology. Define $f{\colon} E^*\to\mathbb{R}$ by

\begin{equation*} f(x^*) = \begin{cases} t\ &\text{if}\ x^* = tx_0^*\ \text{for some}\ t \gt 0,\\ 0\ &\text{otherwise.} \end{cases} \end{equation*}

It is easy to see that f is positively homogeneous. Choose $x_0\in E$ such that $x_0^*(x_0)=1$. Then $| f| \leqslant |{\delta}_{x_0}|$, so $f\in I[E]$. On the other hand, $f(x_0^* + \frac1n y_0^*)=0$ for each $n\in\mathbb{N}$ by definition, but $f(x_0^*) =1$, so $f{\upharpoonright_{B_{E^*}}}$ is discontinuous at  $x_0^*$, and therefore $f\notin H_{w^*}[E]$.

Proof. Lemma 2.1(vi) implies that $R(\textrm{FBL}^p[E])\subseteq R(H_{w^*}[E])$, while the inclusion $R(H_{w^*}[E])\subseteq C(S_{E^*})$ follows from the fact that the weak* and norm topologies on  $E^*$ coincide. Therefore, to prove the left-hand identity, it only remains to verify that $C(S_{E^*})\subseteq R(\textrm{FBL}^p[E])$, which can be done by arguing as in the proof of [Reference de Pagter and Wickstead14, Proposition 5.3]; see also [Reference Avilés, Rodríguez and Tradacete6, Corollary 2.9(iii)] for the connection between de Pagter and Wickstead’s notion of the free Banach lattice over a set studied in [Reference de Pagter and Wickstead14] and the free Banach lattice over a Banach space.

To prove the right-hand identity, we begin by observing that $R(J_{w^*}^p[E])\subseteq R(H^p[E])$ because $J_{w^*}^p[E]\subseteq H^p[E]$ by definition, and $R(H^p[E])\subseteq{\ell}_\infty(S_{E^*})$ because $| f(x^*)| \leqslant \| f\|_{\textrm{FBL}^p[E]}$ for every $x^*\in B_{E^*}$ and $f\in H^p[E]$. Finally, to verify that ${\ell}_\infty(S_{E^*})\subseteq R(J_{w^*}^p[E])$, take $f_0\in {\ell}_\infty(S_{E^*})$. In view of the remarks we made before the statement of the result, we can find $f\in H[E]$ such that $Rf = f_0$. Let $\nu\in H[E]$ be the “norm function” defined by $\nu(x^\ast)=\|x^\ast\|$ for every $x^*\in E^*$. Then $R\nu$ is the constant function 1, which obviously belongs to  $C(S_{E^*})$, so $\nu\in\textrm{FBL}^p[E]\subseteq H_{w^*}^p[E]$ by the first part of the proof and lemma 2.1(vi). Since f is positively homogeneous, we have $\vert f\vert\leqslant\| f\upharpoonright_{S_{E^\ast}}\|_\infty\,\nu$, which shows that $f\in J_{w^*}^p[E]$, and the conclusion follows.

Proof. Set $f = {\bigvee}_{n\in\mathbb{N}} |{\delta}_{x_n}|$, which is clearly a positively homogeneous function, and consider the functions $f_m = {\bigvee}_{n=1}^m |{\delta}_{x_n}|\in H_{w^*}[E]$ for $m\in\mathbb{N}$. They satisfy

\begin{align*} \| (f-f_m){\upharpoonright_{B_{E^*}}}\|_\infty &= \text{sup}_{x^*\in B_{E^*}} \bigg|{\bigvee}_{n=1}^\infty| x^*(x_n)| - {\bigvee}_{n=1}^m| x^*(x_n)|\bigg|\\ & \leqslant \text{sup}_{x^*\in B_{E^*}} {\bigvee}_{n=m+1}^\infty | x^*(x_n)| = {\bigvee}_{n=m+1}^\infty \|x_n\|\underset{m\to\infty}\longrightarrow 0, \end{align*}

so $f\in H_{w^*}[E]$ by lemma 2.1(iv).

Proof. The proof has three parts: first we show that conditions (a)–(c) are equivalent, then we show that conditions (a) and (d)–(g) are equivalent, and finally we show that conditions (h)–(j) are equivalent to the other conditions.

(a) $\Rightarrow$(b). This follows from the first identity in proposition 3.2.

(b) $\Rightarrow$(c). This is obvious because $\textrm{FBL}^p[E]\subseteq H^p[E]$ by definition.

(c) $\Rightarrow$(a). To prove the contrapositive, suppose that E is infinite-dimensional. Then, by the weak Dvoretzky–Rogers theorem [Reference Diestel, Jarchow and Tonge15, Theorem 2.18], $E^*$ contains a sequence $(x_n^*)$ which is weakly p-summable, but not strongly p-summable. By scaling this sequence by a suitable constant, we may suppose that

\begin{equation*} \text{sup}_{x\in B_E}\sum_{n=1}^\infty|x_n^*(x)|^p \leqslant1\qquad\text{and}\qquad \sum_{n=1}^\infty\|x_n^*\|^p=\infty. \end{equation*}

For each $n\in\mathbb N$, choose $x_n\in B_E$ such that $x_n^*(x_n)\geqslant \frac{\|x_n^*\|}{2}$, and choose a decreasing sequence $(s_n)$ in $(0,1)$ such that

\begin{equation*} \lim_{n\to\infty} s_n=0\qquad\text{and}\qquad \sum_{n=1}^\infty s_n^p\|x_n^*\|^p=\infty. \end{equation*}

Lemma 3.4 shows that the function $f = {\bigvee}_{n\in\mathbb N} |{\delta}_{s_nx_n}|$ belongs to $H_{w^*}[E]$. However, $f\notin H^p[E]$ because the fact that $\|(x_k^*)_{k=1}^m\|_{p,{{\text{weak}}}} \leqslant 1$ for every $m\in\mathbb{N}$ implies that

\begin{align*} \|f\|^p_{\textrm{FBL}^p[E]} &\geqslant \sum_{k=1}^m |f(x_k^*)|^p = \sum_{k=1}^m \bigg({\bigvee}_{n\in\mathbb N} s_n |x_k^*(x_n)|\bigg)^p\\ &\geqslant \sum_{k=1}^m s_k^p|x_k^*(x_k)|^p \geqslant \frac1{2^p}\sum_{k=1}^m s_k^p\|x_k^*\|^p\underset{m\rightarrow\infty}\longrightarrow\infty. \end{align*}

Consequently $H_{w^*}[E]\nsubseteq H^p[E]$.

(a) $\Rightarrow$(d). Suppose that E has dimension  $n\in\mathbb{N}$, and let $(b_j)_{j=1}^n$ be a basis for E with coordinate functionals $(b_j^*)_{j=1}^n$, which form a basis for  $E^*$. We may suppose that $\| b_j^*\| = 1$ for each $j\in\{1,\ldots,n\}$. Given $f\in H^p[E]$, the number $t = \| f{\upharpoonright_{B_{E^*}}}\|_\infty$ is finite by lemma 2.1(iii), and for each non-zero $x^*\in E^*$, we have

\begin{align*} | f(x^*)| &= \| x^*\|\,\bigg| f\Big(\frac{x^*}{\| x^*\|}\Big)\bigg| \leqslant \| x^*\|\cdot t = \bigg\|\sum_{j=1}^n x^*(tb_j)b_j^*\bigg\| \leqslant \sum_{j=1}^n| x^*(tb_j)|. \end{align*}

This inequality is trivially true for $x^*=0$, so $| f| \leqslant \sum_{j=1}^n| {\delta}_{tb_j}|$, and therefore $f\in I[E]$ by lemma 2.1(i). This shows that $H^p[E]\subseteq I[E]$, while the opposite inclusion follows from lemma 2.1(vii).

It is clear that (d) $\Rightarrow$(f) $\Rightarrow$(g). We also have (d) $\Rightarrow$(e) $\Rightarrow$(g) because lemma 2.1(vii) and (vi) show that the following inclusions always hold:

\begin{equation*} I[E]\subseteq J_{w^*}^p[E]\subseteq H^p[E]\quad \text{and}\quad \textrm{FBL}^p[E]\subseteq H_{w^*}^p[E]\subseteq J_{w^*}^p[E]. \end{equation*}

We prove that (g) $\Rightarrow$(a) by contraposition. Suppose that E is infinite-dimensional, and take a normalized basic sequence $(b_n)_{n\in\mathbb{N}}$ in E. Let  $b_n^*\in E^*$ be a Hahn–Banach extension of the n ^th coordinate functional of this basic sequence for each $n\in\mathbb{N}$, and define $f = \sum_{n=1}^\infty|{\delta}_{b_n}|/2^n$, which belongs to  $\textrm{FBL}^p[E]$ because the series converges absolutely in the norm  $\|\,\cdot\,\|_{\textrm{FBL}^p[E]}$.

We claim that $f\notin I[E]$. By lemma 2.1(i), this amounts to showing that for every $m\in\mathbb{N}$ and $x_1,\ldots,x_m\in E$, we can find $x^*\in E^*$ such that $f(x^*) \gt \sum_{j=1}^m| x^*(x_j)|$. Since $m+1 \gt m$, the system of m homogeneous linear equations

\begin{equation*} \sum_{k=1}^{m+1}b_k^*(x_j)t_k = 0\qquad (j=1,\ldots,m) \end{equation*}

in the m + 1 real variables $t_1,\ldots,t_{m+1}$ has a non-trivial solution, which we denote $(t_1,\ldots,t_{m+1})\in\mathbb{R}^{m+1}\setminus\{(0,\ldots,0)\}$. Then the functional $x^*= \sum_{k=1}^{m+1} t_kb_k^*\in E^*$ satisfies $x^*(x_j)=0$ for each $j=1,\ldots,m$, and therefore

\begin{equation*} f(x^*) =\sum_{n=1}^{m+1}\frac{| t_n|}{2^n} \gt 0 = \sum_{j=1}^m| x^*(x_j)|. \end{equation*}

We have thus shown that conditions (a)–(g) are equivalent. Combining conditions (c) and (d), we see that they imply condition (h).

