We introduce the class of ψ-convex functions f:[0,∞)→R, where ψ∈C([0,1]) satisfies ψ≥0 and ψ(0)≠ψ(1). This class includes several types of convex functions introduced in previous works. We first study some properties of such functions. Next, we establish a double Hermite-Hadamard-type inequality involving ψ-convex functions and a Simpson-type inequality for functions f∈C1([0,∞)) such that |f′| is ψ-convex. Our obtained results are new and recover several existing results from the literature.
Citation: Hassen Aydi, Bessem Samet, Manuel De la Sen. On ψ-convex functions and related inequalities[J]. AIMS Mathematics, 2024, 9(5): 11139-11155. doi: 10.3934/math.2024546
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We introduce the class of ψ-convex functions f:[0,∞)→R, where ψ∈C([0,1]) satisfies ψ≥0 and ψ(0)≠ψ(1). This class includes several types of convex functions introduced in previous works. We first study some properties of such functions. Next, we establish a double Hermite-Hadamard-type inequality involving ψ-convex functions and a Simpson-type inequality for functions f∈C1([0,∞)) such that |f′| is ψ-convex. Our obtained results are new and recover several existing results from the literature.
The class of convex functions is widely used in many branches of pure and applied mathematics. Due to this reason, we find in the literature several studies related to convex functions, see e.g., [1,2,3,4,5,6,7]. One of the most famous inequalities involving convex functions is the double Hermite-Hadamard inequality [8,9]:
f(a+b2)≤1b−a∫baf(x)dx≤f(a)+f(b)2, | (1.1) |
which holds for any convex function f:I→R and a,b∈I, where a<b and I is an interval of R. For more details about (1.1), see e.g., [10]. The double inequality (1.1) has been refined and extended to various classes of functions such as log-convex functions [11], s-logarithmically convex functions [12], hyperbolic p-convex functions [13], s-convex functions [14], convex functions on the co-ordinates [15], F-convex functions [16,17], h-convex and harmonically h-convex interval-valued functions [18], m-convex functions [19], (m1,m2)-convex functions [20], (α,m)-convex functions [21], (α,m1,m2)-convex functions [22], etc. In particular, when f:[0,∞)→R is m-convex (see Definition 2), where 0<m≤1, Dragomir [19] proved that for all a,b≥0 with 0≤a<b, we have
1b−a∫baf(x)dx≤12min{f(a)+mf(bm),f(b)+mf(am)} |
and
f(a+b2)≤1b−a∫baf(x)+mf(xm)2dx≤m+14(f(a)+f(b)2+mf(am)+f(bm)2). |
Notice that, if f is convex (so m=1, see Definition 2.2), the above double inequality reduces to (1.1).
Another important inequality, which is very useful in numerical integrations, is the Simpson's inequality
|1b−a∫baf(x)dx−13[f(a)+f(b)2+2f(a+b2)]|≤(b−a)42880‖f(4)‖∞, |
where f∈C4([a,b]) and ‖f(4)‖∞=maxa≤x≤b|f(4)(x)|. The above inequality has been studied in several papers for different classes of functions, see e.g., [23,24,25,26,27,28]. For instance, Dragomir [25], proved that, if f:[a,b]→R is a function of bounded variation, then
|1b−a∫baf(x)dx−13[f(a)+f(b)2+2f(a+b2)]|≤13Vba, |
where Vba denotes the total variation of f on [a,b].
Notice that it is always interesting to extend the above important inequalities to other classes of functions. Such extensions will be useful for example in numerical integrations and many other applications. Motivated by this fact, we introduce in this paper the class of ψ-convex functions f:[0,∞)→R, where ψ∈C([0,1]) is a function satisfying certain conditions. This class includes several types of convex functions from the literature: m-convex functions, (m1,m2)-convex functions, (α,m)-convex functions and (α,m1,m2)-convex functions. Moreover, after studying some properties of this introduced class of functions, we establish a double Hermite-Hadamard-type inequality involving ψ-convex functions and a Simpson-type inequality for functions f∈C1([0,∞)), where |f′| is ψ-convex. Our obtained results are new and recover several results from the literature.
