Research article

A variant of Jensen-type inequality and related results for harmonic convex functions

  • Received: 21 June 2020 Accepted: 03 August 2020 Published: 13 August 2020
  • MSC : 26D15, 26A51, 26D10, 26A15

  • In this article, we present a variant of discrete Jensen-type inequality for harmonic convex functions and establish a Jensen-type inequality for harmonic h-convex functions. Furthermore, we found a variant of Jensen-type inequality for harmonic h-convex functions.

    Citation: Imran Abbas Baloch, Aqeel Ahmad Mughal, Yu-Ming Chu, Absar Ul Haq, Manuel De La Sen. A variant of Jensen-type inequality and related results for harmonic convex functions[J]. AIMS Mathematics, 2020, 5(6): 6404-6418. doi: 10.3934/math.2020412

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  • In this article, we present a variant of discrete Jensen-type inequality for harmonic convex functions and establish a Jensen-type inequality for harmonic h-convex functions. Furthermore, we found a variant of Jensen-type inequality for harmonic h-convex functions.


    Let V be an -dimensional vector space over a field K of characteristic 0. An arrangement of hyperplanes A is a finite collection of codimension one affine subspaces in V. An arrangement A is called central if every hyperplane HA goes through the origin.

    Let V be the dual space of V, and S=S(V) be the symmetric algebra over V. A K-linear map θ:SS is called a derivation if for f,gS,

    θ(fg)=fθ(g)+gθ(f).

    Let DerK(S) be the S-module of derivations. When A is central, for each HA, choose αHV with ker(αH)=H. Define an S-submodule of DerK(S), called the module of A-derivations by

    D(A):={θDerK(S)|θ(αH)αHSforallHA}.

    The arrangement A is called free if D(A) is a free S-module. Then, D(A) has a basis comprising of homogeneous elements. For an affine arrangement A in V, cA denotes the cone over A [7], which is a central arrangement in an (+1)-dimensional vector space by adding the new coordinate z.

    Let E=R be an -dimensional Euclidean space with a coordinate system x1,,x, and let Φ be a crystallographic irreducible root system in the dual space E. Let Φ+ be a positive system of Φ. For αΦ+ and kZ, define an affine hyperplane Hα,k by

    Hα,k:={vE(α,v)=k}.

    The Shi arrangement Shi() was introduced by Shi in the study of the Kazhdan-Lusztig representation theory of the affine Weyl groups in [9] by

    Shi():={Hα,kαΦ+,0k1},

    when the root system is of type A1.

    For mZ0, the extended Shi arrangement Shik of the type Φ is an affine arrangement defined by

    Shik:={Hα,kαΦ+,m+1km}.

    There are a lot of researches on the freeness of the cones over the extended Shi arrangements [1,3,4]. The first breakthrough was the proof of the freeness of multi-Coxeter arrangements with constant multiplicities by Terao in [13]. Combining it with algebro-geometric method, Yoshinaga proved the freeness of the extended Shi arrangements in [15]. Nevertheless, there has been limited progress in constructing their bases, and a universal method for determining these bases remains elusive. For types A1, B, C, and D, explicit bases for the cones over the Shi arrangements were constructed in [6,10,11]. Notably, a basis for the extended Shi arrangements of type A2 was established in [2]. Recently, Suyama and Yoshinaga constructed explicit bases for the extended Shi arrangements of type A1 using discrete integrals in [12]. Feigin et al. presented integral expressions for specific bases of certain multiarrangements in [5]. In these studies, Suyama and Terao first constructed a basis for the derivation module of the cone over the Shi arrangement, as detailed in [11], with Bernoulli polynomials playing a central role in their approach. The following definitions are pertinent to this result.

    For (k1,k2)(Z0)2, the homogenization polynomial of degree k1+k2+1 is defined by

    ¯Bk1,k2(x,z):=zk1+k2+1k1i=01k2+i+1(k1i){Bk2+i+1(xz)Bk2+i+1},

    where Bk(x) denotes the k-th Bernoulli polynomial and Bk(0)=Bk denotes the k-th Bernoulli number. Using this polynomial, the basis for D(cShi()) was constructed as follows.

