The boundedness and compactness of a new class of linear operators from the weighted Bergman space to the weighted-type spaces on the unit ball are characterized.
Citation: Stevo Stević. Note on a new class of operators between some spaces of holomorphic functions[J]. AIMS Mathematics, 2023, 8(2): 4153-4167. doi: 10.3934/math.2023207
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The boundedness and compactness of a new class of linear operators from the weighted Bergman space to the weighted-type spaces on the unit ball are characterized.
By B we denote the open unit ball in Cn, S is the unit sphere in Cn, B(z,r) is the open ball centered at z and with radius r, dσ is the normalized rotation invariant measure on S, dV(z) is the Lebesgue measure, and dVα(z):=cα,n(1−|z|2)αdV(z), α>−1, where cα,n is the normalization constant such that Vα(B)=1. The linear space of holomorphic functions on B we denote by H(B), whereas S(B) denotes the class of holomorphic self-maps of B. The standard inner product between the vectors z,w∈Cn is denoted by ⟨z,w⟩, whereas |z|=√⟨z,z⟩ is the Euclidean norm in Cn. Many classical results on functions in H(B) can be found in [1]. If f∈C(B) is a positive function, then we call it a weight function, and the class of functions is denoted by W(B). If p,q∈N0, p≤q, then the notation j=¯p,q is an abbreviation for the notation j=p,p+1,…,q. If X is a Banach space, then by BX we denote the unit ball in X.
Each φ∈S(B) induces the composition operator Cφf(z)=f(φ(z)), whereas each u∈H(B) induces the multiplication operator Muf(z)=u(z)f(z). The radial derivative of f∈H(B) is defined by
ℜf(z)=n∑j=1zjDjf(z), |
where Djf(z)=∂f∂zj(z),j=¯1,n (if n=1, then we regard D1f:=Df=f′). There has been a huge interest in the operators and their products on subspaces of H(B). The first investigations have been mostly devoted to the case n=1. Beside the products of the operators Cφ and Mu, which have been studied a lot, there have been some investigations of the products of the operators D and Cφ. For some products of these and other concrete operators, see, for example, [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25] and the related references therein. The boundedness and compactness [26,27] of the operators have been predominately studied so far.
The weighted Bergman space Apα=Apα(B), p>0, α>−1, consists of all f∈H(B) such that
‖f‖Apα=(∫B|f(z)|pdVα(z))1/p<+∞, |
which for p≥1 is a norm on Apα. With the norm the space is Banach. For some results on the space and operators on it, see, e.g., [4,6,14,15,22,28,29,30,31].
If μ is a weight function, then the space of all f∈H(B) such that
‖f‖H∞μ=sup |
is called the weighted-type space and denoted by H_{\mu}^{\infty}(\mathbb{B}) = H_{\mu}^{\infty} , whereas the little weighted-type space is its closed subspace consisting of all f\in H({\mathbb{B}}) such that \lim_{|z|\rightarrow 1}\mu(z)|f(z)| = 0, and is denoted by H_{\mu, 0}^{\infty}(\mathbb{B}) = H_{\mu, 0}^{\infty} . There has been a huge interest in investigating the spaces, their generalizations, and linear operators on them, especially in the boundedness and compactness [2,11,13,19,23,31,32,33,34].
The product operator \Re^m_{u, {\varphi}} = M_{u}C_{\varphi}\Re^m was introduced in [35]. For some investigations in the direction, see also [36]. Motivated, among others, by our investigations in [14,15,16,35], I have introduced the operator
\begin{eqnarray} {\mathfrak S}^m_{\vec u, {\varphi}} = \sum\limits_{j = 0}^m M_{u_j}C_{\varphi}\Re^j = \sum\limits_{j = 0}^m\Re^j_{u_j, {\varphi}}, \end{eqnarray} | (1.1) |
where m\in {\mathbb N} , u_j\in H({\mathbb{B}}) , j = \overline{0, m} , and {\varphi}\in S({\mathbb{B}}) , and studied it, for example, in [37]. For some related studies see also [2,3].
This note continues some of our previous investigations (for example, the ones in [13,14,15,16,35,37]), by studying the boundedness and compactness of the operators {\mathfrak S}^m_{\vec u, {\varphi}}: A^p_{\alpha}\to H_\mu^\infty (or H_{\mu, 0}^{\infty} ), where p\ge 1 and {\alpha} > -1 .
By C we denote some positive constants independent of essential variables and functions which may differ from line to line, whereas a\lesssim b (resp. a\gtrsim b ) means that there is C > 0 such that a\leq Cb (resp. a\geq Cb ). If a\lesssim b and b\lesssim a , then we use the notation a\asymp b .
The first result is a standard Schwartz-type lemma [38].
Lemma 2.1. Assume p\ge 1 , {\alpha} > -1 , \mu\in W({\mathbb{B}}) , u_j\in H(\mathbb{B}) , j = \overline{0, m} , m\in\mathbb{N} , \varphi\in S(\mathbb{B}) , and that the operator {\mathfrak S}^m_{\vec u, {\varphi}}: A^p_{\alpha}\to H_\mu^\infty is bounded. Then, the operator is compact if and only if for every bounded sequence (f_k)_{k\in {\mathbb N}}\subset A^p_{\alpha} uniformly converging to zero on compacts of \mathbb{B} , we have
\lim\limits_{k\to+\infty}\|{\mathfrak S}^m_{\vec u, {\varphi}}f_{k}\|_{H_{\mu}^{\infty}} = 0. |
The following lemma was essentially proved in [39], so we omit the proof.
Lemma 2.2. A closed set K in H_{\mu, 0}^{\infty} is compact if and only if it is bounded and
\begin{align*} \lim\limits_{|z|\rightarrow 1} \sup\limits_{f\in K} \mu(z)|f(z)| = 0. \end{align*} |
The following lemma is well known (see [29]; for a less precise version see also [1]).
Lemma 2.3. Assume p\in(0, \infty) , {\alpha} > -1 , and f\in A_{\alpha}^p({\mathbb{B}}) ; Then,
\begin{align} |f(z)|\leq \frac {\|f\|_{A^p_{\alpha}}}{(1-|z|^2)^{\frac{n+{\alpha}+1}p}}, \; z\in{\mathbb{B}}. \end{align} | (2.1) |
Lemma 2.4. Assume p\in(0, \infty) , {\alpha} > -1 , and m\in\mathbb{N} . Then,
\begin{align} \big|\Re^{m}f(z)\big|\lesssim\frac{|z|}{(1-|z|^{2})^{\frac{n+{\alpha}+1}{p}+m}}\|f\|_{A_{\alpha}^p}, \end{align} | (2.2) |
for every f\in A_{\alpha}^p and z\in\mathbb{B} .
