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Boundedness of sublinear operators on weighted grand Herz-Morrey spaces

  • In this paper, we introduce weighted grand Herz-Morrey type spaces and prove the boundedness of sublinear operators and their multilinear commutators on these spaces. The results are still new even in the unweighted setting.

    Citation: Wanjing Zhang, Suixin He, Jing Zhang. Boundedness of sublinear operators on weighted grand Herz-Morrey spaces[J]. AIMS Mathematics, 2023, 8(8): 17381-17401. doi: 10.3934/math.2023888

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  • In this paper, we introduce weighted grand Herz-Morrey type spaces and prove the boundedness of sublinear operators and their multilinear commutators on these spaces. The results are still new even in the unweighted setting.



    Graph categories considered in this study are simple, finite, and linked. Allow the graph G to be made up of VG and edge set EG, i.e., G=(VG,EG). For more graph notations, readers should refer to [1].

    If and only if two neighbouring verticies i and j of G, the adjacency matrix A(G)=(aij) is a (0, 1)-matrix, aij=1. The diagonal degree matrix of G is

    DG=diag(d1,d2,,dn),

    where di represents the degree of vertex i in G. The difference between the degree matrix DG and adjacency matrix AG of G gives rise to the Laplace matrix, denoted as LG. The normalized Laplacian [2] is defined as

    L(G)=ID(G)12(D(G)1A(G))D(G)12=D(G)12L(G)D(G)12.

    The (m,n)th-entry of L(G), which is designated

    (L(G))mn={1,m=n;1dmdn,mn,vmisadjacenttovn;0,otherwise. (1.1)

    The distance between both vi and vj, known as dij=dG(vi,vj), represents the length of the smallest path in question. Wiener and Dobrynin [3,4] introduced the Wiener index for the first time in 1947. In addition, the Wiener index is denoted as

    W(G)=i<jdij.

    For further information on the Wiener index, please refer to [5,6,7,8,9].

    The Gutman index of a simple graph G is introduced [10] and denoted as

    Gut(G)=i<jdidjdij,

    taking into account the degree di of vertex vi.

    The Kirchhoff index[11,12] characterizes graph G by summing the resistance distances between every pair of vertices, similar to the Wiener index, namely

    Kf(G)=i<jrij.

    The multiplicative degree-Kirchhoff index, initially introduced by Chen and Zhang[13] in 2007. It is an extension of the traditional Kirchhoff index, which is expressed as

    Kf(G)=i<jdidjrij.

    The techniques of the Kirchhoff index and multiplication degree-Kirchhoff index can be found in [14,15,16,17,18]. The multiplication degree-Kirchhoff index has garnered significant attention due to its remarkable contributions in academia and practical applications in computer network science, epidemiology, social economics, and other fields. Further research results on the Kirchhoff index and multiplication degree-Kirchhoff index can be explored through [19,20,21,22,23].

    The spanning tree of a graph G, also known as complexity, denoted as τ(G), refers to the number of subgraphs that encompass all vertices in G. This measure serves as a crucial indicator for network stability and plays a significant role in assessing the structural characteristics of graphs. For further insights into related topics such as the count of spanning trees, interested readers are encouraged to consult [24,25,26].

    With the rapid advancement of scientific research and the successful application of topology in practical scenarios, topological theory has gained increasing recognition worldwide. The calculation problem concerning the phenylene Wiener index has been effectively resolved by Pavlović and Gutman [27]. Chen and Zhang [28] have developed a precise equation for predicting the Wiener index of random phenylene chains. Additionally, Liu et al. [29] have identified both the degree-Kirchhoff index and the number of spanning trees for Ln dicyclobutadieno derivatives of [n] phenylenes.

