$(p,\sigma )$-Absolute continuity of Bloch maps

Abstract

Motivated by new progress in the theory of ideals of Bloch maps, we introduce $(p,\sigma )$-absolutely continuous Bloch maps with $p\in [1,\infty )$ and $\sigma \in [0,1)$ from the complex unit open disc $\mathbb {D}$ into a complex Banach space X. We prove a Pietsch domination/factorization theorem for such Bloch maps that provides a reformulation of some results on both absolutely continuous (multilinear) operators and Lipschitz operators. We also identify the spaces of $(p,\sigma )$-absolutely continuous Bloch zero-preserving maps from $\mathbb {D}$ into $X^*$ under a suitable norm $\pi ^{\mathcal {B}}_{p,\sigma }$ with the duals of the spaces of X-valued Bloch molecules on $\mathbb {D}$ equipped with the Bloch version of the $(p^*,\sigma )$-Chevet–Saphar tensor norms.

1 Introduction and preliminaries

For any Banach spaces X and Y, $\mathcal {L}(X,Y)$ denotes the Banach space of all continuous linear operators from X into Y, under the operator norm. In particular, $\mathcal {L}(X,\mathbb {K})$ is denoted by $X^*$. As usual, $B_{X}$ stands for the closed unit ball of X.

Recall that $T\in \mathcal {L}(X,Y)$ is called p-summing with $p\in [1,\infty )$ if there exists $C\ge 0$ so that

$$\begin{aligned} \left( \sum _{i=1}^n\left\| T(x_i)\right\| ^p\right) ^{\frac{1}{p}}\le C \sup _{x^*\in B_{X^*}}\left( \sum _{i=1}^n\left| x^*(x_i)\right| ^p\right) ^{\frac{1}{p}} \end{aligned}$$

for any $n\in \mathbb {N}$ and $x_1,\ldots ,x_n\in X$. The infimum of such constants C is denoted by $\pi _p(T)$, and the Banach space of all p-summing operators from X into Y, under the norm $\pi _p$, by $\Pi _p(X,Y)$.

In the eighties, Matter considered the ideal of $(p,\sigma )$-absolutely continuous linear operators for any $p\in [1,\infty )$ and $\sigma \in [0,1)$, with the aim of analysing super-reflexive Banach spaces, providing its main properties in the papers [13, 14].

Let us recall that a linear map $T:X\rightarrow Y$ is called $(p,\sigma )$-absolutely continuous for $p\in [1,\infty )$ and $\sigma \in [0,1)$ if there exist a Banach space Z and a p-summing operator $S\in \Pi _p(X,Z)$ for which

$$\begin{aligned} \left\| T(x)\right\| \le \left\| x\right\| ^\sigma \left\| S(x)\right\| ^{1-\sigma }\qquad (x\in X). \end{aligned}$$

We set $\pi _{p,\sigma }(T)=\inf \{\pi _p(S)^{1-\sigma }\}$, where the infimum is taken over all Banach spaces Z and $S\in \Pi _p(X,Z)$ such that the above inequality holds. Let $\Pi _{p,\sigma }(X,Y)$ be the Banach space of all $(p,\sigma )$-absolutely continuous operators from X into Y, under the norm $\pi _{p,\sigma }$.

In the nineties, López Molina and Sánchez Pérez investigated on the factorization properties and the tensor norms related to these operator ideals in the papers [11, 12, 19]. Roughly speaking, the ideal of $(p,\sigma )$-absolutely continuous operators can be considered as an interpolating ideal between the p-summing operators and the continuous operators since

$$\begin{aligned} \Pi _{p}(X,Y)\subseteq \Pi _{p,\sigma }(X,Y)\subseteq \mathcal {L}(X,Y) \end{aligned}$$

with

$$\begin{aligned} \left\| T\right\| \le \pi _{p,\sigma }(T)\le \pi _p(T)\qquad (T\in \Pi _{p}(X,Y)). \end{aligned}$$

We refer the reader to the book [9] for a complete study on p-summing operators.

In the second decade of the twentieth century, Achour, Dahia, Rueda and Sánchez Pérez dealt with the factorization of both absolutely continuous polynomials and strongly $(p,\sigma )$-continuous multilinear operators in [1, 2]. Besides, Achour, Rueda and Yahi [3] extended these studies for Lipschitz maps from a metric space into a Banach space.

Our main purpose in this paper is to introduce and establish the most notable properties of a notion of $(p,\sigma )$-absolutely continuous Bloch map on the open unit disc $\mathbb {D}\subseteq \mathbb {C}$, in terms of the concept of p-summing Bloch map. From now on, unless otherwise stated, X will denote a complex Banach space.

If $\mathcal {H}(\mathbb {D},X)$ represents the space of all holomorphic maps from $\mathbb {D}$ into X, a map $f\in \mathcal {H}(\mathbb {D},X)$ is called Bloch if

$$\begin{aligned} \rho _{\mathcal {B}}\left( f\right) :=\sup \left\{ (1-|z|^{2})\left\| f'(z)\right\| :z\in \mathbb {D}\right\} <\infty . \end{aligned}$$

The linear space of all Bloch maps from $\mathbb {D}$ into X, under the Bloch seminorm $\rho _{\mathcal {B}}$, is denoted by $\mathcal {B}(\mathbb {D},X)$. The normalized Bloch space $\widehat{\mathcal {B}}(\mathbb {D},X)$ is the closed subspace of $\mathcal {B}(\mathbb {D},X)$ formed by all those maps f for which $f(0)=0$, under the Bloch norm $\rho _{\mathcal {B}}$. For simplicity, we write $\widehat{\mathcal {B}}\left( \mathbb {D}\right)$ instead of $\widehat{\mathcal {B}}(\mathbb {D},\mathbb {C)}$. Numerous authors have studied these function spaces (see, for example, the monographs [4] for the complex-valued case, and [20] for the vector-valued case).

In a recent paper [6], the p-summability of operators was adapted to address the property of p-summability in the setting of Bloch maps, as follows.

For any $p\in [1,\infty )$, we say that a map $f\in \mathcal {H}(\mathbb {D},X)$ is p-summing Bloch if there exists $C\ge 0$ such that for any n in $\mathbb {N}$, $\lambda _1,\ldots ,\lambda _n$ in $\mathbb {C}$ and $z_1,\ldots ,z_n$ in $\mathbb {D}$, one has

$$\begin{aligned} \left( \sum _{i=1}^n\left| \lambda _i\right| ^p\left\| f'(z_i)\right\| ^p\right) ^{\frac{1}{p}}\le C\sup _{g\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \sum _{i=1}^n\left| \lambda _i\right| ^p\left| g'(z_i)\right| ^p\right) ^{\frac{1}{p}}. \end{aligned}$$

The infimum of the constants C for which this inequality holds, denoted by $\pi _{p}^{\mathcal {B}}$, defines a seminorm on the linear space $\Pi _{p}^{\mathcal {B}}(\mathbb {D},X)$ of all p-absolutely continuous Bloch maps from $\mathbb {D}$ into X. Furthermore, this seminorm becomes a norm on the subspace $\Pi _{p}^{\widehat{\mathcal {B}}}(\mathbb {D},X)$ consisting of all those maps $f\in \Pi _{p}^{\mathcal {B}}\left( \mathbb {D},X\right)$ so that $f(0)=0$. A complete study on these spaces can be consulted in [6].

Now, we introduce the Bloch analogue of the notion of $(p,\sigma )$-absolutely continuous operator.

Definition 1.1

For any $p\in [1,\infty )$ and $\sigma \in [0,1)$, we say that a map $f\in \mathcal {H}\left( \mathbb {D},X\right)$ is $(p,\sigma )$-absolutely continuous Bloch if there exist a complex Banach space Y and a map $g\in \Pi _{p}^{\mathcal {B}}\left( \mathbb {D},Y\right)$ such that

$$\begin{aligned} \left\| f'\left( z\right) \right\| \le \left( \frac{1}{1-\left| z\right| ^{2}}\right) ^{\sigma }\left\| g'\left( z\right) \right\| ^{1-\sigma }\qquad (z\in \mathbb {D}). \end{aligned}$$

In such case, we put

$$\begin{aligned} \pi _{p,\sigma }^{\mathcal {B}}\left( f\right) =\inf \left\{ \pi _{p}^{\mathcal {B}}\left( g\right) ^{1-\sigma }\right\} , \end{aligned}$$

taking the infimum over all complex Banach spaces Y and all $g\in \Pi _{p}^{\mathcal {B}}\left( \mathbb {D},Y\right)$ such that the above inequality holds. $\Pi _{p,\sigma }^{\mathcal {B}}\left( \mathbb {D},X\right)$ stands for the linear space of all $(p,\sigma )$-absolutely continuous Bloch maps $f:\mathbb {D}\rightarrow X$. The linear subspace of $\Pi _{p,\sigma }^{\mathcal {B}}\left( \mathbb {D},X\right)$ consisting of all those maps f for which $f(0)=0$ is denoted by $\Pi _{p,\sigma }^{\widehat{\mathcal {B}}}\left( \mathbb {D},X\right)$.

We divide the contents of this paper into some sections. We start by showing that $(\Pi _{p,0}^{\mathcal {B}}(\mathbb {D},X),\pi _{p,0}^{\mathcal {B}})$ can be identified with $(\Pi _p^{\mathcal {B}}(\mathbb {D},X),\pi _{p}^{\mathcal {B}})$. For this reason, the results that we establish in this paper extend some obtained in [6]. In a clear parallel with the linear setting, the class $\Pi _{p,\sigma }^{\mathcal {B}}$ can be considered as an interpolating class between the classes $\Pi _{p}^{\mathcal {B}}$ and $\mathcal {B}$.

In Sects. 2 and 5, we prove that $[\Pi _{p,\sigma }^{\widehat{\mathcal {B}}},\pi _{p,\sigma }^{\mathcal {B}}]$ is an injective Banach normalized Bloch ideal. Sections 3 and 4 are devoted to both versions of Pietsch domination theorem and Pietsch factorization theorem for $(p,\sigma )$-absolutely continuous Bloch maps on $\mathbb {D}$. We also address the invariance of the space $(\Pi ^{\mathcal {B}}_p(\mathbb {D},X),\pi ^{\mathcal {B}}_p)$ by Möbius transformations of $\mathbb {D}$. In Sect. 6, we introduce and analyse the so-called $(p,\sigma )$-Chevet–Saphar Bloch norms $d^{\widehat{\mathcal {B}}}_{p,\sigma }$ on the tensor product space $\mathcal {G}(\mathbb {D})\otimes X$, where $\mathcal {G}(\mathbb {D})$ is the Bloch-free Banach space. If $p^*=\infty$ for $p=1$, and $p^*=p/(p-1)$ for $1<p<\infty$, we show that $(\Pi _{p,\sigma }^{\widehat{\mathcal {B}}}(\mathbb {D},X^*),\pi _{p,\sigma }^{\mathcal {B}})$ can be canonically identified with the dual of the completion of the space $\mathcal {G}(\mathbb {D})\otimes _{d^{\widehat{\mathcal {B}}}_{p^*,\sigma }}X$.

2 Banach structure

We begin with an easy result on interpolation which can be compared to [13, Proposition 3.3]. We will need the following class of Bloch functions. For each $z\in \mathbb {D}$, the map $f_z:\mathbb {D}\rightarrow \mathbb {C}$ defined by

$$\begin{aligned} f_z(w)=\frac{(1-|z|^2)w}{1-\overline{z}w}\qquad (w\in \mathbb {D}), \end{aligned}$$

is in $\widehat{\mathcal {B}}(\mathbb {D})$ and $\rho _{\mathcal {B}}(f_z)=1=(1-|z|^2)f_z'(z)$ (see [10, Proposition 2.2]). Clearly, $f_z\in \Pi _p^{\widehat{\mathcal {B}}}(\mathbb {D},\mathbb {C})$ with $\pi _{p}^{\mathcal {B}}(f_z)\le 1$ for any $p\in [1,\infty )$.

Given two semi-normed spaces $(X,\rho _X)$ and $(Y,\rho _Y)$, we will write $(X,\rho _X)\le (Y,\rho _Y)$ to indicate that $X\subseteq Y$ and $\rho _Y(x)\le \rho _X(x)$ for all $x\in X$.

Proposition 2.1

If $p,q\in [1,\infty )$ with $p<q$ and $\sigma \in [0,1)$, then

$$\begin{aligned} (\Pi _{p,0}^{\mathcal {B}}(\mathbb {D},X),\pi _{p,0}^{\mathcal {B}})&=(\Pi _p^{\mathcal {B}}(\mathbb {D},X),\pi _p^{\mathcal {B}})\le (\Pi _{p,\sigma }^{\mathcal {B}}(\mathbb {D},X),\pi _{p,\sigma }^{\mathcal {B}})\\&\le (\Pi _{q,\sigma }^{\mathcal {B}}(\mathbb {D},X),\pi _{q,\sigma }^{\mathcal {B}})\le (\mathcal {B}(\mathbb {D},X),\rho _{\mathcal {B}}). \end{aligned}$$

Proof

If $f\in \Pi _{p,0}^{\mathcal {B}}(\mathbb {D},X)$, there is a map $g\in \Pi _{p}^{\mathcal {B}}(\mathbb {D},Y)$ for some complex Banach space Y such that $\left\| f'(z)\right\| \le \left\| g'(z)\right\|$ for all $z\in \mathbb {D}$. Given $n\in \mathbb {N}$, $\lambda _i\in \mathbb {C}$ and $z_i\in \mathbb {D}$ for all $i\in \{1,\ldots ,n\}$, we get

$$\begin{aligned} \left( \sum _{i=1}^n\left| \lambda _i\right| ^p\left\| f'(z_i)\right\| ^p\right) ^{\frac{1}{p}}&\le \left( \sum _{i=1}^n\left| \lambda _i\right| ^p\left\| g'(z_i)\right\| ^p\right) ^{\frac{1}{p}}\\&\le \pi _p^{\mathcal {B}}(g)\sup _{h\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \sum _{i=1}^n\left| \lambda _i\right| ^p\left| h'(z_i)\right| ^p\right) ^{\frac{1}{p}}, \end{aligned}$$

hence $f\in \Pi _p^{\mathcal {B}}(\mathbb {D},X)$ with $\pi _p^{\mathcal {B}}(f)\le \pi _p^{\mathcal {B}}(g)$, and passing to the infimum over all such complex Banach spaces Y and all such maps g, one has $\rho _{\mathcal {B}}(f)\le \pi ^\mathcal {B}_{p,0}(f)$.