Proof. Note that it is enough to prove that for all f\in A_{\alpha}^p and z\in\mathbb{B} ,
\begin{align} |\Re^{m}f(z)|\lesssim\frac{|z|}{(1-|z|)^{\frac{n+{\alpha}+1}{p}+m}}\|f\|_{A_{\alpha}^p}. \end{align} | (2.3) |
Let r\in(0, 1) be fixed. Then, the Cauchy-Schwartz and Cauchy inequalities imply
\begin{align} |\Re f(z)|\lesssim|z|\frac{\sup\limits_{w\in B(z, r(1-|z|))}|f(w)|}{1-|z|}, \; z\in{\mathbb{B}}, \; f\in H({\mathbb{B}}). \end{align} | (2.4) |
Inequality (2.1) implies that
\begin{align} \sup\limits_{w\in B(z, r(1-|z|))}|f(w)|\lesssim\frac{\|f\|_{A_{\alpha}^p}}{[(1-r)(1-|z|)]^{\frac{n+{\alpha}+1}p}}. \end{align} | (2.5) |
Since r is fixed, by (2.4) and (2.5) we get
\begin{align} |\Re f(z)|\lesssim\frac{|z|}{(1-|z|)^{\frac{n+{\alpha}+1}p+1}}\|f\|_{A_{\alpha}^p}, \end{align} | (2.6) |
that is, (2.3) holds when m = 1 .
Assume that for a k\in {\mathbb N}\setminus\{1\} and all f\in A_{\alpha}^p and z\in\mathbb{B} holds,
\begin{align} |\Re^{k-1} f(z)|\lesssim\frac{|z|}{(1-|z|)^{\frac{n+{\alpha}+1}p+k-1}}\|f\|_{A_{\alpha}^p}. \end{align} | (2.7) |
Then, since for w\in B(z, r(1-|z|)) we have (1-r)^{\frac{n+{\alpha}+1}p+k-1}(1-|z|)^{\frac{n+{\alpha}+1}p+k-1}\leq (1-|w|)^{\frac{n+{\alpha}+1}p+k-1}, from (2.7) we have
\begin{align} \sup\limits_{w\in B(z, r(1-|z|))}|\Re^{k-1}f(w)|\lesssim\frac1{(1-|z|)^{\frac{n+{\alpha}+1}p+k-1}}\|f\|_{A_{\alpha}^p}. \end{align} | (2.8) |
If in (2.4) we replace f by \Re^{k-1}f , we get
\begin{align} |\Re^{k} f(z)|\lesssim|z|\frac{\sup\nolimits_{w\in B(z, r(1-|z|))}|\Re^{k-1}f(w)|}{1-|z|}. \end{align} | (2.9) |
Combining (2.8) and (2.9), we have
\begin{align*} |\Re^{k} f(z)|\lesssim\frac{|z|}{(1-|z|)^{\frac{n+{\alpha}+1}p+k}}\|f\|_{A_{\alpha}^p}. \end{align*} |
Thus, (2.3) holds for each m\in {\mathbb N}, implying (2.2).
The following lemma is well known.
Lemma 2.5. Let p\ge 1 and {\alpha} > -1 . Then, for any t\geq0 and w\in\mathbb{B} ,
\begin{eqnarray} f_{w, t}(z): = \frac{(1-|w|^{2})^{t+1}}{(1-\langle z, w\rangle)^{\frac{n+{\alpha}+1}p+t+1}}, \end{eqnarray} | (2.10) |
belongs to A_{\alpha}^p and \sup_{w\in\mathbb{B}}\|f_{w, t}\|_{A_{\alpha}^p}\lesssim 1.
The following lemma is from [34] and [35].
Lemma 2.6. Let s\geq 0 , w\in{\mathbb{B}} and g_{w, s}(z) = (1-\langle z, w\rangle)^{-s}. Then,
\begin{align} \Re^{k}g_{w, s}(z) = s\frac{P_k(\langle z, w\rangle)}{(1-\langle z, w\rangle)^{s+k}}, \end{align} | (2.11) |
where P_k(w) = s^{k-1}w^k+p_{k-1}^{(k)}(s)w^{k-1}+\cdots+p_2^{(k)}(s)w^2+w, and where p^{(k)}_j(s) , j = \overline{2, k-1} , are nonnegative polynomials for s > 0 ;
\begin{align} \Re^{k}g_{w, s}(z) = \sum\limits_{t = 1}^{k}a_{t}^{(k)}\bigg(\prod\limits_{j = 0}^{t-1}(s+j)\bigg)\frac{\langle z, w\rangle^{t}}{(1-\langle z, w\rangle)^{s+t}}, \end{align} | (2.12) |
where (a_{t}^{(k)}) , t = \overline{1, k} , k\in\mathbb{N}, are defined as
\begin{eqnarray} a_{1}^{(k)} = a_{k}^{(k)} = 1, \; k\in {\mathbb N}; \end{eqnarray} | (2.13) |
and for 2\leq t\leq k-1, k\geq3 ,
\begin{eqnarray} a_{t}^{(k)} = ta_{t}^{(k-1)}+a_{t-1}^{(k-1)}. \end{eqnarray} | (2.14) |
Lemma 2.7. Assume p\ge 1 , {\alpha} > -1 , m\in\mathbb{N} , w\in{\mathbb{B}} , f_{w, t} is defined in (2.10), and (a_{t}^{(k)})_{t = \overline{1, k}} , k = \overline{1, m} , are defined in (2.13) and (2.14). Then,
(a) for each l\in\{1, \ldots, m\} , there is
\begin{eqnarray} h_{w}^{(l)}(z) = \sum\limits_{k = 0}^{m}c_{k}^{(l)}f_{w, k}(z), \end{eqnarray} | (2.15) |
where c_{k}^{(l)} , k = \overline{0, m} , are numbers, such that
\begin{equation} \Re^{j}h_{w}^{(l)}(w) = 0, \; 0\leq j < l, \end{equation} | (2.16) |
\begin{equation} \Re^{j}h_{w}^{(l)}(w) = a_l^{(j)}\frac{|w|^{2l}}{(1-|w|^{2})^{\frac{n+{\alpha}+1}p+l}}, \; l\leq j\leq m, \end{equation} | (2.17) |
hold. Moreover, we have \sup_{w\in\mathbb{B}}\|h_{w}^{(l)}\|_{A_{\alpha}^p} < +\infty;
(b) there is
\begin{eqnarray} h_{w}^{(0)}(z) = \sum\limits_{k = 0}^{m}c_{k}^{(0)}f_{w, k}(z), \end{eqnarray} | (2.18) |
where c_k^{(0)} , k = \overline{0, m} , are numbers, such that
h_{w}^{(0)}(w) = \frac1{(1-|w|^{2})^{\frac{n+{\alpha}+1}p}}, \; \Re^{j}h_{w}^{(0)}(w) = 0, \; j = \overline{1, m}, |
hold. Moreover, we have \sup_{w\in\mathbb{B}}\|h_{w}^{(0)}\|_{A_{\alpha}^p} < +\infty.
