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:I⊆R→R is said convex if
f(ta+(1−t)b)≤tf(a)+(1−t)f(b) |
for all a,b∈Iand t∈[0,1].
In [3] Toader defines the m−convexity:
Definition 1.2. A functionf:[0,d]→R0=[0,∞) is said m−convex on [0,d] for some m∈(0,1],if
f(ta+m(1−t)b)≤tf(a)+m(1−t)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(1−t)b)≤tαf(a)+m(1−tα)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(1−t)b)≤tsf(a)+m(1−t)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:I⊆R→R be a twice differentiable mapping on I∘(interior of I), a≠b∈I and f′′∈L[a,b], then the following equality holds:
f(a)+f(b)2−1b−a∫baf(x)dx=(b−a)22∫10t(1−t)f′′(ta+(1−t)b)dt. |
Lemma 1.2. Let f:I⊆R→R be a twice differentiable mapping on I∘, where a,b∈I with a<mb and m∈(0,1].If f′′∈L[a,b],then the following equality holds:
f(a)+f(mb)2−1mb−a∫mbaf(x)dx=(mb−a)22∫10t(1−t)f′′(ta+m(1−t)b)dt. | (1.1) |
In [7], M. Z. Sarıkaya and N. Aktan obtained the following Trapezoid inequality for convex functions
|f(a)+f(b)2−1b−a∫baf(t)dt|≤(b−a)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 q−1+p−1=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≤ 1b−a(∫ba(b−t)|f(t)|pdt )1p(∫ba(b−t)|g(t)|qdt )1q+1b−a(∫ba(t−a)|f(t)|pdt )1p(∫ba(t−a)|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 q≥1. 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)|dt≤1b−a(∫ba(b−t)|f(t)|dt )1−1q(∫ba(b−t)|f(t)||g(t)|qdt )1q+1b−a(∫ba(t−a)|f(t)|dt )1−1q(∫ba(t−a)|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(1−t)b)≤tαsf(a)+m(1−tα)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 f∈Kmα, we have
f(ta+m(1−t)b)≤tαf(a)+m(1−tα)f(b) |
∀a,b∈[0,d],t∈[0,1]and (α,m)∈(0,1]2. On the other hand, we have tα≤tαsand 1−tα≤(1−tα)s,for s∈[−1,1],thus
f(ta+m(1−t)b)≤tαf(a)+m(1−tα)f(b)≤tαsf(a)+m(1−tα)sf(b), |
i.ef∈Ks,mα.
Proposition 2.2. Let α∈(0,1]and s∈[−1,1],then
(i)when s∈(−1,1],we have
Mβ(α,s)≈∫10t(1−t)(1−tα)sdt=1α{β(2α,s+1)−β(3α,s+1)}, |
(ii)whens=−1, we have ∫10t(1−t)(1−tα)dt=1α{Ψ(3α)−Ψ(2α)},
where Γ(.),β(.,.),Ψ(.)are the classical Euler-Gamma, Beta and Psi-Gamma functions, respectively.
Γ(x)=∫∞0zx−1ezdz, β(x,y)=∫10zx−1(1−z)y−1dz(x>0,y>0),Ψ(x)=dlnΓ(x)dx=−∫∞011−e−ze−zxdz. |
Proof. (i)
Mβ(α,s)≈∫10t(1−t)(1−tα)sdt. |
If we change the variable tα=z⇒t=z1α, we get dt=1αz1α−1dz. Hence
∫10z1α(1−z)s1αz1α−1dz−∫10z2α(1−z)s1αz1α−1dz=1α∫10z2α−1(1−z)sdz−1α∫10z3α−1(1−z)sdz=1α{β(2α,s+1)−β(3α,s+1)}, |
which completes the proof of proposition (i).
(ii)Whens=−1
∫10t(1−t)(1−tα)−1dt=∫10t(1−t)1−tαdt. |
Again using the change of variable tα=z⇒t=z1α, we have
∫10t(1−t)1−tαdt=∫10z1α−z2α1−z1αz1α−1dz=1α∫10(1−z3α−1)−(1−z2α−1)1−zdz=1α∫10(1−z3α−1)1−zdz−1α∫10(1−z2α−1)1−zdz=1α{[Ψ(3α)+γ]−[Ψ(2α)+γ]}=1α{[Ψ(3α)]−[Ψ(2α)]}, |
where Ψ(x)+γ=∫101−tx−11−tdt and γ=∫∞0(11+t−e−t)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)≤1mb−a∫mbaf(t)dt≤1αs+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:
1mb−a∫mbaf(t)dt=1mb−a(∫a+mb2af(t)dt+∫mba+mb2f(t)dt)=12∫10[f(a+mb−t(mb−a)2)+f(a+mb+t(mb−a)2)]dt≥f(a+mb2). |
The proof of the right-hand side (3.1).