(h) $\Rightarrow$(i). Suppose that $I_{w^*}[E]=H_{w^*}[E]$. Then $I_{w^*}[E]=H_{w^*}^p[E]$, which is closed by lemma 2.1(v).

(i) $\Rightarrow$(j). This follows from the general fact that $\textrm{FBL}^p[E]\subseteq\overline{I_{w^*}[E]}$.

Finally, we can complete the proof by observing that the implication (j) $\Rightarrow$(g) is trivial.

Proof. The equalities follow from theorem 3.5 and lemma 2.1(ix), while proposition 3.2 implies that the inclusion in the middle is proper.

Question 4.1.

For which Banach spaces E is it true that $\textrm{FBL}^p[E] = \overline{I_{w^*}[E]}$ for some/all $1 \leqslant p \lt \infty$?

Question 4.2.

For which Banach spaces E is it true that $\overline{I_{w^*}[E]} = H_{w^*}^p[E]$ for some/all $1 \leqslant p \lt \infty$?

Proof. We begin by showing that conditions (c), (d) and (e) are equivalent. It is clear that (c) implies (d).

(d) $\Rightarrow$(e). Suppose that $(x_n^*)$ is a sequence in  $E^*\setminus\{0\}$ such that $\bigcup_{m\in\mathbb{N}} W_m$ is dense in E, where $W_m=\bigcap_{n=m}^\infty\textrm{ker} x^*_n$. Clearly $W_m\subseteq W_{m+1}$ for each $m\in\mathbb{N}$, and either $W_m=W_{m+1}$ or W_m has codimension one in  $W_{m+1}$. We cannot have $W_m=W_{m+1}$ for all but finitely many $m\in\mathbb{N}$ because if we did, we would have $\bigcup_{m\in\mathbb{N}} W_m = W_{m_0}$ for some $m_0\in\mathbb{N}$, so $E=W_{m_0}$ because  $\bigcup_{m\in\mathbb{N}} W_m$ is dense in E and  $W_{m_0}$ is closed. This would imply that $x_n^*=0$ for each $n\geqslant m_0$, which contradicts our assumption. Hence, by passing to a subsequence, we may suppose that $W_m\subsetneq W_{m+1}$ for each $m\in\mathbb{N}$. According to a theorem of Mujica (see [Reference Hájek, Santalucía, Vanderwerff and Zizler24, page 199]), this implies that (e) is satisfied.

Alternatively, we can easily verify (e) directly, showing that the quotient space $E/W_1$ is separable and infinite-dimensional. To this end, take $w_m\in W_{m+1}\setminus W_m$ such that $W_{m+1} = W_m + \mathbb{R} w_m$ for each $m\in\mathbb{N}$. Then the subspace

\begin{equation*} W_1+\textrm{span}\{w_m : m\in\mathbb{N}\} \end{equation*}

is dense in E, and therefore the subspace $\textrm{span}\{Qw_m : m\in\mathbb{N}\}$ is dense in  $E/W_1$, where $Q{\colon} E\to E/W_1$ denotes the quotient map. Hence  $E/W_1$ is separable. Furthermore, $E/W_1$ is infinite-dimensional because the sequence $(Qw_m)_{m\in\mathbb{N}}$ is linearly independent. Indeed, assuming the contrary, we can write $Qw_{m+1} = \sum_{j=1}^ms_jQw_j$ for some $m\in\mathbb{N}$ and some $s_1,\ldots,s_m\in\mathbb{R}$. Consequently $w_{m+1}- \sum_{j=1}^ms_jw_j\in W_1$, so $w_{m+1}\in W_{m+1}$, which is a contradiction.

(e) $\Rightarrow$(c). Suppose that W is a closed subspace of E such that $E/W$ is infinite-dimensional and separable. Then $E/W$ has a Markushevich basis (see for instance [Reference Fabian, Habala, Hájek, Montesinos and Zizler18, Theorem 4.59]), say $(Qx_m)_{m\in\mathbb{N}}$, where $x_m\in E$ and $Q{\colon} E\to E/W$ denotes the quotient map. Set $x_n^* = Q^*(y_n^*)\in E^*$ for each $n\in\mathbb{N}$, where $y_n^*\in (E/W)^*$ is the n ^th biorthogonal functional of $(Qx_m)_{m\in\mathbb{N}}$, and observe that $(x_n^*)_{n\in\mathbb{N}}$ is linearly independent because  $Q^*$ is injective. We claim that $\bigcup_{m\in\mathbb{N}}\bigcap_{n=m}^\infty\textrm{ker} x_n^*$ is dense in E. Indeed, given $x\in E$ and ɛ > 0, we can find $m\in\mathbb{N}$, $s_1,\ldots,s_m\in\mathbb{R}$ and $w\in W$ such that $\| x-\sum_{j=1}^ms_jx_j-w\| \lt \varepsilon$ because $\textrm{span}\{Qx_n : n\in\mathbb{N}\}$ is dense in  $E/W$. Now the conclusion follows because

\begin{equation*} \bigg\langle \sum_{j=1}^ms_jx_j+w, x_n^*\bigg\rangle = \sum_{j=1}^ms_j\langle Qx_j, y_n^*\rangle = 0 \end{equation*}

for each n > m, so $\sum_{j=1}^ms_jx_j+w\in \bigcap_{n=m+1}^\infty\textrm{ker} x_n^*$.

Clearly (a) implies (d) and (b) implies (d). We shall now complete the proof by showing that (c) implies (a) and (c) implies (b). This is the harder part of the proof, but it is actually quite simple, as the former consists of making appropriate adjustments to the proof of the Dvoretzky–Rogers theorem given in [Reference Diestel, Jarchow and Tonge15, Theorem 1.2], and the latter is a minor variation, using ideas from [Reference Grothendieck23, Theorem 2]. We include the details for the convenience of the reader.

(c) $\Rightarrow$(a). Let $(\lambda_n)\in{\ell}_2^+$, and set $n_0=0$. As in the proof of [Reference Diestel, Jarchow and Tonge15, Theorem 1.2], we choose integers $1 \leqslant n_1 \lt n_2 \lt \cdots$ such that

(4.1)\begin{equation} \sum_{j=n_k+1}^\infty\lambda_j^2 \leqslant 2^{-2k}\qquad (k\in\mathbb{N}). \end{equation}

Suppose that $(y_n^*)$ is a linearly independent sequence in  $E^*$ such that the subspace $\bigcup_{m\in\mathbb{N}}\bigcap_{n=m}^\infty\textrm{ker} y^*_n$ is dense in E. For each $k\in\mathbb{N}$, we can apply [Reference Diestel, Jarchow and Tonge15, Lemma 1.3] to the $2(n_k-n_{k-1})$-dimensional space $\textrm{span} \{y_j^* : 2n_{k-1} \lt j \leqslant 2n_k\}$ to find functionals

\begin{equation*} z_{n_{k-1}+1}^*,z_{n_{k-1}+2}^*,\ldots,z_{n_k}^*\in\textrm{span} \{y_j^* : 2n_{k-1} \lt j \leqslant 2n_k\}, \end{equation*}

each having norm between  $1/2$ and 1, such that

(4.2)\begin{equation} \bigg\|\sum_{j=n_{k-1}+1}^{n_k}\alpha_jz_j^*\bigg\| \leqslant \bigg(\sum_{j=n_{k-1}+1}^{n_k}\alpha_j^2\bigg)^{1/2}\qquad (\alpha_{n_{k-1}+1},\ldots,\alpha_{n_k}\in\mathbb{R}). \end{equation}

Define $x_j^* = \lambda_jz_j^*/\| z_j^*\|\in E^*$ for every $j\in\{n_{k-1}+1,\ldots,n_k\}$, so that $\| x_j^*\| = \lambda_j$, and observe that

(4.3)\begin{equation} \bigcap_{i=2n_{k-1}+1}^{2n_k}\textrm{ker} y_i^*\subseteq\textrm{ker} z_j^*\subseteq\textrm{ker} x_j^*\qquad (j\in\{n_{k-1}+1,\ldots,n_k\}). \end{equation}

The argument given in [Reference Diestel, Jarchow and Tonge15] shows that the sequence $(x_j^*)$ is weakly 1-summable.

Hence it only remains to show that $\bigcup_{m\in\mathbb{N}}\bigcap_{n=m}^\infty\textrm{ker} x_n^*$ is dense in E. Given $x\in E$ and ɛ > 0, choose $m\in\mathbb{N}$ and $y\in\bigcap_{n=m}^\infty\textrm{ker} y_n^*$ such that $\| x-y\| \lt \varepsilon$. Then, choosing $\ell\in\mathbb{N}_0$ such that $2n_\ell+1\geqslant m$, we obtain

\begin{equation*} y\in \bigcap_{i=2n_\ell+1}^\infty\textrm{ker} y_i^* = \bigcap_{k=\ell}^\infty\bigg(\bigcap_{i=2n_k+1}^{2n_{k+1}}\textrm{ker} y_i^*\bigg)\subseteq \bigcap_{k=\ell}^\infty\bigg(\bigcap_{j=n_k+1}^{n_{k+1}}\textrm{ker} x_j^*\bigg) = \bigcap_{j=n_\ell+1}^\infty\textrm{ker} x_j^*, \end{equation*}

where the inclusion in the middle follows from (4.3).

