Integr. Equ. Oper. Theory (2019) 91:30 https://doi.org/10.1007/s00020-019-2528-0 Published online June 14, 2019 Integral Equations c Springer Nature Switzerland AG 2019 and Operator Theory
Singular Integral Operators with Bergman–Besov Kernels on the Ball
H. Turgay Kaptano˘glu and A. Ersin Ureyen¨
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 CN . The integral oper- ators 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. 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 frac- tional derivative, Schur test, Forelli–Rudin estimate, Inclusion relation.
1. Introduction The Bergman projection is known to be a bounded operator on Lp 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 differ- ing 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 gen- eralizing 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. 30 Page 2 of 30 H. T. Kaptano˘glu, A. E. Ureyen¨ IEOT
We present our results after giving a minimal amount of notation. Let B be the unit ball in CN with respect to the norm |z| = z,z induced by the inner product w, z = w1z1 + ···+ wN zN , which is the unit disc D for N =1.LetH(B)andH∞ 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 2 q dνq(z):=(1−|z| ) dν(z). These measures are finite for q>−1andσ-finite otherwise. For 0
The kernels Kq for q>−(1+N) can also be written as 2F1(1, 1+N+q;1) to complete the picture. The kernels Kq for q<−(1 + N) appear in the literature first in [1, p. 13]. Notice that 1 1 K− N (w, z)= log . (1+ ) w, z 1 −w, z p P For a, b ∈ R, the operators acting from Lq to LQ that we investigate are 2 b Tabf(w)= Ka(w, z)f(z)(1 −|z| ) dν(z) B and 2 b Sabf(w)= |Ka(w, z)| f(z)(1 −|z| ) dν(z). B Our main results are the following two theorems that describe their bound- edness 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 Sab solely because we need operators with positive kernels in Schur tests. Theorem 1.2. Let a, b, q, Q ∈ R, 1 ≤ p ≤ P ≤∞,andassumeQ>−1 when P<∞. Then the conditions (I), (II), (III) are equivalent. p P (I) Tab : Lq → LQ is bounded. p P (II) Sab : Lq → LQ is bounded. IEOT Singular Integral Operators with Bergman–Besov Kernels Page 3 of 30 30
1+q 1+N+Q 1+N+q (III) p < 1+b and a ≤ b + P − p for 1
The very particular case of Theorem 1.8 in which a = −N, b =0,q =0, Q = −N,andP = 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.
p ∞ Theorem 1.9. Supppose a, b, q ∈ R and 1 ≤ p ≤∞.ThenTab : Lq → 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 through- out the ten cases. The sufficiency proofs are either by Schur tests or by direct H¨older or Minkowski type inequalities which also make use of growth rate estimates of Forelli–Rudin type integrals. The necessity proofs are by an orig- inal 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 Sect. 6 and sufficiency parts in Sect. 7. The proofs of Theorems 1.8 and 1.9 occupy Sect. 8. Before the proofs, we list the ma- jor standard results we use in Sect. 5. Earlier in Sect. 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 back- ground on Bergman–Besov spaces. In the next Sect. 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 p P region of boundedness of Tab : Lq → LQ can be described geometrically with fewer variables. In this direction, we let 1+q 1+Q x = − b and y = − a. p P 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 Tab 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 30 Page 6 of 30 H. T. Kaptano˘glu, A. E. Ureyen¨ IEOT
y
U x = y
R ∩ U
R
1 x
Figure 1. A typical region R ∩ U move in the xy-plane. Note that R degenerates to a vertical line segment at x = −b when q = −1. In Theorem 1.3, the inequalities of (III) can now be written in the form x<1andx
y
V
R ∩ V
R
d 1 x + x = y
Figure 2. A typical region R ∩ V 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 N equivalently, p = p/(p − 1). In multi-index notation, γ =(γ1,...,γN ) ∈ N is an N-tuple of nonnegative integers, |γ| = γ1 + ···+ γN , γ!=γ1! ···γN !, 0 γ γ1 ··· γN 0 =1,andz = z1 zN . Let S be the unit sphere in CN , 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 1 f(z) dν(z)=2N r2N−1 f(rζ) dσ(ζ) dr, B 0 S 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
∞ 2 α ∞ Lα := { ϕ measurable on B :(1−|z| ) ϕ(z) ∈ L } L∞ ∞ so that 0 = L , which are normed by α ∞ −| |2 | | ϕ Lα := ess sup (1 z ) ϕ(z) . z∈B The norm on L∞ carries over to H∞ with sup. We show an integral inner product on a space X of functions by [ · , · ]X . 30 Page 8 of 30 H. T. Kaptano˘glu, A. E. Ureyen¨ IEOT
Let’s explain the notation used in Definition 1.1.ThePochhammer sym- bol (u)v is defined by Γ(u + v) (u)v := Γ(u) when u and u + v are off the pole set −N of the gamma function Γ. In particular, (u)0 =1and(u)k = u(u +1)···(u + k − 1) for a positive integer k. The Stirling formula yields
Γ(t + u) (u)t (t)u ∼ tu−v, ∼ tu−v, ∼ tu−v (Re t →∞), (1) Γ(t + v) (v)t (t)v where x ∼ y means both x = O(y)andy = 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 occurrence. The Gauss hypergeometric function 2F1 ∈ H(D) is defined by ∞ (u)k(v)k k 2F1(u, v; t; z)= z . (t)k(1)k k=0 A large part of this work depends on the interactions between the Lebesgue classes and the Bergman–Besov spaces. Given q ∈ R and 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 par- tial derivatives or radial fractional differential operators contain the same functions. The same is true also for a Bloch–Lipschitz space. t p Remark 3.4. Definitions 3.2 and 3.3 imply that Is imbeds Bq isometrically p t ∞ ∞ into Lq if and only if (6) holds, and Is imbeds Bα isometrically into Lα if and only if (8) holds. p ∞ Much more information about the spaces Bq and Bα and their inter- connections can be found in [12].
