In this paper, we introduce the concept of hyper homomorphisms and hyper derivations in Banach algebras and we establish the stability of hyper homomorphisms and hyper derivations in Banach algebras for the following 3-additive functional equation:
g(x1+x2,y1+y2,z1+z2)=2∑i,j,k=1g(xi,yj,zk).
Citation: Yamin Sayyari, Mehdi Dehghanian, Choonkil Park, Jung Rye Lee. Stability of hyper homomorphisms and hyper derivations in complex Banach algebras[J]. AIMS Mathematics, 2022, 7(6): 10700-10710. doi: 10.3934/math.2022597
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In this paper, we introduce the concept of hyper homomorphisms and hyper derivations in Banach algebras and we establish the stability of hyper homomorphisms and hyper derivations in Banach algebras for the following 3-additive functional equation:
g(x1+x2,y1+y2,z1+z2)=2∑i,j,k=1g(xi,yj,zk).
Let I⊆R be an interval. Then a real-valued function f:I→R is said to be convex (concave) if the inequality
f[λx+(1−λ)y]≤(≥)λf(x)+(1−λ)f(y) |
holds whenever x,y∈I and λ∈[0,1]. It is well-know that the convexity (concavity) has wild applications in pure and applied mathematics [1,2,3,4,5], and many inequalities [6,7,8,9,10,11,12,13,14,15,16,17,18,19] can be found in the literature via the convexity theory. Recently, the generalizations and variants for the convexity have attracted the attention of the researchers, for example, the GG- and GA-convexity [20], h-convexity [21], qausi-convexity [22], ρ-convexity [23], exponential convexity [24], harmonic convexity [25], s-convexity [26,27] and others.
The classical Hermite-Hadamard inequality [28,29,30,31,32,33] is one of the most famous inequalities in convex function theory, which can be stated as follows:
A real-valued function ψ:[b1,b2]→R is convex if and only if
ψ(b1+b22)≤1b2−b1∫b2b1ψ(x)dx≤ψ(b1)+ψ(b2)2. | (1.1) |
If ψ is concave, then the inequalities in (1.1) remain valid in the reversed direction. The Hermite-Hadamard inequality (1.1) provides both upper and lower estimates of the integral mean of any convex function defined on a closed and bounded interval involving the endpoints and midpoints of the function's domain. Because of the excellent significance of the Hermite-Hadamard inequality, the literature is replete with ample amount of research articles dedicated to the generalizations, refinements, and extensions for the Hermite-Hadamard inequality for various families of convexity.
Besides generalization via convexity, great effort has gone into extending (1.1) by means of fractional integral operators. Most popular of them is the Riemann-Liouville fractional integral operators given in the following definition.
Definition 1. Let α>0, b1,b2∈R with b1<b2 and ψ∈L[b1,b2]. Then the left and right Riemann-Liouville fractional integrals Jαb1+ψ and Jαb2−ψ of order α are defined by
Jαb1+ψ(x)=1Γ(α)∫xb1(x−t)α−1ψ(t)dt(x>b1) |
and
Jαb2−ψ(x)=1Γ(α)∫b2x(t−x)α−1ψ(t)dt(x<b2) |
respectively where Γ(α)=∫∞0e−ttα−1dt is the gamma function.
Sarikaya et al. [34] established the following fractional version of the Hermite-Hadamard inequality:
Theorem 2. Let 0≤b1<b2 and ψ:[b1,b2]→R be a positive convex function such that ψ∈L[b1,b2]. Then the fractional integrals inequality
ψ(b1+b22)≤Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]≤ψ(b1)+ψ(b2)2 |
holds for α>0.
In view of Theorem 2, it is pertinent to note that the positivity of the function ψ and the numbers b1 and b2 is not necessary. From Definition 1, it is clear that b1 and b2 are any real numbers such that b1<b2.
Our main contribution in this article is to use a similar technique used in [35] to obtain Theorem 2. This time, our main result contains both sided fractional integral operators in the Riemann-Liouville sense. In this method, we use a green function and in the process, we obtain some identities involving the left and right Riemann-Liouville fractional integral operators. These identities are subsequently employed to establish some new results for the class of convex, concave and monotone functions.
Our main result will be anchored on the succeeding lemma.
Lemma 3. (See [36,37]) Let G be the green function defined on [b1,b2]×[b1,b2] by
G(λ,μ)={b1−μ,b1≤μ≤λ;b1−λ,λ≤μ≤b2. |
Then any ψ∈C2([b1,b2]) can be expressed as
ψ(x)=ψ(b1)+(x−b1)ψ′(b2)+∫b2b1G(x,μ)ψ″(μ)dμ. | (2.1) |
We are now in a position to frame and prove our results.
