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

We consider the classes M p (1 < p < ∞) of holomorphic functions on the open unit disk 𝔻 in the complex plane. These classes are in fact generalizations of the class M introduced by Kim (1986). The space M p equipped with the topology given by the metric ρ p defined by ρ p(f, g) = ||f − g||p = (∫ 0 2πlogp(1 + M(f − g)(θ))(dθ/2π))1/p, with f, g∈M p and Mf(θ) = sup0⩽r<1 ⁡|f(re iθ)|, becomes an F-space. By a result of Stoll (1977), the Privalov space N p (1 < p < ∞) with the topology given by the Stoll metric d p is an F-algebra. By using these two facts, we prove that the spaces M p and N p coincide and have the same topological structure. Consequently, we describe a general form of continuous linear functionals on M p (with respect to the metric ρ p). Furthermore, we give a characterization of bounded subsets of the spaces M p. Moreover, we give the examples of bounded subsets of M p that are not relatively compact.


Introduction and Preliminaries
Let D denote the open unit disk in the complex plane and let T denote the boundary of D. Let (T) (0 < ≤ ∞) be the familiar Lebesgue spaces on the unit circle T.
These classes were firstly considered by Privalov in [3, page 93], where is denoted as .
It is well known [16, page 26] that a function ∈ + if and only if = , where is an inner function on D and is an outer function given by where log | * | ∈ 1 (T).
Privalov [3, page 98] showed that ∈ if and only if = , where is an inner function on D and is an outer function as given above with log + | * | ∈ (T).
Stoll [6,Theorem 4.2] showed that the space (with the notation (log + ) in [6]) with the topology given by the metric defined by becomes an -algebra. Recall that the function 1 = defined on the Smirnov class + by (8) with = 1 induces the metric topology on + . Yanagihara [17] showed that, under this topology, + is an -space.
Furthermore, in connection with the spaces (1 < < ∞), Stoll [6] (also see [7] and [12,Section 3]) also studied the spaces (0 < < ∞) (with the notation 1/ in [6]), consisting of those functions holomorphic on D for which lim → 1 where Stoll [6,Theorem 3.2] proved that the space with the topology given by the family of seminorms {‖ ⋅ ‖ , } >0 defined for ∈ as for each > 0, wherê( ) 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 1 < < ∞, and hence we will assume that = > 1 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 D such that for each > 0, wherê( ) is the th Taylor coefficient of . Notice that + coincides with the space 1 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 (1 < < ∞) are generalizations of the Smirnov class + , in Section 2, we consider the classes (1 < < ∞) as generalizations of the class . Accordingly, the class (1 < < ∞) consists of all holomorphic functions on D 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 D 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 D and the convergences induced by the metrics and in the space , respectively. It is proved (Theorem 11) that = for each > 1.
The Scientific World Journal 3 It is proved in Section 4 that the space of all polynomials on C 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 = 1 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).

The Classes
(1 < < ∞) Recall that, for a fixed 1 < < ∞, the class consists of all holomorphic functions on D for which Combining the inequalities log(| | + 1) ≤ log + | | + log 2 and 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 ( , ) = 0, for some , ∈ , then by (23) it follows that ( − )( ) = 0 for almost every ∈ [0, 2 ]. Hence, * ( ) = * ( ) for almost every ∈ T, and, by Riesz uniqueness theorem, we infer that ( ) = ( ) for all ∈ D. 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 D and for any fixed 0 ≤ < 1, denote by the function defined on D as ( ) = ( ), ∈ D. Furthermore, for a given holomorphic function on D, let Proof. The condition (26) implies that ∈ . Conversely, assume that ∈ . Then Since, by the assumption, ∈ ; that is, ∫ 2 0 (log + ( )) ( /2 ) < ∞, using (27) and applying the Lebesgue dominated convergence theorem, we obtain lim → 1 which completes the proof. Proof. For a given function ∈ , define It follows that | ( )| ≤ ( ), and thus ( ) ≤ ( ) for almost every ∈ [0, 2 ]. Therefore ∈ , as desired.

Convergences in the Space
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, For the proof of completeness of the metric space ( , ) we will need the following lemmas.
This completes proof of Lemma 6.

