Embedding of Besov Spaces and the Volterra Integral Operator
The boundedness and compactness of the inclusion mapping from Besov spaces to tent spaces are studied in this paper. Meanwhile, the boundedness, compactness, and essential norm of the Volterra integral operator from Besov spaces to a class of general function spaces are also investigated.
Let be the unit disk in the complex plane and be the class of functions analytic in . For , the Besov space, denoted by , is the space of all functions satisfying
Let , , and . The space is the space consisting of all such thatwhere . This space was first introduced by Zhao in . is the space (see ). is the space. is called the Dirichlet-type space, denoted by . In particular, is the Besov space . is just the classical Bergman space . When , from , we see that is equivalent to the Bloch space, denoted by , which consists of all such that
For and , let denote the space of all such that
The norm for is given by . The space is a Banach space when .
The Volterra integral operator was introduced by Pommerenke in . Here,
In , Pommerenke showed that is bounded on if and only if . Aleman and Siskakis showed that is bounded on () if and only if in . In , Aleman and Siskakis proved that is bounded (compact) if and only if . Recently, the operator has been receiving much attention. See [4–18] and the references therein for more study of the operator .
For any arc , the boundary of , let denote the normalized length of and be the Carleson box defined by
Let and be a positive Borel measure on . We say that is a -Carleson measure if
When , it gives the classical Carleson measure. is said to be a vanishing -Carleson measure if . The Carleson measure is very useful in the theory of function spaces and operator theory. The famous embedding theorem says that the inclusion mapping is bounded if and only if is a Carleson measure (see ). See [7, 20] for the study of the inclusion mapping .
Let , and be a positive Borel measure on . Let denote the space of all -measurable functions such that (see, e.g., )
The tent space was introduced by Liu et al. in . When , will be denoted by for the simplicity. In , Liu et al. studied the embedding of some Möbius invariant spaces, such as the Bloch space and the space, into .
In , Pau and Zhao showed that the inclusion mapping is bounded if and only if is a -logarithmic -Carleson measure. In , Li et al. proved that the inclusion mapping is bounded if and only if is a -Carleson measure. In , Qian and Li proved that the inclusion mapping is bounded (resp. compact) if and only if is a -logarithmic -Carleson measure (resp. vanishing -logarithmic -Carleson measure) under the assumption that and .
Motivated by [14, 21], in this paper, we study the boundedness and compactness of the inclusion mapping . More precisely, we show that is bounded (resp. compact) if and only if is an -Carleson measure (resp. vanishing -Carleson measure) under the assumption that and . As an application, we study the boundedness of the operator . Moreover, the compactness and essential norm of the operator are also investigated.
In this paper, the symbol means that . We say that if there exists a constant such that .
2. Embedding the Besov Space into
We need the following equivalent description of -Carleson measure (see Lemma 2.2 in ).
Lemma 1. Let and be a positive Borel measure on . Then, is an -Carleson measure if and only ifMoreover,
Using Lemma 3.10 in , we can easily obtain the following result.
Lemma 2. Let and . SetThen, .
Theorem 1. Let , , and be a positive Borel measure on . Then, the inclusion mapping is bounded if and only if is an -Carleson measure.
Proof. First, we assume that is bounded. For any given arc , set , and is the center point of . It is easy to see thatLetBy Lemma 2, we see that . From the boundedness of , we haveBy the fact that when , we getHence, is an -Carleson measure.
Conversely, assume that is an -Carleson measure. Let . For any given arc , set , and is the center point of . Then,whereSincewe getNow, we turn to estimate . By Theorem 1 in , we see that is an -Carleson measure if and only if is bounded. Note thatThen,Sincewe deduce that , whereSincewe get thatMaking the change of variable and combining with Proposition 4.2 in , we haveTherefore,which implies the desired result. The proof is completed.
We say that the inclusion mapping is compact ifwhenever and is a bounded sequence in that converges to 0 uniformly on compact subsets of .
Theorem 2. Let and . Let be a nonnegative Borel measure on such that point evaluation is a bounded functional on . Then, the inclusion mapping is compact if and only if is a vanishing -Carleson measure.
Proof. First, we assume that is compact. Let be a sequence arc with . Set , where is the midpoint of arc . TakeWe see that , and converges to 0 uniformly on compact subsets of when . Then, we getas , which implies that is a vanishing -Carleson measure.
Conversely, assume that is a vanishing -Carleson measure. From , we see thatHere, for and for . Let and converge to 0 uniformly on compact subsets of . Then,Letting and then , we haveTherefore, is compact. The proof is completed.
3. Volterra Integral Operator
Lemma 3. Let and . Then, if and only if
Proof. The proof is similar to that of Proposition 1 in . Thus, we omit the details of the proof.
Theorem 3. Let and . Then, is bounded if and only if .
Proof. Assume that is bounded. For any fixed arc , let denote the center of and . SetBy Lemma 2, we have for . In addition, it is easy to see thatwhen . By the boundedness of , we getwhich implies that by .
Conversely, suppose that . By , we see that is an -Carleson measure. Let . By Theorem 1, we see that is bounded, i.e.,By Lemma 3, we get thatTherefore, is bounded.
Next, we give an estimation for the essential norm of . First, we recall some definitions. The essential norm of , denoted by , is defined byHere, and are Banach spaces, and is a bounded linear operator. It is easy to see that is compact if and only if . Let be a closed subspace of . Given , the distance from to , denoted by , is defined by
Lemma 4. (see ). Let and . If , thenHere, , .
Lemma 5. Let and . If and , then is compact.
Proof. Given such that converges to zero uniformly on any compact subset of and . Since , we get thatHence,By the dominated convergence theorem, we get the desired result. The proof is completed.
Lemma 6. (see ). Let be two Banach spaces of analytic functions on . Suppose that(1)The point evaluation functionals on are continuous(2)The closed unit ball of is a compact subset of in the topology of uniform convergence on compact sets(3) is continuous when and are given the topology of uniform convergence on compact setsThen, is a compact operator if and only if for any bounded sequence in such that converges to zero uniformly on every compact set of , the sequence converges to zero in the norm of .
Theorem 4. Let and . If is bounded, then
Proof. Let and as . Suppose is the center of and . For each , letThen, when , and is bounded in . Furthermore, converges to zero uniformly on every compact subset of . Given a compact operator , by Lemma 6, we haveSo,which implies thatIt follows from Lemma 4 thatOn the contrary, by Lemma 5, is compact. Then,Using Lemma 4 again, we haveThe proof is completed.
The following result can be deduced by Theorem 4 directly.
Corollary 1. Let and . If is bounded, then is compact if and only if
No data were used to support this study.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
The first author was supported by the Science Foundation of Hanshan Normal University (no. XN202029). The third author was supported by NNSF of China (nos. 11801250 and 11871257), Overseas Scholarship Program for Elite Young and Middle-aged Teachers of Lingnan Normal University, Yanling Youqing Program of Lingnan Normal University (no. YL20200202), the Key Subject Program of Lingnan Normal University (nos. 1171518004 and LZ1905), and the Department of Education of Guangdong Province (no. 2018KTSCX133).
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