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Research article

Commuting Toeplitz operators on weighted harmonic Bergman spaces and hyponormality on the Bergman space of the punctured unit disk

  • Received: 16 April 2024 Revised: 08 June 2024 Accepted: 14 June 2024 Published: 20 June 2024
  • MSC : 15B05, 15B48, 47B20, 47B35, 47B47

  • We first describe commuting Toeplitz operators with harmonic symbols on weighted harmonic Bergman spaces. Then, a sufficient condition for hyponormality on weighted Bergman spaces of the punctured unit disk, when the analytic part of the symbol is a monomial, is shown.

    Citation: Houcine Sadraoui, Borhen Halouani. Commuting Toeplitz operators on weighted harmonic Bergman spaces and hyponormality on the Bergman space of the punctured unit disk[J]. AIMS Mathematics, 2024, 9(8): 20043-20057. doi: 10.3934/math.2024977

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  • We first describe commuting Toeplitz operators with harmonic symbols on weighted harmonic Bergman spaces. Then, a sufficient condition for hyponormality on weighted Bergman spaces of the punctured unit disk, when the analytic part of the symbol is a monomial, is shown.



    The general theory of convex functions is the origin of powerful tools for the study of problems in analysis. Convex functions are thoroughly associated to the theory of inequalities. Inequalities involving convex functions are the most efficient tools in the development of several branches of mathematics and has been given considerable attention in the literature (see [1,2]) and references therein). We start by giving some renowned definitions:

    Definition 1.1. A function f:IRR is said convex if

    f(ta+(1t)b)tf(a)+(1t)f(b)

    for all a,bIand t[0,1].

    In [3] Toader defines the mconvexity:

    Definition 1.2. A functionf:[0,d]R0=[0,) is said mconvex on [0,d] for some m(0,1],if

    f(ta+m(1t)b)tf(a)+m(1t)f(b), a,b[0,d] andt[0,1].

    In [4] Miheşan introduced the following class of functions:

    Definition 1.3. A functionf:[0,d]R0=[0,) is said (α,m) convex on [0,d],(α,m)(0,1]2, if

    f(ta+m(1t)b)tαf(a)+m(1tα)f(b), a,b[0,d]andt[0,1].

    Let Kmα be the set of (α,m)convex functions on [0,d].

    In [5] Park J. introduced the following class of (s,m)convex functions in the second sense:

    Definition 1.4. For some fixed s(0,1] and m[0,1] a mapping f:[0,d]R is said (s,m) convex in the second sense on [0,d], if

    f(ta+m(1t)b)tsf(a)+m(1t)sf(b)

    holds for all a,b[0,d] and t[0,1].

    In [6], Özdemir et al. gave the following lemmas for twice differentiable functions.

    Lemma 1.1. Let f:IRR be a twice differentiable mapping on I(interior of I), abI and fL[a,b], then the following equality holds:

    f(a)+f(b)21babaf(x)dx=(ba)2210t(1t)f(ta+(1t)b)dt.

    Lemma 1.2. Let f:IRR be a twice differentiable mapping on I, where a,bI with a<mb and m(0,1].If fL[a,b],then the following equality holds:

    f(a)+f(mb)21mbambaf(x)dx=(mba)2210t(1t)f(ta+m(1t)b)dt. (1.1)

    In [7], M. Z. Sarıkaya and N. Aktan obtained the following Trapezoid inequality for convex functions

    |f(a)+f(b)21babaf(t)dt|(ba)224[|f(a)|+|f(b)|]. (1.2)

    In [8], I. Işcan gave a refinement of the Hölder's integral inequality as follows:

    Theorem 1.1. (Hölder- İşcan integral inequality) Let p>1 and q1+p1=1. If f and g are real functions defined on interval [a,b] and if |f|p and |g|q are integrable functions on [a,b], then

    ba|f(t)g(t)|dt 1ba(ba(bt)|f(t)|pdt )1p(ba(bt)|g(t)|qdt )1q+1ba(ba(ta)|f(t)|pdt )1p(ba(ta)|g(t)|qdt )1q.

    An refinement of power-mean integral inequality as a different version of the Hölder- İşcan integral inequality can be given as follows:

    Theorem 1.2. (Improved power-mean integral inequality [9]) Let q1. If f and g are real functions defined on [a,b] and if |f|,|f||g|q are integrable functions on [a,b], then

    ba|f(t)g(t)|dt1ba(ba(bt)|f(t)|dt )11q(ba(bt)|f(t)||g(t)|qdt )1q+1ba(ba(ta)|f(t)|dt )11q(ba(ta)|f(t)||g(t)|qdt )1q.

    There is a massive literature on Hermite – Hadamard type inequalities via different kind of convexities. The most notable are m–convex, (α,m)convex, s –convex and extended s convex (see [10,11,12,13,14,15,16,17,18,19]). Besides this, we mention recent results related to the Hermite–Hadamard type inequalities, for example, see [20,21,22] and the references therein. In the next section we consider the most recent generalized convex functions and its properties which covers all the above as particular cases.