The inequality $(\Pi _p^{\mathcal {B}}(\mathbb {D},X),\pi _p^{\mathcal {B}})\le (\Pi _{p,0}^{\mathcal {B}}(\mathbb {D},X),\pi _{p,0}^{\mathcal {B}})$ is a particular case of the following. If $f\in \Pi _p^{\mathcal {B}}(\mathbb {D},X)$, then

$$\begin{aligned} \left\| f'(z)\right\|&\le \pi _{p}^{\mathcal {B}}(f)\sup _{g\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left| g'(z)\right| \le \pi _{p}^{\mathcal {B}}(f)\frac{1}{1-|z|^2}\\&=\pi _{p}^{\mathcal {B}}(f)\left( \frac{1}{1-|z|^2}\right) ^\sigma \left| f'_z(z)\right| ^{1-\sigma } \qquad (z\in \mathbb {D}). \end{aligned}$$

as for the second inequality we use the supremum is taken over $g's$ in $B_{\widehat{\mathcal {B}}(\mathbb {D})}$ and that $\rho _{\mathcal {B}}(g)\ge (1-|z|^{2})\left\| g'(z)\right\|$. Hence $f\in \Pi _{p,\sigma }^{\mathcal {B}}(\mathbb {D},X)$ with

$$\begin{aligned} \pi _{p,\sigma }^{\mathcal {B}}(f)\le \pi _{p}^{\mathcal {B}}(\pi _{p}^{\mathcal {B}}(f)^{\frac{1}{1-\sigma }}f_z)^{1-\sigma } =\pi _{p}^{\mathcal {B}}(f)\pi _{p}^{\mathcal {B}}(f_z)^{1-\sigma }\le \pi _{p}^{\mathcal {B}}(f). \end{aligned}$$

If $f\in \Pi _{p,\sigma }^{\mathcal {B}}(\mathbb {D},X)$, then $f\in \Pi _{q,\sigma }^{\mathcal {B}}(\mathbb {D},X)$ with $\pi _{q,\sigma }^{\mathcal {B}}(f)\le \pi _{p,\sigma }^{\mathcal {B}}(f)$ follows readily by applying [6, Proposition 1.1].

If $f\in \Pi _{q,\sigma }^{\mathcal {B}}(\mathbb {D},X)$, we can take a complex Banach space Y and a map $g\in \Pi _{q}^{\mathcal {B}}(\mathbb {D},Y)$ such that

$$\begin{aligned} \left\| f'(z)\right\| \le \left( \frac{1}{1-|z|^2}\right) ^\sigma \left\| g'(z)\right\| ^{1-\sigma } \qquad (z\in \mathbb {D}). \end{aligned}$$

It follows that

$$\begin{aligned} (1-|z|^2)\left\| f'(z)\right\| \le \left( (1-|z|^2)\left\| g'(z)\right\| \right) ^{1-\sigma }\le \rho _\mathcal {B}(g)^{1-\sigma } \qquad (z\in \mathbb {D}), \end{aligned}$$

hence $f\in \mathcal {B}(\mathbb {D},X)$ with $\rho _{\mathcal {B}}(f)\le \rho _\mathcal {B}(g)^{1-\sigma }$, and taking infimum over all such complex Banach spaces Y and such maps g, we conclude that $\rho _{\mathcal {B}}(f)\le \pi ^\mathcal {B}_{q,\sigma }(f)$. $\square$

The case $\sigma =0$ in the next result follows from Proposition 2.1 and [6, Proposition 1.2]. In fact, we can adapt the proof of [6, Proposition 1.2] to yield a more general result.

Proposition 2.2

$\left( \Pi _{p,\sigma }^{\widehat{\mathcal {B}}}\left( \mathbb {D},X\right) ,\pi _{p,\sigma }^{\mathcal {B}}\right)$ is a Banach space for any $p\in [1,\infty )$ and $\sigma \in [0,1)$.

Proof

Assume that $\sigma \in (0,1)$. If $f\in \Pi _{p,\sigma }^{\widehat{\mathcal {B}}}\left( \mathbb {D},X\right)$ and $\pi _{p,\sigma }^{\mathcal {B}}\left( f\right) =0$, then $\rho _{\mathcal {B}}\left( f\right) =0$ by Proposition 2.1, and so $f=0$. We now prove the triangle inequality. For $i=1,2$, consider $f_i\in \Pi _{p,\sigma }^{\widehat{\mathcal {B}}}\left( \mathbb {D},X\right)$, a complex Banach space $Y_i$, and $g_{i}\in \Pi _{p}^{\widehat{\mathcal {B}}}\left( \mathbb {D},Y_{i}\right)$ such that

$$\begin{aligned} \left\| f_i'\left( z\right) \right\| \le \left( \frac{1}{1-\left| z\right| ^{2}}\right) ^{\sigma }\left\| g'_{i}\left( z\right) \right\| _{Y_i} ^{1-\sigma }\qquad (z\in \mathbb {D}). \end{aligned}$$

Let Y be the $\ell _{1}$-sum of $Y_{1}$ and $Y_{2}$, and let $I_{i}:Y_{i}\rightarrow Y$ be the canonical injection. The map $g=\sum \nolimits _{i=1}^2\pi _{p}^{\mathcal {B}}\left( g_{i}\right) ^{-\sigma }(I_{i}\circ g_{i})$ belongs to $\Pi _{p}^{\widehat{\mathcal {B}}}\left( \mathbb {D},Y\right)$ and $\pi _{p}^{\mathcal {B}}\left( g\right) \le \sum _{i=1}^2\pi _{p}^{\mathcal {B}}\left( g_{i}\right) ^{1-\sigma }$. Using Holder’s Inequality, we get

$$\begin{aligned} \left\| \sum _{i=1}^2 f'_i\left( z\right) \right\|&\le \sum _{i=1}^2 \left\| f'_i\left( z\right) \right\| \le \left( \frac{1}{1-\left| z\right| ^{2}}\right) ^{\sigma }\sum _{i=1}^2\left\| g'_{i}\left( z\right) \right\| _{Y_i} ^{1-\sigma }\\&= \left( \frac{1}{1-\left| z\right| ^{2}}\right) ^{\sigma }\sum _{i=1}^2\left\| \pi _{p}^{\mathcal {B}}\left( g_i\right) ^{-\sigma }g'_{i}\left( z\right) \right\| _{Y_{i}} ^{1-\sigma }\pi _{p}^{\mathcal {B}}\left( g_{i}\right) ^{\sigma (1-\sigma )}\\&\le \left( \frac{1}{1-\left| z\right| ^{2}}\right) ^{\sigma }\left( \underset{i=1}{\overset{2}{\sum }}\left\| \pi _{p}^{\mathcal {B}}\left( g_{i}\right) ^{-\sigma }g_{i}'\left( z\right) \right\| _{Y_{i}}\right) ^{1-\sigma }\left( \sum _{i=1}^2\pi _{p}^{\mathcal {B}}\left( g_{i}\right) ^{1-\sigma }\right) ^{\sigma } \\&=\left( \frac{1}{1-\left| z\right| ^{2}}\right) ^{\sigma }\left\| g'\left( z\right) \right\| ^{1-\sigma } _{Y}\left( \pi _{p}^{\mathcal {B}}\left( g_{1}\right) ^{1-\sigma }+\pi _{p}^{\mathcal {B}}\left( g_{2}\right) ^{1-\sigma }\right) ^{\sigma }\qquad (z\in \mathbb {D}). \end{aligned}$$

Thus $\sum _{i=1}^2 f_i\in \Pi _{p,\sigma }^{\widehat{\mathcal {B}}}\left( \mathbb {D},X\right)$ with

$$\begin{aligned} \pi _{p,\sigma }^{\mathcal {B}}\left( \sum _{i=1}^2 f_i \right) \le \left( \sum _{i=1}^2\pi _{p}^{\mathcal {B}}\left( g_{i}\right) ^{1-\sigma }\right) ^{\sigma }\pi _{p}^{\mathcal {B}}\left( g\right) ^{1-\sigma } \le \sum _{i=1}^2\pi _{p}^{\mathcal {B}}\left( g_{i}\right) ^{1-\sigma }. \end{aligned}$$

Passing to the infimum over all such complex Banach spaces Y and such maps $g_1$ and $g_2$, we deduce that $\pi _{p,\sigma }^{\mathcal {B}}\left( \sum _{i=1}^2f_i\right) \le \sum _{i=1}^2\pi _{p,\sigma }^{\mathcal {B}}\left( f_i\right)$.

Let $\lambda \in \mathbb {C}$ and $f\in \Pi _{p,\sigma }^{\widehat{\mathcal {B}}}\left( \mathbb {D},X\right)$. We have a complex Banach space Y and $g\in \Pi _{p}^{\widehat{\mathcal {B}}}\left( \mathbb {D},Y\right)$ such that

$$\begin{aligned} \left\| f'\left( z\right) \right\| \le \left( \frac{1}{1-\left| z\right| ^{2}}\right) ^{\sigma }\left\| g'\left( z\right) \right\| ^{1-\sigma }\qquad (z\in \mathbb {D}). \end{aligned}$$

Therefore,

$$\begin{aligned} \left\| \left( \lambda f\right) '\left( z\right) \right\| \le \left| \lambda \right| \left( \frac{1}{1-\left| z\right| ^{2}}\right) ^{\sigma }\left\| g'\left( z\right) \right\| ^{1-\sigma } =\left( \frac{1}{1-|z|^2}\right) ^{\sigma }\left\| \left( \lambda ^{\frac{1}{1-\sigma }}g\right) '(z)\right\| ^{1-\sigma } \qquad (z\in \mathbb {D}). \end{aligned}$$

Since $\lambda ^{\frac{1}{1-\sigma }}g\in \Pi _{p}^{\widehat{\mathcal {B}}}\left( \mathbb {D},X\right)$, we have $\lambda f\in \Pi _{p,\sigma }^{\widehat{\mathcal {B}}}\left( \mathbb {D},X\right)$ with $\pi _{p,\sigma }^{\mathcal {B}}\left( \lambda f\right) \le \pi _{p}^{\mathcal {B}}\left( \lambda ^{\frac{1}{1-\sigma }}g\right) ^{1-\sigma }=\left| \lambda \right| \pi _{p}^{\mathcal {B}}(g)^{1-\sigma }$. For $\lambda =0$, we obtain $\pi _{p,\sigma }^{\mathcal {B}}\left( \lambda f\right) =0=\left| \lambda \right| \pi _{p,\sigma }^{\mathcal {B}}\left( f\right)$. For $\lambda \ne 0$, we deduce that $\pi _{p,\sigma }^{\mathcal {B}}\left( \lambda f\right) \le \left| \lambda \right| \pi _{p,\sigma }^{\mathcal {B}}\left( f\right)$. Hence $\pi _{p,\sigma }^{\mathcal {B}}\left( f\right) \le \left| \lambda \right| ^{-1}\pi _{p,\sigma }^{\mathcal {B}}\left( \lambda f\right)$, then $\left| \lambda \right| \pi _{p,\sigma }^{\mathcal {B}}\left( f\right) \le \pi _{p,\sigma }^{\mathcal {B}}\left( \lambda f\right)$, and thus $\pi _{p,\sigma }^{\mathcal {B}}\left( \lambda f\right) =\left| \lambda \right| \pi _{p,\sigma }^{\mathcal {B}}\left( f\right)$. So $(\Pi _{p,\sigma }^{\widehat{\mathcal {B}}}\left( \mathbb {D},X\right) ,\pi _{p,\sigma }^{\mathcal {B}})$ is a complex normed space.