(c) $\Rightarrow$(b). As already stated, this proof is a minor variation of the argument we have just given to show that (c) implies (a), so we shall focus on explaining the differences. It suffices to consider $(\lambda_n)\in c_0^+$ with norm at most 1. We can no longer ensure that (4.1) is satisfied; instead, we choose integers $0=n_0 \lt n_1 \lt n_2 \lt \cdots$ such that

(4.4)\begin{equation} \text{sup}_{j \gt n_k}\lambda_j \leqslant 2^{-k}\qquad (k\in\mathbb{N}). \end{equation}

Our assumption that $\|(\lambda_n)\|_\infty\leqslant1$ implies that this estimate is also valid for k = 0. Now, for $(y_n^*)$ chosen as before, we can find the sequence $(z_j^*)$ and define $(x_j^*)$ as above. Then the proof that $\bigcup_{m\in\mathbb{N}}\bigcap_{n=m}^\infty\textrm{ker} x_n^*$ is dense in E carries over verbatim, so we only need to verify that $(x_j^*)$ is weakly 2-summable. For every $(a_j)\in B_{{\ell}_2}$ and $m\in\mathbb{N}$, we can apply (4.2), (4.4) and the fact that $\|z_j^\ast\|\geqslant\frac12$ to obtain

\begin{align*} \bigg\|\sum_{j=1}^{n_m} a_jx_j^*\bigg\|& \leqslant \sum_{k=1}^m\bigg\|\sum_{j=n_{k-1}+1}^{n_{k}}a_jx_j^*\bigg\| = \sum_{k=1}^m\bigg\|\sum_{j=n_{k-1}+1}^{n_{k}}\frac{a_j\lambda_j}{\| z_j^*\|}z_j^*\bigg\|\\ & \leqslant \sum_{k=1}^m\bigg(\sum_{j=n_{k-1}+1}^{n_{k}}\frac{a_j^2\lambda_j^2}{\| z_j^*\|^2}\bigg)^{\frac12} \leqslant \sum_{k=1}^m 2^{2-k}\bigg(\sum_{j=n_{k-1}+1}^{n_{k}}a_j^2\bigg)^{\frac12} \leqslant4. \end{align*}

This shows that $(x_j^*)$ is weakly 2-summable because $n_m\to\infty$ as $m\to\infty$.

Proof of theorem 4.3.

This proof is a more subtle variant of the proof of the implication (c) $\Rightarrow$(a) in theorem 3.5.

Recall that we seek a function $f\in I_{w^*}[E]\setminus\textrm{FBL}^p[E]$. It turns out to be convenient to decompose E as $E=F\oplus\mathbb{R} x_1$ and then carry out the main technical part of the construction of f within the subspace F. Consequently we choose $x_1\in E$ and $x_1^*\in E^*$ such that $\| x_1\| = \| x_1^*\| = 1 = x_1^*(x_1)$, and we then define $F=\textrm{ker} x_1^*$.

Since ${E}$ admits an infinite-dimensional, separable quotient space, so does F; that is, F satisfies one, and hence all, of the equivalent conditions in lemma 4.4. Using this, we shall prove that, for a suitably chosen real number q > p, $F^*$ contains a weakly p-summable sequence $(y_n^*)$ such that

(4.5)\begin{equation} \| y^*_n\| = n^{-\frac1q}\qquad (n\in\mathbb{N})\qquad\text{and}\qquad \overline{\bigcup_{m\in\mathbb{N}}\bigcap_{n\geqslant m}\textrm{ker} y^*_n} = F. \end{equation}

We split the proof of this claim in two cases:

• • If p < 2, we choose $q\in(p,2)$ and observe that $(n^{-\frac1q})_{n\in\mathbb{N}}\in{\ell}_2^+$. Therefore lemma 4.4(a) implies that  $F^*$ contains a weakly 1-summable sequence $(y_n^*)$ which satisfies (4.5). Since $p\geqslant1$, it follows that $(y_n^*)$ is weakly p-summable.

• • Otherwise $p\geqslant 2$, in which case we choose $q\in(p,\infty)$. Since $(n^{-\frac1q})_{n\in\mathbb{N}}\in c_0^+$, lemma 4.4(b) shows that  $F^*$ contains a weakly 2-summable sequence $(y_n^*)$ which satisfies (4.5). The fact that $p\geqslant2$ ensures that $(y_n^*)$ is weakly p-summable.

Let  ${P}$ be the projection of  ${E}$ onto  ${F}$ given by $Px = x- x_1^*(x)x_1$ for $x\in E$. For each $n\in\mathbb{N}$, set $x_{n+1}^* = P^*y_n^*\in E^*$, and take a unit vector $x_{n+1}\in F$ such that $y^*_n(x_{n+1})\geqslant \| y_n^*\|/2$. Then we have:

• • $x_{n+1}^*(x_1) =0$ and $x_{n+1}^*(x_{n+1})\geqslant \| y_{n}^*\|/2$ for every $n\in\mathbb{N}$;

• • the subspace $W=\bigcup_{m\in\mathbb{N}}W_m$ is dense in E, where $W_m = \bigcap_{n=m}^\infty\textrm{ker} x^*_n$;

• • the sequence $(x_n^*)$ is weakly p-summable, so

  (4.6)\begin{equation} K := \text{sup}_{x\in B_E}\bigg(\sum_{n=1}^\infty | x_n^*(x)|^p\bigg)^{\frac1p} \lt \infty. \end{equation}

Set $s_n = n^{\frac1q-\frac1p}\in(0,1]$ for each $n\in\mathbb{N}$. Then $s_n\to 0$ as $n\to\infty$, so lemma 3.4 shows that the function $g = {\bigvee}_{n=1}^\infty | {\delta}_{s_nx_{n+1}}|$ belongs to  $H_{w^*}[E]$, and therefore $f =|{\delta}_{x_1}|\wedge g$ belongs to  $I_{w^*}[E]$ by lemma 2.1(ii). We shall now complete the proof by showing that $f\notin \textrm{FBL}^p[E]$, using $(x_n^*)$ to produce a suitable “witness sequence”.

The sublattice L of  $H[E]$ generated by $\{{\delta}_w : w\in W\}$ is dense in $\textrm{FBL}^p[E]$ because W is dense in E. (This follows easily from the construction of  $\textrm{FBL}^p[E]$ in [Reference Jardón-Sánchez, Laustsen, Taylor, Tradacete and Troitsky25, Theorem 6.1].) Therefore it will suffice to show that

(4.7)\begin{equation} \inf\Big\{\| f - h\|_{\textrm{FBL}^p[E]} : h\in L\Big\} \gt 0. \end{equation}

Since W is a subspace of E, every function $h\in L$ can be expressed in the form $h = {\bigvee}_{i=1}^k {\delta}_{v_i} - {\bigvee}_{i=1}^k {\delta}_{w_i}$ for some $k\in\mathbb{N}$ and some $v_1,\ldots,v_k,w_1,\ldots,w_k\in W$. As W is the union of the increasing sequence $(W_m)$ of subspaces, we can find $m\in\mathbb{N}$ such that $v_1,\ldots,v_k,w_1,\ldots,w_k\in W_m$. Then, for each $j\in\{1,\ldots,m\}$, we see that the functional $z_j^* = m^{-\frac1p} x_1^* + x_{m+j}^*\in E^*$ satisfies

(4.8)\begin{align} h(z_j^*) &= {\bigvee}_{i=1}^k \Big(m^{-\frac1p} x_1^*(v_i) + x_{m+j}^*(v_i)\Big) - {\bigvee}_{i=1}^k \Big(m^{-\frac1p}x_1^*(w_i) + x_{m+j}^*(w_i)\Big)\nonumber\\ &= {\bigvee}_{i=1}^k m^{-\frac1p} x_1^*(v_i) - {\bigvee}_{i=1}^k m^{-\frac1p}x_1^*(w_i) = m^{-\frac1p}h(x_1^*) \end{align}

because $v_1,\ldots,v_k,w_1,\ldots,w_k\in W_m\subseteq\textrm{ker} x_{n}^*$ for every $n\geqslant m$.

It follows from (4.6) that $\|(x_j^\ast)_{j=m+1}^{2m}\|_{p,\text{weak}}\leqslant K$. Combining this estimate with the easy observation that

(4.9)\begin{equation} \big\| (\underbrace{m^{-\frac1p} x_1^*,\ldots,m^{-\frac1p} x_1^*}_{m})\big\|_{p,{{\text{weak}}}} = \| x^*_1\| = 1 \end{equation}

and the subadditivity of  $\|\cdot\|_{p,\mathrm{weak}}$, we see that $\|(z_j^*)_{j=1}^m\|_{p,{{\text{weak}}}} \leqslant K+1$, so

(4.10)\begin{equation} (K+1)\| f-h\|_{\textrm{FBL}^p[E]}\geqslant \bigg(\sum_{j=1}^{m}\mid f(z_j^*)-h(z_j^*)\mid^p\bigg)^\frac1p. \end{equation}

Another application of (4.9) shows that

(4.11)\begin{equation} \| f-h\|_{\textrm{FBL}^p[E]}\geqslant \bigg(\sum_{j=1}^{m}\mid m^{-\frac1p}(f(x_1^*)-h(x_1^*))\mid^p\bigg)^{\frac1p}. \end{equation}

Now we combine (4.10) and (4.11) with the subadditivity of the norm on  ${\ell}_p^{m}$ to obtain

(4.12)\begin{align} (K+2)\| f-h\|_{\textrm{FBL}^p[E]}&\geqslant \bigg(\sum_{j=1}^{m}\mid f(z_j^*)-h(z_j^*) - m^{-\frac1p}(f(x_1^*) - h(x_1^*))\mid^p\bigg)^{\frac1p}\nonumber\\ &= \bigg(\sum_{j=1}^{m}\mid f(z_j^*) - m^{-\frac1p}f(x_1^*)\mid^p\bigg)^\frac1p, \end{align}

where the final equality follows from (4.8).