Lemma 7.
For any > 1, ⊆ and where is the metric of defined by (8).
Proof. The inclusion ⊆ is obvious, and (38) follows by the definition of the metrics and . Proof. The assertion immediately follows from the inequality on [5, page 898], which implies that, for any function ∈ and 0 ≤ < 1, we have 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 D.
The following result is a maximal theorem of Hardy and Littlewood.
The Scientific World Journal 5 Lemma 10 (see [16, page 11]). Let 1 < ≤ +∞ and let be a function in the Lebesgue space (T). Let be the Poisson integral of the function . Define Then ∈ (T) and there is a constant depending only on such that where ‖ ⋅ ‖ is the usual norm of the space (T).
We are now ready to state the following result.
Theorem 11. = for each > 1; that is, the spaces and coincide.
Proof. By Lemma 7, ⊆ for each > 1. 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, log + | * | ∈ (T). As | ( )| ≤ 1, for each ∈ D, the previous factorization and the fact that ∈ immediately imply that ∈ . Since from (44), we immediately obtain log ( ) whence it follows that, for 0 ≤ < 1, The above inequality yields From the above inequality and the fact that log + | * | ∈ (T), we conclude by Lemma 10 that log + ( ) ∈ (T). This means that ∈ and therefore ∈ . Thus ⊆ , and therefore = . This completes the proof.

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

as an -Algebra
Theorem 13. The space of all polynomials over C is a dense subset of . Hence, is a separable metric space.
Proof. Suppose that ∈ . Since, for a fixed 0 ≤ < 1, is a holomorphic function on the closed unit disk D : | | ≤ 1, by Runge's theorem, can be uniformly approximated by polynomials on D. This together with the fact that, by Theorem 5, → in as → 1− yields that the space of all polynomials over C 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 0 < < 1 and 1 ∈ N such that ( , ) < 3 , ( , ( ) ) < 3 for each ≥ 1 .
(50) 6 The Scientific World Journal Since, by Lemma 9, a sequence { } converges uniformly on each closed disk | | ≤ < 1 to some function , it follows that there exists 2 ∈ N such that Taking 0 = max{ 1 , 2 }, 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 C into , (iii) for any fixed ∈ C, → 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 ∈ N 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.  1/( +1) )), for some > 0, such that where ( ) = ∑ ∞ =0 ∈ , with convergence being absolute. Conversely, if { } is a sequence of complex numbers for which then (55) defines a continuous linear functional on .
Proof. By Theorem 15, becomes an -space. As is an -algebra, by Theorem 16, the multiplication is also continuous on . Hence, is an -algebra.

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 * ( ) = lim → 1− ( ) which exists for almost every ∈ T. 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: (ii) for all > 0 there exists > 0 such that for every measurable set ⊂ T with the Lebesgue measure | | < ; (iii) for all > 0 there exists > 0 such that for each measurable set ⊂ T with the Lebesgue measure | | < . be an arbitrary neighborhood of zero in . Choose sufficiently small > 0 such that Now it follows that there exists , 0 < < , such that (iii) holds. Choose an ∈ N for which 1/ < . Set Then | | = 1/ < , and thus by (iii) we have By (62) and Chebyshev's inequality, we conclude that for every function ∈ there exists a measurable set ⊂ T depending on such that Choose such that 0 < < / . Then using the inequality Therefore, ( , 0) < , from which it follows that ⊂ . Hence, is a bounded subset of . (i)⇒(ii). Assume that is a bounded subset of . Then for any given > 0 there is a 0 = 0 ( ), 0 < 0 < 1, such that for each ∈ and | | ≤ 0 . It follows that Since we obtain For given > 0, choose > 0 satisfying and 0 = 0 ( ) satisfying (67) and so also satisfying (68). Next, take > 0 such that Then for each set ⊂ T with | | < , by (68)-(72), for every ∈ , we obtain ∫ (log + ( )) 2 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 {(log + ( )) : ∈ } is uniformly integrable on T. 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 {(log + ( )) : ∈ } forms a bounded subset of the space 1 (T); that is, there holds lim sup ∈ ∫ 2 0 (log + ( )) 2 < +∞. 8 The Scientific World Journal Similarly, it follows from (iii) that the family {(log + | * ( )|) : ∈ } is bounded in 1 (T).

Corollary 21. If is a subset of for which the family
is uniformly integrable, then the family is also uniformly integrable.
The following result gives a necessary condition for a subset of (= ) to be bounded.
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 > 0 there exists = ( ) > 0, such that and for every measurable set ⊂ T with the Lebesgue measure | | < .
Further, from the proof of (iii)⇒(i) of Theorem 19, we see that for each ∈ there is a measurable set ⊂ T depending on for which for almost every ∈ . From (78)-(80), we obtain (log + ( ) ) = ∫ + ∫ whence it follows that Choose a sequence { } of positive numbers such that ↓ 0.
Remark 23. The condition of Theorem 22 is not a sufficient condition for a set ⊂ to be bounded. To show this, define The Scientific World Journal 9 where = −1/2( +1) .
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 log * ( ) = 1/2( +1) , we see that is not bounded in .
Obviously, { } ⊂ and for each measurable set ⊂ T 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 ( , ) → 0 as → ∞, and thus ( ) → ( ) , uniformly on each closed disk | | ≤ < 1.
This contradiction shows that is not relatively compact in .