Th rest of the paper is as follows. In Section 2, we introduce the class of ψ-convex functions and we study some properties of such functions. We also provide several examples of functions that belong to the introduced class. In Section 3, we establish Hermite-Hadamard-type inequalities involving ψ-convex functions. In Section 4, a Simpson-type inequality is proved for functions f∈C1([0,∞)), where |f′| is ψ-convex.
In this section, we introduce the class of ψ-convex functions. We first introduce the set
Ψ={ψ∈C([0,1]):ψ≥0,ψ(0)≠ψ(1)}. |
Definition 2.1. Let ψ∈Ψ. A function f:[0,∞)→R is said to be ψ-convex, if
f(ψ(0)tx+ψ(1)(1−t)y)≤ψ(0)ψ(t)−ψ(0)ψ(1)−ψ(0)f(x)+ψ(1)ψ(1)−ψ(t)ψ(1)−ψ(0)f(y) | (2.1) |
for all t∈[0,1] and x,y≥0.
We will show below that the introduced class of functions includes several classes of functions from the literature. Let us first start with some simple examples of ψ-convex functions.
Example 2.1. Let us consider the function ψ:[0,1]→R defined by
ψ(t)=at,0≤t≤1, |
where 0<a≤1 is a constant. Clearly, the function ψ∈Ψ and ψ(0)=0<a=ψ(1). On the other hand, for all t∈[0,1], we have
ψ(0)ψ(t)−ψ(0)ψ(1)−ψ(0)=0 |
and
ψ(1)ψ(1)−ψ(t)ψ(1)−ψ(0)=a(1−t). |
Let f:[0,∞)→R be the function defined by
f(x)=Ax2+Bx,x≥0, |
where A≥0 and B∈R are constants. For all t∈[0,1] and x,y≥0, we have
ψ(0)ψ(t)−ψ(0)ψ(1)−ψ(0)f(x)+ψ(1)ψ(1)−ψ(t)ψ(1)−ψ(0)f(y)−f(ψ(0)tx+ψ(1)(1−t)y)=a(1−t)f(y)−f(a(1−t)y)=a(1−t)(Ay2+By)−A(a(1−t))2−B(a(1−t))=aA(1−t)y2(at+1−a)≥0, |
which shows that f is ψ-convex.
Example 2.2. We consider the function ψ:[0,1]→R defined by
ψ(t)=t2,0≤t≤1. |
Clearly, the function ψ∈Ψ. On the other hand, for all t∈[0,1], we have
ψ(0)ψ(t)−ψ(0)ψ(1)−ψ(0)=0 |
and
ψ(1)ψ(1)−ψ(t)ψ(1)−ψ(0)=1−t2. |
Let f:[0,∞)→R be the function defined by
f(x)=x3−x2+x,x≥0. |
For all t∈[0,1] and x,y≥0, we have
ψ(0)ψ(t)−ψ(0)ψ(1)−ψ(0)f(x)+ψ(1)ψ(1)−ψ(t)ψ(1)−ψ(0)f(y)−f(ψ(0)tx+ψ(1)(1−t)y)=(1−t2)(y3−y2+y)−[(1−t)y]3+[(1−t)y]2−[(1−t)y]=ty(1−t)Pt(y), |
where Pt(y) is the second order polynomial function (with respect to y) given by
Pt(y)=(3−t)y2−2y+1. |
Observe that for all t∈[0,1], the the discriminant of Pt is given by
Δ=4(t−2)<0, |
which implies (since 3−t>0) that Pt(y)≥0. Consequently, for all t∈[0,1] and x,y≥0,
ψ(0)ψ(t)−ψ(0)ψ(1)−ψ(0)f(x)+ψ(1)ψ(1)−ψ(t)ψ(1)−ψ(0)f(y)−f(ψ(0)tx+ψ(1)(1−t)y)≥0, |
which shows that f is ψ-convex.
We now recall the following notion introduced by Toader [29].