Theorem 4. Let ψ∈C2([b1,b2]) be a convex function. Then, for any α>0, the following fractional integral inequalities hold:
ψ(b1+b22)≤Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]≤ψ(b1)+ψ(b2)2. |
Proof. Setting x=b1+b22 in (2.1), we get
ψ(b1+b22)=ψ(b1)+(b1+b22−b1)ψ′(b2)+∫b2b1G(b1+b22,μ)ψ″(μ)dμ. |
Equivalently,
ψ(b1+b22)=ψ(b1)+(b2−b12)ψ′(b2)+∫b2b1G(b1+b22,μ)ψ″(μ)dμ. | (2.2) |
Using (2.1), we do the following computations:
Jαb1+ψ(b2)=1Γ(α)∫b2b1(b2−x)α−1ψ(x)dx=1Γ(α)∫b2b1(b2−x)α−1{ψ(b1)+(x−b1)ψ′(b2)+∫b2b1G(x,μ)ψ″(μ)dμ}dx=1Γ(α){ψ(b1)∫b2b1(b2−x)α−1dx+ψ′(b2)∫b2b1(b2−x)α−1(x−b1)dx+∫b2b1∫b2b1(b2−x)α−1G(x,μ)ψ″(μ)dμdx}=1Γ(α)[ψ(b1)(b2−x)α−α|b2b1+ψ′(b2){(x−b1)(b2−x)α−α|b2b1−∫b2b1(b2−x)α−αdx}+∫b2b1∫b2b1(b2−x)α−1G(x,μ)ψ″(μ)dμdx]=1Γ(α)[ψ(b1)(b2−b1)αα+ψ′(b2){0−(b2−x)α+1α(α+1)|b2b1}+∫b2b1∫b2b1(b2−x)α−1G(x,μ)ψ″(μ)dμdx]. |
Therefore,
Jαb1+ψ(b2)=1Γ(α)[(b2−b1)ααψ(b1)+ψ′(b2)(b2−b1)α+1α(α+1)+∫b2b1∫b2b1(b2−x)α−1G(x,μ)ψ″(μ)dμdx]. | (2.3) |
Similarly,
Jαb2−ψ(b1)=1Γ(α)∫b2b1(x−b1)α−1ψ(x)dx=1Γ(α)∫b2b1(x−b1)α−1{ψ(b1)+(x−b1)ψ′(b2)+∫b2b1G(x,μ)ψ″(μ)dμ}dx=1Γ(α){ψ(b1)∫b2b1(x−b1)α−1dx+ψ′(b2)∫b2b1(x−b1)α−1(x−b1)dx+∫b2b1∫b2b1(x−b1)α−1G(x,μ)ψ″(μ)dμdx}=1Γ(α)[ψ(b1)(x−b1)αα|b2b1+ψ′(b2)∫b2b1(x−b1)αdx+∫b2b1∫b2b1(x−b1)α−1G(x,μ)ψ″(μ)dμdx]=1Γ(α)[ψ(b1)(b2−b1)αα+ψ′(b2)(x−b1)α+1α+1|b2b1+∫b2b1∫b2b1(x−b1)α−1G(x,μ)ψ″(μ)dμdx]. |
So,
Jαb2−ψ(b1)=1Γ(α)[(b2−b1)ααψ(b1)+ψ′(b2)(b2−b1)α+1α+1+∫b2b1∫b2b1(x−b1)α−1G(x,μ)ψ″(μ)dμdx]. | (2.4) |
Now, adding (2.3) and (2.4) and then multiplying the resultant sum by Γ(α+1)2(b2−b1)α to get:
Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]=Γ(α+1)2(b2−b1)α1Γ(α)[(b2−b1)ααψ(b1)+ψ′(b2)(b2−b1)α+1α(α+1)+∫b2b1∫b2b1(b2−x)α−1G(x,μ)ψ″(μ)dμdx+(b2−b1)ααψ(b1)+ψ′(b2)(b2−b1)α+1α+1+∫b2b1∫b2b1(x−b1)α−1G(x,μ)ψ″(μ)dμdx]=α2(b2−b1)α[2(b2−b1)ααψ(b1)+ψ′(b2)(b2−b1)α+1(α+1){1α+1}+∫b2b1∫b2b1(b2−x)α−1G(x,μ)ψ″(μ)dμdx+∫b2b1∫b2b1(x−b1)α−1G(x,μ)ψ″(μ)dμdx]=ψ(b1)+ψ′(b2)(b2−b1)2+α2(b2−b1)α[∫b2b1∫b2b1(b2−x)α−1G(x,μ)ψ″(μ)dμdx+∫b2b1∫b2b1(x−b1)α−1G(x,μ)ψ″(μ)dμdx]. | (2.5) |
Subtracting (2.5) from (2.2), we obtain
ψ(b1+b22)−Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]=ψ(b1)+(b2−b12)ψ′(b2)+∫b2b1G(b1+b22,μ)ψ″(μ)dμ−ψ(b1)−ψ′(b2)(b2−b1)2−α2(b2−b1)α[∫b2b1∫b2b1(b2−x)α−1G(x,μ)ψ″(μ)dμdx+∫b2b1∫b2b1(x−b1)α−1G(x,μ)ψ″(μ)dμdx]=∫b2b1[G(b1+b22,μ)−α2(b2−b1)α{∫b2b1(b2−x)α−1G(x,μ)dx+∫b2b1(x−b1)α−1G(x,μ)dx}]ψ″(μ)dμ. |
By the definition of the Green function,
G(x,μ)={b1−μ,if b1≤μ≤xb1−x,if x≤μ≤b2, |
we obtain
∫b2b1(b2−x)α−1G(x,μ)dx=1α(α+1)[(b2−μ)α+1−(b2−b1)α+1] | (2.6) |
and
