Two-sided norm estimates for Bergman-type projections with an asymptotically sharp lower bound

0.18 MB·

by  Congwen Liu

TWO-SIDED NORM ESTIMATES FOR BERGMAN-TYPE
PROJECTIONS WITH AN ASYMPTOTICALLY SHARP LOWER
BOUND
7
1
0 CONGWENLIU,ANTTIPERA¨LA¨,ANDLIFANGZHOU
2

n
Abstract. Weobtainnewtwo-sidednormestimatesforthefamilyofBergman-
a typeprojectionsarisingfromthestandardweights (1−|z|2)α whereα>−1.
J
Asα→−1,thelowerboundissharpinthesensethatitasymptoticallyagrees
8
withthenormoftheRieszprojection. Theupperboundisestimatedinterms

ofthemaximalBergmanprojection,whoseexactoperatornormwecalculate.
]
V Theresultsprovideevidencetowardsaconjecturethatwasposedveryrecently
bythefirstauthor.
C
.
h
t
a
m 1. Introduction
[ Let D denote the unit disk in the complex plane and write dA for the normalized

Lebesgue area measure dA(z) = π−1dxdy (where z = x+iy). For α > −1, the
1
v standard weighted area measure dAα is given by
8
dA (z)=(1+α)(1−|z|2)αdA(z).
8 α
9
Because of the normalization, dA is also a probability measure. As usual, for
1 p > 0, the space Lp(D) consists oαf all Lebesgue measurable functions f on D for
0 α
. which
1 1
p
0 kfk := |f(z)|pdA (z)
p,α α
7 D
(cid:26)Z (cid:27)
1 is finite. For 0<p<∞, the Bergman space Ap consists of analytic functions f in
: α
v Lp(D), while H∞(D) denotes the space of bounded analytic functions. Under the
α
i weightedLp-topologies,the spacesaboveare complete, andBanachwheneverthey
X
are normed.
r
a We consider the natural projection onto these spaces, i.e., the orthogonal pro-
jection from L2(D) onto A2, also known as the Bergman projection. It can be
α α
expressed as an integral operator:
f(w)
P f(z)= dA (w).
α D (1−zw¯)2+α α
Z
As usual, dA =dA, Lp(D)=Lp(D), Ap(D)=Ap(D), k·k =k·k and P =P.
0 0 0 p,0 p 0
2010 Mathematics Subject Classification. Primary32A25;Secondary32A36, 47G10.
Key words and phrases. Bergmanprojection,Weighted Bergmanspace,Operatornorm.
The first author was supported by the National Natural Science Foundation of China grant
11171318andOATF,USTC;thesecondauthorwassupportedbytheAcademyofFinlandproject
number 79999201 and the Emil Aaltonen Foundation; the third author was supported by the
NationalNaturalScienceFoundationofChinagrant11271124,11301136,andtheNaturalScience
FoundationofZhejiangprovincegrantLQ13A010005.
1
2 CONGWENLIU,ANTTIPERA¨LA¨,ANDLIFANGZHOU
The Bergman projection is a central object in the study of analytic function
spaces. It naturallyrelatesto fundamentalquestions suchas duality andharmonic
conjugates, and it is also a building block for Toeplitz operators. Understanding
its behaviour and estimating its size is therefore of vital importance on several oc-
casions. There areseveraltextbooks onBergmanspacesandBergmanprojections.
For the interested reader, we recommend [5, 9, 20, 22].
Let
kP fk
kP k :=sup α p,α : f ∈Lp,f 6=0
α p,α kfk α
(cid:26) p,α (cid:27)
be the operator norm of P . For p=2, P is an orthogonalprojection on L2, and
α α α
hencekP k =1. Thepurposeofthepresentworkistoobtainnewandimproved
α 2,α
two-sidedestimates for kP k for the full rangeof p∈(1,∞), α>−1. Our main
α p,α
result reads as follows.