To find a lower bound on the right-hand side of (4.12), we recall that $f =|{\delta}_{x_1}|\wedge g$, where $g(x_1^*) =0$ because $x_{n+1}\in F= \textrm{ker} x_1^*$ for every $n\in\mathbb{N}$, so $f(x_1^*) = 0$, while

\begin{align*} f(z_j^*) &= \mid m^{-\frac1p} x_1^*(x_1) + x_{m+j}^*(x_1)\mid\wedge {\bigvee}_{n=1}^\infty s_n\mid m^{-\frac1p}x_1^*(x_{n+1}) + x_{m+j}^*(x_{n+1})\mid\\ &= m^{-\frac1p}\wedge {\bigvee}_{n=1}^\infty s_n\mid x_{m+j}^*(x_{n+1})\mid\geqslant m^{-\frac1p}\wedge s_{m+j-1}\mid x_{m+j}^*(x_{m+j})\mid\\ &\geqslant m^{-\frac1p}\wedge \frac{s_{m+j-1}\| y_{m+j-1}^*\|}2 = m^{-\frac1p}\wedge \frac12 (m+j-1)^{-\frac1p} \geqslant \frac{1}{2}(2m)^{-\frac1p} \end{align*}

for each $j\in\{1,\ldots,m\}$. Substituting these estimates into (4.12), we conclude that

\begin{align*} (K+2)\| f-h\|_{\textrm{FBL}^p[E]} &\geqslant 2^{-\frac{p+1}{p}}, \end{align*}

from which (4.7) follows because the constant K is independent of h.

This completes the proof that $f\in I_{w^*}[E]\setminus\textrm{FBL}^p[E]$. The final clause follows because $\textrm{FBL}^p[E]\subseteq\overline{I_{w^*}[E]}$.

Proof. Set $E = {\ell}_1(A)$, and let $R{\colon} E^*\to C(B_{E^{**}})$ be the bounded linear operator defined by $(Rx^*)(x^{**}) = x^{**}(x^*)$ for $x^*\in E^*$ and $x^{**}\in B_{E^{**}}$; that is, $Rx^*$ is the restriction of the evaluation map  ${\delta}_{x^*}$ to  $B_{E^{**}}$.

For $f\in H_{w^*}^p[E]$, we can choose a probability measure $\mu\in C(B_{E^{**}})_+^*$ such that (4.14) is satisfied. Let $J_\infty{\colon} C(B_{E^{**}}) \to L_\infty(B_{E^{**}},\mu)$ and $J_p{\colon} L_\infty(B_{E^{**}},\mu)\to L_p(B_{E^{**}},\mu)$ be the formal inclusion maps. Then, by [Reference Cardassi11, Theorem I.8] or [Reference Diestel, Jarchow and Tonge15, page 97], the composite operator $T=J_pJ_\infty R{\colon} E^*\to L_p(B_{E^{**}},\mu)$ is strictly p-integral, and we can restate the inequality (4.14) as

(4.15)\begin{equation}\vert f(x^\ast)\vert\leqslant\Arrowvert f\Arrowvert_{\text{FBL}^p\lbrack E\rbrack}\|Tx^\ast\|\qquad(x^\ast\in E^\ast).\end{equation}

Let $J_0{\colon} c_0(A)\to{\ell}_\infty(A)=E^*$ be the inclusion map. Its domain is an Asplund space, so [Reference Cardassi11, Theorem II.3] shows that the operator TJ ₀ is p-nuclear; therefore, using [Reference Cardassi11, Definition I.1] or [Reference Diestel, Jarchow and Tonge15, Proposition 5.23], we can find sequences $(y_j)_{j\in\mathbb{N}}$ in  $E=c_0(A)^*$ and $(z_j)_{j\in \mathbb{N}}$ in  $L_p(B_{E^{**}},\mu)$ such that

1. (i) $TJ_0x^*=\sum_{j=1}^\infty \langle y_j,x^*\rangle z_j$ for every $x^*\in c_0(A)$;

2. (ii) $(y_j)_{j\in\mathbb{N}}$ is strongly p-summable; that is, $\sum_{j=1}^\infty \|y_j\|^p \lt \infty$;

3. (iii) $(z_j)_{j\in \mathbb{N}}$ is weakly $p^*$-summable, where $p^*\in(1,\infty]$ is the conjugate exponent of p; that is, $\text{sup}_{j\in\mathbb{N}}\|z_j\| \lt \infty$ if p = 1, and otherwise

   \begin{equation*} \text{sup}\bigg\{\sum_{j=1}^\infty|\langle z^*,z_j\rangle|^{p^*} :z^*\in L_{p^*}(B_{E^{**}},\mu),\ \|z^*\| \leqslant 1\bigg\} \lt \infty, \end{equation*}

   where $\frac{1}{p}+\frac{1}{p^*}=1$. By rescaling, we may suppose that these suprema equal 1.

Condition (ii) implies that we can define a bounded linear operator $U{\colon} E^*\to{\ell}_p$ by $Ux^* = (\langle y_j,x^*\rangle)_{j\in\mathbb{N}}$. Applying (i) and (iii), as well as Hölder’s inequality if p > 1, we obtain

(4.16)\begin{align} \| TJ_0x^*\| &= \mathop{\rm sup}_{\| z^*\| \leqslant1}|\langle z^*,TJ_0x^*\rangle|=\mathop{\rm sup}_{\| z^*\| \leqslant1}\bigg|\sum_{j=1}^\infty\langle y_j,x^*\rangle\langle z^*,z_j\rangle\bigg|\nonumber\\ & \leqslant \bigg(\sum_{j=1}^\infty|\langle y_j,x^*\rangle|^p\bigg)^{\frac{1}{p}} = \| Ux^*\|\qquad (x^*\in c_0(A)). \end{align}

We now claim that the map $g\in H[E]$ defined by $g(x^*) = \| Ux^*\|$ for $x^*\in E^*$ belongs to  $\textrm{FBL}^p[E]$. For ɛ > 0, take $m\in\mathbb{N}$ such that $\sum_{j=m+1}^\infty\| y_j\|^p \leqslant\varepsilon^p$, and set $g_m = \Big(\sum_{j=1}^m|{\delta}_{y_j}|^p\Big)^{\frac1p}\in\textrm{FBL}^p[E]$. We observe that $g_m(x^*) = \Big(\sum_{j=1}^m|\langle y_j,x^*\rangle|^p\Big)^{\frac1p} = \| P_mUx^*\|$ for $x^*\in E^*$, where $P_m{\colon}{\ell}_p\to{\ell}_p$ denotes the m ^th basis projection. Hence, for $n\in\mathbb{N}$ and $(x_k^*)_{k=1}^n\subset E^*$ with $\|(x_k^*)_{k=1}^n\|_{p,{{\text{weak}}}} \leqslant 1$, we have

\begin{align*} \sum_{k=1}^n| (g-g_m)(x_k^*)|^p &= \sum_{k=1}^n\mid \| Ux_k^*\| - \| P_mUx_k^*\|\mid^p \leqslant \sum_{k=1}^n\| (I_{{\ell}_p} - P_m)Ux_k^*\|^p\\ &=\sum_{k=1}^n\sum_{j=m+1}^\infty|\langle y_j,x_k^*\rangle|^p = \sum_{j=m+1}^\infty\sum_{k=1}^n|\langle y_j,x_k^*\rangle|^p \leqslant \sum_{j=m+1}^\infty\| y_j\|^p \leqslant\varepsilon^p. \end{align*}

This proves that $\| g-g_m\|_{\textrm{FBL}^p[E]} \leqslant\varepsilon$, and the claim follows.

Combining the inequalities (4.15)–(4.16) with the definition of g, we see that $| f(x^*)| \leqslant \| f\|_{\textrm{FBL}^p[E]}\, g(x^*)$ for $x^*\in c_0(A)$. Goldstine’s theorem allows us to extend this inequality to $x^*\in E^*$ because the restrictions of f and g to  $B_{E^*}$ are weak*-continuous. This shows that $f\in\overline{I_{w^*}[E]}$, and therefore $H_{w^*}^p[E]\subseteq \overline{I_{w^*}[E]}$.

The opposite inclusion follows from lemma 2.1(viii).

Proof. (i) Since

\begin{equation*} \bigg(\sum_{j=1}^m | x^{**}(x_j^*)|^p\bigg)^{\frac1p} \leqslant \| x^{**}\|\cdot \|(x_j^*)_{j=1}^m\|_{p,{{\text{weak}}}} \end{equation*}

for $x^{**}\in E^{**}$, $m\in\mathbb{N}$ and $x_1^*,\ldots,x_m^*\in E^*$, we have $\| x^{**}\|_{\textrm{FBL}^p[E]} \leqslant\| x^{**}\|$. The opposite inequality follows from the fact that $\| x^*\|_{p,{{\text{weak}}}}=\| x^*\|$ for every $x^*\in E^*$.

(ii) It is clear that ${\delta}_x\in E^{**}\cap I[E]\cap H_{w^*}[E]$ for every $x\in E$, and lemma 2.1(vii) and (ix) show that $I[E]\subseteq M_{w^*,0}^p[E]$. Hence we can complete the proof by showing that if $x^{**}\in E^{**}$ belongs to  $H_{w^*}[E]$ or  $M_{w^*,0}^p[E]$, then $x^{**}={\delta}_x$ for some $x\in E$. In both cases $x^{**}{\upharpoonright_{B_{E^*}}}$ is weak*-continuous at 0, so $x^{**}{\upharpoonright_{B_{E^*}}^{\,-1}}(\{0\}) = \textrm{ker} x^{**}\cap B_{E^*}$ is weak*-closed. This implies that $x^{**}={\delta}_x$ for some $x\in E$ by an application of the Krein–Smulian theorem (see, e.g., [Reference Fabian, Habala, Hájek, Montesinos and Zizler18, Corollary 3.94]).