To prove its completeness, let $(f_{n})$ be a sequence in $\Pi _{p,\sigma }^{\widehat{\mathcal {B}}}\left( \mathbb {D},X\right)$ for which $\sum _{n=1}^\infty \pi _{p,\sigma }^{\mathcal {B}}\left( f_{n}\right) <\infty$. Since $\rho _{\mathcal {B}}\le \pi _{p,\sigma }^{\mathcal {B}}$ on $\Pi _{p,\sigma }^{\widehat{\mathcal {B}}}\left( \mathbb {D},X\right)$ (by Proposition 2.1) and $\widehat{\mathcal {B}}\left( \mathbb {D},X\right)$ with the norm $\rho _{\mathcal {B}}$ is a Banach space, there exists $f\in \widehat{\mathcal {B}}\left( \mathbb {D},X\right)$ such that $\sum _{n=1}^\infty f_{n}=f$ for $\rho _{\mathcal {B}}$. We will prove that $\sum _{n=1}^\infty f_{n}=f$ for $\pi _{p,\sigma }^{\mathcal {B}}$. Let $\varepsilon >0$, and for each $n\in \mathbb {N}$, we can take a complex Banach space $Y_n$ and a map $g_{n}\in \Pi _{p}^{\widehat{\mathcal {B}}}\left( \mathbb {D},Y_{n}\right)$ for which

$$\begin{aligned} \left\| f_{n}'\left( z\right) \right\| \le \left( \frac{1}{1-\left| z\right| ^{2}}\right) ^{\sigma }\left\| g'_{n}\left( z\right) \right\| _{Y_n} ^{1-\sigma }\qquad (z\in \mathbb {D}), \end{aligned}$$

with

$$\begin{aligned} \pi _{p}^{\mathcal {B}}\left( g_{n}\right) ^{1-\sigma }\le \pi _{p,\sigma }^{\mathcal {B}}\left( f_{n}\right) +\frac{\varepsilon }{2^{n}}. \end{aligned}$$

Then

$$\begin{aligned} \underset{n=1}{\overset{\infty }{\sum }}\pi _{p}^{\mathcal {B}}\left( g_{n}\right) ^{1-\sigma }\le \underset{n=1}{\overset{\infty }{\sum }}\pi _{p,\sigma }^{\mathcal {B}}\left( f_{n}\right) +\varepsilon . \end{aligned}$$

Let $g={{\sum _{n=1}^{\infty }}}\pi _{p}^{\mathcal {B}}\left( g_{n}\right) ^{-\sigma }\left( I_{n}\circ g_{n}\right) \in \Pi _{p}^{\widehat{\mathcal {B}}}\left( \mathbb {D},Y\right)$, where Y is the $\ell _{1}$-sum of all $Y_{n}$ and $I_{n}:Y_{n}\rightarrow Y$ is the canonical injection. Hence

$$\begin{aligned} \left\| f'\left( z\right) \right\|&\le \underset{n=1}{\overset{\infty }{\sum }}\left\| f_{n}'\left( z\right) \right\| \le \underset{n=1}{\overset{\infty }{\sum }}\left( \frac{1}{1-\left| z\right| ^{2}}\right) ^{\sigma }\left\| g_{n}'\left( z\right) \right\| _{Y_{n}}^{1-\sigma } \\&\le \left( \frac{1}{1-\left| z\right| ^{2}}\right) ^{\sigma }\left\| g'\left( z\right) \right\| _{Y}^{1-\sigma }\left( \underset{n=1}{\overset{\infty }{\sum }}\pi _{p}^{\mathcal {B}}\left( g_{n}\right) ^{1-\sigma }\right) ^{\sigma }\qquad (z\in \mathbb {D}). \end{aligned}$$

This implies that $f\in \Pi _{p,\sigma }^{\widehat{\mathcal {B}}}\left( \mathbb {D},X\right)$ with

$$\begin{aligned} \pi _{p,\sigma }^{\mathcal {B}}\left( f\right)&\le \left( \underset{n=1}{\overset{\infty }{\sum }}\pi _{p}^{\mathcal {B}}\left( g_{n}\right) ^{1-\sigma }\right) ^{\sigma }\pi _p(g)^{1-\sigma }\\&\le \left( \underset{n=1}{\overset{\infty }{\sum }}\pi _{p}^{\mathcal {B}}\left( g_{n}\right) ^{1-\sigma }\right) ^{\sigma }\left( \underset{n=1}{\overset{\infty }{\sum }}\pi _{p}^{\mathcal {B}}\left( g_{n}\right) ^{1-\sigma }\right) ^{1-\sigma }\\&= \underset{n=1}{\overset{\infty }{\sum }}\pi _{p}^{\mathcal {B}}\left( g_{n}\right) ^{1-\sigma }\le \underset{n=1}{\overset{\infty }{\sum }}\pi _{p,\sigma }^{\mathcal {B}}\left( f_{n}\right) +\varepsilon . \end{aligned}$$

Moreover, we have

$$\begin{aligned} \pi _{p,\sigma }^{\mathcal {B}}\left( f-\underset{k=1}{\overset{n}{\sum }}f_{k}\right) =\pi _{p,\sigma }^{\mathcal {B}}\left( \underset{k=n+1}{\overset{\infty }{\sum }}f_{k}\right) \le \underset{k=n+1}{\overset{\infty }{\sum }}\pi _{p}^{\mathcal {B}}\left( f_{k}\right) \qquad (n\in \mathbb {N}), \end{aligned}$$

and thus $\sum \nolimits _{n=1}^{\infty } f_{n}=f$ for $\pi _{p,\sigma }^{\mathcal {B}}$. $\square$

3 Pietsch domination

Our next result is a reformulation for $(p,\sigma )$-absolutely continuous Bloch maps of Pietsch domination theorem for $(p,\sigma )$-absolutely continuous operators stated by Matter in [13, Theorem 4.1]. However, to prove our result, we will apply an unified abstract version of the Pietsch domination theorem established by Pellegrino and Santos in [15, Theorem 3.1] (see also [5, 16]). Our proof is based on [6, Theorem 1.4 and Lemma 1.5].

Let us recall that $\widehat{\mathcal {B}}\left( \mathbb {D}\right)$ is a dual Banach space (see, for example, [20]) and therefore we can consider this space equipped with its weak* topology. Let $\mathcal {P}(B_{\widehat{\mathcal {B}}\left( \mathbb {D}\right) })$ be the set of all Borel regular probability measures $\mu$ on $(B_{\widehat{\mathcal {B}}\left( \mathbb {D}\right) },w^{*})$.

Given $\mu \in \mathcal {P}(B_{\widehat{\mathcal {B}}(\mathbb {D})})$, $p\in [1,\infty )$ and $\sigma \in [0,1)$, consider the inclusion operators

$$\begin{aligned} I_{\infty ,p/(1-\sigma )}:L_\infty (\mu )\rightarrow L_{p/(1-\sigma )}(\mu ) \end{aligned}$$

and

$$\begin{aligned} j_{\infty }:C(B_{\widehat{\mathcal {B}}(\mathbb {D})})\rightarrow L_\infty (\mu ). \end{aligned}$$

We will also use the map

$$\begin{aligned} \iota _\mathbb {D}:\mathbb {D}\rightarrow C(B_{\widehat{\mathcal {B}}(\mathbb {D})}) \end{aligned}$$

defined by

$$\begin{aligned} \iota _\mathbb {D}(z)(g)=g'(z)\qquad \left( g\in B_{\widehat{\mathcal {B}}(\mathbb {D})},\; z\in \mathbb {D}\right) , \end{aligned}$$

and, for a complex Banach space X, the isometric linear embedding

$$\begin{aligned} \iota _X:X\rightarrow \ell _\infty (B_{X^*}) \end{aligned}$$

given by

$$\begin{aligned} \iota _X(x)(x^*)=x^*(x)\qquad (x^*\in B_{X^*},\; x\in X). \end{aligned}$$

Theorem 3.1

(Pietsch domination). Let $p\in [1,\infty )$, $\sigma \in [0,1)$ and $f\in \widehat{\mathcal {B}}\left( \mathbb {D},X\right)$. The following are equivalent:

• (1) $f\in \Pi _{p,\sigma }^{\widehat{\mathcal {B}}}\left( \mathbb {D},X\right)$.

• (2) There is a constant $C\ge 0$ and a measure $\mu \in \mathcal {P}\left( B_{\widehat{\mathcal {B}}\left( \mathbb {D}\right) }\right)$ such that

  $$\begin{aligned} \left\| f'\left( z\right) \right\| \le C\left( \int _{B_{\widehat{\mathcal {B}}\left( \mathbb {D}\right) }}\left( \left( \frac{1}{1-\left| z\right| ^{2}}\right) ^{\sigma } \left| h'\left( z\right) \right| ^{1-\sigma }\right) ^{\frac{p}{1-\sigma }}\textrm{d}\mu \left( h\right) \right) ^{\frac{1-\sigma }{p}} \end{aligned}$$

  for all $z\in \mathbb {D}$.

• (3) There is a constant $C\ge 0$ such that

  $$\begin{aligned}{} & {} \left( \underset{i=1}{\overset{n}{\sum }}\left| \lambda _{i}\right| ^{\frac{p}{1-\sigma }}\left\| f'\left( z_{i}\right) \right\| ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}}\le C \\{} & {} \qquad \underset{h\in B_{\widehat{\mathcal {B}}\left( \mathbb {D}\right) }}{\sup }\left( \underset{i=1}{\overset{n}{\sum }}\left( \left| \lambda _{i}\right| \left( \frac{1}{1-\left| z_{i}\right| ^{2}}\right) ^{\sigma }\left| h'\left( z_{i}\right) \right| ^{1-\sigma }\right) ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}} \end{aligned}$$

  for all $n\in \mathbb {N}$, $\lambda _i\in \mathbb {C}$ and $z_i\in \mathbb {D}$ for all $i\in \{1,\ldots ,n\}$.

Furthermore, the infimum of the constants $C\ge 0$ in (2) (and in (3)) is $\pi _{p,\sigma }^{\mathcal {B}}\left( f\right)$.

Proof

$\left( 1\right) \Rightarrow \left( 2\right)$: If $f\in \Pi _{p,\sigma }^{\widehat{\mathcal {B}}}\left( \mathbb {D},X\right)$, then there exist a complex Banach space Y and a map $g\in \Pi _{p}^{\widehat{\mathcal {B}}}\left( \mathbb {D},Y\right)$ such that

$$\begin{aligned} \left\| f'\left( z\right) \right\| \le \left( \frac{1}{1-\left| z\right| ^{2}}\right) ^{\sigma }\left\| g\left( z\right) \right\| ^{1-\sigma }\qquad (z\in \mathbb {D}). \end{aligned}$$

By [6, Theorem 1.4], there is a measure $\mu \in \mathcal {P}\left( B_{\widehat{\mathcal {B}}\left( \mathbb {D}\right) }\right)$ such that

$$\begin{aligned} \left\| g'(z)\right\| \le \pi _{p}^{\mathcal {B}}(g)\left( \int _{B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left| h'(z)\right| ^{p}\textrm{d}\mu (h)\right) ^{\frac{1}{p}}\qquad (z\in \mathbb {D}), \end{aligned}$$

and therefore

$$\begin{aligned} \left\| f'\left( z\right) \right\|&\le \left( \frac{1}{1-\left| z\right| ^{2}}\right) ^{\sigma }\left\| g'\left( z\right) \right\| ^{1-\sigma } \\&\le \pi _{p}^{\mathcal {B}}\left( g\right) ^{1-\sigma }\left( \int _{B_{\widehat{\mathcal {B}}\left( \mathbb {D}\right) }}\left( \left( \frac{1}{1-\left| z\right| ^{2}}\right) ^{\sigma }\left| h'\left( z\right) \right| ^{1-\sigma }\right) ^{\frac{p}{1-\sigma }}\textrm{d}\mu \left( h\right) \right) ^{\frac{1-\sigma }{p}}(z\in \mathbb {D}). \end{aligned}$$

$\left( 2\right) \Rightarrow \left( 1\right)$: By [6, Lemma 1.5], there exists a map $k\in \widehat{\mathcal {B}}\left( \mathbb {D},L_{\infty }\left( \mu \right) \right)$ with $\rho _{\mathcal {B}}\left( k\right) =1$ such that $k'=j_{\infty }\circ \iota _{\mathbb {D}}$. In fact, $k\in \Pi _{p}^{\widehat{\mathcal {B}}}\left( \mathbb {D},L_{\infty }\left( \mu \right) \right)$ with $\pi _{p}^{\mathcal {B}}\left( k\right) =1$. By (2), we can write

$$\begin{aligned} \left\| f'(z)\right\|&\le C\left( \int _{B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \left( \frac{1}{1-|z|^{2}}\right) ^{\sigma }\left| h'(z)\right| ^{1-\sigma }\right) ^{\frac{p}{1-\sigma }}\textrm{d}\mu (h)\right) ^{\frac{1-\sigma }{p}} \\&=\left( \frac{1}{1-\left| z\right| ^{2}}\right) ^{\sigma }\left( \int _{B_{\widehat{\mathcal {B}}\left( \mathbb {D}\right) }}\left| C^{\frac{1}{1-\sigma }}(I_{\infty ,p}\circ j_{\infty }\circ \iota _{\mathbb {D}})\left( z\right) \left( h\right) \right| ^{p}\textrm{d}\mu \left( h\right) \right) ^{\frac{1-\sigma }{p}} \\&=\left( \frac{1}{1-\left| z\right| ^{2}}\right) ^{\sigma }\left( \int _{B_{\widehat{\mathcal {B}}(\mathbb {D}}}\left| C^{\frac{1}{1-\sigma }}(I_{\infty ,p}\circ k)'\left( z\right) \left( h\right) \right| ^{p}\textrm{d}\mu \left( h\right) \right) ^{\frac{1-\sigma }{p}}\\&=\left( \frac{1}{1-\left| z\right| ^{2}}\right) ^{\sigma }\left\| g'\left( z\right) \right\| _{L_p(\mu )} ^{1-\sigma }, \end{aligned}$$

where $g=C^{\frac{1}{1-\sigma }}(I_{\infty ,p}\circ k)\in \Pi _{p}^{\widehat{\mathcal {B}}}(\mathbb {D},L_p(\mu ))$.

