Singular Integral Operators with Bergman-Besov Kernels on the Ball

H. Turgay Kaptanoğlu and A. Ersin Üreyen

Abstract. We completely characterize in terms of the six parameters involved the boundedness of all standard weighted integral operators induced by Bergman-Besov kernels acting between different Lebesgue classes with standard weights on the unit ball of ℂ^{N}. The integral operators generalize the Bergman-Besov projections. To find the necessary conditions for boundedness, we employ a new versatile method that depends on precise imbedding and inclusion relations among various holomorphic function spaces. The sufficiency proofs are by Schur tests or integral inequalities.

Mathematics Subject Classification (2010). Primary 47B34, 47G10; Secondary 32A55, 45P05, 46E15, 32A37, 32A36, 30H25, 30H20.

Keywords. Integral operator, Bergman-Besov kernel, Bergman-Besov space, Bloch-Lipschitz space, Bergman-Besov projection, radial fractional derivative, Schur test, Forelli-Rudin estimate, inclusion relation.

1. Introduction

The Bergman projection is known to be a bounded operator on L^{p} of the disc for all p > 1 ever since [17]. Weighted versions in several variables are considered with the help of the Schur test in [7] resulting in projections also for p = 1. After many modifications, integral operators similar to Bergman projections are investigated between different Lebesgue classes on the ball in several publications, such as the more recent [18].

Generalizations to other types of spaces on various domains with differing kernels are too numerous to mention here. But a complete analysis of the integral operators arising from Bergman kernels between Lebesgue classes is rather new and is attempted in [3] on the disc and for one single kernel in [2] on the ball. Here we undertake and complete the task of extending and generalizing their work to the ball, to weighted operators, to all Bergman-Besov kernels, and to Lebesgue classes with standard weights but with different exponents. Many of our results are new even in the disc.

We present our results after giving a minimal amount of notation. Let 𝔹 be the unit ball in ℂ^{N} with respect to the norm |z| = \sqrt{⟨z, z⟩} induced by the inner product ⟨w, z⟩ = w_{1}\overline{z}_{1} + ⋯ + w_{N} \overline{z}_{N}, which is the unit disc 𝔻 for N = 1. Let H(𝔹) and H^{∞} denote the spaces of all and bounded holomorphic functions on 𝔹, respectively.

We let ν be the Lebesgue measure on 𝔹 normalized so that ν(𝔹) = 1. For q ∈ ℝ, we also define on 𝔹 the measures

dν_{q}(z) := (1 − |z|^{2})^{q} dν(z).

These measures are finite for q > −1 and σ-finite otherwise. For 0 < p < ∞, we denote the Lebesgue classes with respect to ν_{q} by L_{q}^{p}, writing also L^{p} = L_{0}^{p}. The Lebesgue class L_{q}^{∞} of essentially bounded functions on 𝔹 with respect to any ν_{q} is the same (see [10, Proposition 2.3]), so we denote them all by L^{∞}.

Definition 1.1. For q ∈ ℝ and w, z ∈ 𝔹, the Bergman-Besov kernels are

K_{q}(w, z) :=\begin{cases}\frac{1}{(1−⟨w, z⟩)^{1+N+q}} =∑_{k=0}^{∞} \frac{(1+N +q)_{k}}{k!} ⟨w, z⟩^{k}, & q > −(1+N), \\ _{2}F_{1}(1, 1; 1−(N +q); ⟨w, z⟩) =∑_{k=0}^{∞} \frac{k! ⟨w, z⟩^{k}}{(1−(N +q))_{k}} , & q ≤ −(1+N), \end{cases}

where _{2}F_{1} ∈ H(𝔻) is the Gauss hypergeometric function and (u)_{v} is the Pochhammer symbol. They are the reproducing kernels of Hilbert Bergman-Besov spaces.

The kernels K_{q} for q > −(1+N) can also be written as_{2}F_{1}(1, 1+N+q; 1) to complete the picture. The kernels K_{q} for q < −(1 + N) appear in the literature first in [1, p. 13]. Notice that

K_{−(1+N)}(w, z) = \frac{1}{⟨w, z⟩} log \frac{1}{1 − ⟨w, z⟩}.

For a, b ∈ ℝ, the operators acting from L_{q}^{p} to L_{Q}^{P} that we investigate are

T_{ab}f(w) = ∫_{𝔹} K_{a}(w, z)f(z)(1 − |z|^{2})^{b} dν(z)

and

S_{ab}f(w) = ∫_{𝔹} |K_{a}(w, z)| f(z)(1 − |z|^{2})^{b} dν(z).

Our main results are the following two theorems that describe their boundedness in terms of the 6 parameters (a, b, p, q, P, Q) involved. Note that at the endpoints when p or P is 1 or ∞ in Theorem 1.2, the inequalities of (III) take several different forms. They are described in more detail in Remark 1.4 immediately following. We use S_{ab} solely because we need operators with positive kernels in Schur tests.

Theorem 1.2. Let a, b, q, Q ∈ ℝ, 1 ≤ p ≤ P ≤ ∞, and assume Q > −1 when P < ∞. Then the conditions (I), (II), (III) are equivalent.

(I) T_{ab} : L_{q}^{p} → L_{Q}^{P} is bounded. (II) S_{ab} : L_{q}^{p} → L_{Q}^{P} is bounded.

(III) \frac{1 + q}{p} < 1+b and a ≤ b+ \frac{1 + N + Q}{P} − \frac{1 + N + q}{p} for 1 < p ≤ P < ∞;

\frac{1 + q}{p} ≤ 1+b and a ≤ b+ \frac{1 + N + Q}{P} − \frac{1 + N + q}{p} for 1 = p ≤ P ≤ ∞,

but at least one inequality must be strict;

\frac{1 + q}{p} < 1+b and a < b+ \frac{1 + N + Q}{P} − \frac{1 + N + q}{p} for 1 < p ≤ P = ∞.

Theorem 1.3. Let a, b, q, Q ∈ ℝ, 1 ≤ P < p ≤ ∞, and assume Q > −1. Then the conditions (I), (II), (III) are equivalent.

(I) T_{ab} : L_{q}^{p} → L_{Q}^{P} is bounded. (II) S_{ab} : L_{q}^{p} → L_{Q}^{P} is bounded.

(III) \frac{1 + q}{p} < 1 + b and a < b + \frac{1 + Q}{P} − \frac{1 + q}{p}.

Remark 1.4. When p or P is 1 or ∞, clearly the inequalities in (III) get simplified by cancellation or by 1/∞ = 0. Considering all possible relative values of p and P, there are 10 distinct cases each of which requiring a somewhat different proof, 6 cases for Theorem 1.2 and 4 cases for Theorem 1.3. We list below all ten of them and the exact form of (III) for each, while (I) and (II) staying the same as above. The cases ⟨1⟩ and ⟨7⟩ are the generic cases for p ≤ P and P < p, respectively, and the other 8 cases are the endpoints at 1 or ∞.

⟨1⟩ 1 < p ≤ P < ∞: (III) \frac{1+q}{p} < 1+b and a ≤ b + \frac{1+N +Q}{P} − \frac{1+N +q}{p}.

⟨2⟩ 1 = p = P : (III) q ≤ b and a ≤ b+Q−q, but not both = simultaneously.

⟨3⟩ 1 = p < P < ∞: (III) q ≤ b and a ≤ b + \frac{1 + N + Q}{P} − (1 + N + q),

but not both = simultaneously. ⟨4⟩ 1 = p < P = ∞: (III) q ≤ b and a ≤ b − (1 + N + q), but not both =

simultaneously.

⟨5⟩ 1 < p < P = ∞: (III) \frac{1 + q}{p} < 1 + b and a < b − \frac{1 + N + q}{p}.

⟨6⟩ p = P = ∞: (III) 0 < 1 + b and a < b.

⟨7⟩ 1 < P < p < ∞: (III) \frac{1 + q}{p} < 1 + b and a < b + \frac{1 + Q}{P} − \frac{1 + q}{p}.

⟨8⟩ 1 = P < p < ∞: (III) \frac{1 + q}{p} < 1 + b and a < b + (1 + Q) − \frac{1 + q}{p}.

⟨9⟩ 1 = P < p = ∞: (III) 0 < 1 + b and a < b + (1 + Q).

⟨10⟩ 1 < P < p = ∞: (III) 0 < 1 + b and a < b + \frac{1 + Q}{P}.

In the proofs, “Necessity” refers to the implication (I) ⇒ (III), and “Sufficiency” to the implication (III) ⇒ (II). The implication (II) ⇒ (I) is obvious.

Remark 1.5. The condition Q > −1 when P < ∞ in Theorems 1.2 and 1.3 cannot be removed as we explain in Corollary 4.11 below. This condition arises from the fact that T_{ab} generates holomorphic functions and |T_{ab}f|^{P} is subharmonic for P < ∞. It is important to note that this condition does not put any extra constraint when P = ∞ since L_{Q}^{∞} = L^{∞} for any Q. It is no surprise that those terms in the inequalities in (III) that contain Q disappear when P = ∞. This phenomenon occurs in the cases ⟨4⟩, ⟨5⟩, ⟨6⟩, in which Q ∈ ℝ. So Q > −1 is meaningful in the remaining 7 cases.

Remark 1.6. When a = Q in the case ⟨2⟩, when a = (1+N +Q)/P −(1+N) in the case ⟨3⟩, and when a = −(1 + N) in the case ⟨4⟩, the two inequalities in (III) are the same and are q ≤ b. In such cases, q = b cannot hold as stated above and proved in Theorem 6.3 below. So for these special values of a, we have (III) q < b in the cases ⟨2⟩, ⟨3⟩, ⟨4⟩.

Remark 1.7. If a ≤ −(1 + N), then the first inequality in Theorem 1.2 (III) implies the second. Indeed, the given conditions and the first inequality imply

\frac{1 + N + Q}{P} − a ≥ \frac{N}{P} + 1 + N > 1 + N ≥ \frac{1 + q}{p} − b + \frac{N}{p} ,

in which the last inequality is strict in the cases ⟨5⟩, ⟨6⟩, whence the second inequality. Similarly, if a ≤ −1, then the first inequality in Theorem 1.3 (III) implies the second. Indeed, the given conditions and the first inequality imply (1 + Q)/P − a > 0 + 1 > (1 + q)/p − b, whence the second inequality.

The special case with N = 1, a > −2, b = 0, and q = Q = 0 of both Theorems 1.2 and 1.3 appear in [3, Theorems 1, 2, 3, 4]; also the more restricted case with a = −N, b = 0, and q = Q = 0 in [2, Theorem 2]. These two references do not consider any kernels of ours for a ≤ −(1 + N). Those parts of only Theorem 1.2 with a > −(1 + N), q > −1, and P < ∞ appear in [18, Theorems 3 and 4], and its parts with a > −(1 + N) and p = P = ∞ in [11, Theorem 7.2].

Everything else in Theorems 1.2 and 1.3 is new, even for N = 1. Thus we present a complete picture on 𝔹 as far as the standard weights are concerned.

In [18], the kernels considered for q ≤ −(1 + N) are simply the binomial form of K_{q}(w, z) with powers negated, but these kernels are not natural in the sense that they are not positive definite, and hence cannot be reproducing kernels of Hilbert spaces; see [6, Lemma 5.1] or [9, Corollary 6.3]. Our kernels are the reproducing kernels of the Hilbert Bergman-Besov spaces B_{q}^{2} and thus are positive definite. In particular, [18] does not consider a logarithmic kernel. But see [18] also for further references to earlier results.