Proof. (a) Let d_{k} = \frac{n+{\alpha}+1}p+k+1 , k\in {\mathbb N}_0 . Replace the constants c_{k}^{(l)} in (2.15) by c_k . Then, from (2.12) we get
\begin{align} h_{w}^{(l)}(w)& = \frac{c_0+c_1+\cdots+c_m}{(1-|w|^2)^{\frac{n+{\alpha}+1}p}}, \\ \Re h_{w}^{(l)}(w)& = \frac{(d_0c_0+d_1c_1+\cdots+d_mc_m)|w|^2}{(1-|w|^2)^{\frac{n+{\alpha}+1}p+1}}, \\ &\qquad\qquad\vdots \\ \Re^m h_{w}^{(l)}(w)& = a_1^{(m)}\frac{(d_0c_0+d_1c_1+\cdots+d_mc_m)|w|^2}{(1-|w|^2)^{\frac{n+{\alpha}+1}p+1}}+\cdots\\ &+a_l^{(m)}\frac{(d_0\cdots d_{l-1}c_0+d_1\cdots d_lc_1+\cdots+d_m\cdots d_{m+l-1}c_m)|w|^{2l}}{(1-|w|^2)^{\frac{n+{\alpha}+1}p+l}}+\cdots\\ &+a_m^{(m)}\frac{(d_0\cdots d_{m-1}c_0+d_1\cdots d_mc_1+\cdots+d_m\cdots d_{2m-1}c_m)|w|^{2m}}{(1-|w|^2)^{\frac{n+{\alpha}+1}p+m}}. \end{align} | (2.19) |
Lemma 2.5 in [11] shows that the determinant of the system,
\begin{equation} \left[ \begin{array}{cccc} 1 & 1 &\cdots & 1 \\ d_{0} & d_{1} &\cdots & d_{m} \\ \vdots &\vdots & &\vdots \\ \prod\limits_{k = 0}^ld_{k}& \prod\limits_{k = 0}^ld_{k+1}&\cdots & \prod\limits_{k = 0}^ld_{m+k} \\ \vdots &\vdots & &\vdots \\ \prod\limits_{k = 0}^{m-1}d_{k} & \prod\limits_{k = 0}^{m-1}d_{k+1} &\cdots & \prod\limits_{k = 0}^{m-1}d_{m+k} \end{array} \right] \left[ \begin{array}{cccc} c_{0}\\ c_{1}\\ \qquad\\ \qquad\\ \vdots\\ \qquad\\ \qquad\\ \qquad\\ c_{m} \end{array} \right] = \left[ \begin{array}{cccc} 0\\ 0\\ \vdots\\ 0\\ 1\\ 0\\ \vdots\\ 0 \end{array} \right], \end{equation} | (2.20) |
is different from zero (on the right-hand side of (2.20), the unit is in the (l+1) th position). Thus, there is a unique solution c_k = c_k^{(l)} , k = \overline{0, m}, to (2.20). For these c_k -s, function (2.15) satisfies (2.16) and (2.17). By Lemma 2.5 we have \sup_{w\in\mathbb{B}}\|h_{w}^{(l)}\|_{A_{\alpha}^p} < +\infty.
(b) The proof is similar, so it is omitted.
Our main results are formulated and proved in this section.
Theorem 3.1. Let p\ge 1 , {\alpha} > -1 , k\in {\mathbb N} , u\in H(\mathbb{B}) , \varphi\in S(\mathbb{B}) and \mu\in W({\mathbb{B}}) . Then, the operator \Re^k_{u, {\varphi}}: A_{\alpha}^p\to H_{\mu}^{\infty} is bounded if and only if
\begin{align} J_k: = \sup\limits_{z\in\mathbb{B}}\frac{\mu(z)|u(z)||{\varphi}(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+{\alpha}+1}p+k}} < +\infty, \end{align} | (3.1) |
and if it is bounded, then we have
\begin{align} \|\Re^k_{u, {\varphi}}\|_{A_{\alpha}^p\rightarrow H_{\mu}^{\infty}}\asymp J_k. \end{align} | (3.2) |
Proof. Assume \Re^k_{u, {\varphi}}: A_{\alpha}^p\to H_{\mu}^{\infty} is bounded. Let g_w(z) = f_{{\varphi}(w), 1}(z) . By Lemma 2.6 the coefficients of the polynomial P_k therein are nonnegative, so we have
\begin{align} s\frac{\mu(w)|u(w)||{\varphi}(w)|^2}{(1-|{\varphi}(w)|^2)^{\frac{n+{\alpha}+1}p+k}}\le s\frac{\mu(w)|u(w)|P_k(|{\varphi}(w)|^2)}{(1-|{\varphi}(w)|^2)^{\frac{n+{\alpha}+1}p+k}}\le \|\Re^k_{u, {\varphi}}g_w\|_{H^\infty_\mu}. \end{align} | (3.3) |
The boundedness, (3.3) and the fact \sup_{w\in{\mathbb{B}}}\|g_w\|_{A_{\alpha}^p} < +\infty , imply
\begin{align} \sup\limits_{|{\varphi}(z)| > 1/2}\frac{\mu(z)|u(z)||{\varphi}(z)|}{(1-|{\varphi}(z)|^2)^{\frac{n+{\alpha}+1}p+k}}\lesssim\|\Re^k_{u, {\varphi}}\|_{A_{\alpha}^p\to H^\infty_\mu}. \end{align} | (3.4) |
Further, the fact f_{j}(z) = z_{j}\in A_{\alpha}^p , j = \overline{1, n} , implies \Re^k_{u, {\varphi}}f_j\in H_\mu^\infty , j = \overline{1, n} , from which, together with \Re f_j = f_j, j = \overline{1, n}, we get
\begin{align*} \sup\limits_{z\in{\mathbb{B}}}\mu(z)|u(z)||{\varphi}_j(z)| = \|\Re^k_{u, {\varphi}}f_j\|_{H_\mu^\infty} \leq\|\Re^k_{u, {\varphi}}\|_{A_{\alpha}^p\to H^\infty_\mu}\|z_j\|_{A_{\alpha}^p}, \; j = \overline{1, n}, \end{align*} |
from which we get
\begin{align} \sup\limits_{z\in{\mathbb{B}}}\mu(z)|u(z)||{\varphi}(z)|\lesssim\|\Re^k_{u, {\varphi}}\|_{A_{\alpha}^p\to H^\infty_\mu}. \end{align} | (3.5) |
Inequality (3.5) together with
\sup\limits_{|{\varphi}(z)|\leq1/2}\frac{\mu(z)|u(z)||{\varphi}(z)|}{(1-|{\varphi}(z)|^2)^{\frac{n+{\alpha}+1}p+k}} \lesssim\sup\limits_{|{\varphi}(z)|\leq1/2}\mu(z)|u(z)||{\varphi}(z)|, |
implies
\begin{align} \sup\limits_{|{\varphi}(z)|\leq1/2}\frac{\mu(z)|u(z)||{\varphi}(z)|}{(1-|{\varphi}(z)|^2)^{\frac{n+{\alpha}+1}p+k}} \lesssim\|\Re^k_{u, {\varphi}}\|_{A_{\alpha}^p\to H^\infty_\mu}. \end{align} | (3.6) |
Combining (3.4) and (3.6), we get (3.1) and J_k\lesssim\|\Re^k_{u, {\varphi}}\|_{A_{\alpha}^p\to H^\infty_\mu}.
