Proof. Let \(e=\intoo[0]{\sum_{i=1}^n |x_i|^p}^\frac{1}{p}\in X\) and \(A=\cbr{\sum_{i=1}^n a_ix_i: (a_i)_{i=1}^n\in B_{\ell_{p^*}^n}}\). We need to show that \(e=\sup A\). To do so, consider the ideal \(I_e\) generated by \(e\) and represent it as a \(C(K)\) space. Since the equality \[\intoo[3]{\sum_{i=1}^n |t_i|^p}^\frac{1}{p}=\sup \cbr{\sum_{i=1}^n a_it_i: (a_i)_{i=1}^n\in B_{\ell_{p^*}^n}}\] holds for any choice \(t_1,\ldots,t_n\in\mathbb{R}\), and in \(C(K)\) the value at \(s\in K\) of the function \(e=\intoo[0]{\sum_{i=1}^n |x_i|^p}^\frac{1}{p}\) is given by \(\intoo[0]{\sum_{i=1}^n |x_i(s)|^p}^\frac{1}{p}\), it follows that the proposition holds in \(C(K)\). Since lattice isomorphisms preserve arbitrary suprema, the same equality holds in \(X\). ◻

Proof. Note that, by 2, for every \((a_i)_{i=1}^n\in B_{\ell_{p^*}^n}\) we have \[\sum_{i=1}^n a_i Tx_i= T\intoo[3]{\sum_{i=1}^n a_i x_i}\leq T \intoo[3]{\sum_{i=1}^n |x_i|^p}^\frac{1}{p},\] so again by 2 we conclude that \[\intoo[3]{\sum_{i=1}^n |Tx_i|^p}^\frac{1}{p}\leq T \intoo[3]{\sum_{i=1}^n |x_i|^p}^\frac{1}{p}. \qedhere\] ◻

Proof. The norm one inclusions \(\ell_1^n\subseteq \ell_p^n \subseteq \ell_\infty^n\) imply that \[\sum_{i=1}^n |x_i| \geq \intoo[3]{\sum_{i=1}^n |x_i|^p}^\frac{1}{p} \geq \bigvee_{i=1}^n |x_i|.\] Now, since the vectors \(x_1,\ldots,x_n\) are disjoint, it follows that \(\bigvee_{i=1}^n |x_i|=\sum_{i=1}^n |x_i|\), so the statement follows. ◻

Proof. Observe that the function of real variables \(f(t_1,\ldots,t_n)=\intoo{\sum_{i=1}^n |t_i|^p}^\frac{1}{p}\) satisfies \(f(s_1,\ldots,s_n)\leq f(t_1,\ldots,t_n)\) whenever \(0\leq s_i\leq t_i\) for every \(i=1,\ldots,n\), so the proposition is true in every \(C(K)\) space as well. Now, let \(e=\bigvee_{i=1}^n |y_i|\) and note that every \(x_i\) and \(y_i\) belongs to the ideal \(I_e\) generated by \(e\). Kakutani’s Representation Theorem for \(AM\)-spaces and the uniqueness of the functional calculus yield the result. ◻

Proof. Let \(e\in X\) be such that \(x_1,\ldots,x_n\in I_e\), and let us endow \(I_e\) with the principal ideal norm \(\|x\|_e=\inf\{\lambda>0:|x|\leq \lambda e\) for every \(x\in I_e\). Note that \(\|x\|\leq \lambda \|e\|\) whenever \(|x|\leq \lambda e\), so \(\|x\|\leq \|x\|_e \|e\|\). This implies that the restrictions of any functional \(x^*\in X^*\) to \(I_e\) is continuous with respect to the principal ideal norm: \[|x^*(x)|\leq \|x^*\|\|x\|\leq \|e\|\|x^*\|\|x\|_e.\]

We claim that without loss of generality we can assume \(X=(I_e,\|\cdot\|_e)\). Indeed, let \(J:C(K)\rightarrow I_e\subseteq X\) be the isometric lattice identification of \(I_e\) with a \(C(K)\) space for some Hausdorff compact space \(K\). By the above, we have that for every \(f\in C(K)\) \[\|Jf\|\leq \|e\|\|Jf\|_e =\|e\|\|f\|_\infty,\] so \(J\) is a bounded operator from \(C(K)\) into \(X\). Note that \(J^*:X^*\rightarrow C(K)^*\) behaves as the restriction operator to \(I_e\), since \(J^*x^*(f)=x^*(Jf)\) for every \(f\in C(K)\) and \(x^*\in X^*\). Moreover, since \(J\) is an interval preserving lattice homomorphism (\(J(C(K))=I_e\) is an ideal of \(X\)), \(J^*\) is again an interval preservin lattice homomorphism. In particular, both \(J\) and \(J^*\) preserve the functional calculus of \(C(K)\) and \(X^*\), respectively. Let \(f_1,\ldots,f_n\in C(K)\) be such that \(Jf_i=x_i\), and denote by \(\mu_i=J^*x_i^*\) for every \(i=1,\ldots,n\). Then \[\begin{align} \intoo[3]{\sum_{i=1}^n |x_i^*|^{p^*}}^\frac{1}{p^*} \intoo[4]{\intoo[3]{\sum_{i=1}^n |x_i|^p}^\frac{1}{p}} & = \intoo[3]{\sum_{i=1}^n |x_i^*|^{p^*}}^\frac{1}{p^*} \intoo[4]{\intoo[3]{\sum_{i=1}^n |J f_i|^p}^\frac{1}{p}} \\ = \intoo[3]{\sum_{i=1}^n |x_i^*|^{p^*}}^\frac{1}{p^*} \intoo[4]{J\intoo[3]{\sum_{i=1}^n |f_i|^p}^\frac{1}{p}} & = J^*\intoo[3]{\sum_{i=1}^n |x_i^*|^{p^*}}^\frac{1}{p^*} \intoo[4]{\intoo[3]{\sum_{i=1}^n |f_i|^p}^\frac{1}{p}}\\ = \intoo[3]{\sum_{i=1}^n |J^*x_i^*|^{p^*}}^\frac{1}{p^*} \intoo[4]{\intoo[3]{\sum_{i=1}^n |f_i|^p}^\frac{1}{p}} &=\intoo[3]{\sum_{i=1}^n |\mu_i|^{p^*}}^\frac{1}{p^*} \intoo[4]{\intoo[3]{\sum_{i=1}^n |f_i|^p}^\frac{1}{p}}. \end{align}\] Therefore, it suffices to show the statement when \(X=C(K)\). Let \(f_1,\ldots,f_n\in C(K)\) and \(\mu_1,\ldots,\mu_n \in C(K)^*\), which we can identify with measures over \(C(K)\) by Riesz Representation Theorem. Let \(\mu\in C(K)^*_+\) be a measure such that \(|\mu_i|\leq \mu\) for every \(i=1,\ldots,n\), so in particular \(\mu_i\ll \mu\), and we can apply Radon–Nikodym Theorem to obtain a function \(h_i\in L_1(\mu)\), the Radon–Nikodym derivative of \(\mu_i\) with respect to \(\mu\), such that \(\mu_i(f)=\int_K f \,d\mu_i=\int_K f h_i \,d\mu\) for every \(f\in C(K)\). Moreover, the identification of a measure in \(B_\mu\), the band generated by \(\mu\) in \(C(K)^*\), with its Radon–Nikodym derivative in \(L_1(\mu)\) is a lattice isometry. In particular, the derivative of \(\intoo{\sum_{i=1}^n |\mu_i|^{p^*}}^\frac{1}{p^*}\) corresponds to \(\intoo[3]{\sum_{i=1}^n |h_i|^{p^*}}^\frac{1}{p^*}\), so \[\begin{align} \left|\sum_{i=1}^n \mu_i^*(f_i) \right| &= \left|\sum_{i=1}^n\int_K f_i h_i \,d\mu \right| \leq \int_K \sum_{i=1}^n |f_i h_i| \,d\mu \\ & \leq \int_K \intoo[3]{\sum_{i=1}^n |h_i|^{p^*}}^\frac{1}{p^*} \intoo[3]{\sum_{i=1}^n |f_i|^p}^\frac{1}{p} \,d\mu \\ & = \intoo[3]{\sum_{i=1}^n |\mu_i|^{p^*}}^\frac{1}{p^*} \intoo[4]{\intoo[3]{\sum_{i=1}^n |f_i|^p}^\frac{1}{p}}, \end{align}\] as we wanted to show. ◻

Proof. Let \(x\in X_+\). First, assume that \(x_1,\ldots,x_n\in X\) are such that \(\intoo{\sum_{i=1}^n |x_i|^{p}}^\frac{1}{p}\leq x\). Then, by 6 it follows that \[\left|\sum_{i=1}^n x_i^*(x_i)\right| \leq \intoo[3]{\sum_{i=1}^n |x_i^*|^{p^*}}^\frac{1}{p^*} \intoo[4]{\intoo[3]{\sum_{i=1}^n |x_i|^p}^\frac{1}{p}} \leq \intoo[3]{\sum_{i=1}^n |x_i^*|^{p^*}}^\frac{1}{p^*}(x).\] To prove the reverse inequality, we apply 2 and the Riesz–Kantorovich formula (cf. [8]) to obtain \[\begin{align} \intoo[3]{\sum_{i=1}^n |x_i^*|^{p^*}}^\frac{1}{p^*}(x) &= \sup\cbr{\intoo[3]{\sum_{i=1}^n a_{i}x_i^*}: (a_{i})_{i=1}^n\in B_{\ell_p^n}}(x)\\ &= \sup\cbr{\bigvee_{j=1}^m \intoo[3]{\sum_{i=1}^n a_{ij}x_i^*}: (a_{ij})_{i=1}^n\in B_{\ell_p^n}, j=1,\ldots,m, m\in \mathbb{N}}(x)\\ & = \sup\cbr{\intoo[3]{\bigvee_{j=1}^m \intoo[3]{\sum_{i=1}^n a_{ij}x_i^*}}(x): (a_{ij})_{i=1}^n\in B_{\ell_p^n}}\\ & = \sup\cbr{\sum_{j=1}^m \intoo[3]{\sum_{i=1}^n a_{ij}x_i^*}(z_j): z_j\geq 0,\sum_{j=1}^m z_j=x, (a_{ij})_{i=1}^n\in B_{\ell_p^n}}\\ & = \sup\cbr{\sum_{i=1}^n x_i^*\intoo{\sum_{j=1}^m a_{ij}z_j}: z_j\geq 0,\sum_{j=1}^m z_j=x, (a_{ij})_{i=1}^n\in B_{\ell_p^n}}. \end{align}\] Note that the vectors \(\sum_{j=1}^m a_{ij}z_j\), \(1\leq i\leq n\) with \(z_j\geq 0,\sum_{j=1}^m z_j=x, (a_{ij})_{i=1}^n\in B_{\ell_p^n}\) satisfy \[\intoo{\sum_{i=1}^n \left|\sum_{j=1}^m a_{ij}z_j\right|^{p}}^\frac{1}{p}\leq \sum_{j=1}^m \intoo{\sum_{i=1}^n | a_{ij}z_j|^{p}}^\frac{1}{p} =\sum_{j=1}^m z_j \intoo{\sum_{i=1}^n | a_{ij}|^{p}}^\frac{1}{p} \leq \sum_{j=1}^m z_j = x,\] where the first inequality follows from the triangular inequality of the norm of \(\ell_p^n\). Therefore, we conclude that \[\intoo[3]{\sum_{i=1}^n |x_i^*|^{p^*}}^\frac{1}{p^*}(x) \leq \sup\cbr{\sum_{i=1}^n x_i^*(x_i): x_1,\ldots,x_n\in X, \intoo[3]{\sum_{i=1}^n |x_i|^{p}}^\frac{1}{p}\leq x},\] so the statement holds. ◻

Proof. \((1)\) From Young’s inequality in \(\mathbb{R}^2\) \[|s|^\theta|t|^{1-\theta}\leq \theta |s|+(1-\theta)|t|\] it follows that for every \(c>0\) we have \[\||x|^\theta|y|^{1-\theta}\|=\||c^\frac{1}{\theta}x|^\theta|c^{-\frac{1}{1-\theta}}y|^{1-\theta}\|\leq \theta c^\frac{1}{\theta}\|x\|+(1-\theta) c^{-\frac{1}{1-\theta}}\|y\|.\] In particular, evaluating at \(c=\intoo[1]{\frac{\|y\|}{\|x\|}}^{\theta(1-\theta)}\) we obtain the desired inequality.
\((2)\) Hölder’s inequality with exponent \(s=\frac{p}{\theta r}\) yields \[\intoo[3]{\sum_{i=1}^n a_i|t_i|^{r}}^\frac{1}{r}=\intoo[3]{\sum_{i=1}^n a_i^{\frac{1}{s}+\frac{1}{s^*}}|t_i|^{r\theta +r(1-\theta)}}^\frac{1}{r}\leq \intoo[3]{\sum_{i=1}^n a_i|t_i|^{p}}^\frac{\theta}{p} \intoo[3]{\sum_{i=1}^n a_i|t_i|^{q}}^\frac{1-\theta}{q}\] for every \(t_1,\ldots,t_n\in \mathbb{R}\). Therefore, the same inequality holds in \(X\), writing \(x_1,\ldots,x_n\) instead. Applying \((1)\) we obtain \((2)\). ◻

Proof. The first inequality follows from observing that for every \(t>0\) we have \[\begin{align} t\mu(\{|f|>t\})^{\frac{1}{p}} & =\mu(\{|f|>t\})^{\frac{1}{p}-\frac{1}{r}}\intoo[3]{\int_{\{|f|>t\}}t^r \,d\mu}^\frac{1}{r}\\ & \leq \mu(\{|f|>t\})^{\frac{1}{p}-\frac{1}{r}}\intoo[3]{\int_{\{|f|>t\}}|f|^r \,d\mu}^\frac{1}{r} \leq \|f\|_{L_{p,\infty}^{[r]}}. \end{align}\] In order to prove the second inequality, we start by proving that for any measurable set \(A\) with \(0<\mu(A)<\infty\) we have \[\int_A |f|^r \,d\mu \leq \frac{p}{p-r}\mu(A)^{1-\frac{r}{p}}\vvvert f\vvvert_{L_{p,\infty}}^r.\] To do so, observe first that, by the definition of the quasinorm \(\vvvert \cdot \vvvert_{L_{p,\infty}}\), it follows that \[\mu (A\cap\{|f|>\alpha\})\leq \min \{\mu(A), \alpha^{-p}\vvvert f\vvvert_{L_{p,\infty}}^p\}\] for every \(\alpha>0\). Now, we decompose the integral over \(A\) as \[\int_A |f|^r \,d\mu=\int_{A\cap\{|f|\leq\alpha\}} |f|^r \,d\mu+\int_{A\cap\{|f|>\alpha\}} |f|^r \,d\mu.\] The first term is easy to bound: \[\int_{A\cap\{|f|\leq\alpha\}} |f|^r \,d\mu\leq \alpha^r \mu(A\cap\{|f|\leq\alpha\})\leq \alpha^r \mu(A).\] In order to estimate the second term, we use a Fubini-type argument: \[\begin{align} \int_{A\cap\{|f|>\alpha\}} |f|^r \,d\mu& =\int_{\{|f|>\alpha\}} \chi_A(\omega) \int_0^{|f(\omega)|}rt^{r-1} \,dt \,d\mu(\omega)\\ & =\int_0^\infty rt^{r-1} \int_{\{|f|>\max\{t,\alpha\}\}} \chi_A(\omega) \,d\mu(\omega) \,dt\\ & =\int_0^\infty rt^{r-1} \mu(A\cap\{|f|>\max\{t,\alpha\}\}) \,dt\\ & \leq \int_0^\infty rt^{r-1} \min \{\mu(A), \max\{t,\alpha\}^{-p}\vvvert f\vvvert_{L_{p,\infty}}^p\} \,dt. \end{align}\] If \(\alpha<\mu(A)^{-\frac{1}{p}}\vvvert f\vvvert_{L_{p,\infty}}\), we can split this integral in two terms to obtain the desired inequality: \[\begin{align} \int_{A\cap\{|f|>\alpha\}} |f|^r \,d\mu& \leq \int_0^\infty rt^{r-1} \min \{\mu(A), \max\{t,\alpha\}^{-p}\vvvert f\vvvert_{L_{p,\infty}}^p\} \,dt\\ & = \int_0^{\mu(A)^{-\frac{1}{p}}\vvvert f\vvvert_{L_{p,\infty}}} rt^{r-1}\mu(A) \,dt +\int_{\mu(A)^{-\frac{1}{p}}\vvvert f\vvvert_{L_{p,\infty}}}^\infty rt^{r-1} t^{-p}\vvvert f\vvvert_{L_{p,\infty}}^p \,dt \\ &= \mu(A)^{1-\frac{r}{p}} \vvvert f\vvvert_{L_{p,\infty}}^r+\frac{r}{p-r} \mu(A)^{1-\frac{r}{p}} \vvvert f\vvvert_{L_{p,\infty}}^r = \frac{p}{p-r} \mu(A)^{1-\frac{r}{p}} \vvvert f\vvvert_{L_{p,\infty}}^r. \end{align}\] Rearranging the terms and taking the supremum over all sets \(A\) of positive finite measure we obtain the second inequality of the statement. ◻

Proof. To check the first assertion, let \(f_1,\ldots,f_n\in L_{p,\infty}(\mu)_+\) be disjoint functions. Given any measurable set \(A\) with \(0<\mu(A)<\infty\), let us denote by \(A_i=A\cap \text{supp}f_i\), \(i=1,\ldots,n\), so that \[\begin{align} &\mu(A)^{\frac{1}{p}-\frac{1}{r}} \left(\int_A \intoo[3]{\sum_{i=1}^n |f_i|}^r\,d\mu\right)^{\frac{1}{r}} = \mu(A)^{\frac{1}{p}-\frac{1}{r}} \left(\sum_{i=1}^n\int_{A_i} |f_i|^r\,d\mu\right)^{\frac{1}{r}}\\ & \leq \left(\sum_{i=1}^n \left(\frac{\mu(A_i)}{\mu(A)}\right)^{1-\frac{r}{p}} \|f\|_{L_{p,\infty}^{[r]}}^r\right)^{\frac{1}{r}} \leq \left(\sum_{i=1}^n \frac{\mu(A_i)}{\mu(A)} \right)^{\frac{p-r}{pr}} \left(\sum_{i=1}^n \|f\|_{L_{p,\infty}^{[r]}}^p\right)^{\frac{1}{p}} \leq \left(\sum_{i=1}^n \|f\|_{L_{p,\infty}^{[r]}}^p\right)^{\frac{1}{p}}, \end{align}\] where in the second inequality we have used Hölder’s inequality with exponent \(\frac{p}{r}\). Therefore, \((L_{p,\infty}(\mu),\|\cdot\|_{L_{p,\infty}^{[r]}})\) satisfies an upper \(p\)-estimate with constant one.
Now, given \(f_1,\ldots,f_n \in L_{p,\infty}(\mu)\) and a measurable set \(A\) with \(0<\mu(A)<\infty\), it follows that \[\mu(A)^{\frac{1}{p}-\frac{1}{r}} \intoo[3]{\int_A \sum_{i=1}^n|f_i|^r\, d\mu}^\frac{1}{r}= \intoo[3]{ \sum_{i=1}^n \intoo[3]{\mu(A)^{\frac{1}{p}-\frac{1}{r}}\intoo[3]{\int_A|f_i|^r\, d\mu}^\frac{1}{r}}^r}^\frac{1}{r} \leq \intoo[3]{ \sum_{i=1}^n \|f\|_{L_{p,\infty}^{[r]}}^r}^\frac{1}{r}.\] Taking the supremum over all the sets \(A\), we obtain \[\norm[3]{\intoo[3]{ \sum_{i=1}^n |f_i|^r}^\frac{1}{r}}_{L_{p,\infty}^{[r]}} \leq \intoo[3]{ \sum_{i=1}^n \|f\|_{L_{p,\infty}^{[r]}}^r}^\frac{1}{r},\] so \((L_{p,\infty}(\mu),\|\cdot\|_{L_{p,\infty}^{[r]}})\) is \(r\)-convex with constant one. ◻

Proof. Fix \(n\in \mathbb{N}\) and let \((e_k)_{k=0}^{n-1}\) be the canonical basis of \(\ell_{p,\infty}^n\). For \(j=0,\ldots,n-1\), we write \(a^{(j)}=\sum_{k=1}^n \alpha_k e_{(k+j)_n}\) for the cyclic permutations of the coefficients \(\alpha_k=k^\frac{1}{p^*}-(k-1)^\frac{1}{p^*}\), \(1\leq k\leq n\), over the basis \((e_k)_{k=0}^{n-1}\). Here, by \(i_n\) we mean the remainder of \(i\) divided by \(n\). It is clear that \(\|a^{(j)}\|_{\ell_{p,\infty}^n}=1\) for every \(j\). Moreover, \[\intoo[3]{\sum_{j=0}^{n-1} |a^{(j)}|^p}^\frac{1}{p}=\sum_{j=0}^{n-1} \intoo[3]{\sum_{k=1}^n \alpha_k^p}^\frac{1}{p} e_{j},\] so the \(p\)-sum of the vectors \(a^{(j)}\) is a multiple of the constant vector. Let \(A_n=\intoo{\sum_{k=1}^n \alpha_k^p}^\frac{1}{p}\). We claim that \(A_n\) diverges as \(n \rightarrow \infty\). More specifically, let us show that \(\alpha_k \approx k^{-\frac{1}{p}}\) uniformly on \(k\), so that \(A_n\approx \intoo{\sum_{k=1}^n k^{-1}}^\frac{1}{p}\), which diverges. Indeed, let \(f(t)=t-t^\frac{1}{p}(t-1)^\frac{1}{p^*} = t^\frac{1}{p}(t^\frac{1}{p^*}-(t-1)^\frac{1}{p^*})\), which is continuous on \([1,\infty)\) and satisfies \(f(k)=k^\frac{1}{p}\alpha_k\). We have \(0<f(f)\leq t-(t-1)=1\). Moreover, making use of L’Hôpital rule, it is easy to show that \[\lim_{t\rightarrow \infty}f(t)=\lim_{t\rightarrow \infty}\frac{t^\frac{1}{p^*}-(t-1)^\frac{1}{p^*}}{t^{-\frac{1}{p}}}=\frac{1}{p^*}.\] Therefore, \(f(t)\) is uniformly bounded from below on the interval \([1,\infty)\) by a constant \(c>0\), so \(c k^{-\frac{1}{p}} \leq \alpha_k \leq k^{-\frac{1}{p}}\) for every \(k\geq 1\).
To conclude the argument, assume that \(X\) is \(p\)-convex with constant \(K>0\), and let \(T_n:\ell_{p,\infty}^n \rightarrow X\) be lattice isomorphic embeddings with \(M=\sup_n \|T_n\|\|T_n^{-1}\|<\infty\). Since lattice homomorphisms preserve \(p\)-sums, for every \(n\) we have \[\begin{align} A_n n^\frac{1}{p} &= \norm[3]{\intoo[3]{\sum_{j=0}^{n-1} |a^{(j)}|^p}^\frac{1}{p}}_{\ell_{p,\infty}^n} \leq \|T_n^{-1}\| \norm[3]{\intoo[3]{\sum_{j=0}^{n-1} |T_na^{(j)}|^p}^\frac{1}{p}}_X \\ & \leq \|T_n^{-1}\| K \intoo[3]{\sum_{j=0}^{n-1} \|T_n a^{(j)}\|_X^p}^\frac{1}{p} \leq \|T_n\|\|T_n^{-1}\| K \intoo[3]{\sum_{j=0}^{n-1} \|a^{(j)}\|_{\ell_{p,\infty}^n}^p}^\frac{1}{p} \leq M K n^\frac{1}{p}, \end{align}\] so \(A_n\leq MK\) for every \(n\), which is false. ◻

Proof. Let \(w_i=\mu(B_i)\) and \(g_i=w_i^{-\frac{1}{p}}\chi_{B_i}\), so that \((g_i)_{i=1}^n\) is a normalized basis of \(\mathop{\mathrm{span}}\cbr{\chi_{B_i}}_{i=1}^n\). Given \(f=\sum_{i=1}^n a_i g_i\), we have that \[\begin{align} \|f\|_{L_{p,\infty}}&=\sup_{0<\mu(A)<\infty} \mu(A)^{-\frac{1}{p^*}} \sum_{i=1}^n |a_i| \frac{\mu(A\cap B_i)}{w_i^\frac{1}{p}}\\ &= \sup_{\substack{0<\mu(A)<\infty\\ A\subseteq \bigcup_i B_i}} \intoo[3]{\sum_{i=1}^n\mu(A\cap B_i)}^{-\frac{1}{p^*}} \sum_{i=1}^n |a_i| \frac{\mu(A\cap B_i)}{w_i^\frac{1}{p}}. \end{align}\]