Definition 2.2. Let m∈[0,1] and f:[0,∞)→R. The function f is said to be m-convex, if
f(tx+m(1−t)y)≤tf(x)+m(1−t)f(y) |
for all t∈[0,1] and x,y≥0.
Proposition 2.1. If f:[0,∞)→R is m-convex, where 0≤m<1, then f is ψ-convex for some ψ∈Ψ.
Proof. Let
ψ(t)=(m−1)t+1,0≤t≤1. |
Clearly, ψ is a nonnegative and continuous function and ψ(0)−ψ(1)=1−m≠0, which shows that ψ∈Ψ. We also have
ψ(0)ψ(t)−ψ(0)ψ(1)−ψ(0)=(m−1)tm−1=t |
and
ψ(1)ψ(1)−ψ(t)ψ(1)−ψ(0)=m(m−1)(1−t)m−1=m(1−t). |
Consequently, we get
f(ψ(0)tx+ψ(1)(1−t)y)=f(tx+m(1−t)y)≤tf(x)+m(1−t)f(y)=ψ(0)ψ(t)−ψ(0)ψ(1)−ψ(0)f(x)+ψ(1)ψ(1)−ψ(t)ψ(1)−ψ(0)f(y), |
which shows that f is ψ-convex.
The following class of functions was introduced by Kadakal [20].
Definition 2.3. Let m1,m2∈[0,1] and f:[0,∞)→R. The function f is said to be (m1,m2)-convex, if
f(m1tx+m2(1−t)y)≤m1tf(x)+m2(1−t)f(y) |
for all t∈[0,1] and x,y≥0.
Proposition 2.2. If f:[0,∞)→R is (m1,m2)-convex, where m1,m2∈[0,1] with m1≠m2, then f is ψ-convex for some ψ∈Ψ.
Proof. Let
ψ(t)=(m2−m1)t+m1,0≤t≤1. |
Clearly, ψ is a nonnegative and continuous function and ψ(0)−ψ(1)=m1−m2≠0, which shows that ψ∈Ψ. We also have
ψ(0)ψ(t)−ψ(0)ψ(1)−ψ(0)=m1(m2−m1)tm2−m1=m1t |
and
ψ(1)ψ(1)−ψ(t)ψ(1)−ψ(0)=m2(m2−m1)(1−t)m2−m1=m2(1−t). |
Consequently, we get
f(ψ(0)tx+ψ(1)(1−t)y)=f(m1tx+m2(1−t)y)≤m1tf(x)+m2(1−t)f(y)=ψ(0)ψ(t)−ψ(0)ψ(1)−ψ(0)f(x)+ψ(1)ψ(1)−ψ(t)ψ(1)−ψ(0)f(y), |
which shows that f is ψ-convex.
Miheşan [30] was introduced the following concept.
Definition 2.4. Let α,m∈[0,1] and f:[0,∞)→R. The function f is said to be (α,m)-convex, if
f(tx+m(1−t)y)≤tαf(x)+m(1−tα)f(y) |
for all t∈[0,1] and x,y≥0.
Proposition 2.3. If f:[0,∞)→R is a (α,m)-convex function, where α∈(0,1] and 0≤m<1, then f is ψ-convex for some ψ∈Ψ.
Proof. Let
ψ(t)=(m−1)tα+1,0≤t≤1. |
It is clear that ψ∈Ψ, ψ(0)=1 and ψ(1)=m. We also have
ψ(0)ψ(t)−ψ(0)ψ(1)−ψ(0)=(m−1)tαm−1=tα |
and
ψ(1)ψ(1)−ψ(t)ψ(1)−ψ(0)=m(m−1)(1−tα)m−1=m(1−tα). |
Consequently, we get
f(ψ(0)tx+ψ(1)(1−t)y)=f(tx+m(1−t)y)≤tαf(x)+m(1−tα)f(y)=ψ(0)ψ(t)−ψ(0)ψ(1)−ψ(0)f(x)+ψ(1)ψ(1)−ψ(t)ψ(1)−ψ(0)f(y), |
which shows that f is ψ-convex.
In [22], Kadakal introduced the following notion.