    Given two automorphic graphs S and K, we define the symbol SK to represent the strong product of these two graphs with V(S)×V(K), which is commonly referred to as the strong product in graph theory literature. Readers can refer to [30] for more comprehensive definitions and concepts. Recently, Pan et al.[25] utilized the resistance distance of a strong prism formed by Pn and Cn to determine the Kirchhoff index. Similarly, Li et al.[31] derived graph invariants and spanning trees from the strong prism of a star Sn. Motivated by [30,31,32,33], we obtain the pentagonal network Rn and its strong product P2n. The pentagonal network consists of numerous adjacent pentagons and quadrilaterals with each quadrangle having a maximum of two non-adjacent pentagons, as shown in Figure 1. The P2n is the strong product of Rn, as depicted in Figure 2. It obviously that

    |V(P2n)|=14nand|E(P2n)|=47n8.
    Figure 1.  Linear pentagonal network Rn.
    Figure 2.  The strong product P2n of the linear pentagonal.

    In this paper, we focus on the strong product of pentagonal networks, specifically examining the graph P2n with n1. The subsequent sections are organized as follows: Section 2 provides a comprehensive review of relevant research materials, presenting illustrations, concepts and lemmas. In Section 3, we derive the normalized Laplacian spectrum and present an explicit closed formula for the multiplicative degree-Kirchhoff index. Additionally, we calculate the complexity of P2n. In Section 4, we conclude the paper.

    In this section, let Rn represent the penagonal-quadrilateral networks, as illustrated in Figure 1. P2n is composed of Rn and its copy Rn, positioned one in front and one behind, as shown in Figure 2. Moreover,

    ΦA(x)=det(xInA)

    represents the characteristic polynomial of matrix A.

    The fact that

    π=(1o,2o,,no)(1,1)(2,2)((3n),(3n))

    is an automorphism deserves attention. Let

    V1={1o,2o,,no,u1,u2,,u3n,v1,,v3n},
    V2={1o,2o,,no,u1,u2,,u3n,v1,,v3n},
    |V(P2n)|=14nand|E(P2n)|=47n8.

    Subsequently, the normalized Laplacian matrix can be represented as a block matrix, that is

    L(P2n)=(LV1V1LV1V2LV2V1LV2V2),

    in which

    LV1V1=LV2V2,LV1V2=LV2V1.

    Let

    W=(12I6n12I6n12I6n12I6n),

    then,

    WL(P2n)W=(LA00LS),

    where

    LA=LV1V1+LV1V2andLS=LV1V1LV1V2.

    Observe that W and W are transpose matrices of each other.

    The characteristic polynomial of the matrix R, is denoted as

    Φ(R):=det(xIR).

    The decomposition theorem process for P2n is obtained in a similar manner to Pan and Li[34], thus we omit this proof and present it as follows:

    Lemma 2.1. [35] Assuming that the determination of LA and LS has been previously described, then

    ΦL(Ln)(x)=ΦLA(x)ΦLS(x).

    Lemma 2.2. [13] The graph G is an undirected connected graph with n vertices and m edges. Then

    Kf(G)=2mnk=21λk.

    Lemma 2.3. [2] The number of spanning trees in G, referred to as the graph's complexity, can be considered a fundamental measure in graph theory. Then

    τ(G)=12mni=1dinj=2λj.

    In this section, we explore the methodology for deriving an explicit analytical expression of the multiplicative Kirchhoff index by traversing the normalized Laplacian matrix. Meanwhile, we determine the computational complexity of P2n. Subsequently, employing the normalized Laplacian, we derive matrices of order 6n as

    LV1V1=(1135000150000135117000170000171000017000001135000170000135100001515000011350000170001351170000170001710000017000011350000150001351)

    and

    LV1V2=(15135000150000135171700017000017170000170000017135000170000135150000151500001513500001700013517170000170001717000001700001713500001500013515).

    Due to

    LA=LV1V1(P2n)+LV1V2(P2n)

    and

    LS=LV1V1(P2n)LV1V2(P2n),

    it can be convincingly argued that

    LA=2(2513500015000013537170001700001737000000000037135000170000135250000151500002513500001700013537170000170001737000001700003713500001500013525)

    and

    LS=diag(65,87,87,,87,65,65,87,87,,87,65).