By the (α,s,m)−convexity of fwe observe that
f(ta+m(1−t)b)≤tαsf(a)+m(1−tα)sf(b),∀t∈[0,1]. |
Integrating the resulting inequality with respect to t, we get
∫10f(ta+m(1−t)b)dt≤∫10[tαsf(a)+m(1−tα)sf(b)]dt |
as
∫10f(ta+m(1−t)b)dt=1mb−a∫mbaf(t)dt |
and
∫10[tαsf(a)+m(1−tα)sf(b)]dt=∫10tαsf(a)dt+∫10m(1−tα)sf(b)dt≤∫10tαsf(a)dt+mf(b)∫10(1−t)αsdt=1αs+1(f(a)+mf(b)), |
which proved the theorem as we used the fact
∫10(1−tα)sdt≤∫10(1−t)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)≤1b−a∫baf(x)dx≤12(f(a)+f(b)). |
Theorem 3.2. Let f:I⊂(0,d]→R be a differentiable function on I∘(I∘is interior of I) for a,b∈I∘ such that f′′∈L[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)2−1mb−a∫mbaf(x)dx|≤(mb−a)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)2−1mb−a∫mbaf(x)dx|≤(mb−a)22∫10t(1−t)[tαs|f′′(a)|+m(1−tα)s|f′′(b)|]dt=(mb−a)22[|f′′(a)|∫10tαs+1(1−t)dt+m|f′′(b)|∫10t(1−t)(1−tα)sdt]≤(mb−a)22[|f′′(a)|∫10tαs+1(1−t)dt+m|f′′(b)|∫10t(1−t)(1−t)αsdt]=(mb−a)22|f′′(a)|+m|f′′(b)|(αs+2)(αs+3), |
where we used the facts that, since tα≥tfor all α∈(0,1]and t∈[0,1],we have −tα≤−tor1−tα≤1−t, 1−tα≤(1−t)αand (1−tα)s≤(1−t)αs,s∈(0,1].If we multiply the resulting inequality with t(1−t)and taking integral on [0,1],we have
∫10t(1−t)(1−tα)sdt≤∫10t(1−t)(1−t)αsdt=1(αs+2)(αs+3) |
and
∫10tαs+1(1−t)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)2−1mb−a∫mbaf(x)dx|≤M(mb−a)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∘(I∘is interior of I) for a,b∈I∘, f′′∈L[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 q≥1, then the following inequality holds:
|f(a)+f(b)2−1b−a∫baf(x)dx|≤(b−a)22(16)1−1q× [|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)2−1b−a∫baf(x)dx|≤(b−a)22∫10t(1−t)|f′′(ta+m(1−t)bm)|dt≤(b−a)22∫10t(1−t)(tαs|f′′(a)|+m(1−tα)s|f′′(bm)|)dt=(b−a)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(t−t2)|f′′(ta+m(1−t)bm)|dt=∫10(t−t2)1−1q(t−t2)1q|f′′(ta+m(1−t)bm)|dt≤[∫10(t−t2)dt]1−1q[∫10(t−t2)|f′′(ta+m(1−t)bm)|qdt]1q, | (3.4) |
where p−1+q−1=1.
Since |f′′|qis (α,s,m)−convex on [a,b], we know that for every t∈[0,1].
