A note on a two-parameter family of operators Ab,c on weighted Bergman spaces Текст научной статьи по специальности «Математика»

Probl. Anal. Issues Anal. Vol. 8 (26), No3, 2019, pp. 125-136

DOI: 10.15393/j3.art.2019.6570

125

UDC 517.547, 517.588

S. Naik, P. K. Nath

A NOTE ON A TWO-PARAMETER FAMILY OF OPERATORS c ON WEIGHTED BERGMAN SPACES

Abstract. In this article, we prove that the two-parameter family of operators Ab'c is bounded on the weighted Bergman spaces Bpa+c-1 if a + 2 < p and unbounded if a + 2 = p.

Key words: Generalized Cesaro operator, weighted Bergman space, boundedness

2010 Mathematical Subject Classification: 47B38, 46E15

1. Introduction. Let D denote the unit disc in the complex plane C, dD its boundary, H(D) the set of all analytic functions on D and dm(-) = 1/n r drdd the normalized Lebesgue area measure on D. For 0 < p < <x, the weighted Bergman space for —1 < a < consists of functions f E H(D), such that

Bp = (a + 1)/ |f (z)|p(1 — |z|2)adm(z) =

D 1

a +W, ,w, a.

n J 0

2n

id\

Mp(r, fX1 - r2<

where Mp(r,f) = / |f(re )|vdB.

0

To define the adjoint of the generalized Cesaro operator, we need the Gaussian hypergeometric function. Let (a,n) be the shifted factorial defined by Appel's symbol

r(a + n)

(a, n) = a(a + 1)... (a + n — 1) = ^ . ;, n E N = {1, 2, 3,...}

r(a)

© Petrozavodsk State University, 2019

and (a, 0) = 1 for a = 0. The Gaussain hypergeometric function is defined by the power series expansion

F(a,b; c; z) =

n=0

(a, n)(b, n) zn (c,n)

n!

(|z| < 1),

where a, b, c are complex numbers, such that c = —m, m = 0,1, 2, 3,..., and we assume c = —m, m = 0,1, 2, 3,..., to avoid zero denominators. Clearly, F(a,b, c, z) belongs to H(D). Many properties of the hypergeometric functions are found in [1]. Asymptotic behavior of the zero-balanced (i. e., the c = a+b case) is well-known. For the non-zero balanced case, improved formulation is obtained in [4,12], whereas the geometric properties of Gaussian hypergeometric functions are considered, for example, in [9,10]. The same problems for linear and convolution operator are dealt with in [11].

Let b, c E C with Reb > 0, Re c > 0. For a function f (z) = anzn, f (z) E H(D), the two-parameter family of Cesaro averaging operators Pb'c is given by

Pb,cf (z) = E( TbTIC E bn-kak

An k=n

n=0

where

and bk are given by b0 = 1,

A!

b;c

(b,n) n)

bk

1 + b - c

Ab+l;c+l Ak-1

1 + b - c b

Ak;c

for k ^ 1. The operators Pb'c were introduced in [2] and have been studied for boundedness on various function spaces, such as Hp, BMOA, Ba [7,8], on mixed norm spaces [6], as well as on the Dirichlet space [7]. For b = 1 + a and c = 1, we obtain the generalized Cesaro operators p 1+7,1 = c7 introduced in [14]. It is known that operators CY are bounded on the Hardy space for 0 < p < x>, BMOA, and Bloch space [17] and on the Dirichlet space [16].

For b,c E C, such that Reb > 0, Rec > 0, let Ab,c be the adjoint operator of Pb,c, given by

Abcf (z) = EE

n=0

bk -nak Ab+l;c k=n Ak

c

n

z

where f (z) = ^=0 anzn E H(D) and Ablkc and bb are the same as defined for Pb'c. These operators were formally introduced in [3] and studied for boundedness on the space of Cauchy transforms. In the notation of Stempak [14], we find that

A1+7>1f = AY f.