Definition 2.5. Let α,m1,m2∈[0,1] and f:[0,∞)→R. The function f is said to be (α,m1,m2)-convex, if
f(m1tx+m2(1−t)y)≤m1tαf(x)+m2(1−tα)f(y) |
for all t∈[0,1] and x,y≥0.
Proposition 2.4. If f:[0,∞)→R is a (α,m1,m2)-convex function, where α∈(0,1] and m1,m2∈[0,1] with m1≠m2, then f is ψ-convex for some ψ∈Ψ.
Proof. Let
ψ(t)=(m2−m1)tα+m1,0≤t≤1. |
It is clear that ψ∈Ψ, ψ(0)=m1 and ψ(1)=m2. We also have
ψ(0)ψ(t)−ψ(0)ψ(1)−ψ(0)=m1tα |
and
ψ(1)ψ(1)−ψ(t)ψ(1)−ψ(0)=m2(1−tα). |
Consequently, we get
f(ψ(0)tx+ψ(1)(1−t)y)=f(m1tx+m2(1−t)y)≤m1tαf(x)+m2(1−tα)f(y)=ψ(0)ψ(t)−ψ(0)ψ(1)−ψ(0)f(x)+ψ(1)ψ(1)−ψ(t)ψ(1)−ψ(0)f(y), |
which shows that f is ψ-convex.
Remark 2.1. Let f:[0,∞)→R be (α,m1,m2)-convex, where α,m1,m2∈[0,1].
(i) If α=1, then f is (m1,m2)-convex.
(ii) If m1=1 and m2=m, then f is (α,m)-convex.
(iii) If α=1, m1=1 and m2=m, then f is m-convex.
We provide below some properties of ψ-convex functions.
Proposition 2.5.
(i) Let σ,μ≥0 and ψ∈Ψ. If f,g:[0,∞)→R are ψ-convex, then σf+μg is ψ-convex.
(ii) Let ψ∈Ψ. If f:[0,∞)→R is ψ-convex, then for all x≥0,
f(ψ(0)x)≤ψ(0)f(x),f(ψ(1)x)≤ψ(1)f(x),f(ψ(0)x)+f(ψ(1)x)ψ(0)+ψ(1)≤f(x). |
(iii) Let ψ∈Ψ and f:[0,∞)→R be ψ-convex.
● If f(0)>0, then for all t∈[0,1],
ψ(t)≤ψ(0)+ψ(1)−1. |
● If f(0)<0, then for all t∈[0,1],
ψ(t)≥ψ(0)+ψ(1)−1. |
Proof. (ⅰ) Let f,g:[0,∞)→R be two ψ-convex functions. Let h=σf+μg. For all t∈[0,1] and x,y≥0, we have
f(ψ(0)tx+ψ(1)(1−t)y)≤ψ(0)ψ(t)−ψ(0)ψ(1)−ψ(0)f(x)+ψ(1)ψ(1)−ψ(t)ψ(1)−ψ(0)f(y) |
and
g(ψ(0)tx+ψ(1)(1−t)y)≤ψ(0)ψ(t)−ψ(0)ψ(1)−ψ(0)g(x)+ψ(1)ψ(1)−ψ(t)ψ(1)−ψ(0)g(y). |
Multiplying the first inequality (resp. second inequality) by σ≥0 (resp. μ≥0), we obtain
h(ψ(0)tx+ψ(1)(1−t)y)≤ψ(0)ψ(t)−ψ(0)ψ(1)−ψ(0)σf(x)+ψ(1)ψ(1)−ψ(t)ψ(1)−ψ(0)σf(y)+ψ(0)ψ(t)−ψ(0)ψ(1)−ψ(0)μg(x)+ψ(1)ψ(1)−ψ(t)ψ(1)−ψ(0)μg(y)=ψ(0)ψ(t)−ψ(0)ψ(1)−ψ(0)h(x)+ψ(1)ψ(1)−ψ(t)ψ(1)−ψ(0)h(y), |
which shows that h is ψ-convex.
