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Research article Special Issues

Geopolitics of Covid-19: global challenge at national borders

  • Received: 03 November 2020 Accepted: 19 November 2020 Published: 27 November 2020
  • The current Covid-19 pandemic seems to confirm that glocalism, a phenomenon that theorizes a close correlation between the sphere of the local and the global, is now the new normal. The rapid spread of the virus seems to show how the international order, focused on borders and political-territorial spheres, is currently struggling to manage complex problems caused by factors such as innovation and the mobility of people, goods and information. In a short space of time, the virus spread quickly from the market in the Chinese metropolis of Wuhan to the rest of the world, thanks to the infection being spread from person to person. To this global change, states have responded with national solutions. The perception that the risks caused by the spread of the virus could undermine national security and sovereignty has led most states to isolate themselves by refraining from multilateral cooperation. At the peak of the pandemic, contrary to the instructions of the World Health Organization, more than one hundred and thirty countries closed borders or imposed strict border controls and banned from entering a selection of citizens from the outbreak areas of contagion. These "cures" are often worse than the disease. The international health emergency seems to have accelerated the beginning of a new glocal era, based on a close correlation between the local and the global sphere. This article aims to analyse the causes, consequences and possible geopolitical scenarios of this phenomenon.

    Citation: Giuseppe Terranova. Geopolitics of Covid-19: global challenge at national borders[J]. AIMS Geosciences, 2020, 6(4): 515-524. doi: 10.3934/geosci.2020029

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  • The current Covid-19 pandemic seems to confirm that glocalism, a phenomenon that theorizes a close correlation between the sphere of the local and the global, is now the new normal. The rapid spread of the virus seems to show how the international order, focused on borders and political-territorial spheres, is currently struggling to manage complex problems caused by factors such as innovation and the mobility of people, goods and information. In a short space of time, the virus spread quickly from the market in the Chinese metropolis of Wuhan to the rest of the world, thanks to the infection being spread from person to person. To this global change, states have responded with national solutions. The perception that the risks caused by the spread of the virus could undermine national security and sovereignty has led most states to isolate themselves by refraining from multilateral cooperation. At the peak of the pandemic, contrary to the instructions of the World Health Organization, more than one hundred and thirty countries closed borders or imposed strict border controls and banned from entering a selection of citizens from the outbreak areas of contagion. These "cures" are often worse than the disease. The international health emergency seems to have accelerated the beginning of a new glocal era, based on a close correlation between the local and the global sphere. This article aims to analyse the causes, consequences and possible geopolitical scenarios of this phenomenon.


    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)1babaf(x)dxf(a)+f(b)2, (1.1)

    which holds for any convex function f:IR and a,bI, 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<m1, Dragomir [19] proved that for all a,b0 with 0a<b, we have

    1babaf(x)dx12min{f(a)+mf(bm),f(b)+mf(am)}

    and

    f(a+b2)1babaf(x)+mf(xm)2dxm+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

    |1babaf(x)dx13[f(a)+f(b)2+2f(a+b2)]|(ba)42880f(4),

    where fC4([a,b]) and f(4)=maxaxb|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

    |1babaf(x)dx13[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 ψC([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 fC1([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 fC1([0,)), where |f| is ψ-convex.

    In this section, we introduce the class of ψ-convex functions. We first introduce the set

    Ψ={ψC([0,1]):ψ0,ψ(0)ψ(1)}.

    Definition 2.1. Let ψΨ. A function f:[0,)R is said to be ψ-convex, if

    f(ψ(0)tx+ψ(1)(1t)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,y0.

    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,0t1,

    where 0<a1 is a constant. Clearly, the function ψΨ 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(1t).

    Let f:[0,)R be the function defined by

    f(x)=Ax2+Bx,x0,

    where A0 and BR are constants. For all t[0,1] and x,y0, 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)(1t)y)=a(1t)f(y)f(a(1t)y)=a(1t)(Ay2+By)A(a(1t))2B(a(1t))=aA(1t)y2(at+1a)0,

    which shows that f is ψ-convex.