In particular, for y = 0

00

A"f = Af = £ Ei+I a z'"

n=0 \b=n /

where AY is the adjoint operator of the generalized Cesaro operator C1 (see [17]). If y = 0, the AY is simply adjoint of the classical Cesaro operator C (see [13]). Now we recall a known result that gives an integral representation of the operator Ab'c.

Lemma 1. [3] Let b,c E C with Re b > 0, Re c > 1 and function ft,s(z) = 1 — t — s + st + tz. Then

1 1

Ab,cf (z) = mJJ sc-2(1 — t)b-1f (^t,s(z)) * F(c,1; 1; ^t,s(z))dsdt,

00

where M = (1 + b — c)(c — 1).

Here * denotes the Hadamard product (or convolution) of power series. That is, if f (z) = Y1 ^=0 anzn and g(z) = Y1 ^=0 bnzn are two analytic functions in |z| < R, then convolution between f and g is denoted by f * g and is defined by (f * g)(z) = ^=0 anbnzn. This series converges for |z| < R2. Moreover,

(f * g)(z) = 2ni i f (w)g(z/w)|z| < pR < R2. ln% J w

|w| =p

2. Preliminary results. In this section, we recall few preliminary results, which are used to state and prove the main results of this article. The adjoint operator was considered in [13] for the case (y = 0) and in [17]. In [13], Siskakis proved the following result.

Theorem 1. The operator A is bounded on the weighted Bergman space BOa if and only if a + 2 < p.

Stevic proved, in [15], a generalization of Theorem 1 for the operators AY, when y = 0:

Theorem 2. The operator AY is bounded on the weighted Bergman space Bp if and only if a + 2 < p.

The main aim of this article is to generalize Theorem 2 by finding conditions on the parameters b and c for which the operators Ab'c are bounded on the weighted Bergman spaces.

We will use the following lemma in the sequel.

Lemma 2. [5, p. 65] For each 1 < a < ro there is a positive constant C = C(a), such that

f |1 - reie|-a d0 ^ C(1 - r)-(a-1),

if 0 ^ r < 1.

Henceforth, C, K, and C1 denote positive constants, whose values are different at different occurrences.

3. Main Results. In this section, we consider the so-called convolution operator and prove its boundedness on the weighted Bergman space Ba+c-1 for c ^ 1. Also, we state and prove the main result of this paper. From now onwards, we denote F(z) = F(1, c; 1; z) for all z E D.

Lemma 3. If p E [1, ro),a> -1, c ^ 1, f E Bpa, then f * F E Bpa+{c-1)p.

Proof. Let f E Bpa. Then

1

a + 1 f , , W, „2Na

n J

0

Mp(f, r)(1 - r2)ardr < ro. (2)

Using the definition of convolution and the fact that F(1, c; 1; z) = (1-z)-for 0 < r < p < 1, we have

2n

MP(f * F, r) = 2n f |(f * F)(peid)|pdd =

1

2n

0

2n 2n

P

de

21n f F(pe^f p^) dt

2n 2n

1

2- (1 - pe«)-c/(!>-t>)dt

2n

0 0

de.

Applying Minkowski's integral inequality and Lemma 2 above, we have

2n 2n

1 Ç/ 1

Mp(f * F,r) ^ 2ny (2ny |(1 - pe^fPe^pe) p dt ^

2n J V2n

00 2n 2n

^ 2n/ t/

00

1 — pe

it

—cp

-e P

¿(0—t)

de p dt

2n 2n

— i |1 — peit|—cdt(— 2n J 1 ^ 1 V2n

0

¿c (1—p)—c+iMp(/,r :

rei(0—t)

de

From the above inequality, we find

(1 — p)(c—1)pM^(/ * F, r)(1 — r)adr ^K^ M^f,r) (1 — r)adr

where K = C/2n.

Now, taking r = p2, we have

Mp(f * F, p2)(1 — p2)a+(c—1)pdp2 ^

^ K(1 + p)(c—1)p MP (f,p)(1 — p2)adp2.

A simple calculation shows:

MPf * F, p2)(1 — p4)a+(c—1)q p2dp2 ^ K Mp (f,p) (1 — p2)ap2dp2

p

r

p

p

1

1

1

1

1

The last inequality and (2) give i

y * F,p2)(1 - p4)a+(c-1)qp2dp2 < œ. 0

This completes the proof. □

Now we give an estimate on the norm of the convolution operator )

on Ba.

