On 𝐹-Algebras 𝑀^𝑝(1<𝑝<∞) of Holomorphic Functions

Abstract

We consider the classes of holomorphic functions on the open unit disk in the complex plane. These classes are in fact generalizations of the class introduced by Kim (1986). The space equipped with the topology given by the metric defined by , with and , becomes an -space. By a result of Stoll (1977), the Privalov space with the topology given by the Stoll metric is an -algebra. By using these two facts, we prove that the spaces and coincide and have the same topological structure. Consequently, we describe a general form of continuous linear functionals on (with respect to the metric ). Furthermore, we give a characterization of bounded subsets of the spaces . Moreover, we give the examples of bounded subsets of that are not relatively compact.

1. Introduction and Preliminaries

Let denote the open unit disk in the complex plane and let denote the boundary of . Let    be the familiar Lebesgue spaces on the unit circle .

Following Kim ([1, 2]), the class   consists of all holomorphic functions on for which where and is the maximal radial function of . The Privalov class   consists of all holomorphic functions on for which These classes were firstly considered by Privalov in [3, page 93], where is denoted as .

Notice that for , the condition (3) defines the Nevanlinna class   of holomorphic functions in . Recall that the Smirnov class   is the set of all functions holomorphic on such that where is the boundary function of on ; that is, is the radial limit of which exists for almost every . We denote by the classical Hardy space on . It is known (see [4, 5]) that where the above containment relations are proper.

The study of the spaces was continued in 1977 by Stoll [6] (with the notation in [6]). Further, the topological and functional properties of these spaces were studied in [4, 5, 7–14]; typically, the notation varied and these spaces are called the Privalov spaces in [12–15].

It is well known [16, page 26] that a function if and only if , where is an inner function on and is an outer function given by where .

Privalov [3, page 98] showed that if and only if , where is an inner function on and is an outer function as given above with .

Stoll [6, Theorem 4.2] showed that the space (with the notation in [6]) with the topology given by the metric defined by becomes an -algebra. Recall that the function defined on the Smirnov class by (8) with induces the metric topology on . Yanagihara [17] showed that, under this topology, is an -space.

Furthermore, in connection with the spaces , Stoll [6] (also see [7] and [12, Section 3]) also studied the spaces (with the notation in [6]), consisting of those functions holomorphic on for which where Stoll [6, Theorem 3.2] proved that the space with the topology given by the family of seminorms defined for as for each , where is the th Taylor coefficient of , becomes a countably normed Fréchet algebra. By a result of Eoff [7, Theorem 4.2], is the Fréchet envelope  of  , and hence and have the same topological duals.

Here, as always in the sequel, we will need some of Stoll’s results concerning the spaces only with , and hence we will assume that is any fixed number.

The study of the class has been extensively investigated by Kim in [1, 2], Gavrilov and Zaharyan [18], and Nawrocky [19]. Kim [2, Theorems 3.1 and 6.1] showed that the space with the topology given by the metric defined by becomes an -algebra. Furthermore, Kim [2, Theorems 5.2 and 5.3] gave an incomplete characterization of multipliers of into . Consequently, the topological dual of is not exactly determined in [2], but, as an application, it was proved in [2, Theorem 5.4] (also cf. [19, Corollary 4]) that is not locally convex space. Furthermore, the space is not locally bounded ([2, Theorem 4.5] and [19, Corollary 5]).

Although the class is essentially smaller than the class , Nawrocky [19] showed that the class and the Smirnov class have the same corresponding locally convex structure which was already established by Yanagihara for the Smirnov class in [17, 20]. More precisely, it was proved in [19, Theorem 1] that the Fréchet envelope of the class can be identified with the space of holomorphic functions on the open unit disk such that for each , where is the th Taylor coefficient of . Notice that coincides with the space defined above. It was shown in [17, 21] that is actually the containing Fréchet space for . Moreover, Nawrocky [19, Theorem 1] characterized the set of all continuous linear functionals on which by a result of Yanagihara [17] coincides with those on the Smirnov class .