$\left( 2\right) \Rightarrow \left( 3\right)$: If (2) holds, then

$$\begin{aligned}&\left( \underset{i=1}{\overset{n}{\sum }}\left| \lambda _{i}\right| ^{\frac{p}{1-\sigma }}\left\| f'\left( z_{i}\right) \right\| ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}}\\&\quad \le C\left( \sum _{i=1}^n\int _{B_{\widehat{\mathcal {B}}\left( \mathbb {D}\right) }}\left( \left| \lambda _i\right| \left( \frac{1}{1-\left| z_{i}\right| ^{2}}\right) ^{\sigma } \left| h'\left( z_{i}\right) \right| ^{1-\sigma }\right) ^{\frac{p}{1-\sigma }} \textrm{d}\mu \left( h\right) \right) ^{\frac{1-\sigma }{p}}\\&\quad \le C\left( \sum _{i=1}^n\int _{B_{\widehat{\mathcal {B}}\left( \mathbb {D}\right) }}\left( \left| \lambda _i\right| \left( \frac{1}{1-\left| z_{i}\right| ^{2}}\right) ^{\sigma }\left( \frac{1}{1-\left| z_{i}\right| ^{2}}\right) ^{1-\sigma }\right) ^{\frac{p}{1-\sigma }}\textrm{d}\mu \left( h\right) \right) ^{\frac{1-\sigma }{p}}\\&\quad = C \left( \sum _{i=1}^n\int _{B_{\widehat{\mathcal {B}}\left( \mathbb {D}\right) }}\left( \left| \lambda _i\right| \left( \frac{1}{1-\left| z_{i}\right| ^{2}}\right) ^{\sigma }\left| f'_{z_i}(z_i)\right| ^{1-\sigma }\right) ^{\frac{p}{1-\sigma }}\textrm{d}\mu \left( h\right) \right) ^{\frac{1-\sigma }{p}}\\&\quad \le C\left( \sum _{i=1}^n\left( \left| \lambda _{i}\right| \left( \frac{1}{1-\left| z_{i}\right| ^{2}}\right) ^{\sigma }\left| f'_{z_i}(z_i)\right| ^{1-\sigma }\right) ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}}\\&\quad \le C\sup _{h\in B_{\widehat{\mathcal {B}}\left( \mathbb {D}\right) }}\left( \sum _{i=1}^n\left( \left| \lambda _{i}\right| \left( \frac{1}{1-\left| z_{i}\right| ^{2}}\right) ^{\sigma }\left| h'\left( z_{i}\right) \right| ^{1-\sigma }\right) ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}} \end{aligned}$$

for all $n\in \mathbb {N}$ $\lambda _i\in \mathbb {C}$ and $z_i\in \mathbb {D}$ for all $i\in \{1,\ldots ,n\}$, and this proves (3).

$\left( 3\right) \Rightarrow \left( 2\right)$: Let $R:B_{\widehat{\mathcal {B}}\left( \mathbb {D}\right) }\times \left( \mathbb {D}\times \mathbb {R}\right) \times \mathbb {R}\rightarrow \bigg [ 0,+\infty \bigg [$ be given by

$$\begin{aligned} R\left( h,\left( z,\lambda \right) ,b\right) =\left| \lambda \right| \left( \frac{1}{1-\left| z\right| ^{2}}\right) ^{\sigma }\left| h'\left( z\right) \right| ^{1-\sigma }\left| b\right| , \end{aligned}$$

and let $S:\widehat{\mathcal {B}}\left( \mathbb {D},X\right) \times \left( \mathbb {D}\times \mathbb {R}\right) \times \mathbb {R}\rightarrow \bigg [ 0,+\infty \bigg [$ be defined by

$$\begin{aligned} S\left( f,\left( z,\lambda \right) ,b\right) =\left| \lambda \right| \left\| f'\left( z\right) \right\| \left| b\right| . \end{aligned}$$

Then f is R-S-abstract $p/(1-\sigma )$-summing (see definition in [15]) since

$$\begin{aligned}&\left( \underset{i=1}{\overset{n}{\sum }}S\left( f,\left( z_{i},\lambda _{i}\right) ,b_{i}\right) ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}} =\left( \underset{i=1}{\overset{n}{\sum }}\left( \left| \lambda _{i}\right| \left\| f'\left( z_{i}\right) \right\| \left| b_{i}\right| \right) ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}} \\&\quad \le C\underset{h\in B_{\widehat{\mathcal {B}}\left( \mathbb {D}\right) }}{\sup }\left( \underset{i=1}{\overset{n}{\sum }}\left( \left| \lambda _{i}\right| \left( \frac{1}{1-\left| z_i\right| ^{2}}\right) ^{\sigma } \left| h'\left( z_{i}\right) \right| ^{1-\sigma }\left| b_{i}\right| \right) ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}} \\&\quad = C\underset{h\in B_{\widehat{\mathcal {B}}\left( \mathbb {D}\right) }}{\sup }\left( \underset{i=1}{\overset{n}{\sum }}R\left( h,\left( z_{i},\lambda _{i}\right) ,b_{i}\right) ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}}. \end{aligned}$$

Then, by [15, Theorem 3.1], there are $C>0$ and $\mu \in \mathcal {P}\left( B_{\widehat{\mathcal {B}}\left( \mathbb {D}\right) }\right)$ such that

$$\begin{aligned} S\left( f,\left( z,\lambda \right) ,b\right) \le C\left( \int _{B_{\widehat{\mathcal {B}}\left( \mathbb {D}\right) }}R\left( h,\left( z,\lambda \right) ,b\right) ^{\frac{p}{1-\sigma }}\textrm{d}\mu \left( h\right) \right) ^{\frac{1-\sigma }{p}} \end{aligned}$$

for all $\left( z,\lambda \right) \in \mathbb {D\times R}$ and $b\in \mathbb {R}$. In particular, we have

$$\begin{aligned} \left\| f'\left( z\right) \right\| \le C\left( \int _{B_{\widehat{\mathcal {B}}\left( \mathbb {D}\right) }}\left( \left( \frac{1}{1-\left| z\right| ^{2}}\right) ^{\sigma } \left| h'\left( z\right) \right| ^{1-\sigma }\right) ^{\frac{p}{1-\sigma }}\textrm{d}\mu \left( h\right) \right) ^{\frac{1-\sigma }{p}}\qquad (z\in \mathbb {D}). \end{aligned}$$

$\square$

4 Pietsch factorization

We now present the analogue for $(p,\sigma )$-absolutely continuous Bloch maps of Pietsch factorization theorem for $(p,\sigma )$-summing operators. Its proof is based on those of [6, Theorem 1.6] and [7, Theorem 3.5].

Theorem 4.1

(Pietsch factorization). Let $p\in [1,\infty )$, $\sigma \in [0,1)$ and $f\in \widehat{\mathcal {B}}(\mathbb {D},X)$. The following are equivalent:

• (1) $f\in \Pi _{p,\sigma }^{\widehat{\mathcal {B}}}\left( \mathbb {D},X\right)$.

• (2) There exist a measure $\mu \in \mathcal {P}(B_{\widehat{\mathcal {B}}(\mathbb {D})})$, a map $h\in \widehat{\mathcal {B}}(\mathbb {D},L_\infty (\mu ))$ and an operator $T\in \mathcal {L}(L_{p/(1-\sigma )}(\mu ),\ell _\infty (B_{X^*}))$ such that the following diagram commutes:

  

Furthermore, $\pi ^{\mathcal {B}}_{p,\sigma }(f)=\inf \left\{ \left\| T\right\| \rho _\mathcal {B}(h)\right\}$, the infimum being extended over all such decompositions of $\iota _X\circ f'$ as above, and this infimum is attained.

Proof

If (1) holds, then Theorem 3.1 provides a measure $\mu \in \mathcal {P}(B_{\widehat{\mathcal {B}}(\mathbb {D})})$ such that

$$\begin{aligned} \left\| f'(z)\right\| \le \pi ^{\mathcal {B}}_{p,\sigma }(f)\left( \int _{B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \left( \frac{1}{1-|z|^2}\right) ^\sigma \left| g'(z)\right| ^{1-\sigma }\right) ^{\frac{p}{1-\sigma }}\textrm{d}\mu (g)\right) ^{\frac{1-\sigma }{p}}\qquad (z\in \mathbb {D}). \end{aligned}$$

By [6, Lemma 1.5], there exists a map $h\in \widehat{\mathcal {B}}(\mathbb {D},L_\infty (\mu ))$ with $\rho _\mathcal {B}(h)=1$ such that $h'=j_{\infty }\circ \iota _\mathbb {D}$. Denote the closed linear subspace

$$\begin{aligned} S_{p/(1-\sigma )}:=\overline{\textrm{lin}}(I_{\infty ,p/(1-\sigma )}(h'(\mathbb {D})))\subseteq L_{p/(1-\sigma )}(\mu ), \end{aligned}$$

and define $T_0\in \mathcal {L}(S_{p/(1-\sigma )},\ell _\infty (B_{X^*}))$ by

$$\begin{aligned} T_0(I_{\infty ,p/(1-\sigma )}(h'(z)))=\iota _X(f'(z))\qquad (z\in \mathbb {D}). \end{aligned}$$

Notice that $\left\| T_0\right\| \le \pi ^{\mathcal {B}}_{p,\sigma }(f)$ since

$$\begin{aligned}&\left\| T_0\left( \sum _{i=1}^n\alpha _iI_{\infty ,p/(1-\sigma )}(h'(z_i))\right) \right\| _\infty =\left\| \sum _{i=1}^n\alpha _i T_0(I_{\infty ,p/(1-\sigma )}(h'(z_i)))\right\| _\infty \\&\quad =\left\| \sum _{i=1}^n\alpha _i \iota _X(f'(z_i))\right\| _\infty \le \sum _{i=1}^n\left| \alpha _i\right| \left\| \iota _X(f'(z_i))\right\| _\infty =\sum _{i=1}^n\left| \alpha _i\right| \left\| f'(z_i)\right\| \\&\quad \le \pi ^{\mathcal {B}}_{p,\sigma }(f)\sum _{i=1}^n\left| \alpha _i\right| \left( \int _{B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \left( \frac{1}{1-|z_i|^2}\right) ^\sigma \left| g'(z_i)\right| ^{1-\sigma }\right) ^{\frac{p}{1-\sigma }}\textrm{d}\mu (g)\right) ^{\frac{1-\sigma }{p}}\\&\quad \le \pi ^{\mathcal {B}}_{p,\sigma }(f)\sum _{i=1}^n\frac{\left| \alpha _i\right| }{1-|z_i|^2} \end{aligned}$$

and

$$\begin{aligned} \sum _{i=1}^n\frac{\left| \alpha _i\right| }{1-|z_i|^2}&=\left| \sum _{i=1}^n\alpha _i\frac{\overline{\alpha _i}}{\left| \alpha _i\right| }f'_{z_i}(z_i)\right| =\sup _{g\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left| \sum _{i=1}^n\alpha _i g'(z_i)\right| =\sup _{g\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left| \sum _{i=1}^n\alpha _i \iota _\mathbb {D}(z_i)(g)\right| \\&=\left\| \sum _{i=1}^n\alpha _i\iota _\mathbb {D}(z_i)\right\| _\infty =\left\| \sum _{i=1}^n\alpha _i j_{\infty }(\iota _\mathbb {D}(z_i))\right\| _{L_\infty (\mu )}=\left\| \sum _{i=1}^n\alpha _i h'(z_i)\right\| _{L_\infty (\mu )}\\&=\left\| I_{\infty ,p/(1-\sigma )}\left( \sum _{i=1}^n\alpha _ih'(z_i)\right) \right\| _{L_{p/(1-\sigma )}}=\left\| \sum _{i=1}^n\alpha _iI_{\infty ,p/(1-\sigma )}(h'(z_i))\right\| _{L_{p/(1-\sigma )}} \end{aligned}$$

for any $n\in \mathbb {N}$, $\alpha _i\in \mathbb {C}^*$ and $z_i\in \mathbb {D}$ for all $i\in \{1,\ldots ,n\}$. By the injectivity of the Banach space $\ell _\infty (B_{X^*})$ (see [9, p. 45]), there exists $T\in \mathcal {L}(L_{p/(1-\sigma )}(\mu ),\ell _\infty (B_{X^*}))$ such that $\left. T\right| _{S_{p/(1-\sigma )}}=T_0$ with $\left\| T\right\| =\left\| T_0\right\|$. This allows us to conclude that $\iota _X\circ f'=T\circ I_{\infty ,p/(1-\sigma )}\circ h'$ with $\left\| T\right\| \rho _\mathcal {B}(h)\le \pi ^{\mathcal {B}}_{p,\sigma }(f)$.

Conversely, assume that $\iota _X\circ f'=T\circ I_{\infty ,p/(1-\sigma )}\circ h'$ as in (2). We have

$$\begin{aligned}&\left( \sum _{i=1}^n\left| \lambda _i\right| ^{\frac{p}{1-\sigma }}\left\| f'(z_i)\right\| ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}} =\left( \sum _{i=1}^n\left| \lambda _i\right| ^{\frac{p}{1-\sigma }}\left\| \iota _X(f'(z_i))\right\| _\infty ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}}\\&\quad =\left( \sum _{i=1}^n\left| \lambda _i\right| ^{\frac{p}{1-\sigma }}\left\| T(I_{\infty ,p/(1-\sigma )}(h'(z_i)))\right\| _\infty ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}}\\&\quad \le \left\| T\right\| \left( \sum _{i=1}^n\left| \lambda _i\right| ^{\frac{p}{1-\sigma }}\left\| I_{\infty ,p/(1-\sigma )}(h'(z_i))\right\| _{L_{p/(1-\sigma )}(\mu )}^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}}\\&\quad =\left\| T\right\| \left( \sum _{i=1}^n\left| \lambda _i\right| ^{\frac{p}{1-\sigma }}\left\| h'(z_i)\right\| _{L_\infty (\mu )}^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}}\\&\quad \le \left\| T\right\| \rho _\mathcal {B}(h)\left( \sum _{i=1}^n\frac{\left| \lambda _i\right| ^{\frac{p}{1-\sigma }}}{(1-|z_i|^2)^{\frac{p}{1-\sigma }}}\right) ^{\frac{1-\sigma }{p}}\\&\quad =\left\| T\right\| \rho _\mathcal {B}(h)\left( \sum _{i=1}^n\left( \left| \lambda _i\right| \left( \frac{1}{1-|z_i|^2}\right) ^\sigma \left| f'_{z_i}(z_i)\right| ^{1-\sigma }\right) ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}}\\&\quad \le \left\| T\right\| \rho _\mathcal {B}(h)\sup _{g\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \sum _{i=1}^n\left( \left| \lambda _i\right| \left( \frac{1}{1-|z_i|^2}\right) ^\sigma \left| g'(z_i)\right| ^{1-\sigma }\right) ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}} \end{aligned}$$

for any $n\in \mathbb {N}$, $\lambda _i\in \mathbb {C}$ and $z_i\in \mathbb {D}$ for all $i\in \{1,\ldots ,n\}$. Hence $f\in \Pi ^{\widehat{\mathcal {B}}}_{p,\sigma }(\mathbb {D},X)$ with $\pi ^{\mathcal {B}}_{p,\sigma }(f)\le \left\| T\right\| \rho _\mathcal {B}(h)$ by Theorem 3.1. $\square$

We now relate $(p,\sigma )$-absolutely continuous Bloch maps with (weakly) compact Bloch maps which were introduced in [10].