Proof. Lemma 5.1 shows that $E^{**}\setminus\{{\delta}_x:x\in E\}\subseteq H^p[E]\setminus (M_{w^*,0}^p[E]\cup H_{w^*}[E])$.

Proof. We begin by showing that ψ_p is weakly sequentially continuous. Take a sequence $(x_n^*)_{n\in\mathbb N}$ in  $B_{E^*}$ which converges weakly to some  $x^*\in B_{E^*}$. Then ${\text{sup}}_{n\in\mathbb{N}}\|\psi_p(x_n^\ast)\|_\infty\leqslant1$ and

\begin{equation*} \psi_p(x^*_n)(x)= | x_n^*(x)|^p\underset{n\to\infty}\longrightarrow | x^*(x)|^p = \psi_p(x^*)\qquad (x\in B_E), \end{equation*}

so theorem 5.3 shows that $(\psi_p(x_n^*))_{n\in\mathbb{N}}$ converges weakly to  $\psi_p(x^*)$.

Now, in order to complete the proof, take a weakly closed subset C of $C(B_E)$, and let us prove that $\psi_p^{-1}(C)$ is weakly closed in $B_{E^*}$. By the reflexivity of E and the Eberlein–Smulian theorem, this is equivalent to showing that $\psi_p^{-1}(C)$ is weakly sequentially compact. To verify this, take a sequence $(x_n^*)_{n\in\mathbb N}$ in $\psi_p^{-1}(C)$. Due to the weak sequential compactness of $B_{E^*}$, $(x_n^*)_{n\in\mathbb{N}}$ has a subsequence $(x_{n_j}^*)_{j\in\mathbb{N}}$ which converges weakly to some $x^*\in B_{E^*}$. The first part of the proof implies that $(\psi_p(x_{n_j}^*))_{j\in\mathbb{N}}$ converges weakly to $\psi_p(x^*)$. Since $\psi_p(x_{n_j}^*)$ belongs to the weakly closed set C for each $j\in\mathbb{N}$, we see that $\psi_p(x^*)\in C$. In conclusion, we have shown that $(x_n^*)_{n\in\mathbb N}$ has a subsequence which converges weakly to $x^*\in\psi_p^{-1}(C)$, and the result follows.

Proof. (a) $\Rightarrow$(b). Suppose that E is reflexive, so that $E^{**}=E$ with the weak* and weak topologies coinciding, and take $\mu\in C(B_E)^*_+$. For $x^*\in B_{E^*}$, we have

\begin{equation*} f_\mu^p(x^*)^p = \int_{B_{E}} |x^{*}(x)|^p\, d\mu(x) =\langle \psi_p(x^*),\mu\rangle = (\mu\circ\psi_p)(x^*), \end{equation*}

so that $(f_\mu^p{\upharpoonright_{B_{E^*}}})^p=\mu\circ\psi_p$, where ψ_p is the map defined in lemma 5.4. Take a net $(x^*_\alpha)$ in $B_{E^*}$ which weak*-converges to  $x^*$. Then, by reflexivity, $(x^*_\alpha)$ converges weakly to  $x^*$, so lemma 5.4 implies that $(\psi_p(x^*_\alpha))$ converges weakly to $\psi_p(x^*)$, which means that $(\langle \psi_p(x_\alpha^*),\mu\rangle)=(f_\mu^p(x^*_\alpha)^p)$ converges to $\langle \psi_p(x^*),\mu\rangle = f_\mu^p(x^*)^p$. This shows that $f_\mu^p\in H_{w^*}[E]$.

(b) $\Rightarrow$(c). Suppose that $f_\mu^p\in H_{w^*}[E]$ for every $\mu\in C(B_{E^{**}})^*_+$. Then $f_\mu^p\in H_{w^*}^p[E]$ because $f_\mu^p$ always belongs to  $H^p[E]$. As mentioned above, for every $f\in H^p[E]$, we can find a probability measure $\mu\in C(B_{E^{**}})^*_+$ such that (4.14) is satisfied. This implies that $f\in J_{w^*}^p[E]$, and consequently $H^p[E] = J_{w^*}^p[E]$.

(c) $\Rightarrow$(d) is trivial.

(d) $\Rightarrow$(e). This follows from lemma 2.1(ix) because $f_\mu^p\in H^p[E]$ for every $\mu\in C(B_{E^{**}})^*_+$.

Finally, we prove the implication (e) $\Rightarrow$(a) by contraposition. Suppose that E is non-reflexive. Then $B_{E^*}$ contains a net $(x^*_\alpha)$ which weak*-converges to 0, but does not converge weakly to 0. Take $x_0^{**}\in E^{**}$ such that $(x_0^{**}(x^*_\alpha))$ does not converge to 0, and consider the Dirac measure $\mu= {\delta}_{x^{**}_0}\in C(B_{E^{**}})_+^*$; it satisfies

\begin{align*} f_\mu^p(x_\alpha^*) &= \bigg(\int_{B_{E^{**}}} |x^{**}(x^*_\alpha)|^p\,d\mu(x^{**})\bigg)^{\frac1p} = \mid x^{**}_0(x_\alpha^*)\mid\not\rightarrow 0 = f_\mu^p(0). \end{align*}

This shows that $f_{\mu}^p{\upharpoonright_{B_{E^*}}}$ fails to be weak*-continuous at 0, so $f_\mu^p\notin M_{w^*,0}^p[E]$.

Proof. (i) The composite map $f\circ\Phi$ is clearly positively homogeneous when f and Φ are, so $C_\Phi(H[E])\subseteq H[F]$, and $C_\Phi$ is a lattice homomorphism because the vector lattice operations are defined coordinatewise.

(ii) Suppose that $\Phi\in\textrm{PH}^p[F^*,E^*]$. Then  $\|\Phi(y^\ast)\|\leqslant\Arrowvert\Phi\Arrowvert_{\text{PH}^p\lbrack F^\ast,E^\ast\rbrack}$ for every $y^*\in B_{F^*}$ because $\|z^\ast\|_{p,\text{weak}}=\|z^\ast\|$ for every functional  $z^*$ (in either  $E^*$ or  $F^*)$, and therefore $\Phi(B_{F^*})\subseteq \|\Phi\|_{\textrm{PH}^p[F^*,E^*]} B_{E^*}$. Furthermore, working straight from the definitions of the norms, we obtain

\begin{multline*} \| C_\Phi (f)\|_{\textrm{FBL}^p[F]} = \text{sup}\bigg\{\Big(\sum_{j=1}^m| f(\Phi y_j^*)|^p\Big)^{\frac1p} : m\in\mathbb N,\, (y_j^*)_{j=1}^m\subset F^*,\bigg.\\[-1mm] \bigg.\| (y_j^*)_{j=1}^m\|_{p,{{\text{weak}}}} \leqslant1\bigg\} \leqslant\|\Phi\|_{\textrm{PH}^p[F^*,E^*]}\, \|f\|_{\textrm{FBL}^p[E]} \qquad (f\in H^p[E]). \end{multline*}

This shows that  $C_{\Phi}$ maps  $H^p[E]$ into  $H^p[F]$ and $\|C_\Phi\| \leqslant \|\Phi\|_{\textrm{PH}^p[F^*,E^*]}$.

On the other hand, given $x\in B_E$ and $(y_j^*)_{j=1}^m\subset F^*$ for some $m\in\mathbb{N}$, we have

\begin{align*} \Big(\sum_{j=1}^m |(\Phi y_j^*)(x)|^p\Big)^{\frac{1}{p}}&=\Big(\sum_{j=1}^m |{\delta}_x(\Phi y_j^*)|^p\Big)^{\frac{1}{p}}=\Big(\sum_{j=1}^m |(C_\Phi {\delta}_x) (y_j^*)|^p\Big)^{\frac{1}{p}}\\ & \leqslant\|C_{\Phi} {\delta}_x\|_{\textrm{FBL}^p[F]}\, \| (y_j^*)_{j=1}^m\|_{p,{{\text{weak}}}} \leqslant\|C_{\Phi}\|\, \| (y_j^*)_{j=1}^m\|_{p,{{\text{weak}}}}, \end{align*}

which shows that $\|\Phi\|_{\textrm{PH}^p[F^*,E^*]} \leqslant \|C_{\Phi}\|$, and therefore $\|C_{\Phi}\|=\|\Phi\|_{\textrm{PH}^p[F^*,E^*]}$.

Now suppose that the restriction of Φ to  $B_{F^*}$ is weak*-to-weak* continuous (as well as $\Phi\in\textrm{PH}^p[F^*,E^*]$). Since $\Phi(B_{F^*})\subseteq \|\Phi\|_{\textrm{PH}^p[F^*,E^*]} B_{E^*}$ and the functions in  $H_{w^*}[E]$ are positively homogeneous, we see that  $C_\Phi$ maps  $H_{w^*}[E]$ into  $H_{w^*}[F]$ and  $M^p_{w^*,0}[E]$ into  $M^p_{w^*,0}[F]$. The inclusion $C_{\Phi}(H^p_{w^*}[E])\subseteq H^p_{w^*}[F]$ follows by combining the two inclusions $C_\Phi(H_{w^*}[E])\subseteq H_{w^*}[F]$ and $C_{\Phi}(H^p[E])\subseteq H^p[F]$.