    Utilizing Lemma 2.1, it is revealed that the P2n normalized Laplacian spectrum consists of the eigenvalues from LA and LS. It is established that the LS possesses eigenvalues 65 and 87 with multiplicities of 4 and (6n4), respectively.

    Let

    M=(2513500000135371700000173717000001737000000037170000017371350000013525)(3n)×(3n)

    and

    N=diag(15,17,17,17,17,,17,17,15),

    where the matrices M and N are both of order 3n.

    The matrices M and N are combined to form a block matrix, denoted as 12LA, in the following manner:

    12LA=(MNNM).

    Suppose that

    W=(12I3n12I3n12I3n12I3n)

    is a block matrix. Hence, we can obtain

    W(12LA)W=(M+N00MN).

    Let J=M+N and K=MN. Then,

    J=(1513500000135271700000172717000001727000000027170000017471350000013535)(3n)×(3n)

    and

    K=(3513500000135471700000174717000001747000000047170000017471350000013535)(3n)×(3n),

    in which the diagonal elements are

    (35,47,47,47,47,47,35).

    Based on Lemma 2.1, it is evident and demonstrable that the eigenvalues of 12LA are identical to those of J and K. Assume that the eigenvalues of J and K are σi and ςj(i,j=1,2,,3n), respectively, with

    σ1σ2σ3σ3n,ς1ς2ς3ς3n.

    We verify σ10 and ς10. In addition, it is easy to know that the normalized Laplacian spectrum of P2n is {2σ1,2σ2,,2σ3n,2ς1,2ς2,,2ς3n}. Note that

    |E(P2n)|=47n8,

    we can obtain Lemma 3.1 according to Lemma 2.2.

    Lemma 3.1. Assume that P2n is the strong product of the pentagonal network. Then,

    Kf(P2n)=2(47n8)(2×56+(6n2)78+123ni=21σi+123nj=11ςj)=(47n8)(63n16+3ni=21σi+3nj=11ςj).

    Subsequently, we partition the computation of the aforementioned equation into two distinct components and prioritize the initial calculation of 3ni=21σi.

    Lemma 3.2. Suppose that σi(i=1,2,,3n) is defined as described previously.

    3ni=21σi=1035n3+142n2+617n2(81n+490).

    Proof. Suppose that

    Φ(J)=x3n+a1x3n1++a3nx2+a3n+1x=x(x3n1+a1x3n2++a3n1x+a3n2).

    Then σ2,σ3,,σ3n fulfil the following equation

    x3n1+a1x3n2++a3n2x+a3n1=0,

    and we observe that 1σ2,1σ3,,1σ3n are the roots of the following equation

    a3n1x3n1+a3n2x3n2++a1x+1=0.

    By Vieta's Theorem, one has

    3ni=21σi=(1)3n2a3n2(1)3n1a3n1. (3.1)

    For each value of i from 1 to 3n+1, we consider Ji and assign ji as the determinant of Ji. We will derive the formula for ji, which can be utilized to calculate (1)3n2a3n2 and (1)3n1a3n1. Then one has

    j1=15,j2=135,j3=1245,j4=11715,j5=112005,j6=184035,j7=1588245,j8=14117715,

    and

    {j3i=27j3i1149j3i2,1in;j3i+1=27j3i149j3i1,0in1;j3i+2=27j3i+1149j3i,0in1.

    Through a straightforward computation, one can derive the following general formulas:

    {j3i=75(1343)i,1in;j3i+1=15(1343)i,0in1;j3i+2=135(1343)i,0in1. (3.2)

    According to Eq (3.1), we divide the numerator and denominator into two facts and reveal them later. For the sake of convenience, we represent the diagonal elements of J as lii in a simplified manner.

    Fact 3.3.

    (1)3n1a3n1=490+81n25(1343)n.

    Proof. Given that J is a left-right symmetric matrix, the sum of all principal minors can be represented by the number (1)3n1a3n1, where these minors correspond to rows and columns with indices equal to (3n1).