Next, let us consider the family of functions of the form \[F(t)=(t+a)^{-\frac{1}{p^*}}(bt+c)\] defined on \([0,d]\), with \(a,b,c\geq 0\) (with the additional restriction that if \(a=0\), then \(c=0\) too) and \(d>0\). We claim that all these functions attain their maximum in the interval \([0,d]\) either at \(0\) or at \(d\). On the one hand, if \(a=0=c\), it is clear that \(F(t)=bt^\frac{1}{p}\) is increasing, so its maximum is attained at \(d\). On the other hand, if \(a>0\), then \(F\) is differentiable on \([0,d]\), so a standard optimization argument yields that either \(F'\) is non-zero in the whole interval \([0,d]\), which means that \(F\) is strictly monotone, or \(F\) has only a local minimum in \([0,d]\). In both cases, it follows that \(F\) attains its maximum at one of the extreme points of the interval.
Now that the claim has been proved, we can simplify the expression for the norm of \(f=\sum_{i=1}^n a_i g_i\). More specifically, given any measurable subset \(A\subseteq \bigcup_{i=1}^n B_i\) with finite measure, we can find a set of indices \(I\subseteq \{1,\ldots,n\}\) such that \(\bigcup_{i\in I} B_i\) approximates the norm of \(f\) better than \(A\). Indeed, fixing the parameters \(a=\sum_{i=2}^n\mu(A\cap B_i)\), \(b=\frac{|a_1|}{w_1}\), \(c= \sum_{i=2}^n |a_i| \frac{\mu(A\cap B_i)}{w_i^\frac{1}{p}}\) and \(d=w_1\), it follows that \[\intoo[3]{\sum_{i=1}^n\mu(A\cap B_i)}^{-\frac{1}{p^*}} \sum_{i=1}^n |a_i| \frac{\mu(A\cap B_i)}{w_i^\frac{1}{p}} =F(\mu(A\cap B_1))\leq \max \{F(0),F(w_1)\}=F(\theta_1w_1)\] for some \(\theta_1\in \{0,1\}\). We can inductively apply this argument to obtain a sequence \((\theta_i)_{i=1}^n\subseteq \{0,1\}\) such that \[\begin{align} \intoo[3]{\sum_{i=1}^n\mu(A\cap B_i)}^{-\frac{1}{p^*}} & \sum_{i=1}^n |a_i| \frac{\mu(A\cap B_i)}{w_i^\frac{1}{p}} \\ & \leq \intoo[3]{\theta_1 w_1+\sum_{i=2}^n\mu(A\cap B_i)}^{-\frac{1}{p^*}} \intoo[3]{\theta_1 |a_1|w_1^\frac{1}{p^*}+\sum_{i=2}^n |a_i| \frac{\mu(A\cap B_i)}{w_i^\frac{1}{p}}}\\ & \leq \intoo[3]{\sum_{i=1}^k\theta_i w_i+\sum_{i=k+1}^n\mu(A\cap B_i)}^{-\frac{1}{p^*}} \intoo[3]{\sum_{i=1}^k\theta_i |a_i|w_i^\frac{1}{p^*}+\sum_{i=k+1}^n |a_i| \frac{\mu(A\cap B_i)}{w_i^\frac{1}{p}}} \\ & \leq \intoo[3]{\sum_{i=1}^n\theta_i w_i}^{-\frac{1}{p^*}} \intoo[3]{\sum_{i=1}^n\theta_i |a_i|w_i^\frac{1}{p^*}}\\ & \leq \intoo[3]{\sum_{i\in I}w_i}^{-\frac{1}{p^*}} \intoo[3]{\sum_{i\in I } |a_i|w_i^\frac{1}{p^*}}, \end{align}\] where \(I=\{i:\theta_i=1\}\). Therefore, we conclude that \[\|f\|_{L_{p,\infty}} = \sup_{I\subseteq \{1,\ldots,n\}} \intoo[3]{\sum_{i\in I}w_i}^{-\frac{1}{p^*}} \intoo[3]{\sum_{i\in I } |a_i|w_i^\frac{1}{p^*}},\] which is precisely the norm in the weighted finite-dimensional \(\ell_{p,\infty}^n(w_i)_{i=1}^n\) of the sequence \(\sum_{i=1}^n a_i e_i\), where \(e_i=w_i^{-\frac{1}{p}}\chi_{\{i\}}\) is the normalized canonical basis. Therefore, \(\mathop{\mathrm{span}}\cbr{\chi_{B_i}}_{i=1}^n\) is lattice isometric to \(\ell_{p,\infty}^n((w_i)_{i=1}^n)\). Finally, if we denote by \(w_-=\min_i w_i\) and \(w_+=\max_i w_i\), it is easy to see that \[\intoo[2]{\frac{w_-}{w_+}}^\frac{1}{p^*}\|(a_i)_{i=1}^n\|_{\ell_{p,\infty}^n} \leq \|f\|_{L_{p,\infty}} \leq \intoo[2]{\frac{w_+}{w_-}}^\frac{1}{p^*}\|(a_i)_{i=1}^n\|_{\ell_{p,\infty}^n},\] so \(\mathop{\mathrm{span}}\cbr{\chi_{B_i}}_{i=1}^n\) is lattice isomorphic to \(\ell_{p,\infty}^n\). ◻

Proof. Let us consider first a \((p,q)\)-convex operator \(T:E\rightarrow X\), and fix \(x_1^*,\ldots, x_n^*\in X^*\). Then, for every sequence \((u_i)_{i=1}^n\in S_{\ell_p^n(E)}\), using 6 we have that \[\begin{align} \sum_{i=1}^nT^*x_i^*(u_i)&= \sum_{i=1}^nx_i^*(Tu_i)\leq \intoo[3]{\sum_{i=1}^n |x_i^*|^{q^*}}^\frac{1}{q^*} \intoo[4]{\intoo[3]{\sum_{i=1}^n |Tu_i|^q}^\frac{1}{q}}\\ & \leq \norm[3]{\intoo[3]{\sum_{i=1}^n |x_i^*|^{q^*}}^\frac{1}{q^*}} \norm[3]{\intoo[3]{\sum_{i=1}^n |Tu_i|^q}^\frac{1}{q}}\\ & \leq \norm[3]{\intoo[3]{\sum_{i=1}^n |x_i^*|^{q^*}}^\frac{1}{q^*}} K^{(p,q)}(T) \intoo[3]{\sum_{i=1}^n \|u_i\|^p}^\frac{1}{p} = K^{(p,q)}(T) \norm[3]{\intoo[3]{\sum_{i=1}^n |x_i^*|^{q^*}}^\frac{1}{q^*}}. \end{align}\] Taking the supremum over all \((u_i)_{i=1}^n\in S_{\ell_p^n(E)}\) we conclude that \(T^*\) is \((p^*,q^*)\)-concave with constant \(K^{(p,q)}(T)\).
Next, let us consider a \((p,q)\)-concave operator \(S:X\rightarrow E\) and \(u_1^*,\ldots, u_n^*\in E^*\). Then, for any \(x\in X_+\), we know from 7 that \[\intoo[3]{\sum_{i=1}^n |S^*u_i^*|^{q^*}}^\frac{1}{q^*}(x) = \sup\cbr{\sum_{i=1}^n S^*u_i^*(x_i): x_1,\ldots,x_n\in X, \intoo[3]{\sum_{i=1}^n |x_i|^{q}}^\frac{1}{q}\leq x}.\] Therefore, let us fix \(x_1,\ldots,x_n\in X\) such that \(\intoo{\sum_{i=1}^n |x_i|^{q}}^\frac{1}{q}\leq x\). Then, \[\begin{align} \sum_{i=1}^n S^*u_i^*(x_i) & = \sum_{i=1}^n u_i^*(Sx_i) \leq \intoo[3]{\sum_{i=1}^n \|u_i^*\|^{p^*}}^\frac{1}{p^*} \intoo[3]{\sum_{i=1}^n \|Sx_i\|^p}^\frac{1}{p}\\ &\leq \intoo[3]{\sum_{i=1}^n \|u_i^*\|^{p^*}}^\frac{1}{p^*} K_{(p,q)}(S) \norm[3]{\intoo[3]{\sum_{i=1}^n |x_i|^q}^\frac{1}{q}} \\ &\leq K_{(p,q)}(S) \intoo[3]{\sum_{i=1}^n \|u_i^*\|^{p^*}}^\frac{1}{p^*} \|x\|. \end{align}\] Taking the supremum over all the possible choices of \(x_1,\ldots,x_n\in X\) we conclude that \(S^*\) is \((p^*,q^*)\)-convex with constant \(K_{(p,q)}(S)\).
Now, let \(T:E\rightarrow X\) and assume that \(T^*\) is \((p^*,q^*)\)-concave. Then, \(T^{**}\) is \((p,q)\)-convex with constant \(K_{(p^*,q^*)}(T^*)\), and so is \(T^{**}J_E\), where \(J_E:E\rightarrow E^{**}\) denotes the canonical isometric embedding of \(E\) into its bidual. Since \(T^{**}J_E=J_XT\) and \(J_X\) is a lattice isometry from \(X\) into \(J_X(X)\subseteq X\), it follows from 12(7) that \(T=J_X^{-1}T^{**}J_E\) is \((p,q)\)-convex with constant \(K_{(p^*,q^*)}(T^*)\).
Similarly, if \(S:X\rightarrow E\) is such that \(S^*\) is \((p^*,q^*)\)-convex, then \(S^{**}\) is \((p,q)\)-concave with constant \(K^{(p^*q^*)}(S^*)\), and so is \(S=J_E^{-1}S^{**}J_X\). ◻

Proof. Since \(x_i^*\wedge (\sum_{j\neq i}x_j^*)=0\) for every \(i\), Riesz–Kantorovich formulas imply that there exists a vector \(y_i\in [0,x_i]\) such that \(x_i^*(x_i-y_i)+ \sum_{j\neq i}x_j^*(y_i)\leq \frac{\delta}{n}\), so \(x_i^*(y_i)\geq x_i^*(x_i)-\frac{\delta}{n}\), and \(x_k^*(y_i)\leq \sum_{j\neq i}x_j^*(y_i)\leq \frac{\delta}{n}\) for every \(k\neq i\). Put \(z_i=(y_i-\sum_{j\neq i}y_j)_+\in [0,x_i]\) for every \(i\), so that \(y_i-\sum_{j\neq i}y_j \leq z_i\leq y_i\) and thus \(x_i^*(z_i)\geq x_i^*(x_i)-\delta\) and \(x_k^*(z_i)\leq x_k^*(y_i)\leq \delta\) for every \(k\neq i\). Note that if \(i\neq k\), then \(0\leq z_i\wedge z_k\leq (y_i-y_k)_+\wedge (y_k-y_i)_+=0\), so the vectors \(z_1,\ldots,z_n\) are pairwise disjoint. ◻

Proof. We start proving that if \(X\) satisfies an upper \(p\)-estimate, then \(X^*\) satisfies a lower \(p^*\)-estimate. To do so, let \(x_1^*,\ldots,x_n^*\in X^*_+\) be disjoint non-zero functionals. Fix \(\varepsilon>0\) and choose positive \(x_1,\ldots,x_n\in B_X\) such that \(x_i^*(x_i)\geq (1-\varepsilon)\|x_i^*\|\) for every \(i=1,\ldots,n\). Applying 19 with \(\delta = \varepsilon \min_{1\leq j\leq n}\|x_j^*\|\) we can find pairwise disjoint \(z_i\in [0,x_i]\) such that \(x_i^*(z_i)\geq (1-2\varepsilon)\|x_i^*\|\). Let us consider arbitrary positive scalars \((a_i)_{i=1}^n\in B_{\ell_{p}^n}\). Since \(X\) satisfies an upper \(p\)-estimate, it follows that \[\begin{align} \sum_{i=1}^n a_i\|x_i^*\| & \leq \frac{1}{1-2\varepsilon} \sum_{i=1}^n a_i x_i^*(z_i)\leq \frac{1}{1-2\varepsilon} \sum_{i=1}^n \intoo[3]{\sum_{j=1}^n x_j^*}(a_iz_i)\\ & \leq \frac{1}{1-2\varepsilon} \norm[3]{\sum_{j=1}^n x_j^*}\norm[3]{\sum_{i=1}^na_iz_i} \leq \frac{K^{(\uparrow p)}(X)}{1-2\varepsilon} \norm[3]{\sum_{j=1}^n x_j^*}. \end{align}\] Recall that \[\intoo[3]{\sum_{i=1}^n \|x_i^*\|^{p^*}}^\frac{1}{p^*}=\sup \cbr{\sum_{i=1}^n a_i\|x_i^*\|: a_i\geq 0, (a_i)_{i=1}^n\in B_{\ell_{p}^n}},\] so \[\intoo[3]{\sum_{i=1}^n \|x_i^*\|^{p^*}}^\frac{1}{p^*}\leq \frac{K^{(\uparrow p)}(X)}{1-2\varepsilon} \norm[3]{\sum_{j=1}^n x_j^*}\] for every \(\varepsilon>0\), and therefore \(X^*\) satisfies a lower \(p^*\)-estimate with \(K_{(\downarrow p^*)}(X^*)\leq K^{(\uparrow p)}(X)\).
In particular, if \(X^*\) satisfies an upper \(p^*\)-estimate, it follows that \(X^{**}\) satisfies a lower \(p\)-estimate, and therefore \(X\) satisfies a lower \(p\)-estimate too. Moreover, \(K_{(\downarrow p)}(X)\leq K^{(\uparrow p^*)}(X^*)\).
Next, we prove that if \(X\) satisfies a lower \(p\)-estimate, then \(X^*\) satisfies an upper \(p^*\)-estimate. Again, fix disjoint non-zero functionals \(x_1^*,\ldots,x_n^*\in X^*_+\) and \(\varepsilon>0\), and choose a positive \(z\in B_X\) such that \(\|\sum_{i=1}^n x_i^*\|\leq \sum_{i=1}^n x_i^*(z) +\varepsilon\). Using 19 with \(x_i=z\) for every \(i=1,\ldots,n\) and \(\delta=\frac{\varepsilon}{n}\), we obtain pairwise disjoint vectors \(z_1,\ldots,z_n\in [0,z]\) such that \(x_i^*(z_i)\geq x_i^*(z)-\frac{\varepsilon}{n}\). Therefore, the lower \(p\)-estimate of \(X\) yields \[\begin{align} \norm[3]{\sum_{j=1}^n x_j^*} & \leq \sum_{i=1}^n x_i^*(z) +\varepsilon\leq \sum_{i=1}^n x_i^*(z_i) +2\varepsilon\leq \sum_{i=1}^n \|x_i^*\|\|z_i\| +2\varepsilon\\ & \leq \intoo[3]{\sum_{i=1}^n \|x_i^*\|^{p^*}}^\frac{1}{p^*} \intoo[3]{\sum_{i=1}^n \|z_i\|^{p}}^\frac{1}{p} + 2\varepsilon\\ & \leq K_{(\downarrow p)}(X) \intoo[3]{\sum_{i=1}^n \|x_i^*\|^{p^*}}^\frac{1}{p^*}\norm[3]{\sum_{j=1}^n z_i} + 2\varepsilon\\ & \leq K_{(\downarrow p)}(X) \intoo[3]{\sum_{i=1}^n \|x_i^*\|^{p^*}}^\frac{1}{p^*} + 2\varepsilon. \end{align}\] Taking \(\varepsilon\) arbitrarily small we recover that \(X^*\) satisfies an upper \(p^*\)-estimate with \(K^{(\uparrow p^*)}(X^*)\leq K_{(\downarrow p)}(X)\).
Finally, when \(X^*\) satisfies a lower \(p^*\)-estimate, we get that \(X^{**}\) satisfies an upper \(p\)-estimate, and it passes to \(X\). Additionally, \(K^{(\uparrow p)}(X)\leq K_{(\downarrow p^*)}(X^*)\). ◻

Proof. By 18, it suffices to prove \((1)\), since \((2)\) follows by duality.
Let \(T:E\rightarrow X\) be an operator, and note that if \(K^{(s)}(T)<\infty\), then \[\begin{align} K^{(s)}(T)&=\sup \cbr[3]{ \norm[3]{\intoo[3]{\sum_{i=1}^n |Tx_i|^{s}}^\frac{1}{s}}: n\in \mathbb{N}, (x_i)_{i=1}^n\subseteq E, \sum_{i=1}^n \|x_i\|^s\leq 1}\\ &=\sup \cbr[3]{ \norm[3]{\intoo[3]{\sum_{i=1}^n a_i|Ty_i|^{s}}^\frac{1}{s}}: n\in \mathbb{N}, (y_i)_{i=1}^n\subseteq E, \|y_i\|=1, a_i\geq 0, \sum_{i=1}^n a_i\leq 1}. \end{align}\] Applying 8(2) for any choice \((y_i)_{i=1}^n\subseteq S_E\) and scalars \(a_i\geq 0\) with \(\sum_{i=1}^n a_i\leq 1\), and taking the supremum of all such quantities, we conclude that \(K^{(s)}(T)\leq K^{(r)}(T)^\theta K^{(q)}(T)^{1-\theta}\) whenever \(1\leq r<s<q\leq \infty\) and \(\frac{1}{s}=\frac{\theta}{r}+\frac{1-\theta}{q}\), provided \(K^{(r)}(T), K^{(q)}(T)<\infty\). In particular, we know that \(K^{(1)}(T)=\|T\|\) and \(K^{(p)}(T)<\infty\), so \(K^{(s)}(T)<\infty\) for every \(1\leq s \leq p\). Moreover, since \(K^{(1)}(T)=\|T\|\leq K^{(q)}(T)\) for every \(1<q\leq p\), we see that \(K^{(s)}(T)\leq K^{(q)}(T)\) whenever \(s<q\). Finally, we observe that the function \(\varphi(\alpha)=\log K^{\intoo[0]{\frac{1}{\alpha}}}(T)\) is convex (and hence, continuous) in \(\intcc[1]{\frac{1}{p},1}\), so \(K^{(q)}(T)\) is continuous in \([1,p]\). ◻

Proof. We begin by proving \((1)\).
\((\Leftarrow)\) Let \(x_1,\ldots,x_n\in X_+\) be disjoint elements. Then \(\sum_{i=1}^n |x_i| = \bigvee_{i=1}^n |x_i|\) by 4, so the upper \(p\)-estimates inequality is just a particular instance of the \((p,\infty)\)-convexity inequality, and \(K^{(\uparrow p)}(X)\leq K^{(p,\infty)}(X)\).
\((\Rightarrow)\) Let us first assume that \(X\) is order complete, that is, that every order bounded set has a supremum in \(X\). Let \(x_1,\ldots,x_n\in X\). We can assume without loss of generality that they are non-negative. Recall that in an order complete Banach lattice every band is a projection band (cf. [8]), so let us denote by \(z_i=\intoo{\bigvee_{j\neq i}x_j - x_i}_+\) and by \(P_i=P_{z_i}\) the band projection over \(B_{z_i}\), the band generated by \(z_i\), for \(i=1,\ldots, n\). Let \(z=\bigvee_{j=1}^nx_j\), \(y_1=(i_X-P_1)z\) and \(y_i=P_1\ldots P_{i-1}(i_X-P_i)z\) for \(2\leq i\leq n\). Note that \[\bigwedge_{i=1}^n z_i= \bigwedge_{i=1}^n\intoo[3]{x_i\vee \bigvee_{j\neq i}x_j - x_i}=\bigvee_{j=1}^nx_j+\bigwedge_{i=1}^n(- x_i)=0,\] so \(P_1\ldots P_n=P_{\bigwedge_{i} z_i}=0\). Therefore, we can iteratively check that \(\sum_{i=1}^n y_i=z\). Moreover, the vectors \(y_i\) are pairwise disjoint. Indeed, let \(i<k\), and recall that band projections commute and the ranges of \(P_i\) and \(i_X-P_i\) are disjoint, so \(y_i\in (i_X-P_i)(X)\) and \(y_k\in P_i(X)\) must be disjoint. Finally, we want to check that \(y_i\in [0, x_i]\) for every \(i=1,\ldots,n\). Since \(0\leq P_j\leq i_X\) for every \(j\), it suffices to show that \((i_X-P_i)z\leq x_i\). Recall that \(P_i\) has an explicit expression given by \(P_ix=\sup_{m\in \mathbb{N}} x\wedge (mz_i)\) for any \(x\geq 0\), so \(z-x_i=z_i= z\wedge z_i\leq P_i z\), as we wanted to show. Therefore, we can apply the upper \(p\)-estimate condition to conclude that \[\norm[3]{\bigvee_{i=1}^nx_i}=\|z\|=\norm[3]{\sum_{i=1}^n y_i}\leq K^{(\uparrow p)}(X) \intoo[3]{\sum_{i=1}^n \|y_i\|^{p}}^\frac{1}{p} \leq K^{(\uparrow p)}(X) \intoo[3]{\sum_{i=1}^n \|x_i\|^{p}}^\frac{1}{p},\] and thus \(K^{(p,\infty)}(X)\leq K^{(\uparrow p)}(X)\).
Now, if \(X\) is an arbitrary Banach lattice with upper \(p\)-estimates, then by 20 \(X^{**}\) also has upper \(p\)-estimates with the same constant. Since \(X^{**}\) is order complete, by the previous part it is \((p,\infty)\)-convex. As \(X\) is a closed sublattice of \(X^{**}\), it inherits the \((p,\infty)\)-convexity.
Statement \((2)\) now follows automatically by duality (Theorems 18 and 20): \(X\) satisfies a lower \(p\)-estimate if and only if \(X^*\) satisfies an upper \(p^*\)-estimate, if and only if \(X^*\) is \((p^*,\infty)\)-convex, if and only if \(X\) is \((p,1)\)-concave, and the constant remains the same throughout the process. ◻

Proof. It suffices to prove that the left-hand side term is smaller than the right-hand side term. Since the evaluation functionals \(\delta_t\) in \(C(K)^*\), with \(t\in K\), are normalized lattice homomorphism, they preserve functional calculus, so \[\begin{align} \norm[3]{\intoo[3]{\sum_{i=1}^n |f_i|^q}^\frac{1}{q}} & =\sup_{t\in K} \intoo[3]{\sum_{i=1}^n |f_i|^q}^\frac{1}{q}(t) = \sup_{t\in K} \delta_t \intoo[3]{\intoo[3]{\sum_{i=1}^n |f_i|^q}^\frac{1}{q}}\\ & = \sup_{t\in K} \intoo[3]{\sum_{i=1}^n |\delta_t (f_i)|^q}^\frac{1}{q} \leq \sup_{\mu\in B_{C(K)^*}} \intoo[3]{\sum_{i=1}^n |\mu (f_i)|^q}^\frac{1}{q} =\|(f_i)_{i=1}^n\|_{q,w}. \qedhere \end{align}\] ◻

Proof. \((\Rightarrow)\) Let \(K\) be a compact Hausdorff space and \(S:C(K)\rightarrow X\) be a positive operator. Then, using that, by 12(7), \(TS\) is also \((p,q)\)-concave with constant \(M\|S\|\) of \(T\), so using 24 it follows that for every \(f_1,\ldots,f_n \in C(K)\) we have that \[\intoo[3]{\sum_{i=1}^n \|TSf_i\|^p}^\frac{1}{p} \leq M\|S\| \norm[3]{\intoo[3]{\sum_{i=1}^n |f_i|^q}^\frac{1}{q}} = M\|S\| \|(f_i)_{i=1}^n\|_{q,w},\] and hence \(TS\) is \((p,q)\)-summing.
\((\Leftarrow)\) Let \(x_1,\ldots,x_n\in X\), and assume without loss of generality that \(e=\intoo{\sum_{i=1}^n |x_i|^q}^\frac{1}{q}\) has norm one. We identify \(I_e\) with some \(C(K)\) space using Kakutani’s Representation Theorem for \(AM\)-spaces, and let \(j:C(K)\rightarrow I_e\subseteq X\) be the corresponding lattice injection, which has norm one, since \(\|jf\|\leq \|e\| \|jf\|_e=\|f\|_\infty\) for every \(f\in C(K)\), and satisfies \(j \mathbb{1}=e\), where \(\mathbb{1}\) denotes the constant one function. By the assumption, \(Tj\) is \((p,q)\)-summing, so if \(f_i\in C(K)\) are such that \(x_i=jf_i\), it follows that \[\intoo[3]{\sum_{i=1}^n \|Tx_i\|^p}^\frac{1}{p} =\intoo[3]{\sum_{i=1}^n \|Tjf_i\|^p}^\frac{1}{p} \leq \pi_{p,q}(Tj) \norm[3]{\intoo[3]{\sum_{i=1}^n |f_i|^q}^\frac{1}{q}}_\infty = \pi_{p,q}(Tj) \|\mathbb{1}\|_\infty \leq M\|j\| =M.\] We conclude that \(T\) is \((p,q)\)-concave. ◻

Proof. By [8], \(X\) is a KB-space if and only if \(X\) does not contain any sublattice lattice isomorphic to \(c_0\). Assume there exists a positive disjoint sequence \((x_k)_{k=1}^\infty\) that is equivalent to the canonical basis of \(c_0\). Then, there exists a constant \(C>0\) such that for every sequence \((a_k)_{k=1}^\infty\in c_0\), \[C^{-1}\sup_k |a_k|\leq \norm[3]{\sum_{k=1}^\infty a_kx_k}\leq C\sup_k |a_k|.\] In particular, the lower \(p\)-estimate of \(X\) yields that \[C^{-1} n^\frac{1}{p}\leq \intoo[3]{\sum_{k=1}^n \|x_k\|^p}^\frac{1}{p}\leq K_{(\downarrow p)}(X)\norm[3]{\sum_{k=1}^n x_k}\leq K_{(\downarrow p)}(X)C\] for every \(n\in \mathbb{N}\), which leads to a contradiction. ◻