    Theorem 1.1. [11, Theorem 3.5] The arrangement cShi() is free with the exponents (0,1,1). The homogeneous derivations

    η1:=1+2++,η2:=x11+x22++x+zz,ψ()j:=(xjxj+1z)i=10k1j10k2j1(1)k1+k2Ijk11[1,j1]Ijk21[j+2,]¯Bk1,k2(xi,z)i,

    form a basis for D(cShi()), where 1j1 and i(1i), z denote xi, z respectively. Ik[u,v] represents the elementary symmetric function in the variables {xu,xu+1,,xv} of degree k for 1uv.

    The above conclusion was reached by using Saito's criterion, which is a crucial theorem for determining the basis of a free arrangement.

    Theorem 1.2. [8, Saito's criterion] Let A be a central arrangement, and Q(A) be the defining polynomial of A. Given θ1,,θD(A), the following two conditions are equivalent:

    (1)detM(θ1,,θ)Q(A),

    (2)θ1,,θ form a basis for D(A) over S,

    where M(θ1,,θ)=(θj(xi))× denotes the coefficient matrix, and AB means that A=cB, cK{0}.

    This theorem provides a useful tool for determining when a set of derivations forms a basis for the module of derivations associated with a central arrangement.

    Let α=(1,,1)T and β=(x1,,x)T be the ×1 column vectors, and define ψ()i,j:=ψ()j(xi) for 1i, 1j1. Suyuma and Terao in [11] proved the equality

    detM(η1,η2,ψ()1,,ψ()1)=det(αβ(ψ()i,j)×(1)0z01×(1))(+1)×(+1)z1m<n(xmxn)(xmxnz),

    which yields

    det(α(ψ()i,j)×(1))1m<n(xmxn)(xmxnz). (1.1)

    A graph G=(V,E) is defined as an ordered pair, where the set V={1,2,,} represents the vertex set, and E is a collection of two-element subsets of V. If {i,j}E for some i,jV, then {i,j} is referred to as an edge. Writing {i,j}G implies {i,j}E. Let UV, and define E(U)={{i,j}i,jU,{i,j}E}. We say U induces a subgraph GU=(U,E(U)). Specifically, we use the symbol KU for the induced subgraph of the complete graph K. For i<j, the interval notation [i,j] represents {i,i+1,,j}.

    For a graph G on the vertex set {1,2,,}, the arrangement Shi(G) was defined in [14] by

    Shi(G):={{xmxn=0}|{m,n}G}{{xmxn=1}|1m<n}.

    Then, Shi(G) is an arrangement between the Linial arrangement

    {{xmxn=1}1m<n},

    and the Shi arrangement Shi(). Write A(G):=cShi(G). It was classified to be free according to the following theorem.

    Theorem 1.3. [14, Theorem 3] The arrangement A(G) is free if and only if G consists of all edges of three complete induced subgraphs G[1,s],G[t,],G[2,1], where 1s, ts+1. The free arrangement A(G) has exponents (0,1,(1)+ts2,st+1) for s< and t>1, and (0,1,1) for s= or t=1.

    For s,tZ+, we may define the arrangement

    A[s,t]:=A(K[2,1]){{x1xn=0}2ns}{{xmx=0}1tm1}.

    By Theorem 1.3, for 1s and ts+1, the arrangement A[s,t] is free with exponents (0,1,(1)+ts2,st+1) for s< and t>1, and (0,1,1) for s= or t=1.

    For 0q2, we write A[q]:=A[1,q], then A[q] is free with exponents (0,1,(1)q1,q).

    In this section, based on the conclusions of Suyama and Terao, we provide an explicit construction of the basis for D(A[q]), 0q2. First, we shall establish a basis for D(A[0]), which is the ingredient of the basis for D(A[q]).

    Theorem 2.1. For 1j2, define homogeneous derivations

    φ(0)j:=(xjxj+1z)i=10k1j10k2j2(1)k1+k2Ijk11[1,j1]Ijk22[j+2,1]¯Bk1,k2(yi,z)i,φ(0)1:=1s=1(xsxz)D(A[0]),

    where

    yi={xi,1i1,x+z,i=.

    Then, the derivations η1,η2,φ(0)1,,φ(0)1 form a basis for D(A[0]).