Assume (3.1) holds. Then, Lemma 2.4 implies that for any f\in A_{\alpha}^p(\mathbb{B}) and z\in\mathbb{B} ,
\begin{align} \mu(z)\big|\Re^k_{u, {\varphi}}f(z)\big|\lesssim\frac{\mu(z)|u(z)||\varphi(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+{\alpha}+1}p+k}}\|f\|_{A_{\alpha}^p}. \end{align} | (3.7) |
Taking the supremum in (3.7) over B_{A_{\alpha}^p} , and employing (3.1), the boundedness of \Re^k_{u, {\varphi}}:A_{\alpha}^p\rightarrow H_{\mu}^{\infty} and the relation \|\Re^k_{u, {\varphi}}\|_{A_{\alpha}^p\rightarrow H_{\mu}^{\infty}}\lesssim J_k follow, implying (3.2).
The following result is known. For a more general result, see [31].
Theorem 3.2. Let p\ge 1 , {\alpha} > -1 , \mu\in W({\mathbb{B}}) , u\in H(\mathbb{B}) and \varphi\in S(\mathbb{B}) . Then, the operator \Re^0_{u, {\varphi}}: A_{\alpha}^p\to H_{\mu}^{\infty} is bounded if and only if
\begin{align} J_0 = :\sup\limits_{z\in\mathbb{B}}\frac{\mu(z)|u(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+{\alpha}+1}p}} < +\infty, \end{align} | (3.8) |
and if it is bounded, then \|\Re^0_{u, {\varphi}}\|_{A_{\alpha}^p\rightarrow H_{\mu}^{\infty}}\asymp J_0.
Theorem 3.3. Let p\ge 1 , {\alpha} > -1 , m\in\mathbb{N} , u_j\in H(\mathbb{B}) , j = \overline{0, m} , \varphi\in S(\mathbb{B}) and \mu\in W({\mathbb{B}}) . Then, the operators \Re^j_{u_j, {\varphi}}:A_{\alpha}^p\to H_{\mu}^{\infty} , j = \overline{0, m} , are bounded if and only if {\mathfrak S}^m_{\vec u, {\varphi}}:A_{\alpha}^p\rightarrow H_{\mu}^{\infty} is bounded and
\begin{eqnarray} \sup\limits_{z\in{\mathbb{B}}}\mu(z)|u_j(z)||{\varphi}(z)| < +\infty, \; j = \overline{1, m}. \end{eqnarray} | (3.9) |
Proof. Assume {\mathfrak S}^m_{\vec u, {\varphi}}:A_{\alpha}^p\rightarrow H_{\mu}^{\infty} is bounded and (3.9) holds. We need to prove
\begin{align} I_j = \sup\limits_{z\in\mathbb{B}}\frac{\mu(z)|u_j(z)||{\varphi}(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+{\alpha}+1}p+j}} < +\infty, \; j = \overline{1, m}, \end{align} | (3.10) |
and
\begin{align} I_0 = \sup\limits_{z\in\mathbb{B}}\frac{\mu(z)|u_0(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+{\alpha}+1}p}} < +\infty. \end{align} | (3.11) |
If {\varphi}(w)\ne0 , then there is h_{{\varphi}(w)}^{(m)}\in A_{\alpha}^p such that
\Re^{j}h_{{\varphi}(w)}^{(m)}({\varphi}(w)) = 0, \; 0\leq j < m, \; \Re^{m}h_{{\varphi}(w)}^{(m)}({\varphi}(w)) = \frac{|{\varphi}(w)|^{2m}}{(1-|{\varphi}(w)|^{2})^{\frac{n+{\alpha}+1}p+m}}, |
and \sup_{w\in\mathbb{B}}\|h_{{\varphi}(w)}^{(m)}\|_{A_{\alpha}^p} < +\infty (see Lemma 2.7 (a)). This, together with the boundedness of {\mathfrak S}^m_{\vec u, {\varphi}}:A_{\alpha}^p\rightarrow H_{\mu}^{\infty} , implies
\begin{align} \|{\mathfrak S}^m_{\vec u, {\varphi}}\|_{A_{\alpha}^p\rightarrow H_{\mu}^{\infty}} &\gtrsim\|{\mathfrak S}^m_{\vec u, {\varphi}}h_{\varphi(w)}^{(m)}\|_{H_{\mu}^{\infty}} \geq\mu(w)\bigg|\sum\limits_{j = 0}^{m}u_j(w)\Re^{j}h_{\varphi(w)}^{(m)}(\varphi(w))\bigg|\\ & = \frac{\mu(w)|u_m(w)||{\varphi}(w)|^{2m}}{(1-|{\varphi}(w)|^{2})^{\frac{n+{\alpha}+1}p+m}}, \end{align} | (3.12) |
from which it follows that
\sup\limits_{|{\varphi}(z)| > 1/2}\frac{\mu(z)|u_m(z)||{\varphi}(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+{\alpha}+1}p+m}}\lesssim\|{\mathfrak S}^m_{\vec u, {\varphi}}\|_{A_{\alpha}^p\rightarrow H_{\mu}^{\infty}}, |
and along with
\begin{align*} \sup\limits_{|{\varphi}(z)|\leq1/2}\frac{\mu(z)|u_m(z)||\varphi(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+{\alpha}+1}p+m}}\lesssim\sup\limits_{z\in{\mathbb{B}}}\mu(z)|u_m(z)||\varphi(z)| < +\infty, \end{align*} |
implies I_m < +\infty .