In [4, Theorem 1.2], the authors prove a result similar to the sufficiency of Theorem 1.2 for T_{ab} with parameters corresponding to a = 0, b ≥ 0, and q = Q = 0 on the more general smoothly bounded, strongly pseudoconvex domains. When they further restrict to N = 1, to 𝔻, and to 1 < p < ∞ keeping a = 0, b ≥ 0, and q = Q = 0, they also obtain the necessity result of Theorem 1.2. They further discuss that the second inequality in Theorem 1.2 (III) may not depend on N, but it turns out here that it does. We do not attempt to survey the large literature on more general domains or on more general weights. We do not also try to estimate the norms of the main operators.

However, we do consider a variation of Theorems 1.2 and 1.3 that removes the annoying condition Q > −1 when P < ∞. We achieve this by mapping T_{ab} into the Bergman-Besov spaces B_{Q}^{P} (see Section 3 and in particular Definition 3.2) instead of the Lebesgue classes. In conjuction with the claim T_{ab}f ∈ H(𝔹) of Corollary 4.11, this variation seems quite natural.

Theorem 1.8. Supppose a, b, q, Q ∈ ℝ, 1 ≤ p ≤ ∞, and 1 ≤ P < ∞. Then T_{ab} : L_{q}^{p} → B_{Q}^{P} is bounded if and only if the two inequalities in (III) hold in the 7 cases with P < ∞, that is, excluding the cases ⟨4⟩, ⟨5⟩, ⟨6⟩.

The very particular case of Theorem 1.8 in which a = −N, b = 0, q = 0, Q = −N, and P = 2 is in [2, Theorem 1]. The following can be considered as the P = ∞ version of Theorem 1.8, but it hardly says anything new in view of Corollary 4.11.

Theorem 1.9. Supppose a, b, q ∈ ℝ and 1 ≤ p ≤ ∞. Then T_{ab} : L_{q}^{p} → H^{∞} is bounded if and only if the two inequalities in (III) hold in the 3 cases with P = ∞, that is, in the cases ⟨4⟩, ⟨5⟩, ⟨6⟩.

Unlike earlier work, methods of proof we employ are uniform throughout the ten cases. The sufficiency proofs are either by Schur tests or by direct Hölder or Minkowski type inequalities which also make use of growth rate estimates of Forelli-Rudin type integrals. The necessity proofs are by an original technique that heavily depends on the precise imbedding and inclusion relations among holomorphic function spaces on 𝔹. This technique has the potential to be used also with other kernels and spaces. By contrast, we do not use any results on Carleson measures or coefficient multipliers employed in earlier works. Our new technique is also the reason why we give all the proofs in detail, including those particular cases that are proved elsewhere by other means. It makes this paper more or less self-contained apart from some standard results and the inclusion relations.

The proofs of Theorems 1.2 and 1.3 are rather long and are presented late, necessity parts in Section 6 and sufficiency parts in Section 7. The proofs of Theorems 1.8 and 1.9 occupy Section 8. Before the proofs, we list the major standard results we use in Section 5. Earlier in Section 4, we place the main operators in context and develop their elementary properties. It is also here that we obtain the condition Q > −1. Section 3 covers the necessary background on Bergman-Besov spaces. In the next Section 2, we exhibit the regions of boundedness of the main operators graphically.

2. Graphical Representation

The repeated terms in the inequalities in (III) of Theorems 1.2 and 1.3 suggest that the 6 parameters in them can be combined in interesting ways and the region of boundedness of T_{ab} : L_{q}^{p} → L_{Q}^{P} can be described geometrically with fewer variables. In this direction, we let

x = \frac{1 + q}{p} − b  and  y = \frac{1 + Q}{P} − a.

In [3], such a region of boundedness is graphed in the 1/p-1/P -plane and it is almost the same as our xy-plane in the absence of b, q, Q and with N = 1. With 4 extra parameters and more freedom for a, something similar is still possible.

There are natural bounds for x, y imposed by the bounds 1 ≤ p, P ≤ ∞. Since also Q > −1, the region of boundedness of T_{ab} lies in the rectangle R determined by −a ≤ y ≤ 1 + Q − a and one of −b ≤ x ≤ 1 + q − b and 1 + q − b ≤ x ≤ −b depending on the sign of 1 + q. Otherwise R is free to move in the xy-plane. Note that R degenerates to a vertical line segment at x = −b when q = −1.

Figure 1. A typical region R ∩ U.

In Theorem 1.3, the inequalities of (III) can now be written in the form x < 1 and x < y. Each inequality determines a half plane whose intersection we call U. For P < p, the operator T_{ab} : L_{q}^{p} → L_{Q}^{P} is bounded precisely when (x, y) ∈ R ∩ U. The intersection R ∩ U can be triangular, quadrilateral, or pentagonal, or as simple as a vertical line segment. Part of Remark 1.7 is clearer now since once y > 1, if R is to the left of x = 1, then it is also above y = x. A typical R ∩ U is illustrated in Figure 1.

In Theorem 1.2, the inequalities of (III), say in the case ⟨1⟩, can now be

written in the form x < 1 and x + d ≤ y, where d = N(\frac{1}{p} − \frac{1}{P} ) ≥ 0. Each inequality again determines a half plane whose intersection we call V. For p ≤ P, the operator T_{ab} : L_{q}^{p} → L_{Q}^{P} is bounded precisely when (x, y) ∈ R∩V. The shape of R ∩ V is like that of R ∩ U, but now R ∩ V can also be a single point when the upper left corner of R lies on the line y = x+d. This happens only in the case ⟨2⟩ with 1 = p = P and hence d = 0, and the minimum value of x equaling the maximum value of y in R. So with 1 + q < 0, we have 1 + q − b = 1 + Q − a, which is actually the equality in the second inequality of (III). Then the first inequality in (III) must be strict and be q < b. Of course Q > −1. So a single point in the xy-plane need not correspond to a single set of values for the six parameters. This phenomenon does not occur for P < p since both inequalities in (III) are then strict. The other part of Remark 1.7 is clearer now since once y > 1 + N ≥ 1 + d, if R is to the left of x = 1, then it is also above y = x + d. A typical R ∩ V is illustrated in Figure 2.

Figure 2. A typical region R ∩ V.

The form of the variables x, y brings to mind whether or not our results on weighted spaces can be obtained from those on unweighted spaces with q = Q = 0. It turns out that they can and we explain how in Remark 8.1. However, our proofs are not simplified significantly with q = Q = 0. The classification in Remark 1.4 is according to p, P and there seems to be no simple way of reducing it to fewer cases, because the inequalities in (III) can change between < and ≤ without any apparent reason with p, P and the norms on the spaces are different when p or P is ∞. Proving everything in full generality in one pass is a good idea.

3. Preliminaries on Spaces

We let 1 ≤ p, p′ ≤ ∞ be conjugate exponents, that is, 1/p + 1/p′ = 1, or equivalently, p′ = p/(p − 1). In multi-index notation, γ = (γ_{1}, …, γ_{N}) ∈ ℕ^{N} is an N-tuple of nonnegative integers, |γ| = γ_{1} + ⋯ + γ_{N}, γ! = γ_{1}! ⋯ γ_{N}!, 0^{0} = 1, and z^{γ} = z_{1}^{γ_{1}} ⋯ z_{N}^{γ_{N}}.

Let 𝕊 be the unit sphere in ℂ^{N}, which is the unit circle 𝕋 when N = 1. We let σ be the Lebesgue measure on 𝕊 normalized so that σ(𝕊) = 1. The polar coordinates formula that relates σ and ν as given in [16, § 1.4.3] is

∫_{𝔹} f(z) dν(z) = 2N ∫_{0}^{1} r^{2N−1} ∫_{𝕊} f(rζ) dσ(ζ) dr,

in which z = rζ, and we also use w = ρη with ζ, η ∈ 𝕊 and r, ρ ≥ 0. For α ∈ ℝ, we also define the weighted classes

L_{α}^{∞} := { φ measurable on 𝔹 : (1 − |z|^{2})^{α} φ(z) ∈ L^{∞} }

so that L_{0}^{∞} = L^{∞}, which are normed by

∥φ∥_{L_{α}^{∞}} := ess sup_{z∈𝔹} (1 − |z|^{2})^{α} |φ(z)|.

The norm on L^{∞} carries over to H^{∞} with sup.

We show an integral inner product on a space X of functions by [ · , ·]_{X}. Let’s explain the notation used in Definition 1.1. The Pochhammer sym-

bol (u)_{v} is defined by

(u)_{v} := \frac{Γ(u + v)}{Γ(u)}

when u and u + v are off the pole set −ℕ of the gamma function Γ. In particular, (u)_{0} = 1 and (u)_{k} = u(u + 1) ⋯ (u + k − 1) for a positive integer k. The Stirling formula yields

\frac{Γ(t + u)}{Γ(t + v)} ∼ t^{u−v},  \frac{(u)_{t}}{(v)_{t}} ∼ t^{u−v},  \frac{(t)_{u}}{(t)_{v}} ∼ t^{u−v}  (Re t → ∞), (1)

where x ∼ y means both x = O(y) and y = O(x) for all x, y in question. If only x = O(y), we write x ≲ y. A constant C may attain a different value at each occurence. The Gauss hypergeometric function _{2}F_{1} ∈ H(𝔻) is defined by

2F_{1}(u, v; t; z) = ∑_{k=0}^{∞} \frac{(u)_{k}(v)_{k}}{(t)_{k}(1)_{k}} z^{k}.

A large part of this work depends on the interactions between the Lebesgue classes and the Bergman-Besov spaces. Given q ∈ ℝ and 0 < p < ∞, let m be a nonnegative integer such that q + pm > −1. In more common notation, the Bergman-Besov space B_{q}^{p} consists of all f ∈ H(𝔹) for which

(1 − |z|^{2})^{m} \frac{∂^{m}f}{∂z_{1}^{γ_{1}} ⋯ ∂z_{N}^{γ_{N}}} ∈ L_{q}^{p}

for every multi-index γ = (γ_{1}, …, γ_{N}) with γ_{1} + ⋯ + γ_{N} = m.

Likewise, given α ∈ ℝ, let m be a nonnegative integer so that α+m > 0. The Bloch-Lipschitz space B_{α}^{∞} consists of all f ∈ H(𝔹) for which

(1 − |z|^{2})^{m} \frac{∂^{m}f}{∂z_{1}^{γ_{1}} ⋯ ∂z_{N}^{γ_{N}}} ∈ L_{α}^{∞}

for every multi-index γ = (γ_{1}, …, γ_{N}) with γ_{1} + ⋯ + γ_{N} = m.

However, partial derivatives are more difficult to use in the context of this paper and we follow an equivalent path. So we now introduce the radial fractional derivatives that not only allow us to define the holomorphic variants of the L_{q}^{p} spaces more easily, but also form some of the most useful operators in this paper.

First let the coefficient of ⟨w, z⟩^{k} in the series expansion of K_{q}(w, z) in Definition 1.1 be c_{k}(q). So

K_{q}(w, z) = ∑_{k=0}^{∞} c_{k}(q)⟨w, z⟩^{k}  (q ∈ ℝ),  (2)

where evidently the series converges absolutely and uniformly when one of the variables w, z lies in a compact subset of 𝔹. Note that c_{0}(q) = 1, c_{k}(q) > 0 for any k, and by (1),

c_{k}(q) ∼ k^{N+q}  (k → ∞),  (3)

for every q. This explains the choice of the parameters of the hypergeometric function in K_{q}.