(ⅱ) Let x≥0. Taking t=1 in (2.1), we get
f(ψ(0)x)≤ψ(0)f(x). | (2.2) |
Taking t=1 in (2.1), we get
f(ψ(1)x)≤ψ(1)f(x). | (2.3) |
Summing (2.2) and (2.3), we obtain
f(ψ(0)x)+f(ψ(1)x)≤(ψ(0)+ψ(1))f(x), |
that is,
f(ψ(0)x)+f(ψ(1)x)ψ(0)+ψ(1)≤f(x). |
(ⅲ) Let t∈[0,1]. Taking x=y=0 in (2.1), we obtain
f(0)≤1ψ(1)−ψ(0)(ψ(0)ψ(t)−ψ2(0)+ψ2(1)−ψ(1)ψ(t))f(0)=(ψ(1)+ψ(0)−ψ(t))f(0). |
Hence, if f(0)>0, the above inequality yields
ψ(t)≤ψ(1)+ψ(0)−1. |
Similarly, if f(0)<0, we get
ψ(t)≥ψ(1)+ψ(0)−1. |
In this section, we extend the Hermite-Hadamard double inequality (1.1) to the class of ψ-convex functions. We first fix some notations.
For all ψ∈ψ, let
Aψ=ψ(0)ψ(1)−ψ(0)∫10(ψ(t)−ψ(0))dt,Bψ=ψ(1)ψ(1)−ψ(0)∫10(ψ(1)−ψ(t))dt,Eψ=ψ(1)(ψ(1)−ψ(12))ψ(1)−ψ(0),Fψ=ψ(0)(ψ(12)−ψ(0))ψ(1)−ψ(0). |
Theorem 3.1. Let ψ∈Ψ be such that ψ(0)ψ(1)≠0. If f:[0,∞)→R is a ψ-convex function, then for all a,b≥0 with a<b, we have
1b−a∫baf(x)dx≤min{f(aψ(0))Aψ+f(bψ(1))Bψ,f(bψ(0))Aψ+f(aψ(1))Bψ}. | (3.1) |
Proof. Let 0≤a<b. For all t∈(0,1), we have
f(ta+(1−t)b)=f(ψ(0)t[aψ(0)]+ψ(1)(1−t)[bψ(1)])≤ψ(0)ψ(t)−ψ(0)ψ(1)−ψ(0)f(aψ(0))+ψ(1)ψ(1)−ψ(t)ψ(1)−ψ(0)f(bψ(1)), |
which implies after integration over t∈(0,1) that
∫10f(ta+(1−t)b)dt≤f(aψ(0))Aψ+f(bψ(1))Bψ. | (3.2) |
Similarly, we have
f(tb+(1−t)a)≤ψ(0)ψ(t)−ψ(0)ψ(1)−ψ(0)f(bψ(0))+ψ(1)ψ(1)−ψ(t)ψ(1)−ψ(0)f(aψ(1)), |
which implies after integration over t∈(0,1) that
∫10f(tb+(1−t)a)dt≤f(bψ(0))Aψ+f(aψ(1))Bψ. | (3.3) |
Finally, using (3.2), (3.3) and
∫10f(ta+(1−t)b)dt=∫10f(tb+(1−t)a)dt=1b−a∫baf(x)dx, |
we obtain (3.1).
Remark 3.1. Let us consider the case when f:[0,∞)→R is (α,m1,m2) convex, where α,m1,m1∈(0,1] with m1≠m2. From Proposition 2.4, f is ψ-convex, where
ψ(t)=(m2−m1)tα+m1,0≤t≤1. |
In this case, elementary calculations give us that
Aψ=ψ(0)ψ(1)−ψ(0)∫10(ψ(t)−ψ(0))dt=m1m2−m1∫10(m2−m1)tαdt=m1α+1 |
and
Bψ=ψ(1)ψ(1)−ψ(0)∫10(ψ(1)−ψ(t))dt=m2m2−m1∫10(m2−m1)(1−tα)dt=m2αα+1. |
Hence, (3.1) reduces to
1b−a∫baf(x)dx≤min{m1α+1f(am1)+m2αα+1f(bm2),m1α+1f(bm1)+m2αα+1f(am2)} | (3.4) |
and we recover the obtained inequality in [22].