Assume (3.10) holds for j = \overline{s+1, m} , for an s\in\{1, 2, \ldots, m-1\} . Let h_{{\varphi}(w)}^{(s)}(z) be as in Lemma 2.7 (a). Then, \sup_{w\in{\mathbb{B}}}\|h_{{\varphi}(w)}^{(s)}\|_{A_{\alpha}^p} < +\infty , and
\begin{align} \mu(w)\bigg|\sum\limits_{j = s}^{m}a_s^{(j)}u_j(w)\frac{|{\varphi}(w)|^{2s}}{(1-|\varphi(w)|^{2})^{\frac{n+{\alpha}+1}p+s}}\bigg|&\le \sup\limits_{z\in\mathbb{B}}\mu(z)\Big|\sum\limits_{j = 0}^{m}u_j(z)\Re^jh_{\varphi(w)}^{(s)}(\varphi(z))\Big|\\ &\lesssim\|{\mathfrak S}^m_{\vec u, {\varphi}}\|_{A_{\alpha}^p\rightarrow H_{\mu}^{\infty}}, \end{align} |
from which we easily get
\begin{eqnarray} \frac{\mu(w)|u_s(w)||{\varphi}(w)|^{2s}}{(1-|\varphi(w)|^{2})^{\frac{n+{\alpha}+1}p+s}}\lesssim\|{\mathfrak S}^m_{\vec u, {\varphi}}\|_{A_{\alpha}^p\rightarrow H_{\mu}^{\infty}} +\sum\limits_{j = s+1}^m\frac{\mu(w)|u_j(w)||{\varphi}(w)|^{2s}}{(1-|\varphi(w)|^{2})^{\frac{n+{\alpha}+1}p+s}}. \end{eqnarray} | (3.13) |
From (3.13) and the fact s\ge 1 , we have
\begin{align} \sup\limits_{|{\varphi}(z)| > 1/2}\frac{\mu(z)|u_s(z)||{\varphi}(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+{\alpha}+1}p+s}} &\lesssim\|{\mathfrak S}^m_{\vec u, {\varphi}}\|_{A_{\alpha}^p\rightarrow H_{\mu}^{\infty}} +\sum\limits_{j = s+1}^m\sup\limits_{|{\varphi}(z)| > 1/2}\frac{\mu(z)|u_j(z)||{\varphi}(z)|^{2s}}{(1-|\varphi(z)|^{2})^{\frac{n+{\alpha}+1}p+j}}\\ &\le \|{\mathfrak S}^m_{\vec u, {\varphi}}\|_{A_{\alpha}^p\rightarrow H_{\mu}^{\infty}} +\sum\limits_{j = s+1}^mI_j. \end{align} |
This, together with the fact
\begin{align*} \sup\limits_{|{\varphi}(z)|\leq 1/2}\frac{\mu(z)|u_s(z)||{\varphi}(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+{\alpha}+1}p+s}} \lesssim\sup\limits_{z\in{\mathbb{B}}}\mu(z)|u_s(z)||{\varphi}(z)| < +\infty, \end{align*} |
implies (3.10) for j = s . Thus, (3.10) holds for any j\in\{1, \ldots, m\} .
For any w\in{\mathbb{B}} , there is h_{{\varphi}(w)}^{(0)}\in A_{\alpha}^p such that
h_{{\varphi}(w)}^{(0)}({\varphi}(w)) = \frac1{(1-|{\varphi}(w)|^2)^{\frac{n+{\alpha}+1}p}}, \; \Re^j h_{{\varphi}(w)}^{(0)}({\varphi}(w)) = 0, \; j = \overline{1, m}, |
and \sup_{w\in\mathbb{B}}\|h_{{\varphi}(w)}^{(0)}\|_{A_{\alpha}^p} < +\infty (see Lemma 2.7 (b)).
This together with the boundedness of {\mathfrak S}^m_{\vec u, {\varphi}}:A_{\alpha}^p\rightarrow H_{\mu}^{\infty} implies
\begin{align} \frac{\mu(w)|u_0(w)|}{(1-|{\varphi}(w)|^{2})^{\frac{n+{\alpha}+1}p}}\le\|{\mathfrak S}^m_{\vec u, {\varphi}}h_{\varphi(w)}^{(0)}\|_{H_{\mu}^{\infty}}\lesssim \|{\mathfrak S}^m_{\vec u, {\varphi}}\|_{A_{\alpha}^p\rightarrow H_{\mu}^{\infty}}, \end{align} | (3.14) |
from which (3.11) follows, as claimed.
Assume \Re^j_{u_j, {\varphi}}:A_{\alpha}^p\to H_{\mu}^{\infty} , j = \overline{0, m} , are bounded. Then, {\mathfrak S}^m_{\vec u, {\varphi}}:A_{\alpha}^p\to H_{\mu}^{\infty} is also bounded. If u in (3.5) is replaced by u_j , we get (3.9).
Theorem 3.4. Let p\ge 1 , {\alpha} > -1 , k\in {\mathbb N} , u\in H(\mathbb{B}) , \varphi\in S(\mathbb{B}) and \mu\in W({\mathbb{B}}) . Then, the operator \Re^k_{u, {\varphi}}: A_{\alpha}^p\to H_{\mu}^{\infty} is compact if and only if it is bounded and
\begin{align} \lim\limits_{|{\varphi}(z)|\to1}\frac{\mu(z)|u(z)||{\varphi}(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+{\alpha}+1}p+k}} = 0. \end{align} | (3.15) |
Proof. If \Re^k_{u, {\varphi}}: A_{\alpha}^p\to H_{\mu}^{\infty} is compact, it is also bounded. If \|\varphi\|_{\infty} < 1 , (3.15) automatically/vacuously holds. If \|\varphi\|_{\infty} = 1 and (z_j)_{j\in {\mathbb N}}\subset\mathbb{B} is such that |\varphi(z_j)|\rightarrow 1 as j\to+\infty , and h_j(z) = f_{\varphi(z_j), t}(z) , then \sup_{j\in\mathbb{N}}\|h_j\|_{A_{\alpha}^p} < +\infty . From \lim_{j\to+\infty}(1-|\varphi(z_j)|^{2})^{t+1} = 0 , we have h_j\to 0 as j\to+\infty , uniformly on compacta of \mathbb{B} . Using Lemma 2.1, it follows that \lim_{j\to+\infty}\|\Re^k_{u, {\varphi}}h_j\|_{H_{\mu}^{\infty}} = 0, from which, along with the consequence of (3.3),
\frac{\mu(z_j)|u(z_j)||{\varphi}(z_j)|}{(1-|\varphi(z_j)|^{2})^{\frac{n+{\alpha}+1}p+k}}\leq C\|\Re^k_{u, {\varphi}}h_j\|_{H_{\mu}^{\infty}}, |
which holds for sufficiently large j , and we easily get (3.15).