Theorem 3. Let p G (0, œ),a > — 1,c ^ 1,0 : D ^ D be a non-constant analytic function. Then the operator ) = (/ * F)(0), where F = F(1,c; 1, z) on Ba+c-1(D), satisfies the following inequality:

a + 2

IIW)Ik«_. S C (MKMV II/IIBp,

C = () if a > 0 and C = () (||0||» + |0(0)|)p +3|0(0)|)-p.

Proof. We will use the method of Siskakis [13]. Let a = |0(0)| and b = «0«» and fix 0 < r < 1. By the well known consequence of the Schwarz-pick lemma on the map 01 = b-10, we have |0(z)| ^ ^o+O^, for Izl ^ r. Since a ^ b, we have (o+br) ^ ((b-a)r+2a) for all 0 < r < 1,

^ ' (b+or) ^ (b+o) '

so |0(z)| ^ bR ^ R, where R = R(r) = ((b-b+r0+2o) for all 0 < r < 1. If f * F G Ba+c-1(D), let |(u * F)(z)| be the harmonic extension |(f * F)(Rei6>)|p on |z| ^ bR; |(u * F)(z)| is continuous on |z| ^ bR and majorizes |(f * F)(z)|p there, so

|(f * F)(0(z))|p ^ |(u * F)(0(z))| for |z| ^ r.

It follows that

2n 2n

M ((f * F)0,r) = J |(f * F)(0(rei0))|pd£ |(u * F)(0(rei0))|d0. (3) 0 0

Now, for 0 < p < 1,

2n

(u * F)(p0(0)) = ~n/ F(pei0)«(0(0)e-i0)d0,

DO

2n

|(u * F)(p0(O))| = 12-/ F(pei0)u#(0)e-i0)de

0

2n

^ 2b/|F(Pei6)||u(0(O)e-ie)|de = 0

2n

= 2n f |1 - pei0|-Cu(0(O)e-i0)de.

Finally, by Harnack's inequality and the Mean Value Theorem, we have

2n

'"«'»e-) < ^"<°> = ^2n / |f <bRe")|pde- (5)

0

From (3), (4), and (5) and using Lemma 2, we obtain

2n 2n

de <

MP((f * F)0,r) ^ ( ) I |1 - pei0|-C I |f (dReit:)|Pdt

id I-c

. I f I I — I

(2n)HbR - a

00

2n

< ^^ (bR-a)^|f)|Pde =

0

D(1 - p)-c+1 (bR + a) Mp(f R) (6)

= (2n)2 ^bR-^JMpP(f,R). (6)

Now multiply both sides of (6) by (1 - r2)ar and integrate with respect to r from 0 to 1 to get

1

J MpP((f * F)<£, r)(1 - p)c-1(1 - r2)ardr ^ 0

1

^ 77^2 i ) M(f, R)(1 - r2)ardr.

(2n)27 \bR - a

Taking p = r2, we get

1

J M((f * F)0,r)(1 - r2)a+c-1rdr * 0

1

* (2^/ (^)m(f,R)(1 - r2)"rdr.

0

Proceeding as in [13], we get i

i Mp((/ * F)0, r)(1 - r2)a+c-1rdr *

a+2 1

i /(1 - u2)aMp(/,u)udu

(2n)2 Vb + 3a/ V&- a) J K J p

0

for -1 < a < 0.

™* (^ )a( b-a )"+2»/

Hence, the conclusion follows. □

The following results, regarding the boundedness of the composition operator on the Weighted Bergman space, was proved in [13], which is a particular case of c =1 of our result, as given in Theorem 3.

Corollary. Let 0 : D ^ D be a non-constant analytic function. Then the operator T^(/) = /o0 on Bp satisfies the following inequality:

a + 2

* c, u-r + |0(O)|\ p

where C =1 if a > 0 and C = (||0|U + |0(O)|)P(||0|U + 3|0(O)|)-P, -1 < a < 0.