Theorem 1.1. For 1<p<∞, we have
Γ(2+α)Γ(2+α) Γ(1+α)Γ(1+α)
p q p q
(1.1) ≤ kP k ≤ (1+α) ,
Γ2(2+α) α p,α Γ2(2+α)
2 2
where q := p is the conjugate exponent of p.
p−1
TheboundednessofP whenα=0datesbackfiftyyearsandisduetoZaharjuta
α
and Judoviˇc [19]. About ten years later, the result was established for the whole
scale α > −1 by Forelli and Rudin [7]. However, calculating the exact value of
kP k has turned out to be very challenging. Zhu [21] first obtained the right
α p,α
asymptoticsofkP k ,andlaterDostani´c[4]foundaquantitativeversionofZhu’s
α p,α
result for α=0. Dostani´c also conjectured that
(1.2) kPk =csc(π/p).
p
Notethattheconjectureagreeswiththealreadyestablishedvalueforthenormof
the Riesz projectiondue to Hollenbeck and Verbitsky [10]. However,very recently,
in [13], the first author of the present paper obtained
kPk ≥Γ(2/p)Γ(2/q),
p
which disproves (1.2). However, this motivated the following conjecture.
Conjecture (Liu). For 1<p<∞, we have
kPk =Γ(2/p)Γ(2/q).
p
One of the reasons for writing this paper is to provide evidence supporting the
above conjecture. Namely, observe that the lower bound in (1.1) tends to csc(π/p)
as α→−1. Since α=−1 can often be viewed as the case of the Hardy spaces and
the Riesz projection, we have convincing evidence that
Γ(2+α)Γ(2+α)
p q
=kP k ,
Γ2(2+α) α p,α
2
which we conjecture to be true. When α = 0, we then recover the conjecture
from [13]. The lower bound is obtained by using a suitable choice of test functions
formed from Bergman-type kernels along with some interpolation, manipulation
of the classical Forelli-Rudin estimates [7], and a Hausdorff-Young type inequality.
We remark that in our argument there is a cut-off at α= (2−p)/(1−p), and we
NORM OF THE BERGMAN PROJECTION 3
needtwoseparatemethods. Itwouldbe ofsomeinteresttofindaunifiedapproach
that directly covers all cases.
For the upper bound, we consider the maximal Bergman projection, which is
arguably a quite standard approach in this direction. However, we manage to
calculate its exact operator norm. For the unweighted case, this result can be
deduced from the work of Dostani´c [3] – the weighted case is probably new.
In the recent years, there has been increasing interest in the study of the size
of Bergman projections in various context other than Ap. For the Bloch space,
α
we mention the works of the second author [14, 15], as well as the work of Kalaj-
Markovi´c[11]. Forthe Besovspaceswe mentionthe papersofKaptano˘glu-U¨reyen,
Per¨al¨a and Vujadinovi´c [12, 16, 18].
2. Preliminaries
We use the classical notation for the functions F
2 1
a, b ∞ (a) (b) λk
k k
F ; λ =
2 1 c (c) k!
(cid:20) (cid:21) k=0 k
X
with c6=0,−1,−2,..., where
(a) =1, (a) =a(a+1)...(a+k−1) for k≥1.
0 k
denotes the Pochhammer symbol of a. This series gives an analytic function for
|λ|<1, called the Gauss hypergeometric function associated to (a,b,c).
We refer to [6, Chapter II] for the properties of these functions. Here, we only
list some of them for later reference.
a, b Γ(c)Γ(c−a−b)
(2.1) F ; 1− = , Re(c−a−b)>0.
2 1 c Γ(c−a)Γ(c−b)
(cid:20) (cid:21)
a, b c−a, c−b
(2.2) F ; λ =(1−λ)c−a−b F ; λ .
2 1 c 2 1 c
(cid:20) (cid:21) (cid:20) (cid:21)
a, b Γ(c) 1
(2.3) F ; λ = tb−1(1−t)c−b−1(1−tλ)−adt,
2 1 c Γ(b)Γ(c−b)
(cid:20) (cid:21) Z0
Rec>Reb>0; |arg(1−λ)|<π; λ6=1.
a, b Γ(c) 1 a, b
(2.4) F ; λ = tγ−1(1−t)c−γ−1 F ; tλ dt,
2 1 c Γ(γ)Γ(c−γ) 2 1 γ
(cid:20) (cid:21) Z0 (cid:20) (cid:21)
Rec>Reγ >0; |arg(1−λ)|<π; λ6=1.