Definition 3.1. Let f ∈ H(𝔹) be given on 𝔹 by its convergent homogeneous

expansion f = ∑_{k=0}^{∞} f_{k} in which f_{k} is a homogeneous polynomial in z_{1}, …, z_{N}

of degree k. For any s, t ∈ ℝ, we define the radial fractional differential operator D_{s}^{t} on H(𝔹) by

D_{s}^{t}f := ∑_{k=0}^{∞} d_{k}(s, t)f_{k} := ∑_{k=0}^{∞} \frac{c_{k}(s + t)}{c_{k}(s)}f_{k}.

Note that d_{0}(s, t) = 1 so that D_{s}^{t}(1) = 1, d_{k}(s, t) > 0 for any k, and

d_{k}(s, t) ∼ k^{t} (k → ∞),

for any s, t by (3). So D_{s}^{t} is a continuous operator on H(𝔹) and is of order t. In particular, D_{s}^{t}z^{γ} = d_{|γ|}(s, t)z^{γ} for any multi-index γ. More importantly,

D_{s}^{0} = I,  D_{s+t}^{u}D_{s}^{t} = D_{s}^{t+u},  and  (D_{s}^{t})^{−1} = D_{s+t}^{−t}  (4)

for any s, t, u, where the inverse is two-sided. Thus any D_{s}^{t} maps H(𝔹) onto itself.

Consider now the linear transformation I_{s}^{t} defined for f ∈ H(𝔹) by

I_{s}^{t}f(z) := (1 − |z|^{2})^{t} D_{s}^{t}f(z).  (5)

Definition 3.2. For q ∈ ℝ and 0 < p < ∞, we define the Bergman-Besov space B_{q}^{p} to consist of all f ∈ H(𝔹) for which I_{s}^{t}f belongs to L_{q}^{p} for some s, t satisfying

q + pt > −1.  (6)

It is well-known that under (6), Definition 3.2 is independent of s, t and the norms ∥f∥_{B_{q}^{p}} := ∥I_{s}^{t}f∥_{L_{q}^{p}} are all equivalent. Explicitly,

∥f∥_{B_{q}^{p}}^{p} = ∫_{𝔹} |D_{s}^{t}f(z)|^{p}(1 − |z|^{2})^{q+pt} dν(z)  (q + pt > −1).  (7)

When q > −1, we can take t = 0 in (6) and obtain the weighted Bergman spaces A_{q}^{p} = B_{q}^{p}. We also wite A^{p} = A_{0}^{p}. For 0 < p < 1, what we call norms are actually quasinorms.

Definition 3.3. For α ∈ ℝ, we define the Bloch-Lipschitz space B_{α}^{∞} to consist of all f ∈ H(𝔹) for which I_{s}^{t}f belongs to L_{α}^{∞} for some s, t satisfying

α + t > 0.  (8)

It is well-known that under (8), Definition 3.3 is independent of s, t and the norms ∥f∥_{B_{α}^{∞}} := ∥I_{s}^{t}f∥_{L_{α}^{∞}} are all equivalent. Explicitly,

∥f∥_{B_{α}^{∞}} = sup_{z∈𝔹} |D_{s}^{t}f(z)|(1 − |z|^{2})^{α+t} (α + t > 0).

If α > 0, we can take t = 0 in (8) and obtain the weighted Bloch spaces. When α < 0, then the corresponding spaces are the holomorphic Lipschitz spaces Λ_{−α} = B_{α}^{∞}. The usual Bloch space B_{0}^{∞} = B^{∞} corresponds to α = 0. Admittedly this notation is a bit unusual, but there is no mention of little Bloch spaces in this paper.

It is also well-known that a Bergman-Besov space defined using partial derivatives or radial fractional differential operators contain the same functions. The same is true also for a Bloch-Lipschitz space.

Remark 3.4. Definitions 3.2 and 3.3 imply that I_{s}^{t} imbeds B_{q}^{p} isometrically into L_{q}^{p} if and only if (6) holds, and I_{s}^{t} imbeds B_{α}^{∞} isometrically into L_{α}^{∞} if and only if (8) holds.

Much more information about the spaces B_{q}^{p} and B_{α}^{∞} and their interconnections can be found in [12].

4. Properties of Kernels and Operators

It is well-known that every B_{q}^{2} space is a reproducing kernel Hilbert space and its reproducing kernel is K_{q}. Thus K_{q}(w, z) is positive definite for w, z ∈ 𝔹. Further, K_{q}(w, z) is holomorphic in w ∈ 𝔹 and satisfy \overline{K_{q}(z, w)} = K_{q}(w, z). In particular, for q > −1, the B_{q}^{2} are weighted Bergman spaces A_{q}^{2}, B_{−1}^{2} is the Hardy space H^{2}, B_{−N}^{2} is the Drury-Arveson space, and B_{−(1+N)}^{2} is the Dirichlet space. Moreover, the norm on B_{q}^{2} obtained from its reproducing kernel K_{q} is equivalent to its integral norms given in (7).

For ease of writing, put ω = ⟨w, z⟩ ∈ 𝔻 and let k_{q}(ω) = K_{q}(w, z), which is holomorphic in ω ∈ 𝔻.

Lemma 4.1. For q < −(1 + N), each |K_{q}(w, z)| is bounded above as w, z vary in 𝔹. More importantly, each |K_{q}(w, z)| with q ∈ ℝ is bounded below by a positive constant as w, z vary in 𝔹. Therefore no K_{q} has a zero in 𝔹 × 𝔹.

Proof. When q < −(1 + N), the growth rate (3) of the coefficient c_{k} of ω^{k} shows that the power series of k_{q}(ω) converges uniformly for ω ∈ \overline{𝔻}. This shows boundedness. Next we consult [15, Equation 15.13.1] and see that k_{q} is not 0 on a set containing \overline{𝔻} \ \{1\}. The reason for this is that the first parameter 1 of the hypergeometric function defining k_{q} is positive. But also k_{q}(1) ≠ 0 clearly. Thus |k_{q}| is bounded below on \overline{𝔻}.

If q = −(1 + N), then k_{−(1+N)}(ω) = ω^{−1} log(1 − ω)^{−1}. On \overline{𝔻} \ \{1\}, k_{−(1+N)} is not zero and |k_{q}(ω)| blows up as ω → 1 within \overline{𝔻}. So |k_{q}| is bounded below on \overline{𝔻}.

The claim about |K_{q}| for q > −(1 + N) is obvious and the lower bound can be taken as 2^{−(1+N+q)}. □

One of the best things about the radial differential operators D_{s}^{t} is that they allow us to pass easily from one kernel to another and from one space to another in the same family. First, it is immediate that

D_{q}^{t}K_{q}(w, z) = K_{q+t}(w, z)  (q, t ∈ ℝ),  (9)

where differentiation is performed on the holomorphic variable w. But the more versatile result is the following, which is a combination of [11, Proposition 3.2 and Corollary 8.5].

Theorem 4.2. Let q, α, s, t ∈ ℝ and 0 < p < ∞ be arbitrary. Then the maps D_{s}^{t} : B_{q}^{p} → B_{q+pt}^{p} and D_{s}^{t} : B_{α}^{∞} → B_{α+t}^{∞} are isomorphisms, and isometries when the parameters of the norms of the spaces are chosen appropriately.

Definition 4.3. For b ∈ ℝ, the Bergman-Besov projections are the operators T_{bb} acting on a suitable Lebesgue class with range in a holomorphic function space, both on 𝔹.

The following general result describes the boundedness of Bergman-Besov projections on all two-parameter Bergman-Besov spaces and the usual Bloch space, and is [10, Theorem 1.2]. It is also an indicator of the importance of the operators I_{s}^{t} which we use repeatedly in this paper.

Theorem 4.4. For q ∈ ℝ and 1 ≤ p < ∞, the T_{bb} : L_{q}^{p} → B_{q}^{p} is bounded if and only if

\frac{1 + q}{p} < 1 + b.  (10)

And T_{bb} : L^{∞} → B^{∞} is bounded if and only if

0 < 1 + b.  (11)

Moreover, given b satisfying (10), if t satisfies (6), then T_{bb}I_{b}^{t} = Cf for f ∈ B_{q}^{p}. And given b satisfying (11), if t satisfies (8) with α = 0, then T_{bb}I_{b}^{t}f = Cf for f ∈ B^{∞}. Here C is a constant that depends on N, b, t, but not on f.

The next lemma is adapted from [5, Lemma 3.2]. To see what it means, first check that K_{q}(0, z) = 1 for all z ∈ 𝔹.

Lemma 4.5. For each q ∈ ℝ, there is a ρ_{0} < 1 such that for |w| ≤ ρ_{0} and all z ∈ 𝔹, we have Re K_{q}(w, z) ≥ 1/2.

Proof. By (3), |K_{q}(w, z)| ≤ 1 + C ∑_{k=1}^{∞} k^{N+q}|⟨w, z⟩|^{k} for some constant C and

C ∑_{k=1}^{∞} k^{N+q}|⟨w, z⟩|^{k} ≤ C ∑_{k=1}^{∞} k^{N+q}|w|^{k}|z|^{k} ≤ C|w| ∑_{k=1}^{∞} k^{N+q}|w|^{k−1}

for all z, w ∈ 𝔹. The last series converges, say, for |w| = 1/2; call its sum W and set ρ_{0} = min{1/2, 1/2CW }. If |w| ≤ ρ_{0}, then

|C ∑_{k=1}^{∞} k^{N+q}⟨w, z⟩^{k}| ≤ CW |w| ≤ \frac{1}{2} (z ∈ 𝔹).

That is, |K_{q}(w, z) − 1| ≤ 1/2 for |w| ≤ ρ_{0} and all z ∈ 𝔹. This implies the desired result. □

We turn to the operators T_{ab} and formulate their behavior in many important situations. But first we insert some obvious inequalities we use many times in the proofs. If c < d, u > 0, and v ∈ ℝ, then for 0 ≤ r < 1,

(1 − r^{2})^{d} ≤ (1 − r^{2})^{c}  and  (1 − r^{2})^{u}(\frac{1}{r^{2}} log \frac{1}{1 − r^{2}} )^{−v} ≲ 1. (12)

The second leads to an estimate we need several times.

Lemma 4.6. For u, v ∈ ℝ,

∫_{0}^{1}(1 − r^{2})^{u}(\frac{1}{r^{2}} log \frac{1}{1 − r^{2}} )^{−v} dr < ∞

if u > −1 or if u = −1 and v > 1, and the integral diverges otherwise.

Proof. The only singularity of the integrand is at r = 1. For u ≠ −1, polynomial growth dominates a logarithmic one. For u = −1, we reduce the integral into one studied in calculus after changes of variables. □

We call

f_{uv}(z) = (1 − |z|^{2})^{u}(\frac{1}{|z|^{2}} log \frac{1}{1 − |z|^{2}} )^{−v}

test functions, because we derive half the necessary conditions of Theorems 1.2 and 1.3 from the action of T_{ab} on them. Here u, v ∈ ℝ and f_{00} = 1. When we apply Lemma 4.6 to the f_{uv}, we obtain the next result.

Lemma 4.7. For 1 ≤ p < ∞, we have f_{uv} ∈ L_{q}^{p} if and only if q + pu > −1, or q + pu = −1 and pv > 1. For p = ∞, we have f_{uv} ∈ L^{∞} if and only if u > 0, or u = 0 and v ≥ 0.

Lemma 4.8. If b + u > −1 or if b + u = −1 and v > 1, then T_{ab}f_{uv} is a finite positive constant. Otherwise, |T_{ab}f_{uv}(w)| = ∞ for |w| ≤ ρ_{0}, where ρ_{0} is as in Lemma 4.5.