    Definition 2.1. [23] The function f:[0,d]R is said to be (α,s,m)convex, if we have

    f(ta+m(1t)b)tαsf(a)+m(1tα)sf(b),

    where a,b[0,d], t(0,1)and for some s[1,1],(α,m)(0,1]2.

    Here Ks,mα is the set of (α,s,m) convex functions on [0,d].

    Remark 2.1. (i) If s=1, then f is an (α,m) - convex function on (0,d].

    (ii) If α=1, then f is an extended (s,m) - convex function on (0,d].

    (iii) If α=m=1, then f is an extended s -convex function on (0,d].

    (iv) If  α=s=m=1, then f is a convex function on (0,d].

    Proposition 2.1. If a function f is (α,m) – convex for all s[1,1], then it is also (α,s,m) – convex.

    Proof. Since fKmα, we have

    f(ta+m(1t)b)tαf(a)+m(1tα)f(b)

    a,b[0,d],t[0,1]and (α,m)(0,1]2. On the other hand, we have tαtαsand 1tα(1tα)s,for s[1,1],thus

    f(ta+m(1t)b)tαf(a)+m(1tα)f(b)tαsf(a)+m(1tα)sf(b),

    i.efKs,mα.

    Proposition 2.2. Let α(0,1]and s[1,1],then

    (i)when s(1,1],we have

    Mβ(α,s)10t(1t)(1tα)sdt=1α{β(2α,s+1)β(3α,s+1)},

    (ii)whens=1, we have 10t(1t)(1tα)dt=1α{Ψ(3α)Ψ(2α)},

    where Γ(.),β(.,.),Ψ(.)are the classical Euler-Gamma, Beta and Psi-Gamma functions, respectively.

    Γ(x)=0zx1ezdz,  β(x,y)=10zx1(1z)y1dz(x>0,y>0),Ψ(x)=dlnΓ(x)dx=011ezezxdz.

    Proof. (i)

    Mβ(α,s)10t(1t)(1tα)sdt.

    If we change the variable tα=zt=z1α, we get dt=1αz1α1dz. Hence

    10z1α(1z)s1αz1α1dz10z2α(1z)s1αz1α1dz=1α10z2α1(1z)sdz1α10z3α1(1z)sdz=1α{β(2α,s+1)β(3α,s+1)},

    which completes the proof of proposition (i).

    (ii)Whens=1

    10t(1t)(1tα)1dt=10t(1t)1tαdt.

    Again using the change of variable tα=zt=z1α, we have

    10t(1t)1tαdt=10z1αz2α1z1αz1α1dz=1α10(1z3α1)(1z2α1)1zdz=1α10(1z3α1)1zdz1α10(1z2α1)1zdz=1α{[Ψ(3α)+γ][Ψ(2α)+γ]}=1α{[Ψ(3α)][Ψ(2α)]},

    where Ψ(x)+γ=101tx11tdt and γ=0(11+tet)1t dtis Euler–Mascheroini constant (see p.258 [24]). Which completes the proof of proposition (ii).

    Theorem 3.1. If f:[0,d]Ris (α,s,m)convex, we have

    f(a+mb2)1mbambaf(t)dt1αs+1(f(a)+mf(b)), (3.1)

    where a,b[0,d], a<mbandforsomes[0,1],(α,m)(0,1]2.

    Proof. The left-hand side of (3.1) is easy to prove:

    1mbambaf(t)dt=1mba(a+mb2af(t)dt+mba+mb2f(t)dt)=1210[f(a+mbt(mba)2)+f(a+mb+t(mba)2)]dtf(a+mb2).

    The proof of the right-hand side (3.1).

    By the (α,s,m)convexity of fwe observe that

    f(ta+m(1t)b)tαsf(a)+m(1tα)sf(b),t[0,1].

    Integrating the resulting inequality with respect to t, we get

    10f(ta+m(1t)b)dt10[tαsf(a)+m(1tα)sf(b)]dt

    as

    10f(ta+m(1t)b)dt=1mbambaf(t)dt

    and

    10[tαsf(a)+m(1tα)sf(b)]dt=10tαsf(a)dt+10m(1tα)sf(b)dt10tαsf(a)dt+mf(b)10(1t)αsdt=1αs+1(f(a)+mf(b)),

    which proved the theorem as we used the fact

    10(1tα)sdt10(1t)αsdt=1αs+1.

    Remark 3.1. If we take α=m=s=1.The function fbecomes (1,1,1)convex. Hence the inequality (3.1) reduces to the Hermite- Hadamard inequality :

    f(a+b2)1babaf(x)dx12(f(a)+f(b)).

