Weighted holomorphic Besov spaces on the polydisk

This work is an introduction of weighted Besov spaces of holomorphic functions on the polydisk. Let U be the unit polydisk in C and S be the space of functions of regular variation. Let 1 ≤ p < ∞, ω = (ω1, . . . , ωn) , ωj ∈ S(1 ≤ j ≤ n) and f ∈ H(U). The function f is said to be an element of the holomorphic Besov space Bp(ω) if ‖f‖pBp(ω) = ∫ Un |Df(z)| n ∏ j=1 ωj(1− |zj |) (1− |zj |2)2−p dm2n(z) < +∞, where dm2n(z) is the 2n -dimensional Lebesgue measure on U n and D stands for a special fractional derivative of f defined in the paper. For example, if n = 1 then Df is the derivative of the function zf(z) . We describe the holomorphic Besov space in terms of Lp(ω) space. Moreover projection theorems and theorems of the existence of a right inverse are proved.


Introduction
In the present article holomorphic weighted Besov spaces on the polydisk are introduced.Numerous authors have contributed to holomorphic Besov spaces in the unit disk in C and in the unit ball in C n , see K. Zhu [15], K. Stroethoff [13], Arazy-Fisher-Peetre [2], O. Blasco [3], A. Karapetyants [9].
The investigation of holomorphic Besov spaces on the polydisk is of special interest.The polydisk is a product of n disks.So one would expect that the results of the one-dimensional case are also valid in the n− dimensional case.But it turns out that this is not true.The polydisk case is different from the one-dimensional case and from the case of the n− dimensional ball.For example, let us consider the classical theorem of Privalov: if f ∈ Lip α , then Kf ∈ Lip α, where Kf is a Cauchy type integral.It is known that the analogue of this theorem for multidimensional Lipschitz classes is not true ( [14]), even though the analogue of this theorem for a sphere is valid ( [10]).In many cases, especially, when the class is defined by means of derivatives, the generalisation of function spaces in the polydisk is different from that on the unit ball.Our aim is to generalize the notion of Besov space in such way that all the known classical results of the one-variable case remain true here (for instance the description of continuous linear functionals, the projection theorems, etc.).Moreover, our generalisation will give results which are new also for functions of one variable.
In section 2, we give definitions and prove auxiliary lemmas.Note that in contrast to the one-dimensional non-weighted case, the use of weights allows us to define the Besov space as well for p = 1.
In section 3, we describe the holomorphic Besov space in terms of L p (ω)norms.
In section 4, projection theorems and theorems of the existence of a right inverse are proved.
Based on the results of this paper we shall characterize the duals of our Besov spaces in [7].

Definitions and auxiliary constructions
Denote by the unit polydisk in the n-dimensional complex plane C n , and by Definition 1.Let S be the class of all non-negative measurable functions ω on (0, 1) with where ε and η are measurable, bounded function on (0, 1).We assume that there are constants β ω < 0 and α ω with The functions in S are called functions of regular variation (see [11]).
Note that ω ∈ S always satisfies Let ω(t) = (ω 1 (t), . . ., ω n (t)).Throughout the paperwe assume that Besides, for any function f and g the notation f g (f g) We denote by A p (α), the set of all f ∈ H(U n ), for which where dm 2n (z) is the 2n− dimensional Lebesgue measure on U n .
The integral representation formula for the class A p (α) is a trivial consequence of the well known one-dimensional case: (1) for details see [4], [12].Note that the generalization of A p α spaces in terms of ω weights was used for the first time by F. A. Shamoyan [12] who greatly invested in the theory of weighted classes of functions in the polydisk.Definition 3. Let p ≥ 1.We denote by L p (ω) the set of all measurable functions on U n , for which Notice that L p (ω) is the L p − space with respect to the measure ω(1 − |z|)(1 − |z| 2 ) −2 dm 2n (z).In view on our conditions on ω (ω j ∈ S ) this measure is bounded.
The following definition gives the notion of the fractional differential.
Definition 4. For a holomorphic function where Γ(.) is the Gamma function and In particular, if the β j are positive integers, then We have Now we define holomorphic Besov spaces on the polydisk.
is indeed a norm.(We do not have to add |f (0)|).This follows from the fact that here Df = 0 implies f = 0 for holomorphic f .Definition 5 is closely related to the classical Besov space where p > 1, n = 1, and ω(t) = 1 (see [15]) and to the classical Dirichlet space where p = 2 ([1]).Indeed, recall that the norm of the classical Besov space is defined by and p > 1 .Hence, according to Definition 5, |||zf ||| p = ||f || Bp(ω) where ω(t) = 1 .However, in our paper we require ω j ∈ S and β ω < −1.This means in particular that ω(1 − |z|) tends to 0 as |z| tends to 1.We clearly have that B p (ω) is a subset of B 1 (ω) for any p > 1.
As in the one dimensional case, B p (ω) is a Banach space with respect to the norm (3).
To prove the main results we need the following auxiliary lemma: Then, using (1) for D β f and applying DD −β on both sides, we obtain A calculation shows that By partial integration we get where P (ζz) is some Polynom of order m + 1 − β.Then using (2) we get For the proof of ( 5) use the fact that Df ∈ A 1 (m) if m j ≥ α ω − 2 and 1 ≤ j ≤ n.Apply (1) to Df and then apply D β D −1 on both sides.Similarly to the proof of the first part of this lemma we obtain (5).
The proofs of the main theorems are also based on the following lemma.

