Litte Hankel Operators Between Vector-Valued Bergman Spaces on the Unit Ball

Abstract

In this paper, we study the boundedness and the compactness of the little Hankel operators $h_b$ with operator-valued symbols b between different weighted vector-valued Bergman spaces on the open unit ball $\mathbb {B}_n$ in $\mathbb {C}^n.$ More precisely, given two complex Banach spaces X, Y,  and $0 < p,q \le 1,$ we characterize those operator-valued symbols $b: \mathbb {B}_{n}\rightarrow \mathcal {L}(\overline{X},Y)$ for which the little Hankel operator $h_{b}: A^p_{\alpha }(\mathbb {B}_{n},X) \longrightarrow A^q_{\alpha }(\mathbb {B}_{n},Y),$ is a bounded operator. Also, given two reflexive complex Banach spaces X, Y and $1< p \le q < \infty ,$ we characterize those operator-valued symbols $b: \mathbb {B}_{n}\rightarrow \mathcal {L}(\overline{X},Y)$ for which the little Hankel operator $h_{b}: A^p_{\alpha }(\mathbb {B}_{n},X) \longrightarrow A^q_{\alpha }(\mathbb {B}_{n},Y),$ is a compact operator.

1 Introducton

It is well known that Hankel operators constitute a very important class of operators in spaces of analytic functions. The study of these operators on different analytic spaces is not only motivated by the mathematical challenges it raises, but also by many applications on applied mathematics and in physics (see for example [13] for more information). In this paper, we are interested on the boundedness and the compactness problem of the little Hankel operator with operator-valued symbols on weighted vector-valued Bergman spaces on the unit ball.

Throughout this paper, we fix a nonnegative integer n and let

$$\begin{aligned} \mathbb {C}^n = \mathbb {C} \times \cdots \times \mathbb {C} \end{aligned}$$

denote the n-dimensional Euclidean space. For

$$\begin{aligned} z = (z_{1},\ldots ,z_{n}), \qquad w = (w_{1},\ldots ,w_{n}), \end{aligned}$$

in $\mathbb {C}^{n},$ we define the inner product of z and w by

$$\begin{aligned} \langle z,w \rangle = z_{1}\overline{w_{1}} + \cdots + z_{n}\overline{w_{n}}, \end{aligned}$$

where $\overline{w_{k}}$ is the complex conjugate of $w_{k}.$ The resulting norm is then

$$\begin{aligned} |z| = \sqrt{\langle z,z \rangle } = \sqrt{|z_{1}|^2 + \cdots + |z_{n}|^2 }. \end{aligned}$$

Endowed with the above inner product, $\mathbb {C}^n$ become a Hilbert space whose canonical basis consists of the following vectors

$$\begin{aligned} e_{1} = (1,0,\ldots ,0),~~e_{2}=(0,1,0,\ldots ,0),~~ \ldots ,~~e_{n} =(0,\ldots ,0,1). \end{aligned}$$

The open unit ball in $\mathbb {C}^n$ is the set

$$\begin{aligned} \mathbb {B}_n = \lbrace z \in \mathbb {C}^{n}: |z| < 1 \rbrace . \end{aligned}$$

When $\alpha >-1,$ the weighted Lebesgue measure $\mathrm {d}\nu _{\alpha }$ in $\mathbb {B}_n$ is defined by

$$\begin{aligned} \mathrm {d}\nu _{\alpha }(z) = c_{\alpha }(1-|z|^2)^{\alpha }\mathrm {d}\nu (z), \qquad z \in \mathbb {B}_{n} \end{aligned}$$

where $\mathrm {d}\nu$ is the Lebesgue measure in $\mathbb {C}^n$ and

$$\begin{aligned} c_{\alpha } = \dfrac{\Gamma (n+\alpha +1)}{n!\Gamma (\alpha +1)} \end{aligned}$$

is the normalizing constant so that $\mathrm {d}\nu _{\alpha }$ becomes a probability measure on $\mathbb {B}_n.$ A function defined on the unit ball $\mathbb {B}_n$ will be called a vector-valued function when it takes its values in some vector space. If X is a complex Banach space, a vector-valued function $f:\mathbb {B}_{n} \longrightarrow X$ (a X-valued function) is said to be strongly holomorphic in $\mathbb {B}_n$ if for every $z \in \mathbb {B}_{n}$ and for every $k \in \lbrace 1,\ldots , n \rbrace ,$ the limit

$$\begin{aligned} \displaystyle \lim _{\lambda \longrightarrow 0}\dfrac{f(z+\lambda e_{k})-f(z)}{\lambda } \end{aligned}$$

exists in X,  where $\lambda \in \mathbb {C}-\lbrace 0 \rbrace$. The space of all X-valued strongly holomorphic functions on $\mathbb {B}_n$ will be denoted by $\mathcal {H}(\mathbb {B}_{n},X).$ We will also denote by $H^{\infty }(\mathbb {B}_{n},X)$ the space of all bounded X-valued holomorphic functions. Let $X^{\star }$ denotes the space of all bounded linear functionals $x^{\star }:X \longrightarrow \mathbb {C}$ (the topological dual space of X). We say that a vector-valued function $f:\mathbb {B}_{n} \longrightarrow X$ is weakly holomorphic if for every $x^{\star } \in X^{\star },$ the scalar-valued function $x^{\star }(f): \mathbb {B}_{n} \longrightarrow \mathbb {C}$ is holomorphic in the usual sense. An important result by Dunford [7] shows that a vector-valued function is strongly holomorphic if and only if it is weakly holomorphic.

1.1 The Conjugate $\overline{X}$ of the Complex Banach Space X

In the sequel, we will need the notion of “conjugate” of a complex Banach space [11].

We will use the following definition and notation which can be found in [11]. Let $x \in X,$ $x^{\star } \in X^{\star }$ and $\lambda \in \mathbb {C}.$ We define

$$\begin{aligned} (\lambda x^{\star })(x): = \overline{\lambda } x^{\star }(x). \end{aligned}$$

We also use the notation

$$\begin{aligned} \langle x,x^{\star } \rangle _{X,X^{\star }} = x^{\star }(x) \end{aligned}$$

to represent the ‘inner product’ in the complex Banach space X. We have the following identities

$$\begin{aligned} \langle \lambda x,x^{\star } \rangle _{X,X^{\star }} = \lambda \langle x,x^{\star } \rangle _{X,X^{\star }} = \langle x,\overline{\lambda }x^{\star } \rangle _{X,X^{\star }}, \end{aligned}$$

so that we have a regular rule of an inner product. The complex conjugate $\overline{x}$ of $x \in X,$ is the linear functional on $X^{\star }$ defined by

$$\begin{aligned} \overline{x}(x^{\star }) = \overline{\langle x,x^{\star } \rangle }_{X,X^{\star }}, \end{aligned}$$

for every $x^{\star } \in X^{\star }.$ Therefore,

$$\begin{aligned} \overline{X} = \lbrace \overline{x}: x \in X \rbrace \end{aligned}$$

is called the complex conjugate of the Banach space X. With the norm defined by

$$\begin{aligned} \Vert \overline{x}\Vert := \sup _{\Vert x^{\star }\Vert _{X^{\star }} = 1}|\overline{x}(x^{\star })|, \end{aligned}$$

$\overline{X}$ becomes a Banach space. Moreover, we have that $\Vert x\Vert _{X} = \Vert \overline{x}\Vert _{\overline{X}}$ for any $x \in X,$ so that X and $\overline{X}$ are isometrically anti-isomorphic.

1.2 Vector-Valued Bergman Space

In the sequel, we will integrate vector-valued measurable functions in the sense of Bochner (see [7] for more information). Let X be a complex Banach space. A measurable function $f: \mathbb {B}_{n} \longrightarrow X$ is Bochner-integrable with respect to the measure $\nu _{\alpha }$ in the unit ball $\mathbb {B}_n$ if and only if the Lebesgue integral

$$\begin{aligned} \displaystyle \Vert f\Vert _{1,\alpha ,X} = \int _{\mathbb {B}_n}\Vert f(z)\Vert _{X}\mathrm {d}\nu _{\alpha }(z) \end{aligned}$$

is finite. For $0< p < \infty ,$ the Bochner-Lebesgue space $L^{p}_{\nu _{\alpha }}(\mathbb {B}_{n},X)$ consists of all vector-valued measurable functions $f:\mathbb {B}_{n} \longrightarrow X$ such that

$$\begin{aligned} \displaystyle \Vert f\Vert ^p_{p,\alpha ,X} = \int _{\mathbb {B}_n}\Vert f(z)\Vert ^p_{X}\mathrm {d}\nu _{\alpha }(z) < \infty . \end{aligned}$$

The vector-valued Bergman space $A^{p}_{\alpha }(\mathbb {B}_{n},X)$ is defined by

$$\begin{aligned} A^{p}_{\alpha }(\mathbb {B}_{n},X) = L^{p}_{\nu _{\alpha }}(\mathbb {B}_{n},X) \cap \mathcal {H}(\mathbb {B}_{n},X). \end{aligned}$$

The weak Bochner-Lebesgue space $L^{p,\infty }_{\alpha }(\mathbb {B}_{n},X)$ consists of all vector-valued measurable functions $f:\mathbb {B}_n \longrightarrow X$ for which

$$\begin{aligned} \Vert f\Vert _{L^{p,\infty }_{\alpha }(\mathbb {B}_{n},X)} = \left( \sup _{\lambda> 0}\lambda ^{p} \nu _{\alpha }\left( \lbrace z \in \mathbb {B}_{n}: \Vert f(z)\Vert _{X} > \lambda \rbrace \right) \right) ^{1/p} < \infty . \end{aligned}$$

The weak vector-valued Bergman space $A^{p,\infty }_{\alpha }(\mathbb {B}_{n},X)$ is defined by

$$\begin{aligned} A^{p,\infty }_{\alpha }(\mathbb {B}_{n},X) = \mathcal {H}(\mathbb {B}_{n},X) \cap L^{p,\infty }_{\alpha }(\mathbb {B}_{n},X). \end{aligned}$$

Let X, Y be two complex Banach spaces and $\alpha > -1.$ We have the following two lemmas whose proofs can be found in [11].

Lemma 1

Let $T: X \longrightarrow Y$ be a bounded linear operator. If $f:\mathbb {B}_{n} \longrightarrow X$ is $\nu _{\alpha }$-Bochner integrable in the unit ball, then $Tf:\mathbb {B}_{n} \longrightarrow Y$ is $\nu _{\alpha }$-Bochner integrable in the unit ball and we have

$$\begin{aligned} \displaystyle \int _{\mathbb {B}_n} Tf(z)\mathrm {d}\nu _{\alpha }(z) = T\left( \int _{\mathbb {B}_n}f(z)\mathrm {d}\nu _{\alpha }(z)\right) . \end{aligned}$$

Lemma 2

If $f:\mathbb {B}_{n} \longrightarrow X$ is a $\nu _{\alpha }$-Bochner integrable vector-valued function in the unit ball, then the following inequality holds

$$\begin{aligned} \displaystyle \left\| \int _{\mathbb {B}_n}f(z) \mathrm {d}\nu _{\alpha }(z)\right\| _{X} \le \int _{\mathbb {B}_n}\Vert f(z)\Vert _{X}\mathrm {d}\nu _{\alpha }(z). \end{aligned}$$

1.3 Vector-Valued Lipschitz Spaces and Vector-Valued $\gamma$-Bloch Spaces

The radial derivative of a vector-valued holomorphic function $f: \mathbb {B}_{n} \longrightarrow X$ denoted Nf is defined for $z \in \mathbb {B}_n$ by

$$\begin{aligned} \displaystyle Nf(z): = \sum _{j=1}^{n}z_{j}\frac{\partial f}{\partial z_{j}}(z). \end{aligned}$$

(1.1)

Let $f \in \mathcal {H}(\mathbb {B}_{n},X)$ and

$$\begin{aligned} \displaystyle f(z) = \sum _{k=0}^{\infty }f_{k}(z), \qquad z \in \mathbb {B}_n \end{aligned}$$

the homogeneous expansion of the function f where $f_k$ are homogeneous holomorphic polynomials of degree k with coefficients in X. For any two real parameters $\alpha$ and t such that neither $n+\alpha$ nor $n+\alpha +t$ is a negative integer, we define an invertible operator $R^{\alpha ,t} : \mathcal {H}(\mathbb {B}_{n},X) \rightarrow \mathcal {H}(\mathbb {B}_{n},X)$ as

$$\begin{aligned} \displaystyle R^{\alpha ,t}f(z): = \sum _{k=0}^{\infty }\dfrac{\Gamma (n+1+\alpha )\Gamma (n+1+k+\alpha +t)}{\Gamma (n+1+\alpha +t)\Gamma (n+1+k+\alpha )}f_{k}(z), \end{aligned}$$

(1.2)

where $z \in \mathbb {B}_{n}$ and $\Gamma$ is the classical Euler Gamma function. For $\gamma \ge 0,$ we denote by $\Gamma _{\gamma }(\mathbb {B}_{n},X)$ the space of vector-valued holomorphic functions $f:\mathbb {B}_{n} \longrightarrow X$ for which there exists an integer $k > \gamma$ such that

$$\begin{aligned} \Vert f\Vert _{\gamma ,X} = \Vert f(0)\Vert _{X} + \sup _{z \in \mathbb {B}_{n}}(1 - |z|^2)^{k-\gamma }\Vert N^{k}f(z)\Vert _{X} < \infty , \end{aligned}$$

where $N^{k} = N \circ N \circ \cdots \circ N$ k-times. The definition of the space $\Gamma _{\gamma }(\mathbb {B}_{n},X)$ is independent of the integer k used. The space $\Gamma _{\gamma }(\mathbb {B}_{n},X)$ will be called the vector-valued holomorphic Lipschitz space and for $\gamma = 0,$ we write $\mathcal {B}(\mathbb {B}_{n},X) = \Gamma _{0}(\mathbb {B}_{n},X).$ It is clear that $f \in \mathcal {B}(\mathbb {B}_{n},X)$ if and only if f is a vector-valued holomorphic function and

$$\begin{aligned} \Vert f\Vert _{\mathcal {B}(\mathbb {B}_{n},X)} = \Vert f(0)\Vert _{X} + \sup _{z \in \mathbb {B}_{n}}(1 - |z|^2)\Vert Nf(z)\Vert _{X} < \infty . \end{aligned}$$

That is, $\mathcal {B}(\mathbb {B}_{n},X) = \Gamma _{0}(\mathbb {B}_{n},X)$ is the vector-valued Bloch space. The vector-valued $\gamma$-Bloch space $\mathcal {B}_{\gamma }(\mathbb {B}_{n},X)$ for $\gamma > 0,$ is defined as the space of vector-valued holomorphic functions $f \in \mathcal {H}(\mathbb {B}_{n},X)$ such that

$$\begin{aligned} \sup _{z \in \mathbb {B}_{n}}(1 - |z|^2)^{\gamma } \Vert Nf(z)\Vert _{X} < \infty . \end{aligned}$$

The little vector-valued $\gamma$-Bloch space $\mathcal {B}_{\gamma ,0}(\mathbb {B}_{n},X)$ for $\gamma > 0,$ is the subspace of $\mathcal {B}_{\gamma }(\mathbb {B}_{n},X)$ consisting of functions f such that

$$\begin{aligned} \lim _{|z| \rightarrow 1^{-}}(1 - |z|^2)^{\gamma } \Vert Nf(z)\Vert _{X} = 0. \end{aligned}$$

It is easy to see that $\mathcal {B}_{1}(\mathbb {B}_{n},X) = \mathcal {B}(\mathbb {B}_{n},X).$ Therefore, the vector-valued $\gamma$-Bloch spaces with $\gamma > 0$ generalize the vector-valued Bloch space. Let $\gamma \ge 0.$ The generalized vector-valued Lipschitz space $\Lambda _{\gamma }(\mathbb {B}_{n},X)$ consists of vector-valued holomorphic functions f in $\mathbb {B}_{n}$ such that for some nonnegative integer $k > \gamma ,$ we have

$$\begin{aligned} \Vert f\Vert _{\Lambda _{\gamma }(\mathbb {B}_{n},X)} = \sup _{z \in \mathbb {B}_n} (1 - |z|^2)^{k-\gamma }\Vert R^{\alpha ,k}f(z) \Vert _{X} < \infty . \end{aligned}$$

We consider the following norm on the generalized vector-valued Lipschitz space $\Lambda _{\gamma }(\mathbb {B}_{n},X)$ by

$$\begin{aligned} \Vert f\Vert _{\Lambda _{\gamma }(\mathbb {B}_{n},X)} = \sup _{z \in \mathbb {B}_{n}}(1 - |z|^2)^{k-\gamma }\Vert R^{\alpha ,k}f(z)\Vert _{X}, \end{aligned}$$

where $k > \gamma$ is a nonnegative integer. Equipped with this norm, the generalized vector-valued Lipschitz space $\Lambda _{\gamma }(\mathbb {B}_{n},X)$ becomes a Banach space. The generalized little vector-valued Lipschitz space $\Lambda _{\gamma ,0}(\mathbb {B}_{n},X)$ is the subspace of $\Lambda _{\gamma }(\mathbb {B}_{n},X),$ which consists of functions $f \in \Lambda _{\gamma }(\mathbb {B}_{n},X)$ such that

$$\begin{aligned} \lim _{|z| \rightarrow 1^{-}}(1 - |z|^2)^{k-\gamma }\Vert R^{\alpha ,k}f(z)\Vert _{X} = 0. \end{aligned}$$

(1.3)

When $\gamma = 0$ and $k =1,$ then $\Lambda _{0}(\mathbb {B}_{n},X) = \mathcal {B}(\mathbb {B}_{n},X).$ It is also important to note that as in the classical case, when $0< \gamma < 1,$ we have $\Lambda _{\gamma }(\mathbb {B}_{n},X) = \mathcal {B}_{1-\gamma }(\mathbb {B}_{n},X).$

1.4 Little Hankel Operator with Operator-Valued Symbol

Given two complex Banach spaces X and Y,  we denote by $\mathcal {L}(X,Y)$ the space of all bounded linear operators $T : X \longrightarrow Y$ endowed with the following norm

$$\begin{aligned} \Vert T\Vert _{\mathcal {L}(X,Y)} = \sup _{\Vert x\Vert _{X}=1} \Vert Tx\Vert _{Y} = \sup _{\Vert x\Vert _{X}=1, \Vert y^{\star }\Vert _{Y^{\star }}=1} |\langle Tx,y^{\star } \rangle _{Y,Y^{\star }}|, \end{aligned}$$

where $T \in \mathcal {L}(X,Y).$ Then $\mathcal {L}(X,Y)$ is a Banach space. We consider an operator-valued function $b:\mathbb {B}_{n} \longrightarrow \mathcal {L}(\overline{X},Y)$ and we suppose that $b \in \mathcal {H}(\mathbb {B}_{n}, \mathcal {L}(\overline{X},Y)).$ The little Hankel operator with operator-valued symbol b,  denoted $h_b$ is defined for $z \in \mathbb {B}_n$ by

$$\begin{aligned} \displaystyle h_{b}f(z) := \int _{\mathbb {B}_n}\dfrac{b(w)\overline{f(w)}}{(1- \langle z,w\rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(w), \quad f \in H^{\infty }(\mathbb {B}_{n},X). \end{aligned}$$

In the sequel, we will assume that the symbol b satisfies the following condition

$$\begin{aligned} \displaystyle \int _{\mathbb {B}_n} \dfrac{\Vert b(w)\Vert _{\mathcal {L}(\overline{X},Y)}}{|1-\langle z,w \rangle |^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(w) < \infty , \quad \text{ for } \text{ every }~~z \in \mathbb {B}_{n}. \end{aligned}$$

(1.4)

It is easy to check that if b satisfies (1.4), then the little Hankel operator $h_{b}$ is well defined on $H^{\infty }(\mathbb {B}_{n},X).$

1.5 Problems and Known Results

The boundedness properties of the little Hankel operator in the classical case (that is, when $X = Y = \mathbb {C}$) have been extensively studied and many results are now well known. For the case $n = 1,$ important references are [6, 15]. For $n>1,$ a complete characterization has been obtained by Aline Bonami and Luo Luo in [4] when $p \le q.$ In 2015, Pau and Zhao [12] solved the case $1< q< p <\infty .$ Indeed, they showed that if b is a holomorphic symbol, the little Hankel operator $h_{b}$ extends to a bounded operator from $A^p_\alpha (\mathbb {B}_{n},\mathbb {C})$ into $A^q_\alpha (\mathbb {B}_{n},\mathbb {C}),$ with $1< q< p < \infty ,$ if and only if the symbol b belongs to the weighted Bergman space $A^t_\alpha (\mathbb {B}_{n},\mathbb {C})$ where $1/t = 1/q - 1/p.$ We are here concerned with the question of characterizing the operator-valued holomorphic symbols b for which the little Hankel operator $h_{b}$ extends into a bounded operator from $A^p_{\alpha }(\mathbb {B}_{n},X)$ into $A^q_{\alpha }(\mathbb {B}_{n},Y)$ where $0<p,q < \infty .$ In [1] Aleman and Constantin solved this problem for the particular case $n = 1,$ $p = q = 2$ and $X = Y = \mathcal {H}$ where $\mathcal {H}$ is a separable Hilbert space. They showed that the little Hankel operator $h_b$ extends into a bounded operator from $A^2_{\alpha }(\mathbb {B}_{n},\mathcal {H})$ into $A^2_{\alpha }(\mathbb {B}_{n},\mathcal {H})$ if and only if the symbol b belongs to the Bloch space $\mathcal {B}(\mathbb {B}_{n},\mathcal {L}(\mathcal {H})).$ Constantin also obtained in [5] that the little Hankel operator $h_b$ is a compact operator from $A^2_{\alpha }(\mathbb {B}_{n},\mathcal {H})$ into $A^2_{\alpha }(\mathbb {B}_{n},\mathcal {H})$ if and only if the symbol b belongs to the little vector-valued Bloch space $\mathcal {B}_{0}(\mathbb {B}_{n},\mathcal {K}(\mathcal {H})).$ Their results extend clearly the one known in the classical case (when $\mathcal {H} = \mathbb {C}$). In [11], Oliver solved this problem in the case $1< p,q <\infty .$ Mainly, he showed that for $1<p < \infty ,$ the little Hankel operator $h_{b}$ is bounded from $A^p_{\alpha }(\mathbb {B}_{n},X)$ into $A^p_{\alpha }(\mathbb {B}_{n},Y)$ if and only if the symbol b belongs to the vector-valued Bloch space $\mathcal {B}(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y))$ and this result clearly generalizes the one obtained by Aleman and Constantin in [1]. Moreover, for $1<p \le q < \infty ,$ Oliver showed that the little Hankel operator $h_{b}$ is bounded from $A^p_{\alpha }(\mathbb {B}_{n},X)$ into $A^q_{\alpha }(\mathbb {B}_{n},Y)$ if and only if the symbol b belongs to the $\gamma$-Bloch space $\mathcal {B}_{\gamma }(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y))$ with $\gamma = 1+(n+1+\alpha )\left( \frac{1}{q} - \frac{1}{p} \right) .$ Also for $1< q< p < \infty ,$ Oliver showed that the little Hankel operator $h_{b}$ is bounded from $A^p_{\alpha }(\mathbb {B}_{n},X)$ into $A^q_{\alpha }(\mathbb {B}_{n},Y)$ if and only if $b \in A^{t}_{\alpha }(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y)),$ with $1/t = 1/q - 1/p,$ which generalizes the main result in [12]. We are also concerned here with the question of characterizing the operator-valued holomorphic symbols for which $h_b$ extends into a compact operator from $A^p_{\alpha }(\mathbb {B}_{n},X)$ into $A^q_{\alpha }(\mathbb {B}_{n},Y)$ where $1<p\le q<\infty .$

1.6 Statement of Results

Let X be a complex Banach space and $0 <p \le 1.$ The topological dual of the Bergman space $A^p_\alpha (\mathbb {B}_{n},X)$ can be identified with the Lipschitz space $\Gamma _{\gamma }(\mathbb {B}_{n},X^{\star })$ as follows:

Theorem 3

Let $0 < p \le 1.$ The space $(A^p_\alpha (\mathbb {B}_{n},X))^{\star }$ can be identified with $\Gamma _{\gamma }(\mathbb {B}_{n},X^{\star })$ with $\gamma = (n+1+\alpha )\left( \frac{1}{p}-1\right)$ under the pairing

$$\begin{aligned} \displaystyle \langle f,g \rangle _{\alpha ,X} = c_{k}\int _{\mathbb {B}_n} \langle f(z),D_{k}g(z) \rangle _{X,X^{\star }}(1 - |z|^2)^{k} \mathrm {d}\nu _{\alpha }(z), \end{aligned}$$

(1.5)

where $D_{k}$ is defined by (2.3), $k > \gamma ,$ is an integer, $g \in \Gamma _{\gamma }(\mathbb {B}_{n},X^{\star })$ and $f \in A^p_\alpha (\mathbb {B}_{n},X).$ Moreover,

$$\begin{aligned} \Vert g\Vert _{\Gamma _{\gamma }(\mathbb {B}_{n},X^{\star })} \simeq \sup _{\Vert f\Vert _{A^p_\alpha (\mathbb {B}_{n},X)} =1} |\langle f,g \rangle _{\alpha ,X}|. \end{aligned}$$

Before stating the next results, we need to make another assumption on the operator-valued symbol b. More precisely, we assume that the operator-valued holomorphic symbol b satisfies the following condition:

$$\begin{aligned} \displaystyle \int _{\mathbb {B}_{n}} \Vert b(z)\Vert _{\mathcal {L}(\overline{X},Y)} \log \left( \dfrac{1}{1 - |z|^2} \right) \mathrm {d}\nu _{\alpha }(z) < \infty . \end{aligned}$$

(1.6)

Let X and Y be two complex Banach spaces. Our contributions to the boundedness problem of the little Hankel operator with operator-valued symbol for $0 <p,q \le 1$ are the following :

Theorem 4

Suppose $0 < p \le 1,$ and $\alpha > -1.$ If the little Hankel operator $h_{b}$ extends to a bounded operator from $A^p_{\alpha }(\mathbb {B}_{n},X)$ into $A^q_{\alpha }(\mathbb {B}_{n},Y)$ for some positive $q<1,$ then the symbol b is in $\Gamma _{\gamma }(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y))$ with $\gamma = (n+1+\alpha )\left( \frac{1}{p}-1 \right) .$ Conversely, if b is in $\Gamma _{\gamma }(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y))$ with $\gamma = (n+1+\alpha )\left( \frac{1}{p}-1 \right) ,$ then the little Hankel operator $h_{b}: A^p_{\alpha }(\mathbb {B}_{n},X) \longrightarrow A^{1,\infty }_{\alpha }(\mathbb {B}_{n},Y)$ is a bounded operator.