    Theorem 3.2. Let  f:I(0,d]R be a differentiable function on I(Iis interior of I) for a,bI such that fL[a,b] with 0<a<mb. If |f| is an (α,s,m)convex function on [a,b], then the following inequality holds:

    |f(a)+f(mb)21mbambaf(x)dx|(mba)22|f(a)|+m|f(b)|(αs+2)(αs+3), (3.2)

    where(α,m)(0,1]2,s(0,1].

    Proof. From Lemma 1.2 and using the (α,s,m)convexity of |f|, we have

    |f(a)+f(mb)21mbambaf(x)dx|(mba)2210t(1t)[tαs|f(a)|+m(1tα)s|f(b)|]dt=(mba)22[|f(a)|10tαs+1(1t)dt+m|f(b)|10t(1t)(1tα)sdt](mba)22[|f(a)|10tαs+1(1t)dt+m|f(b)|10t(1t)(1t)αsdt]=(mba)22|f(a)|+m|f(b)|(αs+2)(αs+3),

    where we used the facts that, since tαtfor all α(0,1]and t[0,1],we have tαtor1tα1t, 1tα(1t)αand (1tα)s(1t)αs,s(0,1].If we multiply the resulting inequality with t(1t)and taking integral on [0,1],we have

    10t(1t)(1tα)sdt10t(1t)(1t)αsdt=1(αs+2)(αs+3)

    and

    10tαs+1(1t)dt=1(αs+2)(αs+3).

    Corollary 3.1. From Remark 2.1 (iv), since f(.) is a convex function on [a,b] for m=α=s=1, we obtain the inequality (1.2).

    Corollary 3.2. Under the same conditions in Theorem 3.2 with |f(x)|M for all x[a,b], we have

    |f(a)+f(mb)21mbambaf(x)dx|M(mba)22×(m+1)(αs+2)(αs+3).

    The following Theorem presents a upper bound for (α,s,m)-convex functions.

    Theorem 3.3. Let  f:I(0,d]R be a differentiable function on I(Iis interior of I) for a,bI, fL[a,b],(α,m)(0,1]2,s(1,1] with 0<a<mb. If |f|q is an (α,s,m)convex function on [a,b] for q1, then the following inequality holds:

    |f(a)+f(b)21babaf(x)dx|(ba)22(16)11q× [|f(a)|q(αs+2)(αs+3)+mMβ(α,s)|f(bm)|q]1q, (3.3)

    where Mβ(α,s) is given in Proposition 2.2(i).

    Proof. First, we assume that q=1.From Lemma 1.1 and using the (α,s,m)convexity with properties of modulus we have

    |f(a)+f(b)21babaf(x)dx|(ba)2210t(1t)|f(ta+m(1t)bm)|dt(ba)2210t(1t)(tαs|f(a)|+m(1tα)s|f(bm)|)dt=(ba)22[|f(a)|(αs+2)(αs+3)+mMβ(α,s)|f(bm)|],

    which completes the proof for q = 1. Here we also used the Proposition 2.2 (ⅰ).

    Secondly, suppose now that q>1. From Lemma 1 and using the Hölder's integral inequality for q>1 with properties of modulus, we have

    10(tt2)|f(ta+m(1t)bm)|dt=10(tt2)11q(tt2)1q|f(ta+m(1t)bm)|dt[10(tt2)dt]11q[10(tt2)|f(ta+m(1t)bm)|qdt]1q, (3.4)

    where p1+q1=1.

    Since  |f|qis (α,s,m)convex on [a,b], we know that for every t[0,1].

    |f(ta+m(1t)bm)|qtαs|f(a)|q+m(1tα)s|f(bm)|q. (3.5)

    From (3.4), (3.5) and Proposition 2.2 (ⅰ)

    |f(a)+f(b)21babaf(x)dx|(ba)22[10(tt2)dt]11q[10(tt2)|f(ta+m(1t)bm)|qdt]1q(ba)22[10(tt2)dt]11q[10(tt2)(tαs|f(a)|q            +m(1tα)s|f(bm)|q)dt]1q=(ba)22(16)11q[|f(a)|q(αs+2)(αs+3)+mMβ(α,s)|f(bm)|q]1q,

    which completes the inequality (3.3).

    Corollary 3.3. Under the same conditions in Theorem 3.3 with |f(x)|M and m=1 for all x[a,b], we have

    |f(a)+f(b)21babaf(x)dx|M(ba)22(16)11q[1(αs+2)(αs+3)+Mβ(α,s)]1q.

    Theorem 3.4. When s=1, under the same conditions of Theorem 3.3, we obtain the following result

    |f(a)+f(b)21babaf(x)dx|(ba)22(16)11q (3.6)
    ×[|f(a)|q(2α)(3α)+m|f(bm)|q1α{Ψ(3α)Ψ(2α)}]1q.