    Example 2.2. We consider the function ψ:[0,1]R defined by

    ψ(t)=t2,0t1.

    Clearly, the function ψΨ. 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)=1t2.

    Let f:[0,)R be the function defined by

    f(x)=x3x2+x,x0.

    For all t[0,1] and x,y0, 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)(1t)y)=(1t2)(y3y2+y)[(1t)y]3+[(1t)y]2[(1t)y]=ty(1t)Pt(y),

    where Pt(y) is the second order polynomial function (with respect to y) given by

    Pt(y)=(3t)y22y+1.

    Observe that for all t[0,1], the the discriminant of Pt is given by

    Δ=4(t2)<0,

    which implies (since 3t>0) that Pt(y)0. Consequently, for all t[0,1] and x,y0,

    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)(1t)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(1t)y)tf(x)+m(1t)f(y)

    for all t[0,1] and x,y0.

    Proposition 2.1. If f:[0,)R is m-convex, where 0m<1, then f is ψ-convex for some ψΨ.

    Proof. Let

    ψ(t)=(m1)t+1,0t1.

    Clearly, ψ is a nonnegative and continuous function and ψ(0)ψ(1)=1m0, which shows that ψΨ. We also have

    ψ(0)ψ(t)ψ(0)ψ(1)ψ(0)=(m1)tm1=t

    and

    ψ(1)ψ(1)ψ(t)ψ(1)ψ(0)=m(m1)(1t)m1=m(1t).

    Consequently, we get

    f(ψ(0)tx+ψ(1)(1t)y)=f(tx+m(1t)y)tf(x)+m(1t)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(1t)y)m1tf(x)+m2(1t)f(y)

    for all t[0,1] and x,y0.

    Proposition 2.2. If f:[0,)R is (m1,m2)-convex, where m1,m2[0,1] with m1m2, then f is ψ-convex for some ψΨ.

    Proof. Let

    ψ(t)=(m2m1)t+m1,0t1.

    Clearly, ψ is a nonnegative and continuous function and ψ(0)ψ(1)=m1m20, which shows that ψΨ. We also have

    ψ(0)ψ(t)ψ(0)ψ(1)ψ(0)=m1(m2m1)tm2m1=m1t

    and

    ψ(1)ψ(1)ψ(t)ψ(1)ψ(0)=m2(m2m1)(1t)m2m1=m2(1t).

    Consequently, we get

    f(ψ(0)tx+ψ(1)(1t)y)=f(m1tx+m2(1t)y)m1tf(x)+m2(1t)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(1t)y)tαf(x)+m(1tα)f(y)

    for all t[0,1] and x,y0.

    Proposition 2.3. If f:[0,)R is a (α,m)-convex function, where α(0,1] and 0m<1, then f is ψ-convex for some ψΨ.

    Proof. Let

    ψ(t)=(m1)tα+1,0t1.

    It is clear that ψΨ, ψ(0)=1 and ψ(1)=m. We also have

    ψ(0)ψ(t)ψ(0)ψ(1)ψ(0)=(m1)tαm1=tα

    and

    ψ(1)ψ(1)ψ(t)ψ(1)ψ(0)=m(m1)(1tα)m1=m(1tα).

    Consequently, we get

    f(ψ(0)tx+ψ(1)(1t)y)=f(tx+m(1t)y)tαf(x)+m(1tα)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(1t)y)m1tαf(x)+m2(1tα)f(y)

    for all t[0,1] and x,y0.

    Proposition 2.4. If f:[0,)R is a (α,m1,m2)-convex function, where α(0,1] and m1,m2[0,1] with m1m2, then f is ψ-convex for some ψΨ.

    Proof. Let

    ψ(t)=(m2m1)tα+m1,0t1.

    It is clear that ψΨ, 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(1tα).

    Consequently, we get

    f(ψ(0)tx+ψ(1)(1t)y)=f(m1tx+m2(1t)y)m1tαf(x)+m2(1tα)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,μ0 and ψΨ. If f,g:[0,)R are ψ-convex, then σf+μg is ψ-convex.