As a direct consequence, we have the following result:

Corollary 5

Suppose $0 < p \le 1,$ and $\alpha > -1.$ The little Hankel operator $h_{b}$ extends to a bounded operator from $A^p_{\alpha }(\mathbb {B}_{n},X)$ into $A^q_{\alpha }(\mathbb {B}_{n},Y)$ for some positive $q<1$ if and only if its symbol b belongs to $\Gamma _{\gamma }(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y)),$ where $\gamma = (n+1+\alpha )\left( \frac{1}{p}-1\right) .$

Theorem 6

Let $0 < p \le 1,$ $\alpha >-1$ and $\gamma = (n+1+\alpha )\left( \frac{1}{p}-1\right) .$ The little Hankel operator extends to a bounded operator from $A^p_{\alpha }(\mathbb {B}_{n},X)$ into $A^{1}_{\alpha }(\mathbb {B}_{n},Y)$ if and only if for some integer $k > \gamma ,$

$$\begin{aligned} \Vert N^{k}b(w)\Vert _{\mathcal {L}(\overline{X},Y)} \le \dfrac{C}{(1-|w|^2)^{k-\gamma }}\left( \log \dfrac{1}{1-|w|^2} \right) ^{-1} \qquad w \in \mathbb {B}_{n}. \end{aligned}$$

(1.7)

Theorem 7

Suppose $1< p \le q <\infty .$ The little Hankel operator $h_{b}: A^p_{\alpha }(\mathbb {B}_{n},X) \rightarrow A^{q}_{\alpha }(\mathbb {B}_{n},Y)$ is a bounded operator if and only if

$b \in \Lambda _{\gamma _{0}}(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y)),$ where $\gamma _{0} = (n+1+\alpha )\left( \frac{1}{p} - \frac{1}{q}\right) .$ Moreover,

$$\begin{aligned} \Vert h_{b}\Vert _{A^p_{\alpha }(\mathbb {B}_{n},X) \rightarrow A^{q}_{\alpha }(\mathbb {B}_{n},Y)} \simeq \Vert b\Vert _{ \Lambda _{\gamma _{0}}(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y))}. \end{aligned}$$

If X, Y are reflexive complex Banach spaces, then we have the following theorem

Theorem 8

Suppose that $1< p \le q < \infty ,$ and $\alpha >-1$ The little Hankel operator $h_b : A^{p}_{\alpha }(\mathbb {B}_{n},X) \longrightarrow A^{q}_{\alpha }(\mathbb {B}_{n},Y)$ is a compact operator if and only if

$$\begin{aligned} b \in \Lambda _{\gamma _{0},0}(\mathbb {B}_{n},\mathcal {K}(\overline{X},Y)), \end{aligned}$$

where $\Lambda _{\gamma _{0},0}(\mathbb {B}_{n},\mathcal {K}(\overline{X},Y))$ denotes the generalized little vector-valued Lipschitz space and $\gamma _{0} = (n+1+\alpha )\left( \frac{1}{p} - \frac{1}{q}\right) ,$ see (1.3).

1.7 Plan of the Paper

The paper is divided into six sections. In Sect. 2, we recall some preliminary notions on vector-valued holomorphic functions and we also give the proofs of some important results. Sect. 3 contains the proof of Theorem 3 on the dual of the vector-valued Bergman space $A^p_\alpha (\mathbb {B}_{n},X)$ for $0 < p \le 1.$ In Sect. 4, we give the proof of Theorem 4 and Corollary 5. In Sect. 5, we give the proof of Theorem 6. In Sect. 6, We first give some preliminaries results to prepare the proof of Theorem 8. We recall the result by Oliver [11] of the boundedness of the little Hankel operator with operator-valued symbol $h_b$ from $A^p_\alpha (\mathbb {B}_{n},X)$ into $A^{q}_{\alpha }(\mathbb {B}_{n},Y),$ with $1< p \le q < \infty$ and we generalize it. In the same section, we give the proof of Theorem 8.

Throughout this paper, when there is no additional condition, X and Y will denotes two complex Banach spaces, the real parameter $\alpha$ will be chosen such that $\alpha >-1$ and c will be a positive constant whose value may change from one occurrence to the next. We will also adopt the following notation: we will write $A \lesssim B$ whenever there exists a positive constant c such that $A \le c B.$ We also write $A \simeq B$ when $A \lesssim B$ and $B \lesssim A.$

2 Preliminaries

2.1 Vector-Valued Bergman Projection and Integral Estimates

Here we give some definitions and notations which will be used later and can be found in [4, 11].

For $f \in L^{1}_{\alpha }(\mathbb {B}_{n},X)$ and $z \in \mathbb {B}_n,$ the Bergman projection $P_{\alpha }f$ of f is the integral operator defined by

$$\begin{aligned} \displaystyle P_{\alpha }f(z) := \int _{\mathbb {B}_n} K_{\alpha }(z,w) f(w)\mathrm {d}\nu _{\alpha }(w), \end{aligned}$$

where $K_{\alpha }(z,w) := \dfrac{1}{(1-\langle z,w \rangle )^{n+1+\alpha }}$ is the Bergman reproducing kernel of $\mathbb {B}_n.$ In this situation, $P_{\alpha }f$ is also a X-valued holomorphic function.

Lemma 9

(Density) Suppose that $0< p < \infty .$ Then the space of all bounded vector-valued holomorphic functions $H^{\infty }(\mathbb {B}_{n},X)$ is dense in $A^p_{\alpha }(\mathbb {B}_{n},X).$

Proof

We are going to give the proof for $0< p < 1,$ since the case $1 \le p < \infty$ is [11, Lemma 2.1.4]. Given a function $f \in A^p_{\alpha }(\mathbb {B}_{n},X),$ let $f_{\rho }$ defined for $z \in \mathbb {B}_n$ by $f_{\rho }(z): = f(\rho z),$ where $0< \rho < 1.$ The function $f_{\rho }$ is holomorphic in the set $\lbrace z\in \mathbb {B}_{n} : |z| < 1/\rho \rbrace$ hence is bounded on $\mathbb {B}_{n}.$ We first recall that the integral means

$$\begin{aligned} M_{p}(r,f):= \int _{\mathbb {S}_n} \Vert f(r \zeta ) \Vert ^p_{X} \mathrm {d}\sigma (\zeta ), \qquad 0 \le r < 1 \end{aligned}$$

are increasing with r,  see [14, Corollary 4.21]. Since $M_{p}(r,f_{\rho }) = M_{p}(\rho r,f),$ we have by Minkowski’s inequality that

$$\begin{aligned} M^p_{p}(r,f_{\rho }-f) \le M^p_{p}(r,f) + M^p_{p}(r,f_{\rho }) \le 2M^p_{p}(r,f). \end{aligned}$$

By the formula of [11, (1.1.1)], (integration in polar coordinates formula) we get

$$\begin{aligned} \Vert f - f_{\rho }\Vert ^p_{p,\alpha ,X} = 2nc_{\alpha }\int _{0}^{1}M^p_{p}(r,f_{\rho }-f)(1 - r^2)^{\alpha }r^{2n-1}\mathrm {d}r. \end{aligned}$$

(2.1)

Since $f \in A^p_{\alpha }(\mathbb {B}_{n},X),$ we have that the function $M^p_{p}(r,f)$ is integrable over the interval [0, 1) with respect to the measure $2n(1 - r^2)^{\alpha }r^{2n-1}\mathrm {d}r.$ It is also clear that $f_\rho \rightarrow f$ on any compact subsets of $\mathbb {B}_n$ which implies that $M^p_{p}(r,f_{\rho }-f) \rightarrow 0$ for each $r \in [0, 1)$ as $\rho \rightarrow 1.$ Applying the dominated convergence theorem in (2.1), we obtain that $\Vert f - f_{\rho }\Vert ^p_{p,\alpha ,X} \longrightarrow 0,$ as $\rho \rightarrow 1. \square$

Corollary 10

For $0 < p \le 1,$ the following inclusion is dense

$$\begin{aligned} A^2_\alpha (\mathbb {B}_{n},X) \subset A^p_\alpha (\mathbb {B}_{n},X). \end{aligned}$$

Proof

The proof follows directly from Lemma 9. $\square$

In [3], Oscar Blasco obtained the duality theorem for the vector-valued Bergman spaces in the unit disc $\mathbb {B}_{1}$ without any restriction on the Banach space. The proof also works for the unit ball $\mathbb {B}_n.$ The result is stated as follows:

Theorem 11

(Duality). Suppose $1< p < \infty .$ The dual space $(A^p_\alpha (\mathbb {B}_{n},X))^{\star }$ can be identified with $A^{p'}_\alpha (\mathbb {B}_{n},X^{\star }),$ where $p'$ is the conjugate exponent of p given by $\frac{1}{p}+\frac{1}{p'} =1,$ under the integral pairing defined by

$$\begin{aligned} \displaystyle \langle f,g \rangle _{\alpha ,X} := \int _{\mathbb {B}_n}\langle f(z),g(z) \rangle _{X,X^{\star }}\mathrm {d}\nu _{\alpha }(z), \end{aligned}$$

(2.2)

for any $f \in A^p_\alpha (\mathbb {B}_{n},X),$ $g \in A^{p'}_\alpha (\mathbb {B}_{n},X^{\star }).$

Remark 12

Suppose $1<p < \infty .$ If X is a reflexive complex Banach space, then the vector-valued Bergman space $A^p_\alpha (\mathbb {B}_{n},X)$ is a reflexive Banach space.

The following reproducing kernel formula also holds for vector-valued Bergman spaces. The proof can be found in [11, Proposition 2.1.2].

Proposition 13

Let $f \in A^1_\alpha (\mathbb {B}_{n},X).$ We have

$$\begin{aligned} \displaystyle f(z) := \int _{\mathbb {B}_n} \dfrac{f(w)}{(1-\langle z,w \rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(w), \end{aligned}$$

for any $z \in \mathbb {B}_n.$

We have the following pointwise estimate on the vector-valued Bergman spaces. The proof can be found in [11].

Theorem 14

Let $0< p < \infty .$ Then

$$\begin{aligned} \Vert f(z)\Vert _{X} \le \dfrac{\Vert f\Vert _{p,\alpha ,X}}{(1-|z|^2)^{(n+1+\alpha )/p}}, \end{aligned}$$

for any $f \in A^p_{\alpha }(\mathbb {B}_{n},X)$ and $z \in \mathbb {B}_{n}.$

The following lemma is critical for many problems concerning the weighted vector-valued Bergman spaces $A^p_{\alpha }(\mathbb {B}_{n},X)$ whenever $0 < p \le 1$ and will be extensively used.

Lemma 15

Let $0 < p \le 1.$ Then

$$\begin{aligned} \displaystyle \int _{\mathbb {B}_n}\Vert f(z)\Vert _{X}(1-|z|^2)^{(\frac{1}{p}- 1)(n+1+\alpha )}\mathrm {d}\nu _{\alpha }(z) \le \Vert f\Vert _{p,\alpha ,X}, \end{aligned}$$

for all $f \in A^p_{\alpha }(\mathbb {B}_{n},X).$

Proof

Write

$$\begin{aligned} \Vert f(z)\Vert _{X} = \Vert f(z)\Vert ^p_{X} \Vert f(z)\Vert ^{1-p}_{X}, \end{aligned}$$

and estimate the second factor using Theorem 14. The desired result follows. $\square$

The following technical result is proved in [4, Lemma 3.1]

Lemma 16

Let $\beta ,\delta > 0.$ For all $w \in \mathbb {B}_{n},$ we have

$$\begin{aligned} \displaystyle I_{\alpha }(w) := \int _{\mathbb {B}_n}\left| \log \left( \dfrac{1-\langle z,w \rangle }{1-|w|^2} \right) \right| ^{\delta }\dfrac{(1-|w|^2)^{\beta }}{|1-\langle z,w \rangle |^{n+1+\alpha +\beta }}\mathrm {d}\nu _{\alpha }(z) \le C, \end{aligned}$$

where C is independent of w and $\log$ is the principal branch of the logarithm.

In the sequel, we will also need the following lemma which the scalar version can be found in [8].

Lemma 17

If $0< q < 1,$ then the identity $i : L^{1,\infty }_{\alpha }(\mathbb {B}_{n},X) \hookrightarrow L^{q}_{\alpha }(\mathbb {B}_{n},X)$ is continuous in the sense that there exists a constant $C(q) >0$ such that for every $f\in L^{1,\infty }_{\alpha }(\mathbb {B}_{n},X),$ we have

$$\begin{aligned} \Vert f\Vert _{q,\alpha ,X} \le C(q) \Vert f\Vert _{L^{1,\infty }_{\alpha }(\mathbb {B}_{n},X)}. \end{aligned}$$

The following result will be very useful in many situations. A proof can be found in [14].

Theorem 18

For $\beta \in \mathbb {R},$ let

$$\begin{aligned} \displaystyle I_{\alpha ,\beta }(z) : = \int _{\mathbb {B}_n}\dfrac{(1-|w|^2)^{\alpha }\mathrm {d}\nu (w)}{|1- \langle z,w \rangle |^{n+1+\alpha +\beta }}, \quad z \in \mathbb {B}_{n}. \end{aligned}$$

• (i) If $\beta = 0,$ there exists a constant $C >0$ such that

  $$\begin{aligned} I_{\alpha ,\beta }(z) \le C \log \dfrac{1}{1- |z|^2},\quad z\in \mathbb {B}_n. \end{aligned}$$

• (ii) If $\beta > 0,$ there exists a constant $C >0$ such that

  $$\begin{aligned} I_{\alpha ,\beta }(z) \le C \dfrac{1}{(1- |z|^2)^\beta },\quad z\in \mathbb {B}_n. \end{aligned}$$

• (iii) If $\beta < 0,$ there exists a constant $C >0$ such that

  $$\begin{aligned} I_{\alpha ,\beta }(z) \le C. \end{aligned}$$

2.2 Differential Operators and Equivalent Norms for $\Gamma _\gamma$

Given a positive integer k,  we define the differential operator $D_k$ by

$$\begin{aligned} D_{k}:= (2I+N) \circ (3I+N) \circ \ldots \circ ((k+1)I+N), \end{aligned}$$

(2.3)

where I is the identity operator and N is the differential operator given in (1.1).

In the sequel, we denote by $\mathcal {P}(\mathbb {B}_{n},X)$ the space of all vector-valued holomorphic polynomials. The proof of the following lemma is similar as in the scalar case in [10].

Lemma 19

For all $f \in \mathcal {P}(\mathbb {B}_{n},X)$ and $g \in \mathcal {P}(\mathbb {B}_{n},X^{\star }),$ we have the following identity

$$\begin{aligned} \displaystyle \int _{\mathbb {B}_n} \langle f(z),g(z) \rangle _{X,X^{\star }} \mathrm {d}\nu _{\alpha }(z)&= \displaystyle c_{k}\int _{\mathbb {B}_n} \langle f(z),D_{k}g(z) \rangle _{X,X^{\star }} (1-|z|^2)^{k} \mathrm {d}\nu _{\alpha }(z), \end{aligned}$$

where $c_k$ is a positive constant depending only on the integer k. The above identities are valid for vector-valued holomorphic functions when both sides make sense.

The following lemma will be very useful in the sequel.

Lemma 20

Let $\lbrace a_k \rbrace$ a sequence of positive numbers. For any positive integer k,  let $M_{k}$ the differential operator of order k defined by

$$\begin{aligned} M_{k} := (a_{0}I+N) \circ (a_{1}I+N) \circ \ldots \circ (a_{k-1}I+N). \end{aligned}$$

Then a vector-valued holomorphic function f belongs to $\Gamma _{\gamma }(\mathbb {B}_{n},X)$ if and only if there exists an integer $k > \gamma$ such that

$$\begin{aligned} \sup _{z \in \mathbb {B}_{n}}(1-|z|^2)^{k-\gamma }\Vert M_{k}f(z)\Vert _{X} < \infty . \end{aligned}$$

Proof

Let us assume first that $f \in \Gamma _{\gamma }(\mathbb {B}_{n},X),$ and we prove the desired estimate on $M_k.$ By assumption, there exists an integer $k > \gamma$ and a positive constant C such that

$$\begin{aligned} \Vert N^{k}f(z)\Vert _{X} \le C (1 - |z|^2)^{\gamma -k}, \end{aligned}$$

for any $z \in \mathbb {B}_{n}.$ It is enough to prove that the following inequality

$$\begin{aligned} \Vert N^{j}f(z) \Vert _{X} < C (1 - |z|^2)^{\gamma -k} , \end{aligned}$$

holds for $0 \le j < k,$ since the assumption give the case $j = k.$ For $g \in \mathcal {H}(\mathbb {B}_{n},X)$ and $z = rz',$ where $r = |z|,$ and $z'$ is in the unit sphere. We have

$$\begin{aligned} N g(r z') = r \partial _{r} g(rz'). \end{aligned}$$

Thus,

$$\begin{aligned} g(rz') - g(z'/2) = \int _{\frac{1}{2}}^{r} Ng(sz')\frac{ds}{s}. \end{aligned}$$

Now, for $g \in \mathcal {H}(\mathbb {B}_{n},X)$ such that $\Vert Ng(z)\Vert _{X} \le C (1 - |z|^2)^{\gamma -k}.$ We have that

$$\begin{aligned} \Vert g(rz') - g(z'/2) \Vert _{X}&\le 2 \int _{\frac{1}{2}}^{r} \Vert Ng(sz')\Vert _{X} \mathrm {d}s\\&\le 4C \int _{\frac{1}{2}}^{r} (1 - s^2)^{\gamma -k}s \mathrm {d}s \\&= -2C\int _{\frac{1}{2}}^{r} -2s(1 - s^2)^{\gamma -k} \mathrm {d}s\\&= \left[ \dfrac{-2C}{\gamma -k+1} (1 - s^2)^{\gamma -k+1} \right] _{\frac{1}{2}}^{r}\\&= \dfrac{-2C}{\gamma -k+1} \left\{ (1- r^2)^{\gamma -k+1} - (1 - \frac{1}{4})^{\gamma -k+1} \right\} . \end{aligned}$$

Now, if $\gamma -k+1 < 0,$ then

$$\begin{aligned} \Vert g(rz') - g(z'/2) \Vert _{X} \le \dfrac{-2C}{\gamma -k+1} (1 - r^2)^{\gamma -k} = C_{k,\gamma }(1 - |z|^2)^{\gamma -k}. \end{aligned}$$

If $\gamma -k+1 > 0,$ then

$$\begin{aligned} \Vert g(rz') - g(z'/2) \Vert _{X}&\le \dfrac{2C}{\gamma -k+1} \left\{ (1 - \frac{1}{4})^{\gamma -k+1} - (1 - r^2)^{\gamma -k+1} \right\} \\&\le \dfrac{2C}{\gamma -k+1} (1 - \frac{1}{4})^{\gamma -k+1} = C'_{k,\gamma }\\&\le C'_{k,\gamma } (1 - |z|^2)^{\gamma -k}, \end{aligned}$$

where the last inequality is justified using the fact that $(1 - |z|^2)^{\gamma -k} > 1.$ It then follows that

$$\begin{aligned} \Vert g(z)\Vert _{X} \le C (1 - |z|^2)^{\gamma -k}. \end{aligned}$$

Now, we use this fact inductively for $g = N^{k} f,$ then $g = N^{k-1} f,~~\ldots$ to conclude. Conversely, assume that there exists an integer $k > \gamma$ and a positive constant C such that

$$\begin{aligned} \Vert M_{k} f(z) \Vert _{X} \le C (1 - |z|^2)^{\gamma -k}, \end{aligned}$$

for any $z \in \mathbb {B}_{n}.$ To conclude, it is sufficient to prove that for a fixed positive real a, the inequality

$$\begin{aligned} \Vert ag(z) + Ng(z) \Vert _{X} \le C (1 - |z|^2)^{\gamma -k} \end{aligned}$$

(2.4)

implies the inequality

$$\begin{aligned} \Vert Ng(z) \Vert _{X} \le C (1 - |z|^2)^{\gamma -k}, \end{aligned}$$

for any function $g \in \mathcal {H}(\mathbb {B}_{n},X).$ Choose a real $\beta$ such that $\beta +\gamma - k > -1.$ By the assumption (2.4), we have that

$$\begin{aligned} \displaystyle \int _{\mathbb {B}_n} \Vert ag(z) + Ng(z) \Vert _{X} (1 - |z|^2)^{\beta } \mathrm {d}\nu (z) < \infty . \end{aligned}$$

Thus, for any $z \in \mathbb {B}_{n},$ we have

$$\begin{aligned} \displaystyle ag(z) + Ng(z) = c_{\beta } \int _{\mathbb {B}_{n}}\dfrac{[ag(w) + Ng(w)]}{(1 - \langle z,w \rangle )^{n+1+\beta }} (1 - |w|^2)^{\beta } \mathrm {d}\nu (w). \end{aligned}$$

Then, differentiating under the integral sign, we obtain that for all $1 \le i \le n,$ we get

$$\begin{aligned}&\displaystyle \partial _{z_i} \left[ ag(z) + Ng(z) \right] \\&\quad = (n+1+\beta )c_{\beta } \int _{\mathbb {B}_{n}}\dfrac{[ag(w) + Ng(w)]\overline{w}_{i}}{(1 - \langle z,w \rangle )^{n+2+\beta }} (1 - |w|^2)^{\beta } \mathrm {d}\nu (w). \end{aligned}$$

Therefore,

$$\begin{aligned}&\displaystyle N \left( ag(z) + Ng(z) \right) \\&\quad = (n+1+\beta )c_{\beta } \int _{\mathbb {B}_{n}}\dfrac{[ag(w) + Ng(w)]\langle z,w \rangle }{(1 - \langle z,w \rangle )^{n+2+\beta }} (1 - |w|^2)^{\beta } \mathrm {d}\nu (w). \end{aligned}$$

Applying (2.4), and Theorem 18, we get that for all $1 \le i \le n,$

$$\begin{aligned} \displaystyle \Vert N \left( ag(z) + Ng(z) \right) \Vert _{X}&\le C c_{\beta }\int _{\mathbb {B}_{n}} \dfrac{(1 - |w|^2)^{\gamma -k+\beta }}{|1 - \langle z,w \rangle |^{n+1+\gamma -k+\beta +(k-\gamma +1)}}\mathrm {d}\nu (w)\\&\le C (1 - |z|^2)^{\gamma -k-1}. \end{aligned}$$

Thus, the derivative of $ag(z) + Ng(z)$ is bounded by $(1 -|z|^2)^{\gamma -k-1}.$ So, to prove the inequality above, we are reduced to consider smooth functions $\phi$ of one variable $r \in [0,1),$ and to prove that the inequality

$$\begin{aligned} \Vert \psi '(r) \Vert _{X} \le C(1-r)^{\gamma -k-1}, \end{aligned}$$

with $\psi (r) = a\phi (r) + r \phi '(r),$ implies that

$$\begin{aligned} \Vert r\phi '(r)\Vert _{X} \le C (1-r)^{\gamma -k} \end{aligned}$$

(here, $\phi (r) = g(rz')$). Now, differentiating $\psi ,$ we obtain $\psi '(s) = (a+1)\phi '(s) + s\phi ''(s).$ Multiplying both sides of the previous inequality by $s^{a},$ we obtain that $s^{a}\psi '(s) = (a+1)s^{a}\phi '(s) + s^{a+1}\phi ''(s) = \left[ s^{a+1}\phi '(s)\right] '.$ Then integrating the equality above on [0, r],  we obtain that

$$\begin{aligned} \phi '(r) = \dfrac{1}{r^{a+1}}\int _{0}^{r} s^{a}\psi '(s)\mathrm {d}s. \end{aligned}$$

Therefore, the desired estimate follows at once, since $k > \gamma . \square$

Remark 21

We shall use extensively this lemma for two particular classes of differential operators: first the class $D_{k},$ then the class $L_{k},$ corresponding to the choice $a_{j} = n+\alpha +j+1.$ For this choice, we have

$$\begin{aligned} (a_{j}I+N)(1-\langle z,w \rangle )^{-n-\alpha -j-1} = \dfrac{n+\alpha +j+1}{(1-\langle z,w \rangle )^{n+\alpha +j+2}}, \end{aligned}$$

and inductively,

$$\begin{aligned} L_{k}(1-\langle z,w \rangle )^{-n-\alpha -1} = \dfrac{c_{k}}{(1-\langle z,w \rangle )^{n+\alpha +k+1}}. \end{aligned}$$

The proof of Lemma 20 allows us to define an equivalent norm of f in terms of $M_{k}f.$ Particularly, we will write the equivalent norms of f in terms of $D_{k}f$ and $L_{k}f.$ More precisely, we have the following result:

Corollary 22

Let $D_{k}$ a differential operator of order k defined in (2.3) and $L_{k}$ a differential operator of order k defined in Remark 21. For vector-valued holomorphic functions, the following assertions are equivalent:

• (1) $f \in \Gamma _{\gamma }(\mathbb {B}_{n},X).$

• (2) There exists an integer $k > \gamma$ such that

  $$\begin{aligned} \sup _{z \in \mathbb {B}_{n}}(1-|z|^2)^{k-\gamma }\Vert D_{k}f(z)\Vert _{X} < \infty . \end{aligned}$$

• (3) There exists an integer $k > \gamma$ such that

  $$\begin{aligned} \sup _{z \in \mathbb {B}_{n}}(1-|z|^2)^{k-\gamma }\Vert L_{k}f(z)\Vert _{X} < \infty . \end{aligned}$$

  Moreover, the following are equivalent

  $$\begin{aligned} \Vert f\Vert _{ \Gamma _{\gamma }(\mathbb {B}_{n},X)}&\simeq \Vert f(0)\Vert _{X} + \sup _{z \in \mathbb {B}_{n}}(1-|z|^2)^{k-\gamma }\Vert D_{k}f(z)\Vert _{X} \\&\simeq \Vert f(0)\Vert _{X} + \sup _{z \in \mathbb {B}_{n}}(1-|z|^2)^{k-\gamma }\Vert L_{k}f(z)\Vert _{X}. \end{aligned}$$

The proof of some of the results obtained in this paper will be based on the following lemma. A proof is in [11], but for the sake of completeness, we will recall the proof.