Let us recall that the Bloch range of a function $f\in \mathcal {H}(\mathbb {D},X)$, denoted by $\textrm{rang}_{\mathcal {B}}(f)$, is the set

$$\begin{aligned} \left\{ (1-|z|^2)f'(z)\in X:z\in \mathbb {D}\right\} . \end{aligned}$$

A map $f\in \mathcal {H}(\mathbb {D},X)$ is called (weakly) compact Bloch if $\textrm{rang}_{\mathcal {B}}(f)$ is a relatively (weakly) compact set in X.

Proposition 4.2

If $p\in [1,\infty )$ and $\sigma \in [0,1)$, then every $(p,\sigma )$-absolutely continuous Bloch map $f:\mathbb {D}\rightarrow X$ is weakly compact Bloch, and if in addition X is reflexive, then f is compact Bloch.

Proof

Let $f\in \Pi _{p,\sigma }^{\widehat{\mathcal {B}}}(\mathbb {D},X)$. Hence Theorem 4.1 guarantees that

$$\begin{aligned} (\iota _X\circ f)'=\iota _X\circ f'=T\circ I_{\infty ,p/(1-\sigma )}\circ h'=T\circ (I_{\infty ,p/(1-\sigma )}\circ h)', \end{aligned}$$

for some measure $\mu \in \mathcal {P}(B_{\widehat{\mathcal {B}}(\mathbb {D})})$, an operator $T\in \mathcal {L}(L_{p/(1-\sigma )}(\mu ),\ell _{\infty }(B_{X^*}))$ and a map $h\in \widehat{\mathcal {B}}(\mathbb {D},L_{\infty }(\mu ))$. Assume first $p>1$ and then the reflexivity of $L_{p/(1-\sigma )}(\mu )$ shows that $\iota _X\circ f\in \widehat{\mathcal {B}}(\mathbb {D},\ell _{\infty }(B_{X^*}))$ is weakly compact Bloch by [10, Theorem 5.6]. Now, the equality $\textrm{rang}_{\mathcal {B}}(\iota _X\circ f)=\iota _X(\textrm{rang}_{\mathcal {B}}(f))$ yields that f is weakly compact Bloch. The case $p=1$ follows from the previous case when $\sigma \in (0,1)$ and from Proposition 2.1 and [6, Corollary 1.7] when $\sigma =0$.

So we have proved that $\textrm{rang}_{\mathcal {B}}(f)$ is relatively weakly compact in X, and therefore relatively compact in X whenever X is reflexive. $\square$

5 Injective Banach normalized Bloch ideal

Motivated by the theory of operator ideals between Banach spaces [17], the concept of a Banach normalized Bloch ideal on $\mathbb {D}$ was introduced in [10, Definition 5.11]. Proposition 1.2 in [6] asserts that $[\Pi _{p}^{\widehat{\mathcal {B}}},\pi _{p}^{\mathcal {B}}]$ is an injective Banach normalized Bloch ideal for any $p\in [1,\infty )$. We now show that $[\Pi _{p,\sigma }^{\widehat{\mathcal {B}}},\pi _{p,\sigma }^{\mathcal {B}}]$ enjoys the same property using [10].

Proposition 5.1

$[\Pi _{p,\sigma }^{\widehat{\mathcal {B}}},\pi _{p,\sigma }^{\mathcal {B}}]$ is an injective Banach normalized Bloch ideal for any $p\in [1,\infty )$ and $\sigma \in [0,1)$.

Proof

Note that we only need to prove the case $\sigma \in (0,1)$.

(N1): By Proposition 2.2, $(\Pi _{p,\sigma }^{\widehat{\mathcal {B}}}(\mathbb {D},X),\pi _{p,\sigma }^{\mathcal {B}})$ is a Banach space with $\rho _{\mathcal {B}}(f)\le \pi _{p,\sigma }^{\mathcal {B}}(f)$ for all $f\in \Pi _{p,\sigma }^{\widehat{\mathcal {B}}}(\mathbb {D},X)$.

(N2): Let $g\in \widehat{\mathcal {B}}(\mathbb {D})$ and $x\in X$. Let us recall that $g\cdot x\in \widehat{\mathcal {B}}(\mathbb {D},X)$ with $\rho _{\mathcal {B}}(g\cdot x)=\rho _{\mathcal {B}}(g)\left\| x\right\|$ by [10, Proposition 5.13]. Assume $g\ne 0$. For all $n\in \mathbb {N}$, $\lambda _i\in \mathbb {C}$ and $z_i\in \mathbb {D}$ for all $i\in \{1,\ldots ,n\}$, it holds

$$\begin{aligned}&\left( \underset{i=1}{\overset{n}{\sum }}\left| \lambda _{i}\right| ^{\frac{p}{1-\sigma }}\left\| \left( g\cdot x\right) '\left( z_{i}\right) \right\| ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}} =\rho _{\mathcal {B}}(g)\left\| x\right\| \left( \underset{i=1}{\overset{n}{\sum }}\left| \lambda _{i}\right| ^{\frac{p}{1-\sigma }}\left| \left( \frac{g}{\rho _{\mathcal {B}}(g)}\right) ^{^{\prime }}\left( z_{i}\right) \right| ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}} \\&\quad \le \rho _{\mathcal {B}}(g)\left\| x\right\| \underset{h\in B_{\widehat{\mathcal {B}}\left( \mathbb {D}\right) }}{\sup }\left( \underset{i=1}{\overset{n}{\sum }}\left( \left| \lambda _{i}\right| \left( \frac{1}{1-\left| z\right| ^{2}}\right) ^{\sigma }\left| h'\left( z_{i}\right) \right| ^{1-\sigma }\right) ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}}, \end{aligned}$$

and so $g\cdot x\in \Pi _{p,\sigma }^{\widehat{\mathcal {B}}}\left( \mathbb { D},X\right)$ with $\pi _{p,\sigma }^{\mathcal {B}}\left( g\cdot x\right) \le \rho _{\mathcal {B}}(g)\left\| x\right\|$. Since $\rho _{\mathcal {B}}(g)\left\| x\right\| =\rho _{\mathcal {B}}(g\cdot x)\le \pi _{p}^{\mathcal {B}}\left( g\cdot x\right)$, we have $\pi _{p,\sigma }^{\mathcal {B}}\left( g\cdot x\right) =\rho _{\mathcal {B}}(g)\left\| x\right\|$.

(N3): Let $f\in \Pi _{p,\sigma }^{\widehat{\mathcal {B}}}\left( \mathbb {D},X\right) ,T\in \mathcal {L}\left( X,Y\right)$ and let $g:\mathbb {D}\rightarrow \mathbb {D}$ be a holomorphic function with $g(0)=0$. The Pick–Schwarz Lemma assures that

$$\begin{aligned} (1-|z|^2)|g'(z)|\le 1-|g(z)|^2\qquad (z\in \mathbb {D}). \end{aligned}$$

Let us recall that $T\circ f\circ g\in \widehat{\mathcal {B}}\left( \mathbb {D},Y\right)$ by [10, Proposition 5.13]. We have

$$\begin{aligned}&\left( \underset{i=1}{\overset{n}{\sum }}\left| \lambda _{i}\right| ^{\frac{p}{1-\sigma }}\left\| \left( T\circ f\circ g\right) '\left( z_{i}\right) \right\| ^{\frac{p}{1-\sigma } }\right) ^{\frac{1-\sigma }{p}} \\&\quad =\left( \underset{i=1}{\overset{n}{\sum }}\left| \lambda _{i}\right| ^{\frac{p}{1-\sigma }}\left\| T\left( f'\left( g\left( z_{i}\right) \right) g'\left( z_{i}\right) \right) \right\| ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}} \\&\quad \le \left\| T\right\| \left( \underset{i=1}{\overset{n}{\sum }}\left( \left| \lambda _{i}\right| \left| g'\left( z_{i}\right) \right| \left\| f'\left( g\left( z_{i}\right) \right) \right\| \right) ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}} \\&\quad \le \left\| T\right\| \pi _{p,\sigma }^{\mathcal {B}}\left( f\right) \underset{h\in B_{\widehat{\mathcal {B}}\left( \mathbb {D}\right) }}{\sup }\left( \underset{i=1}{\overset{n}{\sum }}\left( \left| \lambda _{i}\right| \left| g'\left( z_{i}\right) \right| \left| h'\left( g\left( z_{i}\right) \right) \right| ^{1-\sigma }\left( \frac{1}{1-\left| g\left( z_{i}\right) \right| ^{2}}\right) ^{\sigma }\right) ^{\frac{p}{1-\sigma }}\right) ^{\frac{ 1-\sigma }{p}} \\&\quad =\left\| T\right\| \pi _{p,\sigma }^{\mathcal {B}}\left( f\right) \underset{h\in B_{\widehat{\mathcal {B}}\left( \mathbb {D}\right) }}{\sup }\left( \underset{i=1}{\overset{n}{\sum }}\left( \left| \lambda _{i}\right| \left| g'\left( z_{i}\right) h'\left( g\left( z_{i}\right) \right) \right| ^{1-\sigma }\left( \frac{\left| g^{\prime }\left( z_{i}\right) \right| }{1-\left| g\left( z_{i}\right) \right| ^{2}} \right) ^{\sigma }\right) ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p} } \\&\quad \le \left\| T\right\| \pi _{p,\sigma }^{\mathcal {B}}\left( f\right) \underset{h\in B_{\widehat{\mathcal {B}}\left( \mathbb {D}\right) }}{\sup } \left( \underset{i=1}{\overset{n}{\sum }}\left( \left| \lambda _{i}\right| \left( \frac{1}{1-\left| z_{i}\right| ^{2}}\right) ^{\sigma }\left| \left( h\circ g\right) '\left( z_{i}\right) \right| ^{1-\sigma }\right) ^{\frac{p}{1-\sigma } }\right) ^{\frac{1-\sigma }{p}} \\&\quad \le \left\| T\right\| \pi _{p,\sigma }^{\mathcal {B}}\left( f\right) \underset{k \in B_{\widehat{\mathcal {B}}\left( \mathbb {D}\right) }}{\sup } \left( \underset{i=1}{\overset{n}{\sum }}\left( \left| \lambda _{i}\right| \left( \frac{1}{1-\left| z_{i}\right| ^{2}}\right) ^{\sigma }\left| k '\left( z_{i}\right) \right| ^{1-\sigma }\right) ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}}. \end{aligned}$$

where $\rho _{\mathcal {B}}(h\circ g)\le \rho _{\mathcal {B}}(h)$ by [10, Proposition 3.6]. Therefore, $T\circ f\circ g\in \Pi _{p,\sigma }^{\widehat{\mathcal {B}}}\left( \mathbb {D},X\right)$ with $\pi _{p,\sigma }^{\mathcal {B}}\left( T\circ f\circ g\right) \le \left\| T\right\| \pi _{p,\sigma }^{\mathcal {B}}\left( f\right)$.

(I): Let $f\in \widehat{\mathcal {B}}\left( \mathbb {D},X\right)$ and let $\iota :X\rightarrow Y$ be a linear isometry so that $\iota \circ f\in \Pi _{p,\sigma }^{\widehat{\mathcal {B}}}\left( \mathbb {D},Y\right)$. We have

$$\begin{aligned}&\left( \underset{i=1}{\overset{n}{\sum }}\left| \lambda _{i}\right| ^{\frac{p}{1-\sigma }}\left\| f'\left( z_{i}\right) \right\| ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}} =\left( \underset{i=1}{\overset{n}{\sum }}\left| \lambda _{i}\right| ^{\frac{p}{1-\sigma }}\left\| \iota \left( f'\left( z_{i}\right) \right) \right\| ^{\frac{p}{1-\sigma }}\right) ^{\frac{ 1-\sigma }{p}} \\&\quad =\left( \underset{i=1}{\overset{n}{\sum }}\left| \lambda _{i}\right| ^{\frac{p}{1-\sigma }}\left\| \left( \iota \circ f\right) '\left( z_{i}\right) \right\| ^{\frac{p}{1-\sigma }}\right) ^{ \frac{1-\sigma }{p}} \\&\quad \le \pi _{p,\sigma }^{\mathcal {B}}\left( \iota \circ f\right) \underset{h\in B_{\widehat{\mathcal {B}}\left( \mathbb {D}\right) }}{\sup }\left( \underset{i=1}{\overset{n}{\sum }}\left( \left| \lambda _{i}\right| \left( \frac{1}{1-\left| z_{i}\right| ^{2}}\right) ^{\sigma }\left| h'\left( z_{i}\right) \right| ^{1-\sigma }\right) ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}} \end{aligned}$$

and thus $f\in \Pi _{p,\sigma }^{\widehat{\mathcal {B}}}\left( \mathbb {D},X\right)$ with $\pi _{p,\sigma }^{\mathcal {B}}\left( f\right) \le \pi _{p,\sigma }^{\mathcal {B}}\left( \iota \circ f\right)$. The reverse inequality follows from (N3). $\square$

The Möbius group of $\mathbb {D}$, designated $\textrm{Aut}(\mathbb {D})$, consists of all biholomorphic bijections from $\mathbb {D}$ onto itself. Let us recall that a linear space $\mathcal {A}(\mathbb {D},X)\subseteq \mathcal {B}(\mathbb {D},X)$, under a seminorm $\rho _\mathcal {A}$, is Möbius-invariant if: (i) there is $C>0$ such that $\rho _{\mathcal {B}}(f)\le C \rho _{\mathcal {A}}(f)$ for all $f\in \mathcal {A}(\mathbb {D},X)$; and (ii) $f\circ \phi \in \mathcal {A}(\mathbb {D},X)$ with $\rho _{\mathcal {A}}(f\circ \phi )=\rho _{\mathcal {A}}(f)$ for all $\phi \in \textrm{Aut}(\mathbb {D})$ and $f\in \mathcal {A}(\mathbb {D},X)$.

