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Research article

A Derivative Hilbert operator acting from Bergman spaces to Hardy spaces

  • Received: 29 September 2022 Revised: 01 February 2023 Accepted: 07 February 2023 Published: 15 February 2023
  • MSC : 47B35, 30H10, 30H20

  • Let μ be a positive Borel measure on the interval [0,1). The Hankel matrix Hμ=(μn,k)n,k0 with entries μn,k=μn+k, where μn=[0,1)tndμ(t), formally induces the operator as follows:

    DHμ(f)(z)=n=0(k=0μn,kak)(n+1)zn,zD,

    where f(z)=n=0anzn is an analytic function in D. In this article, we characterize those positive Borel measures on [0,1) such that DHμ is bounded (resp., compact) from Bergman spaces Ap into Hardy spaces Hq, where 0<p,q<.

    Citation: Yun Xu, Shanli Ye. A Derivative Hilbert operator acting from Bergman spaces to Hardy spaces[J]. AIMS Mathematics, 2023, 8(4): 9290-9302. doi: 10.3934/math.2023466

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  • Let μ be a positive Borel measure on the interval [0,1). The Hankel matrix Hμ=(μn,k)n,k0 with entries μn,k=μn+k, where μn=[0,1)tndμ(t), formally induces the operator as follows:

    DHμ(f)(z)=n=0(k=0μn,kak)(n+1)zn,zD,

    where f(z)=n=0anzn is an analytic function in D. In this article, we characterize those positive Borel measures on [0,1) such that DHμ is bounded (resp., compact) from Bergman spaces Ap into Hardy spaces Hq, where 0<p,q<.



    Ramanujan's last letter to Hardy is one of the most mysterious and important mathematical letters in the history of mathematics. He introduced a class of functions that he called mock theta functions in his letter. For nearly a century, properties of these functions have been widely studied by different mathematicians. The important direction involves the arithmetic properties (see [1,2]), combinatorics (see [3,4]), identities between these functions, and generalized Lambert series (see [5,6]). For the interested reader, regarding the history and new developments in the study of mock theta functions, we refer to [7].

    In 2007, McIntosh studied two second order mock theta functions in reference [8]; more details are given in reference [9]. These mock theta functions are:

    A(q)=n=0q(n+1)2(q;q2)n(q;q2)2(n+1)=n=0qn+1(q2;q2)n(q;q2)n+1, (1.1)
    B(q)=n=0qn(q;q2)n(q;q2)n+1=n=0qn(n+1)(q2;q2)n(q;q2)2n+1, (1.2)

    where

    (a;q)n=n1i=0(1aqi),(a;q)=i=0(1aqi),
    (a1,a2,,am;q)=(a1;q)(a2;q)(am;q),

    for |q|<1.

    The functions A(q) and B(q) have been combinatorially interpreted in terms of overpartitions in [3] using the odd Ferrers diagram. In this paper, we study some arithmetic properties of one of the second order mock theta functions B(q). We start by noting, Bringmann, Ono and Rhoades [10] obtained the following identity:

    B(q)+B(q)2=f54f42, (1.3)

    where

    fkm:=(qm;qm)k,

    for positive integers m and k. We consider the function

    B(q):=n=0b(n)qn. (1.4)

    Followed by Eq (1.3), the even part of B(q) is given by:

    n=0b(2n)qn=f52f41. (1.5)

    In 2012, applying the theory of (mock) modular forms and Zwegers' results, Chan and Mao [5] established two identities for b(n), shown as:

    n=0b(4n+1)qn=2f82f71, (1.6)
    n=0b(4n+2)qn=4f22f44f51. (1.7)

    In a sequel, Qu, Wang and Yao [6] found that all the coefficients for odd powers of q in B(q) are even. Recently, Mao [11] gave analogues of Eqs (1.6) and (1.7) modulo 6

    n=0b(6n+2)qn=4f102f23f101f6, (1.8)
    n=0b(6n+4)qn=9f42f43f6f81, (1.9)

    and proved several congruences for the coefficients of B(q). Motivated from this, we prove similar results for b(n) by applying identities on the coefficients in arithmetic progressions. We present some congruence relations for the coefficients of B(q) modulo certain numbers of the form 2α3β,2α5β,2α7β where α,β0. Our main theorems are given below:

    Theorem 1.1. For n0, we have

    n=0b(12n+9)qn=18[f92f123f171f36+2f52f43f6f91+28f62f33f66f141], (1.10)
    n=0b(12n+10)qn=36[2f162f106f201f3f412qf282f33f212f241f84f2616q2f22f33f84f212f161f26]. (1.11)

    In particular, b(12n+9)0(mod18),b(12n+10)0(mod36).