Proof. First, we observe that the properties of \(F_\sigma\) described above, together with the fact that \(G_\frac{1}{\sigma}(-1)=-1\), imply that \(\oplus_\sigma\) is an associative and commutative operation on \(X\) that has a zero element given by the zero element of \((X,+,\cdot)\), \(0\in X\), and a opposite element for every \(x\in X\) given by the opposite element of \((X,+,\cdot)\), \(-x=(-1)\cdot x=(-1)\odot_\sigma x\). Moreover, \(F_\sigma\) is homogeneous, so in particular, if we fix \(\alpha \in R\), we have \[F_\sigma(G_\frac{1}{\sigma}(\alpha) t_1,G_\frac{1}{\sigma}(\alpha) t_2)= G_\frac{1}{\sigma}(\alpha)F_\sigma(t_1,t_2)\] for every \(t_1,t_2 \in R\). Therefore, for every \(x,y\in X\) \[(\alpha\odot_\sigma x)\oplus_\sigma (\alpha\odot_\sigma y)=F_\sigma(G_\frac{1}{\sigma}(\alpha) \cdot x,G_\frac{1}{\sigma}(\alpha) \cdot y)= G_\frac{1}{\sigma}(\alpha)\cdot F_\sigma(x,y)=\alpha \odot_\sigma (x\oplus_\sigma y),\] so the product \(\odot_\sigma\) is distributive and \((X,\oplus_\sigma, \odot_\sigma)\) is a vector space. Next, let us check that the order \(\leq\) is compatible with the linear structure. Proceeding as in the proof of 5, we can show that given \(x,y,z\in X\) with \(x\leq y\), we have \(x\oplus_\sigma z\leq y\oplus_\sigma z\). Moreover, if \(x,y\in X\) are such that \(x\leq y\) and \(\alpha \in [0,\infty)\), then \[\alpha \odot_\sigma x=|\alpha|^\frac{1}{\sigma}\cdot x\leq |\alpha|^\frac{1}{\sigma}\cdot y=\alpha \odot_\sigma y.\] Finally, the order \(\leq\) is a lattice order, with lattice operations \(\vee\) and \(\wedge\), so \((X,\oplus_\sigma, \odot_\sigma,\leq)\) is a vector lattice. ◻

Proof. \((1)\) Without loss of generality, we can suppose that \(t_2<t_1\). Assume first that \(0\leq t_2 < t_1\). If \(t_2<\frac{t_1}{2}\), then \[0\leq t_1^\frac{1}{p}-t_2^\frac{1}{p}\leq t_1^\frac{1}{p}=2^\frac{1}{p}\intoo[2]{\frac{t_1}{2}}^\frac{1}{p}\leq 2^\frac{1}{p}|t_1-t_2|^\frac{1}{p}.\] On the other hand, if \(t_2\geq\frac{t_1}{2}\), by the Mean Value Theorem there is some \(t_2<s<t_1\) such that \[|t_1^\frac{1}{p}-t_2^\frac{1}{p}|=\frac{1}{p}s^{-\frac{1}{p^*}}|t_1-t_2|=\frac{1}{p}\intoo[2]{\frac{|t_1-t_2|}{s}}^{\frac{1}{p^*}}|t_1-t_2|^\frac{1}{p}\leq \frac{1}{p}|t_1-t_2|^\frac{1}{p}.\] If \(t_2<t_1\leq 0\), the situation is similar. Finally, if \(t_2<0<t_1\), then \[|t_1^\frac{1}{p}-t_2^\frac{1}{p}|=t_1^\frac{1}{p}+|t_2|^\frac{1}{p}\leq 2^\frac{1}{p^*}(t_1+|t_2|)^\frac{1}{p}=2^\frac{1}{p^*}|t_1-t_2|^\frac{1}{p}.\]

\((2)\) Again, we can assume that \(t_2<t_1\). Since \(G_p\) is differentiable in \(\mathbb{R}\), by the Mean Value Theorem there exists \(t_2<s<t_1\) such that \[|t_1^p-t_2^p|^\frac{1}{p}=p^\frac{1}{p} |s|^\frac{p-1}{p} |t_1-t_2|^\frac{1}{p}\leq p^\frac{1}{p} (|t_1|\vee |t_2|)^\frac{1}{p^*} |t_1-t_2|^\frac{1}{p}.\qedhere\] ◻

Proof. We first prove that \(\normiii{\cdot}\) is a norm on \(X_\frac{1}{p}\). Indeed, given \(x\in X\) and \(\alpha\in \mathbb{R}\), it is clear that \(\normiii{x}=0\) if and only if \(x=0\), and \[\normiii{\alpha \odot_\frac{1}{p} x}=\|\alpha^p \cdot x\|^\frac{1}{p}=|\alpha|\|x\|^\frac{1}{p}=|\alpha|\normiii{x}.\] To prove the triangular inequality, note that for every \(t_1,t_2\in \mathbb{R}\) and \(\alpha,\beta\in (0,\infty)\) with \(\alpha^{p^*}+\beta^{p^*}=1\), we have \[|F_\frac{1}{p}(t_1,t_2)|\leq (|t_1|^\frac{1}{p}+|t_2|^\frac{1}{p})^p\leq \frac{|t_1|}{\alpha^p}+\frac{|t_2|}{\beta^p},\] therefore the same inequality holds when we replace \(t_1,t_2\in \mathbb{R}\) by \(x,y\in X\). In particular, taking \[\alpha^p=\frac{\|x\|^\frac{1}{p^*}}{(\|x\|^\frac{1}{p}+\|y\|^\frac{1}{p})^\frac{p}{p^*}}\quad \text{and} \quad \beta^p=\frac{\|y\|^\frac{1}{p^*}}{(\|x\|^\frac{1}{p}+\|y\|^\frac{1}{p})^\frac{p}{p^*}},\] we obtain \[\normiii{x\oplus_\frac{1}{p} y}=\| F_\frac{1}{p}(x,y)\|^\frac{1}{p}\leq \norm[3]{\frac{|x|}{\alpha^p}+\frac{|y|}{\beta^p}}^\frac{1}{p} \leq \intoo[3]{\frac{\|x\|}{\alpha^p}+\frac{\|y\|}{\beta^p}}^\frac{1}{p}= \normiii{x}+\normiii{y}.\] Moreover, \(\normiii{\cdot}\) is a lattice norm. Indeed, if \(|x|\leq |y|\), then \(\normiii{x}=\|x\|^\frac{1}{p}\leq \|y\|^\frac{1}{p}=\normiii{y}\). Therefore, \(X^{(p)}=(X_\frac{1}{p},\normiii{\cdot})\) is a normed vector lattice.
Finally, let us check that \(\normiii{\cdot}\) is a complete norm. Let \((x_n)_{n=1}^\infty\subseteq X\) be a Cauchy sequence in \(X^{(p)}\). In particular, it is bounded, so there is some \(M>0\) such that \(\normiii{x_n}=\|x_n\|^\frac{1}{p}\leq M\) for every \(n\in \mathbb{N}\). Our aim is to show that \((x_n)_{n=1}^\infty\) converges to some \(x\in X\) with respect to the norm \(\normiii{\cdot}\). To do so, we first prove that \((x_n)_{n=1}^\infty\) is a Cauchy sequence in \(X\) (with its usual norm). Observe that, by 28(2), for every \(t_1,t_2\in \mathbb{R}\) we have \[|t_1-t_2|=|(t_1^\frac{1}{p})^p-(t_2^\frac{1}{p})^p|\leq D_p^p(|t_1|\vee |t_2|)^\frac{1}{p^*}|(t_1^\frac{1}{p}-t_2^\frac{1}{p})^p|^\frac{1}{p}.\] In particular, using 8(1), we get that for any \(n,m\in \mathbb{N}\) \[\|x_n-x_m\|\leq D_p^p \||x_n|\vee |x_m|\|^\frac{1}{p^*}\|x_n\ominus_\frac{1}{p} x_m\|^\frac{1}{p}\leq D_p^p (2M^p)^\frac{1}{p^*} \normiii{x_n\ominus_\frac{1}{p} x_m},\] so \((x_n)_{n=1}^\infty\) is a Cauchy sequence in \(X\) for the norm \(\|\cdot\|\). Since \(X\) is a Banach lattice, there exists a vector \(x\in X\) such that \(\|x-x_n\|\rightarrow 0\) as \(n \rightarrow\infty\). By 28(1), it follows that \[|x\ominus_\frac{1}{p}x_n|=|F_\frac{1}{p}(x,-x_n)|\leq C_p |x-x_n|,\] so \[\normiii{x\ominus_\frac{1}{p}x_n}=\|x\ominus_\frac{1}{p}x_n\|^\frac{1}{p}\leq C_p^\frac{1}{p} \|x-x_n\|^\frac{1}{p} \xrightarrow[n\rightarrow \infty]{}0\] and \(X^{(p)}\) is complete. ◻

Proof. Given \(x\in X\), let us define \(\normiii{x}_0=\|x\|^p\) and \[\normiii{x}=\inf \cbr[3]{\sum_{i=1}^n \normiii{x_i}_0: n\in \mathbb{N}, (x_i)_{i=1}^n\subseteq X,|x|=\bigoplus_{i=1}^n |x_i| },\] where \(\bigoplus_{i=1}^n |x_i|= |x_1|\oplus_p\ldots\oplus_p |x_n|\). It is clear that \(\normiii{x}\leq \normiii{x}_0\). Moreover, the \(p\)-convexity of \(X\) yields that for every \((x_i)_{i=1}^n\subseteq X\) such that \(|x|=\bigoplus_{i=1}^n |x_i|\), \[\normiii{x}_0=\|x\|^p \leq K^{(p)}(X)^p \sum_{i=1}^n \|x_i\|^p,\] so taking the infimum over all possible decompositions we get that \(\normiii{x}_0\leq K^{(p)}(X)^p\normiii{x}\). Let us check that \(\normiii{\cdot}\) is a norm. Clearly, \(\normiii{x}=0\) if and only if \(\|x\|=0\), if and only if \(x=0\). Moreover, for every \(\alpha\in \mathbb{R}\) and \(x\in X\), \[\normiii{\alpha \odot_p}_0=\|\alpha^\frac{1}{p}\cdot x\|^p=|\alpha|\|x\|^p=|\alpha|\normiii{x}_0,\] and from the definition of \(\normiii{\cdot}\) we see that \(\normiii{\alpha \odot_p}=|\alpha|\normiii{x}\). Before proving the triangular inequality, let us check that if \(|x|\leq |y|\), then \(\normiii{x}\leq \normiii{y}\). Indeed, if \(|y|=\bigoplus_{i=1}^n |y_i|\) is any decomposition, then, since \(X_p\) is a vector lattice, it has the Riesz decomposition property, so there exist \((x_i)_{i=1}^n\subseteq X\) such that \(0\leq x_i\leq |y_i|\) and \(|x|=\bigoplus_{i=1}^n x_i\). In particular, since \(\|\cdot\|\), and thus \(\normiii{\cdot}_0\), is monotone, we get \[\normiii{x}\leq \sum_{i=1}^n \normiii{x_i}_0\leq \sum_{i=1}^n \normiii{y_i}_0,\] so taking the infimum over all decompositions of \(|y|\) we obtain the desired inequality. Now, in order to prove the triangular inequality, note that for every \(t_1,t_2\in \mathbb{R}\), \[|F_p(t_1,t_2)|\leq F_p(|t_1|,|t_2|),\] and for any \(s_1,\ldots,s_m,t_1,\ldots,t_n\in\mathbb{R}\), \[F_p(F_p(s_1,\ldots,s_m),F_p(t_1,\ldots,t_n))= F_p(s_1,\ldots,s_m,t_1,\ldots,t_n),\] where \(F_p(r_1,\ldots,r_k)=G_\frac{1}{p}\intoo{\sum_{i=1}^k G_p(r_i)}\) is defined for any number \(k\in \mathbb{N}\) of arguments. In particular, for any \(x, y\in X\) and any decompositions \(|x|=\bigoplus_{i=1}^m |x_i|\) and \(|y|=\bigoplus_{j=1}^n |y_j|\) we have \[\begin{align} |x\oplus_p y| &\leq |x|\oplus_p |y| =\intoo[3]{\bigoplus_{i=1}^m |x_i|}\oplus_p \intoo[3]{\bigoplus_{j=1}^n |y_j|}\\ & =|x_1|\oplus_p\ldots \oplus_p |x_m|\oplus_p |y_1|\oplus_p\ldots\oplus_p |y_n|, \end{align}\] so \[\normiii{x\oplus_p y}\leq \sum_{i=1}^m \normiii{x_i}_0+\sum_{j=1}^n \normiii{y_j}_0.\] Taking the infimum over all possible decompositions for \(x\) and \(y\) we obtain the triangular inequality. Therefore, \(X_{(p)}=(X_p,\normiii{\cdot})\) is a normed vector lattice.
Finally, let us check that the norm \(\normiii{\cdot}\) is complete. As in the proof of 29, let \((x_n)_{n=1}^\infty\subseteq X\) be a Cauchy sequence for the norm \(\normiii{\cdot}\). We will show that \((x_n)_{n=1}^\infty\) converges in \(X_{(p)}\) to some \(x\in X\). First, note that applying 28(1) we get that for every \(t_1,t_2\in \mathbb{R}\), \[|t_1-t_2|=|(t_1^p)^\frac{1}{p}-(t_2^p)^\frac{1}{p}|\leq C_p^\frac{1}{p}|t_1^p-t_2^p|^\frac{1}{p},\] so \[\|x_n-x_m\|\leq C_p^\frac{1}{p}\|x_n\ominus_p x_m\|\] for every \(n,m\in \mathbb{N}\). Thus, \((x_n)_{n=1}^\infty\) is a Cauchy sequence in \(X\) (with its usual norm). In particular, it is convergent and bounded. Let \(x\in X\) be the limit of the sequence \((x_n)_{n=1}^\infty\), and let \(M>0\) be such that \(\|x\|,\|x_n\|\leq M\) for every \(n\in \mathbb{N}\). Then, by 28(2) and 8 we get that \[\begin{align} \normiii{x\ominus_p x_n}& \leq \normiii{x\ominus_p x_n}_0=\|x\ominus_p x_n\|^p \leq D_p^p \| (|x|\vee|x_n|)^\frac{1}{p^*}|x-x_n|^\frac{1}{p}\|^p \\ & \leq D_p^p \| |x|\vee|x_n|\|^\frac{p}{p^*}\|x-x_n\| \leq D_p^p (2M)^\frac{p}{p^*}\|x-x_n\| \xrightarrow[n\rightarrow \infty]{} 0, \end{align}\] as we wanted to show. ◻

Proof. \((1)\) If \(\sigma=1\), there is nothing to prove, as \(X_1\) coincides with \(X\) as a vector lattice. If \(0<\sigma<1\), by 29 \(\normiii{\cdot}\) is the norm of the \(\frac{1}{\sigma}\)-convexification of \(\mathcal{H}^n\). If \(1<\sigma<\infty\), note that \(\mathcal{H}^n\) is \(\infty\)-convex with constant one, so by 21 \(\mathcal{H}^n\) is \(\sigma\)-convex with constant one. In particular, by 30 \(\normiii{\cdot}\) is the norm of the \(\sigma\)-concavification of \(\mathcal{H}^n\). In every case, \(\normiii{\cdot}\) is a complete lattice norm on \(X_\sigma\). Therefore, 1 provides a functional calculus operator \(U_{\overline{f}}^\sigma:\mathcal{H}\rightarrow \mathcal{H}^n_\sigma\) for every finite sequence \(\overline{f}\).
\((2)\) Note that \(\phi_\sigma\) is continuous and bijective, with inverse given by \(\phi_\frac{1}{\sigma}\), which is also continuous. Moreover, \(\phi_\sigma(S)=S\): if \((t_i)_{i=1}^n\in S\), there exists some index \(j\) such that \(|t_i|\leq 1=|t_j|\) for every \(i=1,\ldots,n\), so \(|t_i^\sigma|\leq 1=|t_j^\sigma|\). Hence, \(\phi_\sigma(t_1,\ldots,t_n)\in S\). Lastly, for every \(\alpha \geq 0\), \(\phi_\sigma(\alpha t_1,\ldots,\alpha t_n)=\alpha^\sigma \phi_\sigma(t_1,\ldots,t_n)\). Therefore, for every \(f\in \mathcal{H}^n\) the function \(Uf\) is continuous and positively homogeneous, so \(U\) is well defined. Next, we check that \(U\) is linear: \[U(\alpha \cdot f)=G_\frac{1}{\sigma}((\alpha \cdot f)\circ \phi_\sigma)= G_\frac{1}{\sigma}(\alpha \cdot (f\circ \phi_\sigma))=G_\frac{1}{\sigma}(\alpha)\cdot G_\frac{1}{\sigma}(f\circ \phi_\sigma)= \alpha \odot_\sigma Uf\] and \[U(f+g)=G_\frac{1}{\sigma}((f+g)\circ \phi_\sigma)= G_\frac{1}{\sigma}(G_\sigma\circ G_\frac{1}{\sigma}\circ f\circ \phi_\sigma +G_\sigma\circ G_\frac{1}{\sigma}\circ g\circ \phi_\sigma)=(Uf)\oplus_\sigma (Ug).\] Clearly, \[|Uf|=|f\circ\phi_\sigma|^\frac{1}{\sigma}=|f|^\frac{1}{\sigma}\circ\phi_\sigma = U|f|\] and, since \(S=\phi_\frac{1}{\sigma}(S)\), \[\normiii{Uf}=\left(\sup_S |G_\frac{1}{\sigma}\circ f\circ \phi_\sigma|\right)^\sigma = \sup_{\phi_\frac{1}{\sigma}(S)} |G_\frac{1}{\sigma}\circ f\circ \phi_\sigma|^\sigma =\sup_S | f|=\|f\|,\] so \(U\) is an isometric lattice homomorphism. Finally, \[Up_i=G_\frac{1}{\sigma}\circ p_i\circ \phi_\sigma = G_\frac{1}{\sigma}\circ G_\sigma \circ p_i = p_i,\] so \(U\) satisfies the properties that uniquely define the functional calculus associated to \(U_{\overline{p}}^\sigma:\mathcal{H}\rightarrow \mathcal{H}^n_\sigma\), according to 1. Therefore, both operators are the same. ◻

Proof. First note that \(X_\sigma=X\) and \(Y_\sigma=Y\) as sets, so the map \(T:X_\sigma\rightarrow Y_\sigma\) is well defined. Now, if \(x,y\in X\) and \(\alpha\in \mathbb{R}\), we get that \[T(\alpha\odot_\sigma x)= T(\alpha^\frac{1}{\sigma}\cdot x)=\alpha^\frac{1}{\sigma}\cdot Tx=\alpha\odot_\sigma Tx,\] and, since \(T\) preserves functional calculus, \[T(x\oplus_\sigma y)=T(F_\sigma(x,y))=F_\sigma(Tx,Ty)=Tx\oplus_\sigma Ty.\] Therefore, \(T\) is linear. Moreover, \(T\) preserves the order and the lattice operations of \(X_\sigma\) and \(Y_\sigma\), since they coincide with the ones from \(X\) and \(Y\). ◻

Proof. The statement follows from the fact that \(V_{\overline{x}}:\mathcal{H}^n_\sigma\rightarrow X_\sigma\) is by 32 a lattice homomorphism such that \(V_{\overline{x}}\circ U^\sigma_{\overline{p}}p_i=V_{\overline{x}}p_i=x_i\). Therefore, the uniqueness of functional calculus implies that \(V_{\overline{x}}^\sigma=V_{\overline{x}}\circ U^\sigma_{\overline{p}}\). ◻

Proof. Recall that the norm in \(X^{(p)}=(X_\frac{1}{p},\normiii{\cdot})\) is given by \(\normiii{\cdot}=\|\cdot\|^\frac{1}{p}\). Therefore, by 8 we get that \[\begin{align} \normiii[\Bigg]{\intoo[3]{\bigoplus_{i=1}^n |x_i|^{ps}}^\frac{1}{ps}}& = \norm[3]{\intoo[3]{\sum_{i=1}^n |x_i|^{s}}^\frac{1}{s}}^\frac{1}{p}\leq K^{(r,s)}(X)^\frac{1}{p} \intoo[3]{\sum_{i=1}^n \|x_i\|^{r}}^\frac{1}{pr}\\ & =K^{(r,s)}(X)^\frac{1}{p} \intoo[3]{\sum_{i=1}^n \normiii{x_i}^{pr}}^\frac{1}{pr} \end{align}\] and \[\begin{align} \normiii[\Bigg]{\intoo[3]{\bigoplus_{i=1}^n |x_i|^{pv}}^\frac{1}{pv}}& = \norm[3]{\intoo[3]{\sum_{i=1}^n |x_i|^{v}}^\frac{1}{v}}^\frac{1}{p}\geq K_{(u,v)}(X)^{-\frac{1}{p}} \intoo[3]{\sum_{i=1}^n \|x_i\|^{u}}^\frac{1}{pu}\\ & =K_{(u,v)}(X)^{-\frac{1}{p}} \intoo[3]{\sum_{i=1}^n \normiii{x_i}^{pu}}^\frac{1}{pu}. \qedhere \end{align}\] ◻

Proof. Recall that the norm in \(X_{(p)}=(X_\frac{1}{p},\normiii{\cdot})\) satisfies \(\normiii{\cdot}\leq \|\cdot\|^p\leq K^{(p)}(X)^p\normiii{\cdot}\). Again, by 8 we get that \[\begin{align} \normiii[\Bigg]{\intoo[3]{\bigoplus_{i=1}^n |x_i|^{\frac{s}{p}}}^\frac{p}{s}}& \leq \norm[3]{\intoo[3]{\sum_{i=1}^n |x_i|^{s}}^\frac{1}{s}}^p\leq K^{(r,s)}(X)^p \intoo[3]{\sum_{i=1}^n \|x_i\|^{r}}^\frac{p}{r}\\ & \leq K^{(p)}(X)^p K^{(r,s)}(X)^p \intoo[3]{\sum_{i=1}^n \normiii{x_i}^{\frac{r}{p}}}^\frac{p}{r} \end{align}\] and \[\begin{align} \normiii[\Bigg]{\intoo[3]{\bigoplus_{i=1}^n |x_i|^{\frac{v}{p}}}^\frac{p}{v}}& \geq K^{(p)}(X)^{-p} \norm[3]{\intoo[3]{\sum_{i=1}^n |x_i|^{v}}^\frac{1}{v}}^p\geq K^{(p)}(X)^{-p} K_{(u,v)}(X)^{-p} \intoo[3]{\sum_{i=1}^n \|x_i\|^{u}}^\frac{p}{u}\\ & \geq K^{(p)}(X)^{-p} K_{(u,v)}(X)^{-p} \intoo[3]{\sum_{i=1}^n \normiii{x_i}^{\frac{u}{p}}}^\frac{p}{u} \end{align}\] (note that all the exponents involved are greater or equal than one). If \(X\) satisfies a lower \(u\)-estimate instead, then for every pairwise disjoint vectors \(x_1,\ldots,x_n\in X_{(p)}\), we observe that they are also pairwise disjoint in \(X\), so \[\begin{align} \normiii[\Bigg]{\bigoplus_{i=1}^n |x_i|}& \geq K^{(p)}(X)^{-p} \norm[3]{\intoo[3]{\sum_{i=1}^n |x_i|^{p}}^\frac{1}{p}}^p = K^{(p)}(X)^{-p} \norm[3]{\sum_{i=1}^n |x_i|}^p \\ & \geq K^{(p)}(X)^{-p} K_{(\downarrow u)}(X)^{-p} \intoo[3]{\sum_{i=1}^n \|x_i\|^{u}}^\frac{p}{u}\\ & \geq K^{(p)}(X)^{-p} K_{(\downarrow u)}(X)^{-p} \intoo[3]{\sum_{i=1}^n \normiii{x_i}^{\frac{u}{p}}}^\frac{p}{u}.\qedhere \end{align}\] ◻