If \Re^k_{u, {\varphi}}:A_{\alpha}^p\to H_{\mu}^{\infty} is bounded and (3.15) holds, then Theorem 3.1 implies \mu(z)|u(z)||{\varphi}(z)|\leq J_k < +\infty, z\in{\mathbb{B}}, and (3.15) implies that for any \varepsilon > 0 there is \delta\in(0, 1) such that when \delta < |\varphi(z)| < 1 ,
\begin{align} \frac{\mu(z)|u(z)||{\varphi}(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+{\alpha}+1}p+k}} < \varepsilon. \end{align} | (3.16) |
Suppose (f_j)_{j\in {\mathbb N}} is a bounded sequence in A_{\alpha}^p converging to zero uniformly on compacts of \mathbb{B} . Let s_{\delta} = \{z\in\mathbb{B}:|\varphi(z)|\leq\delta\} . Then, Lemma 2.4, together with the fact \sup_{z\in{\mathbb{B}}}\mu(z)|u(z)||{\varphi}(z)| < +\infty, and (3.16), implies
\begin{align} \|\Re^k_{u, {\varphi}}f_j\|_{H_\mu^\infty} &\!\leq\!\sup\limits_{z\in s_{\delta}}\mu(z)\big|u(z)\Re^{k}f_j(\varphi(z))\big|+\sup\limits_{z\in\mathbb{B}\setminus s_{\delta}}\mu(z)\big|u(z)\Re^{k}f_j(\varphi(z))\big|\\ &\!\lesssim\!\sup\limits_{z\in s_{\delta}}\mu(z)\big|u(z)\big||\varphi(z)|\big|\nabla\Re^{k-1}f_j(\varphi(z))\big|\!+\!\sup\limits_{z\in\mathbb{B}\setminus s_{\delta}} \frac{\mu(z)|u(z)||{\varphi}(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+{\alpha}+1}p+k}}\\ &\lesssim \sup\limits_{|w|\leq\delta}\big|\nabla\Re^{k-1}f_j(w)\big|+\varepsilon. \end{align} | (3.17) |
The assumption f_j\to 0 on compacts along with Cauchy's estimate implies \lim_{j\to+\infty}|\nabla\Re^{k-1}f_j| = 0 uniformly on compacts of \mathbb{B} . The set \{w:|w|\leq\delta\} is compact, so by letting j\to+\infty in (3.17), it follows that \limsup_{j\to+\infty}\|\Re^k_{u, {\varphi}}f_j\|_{H_{\mu}^{\infty}}\lesssim\varepsilon, from which it follows that \lim_{j\to+\infty}\|\Re^k_{u, {\varphi}}f_j\|_{H_{\mu}^{\infty}} = 0. From this and Lemma 2.1, the compactness of \Re^k_{u, {\varphi}}:A_{\alpha}^p\to H^\infty_\mu follows.
The following theorem is known. For a more general result, see [31].
Theorem 3.5. Let p\ge 1 , {\alpha} > -1 , u\in H(\mathbb{B}) , \varphi\in S(\mathbb{B}) and \mu\in W({\mathbb{B}}) . Then, the operator \Re^0_{u, {\varphi}}: A_{\alpha}^p\to H_{\mu}^{\infty} is compact if and only if it is bounded and
\begin{align} \lim\limits_{|{\varphi}(z)|\to1}\frac{\mu(z)|u(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+{\alpha}+1}p}} = 0. \end{align} | (3.18) |
Theorem 3.6. Let p\ge 1 , {\alpha} > -1 , m\in\mathbb{N} , u_j\in H(\mathbb{B}) , j = \overline{0, m} , \varphi\in S(\mathbb{B}) and \mu\in W({\mathbb{B}}) . Then, the operator {\mathfrak S}^m_{\vec u, {\varphi}}:A_{\alpha}^p\rightarrow H_{\mu}^{\infty} is compact and (3.9) holds if and only if the operators \Re^j_{u_j, {\varphi}}:A_{\alpha}^p\to H_{\mu}^{\infty} are compact for j = \overline{0, m} .
Proof. If {\mathfrak S}^m_{\vec u, {\varphi}}:A_{\alpha}^p\rightarrow H_{\mu}^{\infty} is compact and (3.9) holds, then the operator is bounded, from which, together with Theorem 3.3, the boundedness of \Re^j_{u_j, {\varphi}}:A_{\alpha}^p\to H_{\mu}^{\infty} , j = \overline{0, m} , follows. The previous two theorems show that it is enough to prove
\begin{align} \lim\limits_{|\varphi(z)|\rightarrow 1}\frac{\mu(z)|u_j(z)||{\varphi}(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+{\alpha}+1}p+j}} = 0, \; j = \overline{1, m}, \end{align} | (3.19) |
and
\begin{align} \lim\limits_{|\varphi(z)|\rightarrow 1}\frac{\mu(z)|u_0(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+{\alpha}+1}p}} = 0. \end{align} | (3.20) |
If \|\varphi\|_{\infty} < 1 , then (3.19) and (3.20) hold. Assume \|\varphi\|_{\infty} = 1 . Let (z_k)_{k\in {\mathbb N}}\subset\mathbb{B} be such that \lim_{k\to+\infty}|\varphi(z_{k})| = 1 , and h_{k}^{(s)}(z) = h_{\varphi(z_{k})}^{(s)}(z) for an s\in\{1, \ldots, m\} (see (2.15)). Then, \sup_{k\in\mathbb{N}}\|h_{k}^{(s)}\|_{A_{\alpha}^p} < +\infty. The fact \lim_{k\to+\infty}(1-|\varphi(z_{k})|^{2})^{t+1} = 0, implies \lim_{k\to+\infty}h_k^{(s)} = 0 uniformly on any compact of \mathbb{B} . So, Lemma 2.1 implies
\begin{align} \lim\limits_{k\to+\infty}\|{\mathfrak S}^m_{\vec u, {\varphi}}h_{k}^{(s)}\|_{H_{\mu}^{\infty}} = 0. \end{align} | (3.21) |
Relation (3.12) implies
\begin{align} \frac{\mu(z_{k})|u_m(z_k)||{\varphi}(z_k)|}{(1-|\varphi(z_{k})|^{2})^{\frac{n+{\alpha}+1}p+m}}\lesssim\|{\mathfrak S}^m_{\vec u, {\varphi}}h_{k}^{(m)}\|_{H_{\mu}^{\infty}}, \end{align} | (3.22) |
for sufficiently large k . From (3.22) and (3.21) with s = m , relation (3.19) with j = m follows.
If (3.19) holds for j = \overline{s+1, m}, for a fixed s\in\{1, \ldots, m-1\} , (3.13) implies
\begin{eqnarray} \frac{\mu(w)|u_s(z_k)||{\varphi}(z_k)|}{(1-|\varphi(z_k)|^{2})^{\frac{n+{\alpha}+1}p+s}}\lesssim\|{\mathfrak S}^m_{\vec u, {\varphi}}h_{k}^{(s)}\|_{A_{\alpha}^p\rightarrow H_{\mu}^{\infty}} +\sum\limits_{j = s+1}^m\frac{\mu(w)|u_j(z_k)||{\varphi}(z_k)|}{(1-|\varphi(z_k)|^{2})^{\frac{n+{\alpha}+1}p+j}}, \end{eqnarray} |
for k large, from which, along with (3.21) and the hypothesis, the relation (3.19) with j = s follows. Thus, (3.19) holds for any s\in\{1, \ldots, m\} .