    Proof. Using the (α,s,m)convexity of |f|q and Proposition 2.2(ⅱ) with the properties of modulus, we have

    |f(a)+f(b)21babaf(x)dx|(ba)22[10(tt2)]11q ×[10(tt2)(tα|f(a)|q+m(1tα)1|f(bm)|q)dt]1q=(ba)22(16)11q[10(tt2)(tα|f(a)|q+m(1tα)1|f(bm)|q)dt]1q=(ba)22(16)11q[|f(a)|q10(t1αt2α)dt+m|f(bm)|q10tt21tαdt]1q=(ba)22(16)11q[|f(a)|q(2α)(3α)+m|f(bm)|q1α{Ψ(3α)Ψ(2α)}]1q,

    which completes the inequality (3.6).

    Corollary 3.4. Under the same conditions in Theorem 3.4 with |f(x)|M and m=1 for all x[a,b], we have

    |f(a)+f(b)21babaf(x)dx| M(ba)22(16)11q×[1(2α)(3α)+1α{Ψ(3α)Ψ(2α)}]1q.

    Theorem 3.5. Let  f:I(0,d]R be a twice differentiable function on I for a,bI such that fL[a,b],(α,m)(0,1]2,s(1,1]with 0<a<mb. If |f|q is an (α,s,m)convex function on [a,b]for q1 andqr>0, then the following inequality holds:

    |f(a)+f(b)21babaf(x)dx|(ba)22(β(2qr1q1,2q1q1))11q (3.7)
    ×[|f(a)|qαs+r+1+mα|f(bm)|q(β(r+1α,s+1))]1q.

    Proof. Using the Lemma 1.1, Hölder's inequality and (α,s,m)convexity of |f|q we have

    |f(a)+f(b)21babaf(x)dx|(ba)22[10(tqrq1(1t)qq1)dt]11q×[10tr(tαs|f(a)|q+m(1tα)s|f(bm)|q)dt]1q=(ba)22[β(2qr1q1,2q1q1)]11q×[|f(a)|q10tαs+rdt+m|f(bm)|q10tr(1tα)sdt]1q.

    After simplifying we get inequality (3.7).

    An immediate consequences of Theorem 3.5 by considering special cases for (r=0) and (m=1) can be given as:

    Corollary 3.5. Under the assumptions of Theorem 3.5, the following inequality holds

    |f(a)+f(b)21babaf(x)dx|(ba)22(β(2q1q1,2q1q1))11q (3.8)
    ×[|f(a)|qαs+1+mα|f(bm)|q(β(1α,s+1))]1q.

    Corollary 3.6. Under the assumptions of Theorem 3.5, the following inequality holds

    |f(a)+f(b)21babaf(x)dx|(ba)22(β(2q1q1,2q1q1))11q (3.9)
    ×[|f(a)|qαs+1+1α |f(b)|q(β(1α,s+1))]1q.

    Corollary 3.7. Under the same conditions in Theorem 3.5 with |f(x)|M and m=1, for all x[a,b], we have

    |f(a)+f(b)21babaf(x)dx|M(ba)22[β(2qr1q1,2q1q1)]11q×[1αs+r+1+1aβ(r+1α,s+1)]1q.

    Theorem 3.6. Let  f:I(0,d]R be a twice differentiable function on I such that fL[a,b] for a,bI,(α,m)(0,1]2,s(1,1] with 0<a<mb. If |f|qis an (α,s,m)convex function on [a,b]for q1,qr>0, then the following inequality holds:

    |f(a)+f(b)21babaf(t)dt|K[β(2qr1q1,2q1q1)]11q×[1αs+r+1|f(a)|q+mαβ(r+1α,s+1)|f(bm)|q]1q, (3.10)

    where  K=(ba)22.

    Proof. By using Lemma 1.1 and Hölder's inequality with (α,s,m)convexity of |f|q, we have

    |f(a)+f(b)21babaf(t)dt|K[10tqrq1(1t)qq1dt]11q[10tr(tαs|f(a)|q+m(1tα)s|f(bm)|q)dt]1qK{[10tqrq1(1t)qq1dt]11q[|f(a)|q10tr+αsdt+m|f(bm)|q10tr(1tα)sdt]1q}=K{[B(2qr1q1,2q1q1)]11q[|f(a)|q10tr+αsdt+m|f(bm)|q10tr(1tα)sdt]1q}.

    Here we used the fact that

    \begin{eqnarray*} \int_{0}^{1}t^{\frac{q-r}{q-1}}\left( 1-t\right) ^{\frac{q}{q-1}}dt = B\left( \frac{2q-r-1}{q-1},\frac{2q-1}{q-1}\right) \end{eqnarray*}

    and with a formal change of variables t^{\alpha } = z

    \begin{eqnarray*} \int_{0}^{1}t^{r}\left( 1-t^{\alpha }\right) ^{s}dt = \frac{1}{\alpha } \int_{0}^{1}z^{\frac{r}{\alpha }}\left( 1-z\right) ^{s}dz = \frac{1}{\alpha } B\left( \frac{r+1}{\alpha },\;s+1\right). \end{eqnarray*}

    If we write the above integrals places we obtain the inequality (3.10).