Finally, for each $f\in J^p_{w^*}[E]$, we can find $g\in H^p_{w^*}[E]$ such that $|f| \leqslant g$. Then we have $|C_{\Phi}(f)| \leqslant C_{\Phi}(g)\in H^p_{w^*}[F]$ by the previous inclusion, and consequently $C_{\Phi}(f)\in J^p_{w^*}[F]$. This implies that $C_{\Phi}(\overline{J^p_{w^*}[E]})\subseteq \overline{J^p_{w^*}[F]}$ because  $C_\Phi$ is continuous with respect to the $\textrm{FBL}^p$-norms, as we have shown above.

(iii) This proof is similar to the proof of [Reference Jardón-Sánchez, Laustsen, Taylor, Tradacete and Troitsky25, Proposition 7.3]. We may suppose that $\|\Phi\|_{\textrm{PH}^p[F^*,E^*]} \lt \infty$, as otherwise there is nothing to prove. Take $x\in B_E$ and $(y_j^*)_{j=1}^m\subset F^*$ with $\|(y_j^\ast)_{j=1}^m\|_{q,\text{weak}}\leqslant1$ for some $m\in\mathbb{N}$, and set $\lambda_j = |\langle x,\Phi(y_j^*)\rangle|^{q/r}\geqslant 0$ for $j\in\{1,\ldots,m\}$, where $r = pq/(q-p)\in(p,\infty)$. Then $(1+r/q)p=r$, which together with the positive homogeneity of Φ implies that

(6.1)\begin{align} \sum_{j=1}^m\lambda_j^r &= \sum_{j=1}^m\Big(\lambda_j|\langle x,\Phi(y_j^*)\rangle|\Big)^p = \sum_{j=1}^m|\langle x,\Phi(\lambda_jy_j^*)\rangle|^p\nonumber\\ & \leqslant \| (\Phi(\lambda_jy_j^*))_{j=1}^m\|_{p,\text{weak}}^p \leqslant \|\Phi\|_{\textrm{PH}^p[F^*,E^*]}^p\, \|(\lambda_jy_j^*)_{j=1}^m\|_{p,\text{weak}}^p. \end{align}

The choice of r means that the conjugate exponent of  $r/p\in(1,\infty)$ is  $q/p$. Hence, for $y\in B_F$, Hölder’s inequality gives

\begin{align*} \sum_{j=1}^m |\langle y, \lambda_jy_j^*\rangle|^p &= \sum_{j=1}^m \lambda_j^p|\langle y,y_j^*\rangle|^p \leqslant \Big(\sum_{j=1}^m \lambda_j^r\Big)^{\frac{p}{r}} \Big(\sum_{j=1}^m|\langle y,y_j^*\rangle|^q\Big)^{\frac{p}{q}} \leqslant \Big(\sum_{j=1}^m\lambda_j^r\Big)^{\frac{p}{r}}. \end{align*}

This shows that $\|(\lambda_jy_j^*)_{j=1}^m\|_{p,\text{weak}}^p \leqslant \Big(\sum_{j=1}^m\lambda_j^r\Big)^{p/r}$, so by (6.1), we have

\begin{equation*} \sum_{j=1}^m\lambda_j^r \leqslant \|\Phi\|_{\textrm{PH}^p[F^*,E^*]}^p\,\Big(\sum_{j=1}^m\lambda_j^r\Big)^{\frac{p}{r}}. \end{equation*}

Rearranging this inequality and using once more that $1-p/r=p/q$, we obtain

\begin{equation*} \|\Phi\|_{\textrm{PH}^p[F^*,E^*]}^p\geqslant \Big(\sum_{j=1}^m\lambda_j^r\Big)^{\frac{p}{q}} = \Big(\sum_{j=1}^m|\langle x,\Phi(y_j^*)\rangle|^q\Big)^{\frac{p}{q}}. \end{equation*}

Now the conclusion follows by taking the $p{\text{th}}$ roots and the supremum over x and $(y_j^*)_{j=1}^m$.

Question 6.3.

Let $\Phi\in\textrm{PH}^p[F^*,E^*]$ for some $1 \leqslant p \lt \infty$ and infinite-dimensional Banach spaces E and F, and suppose that $\Phi{\upharpoonright_{B_{F^*}}}$ is weak*-to-weak* continuous. Is

\begin{equation*} C_\Phi(\overline{I_{w^*}[E]})\subseteq\overline{I_{w^*}[F]}\,? \end{equation*}

Proof. We begin with the uniqueness. Suppose that $\Phi_T{\colon} F^*\rightarrow E^*$ is a map which satisfies (6.2). Substituting $f={\delta}_x$ for some $x\in E$ into this identity and evaluating it at some $y^*\in F^*$, we obtain

(6.3)\begin{equation} (T{\delta}_x)(y^*) = ({\delta}_x\circ\Phi_T)(y^*) = \Phi_T (y^*)(x). \end{equation}

Reading this equation from the right to the left, we conclude that there is only one possible way to define  $\Phi_T$. Taking this as our definition, we must check that $\Phi_T(y^*)\in E^*$, which is easy: Linearity follows from the fact that ${\delta}_{\lambda w+x} = \lambda{\delta}_w+{\delta}_x$ for $\lambda\in\mathbb{R}$ and $w,x\in E$, and continuity from the estimate $\vert(T\delta_x)(y^\ast)\vert\leqslant\|T\|\,\|x\|\,\|y^\ast\|$. This estimate also shows that  $\Phi_T$ maps  $B_{F^*}$ into  $\| T\| B_{E^*}$.

It is equally easy to see that $\Phi_T$ is positively homogeneous because

\begin{equation*} \Phi_T (\lambda y^*) (x)=(T{\delta}_{x})(\lambda y^*)=\lambda (T{\delta}_{x})(y^*)=\lambda \Phi_T(y^*)(x)\quad (\lambda\geqslant0,\,y^*\in F^*,\,x\in E). \end{equation*}

Next, we shall prove (i)–(ii), beginning with the latter. Suppose that $(y^*_\alpha)$ is a net in  $B_{F^*}$ which weak*-converges to some  $y^*\in B_{F^*}$, and take $x\in E$. Lemma 2.1(vi) implies that the net $((T{\delta}_{x})(y^*_\alpha))$ converges to $(T{\delta}_{x})(y^*)$, which in view of the definition (6.3) means that the net $(\Phi_T (y^*_\alpha) (x))$ converges to  $\Phi_T (y^*)(x)$.

To verify (i), we observe that the uniqueness part of the universal property of  $\textrm{FBL}^p[E]$ implies that T is the lattice homomorphism induced by the bounded linear operator $T\circ{\delta}^E{\colon} E\to\textrm{FBL}^p[F]$; that is, $T=\widehat{T\circ{\delta}^E}$, and therefore

\begin{align*} \| T\| &= \| T\circ{\delta}^E\| = \text{sup}\Big\{\| T{\delta}_x\|_{\textrm{FBL}^p[F]} : x\in B_E\Big\} =\text{sup}\bigg\{\Big(\sum_{j=1}^{m}| (T{\delta}_x)(y^*_j)|^p\Big)^{\frac1p} :\bigg.\\[-2mm] &\qquad \qquad \qquad \qquad \qquad \bigg. x\in B_E,\, m\in\mathbb N,\, (y_j^*)_{j=1}^m\subset F^*,\,\| (y^*_j)_{j=1}^{m}\|_{p,{{\text{weak}}}} \leqslant 1\bigg\}\\ &=\text{sup}\left\{\| (\Phi_T (y^*_j)\Big)_{j=1}^m{\|}_{p,{{\text{weak}}}} : m\in\mathbb N,\, (y_j^*)_{j=1}^m\subset F^*,\, \| (y^*_j)_{j=1}^{m}\|_{p,{{\text{weak}}}} \leqslant 1\right\}\\ &=\| \Phi_T\|_{\textrm{PH}^p[F^*,E^*]}. \end{align*}

It remains to verify that the map  $\Phi_T$ defined by (6.3) satisfies (6.2). We have already shown that $\Phi_T\in\textrm{PH}^p[F^*,E^*]$, so lemma 6.1(ii) implies that it induces a lattice homomorphism $C_{\Phi_T}{\colon} H^p[E]\to H^p[F]$ by composition; that is, $C_{\Phi_T}(f) = f\circ\Phi_T$ for $f\in H^p[E]$. The fact that

\begin{equation*} C_{\Phi_T}({\delta}_x)(y^*)={\delta}_x(\Phi_T (y^*)) = \Phi_T(y^*)(x) = (T{\delta}_{x})( y^*)\qquad (x\in E,\, y^*\in F^*) \end{equation*}

shows that the restriction of  $C_{\Phi_T}$ to  $\textrm{FBL}^p[E]$ satisfies $C_{\Phi_T}{\!\upharpoonright_{\textrm{FBL}^p[E]}}\circ{\delta}^E=\iota\circ T\circ{\delta}^E$, where $\iota{\colon}\textrm{FBL}^p[F]\to H^p[F]$ denotes the inclusion map. This in turn implies that $C_{\Phi_T}{\!\upharpoonright_{\textrm{FBL}^p[E]}}=\iota\circ T$ by [Reference Jardón-Sánchez, Laustsen, Taylor, Tradacete and Troitsky25, Corollary 3.5(ii)], bearing in mind that the map δ ^E is denoted  $\phi_E^{\mathcal{D}}$ in [Reference Jardón-Sánchez, Laustsen, Taylor, Tradacete and Troitsky25]. Now the conclusion follows because

\begin{equation*} Tf = C_{\Phi_T}(f) = f\circ\Phi_T\qquad (f\in\textrm{FBL}^p[E]). \end{equation*}

Proof. (i) It suffices to show that $C_{\Phi_T}(f)\in\overline{I_{w^*}[F]}$ for every $f\in I_{w^*}[E]_+$ because  $C_{\Phi_T}$ is a lattice homomorphism (in particular continuous). Take $m\in\mathbb{N}$ and $x_1,\ldots,x_m\in E$ such that $f \leqslant\sum_{j=1}^m|{\delta}_{x_j}|$. Using that  $C_{\Phi_T}$ is a lattice homomorphism together with (6.5), we obtain

\begin{equation*} 0 \leqslant C_{\Phi_T}(f) \leqslant C_{\Phi_T}\Big(\sum_{j=1}^m|{\delta}_{x_j}|\Big) = \sum_{j=1}^m| C_{\Phi_T}({\delta}_{x_j})| = \sum_{j=1}^m| T{\delta}_{x_j}|\in\textrm{FBL}^p[F]\subseteq\overline{I_{w^*}[F]}. \end{equation*}

Since $\overline{I_{w^*}[F]}$ is an ideal of  $H_{w^*}^p[F]$ and $C_{\Phi_T}(f)\in H_{w^*}^p[F]$ by (6.4), we conclude that $C_{\Phi_T}(f)\in \overline{I_{w^*}[F]}$, as desired.

