In this paper, we consider the symmetry properties of the positive solutions of a p-Laplacian problem of the form
{−Δpu=f(x,u), in Ω, u=g(x), on ∂Ω,
where Ω is an open smooth bounded domain in RN,N≥2, and symmetric w.r.t. the hyperplane Tν0(ν is a direction vector in RN,|ν|=1), f: Ω×R+→R+ is a continuous function of class C1 w.r.t. the second variable, g≥0 is continuous, and both f and g are symmetric w.r.t. Tν0, respectively. Introducing some assumptions on nonlinearities, we get that the positive solutions of the problem above are symmetric w.r.t. the direction ν by a new simple idea even if Ω is not convex in the direction ν.
Citation: Keqiang Li, Shangjiu Wang. Symmetry of positive solutions of a p-Laplace equation with convex nonlinearites[J]. AIMS Mathematics, 2023, 8(6): 13425-13431. doi: 10.3934/math.2023680
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In this paper, we consider the symmetry properties of the positive solutions of a p-Laplacian problem of the form
{−Δpu=f(x,u), in Ω, u=g(x), on ∂Ω,
where Ω is an open smooth bounded domain in RN,N≥2, and symmetric w.r.t. the hyperplane Tν0(ν is a direction vector in RN,|ν|=1), f: Ω×R+→R+ is a continuous function of class C1 w.r.t. the second variable, g≥0 is continuous, and both f and g are symmetric w.r.t. Tν0, respectively. Introducing some assumptions on nonlinearities, we get that the positive solutions of the problem above are symmetric w.r.t. the direction ν by a new simple idea even if Ω is not convex in the direction ν.
Hermite-Hadamard inequality is a double inequality for convex functions that has a lot of literary value (please see [16]).
Let ζ:I⟶R, ∅≠I⊆R, ς,τ∈I with ς<τ, be a convex function. Then
ζ(ς+τ2)≤1τ−ςτ∫ςζ(w)dw≤ζ(ς)+ζ(τ)2, | (1.1) |
the inequality holds in reversed direction if ζ is concave.
Fejér [15] established the following double inequality as a weighted generalization of (1.1):
ζ(ς+τ2)τ∫ςϑ(w)dw≤τ∫ςζ(w)ϑ(w)dw≤ζ(ς)+ζ(τ)2τ∫ςϑ(w)dw, | (1.2) |
where ζ:I⟶R, ∅≠I⊆R, ς,τ∈I with ς<τ is any convex function and ϑ:[ς,τ]→R is non-negative integrable and symmetric with respect to w=ς+τ2.
These inequalities have many extensions and generalizations, see [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18] and [19,20,21,22,23,24,25,26,27,28,29,30,31].
Consider the following mappings on [0,1]:
˘G(ι)=12τ∫ς[ζ(ις+(1−ι)ς+τ2)+ζ(ιτ+(1−ι)ς+τ2)]dw, |
H(ι)=1τ−ςτ∫ςζ(ιw+(1−ι)ς+τ2)dw, |
Hϑ(ι)=1τ−ςτ∫ςζ(ιw+(1−ι)ς+τ2)ϑ(w)dw, |
L(ι)=12(τ−ς)τ∫ς[ζ(ις+(1−ι)w)+ζ(ιτ+(1−ι)w)]dw |
and
Lϑ(ι)=12τ∫ς[ζ(ις+(1−ι)w)+ζ(ιτ+(1−ι)w)]ϑ(w)dw, |
where ζ:[ς,τ]→R is a convex function and ϑ:[ς,τ]→R is non-negative integrable and symmetric with respect to w=ς+τ2.
The important results that characterize the properties of the above mappings and inequalities are discussed by a number of mathematicians.
Dragomir [2] established the theorem which refines the first inequality of (1.1).
Theorem 1. [2] Let ζ:[ς,τ]→R be a convex function on [ς,τ]. Then H is monotonically increasing and convex on [0,1]. Moreover, one has thefollowing inequalities
ζ(ς+τ2)=H(0)≤H(ι)≤H(1)=1τ−ς∫τςζ(w)dw. |
Dragomir et al. [7] obtained the refinements of (1.1).
Theorem 2. [2] Let ζ, H be defined asabove. Then
(i) The following inequality holds
ζ(ς+τ2)≤2τ−ς∫ς+3τ43ς+τ4ζ(w)dw≤∫10H(ι)dι≤12[ζ(ς+τ2)+1τ−ς∫τςζ(w)dw]. |
(ii) If ζ is differentiable on [ς,τ], then for all ι∈[0,1], one has
0≤1τ−ς∫τςζ(w)dw−H(ι)≤(1−ι)[ζ(ς)+ζ(τ)2−1τ−ς∫τςζ(w)dw] |
and
0≤ζ(ς)+ζ(τ)2−H(ι)≤(ζ′(τ)−ζ′(ς))(τ−ς)4. |
Theorem 3. [7] Let ζ, H, ˘G be defined as above. We have
(i) ˘G is convex and increasing on [0,1].
(ii) The following hold
infι∈[0,1]˘G(ι)=˘G(0)=ζ(ς+τ2) |
and
supι∈[0,1]˘G(ι)=˘G(1)=ζ(ς)+ζ(τ)2. |
(iii) The inequality
H(ι)≤˘G(ι) |
holds for all ι∈[0,1].
(iv) Then
2τ−ς∫ς+3τ43ς+τ4ζ(w)dw≤12[ζ(3ς+τ4)+ζ(ς+3τ4)]≤∫10˘G(ι)dι≤12[ζ(ς+τ2)+ζ(ς)+ζ(τ)2]. |
(v) If ζ is differentiable on [ς,τ], then for all ι∈[0,1], one has
0≤H(ι)−ζ(ς+τ2)≤˘G(ι)−H(ι). |
Theorem 4. [7] Let ζ, H, ˘G, L be defined as above. Then
(i) L is convex on [0,1].
(ii) The inequalities
˘G(ι)≤L(ι)≤1−ιτ−ς∫τςζ(w)dw+ιζ(ς)+ζ(τ)2≤ζ(ς)+ζ(τ)2 |
hold for all ι∈[0,1] and
supι∈[0,1]L(ι)=L(1)=ζ(ς)+ζ(τ)2. |
(iii) The inequalities
H(1−ι)≤L(ι)andH(ι)+H(1−ι)2≤L(ι) |
hold for all ι∈[0,1].
Teseng et al. [24] proved the following result.
Lemma 1. [24] Let ζ: [ς,τ]→R be a convex function and let ς≤˘ϰ1≤w1≤w2≤˘ϰ2≤τ with w1+w2=˘ϰ1+˘ϰ2. Then
ζ(w1)+ζ(w2)≤ζ(˘ϰ1)+ζ(˘ϰ2). |
Yang and Tseng [28] proven the theorem by using Lemma 1 which refines the first inequality of (1.2) and generalizes Theorem 1.
Theorem 5. [28] Let ζ:[ς,τ]→R be a convex function and ϑ:[ς,τ]→R is non-negative integrable and symmetric with respect to w=ς+τ2. Then Hϑ is convex, increasing on [0,1], and for all ι∈[0,1], we have
ζ(ς+τ2)∫τςϑ(w)dw=Hϑ(0)≤Hϑ(ι)≤Hϑ(1)=∫τςζ(w)ϑ(w)dw. | (1.3) |
One of the generalizations of the convex functions is harmonic functions:
Definition 1. [17] Define I⊆R∖{0} as an interval of real numbers. A function ζ from I to the real numbers is considered to be harmonically convex, if
ζ(w˘ϰιw+(1−ι)˘ϰ)≤ιζ(˘ϰ)+(1−ι)ζ(w) | (1.4) |
for all w,˘ϰ∈I and ι∈[0,1]. Harmonically concave ζ is defined as the inequality in (1.4) reversed.