Proof. We first note that \(X\), \(X_{(p)}\) and \(Y\) coincide as sets, and the order and lattice operations are the same. Next, recall that \[\|\cdot\|_{X_{(p)}}\leq \|\cdot\|^p\leq K^{(p)}(X)^p \|\cdot\|_{X_{(p)}},\] so the norm of \(Y\) is equivalent to the norm of \(X\): \[K^{(p)}(X)^{-1}\|\cdot\|\leq \|\cdot\|_Y=\|\cdot\|_{X_{(p)}}^\frac{1}{p}\leq \|\cdot\|.\] Now, given \(\overline{x}=(x_i)_{i=1}^n\) in the set \(X\), let us denote by \(V_{\overline{x}}:\mathcal{H}_n\rightarrow X\) the functional calculus in the Banach lattice \(X\) associated to \(\overline{x}\), and by \(U^\sigma_{\overline{p}}:\mathcal{H}_n\rightarrow \mathcal{H}_n^\sigma\) the functional calculus associated to the projections \(\overline{p}=(p_i)_{i=1}^n\) on the vector lattice \(\mathcal{H}_n^\sigma\) for any \(0<\sigma<\infty\), as we did in 33. Then, we know by this same result that the functional calculus on \(X_{(p)}\), \(V^p_{\overline{x}}:\mathcal{H}_n\rightarrow X_{(p)}\), is given by \(V^p_{\overline{x}}=V_{\overline{x}}\circ U^p_{\overline{p}}\). Applying 33 again, this time to \(X_{(p)}\) and its \(p\)-convexification \(Y\), we obtain that the functional calculus on \(Y\), \((V_{\overline{x}}^p)^\frac{1}{p}:\mathcal{H}_n\rightarrow Y=(X_{(p)})^{(p)}\), is given by \((V^p_{\overline{x}})^\frac{1}{p}=V^p_{\overline{x}}\circ U^\frac{1}{p}_{\overline{p}}=V_{\overline{x}}\circ U^p_{\overline{p}}\circ U^\frac{1}{p}_{\overline{p}}=V_{\overline{x}}\), that is, it coincides with the original functional calculus on \(X\). In particular, the sum and multiplication by scalars in \(Y\) coincide with the original sum and multiplication by scalars in \(X\). In other words, the vector lattice structures in \(Y\) and \(X\) coincide. Since the norms are equivalent, the identity on \(X\) is a lattice isomorphism from \(X\) to \(Y\) with distortion \(K^{(p)}(X)\), as we wanted to show. Finally, the last statement follows from 35. ◻

Proof. Let \(X_{i(p)}\) be the \(p\)-concavification of \(X_i\) for \(i=0,1,2\). Then, by 32 we get that the maps \(T_i:X_{0(p)}\rightarrow X_{i(p)}\), \(i=1,2\), are lattice homomorphisms. Let us apply 38 to obtain a push-out \(PO\) in the category of all Banach lattices for the \(p\)-concavified diagram: \[\xymatrix{ X_{1(p)} \ar@{->}[r]^{R_1}& PO \\ X_{0(p)} \ar@{->}[r]_{T_2} \ar@{->}[u]^{T_1} & X_{2(p)} \ar@{->}[u]_{R_2} }\] If we \(p\)-convexify this diagram, we find that for every \(i=0,1,2\), \((X_{i(p)})^{(p)}=X_i\) by 37. Arguing as before, the lattice homomorphisms \(R_i:X_i\rightarrow PO^{(p)}\), \(i=1,2\), are contractive. We claim that \(PO^{(p)}\) satisfies the universal property of the statement. Consider a \(p\)-convex Banach lattice \(Y\) with constant one and lattice homomorphisms \(R_i:X_i \rightarrow Y\), \(i=1,2\), satisfying that \(R_1T_1=R_2T_2\). By \(p\)-concavifying the new diagram and applying the universal property of \(PO\), we get that there exists a unique lattice homomorphism \(\gamma : PO\rightarrow Y_{(p)}\) such that \(\|\gamma\|\leq \max_{i=1,2}\|R_i:X_{i(p)}\rightarrow Y_{(p)}\|\) and \(R_i=\gamma S_i\), \(i=1,2\). If we \(p\)-convexify back, we get a unique lattice homomorphism from \(PO^{(p)}\) to \(Y\) satisfying the properties of the statement.
Finally, if \(T_2\) was originally an isometric embedding, then it remains an isometric embedding after the \(p\)-concavification, since \(\|\cdot\|_{X_{i(p)}}=\|\cdot\|_{X_{i}}^p\). 38 then yields that \(R_1:X_{1(p)} \rightarrow PO\) is an isometric embedding, so it remains an isometry after \(p\)-convexifying. ◻

Proof. Recall from 2(3) that \(C(\Delta, L_p[0,1])\) can be identified with the \(p\)-convexification of \(C(\Delta, L_1[0,1])\), so it suffices to show that \(C(\Delta, L_1[0,1])^{(p)}\) is universal for the class of separable \(p\)-convex Banach lattices. Let \(X\) be a \(p\)-convex Banach lattice with constant 1, and consider its \(p\)-concavification \(X_{(p)}\), which is also a Banach lattice. By 40, there exists a lattice isometric embedding \(T:X_{(p)}\rightarrow C(\Delta, L_1[0,1])\). If we \(p\)-convexify both spaces, we find that \(T\) is again a lattice isometry from \(X=(X_{(p)})^{(p)}\) into \(C(\Delta, L_1[0,1])^{(p)}\), as we wanted to show. ◻

Proof. Note that if \(p=\infty\), the result is trivially true considering \(Z^0=X\) (every Banach lattice is \((\infty,q)\)-concave with constant one for every \(1\leq q\leq \infty\)) and \(\phi^0\) the identity operator on \(X\). Therefore, we assume that \(p\) (and hence, \(q\)) is finite. Given \(x\in X\), we define \[\rho(x)=\sup \cbr[3]{\intoo[3]{\sum_{i=1}^n \|Tx_i\|^{p}}^\frac{1}{p} : x_1,\ldots,x_n \in X,\intoo[3]{\sum_{i=1}^n |x_i|^{q}}^\frac{1}{q} \leq |x|}.\] Note that for every possible choice of \(x_1,\ldots,x_n \in X\), the \((p,q)\)-concavity of \(T\) yields that \[\intoo[3]{\sum_{i=1}^n \|Tx_i\|^{p}}^\frac{1}{p} \leq K_{(p,q)}(T)\norm[3]{\intoo[3]{\sum_{i=1}^n |x_i|^{q}}^\frac{1}{q}} \leq K_{(p,q)}(T)\|x\|,\] so in particular \(\rho(x)\) is finite for every \(x\in X\) and \(\rho(x)\leq K_{(p,q)}(T)\|x\|\). It is also clear that \(\|Tx\|\leq \rho(x)\). Let us show that \(\rho\) is a lattice seminorm on \(X\). It is easy to check that \(\rho\) is positively homogeneous. Moreover, if \(|x|\leq |y|\), then every finite sequence \(x_1,\ldots,x_n \in X\) such that \(\intoo{\sum_{i=1}^n |x_i|^{q}}^\frac{1}{q} \leq |x|\) can also be used to compute \(\rho(y)\), so \(\rho(x)\leq \rho(y)\). To prove the triangular inequality, we fix \(x,y\in X\), and let \(z=|x|+|y|\) and \(I_z\) the principal ideal generated by \(z\), that we can identify with a certain \(C(K)\) through an onto lattice isomorphism \(J:I_z\rightarrow C(K)\) such that \(Jz=\mathbb{1}\), the constant one function. For any choice \(z_1,\ldots,z_n\in X\) such that \(\intoo{\sum_{i=1}^n |z_i|^{q}}^\frac{1}{q}\leq |x+y|\), it is clear that \(z_1,\ldots,z_n\in I_z\). Let us denote by \(f=Jx\), \(g=Jy\) and \(h_i=Jz_i\). It follows that \(|f|+|g|=\mathbb{1}\) and \[\intoo{\sum_{i=1}^n |h_i|^{q}}^\frac{1}{q}=J\intoo{\sum_{i=1}^n |z_i|^{q}}^\frac{1}{q}\leq J|x+y|\leq Jz=\mathbb{1}.\] We construct \(f_i=|f|h_i\) and \(g_i=|g|h_i\), \(i=1,\ldots,n\), so that \[\intoo[3]{\sum_{i=1}^n |f_i|^{q}}^\frac{1}{q}=|f|\intoo[3]{\sum_{i=1}^n |h_i|^{q}}^\frac{1}{q}\leq |f|, \quad \intoo[3]{\sum_{i=1}^n |g_i|^{q}}^\frac{1}{q}=|g|\intoo[3]{\sum_{i=1}^n |h_i|^{q}}^\frac{1}{q}\leq |g|\] and \(h_i=f_i+g_i\) for every \(i=1,\ldots,n\). It follows that \(x_i=J^{-1}f_i\) and \(y_i=J^{-1}g_i\) satisfy \(\intoo{\sum_{i=1}^n |x_i|^{q}}^\frac{1}{q}\leq |x|\) and \(\intoo{\sum_{i=1}^n |y_i|^{q}}^\frac{1}{q}\leq |y|\), so \[\intoo[3]{\sum_{i=1}^n \|Tz_i\|^{p}}^\frac{1}{p} \leq \intoo[3]{\sum_{i=1}^n (\|Tx_i\|+\|Ty_i\|)^{p}}^\frac{1}{p} \leq \intoo[3]{\sum_{i=1}^n \|Tx_i\|^{p}}^\frac{1}{p} +\intoo[3]{\sum_{i=1}^n \|Ty_i\|^{p}}^\frac{1}{p} \leq \rho(x)+\rho(y).\] Since the choice of \((z_i)_{i=1}^n\) was arbitrary, we conclude that \(\rho\) satisfies the triangular inequality.
Now, let \(Z^0\) be the completion of \(X/\ker\rho\) with \(\hat{\rho}\), the norm induced by \(\rho\) on the quotient. Let \(\phi^0:X\rightarrow Z^0\) be the composition of the quotient map from \(X\) to \(X/\ker\rho\), which is an onto interval preserving lattice homomorphism, and the formal inclusion from \(X/\ker\rho\) into the completion \(Z^0\), which is an injective lattice homomorphism with dense range, so that \(\phi^0\) is an almost interval preserving lattice homomorphism with dense range, and \(\|\phi^0\|\leq K_{(p,q)}(T)\). Let us define for every equivalence class \(\phi x \in X/\ker\rho\), with \(x\in X\), the map \(S^0(\phi^0 x)=Tx\). It is well defined, since for every \(y\in X\) such that \(\phi^0 y=\phi^0 x\) we have that \(\|Tx-Ty\|\leq \rho(x-y)=0\), so \(S^0 (\phi^0 x)\) does not depend on the representative of the class. Moreover, \(\|S^0(\phi^0 x)\|=\|Ty\|\leq \rho(y)\) for every \(y\) in the class \(\phi^0 x\), so \(\|S^0(\phi^0 x)\|\leq \hat{\rho}(\phi^0 x)\) and \(S^0: X/\ker\rho\rightarrow E\) is bounded. We denote its unique extension to the completion \(Z^0\) by \(S^0\) again. The identity \(S^0 \phi^0=T\) follows from the definition.
Next, we need to show that \(Z^0\) is \((p,q)\)-concave with constant one. Let \(z_1,\ldots,z_n\in Z^0\), and fix \(\varepsilon>0\). We can find \(x_1,\ldots, x_n \in X\) such that \(\hat{\rho}(z_i-\phi^0 x_i)<\varepsilon/n^\frac{1}{p}\), and for each \(x_i\) choose \((y_{i,j})_{j=1}^{k_i}\subseteq X\) such that \[\intoo[3]{\sum_{j=1}^{k_i} |y_{i,j}|^{q}}^\frac{1}{q} \leq |x_i| \quad \text{and} \quad \hat{\rho}(\phi^0 x_i)^p=\rho(x_i)^p\leq \sum_{j=1}^{k_i} \|Ty_{i,j}\|^{p}+ \frac{\varepsilon^p}{n}.\] It follows using 5 that \(\intoo[1]{\sum_{i=1}^n \sum_{j=1}^{k_i} |y_{i,j}|^{q}}^\frac{1}{q} \leq \intoo{\sum_{i=1}^n |x_i|^{q}}^\frac{1}{q}\). Moreover, since \(\phi^0\) is a lattice homomorphism, it preserves the functional calculus, so \[\begin{align} \phi^0 \intoo[3]{\sum_{i=1}^n |x_i|^{q}}^\frac{1}{q}= \intoo[3]{\sum_{i=1}^n |\phi^0 x_i|^{q}}^\frac{1}{q} & \leq \intoo[3]{\sum_{i=1}^n |z_i|^{q}}^\frac{1}{q}+ \intoo[3]{\sum_{i=1}^n |z_i-\phi^0 x_i|^{q}}^\frac{1}{q} \\ & \leq \intoo[3]{\sum_{i=1}^n |z_i|^{q}}^\frac{1}{q}+\sum_{i=1}^n |z_i-\phi^0 x_i| . \end{align}\] To conclude the argument, we just observe that \[\begin{align} \intoo[3]{\sum_{i=1}^n \hat{\rho}(z_i)^{p}}^\frac{1}{p} & \leq \intoo[3]{\sum_{i=1}^n \hat{\rho}(\phi^0 x_i)^{p}}^\frac{1}{p} + \intoo[3]{\sum_{i=1}^n \hat{\rho}(z_i-\phi^0 x_i)^{p}}^\frac{1}{p} \leq \intoo[3]{\sum_{i=1}^n \intoo[3]{\sum_{j=1}^{k_i} \|Ty_{i,j}\|^{p}+ \frac{\varepsilon^p}{n}}}^\frac{1}{p} + \varepsilon\\ & \leq \rho \intoo[3]{\intoo[3]{\sum_{i=1}^n |x_i|^{q}}^\frac{1}{q}} + 2\varepsilon= \hat{\rho} \intoo[3]{\phi^0 \intoo[3]{\sum_{i=1}^n |x_i|^{q}}^\frac{1}{q}} + 2\varepsilon\\ & \leq \hat{\rho} \intoo[3]{ \intoo[3]{\sum_{i=1}^n |z_i|^{q}}^\frac{1}{q}} + \sum_{i=1}^n \hat{\rho}(z_i-\phi^0 x_i)+ 2\varepsilon\leq \hat{\rho} \intoo[3]{ \intoo[3]{\sum_{i=1}^n |z_i|^{q}}^\frac{1}{q}} + (2+n^\frac{1}{p^*})\varepsilon, \end{align}\] so \(Z\) is \((p,q)\)-concave with constant one.
Finally, let us show the last claim. Given \(Z\), \(\phi\) and \(S\) as in the statement, we can define the map \(\psi\) that sends every \(\phi x\in \phi(X)\subseteq Z\) to \(\phi^0 x\in Z^0\). This map is well defined and bounded. Indeed, if \(x_1,\ldots, x_n\in X\) are such that \(\intoo{\sum_{i=1}^n |x_i|^{q}}^\frac{1}{q}\leq |x|\), since \(\phi\) is a lattice homomorphism we have \[\begin{align} \intoo[3]{\sum_{i=1}^n \|Tx_i\|^{p}}^\frac{1}{p}& = \intoo[3]{\sum_{i=1}^n \|S\phi x_i\|^{p}}^\frac{1}{p} \leq \|S\| \intoo[3]{\sum_{i=1}^n \|\phi x_i\|^{p}}^\frac{1}{p} \leq \|S\| K_{(p,q)}(Z) \norm[3]{\intoo[3]{\sum_{i=1}^n |\phi x_i|^{q}}^\frac{1}{q}}\\ & \leq \|S\| K_{(p,q)}(Z) \norm[3]{\phi \intoo[3]{\intoo[3]{\sum_{i=1}^n | x_i|^{q}}^\frac{1}{q} }} \leq \|S\| K_{(p,q)}(Z) \|\phi x\|. \end{align}\] Taking the supremum over all possible choices of \(x_1,\ldots, x_n\in X\) we get that \(\hat{\rho}(\phi^0 x) = \rho (x)\leq \|S\| K_{(p,q)}(Z) \|\phi x\|\). This implies, in particular, that \(\phi^0 x=0\) whenever \(\phi x=0\), so \(\psi\) is well defined, and then the previous inequality reads \(\hat{\rho}(\psi(\phi x)) \leq \|S\| K_{(p,q)}(Z) \|\phi x\|\), so \(\psi\) is bounded. Since \(\phi\) has dense range, \(\psi\) can be extended to the whole \(Z\). It is also clear that \(\psi\) is a lattice homomorphism, and that \(\phi^0=\psi \phi\). Since \(\phi^0\) has dense range, \(\psi\) does too. Finally, for every \(x\in X\) we have \(S^0\psi (\phi x)=S^0\phi^0 x =Tx =S(\phi x)\), so we conclude that \(S=S^0\psi\). Therefore, the proof is concluded. ◻

Proof. One implication follows from 42. To prove the reverse, let \(Z\), \(\phi: X\rightarrow Z\) and \(S:Z\rightarrow E\) be as in the statement. Since \(\phi\) is positive, by 3 \[\begin{align} \intoo[3]{\sum_{i=1}^n \|Tx_i\|^{p}}^\frac{1}{p} &\leq \|S\|\intoo[3]{\sum_{i=1}^n \|\phi x_i\|^{p}}^\frac{1}{p} \leq \|S\| K_{(p,q)}(Z) \norm[3]{\intoo[3]{\sum_{i=1}^n |\phi x_i|^{q}}^\frac{1}{q}}\\ & \leq \|S\| K_{(p,q)}(Z) \norm[3]{\phi \intoo[3]{\sum_{i=1}^n |x_i|^{q}}^\frac{1}{q}} \leq \|S\| \|\phi\| K_{(p,q)}(Z) \norm[3]{ \intoo[3]{\sum_{i=1}^n |x_i|^{q}}^\frac{1}{q}} \end{align}\] for every \(x_1,\ldots, x_n \in X\), so \(T\) is \((p,q)\)-concave. ◻

Proof. \((1)\) Let us consider for every \(x\in X\) the expression from the proof of 42 when \(T\) is the identity operator on \(X\): \[\rho(x)=\sup \cbr[3]{\intoo[3]{\sum_{i=1}^n \|x_i\|^{p_2}}^\frac{1}{p_2} : x_1,\ldots,x_n \in X,\intoo[3]{\sum_{i=1}^n |x_i|^{q_2}}^\frac{1}{q_2} \leq |x|}.\] It was established that \(\rho\) is a lattice seminorm satisfying \(\rho(x)\leq K_{(p_2,q_2)}(X) \|x\|\). However, in this particular case, \(\|x\|\leq \rho(x)\), so it defines an equivalent renorming of \(X\). Moreover, for every \(x_1,\ldots, x_n\in X\), it was shown that \[\intoo[3]{\sum_{i=1}^n \rho(x_i)^{p_2}}^\frac{1}{p_2} \leq \rho \intoo[3]{ \intoo[3]{\sum_{i=1}^n |x_i|^{q_2}}^\frac{1}{q_2}},\] so \((X,\rho)\) is \((p_2,q_2)\)-concave with constant one.
Now, let us fix \(x_1,\ldots, x_n\in X\) and \(\varepsilon>0\), and find \(y_1,\ldots, y_m\in X\) such that \[\intoo[3]{\sum_{j=1}^m |y_j|^{q_2}}^\frac{1}{q_2} \leq \intoo[3]{\sum_{i=1}^n |x_i|^{q_1}}^\frac{1}{q_1} \quad \text{and} \quad \rho \intoo[3]{\intoo[3]{\sum_{i=1}^n |x_i|^{q_1}}^\frac{1}{q_1}} \leq \intoo[3]{\sum_{j=1}^m \|y_j\|^{p_2}}^\frac{1}{p_2} + \varepsilon.\] Define \(e= \intoo{\sum_{i=1}^n |x_i|^{q_1}}^\frac{1}{q_1}\), and let us represent \(I_e\), the principal ideal generated by \(e\), as a \(C(K)\)-space using Kakutani’s Representation Theorem for \(AM\)-spaces. Under this lattice isomorphism, \(e\) corresponds to \(\mathbb{1}\), the constant one function over \(K\), and the vectors \(x_i\) and \(y_j\) can be identified with some \(f_i\) and \(g_j\) in \(C(K)\) such that \[\intoo[3]{\sum_{j=1}^m |g_j|^{q_2}}^\frac{1}{q_2} \leq \intoo[3]{\sum_{i=1}^n |f_i|^{q_1}}^\frac{1}{q_1} =\mathbb{1}.\] Let \(h_{ij}=f_ig_j\), which satisfy that \[\intoo[3]{\sum_{i=1}^n |h_{ij}|^{q_1}}^\frac{1}{q_1} = |g_j|\intoo[3]{\sum_{i=1}^n |f_i|^{q_1}}^\frac{1}{q_1} = |g_j| \quad \text{and} \quad \intoo[3]{\sum_{j=1}^m |h_{ij}|^{q_2}}^\frac{1}{q_2} = |f_i|\intoo[3]{\sum_{j=1}^m |g_j|^{q_2}}^\frac{1}{q_2} \leq |f_i|\] for every \(j=1,\ldots, m\) and \(i=1,\ldots,n\), respectively. Let \(z_{ij}\in I_e\) be the corresponding image of \(h_{ij}\). Since \(I_e\) and \(C(K)\) are lattice isomorphic, \[\intoo[3]{\sum_{i=1}^n |z_{ij}|^{q_1}}^\frac{1}{q_1} = |y_j| \quad \text{and} \quad \intoo[3]{\sum_{j=1}^m |z_{ij}|^{q_2}}^\frac{1}{q_2} \leq |x_i|.\] Therefore, applying the \((p_1,q_1)\)-convexity of \(X\), Minkowski’s (integral) inequality with exponent \(\frac{p_2}{p_1}\) and the definition of \(\rho\), we obtain that \[\begin{align} \rho \intoo[3]{ \intoo[3]{\sum_{i=1}^n |x_i|^{q_1}}^\frac{1}{q_1}} -\varepsilon& \leq \intoo[3]{\sum_{j=1}^m \|y_j\|^{p_2}}^\frac{1}{p_2} = \intoo[3]{\sum_{j=1}^m \norm[3]{\intoo[3]{\sum_{i=1}^n |z_{ij}|^{q_1}}^\frac{1}{q_1} }^{p_2}}^\frac{1}{p_2}\\ & \leq K^{(p_1,q_1)}(X) \intoo[3]{\sum_{j=1}^m \intoo[3]{\sum_{i=1}^n \|z_{ij}\|^{p_1}}^\frac{p_2}{p_1} }^\frac{1}{p_2}\\ & \leq K^{(p_1,q_1)}(X) \intoo[3]{\sum_{i=1}^n \intoo[3]{\sum_{j=1}^m \|z_{ij}\|^{p_2}}^\frac{p_1}{p_2} }^\frac{1}{p_1}\\ & \leq K^{(p_1,q_1)}(X) \intoo[3]{\sum_{i=1}^n \rho(x_i)^{p_1} }^\frac{1}{p_1}, \end{align}\] so \((X,\rho)\) is \((p_1,q_1)\)-convex with constant \(K^{(p_1,q_1)}(X)\), as we wanted to show.
\((2)\) It follows from applying \((1)\) twice. First, we apply it to \(X\), obtaining \(Y=(X,\rho)\). Next, we apply it to \(Y^*\), which, by 18, is \((p_1^*,q_1^*)\)-concave with constant \(K^{(p_1,q_1)}(X)\) and \((p_2^*,q_2^*)\)-convex with constant one. We then obtain an isomorphic Banach lattice \(Z\) whose \((p_1^*,q_1^*)\)-concavity and \((p_2^*,q_2^*)\)-convexity constants are both one. Taking duals again, it is clear that \(X^{**}\) is lattice isomorphic to \(Z^*\), so the result follows. ◻

Proof. \((3)\Rightarrow (1) \Rightarrow (2)\) Trivial.
\((2)\Rightarrow (3)\) Recall from 22 that upper and lower \(p\)-estimates are equivalent to \((p,\infty)\)-convexity and \((p,1)\)-concavity, respectively. Therefore, if \(Y\) denotes the equivalent renorming of \(X\) given by 45, then \(Y\) satisfies an upper and a lower \(p\)-estimate with constant one, and hence its norm is \(p\)-additive on disjoint elements. If \(p<\infty\), by Kakutani’s Representation Theorem for \(AL_p\)-spaces (cf. [8]), it follows that \(Y\) is lattice isometric to an \(L_p(\mu)\) space for some measure \(\mu\), as we wanted to show. If \(p=\infty\), we observe that \(Y^*\) satisfies an upper and a lower \(1\)-estimate with constant one, so it is lattice isometric to an \(L_1(\mu)\)-space. Consequently, \(X\) is lattice isomorphic to a sublattice of \(Y^{**}\), which is a \(C(K)\)-space, so the proof is concluded. ◻