    Theorem 1.2. For n0, we have

    n=0b(18n+10)qn=72[f162f213f271f96+38qf132f123f241+64q2f102f33f96f211], (1.12)
    n=0b(18n+16)qn=72[5f152f183f261f66+64qf122f93f36f231+32q2f92f126f201]. (1.13)

    In particular, b(18n+10)0(mod72),b(18n+16)0(mod72).

    Apart from these congruences, we find some relations between b(n) and restricted partition functions. Here we recall, Partition of a positive integer ν, is a representation of ν as a sum of non-increasing sequence of positive integers μ1,μ2,,μn. The number of partitions of ν is denoted by p(ν) which is called the partition function. If certain conditions are imposed on parts of the partition, is called the restricted partition and corresponding partition function is named as restricted partition function. Euler proved the following recurrence for p(n) [12] [p. 12, Cor. 1.8]:

    (n)p(n1)p(n2)+p(n5)+p(n7)p(n12)p(n15)++(1)kp(nk(3k1)/2)+(1)kp(nk(3k+1)/2)+={1, if n=0,0, otherwise. 

    The numbers k(3k±1)/2 are pentagonal numbers. Following the same idea, different recurrence relations have been found by some researchers for restricted partition functions. For instance, Ewell [13] presented the recurrence for p(n) involving the triangular numbers. For more study of recurrences, see [14,15,16]. Under the influence of these efforts, we express the coefficients of mock theta function B(q) which are in arithmetic progression in terms of recurrence of some restricted partition functions.

    This paper is organized as follows: Section 2, here we recall some preliminary lemmas and present the proof of Theorems 1.1 and 1.2. Section 3 includes some more congruences based on the above results. Section 4 depicts the links between b(n) and some of the restricted partition functions.

    Before proving the results, we recall Ramanujan's theta function:

    j(a,b)=n=an(n+1)2bn(n1)2, for|ab|<1.

    Some special cases of j(a,b) are:

    ϕ(q):=j(q,q)=n=qn2=f52f21f24,ψ(q):=j(q,q3)=n=0qn(n+1)/2=f22f1.

    Also,

    ϕ(q)=f21f2.

    The above function satisfy the following properties (see Entries 19, 20 in [17]).

    j(a,b)=(a,b,ab;ab),(Jacobi's triple product identity),
    j(q,q2)=(q;q),(Euler's pentagonal number theorem).

    We note the following identities which will be used below.

    Lemma 2.1. [[18], Eq (3.1)] We have

    f32f31=f6f3+3qf46f59f83f18+6q2f36f29f218f73+12q3f26f518f63f9. (2.1)

    Lemma 2.2. We have

    f22f1=f6f29f3f18+qf218f9, (2.2)
    f2f21=f46f69f83f318+2qf36f39f73+4q2f26f318f63. (2.3)

    Proof. The first identity follows from [[19] Eq (14.3.3)]. The proof of second identity can be seen from [20].

    Lemma 2.3. We have

    1f41=f144f142f48+4qf24f48f102, (2.4)
    f41=f104f22f484qf22f48f24. (2.5)

    Proof. Identity (2.4) is Eq (1.10.1) from [19]. To obtain (2.5), replacing q by q and then using

    (q;q)=f32f1f4.

    Now, we present the proof of Theorems 1.1 and 1.2.

    Proof of Theorems 1.1 and 1.2. From Eq (1.6), we have

    n=0b(4n+1)qn=2(f32f31)3f22f1.

    Substituting the values from Eqs (2.1) and (2.2) in above, we get

    n=0b(4n+1)qn=2f36f29f33f18+2qf26f218f23f9+12qf66f79f103f218+18q2f96f129f173f318+36q2f56f49f18f93+90q3f86f99f163+72q3f46f9f418f83+48q4f36f718f73f29+288q4f76f69f318f153+504q5f66f39f618f143+576q6f56f918f133. (2.6)

    Bringing out the terms involving q3n+2, dividing by q2 and replacing q3 by q, we get (1.10). Considering Eq (1.5), we have

    n=0b(2n)qn=f32f31f22f1.

    Substituting the values from Eqs (2.1) and (2.2), we obtain

    n=0b(2n)qn=(f6f3+3qf46f59f83f18+6q2f36f29f218f73+12q3f26f518f63f9)(f6f29f3f18+qf218f9).