Notice that,
● if α=1, then (3.4) reduces to the right Hermite-Hadamard inequality for (m1,m2)-convex functions [20],
● if m1=1 and m2=m, then (3.4) reduces to the right Hermite-Hadamard inequality for (α,m)-convex functions [21],
● if α=1, m1=1 and m2=m, then (3.4) reduces to the right Hermite-Hadamard inequality for m-convex functions [19].
Let us denote by L1loc([0,∞)) the set of functions f:[0,∞)→R such that
∫I|f(x)|dx<∞ |
for every closed and bounded interval I⊂[0,∞).
Our second main result is the following.
Theorem 3.2. Let ψ∈Ψ be such that ψ(0)ψ(1)≠0. If f:[0,∞)→R is a ψ-convex function and f∈L1loc([0,∞)), then for all a,b≥0 with a<b, we have
f(a+b2)≤Fψb−a∫baf(xψ(0))dx+Eψb−a∫baf(xψ(1))dx. | (3.5) |
Proof. For all x,y≥0, writing
x+y2=ψ(0)12(xψ(0))+ψ(1)(1−12)(yψ(1)) |
and using the ψ-convexity of f, we obtain
f(x+y2)≤ψ(0)ψ(12)−ψ(0)ψ(1)−ψ(0)f(xψ(0))+ψ(1)ψ(1)−ψ(12)ψ(1)−ψ(0)f(yψ(1)), |
that is,
f(x+y2)≤Fψf(xψ(0))+Eψf(yψ(1)). | (3.6) |
In particular, for x=ta+(1−t)b and y=(1−t)a+tb, where t∈(0,1), (3.6) reduces to
f(a+b2)≤Fψf(ta+(1−t)bψ(0))+Eψf(1−t)a+tbψ(1)). |
Integrating the above inequality over t∈(0,1), we obtain
f(a+b2)≤Fψ∫10f(ta+(1−t)bψ(0))dt+Eψ∫10f(1−t)a+tbψ(1))dt. | (3.7) |
On the other hand, one has
∫10f(ta+(1−t)bψ(0))dt=1b−a∫baf(zψ(0))dz | (3.8) |
and
∫10f(1−t)a+tbψ(1))dt=1b−a∫baf(zψ(1))dz. | (3.9) |
Thus, (3.5) follows from (3.7)–(3.9).
Remark 3.2. Let us consider the case when f:[0,∞)→R is (α,m1,m2) convex, where α,m1,m1∈(0,1] with m1≠m2. Then f is ψ-convex, where
ψ(t)=(m2−m1)tα+m1,0≤t≤1. |
In this case, elementary calculations give us that
Eψ=ψ(1)(ψ(1)−ψ(12))ψ(1)−ψ(0)=2α−12αm2 |
and
Fψ=ψ(0)(ψ(12)−ψ(0))ψ(1)−ψ(0)=m12α. |
Hence, (3.5) reduces to
f(a+b2)≤1b−a(m12α∫baf(xm1)dx+(2α−1)m22α∫baf(xm2)dx) |
and we recover the obtained inequality in [22].
In this section, we establish Simpson-type inequalities for the class of functions f∈C1([0,∞)) such that |f′| is ψ-convex. We first need the following lemma.