Proof. Let us consider the adjoint operator \(T^*:X^*\rightarrow E^*\), which is \((p^*,q^*)\)-concave with \(K_{(p^*,q^*)}(T^*)=K^{(p,q)}(T)\) by 18. As in the proof of 42, we define for every \(x^*\in X^*\) the lattice seminorm \[\rho(x^*)=\sup \cbr[3]{\intoo[3]{\sum_{i=1}^n \|T^*x^*_i\|^{p^*}}^\frac{1}{p^*} : x^*_1,\ldots,x^*_n \in X^*,\intoo[3]{\sum_{i=1}^n |x^*_i|^{q^*}}^\frac{1}{q^*} \leq |x^*|}.\] Recall that \(\|T^*x^*\|\leq\rho(x^*)\leq K^{(p,q)}(T)\|x^*\|\) for every \(x^*\in X^*\). Let us define \[A=\{x^*\in X^*:\rho(x^*)\leq 1\},\] which is a convex balanced solid subset of \(X^*\), and consider its polar set \[A^\circ=\{x\in X: x^*(x)\leq 1 \,\,\forall x^*\in A\}\subseteq X\] (throughout the proof, we will take polars with respect to the pair \(\langle X^*,X\rangle\)). Let \(\tau\) be the Minkowski of \(A^\circ\), which can also be computed using the expression \[\tau(x)=\sup \{x^*(x):x^*\in A\} \quad \text{ for every }\quad x\in X.\] We claim that \(\tau\) is a complete lattice norm in \(Z_0=\mathop{\mathrm{span}}A^\circ\). Clearly, \(\tau\) is positively homogeneous and satisfies the triangular inequality. Next, note that if \(x\geq 0\), \[\tau(x)=\sup\{x^*(x):x^*\in X^*_+\cap A\}.\] From this we get that \(\tau(x)\leq\tau(y)\) whenever \(0\leq x\leq y\). Additionally, using Riesz–Kantorovich formulas and the fact that \(A\) is solid we can show that \(\tau(x)= \tau(|x|)\): \[\begin{align} \tau(|x|)&=\sup \{x^*(|x|):x^*\in X^*_+\cap A\}= \sup \{|y^*(x)|:0\leq |y^*|\leq x^*, x^*\in A\}\\ &= \sup \{y^*(x):y^*\in A\} = \tau(x). \end{align}\] We conclude that \(\tau(x)\leq \tau(y)\) when \(|x|\leq |y|\). Since \(B_{X^*}\subseteq K^{(p,q)}(T) A\), for every \(x\in Z_0\) we get that \[\|x\|\leq K^{(p,q)}(T) \tau(x).\] In particular, \(\tau(x)=0\) if and only if \(x=0\). From all these facts, we conclude that \(Z_0\) is a (not necessarily closed) ideal of \(X\) and that \(\tau\) is a lattice norm in \(Z_0\). It remains to show that \(\tau\) is complete. To this end, note that \(A^\circ\) is closed in \(X\). Now, take a sequence \((x_n)_n\subseteq Z_0\) such that \(\sum_n \tau(x_n) <1\). In particular, \(\sum_n \|x_n\| < K^{(p,q)}(T)\), so \(\sum_{n=1}^m x_n\) converges in \(X\) to some \(x\). By the definition of the Minkowski functional, we can find scalars \((\lambda_n)_n\in \mathbb{R}_+\) so that \(x_n \in \lambda_n A^\circ\) for all \(n\) and \(\sum_n \lambda_n <1\). Since \[\sum^m_{n=1}x_n \in \sum^m_{n=1}\lambda_n\cdot A^\circ \subseteq A^\circ\] for all \(m\) and \(A^\circ\) is closed, \(x\in A^\circ\). Now, if \(k >m\), \[\sum^k_{n=m+1}x_n \in \sum^k_{n=m+1}\lambda_n\cdot A^\circ \subseteq \sum^\infty_{n=m+1}\lambda_n\cdot A^\circ.\] Taking the limit \(k\to \infty\) in \(X\), we see that \(x - \sum^m_{n=1}x_n\in \sum^\infty_{n=m+1}\lambda_n\cdot A^\circ\). Hence, \(\rho(x - \sum^m_{n=1}x_n)\to 0\) as \(m\to \infty\), so \(\rho\) is complete and \((Z_0,\tau)\) is a Banach lattice.
Now, consider the formal identity \(\phi_0:Z_0\rightarrow X\), which is an injective interval preserving lattice homomorphism with \(\phi_0(B_{Z_0})=A^\circ\), which is closed. In other words, \(\phi_0\) is a class \(\mathcal{C}\) operator. Moreover, \(\|\phi_0\|\leq K^{(p,q)}(T)\). Next, we note that \(T(E)\subseteq Z_0\) and \(\tau(Tu)\leq \|u\|\) for every \(u\in E\). Indeed, given \(x^*\in A\), \[|x^*(Tu)|=|T^*x^*(u)|\leq \|T^*x^*\|\|u\|\leq \rho(x^*)\|u\|\leq \|u\|.\] Therefore, the operator \(S_0:E\rightarrow Z_0\) given by \(S_0u=Tu\) is well defined and bounded, with \(\|S_0\|\leq 1\), and \(T\) factors as \(T=\phi_0S_0\).
Next, we need to show that \(Z_0\) is \((p,q)\)-convex with constant one. Let us fix \((x_i)_{i=1}^n\subseteq Z_0\), \(x^*\in A\) and \(\varepsilon>0\). By 7 (applied in \(X^*\) instead of \(X\)), we can find \((x_i^*)_{i=1}^n\subseteq X^*\) such that \[\intoo[3]{\sum_{i=1}^n |x^*_i|^{q^*}}^\frac{1}{q^*} \leq |x^*|\quad \text{and} \quad |x^*|\intoo[3]{\intoo[3]{\sum_{i=1}^n |x_i|^{q}}^\frac{1}{q}}\leq \sum_{i=1}^n |x_i^*|(|x_i|)+\varepsilon.\] In the proof of 42 it was proven that \(\rho\) satisfies the \((p^*,q^*)\)-concavity inequality, so we know that \[\intoo[3]{\sum_{i=1}^n \rho(x^*_i)^{p^*}}^\frac{1}{p^*} \leq \rho \intoo[3]{\intoo[3]{\sum_{i=1}^n |x^*_i|^{q^*}}^\frac{1}{q^*}}\leq \rho (x^*)\leq 1.\] Therefore, we obtain that \[\begin{align} |x^*|\intoo[3]{\intoo[3]{\sum_{i=1}^n |x_i|^{q}}^\frac{1}{q}}&\leq \sum_{i=1}^n |x_i^*|(|x_i|)+\varepsilon\leq \sum_{i=1}^n \rho(x_i^*)\tau(x_i)+\varepsilon\\ &\leq \intoo[3]{\sum_{i=1}^n \rho(x^*_i)^{p^*}}^\frac{1}{p^*} \intoo[3]{\sum_{i=1}^n \tau(x_i)^{p}}^\frac{1}{p}+\varepsilon\leq \intoo[3]{\sum_{i=1}^n \tau(x_i)^{p}}^\frac{1}{p}+\varepsilon. \end{align}\] Making \(\varepsilon\) tend to \(0\) and taking the supremum over all \(x^*\in A\) we conclude that \[\tau \intoo[3]{\intoo[3]{\sum_{i=1}^n |x_i|^{q}}^\frac{1}{q}}\leq \intoo[3]{\sum_{i=1}^n \tau(x_i)^{p}}^\frac{1}{p},\] as we wanted to show.
To conclude the proof, let \(Z\) be a \((p,q)\)-convex Banach lattice, \(\phi:Z\rightarrow X\) a class \(\mathcal{C}\) operator and \(S:E\rightarrow Z\) an operator, so that \(T=\phi S\). Let us write \(K=K^{(p,q)}(Z)\) and, without loss of generality, assume that \(S\) is contractive. We claim that \((K\phi(B_Z))^\circ \subseteq A\). Indeed, assume that \(x^*\notin A\), i.e., \(\rho(x^*)>1\). Then, there exists a sequence \((x_i^*)_{i=1}^n\subseteq X^*\) such that \[\intoo[3]{\sum_{i=1}^n |x^*_i|^{q^*}}^\frac{1}{q^*} \leq |x^*|\quad \text{but} \quad \intoo[3]{\sum_{i=1}^n \|T^*x^*_i\|^{p^*}}^\frac{1}{p^*} >1.\] We can find \((u_i)_{i=1}^n\subseteq E\) such that \(\intoo{\sum_{i=1}^n \|u_i\|^{p}}^\frac{1}{p}\leq 1\) and \[\begin{align} 1&< \sum_{i=1}^n T^*x^*_i(u_i)=\sum_{i=1}^n x^*_i(Tu_i)\leq \intoo[3]{\sum_{i=1}^n |x^*_i|^{q^*}}^\frac{1}{q^*}\intoo[3]{\intoo[3]{\sum_{i=1}^n |Tx_i|^{q}}^\frac{1}{q}}\\ &\leq |x^*|\intoo[3]{\intoo[3]{\sum_{i=1}^n |Tx_i|^{q}}^\frac{1}{q}} = |x^*|\intoo[3]{\phi \intoo[3]{\sum_{i=1}^n |Sx_i|^{q}}^\frac{1}{q}}, \end{align}\] where we have used that \(\phi\) is a lattice homomorphism, combined with 6. Note that the \((p,q)\)-convexity of \(Z\) and the contractivity of \(S\) yield that \(\intoo{\sum_{i=1}^n |Sx_i|^{q}}^\frac{1}{q} \in KB_Z\), so \(\phi \intoo{\sum_{i=1}^n |Sx_i|^{q}}^\frac{1}{q} \in K\phi(B_Z)\). It follows that \(x^*\notin (K^{(p,q)}(Z) \phi(B_Z))^\circ\), as we claimed. Taking polars and noting that \(\phi(B_Z)\) is closed and convex, so \(\phi(B_Z) = (\phi(B_Z))^{\circ\circ}\), we obtain that \[\phi_0(B_{Z_0})=A^\circ \subseteq (K \phi(B_Z))^{\circ\circ} =K \phi(B_Z).\] Thus, for every \(x\in Z_0\), there exists \(z\in K \|x\| B_Z\) such that \(\phi_0 x=\phi z\). We define \(\psi:Z_0\rightarrow Z\) as \(\psi x= z\). Clearly, \(\psi\) is an injective interval preserving lattice homomorphism that satisfies \(\phi_0=\phi \psi\) and \(S=\psi S_0\). Finally, if \((x_n)_n\subseteq B_{Z_0}\) is a sequence such that \(\psi x_n\rightarrow z\), it follows that \(\phi_0 x_n=\phi\psi x_n\rightarrow \phi z\). Since \(\phi_0(B_{Z_0})\) is closed, we conclude that \(\phi z=\phi_0x\) for some \(x\in B_{Z_0}\). Therefore, \(z=\psi x\) and \(\psi(B_{Z_0})\) is closed in \(Z\), so \(\psi\) is of class \(\mathcal{C}\) and the proof is concluded. ◻

Proof. Again, one implication is just 47. To obtain the converse, assume that \(Z\), \(\phi\) and \(S\) are as in the statement. Since \(\phi\) is positive, by 3 \[\begin{align} \norm[3]{ \intoo[3]{\sum_{i=1}^n |Tu_i|^{q}}^\frac{1}{q}} & \leq \norm[3]{ \phi \intoo[3]{\sum_{i=1}^n |Su_i|^{q}}^\frac{1}{q}}\leq \|\phi\| \norm[3]{ \intoo[3]{\sum_{i=1}^n |Su_i|^{q}}^\frac{1}{q}}\\ & \leq \|\phi\| K^{(p,q)}(Z) \intoo[3]{\sum_{i=1}^n \|Su_i\|^{p}}^\frac{1}{p} \leq \|\phi\| \|S\| K^{(p,q)}(Z) \intoo[3]{\sum_{i=1}^n \|u_i\|^{p}}^\frac{1}{p} \end{align}\] for every \(u_1,\ldots, u_n \in E\), so \(T\) is \((p,q)\)-convex. ◻

Proof. It suffices to show the following claim:
Claim: Every contractive lattice homomorphism \(T: Y\rightarrow Z\), where \(Y\) is \(p\)-convex and \(Z\) is \(p\)-concave, both with constant one, factors through \(L_p(\mu)\) for some measure, with the factors being contractive lattice homomorphisms.
Indeed, by Corollaries 44 and 49 there exists a \(p\)-convex Banach lattice \(Y\) and a \(p\)-concave Banach lattice \(Z\), both with constant one, lattice homomorphisms \(\psi:Y\rightarrow X\) and \(\phi:X\rightarrow Z\) and operators \(R:E\rightarrow Y\) and \(S: Z\rightarrow F\), such that the following diagram commutes, where \(T=\phi \psi\): \[\xymatrix{ E \ar@{->}[rd]_{R} \ar@{->}[rr]^{U} & & X \ar@{->}[rd]_{\phi} \ar@{->}[rr]^{V} & & F \\ & Y \ar@{->}[ru]_{\psi} \ar@{->}[rr]_{T} & & Z \ar@{->}[ru]_{S} & }\] Moreover, by replacing the norm in \(Y\) by \(K^{(p)}(U)\) times the same norm, we get that \(\|R\|\leq K^{(p)}(U)\) and \(\psi\) is contractive. Similarly, we can assume that \(\|S\|\leq K_{(p)}(V)\) and \(\phi\) is contractive. Therefore, we just need to factor the contractive lattice homomorphism \(T=\phi \psi\) through some \(L_p(\mu)\) space.
Proof of the claim. First we need to reduce \(T:Y\rightarrow Z\) to an injective lattice homomorphism. It is clear that \(\ker T\) is a closed ideal of \(Y\), so \(\widetilde{Y}=Y/\ker T\) is a Banach lattice and \(T\) can be factored as \(T=\widetilde{T} Q\), where \(Q:Y\rightarrow \widetilde{Y}\) is the canonical lattice quotient and \(\widetilde{T}:\widetilde{Y}\rightarrow Z\) is an injective and contractive lattice homomorphism. Moreover, \(\widetilde{Y}\) is \(p\)-convex with constant one. Next, we apply a \(p\)-concavification procedure (see 30) to both \(\widetilde{Y}\) and \(Z\). Since \(\widetilde{Y}\) is \(p\)-convex with constant one, \(\widetilde{Y}_{(p)}\) is a Banach lattice with the norm \(\normiii{\cdot}=\normiii{\cdot}_0=\|\cdot\|^p\). In the case of \(Z\), we cannot find a priori a lattice norm in the vector lattice \(Z_p\) due to the lack of convexity assumptions on \(Z\). However, \(\normiii{\cdot}=\|\cdot\|^p\) induces a lattice quasi-norm in \(Z_p\), meaning that it is positively homogeneous, it vanishes only at zero, \(\normiii{x}\leq \normiii{z}\) for every \(x,z\in Z\) with \(|x|\leq |z|\), and \[\normiii{x\oplus_p z}\leq 2^\frac{p}{p^*}(\normiii{x}+\normiii{z})\] for every \(x,z\in Z\). Moreover, \(\normiii{\cdot}\) is complete, and the \(p\)-concavity of \(Z\) implies that \(\normiii{\cdot}\) satisfies that for every \(z_1,\ldots, z_n\in Z\) \[\normiii{|z_1|\oplus_p\ldots\oplus_p |z_n|}=\norm[3]{\intoo[3]{\sum_{i=1}^n |z_i|^{p}}^\frac{1}{p}}^p\geq \sum_{i=1}^n \|z_i\|^{p}= \sum_{i=1}^n \normiii{z_i}.\] By 32, \(\widetilde{T}:\widetilde{Y}_{(p)}\rightarrow (Z_p,\normiii{\cdot})\) is lattice homomorphism with norm \(\normiii{\widetilde{T}}=\|\widetilde{T}\|^p\leq 1\). We define in \(\widetilde{Y}_{(p)}\) the sets \[C_1=\{y\in \widetilde{Y}_{(p)}: \normiii{y}<1\}\quad \text{and}\quad C_2= \{y\in \widetilde{Y}_{(p)}: \widetilde{T}y\geq 0, \normiii{\widetilde{T}y}\geq \normiii{\widetilde{T}}\}.\] \(C_1\) is clearly convex and open, as it is the open unit ball of \(\widetilde{Y}_{(p)}\), and \(C_2\) is convex due to the \(1\)-concavity of \(\normiii{\cdot}\) in \(Z_p\). Moreover, \(C_1\cap C_2=\varnothing\), so by Hahn–Banach Theorem there exists a functional \(y^*\in (\widetilde{Y}_{(p)})^*\) such that \(y^*(y)\leq 1\) for every \(y\in C_1\) and \(y^*(y)\geq 1\) for every \(y\in C_2\). From this we learn that \(y^*\in B_{(\widetilde{Y}_{(p)})^*}\), and that \(y^*\) is positive. Indeed, if \(0<y\in \widetilde{Y}_{(p)}\), then \(\widetilde{T}y>0\), since \(\widetilde{T}\) is positive and injective, so \(\normiii{\widetilde{T}}y/\normiii{\widetilde{T}y}\in C_2\) and thus \(y^*(y)\geq \normiii{\widetilde{T}y}/\normiii{\widetilde{T}}>0\). It follows that \[\frac{\normiii{\widetilde{T}y}}{\normiii{\widetilde{T}}}\leq y^*(|y|)\leq \normiii{y}\] for every \(y\in \widetilde{Y}_{(p)}\), so \(\sigma(y)=y^*(|y|)\) is a lattice seminorm that induces an \(AL\)-norm in \(\widetilde{Y}_{(p)}/\ker \sigma\). By Kakutani’s Representation Theorem for \(AL\)-spaces, the completion of this quotient can be isometrically identified with \(L_1(\mu)\) for some measure \(\mu\). Let \(P:\widetilde{Y}_{(p)} \rightarrow L_1(\mu)\) be the composition of the quotient, the inclusion into the completion and the isometry into \(L_1(\mu)\), and \(\widehat{T}:L_1(\mu)\rightarrow (Z_p,\normiii{\cdot})\) be the extension to \(L_1(\mu)\) of the lattice homomorphism induced on \(\widetilde{Y}_{(p)}/\ker \sigma\) by \(\widetilde{T}\), so that \(\widetilde{T}=\widehat{T} P\). Applying now a \(p\)-convexification procedure to each space we recover \(\widetilde{Y}\) and \(Z\) with the same norm and vector lattice structure, and we can lattice isometrically identify \(L_1(\mu)^{(p)}\) with \(L_p(\mu)\). Morevoer, by 32, \(P:\widetilde{Y} \rightarrow L_p(\mu)\) and \(\widehat{T}: L_p(\mu)\rightarrow Z\) are contractive lattice homomorphisms such that \(\widetilde{T}=\widehat{T} P\). It follows that \(T=\widetilde{T} Q=\widehat{T} P Q\) factors through \(L_p(\mu)\), as we wanted to show. ◻

Proof. Without loss of generality, we can assume that \(\mu\) is a probability measure. Indeed, the function \(e=\sum_{n=1}^\infty 2^{-n}g_n \in L_1(\mu)_+\) has norm one, so the measure \(e\cdot \mu\) is a probability measure, and satisfies that the band generated by \(e\), \(B_e\), coincides with the band generated by the whole sequence \((g_n)_{n=1}^\infty\). Radon–Nikodym Theorem establishes that \(B_e\) is lattice isometric to \(L_1(e\cdot \mu)\), so \((g_n)_{n=1}^\infty\) satisfies the inequality of the statement in \(L_1(\mu)\) if and only if \((g_n/e)_{n=1}^\infty\) satisfies the corresponding property in \(L_1(e\cdot \mu)\). Since linear operators are weak-to-weak continuous, it suffices to show that \((g_n/e)_{n=1}^\infty\) is relatively weakly compact in \(L_1(e\cdot\mu)\) in order to conclude that \((g_n)_{n=1}^\infty\) is relatively weakly compact in \(L_1(\mu)\).
Hereby, let \(\mu\) be a probability measure. First, we show that for every sequence of pairwise disjoint measurable subsets \((B_n)_{n=1}^\infty\subseteq \Omega\) we have \[\lim_n \int_{B_n} g_n\,d\mu =0.\] To do so, we define \(\lambda_n=\intoo{\int_{B_n} g_n\,d\mu}^{q^*-1}\), so for every \(m\in \mathbb{N}\) we have \[\begin{align} \sum_{n=1}^m \intoo[3]{\int_{B_n} g_n\,d\mu}^{q^*} & = \sum_{n=1}^m \lambda_n \int_{B_n} g_n\,d\mu \leq \sum_{n=1}^m \int_{B_n} \bigvee_{k=1}^m |\lambda_k g_k|\,d\mu \leq \int_{\Omega} \bigvee_{k=1}^m |\lambda_k g_k|\,d\mu \\ &\leq K \intoo[3]{\sum_{n=1}^m |\lambda_n|^{q}}^\frac{1}{q} = K \intoo[3]{\sum_{n=1}^m \intoo[3]{\int_{B_n} g_n\,d\mu}^{q^*}}^\frac{1}{q}. \end{align}\] Rearranging the terms we conclude that \(\sum_{n=1}^\infty \intoo{\int_{B_n} g_n\,d\mu}^{q^*}\leq K^{q^*}\), so in particular \[\lim_n \int_{B_n} g_n\,d\mu =0.\] Next we prove that for every sequence of measurable subsets \((C_n)_{n=1}^\infty\subseteq \Omega\) such that \(\lim_n\mu(C_n)=0\) we have \[\lim_n \int_{C_n} g_n\,d\mu =0.\] Assume the contrary. Then, passing to a subsequence we can find \(\varepsilon>0\) and a sequence \((C_n)_{n=1}^\infty\subseteq \Omega\) such that \(\lim_n\mu(C_n)=0\) but \(\int_{C_n} g_n\,d\mu \geq \varepsilon\). Note that, for every fixed \(m\in \mathbb{N}\), \(g_m\chi_{C_n}\xrightarrow[n]{}0\) \(\mu\)-a.e., so by the Dominated Convergence Theorem \(\lim_n \int_{C_n} g_m\,d\mu =0\). Let \(n_1=1\) and \(j\geq 2\), and assume that \(n_1<\ldots<n_{j-1}\) have been selected. We can choose \(n_j>n_{j-1}\) such that \(\int_{C_{n_j}} g_{n_i}\,d\mu <2^{-j}\varepsilon\) for every \(i=1,\ldots, j-1\). Let \(B_i=C_{n_i}\setminus \bigcup_{j=i+1}^\infty C_{n_j}\). If \(i<j\), \(B_i\cap B_j\subseteq B_i\cap C_{n_j}= \varnothing\), so the new collection is pairwise disjoint. Moreover, \[\int_{B_i} g_{n_i}\,d\mu \geq \int_{C_{n_i}} g_{n_i}\,d\mu-\int_{\bigcup_{j=i+1}^\infty C_{n_j}} g_{n_i}\,d\mu \geq \int_{C_{n_i}} g_{n_i}\,d\mu-\sum_{j=i+1}^\infty\int_{C_{n_j}} g_{n_i}\,d\mu \geq \frac{\varepsilon}{2}.\] This contradicts the fact that \(\lim_n \int_{B_i} g_{n_i}\,d\mu =0\).
This last property implies that \((g_n)_{n=1}^\infty\) is an equi-integrable set, i.e., for every \(\varepsilon>0\) there exists some \(\delta>0\) such that \(\sup_m \int_{B} g_{m}\,d\mu<\varepsilon\) for every measurable subset \(C\) with \(\mu(C)<\delta\). Otherwise, we can find some \(\varepsilon>0\) and a sequence of subsets \((C_n)_{n=1}^\infty\) such that \(\mu(C_n)\leq \frac{1}{n}\) but \(\sup_m \int_{C_n} g_{m}\,d\mu>\varepsilon\) for every \(n\in \mathbb{N}\). For each \(n\) there is some \(m_n\in \mathbb{N}\) such that \(\int_{C_n} g_{m_n}\,d\mu>\varepsilon\). The fact that \(\lim_n \int_{C_n} g_m\,d\mu =0\) for every \(m\in \mathbb{N}\) implies that the set \(\{m_n:n\in \mathbb{N}\}\) contains an increasing subsequence. We can assume that \((m_n)_{n=1}^\infty\) itself is strictly increasing, so by the above property \(\lim_n \int_{C_n} g_{m_n}\,d\mu =0\), which leads to a contradiction. Therefore, \((g_n)_{n=1}^\infty\) is equi-integrable, so by [11] it is relatively weakly bounded. ◻