Lemma 23

Let $f \in H^{\infty }(\mathbb {B}_{n},X)$ and $g \in H^{\infty }(\mathbb {B}_{n},Y^{\star }).$ If $b \in \mathcal {H}(\mathbb {B}_{n}, \mathcal {L}(\overline{X},Y))$ is such that (1.4) and (1.6) hold. Then we have

$$\begin{aligned} \displaystyle \langle h_{b}f,g \rangle _{\alpha ,Y} = \int _{\mathbb {B}_n} \langle b(z)\overline{f(z)}, g(z) \rangle _{Y,Y^{\star }} \mathrm {d}\nu _{\alpha }(z). \end{aligned}$$

(2.5)

Proof

Let $f\in H^{\infty }(\mathbb {B}_{n},X)$ and $g \in H^{\infty }(\mathbb {B}_{n},Y^{\star }).$ By the definition of $\langle \cdot ,\cdot \rangle _{\alpha ,Y},$ Fubini’s theorem, Lemma 1 and the reproducing kernel property, we have:

$$\begin{aligned} \langle h_{b}(f),g \rangle _{\alpha ,Y}&= \int _{\mathbb {B}_{n}} \langle h_{b}(f)(z),g(z) \rangle _{Y,Y^{\star }}\mathrm {d}\nu _{\alpha }(z)\\&= \int _{\mathbb {B}_{n}} \langle \int _{\mathbb {B}_{n}} \dfrac{b(w)(\overline{f(w)})\mathrm {d}\nu _{\alpha }(w)}{(1 - \langle z,w \rangle )^{n+1+\alpha }}, g(z) \rangle _{Y,Y^{\star }}\mathrm {d}\nu _{\alpha }(z)\\&= \int _{\mathbb {B}_{n}}g(z)\left( \int _{\mathbb {B}_{n}} \dfrac{b(w)(\overline{f(w)})\mathrm {d}\nu _{\alpha }(w)}{(1 - \langle z,w \rangle )^{n+1+\alpha }}\right) \mathrm {d}\nu _{\alpha }(z)\\&= \int _{\mathbb {B}_{n}}\int _{\mathbb {B}_{n}}g(z)\left( \dfrac{b(w)(\overline{f(w)})}{(1 - \langle z,w \rangle )^{n+1+\alpha }}\right) \mathrm {d}\nu _{\alpha }(w)\mathrm {d}\nu _{\alpha }(z)\\&= \int _{\mathbb {B}_{n}}\left( \int _{\mathbb {B}_{n}}\dfrac{g(z)}{(1 - \langle w,z \rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(z)\right) \left( b(w)(\overline{f(w)})\right) \mathrm {d}\nu _{\alpha }(w)\\&= \int _{\mathbb {B}_{n}} g(w)\left( b(w)(\overline{f(w)})\right) \mathrm {d}\nu _{\alpha }(w)\\&= \int _{\mathbb {B}_n} \langle b(w)\overline{f(w)}, g(w) \rangle _{Y,Y^{\star }} \mathrm {d}\nu _{\alpha }(w). \end{aligned}$$

It remains to show that the assumption of Fubini’s theorem is fulfilled. Indeed, since $f\in H^{\infty }(\mathbb {B}_{n},X)$ and $g\in H^{\infty }(\mathbb {B}_{n},Y^{\star }),$ by Tonelli’s theorem, Theorem 18 and relation (1.6) we have that

$$\begin{aligned}&\int _{\mathbb {B}_{n}} \int _{\mathbb {B}_{n}} \left| \dfrac{g(z)\left( b(w)(\overline{f(w)})\right) }{(1 - \langle z,w \rangle )^{n+1+\alpha }} \right| \mathrm {d}\nu _{\alpha }(w)\mathrm {d}\nu _{\alpha }(z) \\&\lesssim \displaystyle \int _{\mathbb {B}_{n}} \int _{\mathbb {B}_{n}} \dfrac{\Vert b(w)\Vert _{\mathcal {L}(\overline{X},Y)}}{|1 - \langle z,w \rangle |^{n+1+\alpha }} \mathrm {d}\nu _{\alpha }(w)\mathrm {d}\nu _{\alpha }(z)\\&\lesssim \displaystyle \int _{\mathbb {B}_{n}} \Vert b(w)\Vert _{\mathcal {L}(\overline{X},Y)} \log \left( \dfrac{1}{1-|w|^2} \right) \mathrm {d}\nu _{\alpha }(w) < \infty . \end{aligned}$$

$\square$

Lemma 24

Let $f \in H^{\infty }(\mathbb {B}_{n},X)$ and $z \in \mathbb {B}_n.$ For $b \in \mathcal {H}(\mathbb {B}_{n}, \mathcal {L}(\overline{X},Y))$ satisfying (1.4) and (1.6), the function

$$\begin{aligned} g_{z}(w):= \dfrac{f(w)}{(1 - \langle w,z \rangle )^{n+1+\alpha }}, \qquad w \in \mathbb {B}_n \end{aligned}$$

belongs to $H^{\infty }(\mathbb {B}_{n},X)$ and the following identity holds:

$$\begin{aligned} \displaystyle h_{b}(f)(z) = C_{k}\int _{\mathbb {B}_n} L_{k}\left( b(w)(\overline{g_{z}(w)})\right) \mathrm {d}\nu _{\alpha +k}(w), \end{aligned}$$

where k is any positive integer and $C_{k}$ is a positive constant depending only on k.

Proof

It is clear that $g_{z} \in H^{\infty }(\mathbb {B}_{n},X).$ By the definition of the little Hankel operator and the reproducing kernel property, we have

$$\begin{aligned} h_b(f)(z)&= \int _{\mathbb {B}_n} \dfrac{b(w)\overline{f(w)}}{(1 - \langle z,w \rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(w)\\&= \displaystyle \int _{\mathbb {B}_n} b(w)\left( \overline{ \dfrac{f(w)}{(1 - \langle w,z \rangle )^{n+1+\alpha }}}\right) \mathrm {d}\nu _{\alpha }(w)\\&= \int _{\mathbb {B}_n} b(w)(\overline{g_{z}(w)})\mathrm {d}\nu _{\alpha }(w)\\&= \int _{\mathbb {B}_n} b(w)\left( \overline{\int _{\mathbb {B}_n} \dfrac{g_{z}(\zeta )}{(1 - \langle w,\zeta \rangle )^{n+1+\alpha +k}}\mathrm {d}\nu _{\alpha +k}(\zeta )} \right) \mathrm {d}\nu _{\alpha }(w)\\&= \int _{\mathbb {B}_n} \left( \int _{\mathbb {B}_n} \dfrac{b(w)(\overline{g_{z}(\zeta )})}{(1 - \langle \zeta ,w \rangle )^{n+1+\alpha +k}} \mathrm {d}\nu _{\alpha }(w) \right) \mathrm {d}\nu _{\alpha +k}(\zeta )\\&= c_{k}^{-1} \int _{\mathbb {B}_n}L_{k}\left( \int _{\mathbb {B}_n} \dfrac{b(w)(\overline{g_{z}(\zeta )})}{(1 - \langle \zeta ,w \rangle )^{n+1+\alpha }} \mathrm {d}\nu _{\alpha }(w) \right) \mathrm {d}\nu _{\alpha +k}(\zeta )\\&= c_{k}^{-1} \int _{\mathbb {B}_n}L_{k}\left( b(\zeta )(\overline{g_{z}(\zeta )})\right) \mathrm {d}\nu _{\alpha +k}(\zeta ). \end{aligned}$$

The assumption of Fubini’s theorem is fulfilled. Indeed by (1.6), we have that

$$\begin{aligned}&\int _{\mathbb {B}_n} \left\| \int _{\mathbb {B}_n} \dfrac{b(w)(\overline{g_{z}(\zeta )})}{(1 - \langle \zeta ,w \rangle )^{n+1+\alpha +k}}\mathrm {d}\nu _{\alpha }(w)\right\| _{Y}\mathrm {d}\nu _{\alpha +k}(\zeta )\\&\le \Vert g_{z}\Vert _{\infty ,X} \int _{\mathbb {B}_n} \int _{\mathbb {B}_n} \dfrac{\Vert b(w)\Vert _{\mathcal {L}(\overline{X},Y)}}{|1 - \langle w,\zeta \rangle |^{n+1+\alpha +k}} \mathrm {d}\nu _{\alpha }(w)\mathrm {d}\nu _{\alpha +k}(\zeta )\\&= \Vert g_{z}\Vert _{\infty ,X} \int _{\mathbb {B}_n} \Vert b(w)\Vert _{\mathcal {L}(\overline{X},Y)} \times \left( \int _{\mathbb {B}_n} \dfrac{\mathrm {d}\nu _{\alpha +k}(\zeta )}{|1 - \langle w,\zeta \rangle |^{n+1+\alpha +k}}\right) \mathrm {d}\nu _{\alpha }(w)\\&\le \Vert g_{z}\Vert _{\infty ,X}\int _{\mathbb {B}_n} \Vert b(w)\Vert _{\mathcal {L}(\overline{X},Y)}\left( \log \dfrac{1}{1 - |w|^2} \right) \mathrm {d}\nu _{\alpha }(w) < \infty . \end{aligned}$$

$\square$

3 The Proof of Theorem 3

Proof

We first suppose that $g \in \Gamma _{\gamma }(\mathbb {B}_{n},X^{\star }),$ with $\gamma = (n+1+\alpha )\left( \frac{1}{p}-1\right) .$ Given a positive integer $k > \gamma ,$ we define the functional

$$\begin{aligned}&\displaystyle \wedge _{g}: A^p_{\alpha }(\mathbb {B}_{n},X) \longrightarrow \mathbb {C} \\&f \mapsto \wedge _{g}(f) = c_{k}\int _{\mathbb {B}_n} \langle f(z),D_{k}g(z) \rangle _{X,X^{\star }}(1 - |z|^2)^{k}\mathrm {d}\nu _{\alpha }(z), \end{aligned}$$

where $c_k$ is the positive constant in Lemma 19. It is clear that $\wedge _{g}$ is linear and is well defined on $A^p_{\alpha }(\mathbb {B}_{n},X).$ Indeed, let $f \in A^p_{\alpha }(\mathbb {B}_{n},X).$ By Lemma 15, we have

$$\begin{aligned} |\wedge _{g}(f)|&= \displaystyle c_{k}\left| \int _{\mathbb {B}_n} \langle f(z),D_{k}g(z) \rangle _{X,X^{\star }}(1-|z|^2)^k\mathrm {d}\nu _{\alpha }(z) \right| \\&\le \displaystyle c_{k} \int _{\mathbb {B}_n} \Vert f(z)\Vert _{X}\Vert D_{k}g(z)\Vert _{X^{\star }}(1-|z|^2)^k \mathrm {d}\nu _{\alpha }(z)\\&= \displaystyle c_{k} \int _{\mathbb {B}_n} (1-|z|^2)^{k-\gamma }\Vert D_{k}g(z)\Vert _{X^{\star }}(1-|z|^2)^{\gamma }\Vert f(z)\Vert _{X} \mathrm {d}\nu _{\alpha }(z)\\&\le \displaystyle c_{k} \sup _{z \in \mathbb {B}_{n}}(1-|z|^2)^{k-\gamma } \Vert D_{k}g(z)\Vert _{X^{\star }}\int _{\mathbb {B}_n} (1-|z|^2)^{\gamma } \Vert f(z)\Vert _{X}\mathrm {d}\nu _{\alpha }(z)\\&\lesssim \displaystyle \Vert g\Vert _{\Gamma _{\gamma }(\mathbb {B}_{n},X^{\star })}\int _{\mathbb {B}_n} (1-|z|^2)^{(\frac{1}{p}-1)(n+1+\alpha )} \Vert f(z)\Vert _{X}\mathrm {d}\nu _{\alpha }(z)\\&\lesssim \Vert g\Vert _{\Gamma _{\gamma }(\mathbb {B}_{n},X^{\star })}\Vert f\Vert _{p,\alpha ,X}. \end{aligned}$$

We conclude that $\wedge _{g}$ is bounded on $A^p_{\alpha }(\mathbb {B}_{n},X)$ and $\Vert \wedge _{g}\Vert \lesssim \Vert g\Vert _{\Gamma _{\gamma }(\mathbb {B}_{n},X^{\star })}.$

Conversely, let $\wedge$ be a bounded linear functional on $A^p_{\alpha }(\mathbb {B}_{n},X).$ Let us show that there exists $g \in \Gamma _{\gamma }(\mathbb {B}_{n},X^{\star }),$ with $\gamma = (n+1+\alpha )\left( \frac{1}{p}-1\right)$ such that $\wedge = \wedge _{g}.$ Since $A^2_{\alpha }(\mathbb {B}_{n},X) \subset A^p_{\alpha }(\mathbb {B}_{n},X)$ and $\wedge$ is bounded on $A^p_{\alpha }(\mathbb {B}_{n},X),$ $\wedge$ is also bounded on $A^2_{\alpha }(\mathbb {B}_{n},X).$ Then by Theorem 11, there exists $g \in A^2_{\alpha }(\mathbb {B}_{n},X^{\star })$ such that

$$\begin{aligned} \displaystyle \wedge (f) = \int _{\mathbb {B}_n} \langle f(z),g(z) \rangle _{X,X^{\star }} \mathrm {d}\nu _{\alpha }(z), \end{aligned}$$

(3.1)

for all $f \in A^2_{\alpha }(\mathbb {B}_{n},X).$ Since $g \in A^2_{\alpha }(\mathbb {B}_{n},X^{\star }),$ for any positive integer k,  we have $D_{k}g \in A^2_{\alpha +k}(\mathbb {B}_{n},X^{\star }).$ Applying Lemma 19 in (3.1), we obtain that

$$\begin{aligned} \displaystyle \wedge (f) = c_{k}\int _{\mathbb {B}_n} \langle f(z),D_{k}g(z) \rangle _{X,X^{\star }}(1 - |z|^2)^k \mathrm {d}\nu _{\alpha }(z), \end{aligned}$$

(3.2)

for all $f \in A^2_{\alpha }(\mathbb {B}_{n},X).$ Now, we fix $x \in X,$ $w \in \mathbb {B}_n$ and an integer $k > \gamma .$ Let

$$\begin{aligned} f(z) = \dfrac{(1-|w|^2)^{k-\gamma }}{(1-\langle z,w \rangle )^{n+1+\alpha +k}}x, \quad z\in \mathbb {B}_{n}. \end{aligned}$$

By Theorem 18, we have that $f \in A^2_{\alpha }(\mathbb {B}_{n},X).$ Proposition 13 and (3.2), give us

$$\begin{aligned} \wedge (f)&= \displaystyle c_{k}\int _{\mathbb {B}_n} \langle f(z),D_{k}g(z) \rangle _{X,X^{\star }}(1 - |z|^2)^k\mathrm {d}\nu _{\alpha }(z)\\&= c_{k}\int _{\mathbb {B}_n} \big \langle \dfrac{(1-|w|^2)^{k-\gamma }}{(1-\langle z,w \rangle )^{n+1+\alpha +k}}x, D_{k}g(z) \big \rangle _{X,X^{\star }}(1 - |z|^2)^k\mathrm {d}\nu _{\alpha }(z)\\&= \dfrac{c_{\alpha }c_{k}}{c_{\alpha +k}}(1-|w|^2)^{k-\gamma } \big \langle x, \int _{\mathbb {B}_n}\dfrac{D_{k}g(z)}{(1-\langle w,z \rangle )^{n+1+\alpha +k}}\mathrm {d}\nu _{\alpha +k}(z)\big \rangle _{X,X^{\star }}\\&= \dfrac{c_{\alpha }c_{k}}{c_{\alpha +k}}(1-|w|^2)^{k-\gamma } \big \langle x,D_{k}g(w)\big \rangle _{X,X^{\star }}. \end{aligned}$$

By Theorem 18, $f \in A^p_{\alpha }(\mathbb {B}_{n},X)$ and $\Vert f\Vert _{p,\alpha ,X} \lesssim \Vert x\Vert _{X}.$ Since x is arbitrary, by duality, we have that

$$\begin{aligned} \Vert D_{k}g(w)\Vert _{X^{\star }}&= \sup _{\Vert x\Vert _{X} = 1} |\langle x,D_{k}g(w) \rangle _{X,X^{\star }}|\\&= \dfrac{c_{\alpha +k}}{c_{\alpha }c_{k}}\sup _{\Vert x\Vert _{X} = 1} \dfrac{1}{(1-|w|^2)^{k-\gamma }}|\wedge (f)| \\&\lesssim \displaystyle \sup _{\Vert x\Vert _{X} = 1} \dfrac{1}{(1-|w|^2)^{k-\gamma }} \Vert \wedge \Vert \Vert f\Vert _{p,\alpha ,X}\\&\lesssim \sup _{\Vert x\Vert _{X} = 1} \dfrac{\Vert \wedge \Vert }{(1-|w|^2)^{k-\gamma }}\Vert x\Vert _{X}\\&\lesssim \dfrac{\Vert \wedge \Vert }{(1-|w|^2)^{k-\gamma }}. \end{aligned}$$

According to Corollary 22, we conclude that

$$\begin{aligned} g \in \Gamma _{\gamma }(\mathbb {B}_{n},X^{\star }) ~~~~\text{ and }~~~~ \Vert g\Vert _{\Gamma _{\gamma }(\mathbb {B}_{n},X^{\star })} \lesssim \Vert \wedge \Vert , \end{aligned}$$

with $\gamma = (n+1+\alpha )\left( \frac{1}{p}-1\right) .$ To finish the proof, it remains to show that (3.1) remains true for functions in $A^p_{\alpha }(\mathbb {B}_{n},X)$ which is a direct consequence of the density in Corollary 10. $\square$

4 The Proofs of Theorem 4 and Corollary 5

In this section, we will give the proofs of Theorem 4 and Corollary 5.

4.1 Proof of Theorem 4

Proof

First assume that $h_{b}$ extends to a bounded operator from $A^p_{\alpha }(\mathbb {B}_{n},X)$ to $A^q_{\alpha }(\mathbb {B}_{n},Y),$ with $q < 1.$ Let $\Vert h_{b}\Vert := \Vert h_{b}\Vert _{A^{p}_{\alpha }(\mathbb {B}_{n},X) \longrightarrow A^{q}_{\alpha }(\mathbb {B}_{n},Y)}.$ We want to show that $b \in \Gamma _{\gamma }(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y)).$ Since $h_{b}: A^p_{\alpha }(\mathbb {B}_{n},X) \longrightarrow A^{q}_{\alpha }(\mathbb {B}_{n},Y)$ is a bounded operator, we have by Theorem 3 that

$$\begin{aligned} |\langle h_{b}(f),g \rangle _{\alpha ,Y}| \lesssim \Vert h_{b}\Vert \Vert f\Vert _{p,\alpha ,X} \Vert g\Vert _{\Gamma _{\beta }(\mathbb {B}_{n},Y^{\star })}, \end{aligned}$$

for every $f \in A^p_{\alpha }(\mathbb {B}_{n},X)$ and $g \in \Gamma _{\beta }(\mathbb {B}_{n},Y^{\star }),$ with $\beta = (n+1+\alpha )\left( \frac{1}{q} -1\right) .$ Let $x \in X,$ $y^{\star } \in Y^{\star },$ $w \in \mathbb {B}_{n}$ and an integer k such that $k >\gamma = (n+1+\alpha )\left( \frac{1}{p}-1\right) .$ Let $g(z) = y^{\star },$ and $f(z) = \dfrac{(1-|w|^2)^{k-\gamma }}{(1-\langle z,w \rangle )^{n+1+\alpha + k}}x.$ It is clear that $f\in H^{\infty }(\mathbb {B}_{n},X)$ and $g \in \Gamma _{\beta }(\mathbb {B}_{n},Y^{\star }),$ with $\Vert g\Vert _{\Gamma _{\beta }(\mathbb {B}_{n},Y^{\star })} = \Vert y^{\star }\Vert _{Y^{\star }}.$ We also have by Theorem 18 that $f \in A^p_{\alpha }(\mathbb {B}_{n},X),$ with $\Vert f\Vert _{p,\alpha ,X} \lesssim \Vert x\Vert _{X}.$ Hence

$$\begin{aligned} |\langle h_{b}(f),g \rangle _{\alpha ,Y}| \lesssim \Vert h_{b}\Vert \Vert x\Vert _{X} \Vert y{\star }\Vert _{Y^{\star }}, \end{aligned}$$

(4.1)

Applying Lemma 23 and the reproducing kernel property, we have that

$$\begin{aligned}&|\langle h_{b}(f),g \rangle _{\alpha ,Y}|\\&= \displaystyle \left| \int _{\mathbb {B}_n} \big \langle b(z)\left( \overline{\dfrac{(1-|w|^2)^{k-\gamma }}{(1-\langle z,w \rangle )^{n+1+\alpha +k}}x} \right) , y^{\star } \big \rangle _{Y,Y^{\star }} \mathrm {d}\nu _{\alpha }(z) \right| \\&= \displaystyle (1-|w|^2)^{k-\gamma }\left| \int _{\mathbb {B}_n} \big \langle b(z)\left( \dfrac{\overline{x}}{(1-\langle w,z \rangle )^{n+1+\alpha +k}} \right) , y^{\star } \big \rangle _{Y,Y^{\star }} \mathrm {d}\nu _{\alpha }(z) \right| \\&= \displaystyle (1-|w|^2)^{k-\gamma }\left| \int _{\mathbb {B}_n} \big \langle \dfrac{b(z)\left( \overline{x}\right) }{(1-\langle w,z \rangle )^{n+1+\alpha +k}} , y^{\star } \big \rangle _{Y,Y^{\star }} \mathrm {d}\nu _{\alpha }(z) \right| \\&= \displaystyle (1-|w|^2)^{k-\gamma }\left| \big \langle \int _{\mathbb {B}_n}\dfrac{b(z)\left( \overline{x}\right) }{(1-\langle w,z \rangle )^{n+1+\alpha +k}}\mathrm {d}\nu _{\alpha }(z) , y^{\star } \big \rangle _{Y,Y^{\star }} \right| \\&= \displaystyle \frac{(1-|w|^2)^{k-\gamma }}{c_{k}}\left| \big \langle \int _{\mathbb {B}_n}L_{k}\left( \dfrac{b(z)\left( \overline{x}\right) }{(1-\langle w,z \rangle )^{n+1+\alpha }}\right) \mathrm {d}\nu _{\alpha }(z) , y^{\star } \big \rangle _{Y,Y^{\star }} \right| \\&= \displaystyle \frac{(1-|w|^2)^{k-\gamma }}{c_{k}}\left| \big \langle L_{k}\left( \int _{\mathbb {B}_n} \dfrac{b(z)\left( \overline{x}\right) }{(1-\langle w,z \rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(z)\right) , y^{\star } \big \rangle _{Y,Y^{\star }} \right| \\&= \displaystyle \frac{(1-|w|^2)^{k-\gamma }}{c_{k}} \left| \big \langle L_{k}\left( b(w)\left( \overline{x}\right) \right) , y^{\star } \big \rangle _{Y,Y^{\star }} \right| . \end{aligned}$$

Thus,

$$\begin{aligned} |\langle h_{b}(f),g \rangle _{\alpha ,Y}| = \displaystyle \frac{(1-|w|^2)^{k-\gamma }}{c_{k}} \left| \big \langle L_{k}\left( b(w)\left( \overline{x}\right) \right) , y^{\star } \big \rangle _{Y,Y^{\star }} \right| . \end{aligned}$$

(4.2)

From (4.1), (4.2) and the fact that $\Vert x\Vert _{X} = \Vert \overline{x}\Vert _{\overline{X}},$ we deduce that

$$\begin{aligned} (1-|w|^2)^{k-\gamma } \left| \big \langle L_{k}\left( b(w)\left( \overline{x}\right) \right) , y^{\star } \big \rangle _{Y,Y^{\star }} \right| \lesssim \Vert h_{b}\Vert \Vert \overline{x}\Vert _{\overline{X}} \Vert y^{\star }\Vert _{Y^{\star }}. \end{aligned}$$

(4.3)

Since x and $y^{\star }$ are arbitrary, we get that

$$\begin{aligned} \sup _{w \in \mathbb {B}_n}(1-|w|^2)^{k-\gamma }\Vert L_{k}b(w)\Vert _{\mathcal {L}(\overline{X},Y^{\star })} \lesssim \Vert h_{b}\Vert . \end{aligned}$$