İşcan [17] used harmonic-convexity to develop the inequalities of Hermite-Hadamard type.
Theorem 6. [17] Let ζ:I⊆R∖{0}→R be a harmonically convex function and ς,τ∈I with ς<τ. If ζ∈L([ς,τ]) then theinequalities
ζ(2ςτς+τ)≤ςττ−ς∫ςτζ(w)w2dw≤ζ(ς)+ζ(τ)2 | (1.5) |
hold.
Let ζ:[ς,τ]⊂(0,∞)→R be a harmonic convex mapping and let S,V:[0,1]→R be defined by
S(ι)=ςττ−ς∫τς1w2ζ(2ςτw2ςτι+(1−ι)w(ς+τ))dw | (1.6) |
and
V(ι)=ςτ2(τ−ς)∫τς1w2[ζ(2τw(1+ι)w+(1−ι)τ)+ζ(2ςw(1+ι)w+(1−ι)ς)]dw. | (1.7) |
The author obtained the refinement inequalities for (1.5) related to the above mappings:
Theorem 7. [21] Let ζ:[ς,τ]⊂(0,∞)→R be a harmonic convex function on [ς,τ]. Then
(i) S is harmonic convex (0,1] andincreases monotonically on [0,1].
(ii) The following hold:
ζ(2ςτς+τ)=S(0)≤S(ι)≤S(1)=ςττ−ς∫τςζ(w)w2dw. |
Theorem 8. [21] Let ζ:[ς,τ]⊂(0,∞)→R be a harmonic convex function on [ς,τ]. Then
(i) V is harmonic convex (0,1] andincreases monotonically on [0,1].
(ii) The following hold:
ςττ−ς∫τςζ(w)w2dw=V(0)≤V(ι)≤V(1)=ζ(ς)+ζ(τ)2. |
Harmonic symmetricity of a function is given in the definition below.
Definition 2. [22] A function ϑ:[ς,τ]⊆R∖{0}→R is harmonically symmetric with respect to 2ςτς+τ if
ϑ(w)=ϑ(11ς+1τ−1w) |
holds for all w∈[ς,τ].
Fejér type inequalities using harmonic convexity and the notion of harmonic symmetricity were presented in Chan and Wu [1].
Theorem 9. [1] Let ζ:I⊆R∖{0}→R be a harmonically convex function and ς,τ∈I with ς<τ. If ζ∈L([ς,τ]) and ϑ:[ς,τ]⊆R∖{0}→R is nonnegative, integrable and harmonically symmetric with respect to 2ςτς+τ, then
ζ(2ςτς+τ)∫ςτϑ(w)w2dw≤∫ςτζ(w)ϑ(w)w2dw≤ζ(ς)+ζ(τ)2∫ςτϑ(w)w2dw. | (1.8) |
Chan and Wu [1] also defined some mappings related to (1.8) and discussed important properties of these mappings.
Motivated by the studies conducted in [2,21,24,27], we define some new mappings in connection to (1.8) and to prove new Féjer type inequalities which indeed provide refinement inequalities as well.
To prove the major findings of this work, we employ the given important facts about harmonic convex and convex functions.
Theorem 10. [8,9] If [ς,τ]⊂I⊂(0,∞) and if we consider thefunction ˘h:[1τ,1ς]→R defined by ˘h(ι)=ζ(1ι), then ζ is harmonicallyconvex on [ς,τ] if and only if ˘h is convex in the usual sense on [1τ,1ς].
Theorem 11. [8,9] If I⊂(0,∞) and ζ is convex and nondecreasing function then ζ isharmonic convex and if ζ is harmonic convex and nonincreasing function then ζ is convex.
Let us now define some mappings on [0,1] related to (1.8) and prove some refinement inequalities.
˘G1(ι)=12[ζ(2ςτ2ςι+(1−ι)(ς+τ))+ζ(2ςτ2τι+(1−ι)(ς+τ))]dw, |
S(ι)=ςττ−ςτ∫ς1w2ζ(2ςτw2ςτι+(1−ι)(ς+τ)w)dw, |
Sϑ(ι)=τ∫ςζ(2ςτw2ςτι+(1−ι)(ς+τ)w)ϑ(w)w2dw, |
T(ι)=ςτ2(τ−ς)τ∫ς1w2[ζ(τwιw+(1−ι)τ)+ζ(ςwιw+(1−ι)ς)]dw |
and
Tϑ(ι)=12τ∫ς[ζ(τwιw+(1−ι)τ)+ζ(ςwιw+(1−ι)ς)]ϑ(w)w2dw, |
where ζ:[ς,τ]→R is a harmonic convex function and ϑ:[ς,τ]→R is non-negative integrable and symmetric with respect to w=2ςτς+τ.
Lemma 2. Let ζ: [ς,τ]⊂(0,∞)→R be a harmonic convex function and let ς≤˘ϰ1≤w1≤w2≤˘ϰ2≤τ with w1w2w1+w2=˘ϰ1˘ϰ2˘ϰ1+˘ϰ2. Then
ζ(w1)+ζ(w2)≤ζ(˘ϰ1)+ζ(˘ϰ2). |
Proof. For ˘ϰ1=˘ϰ2, the result is obvious. We observe that
w1=˘ϰ1˘ϰ2(w1˘ϰ1−˘ϰ1˘ϰ2w1˘ϰ1−w1˘ϰ2)˘ϰ2+(˘ϰ1˘ϰ2−w1˘ϰ2w1˘ϰ1−w1˘ϰ2)˘ϰ1andw2=˘ϰ1˘ϰ2(w2˘ϰ1−˘ϰ1˘ϰ2w2˘ϰ1−w2˘ϰ2)˘ϰ2+(˘ϰ1˘ϰ2−w2˘ϰ2w2˘ϰ1−w2˘ϰ2)˘ϰ1. |
By applying the harmonic convexity, we obtain
ζ(w1)+ζ(w2)≤(w1˘ϰ1−˘ϰ1˘ϰ2w1˘ϰ1−w1˘ϰ2)ζ(˘ϰ1)+(˘ϰ1˘ϰ2−w1˘ϰ2w1˘ϰ1−w1˘ϰ2)ζ(˘ϰ2)+(w2˘ϰ1−˘ϰ1˘ϰ2w2˘ϰ1−w2˘ϰ2)ζ(˘ϰ1)+(˘ϰ1˘ϰ2−w2˘ϰ2w2˘ϰ1−w2˘ϰ2)ζ(˘ϰ2)=(w1˘ϰ1−˘ϰ1˘ϰ2w1˘ϰ1−w1˘ϰ2+w2˘ϰ1−˘ϰ1˘ϰ2w2˘ϰ1−w2˘ϰ2)ζ(˘ϰ1)+(˘ϰ1˘ϰ2−w1˘ϰ2w1˘ϰ1−w1˘ϰ2+˘ϰ1˘ϰ2−w2˘ϰ2w2˘ϰ1−w2˘ϰ2)ζ(˘ϰ2)=˘ϰ1˘ϰ2−˘ϰ1((w1+w2w1w2)˘ϰ2−2)ζ(˘ϰ1)+˘ϰ2˘ϰ2−˘ϰ1(2−(w1+w2w1w2)˘ϰ1)ζ(˘ϰ2)=ζ(˘ϰ1)+ζ(˘ϰ2). |
We first prove a result similar to (1.3) for harmonically convex functions which provide refinement inequalities for (1.8).