    Proof. Write φ(0)i,j:=φ(0)j(xi),1i,1j1, and from the definitions of φ(0)j and ψ()j, we can get

    φ(0)i,j=ψ(1)i,j, (2.1)

    for 1i1, 1j2. Consequently, for 1m<n1, it follows that φ(0)j(xmxn) is divisible by xmxn, and φ(0)j(xmxnz) is divisible by xmxnz. For 1m1, let the congruence notation (m,k) in the subsequent calculation denote modulo the ideal (xmxkz). We derive

    φ(0)j(xmxz)=(xjxj+1z)0k1j10k2j2(1)k1+k2Ijk11[1,j1]Ijk22[j+2,1][¯Bk1,k2(xm,z)¯Bk1,k2(x+z,z)](m,1)0,

    which implies that φ(0)j(xmxz) is divisible by xmxz. Thus, φ(0)jD(A[0]) for 1j2. Therefore, we have η1,η2,φ(0)1,,φ(0)1D(A[0]). Additionally, we obtain

    detM(η1,η2,φ(0)1,,φ(0)1)=(1)+1zdet(1φ(0)1,1φ(0)1,201φ(0)1,1φ(0)1,201φ(0),1φ(0),21s=1(xsxz))×=(1)+1z1s=1(xsxz)det(1φ(0)1,1φ(0)1,21φ(0)1,1φ(0)1,2)(1)×(1)=(1)+1z1s=1(xsxz)det(α1(φ(0)i,j)(1i1,1j2))(1)×(1).

    According to the equalities (1.1) and (2.1), we have

    det(α1(φ(0)i,j)(1i1,1j2))1m<n1(xmxn)(xmxnz).

    Hence, we obtain

    detM(η1,η2,φ(0)1,,φ(0)1).=z1s=1(xsxz)1m<n1(xmxn)(xmxnz)=z1m<n1(xmxn)1m<n(xmxnz)=Q(A[0]).

    By applying Theorem 1.2, we conclude that the derivations η1,η2,φ(0)1,,φ(0)1 form a basis for D(A[0]).

    Definition 2.1. For 1q2, 1j1, define the homogeneous derivations

    φ(q)j:={φ(0)j,1jq2,(xj+1x)φ(0)j(xjxj+1z)j1a=q1φ(0)a,q1j3,φ(0)j+1+ja=q1(a1)φ(0)a,j=2,(xjx)φ(0)j+(xjxj+1z)j2a=q1(a2)φ(0)a,j=1.

    To prove the derivations η1,η2,φ(q)1,,φ(q)1 form a basis for D(A[q]), first we prove all such derivations belong to the module D(A[q]).

    Theorem 2.2. For 1m1, 1j2, we have

    φ(0)j(xmx)(m,0)(z)(xjxj+1z)j1s=1(xsxmz)1s=j+2(xsxm). (2.2)

    Proof. We have the following congruence relation of polynomials modulo the ideal (xmx).

    φ(0)j(xmx)=(xjxj+1z)0k1j10k2j2(1)k1+k2Ijk11[1,j1]Ijk22[j+2,1][¯Bk1,k2(xm,z)¯Bk1,k2(x+z,z)](m,0)(xjxj+1z)0k1j10k2j2(1)k1+k2+1Ijk11[1,j1]Ijk22[j+2,1][¯Bk1,k2(xm+z,z)¯Bk1,k2(xm,z)]=(xjxj+1z)0k1j10k2j2(1)k1+k2+1Ijk11[1,j1]Ijk22[j+2,1]zk1+k2+1(xm+zz)k1(xmz)k2=(z)(xjxj+1z)j1k1=0Ijk11[1,j1][(xm+z)]k1j2k2=0Ijk22[j+2,1](xm)k2=(z)(xjxj+1z)j1s=1(xsxmz)1s=j+2(xsxm).

    We complete the proof.

    Remark 2.1 In equality (2.2), we observe that 1s=j+2(xsxm)=0 for j+2m1. This implies that φ(0)j(xmx) is divisible by xmx for 1j3 and j+2m1.

    According to Remark 2.1, for 1jq2, we have φ(0)j(xmx) is divisible by xmx for qm1, which implies that φ(q)j=φ(0)jD(A[q]). Therefore, to prove the derivations belong to the module D(A[q]), it suffices to prove φ(q)jD(A[q]) for q1j1.

    For the sake of convenience in the proof, let us introduce the notations for f,g,hZ+,

    A[g,h]f:=hs=g(xsxf),B[g,h]f:=hs=g(xsxfz).

    If g>h, we agree that A[g,h]f=B[g,h]f=1.