    Theorem 3.7. Let \ f:I\subset (0, d]\rightarrow R be a twice differentiable function on I^{\circ} such that \; f^{\prime \prime }\in L[a, b] for a, b\in I^{\circ}, \; (\alpha, m)\in (0, 1]^{2}, s\in [0, 1]\; with 0 < a < mb . Also let p > 1 such that q = \frac{p}{p-1} , if \left\vert f^{\prime \prime }\right\vert ^{q} is an \left(\alpha, s, m\right) - convex on [a, b] , then the following inequality holds:

    \begin{equation} \left\vert \frac{f(a)+f(mb)}{2}-\frac{1}{mb-a}\int_{a}^{mb}f\left( t\right) dt\right\vert \end{equation} (3.11)
    \begin{eqnarray*} \leq G \left\{ \begin{array}{c} \left( \frac{3p}{ 2\left( p+1\right) \left( p+2\right) \left( 2p+1\right) }\right) ^{\frac{1}{p }}\left( \frac{\left\vert f^{{\prime \prime }}\left( a\right) \right\vert ^{q}}{\left( \alpha s+1\right) \left( \alpha s+2\right) }+\frac{m\left\vert f^{{\prime \prime }}\left( b\right) \right\vert ^{q}}{\left( \alpha s+2\right) }\right) ^{\frac{1}{q}} \\ +\left( \frac{p}{2\left( p+1\right) \left( p+2\right) }\right) ^{ \frac{1}{p}}\left( \frac{\left\vert f^{^{\prime \prime }}\left( a\right) \right\vert ^{q}}{\left( \alpha s+2\right) }+\frac{m\left\vert f^{^{\prime \prime }}\left( b\right) \right\vert ^{q}}{\left( \alpha s+1\right) \left( \alpha s+2\right) }\right) ^{\frac{1}{q}} \end{array} \right\}, \end{eqnarray*}

    where G = \frac{\left(mb-a\right) ^{2}}{2}.

    Proof. Using Lemma 1.2 and Hölder- İşcan integral inequality along with \left(\alpha, s, m\right) - convexity of \left\vert f^{\prime \prime }\right\vert ^{q} , we have

    \begin{eqnarray*} &&\left\vert \frac{f(a)+f(mb)}{2}-\frac{1}{mb-a}\int_{a}^{mb}f\left( t\right) dt\right\vert \leq G \int_{0}^{1}\left( t-t^{2}\right) \Big\vert f^{{\prime \prime }}\left( ta+m\left( 1-t\right) b\right) dt\Big\vert \ dt \\ &\leq &G \left\{ \begin{array}{c} \left( \int_{0}^{1}\left( 1-t\right) \left\vert t-t^{2}\right\vert ^{p}dt\right) ^{\frac{1}{p}}\left( \int_{0}^{1}\left( 1-t\right) \Big\vert f^{{\prime \prime }}\left( ta+m\left( 1-t\right) b\right) dt\Big\vert ^{q}dt\right) ^{\frac{1}{q}} \\ +\left( \int_{0}^{1}t\left\vert t-t^{2}\right\vert ^{p}dt\right) ^{\frac{1 }{p}}\left( t\Big\vert f^{{\prime \prime }}\left( ta+m\left( 1-t\right) b\right) dt\Big\vert ^{q}dt\right) ^{\frac{1}{q}}. \end{array} \right\} \\ \end{eqnarray*}

    Using the fact that \left\vert m-n\right\vert ^{c}\leq m^{c}-n^{c} for c > 1 and for m, n > \ 0; \ \ m > n . The right-hand side of above inequality becomes

    \begin{eqnarray*} &\leq &G \left\{ \begin{array}{c} \left( \int_{0}^{1}\left( 1-t\right) \left( t^{p}-t^{2p}\right) dt\right) ^{\frac{1}{p}}\left( \int_{0}^{1}\left( 1-t\right) \left\{ t^{\alpha s}\left\vert f^{{\prime \prime }}\left( a\right) \right\vert ^{q}+m\left\vert f^{{\prime \prime }}\left( b\right) \right\vert ^{q}\left( 1-t\right) ^{\alpha s}\right\} dt\right) ^{\frac{1}{q} } \\ +\ \left( \int_{0}^{1}t\left( t^{p}-t^{2p}\right) dt\right) ^{\frac{1}{p} }\left( t\left\{ t^{\alpha s}\left\vert f^{{\prime \prime }}\left( a\right) \right\vert ^{q}+m\left\vert f^{{\prime \prime }}\left( b\right) \right\vert ^{q}\left( 1-t\right) ^{\alpha s}\right\} dt\right) ^{\frac{1}{q} }. \end{array} \right\} \\ \end{eqnarray*}

    Simplifying above integrals, we obtain the result of the Theorem 3.7.