(2.5) dk F a, b; λ = (a)k(b)k F a+k, b+k; λ , k ∈N.
dλk 2 1 c (c) 2 1 c+k
(cid:20) (cid:21) k (cid:20) (cid:21)
Lemma 2.1. Suppose Rec>0, Reδ >0 and Re(δ+c−a−b)>0. Then
1 a, b Γ(c)Γ(δ)Γ(δ+c−a−b)
(2.6) tc−1(1−t)δ−1 F ; t dt= .
2 1 c Γ(δ+c−a)Γ(δ+c−b)
Z0 (cid:20) (cid:21)
Proof. Note that, under the assumption of the lemma, both sides of (2.4) are con-
tinuousatλ=1. Thelemmathenfollowsbylettingλ→1−andapplying(2.1). (cid:3)
4 CONGWENLIU,ANTTIPERA¨LA¨,ANDLIFANGZHOU
Lemma 2.2. Let a,b,c∈R and t>−1. The identity
(1−|ξ|2)tdA(ξ)
(2.7)
D (1−zξ¯)a(1−wξ¯)b(1−ξw¯)c
Z
∞
= 1 (a)j(c)j F b, c+j ; |w|2 (zw¯)j
1+t (2+t) j! 2 1 2+t+j
j=0 j (cid:20) (cid:21)
X
holds for any z,w∈D.
Proof. Recall that
∞
(γ)
(2.8) (1−λ)−γ = kλk
k!
k=0
X
holds for λ∈C, |λ|<1 and γ ∈R. This leads to
1 2π
(1−ze−iθ)−a(1−we−iθ)−b(1−eiθw¯)−cdθ
2π
Z0
∞ ∞ ∞ (a) (b) (c) 1 2π
= j k ℓ (ze−iθ)j(we−iθ)k(w¯eiθ)ℓdθ
j!k!ℓ! 2π
j=0k=0ℓ=0 Z0
XXX
∞ ∞
(a) (b) (c)
= j k j+k |w|2k(zw¯)j.
j!k!(j+k)!
j=0k=0
XX
Note that (c) =(c+j) (c) and (j+k)!=(1+j) (1) . Then
j+k k j k j
1 2π
(1−ze−iθ)−a(1−we−iθ)−b(1−eiθw¯)−cdθ
2π
Z0
∞ ∞
(a) (c) (b) (c+j)
= j j k k|w|2k (zw¯)j
(1) j! (1+j) k!
j ( k )
j=0 k=0
X X
∞
(a) (c) b, c+j
= j j F ; |w|2 (zw¯)j.
(1) j! 2 1 1+j
j=0 j (cid:20) (cid:21)
X
Therefore
(1−|ξ|2)tdA(ξ)
D (1−zξ¯)a(1−wξ¯)b(1−ξw¯)c
Z
1 1 2π
= 2 (1−zre−iθ)−a(1−wre−iθ)−b(1−reiθw¯)−cdθ (1−r2)trdr
2π
Z0 (cid:26) Z0 (cid:27)
∞ (a) (c) 1 b, c+j
= 2 j j r1+2j(1−r2)t F ; r2|w|2 dr (zw¯)j.
(1) j! 2 1 1+j
j=0 j (cid:26)Z0 (cid:20) (cid:21) (cid:27)
X
By (2.4), the integral in the parentheses equals
1Γ(1+j)Γ(1+t) b, c+j
F ; |w|2 .
2 Γ(2+t+j) 2 1 2+t+j
(cid:20) (cid:21)
Inserting this into the above series yields (2.7). (cid:3)
As an immediate consequence, we have the following.