4. Properties of Kernels and Operators 2 It is well-known that every Bq space is a reproducing kernel Hilbert space and its reproducing kernel is Kq.ThusKq(w, z) is positive definite for w, z ∈ B. Further, Kq(w, z) is holomorphic in w ∈ B and satisfy Kq(z,w)=Kq(w, z). − 2 2 2 In particular, for q> 1, the Bq are weighted Bergman spaces Aq, B−1 is 2 2 2 the Hardy space H , B−N is the Drury-Arveson space, and B−(1+N) is the 2 Dirichlet space. Moreover, the norm on Bq obtained from its reproducing kernel Kq is equivalent to its integral norms given in (7). For ease of writing, put ω = w, z∈D and let kq(ω)=Kq(w, z), which is holomorphic in ω ∈ D.
Lemma 4.1. For q<−(1+N),each|Kq(w, z)| is bounded above as w, z vary in B. More importantly, each |Kq(w, z)| with q ∈ R is bounded below by a positive constant as w, z vary in B. Therefore no Kq has a zero in B × B. k Proof. When q<−(1 + N), the growth rate (3) of the coefficient ck of ω shows that the power series of kq(ω) converges uniformly for ω ∈ D. This shows boundedness. Next we consult [15, Equation 15.13.1] and see that kq is not 0 on a set containing D\{1}. The reason for this is that the first parameter 1 of the hypergeometric function defining kq is positive. But also kq(1) = 0 clearly. Thus |kq| is bounded below on D. −1 −1 If q = −(1 + N), then k−(1+N)(ω)=ω log(1 − ω) .OnD\{1}, k−(1+N) is not zero and |kq(ω)| blows up as ω → 1 within D.So|kq| is bounded below on D. The claim about |Kq| for q>−(1 + N) is obvious and the lower bound can be taken as 2−(1+N+q). t One of the best things about the radial differential operators Ds 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 t DqKq(w, z)=Kq+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,Proposi- tion 3.2 and Corollary 8.5]. IEOT Singular Integral Operators with Bergman–Besov Kernels Page 11 of 30 30
Theorem 4.2. Let q,α,s,t ∈ R and 0
Lemma 4.5. For each q ∈ R, there is a ρ0 < 1 such that for |w|≤ρ0 and all z ∈ B, we have Re Kq(w, z) ≥ 1/2. ∞ N+q k Proof. By (3), |Kq(w, z)|≤1+C k |w, z| for some constant C and k=1 ∞ ∞ ∞ C kN+q|w, z|k ≤ C kN+q|w|k|z|k ≤ C|w| kN+q|w|k−1 k=1 k=1 k=1 for all z,w ∈ B. The last series converges, say, for |w| =1/2; call its sum W and set ρ0 = min{1/2, 1/2CW}.If|w|≤ρ0, then
∞ N+q k 1 C k w, z ≤ CW|w|≤ (z ∈ B). 2 k=1
That is, |Kq(w, z) − 1|≤1/2for|w|≤ρ0 and all z ∈ B. This implies the desired result.
We turn to the operators Tab and formulate their behavior in many important situations. But first we insert some obvious inequalities we use many times in the proofs. If c
Lemma 4.6. For u, v ∈ R, −v 1 1 1 (1 − r2)u log dr < ∞ 2 − 2 0 r 1 r 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.Foru = −1, polyno- mial growth dominates a logarithmic one. For u = −1, we reduce the integral into one studied in calculus after changes of variables.