Proof. \((2)\Rightarrow (1)\) Let us fix \(f_1,\ldots,f_n\in A\) and \(\alpha_1,\ldots,\alpha_n\in \mathbb{R}\). Then, using Hölder’s inequality with exponent \(\frac{p}{r}\) we get \[\begin{align} \norm[3]{ \intoo[3]{\sum_{i=1}^n |\alpha_i f_i|^{p}}^\frac{1}{p}}_{L_r(\mu)} & = \intoo[3]{\int \intoo[3]{\sum_{i=1}^n |\alpha_i f_ig^{-\frac{1}{r}}|^{p}}^\frac{r}{p} g\, d\mu}^\frac{1}{r}\\ & \leq \intoo[3]{\int \sum_{i=1}^n |\alpha_i f_i g^{-\frac{1}{r}}|^{p} g\, d\mu}^\frac{1}{p} \intoo[3]{\int g\, d\mu}^{\frac{1}{r} -\frac{1}{p}} \\ & = \intoo[3]{ \sum_{i=1}^n |\alpha_i|^p \|f_i g^{-\frac{1}{r}}\|_{L_p(g\cdot \mu)}^{p}}^\frac{1}{p} \leq C \intoo[3]{\sum_{i=1}^n |\alpha_i |^{p}}^\frac{1}{p}. \end{align}\] \((1)\Rightarrow (2)\) Let us fix \(n\in \mathbb{N}\) and define \[C_n=\sup\cbr[3]{\norm[3]{ \intoo[3]{\sum_{i=1}^n |\alpha_i f_i|^{p}}^\frac{1}{p}}_{L_r(\mu)}: f_1,\ldots,f_n\in A, \sum_{i=1}^n |\alpha_i |^{p}\leq 1}.\] Clearly \(C_n\uparrow C\) and \(C_n\neq 0\) as long as \(A\neq \{0\}\). We also select a sequence of scalars \(\delta_n\downarrow 1\). Then, we can choose \(f_{n,1},\ldots, f_{n,n}\in A\) and \(\alpha_{n,1},\ldots, \alpha_{n,n}\in \mathbb{R}\) such that \[\norm[3]{ \intoo[3]{\sum_{i=1}^n |\alpha_{n,i} f_{n,i}|^{p}}^\frac{1}{p}}_{L_r(\mu)}=1 \quad \text{and}\quad \intoo[3]{\sum_{i=1}^n |\alpha_{n,i} |^{p}}^\frac{1}{p}\leq \frac{\delta_n}{C_n}.\] Let \(g_n= \intoo[3]{\sum_{i=1}^n |\alpha_{n,i} f_{n,i}|^{p}}^\frac{r}{p}\), which belongs to \(L_1(\mu)_+\) and has norm one. Then, the sequence \((g_n)_{n=1}^\infty\) satisfies that for any finite choice of scalars \(\lambda_1,\ldots,\lambda_m\in \mathbb{R}\), \[\begin{align} \norm[3]{\bigvee_{n=1}^m |\lambda_n g_n|}_{L_1(\mu)}& \leq \norm[3]{\intoo[3]{\sum_{n=1}^m |\lambda_n g_n|^{\frac{p}{r}}}^\frac{r}{p}}_{L_1(\mu)}= \norm[3]{\intoo[3]{\sum_{n=1}^m \sum_{i=1}^n |\lambda_n|^{\frac{p}{r}} |\alpha_{n,i} f_{n,i}|^{p} }^\frac{1}{p}}_{L_r(\mu)}^r \\ & \leq C^r \intoo[3]{\sum_{n=1}^m \sum_{i=1}^n |\lambda_n|^{\frac{p}{r}} |\alpha_{n,i}|^{p} }^\frac{r}{p} \leq C^r \intoo[3]{\sum_{n=1}^m |\lambda_n|^{\frac{p}{r}} \frac{\delta_n^p}{C_n^p} }^\frac{r}{p} \\ & \leq \intoo[3]{\frac{C\delta_1}{C_1}}^r \intoo[3]{\sum_{n=1}^m |\lambda_n|^{\frac{p}{r}} }^\frac{r}{p}, \end{align}\] so by 53 \((g_n)_{n=1}^\infty\) is relatively weakly compact in \(L_1(\mu)\). The Eberlein–Šmulian Theorem (cf. [11]) implies that there exists a subsequence \((g_{n_m})_{m=1}^\infty\) that converges weakly to some \(g\in L_1(\mu)\). Clearly, \(g\geq 0\) and \(\|g\|_{L_1(\mu)}=1\). Next, we observe that for any \(f\in A\), \(\varepsilon>0\) and \(n\in \mathbb{N}\), we have \[\int (g_n^\frac{p}{r}+(\varepsilon C_n^{-1} |f|)^p)^\frac{r}{p}\, d\mu= \norm[3]{ \intoo[3]{\sum_{i=1}^n |\alpha_{n,i} f_{n,i}|^{p} + (\varepsilon C_n^{-1} |f|)^p }^\frac{1}{p}}_{L_r(\mu)}^r \leq \intoo[3]{\frac{C_{n+1}}{C_n}}^r(\delta_n^p+\varepsilon^p)^\frac{r}{p}.\] If we denote by \(q=\frac{p}{r}\), then we know by 2 that \[(g_n^\frac{p}{r}+(\varepsilon C_n^{-1} |f|)^p)^\frac{r}{p}=\sup \cbr[3]{\bigvee_{j=1}^k (a_j g_n+b_j (\varepsilon C_n^{-1} |f|)^r): a_j^{q^*}+b_j^{q^*}\leq 1, j=1,\ldots, k, k\in \mathbb{N}},\] and similarly, \[(g^\frac{p}{r}+(\varepsilon C^{-1} |f|)^p)^\frac{r}{p}=\sup \cbr[3]{\bigvee_{j=1}^k (a_j g+b_j (\varepsilon C^{-1} |f|)^r): a_j^{q^*}+b_j^{q^*}\leq 1, j=1,\ldots, k, k\in \mathbb{N}}.\] In particular, for any choice of scalars such that \(a_j^{q^*}+b_j^{q^*}\leq 1\), \(j=1,\ldots, k\), we can define the sets \[A_i=\cbr[3]{\bigvee_{j=1}^k (a_j g+b_j (\varepsilon C^{-1} |f|)^r)= a_i g+b_i (\varepsilon C^{-1} |f|)^r}\] for \(i=1,\ldots, k\), that can be assumed to be pairwise disjoint without loss of generality. Recall that \((g_{n_m})_{m=1}^\infty\) converges weakly to \(g\), and clearly \((\varepsilon C_{n_m}^{-1} |f|)^r\) converges weakly to \((\varepsilon C^{-1} |f|)^r\) in \(L_1(\mu)\), so \[\begin{align} \int \bigvee_{j=1}^k (a_j g+b_j (\varepsilon C^{-1} |f|)^r) \, d\mu & = \int \sum_{j=1}^k (a_j\chi_{A_j} g+b_j\chi_{A_j} (\varepsilon C^{-1} |f|)^r) \, d\mu \\ & = \lim_m \int \sum_{j=1}^k (a_j\chi_{A_j} g_{n_m}+b_j\chi_{A_j} (\varepsilon C_{n_m}^{-1} |f|)^r) \, d\mu. \end{align}\] Moreover, for every \(m\in \mathbb{N}\), \[\begin{align} &\int \sum_{j=1}^k (a_j\chi_{A_j} g_{n_m}+b_j\chi_{A_j} (\varepsilon C_{n_m}^{-1} |f|)^r) \, d\mu \leq \int \bigvee_{j=1}^k (a_j g_{n_m}+b_j (\varepsilon C_{n_m}^{-1} |f|)^r) \, d\mu\\ & \leq \int (g_{n_m}^\frac{p}{r}+(\varepsilon C_n^{-1} |f|)^p)^\frac{r}{p} \, d\mu \leq \intoo[3]{\frac{C_{n_m+1}}{C_{n_m}}}^r(\delta_{n_m}^p+\varepsilon^p)^\frac{r}{p}, \end{align}\] so \[\int \bigvee_{j=1}^k (a_j g+b_j (\varepsilon C^{-1} |f|)^r) \, d\mu \leq \lim_m \intoo[3]{\frac{C_{n_m+1}}{C_{n_m}}}^r(\delta_{n_m}^p+\varepsilon^p)^\frac{r}{p} = (1+\varepsilon^p)^\frac{r}{p} .\] Since \(L_1(\mu)\) has the Fatou property (\(h_\alpha \uparrow h\) implies \(\|h_\alpha\|\uparrow \|h\|\) for every net \((h_\alpha)_\alpha\) in \(L_1(\mu)_+\)), it follows that \[\int (g^\frac{p}{r}+(\varepsilon C^{-1} |f|)^p)^\frac{r}{p} \, d\mu \leq (1+\varepsilon^p)^\frac{r}{p}\] (recall that \(f\in A\) and \(\varepsilon>0\) are arbitrary). From here, we deduce that \[\begin{align} 1+\varepsilon^r C^{-r} \int_{\{g=0\}}|f|^r \, d\mu & = \int_{\{g>0\}} g \, d\mu + \varepsilon^r C^{-r} \int_{\{g=0\}}|f|^r \, d\mu \\ & \leq \int (g^\frac{p}{r}+(\varepsilon C^{-1} |f|)^p)^\frac{r}{p} \, d\mu \leq (1+\varepsilon^p)^\frac{r}{p}, \end{align}\] so \[0\leq \int_{\{g=0\}}|f|^r \, d\mu \leq C^r \lim_{\varepsilon\rightarrow 0} \frac{(1+\varepsilon^p)^\frac{r}{p}-1}{\varepsilon^r}=0\] and hence \(\{g=0\}\subseteq \{f=0\}\) for every \(f\in A\). On the other hand, it is clear that \[\lim_{\varepsilon\rightarrow 0}\frac{(1+\varepsilon^p (C^{-1} |f| g^{-\frac{1}{r}})^p)^\frac{r}{p}-1}{\varepsilon^p}g =\frac{r}{p} (C^{-1} |f| g^{-\frac{1}{r}})^p g \quad \mu\text{-a.e.},\] so applying Fatou’s Lemma, we obtain that \[\begin{align} \frac{r}{p} C^{-p} \int (|f| g^{-\frac{1}{r}})^p g \,d\mu \leq \liminf_{\varepsilon\rightarrow 0} \frac{1}{\varepsilon^p} \int ( (g^\frac{p}{r}+(\varepsilon C^{-1} |f|)^p)^\frac{r}{p}-g)\,d\mu \leq \lim_{\varepsilon\rightarrow 0} \frac{(1+\varepsilon^p)^\frac{r}{p}-1}{\varepsilon^p} = \frac{r}{p}. \end{align}\] Therefore, we conclude that \[\|g^{-\frac{1}{r}}f\|_{L_p(g\cdot\mu)}\leq C\] for every \(f\in A\). ◻

Proof. \((1)\Leftrightarrow(2)\) Let \(A=T(B_E)\subseteq L_r(\mu)\). Then, it is clear that conditions \((1)\) and \((2)\) from the statement are equivalent to \(A\) satisfying conditions \((1)\) and \((2)\) of 54, respectively.
\((2)\Rightarrow(3)\) Since \(p>r\) and \(g\cdot \mu\) is a probability measure, the operator \[\fullfunction{M}{L_p(g\cdot \mu)}{L_r(\mu)}{h}{g^\frac{1}{r} h}\] always satisfies \(\|Mh\|_{L_r(\mu)}=\|h\|_{L_r(g\cdot\mu)}\leq \|h\|_{L_p(g\cdot \mu)}\). On the other hand, the operator \(R\) is bounded by \((2)\).
\((3)\Rightarrow(2)\) It follows from the boundedness of \(R\). ◻

Proof. \((2)\Rightarrow (3)\) Let us denote \(\nu=g\cdot \mu\), which is a probability measure. For a fixed \(f\in A\), we can consider the set \(B=\{g=0\}\cap \{|f|>0\}\). If \(\mu(B)\) was non-zero, we would have that \[0<\|f\chi_B\|_{L_r(\mu)}\leq C' \intoo[3]{\int_B g\, d\mu}^{\frac{1}{r}-\frac{1}{p}}=0,\] which is not possible. Therefore, \(\{g=0\}\subseteq \{f=0\}\) (up to a set of measure zero). Now, given any measurable subset \(B\subseteq \Omega\) with \(\nu(B)=\int_B g\, d\mu >0\), we observe that \[\nu(B)^{\frac{1}{p}-\frac{1}{r}} \intoo[3]{\int_B |g^{-\frac{1}{r}} f|^r \,d\nu }^\frac{1}{r}= \intoo[3]{\int_B g\, d\mu}^{\frac{1}{p}-\frac{1}{r}} \|f\chi_B\|_{L_r(\mu)}\leq C',\] so taking the supremum over all such subsets \(0<\nu(B)<\infty\) we obtain \((3)\).
\((3)\Rightarrow (1)\) We first note that if \(\nu\) is a probability measure, then \(\|h\|_{L_r(\nu)}\leq \|h\|_{L^{[r]}_{p,\infty}(\nu)}\) for every \(h\in L_{p,\infty}(\nu)\). Moreover, recall that \((L_{p,\infty}(\nu), \|\cdot \|_{L^{[r]}_{p,\infty}(\nu)})\) satisfies an upper \(p\)-estimate with constant one for any measure \(\nu\). Thus, given \(f_1,\ldots,f_n\in A\) and \(\alpha_1,\ldots,\alpha_n\in \mathbb{R}\), since \(g\cdot \mu\) is a probability measure we have that \[\begin{align} \norm[3]{ \bigvee_{i=1}^n |\alpha_i f_i|}_{L_r(\mu)} & = \norm[3]{ \bigvee_{i=1}^n |\alpha_i f_i g^{-\frac{1}{r}}|}_{L_r(g\cdot \mu)} \leq \norm[3]{ \bigvee_{i=1}^n |\alpha_i f_i g^{-\frac{1}{r}}|}_{L^{[r]}_{p,\infty}(g\cdot \mu)}\\ & \leq \intoo[3]{\sum_{i=1}^n \|\alpha_i f_i g^{-\frac{1}{r}}\|_{L^{[r]}_{p,\infty}(g\cdot \mu)}^{p}}^\frac{1}{p} \leq C \intoo[3]{\sum_{i=1}^n |\alpha_i |^{p}}^\frac{1}{p}. \end{align}\]

\((1)\Rightarrow (2)\) As in the proof of 54, for every \(n\in \mathbb{N}\) we define \[C_n=\sup\cbr[3]{\norm[3]{ \bigvee_{i=1}^n |\alpha_i f_i|}_{L_r(\mu)}: f_1,\ldots,f_n\in A, \sum_{i=1}^n |\alpha_i |^{p}\leq 1},\] that satisfy \(C_n\uparrow C\) and \(C_n\neq 0\) whenever \(A\neq \{0\}\). Let \(\delta_n\downarrow 1\) be a decreasing sequence of scalars, and find \(f_{n,1},\ldots, f_{n,n}\in A\) and \(\alpha_{n,1},\ldots, \alpha_{n,n}\in \mathbb{R}\) satisfying that \[\norm[3]{ \bigvee_{i=1}^n|\alpha_{n,i} f_{n,i}|}_{L_r(\mu)}=1 \quad \text{and}\quad \intoo[3]{\sum_{i=1}^n |\alpha_{n,i} |^{p}}^\frac{1}{p}\leq \frac{\delta_n}{C_n}.\] We denote by \(g_n=\intoo{\bigvee_{i=1}^n|\alpha_{n,i} f_{n,i}|}^r\in L_1(\mu)_+\), that has norm one. For any finite choice of scalars \(\lambda_1,\ldots,\lambda_m\in \mathbb{R}\), the sequence \((g_n)_{n=1}^\infty\) satisfies that \[\begin{align} \norm[3]{\bigvee_{n=1}^m |\lambda_n g_n|}_{L_1(\mu)}& =\norm[3]{\bigvee_{n=1}^m |\lambda_n| \intoo{\bigvee_{i=1}^n|\alpha_{n,i} f_{n,i}|}^r}_{L_1(\mu)} =\norm[3]{\bigvee_{n=1}^m \bigvee_{i=1}^n |\lambda_n|^\frac{1}{r}|\alpha_{n,i} f_{n,i}|}^r_{L_r(\mu)}\\ & \leq C^r \intoo[3]{\sum_{n=1}^m \sum_{i=1}^n |\lambda_n|^{\frac{p}{r}} |\alpha_{n,i}|^{p} }^\frac{r}{p} \leq C^r \intoo[3]{\sum_{n=1}^m |\lambda_n|^{\frac{p}{r}} \frac{\delta_n^p}{C_n^p} }^\frac{r}{p} \\ & \leq \intoo[3]{\frac{C\delta_1}{C_1}}^r \intoo[3]{\sum_{n=1}^m |\lambda_n|^{\frac{p}{r}} }^\frac{r}{p}. \end{align}\] 53 then yields that \((g_n)_{n=1}^\infty\) is relatively weakly compact in \(L_1(\mu)\), so by the Eberlein–Šmulian Theorem (cf. [11]) we can extract a subsequence \((g_{n_m})_{m=1}^\infty\) that converges weakly to some normalized and positive \(g\in L_1(\mu)\). On the other hand, for any \(f\in A\), \(B\subseteq \Omega\) measurable, \(\varepsilon>0\) and \(n\in \mathbb{N}\), we have \[\begin{align} \int_{\Omega\setminus B} g_n\, d\mu + \int_B (\varepsilon C_n^{-1} |f|)^r\, d\mu & \leq \int g_n\vee (\varepsilon C_n^{-1} |f|)^r\, d\mu = \| g_n^\frac{1}{r}\vee (\varepsilon C_n^{-1} |f|)\|^r_{L_r(\mu)}\\ & = \norm[3]{ \intoo[3]{\bigvee_{i=1}^n|\alpha_{n,i} f_{n,i}|} \vee (\varepsilon C_n^{-1} |f|) }_{L_r(\mu)} \leq \intoo[3]{\frac{C_{n+1}}{C_n}}^r(\delta_n^p+\varepsilon^p)^\frac{r}{p}. \end{align}\] Since \(\int g_n\, d\mu=1\), we can rearrange the terms to obtain that \[\varepsilon^r C_n^{-r} \int_B |f|^r\, d\mu \leq \int_B g_n\, d\mu + \intoo[3]{\frac{C_{n+1}}{C_n}}^r(\delta_n^p+\varepsilon^p)^\frac{r}{p}-1.\] In particular, taking the limit over the subsequence \((n_m)_{m=1}^\infty\) we get that \[\varepsilon^r C^{-r} \int_B |f|^r\, d\mu \leq \int_B g\, d\mu + (1+\varepsilon^p)^\frac{r}{p}-1\leq \int_B g\, d\mu + \frac{r}{p}\varepsilon^p,\] or equivalently, \[C^{-r} \int_B |f|^r\, d\mu \leq \varepsilon^{-r} \int_B g\, d\mu + \frac{r}{p}\varepsilon^{p-r}\] for every \(\varepsilon>0\). The second term is minimized when \(\varepsilon^p=\frac{p}{p-r}\int_B g \, d\mu\), which yields \[\int_B |f|^r\, d\mu \leq C^r \intoo[3]{1-\frac{r}{p}}^{\frac{r}{p}-1} \intoo[3]{\int_B g\, d\mu}^{1-\frac{r}{p}}\] for any \(f\in A\) and \(B\subseteq \Omega\), as we wanted to show. ◻

Proof. Conditions \((1)\), \((2)\) and \((3)\) are equivalent to the corresponding conditions of 56 taking \(A=T(B_E)\).
\((3)\Rightarrow(4)\) Since \(\int g\, d\mu=1\), the operator \[\fullfunction{M}{L^{[r]}_{p,\infty}(g\cdot\mu)}{L_r(\mu)}{h}{g^\frac{1}{r} h}\] is bounded with norm one: \[\|Mh\|_{L_r(\mu)}=\intoo[3]{\int |h|^r g\,d\mu}^\frac{1}{r}\leq \intoo[3]{\int g\, d\mu}^{\frac{1}{r}-\frac{1}{p}} \|h\|_{L^{[r]}_{p,\infty}(g\cdot\mu)}= \|h\|_{L^{[r]}_{p,\infty}(g\cdot\mu)}.\] On the other hand, the operator \(R\) is by \((3)\).
\((4)\Rightarrow(3)\) It follows from the boundedness of \(R\). ◻

Proof. Let \(p=q^*\) and \(r=s^*\), so that \(1< r<p<\infty\). We will establish the equivalence of the conditions in the statement with the corresponding conditions in 55 for the operator \(T^*:E^*\rightarrow L_r(\mu)\), that will be denoted by \((1^*)\), \((2^*)\) and \((3^*)\).
\((1)\Leftrightarrow (1^*)\) It follows from 18.
\((2)\Rightarrow (3)\) Let \(\nu=g\cdot \mu\). Note that for every \(f\in L_s(\mu)\), \(\nu (\{g=0\}\cap\{f\neq 0\})=0\), so \(\widetilde{M}f=fg^{-\frac{1}{s}}\) is a measurable and finite \(\nu\)-a.e. function. Moreover, since \(q<s\) and \(\nu\) is a probability measure, \[\|\widetilde{M}f\|_{L_q(\nu)}\leq \|\widetilde{M}f\|_{L_s(\nu)}=\intoo[3]{\int_{\{g>0\}} |f|^s\, d\mu}^\frac{1}{s}\leq \|f\|_{L_s(\mu)},\] so \(\widetilde{M}\) is bounded. Observe that for every \(h\in \widetilde{M}(L_s(\mu))\), the operator \(Sh=T(hg^\frac{1}{s})\) is bounded with norm below \(C\). Since \(\widetilde{M}(L_s(\mu))=L_s(\nu)\subseteq L_q(\nu)\) is dense in \(L_q(\nu)\), we can extend it to \(S:L_q(\nu)\rightarrow E\). We see that for every \(f\in L_s(\mu)\), \[S \widetilde{M} f= S(fg^{-\frac{1}{s}})=S(fg^{-\frac{1}{s}}\chi_{\{g>0\}})=T(f\chi_{\{g>0\}}),\] but condition \((2)\) implies that \[\|Tf-T(f\chi_{g>0})\|\leq C\|(f-f\chi_{g>0})g^{-\frac{1}{s}}\|_{L_q(\nu)}=0,\] so \(T=S \widetilde{M}\).
\((3)\Rightarrow (2)\) It follows from the fact that \(Tf=S(f g^{-\frac{1}{s}})\) and \(S\) is bounded.
In order to prove \((3)\Leftrightarrow (3^*)\), we first observe that given a normalized function \(g\in L_1(\mu)_+\), both operators \[\fullfunction{\widetilde{M}}{L_s(\mu)}{L_q(g\cdot \mu)}{f}{fg^{-\frac{1}{s}}}\quad \text{and} \quad \fullfunction{M}{L_p(g\cdot \mu)}{L_r(\mu)}{h}{hg^\frac{1}{r}}\] are contractive. Moreover, for every \(f\in L_s(\mu)\) and \(h\in L_p(g\cdot \mu)\), \[\langle \widetilde{M}^*h, f \rangle_{L_s(\mu)} = \langle h ,\widetilde{M}f \rangle_{L_s(\mu)}= \int h (fg^{-\frac{1}{s}}) g \,d\mu = \int (hg^{\frac{1}{r}}) f \,d\mu = \langle Mh, f \rangle_{L_s(\mu)},\] so \(\widetilde{M}^*=M\). Since both \(L_s(\mu)\) and \(L_q(g\cdot \mu)\) are reflexive, it follows that \(M^*=\widetilde{M}^{**}=\widetilde{M}\).
Now, to check that \((3)\Rightarrow (3^*)\), note that if \(T\) factors through \(L_q(g\cdot \mu)\) as \(T=S \widetilde{M}\), taking adjoints we get that \(T^*\) factors through \(L_p(g\cdot \mu)\) as \(T^*=M S^*\). Moreover, \((3)\) (or equivalently, \((2)\)) implies that \(T(f\chi_{\{g=0\}})=0\) for every \(f\in L_s(\mu)\). By the definition of \(T^*\), this is equivalent to \(\{g=0\}\subseteq \{T^*x^*=0\}\) for every \(x^*\in E^*\), so \((3^*)\) holds.
Conversely, to show \((3^*)\Rightarrow (3)\) we observe that, if \(T^*\) factors through \(L_p(g\cdot \mu)\) as \(T^*=M R\), taking adjoints we get that \(T^{**}:L_s(\mu)\rightarrow E^{**}\) factors as \(T^{**}=R^* \widetilde{M}\). In general, if \(U: F\rightarrow E\) is a bounded operators between Banach spaces, it holds that \(U^{**} J_F=J_E U\), where \(J_F\) and \(J_E\) denote the corresponding canonical embeddings into the biduals. In the case \(U=T\), \(F=L_s(\mu)\) is reflexive, so \(J_F=i_F\) and we get that \(T^{**}=J_E T\), so we can write \(T=J_E^{-1} R^* \widetilde{M}\), as we wanted to show. ◻

Proof. Again, put \(1<r=s^*<p=q^*<\infty\), and denote by \((1^*)\), \((2^*)\), \((3^*)\) and \((4^*)\) the conditions in 57 associated to the operator \(T^*:E^*\rightarrow L_r(\mu)\). Throughout the proof, for any measure \(\nu\), \(L_{q,1}(\nu)\) will be endowed with the equivalent renorming \[\|f\|_{L^{[r]}_{q,1}(\nu)}=\sup \cbr[3]{\int fh \,d\nu : \|h\|_{L^{[r]}_{p,\infty}(\nu)}\leq 1}\] for every \(f\in L_{q,1}(\nu)\).
The equivalence \((1)\Leftrightarrow (1^*)\) follows from 18, and \((2)\Leftrightarrow (2^*)\) is just a consequence of the fact that \[\langle x^*, T(f\chi_B) \rangle_{E}=\langle T^* x^*, f\chi_B \rangle_{L_s(\mu)}=\langle (T^* x^*)\chi_B, f \rangle_{L_s(\mu)}\] for every \(f\in L_s(\mu)\), \(x^*\in E^*\) and measurable \(B\subseteq \Omega\).
\((3)\Rightarrow (4)\) Note that for every \(f\in L_s(\mu)\) and \(h\in L_{p,\infty}(g\cdot \mu)\), using Hölder’s inequality and the fact that \(g\cdot \mu\) is a probability measure, we get that \[\left| \int fg^{-\frac{1}{s}} h g\,d\mu \right| \leq \int |fh|g^{\frac{1}{r}} \,d\mu \leq \intoo[3]{\int |f|^s\, d\mu}^\frac{1}{s} \intoo[3]{\int |h|^r g\, d\mu}^\frac{1}{r} \leq \|f\|_{L_s(\mu)} \|h\|_{L^{[r]}_{p,\infty}(g\cdot\mu)},\] so the operator \[\fullfunction{\widetilde{M}}{L_s(\mu)}{L_{q,1}(g\cdot \mu)}{f}{fg^{-\frac{1}{s}}}\] is a contraction. Moreover, \(\widetilde{M}(L_s(\mu))=L_s(g\cdot\mu)\) is dense in \(L_{q,1}(g\cdot \mu)\). The rest of the proof follows as in 58.
\((4)\Rightarrow (3)\) It is straightforward.
To show \((4)\Leftrightarrow (4^*)\) we observe again that for any normalized function \(g\in L_1(\mu)_+\), both operators \[\fullfunction{\widetilde{M}}{L_s(\mu)}{L_{q,1}(g\cdot \mu)}{f}{fg^{-\frac{1}{s}}}\quad \text{and} \quad \fullfunction{M}{L_{p,\infty}(g\cdot \mu)}{L_r(\mu)}{h}{hg^\frac{1}{r}}\] are contractive and \(\widetilde{M}^*=M\). Even though \(M^*\) no longer coincides formally with \(\widetilde{M}\), as \(L_{q,1}(g\cdot \mu)\) is not reflexive, \(L_s(\mu)\) is reflexive, so \(M^*=\widetilde{M}^{**}=J_{L_{q,1}(g\cdot \mu)} \widetilde{M}\).
Now \((4)\Rightarrow (4^*)\) follows by taking adjoints in the factorization of \(T\) given in \((4)\) and observing that condition \((3)\) implies that \(\{g=0\}\subseteq \{T^*x^*=0\}\). On the other hand, to show \((4^*)\Rightarrow (4)\) we write \(T^*=M R\), so that taking adjoints and using the reflexivity of \(L_s(\mu)\) we get that \[J_E T=T^{**}= R^* M^*=R^* J_{L_{q,1}(g\cdot \mu)} M,\] so \(T = J_E^{-1} R^* J_{L_{q,1}(g\cdot \mu)} M\). ◻