Let h_{k}^{(0)}(z) = h_{\varphi(z_{k})}^{(0)}(z) (see Lemma 2.7 (b)). Then, \sup_{k\in\mathbb{N}}\|h_{k}^{(0)}\|_{A_{\alpha}^p} < +\infty, and \lim_{k\to+\infty}h_k^{(0)}(z) = 0 uniformly on compacts of \mathbb{B} . From Lemma 2.1 we have that \lim_{k\to+\infty}\|{\mathfrak S}^m_{\vec u, {\varphi}}h_{k}^{(0)}\|_{H_{\mu}^{\infty}} = 0, from which, along with the consequence of (3.14),
\frac{\mu(z_{k})|u_0(z_k)|}{(1-|\varphi(z_{k})|^{2})^{\frac{n+{\alpha}+1}p}}\lesssim\|{\mathfrak S}^m_{\vec u, {\varphi}}h_{k}^{(0)}\|_{H_{\mu}^{\infty}}, |
(3.20) follows.
Assume \Re^j_{u_j, {\varphi}}:A_{\alpha}^p\to H_{\mu}^{\infty} , j = \overline{0, m} , are compact. Then, {\mathfrak S}^m_{\vec u, {\varphi}}:A_{\alpha}^p\rightarrow H_{\mu}^{\infty} is also compact, and by Theorem 3.3 is obtained (3.9).
Theorem 3.7. Let p\ge 1 , {\alpha} > -1 , m\in\mathbb{N} , u_j\in H(\mathbb{B}) , j = \overline{0, m} , \varphi\in S(\mathbb{B}) and \mu\in W({\mathbb{B}}) . Then, the operator {\mathfrak S}^m_{\vec u, {\varphi}}:A_{\alpha}^p\rightarrow H_{\mu, 0}^{\infty} is bounded if and only if {\mathfrak S}^m_{\vec u, {\varphi}}:A_{\alpha}^p\to H_{\mu}^{\infty} is bounded and
\begin{align} \lim\limits_{|z|\rightarrow 1}\mu(z)\bigg|\sum\limits_{j = 0}^mu_j(z)l^j\bigg||{\varphi}(z)|^l = 0, \quad l\in {\mathbb N}_0. \end{align} | (3.23) |
Proof. If {\mathfrak S}^m_{\vec u, {\varphi}}:A_{\alpha}^p\rightarrow H_{\mu}^{\infty} is bounded and (3.23) holds, then since any polynomial p is represented as p(z) = \sum_{l = 0}^tp_l(z) , where p_l , l = \overline{0, t} are homogeneous polynomials of degree l , it follows that as |z|\to 1 ,
\begin{align*} \mu(z)\big|({\mathfrak S}^m_{\vec u, {\varphi}}p)(z)\big| &\leq \sum\limits_{l = 0}^t\mu(z)\bigg|\sum\limits_{j = 0}^mu_j(z)l^j\bigg||p_l({\varphi}(z))|\lesssim\sum\limits_{l = 0}^t\mu(z)\bigg|\sum\limits_{j = 0}^mu_j(z)l^j\bigg||{\varphi}(z)|^l\to 0. \end{align*} |
Hence, {\mathfrak S}^m_{\vec u, {\varphi}}p\in H_{\mu, 0}^{\infty} . The density of the set of polynomials in A_{\alpha}^p , implies that for any f\in A_{\alpha}^p there are polynomials (p_k)_{k\in {\mathbb N}} such that \lim_{k\to+\infty}\|f-p_{k}\|_{A_{\alpha}^p} = 0. From the boundedness of {\mathfrak S}^m_{\vec u, {\varphi}}:A_{\alpha}^p\to H_{\mu}^{\infty} we have
\begin{align*} \|{\mathfrak S}^m_{\vec u, {\varphi}}f-{\mathfrak S}^m_{\vec u, {\varphi}}p_{k}\|_{H_{\mu}^{\infty}} \leq\|{\mathfrak S}^m_{\vec u, {\varphi}}\|_{A_{\alpha}^p\to H_{\mu}^{\infty}}\|f-p_{k}\|_{A_{\alpha}^p}\rightarrow 0, \end{align*} |
as k\to+\infty . So, {\mathfrak S}^m_{\vec u, {\varphi}}(A_{\alpha}^p)\subseteq H_{\mu, 0}^{\infty} , implying the boundedness of {\mathfrak S}^m_{\vec u, {\varphi}}:A_{\alpha}^p\rightarrow H_{\mu, 0}^{\infty} .
If {\mathfrak S}^m_{\vec u, {\varphi}}:A_{\alpha}^p\to H_{\mu, 0}^{\infty} is bounded, then {\mathfrak S}^m_{\vec u, {\varphi}}:A_{\alpha}^p\to H_{\mu}^{\infty} is also bounded. The fact f_{s, l}(z) = z_s^l\in A_{\alpha}^p , s = \overline{1, n}, l\in {\mathbb N}_0 , implies {\mathfrak S}^m_{\vec u, {\varphi}}f_{s, l}\in H_{\mu, 0}^\infty , s = \overline{1, n}, l\in {\mathbb N}_0 . Hence, for s = \overline{1, n}, l\in {\mathbb N}_0 , we have
\begin{align} \lim\limits_{|z|\to1}\mu(z)|{\mathfrak S}^m_{\vec u, {\varphi}}f_{s, l}(z)| = &\lim\limits_{|z|\to1}\mu(z)\bigg|\sum\limits_{j = 0}^mu_j(z)l^j\bigg||{\varphi}_s(z)|^l = 0, \end{align} |
from which, along with |{\varphi}(z)|^l\lesssim\sum_{s = 1}^n|{\varphi}_s(z)|^l , (3.23) follows for each l\in {\mathbb N}_0.
Theorem 3.8. Let p\ge 1 , {\alpha} > -1 , k\in {\mathbb N} , u\in H(\mathbb{B}) , \varphi\in S(\mathbb{B}) and \mu\in W({\mathbb{B}}) . Then, the operator \Re^k_{u, {\varphi}}: A_{\alpha}^p\to H_{\mu, 0}^{\infty} is compact if and only if
\begin{align} \lim\limits_{|z|\to1}\frac{\mu(z)|u(z)||{\varphi}(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+{\alpha}+1}p+k}} = 0. \end{align} | (3.24) |
Proof. Relation (3.24) implies (3.1). Taking the supremum in (3.7) over {\mathbb{B}} and B_{A_{\alpha}^p} , and employing (3.1), it follows that
\begin{align} \sup\limits_{f\in B_{A_{\alpha}^p}}\sup\limits_{z\in{\mathbb{B}}}\mu(z)\big|\Re^k_{u, {\varphi}}f(z)\big|\lesssim\sup\limits_{z\in{\mathbb{B}}}\frac{\mu(z)|u(z)||\varphi(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+{\alpha}+1}p+k}} < +\infty. \end{align} | (3.25) |
Hence, the set {\mathcal S} = \{\Re^k_{u, {\varphi}}f\in H^\infty_\mu : f\in B_{A_{\alpha}^p}\} is bounded in H^\infty_\mu . From (3.7) and (3.24) we easily get \Re^k_{u, {\varphi}}f\in H^\infty_{\mu, 0} for any f\in B_{A_{\alpha}^p} , i.e., {\mathcal S}\subset H^\infty_{\mu, 0} . Taking the supremum in (3.7) over B_{A_{\alpha}^p} and employing (3.24), it follows that
\lim\limits_{|z|\to 1}\sup\limits_{f\in B_{A_{\alpha}^p}}\mu(z)\big|\Re^k_{u, {\varphi}}f(z)\big| = 0. |
This fact and Lemma 2.2 imply the compactness of \Re^k_{u, {\varphi}}:A_{\alpha}^p\rightarrow H_{\mu, 0}^{\infty} .