We call −v 2 u 1 1 fuv(z)=(1−|z| ) log |z|2 1 −|z|2 test functions, because we derive half the necessary conditions of Theo- rems 1.2 and 1.3 from the action of Tab on them. Here u, v ∈ R and f00 =1. When we apply Lemma 4.6 to the fuv, we obtain the next result.
p Lemma 4.7. For 1 ≤ p<∞, we have fuv ∈ Lq if and only if q + pu > −1, ∞ or q + pu = −1 and pv > 1.Forp = ∞, we have fuv ∈ L if and only if u>0,oru =0and v ≥ 0.
Lemma 4.8. If b + u>−1 or if b + u = −1 and v>1, then Tabfuv is a finite positive constant. Otherwise, |Tabfuv(w)| = ∞ for |w|≤ρ0,whereρ0 is as in Lemma 4.5.
Proof. If b + u>−1, or b + u = −1andv>1, then by polar coordinates and the mean value property, −v 2 b+u 1 1 ab uv a −| | T f (w)= K (w, z)(1 z ) 2 log 2 dν(z) B |z| 1 −|z| 1 −v 2N−1 2 b+u 1 1 − a = C r (1 r ) 2 log 2 K (rζ, w) dσ(ζ) dr r 1−r S 0 1 −v 2N−1 2 b+u 1 1 = C r (1 − r ) log Ka(0,w) dr 2 − 2 0 r 1 r −v 1 1 1 = C r2N−1(1 − r2)b+u log dr. 2 − 2 0 r 1 r The last integral is finite by Lemma 4.6, and then evidently Tabfuv is a constant. For other values of the parameters, −v 1 2 b+u 1 1 | ab uv |≥ ab uv ≥ −| | T f (w) Re T f (w) (1 z ) 2 log 2 dν(z) 2 B |z| 1 −|z| = ∞ for |w|≤ρ0 by Lemma 4.5 and Lemma 4.6.
The adjoint of Tab is readily computed. IEOT Singular Integral Operators with Bergman–Besov Kernels Page 13 of 30 30
∗ P p p P Proposition 4.9. The formal adjoint Tab : LQ → Lq of Tab : Lq → LQ ∗ for 1 ≤ p, P < ∞ is Tab = Mb−qTaQ,whereMb−q denotes the operator of multiplication by (1 −|z|2)b−q.
p P Proof. Let f ∈ Lq and g ∈ LQ . Then 2 b 2 Q ab 2 a −| | −| | [ T f,g ]LQ = K (w, z)f(z)(1 z ) dν(z)g(w)(1 w ) dν(w) B B b−q Q = f(z)(1 −|z|2) Ka(z,w)g(w)(1 −|w|2) dν(w) B B · (1 −|z|2)q dν(z) ∗ ∗ 2 = f Tab(g) dνq =[f,Tabg ]Lq . B Thus ∗ 2 b−q 2 Q Tabg(z)=(1−|z| ) Ka(z,w)g(w)(1 −|w| ) dν(w). B
The following simple but very helpful result is well known, but we in- clude its proof for completeness.
Lemma 4.10. If 0
p P p Corollary 4.11. If Tab : Lq → LQ is bounded and f ∈ Lq , then g = Tabf p P lies in H(B).IfalsoP<∞, then Q>−1.ThusTab : Lq → AQ when it is bounded with P<∞. Consequently, if P<∞ and Q ≤−1, then p P p P Tab : Lq → LQ is not bounded, and hence Sab : Lq → LQ is not bounded. On p ∞ p ∞ the other hand, if Tab : Lq → L is bounded and f ∈ Lq , then Tabf ∈ H naturally. Proof. That g is holomorphic follows, for example, by differentiation under the integral sign, from the fact that Kq(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 P Q when P = ∞, simply because the space LQ and the inequalities in (III) are independent of Q when P = ∞. 30 Page 14 of 30 H. T. Kaptano˘glu, A. E. Ureyen¨ IEOT
p P p P Lemma 4.12. If Scb : Lq → LQ is bounded and a Proof. This is because |Ka| |Kc|, which we now prove. If a>−(1 + N), as done in [18, p. 525], we write 1 |1 −w, z|c−a 2c−a = ≤ . |1 −w, z|1+N+a |1 −w, z|1+N+c |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 |kc| is bounded below by Lemma 4.1. On that part of D near 1, |k−(1+N)| is still dominated by |kc|, because ω−1 log(1 − ω)−1 lim − N c =0. Dω→1 (1 − ω) (1+ + ) If a<−(1 + N), then |Kc| dominates |Ka| by Lemma 4.1. Thus in all cases |Ka(w, z)| |Kc(w, z)| for all w, z ∈ B if a 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)= G(x, y)f(x) dλ(x). 