Proof. As in the proof of 52, it suffices to show the statement for a contractive lattice homomorphism \(Q:Y\rightarrow Z\), where \(Y\) is \(p\)-convex and \(Z\) is \(q\)-concave, both with constant one.
Since \(p>q\), by 21 \(Y\) is \(q\)-convex with constant one, so, by 52, there exists a measure \(\mu\) and contractive lattice homomorphisms \(W:Y\rightarrow L_q(\mu)\) and \(T:L_q(\mu)\rightarrow Z\) such that \(Q=T W\). It follows that \(W\) is \(p\)-convex with constant one, so we can apply 55 to find a normalized function \(g\in L_1(\mu)_+\) such that the operators \(P:Y\rightarrow L_p(g\cdot \mu)\) and \(M:L_p(g\cdot \mu)\rightarrow L_q(\mu)\), given by \(Py=g^{-\frac{1}{q}}Wy\) and \(Mf=g^\frac{1}{q} f\), are contractive lattice homomorphisms that satisfy \(W=M P\). Let \(h=g^\frac{1}{r}\), which is a positive and normalized function in \(L_r(\mu)\), \(\frac{1}{q}=\frac{1}{p}+\frac{1}{r}\). Then, if we define \(R:Y\rightarrow L_p(\mu)\) and \(S:L_p( \mu)\rightarrow L_q(\mu)\) by \(Ry=h^{-1} Wy\) and \(Df=h f\), we get that \(W=S R\), and moreover \[\|Ry\|_{L_p(\mu)}^p=\int (g^{\frac{1}{p}-\frac{1}{q}}|Wy|)^p\, d\mu = \int (g^{-\frac{1}{q}}|Wy|)^p g\, d\mu = \|Py\|_{L_p(g\cdot \mu)}^p\leq \|y\|^p\] and \[\|Sf\|_{L_q(\mu)}=\intoo[3]{\int (g^{\frac{1}{q}-\frac{1}{p}}|f|)^q\, d\mu}^\frac{1}{q}\leq \intoo[3]{\int g\, d\mu}^\frac{1}{r}\intoo[3]{\int |f|^p\, d\mu}^\frac{1}{p}=\|f\|_{L_p(\mu)}.\] Therefore, \(Q\) factors as \(T S R\), where \(R:E\rightarrow L_p(\mu)\), \(S:L_p(\mu)\rightarrow L_q(\mu)\) and \(T:L_q(\mu)\rightarrow F\) are contractive lattice homomorphisms and \(Sf=hf\) for some \(h\in L_r(\mu)_+\) of norm one. ◻

Proof. Let us fix \(f\neq 0\) and \(b>0\). Then, \[\int_0^b t^{-\frac{1}{p^*}}f^*(t)\, dt\leq \int_0^b t^{-\frac{1}{p^*}}\|f\|_\infty \, dt = pb^\frac{1}{p}\|f\|_\infty,\] and \[\int_b^\infty t^{-\frac{1}{p^*}}f^*(t)\, dt\leq \intoo[3]{\int_b^\infty f^*(t)^q \, dt}^\frac{1}{q} \intoo[3]{\int_b^\infty t^{-\frac{q^*}{p^*}}\, dt}^\frac{1}{q^*}\leq \intoo[2]{\frac{p(q-1)}{p-q}}^\frac{1}{q^*}b^{\frac{1}{p}-\frac{1}{q}}\|f\|_q\] In particular, for \(b=\intoo[1]{\frac{\|f\|_q}{\|f\|_\infty}}^q\) the above inequalities yield \[\vvvert f \vvvert_{L_{p,1}}\leq \intoo[2]{ p+\intoo[2]{\frac{p(q-1)}{p-q}}^\frac{1}{q^*}} \|f\|_q^{\frac{q}{p}}\|f\|_\infty^{1-\frac{q}{p}} \leq \intoo[2]{ p+\intoo[2]{\frac{1}{q}-\frac{1}{p}}^{-\frac{1}{q^*}}} \|f\|_q^{\frac{q}{p}}\|f\|_\infty^{1-\frac{q}{p}}. \qedhere\] ◻

Proof. \((1)\) Let \(f_1,\ldots,f_n\in C(K)\). Then, by 24, \[\intoo[3]{\sum_{i=1}^n \|f_i\|_p^p}^\frac{1}{p} = \intoo[3]{\int_K \sum_{i=1}^n |f_i(t)|^p \, d\mu}^\frac{1}{p} \leq \mu(K)^\frac{1}{p} \intoo[3]{\sup_{t\in K} \sum_{i=1}^n |f_i(t)|^p }^\frac{1}{p} =\mu(K)^\frac{1}{p} \|(f_i)_{i=1}^n\|_{p,w}.\]

\((2)\) Fix \(1\leq q<p\) and let \(f_1,\ldots,f_n\in C(K)\). We denote by \(A=\intoo[2]{ p+\intoo[2]{\frac{1}{q}-\frac{1}{p}}^{-\frac{1}{q^*}}}\). Then, using 61, the first statement of this lemma, and 24, we get \[\begin{align} \intoo[3]{\sum_{i=1}^n \vvvert f_i \vvvert_{L_{p,1}}^p}^\frac{1}{p} &\leq A \intoo[3]{\sum_{i=1}^n \|f_i\|_q^q \|f_i\|_\infty^{p-q}}^\frac{1}{p}\leq A \intoo[3]{\sum_{i=1}^n \|f_i\|_q^q }^\frac{1}{p} \sup_i \|f_i\|_\infty^{1-\frac{q}{p}}\\ & \leq A \pi_q(J_q)^\frac{q}{p} \|(f_i)_{i=1}^n\|_{q,w}^\frac{q}{p} \norm[3]{\intoo[3]{\sum_{i=1}^n |f_i|^q }^\frac{1}{q}}_\infty^{1-\frac{q}{p}}\leq A \mu(K)^\frac{1}{p} \|(f_i)_{i=1}^n\|_{q,w}.\qedhere \end{align}\] ◻

Proof. \((2)\Rightarrow (1)\) It follows easily from the fact that \(J_p\) is \(p\)-summing with constant \(\mu(K)^\frac{1}{p}=1\) by 62.
\((1)\Rightarrow (2)\) Let \[W=\cbr[3]{f\in C(K): f=C^p\sum_{i=1}^n|f_i|^p\text{ for some } (f_i)_{i=1}^n\subseteq C(K)\text{ with }\sum_{i=1}^n \|Tf_i\|^p=1},\] which is clearly a convex subset of \(C(K)_+\). Since \(T\) is \(p\)-concave, it follows that \(\|f\|\geq 1\) for every \(f\in W\), so \(W\) is disjoint from \(B\), the open unit ball of \(C(K)\). Therefore, by Hahn–Banach Theorem there exists a measure \(\nu\in C(K)\) such that \(\nu(g)<1\) for every \(g\in B\) and \(\nu(f)\geq 1\) for every \(f\in W\). In particular, \(\|\nu\|\leq 1\). Let \(\mu=|\nu|/\|\nu\|\). Then, \(\mu\) is a probability measure such that \(\mu(f)\geq \nu(f)/\|\nu\|\geq 1\) for every \(f\in W\). In particular, for every \(f\notin \ker T\), the function \(C^p|f|^p/\|Tf\|^p\) belongs to \(W\), which means that \[\|Tf\|^p\leq C^p\mu(|f|^p)=C^p\int_K|f|^p\, d\mu = C^p\|J_pf\|_{L_p(K,\mu)}^p.\] Therefore, we can define an operator \(\overline{T}\) on \(J_p(C(K))\) by \(\overline{T} J_p f=Tf\) whose norm is bounded by \(C\). Since \(J_p\) has dense range in \(L_p(K,\mu)\), \(\overline{T}\) can be extended to the whole space. ◻

Proof. \((1)\Rightarrow (2)\) Fix \(n\in\mathbb{N}\) and let \(\delta_n=1+\frac{1}{n}\). If we denote by \[\pi^{(n)}_{p,q}(T)=\sup \cbr[3]{\frac{\intoo[0]{\sum_{i=1}^n \|Tf_i\|^p }^\frac{1}{p}}{\norm[0]{\intoo[0]{\sum_{i=1}^n |f_i|^q }^\frac{1}{q}}_\infty}: f_1,\ldots,f_n\in C(K)},\] we can find functions \(f_1,\ldots,f_n\in C(K)\) such that \[\intoo[3]{\sum_{i=1}^n \|Tf_i\|^p }^\frac{1}{p}=1\quad \text{and}\quad \norm[3]{\intoo[3]{\sum_{i=1}^n |f_i|^q }^\frac{1}{q}}_\infty \leq \frac{\delta_n}{\pi^{(n)}_{p,q}(T)}.\] For each \(i=1,\ldots,n\), there exists a normalized functional \(z_i^*\in E^*\) such that \(z_i^*(Tf_i)=\|Tf_i\|\), so \[\begin{align} 1 &= \intoo[3]{\sum_{i=1}^n \|Tf_i\|^p }^\frac{1}{p}= \max \cbr[3]{\sum_{i=1}^n a_i\|Tf_i\|: (a_i)_{i=1}^n\in B_{\ell_{p^*}^n}}\\ & =\max \cbr[3]{\sum_{i=1}^n a_iz_i^*(Tf_i): (a_i)_{i=1}^n\in B_{\ell_{p^*}^n}} =\sum_{i=1}^n a_iz_i^*(Tf_i) \end{align}\] for some \((a_i)_{i=1}^n\in B_{\ell_{p^*}^n}\). If we denote by \(y_i^*=a_iz_i^*\), then it follows that \[\sum_{i=1}^n \|y^*_i\|^{p^*} =1 \quad \text{and} \quad \sum_{i=1}^n y_i^*(Tf_i) =1.\] Let us define for every \(n\in \mathbb{N}\) the functional \(\mu_n\) by \[\mu_n(f)=\sum_{i=1}^n y_i^*(T(ff_i))\] for every \(f\in C(K)\). Clearly, \(\mu_n\) is linear and \(\mu_n(\mathbb{1})=1\) for every \(n\in \mathbb{N}\), where \(\mathbb{1}\) denotes the constant one function over \(K\). Moreover, \[\begin{align} |\mu_n(f)| &\leq \sum_{i=1}^n \|y_i^*\|\|T(ff_i)\| \leq \intoo[3]{\sum_{i=1}^n \|y^*_i\|^{p^*} }^\frac{1}{p^*} \intoo[3]{\sum_{i=1}^n \|T(ff_i)\|^p }^\frac{1}{p}\\ & \leq \pi^{(n)}_{p,q}(T) \norm[3]{\intoo[3]{\sum_{i=1}^n |ff_i|^q }^\frac{1}{q}}_\infty \leq \delta_n \|f\|_\infty \leq 2 \|f\|_\infty. \end{align}\] so \(\mu_n\in C(K)^*\) for every \(n\in \mathbb{N}\) and they are uniformly bounded. Let \((\mu_{n_\alpha})_\alpha\) be a \(w^*\)-convergent subnet of \(\{\mu_n\}_n\) that converges to a certain acumulation point \(\mu\in C(K)^*\). It follows that \[|\mu(f)|=\lim_\alpha |\mu_{n_\alpha}(f)|\leq \lim_\alpha \delta_{n_\alpha} \|f\|_\infty =\|f\|_\infty\] for every \(f\in C(K)\), and in particular \[\mu(\mathbb{1})=\lim_\alpha \mu_{n_\alpha}(\mathbb{1})=1,\] so \(\|\mu\|=1\). Note that, since \(C(K)^*\) is an \(AL\)-space, it follows that \[\mu_+(\mathbb{1})-\mu_-(\mathbb{1})=\mu(\mathbb{1})=1=\|\mu\|=\||\mu|\| =\|\mu_+\|+\|\mu_-\|=\mu_+(\mathbb{1})+\mu_-(\mathbb{1}),\] so \(0=\mu_-(\mathbb{1})=\|\mu_-\|\) and therefore \(\mu\) is a (positive) probability measure. Now, let us fix a non-zero \(f\in C(K)\) and \(n\in \mathbb{N}\). We can define \[h_i=f_i\intoo[3]{\mathbb{1}-\frac{|f|^q}{\|f\|^q_\infty}}^\frac{1}{q}, i=1,\ldots,n\quad \text{and}\quad h_{n+1}=\frac{\delta_n}{\pi^{(n)}_{p,q}(T)}\cdot \frac{f}{\|f\|_\infty},\] so that \[\begin{align} \intoo[3]{\sum_{i=1}^{n+1} |h_i|^q }^\frac{1}{q} &= \intoo[3]{\sum_{i=1}^{n} |f_i|^q \intoo[3]{\mathbb{1}-\frac{|f|^q}{\|f\|^q_\infty}} +\intoo[3]{\frac{\delta_n}{\pi^{(n)}_{p,q}(T)}}^q \frac{|f|^q}{\|f\|^q_\infty}}^\frac{1}{q}\\ & \leq \intoo[3]{\intoo[3]{\frac{\delta_n}{\pi^{(n)}_{p,q}(T)}}^q \intoo[3]{\mathbb{1}-\frac{|f|^q}{\|f\|^q_\infty}} +\intoo[3]{\frac{\delta_n}{\pi^{(n)}_{p,q}(T)}}^q \frac{|f|^q}{\|f\|^q_\infty}}^\frac{1}{q} = \frac{\delta_n}{\pi^{(n)}_{p,q}(T)}\mathbb{1}. \end{align}\] By the definition of \(\pi^{(n+1)}_{p,q}(T)\), \[\intoo[3]{\sum_{i=1}^{n+1} \|Th_i\|^p }^\frac{1}{p}\leq \pi^{(n+1)}_{p,q}(T) \norm[3]{\intoo[3]{\sum_{i=1}^{n+1} |h_i|^q }^\frac{1}{q}}_\infty \leq \frac{\pi^{(n+1)}_{p,q}(T)}{\pi^{(n)}_{p,q}(T)}\delta_n,\] and by Hölder’s inequality \[\abs[3]{\mu_n \intoo[3]{\intoo[3]{\mathbb{1}-\frac{|f|^q}{\|f\|^q_\infty}}^\frac{1}{q}}}=\abs[3]{\sum_{i=1}^n y_i^*(Th_i)} \leq \intoo[3]{\sum_{i=1}^n \|Th_i\|^p }^\frac{1}{p}.\] Therefore \[\begin{align} \|Tf\|^p &= \intoo[3]{\frac{\pi^{(n)}_{p,q}(T)}{\delta_n}}^p \|f\|_\infty^p \|Th_{n+1}\|^p \\ & \leq \intoo[3]{\frac{\pi^{(n)}_{p,q}(T)}{\delta_n}}^p \|f\|_\infty^p \intoo{\intoo[3]{\frac{\pi^{(n+1)}_{p,q}(T)}{\pi^{(n)}_{p,q}(T)}\delta_n}^p - \abs[3]{\mu_n \intoo[3]{\intoo[3]{\mathbb{1}-\frac{|f|^q}{\|f\|^q_\infty}}^\frac{1}{q}}}^p}. \end{align}\] Taking the limit over the net \(n_\alpha\), we get \[\|Tf\|^p \leq \pi_{p,q}(T)^p \|f\|_\infty^p \intoo{1 - \abs[3]{\mu \intoo[3]{\intoo[3]{\mathbb{1}-\frac{|f|^q}{\|f\|^q_\infty}}^\frac{1}{q}}}^p}.\] Now, observe that \(0\leq \mathbb{1}-\frac{|f|^q}{\|f\|^q_\infty}\leq 1\), so \(0\leq \mathbb{1}-\frac{|f|^q}{\|f\|^q_\infty}\leq \intoo[3]{\mathbb{1}-\frac{|f|^q}{\|f\|^q_\infty}}^\frac{1}{q}\), and since \(\mu\) is positive, we get \[\begin{align} \|Tf\|^p &\leq \pi_{p,q}(T)^p \|f\|_\infty^p \intoo{1 - \abs[3]{\mu \intoo[3]{\mathbb{1}-\frac{|f|^q}{\|f\|^q_\infty}}}^p} \\ & = \pi_{p,q}(T)^p \|f\|_\infty^p \intoo{1 - \abs[3]{1-\mu \intoo[3]{\frac{|f|^q}{\|f\|^q_\infty}}}^p} . \end{align}\] Using the inequality \(1-|1-t|^p\leq pt\), that holds for every \(t\in [0,1]\), we get \[\|Tf\|^p \leq \pi_{p,q}(T)^p \|f\|_\infty^p p \mu \intoo[3]{\frac{|f|^q}{\|f\|^q_\infty}} = p \pi_{p,q}(T)^p \|f\|_\infty^p \int \frac{|f|^q}{\|f\|^q_\infty} \,d\mu ,\] or equivalently \[\|Tf\|\leq p^\frac{1}{p} \pi_{p,q}(T) \|f\|_q^{\frac{q}{p}}\|f\|_\infty^{1-\frac{q}{p}}.\]

\((2)\Rightarrow (3)\) Let \(\mu\) be a probability measure on \(K\) satisfying \((2)\). Then, the operator \(T:C(K)\rightarrow E\) can be extended to \(\hat{T}:L_\infty(K,\mu) \rightarrow E\). Indeed, by Lusin’s Theorem, for every \(f\in L_\infty(K,\mu)\) there exists a sequence \((f_n)_{n=1}^\infty\subseteq C(K)\) converging \(\mu\)-a.e. to \(f\) and such that \(\|f_n\|_\infty \leq \|f\|_\infty\) for every \(n\in \mathbb{N}\). Note that \(|f_n-f|^q\leq (2\|f\|_\infty)^q\mathbb{1}\in L_1(K,\mu)\) and \(|f_n-f|^q\rightarrow 0\) \(\mu\)-a.e., so by the Dominated Convergence Theorem we get that \(f_n\rightarrow f\) in \(L_q(K,\mu)\). In particular, \((f_n)_{n=1}^\infty\) is a Cauchy sequence in \(L_q(K,\mu)\), so by the hypothesis we have \[\|Tf_m-Tf_n\|\leq A'\|f_m-f_n\|_q^{\frac{q}{p}} \|f_m-f_n\|_\infty^{1-\frac{q}{p}} \leq A' (2\|f\|_\infty)^{1-\frac{q}{p}}\|f_m-f_n\|_q^{\frac{q}{p}}\] and \((Tf_n)_{n=1}^\infty\) is a Cauchy sequence in \(E\). Let us denote by \(\hat{T}f\) the limit of \((Tf_n)_{n=1}^\infty\). This limit does not depend of the original choice of sequence \((f_n)_{n=1}^\infty\subseteq C(K)\): if \((g_n)_{n=1}^\infty\subseteq C(K)\) is another sequence with \(\|g_n\|_\infty \leq \|f\|_\infty\) that converges \(\mu\)-a.e. to \(f\) then the alternate sequence \((h_n)_{n=1}^\infty\) given by \(h_{2n-1}=f_n\) and \(h_{2n}=g_n\) also satisfies the previous conditions, so repeating the argument we conclude that \((Th_n)_{n=1}^\infty\) is a Cauchy sequence, so in particular \(\lim_n Tf_n=\lim_n Th_n=\lim_n Tg_n\) and \(\hat{T}f\) is well defined. Note that for the above argument to hold, it suffices to consider sequences \((f_n)_{n=1}^\infty\subseteq C(K)\) converging \(\mu\)-a.e. to \(f\) with \(\|f_n\|_\infty \leq C\|f\|_\infty\) for some fixed constant \(C>0\) independent of the sequence. In particular, this implies that \(T\) is additive (and hence, linear): if \(f,g\in L_\infty(K,\mu)\), we can find sequences \((f_n)_{n=1}^\infty, (g_n)_{n=1}^\infty \subseteq C(K)\) satisfying Lusin’s Theorem, so that \(f_n+g_n\rightarrow f+g\) \(\mu\)-a.e. and \[\|f_n+g_n\|_\infty \leq \|f_n\|_\infty +\|g_n\|_\infty\leq \|f\|_\infty+ \|g\|_\infty\leq 2\|f\|_\infty\vee \|g\|_\infty \leq 2\|f+g\|_\infty.\] Therefore, the sequence \((T(f_n+g_n))_{n=1}^\infty\) is a Cauchy sequence in \(E\) that converges to \(\hat{T}(f+g)\) on one hand, and to \(\hat{T}f+\hat{T}g\) on the other, so both limits must coincide. Recall that the sequences that we have considered satisfy \(f_n\rightarrow f\) in \(L_q(K,\mu)\), in particular \(\|f_n\|_q\rightarrow \|f\|_q\), so the operator \(\hat{T}\) satisfies the estimate in \((2)\) too: \[\|\hat{T}f\|=\lim_n \|Tf_n\|\leq A' \limsup_n\|f_n\|_q^{\frac{q}{p}} \|f_n\|_\infty^{1-\frac{q}{p}} \leq A' \lim_n\|f_n\|_q^{\frac{q}{p}} \|f\|_\infty^{1-\frac{q}{p}}= A' \|f\|_q^{\frac{q}{p}} \|f\|_\infty^{1-\frac{q}{p}} .\] Hence, \(\hat{T}\) is a bounded operator.
Next, let us fix a simple function \(f\in L_\infty(K,\mu)\), given by \(f=\sum_{i=1}^m a_i\chi_{K_i}\) with \(K_i\subseteq K\) pairwise disjoint and \((|a_i|)_{i=1}^m\) ordered in a decreasing way. Let us define \(\varepsilon_j=\text{sgn }a_j\), \(f_j=\sum_{i=1}^j \varepsilon_i \chi_{K_i}\), \(B_j=\bigcup_{i=1}^j K_i\) and \(t_j=\mu(B_j)\) for \(j=1,\ldots,m\). Observe that \(|f_j|=\chi_{B_j}\) for every \(j\). We can decompose \(f\) using the functions \(f_j\) in the following way (here, \(a_{m+1}=0\)): \[\begin{align} f&=\sum_{i=1}^m |a_i|\varepsilon_i \chi_{K_i}=\sum_{i=1}^m \sum_{j=i}^m (|a_j|-|a_{j+1}|) \varepsilon_i \chi_{K_i}\\ & =\sum_{j=1}^m (|a_j|-|a_{j+1}|)\sum_{i=1}^j \varepsilon_i \chi_{K_i} = \sum_{j=1}^m (|a_j|-|a_{j+1}|)f_j. \end{align}\] We will need to compute the norm of \(f\) in \(L_{p,1}(K,\mu)\). To do so, first note that the decreasing rearrangement of the simple function \(f\) is given by a simple function in \([0,\infty)\) that takes the value \(|a_j|\) at the interval \([t_{j-1}, t_j)\) (here \(t_0=0\)) of measure \(\mu(K_j)\), i.e., \(f^*=\sum_{j=1}^m |a_j|\chi_{[t_{j-1}, t_j)}\). Therefore, \[\vvvert f \vvvert_{L_{p,1}} = \int_0^\infty t^{-\frac{1}{p^*}}f^*(t) \, dt = \sum_{j=1}^m |a_j| \int_{t_{j-1}}^{t_j} t^{-\frac{1}{p^*}} \, dt = p\sum_{j=1}^m |a_j|(t_j^\frac{1}{p}-t_{j-1}^\frac{1}{p})\] Now, we can use the above estimates to obtain that \[\begin{align} \|\hat{T}f\|&\leq \sum_{j=1}^m (|a_j|-|a_{j+1}|)\|\hat{T}f_j\| \leq \sum_{j=1}^m A'(|a_j|-|a_{j+1}|)\|f_j\|_q^{\frac{q}{p}} \|f_j\|_\infty^{1-\frac{q}{p}} \\ & = A' \sum_{j=1}^m (|a_j|-|a_{j+1}|) t_j^{\frac{1}{p}} = A' \sum_{j=1}^m |a_j|(t_j^\frac{1}{p}-t_{j-1}^\frac{1}{p}) =\frac{A'}{p} \vvvert f \vvvert_{L_{p,1}}. \end{align}\] Since simple functions are dense in \(L_\infty (K,\mu)\), and the identity operator from \(L_\infty (K,\mu)\) into \(L_{p,1} (K,\mu)\) is continuous, we conclude that \[\|\hat{T}f\| \leq \frac{A'}{p} \vvvert f \vvvert_{L_{p,1}}\] for every \(f\in L_\infty(K,\mu)\), so in particular \((3)\) holds for every \(f\in C(K)\).
\((3)\Leftrightarrow (4)\) Since \(J_{p,1}(C(K))\) is dense in \(L_{p,1}(K,\mu)\), condition \((3)\) allows us to define an operator \(\overline{T}:L_{p,1}(K,\mu)\rightarrow E\) such that \(\overline{T}J_{p,1}=T\). For the reverse implication, just apply the boundedness of \(\overline{T}\) to any \(J_{p,1}f\) with \(f\in C(K)\).
\((4)\Rightarrow (1)\) By 62, \(J_{p,1}\) is \((p,q)\)-summing, so \(T=\overline{T}J_{p,1}\) is also \((p,q)\)-summing, with constant \[\pi_{p,q}(T)\leq \|\overline{T}\|\pi_{p,q}(J_{p,1}) \leq \intoo[2]{ p+\intoo[2]{\frac{1}{q}-\frac{1}{p}}^{-\frac{1}{q^*}}} A''. \qedhere\] ◻