∫b2b1(x−b1)α−1G(x,μ)dx=1α(α+1){(α+1)(b1−μ)(b2−b1)α+(μ−b1)α+1}. | (2.7) |
Substituting Eqs (2.6) and (2.7) into (2.6), then we obtain
ψ(b1+b22)−Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]=∫b2b1[G(b1+b22,μ)−(b2−μ)α+12(α+1)(b2−b1)α+b2−b12(α+1)−b1−μ2−(μ−b1)α+12(α+1)(b2−b1)α]ψ″(μ)dμ. | (2.8) |
Let
f(μ)=G(b1+b22,μ)−(b2−μ)α+12(α+1)(b2−b1)α+b2−b12(α+1)−b1−μ2−(μ−b1)α+12(α+1)(b2−b1)α. | (2.9) |
Here,
G(b1+b22,μ)={b1−μ,b1≤μ≤b1+b22;b1−b22,b1+b22≤μ≤b2. | (2.10) |
Now, if b1≤μ≤b1+b22, then from (2.9) and (2.10), we have
f(μ)=b1−μ2−(b2−μ)α+12(α+1)(b2−b1)α+b2−b12(α+1)−(μ−b1)α+12(α+1)(b2−b1)α. |
f′(μ)=−12+(b2−μ)α2(b2−b1)α−(μ−b1)α2(b2−b1)α≤0. |
This shows that f is decreasing and f(b1)=0 then f(μ)≤0 for all μ∈[b1,b1+b22].
If, on the other hand, b1+b22≤μ≤b2, then
f(μ)=b1−b22−(b2−μ)α+12(α+1)(b2−b1)α+b2−b12(α+1)−b1−μ2−(μ−b1)α+12(α+1)(b2−b1)α=μ−b22−(b2−μ)α+12(α+1)(b2−b1)α+b2−b12(α+1)−(μ−b1)α+12(α+1)(b2−b1)α. |
Therefore,
f′(μ)=12+(b2−μ)α2(b2−b1)α−(μ−b1)α2(b2−b1)α,f″(μ)=−α(b2−μ)α−12(b2−b1)α−α(μ−b1)α−12(b2−b1)α≤0, |
which shows that f′ is decreasing and f′(b2)=0 and thus f′(μ)≥0. Therefore, f is increasing and f(b2)=0. Hence, f(μ)≤0 for all μ∈[b1+b22,b2]. Combining the two cases discussed above, we have that
f(μ)≤0for allμ∈[b1,b2]. |
Using (2.8), we deduce the first inequality:
ψ(b1+b22)≤Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]. |
For the right hand side of the inequality, we recall:
ψ(x)=ψ(b1)+(x−b1)ψ′(b2)+∫b2b1G(x,μ)ψ″(μ)dμ,ψ(b2)=ψ(b1)+(b2−b1)ψ′(b2)+∫b2b1G(b2,μ)ψ″(μ)dμ,ψ(b1)+ψ(b2)=2ψ(b1)+(b2−b1)ψ′(b2)+∫b2b1G(b2,μ)ψ″(μ)dμ,ψ(b1)+ψ(b2)2=ψ(b1)+(b2−b1)2ψ′(b2)+12∫b2b1G(b2,μ)ψ″(μ)dμ. | (2.11) |
Subtracting Eq (2.5) from (2.11), we get:
ψ(b1)+ψ(b2)2−Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]=ψ(b1)+(b2−b1)2ψ′(b2)+12∫b2b1G(b2,μ)ψ″(μ)dμ−ψ(b1)−ψ′(b2)(b2−b1)2−α2(b2−b1)α[∫b2b1∫b2b1(b2−x)α−1G(x,μ)ψ″(μ)dμdx+∫b2b1∫b2b1(x−b1)α−1G(x,μ)ψ″(μ)dμdx]=12∫b2b1[G(b2,μ)−α2(b2−b1)α∫b2b1(b2−x)α−1G(x,μ)dx+∫b2b1(x−b1)α−1G(x,μ)dx]ψ″(μ)dμ=12∫b2b1[G(b2,μ)−(b2−μ)α+1(α+1)(b2−b1)α+b2−b1(α+1)−b1+μ−(μ−b1)α+1(α+1)(b2−b1)α]ψ″(μ)dμ. | (2.12) |
If we set
F(μ)=G(b2,μ)−(b2−μ)α+1(α+1)(b2−b1)α+b2−b1(α+1)−b1+μ−(μ−b1)α+1(α+1)(b2−b1)α, | (2.13) |
then for b1≤μ≤b2,
F(μ)=b1−μ−(b2−μ)α+1(α+1)(b2−b1)α+b2−b1(α+1)−b1+μ−(μ−b1)α+1(α+1)(b2−b1)α=−(b2−μ)α+1(α+1)(b2−b1)α+b2−b1(α+1)−(μ−b1)α+1(α+1)(b2−b1)α=(b2−b1)α+1−(b2−μ)α+1−(μ−b1)α+1(α+1)(b2−b1)α. |
If b1≤μ≤b1+b22, then
F′(μ)=(b2−μ)α−(μ−b1)α(b2−b1)α≥0 |
which proves that F is increasing and F(b1)=0, and hence F(μ)≥0.