Proof. If b + u > −1, or b + u = −1 and v > 1, then by polar coordinates and the mean value property,

T_{ab}f_{uv}(w) = ∫_{𝔹} K_{a}(w, z)(1 − |z|^{2})^{b+u}(\frac{1}{|z|^{2}} log \frac{1}{1 − |z|^{2}} )^{−v} dν(z)

= C ∫_{0}^{1} r^{2N−1}(1−r^{2})^{b+u}(\frac{1}{r^{2}} log \frac{1}{1−r^{2}} )^{−v} \overline{∫_{𝕊} K_{a}(rζ, w) dσ(ζ)} dr

= C ∫_{0}^{1} r^{2N−1}(1 − r^{2})^{b+u}(\frac{1}{r^{2}} log \frac{1}{1 − r^{2}} )^{−v} \overline{K_{a}(0, w)} dr

= C ∫_{0}^{1} r^{2N−1}(1 − r^{2})^{b+u}(\frac{1}{r^{2}} log \frac{1}{1 − r^{2}} )^{−v} dr.

The last integral is finite by Lemma 4.6, and then evidently T_{ab}f_{uv} is a constant.

For other values of the parameters,

|T_{ab}f_{uv}(w)| ≥ Re T_{ab}f_{uv}(w) ≥ \frac{1}{2} ∫_{𝔹}(1 − |z|^{2})^{b+u}(\frac{1}{|z|^{2}} log \frac{1}{1 − |z|^{2}} )^{−v} dν(z)
= ∞

for |w| ≤ ρ_{0} by Lemma 4.5 and Lemma 4.6. □

The adjoint of T_{ab} is readily computed.

Proposition 4.9. The formal adjoint T_{ab}^{∗} : L_{Q}^{P′} → L_{q}^{p′} of T_{ab} : L_{q}^{p} → L_{Q}^{P} for 1 ≤ p, P < ∞ is T_{ab}^{∗} = M_{b−q}T_{aQ}, where M_{b−q} denotes the operator of multiplication by (1 − |z|^{2})^{b−q}.

Proof. Let f ∈ L_{q}^{p} and g ∈ L_{Q}^{P′}. Then

[ T_{ab}f, g]_{L_{Q}^{2}} = ∫_{𝔹} ∫_{𝔹} K_{a}(w, z)f(z)(1 − |z|^{2})^{b} dν(z)\overline{g(w)}(1 − |w|^{2})^{Q} dν(w)
= ∫_{𝔹} f(z)\overline{(1 − |z|^{2})^{b−q} ∫_{𝔹} K_{a}(z, w)g(w)(1 − |w|^{2})^{Q} dν(w)}
· (1 − |z|^{2})^{q} dν(z)
= ∫_{𝔹} f \overline{T_{ab}^{∗}(g)} dν_{q} = [ f, T_{ab}^{∗}g]_{L_{q}^{2}}.

Thus

T_{ab}^{∗}g(z) = (1 − |z|^{2})^{b−q} ∫_{𝔹} K_{a}(z, w)g(w)(1 − |w|^{2})^{Q} dν(w).  □

The following simple but very helpful result is well known, but we include its proof for completeness.

Lemma 4.10. If 0 < P < ∞, Q ≤ −1, g ∈ H(𝔹), and g ≢ 0, then

J := ∫_{𝔹} |g(w)|^{P} (1 − |w|^{2})^{Q} dν(w) = ∞.

Proof. We have

J ≥ 2N ∫_{1/2}^{1} ρ^{2N−1}(1 − ρ^{2})^{Q} ∫_{𝕊} |g(ρη)|^{P} dσ(η) dρ

≳ ∫_{1/2}^{1} \frac{1}{2^{2N−1}} (1 − ρ^{2})^{Q} ∫_{𝕊}|g(\frac{η}{2} )|^{P} dσ(η) dρ ≳ ∫_{1/2}^{1}(1 − ρ^{2})^{Q} dρ = ∞

by polar coordinates and the subharmonicity of |g|^{P}. □

Corollary 4.11. If T_{ab} : L_{q}^{p} → L_{Q}^{P} is bounded and f ∈ L_{q}^{p}, then g = T_{ab}f lies in H(𝔹). If also P < ∞, then Q > −1. Thus T_{ab} : L_{q}^{p} → A_{Q}^{P} when it is bounded with P < ∞. Consequently, if P < ∞ and Q ≤ −1, then T_{ab} : L_{q}^{p} → L_{Q}^{P} is not bounded, and hence S_{ab} : L_{q}^{p} → L_{Q}^{P} is not bounded. On the other hand, if T_{ab} : L_{q}^{p} → L^{∞} is bounded and f ∈ L_{q}^{p}, then T_{ab}f ∈ H^{∞} naturally.

Proof. That g is holomorphic follows, for example, by differentiation under the integral sign, from the fact that K_{q}(w, z) is holomorphic in w. That Q > −1 follows from Lemma 4.10. □

Let’s stress again that Corollary 4.11 does not place any restriction on Q when P = ∞, simply because the space L_{Q}^{P} and the inequalities in (III) are independent of Q when P = ∞.

Lemma 4.12. If S_{cb} : L_{q}^{p} → L_{Q}^{P} is bounded and a < c, then S_{ab} : L_{q}^{p} → L_{Q}^{P} is also bounded.

Proof. This is because |K_{a}| ≲ |K_{c}|, which we now prove. If a > −(1 + N), as done in [18, p. 525], we write

\frac{1}{|1 − ⟨w, z⟩|^{1+N+a}} = \frac{|1 − ⟨w, z⟩|^{c−a}}{|1 − ⟨w, z⟩|^{1+N+c}} ≤ \frac{2^{c−a}}{|1 − ⟨w, z⟩|^{1+N+c}} .

Let now a = −(1 + N). On that part of 𝔻 away from 1, |k_{−(1+N)}| is bounded above and |k_{c}| is bounded below by Lemma 4.1. On that part of 𝔻 near 1, |k_{−(1+N)}| is still dominated by |k_{c}|, because

lim_{𝔻∋ω→1} \frac{ω^{−1} log(1 − ω)^{−1}}{(1 − ω)^{−(1+N+c)}} = 0.

If a < −(1 + N), then |K_{c}| dominates |K_{a}| by Lemma 4.1. Thus in all cases |K_{a}(w, z)| ≲ |K_{c}(w, z)| for all w, z ∈ 𝔹 if a < c. □

5. Main Tools

Let (X, A, λ) and (Y, B, μ) be two measure spaces, G(x, y) a nonnegative function on X × Y measurable with respect to A × B, and let Z be given by

Zf(y) = ∫_{X} G(x, y)f(x) dλ(x).

The following two Schur tests are of crucial importance. The first is [13, Theorem 2.1], and is rediscovered in [18, Theorem 2].

Theorem 5.1. Let 1 < p ≤ P < ∞, and suppose that there are c, d ∈ ℝ with c + d = 1 and strictly positive functions ϕ on X and ψ on Y such that

∫_{X} G(x, y)^{cp′}ϕ(x)^{p′} dλ(x) ≲ ψ(y)^{p′} a.e. [μ],
∫_{Y} G(x, y)^{dP} ψ(y)^{P} dμ(y) ≲ ϕ(x)^{P} a.e. [λ].

Then Z : L^{p}(λ) → L^{P} (μ) is bounded.

The second is [8, Theorem 1.I].

Theorem 5.2. Let 1 < P < p < ∞, and suppose that there are strictly positive functions ϕ on X and ψ on Y such that

∫_{X} G(x, y)ϕ(x)^{p′} dλ(x) ≲ ψ(y)^{P′} a.e. [μ],
∫_{Y} G(x, y)ψ(y)^{P} dμ(y) ≲ ϕ(x)^{p} a.e. [λ],
∫∫_{X×Y} G(x, y)ϕ(x)^{p′}ψ(y)^{P} d(λ × μ)(x, y) ≲ 1.

Then Z : L^{p}(λ) → L^{P} (μ) is bounded.

We need in one place the less known Minkowski integral inequality that in effect exchanges the order of integration; for a proof, see [14, Theorem 3.3.5] for example.

Lemma 5.3. If 1 ≤ p ≤ ∞ and f(x, y) is measurable with respect to A × B, then

(∫_{Y} (∫_{X} |f(x, y)| dλ(x))^{p}dμ(y))^{1/p} ≤ ∫_{X}(∫_{Y} |f(x, y)|^{p} dμ(y))^{1/p}dλ(x),

with an appropriate interpretation with the L^{∞} norm when p = ∞.

We cannot do without the Forelli-Rudin estimates of [16, Proposition 1.4.10].

Proposition 5.4. For d > −1 and c ∈ ℝ, we have

∫_{𝔹} \frac{(1 − |w|^{2})^{d}}{|1 − ⟨z, w⟩|^{1+N+c}} dν(w) ∼ \begin{cases}1, & c < d, \\ |z|^{−2} log(1 − |z|^{2})^{−1}, & c = d, \\ (1 − |z|^{2})^{−(c−d)}, & c > d. \end{cases}

The following result from [10, Lemma 5.1] is extremely useful.

Lemma 5.5. If b > −1, a ∈ ℝ, and f ∈ H(𝔹) ∩ L_{b}^{1}, then

T_{ab}f(w) = ∫_{𝔹} K_{a}(w, z)f(z)(1 − |z|^{2})^{b} dν(z) = \frac{N!}{(1 + b)_{N}} D_{b}^{a−b}f(w).

The important result that we prove now is indispensable in our necessity proofs.

Lemma 5.6. If b + t > −1, then T_{ab}I_{b}^{t}h = CD_{b}^{a−b}h for h ∈ B_{b}^{1}, where I_{b}^{t} is as given in (5). As consequences, D_{a}^{b−a}T_{ab}I_{b}^{t}h = Ch for h ∈ B_{b}^{1} and T_{ab}I_{b}^{t}D_{a}^{b−a} = Ch for h ∈ B_{a}^{1}.

Proof. If h ∈ B_{b}^{1} and b + t > −1, then D_{b}^{t}h ∈ B_{b+t}^{1} ⊂ L_{b+t}^{1} by Theorem 4.2. Then by Lemma 5.5 and (4),

T_{ab}I_{b}^{t}h(w) = ∫_{𝔹} K_{a}(w, z)D_{b}^{t}h(z)(1 − |z|^{2})^{b+t} dν(z)
= T_{a,b+t}D_{b}^{t}h(w) = CD_{b+t}^{a−b−t}D_{b}^{t}h(w) = CD_{b}^{a−b}h(w).

The identities on triple compositions are consequences of the identities in (4). □

Here’s another similar result adapted from [5, Lemma 2.3].

Lemma 5.7. If f ∈ L_{b}^{1}, then D_{a}^{t}T_{ab}f = T_{a+t,b}f.

Proof. By (2), if f ∈ L_{b}^{1}, then

T_{ab}f(w) = ∫_{𝔹} K_{a}(w, z)f(z) dν_{b}(z) = ∫_{𝔹} ∑_{k=0}^{∞} c_{k}(a) ⟨w, z⟩^{k}f(z) dν_{b}(z)

= ∑_{k=0}^{∞} c_{k}(a) ∫_{𝔹}⟨w, z⟩^{k}f(z) dν_{b}(z) =: ∑_{k=0}^{∞} c_{k}(a)p_{k}(w),

where p_{k} is a holomorphic homogeneous polynomial of degree k. Then by Definition 3.1 and (2) again,

D_{a}^{t}T_{ab}f(w) = ∑_{k=0}^{∞} d_{k}(a, t)c_{k}(a)p_{k}(w) = ∑_{k=0}^{∞} c_{k}(a + t)p_{k}(w)

= ∑_{k=0}^{∞} c_{k}(a + t) ∫_{𝔹}⟨w, z⟩^{k}f(z) dν_{b}(z)

= ∫_{𝔹} ∑_{k=0}^{∞} c_{k}(a + t)⟨w, z⟩^{k}f(z) dν_{b}(z)

= ∫_{𝔹} K_{a+t}(w, z)f(z) dν_{b}(z) = T_{a+t,b}f(w).