Relations of B p (ω) to weighted Lebesgue spaces
As the first property, we prove that B p (ω) is a subspace of L p (ω) for any p.
Then using the inequality (5) with β = 0 and m + δ instead of m we get where δ > 0 and δp is sufficiently small.The Hölder inequality with 1/q + 1/p = 1 yields Here we used Lemma 2 b).With Lemma 2 we also get If p = 1 then (5) with β = 0 yields Here we used Lemma 2 again.
Proof.Let f ∈ B p (ω).Since then f ∈ B 1 (ω) we can apply (5).We get We consider the cases p > 1 and p = 1 separately.Let p > 1.Using Hölder's inequality we get from (6) with Lemma 2 Therefore Using Lemma 2 we have Then by Fubini theorem and ( 6) we get In the last inequality we have used Lemma 2 again.Summing up, we have proved, that g ∈ L p (ω) and ( 7) Conversely, let f ∈ H(U n ) and g ∈ L p (ω).Then using (4) and Lemma 2, in the case p > 1, we have This proves f Bp(ω) g L p (ω) .In the case p = 1 we have, using Lemma 2, and therefore f B1(ω) g L1(ω) .So we have proved that f ∈ B p (ω) and ( 8) From ( 7) and ( 8) we get f Bp(ω) g L p (ω) .
B 1 (ω i1 . . .ω i k ) will be called holomorphic Besov space with respect to (i 1 , . . ., i k ).The next theorem shows how the two Besov spaces B 1 and B1 are related.
Our aim is to show, that the following function belongs to L 1 (ω) , using the properties of ω j we obtain Df ∈ A 1 (m).Applying (1) to Df and integrating with respect to z, we get where the function (1 − (1 − ζz) m+1 )/ζz is bounded and has bounded derivatives on U n .Differentiating the last equality we get where m 2 = (m 2 , . . ., m n ).Then, in view of Lemma 2, Repeating the same argument, we get similarly

Bounded and right inverse operators on B p (ω)
Let us consider the following linear operator on L p (ω) where The problem is to characterize the functions P α (g) with g ∈ L p (ω).The next theorem shows that P α is bounded on B p (ω).
Let p > 1.Then, with the Hölder inequality (1/p + 1/q = 1 ) and Lemma 2, We have used the fact that α j p > α ωj − 2 and have taken β j so, that Proof of Remark 1.Let P α be bounded from L p (ω) to B p (ω) and let Lemma 2).We consider the domain where (1 ≤ j ≤ n) and U n = U 1 × . . .× U n .Further, let V n be the following polydisk centered at (r 1 , ..., r n ) Obviously V n ⊂ U n .Fix β such that α j +β j −1 > 0 .Then, using Theorem 2 and ([8], Theorem 1.9), we get where P α is the corresponding integral from Theorem 1.9 (see [8]).If Is not difficult to show that (we have used the inequality 1 − r|z| ≤ |1 − rz|).If we assume that α j p ≤ α ωj − 2 for some j , then for the corresponding integral taking ω j (t) = t αω j we get Consequently,

the set of all bounded measurable functions on U n and let T h (f ) be the Toeplitz operator of the form T
The proof of Theorem 5 is similar to the proof of Thoerem 4. We omit the details.
Remark 2. As in the case of P α , one can prove that the condition α j p > α ωj − 2 (1 ≤ j ≤ n) in Theorem 5 is necessary, too( see Remark 1).
We do not know whether the operators P α and R α in Theorem 5 and 6 are surjective for p > 1, too.
We pass to the analysis of the right inverse map and consider the question: does there exist operators R (where ω is the weight of Theorem 5).The answer turns out to be positive.In the next theorem we discuss the boundedness of T β,γ .Theorem 6.Let α j > α ωj − 2, β j > 0 (1 ≤ j ≤ n).Then (1) T β,α is bounded from B p (ω) to L p ( ω) and R α (T β,α )f (z) ≡ f (z), z ∈ U n .(2) T β,0 is bounded from B p (ω) to L p (ω) and P α (T β,0 )f (z) ≡ f (z), z ∈ U n .