Suppose also b1+b22≤μ≤b2. Then,
F′(μ)=(b2−μ)α−(μ−b1)α(b2−b1)α≤0. |
This implies that F is a decreasing function and F(b2)=0, and thus F(μ)≥0. Also, ψ″(μ)≥0 since ψ is convex. Hence, F(μ)≥0 for all μ∈[b1,b2] and using (2.12) amounts to:
Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]≤ψ(b1)+ψ(b2)2. |
That completes the proof.
Next, we present new Hermite-Hadamard type inequalities for the class of monotone and convex function.
Theorem 5. Let ψ∈C2([b1,b2]) and α>0, Then the following statements are true:
(ⅰ) If |ψ″| is an increasing function, then
|ψ(b1)+ψ(b2)2−Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]|≤|ψ″(b2)|α(b2−b1)22(α+1)(α+2). |
(ⅱ) If |ψ″| is a decreasing function, then
|ψ(b1)+ψ(b2)2−Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]|≤|ψ″(b1)|α(b2−b1)22(α+1)(α+2). |
(ⅲ) If |ψ″| is a convex function, then
|ψ(b1)+ψ(b2)2−Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]|≤max{|ψ″(b1)|,|ψ″(b2)|}α(b2−b1)22(α+1)(α+2). |
Proof. To prove (ⅰ), we use the following identity obtained from (2.12):
ψ(b1)+ψ(b2)2−Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]=12∫b2b1[G(b2,μ)−(b2−μ)α+1(α+1)(b2−b1)α+b2−b1(α+1)−b1+μ−(μ−b1)α+1(α+1)(b2−b1)α]ψ″(μ)dμ. | (2.14) |
Taking absolute values on both sides of (2.14) and using the fact that |ψ″| is an increasing function, we have
|ψ(b1)+ψ(b2)2−Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]|≤|ψ″(b2)|12∫b2b1{−(b2−μ)α+1(α+1)(b2−b1)α+b2−b1α+1−(μ−b1)α+1(α+1)(b2−b1)α}dμ=|ψ″(b2)|12{(b2−μ)α+2(α+1)(α+2)(b2−b1)α|b2b1+b2−b1α+1μ|b2b1−(μ−b1)α+2(α+1)(α+2)(b2−b1)α|b2b1}=|ψ″(b2)|12{−(b2−b1)α+2(α+1)(α+2)(b2−b1)α+(b2−b1)2α+1−(b2−b1)α+2(α+1)(α+2)(b2−b1)α}=|ψ″(b2)|12{−2(b2−b1)α+2(α+1)(α+2)(b2−b1)α+(b2−b1)2α+1}=|ψ″(b2)|α(b2−b1)22(α+1)(α+2). |
Therefore,
|ψ(b1)+ψ(b2)2−Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]|≤|ψ″(b2)|α(b2−b1)22(α+1)(α+2). |
This establishes the inequality in (ⅰ).