Theorem 12. Let ζ: [ς,τ]⊂(0,∞)→R be a harmonic convex function, 0<ρ<1, 0<θ<1, λ=ςτρς+(1−ρ)τ, τ0=(ςττ−ς)min{ρ1−θ,1−ρθ} andlet ϑ:[ς,τ]→R be nonnegative and integrable and ϑ(λ1−θιλ)=ϑ(λ1+(1−θ)ιλ), ι∈[0,τ0]. Then
ζ(ςτρς+(1−ρ)τ)∫λ1−θιλλ1+(1−θ)ιλϑ(w)w2dw≤1−θθ∫λλ1−θιλζ(w)ϑ(w)w2dw+θ1−θ∫λ1+(1−θ)ιλλζ(w)ϑ(w)w2dw≤[ρζ(τ)+(1−ρ)ζ(ς)]∫λ1−θιλλ1+(1−θ)ιλϑ(w)w2dw. | (2.1) |
Proof. For every τ∈[0,τ0], we have the identity
∫λ1−θιλλ1+(1−θ)ιλϑ(w)w2dw=∫λλ1+(1−θ)ιλϑ(w)w2dw+∫λ1−θιλλϑ(w)w2dw=θ∫ι0ϑ(λ1−θwλ)w2dw+(1−θ)∫ι0ϑ(λ1−θwλ)w2dw=∫ι0ϑ(λ1−θwλ)w2dw. | (2.2) |
We now prove that the mapping W:[0,τ0]→R defined by
W(ι)=(1−θ)ζ(λ1−θιλ)+θζ(λ1+(1−θ)ιλ) |
is harmonic convex (0,τ0] and monotonically increasing on [0,τ0].
Since the sum of two harmonic convex functions is a harmonic convex, hence W is a harmonic convex on (0,τ0]. Let ι∈(0,τ0], it follows from the harmonic convexity of ζ that
W(ι)=(1−θ)ζ(λ1−θιλ)+θζ(λ1+(1−θ)ιλ)≥ζ((λ1−θιλ)(λ1+(1−θ)ιλ)θ(λ1−θιλ)+(1−θ)(λ1+(1−θ)ιλ))=ζ(λ)=ζ(ςτρς+(1−ρ)τ). | (2.3) |
We observed that 0<ρ≤ρ(τ−ς)+ςτθιτ−ς≤1, 0≤(1−ρ)(τ−ς)−θιςττ−ς≤1−ρ<1, 0≤ρ≤ρ(τ−ς)−(1−θ)ιςττ−ς≤ρ≤1 and 0<1−ρ≤(1−ρ)(τ−ς)+(1−θ)ιςττ−ς≤1. Thus, by using the harmonic convexity, we obtain
W(ι)=(1−θ)ζ(λ1−θιλ)+θζ(λ1+(1−θ)ιλ)=(1−θ)ζ(ςτ(ρ(τ−ς)+ςτιθτ−ς)ς+((1−ρ)(τ−ς)−θιςττ−ς)τ)+θζ(ςτ(ρ(τ−ς)−(1−θ)ιςτ(τ−ς))ς+((1−ρ)(τ−ς)+(1−θ)ιςττ−ς)τ)≤(1−θ)((1−ρ)(τ−ς)−θιςττ−ς)ζ(ς)+(1−θ)×(ρ(τ−ς)+ιςτθτ−ς)ζ(τ)+θ((1−ρ)(τ−ς)+(1−θ)ιςτ(τ−ς))ζ(ς)+θ(ρ(τ−ς)−(1−θ)ιςττ−ς)ζ(τ)=(1−ρ)ζ(ς)+ρζ(τ). | (2.4) |
From (2.3) and (2.4), we obtain
ζ(ςτρς+(1−ρ)τ)≤W(ι)≤(1−ρ)ζ(ς)+ρζ(τ). | (2.5) |
Finally, for ι1, ι2, such that 0<ι1<ι2≤ς0, since W(ι) is harmonic convex, it follows from (2.3) that
W(ι2)−W(ι1)ι2−ι1≥0. |
This shows that W is increasing on [0,τ0].
Since ϑ is nonnegative, multiplying (2.5) by ϑ(λ1−θwλ)w2, integrating the resulting inequalities over [0,ι] and using ϑ(λ1−θwλ)=ϑ(λ1+(1−θ)wλ), we have
ζ(ςτρς+(1−ρ)τ)∫ι0ϑ(λ1−θwλ)w2dw≤(1−θ)∫ι0ζ(λ1−θwλ)ϑ(λ1−θwλ)w2dw+θ∫ι0ζ(λ1+(1−θ)wλ)ϑ(λ1+(1−θ)wλ)w2dw≤[(1−ρ)ζ(ς)+ρζ(τ)]∫ι0ϑ(λ1−θwλ)w2dw. | (2.6) |
By using the identity (2.2) in (2.6), we obtain (2.1).
Remark 1. If we choose ρ=ϑϑ+q, θ=12, ι=2˘ϰ in Theorem 12, then
ζ(ςτ(ϑ+q)ϑς+qτ)∫λ1−λτ˘ϰλ1+λτ˘ϰϑ(w)w2dw≤∫λ1−λτ˘ϰλ1+λτ˘ϰζ(w)ϑ(w)w2dw≤[ϑζ(τ)+qζ(ς)ϑ+q]∫λ1−λτ˘ϰλ1+λτ˘ϰϑ(w)w2dw. | (2.7) |
Remark 2. If we choose ρ= θ=12, ι=τ0=ςττ−ς in Theorem 12, then we get (1.8).
Remark 3. If we choose ρ= θ=12, ι=τ0=ςττ−ς and ϑ(w)=1, w∈[ς,τ] in Theorem 12, then we get (1.5).
Theorem 13. Let ζ, λ and τ0 be defined as in Theorem 12, 0<ρ<1, 0<θ<1, ρ+θ≤1 and let X be defined on [0,1] as
X(ι)=ρςτ(1−θ)(τ−ς)×∫ρςτ(1−θ)(τ−ς)0[(1−θ)ζ(λ1−θιwλ)+θζ(λ1+(1−θ)ιwλ)]dw. | (2.8) |
Then, X is harmonically convex on (0,1] andmonotonically increasing on [0,1], and
ζ(ςτρς+(1−ρ)τ)≤X(ι)≤X(1)=ρςτ(1−θ)(τ−ς)×∫ρςτ(1−θ)(τ−ς)0[(1−θ)ζ(λ1−θwλ)+θζ(λ1+(1−θ)wλ)]dw≤(1−ρ)ζ(ς)+ρζ(τ). |
Proof. Since ζ is harmonically convex on [ς,τ] this prove the harmonic convexity of X on (0,1]. By using the condition ρ+θ≤1 implies that τ0=ρςτ(1−θ)(τ−ς). Since the mapping W:[0,τ0]→R defined by
W(ι)=(1−θ)ζ(λ1−θιλ)+θζ(λ1+(1−θ)ιλ) | (2.9) |
has been proved to be monotonically increasing on [0,τ0], thus the mapping X is also monotonically increasing on [0,1].
Because X is monotonically increasing on [0,1], it follows that the inequalities (2.8) can be deduced from these inequalities (2.5). The proof of the theorem was completed as a result of this.