NORM OF THE BERGMAN PROJECTION 5
Corollary 2.3. For a∈R and t>−1, we have
(1−|ξ|2)t 1 a, a
(2.9) dA(ξ)= F ; |z|2
D |1−zξ¯|2a 1+t 2 1 2+t
Z (cid:20) (cid:21)
holds for all z ∈D.
Corollary 2.4 (Forelli-Rudin estimates, see, for instance, [9, p.7, Theorem 1.7]).
For a∈R and t>−1, we have
1, if c<0;
(1−|ξ|2)t 1
(2.10) dA(ξ) ≈ log , if c=0;
ZD |1−zξ¯|2+t+c (1−1|−z|2|z)−|2c, if c>0.
as |z| → 1−. Here, we use the symbol ≈to indicate that two quantities have the
same behavior asymptotically.
Corollary 2.5. Let t>−1 and a>1+t/2. We have
(1−|w|2)tdA(w) Γ(1+t)Γ(2a−t−2)
(2.11) sup (1−|z|2)2a−t−2 = .
z∈D(cid:26) ZD |1−zw¯|2a (cid:27) Γ2(a)
Proof. By (2.9) and (2.2), we have
(1−|w|2)tdA(w) 1 2+t−a, 2+t−a
(1−|z|2)2a−t−2 = F ; |z|2 .
D |1−zw¯|2a 1+t 2 1 2+t
Z (cid:20) (cid:21)
Note that the last hypergeometricfunction is increasing in the interval [0,1), since
its Taylor coefficients are all positive. It follows that
2+t−a, 2+t−a 2+t−a, 2+t−a
sup F ; |z|2 = F ; 1−
2 1 2+t 2 1 2+t
z∈D (cid:20) (cid:21) (cid:20) (cid:21)
Γ(2+t)Γ(2a−2−t)
= .
Γ2(a)
This gives (2.11). (cid:3)
Corollary 2.6. Suppose that a,b>0, c∈R, and 1+a+b−2c>0. Then
(1−|w|2)b−1
(2.12) |z|2b(1−|z|2)a−1 dA(w) dA(z)
D D |1−zw¯|2c
Z (cid:26)Z (cid:27)
Γ(a)Γ(b)Γ(1+a+b−2c)
= .
Γ2(1+a+b−c)
Proof. Using (2.9) in the inner integral, the left-hand side of (2.12) equals
1 c, c
|z|2b(1−|z|2)a−1 F ; |z|2 dA(z)
b D 2 1 1+b
Z (cid:20) (cid:21)
1 1 c, c
= rb(1−r)a−1 F ; r dr.
b 2 1 1+b
Z0 (cid:20) (cid:21)
Now (2.12) follows from an application of Lemma 2.1. (cid:3)
6 CONGWENLIU,ANTTIPERA¨LA¨,ANDLIFANGZHOU
3. The proof of Theorem 1.1: the upper estimate
We consider the “maximal Bergman projection”
f(w)
P♯f(z)= dA (w).
α D |1−zw¯|2+α α
Z
It is clear that kP k ≤kP♯k , so it suffices to show the following.
α p,α α p,α
Proposition 3.1. For 1<p<∞ and α>−1, we have
(1+α)Γ(1+α)Γ(1+α)
kP♯k = p q .
α p,α Γ2 2+α
2
We appeal to the well known Schur’s test(cid:0)(see,(cid:1)for instance, [22, Theorem 3.6]).
Lemma 3.2. Suppose that (X,µ) is a σ-finite measure space and K(x,y) is a
nonnegative measurable function on X×X and T the associated integral operator
Tf(x)= K(x,y)f(y)dµ(y).
ZX
Let1<p<∞and1/p+1/q=1. Ifthereexistapositive constantC andapositive
measurable function u on X such that
K(x,y)u(y)qdµ(y)≤Cu(x)q
ZX
for almost every x in X and
K(x,y)u(x)pdµ(x)≤Cu(y)p
ZX
for almost every y in X, then T is bounded on Lp(X,dµ) with kTk≤C.