Above, we have used in effect differentiation under the integral sign and (9). □

The next four theorems on inclusions have been developed by various authors culminating in [12], where references to earlier work can be found. We require them in the necessity proofs. All inclusions in them are continuous, strict, and the best possible. As for notation, if X_{v} is a family of spaces indexed by v ∈ ℝ, the symbol X_{<v} denotes any one of the spaces X_{u} with u < v. Let’s also single out the trivial inclusions which are special cases:

B_{q}^{p} ⊂ B_{Q}^{p} (q ≤ Q)  and  B_{α}^{∞} ⊂ B_{β}^{∞} (α ≤ β).  (13)

Theorem 5.8. Let B_{q}^{p} be given.

(i) If p ≤ P, then B_{q}^{p} ⊂ B_{Q}^{P} if and only if \frac{1 + N + q}{p} ≤ \frac{1 + N + Q}{P}.

(ii) If P < p, then B_{q}^{p} ⊂ B_{Q}^{P} if and only if \frac{1 + q}{p} < \frac{1 + Q}{P}.

Theorem 5.9. (i) H^{∞} ⊂ B_{q}^{p} if and only if q > −1, or q = −1 and p ≥ 2.

(ii) B_{q}^{p} ⊂ H^{∞} if and only if q < −(1 + N), or q = −(1 + N) and 0 < p ≤ 1.

Theorem 5.10. Given B_{q}^{p}, we have B_{< \frac{1+q}{p}}^{∞} ⊂ B_{q}^{p} ⊂ B∞_{\frac{1+N+q}{p}}.

Theorem 5.11. B_{<0}^{∞} ⊂ H^{∞} ⊂ B^{∞}.

6. Necessity Proofs

Now we obtain the two inequalities in (III) of Theorems 1.2 and 1.3 from the boundedness of T_{ab}. In this section, we do not need to assume Q > −1 since the boundedness of T_{ab} implies Q > −1 when P < ∞ by Corollary 4.11. We first derive the first inequality in (III) of each of the 10 cases as a separate theorem. We call it the first necessary condition. The reason underlying it is understandably Theorem 4.4.

Theorem 6.1. Let a, b, q, Q ∈ ℝ and 1 ≤ p, P ≤ ∞. If T_{ab} : L_{q}^{p} → L_{Q}^{P} is

bounded, then \frac{1 + q}{p} ≤ 1 + b in the cases ⟨2⟩, ⟨3⟩, ⟨4⟩, and \frac{1 + q}{p} < 1 + b in

the remaining cases.

Proof. The ten cases can be handled in three groups depending on the value of p. The first group consists of the cases ⟨1⟩, ⟨5⟩, ⟨7⟩, ⟨8⟩ in which 1 < p < ∞. Consider f_{uv} with u = −(1 + q)/p and v = 1 so that f_{uv} ∈ L_{q}^{p} by Lemma 4.7. Then T_{ab}f_{uv} ∈ L_{Q}^{P} and Lemma 4.8 implies b + u > −1. With the value of u chosen, this yields (1 + q)/p < 1 + b.

The second group consists of the cases ⟨2⟩, ⟨3⟩, ⟨4⟩; here p = 1. Consider f_{u0} with u > −(1 + q) so that f_{u0} ∈ L_{q}^{1} by Lemma 4.7. Then T_{ab}f_{u0} ∈ L_{Q}^{P} and Lemma 4.8 implies b + u > −1. Writing u = −1 − q + ε with ε > 0, we obtain q < b + ε. This is nothing but q ≤ b.

The third group consists of the cases ⟨6⟩, ⟨9⟩, ⟨10⟩. Now p = ∞ and f_{00} = 1 is in L^{∞}. Then T_{ab}f_{00} ∈ L_{Q}^{P} and hence 0 < 1 + b by Lemma 4.8. □

We next derive the second inequality in (III) of each of the 10 cases also as a separate theorem. We call it the second necessary condition. We do this by a new original method that relies on knowing which Bergman-Besov and Bloch-Lipschitz spaces lie in which others and in H^{∞}. A crucial component of this method is Lemma 5.6.

Theorem 6.2. Let a, b, q, Q ∈ ℝ and 1 ≤ p, P ≤ ∞. Suppose T_{ab} : L_{q}^{p} → L_{Q}^{P}

is bounded. Then a ≤ b + \frac{1 + N + Q}{P} − \frac{1 + N + q}{p} in the cases ⟨1⟩, ⟨2⟩,

⟨3⟩, ⟨4⟩, also a < b + \frac{1 + N + Q}{P} − \frac{1 + N + q}{p} in the cases ⟨5⟩, ⟨6⟩, and

a < b + \frac{1 + Q}{P} − \frac{1 + q}{p} in the cases ⟨7⟩, ⟨8⟩, ⟨9⟩, ⟨10⟩.

Proof. The boundedness of T_{ab} via Corollary 4.11 implies Q > −1 when P < ∞.

This time we form four groups of the ten cases. The first group again consists of the cases ⟨1⟩, ⟨7⟩, ⟨8⟩, and ⟨2⟩, ⟨3⟩. Let h ∈ B_{q}^{p}. In order to be able to use Lemma 5.6, we need to make sure h ∈ B_{b}^{1}. In the cases ⟨1⟩, ⟨7⟩, ⟨8⟩, the first necessary condition gives (1 + q)/p < 1 + b, and then Theorem 5.8 (ii) shows h ∈ B_{b}^{1}. In the cases ⟨2⟩, ⟨3⟩ where p = 1, Theorem 6.1 gives q ≤ b, and then (13) shows h ∈ B_{b}^{1} again. Let t satisfy (6). Together with the first necessary condition, this gives b + t > −1. Now consider the sequence of bounded maps

B_{q}^{p} ⟶^{I_{b}^{t}} L_{q}^{p} ⟶^{T_{ab}} L_{Q}^{P} ∩ H(𝔹) = A_{Q}^{P} ⟶^{D_{a}^{b−a}} B_{Q+P (b−a)}^{P},

where Theorem 4.2 is invoked and Q > −1 is used in writing A_{Q}^{P}. Lemma 5.6 yields that B_{q}^{p} is imbedded in B_{Q+P (b−a)}^{P} by the inclusion map. By Theorem 5.8, this implies a < b + (1 + Q)/P − (1 + q)/p in the cases ⟨7⟩, ⟨8⟩, and it implies a ≤ b + (1 + N + Q)/P − (1 + N + q)/p in the cases ⟨1⟩, ⟨2⟩, ⟨3⟩.

The second group consists of the cases ⟨4⟩, ⟨5⟩. Let H ∈ B_{q+p(a−b)}^{p}; then D_{a}^{b−a}H = h ∈ B_{q}^{p} by Theorem 4.2. Exactly as in the proof of the first group of cases, h ∈ B_{b}^{1}. Let t satisfy (6). Together with the first necessary condition, this gives b + t > −1. Now consider the sequence of bounded maps

B_{q+p(a−b)}^{p} ⟶^{D_{a}^{b−a}} B_{q}^{p} ⟶^{I_{b}^{t}} L_{q}^{p} ⟶^{T_{ab}} L^{∞} ∩ H(𝔹) = H^{∞}.

Lemma 5.6 yields that B_{q+p(a−b)}^{p} is imbedded in H^{∞} by the inclusion map. By Theorem 5.9 (ii), this implies a < b − (1 + N + q)/p in the case ⟨5⟩, and it implies a ≤ b − (1 + N + q) in the case ⟨4⟩.

The third group consists of the cases ⟨9⟩, ⟨10⟩. Let h ∈ B^{∞}. Theorem 6.1 gives 0 < 1 + b, and then Theorem 5.10 yields h ∈ B_{b}^{1}. Let t > 0, that is, satisfy (8) with α = 0. Together with the first necessary condition, this gives b + t > −1. Now consider the sequence of bounded maps

B^{∞} ⟶^{I_{b}^{t}} L^{∞} ⟶^{T_{ab}} L_{Q}^{P} ∩ H(𝔹) = A_{Q}^{P} ⟶^{D_{a}^{b−a}} B_{Q+P (b−a)}^{P},

where Theorem 4.2 is invoked and Q > −1 is used in writing A_{Q}^{P}. Lemma 5.6 yields that B^{∞} is imbedded in B_{Q+P (b−a)}^{p} by the inclusion map. By Theorem 5.10, this implies a < b + (1 + Q)/P.

The fourth group consists of the case ⟨6⟩ only. Let H ∈ B_{a−b}^{∞}; then D_{a}^{b−a}H = h ∈ B^{∞} by Theorem 4.2. Exactly as in the proof of the third group of cases, h ∈ B_{b}^{1}. Let t > 0, that is, satisfy (8) with α = 0. Together with the first necessary condition, this gives b + t > −1. Now consider the sequence of bounded maps

B_{a−b}^{∞} ⟶^{D_{a}^{b−a}} B^{∞} ⟶^{I_{b}^{t}} L^{∞} ⟶^{T_{ab}} L^{∞} ∩ H(𝔹) = H^{∞}.

Lemma 5.6 yields that B_{a−b}^{∞} is imbedded in H^{∞} by the inclusion map. By Theorem 5.11, this implies a < b. □

Thirdly, we prove that in the cases ⟨2⟩, ⟨3⟩, ⟨4⟩, if one of the inequalities in (III) is an equality, then the other must be a strict inequality.

Theorem 6.3. Let a, b, q, Q ∈ ℝ and 1 ≤ p, P ≤ ∞. If T_{ab} : L_{q}^{p} → L_{Q}^{P} is bounded, then equality cannot hold simultaneously in the inequalities of Theorems 6.1 and 6.2 in the cases ⟨2⟩, ⟨3⟩, ⟨4⟩.

Proof. The boundedness of T_{ab} via Corollary 4.11 implies Q > −1 when P < ∞.

If q = b and a = b + Q − q simultaneously in the case ⟨2⟩, then also a = Q > −1 and T_{ab}^{∗} = T_{QQ} : L^{∞} → L^{∞} is bounded. Let

g_{z}(w) = \frac{(1 − ⟨z, w⟩)^{1+N+Q}}{|1 − ⟨z, w⟩|^{1+N+Q}} ,

which is a uniformly bounded family for z ∈ 𝔹. The same is true also of {T_{QQ}g_{z}}. But

T_{QQ}g_{z}(z) = ∫_{𝔹} \frac{(1 − |w|^{2})^{Q}}{|1 − ⟨z, w⟩|^{1+N+Q}} dν(w) ∼ \frac{1}{|z|^{2}} log \frac{1}{1 − |z|^{2}}

by Proposition 5.4, contradicting uniform boundedness.

If q = b and a = b + (1 + N + Q)/P − (1 + N + q) simultaneously in the case ⟨3⟩, then also a = (1 + N + Q)/P − (1 + N) and T_{ab}^{∗} = T_{aQ} : L_{Q}^{P′} → L^{∞} is bounded with P′ > 1. There is an unbounded g ∈ B_{−(1+N)}^{P′} by Theorem

5.9. Then h = D_{Q−(1+N+Q)/P′}^{(1+N+Q)/P′}g ∈ B_{Q}^{P′} ⊂ L_{Q}^{P′} ⊂ L_{Q}^{1}. By Lemma 5.5 and (4), we obtain

T_{ab}^{∗}h = T_{aQ}h = CD_{Q}^{a−Q}h = CD_{Q}^{a−Q}D_{Q−(1+N+Q)/P′}^{(1+N+Q)/P′}g = Cg

since (a − Q) + (1 + N + Q)/P′ = 0 by the conditions on the parameters. But g ∉ L^{∞}, and this contradicts that T_{ab}^{∗} : L_{Q}^{P′} → L^{∞}.