That is, $b \in \Gamma _{\gamma }(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y^{\star }))$   with   $\Vert b\Vert _{\Gamma _{\gamma }(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y))} \lesssim \Vert h_{b}\Vert .$

Conversely, assume that $b \in \Gamma _{\gamma }(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y))$ and let us prove that $h_{b}$ extends to a bounded operator from $A^p_{\alpha }(\mathbb {B}_{n},X)$ to $A^{1,\infty }_{\alpha }(\mathbb {B}_{n},Y).$ Choose a positive integer $k>\gamma ,$ and let $f \in H^{\infty }(\mathbb {B}_{n},X).$ Taking

$$\begin{aligned} g_{z}(w) = \dfrac{f(w)}{(1 - \langle w,z \rangle )^{n+1+\alpha }}, \end{aligned}$$

with $w\in \mathbb {B}_{n}$ and applying Lemma 24, Lemma 2 and the assumption we obtain

$$\begin{aligned} \Vert h_{b}f(z)\Vert _{Y}&= \displaystyle \left\| \int _{\mathbb {B}_n} \dfrac{b(w)(\overline{f(w)})}{(1-\langle z,w \rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(w) \right\| _{Y} \\&= \displaystyle c_{k}\left\| \int _{\mathbb {B}_n} L_{k}\left( b(w)\overline{g_{z}(w)}\right) \mathrm {d}\nu _{\alpha +k}(w) \right\| _{Y} \\&= \displaystyle c_{k}\left\| \int _{\mathbb {B}_n} \dfrac{L_{k}\left( b(w)\overline{f(w)}\right) }{(1-\langle z,w \rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha +k}(w) \right\| _{Y} \\&\le \displaystyle c_{k} \int _{\mathbb {B}_n} \left\| \dfrac{L_{k}\left( b(w)\overline{f(w)}\right) }{(1-\langle z,w \rangle )^{n+1+\alpha }} \right\| _{Y} \mathrm {d}\nu _{\alpha +k}(w)\\&\le \dfrac{c_{k}c_{\alpha +k}}{c_{\alpha }}\int _{\mathbb {B}_n}\dfrac{(1-|w|^2)^{k}\Vert L_{k}b(w)\Vert _{\mathcal {L(\overline{X}, Y)}}\Vert \overline{f(w)}\Vert _{\overline{X}}}{|1 - \langle z,w \rangle |^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(w) \\&\lesssim \Vert b\Vert _{\Gamma _{\gamma }(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y))}\int _{\mathbb {B}_n} \dfrac{(1-|w|^2)^{\gamma }\Vert f(w)\Vert _{X}}{|1 - \langle z,w \rangle |^{n+1+\alpha }} \mathrm {d}\nu _{\alpha }(w)\\&= \Vert b\Vert _{\Gamma _{\gamma }(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y))}P^{+}_{\alpha } g(z), \end{aligned}$$

where the reproducing kernel is justified by (1.4) and

$$\begin{aligned} \displaystyle P^{+}_{\alpha } g(z) = \int _{\mathbb {B}_n} \dfrac{g(w)}{|1 - \langle z,w \rangle |^{n+1+\alpha }} \mathrm {d}\nu _{\alpha }(w) \end{aligned}$$

is the positive Bergman operator of the positive function $\displaystyle g(z) = (1-|z|^2)^{\gamma }\Vert f(z)\Vert _{X}.$

Now, let $\lambda >0.$ We have that

$$\begin{aligned} \nu _{\alpha } (\lbrace z \in \mathbb {B}_n : \Vert h_{b}f(z)\Vert _{Y}> \lambda \rbrace ) \le \nu _{\alpha }(\lbrace z \in \mathbb {B}_n : c_{k}\Vert b\Vert _{\Gamma _{\gamma }(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y))}P^{+}_{\alpha }g(z) > \lambda \rbrace ). \end{aligned}$$

Since the positive Bergman operator $P^{+}_{\alpha } : L^{1}_{\alpha }(\mathbb {B}_n) \longrightarrow L^{1,\infty }_{\alpha }(\mathbb {B}_n)$ is bounded (cf. e.g [2]), there exists a constant c such that

$$\begin{aligned} \nu _{\alpha }(\lbrace z \in \mathbb {B}_n : c_{k}\Vert b\Vert _{\Gamma _{\gamma }(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y)}P^{+}_{\alpha }g(z) > \lambda \rbrace )&\le \displaystyle \dfrac{c}{\frac{\lambda }{c_{k}\Vert b\Vert }_{\Gamma _{\gamma }(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y))}}\Vert g\Vert _{L^{1}_{\alpha }(\mathbb {B}_n)}\\&= \displaystyle \dfrac{cc_{k}}{\lambda }\Vert b\Vert _{\Gamma _{\gamma }(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y))}\Vert g\Vert _{L^{1}_{\alpha }(\mathbb {B}_n)}. \end{aligned}$$

Applying Lemma 15 to the function f,  we get that

$$\begin{aligned} \Vert g\Vert _{L^{1}_{\alpha }(\mathbb {B}_n)}&= \displaystyle \int _{\mathbb {B}_n}(1-|z|^2)^{\gamma }\Vert f(z)\Vert _{X} \mathrm {d}\nu _{\alpha }(z)\\&= \displaystyle \int _{\mathbb {B}_n}(1-|z|^2)^{(\frac{1}{p}-1)(n+1+\alpha )} \Vert f(z)\Vert _{X} \mathrm {d}\nu _{\alpha }(z)\\&\le \Vert f\Vert _{p,\alpha ,X}. \end{aligned}$$

It follows that

$$\begin{aligned} \lambda \nu _{\alpha } (\lbrace z \in \mathbb {B}_n : \Vert h_{b}f(z)\Vert _{Y} > \lambda \rbrace ) \lesssim \Vert b\Vert _{\Gamma _{\gamma }(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y))} \Vert f\Vert _{p,\alpha ,X} \end{aligned}$$

for all $\lambda > 0.$ Therefore, $h_{b}$ extends into a bounded operator from $A^p_{\alpha }(\mathbb {B}_{n},X)$ to $A^{1,\infty }_{\alpha }(\mathbb {B}_{n},Y)$ with

$$\begin{aligned} \Vert h_{b}\Vert _{A^p_\alpha (\mathbb {B}_{n},X) \longrightarrow A^{1,\infty }_{\alpha }(\mathbb {B}_{n},Y)} \lesssim \Vert b\Vert _{\Gamma _{\gamma }(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y))}. \end{aligned}$$

By density of $H^{\infty }(\mathbb {B}_{n},X)$ on $A^p_\alpha (\mathbb {B}_{n},X),$ the proof of the theorem is finished.

$\square$

4.2 Proof of Corollary 5

Proof

Just apply Lemma 17 and the second part of Theorem 4 to conclude.

$\square$

5 The Proof of Theorem 6

This section is devoted to the proof of Theorem 6.

Proof

We first prove the sufficiency of the theorem. We assume that there exists a constant $C' > 0$ such that

$$\begin{aligned} \Vert N^{k}b(w)\Vert _{\mathcal {L}(\overline{X},Y)} \le \dfrac{C'}{(1-|w|^2)^{k-\gamma }}\left( \log \dfrac{1}{1-|w|^2} \right) ^{-1}. \end{aligned}$$

Likewise by Corollary 22, we have that, there exists a constant $C >0$ such that

$$\begin{aligned} \Vert L_{k}b(w)\Vert _{\mathcal {L}(\overline{X},Y)} \le \dfrac{C}{(1-|w|^2)^{k-\gamma }}\left( \log \dfrac{1}{1-|w|^2} \right) ^{-1}. \end{aligned}$$

Applying Lemma 24 for any $f \in H^{\infty }(\mathbb {B}_{n},X),$ we get

$$\begin{aligned} \int _{\mathbb {B}_n} \dfrac{ b(w)(\overline{f(w)})}{(1 - \langle z,w \rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(w) = c_{k}\int _{\mathbb {B}_n} \dfrac{L_{k}b(w)(\overline{f(w)})}{(1 - \langle z,w \rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha +k}(w). \end{aligned}$$

Thus, by the assumption, Lemma 24 and Lemma 15 we have that

$$\begin{aligned}&\Vert h_{b}f\Vert _{A^{1}_{\alpha }(\mathbb {B}_{n},Y)} \\&= \int _{\mathbb {B}_n} \left\| c_{k}\int _{\mathbb {B}_n}\frac{ L_{k}b(w)(\overline{f(w)})}{(1- \langle z,w \rangle )^{n+1+\alpha }} \mathrm {d}\nu _{\alpha +k}(w) \right\| _{Y} \mathrm {d}\nu _{\alpha }(z)\\&\lesssim \int _{\mathbb {B}_n} \int _{\mathbb {B}_n} \left\| \dfrac{ L_{k}b(w)(\overline{f(w)})}{(1- \langle z,w \rangle )^{n+1+\alpha }} \right\| _{Y}(1-|w|^2)^{k} \mathrm {d}\nu _{\alpha }(w) \mathrm {d}\nu _{\alpha }(z)\\&\lesssim \int _{\mathbb {B}_n} \int _{\mathbb {B}_n} \dfrac{\Vert L_{k}b(w) \Vert _{\mathcal {L}(\overline{X},Y)} }{|1 - \langle z,w \rangle |^{n+1+\alpha }}\Vert \overline{f(w)}\Vert _{\overline{X}}(1-|w|^2)^{k} \mathrm {d}\nu _{\alpha }(w) \mathrm {d}\nu _{\alpha }(z)\\&= \int _{\mathbb {B}_n} \left( \int _{\mathbb {B}_n} \dfrac{1}{|1 - \langle z,w \rangle |^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(z)\right) \\&\Vert L_{k}b(w) \Vert _{\mathcal {L}(\overline{X},Y)} \Vert \overline{f(w)}\Vert _{\overline{X}}(1-|w|^2)^{k} \mathrm {d}\nu _{\alpha }(w)\\&\lesssim \displaystyle \int _{\mathbb {B}_n} \left( \log \dfrac{1}{1-|w|^2}\right) \Vert \overline{f(w)}\Vert _{\overline{X}} \dfrac{(1-|w|^2)^{k}}{(1-|w|^2)^{k-\gamma }}\left( \log \dfrac{1}{1-|w|^2} \right) ^{-1}\mathrm {d}\nu _{\alpha }(w)\\&= \int _{\mathbb {B}_n} \Vert f(w)\Vert _{X}(1-|w|^2)^{\gamma }\mathrm {d}\nu _{\alpha }(w)\\&= \int _{\mathbb {B}_n} \Vert f(w)\Vert _{X}(1-|w|^2)^{(\frac{1}{p}-1)(n+1+\alpha )}\mathrm {d}\nu _{\alpha }(w)\\&\lesssim \Vert f\Vert _{p,\alpha ,X}. \end{aligned}$$

Conversely, we assume that $h_{b}$ extends into a bounded operator from $A^p_{\alpha }(\mathbb {B}_{n},X)$ to $A^{1}_{\alpha }(\mathbb {B}_{n},Y).$ Then for all $f \in H^{\infty }(\mathbb {B}_{n},X)$ and $g \in \mathcal {B}(\mathbb {B}_{n},Y^{\star }),$ we have

$$\begin{aligned} |\langle h_{b}(f), g \rangle _{\alpha ,Y}| \le \Vert h_b\Vert \Vert f\Vert _{p,\alpha ,X}\Vert g\Vert _{\mathcal {B}(\mathbb {B}_{n},Y^{\star })}. \end{aligned}$$

(5.1)

We choose the particular function $g(z) = y^{\star },$ with $y^{\star } \in Y^{\star }.$ Applying Lemma 23, relation (5.1) becomes

$$\begin{aligned} \left| \displaystyle \int _{\mathbb {B}_n} \big \langle h_{b}f(z),y^{\star } \big \rangle _{Y,Y^{\star }}\mathrm {d}\nu _{\alpha }(z)\right|&= \displaystyle \left| \big \langle \int _{\mathbb {B}_n} b(z)\overline{f(z)}\mathrm {d}\nu _{\alpha }(z),y^{\star } \big \rangle _{Y,Y^{\star }} \right| \\&\le \Vert h_b\Vert \Vert f\Vert _{p,\alpha ,X}\Vert y^{\star }\Vert _{Y^{\star }}. \end{aligned}$$

Thus

$$\begin{aligned} \left| \displaystyle \int _{\mathbb {B}_n} \big \langle b(z)\overline{f(z)},y^{\star } \big \rangle _{Y,Y^{\star }}\mathrm {d}\nu _{\alpha }(z)\right| \le \Vert h_b\Vert \Vert f\Vert _{p,\alpha ,X}\Vert y^{\star }\Vert _{Y^{\star }} \end{aligned}$$

(5.2)

for all $f \in H^{\infty }(\mathbb {B}_{n},X)$ and $y^{\star } \in Y^{\star }.$ Now, take $x \in X,$ $y^{\star } \in Y^{\star },$ and an integer k such that $k > \gamma .$ Fix $w \in \mathbb {B}_{n}$ and put

$$\begin{aligned} f(z) = \dfrac{(1-|w|^2)^{k-\gamma }}{(1-\langle z,w \rangle )^{n+1+\alpha +k}}x~~; \qquad g(z) = \log (1- \langle z,w \rangle )y^{\star }, \end{aligned}$$

where $\log$ is the principal branch of the logarithm. Since $f \in H^{\infty }(\mathbb {B}_{n},X)$ and $g \in \mathcal {B}(\mathbb {B}_{n},Y^{\star }),$ by relation (5.1), we have that

$$\begin{aligned} |\langle h_{b}f,g \rangle |_{\alpha ,Y} \le \Vert h_b\Vert \Vert x\Vert _{X}\Vert y^{\star }\Vert _{Y^{\star }} . \end{aligned}$$

(5.3)

Applying Lemma 23 for those particular vector-valued holomorphic functions f and g and using the fact that

$$\begin{aligned} \log (1- \langle w,z \rangle ) = \log (1-|w|^2)+\log \left( \dfrac{1- \langle w,z \rangle }{1-|w|^2}\right) , \end{aligned}$$

we obtain

$$\begin{aligned}&\langle h_{b}f,g \rangle _{\alpha ,Y} \\&= \int _{\mathbb {B}_n} \big \langle b(z)\overline{\left( \dfrac{(1-|w|^2)^{k-\gamma }}{(1-\langle z,w \rangle )^{n+1+\alpha +k}}x\right) }, \log (1-\langle z,w \rangle )y^{\star } \big \rangle _{Y,Y^{\star }} \mathrm {d}\nu _{\alpha }(z)\\&= \big \langle \int _{\mathbb {B}_n} b(z)\left[ \dfrac{(1-|w|^2)^{k-\gamma } \log (1-\langle w,z \rangle )}{(1-\langle w,z \rangle )^{n+1+\alpha +k}}\overline{x}\right] \mathrm {d}\nu _{\alpha }(z),y^{\star } \big \rangle _{Y,Y^{\star }} \\&= \big \langle \int _{\mathbb {B}_n} \dfrac{b(z)(\overline{x})(1-|w|^2)^{k-\gamma }\log (1-|w|^2)}{(1-\langle w,z \rangle )^{n+1+\alpha +k}}\mathrm {d}\nu _{\alpha }(z),y^{\star } \big \rangle _{Y,Y^{\star }} \\&+ \big \langle \int _{\mathbb {B}_n} b(z)\left[ \dfrac{(1-|w|^2)^{k-\gamma }}{(1-\langle w,z \rangle )^{n+1+\alpha +k}}\log \left( \dfrac{1-\langle w,z \rangle }{1-|w|^2} \right) \overline{x}\right] \mathrm {d}\nu _{\alpha }(z),y^{\star } \big \rangle _{Y,Y^{\star }}\\&= \big \langle (1-|w|^2)^{k-\gamma }\log (1-|w|^2)\int _{\mathbb {B}_n} \dfrac{b(z)(\overline{x})\mathrm {d}\nu _{\alpha }(z)}{(1-\langle w,z \rangle )^{n+1+\alpha +k}},y^{\star } \big \rangle _{Y,Y^{\star }} \\&\quad + \big \langle \int _{\mathbb {B}_n} b(z)\left( \overline{\dfrac{(1-|w|^2)^{k-\gamma }}{(1-\langle z,w \rangle )^{n+1+\alpha +k}}\log \left( \dfrac{1- \langle z,w \rangle }{1-|w|^2}\right) x}\right) \mathrm {d}\nu _{\alpha }(z),y^{\star } \big \rangle _{Y,Y^{\star }}\\&= (1-|w|^2)^{k-\gamma }\log (1-|w|^2) \big \langle L_{k}\left( \int _{\mathbb {B}_n} \dfrac{b(z)(\overline{x})}{(1-\langle w,z \rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(z)\right) ,y^{\star } \big \rangle _{Y,Y^{\star }} \\&\quad + \big \langle \int _{\mathbb {B}_n} b(z)\left( \overline{f(z)\log \left( \dfrac{1- \langle z,w \rangle }{1-|w|^2}\right) }\right) \mathrm {d}\nu _{\alpha }(z),y^{\star } \big \rangle _{Y,Y^{\star }}\\&= (1-|w|^2)^{k-\gamma }\log (1-|w|^2) \langle L_{k}(b(w)(\overline{x})),y^{\star } \rangle _{Y,Y^{\star }} \\&\quad + \langle \int _{\mathbb {B}_n} b(z)(\overline{\varphi (z)}) \mathrm {d}\nu _{\alpha }(z),y^{\star } \rangle _{Y,Y^{\star }}, \end{aligned}$$

where $\varphi (z) = f(z)\log \left( \dfrac{1- \langle z,w \rangle }{1-|w|^2}\right) .$ Therefore, we can write $\langle h_{b}f,g \rangle _{\alpha ,Y} = I_{1}+I_{2},$ with

$$\begin{aligned} \displaystyle I_{1} = (1-|w|^2)^{k-\gamma }\log (1-|w|^2) \langle L_{k}(b(w)(\overline{x})),y^{\star } \rangle _{Y,Y^{\star }} \end{aligned}$$

and

$$\begin{aligned} I_{2} = \displaystyle \big \langle \int _{\mathbb {B}_n} b(z)(\overline{\varphi (z)}) \mathrm {d}\nu _{\alpha }(z),y^{\star } \big \rangle _{Y,Y^{\star }}. \end{aligned}$$

Applying Lemma 16 with $\delta = p,$ and $\beta = p(k-\gamma ),$ we obtain that

$$\begin{aligned}&\Vert \varphi \Vert _{p,\alpha ,X} \\&= \displaystyle \left( \int _{\mathbb {B}_n}\left| \log \left( \dfrac{1-\langle z,w \rangle }{1-|w|^2}\right) \right| ^{p}\dfrac{(1-|w|^2)^{p(k-\gamma )}}{|1-\langle z,w \rangle |^{p(n+1+\alpha +k)}}\Vert x\Vert ^p_{X}\mathrm {d}\nu _{\alpha }(z)\right) ^{1/p}\\&= \displaystyle \Vert x\Vert _{X}\left( \int _{\mathbb {B}_n}\left| \log \left( \dfrac{1-\langle z,w \rangle }{1-|w|^2}\right) \right| ^{p}\dfrac{(1-|w|^2)^{p(k-\gamma )}}{|1-\langle z,w \rangle |^{n+1+\alpha +p(k-\gamma )}}\mathrm {d}\nu _{\alpha }(z)\right) ^{1/p} \\&\lesssim \Vert x\Vert _{X}. \end{aligned}$$

According to the relation (5.2), we obtain the following estimation of $I_{2}$

$$\begin{aligned} \displaystyle |I_{2}| \le \Vert h_b\Vert \Vert \varphi \Vert _{p,\alpha ,X}\Vert y^{\star }\Vert _{Y^{\star }} \lesssim \Vert h_b\Vert \Vert x\Vert _{X}\Vert y^{\star }\Vert _{Y^{\star }}. \end{aligned}$$

Since $I_{1} = \langle h_{b}f,g \rangle _{\alpha ,Y} - I_{2},$ by the relation (5.3) and the previous estimates on $I_{2},$ we have that

$$\begin{aligned} |I_{1}| \le |\langle h_{b}f,g \rangle _{\alpha ,Y}| + |I_{2}| \lesssim \Vert h_b\Vert \Vert x\Vert _{X}\Vert y^{\star }\Vert _{Y^{\star }}. \end{aligned}$$

Since $x \in X,$ $y^{\star } \in Y^{\star }$ are arbitrary and $\Vert x\Vert _{X} = \Vert \overline{x}\Vert _{\overline{X}},$ we get that

$$\begin{aligned} |I_{1}|= & {} (1-|w|^2)^{k-\gamma }\log \left( \dfrac{1}{1-|w|^2} \right) |\langle L_{k}(b(w)(\overline{x})),y^{\star } \rangle _{Y,Y^{\star }}| \\&\le C \Vert h_b\Vert \Vert \overline{x}\Vert _{\overline{X}}\Vert y^{\star }\Vert _{Y^{\star }}. \end{aligned}$$

Since $\overline{x} \in \overline{X}$ and $y^{\star } \in Y^{\star }$ are arbitrary, we deduce that :

$$\begin{aligned} \Vert L_{k}b(w)\Vert _{\mathcal {L}(\overline{X},Y)}&= \displaystyle \sup _{\Vert \overline{x}\Vert _{\overline{X}} = 1, \Vert y^{\star }\Vert _{Y^{\star }} = 1} |\langle L_{k}(b(w)(\overline{x})),y^{\star } \rangle _{Y,Y^{\star }}|\\&\le \dfrac{C}{(1-|w|^2)^{k-\gamma }}\left( \log \dfrac{1}{1-|w|^2}\right) ^{-1}. \end{aligned}$$

The desired result follows at once using Corollary 22. $\square$

6 Compactness of the Little Hankel Operator, $h_b$, with Operator-Valued Symbols b From $A^{p}_{\alpha }(\mathbb {B}_{n},X)$ to $A^{q}_{\alpha }(\mathbb {B}_{n},Y),$ With $1< p \le q < \infty$

In this section, we are going to characterize those symbols b for whch the little Hankel operator extends into a bounded compact oparator from $A^{p}_{\alpha }(\mathbb {B}_{n},X)$ to $A^{q}_{\alpha }(\mathbb {B}_{n},Y),$ where $1< p \le q < \infty$ and X, Y are two reflexive complex Banach spaces.

6.1 Preliminaries Notions

The proof of the following remark can be found in [11, Proposition 1.6.1]

Remark 25

Let $t \ge 0.$ Then the operator $R^{\alpha ,t}$ is the unique continuous linear operator on $\mathcal {H}(\mathbb {B}_{n},X)$ satisfying

$$\begin{aligned} \displaystyle R^{\alpha ,t} \left( \dfrac{x}{(1-\langle z,w \rangle )^{n+1+\alpha }}\right) = \dfrac{x}{(1-\langle z,w \rangle )^{n+1+\alpha +t}}, \end{aligned}$$

for every $z \in \mathbb {B}_{n}$ and $x \in X.$

We will use the operator $R^{\alpha ,t},$ for $t > 0,$ in the vector-valued Bergman space $A^1_{\alpha }(\mathbb {B}_{n},X)$ as follows:

Proposition 26

Let $t > 0$ and $f \in A^1_{\alpha }(\mathbb {B}_{n},X).$ Then

$$\begin{aligned} \displaystyle R^{\alpha ,t}f(z) = \int _{\mathbb {B}_n} \dfrac{f(w)}{(1-\langle z,w \rangle )^{n+1+\alpha +t}}\mathrm {d}\nu _{\alpha }(w), \end{aligned}$$

for each $z \in \mathbb {B}_{n}.$

The proof of the following proposition is not quite different to the proof in [14, Proposition 1.15], but for the sake of completeness, we will recall the proof.

Proposition 27

Suppose N is a positive integer and $\alpha$ is a real such that $n+\alpha$ is not a negative integer. Then $R^{\alpha ,N}$ as an operator acting on $\mathcal {H}(\mathbb {B}_{n},X)$ is a linear partial differential operator of order N with polynomial coefficients, that is

$$\begin{aligned} \displaystyle R^{\alpha ,N}f(z) = \sum _{m \in \mathbb {N}^{n},|m| \le N} p_{m}(z)\dfrac{\partial ^{|m|} f}{\partial z^{m}}(z), \end{aligned}$$

where each $p_{m}$ is a polynomial.

Proof

Let $x \in X$ and $w \in \mathbb {B}_n.$ By using the multi-nomial formula

$$\begin{aligned} \langle z,w \rangle ^{k} = \displaystyle \sum _{|m| = k} \dfrac{k!}{m!}z^{m}\overline{w}^{m}, \end{aligned}$$

it follows that

$$\begin{aligned}&\dfrac{x}{(1 - \langle z,w \rangle )^{n+1+\alpha +N}}\\&= \dfrac{x(1 -\langle z,w \rangle + \langle z,w \rangle )^{N}}{(1 - \langle z,w \rangle )^{n+1+\alpha +N}}\\&= \sum _{k=0}^{N}\dfrac{N!}{k!(N-k)!}\dfrac{\langle z,w \rangle ^{k}x\; (1 - \langle z,w \rangle )^{N-k}}{(1 - \langle z,w \rangle )^{n+1+\alpha +N}}\\&= \sum _{k=0}^{N}\dfrac{N!}{k!(N-k)!}\sum _{|m| = k} \dfrac{k!}{m!}z^{m}\dfrac{\overline{w}^{m}x}{(1 - \langle z,w \rangle )^{n+1+\alpha +k}}\\&= \sum _{k=0}^{N}\sum _{|m| = k} \dfrac{N!}{m!(N-k)!}z^{m}\dfrac{\overline{w}^{m}x}{(1 - \langle z,w \rangle )^{n+1+\alpha +k}}\\&= \sum _{k=0}^{N}\sum _{|m| = k} \dfrac{N!}{\prod _{j=0}^{k}(n+1+\alpha +j)m!(N-k)!}z^{m}\dfrac{\partial ^{k} }{\partial z^{m}}\left( \dfrac{x}{(1 - \langle z,w \rangle )^{n+1+\alpha }}\right) . \end{aligned}$$

Therefore, there exists a constant $c_{mk}$ such that

$$\begin{aligned} R^{\alpha ,N}\left( \dfrac{x}{(1 - \langle z,w \rangle )^{n+1+\alpha }}\right) = \displaystyle \sum _{k=0}^{N}\sum _{|m| = k}c_{mk}z^{m}\dfrac{\partial ^{k} }{\partial z^{m}}\left( \dfrac{x}{(1 - \langle z,w \rangle )^{n+1+\alpha }}\right) . \end{aligned}$$

Thus

$$\begin{aligned} R^{\alpha ,N} = \displaystyle \sum _{k=0}^{N}\sum _{|m| = k}c_{mk}z^{m}\dfrac{\partial ^{k} }{\partial z^{m}}.\qquad \qquad \qquad \end{aligned}$$

$\square$

We will also need the following results whose proofs can be found in [11].