    Lemma 2.1. For any u,v,wZ+ that satisfy 4j+1u2, 3v2, and 3w2, we have the following three equalities:

    B[u,j1]u1=A[u+1,j]u1+j1a=u(xaxa+1z)A[a+2,j]u1B[u,a1]u1. (2.3)
    B[v,1]v1=(v+1)A[v,1]v1+2a=v1(a1)(xaxa+1z)A[a+2,1]v1B[v1,a1]v1. (2.4)
    B[w,2]w1=wA[w,2]w1+3a=w1(a2)(xaxa+1z)A[a+2,2]w1B[w1,a1]w1. (2.5)

    Proof. We will only prove equality (2.5) by induction on w. The proofs of equalities (2.3) and (2.4) are similar. For w=3,

    3A[3,2]4+3a=4(a2)(xaxa+1z)A[a+2,2]4B[4,a1]4=3(x3x4)(x2x4)+2(x4x3z)(x2x4)+(x3x2z)(z)=(x3x4z)(x2x4z)=B[3,2]4,

    and the equality holds. Assume that for w=k3, the equality holds. Then, we replace xk1 with xk2, and multiply both sides of the equality by (xk1xk2z) to get

    B[k1,2]k2=(k1)(xk2xkz)(xk1xk2z)A[k+1,2]k2+k(xk1xk2z)A[k,2]k2+3a=k(a2)(xaxa+1z)A[a+2,2]k2B[k2,a1]k2=(k+1)A[k1,2]k2+3a=k2(a2)(xaxa+1z)A[a+2,2]k2B[k2,a1]k2.

    We have completed the induction.

    Lemma 2.2. The derivation φ(q)j belongs to the module D(A[q]) for 2q2 and q1j3.

    Proof. For 2q2 and j=q1, it is evident from Remark 2.1 that

    φ(q)q1=(xqx)φ(0)q1D(A[q]).

    For 3q2 and qj3, we will establish this by induction on q. From Theorem 2.2, for qm1, we have

    φ(q)j(xmx)=(xj+1x)φ(0)j(xmx)(xjxj+1z)j1a=q1φ(0)a(xmx)(m,0)(z)(xjxj+1z)A[j+1,1]mB[1,q2]m[B[q1,j1]mj1a=q1(xaxa+1z)A[a+2,j]mB[q1,a1]m].

    (1) For q=3, we get

    φ(3)3(xmx)(m,0)(z)(x3x2z)A[2,1]mB[1,5]m(x3xm).

    If m=3,2,1, then we have φ(3)3(xmx)(m,0)0, which indicates that φ(3)3(xmx) is divisible by xmx for m=3,2,1. Therefore, φ(3)jD(A[3]).

    (2) For q=k3, assume that φ(k)jD(A[k]), which implies that φ(k)j(xmx) is divisible by xmx for km1.

    For q=k+1, we observe that

    φ(k+1)j=φ(k)j(xjxj+1z)φ(0)k2.

    According to the induction hypothesis and Remark 2.1, it is sufficient to prove that φ(k+1)j(xk1x) is divisible by xk1x. By using the equality (2.3), we obtain

    φ(k+1)j(xk1x)(k1,0)(z)(xjxj+1z)A[j+1,1]k1B[1,k3]k1[B[k2,j1]k1j1a=k2(xaxa+1z)A[a+2,j]k1B[k2,a1]k1]=(z)2(xjxj+1z)A[j+1,1]k1B[1,k2]k1[B[k,j1]k1A[k+1,j]k1j1a=k(xaxa+1z)A[a+2,j]k1B[k,a1]k1]=0.

    Therefore, φ(k+1)j(xk1x) is divisible by xk1x, and it follows that φ(k+1)jD(A[k+1]). Consequently, we can conclude that for any 3q2 and qj3, φ(q)jD(A[q]).

    Lemma 2.3. The derivation φ(q)2 belongs to the module D(A[q]) for 1q2.

    Proof. From Theorem 2.2, we can get the following equality for qm1,

    φ(q)2(xmx)=φ(0)1(xmx)+2a=q1(a1)φ(0)a(xmx)(m,0)B[1,1]m+(z)2a=q1(a1)(xaxa+1z)A[a+2,1]mB[1,a1]m.

    (1) For q=1,2, this conclusion is straightforward to verify.