Proof of Proposition 3.1. With
1
K(z,w)= ,
|1−zw¯|2+α
u(z)=(1−|z|2)−1p+qα,
where q is the conjugate exponent of p, it is clear that
K(z,w)u(w)qdA (w)≤C(p)u(z)q
α
D
Z
for almost every z ∈D and
K(z,w)u(z)pdA (z)≤C(q)u(w)p
α
D
Z
for almost every w ∈D. Here
C(p) = (1+α)sup (1−|z|2)1+pα (1−|w|2)−1+pα+αdA(w) .
z∈D( ZD |1−zw¯|2+α )
In view of (2.11), we find that
Γ(1+α)Γ(1+α)
C(p)=(1+α) p q =C(q).
Γ2 2+α
2
(cid:0) (cid:1)
NORM OF THE BERGMAN PROJECTION 7
Thus, an application of Schur’s test gives
(1+α)Γ(1+α)Γ(1+α)
kP♯k ≤ p q .
α p,α Γ2 2+α
2
To prove the converse inequality, we define(cid:0), for(cid:1)ǫ>0,
g (w) := ǫ1/p(1−|w|2)(ǫ−1)(1+α)/p,
ǫ
1/q
Γ(2+α+ǫ(1+α)q)
h (z) :=
ǫ
(1+α)Γ(ǫ(1+α))Γ(2+α+ǫ(1+α)q/p)
(cid:26) (cid:27)
×|z|2+2α+2(ǫ−1)(1+α)/p(1−|z|2)(ǫ−1)(1+α)/q.
Easy calculations show that kg k =kh k =1.
ǫ p,α ǫ q,α
Applying Corollary2.6, witha=(1+α)[1+(ǫ−1)/q],b=(1+α)[1+(ǫ−1)/p]
and c=1+α/2, we obtain
g (w)
ǫ
dA (w) h (z)dA (z)
D D |1−zw¯|2+α α ǫ α
Z (cid:26)Z (cid:27)
Γ 1+α + ǫ(1+α) Γ 1+α + ǫ(1+α) Γ(ǫ(1+α))
= (1+α)2× p q q p
(cid:16) Γ2(cid:17)2+(cid:16)α +ǫ(1+α) (cid:17)
2
1/q
Γ(2+α(cid:0)+ǫ(1+α)q) (cid:1)
×ǫ1/p .
(1+α)Γ(ǫ(1+α))Γ(2+α+ǫ(1+α)q/p)
(cid:26) (cid:27)
Having in mind that
f(w)
kP♯k = sup dA (w) g(z)dA (z) ,
p kfkp,α=1(cid:12)ZD(cid:18)ZD |1−zw¯|2+α α (cid:19) α (cid:12)
kgkq,α=1(cid:12) (cid:12)
(cid:12) (cid:12)
this implies (cid:12) (cid:12)
Γ 1+α + ǫ(1+α) Γ 1+α + ǫ(1+α) Γ(ǫ(1+α))
kP♯k ≥ (1+α)2× p q q p
p,α (cid:16) Γ2(cid:17)2+(cid:16)α +ǫ(1+α) (cid:17)
2
1/q
Γ(2(cid:0)+α+ǫ(1+α)q)(cid:1)
×ǫ1/p .
(1+α)Γ(ǫ(1+α))Γ(2+α+ǫ(1+α)q/p)
(cid:26) (cid:27)
The proof is completed by letting ǫ→0+. (cid:3)
4. The proof of Theorem 1.1: the lower estimate
We proceed to show
Γ 2+α Γ 2+α
p q
(4.1) kP k ≥ .
α p,α (cid:16) Γ2(cid:17)2+α(cid:16) (cid:17)
2
We only need to consider the case when p >(cid:0)2, an(cid:1)d the case when 1 < p <2 then
followsfromtheduality. Curiously,itturnsoutthatwecannotdealwiththewhole
range α∈(−1,∞) using the same argument. Indeed, we need separate arguments
for the cases
α<(2−p)/(1−p)
and
α>(2−p)/(1−p).