|f′′(ta+m(1−t)bm)|q≤tαs|f′′(a)|q+m(1−tα)s|f′′(bm)|q. | (3.5) |
From (3.4), (3.5) and Proposition 2.2 (ⅰ)
|f(a)+f(b)2−1b−a∫baf(x)dx|≤(b−a)22[∫10(t−t2)dt]1−1q[∫10(t−t2)|f′′(ta+m(1−t)bm)|qdt]1q≤(b−a)22[∫10(t−t2)dt]1−1q[∫10(t−t2)(tαs|f′′(a)|q +m(1−tα)s|f′′(bm)|q)dt]1q=(b−a)22(16)1−1q[|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)2−1b−a∫baf(x)dx|≤M(b−a)22(16)1−1q[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)2−1b−a∫baf(x)dx|≤(b−a)22(16)1−1q | (3.6) |
×[|f′′(a)|q(2−α)(3−α)+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)2−1b−a∫baf(x)dx|≤(b−a)22[∫10(t−t2)]1−1q ×[∫10(t−t2)(t−α|f′′(a)|q+m(1−tα)−1|f′′(bm)|q)dt]1q=(b−a)22(16)1−1q[∫10(t−t2)(t−α|f′′(a)|q+m(1−tα)−1|f′′(bm)|q)dt]1q=(b−a)22(16)1−1q[|f′′(a)|q∫10(t1−α−t2−α)dt+m|f′′(bm)|q∫10t−t21−tαdt]1q=(b−a)22(16)1−1q[|f′′(a)|q(2−α)(3−α)+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)2−1b−a∫baf(x)dx| ≤M(b−a)22(16)1−1q×[1(2−α)(3−α)+1α{Ψ(3α)−Ψ(2α)}]1q. |
Theorem 3.5. Let f:I⊂(0,d]→R be a twice differentiable function on I∘ for a,b∈I∘ such that f′′∈L[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 q≥1 andq≥r>0, then the following inequality holds:
|f(a)+f(b)2−1b−a∫baf(x)dx|≤(b−a)22(β(2q−r−1q−1,2q−1q−1))1−1q | (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)2−1b−a∫baf(x)dx|≤(b−a)22[∫10(tq−rq−1(1−t)qq−1)dt]1−1q×[∫10tr(tαs|f′′(a)|q+m(1−tα)s|f′′(bm)|q)dt]1q=(b−a)22[β(2q−r−1q−1,2q−1q−1)]1−1q×[|f′′(a)|q∫10tαs+rdt+m|f′′(bm)|q∫10tr(1−tα)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)2−1b−a∫baf(x)dx|≤(b−a)22(β(2q−1q−1,2q−1q−1))1−1q | (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)2−1b−a∫baf(x)dx|≤(b−a)22(β(2q−1q−1,2q−1q−1))1−1q | (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)2−1b−a∫baf(x)dx|≤M(b−a)22[β(2q−r−1q−1,2q−1q−1)]1−1q×[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 f′′∈L[a,b] for a,b∈I∘,(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 q≥1,q≥r>0, then the following inequality holds:
|f(a)+f(b)2−1b−a∫baf(t)dt|≤K[β(2q−r−1q−1,2q−1q−1)]1−1q×[1αs+r+1|f′′(a)|q+mα⋅β(r+1α,s+1)|f′′(bm)|q]1q, | (3.10) |
where K=(b−a)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)2−1b−a∫baf(t)dt|≤K[∫10tq−rq−1(1−t)qq−1dt]1−1q[∫10tr(tαs|f′′(a)|q+m(1−tα)s|f′′(bm)|q)dt]1q≤K{[∫10tq−rq−1(1−t)qq−1dt]1−1q[|f′′(a)|q∫10tr+αsdt+m|f′′(bm)|q∫10tr(1−tα)sdt]1q}=K{[B(2q−r−1q−1,2q−1q−1)]1−1q[|f′′(a)|q∫10tr+αsdt+m|f′′(bm)|q∫10tr(1−tα)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.
[1] |
S. Axler, Z. Cuckovic, Commuting Toeplitz operators with harmonic symbols, Integr. Equ. Oper. Theory, 14 (1991), 1–12. https://doi.org/10.1007/BF01194925 doi: 10.1007/BF01194925
![]() |
[2] |
B. R. Choe, Y. J. Lee, Commuting Toeplitz operators on the harmonic Bergman space, Michigan Math. J., 46 (1999), 163–174. https://doi.org/10.1307/mmj/1030132367 doi: 10.1307/mmj/1030132367
![]() |
[3] | H. Sadraoui, M. Guediri, Hyponormality of Toeplitz operators on the Bergman space of an annulus, Rev. Union Mat. Argent., 61 (2020), 303–313. |
[4] | H. Sadraoui, Hyponormality of Toeplitz operators and cohyponormality of composition operators, PhD Thesis, Purdue University, 1992. |
[5] | P. Ahern, Z. Cuckovic, A mean value inequality with applications to Bergman space operators, Pacific J. Math., 173 (1996), 295–305. |
[6] |
H. Sadraoui, M. Garayev, H. Guediri, On Hyponormality of Toeplitz operators, Rocky Mountain J. Math., 51 (2021), 1821–1831. https://doi.org/10.1216/rmj.2021.51.1821 doi: 10.1216/rmj.2021.51.1821
![]() |
[7] | H. Hedenmalm, B. Korenblum, K. H. Zhu, Theory of Bergman spaces, New York: Springer, 2000. https://doi.org/10.1007/978-1-4612-0497-8 |
[8] |
N. Ghiloufi, M. Zaway, Meromorphic Bergman spaces, Ukr. Math. J., 74 (2023), 1209–1224. https://doi.org/10.1007/s11253-023-02130-9 doi: 10.1007/s11253-023-02130-9
![]() |
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