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 Theorem 5.2. Let 1 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 p 1/p 1/p |f(x, y)| dλ(x) dμ(y) ≤ |f(x, y)|p dμ(y) dλ(x), Y X X Y 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 ⎧ ⎨⎪1,c = Ka+t(w, z)f(z) dνb(z)=Ta+t,bf(w). B Above, we have used 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 Xv is a family of spaces indexed by v ∈ R,thesymbolX p Theorem 5.8. Let Bq be given. p P 1+N+q 1+N+Q (i) If p ≤ P , then Bq ⊂ BQ if and only if p ≤ P . p P 1+q 1+Q (ii) If P ∞ p Theorem 5.9. (i) H ⊂ Bq if and only if q>−1,orq = −1 and p ≥ 2. p ∞ (ii) Bq ⊂ H if and only if q<−(1 + N),orq = −(1 + N) and 0 p ∞ p ∞ Theorem 5.10. Given Bq , we have B 1+q ⊂ Bq ⊂B1+N+q . < p p B∞ ⊂ ∞ ⊂B∞ Theorem 5.11. <0 H . 6. Necessity Proofs Now we obtain the two inequalities in (III) of Theorems 1.2 and 1.3 from the boundedness of Tab. In this section, we do not need to assume Q>−1 since the boundedness of Tab 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. p P Theorem 6.1. Let a, b, q, Q ∈ R and 1 ≤ p, P ≤∞.IfTab : Lq → LQ is 1+q 1+q bounded, then p ≤ 1+b in the cases 2, 3, 4,and p < 1+b in the remaining cases. IEOT Singular Integral Operators with Bergman–Besov Kernels Page 17 of 30 30 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 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. p P Theorem 6.2. Let a, b, q, Q ∈ R and 1 ≤ p, P ≤∞.SupposeTab : Lq → LQ 1+N+Q 1+N+q is bounded. Then a ≤ b + P − p in the cases 1, 2, 3, 4,also 1+N+Q 1+N+q 1+Q 1+q a Proof. The boundedness of Tab via Corollary 4.11 implies Q>−1 when P<∞. This time we form four groups of the ten cases. The first group again p consists of the cases 1, 7, 8,and2, 3.Leth ∈ Bq . In order to be able 1 to use Lemma 5.6, we need to make sure h ∈ Bb . In the cases 1, 7, 8, the first necessary condition gives (1 + q)/p < 1+b, and then Theorem 5.8 1 (ii) shows h ∈ Bb . In the cases 2, 3 where p = 1, Theorem 6.1 gives 1 q ≤ b, and then (13)showsh ∈ Bb again. Let t satisfy (6). Together with the first necessary condition, this gives b + t>−1. Now consider the sequence of bounded maps It Db−a p −−−−b → p −−−−Tab→ P ∩ B P −−−−a → P Bq Lq LQ H( )=AQ BQ+P (b−a), P where Theorem 4.2 is invoked and Q>−1 is used in writing AQ. Lemma 5.6 p P yields that Bq is imbedded in BQ+P (b−a) by the inclusion map. By Theo- rem 5.8, this implies a Db−a It p −−−−a → p −−−−b → p −−−−Tab→ ∞ ∩ B ∞ Bq+p(a−b) Bq Lq L H( )=H . 30 Page 18 of 30 H. T. Kaptano˘glu, A. E. Ureyen¨ IEOT p ∞ Lemma 5.6 yields that Bq+p(a−b) is imbedded in H by the inclusion map. By Theorem 5.9 (ii), this implies a Proof. The boundedness of Tab 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>−1andTab = TQQ : L → L is bounded. Let (1 −z,w)1+N+Q gz(w)= , |1 −z,w|1+N+Q which is a uniformly bounded family for z ∈ B. The same is true also of {TQQgz}.But (1 −|w|2)Q 1 1 QQ z ∼ T g (z)= 1+N+Q dν(w) 2 log 2 B |1 −z,w| |z| 1 −|z| by Proposition 5.4, contradicting uniform boundedness. IEOT Singular Integral Operators with Bergman–Besov Kernels Page 19 of 30 30 If q = b and a = b +(1+N + Q)/P − (1 + N + q) simultaneously in the ∗ P ∞ case 3, then also a =(1+N +Q)/P −(1+N)andTab = TaQ : LQ → L is ∈ P bounded with P > 1. There is an unbounded g B−(1+N) by Theorem 5.9. (1+N+Q)/P ∈ P ⊂ P ⊂ 1 Then h = DQ−(1+N+Q)/P g BQ LQ LQ. By Lemma 5.5 and (4), we obtain ∗ a−Q a−Q (1+N+Q)/P Tabh = TaQh = CDQ h = CDQ DQ−(1+N+Q)/P g = Cg since (a − Q)+(1+N + Q)/P = 0 by the conditions on the parameters. But ∞ ∗ P ∞ g/∈ L , and this contradicts that Tab : LQ → L . If q = b and a = b − (1 + N + q) simultaneously in the case 4, then also a = −(1 + N). For j =1, 2,...,letzj =(1− 1/j, 0,...,0) and Ej the ball of radius 1/2j centered at zj, and define χ (z) Ej fj(z)= q . ν(Ej)(1 −|z|2) ∈ 1 1 { } { } Clearly fj Lq and fj Lq =1foreveryj. Then Tabfj = T−(1+N),qfj is a uniformly bounded family. By the mean value property, 1 T−(1+N),qfj(w)= K−(1+N)(z,w) dν(z)=K−(1+N)(w, zj ). ν(Ej) Ej But 1 1 T−(1+N),qfj(zj)=K−(1+N)(zj,zj)= log |zj|2 1 −|zj|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 Sab. 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)andKa 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. 30 Page 20 of 30 H. T. Kaptano˘glu, A. E. Ureyen¨ IEOT 1 1 Case 2.Letf ∈ Lq. We write the LQ norm of Sabf explicitly, then exchange the order of integration by the Fubini theorem, and obtain 2 b 2 Q ab 1 ≤ | a || | −| | −| | S f LQ K (w, z) f(z) (1 z ) dν(z)(1 w ) dν(w) B B (1 −|w|2)Q = |f(z)|(1 −|z|2)b dν(w) dν(z) |1 −z,w|1+N+a B B −| |2 Q 2 b−q (1 w ) | | −| | q = f(z) (1 z ) 1+N+a dν(w) dν (z) B B |1 −z,w| 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.Ifa − Q<0, then the integral in J(z) is bounded and J(z) is also bounded since b − q ≥ 0bythe first inequality in (III). If a − Q = 0, then the integral in J(z) is logarithmic. But now b>qby (III) and Remark 1.6,andJ(z) is bounded this time by (12). If a − Q>0, then J(z) ∼ (1 −|z|2)b−q−a+Q.Butb − q − a + Q ≥ 0 by the second inequality in (III); hence J(z) is bounded once again. Thus 1 1 ab 1 L1 ab → S f LQ f q and S : Lq LQ is bounded. 1 P Case 3.Letf ∈ Bq . We write the LQ norm of Sabf explicitly, then use Lemma 5.3 with the measures νq and νQ, and obtain P 1/P 2 b−q ab P a −| | q Q S f LQ = K (w, z)f(z)(1 z ) dν (z) dν (w) B B 1/P P P 2 (b−q)P ≤ |Ka(w, z)| |f(z)| (1 −|z| ) dνQ(w) dνq(z) B B −| |2 Q 1/P 2 b−q (1 w ) = |f(z)|(1−|z| ) dν(w) dνq(z). (1+N+a)P B B |1−z,w| For z ∈ B,letJ(z) be that part of the integrand of the outer integral multi- plying |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.Ifu<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 u 1+N + Q b − q − = b − q − (1 + N + a)+ ≥ 0 P P IEOT Singular Integral Operators with Bergman–Besov Kernels Page 21 of 30 30 by the second inequality in (III). So J(z) is bounded one more time. Thus 1 P ab P L1 ab → S f LQ f q and S : Lq LQ is bounded. 1 Case 4.Letf ∈ Lq.Ifq = b, then (III) says that a<−(1 + N)andKa is bounded by Lemma 4.1. 2 b | |≤ | || | −| | 1 ∈ B Sabf(w) Ka(w, z) f(z) (1 z ) dν(z) f Lq (w ). B ∞ 1 Then Sabf L f Lq . Otherwise q | | | | 1 ∈ B Sabf(w) f(z) dνq(z)= f Lq (w ) B ∞ ∞ 1 1 → and Sabf L f Lq .ThusSab : Lq L is bounded. p Case 5.Letf ∈ Lq . We write Sabf(w) explicitly, apply the H¨older inequal- ity with the measure νq, and obtain 2 b |Sabf(w)|≤ |Ka(w, z)||f(z)|(1−|z| ) dν(z) B (1 −|z|2)b−q | | q = f(z) 1+N+a dν (z) B |1 −w, z| b−q p q 1/p −| |2 ( ) + (1 z ) 1/p ≤ f Lp dν(z) =: J(w) f Lp . q (1+N+a)p q B |1 −w, z| We show that J is bounded on B by Proposition 5.4.Now (q + 1)(p − 1) 1+q (b − q)p + q +1=p b − q + = p 1+b − > 0 p p by the first inequality in (III), so the power on 1 −|z|2 in J(w)is> −1as required. Note that 1+N +q (1+N +a)p − (1+N) − (b−q)p − q = p 1+N + a − b + q − p 1+N + q = p a − b + < 0 p 30 Page 22 of 30 H. T. Kaptano˘glu, A. E. Ureyen¨ IEOT by the second inequality in (III). This shows that J(w) is bounded for all ∈ B | | p ∈ B ∞ p w . Then Sabf(w) f Lq for all w and Sabf L f Lq .Thus p ∞ Sab : Lq → L is bounded. ∞ ∞ Case 6.Letf ∈ L . We write Sabf(w) explicitly, take the L norm of f out of the integral, and obtain 2 b |Sabf(w)|≤ |Ka(w, z)||f(z)|(1−|z| ) dν(z) B (1 −|z|2)b ≤ L∞ L∞ f 1+N+a dν(z)=:J(w) f . B |1 −w, z| We show that J is bounded on B by Proposition 5.4. Check that b>−1by the first inequality in (III). The second inequality in (III) gives that a p 1 Case 8.Letf ∈ Lq . We write the LQ norm of Sabf explicitly, then exchange the order of integration by the Fubini theorem, afterwards apply the H¨older inequality, and obtain 2 b 2 Q ab 1 ≤ | a || | −| | −| | S f LQ K (w, z) f(z) (1 z ) dν(z)(1 w ) dν(w) B B (1 −|w|2)Q = |f(z)|(1 −|z|2)b dν(w) dν(z) |1 −z,w|1+N+a B B −| |2 Q | | −| |2 q/p (1 w ) dν(w) −| |2 b−q/p = f(z) (1 z ) 1+N+a (1 z ) dν(z) B B |1 −z,w| 2 Q p 1/p (1 −|w| ) dν(w) b−q/p p ≤ p −| |2 ( ) f Lq 1+N+a (1 z ) dν(z) B B |1 −z,w| /p 1 p =: J f Lq We show that J is finite using Proposition 5.4. Check that Q>−1asre- quired. ∼ −| |2 (b−q/p)p If a q 1+q 1 1 b − p = p 1+b − − >p − = −1 p p p p by the first inequality in (III). So J is finite by Lemma 4.6.Ifa = Q, then p ∼ 1 1 −| |2 (b−q/p)p ∞ J 2 log 2 (1 z ) dν(z) < B |z| 1 −|z| also by Lemma 4.6.Ifa>Q, then J ∼ (1 −|z|2)(b−q/p)p −(a−Q)p dν(z). B But q 1+q 1 1 b − p − (a − Q)p = p b − a +1+Q − − >p − = −1 p p p p IEOT Singular Integral Operators with Bergman–Besov Kernels Page 23 of 30 30 by the second inequality in (III). So J is finite one more time. We conclude p 1 that Sab : Lq → LQ is bounded. ∞ 1 Case 9.Letf ∈ L . We write the LQ norm of Sabf explicitly, then ex- change the order of integration by the Fubini theorem, take the L∞ norm of f out of the integral, and obtain 2 b 2 Q ab 1 ≤ | a || | −| | −| | S f LQ K (w, z) f(z) (1 z ) dν(z)(1 w ) dν(w) B B (1 −|w|2)Q = |f(z)|(1 −|z|2)b dν(w) dν(z) |1 −z,w|1+N+a B B −| |2 Q 2 b (1 w ) ≤ L∞ −| | f (1 z ) 1+N+a dν(w) dν(z) B B |1 −z,w| =: J f L∞ . We show that J is finite using Proposition 5.4. Check that Q>−1asre- quired. If a 1+Q Q − (a − b)P = Q − (a − b)P +1− 1=P + b − a − 1 > −1 P by the second inequality in (III), which shows that J is again finite. Conse- ∞ P quently Sab : L → LQ is bounded. Case 7. We employ the Schur test in Theorem 5.2 with the measures (1−|z|2)b−q λ = νq, μ = νQ and the kernel G(z,w)= |1−z,w |1+N+a which together 2 m give us Z = Sab, along with the strictly positive functions φ(z)=(1−|z| ) 30 Page 24 of 30 H. T. Kaptano˘glu, A. E. Ureyen¨ IEOT 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 2 b−q (1 −|z| ) (1 −|z|2)mp (1 −|z|2)q dν(z) (1 −|w|2)nP , |1 −w, z|1+N+a B −| |2 b−q (1 z ) −| |2 nP −| |2 Q −| |2 mp 1+N+a (1 w ) (1 w ) dν(w) (1 z ) . B |1 −w, z| 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 < 0and −nP = a − (b + mp), (15) −mp = a − (nP + Q) − (b − q). But we must also make sure that the conditions of Proposition 5.4 for this to happen are met, that is, b + mp > −1,nP+ Q>−1, (16) a − (b + mp) > 0,a− (nP + Q) > 0. 