    Corollary 3.8. Under the same conditions in Theorem 3.7 with \left\vert f^{\prime \prime }\left(x\right) \right\vert\leq M for all x\in \lbrack a, b] , we have

    \begin{equation*} \left\vert \frac{f\left( a\right) +f\left( mb\right) }{2}-\frac{1}{mb-a} \int_{a}^{mb}f\left( x\right) dx\right\vert \leq G\cdot M\cdot \left\{ \begin{array}{c} \left( \frac{3p}{ 2\left( p+1\right) \left( p+2\right) \left( 2p+1\right) }\right) ^{\frac{1}{p }}\left( \frac{1}{\left( \alpha s+1\right) \left( \alpha s+2\right) }+\frac{m}{\left( \alpha s+2\right) }\right) ^{\frac{1}{q}} \\ +\left( \frac{p}{2\left( p+1\right) \left( p+2\right) }\right) ^{ \frac{1}{p}}\left( \frac{1}{\left( \alpha s+2\right) }+\frac{m}{\left( \alpha s+1\right) \left( \alpha s+2\right) }\right) ^{\frac{1}{q}} \end{array} \right\} \end{equation*}

    Theorem 3.8. Let \ f:I\subset (0, d]\rightarrow R be a twice differentiable function on I^{\circ} such that \; f^{\prime \prime }\in L[a, b] for a, b\in I^{\circ}, \; (\alpha, m)\in (0, 1]^{2}, s\in [0, 1]\; with 0 < a < mb . If \left\vert f^{\prime \prime }\right\vert ^{q} is an \left(\alpha, s, m\right) - convex on [a, b] for q\ge1 , then the following inequality holds:

    \begin{equation} \left\vert \frac{f(a)+f(mb)}{2}-\frac{1}{mb-a}\int_{a}^{mb}f\left( t\right) dt\right\vert \leq \frac{G}{12}\left( \frac{12}{\left( \alpha s+3\right) \left( \alpha s+4\right) }\right) ^{\frac{1}{q}} \end{equation} (3.12)
    \begin{equation*} \times \left\{ \left( \frac{2\left\vert f^{^{\prime \prime }}\left( a\right) \right\vert ^{q}}{\alpha s+2}+m\ \left\vert f^{^{\prime \prime }}\left( b\right) \right\vert ^{q}\right) ^{\frac{1}{q}}+\left( \left\vert f^{^{\prime \prime }}\left( a\right) \right\vert ^{q}+\frac{2m\ \left\vert f^{^{\prime \prime }}\left( b\right) \right\vert ^{q}}{\alpha s+2}\right) ^{ \frac{1}{q}}\right\}, \end{equation*}

    where G = \frac{\left(mb-a\right) ^{2}}{2}.

    Proof. Using Lemma 1.2 and improved power mean integral inequality with \left(\alpha, s, m\right)- convexity of \left\vert f^{\prime \prime }\right\vert ^{q} , we have

    \begin{equation*} \ \ \ \left\vert \frac{f(a)+f(mb)}{2}-\frac{1}{mb-a}\int_{a}^{mb}f\left( t\right) dz\right\vert \end{equation*}
    \begin{equation*} \leq G\cdot\left\{ \begin{array}{c} \left( \int_{0}^{1}\left( 1-t\right) \left\vert t-t^{2}\right\vert \ dt\right) ^{1-\frac{1}{q}}\left( \int_{0}^{1}\left( 1-t\right) \left\vert t-t^{2}\right\vert \ \left\vert f^{^{\prime \prime }}\left( ta+m\left( 1-t\right) b\right) \right\vert ^{q}dt\right) ^{\frac{1}{q}} \\ +\ \left( \int_{0}^{1}t\left\vert t-t^{2}\right\vert \ dt\right) ^{1-\frac{1 }{q}}\left( \int_{0}^{1}t\left\vert t-t^{2}\right\vert \ \left\vert f^{^{\prime \prime }}\left( ta+m\left( 1-t\right) b\right) \right\vert ^{q}dt\right) ^{\frac{1}{q}} \end{array} \right\} \end{equation*}

    By employing \left(\alpha, s, m\right)- convexity and simplifying, we have

    \begin{eqnarray} &\leq & G\cdot\left\{ \begin{array}{c} \left( \int_{0}^{1}\left( 1-t\right) \left\vert t-t^{2}\right\vert dt\right) ^{1-\frac{1}{q}} \\ \times \left[ \int_{0}^{1}\left( 1-t\right) \left\vert t-t^{2}\right\vert \ \left( t^{\alpha s}\left\vert f^{^{\prime \prime }}\left( a\right) \right\vert ^{q}+m\left\vert f^{\prime \prime }\left( b\right) \right\vert ^{q}\left( 1-t^{\alpha }\right) ^{s}dt\right) \right] ^{\frac{1}{q}} \\ +\left( \int_{0}^{1}t\left\vert t-t^{2}\right\vert \ dt\right) ^{1-\frac{1}{q }}\left[ \int_{0}^{1}t\left\vert t-t^{2}\right\vert \ \left( t^{\alpha s}\left\vert f^{\prime \prime }\left( a\right) \right\vert ^{q}+m\left\vert f^{\prime \prime }\left( b\right) \right\vert ^{q}\left( 1-t^{\alpha }\right) ^{s}\right) dt\right] ^{\frac{1}{q}} \end{array} \right\} \end{eqnarray} (3.13)