    (1)3n1a3n1=3ni=1detLA[i]=3ni=1det(Ji100J3ni)=3ni=1ji1j3ni, (3.3)

    where

    J3ni=(li+1,i+1000l3n1,3n11350135l3n,3n).

    By Eqs (3.2) and (3.3), we have

    (1)3n1a3n1=2j3n+1+nl=1j3(l1)+2j3(nl)+2+nl=1j3lj3(nl)+1+n1l=0j3l+1j3(nl)=985(1343)n+343n352(1343)n+7n25(1343)n+n(1343)n=490+81n25(1343)n.

    This is the completion of the proof.

    Fact 3.4.

    (1)3n2a3n2=1035n3+142n2617n50(1343)n.

    Proof. We note that the sum of all principal minors of order 3n2 in J can be expressed as (1)3n2a3n2. After that,

    (1)3n2a3n2=3n1i<j|Ji1000Z000J3nj|,1i<j3n2,

    where

    Z=(li+1,i+100lj1,j1)

    and

    J3nj=(lj+1,j+1000l3n1,3n11350135l3n,3n).

    Note that

    (1)3n2a3n2=3n1i<jdetJi1detZdetJ3nj=3n1i<jdetZsi1s3nj. (3.4)

    According to Eq (3.4), the determinant of Z varies depending on the values of i and j, as well as s and t. Consequently, we can categorize the primary scenarios into six distinct classifications.

    Case 1. i=3s, j=3t for 1s<tn, and

    detZ1=|27170000172717000017271700001727000000271700001727|(3s3t1)=21(st)(1343)st.

    Case 2. i=3s, j=3t+1 for 1stn, and

    detZ2=|27170000172717000017271700001727000000271700001727|(3s3t)=[3(st)+1](1343)st,

    or i=3s+2, j=3t for 0s<tn, and

    detZ3=|27170000172717000017271700001727000000271700001727|(3s3t3)=(3s3t2)(1245)st1.

    Case 3. In the same way, i=3s, j=3t+2 for 1stn, and

    detZ4=|2717000017271700001727170000172700000027135000013515|(3s3t+1)=49(3s3t+2)(1343)st+1,

    or i=3s+1, j=3t for 0s<tn, and

    detZ5=|27170000172717000017271700001727000000271700001727|(3s3t2)=49(3s3t1)(1343)st1.

    Case 4. Similarly, i=3s+1, j=3t+1 for 0s<tn, and

    detZ6=|27170000172717000017271700001727000000271700001727|(3s3t1)=21(st)(1343)st,

    or i=3s+2, j=3t+2 for 0k<ln, and

    detZ7=|2717000017271700001727170000172700000027135000013515|(3s3t1)=detZ6.

    Case 5. i=3s+1, j=3t+2 for 0stn, and

    detZ8=|2717000017271700001727170000172700000025135000013527|(3s3t)=(3s3t+1)(1343)st.

    Case 6. i=3s+2, j=3t+1 for 0k<ln, and

    detZ9=|27170000172717000017271700001727000000271700001727|(3s3t2)=49(3s3t1)(1343)st.

    Therefore, we can obtain

    (1)3n2a3n2=1i<j3ndetZjk1j3nl=1+2+3, (3.5)

    where

    1=1s<tndetZ1j3s1j3n3t+1stndetZ2j3s1j3n3t1+1stn1detZ4j3s1j3n3t+1sndetJ[3s,3n+2]j3s1=7n(n21)50(1343)n+(n2+2n)(n+1)2450(1343)n+n2(n1)2450(1343)n+n(3n+1)490(1343)n=345n3+17n2336n50(1343)n,
    2=1s<tndetZ5j3sj3n3t+2+1s<tndetZ6j3sj3n3t+1+1stn1detZ8j3sj3n3t+1sndetJ[3s+1,3n+2]j3s+1sndetJ[1,3t]j3n3t+2+1tndetS[1,3t+1]s3n3t+1+0tn1detJ[1,3t+2]j3n3t+detJ[1,3n+2]=4950(n3n22n+2)(1343)n1+49n(n21)50(1343)n+49n(n21)50(1343)n+7n(3n1)10(1343)n+7n(3n+1)10(1343)n1+21n(n+1)10(1343)n+7n(3n1)10(1343)n+n(3n+1)(1343)n=345n3+73n292n50(1343)n