The next theorem can be proved similarly:
Theorem 14. Let ζ, λ, τ0, ρ and θ be defined as in Theorem 13. Let X1 be defined on [0,1] as
X1(ι)=ρςτ(1−θ)(τ−ς)×∫ρςτ(1−θ)(τ−ς)0[(1−θ)ζ(λ(1−θ)ςτ(1−θ)ςτ−θ(ρ(τ−ς)−w(1−ι)(1−θ)ςτ)λ)+θζ(λ(1−θ)ςτ(1−θ)ςτ+(1−θ)(ρ(τ−ς)−w(1−ι)(1−θ)ςτ)λ)]dw. | (2.10) |
Then, X1 is harmonically convex monotonically increasing on [0,1], and
(1−θ)2ςτρθ(τ−ς)∫(1−θ)ςτ(1−θ)ς−ρ(τ−ς)λζ(w)w2dw+θςτρ(τ−ς)∫λςζ(w)w2dw≤X1(ι)≤X1(1)=(1−θ)ζ((1−θ)ςτ(ς−τ)ρ+(1−θ)τ)+θζ(ς)≤(1−ρ)ζ(ς)+ρζ(τ). | (2.11) |
Remark 4. Taking ρ=θ=12 in the inequality (2.8) reduces to
S(ι)=ςττ−ς∫τςζ(2ςτw2ςτι+(1−ι)w(ς+τ))dww2. |
Remark 5. Taking ρ=θ=12 in the inequality (2.10) reduces to
V(ι)=ςτ2(τ−ς)∫τς1w2[ζ(2τw(1+ι)w+(1−ι)τ)+ζ(2ςw(1+ι)w+(1−ι)ς)]dw. | (2.12) |
Theorem 15. Let ζ, ρ, θ, λ, τ0 be defined as in Theorem 13 and let ϑ be defined as in Theorem 12. Let Y be a function defined on [0,1] by
Y(ι)=∫s0[(1−θ)ζ(λ1−θλιw)ϑ(λ1−θλw)+θζ(λ1+(1−θ)λιw)ϑ(λ1+(1−θ)λw)]dw | (2.13) |
for some s∈[0,τ0]. Then Y isharmonic convex and monotonically increasing on (0,1] and
ζ(ςτρς+(1−ρ)τ)∫λ1−θλsλ1+(1−θ)λsϑ(w)dw≤Y(ι)≤Y(1)=1−θθ∫λλ1−θλsζ(w)ϑ(w)w2dw+θ1−θ∫λ1+(1−θ)λsλζ(w)ϑ(w)w2dw. | (2.14) |
Proof. Since ζ is harmonic convex and ϑ is nonnegative, we see that Y is harmonic convex on (0,1]. Let w∈[0,s], where s∈[0,τ0], from Theorem 12 we get ˘h(ιw)=(1−θ)ζ(λ1−θλιw)+θζ(λ1+(1−θ)λιw) is increasing for ι∈[0,1]. Therefore the inequalities (2.14) are achieved immediately.
Theorem 16. Let ζ, ρ, θ, λ, τ0 be defined as in Theorem 15 and let ϑ be defined as in Theorem 12. Let Y1 be a function defined on [0,1] by
Y1(ι)=∫s0[(1−θ)ζ(λ1−θs+θw(1−ι)λ)ϑ(λ1−θ(s−w)λ)+θζ(λ1+(1−θ)s−(1−θ)w(1−ι)λ)ϑ(λ1+(1−θ)(s−w)λ)]dw | (2.15) |
for some s∈[0,τ0]. Then Y1 isharmonic convex (0,1] and monotonically increasing on [0,1], and
1−θθ∫λλ1−θλsζ(w)ϑ(w)w2dw+θ1−θ∫λ1+(1−θ)λsλζ(w)ϑ(w)w2dw≤Y(ι)≤Y(1)=[(1−θ)ζ(λ1−θsλ)+(1−θ)ζ(λ1+(1−θ)sλ)]×∫λ1+(1−θ)λsλ1−θλsϑ(w)w2dw≤[(1−ρ)ζ(ς)+ρζ(τ)]∫λ1+(1−θ)λsλ1−θλsϑ(w)w2dw. | (2.16) |
Proof. Since ζ is harmonic convex and ϑ is nonnegative, we see that Y1 is harmonic convex on (0,1]. Next, for each w∈[0,ι], where ι∈[0,τ0], it follows from Theorem 12 that ˘h(ι)=(1−θ)ζ(λ1−θλι)+θζ(λ1+(1−θ)λι) and k(ι)=s−(1−ι)w are increasing on [0,τ0] and [0,1] respectively. Hence
˘h(k(ι))=(1−θ)ζ(λ1−θs+θw(1−ι)λ)ϑ(λ1−θ(s−w)λ)+θζ(λ1+(1−θ)s−(1−θ)w(1−ι)λ)ϑ(λ1+(1−θ)(s−w)λ) |
is increasing on [0,1]. Using the identity ϑ(λ1−θλι)=ϑ(λ1+(1−θ)λι) we see that Y(ι) is increasing on [0,1]. Therefore the inequalities (2.16) follows from
ζ(ςτρς+(1−ρ)τ)≤W(k(ι))≤(1−ρ)ζ(ς)+ρζ(τ) |
and (2.16).
Remark 6. Choose ρ=θ=12, s=τ0=ςττ−ς in Theorems 15 and 16. Then the inequalities (2.14) and (2.16) reduce to
ζ(2ςτς+τ)∫τςϑ(w)w2dw≤Y(ι)≤Y(1)=∫τςζ(w)ϑ(w)w2dw≤Y1(ι)≤Y1(1)=ζ(ς)+ζ(τ)2∫τςϑ(w)w2dw, | (2.17) |
where
Y(ι)=ςττ−ς∫τςζ(2ςτw2ςτι+(1−ι)w(ς+τ))ϑ(w)w2dw |
and
Y1(ι)=12∫τς1w2[ζ(2τw(1+ι)w+(1−ι)τ)ϑ(2ςwς+w)+ζ(2ςw(1+ι)w+(1−ι)ς)ϑ(2wτw+τ)]dw. | (2.18) |
Remark 7. The inequalities (2.17) provide weighted generalizations of Theorems 9 and 15.
In the coming results we provide weighted generalizations of Theorems 2–4 for harmonic convex functions by using Lemma 2.
Theorem 17. Let ζ, ϑ, Sϑ be defined as above. Then
(i) The inequality
ζ(2ςτς+τ)∫τςϑ(w)w2dw≤2∫4ςτ3ς+τ4ςτς+3τζ(w)ϑ(2ςτw4ςτ−(ς+τ)w)dww2≤∫10Sϑ(ι)dι≤12[ζ(2ςτς+τ)∫τςϑ(w)w2dw+∫τςζ(w)ϑ(w)w2dw] | (2.19) |
holds.
(ii) If ζ is differentiable on [ς,τ] and ϑ isbounded on [ς,τ], then theinequalities
0≤∫τςζ(w)ϑ(w)w2dw−Sϑ(ι)≤(1−ι)[(τ−ςςτ)[ζ(ς)+ζ(τ)2]−∫τςζ(w)w2dw]‖ϑ‖∞, | (2.20) |
hold for all ι∈[0,1], where ‖ϑ‖∞=supw∈[ς,τ]ϑ(w).