    It can be noticed that

    \begin{equation} \int_{0}^{1}\left( 1-t\right) \left\vert t-t^{2}\right\vert \ dt = \int_{0}^{1}t\left\vert t-t^{2}\right\vert \ dt = \frac{1}{12} \end{equation} (3.14)

    and

    \begin{equation} \int_{0}^{1}t^{\alpha s}\left( 1-t\right) \left\vert t-t^{2}\right\vert \ dt\ \ = \ \frac{2}{\left( \alpha s+2\right) \left( \alpha s+3\right) \left( \alpha s+4\right) }\ \end{equation} (3.15)

    and

    \begin{eqnarray} \int_{0}^{1}\left( 1-t\right) \left\vert t-t^{2}\right\vert \ \left( 1-t^{\alpha }\right) ^{s}dt &\leq &\int_{0}^{1}\left\vert t-t^{2}\right\vert \ \left( 1-t\right) ^{\alpha s+1}dt \\ & = &\int_{0}^{1}t\left( 1-t\right) ^{\alpha s+2}dt = \frac{1}{\left( \alpha s+3\right) \left( \alpha s+4\right) } \end{eqnarray} (3.16)
    \begin{eqnarray} \int_{0}^{1}t\left\vert t-t^{2}\right\vert \ t^{\alpha s}\ dt & = &\int_{0}^{1}\left( 1-t\right) \ t^{\alpha s+2}\ dt\ = \frac{1}{\left( \alpha s+3\right) \left( \alpha s+4\right) } \\ \int_{0}^{1}t\left\vert t-t^{2}\right\vert \ \left( 1-t^{\alpha }\right) ^{s}dt\ &\leq &\int_{0}^{1}t\left\vert t-t^{2}\right\vert \ \left( 1-t\right) ^{\alpha s}dt \\ & = &\int_{0}^{1}t^{2}\ \left( 1-t\right) ^{\alpha s+1}dt = \frac{2}{\left( \alpha s+2\right) \left( \alpha s+3\right) \left( \alpha s+4\right) } \end{eqnarray} (3.17)

    Replacing the values of the integrals computed in (3.14), (3.15), (3.16) and (3.17) in (3.13), we get (3.12).

    Corollary 3.9. Under the same conditions in Theorem 3.8 with \left\vert f^{^{\prime \prime }}\left(x\right) \right\vert\leq M for all x\in \lbrack a, b] , we have

    \begin{eqnarray*} &&\left\vert \frac{f\left( a\right) +f\left( mb\right) }{2}-\frac{1 }{mb-a}\int_{a}^{mb}f\left( x\right) dx\right\vert \ \\ &\leq &M\cdot \frac{\left( mb-a\right) ^{2}}{24}\left( \frac{12}{\left( \alpha s+3\right) \left( \alpha s+4\right) }\right) ^{\frac{1}{q}}\left[ \left( \frac{2}{\alpha s+2}+m\right) ^{\frac{1}{q}}+\left( 1+\frac{2m}{ \alpha s+2}\right) ^{\frac{1}{q}}\right]. \end{eqnarray*}

    Now we let us consider some special means for arbitrary positive real numbers c and d .

    The\ arithmetic\ mean\ \ \ :

    \ \ \ \ \ \ \ A\left(c, d\right) = \frac{c+d}{2}, \qquad \qquad c, d\in R\ with\ c, d > 0.

    The\ geometric\ mean\ \ \ :

    G\left(c, d\right) = \left(c \cdot d\right) ^{\frac{1}{2}}, \qquad c, d\in R\ with\ c, d > 0.

    The\ harmonic\ mean\ \ \ :

    H\left(c, d\right) = \frac{2\ c\ d}{c + d}, \qquad \qquad c, d\in R\ \backslash \{0\}.

    The\ logarithmic\ mean\ \ \ :

    L\left(c, d\right) = \left\{ \begin{array}{c} c, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ if\ \ c = d \\ \frac {d-c}{ln d - ln c}, \ \ \ \ \ \ \ \ if\ \ c\neq d. \end{array} \ \ \ \ \right.

    The\ Generalized\ logarithmric\ mean\ \ \ :

    L_{n}\left(c, d\right) = \left\{ \begin{array}{c} c, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ if\ \ c = d, \\ \left(\frac{d^{n+1}-c^{n+1}}{\left(n+1\right) \left(d-c\right) }\right)^{1/n}, \ \ \ \ \ \ \ \ if\ \ c\neq d, \ \ n\in Z\backslash \{-1, 0\}; c, d > 0. \end{array} \ \ \ \ \right.