Motivated by the mentioned investigations of the classes and , and the fact that the classes are generalizations of the Smirnov class , in Section 2, we consider the classes as generalizations of the class . Accordingly, the class   consists of all holomorphic functions on for which Obviously, Following [2], by analogy with the space , the space is equipped with the topology induced by the metric defined as with .

In Section 2, we give the integral limit criterion for a function holomorphic on the disk to belong to the class (Lemma 3). Furthermore, we prove that the space is closed under integration (Theorem 4).

In Section 3 we study and compare the uniform convergence on compact subsets of and the convergences induced by the metrics and in the space , respectively. It is proved (Theorem 11) that for each .

It is proved in Section 4 that the space of all polynomials on is a dense subset of (Theorem 13). Hence, is a separable metric space. We show that the space with the topology given by the metric becomes an -space (Theorem 15). As an application, we prove that the metric spaces and have the same topological structure (Theorem 16). Consequently, we obtain a characterization of continuous linear functionals on (Theorem 17). Notice that Theorem 17 with characterizes the set of all continuous linear functionals on the space , which is in fact the Nawrocky result [19, Theorem 1] mentioned above.

In Section 5 we obtain a characterization of bounded subsets of the spaces (Theorem 19). It is also given another necessary condition for a subset of () to be bounded (Theorem 22). Finally, we give the examples of bounded subsets of that are not relatively compact (Theorem 24).

2. The Classes

Recall that, for a fixed , the class consists of all holomorphic functions on for which Combining the inequalities and , we obtain . The last inequality implies the fact that the condition (17) is equivalent to

Lemma 1. The function defined on by (18) satisfies the following conditions: Hence, is an algebra with respect to the pointwise addition and multiplication of functions.

Proof. Combining the inequality with Minkowski’s integral inequality (with the power ), we immediately obtain (19). Similarly, combining the inequality with Minkowski’s integral inequality (with the exponent ), we obtain (20).

Theorem 2. The function defined on as is a translation invariant metric on . Further, the space is a complete metric space with respect to the metric .

Proof. If we suppose that , for some , then by (23) it follows that for almost every . Hence, for almost every , and, by Riesz uniqueness theorem, we infer that for all . As, by (19), the triangle inequality is satisfied, it follows that is a metric on . Finally, by the obvious inequality we see that is a translation invariant metric. This concludes the proof.

For simplicity, here as always in the sequel, we shall write instead of the metric space . For a function holomorphic in and for any fixed , denote by the function defined on as , . Furthermore, for a given holomorphic function on , let

Lemma 3. A function holomorphic on the unit disk belongs to the class if and only if it satisfies

Proof. The condition (26) implies that . Conversely, assume that . Then Since, by the assumption, ; that is, , using (27) and applying the Lebesgue dominated convergence theorem, we obtain which completes the proof.

Theorem 4. The space is closed under integration.

Proof. For a given function , define It follows that , and thus for almost every . Therefore , as desired.

3. Convergences in the Space

Theorem 5. For each function , in as .

Proof. Assume that . Since , by Fatou's theorem, the radial limit exists for almost every . Hence, for such a fixed , the function is a continuous on , and thus it is uniformly continuous on . Therefore, for such a , we have By the inequality in view of the fact that (18) is satisfied for , we obtain From this and (30), by the Lebesgue dominated convergence theorem, we obtain That is, in as .

For the proof of completeness of the metric space we will need the following lemmas.

Lemma 6. If is a Cauchy sequence in , then in as , where this convergence is uniform with respect to .

Proof. Suppose that is an arbitrary Cauchy sequence in . Then for a given there is a such that So by the triangle inequality, for each , we have By Theorem 5, there exists   sufficiently near to , for which Hence, by (35), we immediately obtain This completes proof of Lemma 6.

Lemma 7. For any , and where is the metric of defined by (8).

Proof. The inclusion is obvious, and (38) follows by the definition of the metrics and .

Lemma 8. The convergence with respect to the metric of the space is stronger than the metric of uniform convergence on compact subsets of the disk .