(ii) Suppose that $1 \leqslant p \lt q \lt \infty$. Then $\|f\|_{\text{FBL}^q\lbrack F\rbrack}\leqslant\|f\|_{\text{FBL}^p\lbrack F\rbrack}$ for every $f\in H[F]$ by [Reference Jardón-Sánchez, Laustsen, Taylor, Tradacete and Troitsky25, Proposition 7.3] (or lemma 6.1(iii) in the special case $\Phi=f{\colon} F^*\to\mathbb{R}$). This implies that $\textrm{FBL}^p[F]\subseteq\textrm{FBL}^q[F]$ and therefore, using (6.5), we obtain

\begin{equation*} C_{\Phi_T}({\delta}_x) = T{\delta}_x\in \textrm{FBL}^p[F]\subseteq\textrm{FBL}^q[F]\qquad (x\in E). \end{equation*}

Now the conclusion follows by viewing  $C_{\Phi_T}$ as a lattice homomorphism from  $H^q[E]$ to  $H^q[F]$, as shown in (6.4).

Proof. Suppose that $T{\colon} \textrm{FBL}^p[E]\rightarrow \textrm{FBL}^p[F]$ is a lattice isomorphism with inverse $T^{-1}{\colon}\textrm{FBL}^p[F]\rightarrow \textrm{FBL}^p[E]$. Then, for $x\in E$ and $x^*\in E^*$, we have

\begin{equation*} x^*(x) = (T^{-1}(T{\delta}_x))(x^*) = ({\delta}_x\circ\Phi_T\circ\Phi_{T^{-1}})(x^*) = (\Phi_T\circ\Phi_{T^{-1}})(x^*)(x), \end{equation*}

with a similar calculation showing that $\Phi_{T^{-1}}\circ\Phi_T = I_{F^*}$; that is, the induced maps $\Phi_T{\colon} F^*\to E^*$ and $\Phi_{T^{-1}}{\colon} E^*\to F^*$ are inverses of each other, and therefore the composition operators  $C_{\Phi_T}$ and  $C_{\Phi_{T^{-1}}}$ are also inverses of each other. As we saw in (6.4) and lemma 6.7, they restrict to lattice homomorphisms between each of specified pairs, from which the conclusions follow. (In the case of the $\textrm{FBL}^q$-closure of the ideal  $I_{w^*}$ for q > p, a little extra care is required; before invoking lemma 6.7(i), we use the second part of the same result to deduce that  $C_{\Phi_T}$ is a lattice isomorphism between  $\textrm{FBL}^q[E]$ and  $\textrm{FBL}^q[F]$ with inverse  $C_{\Phi_{T^{-1}}}$.)

Proof. Since $H^p[G_1]$ and $H^p[G_2]$ are lattice isomorphic whenever G ₁ and G ₂ are linearly isomorphic Banach spaces, it suffices to show that  $H^p[E]$ and  $H^p[F]$ are lattice isomorphic for $E= {\ell}_1^m\oplus G$ and $F= {\ell}_1^n\oplus G$, where G is a Banach space and $m,n\in\mathbb{N}\setminus\{1\}$. Take a bijection $\theta{\colon} S_{{\ell}_\infty^n}\rightarrow S_{{\ell}_\infty^m}$, where $S_{{\ell}_\infty^k}$ denotes the unit sphere of ${\ell}_\infty^k = ({\ell}_1^k)^*$, as usual. (Such a bijection exists because both sets have cardinality  $\mathfrak{c}$.) Let $\Theta{\colon} {\ell}_\infty^n\rightarrow {\ell}_\infty^m$ be the positively homogeneous extension of θ, that is, $\Theta(0)=0$ and

\begin{equation*} \Theta(y^*)=\| y^*\|_\infty\,\theta\Big(\frac{y^*}{\| y^*\|_\infty}\Big) = {\bigvee}_{i=1}^n|{\delta}_{e_i}(y^*)|\,\theta\Big(\frac{y^*}{\| y^*\|_\infty}\Big)\qquad (y^*\in{\ell}^n_\infty\setminus\{0\}), \end{equation*}

where $(e_i)_{i=1}^n$ denotes the unit vector basis for  ${\ell}_1^n$.

To verify that $\Theta\in\textrm{PH}^p[{\ell}_\infty^n,{\ell}_\infty^m]$, take $(y_j^*)_{j=1}^k\subset{\ell}_\infty^n$ with $\|(y_j^\ast)_{j=1}^k\|_{p,\text{weak}}\leqslant1$ for some $k\in\mathbb{N}$. We may suppose that $y_j^*\ne 0$ for each $j\in\{1,\ldots,k\}$, and find

\begin{align*} \left\| \Big(\Theta(y_j^*)\Big)_{j=1}^k \right\|_{p,\text{weak}} &= \text{sup}_{x\in B_{{\ell}_1^m}}\bigg(\sum_{j=1}^k\Big({\bigvee}_{i=1}^n\|{\delta}_{e_i}(y^*_j)\|\Big)^p\, \left| \Big\langle \theta\Big(\frac{y^*_j}{\| y^*_j\|_\infty}\Big),x\Big\rangle\right|^p\bigg)^{\frac1p}\\ & \leqslant \bigg(\sum_{j=1}^k\Big({\bigvee}_{i=1}^n\|{\delta}_{e_i}(y^*_j)\|\Big)^p\bigg)^{\frac1p} \leqslant \| {\bigvee}_{i=1}^n|{\delta}_{e_i}|{\|}_{\textrm{FBL}^p[{\ell}_1^n]}, \end{align*}

where the first inequality follows from the fact that $\|\theta(y^*_j/\| y^*_j\|_\infty){\|}_\infty = 1$. Hence $\|\Theta\|_{\textrm{PH}^p[{\ell}_\infty^n,{\ell}_\infty^m]} \leqslant \| {\bigvee}_{i=1}^n|{\delta}_{e_i}|{\|}_{\textrm{FBL}^p[{\ell}_1^n]} \lt \infty$, so $\Theta\in\textrm{PH}^p[{\ell}_\infty^n,{\ell}_\infty^m]$.

We can obviously extend Θ to a positively homogeneous map $\Phi{\colon} F^*\to E^*$ by defining $\Phi(y^*,z^*) = (\Theta(y^*),z^*)$ for $y^*\in{\ell}_\infty^n$ and $z^*\in G^*$. This extension belongs to  $\textrm{PH}^p[F^*,E^*]$ with $\| \Phi\|_{\textrm{PH}^p[F^*,E^*]} \leqslant \| \Theta\|_{\textrm{PH}^p[{\ell}_\infty^n,{\ell}_\infty^m]} +1$ because

\begin{multline*} \| (\Phi(y_j^*,z^*_j))_{j=1}^k{\|}_{p,\text{weak}} = \| \Big(\Theta(y_j^*),z^*_j\Big)_{j=1}^k{\|}_{p,\text{weak}}\\ \leqslant \| \Big(\Theta(y_j^*)\Big)_{j=1}^k{\|}_{p,\text{weak}} + \| (z^*_j)_{j=1}^k\|_{p,\text{weak}} \leqslant \| \Theta\|_{\textrm{PH}^p[{\ell}_\infty^n,{\ell}_\infty^m]} +1 \end{multline*}

for every $k\in\mathbb{N}$ and $(y_j^*,z^*_j)_{j=1}^k\subset F^*$ with $\|(y_j^\ast,z_j^\ast)_{j=1}^k\|_{p,\text{weak}}\leqslant1$.

On the other hand, by repeating the above arguments for the inverse map $\theta^{-1}{\colon} S_{{\ell}_\infty^m}\rightarrow S_{{\ell}_\infty^n}$ instead of θ, we obtain a map $\Phi^{-1}\in\textrm{PH}^p[E^*,F^*]$ which is an inverse of Φ (as our choice of notation indicates). Therefore, lemma 6.1(ii) implies that we may regard the restrictions of  $C_\Phi$ and $C_{\Phi^{-1}}$ as lattice homomorphisms $C_\Phi{\colon} H^p[E]\to H^p[F]$ and $C_{\Phi^{-1}}{\colon} H^p[F]\to H^p[E]$, respectively. They are clearly inverses of each other, from which the conclusion follows.

Proof. (i) follows from the fact that  $\textrm{FBL}^p[E]$ is the closed sublattice of  $H^p[E]$ generated by the image of δ ^E.