If q = b and a = b − (1 + N + q) simultaneously in the case ⟨4⟩, then also a = −(1 + N). For j = 1, 2, …, let z_{j} = (1 − 1/j, 0, …, 0) and E_{j} the ball of radius 1/2j centered at z_{j}, and define

f_{j}(z) = \frac{χ_{E_{j}} (z)}{ν(E_{j})(1 − |z|^{2})^{q}} .

Clearly f_{j} ∈ L_{q}^{1} and ∥f_{j}∥_{L_{q}^{1}} = 1 for every j. Then {T_{ab}f_{j}} = {T_{−(1+N),q}f_{j}} is a uniformly bounded family. By the mean value property,

T_{−(1+N),q}f_{j}(w) = \overline{\frac{1}{ν(E_{j})} ∫_{E_{j}} K_{−(1+N)}(z, w) dν(z)} = K_{−(1+N)}(w, z_{j}).

But

T_{−(1+N),q}f_{j}(z_{j}) = K_{−(1+N)}(z_{j}, z_{j}) = \frac{1}{|z_{j}|^{2}} log \frac{1}{1 − |z_{j}|^{2}}

contradicting uniform boundedness. □

7. Sufficiency Proofs

Conversely, we now present the proofs that the two inequalities in (III) of Theorems 1.2 and 1.3 imply the boundedness of S_{ab}. By Lemma 4.12, it suffices to prove this only for large values of a. In all the cases except ⟨4⟩, there are values of a > −(1+N) satisfying the inequalities in (III). So in this section we make the standing hypothesis

a > −(1 + N) and K_{a} is binomial  except in case ⟨4⟩. (14)

In the case ⟨4⟩, the remaining values of a are handled separately and swiftly.

Proof of Sufficiency. Each of the ten cases has a sufficiently different proof from those of the other cases and we treat each case separately. The cases ⟨7⟩ and ⟨1⟩ are the generic cases of Theorems 1.3 and Theorem 1.2, respectively, and have the most involved proofs, so we leave them to the end. Throughout, our hypothesis is that the two inequalities in (III) hold.

Case ⟨2⟩. Let f ∈ L_{q}^{1}. We write the L_{Q}^{1} norm of S_{ab}f explicitly, then exchange the order of integration by the Fubini theorem, and obtain

∥S_{ab}f∥_{L_{Q}^{1}} ≤ ∫_{𝔹} ∫_{𝔹} |K_{a}(w, z)| |f(z)|(1 − |z|^{2})^{b} dν(z) (1 − |w|^{2})^{Q} dν(w)

= ∫_{𝔹} |f(z)|(1 − |z|^{2})^{b} ∫_{𝔹} \frac{(1 − |w|^{2})^{Q}}{|1 − ⟨z, w⟩|^{1+N+a}} dν(w) dν(z)

= ∫_{𝔹} |f(z)|(1 − |z|^{2})^{b−q} ∫_{𝔹} \frac{(1 − |w|^{2})^{Q}}{|1 − ⟨z, w⟩|^{1+N+a}} dν(w) dν_{q}(z)

Let J(z) be that part of the integrand of the outer integral multiplying |f(z)|. We show that J is bounded on 𝔹 using Proposition 5.4. Check that Q > −1 as required.

Consider (1 + N + a) − (1 + N) − Q = a − Q. If a − Q < 0, then the integral in J(z) is bounded and J(z) is also bounded since b − q ≥ 0 by the first inequality in (III). If a − Q = 0, then the integral in J(z) is logarithmic. But now b > q by (III) and Remark 1.6, and J(z) is bounded this time by (12). If a − Q > 0, then J(z) ∼ (1 − |z|^{2})^{b−q−a+Q}. But b − q − a + Q ≥ 0 by the second inequality in (III); hence J(z) is bounded once again. Thus ∥S_{ab}f∥_{L_{Q}^{1}} ≲ ∥f∥_{L_{q}^{1}} and S_{ab} : L_{q}^{1} → L_{Q}^{1} is bounded.

Case ⟨3⟩. Let f ∈ B_{q}^{1}. We write the L_{Q}^{P} norm of S_{ab}f explicitly, then use Lemma 5.3 with the measures ν_{q} and ν_{Q}, and obtain

∥S_{ab}f∥_{L_{Q}^{P}} = (∫_{𝔹}|∫_{𝔹} K_{a}(w, z)f(z)(1 − |z|^{2})^{b−q} dν_{q}(z)|^{P} dν_{Q}(w))^{1/P}

≤ ∫_{𝔹}(∫_{𝔹} |K_{a}(w, z)|^{P} |f(z)|^{P} (1 − |z|^{2})^{(b−q)P} dν_{Q}(w))^{1/P} dν_{q}(z)

= ∫_{𝔹} |f(z)|(1−|z|^{2})^{b−q}(∫_{𝔹} \frac{(1 − |w|^{2})^{Q}}{|1−⟨z, w⟩|^{(1+N+a)P}} dν(w))^{1/P} dν_{q}(z).

For z ∈ 𝔹, let J(z) be that part of the integrand of the outer integral multiplying |f(z)|. We show that J is bounded on 𝔹 using Proposition 5.4. Check that Q > −1 as required.

Let u = (1 + N + a)P − (1 + N) − Q. If u < 0, then the integral in J(z) is bounded and J(z) is also bounded since b − q ≥ 0 by the first inequality in (III). If u = 0, then the integral in J(z) is logarithmic and J(z) is bounded by (12) since (III) reads q < b by Remark 1.6 with this value of u. If u > 0, then J(z) ∼ (1 − |z|^{2})^{b−q−u/P}. Note that

b − q − \frac{u}{P} = b − q − (1 + N + a) + \frac{1 + N + Q}{P} ≥ 0

by the second inequality in (III). So J(z) is bounded one more time. Thus ∥S_{ab}f∥_{L_{Q}^{P}} ≲ ∥f∥_{L_{q}^{1}} and S_{ab} : L_{q}^{1} → L_{Q}^{P} is bounded.

Case ⟨4⟩. Let f ∈ L_{q}^{1}. If q = b, then (III) says that a < −(1 + N) and K_{a} is bounded by Lemma 4.1.

|S_{ab}f(w)| ≤ ∫_{𝔹} |K_{a}(w, z)| |f(z)|(1 − |z|^{2})^{b} dν(z) ≲ ∥f∥_{L_{q}^{1}} (w ∈ 𝔹).

Then ∥S_{ab}f∥_{L^{∞}} ≲ ∥f∥_{L_{q}^{1}}.

Otherwise q < b and there are values of a > −(1 + N) satisfying the inequalities in (III). So in the rest of this case we can assume a > −(1 + N) and K_{a} is binomial by Lemma 4.12. Then writing S_{ab}f(w) explicitly and simple manipulations give

|S_{ab}f(w)| ≤ ∫_{𝔹} |K_{a}(w, z)| |f(z)|(1 − |z|^{2})^{b} dν(z)

= ∫_{𝔹} |f(z)|\frac{(1 − |z|^{2})^{b−q}}{|1 − ⟨w, z⟩|^{1+N+a}} dν_{q}(z) =: ∫_{𝔹} |f(z)|J(z, w) dν_{q}(z).

Since |⟨w, z⟩| ≤ |w| |z| ≤ |z| for w ∈𝔹, we have J(z, w) ≲ (1−|z|^{2})^{b−q−(1+N+a)} for all such w. Note that the power here is nonnegative by the second inequality in (III) yielding that J(z, w) is bounded for all z, w ∈ 𝔹. Then we obtain

|S_{ab}f(w)| ≲ ∫_{𝔹} |f(z)| dν_{q}(z) = ∥f∥_{L_{q}^{1}} (w ∈ 𝔹)

and ∥S_{ab}f∥_{L^{∞}} ≲ ∥f∥_{L_{q}^{1}}. Thus S_{ab} : L_{q}^{1} → L^{∞} is bounded.

Case ⟨5⟩. Let f ∈ L_{q}^{p}. We write S_{ab}f(w) explicitly, apply the Hölder inequality with the measure ν_{q}, and obtain

|S_{ab}f(w)| ≤ ∫_{𝔹} |K_{a}(w, z)| |f(z)|(1−|z|^{2})^{b} dν(z)

= ∫_{𝔹} |f(z)|\frac{(1 − |z|^{2})^{b−q}}{|1 − ⟨w, z⟩|^{1+N+a}} dν_{q}(z)

≤ ∥f∥_{L_{q}^{p}}(∫_{𝔹} \frac{(1 − |z|^{2})^{(b−q)p′+q}}{|1 − ⟨w, z⟩|^{(1+N+a)p′}} dν(z))^{1/p′} =: J(w)^{1/p′}∥f∥_{L_{q}^{p}}.

We show that J is bounded on 𝔹 by Proposition 5.4. Now

(b − q)p′ + q + 1 = p′(b − q + \frac{(q + 1)(p − 1)}{p}) = p′(1 + b − \frac{1 + q}{p} ) > 0

by the first inequality in (III), so the power on 1 − |z|^{2} in J(w) is > −1 as required.

Note that

(1+N +a)p′ − (1+N) − (b−q)p′ − q = p′(1 + N + a − b + q − \frac{1+N +q}{p′} )

= p′(a − b + \frac{1 + N + q}{p} ) < 0

by the second inequality in (III). This shows that J(w) is bounded for all w ∈ 𝔹. Then |S_{ab}f(w)| ≲ ∥f∥_{L_{q}^{p}} for all w ∈ 𝔹 and ∥S_{ab}f∥_{L^{∞}} ≲ ∥f∥_{L_{q}^{p}}. Thus S_{ab} : L_{q}^{p} → L^{∞} is bounded.

Case ⟨6⟩. Let f ∈ L^{∞}. We write S_{ab}f(w) explicitly, take the L^{∞} norm of f out of the integral, and obtain

|S_{ab}f(w)| ≤ ∫_{𝔹} |K_{a}(w, z)| |f(z)|(1−|z|^{2})^{b} dν(z)

≤ ∥f∥_{L^{∞}} ∫_{𝔹} \frac{(1 − |z|^{2})^{b}}{|1 − ⟨w, z⟩|^{1+N+a}} dν(z) =: J(w)∥f∥_{L^{∞}}.

We show that J is bounded on 𝔹 by Proposition 5.4. Check that b > −1 by the first inequality in (III). The second inequality in (III) gives that a < b and hence J(w) is indeed bounded for w ∈ 𝔹. Then we obtain |S_{ab}f(w)| ≲ ∥f∥_{L^{∞}} for all w ∈ 𝔹. Thus ∥S_{ab}f∥_{L^{∞}} ≲ ∥f∥_{L_{q}^{p}} and S_{ab} : L^{∞} → L^{∞} is bounded.