8 CONGWENLIU,ANTTIPERA¨LA¨,ANDLIFANGZHOU
Note that we can assume that α6=(2−p)/(p−1), since when α=(2−p)/(p−1),
(4.1) can be easily derived from the other case. To see this, we begin with the
following
Lemma 4.1. The function p7→kP k is increasing on [2,∞).
α p,α
Proof. Assumethat2<p <p . Then,bythe Riesz-Thorininterpolationtheorem
1 2
(see [8], p.34, Theorem 1.3.4), we have
kP k ≤kP k1−θkP kθ
α p1,α α 2,α α p2,α
where θ is given by the relation
1 1−θ θ
= + .
p 2 p
1 2
Having in mind that kP k =1, kP k ≥1 and θ ∈(0,1), we get
α 2,α α p2,α
kP k ≤kP kθ ≤kP k .
α p1,α α p2,α α p2,α
(cid:3)
Assume now that we have shown (4.1) for all p>2 and α6=(2−p)/(p−1), or
in other words, (4.1) holds for all 2<p6=p∗ :=(2+α)/(1+α). Thus, by Lemma
4.1, we have
Γ 2+α Γ 2+α− 2+α
p∗−ǫ p∗−ǫ
kPαkp∗,α ≥kPαkp∗−ǫ,α ≥ (cid:16) (cid:17)Γ2(cid:16)2+α (cid:17)
2
for any 0<ǫ<p∗−2, which implies (4.1) is valid fo(cid:0)r p=(cid:1)p∗.
Fromnowon,weassumep>2,α>−1,α6=(2−p)/(p−1)andletβ :=(2+α)/2.
We fix ξ ∈D and define
f (z):=(1−ξz¯)β−2β/p(1−zξ¯)−β, z ∈D.
ξ
Using (2.7) we get
(1−|w|2)2β−2dA(w)
P f (z) = (2β−1)
α ξ D (1−zw¯)2β(1−ξw¯)2β/p−β(1−wξ¯)β
Z
∞
(β) 2β/p−β, β+k
= k F ; |ξ|2 (zξ¯)k.
k! 2 1 2β+k
k=0 (cid:20) (cid:21)
X
We now decompose
P f (z)=Φ (z)+Ψ (z)+Υ (z),
α ξ ξ ξ ξ
where
Γ(2β/p)Γ(2β/q)
(4.2) Φ (z) := (1−zξ¯)−2β/p,
ξ Γ2(β)
∞
Γ(2β/p)Γ(2β/q)
(4.3) Ψ (z) := ǫ ξ¯kzk,
ξ Γ2(β) k
k=0
X
∞
(4.4) Υ (z) := a (ξ)ξ¯kzk,
ξ k
k=0
X
NORM OF THE BERGMAN PROJECTION 9
and
(2β/p) Γ(k+2β)Γ(k+β)
k
ǫ := −1 ,
k
k! Γ(k+β+2β/q)Γ(k+2β/p)
(cid:26) (cid:27)
(β) 2β/p−β, β+k Γ(2β/q)Γ(2β+k)
a (ξ):= k F ; |ξ|2 − .
k k! 2 1 2β+k Γ(β)Γ(β+2β/q+k)
(cid:26) (cid:20) (cid:21) (cid:27)
To see this, just use the formula
Γ(a+k)
(a) =
k Γ(a)
and the formula (2.8). Note that the functions above obviously depend also on p
and α; we suppress the notation to keep the argument readable.
Since obviously
Γ(2β/p)Γ(2β/q)
kΦ k = kf k
ξ p,α Γ2(β) ξ p,α
and by Corollary 2.6,
limsupkf k =∞,
ξ p,α
|ξ|→1−
we are done, once the following lemma is proved.
Lemma 4.2. The functions
ξ 7→kΨ k
ξ p,α
and
ξ 7→kΥ k
ξ p,α
are bounded on D.
To prove this lemma, we shall use the following simple fact (See [1, Lemma 2.2]
or [2, Theorem 2]).