(iii) If ζ is differentiable on [ς,τ], then, for all ι∈[0,1], then
0≤ζ(ς)+ζ(τ)2∫τςϑ(w)w2dw−Sϑ(ι)≤(τ−ς)(τ2ζ′(τ)−ς2ζ′(ς))4ςτ∫τςϑ(w)w2dw. | (2.21) |
Proof. (i) Using techniques of integration and the hypothesis of ϑ, we have the following identities:
ζ(2ςτς+τ)∫τςϑ(w)w2dw=4∫2ςτς+τς∫120ζ(2ςτς+τ)ϑ(w)w2dιdw, | (2.22) |
2∫4ςτ3ς+τ4ςτς+3τζ(w)w2ϑ(2ςτw4ςτ−(ς+τ)w)dw=2∫2ςτς+τς∫120[ζ(4ςτw2ςτ+(ς+τ)w)+ζ(4ςτw3(ς+τ)w−2ςτ)]ϑ(w)w2dιdw, | (2.23) |
∫10Sϑ(ι)dι=∫2ςτς+τς∫120[ζ(2ςτwι(ς+τ)w+2(1−ι)ςτ)+ζ(2ςτw2ςτι+(1−ι)(ς+τ)w)]×ϑ(w)w2dιdw+∫2ςτς+τς∫120[ζ(2ςτw2ι(ςw+τw−ςτ)+(1−ι)(ς+τ)w)+ζ(2ςτw2(1−ι)(ςw+τw−ςτ)+ι(ς+τ)w)]ϑ(w)w2dιdw | (2.24) |
and
12[ζ(2ςτς+τ)∫τςϑ(w)w2dw+∫τςζ(w)ϑ(w)w2dw]=∫2ςτς+τς∫120[ζ(w)+ζ(2ςτς+τ)]ϑ(w)w2dιdw+∫2ςτς+τς∫120[ζ(2ςτς+τ)+ζ(ςτwς+τ−w)]ϑ(w)w2dιdw. | (2.25) |
By using Lemma 2, we observe that the following inequalities hold for all ι∈[0,12] and w∈[ς,2ςτς+τ]:
4ζ(2ςτς+τ)≤2[ζ(4ςτw2ςτ+(ς+τ)w)+ζ(4ςτw3(ς+τ)w−2ςτ)], | (2.26) |
2ζ(4ςτw2ςτ+(ς+τ)w)≤ζ(2ςτwι(ς+τ)w+2(1−ι)ςτ)+ζ(2ςτw2ςτι+(1−ι)(ς+τ)w), | (2.27) |
2ζ(4ςτw3(ς+τ)w−2ςτ)≤ζ(2ςτw2ι(ςw+τw−ςτ)+(1−ι)(ς+τ)w)+ζ(2ςτw2(1−ι)(ςw+τw−ςτ)+ι(ς+τ)w), | (2.28) |
ζ(2ςτwι(ς+τ)w+2(1−ι)ςτ)+ζ(2ςτw2ςτι+(1−ι)(ς+τ)w)≤ζ(w)+ζ(2ςτς+τ) | (2.29) |
and
ζ(2ςτw2ι(ςw+τw−ςτ)+(1−ι)(ς+τ)w)+ζ(2ςτw2(1−ι)(ςw+τw−ςτ)+ι(ς+τ)w)≤ζ(2ςτς+τ)+ζ(ςτwς+τ−w). | (2.30) |
Multiplying the inequalities (2.26)–(2.30) by ϑ(w)w2 and integrating them over ι on [0,12], over w on [ς,2ςτς+τ] and using identities (2.22)–(2.25), we derive (2.19).
(ii) Since ζ:[ς,τ]→R is harmonic convex on [ς,τ], hence ˘h:[1τ,1ς]→R defined by ˘h(w)=ζ(1w) is convex on [1τ,1ς]. Thus, by integration by parts, we get that following identity holds:
∫ς+τ2ςτ1τ(ς+τ2ςτ−w)[˘h′(1ς+1τ−w)−˘h′(w)]dw=(τ−ς2ςτ)[˘h(1ς)+˘h(1τ)]−∫ς+τ2ςτ1τ[˘h(1ς+1τ−w)+˘h(w)]dw. | (2.31) |
The equality (2.31) is equivalent to the equality:
∫2ςτς+τς1w2(1w−ς+τ2ςτ)[ζ′(11ς+1τ−1w)(1ς+1τ−1w)2−w2ζ′(w)]dw=(τ−ςςτ)[ζ(ς)+ζ(τ)2]−∫τςζ(w)w2dw. | (2.32) |
Using substitution rules for integration and the hypothesis of ϑ, we have the following identities:
∫τςζ(w)ϑ(w)w2dw=∫2ςτς+τς[ζ(w)+ζ(11ς+1τ−1w)]ϑ(w)w2dw | (2.33) |
and
Sϑ(ι)=∫2ςτς+τς[ζ(2ςτw2ςτι+(1−ι)(ς+τ)w)+ζ(2ςτw2ι(ςw+τw−ςτ)+(1−ι)(ς+τ)w)]ϑ(w)w2dw. | (2.34) |
Now, using the convexity of ˘h(w)=ζ(1w) on [1τ,1ς] and the hypothesis of ϑ, the following inequality holds for all ι∈[0,1] and w∈[1τ,ς+τ2ςτ]:
[˘h(w)−˘h(ιw+(1−ι)(ς+τ2ςτ))]ϑ(1w)+[˘h(1ς+1τ−w)−˘h(ι(1ς+1τ−w)+(1−ι)(ς+τ2ςτ))]ϑ(1w)≤(1−ι)(w−ς+τ2ςτ)˘h′(w)ϑ(1w)+(1−ι)(ς+τ2ςτ−w)˘h′(1ς+1τ−w)ϑ(1w)=(1−ι)(ς+τ2ςτ−w)[˘h′(1ς+1τ−w)−˘h′(w)]ϑ(1w) | (2.35) |
which is equivalent to
[ζ(1w)−ζ(1ιw+(1−ι)(ς+τ2ςτ))]ϑ(1w)+[ζ(11ς+1τ−w)−ζ(1ι(1ς+1τ−w)+(1−ι)(ς+τ2ςτ))]ϑ(1w)≤(1−ι)(ς+τ2ςτ−w)[ζ′(1w)w2−ζ′(11ς+1τ−w)(1ς+1τ−w)2]ϑ(1w). | (2.36) |
Integrating the above inequalities over w on [1τ,ς+τ2ςτ], we get
∫ς+τ2ςτ1τ[ζ(1w)−ζ(1ιw+(1−ι)(ς+τ2ςτ))]ϑ(1w)dw+∫ς+τ2ςτ1τ[ζ(11ς+1τ−w)−ζ(1ι(1ς+1τ−w)+(1−ι)(ς+τ2ςτ))]ϑ(1w)dw≤(1−ι)∫ς+τ2ςτ1τ(ς+τ2ςτ−w)[ζ′(1w)w2−ζ′(11ς+1τ−w)(1ς+1τ−w)2]ϑ(1w)dw. | (2.37) |
After making use of suitable substitution, the inequality (2.37) takes the form:
∫2ςτς+τς1w2[ζ(1ι1w+(1−ι)(ς+τ2ςτ))−ζ(w)]ϑ(w)dw+∫2ςτς+τς1w2[ζ(1ι(1ς+1τ−1w)+(1−ι)(ς+τ2ςτ))−ζ(11ς+1τ−1w)]ϑ(w)dw≤‖ϑ‖∞(1−ι)∫2ςτς+τς1w2(1w−ς+τ2ςτ)[ζ′(11ς+1τ−1w)(1ς+1τ−1w)2−w2ζ′(w)]dw. | (2.38) |
Inequality (2.20) follows from (2.31)–(2.34) and (2.38).
(iii) We use the fact that ζ:[ς,τ]→R is harmonic convex on [ς,τ], hence ˘h:[1τ,1ς]→R defined by ˘h(w)=ζ(1w) is convex on [1τ,1ς]. Thus
˘h(1τ)−˘h(ς+τ2ςτ)2≤ς−τ4ςτ˘h′(1τ) |
and
˘h(1ς)−˘h(ς+τ2ςτ)2≤τ−ς4ςτ˘h′(1ς). |
Adding the above inequalities
˘h(1ς)+˘h(1τ)2−˘h(ς+τ2ςτ)≤(τ−ς)(˘h′(1ς)−˘h′(1τ))4ςτ. | (2.39) |
The inequality (2.39) becomes
ζ(ς)+ζ(τ)2−ζ(2ςτς+τ)≤(τ−ς)(τ2ζ′(τ)−ς2ζ′(ς))4ςτ. | (2.40) |
Multiplying (2.40) both sides by ϑ(w)w2 and integrating over [ς,τ], we get
ζ(ς)+ζ(τ)2∫τςϑ(w)w2dw−ζ(2ςτς+τ)∫τςϑ(w)w2dw≤(τ−ς)(τ2ζ′(τ)−ς2ζ′(ς))4ςτ∫τςϑ(w)w2dw. | (2.41) |
From (2.17) and (2.41) we get (2.21).