Lemma 28

Let $t > 0.$ Then

$$\begin{aligned} \displaystyle \int _{\mathbb {B}_n} f(z)\overline{g(z)}\mathrm {d}\nu _{\alpha }(z) = \int _{\mathbb {B}_n} R^{\alpha ,t}f(z)\overline{g(z)} \mathrm {d}\nu _{\alpha +t}(z), \end{aligned}$$

for all $f \in A^1_{\alpha }(\mathbb {B}_{n},X)$ and $g \in H^{\infty }(\mathbb {B}_{n},\mathbb {C}).$

Lemma 29

Let $t > 0$ and X a complex Banach space. Then

$$\begin{aligned} \int _{\mathbb {B}_n} \langle f(z),g(z) \rangle _{X,X^{\star }}\mathrm {d}\nu _{\alpha }(z)&= \int _{\mathbb {B}_n} \langle R^{\alpha ,t}f(z),g(z) \rangle _{X,X^{\star }}\mathrm {d}\nu _{\alpha +t}(z)\\&= \int _{\mathbb {B}_n} \langle f(z),R^{\alpha ,t}g(z) \rangle _{X,X^{\star }}\mathrm {d}\nu _{\alpha +t}(z), \end{aligned}$$

for every $f \in A^{1}_{\alpha }(\mathbb {B}_{n},X)$ and $g \in H^{\infty }(\mathbb {B}_{n},X^{\star }).$

Corollary 30

Suppose $t >0$ and $1< p < \infty .$ If $b \in A^{p'}_{\alpha }(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y)),$ where $p'$ is the conjugate exponent of p,  then the following equality holds

$$\begin{aligned} \displaystyle \int _{\mathbb {B}_n} \langle b(z)\overline{f(z)},g(z) \rangle _{Y,Y^{\star }}\mathrm {d}\nu _{\alpha }(z) = \int _{\mathbb {B}_n} \langle R^{\alpha ,t}b(z)\overline{f(z)},g(z) \rangle _{Y,Y^{\star }}\mathrm {d}\nu _{\alpha +t}(z) \end{aligned}$$

for $f \in H^{\infty }(\mathbb {B}_{n},X)$ and $g \in H^{\infty }(\mathbb {B}_{n},Y^{\star }).$

In the sequel, we will need to interchange the position of the summation symbol and the integral symbol in a particular situation. That is why we introduce this lemma.

Lemma 31

Assume $1< t < \infty .$ Let $b(z) = \sum _{\beta \in \mathbb {N}^{n}}\hat{b}(\beta )z^{\beta } \in A^t_{\alpha }(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y)).$ Then

$$\begin{aligned} \int _{\mathbb {B}_n} \langle b(z)\left( \overline{f(z)}\right) ,y^{\star }_{0} \rangle _{Y,Y^{\star }}\mathrm {d}\nu _{\alpha }(z) = \sum _{\beta \in \mathbb {N}^n}\int _{\mathbb {B}_n}z^{\beta } \langle \hat{b}(\beta ) \left( \overline{f(z)}\right) ,y^{\star }_{0} \rangle _{Y,Y^{\star }}\mathrm {d}\nu _{\alpha }(z), \end{aligned}$$

for every $f \in H^{\infty }(\mathbb {B}_{n},X)$ and $y^{\star }_{0} \in Y^{\star }$ with $\Vert y^{\star }_{0}\Vert _{Y^{\star }} = 1.$

Proof

Since $b(z) = \sum _{\beta \in \mathbb {N}^{n}}\hat{b}(\beta )z^{\beta } \in A^{t}_{\alpha }(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y)),$ we have that

$$\begin{aligned} \lim _{N \rightarrow \infty } \int _{\mathbb {B}_n} \left\| b(z) - \sum _{\beta \in \mathbb {N}^{n}, |\beta | \le N }\hat{b}(\beta )z^{\beta } \right\| ^t_{\mathcal {L}(\overline{X},Y)}\mathrm {d}\nu _{\alpha }(z) = 0. \end{aligned}$$

We have

$$\begin{aligned} \displaystyle \left| \int _{\mathbb {B}_n} \left\langle \left( b(z) - \sum _{\beta \in \mathbb {N}^{n}:|\beta | \le N}\hat{b}(\beta )z^{\beta }\right) (\overline{f(z)}),y^{\star }_{0} \right\rangle _{Y,Y^{\star }} \mathrm {d}\nu _{\alpha }(z)\right| \le \\ \displaystyle \int _{\mathbb {B}_n} \left\| b(z) - \sum _{\beta \in \mathbb {N}^{n}:|\beta | \le N}\hat{b}(\beta )z^{\beta } \right\| _{\mathcal {L}(\overline{X},Y)} \Vert \overline{f(z)}\Vert _{\overline{X}} \Vert y^{\star }_{0}\Vert _{Y^{\star }} \mathrm {d}\nu _{\alpha }(z) = \\\displaystyle \int _{\mathbb {B}_n} \left\| b(z) - \sum _{\beta \in \mathbb {N}^{n}:|\beta | \le N}\hat{b}(\beta )z^{\beta } \right\| _{\mathcal {L}(\overline{X},Y)} \Vert f(z)\Vert _{X} \mathrm {d}\nu _{\alpha }(z) \lesssim \\\displaystyle \int _{\mathbb {B}_n} \left\| b(z) - \sum _{\beta \in \mathbb {N}^{n}:|\beta | \le N}\hat{b}(\beta )z^{\beta } \right\| ^t_{\mathcal {L}(\overline{X},Y)} \mathrm {d}\nu _{\alpha }(z) \longrightarrow 0 \end{aligned}$$

as $N \rightarrow \infty .$ Therefore, we have that

$$\begin{aligned}&\int _{\mathbb {B}_n} \langle b(z)\left( \overline{f(z)}\right) ,y^{\star }_{0} \rangle _{Y,Y^{\star }}\mathrm {d}\nu _{\alpha }(z) \\&= \lim _{N \rightarrow \infty }\int _{\mathbb {B}_n} \left\langle \sum _{\beta \in \mathbb {N}^{n}:|\beta | \le N} \hat{b}(\beta )z^{\beta } \left( \overline{f(z)}\right) ,y^{\star }_{0} \right\rangle _{Y,Y^{\star }}\mathrm {d}\nu _{\alpha }(z)\\&= \lim _{N \rightarrow \infty }\int _{\mathbb {B}_n} \sum _{\beta \in \mathbb {N}^{n}:|\beta | \le N} \left\langle \hat{b}(\beta )z^{\beta } \left( \overline{f(z)}\right) ,y^{\star }_{0} \right\rangle _{Y,Y^{\star }}\mathrm {d}\nu _{\alpha }(z)\\&= \lim _{N \rightarrow \infty }\sum _{\beta \in \mathbb {N}^{n}:|\beta | \le N} \int _{\mathbb {B}_n} \left\langle \hat{b}(\beta )z^{\beta } \left( \overline{f(z)}\right) ,y^{\star }_{0} \right\rangle _{Y,Y^{\star }}\mathrm {d}\nu _{\alpha }(z)\\&= \sum _{\beta \in \mathbb {N}^{n}} \int _{\mathbb {B}_n} \left\langle \hat{b}(\beta )z^{\beta } \left( \overline{f(z)}\right) ,y^{\star }_{0} \right\rangle _{Y,Y^{\star }}\mathrm {d}\nu _{\alpha }(z). \qquad \qquad \qquad \end{aligned}$$

$\square$

In the following lemma, we compute the little Hankel operator when the operator-valued symbol is a monomial.

Lemma 32

Suppose $1< p < \infty$ and $\gamma \in \mathbb {N}^{n}.$ If $a_{\gamma } \in \mathcal {L}(\overline{X},Y),$ then for every $\displaystyle f(z) = \sum \nolimits _{\beta \in \mathbb {N}^{n}}c_{\beta }z^{\beta } \in A^{p}_{\alpha }(\mathbb {B}_{n},X),$ we have

$$\begin{aligned} \displaystyle h_{a_{\gamma }z^{\gamma }}f(z) = \sum _{\beta \in \mathbb {N}^{n},\beta \le \gamma } a_{\gamma }(\overline{c_{\beta }})\dfrac{\gamma !\Gamma (n+1+\alpha +|\gamma -\beta |)}{(\gamma -\beta )!\Gamma (n+1+\alpha +|\gamma |)}z^{\gamma -\beta }. \end{aligned}$$

Proof

Since

$$\begin{aligned} \displaystyle f(z) = \sum _{\beta \in \mathbb {N}^{n}}c_{\beta }z^{\beta } \in A^{p}_{\alpha }(\mathbb {B}_{n},X), \end{aligned}$$

and $p>1$ by using [16, Corollary 4], it follows that

$$\begin{aligned} \displaystyle \int _{\mathbb {B}_n} \left\| \sum _{|\beta | \ge N+1}c_{\beta }z^{\beta } \right\| ^p_{X}\mathrm {d}\nu _{\alpha }(z) \rightarrow 0~~~~\text{ as }~~ N \rightarrow \infty . \end{aligned}$$

(6.1)

Firstly, let us prove that

$$\begin{aligned} \displaystyle { \int _{\mathbb {B}_n}\dfrac{\sum _{\beta \in \mathbb {N}^{n}}a_{\gamma }(\overline{c_{\beta }})\overline{w^{\beta }}}{(1 - \langle z,w \rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(w) = \sum _{\beta \in \mathbb {N}^{n}} \int _{\mathbb {B}_n}\dfrac{a_{\gamma }(\overline{c_{\beta }})\overline{w^{\beta }}}{(1 - \langle z,w \rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(w)} \end{aligned}$$

(6.2)

Let $N \in \mathbb {N}.$ We have that

$$\begin{aligned}&\displaystyle { \int _{\mathbb {B}_n}\left\| \dfrac{\sum _{\beta \in \mathbb {N}^{n}}a_{\gamma }(\overline{c_{\beta }})\overline{w^{\beta }} - \sum _{|\beta | \le N}a_{\gamma }(\overline{c_{\beta }})\overline{w^{\beta }}}{(1 - \langle z,w \rangle )^{n+1+\alpha }}\right\| _{Y}\mathrm {d}\nu _{\alpha }(w)} \\&= \int _{\mathbb {B}_n}\left\| \dfrac{\sum _{|\beta | \ge N+1}a_{\gamma }(\overline{c_{\beta }})\overline{w^{\beta }}}{(1 - \langle z,w \rangle )^{n+1+\alpha }} \right\| _{Y}\mathrm {d}\nu _{\alpha }(w)\\&= \int _{\mathbb {B}_n}\left\| \dfrac{a_{\gamma }\left( \sum _{|\beta | \ge N+1}(\overline{c_{\beta }})\overline{w^{\beta }}\right) }{(1 - \langle z,w \rangle )^{n+1+\alpha }}\right\| _{Y}\mathrm {d}\nu _{\alpha }(w)\\&\le \dfrac{\Vert a_{\gamma }\Vert _{\mathcal {L}(\overline{X},Y)}}{(1 - |z|)^{n+1+\alpha }}\int _{\mathbb {B}_n}\left\| \sum _{|\beta | \ge N+1}c_{\beta }w^{\beta }\right\| _{X}\mathrm {d}\nu _{\alpha }(w)\\&\le \dfrac{\Vert a_{\gamma }\Vert _{\mathcal {L}(\overline{X},Y)}}{(1 - |z|)^{n+1+\alpha }}\int _{\mathbb {B}_n}\left\| \sum _{|\beta | \ge N+1}c_{\beta }w^{\beta }\right\| _{X}\mathrm {d}\nu _{\alpha }(w)\\&\le \dfrac{\Vert a_{\gamma }\Vert _{\mathcal {L}(\overline{X},Y)}}{(1 - |z|)^{n+1+\alpha }}\left( \int _{\mathbb {B}_n}\left\| \sum _{|\beta | \ge N+1}c_{\beta }w^{\beta }\right\| ^p_{X}\mathrm {d}\nu _{\alpha }(w)\right) ^{1/p}. \end{aligned}$$

Therefore

$$\begin{aligned} \displaystyle { \left\| \int _{\mathbb {B}_n}\dfrac{\sum _{\beta \in \mathbb {N}^{n}}a_{\gamma }(\overline{c_{\beta }})\overline{w^{\beta }} - \sum _{|\beta | \le N}a_{\gamma }(\overline{c_{\beta }})\overline{w^{\beta }}}{(1 - \langle z,w \rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(w) \right\| _{Y}} \end{aligned}$$

is less than or equal to

$$\begin{aligned} \displaystyle { \dfrac{\Vert a_{\gamma }\Vert _{\mathcal {L}(\overline{X},Y)}}{(1 - |z|)^{n+1+\alpha }}\left( \int _{\mathbb {B}_n}\left\| \sum _{|\beta | \ge N+1}c_{\beta }w^{\beta }\right\| ^p_{X}\mathrm {d}\nu _{\alpha }(w)\right) ^{1/p}.} \end{aligned}$$

(6.3)

By using (6.1) and (6.3), it follows that

$$\begin{aligned} \displaystyle \left\| \int _{\mathbb {B}_n}\dfrac{\displaystyle \sum _{\beta \in \mathbb {N}^{n}}a_{\gamma }(\overline{c_{\beta }})\overline{w^{\beta }} - \sum _{|\beta | \le N}a_{\gamma }(\overline{c_{\beta }})\overline{w^{\beta }}}{(1 - \langle z,w \rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(w) \right\| _{Y} \rightarrow 0 \end{aligned}$$

as $N \rightarrow \infty ,$ and so

$$\begin{aligned} \displaystyle \int _{\mathbb {B}_n}\dfrac{\displaystyle \sum \nolimits _{\beta \in \mathbb {N}^{n}}a_{\gamma }(\overline{c_{\beta }})\overline{w^{\beta }}}{(1 - \langle z,w \rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(w)&= \displaystyle \lim _{N\rightarrow \infty }\int _{\mathbb {B}_n}\dfrac{\displaystyle \sum \nolimits _{|\beta | \le N}a_{\gamma }(\overline{c_{\beta }})\overline{w^{\beta }}}{(1 - \langle z,w \rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(w) \\&= \displaystyle \lim _{N\rightarrow \infty } \sum _{|\beta | \le N}\int _{\mathbb {B}_n}\dfrac{a_{\gamma }(\overline{c_{\beta }})\overline{w^{\beta }}}{(1 - \langle z,w \rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(w)\\&= \displaystyle \sum _{\beta \in \mathbb {N}^{n}}\int _{\mathbb {B}_n}\dfrac{a_{\gamma }(\overline{c_{\beta }})\overline{w^{\beta }}}{(1 - \langle z,w \rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(w), \end{aligned}$$

which is the desired result. Secondly, let us prove that

$$\begin{aligned}&\int _{\mathbb {B}_{n}} \sum _{k=0}^{\infty }\dfrac{\Gamma (n+1+\alpha +k)}{\Gamma (n+1+\alpha )k!} \langle z,w \rangle ^{k} \mathrm {d}\nu _{\alpha }(w) = \nonumber \\&\quad \sum _{k=0}^{\infty } \int _{\mathbb {B}_{n}}\dfrac{\Gamma (n+1+\alpha +k)}{\Gamma (n+1+\alpha )k!} \langle z,w \rangle ^{k} \mathrm {d}\nu _{\alpha }(w). \end{aligned}$$

(6.4)

Let $N \in \mathbb {N}.$ We have

$$\begin{aligned} \displaystyle \left| \sum _{k=0}^{N}\dfrac{\Gamma (n+1+\alpha +k)}{\Gamma (n+1+\alpha )k!} \langle z,w \rangle ^{k}\right|&\le \sum _{k=0}^{N}\dfrac{\Gamma (n+1+\alpha +k)}{\Gamma (n+1+\alpha )k!}|z|^{k}\\&\le \sum _{k=0}^{\infty }\dfrac{\Gamma (n+1+\alpha +k)}{\Gamma (n+1+\alpha )k!}|z|^{k}\\&= \dfrac{1}{(1 - |z|)^{n+1+\alpha }}. \end{aligned}$$

Since $\int _{\mathbb {B}_{n}} \dfrac{1}{(1 - |z|)^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(w) = \dfrac{1}{(1 - |z|)^{n+1+\alpha }},$ by the dominated convergence theorem, we have that

$$\begin{aligned}&\sum _{k=0}^{\infty } \int _{\mathbb {B}_{n}} \dfrac{\Gamma (n+1+\alpha +k)}{\Gamma (n+1+\alpha )k!} \langle z,w \rangle ^{k} \mathrm {d}\nu _{\alpha }(w) \\&= \lim _{N\rightarrow \infty }\sum _{k=0}^{N}\int _{\mathbb {B}_{n}}\dfrac{\Gamma (n+1+\alpha +k)}{\Gamma (n+1+\alpha )k!} \langle z,w \rangle ^{k} \mathrm {d}\nu _{\alpha }(w)\\&= \lim _{N\rightarrow \infty }\int _{\mathbb {B}_{n}} \sum _{k=0}^{N}\dfrac{\Gamma (n+1+\alpha +k)}{\Gamma (n+1+\alpha )k!} \langle z,w \rangle ^{k} \mathrm {d}\nu _{\alpha }(w)\\&= \int _{\mathbb {B}_{n}}\lim _{N\rightarrow \infty }\sum _{k=0}^{N}\dfrac{\Gamma (n+1+\alpha +k)}{\Gamma (n+1+\alpha )k!} \langle z,w \rangle ^{k} \mathrm {d}\nu _{\alpha }(w)\\&= \int _{\mathbb {B}_{n}} \sum _{k=0}^{\infty }\dfrac{\Gamma (n+1+\alpha +k)}{\Gamma (n+1+\alpha )k!} \langle z,w \rangle ^{k} \mathrm {d}\nu _{\alpha }(w). \end{aligned}$$

We are now ready to prove our lemma. For $f(z) = \sum _{\beta \in \mathbb {N}^{n}}c_{\beta }z^{\beta } \in A^{p}_{\alpha }(\mathbb {B}_{n},X),$ by using the following multi-nomial formula [14, (1.1)] and the following formula [14, (1.23)] respectively

$$\begin{aligned} \langle z,w \rangle ^k = \displaystyle \sum _{|m| = k}\frac{k!}{m!}z^{m}\overline{w^m}, \qquad \qquad \int _{\mathbb {B}_n}|z^m|^2\mathbb {d}\nu _{\alpha }(z) = \dfrac{m!\Gamma (n+\alpha +1)}{\Gamma (n+|m|+\alpha +1)}, \end{aligned}$$

we get that, using (6.2) and (6.4)

$$\begin{aligned}&h_{a_{\gamma }z^{\gamma }}f(z)\\&= \int _{\mathbb {B}_n} \dfrac{ a_{\gamma }w^{\gamma }\left( \overline{ \sum _{\beta \in \mathbb {N}^{n}} c_{\beta }w^{\beta }}\right) }{(1-\langle z,w \rangle )^{n+1+\alpha }} \mathrm {d}\nu _{\alpha }(w)\\&= \int _{\mathbb {B}_n}\dfrac{w^{\gamma }\displaystyle \sum \nolimits _{\beta \in \mathbb {N}^{n}}a_{\gamma }(\overline{c_{\beta }})\overline{w^{\beta }}}{(1 - \langle z,w \rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(w)\\&= \sum _{\beta \in \mathbb {N}^{n}} \int _{\mathbb {B}_n}\dfrac{w^{\gamma }a_{\gamma }(\overline{c_{\beta }})\overline{w^{\beta }}}{(1 - \langle z,w \rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(w) \\&= \sum _{\beta \in \mathbb {N}^{n}}a_{\gamma }(\overline{c_{\beta }}) \int _{\mathbb {B}_{n}}w^{\gamma }\overline{w^{\beta }} \sum _{k=0}^{\infty }\dfrac{\Gamma (n+1+\alpha +k)}{\Gamma (n+1+\alpha )k!} \langle z,w \rangle ^{k} \mathrm {d}\nu _{\alpha }(w)\\&= \sum _{\beta \in \mathbb {N}^{n}}a_{\gamma }(\overline{c_{\beta }}) \sum _{k=0}^{\infty }\int _{\mathbb {B}_{n}}w^{\gamma }\overline{w^{\beta }}\dfrac{\Gamma (n+1+\alpha +k)}{\Gamma (n+1+\alpha )k!} \langle z,w \rangle ^{k} \mathrm {d}\nu _{\alpha }(w)\\&= \sum _{\beta \in \mathbb {N}^{n}}a_{\gamma }(\overline{c_{\beta }}) \sum _{k=0}^{\infty }\dfrac{\Gamma (n+1+\alpha +k)}{\Gamma (n+1+\alpha )k!} \int _{\mathbb {B}_{n}}w^{\gamma }\overline{w^{\beta }} \sum _{|m|=k} \dfrac{k!}{m!}z^{m}\overline{w^{m}} \mathrm {d}\nu _{\alpha }(w)\\&= \sum _{\beta \in \mathbb {N}^{n}}a_{\gamma }(\overline{c_{\beta }})\sum _{k=0}^{\infty }\dfrac{\Gamma (n+1+\alpha +k)}{\Gamma (n+1+\alpha )k!} \sum _{|m|=k} \dfrac{k!}{m!} \int _{\mathbb {B}_{n}}w^{\gamma }\overline{w^{\beta }}z^{m}\overline{w^{m}} \mathrm {d}\nu _{\alpha }(w)\\&= \sum _{\beta \in \mathbb {N}^{n}}a_{\gamma }(\overline{c_{\beta }}) \sum _{k=0}^{\infty }\sum _{|m|=k}\dfrac{\Gamma (n+1+\alpha +k)}{\Gamma (n+1+\alpha )m!}\int _{\mathbb {B}_{n}}w^{\gamma } z^{m}\overline{w^{m+\beta }} \mathrm {d}\nu _{\alpha }(w)\\&= \sum _{\beta \in \mathbb {N}^{n}}a_{\gamma }(\overline{c_{\beta }}) \sum _{m \in \mathbb {N}^{n}}\dfrac{\Gamma (n+1+\alpha +|m|)}{\Gamma (n+1+\alpha )m!}z^{m} \int _{\mathbb {B}_{n}}w^{\gamma }\overline{w^{\beta +m}} \mathrm {d}\nu _{\alpha }(w)\\&= \sum _{\beta \in \mathbb {N}^{n},\beta \le \gamma }a_{\gamma }(\overline{c_{\beta }}) \dfrac{\Gamma (n+1+\alpha +|\gamma -\beta |)}{\Gamma (n+1+\alpha )(\gamma -\beta )!}z^{\gamma -\beta } \int _{\mathbb {B}_n}|z^{\gamma }|^{2}\mathrm {d}\nu _{\alpha }(w)\\&= \sum _{\beta \in \mathbb {N}^{n},\beta \le \gamma }a_{\gamma }(\overline{c_{\beta }}) \dfrac{\Gamma (n+1+\alpha +|\gamma -\beta |)}{\Gamma (n+1+\alpha )(\gamma -\beta )!} \dfrac{\gamma !\Gamma (n+1+\alpha )}{\Gamma (n+1+\alpha +|\gamma |)}z^{\gamma -\beta }\\&= \sum _{\beta \in \mathbb {N}^{n},\beta \le \gamma }a_{\gamma }(\overline{c_{\beta }}) \dfrac{\gamma !\Gamma (n+1+\alpha +|\gamma -\beta |)}{(\gamma -\beta )!\Gamma (n+1+\alpha +|\gamma |)} z^{\gamma -\beta }. \qquad \qquad \qquad \qquad \quad \square \end{aligned}$$

The goal of the following lemma is to prove that the linear span of the vector-valued Bergman kernel $\dfrac{x^{\star }}{(1-\langle w,z \rangle )^{n+1+\alpha }},$ where $x^{\star } \in X^{\star }$ and $z,w \in \mathbb {B}_{n}$ form a dense subspace in the vector-valued Bergman space $A^{p'}_{\alpha }(\mathbb {B}_{n},X^{\star }),$ with $1<p < \infty$ and $p'$ is the conjugate exponent of p.