Proof. The assertion immediately follows from the inequality on [5, page 898], which implies that, for any function and , we have

Lemma 9. If is a Cauchy sequence in the space , then converges uniformly on compact subsets of to some holomorphic function on .

Proof. From the inequality (38) of Lemma 7, it follows that is a Cauchy sequence in . Therefore, there exists such that in , and so, by Lemma 8, uniformly on compact subsets of .

The following result is a maximal theorem of Hardy and Littlewood.

Lemma 10 (see [16, page 11]). Let and let be a function in the Lebesgue space . Let be the Poisson integral of the function . Define Then and there is a constant depending only on such that where is the usual norm of the space .

We are now ready to state the following result.

Theorem 11. for each ; that is, the spaces and coincide.

Proof. By Lemma 7, for each . For the proof of the converse of this inclusion, assume that . We will show that . As noticed in Section 1, can be factorized as where is the inner function and is an outer function for the class ; that is, where is a constant of unit modulus. Furthermore, . As , for each , the previous factorization and the fact that immediately imply that . Since from (44), we immediately obtain whence it follows that, for , The above inequality yields From the above inequality and the fact that , we conclude by Lemma 10 that . This means that and therefore . Thus , and therefore . This completes the proof.

Corollary 12. Let . Then where is a nonnegative constant depending only on .

Proof. Let be the outer factor in the canonical factorization of . From the proof of Theorem 11, we see that for the functions and the inequality (42) can be applied from Lemma 10. The obtained inequality is in fact (49) with instead of . Since , for each , it follows that at almost every ; thus (49) is obviously satisfied.

4. as an -Algebra

Theorem 13. The space of all polynomials over is a dense subset of . Hence, is a separable metric space.

Proof. Suppose that . Since, for a fixed , is a holomorphic function on the closed unit disk , by Runge's theorem, can be uniformly approximated by polynomials on . This together with the fact that, by Theorem 5, in as yields that the space of all polynomials over is a dense subset of . Therefore, the set of all polynomials whose coefficients have rational real parts and rational imaginary parts becomes a countable dense subset of . This concludes the proof.

Theorem 14. is a complete metric space.

Proof. Let be a Cauchy sequence in . Then since is complete, there is a such that in . Since, by Theorem 11, , it follows that , and thus it remains to show that in . By Theorem 5 and Lemma 6, there exist and such that Since, by Lemma 9, a sequence converges uniformly on each closed disk to some function , it follows that there exists such that Taking , by (50) and (51), the triangle inequality implies that This shows that in , which completes the proof.

Theorem 15. with the topology given by the metric defined by (23) becomes an -space.

Proof. By [22, page 51], it suffices to show the following properties:(i) is an additive-invariant metric,(ii)for any fixed , is a continuous map from into ,(iii)for any fixed , is a continuous map from into , and(iv) is a complete metric space.The assertion (i) follows from Theorem 2.
By the Lebesgue dominated convergence theorem, we have
Let such that . Then the triangle inequality yields whence we see that is a continuous map from into .
The assertion (iv) is in fact the assertion of Theorem 14.
This concludes the proof.

We are now ready to prove that the (metric) spaces and have the same topological structure.

Theorem 16. For each , the classes and coincide, and the metric spaces and have the same topological structure.

Proof. Consider the identity map . Then, by the inequality (38) of Lemma 7, is continuous. Since, by Theorem 11, , maps onto . Since and are both -spaces, it follows, by the open mapping theorem [23, Corollary 2.12 (b)], that the inverse map of is continuous. Hence, is a homeomorphism, and so the metrics and induce the same topology on and , respectively.

As an application of Theorem 16 and using the characterization of topological dual of the space (which is by [7, Theorem 4.2] the Fréchet envelope of ) given by Stoll [6, Theorem 3.3] (cf. also [12, Theorem 3.5] and [13, Theorem 2]), we immediately get the following result.

Theorem 17. If is a continuous linear functional on , then there exists a sequence of complex numbers with , for some , such that where , with convergence being absolute. Conversely, if is a sequence of complex numbers for which then (55) defines a continuous linear functional on .