Case ⟨8⟩. Let f ∈ L_{q}^{p}. We write the L_{Q}^{1} norm of S_{ab}f explicitly, then exchange the order of integration by the Fubini theorem, afterwards apply the Hölder inequality, and obtain

∥S_{ab}f∥_{L_{Q}^{1}} ≤ ∫_{𝔹} ∫_{𝔹} |K_{a}(w, z)| |f(z)|(1 − |z|^{2})^{b} dν(z) (1 − |w|^{2})^{Q} dν(w)

= ∫_{𝔹} |f(z)|(1 − |z|^{2})^{b} ∫_{𝔹} \frac{(1 − |w|^{2})^{Q}}{|1 − ⟨z, w⟩|^{1+N+a}} dν(w) dν(z)

= ∫_{𝔹} |f(z)|(1 − |z|^{2})^{q/p} ∫_{𝔹} \frac{(1 − |w|^{2})^{Q} dν(w)}{|1 − ⟨z, w⟩|^{1+N+a}} (1 − |z|^{2})^{b−q/p} dν(z)

≤ ∥f∥_{L_{q}^{p}}(∫_{𝔹}(∫_{𝔹} \frac{(1 − |w|^{2})^{Q} dν(w)}{|1 − ⟨z, w⟩|^{1+N+a}})^{p′}(1−|z|^{2})^{(b−q/p)p′} dν(z))^{1/p′}

=: J^{1/p′}∥f∥_{L_{q}^{p}}

We show that J is finite using Proposition 5.4. Check that Q > −1 as required.

If a < Q, then J ∼ ∫_{𝔹}(1 − |z|^{2})^{(b−q/p)p′} dν(z). Note that

(b − \frac{q}{p})p′ = p′(1 + b − \frac{1 + q}{p} − \frac{1}{p′} ) > p′(−\frac{1}{p′} ) = −1

by the first inequality in (III). So J is finite by Lemma 4.6. If a = Q, then

J ∼ ∫_{𝔹}(\frac{1}{|z|^{2}} log \frac{1}{1 − |z|^{2}} )^{p′}(1 − |z|^{2})^{(b−q/p)p′} dν(z) < ∞

also by Lemma 4.6. If a > Q, then J ∼ ∫_{𝔹}(1 − |z|^{2})^{(b−q/p)p′−(a−Q)p′} dν(z).

But

(b − \frac{q}{p})p′ − (a − Q)p′ = p′(b − a + 1 + Q − \frac{1 + q}{p} − \frac{1}{p′} ) > p′(−\frac{1}{p′} ) = −1

by the second inequality in (III). So J is finite one more time. We conclude that S_{ab} : L_{q}^{p} → L_{Q}^{1} is bounded.

Case ⟨9⟩. Let f ∈ L^{∞}. We write the L_{Q}^{1} norm of S_{ab}f explicitly, then exchange the order of integration by the Fubini theorem, take the L^{∞} norm of f out of the integral, and obtain

∥S_{ab}f∥_{L_{Q}^{1}} ≤ ∫_{𝔹} ∫_{𝔹} |K_{a}(w, z)| |f(z)|(1 − |z|^{2})^{b} dν(z) (1 − |w|^{2})^{Q} dν(w)

= ∫_{𝔹} |f(z)|(1 − |z|^{2})^{b} ∫_{𝔹} \frac{(1 − |w|^{2})^{Q}}{|1 − ⟨z, w⟩|^{1+N+a}} dν(w) dν(z)

≤ ∥f∥_{L^{∞}} ∫_{𝔹}(1 − |z|^{2})^{b} ∫_{𝔹} \frac{(1 − |w|^{2})^{Q}}{|1 − ⟨z, w⟩|^{1+N+a}} dν(w) dν(z)

=: J∥f∥_{L^{∞}}.

We show that J is finite using Proposition 5.4. Check that Q > −1 as required.

If a < Q, then the inner integral in J is bounded and hence J is finite by the first inequality in (III). If a = Q, then the inner integral is logarithmic and hence J is finite by the first inequality in (III) and Lemma 4.6. If a > Q,

then J ∼ ∫_{𝔹}(1 − |z|^{2})^{b−a+Q} dν(z) < ∞ since b − a + Q > −1 by the second

inequality in (III). Thus S_{ab} : L^{∞} → L_{Q}^{1} is bounded.

Case ⟨10⟩. Let f ∈ L^{∞}. We write the L_{Q}^{P} norm of S_{ab}f explicitly, take the L^{∞} norm of f out of the integral, and obtain

∥S_{ab}f∥_{L_{Q}^{P}}^{P} = ∫_{𝔹}|∫_{𝔹} K_{a}(w, z)f(z)(1 − |z|^{2})^{b} dν(z)|^{P} (1 − |w|^{2})^{Q} dν(w)

≤ ∥f∥_{L^{∞}}^{P} ∫_{𝔹}(1−|w|^{2})^{Q}(∫_{𝔹} \frac{(1 − |z|^{2})^{b}}{|1−⟨w, z⟩|^{1+N+a}} dν(z))^{P} dν(w)

=: J∥f∥_{L^{∞}}^{P}.

We show that J is finite using Proposition 5.4. Check that b > −1 by the first inequality in (III) as required.

If a < b, then the inner integral in J is bounded and then J is finite since Q > −1. If a = b, then the inner integral in J is logarithmic and then J

is finite by Lemma 4.6. If a > b, then J ∼ ∫_{𝔹}(1 − |w|^{2})^{Q−(a−b)P} dν(w). Note

that

Q − (a − b)P = Q − (a − b)P + 1 − 1 = P (\frac{1 + Q}{P} + b − a) − 1 > −1

by the second inequaliy in (III), which shows that J is again finite. Consequently S_{ab} : L^{∞} → L_{Q}^{P} is bounded.

Case ⟨7⟩. We employ the Schur test in Theorem 5.2 with the measures λ = ν_{q},

μ = ν_{Q} and the kernel G(z, w) = \frac{(1 − |z|^{2})^{b−q}}{|1 − ⟨z, w⟩|^{1+N+a}} which together give

Z = S_{ab}, along with the strictly positive functions ϕ(z) = (1 − |z|^{2})^{m} and ψ(w) = (1 − |w|^{2})^{n} on 𝔹 with m, n ∈ ℝ to be determined. Two of the three conditions that need to be satisfied for the test are

∫_{𝔹} \frac{(1 − |z|^{2})^{b−q}}{|1 − ⟨w, z⟩|^{1+N+a}} (1 − |z|^{2})^{mp′}(1 − |z|^{2})^{q} dν(z) ≲ (1 − |w|^{2})^{nP′},
∫_{𝔹} \frac{(1 − |z|^{2})^{b−q}}{|1 − ⟨w, z⟩|^{1+N+a}} (1 − |w|^{2})^{nP} (1 − |w|^{2})^{Q} dν(w) ≲ (1 − |z|^{2})^{mp}.

One way to satisfy them is by matching the growth rates of their two sides, that is, the powers of the 1 − | · |^{2}. By Proposition 5.4, this is possible if m, n < 0 and

\begin{aligned}−nP′ & = a − (b + mp′), \\ −mp & = a − (nP + Q) − (b − q). \end{aligned}  (15)

But we must also make sure that the conditions of Proposition 5.4 for this to happen are met, that is,

\begin{aligned}b + mp′ & > −1, \\ nP + Q & > −1, \\ a − (b + mp′) & > 0, \\ a − (nP + Q) & > 0. \end{aligned}  (16)

Substituting for p′, P′ in terms of p, P, we can write (15) as a system of two linear equations in the two unknowns m, n as

\begin{aligned}p(P − 1)m − P (p − 1)n & = (a − b)(p − 1)(P − 1), \\ −pm + P n & = a − b + q − Q. \end{aligned}  (17)

This system has the unique solution

\begin{aligned}m & = \frac{(p − 1)(P (a − b) + q − Q)}{p(P − p)}, \\ n & = \frac{(P − 1)(p(a − b) + q − Q)}{P (P − p)} \end{aligned}  (18)

for m, n. The second inequality in (III) can be written in the form

a − b = −ε + \frac{1 + Q}{P} − \frac{1 + q}{p}  (19)

with ε > 0. By Lemma 4.12, it suffices to show that S_{ab} is bounded when (19) holds for small enough ε > 0. Substituting this value of a − b into (18), the solution takes the form

\begin{aligned}m & = \frac{p − 1}{p}(−\frac{1 + q}{p} + \frac{εP}{p − P} ), \\ n & = \frac{P − 1}{P}(−\frac{1 + Q}{P} + \frac{εp}{p − P} ), \end{aligned}  (20)

What is left is to show that this solution satisfies all the required conditions for sufficiently small ε > 0. Bear in mind that Q > −1. First, by the first inequality in (III),

a = −ε + b + \frac{1 + Q}{P} − \frac{1 + q}{p} > −ε + \frac{1 + Q}{P} − 1 > −(1 + ε) > −(1 + N)

provided ε < N. Next we verify the four inequalities in (16). By (20) and the first inequality in (III),

b + mp′ = b − \frac{1 + q}{p} + \frac{εP}{p − P} > −1 + \frac{εP}{p − P} > −1.  (21)

By (20),

nP+ Q = (P −1)(−\frac{1+Q}{P} + \frac{εp}{p−P} )+Q = −1+ \frac{1+Q}{P} + \frac{εp(P −1)}{p − P} > −1. (22)

By (19) and (21),

a − (b + mp′) = −ε + \frac{1 + Q}{P} − \frac{εP}{p − P} = \frac{1 + Q}{P} − \frac{εp}{p − P} > 0  (23)

provided ε < (\frac{1}{P} − \frac{1}{p})(1 + Q). By (19), (22), and the first inequality in (III),

a − (nP + Q) = −ε + b + \frac{1 + Q}{P} − \frac{1 + q}{p} + 1 − \frac{1 + Q}{P} − \frac{εp(P − 1)}{p − P}

= 1 + b − \frac{1 + q}{p} − \frac{εP (p − 1)}{p − P} > 0

provided ε < \frac{p}{p − 1}(\frac{1}{P} − \frac{1}{p})(1 + b − \frac{1 + q}{p} ). Lastly, we check the third

condition of Theorem 5.2, which is the finiteness of

∫_{𝔹}∫_{𝔹} \frac{(1 − |z|^{2})^{b−q}}{|1−⟨w, z⟩|^{1+N+a}} (1−|z|^{2})^{mp′}(1−|w|^{2})^{nP} (1−|z|^{2})^{q}(1−|w|^{2})^{Q}dν(z)dν(w).

Call the double integral J. We estimate first the integral, say, with respect to dν(z) by Proposition 5.4 and obtain

J ∼ ∫_{𝔹}(1 − |w|^{2})^{nP +Q−a+b+mp′} dν(w)

by (23). But by (22) and (23), the power on 1 − |w|^{2} is

nP + Q − (a − (b + mp′)) = −1 + \frac{1 + Q}{P} + \frac{εp(P − 1)}{p − P} − \frac{1 + Q}{P} + \frac{εp}{p − P}

= −1 + \frac{εpP}{p − P} > −1,

making J < ∞. Thus for

0 < ε < min\{N, (\frac{1}{P} − \frac{1}{p})(1 + Q), \frac{p}{p − 1}(\frac{1}{P} − \frac{1}{p})(1 + b − \frac{1 + q}{p} )\},

Theorem 5.2 using the selected ϕ and ψ with the powers in (20) applies and proves that S_{ab} : L_{q}^{p} → L_{Q}^{P} with 1 < P < p < ∞ is bounded when the two inequalities in (III) hold.