    and

    3=0s<tndetZ3j3s+1j3n3t+2+0s<tn1detZ7j3s+1j3n3t+0s<tndetZ9j3s+1j3n3t+1+0sn1detJ[3s+2,3n+2]j3s+1=49n(n2+n4)50(1343)n1+n(n21)50(1343)n+49n(n2+2n1)50(1343)n+21n(n+1)10(1343)n1=345n3+52n2189n50(1343)n.

    By substituting 1, 2, and 3 into Eq (3.5), the desired outcome can be deduced.

    (1)3n2a3n2=1+2+3=1035n3+142n2617n50(1343)n.

    This is the completion of the proof. Let

    0=ς1<ς2ς3ς3n

    represent the eigenvalues of J. By Facts 3.3 and 3.4, we can further investigate Lemma 3.2. According to Eq (3.1), it is evident that

    3ni=21σi=(1)3n2a3n2(1)3n1a3n1=1035n3+142n2+617n2(490+81n).

    Considering Lemma 3.1, we will focus on the calculations of 3nj=11ςj. Let

    δ(n)=10290n(11+230)+3600+124303060000

    and

    ξ(n)=10290n(11230)+3600124303060000.

    Lemma 3.5. The variable ςj (where j ranges from 1 to 3n+2) is assumed to be defined as previously described. One has

    3nj=11ςj=(1)3n1b3n1detK,

    where

    detK=45+1130125(11+430343)n+451130125(11430343)n

    and

    (1)3n1b3n1=δ(n)(11+430343)nξ(n)(11430343)n.

    Proof. The representation of Φ(K) can be expressed as

    y3n+b1y3n1++b3n2y2+b3n1y=y(y3n+1+b1y3n++b3n2y+b3n1),

    where ς1,ς2,,ς3n represent the roots of the equation.

    y3n1+b1y3n2++b3n2y+b3n1=0,

    and the equation is determined to possess 1ς1,1ς2,,1ς3n as its solutions

    b3n1y3n1+b3n2y3n2++b1y+1=0.

    By Vieta's Theorem, one holds that

    3nj=11ςj=(1)3n1b3n1detK. (3.6)

    To simplify the analysis, let Rp denote the p-th order principal minors of matrix K and kp represent the determinant of Rp. We will derive an equation for kp that can be utilized to calculate (1)3n1b3n1 and detK for values of p ranging from 1 to 3n1. Subsequently, we arrive at

    k1=15,k2=135,k3=1245,k4=11715,k5=112005,k6=184035,k7=1588245,k8=14117715

    and

    {k3p=27k3p1149k3p2,1pn;k3p+1=27k3p149k3p1,1pn;k3p+2=27k3p+1149k3p,0pn1.

    After a straightforward computation, the following general formulas can be derived

    {k3p=105+1430150(11+430343)p+1051430150(11430343)p,1pn;k3p+1=45+830150(11+430343)p+45830150(11430343)p,1pn;k3p+2=11+23070(11+430343)p+1123070(11430343)p,0pn1.

    Subsequently, we proceed by examining the following Facts:

    Fact 3.6.

    detK=45+1130125(11+430343)n+451130125(11430343)n.

    Proof. By expanding detK with respect to its last row, we obtain

    detK=35k3n+1135k3n=45+1130125(11+430343)n+451130125(11430343)n.

    The desired outcome has been achieved.

    Fact 3.7.

    (1)3n1b3n1=δ(n)(11+430343)nξ(n)(11430343)n,

    where

    δ(n)=10290n(11+430)+3600+124303060000

    and

    ξ(n)=10290n(11430)+3600124303060000.