Corollary 18. is an -algebra.

Proof. By Theorem 15, becomes an -space. As is an -algebra, by Theorem 16, the multiplication is also continuous on . Hence, is an -algebra.

5. Bounded Subsets of

It is proved in Section 4 (Theorem 16) that the spaces and coincide and have the same topological structure. Since and are not Banach spaces, it is of interest to obtain a characterization of bounded subsets of these spaces in terms of both metrics and .

Recall that, for a function , its boundary function is defined as the radial limit which exists for almost every .

The following result gives a characterization of bounded subsets of . Recall that the assertion (i)(iii) is analogous to Theorem 1 in [21] that describes bounded subsets of .

Theorem 19. For given set , the following conditions are equivalent:(i) is a bounded subset of ;(ii)for all there exists such that for every measurable set with the Lebesgue measure ;(iii)for all there exists such that for each measurable set with the Lebesgue measure .

Proof. (ii)(iii). It follows that from the obvious inequality , , for almost every .
(iii)(i). Let be an arbitrary neighborhood of zero in . Choose sufficiently small such that Now it follows that there exists , , such that (iii) holds. Choose an for which . Set Then , and thus by (iii) we have By (62) and Chebyshev’s inequality, we conclude that for every function there exists a measurable set depending on such that From (63), we obtain Choose such that . Then using the inequality (60) and (iii), for every , we obtain Therefore, , from which it follows that . Hence, is a bounded subset of .
(i)(ii). Assume that is a bounded subset of . Then for any given there is a , , such that for each and . It follows that Since we obtain For given , choose satisfying and satisfying (67) and so also satisfying (68). Next, take such that Then for each set with , by (68)–(72), for every , we obtain Therefore, the condition (ii) of the theorem is satisfied, which concludes the proof.

Remark 20. Note that the condition (ii) from Theorem 19 in fact means that the family is uniformly integrable on . The same assertion is also valid for the condition (iii). On the other hand, from the proof of Theorem 19, we see that (ii) implies that the family forms a bounded subset of the space ; that is, there holds Similarly, it follows from (iii) that the family is bounded in .

Corollary 21. If is a subset of for which the family is uniformly integrable, then the family is also uniformly integrable.

Proof. The condition of Corollary 21 and (iii)(ii) of Theorem 19 immediately yield that the family is uniformly integrable on the circle . This fact and the obvious inequality , , , for almost every , imply that the family is uniformly integrable.

The following result gives a necessary condition for a subset of to be bounded.

Theorem 22. Let be a subset of . If is bounded in , then where , is a positive constant, and , , is a positive continuous function that does not depend on and for which as .

Proof. By the inequqlity (5.4) from the proof of Theorem 5.2 in [4], for all , we have As, by the assumption, is a bounded subset of , by Theorem 19 (iii), for all there exists , such that and for every measurable set with the Lebesgue measure .
Further, from the proof of (iii)(i) of Theorem 19, we see that for each there is a measurable set depending on for which for almost every . From (78)–(80), we obtain whence it follows that Choose a sequence of positive numbers such that . For each , let be a number such that where and Put From (82), (83), and (85) we obtain Since we conclude that there exists a continuous function satisfying Therefore, whence by setting we obtain This concludes the proof.

Remark 23. The condition of Theorem 22 is not a sufficient condition for a set to be bounded. To show this, define where Then as in the proof of Lemma 1 in [21] it is easy to verify that the set satisfies the condition of Theorem 22. Since we see that is not bounded in .

Theorem 24. There exist bounded subsets of that are not relatively compact.

Proof. Define a sequence of functions on as and set Obviously, and for each measurable set we have where denotes the Lebesgue measure of . From this and Theorem 19, we see that the set is bounded in .
Now suppose that is relatively compact. This means that there exists a subsequence of and a function such that and thus Therefore, by (96), it follows that on . On the other hand, from (98), it follows that Therefore, This contradiction shows that is not relatively compact in .

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.