Case ⟨1⟩. The proof of this case follows the same lines as the proof of [18, Lemma 6] and also starts out as in the proof of the case ⟨7⟩. We assume (III) and take a to have its largest value

a = b + \frac{1 + N + Q}{P} − \frac{1 + N + q}{p}  (24)

by Lemma 4.12 with no loss of generality. We have a > −(1 + N) by the first inequality in (III) and Q > −1. Now we employ the Schur test in Theorem 5.1 but with the same test data as in the case ⟨7⟩. So we have the measures

λ = ν_{q}, μ = ν_{Q} and the kernel G(z, w) = \frac{(1 − |z|^{2})^{b−q}}{|1 − ⟨z, w⟩|^{1+N+a}} which together

give Z = S_{ab}, along with the strictly positive functions ϕ(z) = (1−|z|^{2})^{m} and ψ(w) = (1−|w|^{2})^{n} on 𝔹 with m, n ∈ ℝ to be determined. The two conditions that need to be satisfied for the test are

∫_{𝔹} \frac{(1 − |z|^{2})^{(b−q)cp′}}{|1 − ⟨w, z⟩|^{(1+N+a)cp′}} (1 − |z|^{2})^{mp′}(1 − |z|^{2})^{q} dν(z) ≲ (1 − |w|^{2})^{np′},
∫_{𝔹} \frac{(1 − |z|^{2})^{(b−q)dP}}{|1 − ⟨w, z⟩|^{(1+N+a)dP}} (1 − |w|^{2})^{nP} (1 − |w|^{2})^{Q} dν(w) ≲ (1 − |z|^{2})^{mP} .

One way to satisfy them is by matching the growth rates of their two sides, that is, the powers of the 1 − | · |^{2}. By Proposition 5.4, this is possible if m, n < 0 and

\begin{aligned}−np′ & = (1 + N + a)cp′ − (1 + N) − (b − q)cp′ − mp′ − q, \\ −mP & = (1 + N + a)dP − (1 + N) − (nP + Q) − (b − q)dP. \end{aligned} (25)

But we must also make sure that the conditions of Proposition 5.4 for this to happen are met, that is,

\begin{aligned}E := (b − q)cp′ + mp′ + q & > −1, \\ nP + Q & > −1, \\ F := (1 + N + a)cp′ − (1 + N) − (b − q)cp′ − mp′ − q & > 0, \\ (1 + N + a)dP − (1 + N) − (nP + Q) & > 0. \end{aligned}  (26)

If we were to continue as in the proof of the case ⟨7⟩, we would now solve for m, n from the linear equations in (25). However, it turns out that these two equations are linearly dependent, so we follow a different path. We first pick an n to satify the second inequality in (26); so

−\frac{1 + Q}{P} < n < 0,  (27)

which is possible since Q > −1. Next we pick a d to satisfy the fourth inequality in (26) and naturally let c = 1 − d; so we take

\begin{aligned}d & = \frac{1}{1 + N + a}(n + \frac{1 + N + Q}{P} + ε) \\ c & = \frac{1}{1 + N + a}(−n + 1 + N + b − \frac{1 + N + q}{p} − ε) \end{aligned}  (28)

with ε > 0 by (24). Using the chosen values of n, c, d, we then solve for m from, say, the second equation in (25), and simplify it using the definition of d; so

\begin{aligned}m & = (b − q)d − (1 + N + a)d + n + \frac{1 + N + Q}{P} \\ & = (b − q)d − n − \frac{1 + N + Q}{P} − ε + n + \frac{1 + N + Q}{P} = (b − q)d − ε. \end{aligned} (29)

Finally, we make sure the remaining first and third inequalities in (26) hold for some ε > 0. Substituting in the value of m from (29), since c+d = 1,

E + 1 = (b − q)cp′ + (b − q)dp′ − εp′ + q + 1 = (b − q)p′ + q + 1 − εp′

= p′(b − q + \frac{q}{p′} + \frac{1}{p′} ) − εp′ = p′(b − \frac{q}{p} + 1 − \frac{1}{p}) − εp′

= p′(1 + b − \frac{1 + q}{p} − ε) > 0

by the first inequality in (III) provided ε < 1+b− \frac{1 + q}{p}. Again since c+d = 1

and substituting in for m and c from (29) and (28),

F = p′((1 + N + a)c − (b − q)c − (b − q)d + ε − \frac{1 + N + q}{p′} )

= p′(−n + 1 + N + b − \frac{1 + N + q}{p} − ε − b + q + ε − \frac{1 + N + q}{p′} )

= −p′n > 0

by (27). Thus for

0 < ε < 1 + b − \frac{1 + q}{p} ,

Theorem 5.1 using the selected ϕ and ψ with the powers in (29) and (27) applies with the parameters in (28) and proves that S_{ab} : L_{q}^{p} → L_{Q}^{P} with 1 < p ≤ P < ∞ is bounded when the two inequalities in (III) hold.

As added information, from (28) and (27), we see that

d > \frac{1}{1 + N + a}(−\frac{1 + Q}{P} + \frac{1 + N + Q}{P} + ε) = \frac{1}{1 + N + a}(\frac{N}{P} + ε) > 0,

but we do not see any such relation for c = 1 − d. □

8. Mapping into Bergman-Besov Spaces and H^{∞}

We now prove Theorem 1.8. Having proved Theorems 1.2 and 1.3, our task is easy with the technique of composing maps employed in Theorem 6.2. The replacement we need for Lemma 5.6 is Lemma 5.7.

Proof of Theorem 1.8. First Necessary Condition. We imitate the proof of Theorem 6.1 assuming T_{ab} is bounded. We form the same three groups of the ten cases and use the same test functions f_{uv} for each group obtaining T_{ab}f_{uv} ∈ B_{Q}^{P}. Then T_{ab}f_{uv}(0) ∈ ℂ. In the first group excluding the case ⟨5⟩,

T_{ab}f_{uv}(0) = ∫_{𝔹}(1 − |z|^{2})^{b−(1+q)/p}(\frac{1}{|z|^{2}} log \frac{1}{1 − |z|^{2}} )^{−1} dν(z).

Writing the integral in polar coordinates and using Lemma 4.6, we obtain (1 + q)/p < 1 + b. In the second group excluding the case ⟨4⟩,

T_{ab}f_{u0}(0) = ∫_{𝔹}(1 − |z|^{2})^{b+u} dν(z)

with u > −(1 + q). Writing this in polar coordinates and using Lemma 4.6, we obtain q ≤ b. In the third group excluding the case ⟨6⟩,

T_{ab}f_{00}(0) = ∫_{𝔹}(1 − |z|^{2})^{b} dν(z).

Writing this in polar coordinates and using Lemma 4.6, we obtain 0 < 1 + b.

Second Necessary Condition. Assuming T_{ab} is bounded and the first necessary condition, we use Theorems 1.2 and 1.3. Let f ∈ L_{q}^{p}. We first show also f ∈ L_{b}^{1}. In the first group, by the Hölder inequality,

∥f∥_{L_{b}^{1}} = ∫_{𝔹} |f(z)|(1 − |z|^{2})^{b−q} dν_{q}(z)

≤ ∥f∥_{L_{q}^{p}}(∫_{𝔹}(1 − |z|^{2})^{(b−q)p′} dν_{q}(z))^{1/p′} < ∞,

because (b − q)p′ + q = \frac{p(1 + b) − (q + p)}{p − 1} > \frac{1 + q − (q + p)}{p − 1} = −1 by the

already obtained first necessary condition. In the second group, q ≤ b by the first necessary condition and thus f ∈ L_{b}^{1} by (12). In the third group, f ∈ L^{∞} and b > −1 by the first necessary condition; hence f ∈ L_{b}^{1} again.

Now consider the composition of bounded maps

L_{q}^{p} ⟶^{T_{ab}} B_{Q}^{P} ⟶^{D_{a}^{−Q/P}} A^{P} ,

where Theorem 4.2 is put to work. Since we have shown L_{q}^{p} ⊂ L_{b}^{1} in the previous paragraph, by Lemma 5.7, this composition equals T_{a−Q/P,b}. We conclude that T_{a−Q/P,b} : L_{q}^{p} → L^{P} is bounded by Corollary 4.11. Hence the second inequalities in (III) obtain with a − Q/P in place of a and 0 in place of Q. But these are exactly the same second inequalities in (III).

Sufficiency. We assume that the two inequalities in (III) hold excluding the cases ⟨4⟩, ⟨5⟩, ⟨6⟩. As in the previous paragraph, the second inequality is equivalent to that with a replaced by a − Q/P and Q by 0. Now Theorems 1.2 and 1.3 and Corollary 4.11 imply that T_{a−Q/P,b} : L_{q}^{p} → L^{P} ∩ H(𝔹) = A^{P} is bounded. Then the composition of maps

L_{q}^{p} ⟶^{T_{a−Q/P,b}} A^{P} ⟶^{D_{a−Q/P}^{Q/P}} B_{Q}^{P}

is also bounded by Theorem 4.2. The first inequality in (III) as in the proof of the necessity part yields L_{q}^{p} ⊂ L_{b}^{1}. Then by Lemma 5.7, the composition equals T_{ab} : L_{q}^{p} → B_{Q}^{P}. □

Proof of Theorem 1.9. If T_{ab} : L_{q}^{p} → H^{∞} is bounded, then since H^{∞} ⊂ L^{∞} with the same norms, T_{ab} : L_{q}^{p} → L^{∞} is also bounded. Then Theorem 1.2 implies the first and second necessary conditions in the cases ⟨4⟩, ⟨5⟩, ⟨6⟩.

Conversely, if the two inequalities in (III) hold in the cases ⟨4⟩, ⟨5⟩, ⟨6⟩, then T_{ab} : L_{q}^{p} → L^{∞} is bounded by Theorem 1.2. But Corollary 4.11 shows that the range of T_{ab} actually lies in H^{∞} rendering T_{ab} : L_{q}^{p} → H^{∞} bounded. □

30 H. Turgay Kaptanoğlu and A. Ersin Üreyen

Remark 8.1. As a final note, let’s show how our results on weighted spaces can be obtained from those on unweighted spaces. We assume now that Theorems 1.2 and 1.3 are proved in the case q = Q = 0 and we obtain them in the general case with nonzero q, Q. We show the details only in the case ⟨1⟩. We use a few results from Sections 4 and 5 whose proofs are independent of the proofs of our main results. Let M_{u} denote the operator of multiplication by (1 − |z|^{2})^{u} as before. Clearly M_{−q/p} : L^{p} → L_{q}^{p} is an isometric isomorphism. Consider the sequence of bounded maps

L^{p} ⟶^{M_{−q/p}} L_{q}^{p} ⟶^{T_{ab}} L_{Q}^{P} ∩ H(𝔹) = A_{Q}^{P} ⟶^{D_{a}^{−Q/P}} A^{P} ,

where we have A_{Q}^{P} by Corollary 4.11. The last map is also an isomorphism by Theorem 4.2. The composition of the three maps is T_{a−Q/P, b−q/p} : L^{p} → L^{P} by Lemma 5.7, and it is bounded if and only if the middle map T_{ab} : L_{q}^{p} → L_{Q}^{P} is bounded. By assumption, the composition is bounded if and only if

\frac{1}{p} < 1 + b − \frac{q}{p}  and  a − \frac{Q}{P} ≤ b − \frac{q}{p} + \frac{1 + N}{P} − \frac{1 + N}{p} ,

which are nothing but the inequalities in (III) of the case ⟨1⟩.

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Integral Operators with Bergman-Besov Kernels 31

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H. Turgay Kaptanoğlu Bilkent Üniversitesi Matematik Bölümü 06800 Ankara Türkiye

kaptan@fen.bilkent.edu.tr
http://www.fen.bilkent.edu.tr/∼kaptan/

A. Ersin Üreyen Eskişehir Teknik Üniversitesi Fen Fakültesi, Matematik Bölümü 26470 Eskişehir Türkiye

aeureyen@eskisehir.edu.tr
https://akademik.eskisehir.edu.tr/aeureyen/