Essential Norms of Volterra Type Operators between<i> Zygmund</i> Type Spaces

Ye, Shanli; Lin, Caishu

doi:https://doi.org/10.1155/2017/1409642

Journal of Function Spaces

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Research Article | Open Access

Volume 2017 | Article ID 1409642 | https://doi.org/10.1155/2017/1409642

Essential Norms of Volterra Type Operators between Zygmund Type Spaces

Shanli Ye¹and Caishu Lin²

Academic Editor: Ruhan Zhao

Received05 Jan 2017

Revised04 Mar 2017

Accepted07 Mar 2017

Published11 Apr 2017

Abstract

We investigate the boundedness of some Volterra type operators between Zygmund type spaces. Then, we give the essential norms of such operators in terms of , their derivatives, and the nth power of .

1. Introduction

Let be the open unit disk in the complex plane and let be its boundary, and denote the set of all analytic functions on .

For every , we denote by the Bloch type space of all functions satisfying endowed with the norm . The little Bloch type space consists of all satisfying , and is obviously the closed subspace of . When , we get the classical Bloch space and little Bloch space . It is well known that, for , is a subspace of , the Banach space of bounded analytic functions on . Some sources for results and references about the Bloch type functions are the papers of Yoneda [1], Stevic [2, 3], and the first author [4–7].

For we denote by the Zygmund type space of those functions satisfying and the little Zygmund type space consists of all satisfying , and is obviously the closed subspace of . It can easily be proved that is a Banach space under the norm and that is a closed subspace of . When , we get the classical Zygmund space and the little Zygmund space . It is clear that if and only if .

We consider the weighted Banach spaces of analytic functions endowed with norm , where the weight is a continuous, strictly positive, and bounded function. The weight is called radial, if for all For a weight the associated weight is defined by We notice the standard weights , where , and it is well known that . We also consider the logarithmic weight It is straightforward to show that .

For an analytic self-map of and a function , we define the weighted composition operator as for . Weighted composition operators have been extensively studied recently. It is interesting to provide a function theoretic characterization when and induce a bounded or compact composition operator on various function spaces. Some results on the boundedness and compactness of concrete operators between some spaces of analytic functions one of which is of Zygmund type can be found, for example, in [8–19].

Suppose that is an analytic map. Let and denote the Volterra type operators with the analytic symbol on , respectively:

If , then is an integral operator. While , then is Cesàro operator. Pommerenke introduced the type operator and characterized the boundedness of between spaces in [20]. More recently, boundedness and compactness of type operators between several spaces of analytic functions have been studied by many authors; one may see [21, 22].

In this paper, we consider the following integral type operators, which were introduced by Li and Stevic (see, e.g., [10, 23]); they can be defined by We will characterize the boundedness of those integral type operators between Zygmund type spaces and also estimate their essential norms. The boundedness and compactness of these operators on the logarithmic Bloch space have been characterized in [22].

Recall that essential norm of a bounded linear operator is defined as the distance from to , the space of compact operators from to , namely, It provides a measure of noncompactness of . Clearly, is compact if and only if .

Throughout this paper, constants are denoted by , they are positive and may differ from one occurrence to the other. The notation means that there are positive constants such that .

2. Boundedness

In order to prove the main results of this paper. We need some auxiliary results.

Lemma 1 (see [8, 13]). For and let be a bounded sequence in which converges to uniformly on compact subsets of . Then .

Lemma 2 (see [8, 13]). For every , where , one has(i) and for every ;(ii) and for ;(iii), for every ;(iv), for every ;(v), for every ;(vi), for every .

Lemma 3 (see [8]). Let and a radial, nonincreasing weight tending to at boundary of , and let the weighted composition operator be bounded.(i) If , then is a compact operator.(ii) (iii) If , then

The following lemma is due to [24, 25].

Lemma 4. Let and be radial, nonincreasing weights tending to zero at the boundary of . Then(i) The weighted composition operator maps into if and only if with norm comparable to the above supermum.(ii)

Lemma 5 (see [26]). For every , one has

Theorem 6. Let be an analytic self-map of and .(i) If , then is a bounded operator if and only if and (ii) If , then is a bounded operator if and only if (iii) If , then is a bounded operator if and only if

Proof. Suppose that is bounded from to . Using the test functions and , we have Since is a self-map, we get that , .
For every and given nonzero , we take the test functions for every . One can show that , and are in , , and . Since , it follows that for all with , we have Then Now we use (14) and Lemma 4 to conclude that which shows that (16) is necessary for all case.
Conversely, suppose that and (16) holds. Assume that . From Lemma 2, it follows thatwhich implies that is bounded. This completes the proof of (i).
Next we will prove (ii). The necessity in condition (17) has been proved above. Fixing with , we take the functionfor , where Then we have by [11]. Let . It follows that Since (17) holds and is bounded, we obtain that Noting and together with (15) and Lemma 4, we conclude that (18) holds.
The converse implication can be shown as in the proof of (i).
Finally we will prove (iii). We have proved that (19) holds above. To prove (20), we take function defined in (23) for every with and obtain that Since is bounded and (19) holds, we obtain that therefore, we deduce that (20) holds by (14) and Lemma 4.
The converse implication can be shown as in the proof of (i).

Theorem 7. Let be an analytic self-map of and .(i) If , then is a bounded operator if and only if and (ii) If , then is a bounded operator if and only if (iii) If , then is a bounded operator if and only if (36) holds and

The proof is similar to that of Theorem 6, and the details are omitted.