Part (ⅱ) can be proved in a similar way. For (ⅲ), we make use of (2.13) and the fact that every convex function ψ defined on the interval [b1,b2] is bounded above by max{|ψ(b1)|,|ψ(b2)|} to get:
|ψ(b1)+ψ(b2)2−Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]|≤max{|ψ″(b1)|,|ψ″(b2)|}2(α+1)(b2−b1)α|∫b2b1[−(b2−μ)α+1+(b2−b1)α+1−(μ−b1)α+1]dμ|=max{|ψ″(b1)|,|ψ″(b2)|}α(b2−b1)22(α+1)(α+2). |
Remark 6. By setting α=1 in Theorem 5, we get the following inequalities:
|ψ(b1)+ψ(b2)2−1b2−b1∫b2b1ψ(t)dt|≤|ψ″(b2)|(b2−b1)212,|ψ(b1)+ψ(b2)2−1b2−b1∫b2b1ψ(t)dt|≤|ψ″(b1)|(b2−b1)212,|ψ(b1)+ψ(b2)2−1b2−b1∫b2b1ψ(t)dt|≤max{|ψ″(b1)|,|ψ″(b2)|}(b2−b1)212. |
Theorem 7. Let ψ∈C2([b1,b2]) and α>0. Then the following statements are true:
(i) If |ψ″| is an increasing function, then
|ψ(b1+b22)−Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]|≤(b2−b1)2(α2−α+2)16(α+1)(α+2)[|ψ″(b1+b22)|+|ψ″(b2)|]. |
(ii) If |ψ″| is a decreasing function, then
|ψ(b1+b22)−Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]|≤(b2−b1)2(α2−α+2)16(α+1)(α+2)[|ψ″(b1)|+|ψ″(b1+b22)|]. |
(iii) If |ψ″| is a convex function, then
|ψ(b1+b22)−Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]|≤(b2−b1)2(α2−α+2)16(α+1)(α+2)[max{|ψ″(b1)|,|ψ″(b1+b22)|}+{max|ψ″(b1+b22)|,|ψ″(b2)|}]. |
Proof. The inequality in (ⅰ) is obtained by using (2.8):
ψ(b1+b22)−Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]=∫b1+b22b1[G(b1+b22,μ)−(b2−μ)α+12(α+1)(b2−b1)α+b2−b12(α+1)−b1−μ2−(μ−b1)α+12(α+1)(b2−b1)α]ψ″(μ)dμ+∫b2b1+b22[G(b1+b22,μ)−(b2−μ)α+12(α+1)(b2−b1)α+b2−b12(α+1)−b1−μ2−(μ−b1)α+12(α+1)(b2−b1)α]ψ″(μ)dμ. |
Taking absolute values and using triangle inequality, we get
|ψ(b1+b22)−Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]|≤|ψ″(b1+b22)||∫b1+b22b1{b1−μ2−(b2−μ)α+12(α+1)(b2−b1)α+b2−b12(α+1)−(μ−b1)α+12(α+1)(b2−b1)α}dμ|+|ψ″(b2)||∫b2b1+b22{μ−b22−(b2−μ)α+12(α+1)(b2−b1)α+b2−b12(α+1)−(μ−b1)α+12(α+1)(b2−b1)α}dμ|=|ψ″(b1+b22)||{2b1μ−μ24|b1+b22b1+(b2−μ)α+22(α+1)(α+2)(b2−b1)α|b1+b22b1+(b2−b1)2(α+1)μ|b1+b22b1−(μ−b1)α+22(α+1)(α+2)(b2−b1)α|b1+b22b1}|+|ψ″(b2)||{μ2−2b2μ4|b2b1+b22+(b2−μ)α+22(α+1)(α+2)(b2−b1)α|b2b1+b22+b2−b12(α+1)μ|b2b1+b22−(μ−b1)α+22(α+1)(α+2)(b2−b1)α|b2b1+b22}|=|ψ″(b1+b22)||2b1(b1+b22)−(b1+b22)24−2b1(b1)−(b1)24+(b2−b1+b22)α+22(α+1)(α+2)(b2−b1)α−(b2−b1)α+22(α+1)(α+2)(b2−b1)α+(b2−b1)2(α+1)(b1+b22−b1)−(b1+b22−b1)α+22(α+1)(α+2)(b2−b1)α+(b1−b1)α+22(α+1)(α+2)(b2−b1)α|+|ψ″(b2)||(b2)2−2(b2)24−(b1+b22)2−2b2(b1+b22)4−(b2−b1+b22))α+22(α+1)(α+2)(b2−b1)α+b2−b12(α+1)(b2−b1+b22)−(b2−b1)α+22(α+1)(α+2)(b2−b1)α| |
=|ψ″(b1+b22)||−(b2−b1)216+(b2−b1)22α+3(α+1)(α+2)−(b2−b1)22(α+1)(α+2)+(b2−b1)24(α+1)−(b2−b1)22α+3(α+1)(α+2)|+|ψ″(b2)||−(b2−b1)216−(b2−b1)22α+3(α+1)(α+2)+(b2−b1)24(α+1)−(b2−b1)22(α+1)(α+2)+(b2−b1)22α+3(α+1)(α+2)|=|ψ″(b1+b22)||−(b2−b1)216−(b2−b1)22(α+1)(α+2)+(b2−b1)24(α+1)|+|ψ″(b2)||−(b2−b1)216+(b2−b1)24(α+1)−(b2−b1)22(α+1)(α+2)|=|ψ″(b1+b22)|(b2−b1)2|−α2+α−216(α+1)(α+2)|+|ψ″(b2)|(b2−b1)2|−α2+α−216(α+1)(α+2)|={|ψ″(b1+b22)|+|ψ″(b2)|}(b2−b1)2[α2−α+216(α+1)(α+2)]={|ψ″(b1+b22)|+|ψ″(b2)|}(b2−b1)2(α2−α+2)16(α+1)(α+2). |
The second part can be deduced by using the same procedure. For Part (ⅲ), we also employ the fact that the convex function ψ is bounded above by max{|ψ(b1)|,|ψ(b2)|} since it is defined on the interval [b1,b2]. That is, we obtain from (2.8):
|ψ(b1+b22)−Γ(α+1)2(b2−b1)α[Jαb1+(b2)+Jαb2−(b1)]|≤max{|ψ″(b1)|+|ψ″(b1+b22)|,|ψ″(b1+b22)|+|ψ″(b2)|}×(b2−b1)2(α2−α+2)16(α+1)(α+2)=[max{|ψ″(b1)|,|ψ″(b1+b22)|}+max{|ψ″(b1+b22)|,|ψ″(b2)|}]×(b2−b1)2(α2−α+2)16(α+1)(α+2) |
which is the desired inequality.