    Extracting the terms involving q3n,q3n+1,q3n+2 from the above equation, we have

    n=0b(6n)qn=f22f23f21f6+18qf32f3f46f71, (2.7)
    n=0b(6n+2)qn=f2f26f1f3+3f52f73f91f26+12qf22f76f61f23, (2.8)
    n=0b(6n+4)qn=9f42f43f6f81. (2.9)

    Using Eqs (2.4) and (2.5) in Eq (2.9), we get

    n=0b(6n+4)qn=9f42f6(f144f142f48+4qf24f48f102)2(f1012f26f4244q3f26f424f212).

    Extracting the terms involving q2n,q2n+1 from above, we arrive at

    n=0b(12n+4)qn=9(f282f106f241f3f84f412+16qf42f84f106f161f3f41232q2f162f33f412f201f26), (2.10)
    n=0b(12n+10)qn=9(8f162f106f201f3f4124qf282f33f412f241f84f2616q2f42f33f84f412f161f26). (2.11)

    From Eq (2.11), we ultimately arrive at Eq (1.11). To prove Theorem 1.2, consider Eq (2.9) as:

    n=0b(6n+4)qn=9f43f6(f2f21)4.

    Using Eq (2.3) in above, we get

    n=0b(6n+4)qn=9f176f249f283f1218+72qf166f219f273f918+360q2f156f189f263f618+288q3f146f159f253f318+864q3f126f159f193f618+2736q4f136f129f243+4608q5f126f99f318f233+5760q6f116f69f618f223+4608q7f106f39f918f213+2304q8f96f1218f203. (2.12)

    Bringing out the terms involving q3n+1 and q3n+2 from Eq (2.12), we get Eqs (1.12) and (1.13), respectively.

    This segment of the paper contains some more interesting congruence relations for b(n).

    Theorem 3.1. For n0, we have

    b(12n+1){2(1)k(mod6),ifn=3k(3k+1)/2,0(mod6),otherwise. (3.1)

    Theorem 3.2. For n0, we have

    b(2n){(1)k(2k+1)(mod4),ifn=k(k+1),0(mod4),otherwise. (3.2)

    Theorem 3.3. For n0, we have

    b(36n+10)0(mod72), (3.3)
    b(36n+13)0(mod6), (3.4)
    b(36n+25)0(mod12), (3.5)
    b(36n+34)0(mod144), (3.6)
    b(108n+t)0(mod18),for t{49,85}. (3.7)

    Theorem 3.4. For n0, we have

    b(20n+t)0(mod5),for t{8,16} (3.8)
    b(20n+t)0(mod20),for t{6,18} (3.9)
    b(20n+17)0(mod10), (3.10)
    b(28n+t)0(mod14),for t{5,21,25}. (3.11)

    Proof of Theorem 3.1. From Eq (2.6), picking out the terms involving q3n and replacing q3 by q, we have

    n=0b(12n+1)qn=2f32f23f31f6+90qf82f93f161+72qf42f3f46f81+576q2f52f96f31. (3.12)

    Reducing modulo 6, we obtain

    n=0b(12n+1)qn2f3(mod6). (3.13)

    With the help of Euler's pentagonal number theorem,

    n=0b(12n+1)qn2k=(1)kq3k(3k+1)2(mod6),

    which completes the proof of Theorem 3.1.

    Proof of Theorem 3.2. Reducing Eq (1.5) modulo 4, we get

    n=0b(2n)qnf32(mod4). (3.14)

    From Jacobi's triple product identity, we obtain

    n=0b(2n)qnk=0(1)k(2k+1)qk(k+1)(mod4),

    which completes the proof of Theorem 3.2.

    Proof of Theorem 3.3. Consider Eq (1.11), reducing modulo 72

    n=0b(12n+10)qn36qf282f33f412f241f84f26(mod72),
    n=0b(12n+10)qn36qf282f33f412f122f84f12=36qf162f33f312f84(mod72)

    or

    n=0b(12n+10)qn36qf33f312(mod72). (3.15)

    Extracting the terms involving q3n, replacing q3 by q in Eq (3.15), we arrive at Eq (3.3). Similarly, consider Eq (1.13) and reducing modulo 144, we have

    n=0b(18n+16)qn725f152f183f261f66(mod144),72f152f96f132f66=72f22f36(mod144).

    Extracting the terms involving q2n+1, dividing both sides by q and replacing q2 by q, we get Eq (3.6).