Lemma 33

Suppose that $1< p < \infty .$ For each $x^{\star } \in X^{\star }$ and $z \in \mathbb {B}_{n},$ let

$$\begin{aligned} e_{z,x^{\star }}(w) = \dfrac{x^{\star }}{(1-\langle w,z \rangle )^{n+1+\alpha }}; \qquad w \in \mathbb {B}_n. \end{aligned}$$

Then $e_{z,x^{\star }} \in A^{p'}_{\alpha }(\mathbb {B}_{n},X^{\star })$ and the subspace generated by $e_{z,x^{\star }}$ is dense in $A^{p'}_{\alpha }(\mathbb {B}_{n},X^{\star }).$

Proof

Let $\phi \in A^p_{\alpha }(\mathbb {B}_{n},X)$ such that $\langle \phi ,e_{z,x^{\star }} \rangle _{\alpha ,X} = 0$ for all $z \in \mathbb {B}_{n}$ and $x^{\star } \in X^{\star }.$ Let $f^{\star } \in A^{p'}_{\alpha }(\mathbb {B}_{n},X^{\star }).$ According to the Hahn-Banach theorem, it suffices to prove that $\langle \phi ,f^{\star } \rangle _{\alpha ,X} = 0.$ For all $z \in \mathbb {B}_{n}$ and $x^{\star } \in X^{\star },$ using Lemma 1 and the reproducing kernel formula, it follows that

$$\begin{aligned} 0&= \langle \phi ,e_{z,x^{\star }} \rangle _{\alpha ,X} \\&= \displaystyle \int _{\mathbb {B}_n} \langle \phi (w),e_{z,x^{\star }}(w) \rangle _{X,X^{\star }}\mathrm {d}\nu _{\alpha }(w)\\&= \displaystyle \int _{\mathbb {B}_n} \langle \phi (w),\dfrac{x^{\star }}{(1-\langle w,z \rangle )^{n+1+\alpha }} \rangle _{X,X^{\star }}\mathrm {d}\nu _{\alpha }(w)\\&= \displaystyle \int _{\mathbb {B}_n} \langle \dfrac{\phi (w)}{(1-\langle z,w \rangle )^{n+1+\alpha }},x^{\star }\rangle _{X,X^{\star }}\mathrm {d}\nu _{\alpha }(w)\\&= \langle \phi (z),x^{\star } \rangle _{X,X^{\star }}. \end{aligned}$$

Therefore, for all $x^{\star } \in X^{\star },$ we have

$$\begin{aligned} \langle \phi (z),x^{\star } \rangle _{X,X^{\star }} = 0. \end{aligned}$$

Thus $\phi (z)= 0$ for every $z \in \mathbb {B}_{n}.$ It follows that for each $f^{\star } \in A^{p'}_{\alpha }(\mathbb {B}_{n},X^{\star }),$ we have that

$$\begin{aligned} \langle \phi ,f^{\star } \rangle _{\alpha ,X} = \int _{\mathbb {B}_n}\langle \phi (z),f^{\star }(z) \rangle _{X,X^{\star }}\mathrm {d}\nu _{\alpha }(z) = 0. \end{aligned}$$

$\square$

In the proof of the following lemma, we use the fact that when X is a reflexive complex Banach space and $1< p < \infty ,$ the dual of the vector-valued Bergman space $A^{p'}_{\alpha }(\mathbb {B}_{n},X^{\star })$ can be identified with $A^{p}_{\alpha }(\mathbb {B}_{n},X),$ where $p'$ is the conjugate exponent of p.

Lemma 34

Suppose that $1< p < \infty ,$ and X is a reflexive complex Banach space. Let $\lbrace f_{j} \rbrace \subset A^{p}_{\alpha }(\mathbb {B}_{n},X)$ such that $\displaystyle f_{j} \rightarrow 0$ weakly in $A^{p}_{\alpha }(\mathbb {B}_{n},X)$ as $j \rightarrow \infty .$ Then for each $\beta \in \mathbb {N}^{n},$ we have that $\partial ^{\beta }f_{j}(0) \rightarrow 0$ weakly in X as $j \rightarrow \infty ,$ where $\partial ^{\beta } = \frac{\partial ^{|\beta |} }{\partial z^{\beta }}.$

Proof

Since for each $j \in \mathbb {N},$ $f_{j} \in A^{p}_{\alpha }(\mathbb {B}_{n},X),$ using the reproducing kernel formula we have that

$$\begin{aligned} f_{j}(z) = \displaystyle \int _{\mathbb {B}_n}\dfrac{f_{j}(w)}{(1 - \langle z,w \rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(w), \qquad z \in \mathbb {B}_{n}. \end{aligned}$$

Differentiating both sides of the previous relation with respect to z,  we obtain

$$\begin{aligned} \displaystyle \partial ^{\beta } f_{j}(z) = C(n,\alpha ,|\beta |)\int _{\mathbb {B}_n}\dfrac{f_{j}(w)\overline{w}^{\beta }}{(1 - \langle z,w \rangle )^{n+1+\alpha +|\beta |}}\mathrm {d}\nu _{\alpha }(w). \end{aligned}$$

Therefore, we have

$$\begin{aligned} \displaystyle \partial ^{\beta } f_{j}(0) = C(n,\alpha ,|\beta |)\int _{\mathbb {B}_n} f_{j}(w)\overline{w}^{\beta }\mathrm {d}\nu _{\alpha }(w). \end{aligned}$$

Now, let $x^{\star } \in X^{\star }$ and let us show that $\langle \partial ^{\beta } f_{j}(0),x^{\star } \rangle _{X,X^{\star }} \rightarrow 0$ as $j \rightarrow \infty .$ But we have that

$$\begin{aligned} \langle \partial ^{\beta } f_{j}(0),x^{\star } \rangle _{X,X^{\star }}&= \displaystyle C(n,\alpha ,|\beta |) \left\langle \int _{\mathbb {B}_n} f_{j}(w)\overline{w}^{\beta }\mathrm {d}\nu _{\alpha }(w),x^{\star } \right\rangle _{X,X^{\star }}\\&= \displaystyle \int _{\mathbb {B}_n} \langle f_{j}(w),x^{\star } w^{\beta } \rangle _{X,X^{\star }}\mathrm {d}\nu _{\alpha }(w)\\&= \langle f_{j},g \rangle _{\alpha ,X} \rightarrow 0~~~~\text{ as }~~~~ j\rightarrow \infty , \end{aligned}$$

with $g(z) = x^{\star }z^{\beta } \in A^{p'}_{\alpha }(\mathbb {B}_{n},X^{\star }).$ Thus, $\langle \partial ^{\beta } f_{j}(0),x^{\star } \rangle _{X,X^{\star }} \rightarrow 0$ as $j \rightarrow \infty . \square$

We recall that the symbol b used in the following lemma satisfies (1.4) and (1.6).

Lemma 35

Suppose that X is a reflexive complex Banach space and k is a nonnegative integer. If the holomorphic mapping $z \mapsto b(z)$ maps $\mathbb {B}_{n}$ into $\mathcal {K}(\overline{X},Y),$ then the holomorphic mapping $z \mapsto R^{\alpha ,k}b(z)$ also maps $\mathbb {B}_{n}$ into $\mathcal {K}(\overline{X},Y).$

Proof

Let $z \in \mathbb {B}_{n}.$ Let $\lbrace f_{j} \rbrace$ a sequence of elements of X which converges weakly to 0 in X as j tends to infinity. Let us prove that $\lim _{j\rightarrow \infty } \Vert R^{\alpha ,k}b(z)\overline{f_{j}}\Vert _{Y} = 0.$ We know that the sequence $\lbrace f_j \rbrace$ is strongly bounded in X. Let $j \in \mathbb {N},$ by using (1.4) for $z = 0,$ we get that the function $z \mapsto b(z)\overline{f_{j}}\in A^{1}_{\alpha }(\mathbb {B}_{n},Y).$ By the reproducing kernel formula, it follows that

$$\begin{aligned} b(z)\overline{f_{j}} = \displaystyle \int _{\mathbb {B}_n}\dfrac{b(w)\overline{f_{j}}}{(1 - \langle z,w \rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(w). \end{aligned}$$

(6.5)

Applying the partial differential operator $R^{\alpha ,k}$ to (6.5), we have

$$\begin{aligned} R^{\alpha ,k}b(z)\overline{f_{j}} = \displaystyle \int _{\mathbb {B}_n}\dfrac{b(w)\overline{f_{j}}}{(1 - \langle z,w \rangle )^{n+1+\alpha +k}}\mathrm {d}\nu _{\alpha }(w). \end{aligned}$$

We also have

$$\begin{aligned} \displaystyle \dfrac{\Vert b(w)\overline{f_{j}}\Vert _{Y}}{|1 - \langle z,w \rangle |^{n+1+\alpha +k}}&\le \displaystyle \dfrac{\Vert b(w)\Vert _{\mathcal {L}(\overline{X},Y)}\Vert f_{j}\Vert _{X}}{(1 - |z|)^{n+1+\alpha +k}}\\&\le \dfrac{C(n+1+\alpha )}{(1-|z|)^{n+1+\alpha +k}}\Vert b(w)\Vert _{\mathcal {L}(\overline{X},Y)}, \end{aligned}$$

and

$$\begin{aligned} \displaystyle \int _{\mathbb {B}_n}\dfrac{C(n+1+\alpha )}{(1-|z|)^{n+1+\alpha +k}}\Vert b(w)\Vert _{\mathcal {L}(\overline{X},Y)}\mathrm {d}\nu _{\alpha }(w) < \infty . \end{aligned}$$

Therefore, by applying the dominated convergence theorem, we have that

$$\begin{aligned} \limsup _{j\rightarrow \infty }\Vert R^{\alpha ,k}b(z)\overline{f_{j}}\Vert _{Y}&\le \limsup _{j\rightarrow \infty } \displaystyle \int _{\mathbb {B}_n} \dfrac{\Vert b(w)\overline{f_{j}}\Vert _{Y}}{|1 - \langle z,w \rangle |^{n+1+\alpha +k}}\mathrm {d}\nu _{\alpha }(w)\\&= \displaystyle \int _{\mathbb {B}_n}\dfrac{\lim _{j\rightarrow \infty }\Vert b(w)\overline{f_{j}}\Vert _{Y}}{|1 - \langle z,w \rangle |^{n+1+\alpha +k}}\mathrm {d}\nu _{\alpha }(w) = 0. \end{aligned}$$

Thus for each $z \in \mathbb {B}_{n}$

$$\begin{aligned} \lim _{j\rightarrow \infty }\Vert R^{\alpha ,k}b(z)\overline{f_{j}}\Vert _{Y} = 0. \end{aligned}$$

$\square$

The following result will be also important in the sequel.

Lemma 36

Suppose $\beta _{0} \in \mathbb {N}^{n},$ $\lbrace f_j \rbrace$ a sequence of elements of X which converges weakly to 0 as j tends to infinity. For $z \in \mathbb {B}_{n},$ let $x_{j}(z) = z^{\beta _{0}}f_{j}.$ Then $\lbrace x_{j} \rbrace \subset A^p_{\alpha }(\mathbb {B}_{n},X)$ and $\lbrace x_{j} \rbrace$ converges weakly to 0 in $A^p_{\alpha }(\mathbb {B}_{n},X).$

Proof

Let $j \in \mathbb {N}.$ Since $f_j \rightarrow 0$ weakly in X as $j \rightarrow \infty ,$ it follows that $\lbrace f_j \rbrace$ is strongly bounded in X (see [9]). Let $\beta _{0} \in \mathbb {N}^n$ and $x_{j}(z) = z^{\beta _{0}}f_{j}.$ It is clear that $\lbrace x_{j} \rbrace \subset A^{p}_{\alpha }(\mathbb {B}_{n},X).$ For every $g \in A^{p'}_{\alpha }(\mathbb {B}_{n},X^{\star }),$ we have

$$\begin{aligned} \langle x_{j},g \rangle _{\alpha ,X}&= \displaystyle \int _{\mathbb {B}_n} \langle x_{j}(z),g(z) \rangle _{X,X^{\star }} \mathrm {d}\nu _{\alpha }(z)\\&= \displaystyle \int _{\mathbb {B}_n} \langle z^{\beta _{0}}f_{j},g(z) \rangle _{X,X^{\star }}\mathrm {d}\nu _{\alpha }(z)\\&= \displaystyle \int _{\mathbb {B}_n} z^{\beta _{0}} \langle f_{j},g(z) \rangle _{X,X^{\star }}\mathrm {d}\nu _{\alpha }(z). \end{aligned}$$

Since

$$\begin{aligned} \displaystyle \left| z^{\beta _{0}} \langle f_{j},g(z) \rangle _{X,X^{\star }} \right|&\le \displaystyle |z^{\beta _{0}} \langle f_{j},g(z) \rangle _{X,X^{\star }}|\\&\le \Vert f_{j}\Vert _{X}\Vert g(z)\Vert _{X^{\star }} \\&\le C\Vert g(z)\Vert _{X^{\star }}, \end{aligned}$$

and

$$\begin{aligned} \displaystyle \int _{\mathbb {B}_n} \Vert g(z)\Vert _{X^{\star }}\mathrm {d}\nu _{\alpha }(z) \le \left( \int _{\mathbb {B}_n} \Vert g(z)\Vert ^{p'}_{X^{\star }}\mathrm {d}\nu _{\alpha }(z)\right) ^{1/p'} < \infty . \end{aligned}$$

By using the dominated convergence theorem and the assumption, it follows that

$$\begin{aligned} \limsup _{j\longrightarrow \infty }\langle x_{j},g \rangle _{\alpha ,X}&= \displaystyle \int _{\mathbb {B}_n} z^{\beta _{0}} \lim _{j \longrightarrow \infty } \langle f_{j},g(z) \rangle _{X,X^{\star }} \mathrm {d}\nu _{\alpha }(z) = 0. \qquad \qquad \square \end{aligned}$$

6.2 Boundedness of the Little Hankel Operator with Operator-Valued Symbol on Vector-Valued Bergman Spaces

The principal result here is that, the little Hankel operator with operator-valued symbol $h_b$ is a bounded operator form $A^p_{\alpha }(\mathbb {B}_{n},X)$ to $A^q_{\alpha }(\mathbb {B}_{n},Y)$ with $1< p \le q < \infty$ if and only if the symbol b belongs to the generalized vector-valued Lipschitz space $\Lambda _{\gamma _{0}}(\mathbb {B}_{n}, \mathcal {L}(\overline{X},Y)),$ where

$$\begin{aligned} \gamma _{0} = (n+1+\alpha )\left( \frac{1}{p} - \frac{1}{q}\right) . \end{aligned}$$

The result obtained generalize the Oliver’s result [11, Theorem 4.2.2]. In the following lemma, we first prove that the definition of the generalized vector-valued Lipschitz space $\Lambda _{\gamma }(\mathbb {B}_{n},X),$ with $\gamma \ge 0$ is independent of the integer k used.

Lemma 37

Let $f \in \mathcal {H}(\mathbb {B}_{n},X).$ The following conditions are equivalent:

• (a) There exists a nonnegative integer $k > \gamma$ such that

  $$\begin{aligned} \sup _{z \in \mathbb {B}_{n}}(1 - |z|^2)^{k-\gamma }\Vert R^{\alpha ,k}f(z)\Vert _{X} < \infty . \end{aligned}$$

• (b) For every nonnegative integer $k > \gamma$ we have

  $$\begin{aligned} \sup _{z \in \mathbb {B}_{n}}(1 - |z|^2)^{k-\gamma }\Vert R^{\alpha ,k}f(z)\Vert _{X} < \infty . \end{aligned}$$

Proof

It is clear that $(b) \Rightarrow (a).$ So to complete the proof, we will prove that $(a) \Rightarrow (b).$ Suppose that there exists an integer $k > \gamma$ such that

$$\begin{aligned} c:= \sup _{z \in \mathbb {B}_{n}}(1 - |z|^2)^{k-\gamma }\Vert R^{\alpha ,k}f(z)\Vert _{X} < \infty . \end{aligned}$$

We want to prove that

$$\begin{aligned} \sup _{z \in \mathbb {B}_{n}}(1 - |z|^2)^{k+1-\gamma }\Vert R^{\alpha ,k+1}f(z)\Vert _{X} < \infty . \end{aligned}$$

Since $c < \infty ,$ then $f \in A^{1}_{\alpha }(\mathbb {B}_{n},X).$ Indeed, by [11, Theorem 3.1.2], we have that

$$\begin{aligned} \Vert f\Vert _{1,\alpha ,X}&\simeq \displaystyle \int _{\mathbb {B}_n} (1-|z|^2)^{k}\Vert R^{\alpha ,k}f(z)\Vert _{X}\mathrm {d}\nu _{\alpha }(z)\\&= \displaystyle \int _{\mathbb {B}_n} [(1-|z|^2)^{k-\gamma }\Vert R^{\alpha ,k}f(z)\Vert _{X}](1-|z|^2)^{\gamma }\mathrm {d}\nu _{\alpha }(z)\\&\lesssim \displaystyle c \int _{\mathbb {B}_n} (1 - |z|^2)^{\alpha +\gamma } \mathrm {d}\nu (z)\\< & {} \infty . \end{aligned}$$

By using Proposition 26, we have that

$$\begin{aligned} \displaystyle R^{\alpha ,k+1}f(z) = \int _{\mathbb {B}_n}\dfrac{f(w)}{(1 - \langle z,w \rangle )^{n+1+\alpha +k+1}}\mathrm {d}\nu _{\alpha }(w). \end{aligned}$$

Applyng Lemma 28, it follows that

$$\begin{aligned} \displaystyle R^{\alpha ,k+1}f(z) = \int _{\mathbb {B}_n}\dfrac{R^{\alpha ,k}f(w)}{(1 - \langle z,w \rangle )^{n+1+\alpha +k+1}}\mathrm {d}\nu _{\alpha +k}(w). \end{aligned}$$

Thus,

$$\begin{aligned} \Vert R^{\alpha ,k+1}f(z)\Vert _{X}&\lesssim \displaystyle \int _{\mathbb {B}_n} \dfrac{[(1 - |w|^2)^{k-\gamma }\Vert R^{\alpha ,k}f(w)\Vert _{X}](1 - |w|^2)^{\alpha +\gamma }}{|1 - \langle z,w \rangle |^{n+1+\alpha +\gamma +(k+1-\gamma )}}\mathrm {d}\nu (w)\\&\lesssim \dfrac{c}{(1 - |z|^2)^{k+1-\gamma }}. \end{aligned}$$

Therefore, we have that

$$\begin{aligned} \sup _{z \in \mathbb {B}_{n}}(1 - |z|^2)^{k+1-\gamma }\Vert R^{\alpha ,k+1}f(z)\Vert _{X} \lesssim c < \infty . \end{aligned}$$

Also, if k is a nonnegative integer with $k > \gamma$ such that

$$\begin{aligned} c':= \sup _{z \in \mathbb {B}_{n}}(1 - |z|^2)^{k+1-\gamma }\Vert R^{\alpha ,k+1}f(z)\Vert _{X} < \infty , \end{aligned}$$

then

$$\begin{aligned} \sup _{z \in \mathbb {B}_{n}}(1 - |z|^2)^{k-\gamma }\Vert R^{\alpha ,k}f(z)\Vert _{X} < \infty . \end{aligned}$$

Applying Proposition 26 and Lemma 28 we have that

$$\begin{aligned} \displaystyle R^{\alpha ,k} f(z) = \int _{\mathbb {B}_n} \dfrac{f(w)\mathrm {d}\nu _{\alpha }(w)}{(1 - \langle z,w \rangle )^{n+1+\alpha +k}} = \int _{\mathbb {B}_n} \dfrac{R^{\alpha ,k+1} f(w)\mathrm {d}\nu _{\alpha +k+1}(w)}{(1 - \langle z,w \rangle )^{n+1+\alpha +k}}, \end{aligned}$$

where $z \in \mathbb {B}_n.$ By using Theorem 18, it follows that

$$\begin{aligned} \Vert R^{\alpha ,k}f(z)\Vert _{X}&\lesssim \displaystyle \int _{\mathbb {B}_n}\dfrac{[(1 - |w|^2)^{k+1-\gamma }\Vert R^{\alpha ,k+1}f(w)\Vert _{X}](1 - |w|^2)^{\alpha +\gamma }\mathrm {d}\nu (w)}{|1 - \langle z,w \rangle |^{n+1+\alpha +k}}\\&= \displaystyle c'\int _{\mathbb {B}_n} \dfrac{(1-|w|^2)^{\alpha +\gamma }\mathrm {d}\nu (w)}{|1 - \langle z,w \rangle |^{n+1+\alpha +\gamma +(k-\gamma )}}\\&\lesssim \dfrac{c'}{(1 - |z|^2)^{k-\gamma }}. \end{aligned}$$

Since $z \in \mathbb {B}_n$ is arbitrary, we obtain that

$$\begin{aligned} \sup _{z \in \mathbb {B}_{n}}(1 - |z|^2)^{k-\gamma }\Vert R^{\alpha ,k}f(z)\Vert _{X} \lesssim c' < \infty . \end{aligned}$$

$\square$

Proposition 38

Let $\gamma \ge 0$ and $f \in \Lambda _{\gamma }(\mathbb {B}_{n},X).$ The following conditions are equivalent:

• (i) $f \in \Lambda _{\gamma ,0}(\mathbb {B}_{n},X).$

• (ii) $\lim _{s \rightarrow 1^{-}} \Vert f - f_s \Vert _{\Lambda _{\gamma }(\mathbb {B}_{n},X)} = 0,$ where $f_s$ is the dilation function defined for $z \in \mathbb {B}_{n}$ by $f_s(z):=f(sz).$

• (iii) f belongs to the closure of $\mathcal {P}(\mathbb {B}_{n},X),$ where $\mathcal {P}(\mathbb {B}_{n},X)$ is the space of vector-valued holomorphic polynomials.

Proof

$(i) \Rightarrow (ii).$ Suppose that $\frac{1}{2}< r< s < 1,$ and let $f_{s}(z) = f(sz),~~ z\in \mathbb {B}_{n}.$ By the definition, we have:

$$\begin{aligned} \Vert f - f_{s} \Vert _{\Lambda _{\gamma }(\mathbb {B}_{n},X)}&= \displaystyle \sup _{z \in \mathbb {B}_{n}}(1 - |z|^2)^{k-\gamma }\Vert R^{\alpha ,k}(f - f_{s})(z)\Vert _{X}\\&= \sup _{z \in \mathbb {B}_{n}}(1 - |z|^2)^{k-\gamma }\Vert (R^{\alpha ,k}f)(z) - (R^{\alpha ,k}f_{s})(z)\Vert _{X}\\&= \sup _{z \in \mathbb {B}_{n}}(1 - |z|^2)^{k-\gamma }\Vert (R^{\alpha ,k}f)(z) - (R^{\alpha ,k}f)(sz)\Vert _{X}\\&= \sup _{z \in \mathbb {B}_{n}}(1 - |z|^2)^{k-\gamma }\Vert (R^{\alpha ,k}f)(z) - \chi _{r}(z)(R^{\alpha ,k}f)(z)\\&\quad + \chi _{r}(z)(R^{\alpha ,k}f)(z) - (R^{\alpha ,k}f)(sz)\Vert _{X}\\&\le \sup _{z \in \mathbb {B}_{n}}(1 - |z|^2)^{k-\gamma }\Vert (R^{\alpha ,k}f)(z) - \chi _{r}(z)(R^{\alpha ,k}f)(z)\Vert _{X} \\&\quad + \sup _{z \in \mathbb {B}_{n}}(1 - |z|^2)^{k-\gamma }\Vert \chi _{r}(z)(R^{\alpha ,k}f)(z) - (R^{\alpha ,k}f)(sz)\Vert _{X}, \end{aligned}$$

where $\chi _{r}$ is the characteristic function of the set $\lbrace |z| \le r \rbrace .$ We first have the following estimate:

$$\begin{aligned}&\sup _{z \in \mathbb {B}_{n}}(1 - |z|^2)^{k-\gamma }\Vert (R^{\alpha ,k}f)(z) - \chi _{r}(z)(R^{\alpha ,k}f)(z)\Vert _{X} \\\le & {} \sup _{|z| \le r}(1 - |z|^2)^{k-\gamma }\Vert (R^{\alpha ,k}f)(z) - \chi _{r}(z)(R^{\alpha ,k}f)(z)\Vert _{X} \\&+ \sup _{r< |z|< 1}(1 - |z|^2)^{k-\gamma }\Vert (R^{\alpha ,k}f)(z) - \chi _{r}(z)(R^{\alpha ,k}f)(z)\Vert _{X}\\&= \sup _{r< |z|< 1}(1 - |z|^2)^{k-\gamma }\Vert (R^{\alpha ,k}f)(z)\Vert _{X}\\&\le \sup _{r^2< |z| < 1}(1 - |z|^2)^{k-\gamma }\Vert (R^{\alpha ,k}f)(z)\Vert _{X}. \end{aligned}$$

We secondly have the following estimate:

$$\begin{aligned}&\sup _{z \in \mathbb {B}_{n}}(1 - |z|^2)^{k-\gamma }\Vert \chi _{r}(z)(R^{\alpha ,k}f)(z) - (R^{\alpha ,k}f)(sz)\Vert _{X} \\\le & {} \sup _{|z| \le r}(1 - |z|^2)^{k-\gamma }\Vert \chi _{r}(z)(R^{\alpha ,k}f)(z) - (R^{\alpha ,k}f)(sz)\Vert _{X}\\&+ \sup _{r< |z|< 1}(1 - |z|^2)^{k-\gamma }\Vert \chi _{r}(z)(R^{\alpha ,k}f)(z) - (R^{\alpha ,k}f)(sz)\Vert _{X}\\&= \sup _{|z| \le r}(1 - |z|^2)^{k-\gamma }\Vert (R^{\alpha ,k}f)(z) - (R^{\alpha ,k}f)(sz)\Vert _{X}\\&+\sup _{r< |z| < 1}(1 - |z|^2)^{k-\gamma }\Vert (R^{\alpha ,k}f)(sz)\Vert _{X}. \end{aligned}$$