    The\ \ Identric\ mean\ \ \ :

    I\left(c, d\right) = \left\{ \begin{array}{c} c, \qquad if\ \ c = d \\ \frac{1}{e}\left(\frac{d^{d}}{c^{c}}\right) ^{\frac{1}{d-c}}, \qquad if\ \ c\neq d \; \ c, d > 0. \end{array} \ \ \ \ \right.

    Proposition 4.1. Let a, b\in R, \ \ a < b\ , \ a, b > 0\ , then we have the following inequality holds

    \begin{eqnarray*} &&\left\vert A\left( e^{a},\; e^{b}\right) -L\left( e^{a},\; e^{b}\right) \right\vert \\ && \leq \frac{\left( b-a\right) ^{2}}{2}\left( \frac{1}{6}\right) ^{1-\frac{1}{q}}.\left[ \frac{\left\vert e^{a}\right\vert ^{q}}{\left( \alpha s+2\right) \left( \alpha s+3\right) }+mM_{\beta }\left( \alpha ,s\right) \left\vert e^{\frac{b}{m}}\right\vert ^{q}\right] ^{\frac{1 }{q}}. \end{eqnarray*}

    Proof. This assertion follows from Theorem 3.3 for f\left(x\right) = e^{x} and \ x\in R .

    Proposition 4.2. Let a, b\in R, \ \ a < b , a, b > 0 , then we have the following inequality

    \begin{eqnarray*} &&\left\vert A\left( \ln a^{-1},\; \ln b^{-1}\right) +\ln \left( I\right) \right\vert\\ && \leq \frac{\left( b-a\right) ^{2}}{2}\left( \frac{1}{6} \right) ^{1-\frac{1}{q}}\left[ \frac{\left\vert \frac{1}{a^{2}}\right\vert ^{q}}{\left( 2-\alpha \right) \left( 3-\alpha \right) }+\frac{m}{\alpha }\left\vert \frac{m}{b}\right\vert ^{2q}\left\{ \Psi \left( \frac{3}{ \alpha }\right) -\Psi \left( \frac{2}{\alpha }\right) \right\} \right] ^{ \frac{1}{q}}. \end{eqnarray*}

    Proof. This assertion follows from Theorem 3.4 for f\left(x\right) = -ln{x} and \ x > 0 .

    Proposition 4.3. Let a, b\in R, \ \ a < b\ and 0\notin \left[ a, b\right] , then the following inequality holds

    \begin{eqnarray*} &&\left\vert H^{-1}\left( a,b\right) \ -\ L^{-1}\left( a,b\right) \right\vert \leq \frac{\left( b-a\right) ^{2}}{2}\left( \frac{1}{6}\right) ^{1-\frac{1}{q}}\\ && \times\left[ \frac{\left\vert \frac{2}{a^{3}}\right\vert ^{q}}{ \left( 2-\alpha \right) \left( 3-\alpha \right) }+m\; \left\vert \frac{2m^{3}}{ b^{3}}\right\vert ^{q}.\frac{1}{\alpha }\left\{ \Psi \left( \frac{3}{\alpha } \right) -\Psi \left( \frac{2}{\alpha }\right) \right\} \right] ^{\frac{1}{q} }. \end{eqnarray*}

    Proof. This assertion follows from Theorem 3.4 for f\left(x\right) = \frac{1}{x} ; \ x \in[a, b] .

    Proposition 4.4. Let a, b\in R, \ \ a < b and 0\notin \left[ a, b\right] , then the following inequality holds

    \left\vert H^{-1}\left( a,b\right) \ \ -\ \ L^{-1}\left( a,b\right) \right\vert \leq \frac{\left( b-a\right) ^{2}}{2}\left( B\left( \frac{2q-r-1}{q-1},\frac{2q-1}{q-1}\right) \right) ^{1-\frac{1}{q}}\\\times\left[ \frac{\left\vert \frac{2}{a^{3}}\right\vert ^{q}}{\alpha s+r+1}+\frac{m}{ \alpha }\left\vert \frac{2m^{3}}{b^{3}}\right\vert ^{q}B\left( \frac{r+1}{ \alpha },s+1\right) \right] ^{\frac{1}{q}} .

    Proof. This assertion follows from Theorem 3.5 for f\left(x\right) = \frac{1}{x} ; x \in[a, b] .

    In many practical studies, it is necessary to evaluate the difference between the two quantities. From the point of view of programming an algorithm for solving optimization problems, classical inequalities with the smallest upper limit play an important role.

    We have proved several new generalized integral inequalities involving \left(\alpha, s, m\right)- convex functions for extended case of s where s \in[-1, 1] , we obtained Hermite–Hadamrad type inequalities. The most interesting case is extended case for s = -1 , when we get connected to Psi-Gamma functions. We analyze new upper bounds involving special functions by employing different variants of Hölder inequality such as Hölder-Işcan inequality and Improved Power mean inequality.

    The author declares no conflict of interest in this paper.



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