If \Re^k_{u, {\varphi}}: A_{\alpha}^p\to H_{\mu, 0}^{\infty} is compact, then \Re^k_{u, {\varphi}}: A_{\alpha}^p\to H_{\mu}^{\infty} is also compact. From Theorem 3.4 we have that (3.15) and (3.16) hold. The fact f_{j}(z) = z_{j}\in A_{\alpha}^p , j = \overline{1, n} , implies \Re^k_{u, {\varphi}}f_j\in H_{\mu, 0}^\infty , j = \overline{1, n} , from which we have \lim_{|z|\to 1}\mu(z)|u(z)||{\varphi}_j(z)| = 0, j = \overline{1, n}. Hence,
\begin{align} \lim\limits_{|z|\to 1}\mu(z)|u(z)||{\varphi}(z)| = 0. \end{align} | (3.26) |
From (3.26) together with (3.16) we obtain (3.24) in a standard way.
The following result is known. For a more general result, see [31].
Theorem 3.9. Let p\ge 1 , {\alpha} > -1 , u\in H(\mathbb{B}) , \varphi\in S(\mathbb{B}) and \mu\in W({\mathbb{B}}) . Then, the operator \Re^0_{u, {\varphi}}: A_{\alpha}^p\to H_{\mu, 0}^{\infty} is compact if and only if
\begin{align} \lim\limits_{|z|\to1}\frac{\mu(z)|u(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+{\alpha}+1}p}} = 0. \end{align} | (3.27) |
Theorem 3.10. Let p\ge 1 , {\alpha} > -1 , m\in\mathbb{N} , u_j\in H(\mathbb{B}) , j = \overline{0, m} , \varphi\in S(\mathbb{B}) and \mu\in W({\mathbb{B}}) . Then, the operator {\mathfrak S}^m_{\vec u, {\varphi}}:A_{\alpha}^p\rightarrow H_{\mu, 0}^{\infty} is compact and
\begin{eqnarray} \lim\limits_{|z|\to 1}\mu(z)|u_j(z)||{\varphi}(z)| = 0, \; j = \overline{1, m}, \end{eqnarray} | (3.28) |
if and only if \Re^j_{u_j, {\varphi}}:A_{\alpha}^p\to H_{\mu, 0}^{\infty} are compact for j = \overline{0, m} .
Proof. Suppose {\mathfrak S}^m_{\vec u, {\varphi}}:A_{\alpha}^p\rightarrow H_{\mu, 0}^{\infty} is compact and (3.28) holds. For the compactness of \Re^j_{u_j, {\varphi}}:A_{\alpha}^p\to H_{\mu, 0}^{\infty} , j = \overline{0, m}, it is enough to prove (see Theorems 3.8 and 3.9),
\begin{align} \lim\limits_{|z|\to1}\frac{\mu(z)|u_j(z)||{\varphi}(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+{\alpha}+1}p+j}} = 0, \; j = \overline{1, m}, \end{align} | (3.29) |
and
\begin{align} \lim\limits_{|z|\to1}\frac{\mu(z)|u_0(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+{\alpha}+1}p}} = 0. \end{align} | (3.30) |
Note that {\mathfrak S}^m_{\vec u, {\varphi}}:A_{\alpha}^p\rightarrow H_{\mu}^{\infty} is compact, whereas (3.9) follows from (3.28). The compactness of \Re^j_{u_j, {\varphi}}:A_{\alpha}^p\to H_{\mu}^{\infty} , j = \overline{0, m} , follows from Theorem 3.6. Hence, we have (3.19) and (3.20). Therefore, for every {\varepsilon} > 0 there is \delta\in(0, 1) such that for {\delta} < |{\varphi}(z)| < 1 ,
\begin{eqnarray} \frac{\mu(z)|u_j(z)||{\varphi}(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+{\alpha}+1}p+j}} < {\varepsilon}, \; j = \overline{1, m}, \; and\; \frac{\mu(z)|u_0(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+{\alpha}+1}p}} < {\varepsilon}. \end{eqnarray} | (3.31) |
From (3.28) and (3.31), (3.29) easily follows. From the fact f_0(z)\equiv1\in A_{\alpha}^p it follows that {\mathfrak S}^m_{\vec u, {\varphi}}1 = u_0\in H_{\mu, 0}^\infty , from which, together with (3.31), we similarly get (3.30).
If \Re^j_{u_j, {\varphi}}:A_{\alpha}^p\to H_{\mu, 0}^{\infty} , j = \overline{0, m} , are compact, then {\mathfrak S}^m_{\vec u, {\varphi}}:A_{\alpha}^p\rightarrow H_{\mu, 0}^{\infty} is also compact. Beside this (3.26) holds when u is replaced by u_j for each j\in\{1, 2, \ldots, m\}, that is, (3.28) also holds.
Remark 3.1. The quantities J_0 and J_k, k\in {\mathbb N} , in Theorems 3.1 and 3.2, are essentially obtained by using the point evaluations in (2.1) and (2.2), respectively. Since the numerator of the right-hand side in (2.1) does not contain the term |z| , the quantity J_0 does not contain the term |{\varphi}(z)| , unlike the quantities J_k , k\in {\mathbb N}. This is connected with the definition of the radial derivative operator.
Motivated, among others, by our investigations in [14,15,16,35], in 2016 I came up with an idea of studying finite sums of the weighted differentiation composition operators and introduced several operators of this form acting on spaces of holomorphic functions on the unit disk or on the unit ball. One of them was the operator in (1.1). In [37] we have studied the operator from Hardy spaces to weighted-type spaces on the unit ball. Here we complement the main results therein by characterizing the boundedness and compactness of the operator from the weighted Bergman space to the weighted-type spaces on the unit ball. The methods, ideas and tricks presented here, with some modifications, can be used in some other settings, which should lead to some further investigations in the direction.
The paper was made during the investigation supported by the Ministry of Education, Science and Technological Development of Serbia, contract no. 451-03-68/2022-14/200029.
The author declare no conflict of interest.
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