Now we state and prove the main result of this article.

Theorem 4. Let b, c G C with Re(b) > 0 and c ^ 1. Then the operator Ab'c is bounded on the weighted Bergman space Bp+c-1 if a + 2 < p and unbounded for a + 2 = p.

oo

Proof. Case (i) Let a

Here p > 1, because a > -1. Applying Minkowski's inequality twice and taking 0 = in Theorem 3, we obtain

l|A6'c(fm)||BP

a + c-1

1 1

K

((f * F)(<Mz)))(1 -t)b-1sc-2dsdt (1-|z|2r+c-1dm(z)

U 0 0 1 1

^ K

(f * F)(<Mz))(1 -t)b-1 dt"(1 -|z|2)a+c_1dm(zn psc-2ds^

0 U 0 1 1

^ K

(f * F)(<Mz)) " (1 -|z|2)a+c-1dm(z)) p(1 -t)b-1dtsc-2ds

0 0 U

1 1

K I I llW)lBp+c-i(1 - t)b-1dtsc-2ds ^

00

1 1 a + 2

^ KC1|f ||bs // i2 - 2s - t + p (1 - t)b-1dtsc-2ds ^

t

00

^ KCJf ||Bp2^cl^ f _+_(1 - t)b-1dt. c - 1 J t

Here K = (1 + b - c)(c - 1)(a +1). The above integral is convergent for < 1. This completes the proof.

Case (ii) Let a

Suppose f1(z) = . Using Lemma 2, we obtain

/ |i 1Z|P (1 - |z|2)adm(z) =1J(1 - r2)^ |1 - rei0|-Pderdr ^

U 0 -pi

1

2«c /"

^ — (1 - r)a+1-Pdr,

which is finite for a + 2 > p. Hence, f1 E Aja+c_ 1

1

1

n

(a) For c > 1, we have:

1 1

Ab'c(/1(z)) = K ^J(/1 * F)(1 - t - s + ts + tz)(1 - t)b—V-2dsdt:

00 1 1

= K J J F (1,c; 1; 1 - t - s + ts + tz)(1 - t)b-1sc-2dsdt

00 1 1

= K /7 (1 - t)b'1sc'\ dsdt,

00

(t + s - ts - tz)c

where K = (1 + b - c)(c - 1). Now we find 1 1

A/» = K// ^ty dsdt =

00 1 1

OO

K / (1 - t)b-1 t—csc—2 £ ^sn dsdt =

0 0 n=0

~ S 1 1

!n 1 i\ra Wi -Ara+b-1+— (ra+c)^4 / „ra+c—2,

n!

n=0 0 0

1 1

— (c)^ ^n 1 (1 - t)-+b—1

(-1)- /(1 - t)n+b—1t—(n+c)dW sn+c—2ds

= K £ ^(-1)^ (1 -(-+c) dtl sn+c—2ds. n=0 0 0

Since n + c > 1, the first integral diverges. Hence, Ab'c(/1(z)) is unbounded.

(b) For c = 1, from (1) we have

ro ro , ^ ^ / ^ ^ bk—nak \

Ab,1(/(*)) = ^^ A-TTJz-

Ab

г=0 k=n ^fc

For /1(z) = , we find

ro ro , ro , <TO <TO ,

A/)) = £ (£ -r) z- = £ A+ra + £ (£ ^ )z-

n=0 k=n —k k=0 —k n=1 k=n —k

Hence, we have

~ = V (1 + 6 - i)^*1 = b + k - 1

fc=0 k fc=0 ^ k fc=0 which is divergent. □

Remark. It is still an open question, whether the operator Ab'c is bounded on the weighted Bergman space B^+c-1 for a + 2 > p.

Conflicts of Interests: The authors declare that there is no conflict of interests regarding the publication of this paper.

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Received June 27, 2019. In revised form, September 26, 2019. Accepted October 01, 2019. Published online October 16, 2019.

Department of Applied Sciences Gauhati University Guwahati, Assam, India-781 014 S. Naik

E-mail: [email protected]; P. K. Nath

E-mail: [email protected]