Remark 8. In Theorem 7, if we take α = 1, then we obtain
|ψ(b1+b22)−1b2−b1∫b2b1ψ(t)dt|≤(b2−b1)248[|ψ″(b1+b22)|+|ψ″(b2)|],|ψ(b1+b22)−1b2−b1∫b2b1ψ(t)dt|≤(b2−b1)248[|ψ″(b1)|+|ψ″(b1+b22)|],|ψ(b1+b22)−1b2−b1∫b2b1ψ(t)dt|≤(b2−b1)248[max{|ψ″(b1)|,|ψ″(b1+b22)|}+max{|ψ″(b1+b22)|,|ψ″(b2)|}]. |
Theorem 9. Assume that ψ∈C2([b1,b2]) and |ψ″| is a convex function. Then for any α>0 the following inequality holds
|Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]−ψ(b1+b22)|≤(b2−b1)2(α2−α+2)16(α+1)(α+2)[|ψ″(b1)|+|ψ″(b2)|]. |
Proof. The inequality in (ⅰ) is obtained by using (2.8):
Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]−ψ(b1+b22)=∫b1+b22b1[(b2−μ)α+12(α+1)(b2−b1)α−b2−b12(α+1)+μ−b12+(μ−b1)α+12(α+1)(b2−b1)α]ψ″(μ)dμ+∫b2b1+b22[(b2−μ)α+12(α+1)(b2−b1)α−b2−b12(α+1)+b2−μ2+(μ−b1)α+12(α+1)(b2−b1)α]ψ″(μ)dμ. |
Suppose μ=(1−t)b1+tb2 where dμ=(b2−b1)dt,
=∫120[(b2−(1−t)b1−tb2)α+12(α+1)(b2−b1)α−b2−b12(α+1)+(1−t)b1+tb2−b12+((1−t)b1+tb2−b1)α+12(α+1)(b2−b1)α]ψ″((1−t)b1+tb2)(b2−b1)dt+∫112[(b2−(1−t)b1−tb2)α+12(α+1)(b2−b1)α−b2−b12(α+1)+b2−(1−t)b1−tb22+((1−t)b1+tb2−b1)α+12(α+1)(b2−b1)α]ψ″((1−t)b1+tb2)(b2−b1)dt=(b2−b1)22(α+1)[∫120[(1−t)α+1−1+(α+1)t+tα+1]ψ″((1−t)b1+tb2)dt+∫112[(1−t)α+1+α(1−t)−t+tα+1]ψ″((1−t)b1+tb2)dt]. | (2.15) |
Taking absolute and triangular inequality,
≤(b2−b1)22(α+1)[∫120[(1−t)α+1−1+(α+1)t+tα+1](1−t)|ψ″(b1)|dt+∫120[(1−t)α+1−1+(α+1)t+tα+1]t|ψ″(b2)|dt+∫112[(1−t)α+1+α(1−t)−t+tα+1](1−t)|ψ″(b1)|dt+∫112[(1−t)α+1+α(1−t)−t+tα+1]t|ψ″(b2)|]dt]=(b2−b1)22(α+1)[|ψ″(b1)|{∫120((1−t)α+2−(1−t)+(α+1)(t−t2)+tα+1−tα+2)dt+∫112((1−t)α+2+α(1−t)2−(t−t2)+tα+1−tα+2)dt}+|ψ″(b2)|{∫120(t(1−t)α+1−t+(α+1)t2+tα+2)dt+∫112(t(1−t)α+1+α(t−t2)−t2+tα+2)dt}]=(b2−b1)22(α+1)[|ψ″(b1)|{I1}+|ψ″(b2)|{I2}]. | (2.16) |
Putting the values of I1 and I2 in above (2.16), we get:
=(b2−b1)2(α2−α+2)16(α+1)(α+2)[|ψ″(b1)|+|ψ″(b2)|]. |
Remark 10. In Theorem 9, if we take α=1, then we obtain
|ψ(b1+b22)−1b2−b1∫b2b1ψ(t)dt|≤(b2−b1)248[|ψ″(b1)|+|ψ″(b2)|]. |
Theorem 11. Assume that ψ∈C2([b1,b2]) and |ψ″| is a convex function. Then for any α>0 the following inequality holds
|ψ(b1)+ψ(b2)2−Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]|≤α(b2−b1)24(α+1)(α+2)[|ψ″(b1)|+|ψ″(b2)|]. |
Proof. We start by recalling the following identity from (2.12):
ψ(b1)+ψ(b2)2−Γ(α+1)(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]=12∫b2b1[G(b2,μ)−(b2−μ)α+1(α+1)(b2−b1)α+b2−b1(α+1)−b1+μ−(μ−b1)α+1(α+1)(b2−b1)α]ψ″(μ)dμ=12∫b2b1[b1−μ−(b2−μ)α+1(α+1)(b2−b1)α+b2−b1(α+1)−b1+μ−(μ−b1)α+1(α+1)(b2−b1)α]ψ″(μ)dμ=12∫b2b1[−(b2−μ)α+1(α+1)(b2−b1)α+b2−b1(α+1)−(μ−b1)α+1(α+1)(b2−b1)α]ψ″(μ)dμ. |