    From Eq (3.20), we get

    n=0b(12n+1)qn2f3(mod6).

    Bringing out the terms containing q3n+1, dividing both sides by q and replacing q3 by q, we have b(36n+13)0(mod6). Reducing Eq (3.12) modulo 12, we have

    n=0b(12n+1)qn2f32f23f31f6+90qf82f93f161(mod12),
    n=0b(12n+1)qn2f23f6(f6f3+3qf46f59f83f18+6q2f36f29f218f73+12q3f26f518f63f9)+6qf82f93f82.

    Extracting the terms containing q3n+2, dividing by q2 and replacing q3 by q, we obtain Eq (3.5). Reducing Eq (3.12) modulo 18,

    n=0b(12n+1)qn2f32f23f31f6(mod18),=2f23f6(f6f3+3qf46f59f83f18+6q2f36f29f218f73+12q3f26f518f63f9).

    Extracting the terms involving q3n+1, dividing both sides by q and replacing q3 by q, we have

    n=0b(36n+13)qn6f32f53f61f66f6f53f23f6(mod18)

    or

    n=0b(36n+13)qn6f33(mod18).

    Extracting the terms containing q3n+1,q3n+2 from above to get Eq (3.7).

    Proof of Theorem 3.4. From Eqs (1.5) and (2.4), we have

    n=0b(2n)qn=f52(f144f142f48+4qf24f48f102).

    Bringing out the terms containing even powers of q, we obtain

    n=0b(4n)qn=f142f91f44,

    which can be written as:

    n=0b(4n)qn=f152f101f54.f1f4f2f310f25f20.f1f4f2(mod5).

    Here

    f1f4f2=(q;q)(q4;q4)(q2;q2),=(q;q2)(q2;q2)(q4;q4)(q2;q2),
    f1f4f2=(q,q3,q4;q4)=n=(1)nq2n2n, (3.16)

    where the last equality follows from Jacobi's triple product identity. Using the above identity, we have

    n=0b(4n)qnf310f25f20n=(1)nq2n2n(mod5). (3.17)

    Since 2n2n2,4(mod5), it follows that the coefficients of q5n+2,q5n+4 in n=0b(4n)qn are congruent to 0(mod5), which proves that b(20n+t)0(mod5), for t{8,16}.

    Consider Eq (1.7)

    n=0b(4n+2)qn=4f54f51f22f44f20f5f22f4(mod20).

    Now

    f22f4=(q2;q2)2(q4;q4),=(q2;q2)(q2;q4)(q4;q4)(q4;q4),
    f22f4=(q2,q2,q4;q4)=n=(1)nq2n2.

    Using the above identity, we get

    n=0b(4n+2)qn4f20f5n=(1)nq2n2(mod20). (3.18)

    Since 2n21,4(mod5), it follows that the coefficients of q5n+1,q5n+4 in n=0b(4n+2)qn are congruent to 0(mod20), which proves Eq (3.9). For the proof of next part, consider Eq (1.6) as:

    n=0b(4n+1)qn=2f52f101f31f322f10f25f31f32(mod10),
    n=0b(4n+1)qn2f10f25k=0(1)k(2k+1)qk(k+1)2m=0(1)m(2m+1)qm(m+1)(mod10). (3.19)

    Therefore, to contribute the coefficient of q5n+4, (k,m)(2,2)(mod5) and thus the contribution towards the coefficient of q5n+4 is a multiple of 5.

    Consider Eq (1.6) as:

    n=0b(4n+1)qn=2f72f71f22f14f7f2(mod14).

    With the help of Euler's pentagonal number theorem,

    n=0b(4n+1)qn2f14f7n=(1)nqn(3n+1)(mod14). (3.20)

    As n(3n+1)1,5,6(mod7), it readily proves Eq (3.11).

    In this section, we find some recurrence relations connecting b(n) and restricted partition functions. First we define some notations. Let ¯pl(n) denotes the number of overpartitions of n with l copies. Then

    n=0¯pl(n)qn=(f2f21)l.

    Let pld(n) denotes the number of partitions of n into distinct parts with l copies. Then

    n=0pld(n)qn=(f2f1)l.

    Theorem 4.1. We have

    b(2n)=¯p2(n)3¯p2(n)+5¯p2(n)++(1)k(2k+1)¯p2(nk(k+1))+, (4.1)
    (2n)=p4d(n)p4d(n2)p4d(n4)+p4d(n10)+p4d(n14)++(1)kp4d(nk(3k1))+(1)kp4d(nk(3k+1))+. (4.2)

    Theorem 4.2.