(ii) Suppose that the sublattice generated by $(T\circ{\delta}^E)(E)$ is dense in  $\textrm{FBL}^p[F]$. Since  $\Phi_T$ is positively homogeneous, it suffices to verify that its restriction to  $B_{F^*}$ is injective. Take $y^*,z^*\in B_{F^*}$ such that $\Phi_T(y^*)=\Phi_T(z^*)$. By (6.2), we have

\begin{equation*} (T{\delta}_x)(y^*) = {\delta}_x(\Phi_T(y^*)) = {\delta}_x(\Phi_T(z^*)) = (T{\delta}_x)(z^*)\qquad (x\in E). \end{equation*}

This implies that $f(y^*)=f(z^*)$ for every f belonging to the sublattice generated by $(T\circ{\delta}^E)(E)$. By hypothesis, this sublattice is dense in  $\textrm{FBL}^p[F]$ with respect to the  $\textrm{FBL}^p$-norm, which according to lemma 2.1(iii) dominates the uniform norm on  $B_{F^*}$, so we conclude that $f(y^*)=f(z^*)$ for every $f\in \textrm{FBL}^p[F]$. In particular, taking $f={\delta}_y$ for $y\in F$, we see that $y^*(y) = z^*(y)$, which proves that $y^*= z^*$.

Proof. The hypothesis that F is complemented in E means that the inclusion map $J{\colon} F\hookrightarrow E$ has a bounded linear left inverse $L{\colon} E\to F$; that is, $LJ = I_F$. This implies that $(C_{L^*}\circ C_{J^*})f = f\circ J^*\circ L^* = f$ for every $f\in H[F]$, and consequently

\begin{align*} \overline{I_{w^*}[F]} = C_{L^*}(C_{J^*}(\overline{I_{w^*}[F]})) \subseteq C_{L^*}(\overline{I_{w^*}[E]}) &= C_{L^*}(\textrm{FBL}^p[E])\\ &= \overline{L}(\textrm{FBL}^p[E])\subseteq \textrm{FBL}^p[F], \end{align*}

where we have used (6.8), the hypothesis and (6.7). This completes the proof because the opposite inclusion $\textrm{FBL}^p[F]\subseteq\overline{I_{w^*}[F]}$ is always true.

Proof. Let $Q{\colon} E\rightarrow F$ denote the quotient map, and write $q{\colon} B_E\to B_F$ for its restriction to the unit balls. Then q is continuous with respect to the relative weak topologies and surjective (because E and F are reflexive), so we can define an isometric lattice homomorphism $C_q{\colon} C(B_F)\to C(B_E)$ by composition: $C_qf = f\circ q$. Hence, by [Reference Meyer-Nieberg27, Theorem 1.4.19], its adjoint $C_q^*{\colon} C(B_E)^*\to C(B_F)^*$ is a surjective, interval-preserving operator.

As previously mentioned, for every $f\in H_{w^*}^p[F]$, we can find a probability measure $\mu\in C(B_{F})^*_+$ such that (4.14) is satisfied; that is,

(6.9)\begin{equation}\vert f(x^\ast)\vert\leqslant\|f\|_{\text{FBL}^p\lbrack F\rbrack}\,f_\mu^p(x^\ast)\qquad(x^\ast\in F^\ast),\end{equation}

where we recall that the function  $f_\mu^p\in H^p[F]$ is defined by (4.13). In view of the previous paragraph, we can take $\nu\in C(B_E)_+^*$ such that $C_q^*(\nu)=\mu$. Since E is reflexive, $f_\nu^p\in H_{w^*}^p[E] = \overline{I_{w^*}[E]}$ by theorem 5.5 and the hypothesis.

For $y^*\in B_{F^*}$, the function $\psi_p(y^*)\in C(B_F)$ defined in lemma 5.4 satisfies

\begin{equation*} C_q(\psi_p(y^*))(x) = \psi_p(y^*)(q(x)) = | y^*(q(x))|^p = | (Q^*y^*)(x)|^p = \psi_p(Q^*y^*)(x) \end{equation*}

for each $x\in B_E$, so

\begin{align*} \Big(f_\mu^p(y^*)\Big)^p &= \langle \psi_p(y^*),\mu\rangle = \langle C_q(\psi_p(y^*)),\nu\rangle\\ &= \langle\psi_p(Q^*y^*),\nu\rangle = \Big(f_\nu^p(Q^*y^*)\Big)^p = \Big(C_{Q^*}(f_\nu^p)(y^*)\Big)^p. \end{align*}

This proves that $f_\mu^p = C_{Q^*}(f_\nu^p)$, which belongs to  $\overline{I_{w^*}[F]}$ by (6.8), and therefore $f\in\overline{I_{w^*}[F]}$ by (6.9) because  $\overline{I_{w^*}[F]}$ is an ideal of  $H_{w^*}^p[F]$.

Proof. We begin by showing that conditions (a) and (b) are equivalent. It is clear that (b) implies (a). Conversely, suppose that  $\Phi_T$ is linear. By theorem 6.4, its restriction  $\Phi_T{\upharpoonright_{B_{F^*}}}$ is weak*-to-weak* continuous. This implies that $\Phi_T=S^*$ for some bounded linear operator $S{\colon} E\rightarrow F$, as one can prove by adding an application of the Krein–Smulian theorem to the standard proof that a weak*-to-weak* continuous linear operator between dual Banach spaces is the adjoint of a bounded linear operator.

To see that conditions (b) and (c) are also equivalent, we observe that for $x\in E$, $y^*\in F^*$ and a bounded linear operator $S{\colon} E\rightarrow F$, we have

\begin{equation*} ({\delta}^F\circ S)(x)(y^*) = {\delta}_{Sx}(y^*) = y^*(Sx) = (S^*y^*)(x), \end{equation*}

while (6.2) implies that

\begin{equation*} (T\circ{\delta}^E)(x)(y^*) = (T{\delta}_x)(y^*) = ({\delta}_x\circ\Phi_T)(y^*) = \Phi_T(y^*)(x). \end{equation*}

Comparing these two equations, we conclude that $T\circ{\delta}^E={\delta}^F\circ S$ if and only if $\Phi_T=S^*$.

Finally, we verify that (c) and (d) are equivalent. Equation (6.6) shows that (d) implies (c), while the converse implication follows from the fact that  $\overline{S}$ is the unique lattice homomorphism satisfying (6.6).

Proof. We extend R by linearity to the dense subspace of  $F^*$ spanned by $(f_n^*)_{n\in\mathbb{N}}$ and claim that this map is bounded by  $K\,\|\Phi\|_{\text{PH}^1\lbrack F^\ast,E^\ast\rbrack}$, where K denotes the unconditional basis constant of  $(f_n^*)$. Take $(a_j)_{j=1}^m\subset\mathbb{R}$ for some $m\in\mathbb{N}$, and choose $\sigma_j=\pm1$ such that $\sigma_ja_j\geqslant0$ for each $j\in\{1,\ldots,m\}$. Then, by (1.4), we have

\begin{equation*} \|(\sigma_ja_jf_j^*)_{j=1}^m\|_{1,\text{weak}} \leqslant K\,\bigg\|\sum_{j=1}^m a_jf_j^*\bigg\|, \end{equation*}

which implies that

\begin{align*} \|\Phi\|_{\textrm{PH}^1[F^*,E^*]}\,K\,\bigg\|\sum_{j=1}^m a_jf_j^*\bigg\| &\geqslant \| (\Phi(\sigma_ja_jf_j^*))_{j=1}^m{\|}_{1,\text{weak}}\\ &=\mathop{\rm sup}_{\varepsilon_j=\pm1}\bigg\|\sum_{j=1}^m \varepsilon_j\Phi(\sigma_ja_jf_j^*)\bigg\| =\mathop{\rm sup}_{\varepsilon_j=\pm1}\bigg\|\sum_{j=1}^m \varepsilon_j\sigma_ja_j\Phi(f_j^*)\bigg\|\\ &\geqslant \bigg\|\sum_{j=1}^m a_j\Phi(f_j^*)\bigg\| = \bigg\| R\Big(\sum_{j=1}^m a_jf_j^*\Big)\bigg\|. \end{align*}

This proves the claim, and the result follows.

Proof. It suffices to show that $\Phi(y^*)=-\Phi(-y^*)$ for every unit vector $y^*\in F^*$ because Φ is positively homogeneous. Combining the triangle inequality with the fact that $\|\Phi\|_{\textrm{PH}^1[F^*,E^*]} = 1$ and (1.4), we obtain

\begin{align*} 2&\geqslant\max\|\Phi(y^*)\pm\Phi(-y^*)\| = \|(\Phi(y^*),\Phi(-y^*))\|_{1,\text{weak}}\geqslant \|(y^*,-y^*)\|_{1,\text{weak}} \end{align*}

because $\|\Phi^{-1}\|_{\textrm{PH}^1[E^*,F^*]} = 1$. Another application of (1.4) shows that

\begin{equation*} \|(y^*,-y^*)\|_{1,\text{weak}} = \max\| y^*\pm(-y^*)\| = 2, \end{equation*}

so either $\|\Phi(y^*)+\Phi(-y^*)\| = 2$ or $\|\Phi(y^*)-\Phi(-y^*)\| = 2$. Since $\|\pm\Phi(\pm y^*)\| \leqslant \| y^*\| = 1$, the strict convexity of the norm on  $E^*$ implies that $\Phi(y^*)=\Phi(-y^*)$ in the first case and $\Phi(y^*)=-\Phi(-y^*)$ in the second. However, $\Phi(y^*)=\Phi(-y^*)$ is impossible because Φ is injective and $y^*\ne0$, so we must be in the second case, which gives the desired conclusion.

Proof. This follows immediately from lemma 6.22 because the hypothesis that T is an isometric lattice isomorphism means that the induced map $\Phi_T{\colon} F^*\to E^*$ is a positively homogeneous bijection with inverse $\Phi_{T^{-1}}$, and theorem 6.4(i) ensures that $\|\Phi_T\|_{\textrm{PH}^1[F^*,E^*]} = \| T\| = 1$ and $\|\Phi_T^{-1}\|_{\textrm{PH}^1[E^*,F^*]} = \| T^{-1}\| = 1$.