Using the change of variables $w = sz,$ we then obtain

$$\begin{aligned}&\sup _{r< |z|< 1}(1 - |z|^2)^{k-\gamma }\Vert (R^{\alpha ,k}f)(sz)\Vert _{X} \\&= \sup _{rs< |w|< s}\left( 1 - \frac{|w|^2}{s^2}\right) ^{k-\gamma }\Vert (R^{\alpha ,k}f)(w)\Vert _{X}\\&= \sup _{rs< |w|< s} \frac{1}{s^{2(k-\gamma )}}\left( s^{2} - |w|^2\right) ^{k-\gamma }\Vert (R^{\alpha ,k}f)(w)\Vert _{X}\\&\le 2^{2(k-\gamma )}\sup _{r^2< |w| < 1} (1 - |w|^2)^{k-\gamma }\Vert (R^{\alpha ,k}f)(w)\Vert _{X}. \end{aligned}$$

It follows by using the assumption that

$$\begin{aligned} \Vert f - f_{s} \Vert _{\Lambda _{\gamma }(\mathbb {B}_{n},X)}&\le C_{\gamma }\sup _{r^2< |w| < 1} (1 - |w|^2)^{k-\gamma }\Vert (R^{\alpha ,k}f)(w)\Vert _{X} \\&\quad + \sup _{|z| \le r}(1 - |z|^2)^{k-\gamma }\Vert (R^{\alpha ,k}f)(z) - (R^{\alpha ,k}f)(sz)\Vert _{X}, \end{aligned}$$

with $C_{\gamma } = 1+2^{2(k-\gamma )}.$ Since $(R^{\alpha ,k}f)(sz) \rightarrow (R^{\alpha ,k}f)(z)$ in X uniformly on the compact set $\lbrace |z| \le r \rbrace$ as $s\rightarrow 1^{-},$ we have

$$\begin{aligned} \lim _{s \rightarrow 1^{-}}\sup _{|z| \le r}(1 - |z|^2)^{k-\gamma }\Vert (R^{\alpha ,k}f)(z) - R^{\alpha ,k}f(sz)\Vert _{X} = 0. \end{aligned}$$

It follows that

$$\begin{aligned} \lim _{s \rightarrow 1^{-}} \Vert f - f_{s} \Vert _{\Lambda _{\gamma }(\mathbb {B}_{n},X)} \le C_{\gamma }\limsup _{|w|\rightarrow 1^{-}} (1 - |w|^2)^{k-\gamma }\Vert (R^{\alpha ,k}f)(w)\Vert _{X} = 0. \end{aligned}$$

$(ii) \Rightarrow (iii).$ Given $\epsilon > 0,$ by the assumption, there exists $s_{0} \in (0, 1)$ such that

$$\begin{aligned} \Vert f - f_{s_{0}}\Vert _{\Lambda _{\gamma }(\mathbb {B}_{n},X)} < \epsilon . \end{aligned}$$

(6.6)

Further note that $f_{s_{0}} \in \mathcal {H}(\frac{1}{s_{0}}\mathbb {B}_{n},X)$ and $1< \frac{2}{1+s_{0}} < \frac{1}{s_{0}}.$ From this, and by using Taylor’s formula, it follows that for each $m \in \mathbb {N},$ there exists a X-valued polynomial $p_m$ such that

$$\begin{aligned} \lim _{m \rightarrow \infty } \sup _{z \in \frac{2}{1+s_{0}}\overline{\mathbb {B}}_{n} }\Vert f_{s_{0}}(z) - p_{m}(z)\Vert _{X} = 0. \end{aligned}$$

Therefore, there exists $m_{0} \in \mathbb {N}$ such that

$$\begin{aligned} \sup _{z \in \frac{2}{1+s_{0}}\overline{\mathbb {B}}_{n} }\Vert f_{s_{0}}(z) - p_{m}(z)\Vert _{X} < \epsilon , \end{aligned}$$

(6.7)

for $m \ge m_{0}.$ By the Cauchy’s inequality, there exists a constant $c_{s_{0}} > 0$ such that for each $i = 1,\ldots ,n$ we have

$$\begin{aligned} \sup _{z \in \overline{\mathbb {B}}_{n}} \left\| \dfrac{\partial f_{s_{0}}}{\partial z_{i}} - \dfrac{\partial p_{m}}{\partial z_{i}}\right\| _{X} \le c_{s_{0}}\sup _{z \in \frac{2}{1+s_{0}}\overline{\mathbb {B}}_{n} }\Vert f_{s_{0}}(z) - P_{m}(z)\Vert _{X}. \end{aligned}$$

(6.8)

Suppose k is a nonnegative integer with $k > \gamma .$ By using (6.8) and Theorem 27, there is a constant $c = c(s_{0},n,\alpha , k)$ such that

$$\begin{aligned} \sup _{z \in \overline{\mathbb {B}}_{n}}\Vert (R^{\alpha ,k}f_{s_{0}})(z) - (R^{\alpha ,k}p_{m_{0}})(z)\Vert _{X} \le c \sup _{z \in \frac{2}{1+s_{0}}\overline{\mathbb {B}}_{n} }\Vert f_{s_{0}}(z) - P_{m_{0}}(z)\Vert _{X}.\nonumber \\ \end{aligned}$$

(6.9)

It follows by (6.9) and (6.7) that

$$\begin{aligned}&\sup _{z \in \mathbb {B}_{n}}(1- |z|^2)^{k-\gamma }\Vert R^{\alpha ,k}(f_{s_{0}} - p_{m_{0}})(z)\Vert _{X} \\&\le \sup _{z \in \mathbb {B}_{n}}\Vert (R^{\alpha ,k}f_{s_{0}})(z) - (R^{\alpha ,k}p_{m_{0}})(z)\Vert _{X}\\&\le c \sup _{z \in \frac{2}{1+s_{0}}\overline{\mathbb {B}}_{n} }\Vert f_{s_{0}}(z) - p_{m_{0}}(z)\Vert _{X}\\< & {} c\epsilon . \end{aligned}$$

Thus

$$\begin{aligned} \Vert f_{s_{0}} - p_{m_{0}}\Vert _{\Lambda _{\gamma }(\mathbb {B}_{n},X)} < c\epsilon . \end{aligned}$$

(6.10)

Using (6.6) and (6.10), it follows that

$$\begin{aligned} \Vert f - p_{m_{0}}\Vert _{\Lambda _{\gamma }(\mathbb {B}_{n},X)}&\le \Vert f - f_{s_{0}}\Vert _{\Lambda _{\gamma }(\mathbb {B}_{n},X)} + \Vert f_{s_{0}} - p_{m_{0}}\Vert _{\Lambda _{\gamma }(\mathbb {B}_{n},X)}\\< & {} \epsilon + c\epsilon = (1+c)\epsilon . \end{aligned}$$

$(iii) \Rightarrow (i).$ Let f in the closure of the set of vector-valued polynomial $\mathcal {P}(\mathbb {B}_{n},X),$ in $\Lambda _{\gamma }(\mathbb {B}_{n},X).$ There exists a sequence of vector-valued polynomials $\lbrace p_m \rbrace$ in $\mathcal {P}(\mathbb {B}_{n},X)$ such that

$$\begin{aligned} \lim _{m \rightarrow \infty }\Vert f - p_{m}\Vert _{\Lambda _{\gamma }(\mathbb {B}_{n},X)} = 0. \end{aligned}$$

(6.11)

Let us prove that for each $k > \gamma ,$

$$\begin{aligned} \lim _{|z| \rightarrow 1^{-}}(1 - |z|^2)^{k-\gamma }\Vert (R^{\alpha ,k}f)(z)\Vert _{X} = 0. \end{aligned}$$

Let $k > \gamma .$ We have that

$$\begin{aligned} \Vert (R^{\alpha ,k}f)(z)\Vert _{X}&\le \Vert (R^{\alpha ,k}f)(z) - (R^{\alpha ,k}p_{m})(z)\Vert _{X} + \Vert (R^{\alpha ,k}p_{m})(z)\Vert _{X}\\&\le \Vert (R^{\alpha ,k}f)(z) - (R^{\alpha ,k}p_{m})(z)\Vert _{X} + \Vert R^{\alpha ,k}p_{m}\Vert _{\infty ,X}, \end{aligned}$$

where $\Vert R^{\alpha ,k}p_{m}\Vert _{\infty ,X} = \max _{z \in \mathbb {B}_{n}}\Vert (R^{\alpha ,k}p_{m})(z)\Vert _{X}.$ It follows that for each $m \in \mathbb {N},$ we have

$$\begin{aligned}&(1 - |z|^2)^{k-\gamma }\Vert (R^{\alpha ,k}f)(z)\Vert _{X} \\&\le (1 - |z|^2)^{k-\gamma }\Vert (R^{\alpha ,k}f)(z) - (R^{\alpha ,k}p_{m})(z)\Vert _{X} + (1 - |z|^2)^{k-\gamma }\Vert R^{\alpha ,k}p_{m}\Vert _{\infty ,X} \\&\le \Vert f - p_{m} \Vert _{\Lambda _{\gamma }(\mathbb {B}_{n},X)} + (1 - |z|^2)^{k-\gamma }\Vert R^{\alpha ,k}p_{m}\Vert _{\infty ,X}. \end{aligned}$$

Letting $|z| \rightarrow 1^{-},$ we obtain that

$$\begin{aligned} \limsup _{|z| \rightarrow 1^{-}}(1 - |z|^2)^{k-\gamma }\Vert R^{\alpha ,k}f(z)\Vert _{X} \le \Vert f - p_{m} \Vert _{\Lambda _{\gamma }(\mathbb {B}_{n},X)}, \end{aligned}$$

for each $m \in \mathbb {N}.$ Now, letting $m \rightarrow \infty$ on both sides of the previous inequality, it follows by (6.11) that

$$\begin{aligned} \limsup _{|z| \rightarrow 1^{-}}(1 - |z|^2)^{k-\gamma }\Vert R^{\alpha ,k}f(z)\Vert _{X} = 0.\\ \end{aligned}$$

$\square$

Remark 39

One of the consequences of the previous result is that, given $\gamma \ge 0,$ the generalized little vector-valued Lipschitz space $\Lambda _{\gamma ,0}(\mathbb {B}_{n},X)$ is a closed subspace of the generalized vector-valued Lipschitz space $\Lambda _{\gamma }(\mathbb {B}_{n},X).$

From now on, we choose $\gamma _{0} = (n+1+\alpha )\left( \frac{1}{p} - \frac{1}{q}\right) ,$ with $1< p \le q < \infty ,$ and we consider the generalized vector-valued Lipschitz space $\Lambda _{\gamma _{0}}(\mathbb {B}_{n},X).$

Corollary 40

Suppose $1 \le t < \infty .$ Then $\Lambda _{\gamma _{0}}(\mathbb {B}_{n},X) \subset A^{t}_{\alpha }(\mathbb {B}_{n},X).$

Proof

Let $k > \gamma _{0}.$ Applying [11, Theorem 3.1.2], for $f \in \Lambda _{\gamma _{0}}(\mathbb {B}_{n},X),$ we have that

$$\begin{aligned} \Vert f\Vert ^{t}_{t,\alpha ,X}&\simeq \int _{\mathbb {B}_n} [(1 - |z|^2)^{k}\Vert R^{\alpha ,k}f(z)\Vert _{X}]^{t}\mathrm {d}\nu _{\alpha }(z)\\&= \int _{\mathbb {B}_n} [(1 - |z|^2)^{k-\gamma _{0}}\Vert R^{\alpha ,k}f(z)\Vert _{X}]^{t}(1 - |z|^2)^{\gamma _{0}t}\mathrm {d}\nu _{\alpha }(z)\\&\lesssim \Vert f\Vert _{\Lambda _{\gamma _{0}}(\mathbb {B}_{n},X)}\int _{\mathbb {B}_n}(1 - |z|^2)^{\alpha +\gamma _{0}t}\mathrm {d}\nu (z) < \infty . \end{aligned}$$

$\square$

In what follows, we assume that X, Y are reflexives complex Banach spaces. We first introduce the following proposition which will be used in the proof of Theorem 8.

Proposition 41

Suppose $1< p \le q < \infty ,$ $0 \le r < 1$ and $\gamma \in \mathbb {N}^{n}.$ If $a_{\gamma } \in \mathcal {K}(\overline{X},Y),$ then the little Hankel operator $h_{g^{\gamma }_{r}} : A^{p}_{\alpha }(\mathbb {B}_{n},X) \rightarrow A^{q}_{\alpha }(\mathbb {B}_{n},Y)$ is a compact operator, where $g^{\gamma }_{r}(z) = a_{\gamma }(rz)^{\gamma }$ for every $z \in \mathbb {B}_{n}.$

Proof

Let $\lbrace f_{j} \rbrace$ be a sequence in $A^{p}_{\alpha }(\mathbb {B}_{n},X)$ such that $f_{j} \rightarrow 0$ weakly in $A^{p}_{\alpha }(\mathbb {B}_{n},X)$ as j tends to infinity. We want to prove that $\lim _{j\rightarrow \infty }\Vert h_{g^{\gamma }_{r}}f_{j}\Vert _{q,\alpha ,Y} = 0.$ Let the Taylor expansion of $f_j$ given by $\displaystyle f_{j}(z) = \sum _{\beta \in \mathbb {N}^{n}} c^{j}_{\beta }z^{\beta } \in A^{p}_{\alpha }(\mathbb {B}_{n},X).$ Since $f_{j} \rightarrow 0$ weakly in $A^{p}_{\alpha }(\mathbb {B}_{n},X),$ applying Lemma 34, using the fact that $c^{j}_{\beta } = \partial ^{\beta } f_{j}(0)/\beta !,$ we have that for all $\beta \in \mathbb {N}^{n},$ $c^{j}_{\beta } \rightarrow 0$ weakly in X as $j \rightarrow \infty .$ By Lemma 32, for every $z \in \mathbb {B}_{n},$ we have

$$\begin{aligned} \displaystyle h_{g_{r}^{\gamma }}f_{j}(z) = \sum _{\beta \in \mathbb {N}^{n},\beta \le \gamma }a_{\gamma }(\overline{c^{j}_{\beta }}) \dfrac{\gamma !\Gamma (n+1+\alpha +|\gamma -\beta |)}{(\gamma -\beta )!\Gamma (n+1+\alpha +|\gamma |)} r^{|\gamma -\beta |} z^{\gamma -\beta }. \end{aligned}$$

Therefore,

$$\begin{aligned}&\Vert h_{g_{r}^{\gamma }}f_{j}\Vert _{q,\alpha ,Y} \\&= \left( \int _{\mathbb {B}_n} \left\| \sum _{\beta \in \mathbb {N}^{n},\beta \le \gamma }a_{\gamma }(\overline{c^{j}_{\beta }}) \dfrac{\gamma !\Gamma (n+1+\alpha +|\gamma -\beta |)}{(\gamma -\beta )!\Gamma (n+1+\alpha +|\gamma |)}(rz)^{\gamma -\beta }\right\| ^{q}_{Y} \mathrm {d}\nu _{\alpha }(z)\right) ^{1/q}\\&\le \left( \int _{\mathbb {B}_n} \left( \sum _{\beta \in \mathbb {N}^{n},\beta \le \gamma } \dfrac{\gamma !\Gamma (n+1+\alpha +|\gamma -\beta |) \Vert a_{\gamma }(\overline{c^{j}_{\beta }})\Vert _{Y} (r|z|)^{|\gamma -\beta |}}{(\gamma -\beta )!\Gamma (n+1+\alpha +|\gamma |)}\right) ^{q} \mathrm {d}\nu _{\alpha }(z)\right) ^{1/q}\\&\le \sum _{\beta \in \mathbb {N}^{n},\beta \le \gamma } \left( \int _{\mathbb {B}_n}\left( \Vert a_{\gamma }(\overline{c^{j}_{\beta }})\Vert _{Y} \dfrac{\gamma !\Gamma (n+1+\alpha +|\gamma -\beta |)}{(\gamma -\beta )!\Gamma (n+1+\alpha +|\gamma |)}(r|z|)^{|\gamma -\beta |}\right) ^q \mathrm {d}\nu _{\alpha }(z)\right) ^{\frac{1}{q}}\\&= \sum _{\beta \in \mathbb {N}^{n},\beta \le \gamma } \Vert a_{\gamma }(\overline{c^{j}_{\beta }})\Vert _{Y} \dfrac{\gamma !\Gamma (n+1+\alpha +|\gamma -\beta |)}{(\gamma -\beta )!\Gamma (n+1+\alpha +|\gamma |)} \left( \int _{\mathbb {B}_n} (r|z|)^{|\gamma -\beta |q} \mathrm {d}\nu _{\alpha }(z)\right) ^{\frac{1}{q}}\\&\lesssim \displaystyle \sum _{\beta \in \mathbb {N}^{n},\beta \le \gamma } \dfrac{\gamma !\Gamma (n+1+\alpha +|\gamma -\beta |)}{(\gamma -\beta )!\Gamma (n+1+\alpha +|\gamma |)}\Vert a_{\gamma }(\overline{c^{j}_{\beta }})\Vert _{Y}, \end{aligned}$$

where the third line above is justified by the Minkowsky’s inequality for integrals. Thus,

$$\begin{aligned} \displaystyle \Vert h_{g_{r}^{\gamma }}f_{j}\Vert _{q,\alpha ,Y} \lesssim \sum _{\beta \in \mathbb {N}^{n},\beta \le \gamma } \dfrac{\gamma !\Gamma (n+1+\alpha +|\gamma -\beta |)}{(\gamma -\beta )!\Gamma (n+1+\alpha +|\gamma |)} \Vert a_{\gamma }(\overline{c^{j}_{\beta }})\Vert _{Y}. \end{aligned}$$

(6.12)

Now, since $c^{j}_{\beta } \rightarrow 0$ weakly in X as $j\rightarrow \infty ,$ it is clear that $\overline{c^{j}_{\beta }} \rightarrow 0$ weakly in $\overline{X}$ as $j \rightarrow \infty .$ By the assumption, we know that $a_{\gamma } \in \mathcal {K}(\overline{X},Y).$ Since $\overline{c^{j}_{\beta }} \rightarrow 0$ weakly in $\overline{X}$ as $j \rightarrow \infty ,$ we have that $\Vert a_{\gamma }(\overline{c^{j}_{\beta }})\Vert _{Y} \rightarrow 0$ as $j \rightarrow \infty .$ It follows that

$$\begin{aligned}&\limsup _{j\rightarrow \infty } \Vert h_{g_{r}^{\gamma }}f_{j}\Vert _{q,\alpha ,Y} \lesssim \\&\quad \sum _{\beta \in \mathbb {N}^{n},\beta \le \gamma } \dfrac{\gamma !\Gamma (n+1+\alpha +|\gamma -\beta |)}{(\gamma -\beta )!\Gamma (n+1+\alpha +|\gamma |)} \lim _{j\rightarrow \infty } \Vert a_{\gamma }(\overline{c^{j}_{\beta }})\Vert _{Y} = 0. \end{aligned}$$

$\square$

Let us state Oliver’s result on the boundedness of the little Hankel operator with operator-valued symbol between vector-valued Bergman spaces.

Theorem 42

Let $1< p \le q < \infty .$ The little Hankel operator $h_{b}: A^{p}_{\alpha }(\mathbb {B}_{n},X) \rightarrow A^{q}_{\alpha }(\mathbb {B}_{n},Y)$ is a bounded operator if and only if $b \in \mathcal {B}_{\gamma }(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y)),$ where

$$\begin{aligned} \gamma = 1+(n+1+\alpha )\left( \frac{1}{q} - \frac{1}{p}\right) . \end{aligned}$$

Moreover

$$\begin{aligned} \Vert h_b\Vert _{A^{p}_{\alpha }(\mathbb {B}_{n},X) \rightarrow A^{q}_{\alpha }(\mathbb {B}_{n},Y)} \simeq \Vert b\Vert _{\mathcal {B}_{\gamma }(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y))}. \end{aligned}$$

Remark 43

Suppose $1< p< q < \infty ,$ and $\gamma = 1 + (n+1+\alpha )\left( \frac{1}{q} - \frac{1}{p}\right) .$ Then $\gamma$ is not always positive. Indeed, since $1/q-1/p \in (-1,0),$ then $\gamma \in (-n-\alpha , 1).$ It follows that when $\gamma \in (-(n+\alpha ),0),$ the vector-valued $\gamma$-Bloch space $\mathcal {B}_{\gamma }(\mathbb {B}_{n}, \mathcal {L}(\overline{X},Y))$ is not interesting and does not make sense since the definition of the vector-valued $\gamma$-Bloch space introduced by Oliver only takes into account the case where $\gamma > 0.$ In Theorem 7, we correct the problem by replacing the vector-valued $\gamma$-Bloch space with the generalized vector-valued Lipschitz space $\Lambda _{\gamma _{0}}(\mathbb {B}_{n}, \mathcal {L}(\overline{X},Y)),$ where $\gamma _{0} = (n+1+\alpha )\left( \frac{1}{p}-\frac{1}{q}\right) .$ Since $\gamma = 1 - \gamma _{0},$ we see that when $0< \gamma _{0} < 1,$ we have that

$$\begin{aligned} \mathcal {B}_{\gamma }(\mathbb {B}_{n}, \mathcal {L}(\overline{X},Y)) = \Lambda _{\gamma _{0}}(\mathbb {B}_{n}, \mathcal {L}(\overline{X},Y)). \end{aligned}$$

In what follows, we give the proof of Theorem 7 which generalize the Theorem 42 and correct the mistake mentionned in Remark 43.

6.3 Proof of Theorem 7

Let us recall the statement of Theorem 7.

Theorem 44

Suppose $1< p \le q <\infty .$ The little Hankel operator $h_{b}: A^p_{\alpha }(\mathbb {B}_{n},X) \rightarrow A^{q}_{\alpha }(\mathbb {B}_{n},Y)$ is a bounded operator if and only if

$b \in \Lambda _{\gamma _{0}}(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y)),$ where $\gamma _{0} = (n+1+\alpha )\left( \frac{1}{p} - \frac{1}{q}\right) .$ Moreover,

$$\begin{aligned} \Vert h_{b}\Vert _{A^p_{\alpha }(\mathbb {B}_{n},X) \rightarrow A^{q}_{\alpha }(\mathbb {B}_{n},Y)} \simeq \Vert b\Vert _{ \Lambda _{\gamma _{0}}(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y))}. \end{aligned}$$

Proof

Let $p'$ and $q'$ such that $1/p +1/p' = 1$ and $1/q + 1/q' = 1.$ We first assume that $h_{b}$ is a bounded operator from $A^p_{\alpha }(\mathbb {B}_{n},X)$ to $A^q_{\alpha }(\mathbb {B}_{n},Y)$ with norm $\Vert h_b\Vert = \Vert h_b\Vert _{A^p_{\alpha }(\mathbb {B}_{n},X) \rightarrow A^q_{\alpha }(\mathbb {B}_{n},Y)}.$ Let $x \in X$ and $k> (n+1+\alpha )/p.$ Let $z \in \mathbb {B}_{n}$ and put

$$\begin{aligned} f(w) = \dfrac{x}{(1-\langle w,z \rangle )^{k}}, \qquad w \in \mathbb {B}_{n}. \end{aligned}$$

Since $k> (n+1+\alpha )/p,$ by Theorem 18, we have that $f \in A^p_{\alpha }(\mathbb {B}_{n},X)$ and

$$\begin{aligned} \Vert f\Vert _{p,\alpha ,X} \lesssim \dfrac{\Vert x\Vert _{X}}{(1-|z|^2)^{k-(n+1+\alpha )/p}}. \end{aligned}$$

By [11, Proposition 2.1.3 ], we have that

$$\begin{aligned} h_{b}f(z)&= \displaystyle \int _{\mathbb {B}_n}\dfrac{b(w)(\overline{f(w)})}{(1 - \langle z,w \rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(w)\\&= \displaystyle \int _{\mathbb {B}_n}\dfrac{b(w)(\overline{x})}{(1 - \langle z,w \rangle )^{n+1+\alpha +k}}\mathrm {d}\nu _{\alpha }(w)\\&= R^{\alpha ,k}b(z)(\overline{x}). \end{aligned}$$

It follows by Theorem 14 that

$$\begin{aligned} \Vert R^{\alpha ,k}b(z)(\overline{x})\Vert _{Y}&= \Vert h_{b}f(z)\Vert _{Y}\\&\le \dfrac{\Vert h_{b}f \Vert _{q,\alpha ,Y}}{(1-|z|^2)^{(n+1+\alpha )/q}}\\&\le \dfrac{\Vert h_{b}\Vert \Vert f\Vert _{p,\alpha ,X}}{(1-|z|^2)^{(n+1+\alpha )/q}} \\&\lesssim \dfrac{\Vert h_{b}\Vert \Vert x\Vert _{X}}{(1-|z|^2)^{k+(n+1+\alpha )(1/q-1/p)}}\\&= \dfrac{\Vert h_{b}\Vert \Vert x\Vert _{X}}{(1-|z|^2)^{k-\gamma _{0}}}. \end{aligned}$$

Since $x \in X$ is arbitrary and $\Vert x\Vert _{X} = \Vert \overline{x}\Vert _{\overline{X}}$ we get that

$$\begin{aligned} \Vert R^{\alpha ,k}b(z)\Vert _{\mathcal {L}(\overline{X},Y)} \lesssim \dfrac{\Vert h_b\Vert }{(1-|z|^2)^{k-\gamma _{0}}}. \end{aligned}$$

Thus

$$\begin{aligned} \sup _{z \in \mathbb {B}_{n}} (1-|z|^2)^{k-\gamma _{0}}\Vert R^{\alpha ,k}b(z)\Vert _{\mathcal {L}(\overline{X},Y)} \lesssim \Vert h_b\Vert . \end{aligned}$$

By Lemma 37 this means that $b \in \Lambda _{\gamma _{0}}(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y))$ and $\Vert b\Vert _{\Lambda _{\gamma _{0}}(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y))} \lesssim \Vert h_{b}\Vert .$