By Proposition 2.1, $(\Pi _{p,\sigma }^{\mathcal {B}}(\mathbb {D},X),\pi _{p,\sigma }^{\mathcal {B}})\le (\mathcal {B}(\mathbb {D},X),\rho _{\mathcal {B}})$. Moreover, by the proof of (N3) in Proposition 5.1, one has that if $f\in \Pi _{p,\sigma }^{\mathcal {B}}(\mathbb {D},X)$ and $\phi \in \textrm{Aut}(\mathbb {D})$, then $f\circ \phi \in \Pi _{p,\sigma }^{\mathcal {B}}(\mathbb {D},X)$ with $\pi ^{\mathcal {B}}_{p,\sigma }(f\circ \phi )\le \pi ^{\mathcal {B}}_{p,\sigma }(f)$, and this fact also yields $\pi ^{\mathcal {B}}_{p,\sigma }(f)=\pi ^{\mathcal {B}}_{p,\sigma }((f\circ \phi )\circ \phi ^{-1})\le \pi ^{\mathcal {B}}_{p,\sigma }(f\circ \phi )$. So we have stated the following result which extends [6, Proposition 1.3].

Corollary 5.2

$(\Pi ^{\mathcal {B}}_{p,\sigma }(\mathbb {D},X),\pi ^{\mathcal {B}}_{p,\sigma })$ is a Möbius-invariant space for $p\in [1,\infty )$ and $\sigma \in [0,1)$. $\square$

6 Duality

With the aim of studying the duality of the spaces $\Pi ^{\widehat{\mathcal {B}}}_{p,\sigma }(\mathbb {D},X^*)$, we first introduce the Bloch analogues of the $(p,\sigma )$-Chevet–Saphar norms on the tensor product of two Banach spaces. We refer the reader to the references [8, 18] for a complete study on the theory of tensor product. As usual, for any linear spaces E and F, the tensor product $E\otimes F$ equipped with a norm $\alpha$ is denoted by $E\otimes _\alpha F$, and its completion by $E\widehat{\otimes }_\alpha F$.

Towards our aim, we recall some concepts and results of [10]. For each $z\in \mathbb {D}$, a Bloch atom of $\mathbb {D}$ is the functional $\gamma _z\in \widehat{\mathcal {B}}(\mathbb {D})^*$ given by $\gamma _z(f)=f'(z)$ for all $f\in \widehat{\mathcal {B}}(\mathbb {D})$. The named Bloch molecules of $\mathbb {D}$ are the elements of the space

$$\begin{aligned} \textrm{lin}(\{\gamma _z:z\in \mathbb {D}\})\subseteq \widehat{\mathcal {B}}(\mathbb {D})^*, \end{aligned}$$

and the Bloch-free Banach space of $\mathbb {D}$ is the space

$$\begin{aligned} \mathcal {G}(\mathbb {D}):=\overline{\textrm{lin}}(\{\gamma _z:z\in \mathbb {D}\})\subseteq \widehat{\mathcal {B}}(\mathbb {D})^*. \end{aligned}$$

The map $\Gamma :z\in \mathbb {D}\mapsto \gamma _z\in \mathcal {G}(\mathbb {D})$ is holomorphic with $\left\| \gamma _z\right\| =1/(1-|z|^2)$ for all $z\in \mathbb {D}$.

Define now the space of X-valued Bloch molecules of $\mathbb {D}$ by setting

$$\begin{aligned} \textrm{lin}(\Gamma (\mathbb {D}))\otimes X:=\textrm{lin}\left\{ \gamma _z\otimes x:z\in \mathbb {D},\, x\in X\right\} \subseteq \widehat{\mathcal {B}}(\mathbb {D},X^*)^*, \end{aligned}$$

where $\gamma _z\otimes x:\widehat{\mathcal {B}}(\mathbb {D},X^*)\rightarrow \mathbb {C}$ is the functional given by

$$\begin{aligned} \left( \gamma _z\otimes x\right) (f)=\left\langle f'(z),x\right\rangle \qquad \left( f\in \widehat{\mathcal {B}}(\mathbb {D},X^*)\right) . \end{aligned}$$

Plainly, each element $\gamma \in \textrm{lin}(\Gamma (\mathbb {D}))\otimes X$ can be expressed as $\gamma =\sum _{i=1}^n\lambda _i\gamma _{z_i}\otimes x_i$ for some $n\in \mathbb {N}$, $\lambda _i\in \mathbb {C}$, $z_i\in \mathbb {D}$ and $x_i\in X$ for $i=1,\ldots ,n$. Moreover,

$$\begin{aligned} \gamma (f)=\sum _{i=1}^n\lambda _i\left\langle f'(z_i),x_i\right\rangle \qquad (f\in \widehat{\mathcal {B}}(\mathbb {D},X^*)). \end{aligned}$$

The following family of norms contains the p-Chevet–Saphar Bloch norms on $\textrm{lin}(\Gamma (\mathbb {D}))\otimes X$ introduced in [6, Subsection 2.3].

Definition 6.1

Let $p\in (1,\infty )$ and $\sigma \in [0,1)$. We define the $(p,\sigma )$-Chevet–Saphar Bloch norm $d^{\widehat{\mathcal {B}}}_{p,\sigma }$ on $\gamma \in \textrm{lin}(\Gamma (\mathbb {D}))\otimes X$ by

$$\begin{aligned} d^{\widehat{\mathcal {B}}}_{1,\sigma }(\gamma )&=\inf \left\{ \left( \sup _{h\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \max _{1\le i\le n}\left( \left| \lambda _i\right| \left( \frac{1}{1-\left| z_i\right| ^2}\right) ^{\sigma }\left| h'(z_i)\right| ^{1-\sigma }\right) \right) \right) \right. \\&\quad \times \left. \left( \sum _{i=1}^n\left\| x_i\right\| ^{\frac{1}{1-\sigma }}\right) ^{1-\sigma }\right\} ,\\ d^{\widehat{\mathcal {B}}}_{p,\sigma }(\gamma )&=\inf \left\{ \left( \sup _{h\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \sum _{i=1}^n\left( \left| \lambda _i\right| \left( \frac{1}{1-\left| z_i\right| ^2}\right) ^{\sigma }\left| h'(z_i)\right| ^{1-\sigma }\right) ^{\frac{p^*}{1-\sigma }}\right) ^{\frac{1-\sigma }{p^*}}\right) \right. \\&\quad \times \left. \left( \sum _{i=1}^n\left\| x_i\right\| ^{\frac{p^*}{p^*-(1-\sigma )}}\right) ^{\frac{p^*-(1-\sigma )}{p^*}}\right\} \\ d^{\widehat{\mathcal {B}}}_{\infty ,\sigma }(\gamma )&=\inf \left\{ \left( \sup _{h\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \sum _{i=1}^n\left( \left| \lambda _i\right| \left( \frac{1}{1-\left| z_i\right| ^2}\right) ^{\sigma }\left| h'(z_i)\right| ^{1-\sigma }\right) ^{\frac{1}{1-\sigma }}\right) ^{1-\sigma }\right) \right. \\&\quad \times \left. \left( \max _{1\le i\le n}\left\| x_i\right\| ^{\frac{1}{1-\sigma }}\right) ^{1-\sigma }\right\} , \end{aligned}$$

the infimum being taken over all the representations of $\gamma$ as $\sum _{i=1}^n\lambda _i\gamma _{z_i}\otimes x_i$.

The following result concerning Bloch reasonable crossnorms introduced in [6, Definition 2.5] is based on [6, Theorem 2.6].

Given a complex Banach space X, let us recall that a norm $\alpha$ on $\textrm{lin}(\Gamma (\mathbb {D}))\otimes X$ is a Bloch reasonable crossnorm if it satisfies the two conditions: (i) $\alpha (\gamma _z\otimes x)\le \left\| \gamma _z\right\| \left\| x\right\|$ for all $z\in \mathbb {D}$ and $x\in X$; and (ii) Given $g\in \widehat{\mathcal {B}}(\mathbb {D})$ and $x^*\in X^*$, the linear functional $g\otimes x^*:\textrm{lin}(\Gamma (\mathbb {D}))\otimes X\rightarrow \mathbb {C}$ given by $(g\otimes x^*)(\gamma _z\otimes x)=g'(z)x^*(x)$ is bounded on $\textrm{lin}(\Gamma (\mathbb {D}))\otimes _\alpha X$ with $\left\| g\otimes x^*\right\| \le \rho _{\mathcal {B}}(g)\left\| x^*\right\|$.

Theorem 6.2

$d^{\widehat{\mathcal {B}}}_{p,\sigma }$ is a Bloch reasonable crossnorm on $\textrm{lin}(\Gamma (\mathbb {D}))\otimes X$ for $\sigma \in [0,1)$ and $p\in [1,\infty ]$.

Proof

For $\sigma =0$ and $p\in [1,\infty ]$, the result was stated in [6, Theorem 2.6]. We will prove it here for $\sigma \in (0,1)$ and $p\in (1,\infty )$. For $p\in \{1,\infty \}$, the proofs are similar.

Let $\gamma \in \textrm{lin}(\Gamma (\mathbb {D}))\otimes X$ and let $\sum _{i=1}^n\lambda _i\gamma _{z_i}\otimes x_i$ be a representation of $\gamma$. Clearly, $d^{\widehat{\mathcal {B}}}_{p,\sigma }(\gamma )\ge 0$. Given $\lambda \in \mathbb {C}$, it is immediate that

$$\begin{aligned} d^{\widehat{\mathcal {B}}}_{p,\sigma }(\lambda \gamma )\le & {} \left| \lambda \right| \left( \sup _{h\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \sum _{i=1}^n\left( \left| \lambda _i\right| \left( \frac{1}{1-\left| z_i\right| ^2}\right) ^{\sigma }\left| h'(z_i)\right| ^{1-\sigma }\right) ^{\frac{p^*}{1-\sigma }}\right) ^{\frac{1-\sigma }{p^*}}\right) \\{} & {} \times \left( \sum _{i=1}^n\left\| x_i\right\| ^{\frac{p^*}{p^*-(1-\sigma )}}\right) ^{\frac{p^*-(1-\sigma )}{p^*}}. \end{aligned}$$

From this inequality, we infer that $d^{\widehat{\mathcal {B}}}_{p,\sigma }(\lambda \gamma )=0=\left| \lambda \right| d^{\widehat{\mathcal {B}}}_{p,\sigma }(\gamma )$ if $\lambda =0$, and that $d^{\widehat{\mathcal {B}}}_{p,\sigma }(\lambda \gamma )\le \left| \lambda \right| d^{\widehat{\mathcal {B}}}_{p,\sigma }(\gamma )$ if $\lambda \ne 0$. In this case, $d^{\widehat{\mathcal {B}}}_{p,\sigma }(\gamma )=d^{\widehat{\mathcal {B}}}_{p,\sigma }(\lambda ^{-1}(\lambda \gamma ))\le |\lambda ^{-1}|d^{\widehat{\mathcal {B}}}_{p,\sigma }(\lambda \gamma )$, hence $\left| \lambda \right| d^{\widehat{\mathcal {B}}}_{p,\sigma }(\gamma )\le d^{\widehat{\mathcal {B}}}_{p,\sigma }(\lambda \gamma )$, and thus also $d^{\widehat{\mathcal {B}}}_{p,\sigma }(\lambda \gamma )=\left| \lambda \right| d^{\widehat{\mathcal {B}}}_{p,\sigma }(\gamma )$.

To prove the triangular inequality of $d^{\widehat{\mathcal {B}}}_{p,\sigma }$, let $\gamma _1,\gamma _2\in \textrm{lin}(\Gamma (\mathbb {D}))\otimes X$ and $\varepsilon >0$. We can choose representations

$$\begin{aligned} \gamma _1=\sum _{i=1}^n\lambda _{1,i}\gamma _{z_{1,i}}\otimes x_{1,i},\qquad \gamma _2=\sum _{i=1}^m\lambda _{2,i}\gamma _{z_{2,i}}\otimes x_{2,i}, \end{aligned}$$

so that

$$\begin{aligned} \left( \sup _{h\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \sum _{i=1}^n\left( \left| \lambda _{1,i}\right| \left( \frac{1}{1-\left| z_{1,i}\right| ^2}\right) ^{\sigma }\left| h'(z_{1,i})\right| ^{1-\sigma }\right) ^{\frac{p^*}{1-\sigma }}\right) ^{\frac{1-\sigma }{p^*}}\right) \left( \sum _{i=1}^n\left\| x_{1,i}\right\| ^{\frac{p^*}{p^*-(1-\sigma )}}\right) ^{\frac{p^*-(1-\sigma )}{p^*}} \end{aligned}$$

and

$$\begin{aligned} \left( \sup _{h\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \sum _{i=1}^m\left( \left| \lambda _{2,i}\right| \left( \frac{1}{1-\left| z_{2,i}\right| ^2}\right) ^{\sigma }\left| h'(z_{2,i})\right| ^{1-\sigma }\right) ^{\frac{p^*}{1-\sigma }}\right) ^{\frac{1-\sigma }{p^*}}\right) \left( \sum _{i=1}^m\left\| x_{2,i}\right\| ^{\frac{p^*}{p^*-(1-\sigma )}}\right) ^{\frac{p^*-(1-\sigma )}{p^*}} \end{aligned}$$

are less or equal than $d^{\widehat{\mathcal {B}}}_{p,\sigma }(\gamma _1)+\varepsilon$ and $d^{\widehat{\mathcal {B}}}_{p,\sigma }(\gamma _2)+\varepsilon$, respectively.