    Proof. Considering that K has a (3n1)-row and (3n1)-column structure, the sum of all principal minors can be expressed as (1)3n1b3n1. Here, lii represents the diagonal entries of K. It is noteworthy that K exhibits bilateral symmetry, which enables us to derive specific information.

    (1)3n1b3n1=3ni=1detK[i]=3ni=1det(Ri100R3ni)=3ni=1ri1r3ni, (3.7)

    where

    R3ni=(gi+1,i+1000g3n1,3n11350135g3n,3n).

    In line with Eq (3.7), we have

    (1)3n1b3n1=2k3n1+nl=1detK[3l]+n1l=0detK[3l+1]+n1l=0detK[3l+2]=2k3n1+nl=1k3(l1)+2k3(nl)+2+n1l=0k3lk3(nl)+1+n1l=0k3l+1k3(nl). (3.8)

    Immediately, we can get

    nl=1k3(l1)+2k3(nl)+2=245n20(11+430343)n+1(11430343)n+1+30600(11+430343)n30600(11430343)n, (3.9)
    nl=0k3lk3(nl)+1=2391n300(11+430343)n+1(11430343)n+1+1373060000(11+430343)n1373060000(11430343)n, (3.10)
    n1l=0k3l+1k3(nl)=2391n300(11+430343)n+1(11430343)n+1+12713060000(11+430343)n12713060000(11430343)n (3.11)

    and

    2k3n1=90+163075(11+430343)n+90163075(11430343)n. (3.12)

    By incorporating Eqs (3.9)–(3.12) into Eq (3.8), we can achieve Lemma 3.7. Utilizing Eq (3.6), in conjunction with Facts 3.6 and 3.7, Lemma 3.5 can be promptly derived.

    By integrating Lemmas 3.1, 3.2 and 3.5, we can readily deduce the Theorems 3.8 and 3.9.

    Theorem 3.8. Assume that P2n is the strong product of the pentagonal network. One has

    Kf(P2n)=1029n3+5736n2+4275n+8503+(57n+30)(1)3n1b3n1detK,

    where

    (1)3n1b3n1=δ(n)(11+430343)nξ(n)(11430343)n,detK=45+1130125(11+430343)n+451130125(11430343)n,

    additionally,

    δ(n)=10290n(11+230)+3600+124303060000

    and

    ξ(n)=10290n(11230)+3600124303060000.

    Theorem 3.9. Let P2n be the strong product of pentagonal network. Then

    τ(P2n)=35232n+75((45+1130)(11+430)n+(451130)(11430)n).

    Proof. Based on the proof of Lemma 2.2, it is evident that σ1,σ2,σ2n+1 constitute the roots of the equation

    x2n+a1x2n1++a2n1x+a2n=0.

    Accordingly, one has

    3ni=2σi=(1)3n1a3n1.

    By Fact 3.3, we have

    3ni=2σi=12+23n25(1343)n.

    By the same method,

    3nj=1ςj=detK=45+1130375(11+430343)n+451130375((11430)343)n.

    Note that

    vVP2nd(P2n)=54716n4

    and

    |E(P2n)|=48n7.

    In conjunction with Lemma 2.3, we get

    τ(P2n)=12|E(P2n)|((65)2(87)6n23ni=22σi3nj=12ςjvVP2nd(P2n))=35232n+75((45+1130)(11+430)n+(451130)(11430)n).

    This is the completion of the proof.

    In this study, we have derived explicit expressions for the multiplicative degree-Kirchhoff index and complexity of P2n based on the spectrum of the Laplacian matrix, where P2n=RnRn. These two fundamental calculations serve as simple yet reliable graph invariants that effectively capture the stability of diverse networks. Future research should focus on applying our methodology to determine spectra for strong products of automorphic and symmetric networks.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The authors would like to express their sincere gratitude to the editor and anonymous referees for providing valuable suggestions, which significantly contributed to the substantial improvement of the original manuscript. This work was supported by Anhui Jianzhu University Funding under Grant KJ212005, and by Natural Science Fund of Education Department of Anhui Province under Grant KJ2020A0478.

    No potential confilcts of interest were reported by the authors.



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