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 p(P − 1)m − P (p − 1)n =(a − b)(p − 1)(P − 1), (17) −pm + Pn = a − b + q − Q. This system has the unique solution (p − 1)(P (a − b)+q − Q) m = , p(P − p) (18) (P − 1)(p(a − b)+q − Q) n = P (P − p) for m, n. The second inequality in (III) can be written in the form 1+Q 1+q a − b = −ε + − (19) P p with ε>0. By Lemma 4.12, it suffices to show that Sab is bounded when (19) holds for small enough ε>0. Substituting this value of a − b into (18), the solution takes the form p − 1 1+q εP m = − + , p p p − P (20) P − 1 1+Q εp n = − + , P P p − P What is left is to show that this solution satisfies all the required con- ditions for sufficiently small ε>0. Bear in mind that Q>−1. First, by the first inequality in (III), 1+Q 1+q 1+Q a = −ε + b + − > −ε + − 1 > −(1 + ε) > −(1 + N) P p P IEOT Singular Integral Operators with Bergman–Besov Kernels Page 25 of 30 30 provided ε 1+Q εp 1+Q εp(P −1) nP+ Q =(P −1) − + +Q = −1+ + > −1. (22) P p−P P p − P By (19) and (21), 1+Q εP 1+Q εp a − (b + mp)=−ε + − = − > 0 (23) P p − P P p − P 1 1 provided ε< P − p (1 + Q). By (19), (22), and the first inequality in (III), 1+Q 1+q 1+Q εp(P − 1) a − (nP + Q)=−ε + b + − +1− − P p P p − P 1+q εP (p − 1) =1+b − − > 0 p p − P p 1 − 1 − 1+q provided ε< p−1 P p 1+b p . Lastly, we check the third condition of Theorem 5.2, which is the finiteness of b−q −| |2 (1 z ) −| |2 mp −| |2 nP −| |2 q −| |2 Q 1+N+a (1 z ) (1 w ) (1 z ) (1 w ) dν(z)dν(w). B B |1−w, z| Call the double integral J. We estimate first the integral, say, with respect to dν(z) by Proposition 5.4 and obtain J ∼ (1 −|w|2)nP +Q−a+b+mp dν(w) B by (23). But by (22) and (23), the power on 1 −|w|2 is 1+Q εp(P − 1) 1+Q εp nP + Q − (a − (b + mp)) = −1+ + − + P p − P P p − P εpP = −1+ > −1, p − P making J<∞.Thusfor 1 1 p 1 1 1+q 0 <ε 1 1+N + Q d = n + + ε 1+N + a P (28) 1 1+N + q c = −n +1+N + b − − ε 1+N + a p with ε>0by(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 1+N + Q m =(b − q)d − (1 + N + a)d + n + P (29) 1+N + Q 1+N + Q =(b − q)d − n − − ε + n + =(b − q)d − ε. P P IEOT Singular Integral Operators with Bergman–Besov Kernels Page 27 of 30 30 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 q 1 q 1 = p b − q + + − εp = p b − +1− − εp p p p p 1+q = p 1+b − − ε > 0 p 1+q by the first inequality in (III) provided ε<1+b − p . Again since c + d =1 and substituting in for m and c from (29) and (28), 1+N + q F = p (1 + N + a)c − (b − q)c − (b − q)d + ε − p 1+N + q 1+N + q = p −n +1+N + b − − ε − b + q + ε − p p = −pn>0 by (27). Thus for 1+q 0 <ε<1+b − , p Theorem 5.1 using the selected φ and ψ with the powers in (29) and (27) p P applies with the parameters in (28) and proves that Sab : Lq → LQ with 1 1 1+Q 1+N + Q 1 N d> − + + ε = + ε > 0, 1+N + a P P 1+N + a P 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.HavingprovedTheorems1.2 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. ProofofTheorem1.8. First Necessary Condition. We imitate the proof of Theorem 6.1 assuming Tab is bounded. We form the same three groups of the ten cases and use the same test functions fuv for each group obtaining P Tabfuv ∈ BQ. Then Tabfuv(0) ∈ C. In the first group excluding the case 5, −1 2 b−(1+q)/p 1 1 ab uv −| | T f (0) = (1 z ) 2 log 2 dν(z). B |z| 1 −|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, 2 b+u Tabfu0(0) = (1 −|z| ) dν(z) B 30 Page 28 of 30 H. T. Kaptano˘glu, A. E. Ureyen¨ IEOT 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, 2 b Tabf00(0) = (1 −|z| ) dν(z). B Writing this in polar coordinates and using Lemma 4.6, we obtain 0 < 1+b. Second Necessary Condition. Assuming Tab is bounded and the first p necessary condition, we use Theorems 1.2 and 1.3.Letf ∈ Lq . We first show also f ∈ L1. In the first group, by the H¨older