Conversely, assume that $b \in \Lambda _{\gamma _{0}}(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y)).$ Let $f \in A^p_{\alpha }(\mathbb {B}_{n},X),$ $g \in A^{q'}_{\alpha }(\mathbb {B}_{n},Y^{\star })$ and $k > \gamma _{0}.$ By Corollary 40, we have that

$$\begin{aligned} b \in \Lambda _{\gamma _{0}}(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y)) \subset A^{p'}_{\alpha }(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y)), \end{aligned}$$

so by [11, Lemma 4.1.1], Corollary 30, and Lemma 37 it follows that

$$\begin{aligned} |\langle h_{b}f,g \rangle _{\alpha ,Y}|&= \displaystyle \left| \int _{\mathbb {B}_{n}} \langle b(z)\overline{f(z)},g(z) \rangle _{Y} \mathrm {d}\nu _{\alpha }(z)\right| \\&= \displaystyle \left| \int _{\mathbb {B}_n} \langle R^{\alpha ,k+1}b(z)\overline{f(z)},g(z) \rangle _{Y} {d}\nu _{\alpha +k+1}(z)\right| \\&\lesssim \displaystyle \int _{\mathbb {B}_n} \Vert R^{\alpha ,k+1}b(z)\Vert _{\mathcal {L}(\overline{X},Y)}\Vert \overline{f(z)}\Vert _{\overline{X}}\Vert g(z)\Vert _{Y^{\star }}(1-|z|^2)^{k+1+\alpha }\mathrm {d}\nu (z)\\&\lesssim \displaystyle \Vert b\Vert _{\Lambda _{\gamma _{0}}(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y))} \int _{\mathbb {B}_n} \Vert f(z)\Vert _{X}\Vert g(z)\Vert _{Y^{\star }}(1-|z|^2)^{\alpha +\gamma _{0}}\mathrm {d}\nu (z). \end{aligned}$$

By Hölder’s inequality the last integral is less than or equal to

$$\begin{aligned} \displaystyle \left( \int _{\mathbb {B}_{n}}\Vert f(z)\Vert ^q_{X}(1-|z|^2)^{\alpha +q\gamma _{0}}\mathrm {d}\nu (z) \right) ^{1/q} \left( \int _{\mathbb {B}_{n}}\Vert g(z)\Vert ^{q'}_{Y^{\star }}(1-|z|^2)^{\alpha }\mathrm {d}\nu (z) \right) ^{1/q'}. \end{aligned}$$

For $q=p$, we have $\gamma _0=0$ and thus

$$\begin{aligned} |\langle h_{b}f,g \rangle _{\alpha ,Y}|\lesssim \Vert b\Vert _{\Lambda _{\gamma _{0}}(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y))} \Vert f\Vert _{p,\alpha ,X} \Vert g\Vert _{p^\prime ,\alpha ,Y}. \end{aligned}$$

For $q-p > 0,$ using Theorem 14, we have

$$\begin{aligned} \Vert f(z)\Vert ^q_{X} = \Vert f(z)\Vert ^p_{X} \Vert f(z)\Vert ^{q-p}_{X} \le \dfrac{\Vert f(z)\Vert ^p_{X}\Vert f\Vert _{p,\alpha ,X}^{q-p} }{(1-|z|^2)^{(q-p)(n+1+\alpha )/p}} = \\ \dfrac{\Vert f(z)\Vert ^p_{X}\Vert f\Vert ^{q-p}_{p,\alpha ,X}}{(1-|z|^2)^{q\gamma _{0}}}. \end{aligned}$$

It follows that

$$\begin{aligned}&\left( \int _{\mathbb {B}_{n}}\Vert f(z)\Vert ^q_{X}(1-|z|^2)^{\alpha +q\gamma _{0}}\mathrm {d}\nu (z) \right) ^{1/q} \le \\&\quad \Vert f\Vert ^{1-p/q}_{p,\alpha ,X} \left( \int _{\mathbb {B}_{n}}\Vert f(z)\Vert ^{p}_{X}\dfrac{(1-|z|^2)^{\alpha +q\gamma _{0}}}{(1-|z|^2)^{q\gamma _{0}}}\mathrm {d}\nu (z) \right) ^{1/q} = \Vert f\Vert _{p,\alpha ,X}. \end{aligned}$$

Therefore, by duality, we obtain that

$$\begin{aligned}&\Vert h_{b}\Vert _{A^{p}_{\alpha }(\mathbb {B}_{n},X)\rightarrow A^{q}_{\alpha }(\mathbb {B}_{n},Y)} = \\&\quad \sup _{\Vert f\Vert _{p,\alpha ,X} = 1; \Vert g\Vert _{q',\alpha ,Y^{\star }}=1}\left| \langle h_{b}f,g \rangle _{\alpha ,Y} \right| \lesssim \Vert b\Vert _{\Lambda _{\gamma _{0}}(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y))}.\qquad \qquad \qquad \square \end{aligned}$$

6.4 Proof of Theorem 8

We are now ready to give the proof of the main result in this section that is Theorem 8 that we recall here.

Theorem 45

Let X and Y be two reflexive complex Banach spaces. Suppose that $1< p \le q < \infty ,$ and $\alpha >-1$ The little Hankel operator $h_b : A^{p}_{\alpha }(\mathbb {B}_{n},X) \longrightarrow A^{q}_{\alpha }(\mathbb {B}_{n},Y)$ is a compact operator if and only if

$$\begin{aligned} b \in \Lambda _{\gamma _{0},0}(\mathbb {B}_{n},\mathcal {K}(\overline{X},Y)), \end{aligned}$$

where $\Lambda _{\gamma _{0},0}(\mathbb {B}_{n},\mathcal {K}(\overline{X},Y))$ denotes the generalized little vector-valued Lipschitz space and $\gamma _{0} = (n+1+\alpha )\left( \frac{1}{p} - \frac{1}{q}\right) ,$ see (1.3).

Proof

First assume that $b \in \Lambda _{\gamma _{0},0}(\mathbb {B}_{n}, \mathcal {K}(\overline{X},Y))$ and denote by $b_{r}(z):= b(rz)$ with $z\in \mathbb {B}_{n}$ and $0< r < 1.$ Since $b \in \Lambda _{\gamma _{0},0}(\mathbb {B}_{n}, \mathcal {K}(\overline{X},Y)),$ by Theorem 7, we have that

$$\begin{aligned} \Vert h_{b}\Vert _{A^p_{\alpha }(\mathbb {B}_{n},X) \rightarrow A^q_{\alpha }(\mathbb {B}_{n},Y)} \lesssim \Vert b \Vert _{\Lambda _{\gamma _{0}}(\mathbb {B}_{n}, \mathcal {L}(\overline{X},Y))}. \end{aligned}$$

Therefore, we have

$$\begin{aligned} \Vert h_{b}-h_{b_{r}}\Vert _{A^p_{\alpha }(\mathbb {B}_{n},X) \rightarrow A^q_{\alpha }(\mathbb {B}_{n},Y)} \lesssim \Vert b - b_{r} \Vert _{\Lambda _{\gamma _{0}}(\mathbb {B}_{n}, \mathcal {L}(\overline{X},Y))}. \end{aligned}$$

By using Proposition 38, we have that

$$\begin{aligned} \displaystyle \lim _{r \rightarrow 1^{-}} \Vert b - b_r \Vert _{\Lambda _{\gamma _{0}}(\mathbb {B}_{n}, \mathcal {L}(\overline{X},Y))} = 0, \end{aligned}$$

so to prove that $h_b$ is a compact operator, it suffices to prove that $h_{b_{r}}$ is a compact operator. Since $b_r$ is analytic on a neighbourhood of $\overline{\mathbb {B}}_n,$ it can be approximated by its Taylor polynomial in the generalized vector-valued Lipschitz norm. Thus,

$$\begin{aligned} \lim _{N \rightarrow \infty } \Vert b_{r} - P_{N,r} \Vert _{\Lambda _{\gamma _{0}}(\mathbb {B}_{n}, \mathcal {L}(\overline{X},Y))} = 0, \end{aligned}$$

(6.13)

with $\displaystyle P_{N,r}(z) = \sum _{\beta \in \mathbb {N}^{n},|\beta | \le N} \hat{b}(\beta ) r^{|\beta |}z^{\beta },$ where $\hat{b}(\beta ) \in \mathcal {K}(\overline{X},Y)$ are the Taylor coefficients of b. We also have by Theorem 7 that

$$\begin{aligned} \Vert h_{b_{r}}-h_{P_{N,r}}\Vert _{A^p_{\alpha }(\mathbb {B}_{n},X) \rightarrow A^q_{\alpha }(\mathbb {B}_{n},Y)} \lesssim \Vert b_{r} - P_{N,r} \Vert _{\Lambda _{\gamma _{0}}(\mathbb {B}_{n}, \mathcal {L}(\overline{X},Y))}. \end{aligned}$$

So by (6.13), to prove that $h_{b_{r}}$ is a compact operator, it is enough to prove that $h_{P_{N,r}}$ is a compact operator. Since $P_{N,r}$ is a polynomial, it is enough to do the proof for monomials of the form $\hat{b}(\beta ) r^{|\beta |}z^{\beta },$ with $\beta \in \mathbb {N}^{n},$ $z \in \mathbb {B}_{n}$ and $\hat{b}(\beta ) \in K(\overline{X},Y).$ Thus, according to Proposition 41, the proof of this part is complete.

Conversely, for the “only if part”, let us assume that

$$\begin{aligned} h_{b}: A^p_\alpha (\mathbb {B}_{n},X) \longrightarrow A^q_\alpha (\mathbb {B}_{n},Y) \end{aligned}$$

is a compact operator. Since $h_b$ is compact, $h_b$ is then bounded and Theorem 7 yields

$$\begin{aligned} b \in \Lambda _{\gamma _{0}}(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y)). \end{aligned}$$

We shall first prove that the Taylor coefficients $\hat{b}(\beta ),$ $\beta \in \mathbb {N}^{n}$ of b belongs to $\mathcal {K}(\overline{X},Y).$ Let $\lbrace f_j \rbrace \subset X$ such that $f_{j}\longrightarrow 0$ weakly in X as $j\longrightarrow \infty ,$ fix $\beta _{0} \in \mathbb {N}^{n},$ and let $x_{j}(z) = z^{\beta _{0}}f_{j}.$ By Lemma 36, we have $\lbrace x_{j} \rbrace \subset A^p_{\alpha }(\mathbb {B}_{n},X)$ and $\lbrace x_{j} \rbrace$ converges weakly to 0 in $A^p_{\alpha }(\mathbb {B}_{n},X).$ Since

$$\begin{aligned} \Vert \hat{b}(\beta _{0})\overline{f_j}\Vert _{Y} = \sup _{\Vert y^{\star }\Vert _{Y^{\star }} = 1} |\langle \hat{b}(\beta _{0})\overline{f_{j}},y^{\star } \rangle _{Y,Y^{\star }}| \end{aligned}$$

and Y is reflexive, by the Kakutani’s theorem [9, Theorem 3.17] there exists $y^{\star }_{j} \in Y^{\star }$ with $\Vert y^{\star }_{j}\Vert _{Y^{\star }} = 1$ such that

$$\begin{aligned} \Vert \hat{b}(\beta _{0})\overline{f_j}\Vert _{Y} = |\langle \hat{b}(\beta _{0})\overline{f_{j}},y^{\star }_{j} \rangle _{Y,Y^{\star }}|. \end{aligned}$$

But $y^{\star }_{j} \in A^{p'}_{\alpha }(\mathbb {B}_{n},Y^{\star }).$ By Lemma 23, we have

$$\begin{aligned} |\langle h_{b}x_{j} , y^{\star }_{j} \rangle _{\alpha ,Y}|&= \displaystyle \left| \int _{\mathbb {B}_{n}} \langle b(z)\overline{x_{k}}(z),y^{\star }_{j} \rangle _{Y,Y^{\star }}\mathrm {d}\nu _{\alpha }(z)\right| \\&= \left| \displaystyle \int _{\mathbb {B}_{n}} \overline{z}^{\beta _{0}}\langle \sum _{\beta \in \mathbb {N}^{n}} z^{\beta } \hat{b}(\beta )\overline{f_{j}},y^{\star }_{j} \rangle _{Y,Y^{\star }}\mathrm {d}\nu _{\alpha }(z)\right| \\&= \displaystyle \left| \sum _{\beta \in \mathbb {N}^{n}}\langle \hat{b}(\beta )\overline{f_{j}},y^{\star }_{j} \rangle _{Y,Y^{\star }}\int _{\mathbb {B}_{n}}z^{\beta }\overline{z}^{\beta _{0}}\mathrm {d}\nu _{\alpha }(z)\right| \\&= \displaystyle |\langle \hat{b}(\beta _{0})\overline{f_{j}},y^{\star }_{j} \rangle _{Y,Y^{\star }}|\int _{\mathbb {B}_{n}}|z^{\beta _{0}}|^{2}\mathrm {d}\nu _{\alpha }(z)\\&= \dfrac{\beta _{0}!\Gamma (n+\alpha +1)}{\Gamma (n+|\beta _{0}|+\alpha +1)}|\langle \hat{b}(\beta _{0})\overline{f_{j}},y^{\star }_{j} \rangle _{Y,Y^{\star }}|\\&= \dfrac{\beta _{0}!\Gamma (n+\alpha +1)}{\Gamma (n+|\beta _{0}|+\alpha +1)}\Vert \hat{b}(\beta _{0})\overline{f_j}\Vert _{Y}, \end{aligned}$$

where Fubini’s theorem is justified by Lemma 31 with $\lbrace x_{j} \rbrace \subset H^{\infty }(\mathbb {B}_{n},X).$ Since $h_b$ is compact and $\lbrace x_{j} \rbrace$ converges weakly to 0 as j tends to infinity, we have that $\lbrace h_{b}x_{j} \rbrace$ converges strongly to 0 as j tends to infinity, therefore one gets that

$$\begin{aligned} \lim _{j\rightarrow \infty } \langle h_{b}x_{j} , y^{\star }_{j} \rangle _{\alpha ,Y} = 0. \end{aligned}$$

Thus

$$\begin{aligned} \lim _{j \rightarrow \infty }\dfrac{\beta _{0}!\Gamma (n+\alpha +1)}{\Gamma (n+|\beta _{0}|+\alpha +1)}\Vert \hat{b}(\beta _{0})\overline{f_j}\Vert _{Y} = 0. \end{aligned}$$

We then obtain

$$\begin{aligned} \lim _{j \rightarrow \infty } \Vert \hat{b}(\beta _{0})\overline{f_j}\Vert _{Y} = 0. \end{aligned}$$

In fact, we have shown that $\hat{b}(\beta _{0})$ belongs to $\mathcal {K}(\overline{X},Y)$ and as $\beta _{0}$ is arbitrary, this holds for all $\beta \in \mathbb {N}^{n}.$ Let $1< t < \infty .$ Since $b \in \Lambda _{\gamma _{0}}(\mathbb {B}_{n}, \mathcal {L}(\overline{X},Y)),$ we have that $b \in A^t_\alpha (\mathbb {B}_{n},\mathcal {L}(\overline{X},Y))$ and

$$\begin{aligned} \lim _{N \rightarrow \infty } \int _{\mathbb {B}_n} \Vert b(w) - \sum _{|\beta | \le N} \hat{b}(\beta )w^{\beta } \Vert ^t_{\mathcal {L}(\overline{X},Y))} \mathrm {d}\nu _{\alpha }(w) = 0. \end{aligned}$$

Let $z \in \mathbb {B}_n.$ There exists a constant $C_{z} > 0$ such that

$$\begin{aligned} \Vert b(z) - \sum _{|\beta | \le N} \hat{b}(\beta )z^{\beta } \Vert ^t_{\mathcal {L}(\overline{X},Y))} \le C_{z} \int _{\mathbb {B}_n} \Vert b(w) - \sum _{|\beta | \le N} \hat{b}(\beta )w^{\beta } \Vert ^t_{\mathcal {L}(\overline{X},Y))} \mathrm {d}\nu _{\alpha }(w). \end{aligned}$$

Thus,

$$\begin{aligned} \lim _{N \rightarrow \infty } \Vert b(z) - \sum _{|\beta | \le N} \hat{b}(\beta )z^{\beta } \Vert _{\mathcal {L}(\overline{X},Y))} = 0. \end{aligned}$$

Since $z \in \mathbb {B}_n$ is arbitrary, we deduce that $b(z) \in \mathcal {K}(\overline{X},Y),$ for each $z \in \mathbb {B}_{n}.$ It remains to show that b satisfy the “little $\gamma _{0}$- Lipschitz” condition. Let $x \in X$ and $y^{\star } \in Y^{\star }.$ Since $b \in \Lambda _{\gamma _{0}}(\mathbb {B}_{n},\mathcal {L}(\overline{X},Y)),$ then the mapping $z \mapsto \langle b(z)\overline{x},y^{\star } \rangle _{Y,Y^{\star }}$ belongs to $A^1_\alpha (\mathbb {B}_{n},\mathbb {C}).$ By using the reproducing kernel formula, it follows that

$$\begin{aligned} \displaystyle \langle b(z)\overline{x},y^{\star } \rangle _{Y,Y^{\star }} = \int _{\mathbb {B}_{n}} \dfrac{\langle b(w)\overline{x},y^{\star } \rangle _{Y,Y^{\star }}}{(1 - \langle z,w \rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(w). \end{aligned}$$

(6.14)

Let $k > \gamma _{0}.$ Applying the operator $R^{\alpha ,k}$ in (6.14), we obtain that

$$\begin{aligned} \displaystyle \langle R^{\alpha ,k} b(z)\overline{x},y^{\star } \rangle _{Y,Y^{\star }} = \int _{\mathbb {B}_{n}} \dfrac{\langle b(w)\overline{x},y^{\star } \rangle _{Y,Y^{\star }}}{(1 - \langle z,w \rangle )^{n+1+\alpha +k}}\mathrm {d}\nu _{\alpha }(w). \end{aligned}$$

(6.15)

Let $z \in \mathbb {B}_{n}.$ Since $\Vert R^{\alpha ,k} b(z) \Vert _{\mathcal {L}(\overline{X},Y)} = \sup _{\Vert x\Vert _{X} = 1}\Vert R^{\alpha ,k}b(z)(\overline{x})\Vert _{Y},$ and by Lemma 35, the operator $R^{\alpha ,k}b(z)$ is compact. So there exists $x_{0}(z) \in X$ with $\Vert x_{0}(z)\Vert _{X} = 1$ and

$$\begin{aligned} \Vert R^{\alpha ,k}b(z) \Vert _{\mathcal {L}(\overline{X},Y)} = \Vert R^{\alpha ,k}b(z)\overline{x_{0}(z)}\Vert _{Y}. \end{aligned}$$

Also

$$\begin{aligned} \Vert R^{\alpha ,k}b(z)\overline{x_{0}(z)}\Vert _{Y} = \sup _{\Vert y^{\star }\Vert _{Y^{\star }} = 1} |\langle R^{\alpha ,k} b(z)\overline{x_{0}(z)},y^{\star } \rangle _{Y,Y^{\star }}|. \end{aligned}$$

Since Y is reflexive, it follows by the Kakutani’s theorem [9, Theorem 3.17] that there exists $y^{\star }_{0}(z) \in Y^{\star }$ with $\Vert y_{0}^{\star }(z)\Vert _{Y^{\star }} = 1$ such that

$$\begin{aligned} \Vert R^{\alpha ,k}b(z) \Vert _{\mathcal {L}(\overline{X},Y)} = \Vert R^{\alpha ,k}b(z)(\overline{x_{0}(z)})\Vert _{Y} = |\langle R^{\alpha ,k} b(z)\overline{x_{0}(z)}, y_{0}^{\star }(z) \rangle _{Y,Y^{\star }}|.\nonumber \\ \end{aligned}$$

(6.16)

By (6.15) and (6.16) we get

$$\begin{aligned}&(1 - |z|^2)^{k-\gamma _{0}}\Vert R^{\alpha ,k}b(z)\Vert _{\mathcal {L}(\overline{X},Y)} = \\&\quad \left| \int _{\mathbb {B}_n} \langle b(w)\overline{x_{0}(z)},y^{\star }_{0}(z) \rangle _{Y,Y^{\star }} \dfrac{(1-|z|^2)^{k-\gamma _{0}}}{(1 - \langle z,w \rangle )^{n+1+\alpha +k}}\mathrm {d}\nu _{\alpha }(w)\right| = |\langle h_{b}x_{z}, y_{z}^{\star } \rangle _{\alpha ,Y}|, \end{aligned}$$

with

$$\begin{aligned} x_{z}(w) = \dfrac{x_{0}(z)(1 - |z|^2)^{\beta - (n+1+\alpha )/p}}{(1 - \langle w,z \rangle )^{\beta }}, \qquad w \in \mathbb {B}_{n} \end{aligned}$$

and

$$\begin{aligned} y_{z}^{\star }(w) = \dfrac{y^{\star }_{0}(z)(1 - |z|^2)^{k + (n+1+\alpha )/q- \beta }}{(1 - \langle w,z \rangle )^{n+1+\alpha +k-\beta }}, \qquad w \in \mathbb {B}_{n}, \end{aligned}$$

where $\beta$ is chosen such that

$$\begin{aligned} (n+1+\alpha )/p< \beta < k+ (n+1+\alpha )/q . \end{aligned}$$

By Theorem 18, we have $x_{z} \in A^p_{\alpha }(\mathbb {B}_{n},X),$ $y^{\star }_{z} \in A^{q'}_{\alpha }(\mathbb {B}_{n},Y^{\star }),$ and

$$\begin{aligned} \sup _{z \in \mathbb {B}_{n}} \Vert x_{z} \Vert _{p,\alpha ,X}< \infty , \qquad \sup _{z \in \mathbb {B}_{n}} \Vert y^{\star }_{z}\Vert _{q', \alpha , Y^{\star }} < \infty . \end{aligned}$$

Let us prove that

$$\begin{aligned} x_{z} \longrightarrow 0~~\text{ weakly } \text{ in }~~A^p_\alpha (\mathbb {B}_{n},X) ~~\text{ as }~~|z| \longrightarrow 1^{-}. \end{aligned}$$

(6.17)

Since

$$\begin{aligned} \sup _{z \in \mathbb {B}_{n}}\Vert x_{z}\Vert _{p,\alpha ,X} < \infty , \end{aligned}$$

to prove (6.17), by Lemma 33, it suffices to prove that

$$\begin{aligned} \langle x_{z},e_{w,a^{\star }} \rangle _{\alpha ,X} \longrightarrow 0~~\text{ as }~~|z| \longrightarrow 1^{-}, \end{aligned}$$

where for each $a^{\star } \in X^{\star }$ and $w \in \mathbb {B}_{n}$, we have

$$\begin{aligned} e_{w,a^{\star }}(\zeta ) = \dfrac{1}{(1 - \langle \zeta ,w \rangle )^{n+1+\alpha }}a^{\star }, \qquad \zeta \in \mathbb {B}_n. \end{aligned}$$

By using the definition of $e_{w,a^{\star }}$ and the reproducing kernel formula, it follows that

$$\begin{aligned} \displaystyle \langle x_{z},e_{w,a^{\star }} \rangle _{p,\alpha ,X}&= \displaystyle \int _{\mathbb {B}_{n}} \langle x_{z}(\zeta ),e_{w,a^{\star }}(\zeta ) \rangle _{X,X^{\star }}\mathrm {d}\nu _{\alpha }(\zeta )\\&= \displaystyle \int _{\mathbb {B}_{n}} \langle x_{z}(\zeta ),\dfrac{1}{(1-\langle \zeta ,w \rangle )^{n+1+\alpha }}a^{\star } \rangle _{X,X^{\star }}\mathrm {d}\nu _{\alpha }(\zeta )\\&= \displaystyle \big \langle \int _{\mathbb {B}_{n}}\dfrac{ x_{z}(\zeta )}{(1-\langle w,\zeta \rangle )^{n+1+\alpha }}\mathrm {d}\nu _{\alpha }(\zeta ),a^{\star } \big \rangle _{X,X^{\star }}\\&= \displaystyle \langle x_{z}(w),a^{\star } \rangle _{X,X^{\star }}. \end{aligned}$$

Therefore, we have

$$\begin{aligned} |\langle x_{z},e_{w,a^{\star }} \rangle _{p,\alpha ,X}|&= |\langle x_{z}(w),a^{\star } \rangle _{X,X^{\star }}|\\&= \left| \dfrac{(1-|z|^2)^{\beta -(n+1+\alpha )/p}}{(1-\langle w,z \rangle )^{\beta }} \langle x_{0}(z),a^{\star } \rangle _{X,X^{\star }} \right| \\&\le \dfrac{(1-|z|^2)^{\beta -(n+1+\alpha )/p}}{(1- |w|)^{\beta }}\Vert a^{\star }\Vert _{X^{\star }} \longrightarrow 0 \end{aligned}$$

as $|z| \longrightarrow 1^{-}.$ By using (6.17), the compactness of $h_{b}$ and the fact that

$$\begin{aligned} \sup _{z \in \mathbb {B}_{n}}\Vert y_{z}^{\star }\Vert _{q',\alpha ,Y^{\star }}< \infty , \end{aligned}$$

it follows that

$$\begin{aligned} \lim _{|z|\rightarrow 1^{-}} (1 - |z|^2)^{k-\gamma _{0}}\Vert R^{\alpha ,k} b(z)\Vert _{\mathcal {L}(\overline{X},Y)} = \lim _{|z|\rightarrow 1^{-}} |\langle h_{b}x_{z}, y_{z}^{\star } \rangle _{\alpha ,Y}| = 0, \end{aligned}$$

which completes the proof of the theorem. $\square$