For $r,s\in \mathbb {R}^+$ arbitrary, define

$$\begin{aligned} \lambda _{3,i}\gamma _{z_{3,i}}=\left\{ \begin{array}{lll} r^{-1}\lambda _{1,i}\gamma _{z_{1,i}}&{}\quad i=1,\ldots ,n,\\ s^{-1}\lambda _{2,i-n}\gamma _{z_{2,i-n}}&{}\quad i=n+1,\ldots ,n+m, \end{array}\right. \end{aligned}$$

and

$$\begin{aligned} x_{3,i}=\left\{ \begin{array}{lll} rx_{1,i}&{}\quad i=1,\ldots ,n,\\ sx_{2,i-n}&{}\quad i=n+1,\ldots ,n+m. \end{array}\right. \end{aligned}$$

Plainly, $\gamma _1+\gamma _2=\sum _{i=1}^{n+m}\lambda _{3,i}\gamma _{z_{3,i}}\otimes x_{3,i}$ and, in consequence, $d^{\widehat{\mathcal {B}}}_{p,\sigma }(\gamma _1+\gamma _2)$ is less or equal than

$$\begin{aligned}{} & {} \left( \sup _{h\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \sum _{i=1}^{n+m}\left( \left| \lambda _{3,i}\right| \left( \frac{1}{1-\left| z_{3,i}\right| ^2}\right) ^{\sigma }\left| h'(z_{3,i})\right| ^{1-\sigma }\right) ^{\frac{p^*}{1-\sigma }}\right) ^{\frac{1-\sigma }{p^*}}\right) \\{} & {} \quad \times \left( \sum _{i=1}^{n+m}\left\| x_{3,i}\right\| ^{\frac{p^*}{p^*-(1-\sigma )}}\right) ^{\frac{p^*-(1-\sigma )}{p^*}}. \end{aligned}$$

A simple computation produces

$$\begin{aligned}&\left( \sup _{h\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \sum _{i=1}^{n+m}\left( \left| \lambda _{3,i}\right| \left( \frac{1}{1-\left| z_{3,i}\right| ^2}\right) ^{\sigma }\left| h'(z_{3,i})\right| ^{1-\sigma }\right) ^{\frac{p^*}{1-\sigma }}\right) ^{\frac{1-\sigma }{p^*}}\right) ^{\frac{p^*}{1-\sigma }}\\&\quad \le \left( r^{-1}\sup _{h\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \sum _{i=1}^n\left( \left| \lambda _{1,i}\right| \left( \frac{1}{1-\left| z_{1,i}\right| ^2}\right) ^{\sigma }\left| h'(z_{1,i})\right| ^{1-\sigma }\right) ^{\frac{p^*}{1-\sigma }}\right) ^{\frac{1-\sigma }{p^*}}\right) ^{\frac{p^*}{1-\sigma }}\\&\qquad +\left( s^{-1}\sup _{h\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \sum _{i=1}^m\left( \left| \lambda _{2,i}\right| \left( \frac{1}{1-\left| z_{2,i}\right| ^2}\right) ^{\sigma }\left| h'(z_{2,i})\right| ^{1-\sigma }\right) ^{\frac{p^*}{1-\sigma }}\right) ^{\frac{1-\sigma }{p^*}}\right) ^{\frac{p^*}{1-\sigma }} \end{aligned}$$

and

$$\begin{aligned} \sum _{i=1}^{n+m}\left\| x_{3,i}\right\| ^{\frac{p^*}{p^*-(1-\sigma )}} =r^{\frac{p^*}{p^*-(1-\sigma )}}\sum _{i=1}^{n}\left\| x_{1,i}\right\| ^{\frac{p^*}{p^*-(1-\sigma )}} +s^{\frac{p^*}{p^*-(1-\sigma )}}\sum _{i=1}^{m}\left\| x_{2,i}\right\| ^{\frac{p^*}{p^*-(1-\sigma )}}. \end{aligned}$$

Since $p^*/(1-\sigma )>1$ and $(p^*/(1-\sigma ))^*=p^*/(p^*-(1-\sigma ))$, Young’s Inequality gives us

$$\begin{aligned}&d^{\widehat{\mathcal {B}}}_{p,\sigma }(\gamma _1+\gamma _2) \le \frac{1-\sigma }{p^*}\left( \sup _{h\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \sum _{i=1}^{n+m}\left( \left| \lambda _{3,i}\right| \left( \frac{1}{1-\left| z_{3,i}\right| ^2}\right) ^{\sigma }\left| h'(z_{3,i})\right| ^{1-\sigma }\right) ^{\frac{p^*}{1-\sigma }}\right) ^{\frac{1-\sigma }{p^*}}\right) ^{\frac{p^*}{1-\sigma }}\\&\qquad +\frac{p^*-(1-\sigma )}{p^*}\sum _{i=1}^{n+m}\left\| x_{3,i}\right\| ^{\frac{p^*}{p^*-(1-\sigma )}}\\&\quad \le \frac{(1-\sigma )r^{-\frac{p^*}{1-\sigma }}}{p^*}\left( \sup _{h\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \sum _{i=1}^n\left( \left| \lambda _{1,i}\right| \left( \frac{1}{1-\left| z_{1,i}\right| ^2}\right) ^{\sigma }\left| h'(z_{1,i})\right| ^{1-\sigma }\right) ^{\frac{p^*}{1-\sigma }}\right) ^{\frac{1-\sigma }{p^*}}\right) ^{\frac{p^*}{1-\sigma }}\\&\qquad +\frac{(p^*-(1-\sigma ))r^{\frac{p^*}{p^*-(1-\sigma )}}}{p^*}\sum _{i=1}^{n}\left\| x_{1,i}\right\| ^{\frac{p^*}{p^*-(1-\sigma )}}\\&\qquad +\frac{(1-\sigma )s^{-\frac{p^*}{1-\sigma }}}{p^*}\left( \sup _{h\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \sum _{i=1}^m\left( \left| \lambda _{2,i}\right| \left( \frac{1}{1-\left| z_{2,i}\right| ^2}\right) ^{\sigma }\left| h'(z_{2,i})\right| ^{1-\sigma }\right) ^{\frac{p^*}{1-\sigma }}\right) ^{\frac{1-\sigma }{p^*}}\right) ^{\frac{p^*}{1-\sigma }}\\&\qquad +\frac{(p^*-(1-\sigma ))s^{\frac{p^*}{p^*-(1-\sigma )}}}{p^*}\sum _{i=1}^{m}\left\| x_{2,i}\right\| ^{\frac{p^*}{p^*-(1-\sigma )}}. \end{aligned}$$

In particular, taking above

$$\begin{aligned} r&=(d^{\widehat{\mathcal {B}}}_{p,\sigma }(\gamma _1)+\varepsilon )^{-\frac{1-\sigma }{p^*}}\sup _{h\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \sum _{i=1}^n\left( \left| \lambda _{1,i}\right| \left( \frac{1}{1-\left| z_{1,i}\right| ^2}\right) ^{\sigma }\left| h'(z_{1,i})\right| ^{1-\sigma }\right) ^{\frac{p^*}{1-\sigma }}\right) ^{\frac{1-\sigma }{p^*}},\\ s&=(d^{\widehat{\mathcal {B}}}_{p,\sigma }(\gamma _2)+\varepsilon )^{-\frac{1-\sigma }{p^*}}\sup _{h\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \sum _{i=1}^m\left( \left| \lambda _{2,i}\right| \left( \frac{1}{1-\left| z_{2,i}\right| ^2}\right) ^{\sigma }\left| h'(z_{2,i})\right| ^{1-\sigma }\right) ^{\frac{p^*}{1-\sigma }}\right) ^{\frac{1-\sigma }{p^*}}, \end{aligned}$$

one obtains $d^{\widehat{\mathcal {B}}}_{p,\sigma }(\gamma _1+\gamma _2)\le d^{\widehat{\mathcal {B}}}_{p,\sigma }(\gamma _1)+d^{\widehat{\mathcal {B}}}_{p,\sigma }(\gamma _2)+2\varepsilon$, and the arbitrariness of $\varepsilon$ yields

$$\begin{aligned} d^{\widehat{\mathcal {B}}}_{p,\sigma }(\gamma _1+\gamma _2)\le d^{\widehat{\mathcal {B}}}_{p,\sigma }(\gamma _1)+d^{\widehat{\mathcal {B}}}_{p,\sigma }(\gamma _2). \end{aligned}$$

To conclude that $d^{\widehat{\mathcal {B}}}_{p,\sigma }$ is a norm, note that the Hölder’s Inequality gives

$$\begin{aligned}&\left| \sum _{i=1}^n \lambda _i \left( \frac{1}{1-\left| z_i\right| ^2}\right) ^{\sigma }g'(z_i)^{1-\sigma }x^*(x_i)\right| \le \sum _{i=1}^n\left| \lambda _i\right| \left( \frac{1}{1-\left| z_i\right| ^2}\right) ^{\sigma }\left| g'(z_i)\right| ^{1-\sigma }\left\| x_i\right\| \\&\quad \le \left( \sum _{i=1}^n\left( \left| \lambda _i\right| \left( \frac{1}{1-\left| z_i\right| ^2}\right) ^{\sigma }\left| g'(z_i)\right| ^{1-\sigma }\right) ^{\frac{p^*}{1-\sigma }}\right) ^{\frac{1-\sigma }{p^*}}\left( \sum _{i=1}^n\left\| x_i\right\| ^{\frac{p^*}{p^*-(1-\sigma )}}\right) ^{\frac{p^*-(1-\sigma )}{p^*}}\\&\quad \le \sup _{h\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \sum _{i=1}^n\left( \left| \lambda _i\right| \left( \frac{1}{1-\left| z_i\right| ^2}\right) ^{\sigma }\left| h'(z_i)\right| ^{1-\sigma }\right) ^{\frac{p^*}{1-\sigma }}\right) ^{\frac{1-\sigma }{p^*}}\left( \sum _{i=1}^n\left\| x_i\right\| ^{\frac{p^*}{p^*-(1-\sigma )}}\right) ^{\frac{p^*-(1-\sigma )}{p^*}}, \end{aligned}$$

whenever $g\in B_{\widehat{\mathcal {B}}(\mathbb {D})}$ and $x^*\in B_{X^*}$. Note that the value $\left| \sum _{i=1}^n \lambda _i g'(z_i)x^*(x_i)\right|$ is independent on the representation of $\gamma$ seeing as

$$\begin{aligned} \sum _{i=1}^n \lambda _i g'(z_i)x^*(x_i)=\left( \sum _{i=1}^n\lambda _i\gamma _{z_i}\otimes x_i\right) (g\cdot x^*)=\gamma (g\cdot x^*), \end{aligned}$$

and taking infimum over all representations of $\gamma$ produces

$$\begin{aligned} \left| \sum _{i=1}^n \lambda _i g'(z_i)x^*(x_i)\right| \le d^{\widehat{\mathcal {B}}}_{p,\sigma }(\gamma ) \qquad \left( g\in B_{\widehat{\mathcal {B}}(\mathbb {D})},\; x^*\in B_{X^*}\right) . \end{aligned}$$

Now, if $d^{\widehat{\mathcal {B}}}_{p,\sigma }(\gamma )=0$, the above inequality gives

$$\begin{aligned} \left( \sum _{i=1}^n \lambda _i x^*(x_i)\gamma _{z_i}\right) (g)=\sum _{i=1}^n \lambda _i x^*(x_i) g'(z_i)=0\qquad \left( g\in B_{\widehat{\mathcal {B}}(\mathbb {D})},\; x^*\in B_{X^*}\right) . \end{aligned}$$

For each $x^*\in B_{X^*}$, this implies that $\sum _{i=1}^n\lambda _i x^*(x_i)\gamma _{z_i}=0$, and since $\Gamma (\mathbb {D})$ is linearly independent in $\mathcal {G}(\mathbb {D})$ (see [10, Remark 2.8]), we secure that $x^*(x_i)\lambda _i =0$ for all $i\in \{1,\ldots ,n\}$, hence $\lambda _i=0$ for all $i\in \{1,\ldots ,n\}$ since $B_{X^*}$ separate points, and so $\gamma =\sum _{i=1}^n\lambda _i\gamma _{z_i}\otimes x_i=0$.