Lemma 4.1. Let ψ∈Ψ be such that ψ(1)<ψ(0). If f∈C1([0,∞)), then for all a,b≥0 with a<b, we have
16[f(ψ(1)a)+4f(ψ(1)a+ψ(0)b2)+f(ψ(0)b)]−1ψ(0)b−ψ(1)a∫ψ(0)bψ(1)af(x)dx=(ψ(0)b−ψ(1)a)∫10H(t)f′(tψ(0)b+(1−t)ψ(1)a)dt, | (4.1) |
where
H(t)={t−16if0≤t≤12,t−56if12<t≤1. |
Proof. By definition of H and using integrations by parts, we obtain
(ψ(0)b−ψ(1)a)∫10H(t)f′(tψ(0)b+(1−t)ψ(1)a)dt=(ψ(0)b−ψ(1)a)∫120H(t)f′(tψ(0)b+(1−t)ψ(1)a)dt+(ψ(0)b−ψ(1)a)∫112H(t)f′(tψ(0)b+(1−t)ψ(1)a)dt=[H(t)f(tψ(0)b+(1−t)ψ(1)a)]12t=0+[H(t)f(tψ(0)b+(1−t)ψ(1)a)]1t=12−∫10f(tψ(0)b+(1−t)ψ(1)a)dt=13f(ψ(0)b+ψ(1)a2)+16f(ψ(1)a)+16f(ψ(0)b)+13f(ψ(0)b+ψ(1)a2)−1ψ(0)b−ψ(1)a∫ψ(0)bψ(1)af(x)dx, |
which proves (4.1).
We have the following Simpson-type inequality.
Theorem 4.1. Let ψ∈Ψ be a decreasing function. If f∈C1([0,∞)) and |f′| is ψ-convex, then for all a,b≥0 with a<b, we have
|16[f(ψ(1)a)+4f(ψ(1)a+ψ(0)b2)+f(ψ(0)b)]−1ψ(0)b−ψ(1)a∫ψ(0)bψ(1)af(x)dx|≤ψ(0)b−ψ(1)aψ(0)−ψ(1)[(Lψ−536ψ(1))ψ(1)|f′(a)|+(536ψ(0)−Lψ)ψ(0)|f′(b)|], | (4.2) |
where
Lψ=∫120|t−16|ψ(t)dt+∫112|t−56|ψ(t)dt. |
Proof. By Lemma 4.1, we have
|16[f(ψ(1)a)+4f(ψ(1)a+ψ(0)b2)+f(ψ(0)b)]−1ψ(0)b−ψ(1)a∫ψ(0)bψ(1)af(x)dx|≤(ψ(0)b−ψ(1)a)∫10|H(t)||f′(tψ(0)b+(1−t)ψ(1)a)|dt. | (4.3) |
On the other hand, due to the ψ-convexity of |f′|, we have
|f′(tψ(0)b+(1−t)ψ(1)a)|≤ψ(1)ψ(t)−ψ(1)ψ(0)−ψ(1)|f′(a)|+ψ(0)ψ(0)−ψ(t)ψ(0)−ψ(1)|f′(b)| |
for all t∈(0,1), which implies after integration over t∈(0,1) that
(ψ(0)b−ψ(1)a)∫10|H(t)||f′(tψ(0)b+(1−t)ψ(1)a)|dt≤ψ(0)b−ψ(1)aψ(0)−ψ(1)⋅[ψ(1)|f′(a)|∫10|H(t)|(ψ(t)−ψ(1))dt+ψ(0)|f′(b)|∫10|H(t)|(ψ(0)−ψ(t))dt]. | (4.4) |
On the other hand, by the definition of H, we have
∫10|H(t)|(ψ(t)−ψ(1))dt=∫120|t−16|(ψ(t)−ψ(1))dt+∫112|t−56|(ψ(t)−ψ(1))dt=−ψ(1)(∫120|t−16|dt+∫112|t−56|dt)+∫120|t−16|ψ(t)dt+∫112|t−56|ψ(t)dt. |
Notice that ∫120|t−16|dt=∫112|t−56|dt=572. Hence, we get
∫10|H(t)|(ψ(t)−ψ(1))dt=−536ψ(1)+∫120|t−16|ψ(t)dt+∫112|t−56|ψ(t)dt. | (4.5) |
Similarly, we have
∫10|H(t)|(ψ(0)−ψ(t))dt=∫120|t−16|(ψ(0)−ψ(t))dt+∫112|t−56|(ψ(0)−ψ(t))dt=ψ(0)(∫120|t−16|dt+∫112|t−56|dt)−∫120|t−16|ψ(t)dt−∫112|t−56|ψ(t)dt, |
that is,
∫10|H(t)|(ψ(0)−ψ(t))dt=536ψ(0)−∫120|t−16|ψ(t)dt−∫112|t−56|ψ(t)dt. | (4.6) |
Hence, it follows from (4.4)–(4.6) that
(ψ(0)b−ψ(1)a)∫10|H(t)||f′(tψ(0)b+(1−t)ψ(1)a)|dt≤ψ(0)b−ψ(1)aψ(0)−ψ(1)[(Lψ−536ψ(1))ψ(1)|f′(a)|+(536ψ(0)−Lψ)ψ(0)|f′(b)|]. |
Finally, (4.2) follows from (4.3) and the above estimate.