Corollary 1. Suppose that the assumption of Theorem 17 are satisfiedand ϑ(w)=ςττ−ς, w∈[ς,τ], then
(i) The inequalities
ζ(2ςτς+τ)≤2∫4ςτ3ς+τ4ςτς+3τζ(w)w2≤∫10S(ι)dι≤12[ζ(2ςτς+τ)+ςττ−ς∫τςζ(w)w2dw] | (2.42) |
holds.
(ii) The inequalities
0≤ςττ−ς∫τςζ(w)w2dw−S(ι)≤(1−ι)[ζ(ς)+ζ(τ)2−ςττ−ς∫τςζ(w)w2dw] | (2.43) |
hold for all ι∈[0,1].
(iii) The inequalities
0≤ζ(ς)+ζ(τ)2−S(ι)≤(τ−ς)(τ2ζ′(τ)−ς2ζ′(ς))4ςτ | (2.44) |
are valid for all ι∈[0,1].
In the following theorems, we discuss inequalities for the functions S, Sϑ, ˘G1, T and Tϑ as considered above:
Theorem 18. Let ζ, ϑ, ˘G1, Sϑ be defined as above. Then
(i) The inequality
Sϑ(ι)≤˘G1(ι)∫τςϑ(w)w2dw | (2.45) |
holds for all ι∈[0,1].
(ii) The inequalities
2∫4ςτ3ς+τ4ςτς+3τζ(w)ϑ(2ςτw4ςτ−(ς+τ)w)ϑ(w)w2dw≤12[ζ(4ςτ3ς+τ)+ζ(4ςτς+3τ)]∫τςϑ(w)w2dw≤τ−ςςτ∫10λ1(ι)ϑ(ςτ(1−ι)ς+ιτ)dι≤12[ζ(2ςτς+τ)+ζ(ς)+ζ(τ)2]∫τςϑ(w)w2dw | (2.46) |
hold.
(iii) If ζ is differentiable on [ς,τ] and ϑ isbounded on [ς,τ], then, forall ι∈[0,1], then
0≤Sϑ(ι)−ζ(2ςτς+τ)∫τςϑ(w)w2dw≤(τ−ςςτ)[˘G1(ι)−S(ι)]‖ϑ‖∞, | (2.47) |
where ‖ϑ‖∞=supw∈[ς,τ]ϑ(w).
Proof. (i) Using integration by substitution and the assumptions on ϑ, we have that the following identity holds on [0,1]:
˘G1(ι)∫τςϑ(w)w2dw=∫2ςτς+τς[ζ(2ςτ2ςι+(1−ι)(ς+τ))+ζ(2ςτ2τι+(1−ι)(ς+τ))]ϑ(w)w2dw. | (2.48) |
By Lemma 2, the following inequality holds for all w∈[ς,2ςτς+τ] with
w1=2ςτw2ςτι+(1−ι)(ς+τ)w,w2=2ςτw2ι(ςw+τw−ςτ)+(1−ι)(ς+τ)w,˘ϰ1=2ςτ2τι+(1−ι)(ς+τ)and˘ϰ2=2ςτ2ςι+(1−ι)(ς+τ): |
ζ(2ςτw2ςτι+(1−ι)(ς+τ)w)+ζ(2ςτw2ι(ςw+τw−ςτ)+(1−ι)(ς+τ)w)≤ζ(2ςτ2τι+(1−ι)(ς+τ))+ζ(2ςτ2ςι+(1−ι)(ς+τ)). | (2.49) |
Multiplying both sides of (2.49) with ϑ(w)w2, integrating over [ς,2ςτς+τ] and using (2.34) and (2.49), we obtain (2.45).
(ii) We can observe that
12[ζ(4ςτ3ς+τ)+ζ(4ςτς+3τ)]∫τςϑ(w)w2dw=[ζ(4ςτ3ς+τ)+ζ(4ςτς+3τ)]∫2ςτς+τςϑ(w)w2dw. | (2.50) |
By using harmonic symmetric assumption on ϑ, we get
2∫4ςτ3ς+τ4ςτς+3τζ(w)ϑ(2ςτw4ςτ−(ς+τ)w)ϑ(w)w2dw=∫2ςτς+τς[ζ(4ςτw2ςτ+(ς+τ)w)+ζ(4ςτw3(ς+τ)w−2ςτ)]ϑ(w)w2dw. | (2.51) |
We can also see that the following identity holds:
τ−ςςτ∫10˘G1(ι)ϑ(ςτ(1−ι)ς+ιτ)dι=τ−ςςτ×[∫112ζ(12ςι+(1−ι)(ς+τ))ϑ(ςτ(1−ι)ς+ιτ)dι+∫120ζ(12ςι+(1−ι)(ς+τ))ϑ(ςτ(1−ι)τ+ις)dι+∫120ζ(12τι+(1−ι)(ς+τ))ϑ(ςτ(1−ι)τ+ις)dι+∫112ζ(12τι+(1−ι)(ς+τ))ϑ(ςτ(1−ι)ς+ιτ)dι]=12∫2ςτς+τς[ζ(2wςς+w)+ζ(2ςτw2ςw+τw−ςτ)+ζ(2τwτ+w)+ζ(2ςτwςw+2τw−ςτ)]ϑ(w)w2dw. | (2.52) |
Finally we also have
12[ζ(2ςτς+τ)+ζ(ς)+ζ(τ)2]∫τςϑ(w)w2dw=[ζ(2ςτς+τ)+ζ(ς)+ζ(τ)2]∫2ςτς+τςϑ(w)w2dw. | (2.53) |
By Lemma 2, the following inequalities hold for all w∈[ς,2ςτς+τ]:
The inequality
ζ(4ςτw2ςτ+(ς+τ)w)+ζ(4ςτw3(ς+τ)w−2ςτ)≤ζ(4ςτ3ς+τ)+ζ(4ςτς+3τ) | (2.54) |
holds with the choices of w1=4ςτw2ςτ+(ς+τ)w, w2=4ςτw3(ς+τ)w−2ςτ, ˘ϰ1=4ςτ3ς+τ and ˘ϰ2=4ςτς+3τ.
The inequality
ζ(4ςτ3ς+τ)≤12[ζ(2ςτw2ςw+τw−ςτ)+ζ(2ςwς+w)] | (2.55) |
holds with the choices of w1=w2=4ςτ3ς+τ, ˘ϰ1=2ςwς+w, ˘ϰ2=2ςτw2ςw+τw−ςτ.
The inequality
ζ(4ςτς+3τ)≤12[ζ(2ςτwςw+2τw−ςτ)+ζ(2τwτ+w)] | (2.56) |
holds with the choices of w1=w2=4ςτς+3τ, ˘ϰ1=2τwτ+w, ˘ϰ2=2ςτwςw+2τw−ςτ.
The inequality
ζ(2ςτw2ςw+τw−ςτ)+ζ(2ςwς+w)≤ζ(ς)+ζ(2τςς+τ) | (2.57) |
holds with the choices of w1=2ςwς+w, w2=2ςτw2ςw+τw−ςτ, ˘ϰ1=ς, ˘ϰ2=2τςς+τ.