Lemma 4.3 (L’Hˆopital Monotone Rule). Let −∞ < a < b < ∞, and let ϕ,ψ :
[a,b] → R be continuous functions that are differentiable on (a,b), with ϕ(a) =
ψ(a)=0 or ϕ(b)=ψ(b)=0. Assume that ψ′(x)6=0 for each x in (a,b). If ϕ′/ψ′
is increasing (decreasing) on (a,b), then so is ϕ/ψ.
Also, We shall require a result of Hausdorff-Young type for Ap, which is most
α
likely known to the experts. However, we have been unable to find a reference, so
we include a proof, for completeness.
Lemma4.4(Hausdorff-YoungtheoremforAp). Supposethat2≤p<∞, α>−1,
α
and {a }∞ a sequence of complex numbers such that
k k=0
∞ q−1
k!Γ(1+α)
|a |q <∞.
k
Γ(k+2+α)
k=0(cid:26) (cid:27)
X
Then the function ϕ(z)= ∞ a zk is in Ap, and
k=0 k α
P ∞ k!Γ(1+α) q−1
kϕkq ≤ |a |q.
p,α Γ(k+2+α) k
k=0(cid:26) (cid:27)
X
10 CONGWENLIU,ANTTIPERA¨LA¨,ANDLIFANGZHOU
Proof. For simplicity, we denote λ := k!Γ(1+α) . Let µ be the discrete measure
k Γ(k+2+α)
on the set N of nonnegative integers which assigns the mass µ(k) = λ−1 to the
k
integerk =0,1,2,.... Considerthe linear operatorT that maps the sequence b:=
{b }∞ = {λ a }∞ to the formal power series ϕ(z) = a zk. For 2 < p < ∞,
k k=0 k k k=0 k
we want to show that T is bounded as an operator from Lq(N,dµ) to Lp(D,dA ),
α
P
with norm kTk≤1. But for p=2 this follows from the relation
∞ ∞
kϕk2 = λ |a |2 = λ−1|b |2 =kbk2 .
2,α k k k k L2(N,dµ)
k=0 k=0
X X
For p=∞, it is the trivial fact that
∞ ∞
kϕk∞ ≤ |ak|= λ−k1|bk|=kbkL1(N,dµ).
k=0 k=0
X X
ThuswemayinvoketheRiesz-Thorininterpolationtheorem(see[8],p.34,Theorem
1.3.4) to draw the conclusion that kT(b)kp,α ≤kbkLq(N,dµ) for 2≤p≤∞. (cid:3)
Proof of Lemma 4.2. We startwiththe functions ξ 7→kΨ k . Bythe asymptotic
ξ p,α
formula ([6, p.47])
Γ(k+a) 1
≈ ka−b 1+ (a−b)(a+b−1)+O(k−2) ,
Γ(k+b) 2k
(cid:26) (cid:27)
we see that
ǫ =O (k+1)2β−2β/q−2 .
k
In view of Lemma 4.4, this implies(cid:16)that (cid:17)
q ∞ q−1
kΨ kq ≤ Γ(2β/p)Γ(2β/q) k!Γ(2β−1) ǫ |ξ|k q
ξ p,α Γ2(β) Γ(k+2β) k
(cid:26) (cid:27) k=0(cid:26) (cid:27)
X (cid:12) (cid:12)
∞ (cid:12) (cid:12)
. (k+1)−(2β−1)(q−1)+2βq−2β−2q|ξ|qk
k=0
X
∞
. (k+1)−q−1 <+∞.
k=0
X
As for the functions Υ , we proceed as follows. For k∈Z , set
ξ +
β−2β/q, β+k
g (x):= F ; x .
k 2 1 2β+k
(cid:20) (cid:21)
Note by (2.1) that
Γ(2β/q)Γ(2β+k)
g (1−)= .
k
Γ(β)Γ(β+2β/q+k)
Thus we can rewrite (4.4) as
∞
(β)
(4.5) Υ (z)= k g (|ξ|2)−g (1−) (zξ¯)k.
ξ k k
k!
k=0
X (cid:2) (cid:3)
Since we have assume α 6= (2−p)/(p−1), i.e., β 6= q/2, the argument breaks
down into two cases.