Proof. \((1)\) Let us consider first \(T\in \Pi_{p,q}(C(K),E)\). It follows that \[\intoo[3]{\sum_{i=1}^n \|Tf_i\|^p}^\frac{1}{p} \leq \pi_{p,q}(T) \sup_{\mu\in B_{C(K)^*}} \intoo[3]{\sum_{i=1}^n |\mu(f_i)|^q}^\frac{1}{q} \leq \pi_{p,q}(T) \sup_{\mu\in B_{C(K)^*}}\sum_{i=1}^n |\mu(f_i)|\] for every \(f_1,\ldots,f_n\in C(K)\), so the first inequality holds. For the reverse inclusion, we exploit the fact that condition \((4)\) in 64 does not depend on the second exponent \(q\). Therefore, if \(T\in \Pi_{p,1}(C(K),E)\), using \((1)\Rightarrow (4)\) with exponent \(1\) we learn that there exists a probability measure \(\mu\in C(K)^*\) and an operator \(\overline{T}:L_{p,1}(K,\mu)\rightarrow E\) such that \(T=\overline{T}J_{p,1}\) and \(\|\overline{T}\|\leq p^{-\frac{1}{p^*}}\pi_{p,1}(T)\). Therefore, using now \((4)\Rightarrow (1)\) with exponent \(q\) we get that \[\pi_{p,q}(T)\leq \intoo[2]{ p+\intoo[2]{\frac{1}{q}-\frac{1}{p}}^{-\frac{1}{q^*}}}\|\overline{T}\|\leq \intoo[2]{ p^{\frac{1}{p}}+\intoo[2]{\frac{q}{p-q}}^{\frac{1}{q^*}}}\pi_{p,1}(T).\]

\((2)\) Let \(T\in \Pi_{p,1}(C(K),E)\), and apply 64 to find a probability measure \(\mu\in C(K)^*\) and an operator \(\overline{T}:L_{p,1}(K,\mu)\rightarrow E\) such that \(T=\overline{T}J_{p,1}\) and \(\|\overline{T}\|\leq p^{-\frac{1}{p^*}}\pi_{p,1}(T)\). Next, we observe that \(L_q(K,\mu)\subseteq L_{p,1}(K,\mu)\) and the identity operator \(j:L_q(K,\mu)\rightarrow L_{p,1}(K,\mu)\) is bounded. Indeed, if \(f\in L_q(K,\mu)\), then \[\begin{align} \vvvert f \vvvert_{L_{p,1}}& =\int_0^\infty t^{-\frac{1}{p^*}}f^*(t) \, dt = \int_0^{\mu(K)} t^{-\frac{1}{p^*}}f^*(t) \, dt\\ & \leq \intoo[3]{\int_0^1 t^{-\frac{q^*}{p^*}} \, dt}^\frac{1}{q^*} \intoo[3]{\int_0^1 f^*(t)^q \, dt}^\frac{1}{q} = \intoo[2]{ q^*\intoo[2]{\frac{1}{p}-\frac{1}{q}}}^{-\frac{1}{q^*}} \|f\|_q. \end{align}\] In particular, we can write \(J_{p,1}=j J_q\). By 62, \(J_q\) is \(q\)-summing with \(\pi_q(J_q)\leq 1\), so it follows that \[\pi_q(T)\leq \|\overline{T}\|\|j\|\pi_q(J_q) \leq p^{-\frac{1}{p^*}} \intoo[2]{ q^*\intoo[2]{\frac{1}{p}-\frac{1}{q}}}^{-\frac{1}{q^*}} \pi_{p,1}(T).\qedhere\] ◻

Proof. \((1)\) Let us fix \(1\leq q<p\). Then, by 25, \(T\) is \((p,q)\)-concave if and only if for every compact Hausdorff space \(K\) and every positive operator \(S:C(K)\rightarrow X\), \(TS:C(K)\rightarrow E\) is \((p,q)\)-summing. By 65, the operators \(TS\) are \((p,q)\)-summing if and only if they are \((p,1)\)-summing. Using again 25, we conclude that the latter happens if and only if \(T\) is \((p,1)\)-concave. Moreover, if \(T\) is \((p,q)\)-concave with constant \(K_{(p,q)}(T)\), then \(\pi_{p,1}(TS)\leq \pi_{p,q}(TS)\leq \|S\| K_{(p,q)}(T)\), so \(K_{(p,1)}(T) \leq K_{(p,q)}(T)\). On the other hand, if \(T\) is \((p,1)\)-concave with constant \(K_{(p,1)}(T)\), then repeating the argument we get the reverse inequality.
\((2)\) Let us fix \(p< q<\infty\). By 25, if \(T\) is \((p,1)\)-concave with constant \(K_{(p,1)}(T)\), then for every compact Hausdorff space \(K\) and every positive operator \(S:C(K)\rightarrow X\), \(TS:C(K)\rightarrow E\) is \((p,1)\)-summing, and hence \(q\)-summing by 65, with \[\pi_q(T)\leq p^{-\frac{1}{p^*}} \intoo[2]{ q^*\intoo[2]{\frac{1}{p}-\frac{1}{q}}}^{-\frac{1}{q^*}} \pi_{p,1}(TS)\leq p^{-\frac{1}{p^*}} \intoo[2]{ q^*\intoo[2]{\frac{1}{p}-\frac{1}{q}}}^{-\frac{1}{q^*}}\|S\| K_{(p,1)}(T).\] By 25 again, this implies that \(T\) is \(q\)-concave with constant \[K_{(q)}(T) \leq p^{-\frac{1}{p^*}} \intoo[2]{ q^*\intoo[2]{\frac{1}{p}-\frac{1}{q}}}^{-\frac{1}{q^*}} K_{(p,1)}(T). \qedhere\] ◻

Proof. This example is based on the same ideas that we used in 16. We start by showing that \(X\) is \(q\)-convex for any \(1\leq q<p\). Recall from 3.2 that both \(\ell_p\) and \(\ell_{p,\infty}\) are \(q\)-convex with constants 1 and \(\left(\frac{p}{p-q}\right)^{\frac{1}{q}}\), respectively. Given \(\overline{x}^{(i)}=(x_k^{(i)})_{k\in\mathbb{N}}\in X\), \(i=1,\ldots,n\), we have \[\begin{align} \norm[3]{\intoo[3]{\sum_{i=1}^n |\overline{x}^{(i)}|^q}^\frac{1}{q}}_X & = \norm[4]{\intoo[4]{\intoo[3]{\sum_{i=1}^n |x_k^{(i)}|^q}^\frac{1}{q}}_k}_X = \norm[4]{\intoo[4]{\norm[3]{\intoo[3]{\sum_{i=1}^n |x_k^{(i)}|^q}^\frac{1}{q}}_{\ell_p}}_k}_{\ell_{p,\infty}}\\ & \leq \norm[4]{\intoo[4]{\intoo[3]{\sum_{i=1}^n \|x_k^{(i)}\|_{\ell_p}^q}^\frac{1}{q}}_k}_{\ell_{p,\infty}} = \norm[4]{\intoo[3]{\sum_{i=1}^n \abs[2]{\intoo[2]{\|x_k^{(i)}\|_{\ell_p}}_k}^q}^\frac{1}{q}}_{\ell_{p,\infty}} \\ & \leq \left(\frac{p}{p-q}\right)^{\frac{1}{q}} \intoo[3]{\sum_{i=1}^n \norm[2]{\intoo[2]{\|x_k^{(i)}\|_{\ell_p}}_k}_{\ell_{p,\infty}} ^q}^\frac{1}{q} = \left(\frac{p}{p-q}\right)^{\frac{1}{q}} \intoo[3]{\sum_{i=1}^n \|\overline{x}^{(i)}\|_X^q}^\frac{1}{q}, \end{align}\] where the first and second inequalities follow from the \(q\)-convexity of \(\ell_p\) and \(\ell_{p,\infty}\), respectively.

To show that \(X\) does not satisfy an upper \(p\)-estimate, let us assume that the coordinates in \(\ell_p\) and \(\ell_{p,\infty}\) are indexed starting at \(0\). Let us fix \(n\geq 1\) and denote by \((e_j)_{j=0}^\infty\) the canonical basis of \(\ell_p\). For \(k=0,\ldots,n-1\), we write \(a^{(k)}=\sum_{j=0}^{n-1} \alpha_{(k+j)_n} e_j\) for the cyclic permutations of the coefficients \(\alpha_k=(k+1)^\frac{1}{p^*}-k^\frac{1}{p^*}\) over the set \(\{0,\ldots,n-1\}\). Here, by \(i_n\) we mean \(i \mod n\). We consider for \(i=0,\ldots,n-1\) the (finite) sequences \(\overline{x}^{(i)}=(x_k^{(i)})_{k=0}^\infty\in X\) given by \(x_k^{(i)}=\alpha_{(k+i)_n} e_i\) for \(k=0,\ldots,n-1\) and \(x_k^{(i)}=0\) for \(k\geq n\). It is clear that for every \(i\), \[\|\overline{x}^{(i)}\|_X=\|(\alpha_{(k+i)_n})_{k=0}^{n-1}\|_{\ell_{p,\infty}}=\|(\alpha_{k})_{k=0}^{n-1}\|_{\ell_{p,\infty}}=1,\quad \text{so }\intoo[3]{\sum_{i=0}^{n-1} \|\overline{x}^{(i)}\|_X^p}^\frac{1}{p}=n^\frac{1}{p}.\] On the other hand, for every \(k=0,\ldots, n-1\), \[\bigvee_{i=0}^{n-1}|x_k^{(i)}|=\sum_{i=0}^{n-1} \alpha_{(k+i)_n} e_i=a^{(k)},\] so \[\norm[3]{\bigvee_{i=0}^{n-1} |\overline{x}^{(i)}|}_X = \norm[4]{\intoo[4]{\norm[3]{\bigvee_{i=0}^{n-1} |x_k^{(i)}|}_{\ell_p}}_{k=0}^{n-1}}_{\ell_{p,\infty}}= \norm[2]{\intoo[2]{\|a^{(k)}\|_{\ell_p}}_{k=0}^{n-1}}_{\ell_{p,\infty}}= A_n n^\frac{1}{p},\] where \(A_n=\intoo{\sum_{j=0}^{n-1} \alpha_j^p}^\frac{1}{p}\) is the norm of \(a^{(k)}\) in \({\ell_p}\) for every \(k\), and \(n^\frac{1}{p}\) is the norm in \(\ell_{p,\infty}\) of the constant 1 vector on the set \(\{0,\ldots,n-1\}\). As we checked in 16, \(\alpha_k \approx (k+1)^{-\frac{1}{p}}\) uniformly in \(k\), so \(A_n\approx \intoo{\sum_{k=1}^n k^{-1}}^\frac{1}{p}\), which diverges. To conclude the argument, observe that the quotients \(\norm{\bigvee_{i=0}^{n-1} |\overline{x}^{(i)}|}_X /\intoo{\sum_{i=0}^{n-1} \|\overline{x}^{(i)}\|_X^p}^\frac{1}{p}=A_n\) are not uniformly bounded, so \(X\) cannot be \(({p,\infty})\)-convex. ◻

Proof. Let \(x\in S=S_X\cap X_+\). Choose \(x^*\in X^*_+\) such that \(\|x^*\|=1\) and \(x^*(x) = 1\) and define \(\rho_x:X\rightarrow \mathbb{R}\) by \(\rho_x(z) = x^*(|z|)\). Then \(\rho_x\) induces an \(AL\)-norm \(\hat{\rho}_x\) on \(X/\ker \rho_x\). Let \(q_x: X\rightarrow X/\ker\rho_x\) be the quotient map. By Kakutani’s Representation Theorem for \(AL\)-spaces, we know that the completion of \((X/\ker \rho_x,\hat{\rho}_x)\) is lattice isometric to \(L_1(\mu_x)\) for some measure \(\mu_x\), so let us denote by \(i_x: X/\ker\rho_x\rightarrow L_1(\mu_x)\) the formal inclusion, that satisfies \(\|i_x q_x x\|_{L_1(\mu_x)}=\hat{\rho}_x(q_xx)=\rho_x(x)=1\). Since \(X\) is \(p\)-convex or satisfies an upper \(p\)-estimate (i.e., it is \((p,\infty)\)-convex), depending on the case, and \(i_x q_x\) is positive (actually, a lattice homomorphism) and contractive, it follows from 12(7) that \(i_x q_x\) is \(p\)-convex or \((p,\infty)\)-convex, respectively, with the same constant as \(X\). Therefore, 55 and 57 yield that there exists \(g_x\in L_1(\mu_x)_+\) with \(\|g_x\|_{L_1(\mu_x)}= 1\), such that \(i_x q_x\) factors as \(M_x R_x\), with \(R_x z=g_x^{-1}i_x q_x z\) for any \(z\in X\), \(M_x\) is the multiplication by \(g_x\) operator, and \[\|R_x z\|_{L_p(g_x\cdot \mu_x)} \leq K^{(p)}(X)\|x\|, \quad \text{respectively,}\quad \|R_x z\|_{L_{p,\infty} (g_x\cdot \mu_x)} \leq \gamma_p K^{(\uparrow p)}(X)\|x\|.\] Define \[J: X\rightarrow \ell_\infty( L_p(g_x\cdot \mu_x))_{x\in S} , \quad \text{respectively}\quad J : X\rightarrow \ell_\infty( L_{p,\infty}(g_x\cdot \mu_x))_{x\in S}\] by \(Jz = (R_xz)_{x\in S}\) for every \(z\in X\). It is clear that \(R_x\), and thus, \(J\), are lattice homomorphisms. Moreover, \(\|J\|=\sup_{x\in S}\|R_x\|\) is bounded by \(K^{(p)}(X)\) or \(\gamma_p K^{(\uparrow p)}(X)\), respectively. On the other hand, since \(M_x\) is a contraction in both cases, for every \(x\in S\) we have that, in the \(p\)-convex setting, \[\|Jx\| \geq \|R_x x\|_{L_p(g_x\cdot\mu_x)} \geq \|i_x q_xx\|_{L_1(\mu_x)}=1,\] and in the \((p, \infty)\)-convex setting \[\|Jx\| \geq \|R_x x\|_{L_{p,\infty}(g_x\cdot\mu_x)} \geq \|i_x q_xx\|_{L_1(\mu_x)}=1.\] It follows that \(\|Jx\| \geq \|x\|\) for all \(x\in X\) in both cases. ◻

Proof. \((1)\) It is easy to check that \(Y^*=\ell_\infty( Y^*_\gamma)_{\gamma\in\Gamma}\) and that \(T^*x^*=(T^*_\gamma x^*)_{\gamma\in\Gamma}\), which is clearly a lattice homomorphism. Therefore, \(T\) must be almost interval preserving.
\((2)\) \(Y\) is order continuous, so \(J_Y:Y\rightarrow Y^{**}\) is interval preserving [8]. Moreover, \(X\) is a KB-space, so \(J_X(X)\) is a projection band in \(X^{**}\) [8], so there exists an interval preserving operator \(P:X^{**}\rightarrow X\) such that \(P J_X\) is the identity on \(X\). Since \(T\) is almost interval preserving, \(T^{**}:Y^{**}\rightarrow X^{**}\) is interval preserving. But observe that \[T=P J_X T=P T^{**} J_Y,\] so \(T\) must be interval preserving, as it is the composition of interval preserving operators. ◻

Proof of 71. We start as in the proof of 68 by fixing \(x\in S=S_X\cap X_+\) and considering the quotient \(q_x:X^*\rightarrow X^*/\ker \rho_x\), where \(\rho_x(x^*)=|x^*|(x)\), and \(i_x:X^*/\ker \rho_x\rightarrow L_1(\mu_x)\) the inclusion into the completion of \((X^*/\ker \rho_x,\rho_x)\), which is an \(AL\)-space. It is clear that \(i_x q_x\) is a \(q^*\)-convex operator with constant \(K_{(q)}(X)\) (respectively, \((q^*,\infty)\)-convex operator with constant \(K_{(\downarrow q)}(X)\)). By 55 (respectively, 57), there exists a normalized \(g_x\in L_1(g_x)_+\) such that \(i_x=M_x R_x\), where \(R_x x^*=g_x^{-1}i_x q_x x^*\) is defined from \(X^*\) into \(L_{q^*}(g_x\cdot \mu_x)\) with norm \(K_{(q)}(X)\) (respectively, \(L_{q^*,\infty}(g_x\cdot \mu_x)\) with norm \(\gamma_{q^*}K_{(\downarrow q)}(X)\)) and \(M_x\) is the operator of multiplication by \(g_x\).
Recall that in both cases \(X\) is a KB-space (26), so by [8] there exists an interval preserving operator \(P:X^{**}\rightarrow X\) such that \(P J_X\) is the identity on \(X\). Let us define \(Y:=\ell_1(L_q(\mu))_{\mu\in\Gamma}\) and \(Q_x=P R_x^*:L_q(g_x\cdot \mu_x)\rightarrow X\) when \(X\) is \(q\)-concave, and \(Y:=\ell_1(L_{q,1}(\mu))_{\mu\in\Gamma}\) and \(Q_x=P R_x^* J_{L_{q,1}}:L_{q,1}(g_x\cdot \mu_x)\rightarrow X\) when \(X\) satisfies a lower \(q\)-estimate, and denote by \(j_x\) and \(P_x\), \(x\in S\), the embeddings and projections associated to the factors that form \(Y\), which are clearly interval preserving. In both cases, the operators \(Q_x\) are interval preserving (for the second case, use the fact that \(L_{q,1}(g_x\cdot \mu_x)\) is order continuous by 26, so it embeds as an ideal into its bidual [8], which is \(L_{q^*,\infty}(g_x\cdot \mu_x)^*\)), so the compositions \(Q_x P_x\) are interval preserving as well, and their norms are uniformly bounded. Applying 72(1) we obtain an almost interval preserving operator \(Q:Y\rightarrow X\) given by \(Q\overline{f}=\sum_{x\in S} Q_xf_x\) for every \(\overline{f}=(f_x)_{x\in S}\in Y\), with norm bounded by either \(K_{(q)}(X)\) or \(\gamma_{q^*}K_{(\downarrow q)}(X)\). Since \(Y\) is \(q\)-concave or satisfies an upper \(q\)-estimate, it is order continuous, and applying 72(2) we get that \(Q\) is actually interval preserving.
To check the surjectivity of \(Q\), observe that \(Q( j_x\mathbb{1}_x)=x\) for every \(x\in S\), where \(\mathbb{1}_x\) denotes the constant one function in the corresponding \(L_{q}(g_x\cdot \mu_x)\) or \(L_{q,1}(g_x\cdot \mu_x)\) (we only check it in the lower \(q\)-estimate setting, as the \(q\)-concave setting is slightly easier): for every \(x^*\in X^*_+\), \[\begin{align} (R_x^* J_{L_{q,1}}\mathbb{1}_x)(x^*)& =(J_{L_{q,1}}\mathbb{1}_x)(R_x x^*)=(R_x x^*)(\mathbb{1}_x) =\int \frac{i_x q_x x^*}{g_x}\mathbb{1}_x g_x\, d\mu_x\\ &=\|i_x q_x x^*\|_{L_1(\mu_x)}=\rho_x(x^*)=x^*(x)=J_Xx(x^*), \end{align}\] so \(R_x^* J_{L_{q,1}}\mathbb{1}_x=J_X x\) and hence \(Q( j_x\mathbb{1}_x)=PJ_X x=x\). This fact alone already implies that \(Q\) is onto, but actually we can show that \(B_X\subseteq Q(B_Y)\). Indeed, let \(0\neq z\in B_X\). If \(z\geq 0\), then \(z=Q(\|z\|j_x\mathbb{1}_x)\in Q(B_Y)\) for \(x=z/\|z\|\in S\). Otherwise, let \(x=|z|/\|z\|\in S\) and use the fact that \(Q_x\) is interval preserving to find \(f_\pm\in [0,\mathbb{1}_x]\) such that \(Q_xf_\pm=z_\pm\in [0,x]\). Clearly, \(f=f_+-f_-\in [-\mathbb{1}_x,\mathbb{1}_x]\), so \(z=Q(j_xf)\in Q(B_Y)\), and the proof is concluded. ◻

Proof. (1) Without loss of generality, we can assume that the inclusion of \(\widetilde{X}\) into \(L_1(\mu)\) is contractive. Let \(f\in X\) be a simple function. Then, for every \(t\geq 0\) we have \[F(t)=\int (t|f|^p+1)^\frac{1}{p}\,d\mu = \|(t|f|^p+1)^\frac{1}{p}\|_1 \leq \|(t|f|^p+1)^\frac{1}{p}\|_{\widetilde{X}}\leq (t\|f\|_{\widetilde{X}}^p+1)^\frac{1}{p}=G(t).\] Note that \(F(0)=1=G(0)\) and both \(F\) and \(G\) are differentiable, since \(f\) is simple (one can also argue using a derivation under the integral sign argument, cf. [30]), so \[\frac{1}{p}\|f\|_p^p=\frac{1}{p}\int|f|^p=F'(0)\leq G'(0)=\frac{1}{p}\|f\|_{\widetilde{X}}^p,\] that is, \(\|f\|_p\leq \|f\|_{\widetilde{X}}\) for every simple function. Now, recall that if \(f\) is positive measurable function, we can always find an increasing sequence \((f_n)_n\) of simple functions that converges pointwise to \(f\). In particular, for every \(f\in \widetilde{X}_+\), let \((f_n)_n\) be a sequence of this kind, so that \(f_n^p\uparrow f^p\). Using Fatou’s Lemma, we conclude that \[\int f^p\, d\mu \leq \liminf \int f_n^p \, d\mu \leq \liminf \|f_n\|_{\widetilde{X}}^p \leq \|f\|_{\widetilde{X}}^p,\] so \(f\in L_p(\mu)\) and the inclusion is contractive.
(2) If \(X\) is \(q\)-concave, we can assume by rescaling the norm of \(\widetilde{X}\) if needed that \(\widetilde{X}^*\) is contractively contained in \(L_1(\mu)\). Now, repeating the argument from (1), we deduce that \(\widetilde{X}^*\subseteq L_{q^*}(\mu)\) and \(\|g\|_{q^*}\leq \|g\|_{\widetilde{X}^*}\) for every \(g\in \widetilde{X}^*\). Now, given any simple function \(f\in \widetilde{X}_+\), for every \(\varepsilon>0\) we can find \(g\in \widetilde{X}^*\) with \(\|g\|_{\widetilde{X}^*}=1\) such that \[\|f\|_{\widetilde{X}}-\varepsilon\leq \int fg\, d\mu\leq \|f\|_q \|g\|_{q^*}\leq \|f\|_q\|g\|_{\widetilde{X}^*}\leq \|f\|_q,\] so \(\|f\|_{\widetilde{X}}\leq \|f\|_q\) for every simple function. Now, let \(f\in L_q(\mu)_+\) and \((f_n)_n\) be a sequence of simple functions such that \(f_n\uparrow f\). Since \(L_q(\mu)\) is order continuous, \((f_n)_n\) converges to \(f\), so in particular \((f_n)_n\) is a Cauchy sequence in \(L_q(\mu)\). By the inequality above, \(\|f_n-f_m\|_{\widetilde{X}}\leq \|f_n-f_m\|_q\), so \((f_n)_n\) is also Cauchy in \(\widetilde{X}\), so it converges to some \(h\in \widetilde{X}\). Since \((f_n)_n\) is increasing, \(h=\sup_n f_n\), but the suprema of sequences in \(L_q(\mu)\) and \(\widetilde{X}\) coincide, as they are both ideals in \(L_1(\mu)\), so \(h=f\). We conclude that \(\|f\|_{\widetilde{X}}=\lim_n \|f_n\|_{\widetilde{X}}\leq \lim_n \|f_n\|_q= \|f\|_q\). ◻