If μ=(1−t)b1+tb2 with t∈[0,1], then we obtain
ψ(b1)+ψ(b2)2−Γ(α+1)(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]=12∫10[−(b2−(1−t)b1−tb2)α+1(α+1)(b2−b1)α+b2−b1(α+1)−((1−t)b1+tb2−b1)α+1(α+1)(b2−b1)α]×ψ″((1−t)b1+tb2)(b2−b1)dt=12∫10[−(b2−b1)α+1(1−t)α+1(α+1)(b2−b1)α+b2−b1(α+1)−tα+1(b2−b1)α+1(α+1)(b2−b1)α]×ψ″((1−t)b1+tb2)(b2−b1)dt≤(b2−b1)22(α+1)∫10[−(1−t)α+1+1−tα+1][(1−t)|ψ″(b1)|+t|ψ″b2)|]dt=(b2−b1)22(α+1)[|ψ″(b1)|∫10{−(1−t)α+2+1−t−tα+1+tα+2}dt+|ψ″b2)|∫10{−t(1−t)α+1+t−tα+2}dt]. |
Taking absolute values and applying the triangle inequality amounts to the intended result.
Remark 12. Let α=1. Then the inequality in Theorem 11 becomes:
|ψ(b1)+ψ(b2)2−1b2−b1∫b2b1ψ(t)dt|≤(b2−b1)224[|ψ″(b1)|+|ψ″(b2)|]. |
Next, we present results associated with the concave functions.
Theorem 13. Let ψ∈C2([b1,b2]) and |ψ″| be a concave function. Then, for any α>0,
|ψ(b1+b22)−Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]|≤(b2−b1)2(α2−α+2)16(α+1)(α+2)[×|ψ″(b1(2+2α+3(α+2)2α+3(α+2)(α+3)+2α−724)+b2(2α+3−22α+3(α+2)(α+3)+α−224)α2−α+28(α+2))|+|ψ″(b1(α−224+2α+3−22α+3(α+2)(α+3))+b2(2α−724+2α+3(α+2)+22α+3(α+2)(α+3))α2−α+28(α+2))|]. |
Proof. From (2.15), we have:
Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]−ψ(b1+b22)=(b2−b1)22(α+1)[∫120[(1−t)α+1−1+(α+1)t+tα+1]ψ″((1−t)b1+tb2)dt+∫112[(1−t)α+1+α(1−t)−t+tα+1]ψ″((1−t)b1+tb2)dt]. |
Now, we use Jensen's integral inequality to get:
≤(b2−b1)22(α+1)[∫120[(1−t)α+1−1+(α+1)t+tα+1]dt×|ψ″(∫120((1−t)α+1−1+(α+1)t+tα+1)((1−t)b1+tb2)dt∫120(1−t)α+1−1+(α+1)t+tα+1)dt)|+∫112((1−t)α+1+α(1−t)−t+tα+1)dt×|ψ″(∫112((1−t)α+1+α(1−t)−t+tα+1)((1−t)b1+tb2)dt∫112((1−t)α+1+α(1−t)−t+tα+1)dt)|].=(b2−b1)22(α+1)[∫120[I1]dt|ψ″(∫120(I2)dt∫120(I1)dt)|+∫112(A1)dt|ψ″(∫112(A2)dt∫112(A1)dt)|]. | (2.17) |
For this we calculate:
I1=∫120[(1−t)α+1−1+(α+1)t+tα+1]dt=−(1−t)α+2α+2|120−t|120+(α+1)t22|120+tα+2α+2|120=−12α+2(α+2)+1α+2−12+α+18+12α+2(α+2)=α2−α+28(α+2), |
I2=∫120((1−t)α+1−1+(α+1)t+tα+1)((1−t)b1+tb2)dt=b1∫120((1−t)α+2−(1−t)+(α+1)(t−t2)+tα+1−tα+2)dt+b2∫120(t(1−t)α+1−t+(α+1)t2+tα+2)dt=b1(−(1−t)α+3α+3−t+t22+(α+1)(t22−t33)+tα+2α+2−tα+3α+3)|120+b2(−t(1−t)α+2α+2−∫120−(1−t)α+2α+2dt−t22+(α+1)t33+tα+3α+3)|120=b1(2+2α+3(α+2)2α+3(α+2)(α+3)+2α−724)+b2(2α+3−22α+3(α+2)(α+3)+α−224), |
A1=∫112((1−t)α+1+α(1−t)−t+tα+1)dt=−(1−t)α+2α+2|112+α(t−t22)|112−t22|112+tα+2α+2|112=α2−α+28(α+2) |
and