inequality, b 2 b−q 1 | | −| | q f Lb = f(z) (1 z ) dν (z) B 1/p b−q p ≤ p −| |2 ( ) ∞ f Lq (1 z ) dνq(z) < , B − p(1+b)−(q+p) 1+q−(q+p) − because (b q)p + q = p−1 > p−1 = 1 by the already ob- tained first necessary condition. In the second group, q ≤ b by the first 1 ∞ necessary condition and thus f ∈ Lb by (12). In the third group, f ∈ L 1 and b>−1 by the first necessary condition; hence f ∈ Lb again. Now consider the composition of bounded maps D−Q/P p Tab P a P Lq −−−−→ BQ −−−−−→ A , p 1 where Theorem 4.2 is put to work. Since we have shown Lq ⊂ Lb in the previous paragraph, by Lemma 5.7, this composition equals Ta−Q/P,b.We p P conclude that Ta−Q/P,b : Lq → L 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 p P P and 1.3 and Corollary 4.11 imply that Ta−Q/P,b : Lq → L ∩ H(B)=A is bounded. Then the composition of maps Q/P T D p a−Q/P,b P a−Q/P P Lq −−−−−−→ A −−−−−→ BQ is also bounded by Theorem 4.2. The first inequality in (III) as in the proof p 1 of the necessity part yields Lq ⊂ Lb . Then by Lemma 5.7, the composition p P equals Tab : Lq → BQ. p ∞ ∞ ∞ ProofofTheorem1.9. If Tab : Lq → H is bounded, then since H ⊂ L p ∞ with the same norms, Tab : Lq → 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, p ∞ then Tab : Lq → L is bounded by Theorem 1.2. But Corollary 4.11 shows ∞ p ∞ that the range of Tab actually lies in H rendering Tab : Lq → H bounded. 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 The- orems 1.2 and 1.3 areprovedinthecaseq = Q = 0 and we obtain them in IEOT Singular Integral Operators with Bergman–Besov Kernels Page 29 of 30 30 the general case with nonzero q,Q. We show the details only in the case 1. We use a few results from Sects. 4 and 5 whose proofs are independent of the proofs of our main results. Let Mu denote the operator of multiplication by 2 u p p (1 −|z| ) as before. Clearly M−q/p : L → Lq is an isometric isomorphism. Consider the sequence of bounded maps M D−Q/P p −q/p p Tab P P a P L −−−−→ Lq −−−−→ LQ ∩ H(B)=AQ −−−−−→ A , P where we have AQ by Corollary 4.11. The last map is also an isomorphism by p P Theorem 4.2. The composition of the three maps is Ta−Q/P, b−q/p : L → L p P by Lemma 5.7, and it is bounded if and only if the middle map Tab : Lq → LQ is bounded. By assumption, the composition is bounded if and only if 1 q Q q 1+N 1+N < 1+b − and a − ≤ b − + − , p p P p P p which are nothing but the inequalities in (III) of the case 1. Publisher’s Note Springer Nature remains neutral with regard to jurisdic- tional claims in published maps and institutional affiliations. References [1] Beatrous, F., Burbea, J.: Holomorphic Sobolev spaces on the ball. Diss. Math. 276, 57 (1989) [2] Cheng, G., Hou, X., Liu, C.: The singular integral operator induced by Drury– Arveson kernel. Complex Anal. Oper. Theory 12, 917–929 (2018) [3] Cheng, G., Fang, X., Wang, Z., Yu, J.: The hyper-singular cousin of the Bergman projection. Trans. Am. Math. 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Theory 82, 519–532 (2015) H. Turgay Kaptano˘glu(B) Bilkent Universitesi,¨ Matematik B¨ol¨um¨u 06800 Ankara Turkey e-mail: [email protected] URL: http://www.fen.bilkent.edu.tr/∼kaptan/ A. Ersin Ureyen¨ Eski¸sehir Teknik Universitesi,¨ Fen Fak¨ultesi Matematik B¨ol¨um¨u 26470 Eskisehir Turkey e-mail: [email protected] URL: https://akademik.eskisehir.edu.tr/aeureyen/ Received: November 21, 2018. Revised: May 15, 2019.