The inequality
ζ(2ςτwςw+2τw−ςτ)+ζ(2τwτ+w)≤ζ(2τςς+τ)+ζ(τ) | (2.58) |
holds with the choices of w1=2τwτ+w, w2=2ςτwςw+2τw−ςτ, ˘ϰ1=2τςς+τ, ˘ϰ2=τ.
Multiplying (2.54)–(2.58) by ϑ(w), integrating them over [ς,2ςτς+τ] and using (2.50)–(2.53), we get (2.46).
(iii) By integration by parts, we get
ι∫ς+τ2ςτ1τ[(w−ς+τ2ςτ)˘h′(ιw+(1−ι)(ς+τ2ςτ))+(ς+τ2ςτ−w)˘h′(ι(1ς+1τ−w)+(1−ι)(ς+τ2ςτ))]dw=ι∫1ς1τ(w−ς+τ2ςτ)˘h′(ιw+(1−ι)(ς+τ2ςτ))dw=τ−ς2ςτ[ζ(2ςτ2τι+(1−ι)(ς+τ))+ζ(2ςτ2ςι+(1−ι)(ς+τ))]−∫τς1w2ζ(2ςτw2ςτι+(1−ι)(ς+τ))dw=(τ−ςςτ)[˘G1(ι)−S(ι)]. | (2.59) |
Using the convexity of ˘h and the hypothesis of ϑ, the inequality holds for all ι∈[0,1] and w∈[1τ,ς+τ2ςτ]:
[˘h(ιw+(1−ι)(ς+τ2ςτ))−˘h(ς+τ2ςτ)]ϑ(1w)+[˘h(ι(1ς+1τ−w)+(1−ι)(ς+τ2ςτ))−˘h(ς+τ2ςτ)]ϑ(1w)≤ι(w−ς+τ2ςτ)˘h′(ιw+(1−ι)(ς+τ2ςτ))ϑ(1w)+ι(ς+τ2ςτ−w)˘h′(ι(1ς+1τ−w)+(1−ι)(ς+τ2ςτ))ϑ(1w)=ι(ς+τ2ςτ−w)[˘h′(ι(1ς+1τ−w)+(1−ι)(ς+τ2ςτ))−˘h′(ιw+(1−ι)(ς+τ2ςτ))]ϑ(1w)≤ι(ς+τ2ςτ−w)[˘h′(ι(1ς+1τ−w)+(1−ι)(ς+τ2ςτ))−˘h′(ιw+(1−ι)(ς+τ2ςτ))]‖ϑ‖∞. | (2.60) |
Integrating (2.60), using (2.59) and (2.17), we get (2.47).
Corollary 2. According to the assumptions of Theorem 18 with ϑ(w)=ςττ−ς, w∈[ς,τ], then
(i) The inequality
S(ι)≤˘G1(ι) |
holds for all ι∈[0,1].
(ii) The inequalities
2ςττ−ς∫4ςτ3ς+τ4ςτς+3τζ(w)ϑ(2ςτw4ςτ−(ς+τ)w)dww2≤12[ζ(4ςτ3ς+τ)+ζ(4ςτς+3τ)]≤τ−ςςτ∫10˘G1(ι)×ϑ(ςτ(1−ι)ς+ιτ)dι≤12[ζ(2ςτς+τ)+ζ(ς)+ζ(τ)2] | (2.61) |
hold.
(iii) The inequality
0≤S(ι)−ζ(2ςτς+τ)dw≤(τ−ςςτ)[˘G1(ι)−S(ι)] | (2.62) |
holds for all ι∈[0,1].
Theorem 19. Let ζ, ϑ, ˘G1, Sϑ, Tϑ bedefined as above. Then
(i) Tϑ is harmonic convex on (0,1].
(ii) The inequalities
˘G1(ι)∫τςϑ(w)w2dw≤Tϑ(ι)≤(1−ι)∫τςζ(w)ϑ(w)w2dw+ι⋅ζ(ς)+ζ(τ)2∫τςϑ(w)w2dw≤ζ(ς)+ζ(τ)2∫τςϑ(w)w2dw, | (2.63) |
Sϑ(1−ι)≤Tϑ(ι) | (2.64) |
and
Sϑ(ι)+Sϑ(1−ι)2≤Tϑ(ι) | (2.65) |
hold for all \mathfrak{\iota}\in\left[ 0, 1\right] .
(iii) The following bound is true:
\begin{equation} \underset{\mathfrak{\iota}\in\left[ 0,1\right] }{\sup}\mathbb{T} _{\mathbb{\vartheta}}\left( \mathfrak{\iota}\right) = \frac{\mathbb{\zeta }\left( \mathfrak{\varsigma}\right) +\mathbb{\zeta}\left( \mathbb{\tau }\right) }{2}\int_{\mathfrak{\varsigma}}^{\mathbb{\tau}}\frac {\mathbb{\vartheta}\left( \mathfrak{w}\right) }{\mathfrak{w}^{2} }d\mathfrak{w}. \end{equation} | (2.66) |
Proof. (i) Since \mathbb{\zeta} is harmonic convex and \mathbb{\vartheta} is nonnegative, we see that \mathbb{T}_{\mathbb{\vartheta}} is harmonic convex on \left(0, 1\right] .
(ii) We observe that the following identity holds on \left[ 0, 1\right] :
\begin{array}{l} \mathbb{T}_{\mathbb{\vartheta}}\left( \mathfrak{\iota}\right) = \frac{1} {2}\int_{\mathfrak{\varsigma}}^{\frac{2\mathfrak{\varsigma}\mathbb{\tau} }{\mathfrak{\varsigma}+\mathbb{\tau}}}\left[ \mathbb{\zeta}\left( \frac{\mathbb{\tau}\mathfrak{w}}{\mathfrak{\iota w}+\left( 1-\mathfrak{\iota }\right) \mathbb{\tau}}\right) +\mathbb{\zeta}\left( \frac {\mathfrak{\varsigma}\mathbb{\tau}\mathfrak{w}}{\mathfrak{\varsigma w\iota }+\left( 1-\mathfrak{\iota}\right) \left( \mathfrak{\varsigma w}+\mathbb{\tau}\mathfrak{w}-\mathfrak{\varsigma}\mathbb{\tau}\right) }\right) \right. \\ \left. +\mathbb{\zeta}\left( \frac{\mathfrak{\varsigma w}}{\mathfrak{\iota w}+\left( 1-\mathfrak{\iota}\right) \mathfrak{\varsigma}}\right) +\mathbb{\zeta}\left( \frac{\mathfrak{\varsigma}\mathbb{\tau}\mathfrak{w} }{\mathbb{\tau}\mathfrak{w\iota}+\left( 1-\mathfrak{\iota}\right) \left( \mathfrak{\varsigma w}+\mathbb{\tau}\mathfrak{w}-\mathfrak{\varsigma }\mathbb{\tau}\right) }\right) \right] \mathbb{\vartheta}\left( \mathfrak{w}\right) d\mathfrak{w}. \end{array} | (2.67) |
By Lemma 2, the following inequalities hold for all \mathfrak{w} \in\left[ \mathfrak{\varsigma}, \frac{2\mathfrak{\varsigma}\mathbb{\tau} }{\mathfrak{\varsigma}+\mathbb{\tau}}\right] :
\begin{array}{l} 2\mathbb{\zeta}\left( \frac{2\mathfrak{\varsigma}\mathbb{\tau}} {2\mathfrak{\varsigma\iota}+\left( 1-\mathfrak{\iota}\right) \left( \mathfrak{\varsigma}+\mathbb{\tau}\right) }\right) \\ \leq\mathbb{\zeta}\left( \frac{\mathbb{\tau}\mathfrak{w}}{\mathfrak{\iota w}+\left( 1-\mathfrak{\iota}\right) \mathbb{\tau}}\right) +\mathbb{\zeta }\left( \frac{\mathfrak{\varsigma}\mathbb{\tau}\mathfrak{w}} {\mathfrak{\varsigma w\iota}+\left( 1-\mathfrak{\iota}\right) \left( \mathfrak{\varsigma w}+\mathbb{\tau}\mathfrak{w}-\mathfrak{\varsigma }\mathbb{\tau}\right) }\right) \end{array} | (2.68) |
with
\begin{align*} \mathfrak{w}_{1} & = \mathfrak{w}_{2} = \frac{2\mathfrak{\varsigma} \mathbb{\tau}}{2\mathfrak{\varsigma\iota}+\left( 1-\mathfrak{\iota}\right) \left( \mathfrak{\varsigma}+\mathbb{\tau}\right) }{, }\breve{\varkappa }_{1} = \frac{\mathbb{\tau}\mathfrak{w}}{\mathfrak{\iota w}+\left( 1-\mathfrak{\iota}\right) \mathbb{\tau}}\\ {\rm{and }}\;\breve{\varkappa}_{2} & = \frac{\mathfrak{\varsigma}\mathbb{\tau }\mathfrak{w}}{\mathfrak{\varsigma w\iota}+\left( 1-\mathfrak{\iota}\right) \left( \mathfrak{\varsigma w}+\mathbb{\tau}\mathfrak{w}-\mathfrak{\varsigma }\mathbb{\tau}\right) }. \end{align*} |