A2=∫112((1−t)α+1+α(1−t)−t+tα+1)((1−t)b1+tb2)dt=b1∫112((1−t)α+2+α(1−t)2−t+t2+tα+1−tα+2)dt+b2∫112(t(1−t)α+1+α(t−t2)−t2+tα+2)dt=b1[−(1−t)α+3α+3|112−α(1−t)33|112−t22|112+t33|112+tα+2α+2|112−tα+3α+3|112]+b2[−t(1−t)α+2α+2−∫112(1−t)α+2−(α+2)dt+α(t22−t33)−t33+tα+3α+3]112=b1[α−224+2α+3−22α+3(α+2)(α+3)]+b2[2α−724+2α+3(α+2)+22α+3(α+2)(α+3)]. |
Putting the values of I1, I2, A1 and A2 in (2.17), we obtain:
=(b2−b1)22(α+1)[α2−α+28(α+2)|ψ″(b1(2+2α+3(α+2)2α+3(α+2)(α+3)+2α−724)+b2(2α+3−22α+3(α+2)(α+3)+α−224)α2−α+28(α+2))|+α2−α+28(α+2)|ψ″(b1(α−224+2α+3−22α+3(α+2)(α+3))+b2(2α−724+2α+3(α+2)+22α+3(α+2)(α+3))α2−α+28(α+2))|]=(b2−b1)2(α2−α+2)16(α+1)(α+2)×[|ψ″(b1(2+2α+3(α+2)2α+3(α+2)(α+3)+2α−724)+b2(2α+3−22α+3(α+2)(α+3)+α−224)α2−α+28(α+2))|+|ψ″(b1(α−224+2α+3−22α+3(α+2)(α+3))+b2(2α−724+2α+3(α+2)+22α+3(α+2)(α+3))α2−α+28(α+2))|]. |
Remark 14. If we take α=1 in Theorem 13, then we get:
|ψ(b1+b22)−1b2−b1∫b2b1ψ(t)dt|≤(b2−b1)248[|ψ″(5b1+3b28)|+|ψ″(3b1+5b28)|]. |
Theorem 15. Let ψ∈C2([b1,b2]) and |ψ″| be a concave function. Then, for any α>0,
|ψ(b1)+ψ(b2)2−Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]|≤α(b2−b1)22(α+1)(α+2)|ψ″(b1+b22)|. |
Proof. For this, we use (2.14) to obtain the following inequality:
|ψ(b1)+ψ(b2)2−Γ(α+1)2(b2−b1)α[Jαb1+ψ(b2)+Jαb2−ψ(b1)]|=(b2−b1)22(α+1)∫10[−(1−t)α+1+1−tα+1][ψ″((1−t)b1+tb2)]dt. |
We use Jensen's integral inequality to get:
≤(b2−b1)22(α+1){∫10(−(1−t)α+1+1−tα+1)dt×|ψ″(∫10(−(1−t)α+1+1−tα+1)((1−t)b1+tb2)dt∫10(−(1−t)α+1+1−tα+1)dt)|} |
=(b2−b1)22(α+1){αα+2|ψ″(b1(α2(α+2))+b2(α2(α+2))αα+2)|}=(b2−b1)22(α+1){αα+2|ψ″(b1+b22)|}. |
Remark 16. If we set α=1, then Theorem 15 amounts to:
|ψ(b1)+ψ(b2)2−1b2−b1∫b2b1ψ(t)dt|≤(b2−b1)212|ψ″(b1+b22)|. |
By means of a green function, we outlined a new method of proving the Hermite-Hadamard inequality involving the Riemann-Liouville fractional integral operators. In the process of doing this, some new identities were obtained. We applied these identities to prove more results in this direction. Results established in this paper can be recast by using the green function G2(λ,μ) defined on [b1,b2]×[b1,b2] by
G2(λ,μ)={λ−b1,b1≤μ≤λ;μ−b1,λ≤μ≤b2. |
The authors would like to express their sincere thanks to the editor and the anonymous reviewers for their helpful comments and suggestions. The work was supported by the Natural Science Foundation of China (Grant Nos. 61673169, 11301127, 11701176, 11626101, 11601485).
There is no conflict of interest to report.
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