Remark 4.1. Assume that |f′| is (α,m)-convex, where α∈(0,1] and m∈[0,1). Then, by Proposition 2.3, |f′| is ψ-convex, where
ψ(t)=(m−1)tα+1,0≤t≤1. |
Clearly, ψ is a decreasing function. On the other hand, we have
|16[f(ψ(1)a)+4f(ψ(1)a+ψ(0)b2)+f(ψ(0)b)]−1ψ(0)b−ψ(1)a∫ψ(0)bψ(1)af(x)dx|=|16[f(ma)+4f(ma+b2)+f(b)]−1b−ma∫bmaf(x)dx|. |
Furthermore, elementary calculations show that
Lψ=∫120|t−16|ψ(t)dt+∫112|t−56|ψ(t)dt=(m−1)(∫120tα|t−16|dt+∫112tα|t−56|dt)+(∫120|t−16|dt+∫112|t−56|dt)=(m−1)6−α−9×2−α+5α+2×6−α+3α−1218(α+1)(α+2)+536:=(m−1)ν1+536 |
which yields
(Lψ−536ψ(1))ψ(1)=(536−ν1)m(1−m):=ν2m(1−m) |
and
(536ψ(0)−Lψ)ψ(0)=(1−m)ν1. |
We also have
ψ(0)b−ψ(1)aψ(0)−ψ(1)=b−ma1−m. |
Consequently,
ψ(0)b−ψ(1)aψ(0)−ψ(1)[(Lψ−536ψ(1))ψ(1)|f′(a)|+(536ψ(0)−Lψ)ψ(0)|f′(b)|]=(b−ma)(ν2m|f′(a)|+ν1|f′(b)|). |
Hence, (4.2) reduces to
|16[f(ma)+4f(ma+b2)+f(b)]−1b−ma∫bmaf(x)dx|≤(b−ma)(ν2m|f′(a)|+ν1|f′(b)|) |
and we recover the obtained inequality in [23].
We introduced the class of ψ-convex functions, where ψ∈C([0,1]) is nonnegative and satisfies ψ(0)≠ψ(1). The introduced class includes several classes of functions from the literature: m-convex functions, (m1,m2)-convex functions, (α,m)-convex functions and (α,m1,m2)-convex functions. After studying some properties of ψ-convex functions, some known inequalities are extended to this set of functions. Namely, when ψ(0)ψ(1)≠0 and f is ψ-convex, we obtained an upper bound of 1b−a∫baf(x)dx (see Theorem 3.1) and an upper bound of f(a+b2) (see Theorem 3.2). When ψ is nondecreasing and |f′| is ψ-convex, we proved a Simpson-type inequality (see Theorem 4.1), which provides an estimate of
|16[f(ψ(1)a)+4f(ψ(1)a+ψ(0)b2)+f(ψ(0)b)]−1ψ(0)b−ψ(1)a∫ψ(0)bψ(1)af(x)dx|. |
It would be interesting to continue the study of ψ-convex functions in various directions. For instance, in [31], a sandwich like theorem was established for m-convex functions. Recall that any m-convex function is ψ-convex for some ψ∈Ψ (see Proposition 2.1). A natural question is to ask whether it is possible to extend the sandwich like result in [31] to the class of ψ-convex functions.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
B. Samet is supported by Researchers Supporting Project number (RSP2024R4), King Saud University, Riyadh, Saudi Arabia. M. De La Sen is supported by the project: Basque Government IT1555-22.
The authors declare no competing interests.
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