\begin{array}{l} 2\mathbb{\zeta}\left( \frac{2\mathfrak{\varsigma}\mathbb{\tau}} {2\mathbb{\tau}\mathfrak{\iota}+\left( 1-\mathfrak{\iota}\right) \left( \mathfrak{\varsigma}+\mathbb{\tau}\right) }\right) \\ \leq\mathbb{\zeta}\left( \frac{\mathfrak{\varsigma w}}{\mathfrak{\iota w}+\left( 1-\mathfrak{\iota}\right) \mathfrak{\varsigma}}\right) +\mathbb{\zeta}\left( \frac{\mathfrak{\varsigma}\mathbb{\tau}\mathfrak{w} }{\mathbb{\tau}\mathfrak{w\iota}+\left( 1-\mathfrak{\iota}\right) \left( \mathfrak{\varsigma w}+\mathbb{\tau}\mathfrak{w}-\mathfrak{\varsigma }\mathbb{\tau}\right) }\right) \end{array} | (2.69) |
with
\begin{align*} \mathfrak{w}_{1} & = \mathfrak{w}_{2} = \frac{2\mathfrak{\varsigma} \mathbb{\tau}}{2\mathbb{\tau}\mathfrak{\iota}+\left( 1-\mathfrak{\iota }\right) \left( \mathfrak{\varsigma}+\mathbb{\tau}\right) }{, } \breve{\varkappa}_{1} = \frac{\mathfrak{\varsigma w}}{\mathfrak{\iota w}+\left( 1-\mathfrak{\iota}\right) \mathfrak{\varsigma}}\\ {\rm{and }}\;\breve{\varkappa}_{2} & = \frac{\mathfrak{\varsigma}\mathbb{\tau }\mathfrak{w}}{\mathbb{\tau}\mathfrak{w\iota}+\left( 1-\mathfrak{\iota }\right) \left( \mathfrak{\varsigma w}+\mathbb{\tau}\mathfrak{w} -\mathfrak{\varsigma}\mathbb{\tau}\right) }. \end{align*} |
Multiplying the inequalities (2.68) and (2.69) by \mathbb{\vartheta }\left(\mathfrak{w}\right) , integrating them over \mathfrak{w} on \left[ \mathfrak{\varsigma}, \frac{2\mathfrak{\varsigma}\mathbb{\tau} }{\mathfrak{\varsigma}+\mathbb{\tau}}\right] and using identities (2.48) and (2.67), we derive the first inequality of (2.63). Using the harmonic convexity of \mathbb{\zeta} and the inequality (2.17), the last part of (2.63) holds. Using again the harmonic convexity of \mathbb{\zeta} , we get
\begin{array}{l} \mathbb{S}_{\mathbb{\vartheta}}\left( 1-\mathfrak{\iota}\right) = { \int\limits_{\mathfrak{\varsigma}}^{\mathbb{\tau}}} \mathbb{\zeta}\left( \frac{2\mathfrak{\varsigma}\mathbb{\tau}\mathfrak{w} }{2\mathfrak{\varsigma}\mathbb{\tau}\left( 1-\mathfrak{\iota}\right) +\mathfrak{\iota}\left( \mathfrak{\varsigma}+\mathbb{\tau}\right) \mathfrak{w}}\right) \frac{\mathbb{\vartheta}\left( \mathfrak{w}\right) }{\mathfrak{w}^{2}}d\mathfrak{w} \\ = { \int\limits_{\mathfrak{\varsigma}}^{\mathbb{\tau}}} \mathbb{\zeta}\left( \frac{1}{\frac{1}{2}\left( \frac{\mathfrak{\iota w}+\left( 1-\mathfrak{\iota}\right) \mathfrak{\varsigma}} {\mathfrak{\varsigma w}}\right) +\frac{1}{2}\left( \frac{\mathfrak{\iota w}+\left( 1-\mathfrak{\iota}\right) \mathbb{\tau}}{\mathbb{\tau} \mathfrak{w}}\right) }\right) \frac{\mathbb{\vartheta}\left( \mathfrak{w} \right) }{\mathfrak{w}^{2}}d\mathfrak{w}\\ \leq\frac{1}{2} { \int\limits_{\mathfrak{\varsigma}}^{\mathbb{\tau}}} \left[ \mathbb{\zeta}\left( \frac{\mathfrak{\varsigma w}}{\mathfrak{\iota w}+\left( 1-\mathfrak{\iota}\right) \mathfrak{\varsigma}}\right) +\mathbb{\zeta}\left( \frac{\mathfrak{\varsigma w}}{\mathfrak{\iota w}+\left( 1-\mathfrak{\iota}\right) \mathfrak{\varsigma}}\right) \right] \frac{\mathbb{\vartheta}\left( \mathfrak{w}\right) }{\mathfrak{w}^{2} }d\mathfrak{w} = \mathbb{T}_{\mathbb{\vartheta}}\left( \mathfrak{\iota}\right) . \end{array} | (2.70) |
From (2.45), (2.63) and 2.70), we get (2.65).
(iii) (2.66) holds due to the inequality (2.63).
The subject of mathematical inequalities using convex functions has been seen to be an emerging topic during the past more than three decades. The researchers are trying to find new generalizations of convex functions and as a result new results are being adding to the theory of inequalities. In the current research we have used harmonic convex functions to generalize a number of results that hold for convex functions. In order to get the novel results in this study, we defined some new mappings over the interval [0, 1] . We have discussed some interesting properties of these mappings and obtained new refinements of the Hermite-hadamard and Fejér type inequalities already proven for harmonic convex functions. We believe that the results of this paper could be a source of inspiration for mathematicians working in this field and young researchers thinking to start their career in this fascinating field of mathematics.
The author is very thankful to all the anonymous referees for their very useful and constructive comments in order to present the paper in the present form. This work is supported by the Deanship of Scientific Research, King Faisal University under the Nasher Track 2021 (Research Project Number NA000177) which has been converted to Ambitious Researcher Track (Research Project Number GRANT931).
The authors declare that there are no conflicts of interest regarding the publication of this article.
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