To finish, we show that $d^{\widehat{\mathcal {B}}}_{p,\sigma }$ is a Bloch reasonable crossnorm on $\textrm{lin}(\Gamma (\mathbb {D}))\otimes X$:

• (i) Given $z\in \mathbb {D}$ and $x\in X$,

  $$\begin{aligned}{} & {} d^{\widehat{\mathcal {B}}}_{p,\sigma }(\gamma _z\otimes x)\le \left( \sup _{h\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \left( \frac{1}{1-\left| z\right| ^2}\right) ^{\sigma }\left| h'(z)\right| ^{1-\sigma }\right) ^{\frac{p^*}{1-\sigma }}\right) ^{\frac{1-\sigma }{p^*}}\left\| x\right\| \\{} & {} \le \frac{\left\| x\right\| }{1-|z|^2}=\left\| \gamma _z\right\| \left\| x\right\| . \end{aligned}$$

• (ii) For any $g\in \widehat{\mathcal {B}}(\mathbb {D})$ and $x^*\in X^*$,

  $$\begin{aligned}&\left| (g\otimes x^*)(\gamma )\right| =\left| \sum _{i=1}^n\lambda _i(g\otimes x^*)(\gamma _{z_i}\otimes x_i)\right| =\left| \sum _{i=1}^n\lambda _i g'(z_i)x^*(x_i)\right| \\&\quad \le \sum _{i=1}^n\left| \lambda _i\right| \left| g'(z_i)\right| \left| x^*(x_i)\right| \le \rho _{\mathcal {B}}(g)\left\| x^*\right\| \sum _{i=1}^n\frac{\left| \lambda _i\right| }{1-|z_i|^2}\left\| x_i\right\| \\&\quad =\rho _{\mathcal {B}}(g)\left\| x^*\right\| \sum _{i=1}^n\left| \lambda _i\right| \left| f'_{z_i}(z_i)\right| \left\| x_i\right\| =\rho _{\mathcal {B}}(g)\left\| x^*\right\| \sum _{i=1}^n\left| \lambda _i\right| \\&\qquad \times \left( \frac{1}{1-\left| z_i\right| ^2}\right) ^{\sigma }\left| f'_{z_i}(z_i)\right| ^{1-\sigma }\left\| x_i\right\| \\&\quad \le \rho _{\mathcal {B}}(g)\left\| x^*\right\| \left( \sum _{i=1}^n\left( \left| \lambda _i\right| \left( \frac{1}{1-\left| z_i\right| ^2}\right) ^{\sigma }\left| f'_{z_i}(z_i)\right| ^{1-\sigma }\right) ^{\frac{p^*}{1-\sigma }}\right) ^{\frac{1-\sigma }{p^*}}\\&\qquad \times \left( \sum _{i=1}^n\left\| x_i\right\| ^{\frac{p^*}{p^*-(1-\sigma )}}\right) ^{\frac{p^*-(1-\sigma )}{p^*}}\\&\quad \le \rho _{\mathcal {B}}(g)\left\| x^*\right\| \left( \sup _{h\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \sum _{i=1}^n\left( \left| \lambda _i\right| \left( \frac{1}{1-\left| z_i\right| ^2}\right) ^{\sigma }\left| h'(z_i)\right| ^{1-\sigma }\right) ^{\frac{p^*}{1-\sigma }}\right) ^{\frac{1-\sigma }{p^*}}\right) \\&\qquad \times \left( \sum _{i=1}^n\left\| x_i\right\| ^{\frac{p^*}{p^*-(1-\sigma )}}\right) ^{\frac{p^*-(1-\sigma )}{p^*}}. \end{aligned}$$

  Passing to the infimum over all the representations of $\gamma$ yields

  $$\begin{aligned} \left| (g\otimes x^*)(\gamma )\right| \le \rho _{\mathcal {B}}(g)\left\| x^*\right\| d^{\widehat{\mathcal {B}}}_{p,\sigma }(\gamma ). \end{aligned}$$

  Therefore, $g\otimes x^*\in (\textrm{lin}(\Gamma (\mathbb {D}))\otimes _{d^{\widehat{\mathcal {B}}}_{p,\sigma }} X)^*$ and $\left\| g\otimes x^*\right\| \le \rho _{\mathcal {B}}(g)\left\| x^*\right\|$.

$\square$

We are now in a position to address the duality of the space of $(p,\sigma )$-absolutely continuous Bloch maps from $\mathbb {D}$ into the dual space $X^*$ of a complex Banach space X. In the proof of the following result, we will make use of Proposition 2.1 and Theorem 3.1.

Theorem 6.3

Let $p\in [1,\infty )$ and $\sigma \in [0,1)$. Then $\Pi ^{\widehat{\mathcal {B}}}_{p,\sigma }(\mathbb {D},X^*)$ is isometrically isomorphic to $(\mathcal {G}(\mathbb {D})\widehat{\otimes }_{d^{\widehat{\mathcal {B}}}_{p^*,\sigma }} X)^*$, via the canonical pairing

$$\begin{aligned} \Lambda (f)(\gamma )=\sum _{i=1}^n\lambda _i\left\langle f'(z_i),x_i\right\rangle \end{aligned}$$

for all $f\in \Pi ^{\widehat{\mathcal {B}}}_{p,\sigma }(\mathbb {D},X^*)$ and $\gamma =\sum _{i=1}^n\lambda _i\gamma _{z_i}\otimes x_i\in \textrm{lin}(\Gamma (\mathbb {D}))\otimes X$.

Proof

For $\sigma =0$, the result follows from Proposition 2.1 and [6, Theorem 2.8]. Assume $\sigma \in (0,1)$. We are going to prove the case $1<p<\infty$. The case $p=1$ follows similarly.

Let $f\in \Pi ^{\widehat{\mathcal {B}}}_{p,\sigma }(\mathbb {D},X^*)$ and define the linear map $\Lambda _0(f):\textrm{lin}(\Gamma (\mathbb {D}))\otimes X\rightarrow \mathbb {C}$ by setting

$$\begin{aligned} \Lambda _0(f)(\gamma )=\sum _{i=1}^n\lambda _i\left\langle f'(z_i),x_i\right\rangle \qquad (\gamma =\sum _{i=1}^n\lambda _i\gamma _{z_i}\otimes x_i\in \textrm{lin}(\Gamma (\mathbb {D}))\otimes X). \end{aligned}$$

Since $p/(1-\sigma )>1$ and $(p/(1-\sigma ))^*=p/(p-(1-\sigma ))$, Hölder Inequality and Theorem 3.1 provide

$$\begin{aligned}&\left| \Lambda _0(f)(\gamma )\right| =\left| \sum _{i=1}^n\lambda _i\left\langle f'(z_i), x_i\right\rangle \right| \le \sum _{i=1}^n\left| \lambda _i\right| \left\| f'(z_i)\right\| \left\| x_i\right\| \\&\quad \le \left( \sum _{i=1}^n\left( \left| \lambda _i\right| \left\| f'(z_i)\right\| \right) ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}}\left( \sum _{i=1}^n\left\| x_i\right\| ^{\frac{p}{p-(1-\sigma )}}\right) ^{\frac{p-(1-\sigma )}{p}}\\&\quad \le \pi ^{\mathcal {B}}_{p,\sigma }(f)\sup _{h\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \sum _{i=1}^n\left( \left| \lambda _i\right| \left( \frac{1}{1-\left| z_i\right| ^2}\right) ^{\sigma }\left| h'(z_i)\right| ^{1-\sigma }\right) ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}}\\&\quad \left( \sum _{i=1}^n\left\| x_i\right\| ^{\frac{p}{p-(1-\sigma )}}\right) ^{\frac{p-(1-\sigma )}{p}}. \end{aligned}$$

Calculating the infimum on all the representations of $\gamma$ yields

$$\begin{aligned} \left| \Lambda _0(f)(\gamma )\right| \le \pi ^{\mathcal {B}}_{p,\sigma }(f)d^{\widehat{\mathcal {B}}}_{p^*,\sigma }(\gamma ). \end{aligned}$$

Hence $\Lambda _0(f)$ is continuous on $\textrm{lin}(\Gamma (\mathbb {D}))\otimes _{d^{\widehat{\mathcal {B}}}_{p^*,\sigma }} X$ with $\left\| \Lambda _0(f)\right\| \le \pi ^{\mathcal {B}}_{p,\sigma }(f)$.

Clearly, $\mathcal {G}(\mathbb {D})\otimes X$ is a norm-dense linear subspace of $\mathcal {G}(\mathbb {D})\widehat{\otimes }_{d^{\widehat{\mathcal {B}}}_{p^*,\sigma }} X$ and therefore we can find a unique continuous map $\Lambda (f):\mathcal {G}(\mathbb {D})\widehat{\otimes }_{d^{\widehat{\mathcal {B}}}_{p^*,\sigma }} X\rightarrow \mathbb {C}$ extending $\Lambda _0(f)$. Further, $\Lambda (f)$ is linear and $\left\| \Lambda (f)\right\| =\left\| \Lambda _0(f)\right\|$.

In this way, we define a map $\Lambda :\Pi ^{\widehat{\mathcal {B}}}_{p,\sigma }(\mathbb {D},X^*)\rightarrow (\mathcal {G}(\mathbb {D})\widehat{\otimes }_{d^{\widehat{\mathcal {B}}}_{p^*,\sigma }} X)^*$. By [6, Corollary 2.3], $\Lambda$ is linear and injective since $\Pi _{p,\sigma }^{\widehat{\mathcal {B}}}(\mathbb {D},X^*)\subseteq \widehat{\mathcal {B}}(\mathbb {D},X^*)$. We now prove that $\Lambda$ is a surjective isometry. For it, let $\varphi \in (\mathcal {G}(\mathbb {D})\widehat{\otimes }_{d^{\widehat{\mathcal {B}}}_{p^*,\sigma }} X)^*$ and define $F_\varphi :\mathbb {D}\rightarrow X^*$ by

$$\begin{aligned} \left\langle F_\varphi (z),x\right\rangle =\varphi (\gamma _z\otimes x)\qquad \left( z\in \mathbb {D},\; x\in X\right) . \end{aligned}$$

Apparently (see, for example, the proof of [6, Proposition 2.4]), $F_\varphi \in \mathcal {H}(\mathbb {D},X^*)$ and $F_\varphi =f'_\varphi$ for a suitable map $f_\varphi \in \widehat{\mathcal {B}}(\mathbb {D},X^*)$ with $\rho _{\mathcal {B}}(f_\varphi )\le \left\| \varphi \right\|$.

To prove that $f_\varphi \in \Pi ^{\widehat{\mathcal {B}}}_{p,\sigma }(\mathbb {D},X^*)$, let $n\in \mathbb {N}$, $\lambda _i\in \mathbb {C}$ and $z_i\in \mathbb {D}$ for all $i\in \{1,\ldots ,n\}$. Given $\varepsilon >0$, for each $i\in \{1,\ldots ,n\}$, we can find $x_i\in X$ with $\left\| x_i\right\| \le 1+\varepsilon$ so that

$$\begin{aligned} \left\langle f_\varphi '(z_i),x_i\right\rangle =\left\| f_\varphi '(z_i)\right\| . \end{aligned}$$

Obviously, $T:\mathbb {C}^n\rightarrow \mathbb {C}$ defined by

$$\begin{aligned} T(t_1,\ldots ,t_n)=\sum _{i=1}^n t_i \lambda _i\left\| f_\varphi '(z_i)\right\| ,\qquad (t_1,\ldots ,t_n)\in \mathbb {C}^n, \end{aligned}$$

is in $(\mathbb {C}^n,||\cdot ||_{(p/(1-\sigma ))^*})^*$ and

$$\begin{aligned} \left\| T\right\| =\left( \sum _{i=1}^n\left( \left| \lambda _i\right| \left\| f'_\varphi (z_i)\right\| \right) ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}}. \end{aligned}$$

If $||(t_1,\ldots ,t_n)||_{(p/(1-\sigma ))^*}\le 1$, we get

$$\begin{aligned}&\left| T(t_1,\ldots ,t_n)\right| =\left| \varphi \left( \sum _{i=1}^n t_i\lambda _i\gamma _{z_i}\otimes x_i\right) \right| \le \left\| \varphi \right\| d^{\widehat{\mathcal {B}}}_{p^*,\sigma }\left( \sum _{i=1}^n\lambda _i\gamma _{z_i}\otimes t_ix_i\right) \\&\quad \le \left\| \varphi \right\| \sup _{h\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \sum _{i=1}^n\left( \left| \lambda _i\right| \left( \frac{1}{1-\left| z_i\right| ^2}\right) ^{\sigma }\left| h'(z_i)\right| ^{1-\sigma }\right) ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}}\left( \sum _{i=1}^n\left\| t_ix_i\right\| ^{\frac{p}{p-(1-\sigma )}}\right) ^{\frac{p-(1-\sigma )}{p}}\\&\quad \le (1+\varepsilon )\left\| \varphi \right\| \sup _{h\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \sum _{i=1}^n\left( \left| \lambda _i\right| \left( \frac{1}{1-\left| z_i\right| ^2}\right) ^{\sigma }\left| h'(z_i)\right| ^{1-\sigma }\right) ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}}, \end{aligned}$$

therefore

$$\begin{aligned}{} & {} \left( \sum _{i=1}^n\left( \left| \lambda _i\right| \left\| f'_\varphi (z_i)\right\| \right) ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}}\\{} & {} \quad \le \left\| \varphi \right\| \sup _{h\in B_{\widehat{\mathcal {B}}(\mathbb {D})}}\left( \sum _{i=1}^n\left( \left| \lambda _i\right| \left( \frac{1}{1-\left| z_i\right| ^2}\right) ^{\sigma }\left| h'(z_i)\right| ^{1-\sigma }\right) ^{\frac{p}{1-\sigma }}\right) ^{\frac{1-\sigma }{p}}, \end{aligned}$$

and consequently Theorem 3.1 tells us that $f_\varphi \in \Pi ^{\widehat{\mathcal {B}}}_{p,\sigma }(\mathbb {D},X^*)$ with $\pi ^{\mathcal {B}}_{p,\sigma }(f_\varphi )\le \left\| \varphi \right\|$.

Now, for any $\gamma =\sum _{i=1}^n \lambda _i\gamma _{z_i}\otimes x_i\in \textrm{lin}(\Gamma (\mathbb {D}))\otimes X$, one has

$$\begin{aligned} \Lambda (f_\varphi )(\gamma ) =\sum _{i=1}^n\lambda _i\left\langle f'_\varphi (z_i),x_i\right\rangle =\sum _{i=1}^n\lambda _i\varphi (\gamma _{z_i}\otimes x_i) =\varphi \left( \sum _{i=1}^n\lambda _i\gamma _{z_i}\otimes x_i\right) =\varphi (\gamma ), \end{aligned}$$

and $\Lambda (f_\varphi )=\varphi$ on $\mathcal {G}(\mathbb {D})\widehat{\otimes }_{d^{\widehat{\mathcal {B}}}_{p^*,\sigma }}X$. Further, $\pi ^{\mathcal {B}}_{p,\sigma }(f_\varphi )\le \left\| \Lambda (f_\varphi )\right\|$. This completes the proof.$\square$