    (4n+1)=2p8d(n)2p8d(n1)2p8d(n2)+2p8d(n5)+2p8d(n7)++(1)k2p8d(nk(3k1)2)+(1)k2p8d(nk(3k+1)2)+, (4.3)
    b(4n+1)=2nc=0b(2c)p3d(nc). (4.4)

    Theorem 4.3.

    (6n+2)=4p10d(n)8p10d(n3)+8p10d(n12)+8p10d(n27)++8(1)kp10d(n3k2)+. (4.5)

    Proof of Theorem 4.1. Consider (1.5) as:

    n=0b(2n)qn=(f2f21)2f32.

    Then

    n=0b(2n)qn=(n=0¯p2(n)qn)(k=0(1)k(2k+1)qk(k+1)),=n=0k=0(1)k(2k+1)¯p2(n)qn+k(k+1),=n=0(k=0(1)k(2k+1)¯p2(nk(k+1)))qn.

    From the last equality, we readily arrive at (4.1). To prove (4.2), consider (1.5) as:

    n=0b(2n)qn=(f2f1)4f2,=(n=0p4d(n)qn)(k=(1)kqk(3k+1)),=(n=0p4d(n)qn)(1+k=1(1)kqk(3k1)+k=1(1)kqk(3k+1)),
    n=0b(2n)qn=n=0p4d(n)qn+n=0(k=1(1)kp4d(n)qk(3k1)+n)+n=0(k=1(1)kp4d(n)qk(3k+1)+n),
    n=0b(2n)qn=n=0p4d(n)qn+n=0(k=1(1)kp4d(nk(3k1))qn)+n=0(k=1(1)kp4d(nk(3k+1))qn),

    which proves Eq (4.2).

    Proof of Theorem 4.2. Consider Eq (1.6) as:

    n=0b(4n+1)qn=2(f2f1)8f1,=2(n=0p8d(n)qn)(k=(1)kqk(3k+1)2),=2(n=0p8d(n)qn)(1+k=1(1)kqk(3k1)/2+k=1(1)kqk(3k+1)/2),
    n=0b(4n+1)qn=n=0p8d(n)qn+n=0k=1(1)kp8d(n)qk(3k1)/2+n+n=0k=1(1)kp8d(n)qk(3k+1)/2+n,
    n=0b(4n+1)qn=n=0p8d(n)qn+n=0(k=1(1)kp8d(nk(3k1)2))qn+n=0(k=1(1)kp8d(nk(3k+1)2))qn,

    which proves Eq (4.3). To prove Eq (4.4), consider Eq (1.6) as:

    n=0b(4n+1)qn=2(f52f41)f32f31,=2(n=0b(2n)qn)(k=0p3d(k)qk),=2n=0(nc=0b(2c)p3d(nc))qn.

    Comparing the coefficients of qn, we arrive at Eq (4.4).

    Proof of Theorem 4.3. Consider Eq (1.8) as:

    n=0b(6n+2)qn=4(f2f1)10f23f6,=4(n=0p10d(n)qn)(k=(1)kq3k2),=4(n=0p10d(n)qn)(1+2k=1(1)kq3k2),=4n=0p10d(n)qn+8n=0(k=1(1)kp10d(n)q3k2+n),=4n=0p10d(n)qn+8n=0(k=1(1)kp10d(n3k2))qn.

    Comparing the coefficients of qn to obtain Eq (4.5).

    In this paper, we have provided the arithmetic properties of second order mock theta function B(q), introduced by McIntosh. Some congruences are proved for the coefficients of B(q) modulo specific numbers. The questions which arise from this work are:

    (i) Are there exist congruences modulo higher primes for B(q)?

    (ii) Is there exist any other technique (like modular forms) that helps to look for some more arithmetic properties of B(q)?

    (iii) How can we explore the other second order mock theta function A(q)?

    The first author is supported by University Grants Commission (UGC), under grant Ref No. 971/(CSIR-UGC NET JUNE 2018) and the the second author is supported by Science and Engineering Research Research Board (SERB-MATRICS) grant MTR/2019/000123. The authors of this paper are thankful to Dr. Rupam Barman, IIT Guwahati, for his valuable insight during establishing Theorems 3.1 and 3.2. We would like to thank the referee for carefully reading our paper and offering corrections and helpful suggestions.

    The authors declare there is no conflicts of interest.



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