Theorem 8. Let be an analytic self-map of and .(i) If , then is a bounded operator if and only if and .(ii) If , then is a bounded operator if and only if and (iii) If , then is a bounded operator if and only if and (iv) If , then is a bounded operator if and only if (v) If , then is a bounded operator if and only if (40) holds and

Proof. Suppose that is bounded from to space.
(i) Case . Using functions and , we obtain Then we obtain that and are necessary for all case.
For the converse implication, suppose that and . For , it follows from Lemma 2 that Then is bounded. This completes the proof of (i).
(ii) Case . We consider the test function defined in (30) for every with . It follows that Since and , we get Then we have On the other hand, from (15) and Lemma 4, we have Hence (39) holds.
The converse implication can be shown as in the proof of (i).
(iii) Case . has been proved above. We take the test function in (23) for every with ; by the same way as (ii), we can obtain that (40) holds.
The converse implication can be shown as in the proof of (i).
(iv) Case . We have proved that (41) holds above. To prove (42), we consider another test function . Clearly and . For every with , it follows that Applying (41) we get Noting and using Lemma 4 and (15), we conclude that (42) holds.
(v) Case . We have proved that (40) holds above. Applying test function in (23) for every with , we have With the same calculation for test function with , then , and we have that Therefore, Since , we conclude that (43) holds.

Theorem 9. Let be an analytic self-map of and .(i) If , then is a bounded operator if and only if and .(ii) If , then is a bounded operator if and only if and (iii) If , then is a bounded operator if and only if and (iv) If , then is a bounded operator if and only if(v)If , then is a bounded operator if and only if (56) holds and

The proof is similar to that of Theorem 8, and the details are omitted.

3. Essential Norms

In this section we estimate the essential norms of these integral type operators on type spaces in terms of , their derivatives, and the nth power of .

Let and . We note that every compact operator can be extended to a compact operator . In fact, for every , and we can define .

For , we consider the compact operator defined by .

Lemma 10. If is a bounded operator from to space, then

Proof. Clearly . Then we prove .
Let be given. Let be an increasing sequence in converging to and , the closed subspace of . Then Hence Since , we have , and the proof is finished.

Let and be the derivative operators. Then clearly and are linear isometries on and , respectively, and Therefore Similarly,

Theorem 11. Let be a bounded operator from to space.(i) If , then (ii) If , then (iii) If , then

Proof. (i) We start with the upper bound. First we show that is a compact weighted composition operator for into . Suppose that is bounded sequence in . From Lemma .6 in [27], has a subsequence which converges uniformly on to a function, which we can assume to be identically zero. Then it follows from Theorem 6 and Lemma 1 that which shows that is a compact operator and . Applying (63), Lemmas 4, 5, and 10, we get that For the lower bound, let with and as Taking defined in (25), we obtain that is bounded sequence in converging to uniformly on compact subset of and . For every compact operator , Now we use (14) and Lemma 4 to obtain that Hence (70) holds.
(ii) The boundedness of guarantees that and are bounded weighted composition operators. Theorem .4 in [28] ensures that Now we use Lemmas 4, 5 and (63) to conclude that On the other hand, let be a sequence in such that and as . Given where , from [11] we know that is a bounded sequence in which converges to zero uniformly on compact subsets of , and For every compact operator , we have as . By Lemmas 4 and 5, we obtain that Now we take another function From [11] we know that is a bounded sequence in which converges to zero uniformly on compact subsets of , and . It follows from Lemmas 4 and 5 that Noting that , we obtainHence we have
(iii) Let . The proof of the upper bound is similar to that of (ii). From the proof of (i), we get that, for some constant ,Now, let be as before and note that the function given in (23). Then is bounded sequence in converging to zero uniformly on compact subsets of ; therefore By (85), we have and the rest of the proof is similar to that of the previous, and we omit it.

Theorem 12. Let be an analytic self-map of and , and is a bounded operator.(i) If , then (ii) If , then (iii) If , then (iv) If , then (v) If , then

Proof. (i) For the compactness of , the argument is similar to the proof of Theorem 11(i); then we have . Hence by (67) and Lemma 3, we get that
Next we will prove (ii). The boundedness of guarantees that and are bounded weighted composition operators. We know that if is a bounded operator, then is a compact operator by Lemma 3. Hence we consider the boundedness of and just consider that is a bounded operator.
Theorem .4 in [28] ensures that From (67) and Lemmas 4 and 5, we have In order to prove , we take the function where and . From [16] we obtain that is a bounded sequence in which converges to zero uniformly on compact subsets of . By a direct calculation, we have For every compact operator , we have as . Let . It follows from Lemma 5 that This completes the proof.
The proof of (iii) is the same as that of Theorem 11 (iii); we do not prove it again.
(iv) Let . Applying Lemma 3 (ii) and Theorem .2 in [29], we get thatwhich yields the upper bound by (67).
With the same arguments as in the proof of Theorems 8 and 11, for some constant , we haveOn the other hand, let with and as Let the test function From [8] we obtain that is a bounded sequence in which converges to zero uniformly on compact subsets of , and Applying Theorem 8 we get HenceOn the other hand, the lower bound can be easily proved by Lemmas 4 and 5.
If , the proof is similar to that of (iv) except that we now choose the test function instead of . This completes the proof of Theorem 12.

Using the same methods of Theorems 11 and 12, we can have the following results.

Theorem 13. Let be a bounded operator from to space.(i) If , then (ii) If , then (iii) If , then

Theorem 14. Let be an analytic self-map of and , and is a bounded operator.(i) If , then (ii) If , then (iii) If , then (iv) If , then (v) If , then

Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

The first author was partially supported by the National Natural Science Foundation of China (Grant nos. 11671357, 11571217) and the Natural Science Foundation of Fujian Province, China (Grant no. 2015J01005).

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Copyright

Copyright © 2017 Shanli Ye and Caishu Lin. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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