    (2) For q=k3, assume that φ(k)2D(A[k]), which implies that φ(k)2(xmx) is divisible by xmx for km1.

    For q=k+1, we have

    φ(k+1)2=φ(k)2+(k+1)φ(0)k2.

    By using the equality (2.4), we have

    φ(k+1)2(xk1x)(k1,0)B[1,1]k1+(z)2a=k2(a1)(xaxa+1z)A[a+2,1]k1B[1,a1]k1=B[1,k1]k1[B[k,1]k1+(k+1)A[k,1]k1+2a=k1(a1)(xaxa+1z)A[a+2,1]k1B[k1,a1]k1]=0.

    Therefore, φ(k+1)2(xk1x) is divisible by xk1x. According to the induction hypothesis and Remark 2.1, we have φ(k+1)2D(A[k+1]). Hence, we may conclude that for any 1q2, φ(q)2D(A[q]).

    Lemma 2.4. The derivation φ(q)1 belongs to the module D(A[q]) for 1q2.

    Proof. First, from Theorem 2.2, for qm1, we can get

    φ(q)1(xmx)=(x1x)φ(0)1(xmx)+(x1xz)3a=q1(a2)φ(0)a(xmx)(m,0)(x1xm)B[1,1]m+(z)(x1xmz)3a=q1(a2)(xaxa+1z)A[a+2,1]mB[1,a1]m.

    (1) For q=1,2, it is obvious that φ(q)1D(A[q]).

    (2) For q=k3, assume that φ(k)1D(A[k]), which implies that φ(k)1(xmx) is divisible by xmx for km1.

    For q=k+1, we can see

    φ(k+1)1=φ(k)1+k(x1xz)φ(0)k2.

    By using the equality (2.5), we have

    φ(k+1)1(xk1x)(k1,0)(x1xk1)B[1,1]k1+(z)(x1xk1z)3a=k2(a2)(xaxa+1z)A[a+2,1]k1B[1,a1]k1=(x1xk1)(x1xk1z)B[1,k1]k1[B[k,2]k1+kA[k,2]k1+3a=k1(a2)(xaxa+1z)A[a+2,2]k1B[k1,a1]k1]=0.

    Therefore, φ(k+1)1(xk1x) is divisible by xk1x. According to the induction hypothesis and Remark 2.1, we have φ(k+1)1D(A[k+1]). Hence, we may conclude that for any 1q2, φ(q)1D(A[q]).

    From the above proof, we finally conclude that φ(q)1,,φ(q)1 belong to the module D(A[q]).

    Theorem 2.3. For 1q2, the derivations η1,η2,φ(q)1,,φ(q)1 form a basis for D(A[q]).

    Proof. According to Lemmas 2.2–2.4, it suffices to prove that

    detM(η1,η2,φ(q)1,,φ(q)1)Q(A[q]).

    Let

    γ1=(q,q1,,1,1)T

    and

    γ2=((q1)(x1xz),(q2)(x1xz),,x1xz,0,x1x)T

    be the (q+1)×1 column vectors, and define a matrix

    M(q+1)×(q1):=(xqx(xqxq+1z)(x3x2z)0xq+1x(x3x2z)00x2x000000).

    Write ˜M(q+1)×(q+1):=(M(q+1)×(q1),γ1,γ2), then

    det˜M(q+1)×(q+1)=1s=q(xsx).

    Thus, we obtain the following equality

    (η1,η2,φ(q)1,,φ(q)1)(+1)×(+1)=(η1,η2,φ(0)1,,φ(0)1)(Eq0(q)×(q+1)0(q+1)×(q)˜M(q+1)×(q+1)).

    Hence,

    detM(η1,η2,φ(q)1,,φ(q)1)=detM(η1,η2,φ(0)1,,φ(0)1)det˜M(q+1)×(q+1).=z1m<n1(xmxn)1m<n(xmxnz)1s=q(xsx)=Q(A[q]).

    We complete the proof.

    Meihui Jiang: writing-original draft; Ruimei Gao: writing-review and editing, methodology and supervision. All authors have read and approved the final version of the manuscript for publication.

    The authors declare they have used Artificial Intelligence (AI) tools in the creation of this article.

    The work was partially supported by Science and Technology Development Plan Project of Jilin Province, China (No. 20230101186JC) and NSF of China (No. 11501051).

    The authors declare no conflict of interest in this paper.



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