    (ii) Let ψΨ. If f:[0,)R is ψ-convex, then for all x0,

    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 ψΨ 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,y0, we have

    f(ψ(0)tx+ψ(1)(1t)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)(1t)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 σ0 (resp. μ0), we obtain

    h(ψ(0)tx+ψ(1)(1t)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 x0. 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 ψψ, 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 ψΨ be such that ψ(0)ψ(1)0. If f:[0,)R is a ψ-convex function, then for all a,b0 with a<b, we have

    1babaf(x)dxmin{f(aψ(0))Aψ+f(bψ(1))Bψ,f(bψ(0))Aψ+f(aψ(1))Bψ}. (3.1)

    Proof. Let 0a<b. For all t(0,1), we have

    f(ta+(1t)b)=f(ψ(0)t[aψ(0)]+ψ(1)(1t)[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+(1t)b)dtf(aψ(0))Aψ+f(bψ(1))Bψ. (3.2)

    Similarly, we have

    f(tb+(1t)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+(1t)a)dtf(bψ(0))Aψ+f(aψ(1))Bψ. (3.3)

    Finally, using (3.2), (3.3) and

    10f(ta+(1t)b)dt=10f(tb+(1t)a)dt=1babaf(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 m1m2. From Proposition 2.4, f is ψ-convex, where

    ψ(t)=(m2m1)tα+m1,0t1.

    In this case, elementary calculations give us that

    Aψ=ψ(0)ψ(1)ψ(0)10(ψ(t)ψ(0))dt=m1m2m110(m2m1)tαdt=m1α+1

    and

    Bψ=ψ(1)ψ(1)ψ(0)10(ψ(1)ψ(t))dt=m2m2m110(m2m1)(1tα)dt=m2αα+1.

    Hence, (3.1) reduces to

    1babaf(x)dxmin{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 ψΨ be such that ψ(0)ψ(1)0. If f:[0,)R is a ψ-convex function and fL1loc([0,)), then for all a,b0 with a<b, we have

    f(a+b2)Fψbabaf(xψ(0))dx+Eψbabaf(xψ(1))dx. (3.5)

    Proof. For all x,y0, writing

    x+y2=ψ(0)12(xψ(0))+ψ(1)(112)(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+(1t)b and y=(1t)a+tb, where t(0,1), (3.6) reduces to

    f(a+b2)Fψf(ta+(1t)bψ(0))+Eψf(1t)a+tbψ(1)).

    Integrating the above inequality over t(0,1), we obtain

    f(a+b2)Fψ10f(ta+(1t)bψ(0))dt+Eψ10f(1t)a+tbψ(1))dt. (3.7)

    On the other hand, one has

    10f(ta+(1t)bψ(0))dt=1babaf(zψ(0))dz (3.8)

    and

    10f(1t)a+tbψ(1))dt=1babaf(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 m1m2. Then f is ψ-convex, where

    ψ(t)=(m2m1)tα+m1,0t1.

    In this case, elementary calculations give us that

    Eψ=ψ(1)(ψ(1)ψ(12))ψ(1)ψ(0)=2α12αm2

    and

    Fψ=ψ(0)(ψ(12)ψ(0))ψ(1)ψ(0)=m12α.

    Hence, (3.5) reduces to

    f(a+b2)1ba(m12αbaf(xm1)dx+(2α1)m22αbaf(xm2)dx)

    and we recover the obtained inequality in [22].

    In this section, we establish Simpson-type inequalities for the class of functions fC1([0,)) such that |f| is ψ-convex. We first need the following lemma.

    Lemma 4.1. Let ψΨ be such that ψ(1)<ψ(0). If fC1([0,)), then for all a,b0 with a<b, we have

    16[f(ψ(1)a)+4f(ψ(1)a+ψ(0)b2)+f(ψ(0)b)]1ψ(0)bψ(1)aψ(0)bψ(1)af(x)dx=(ψ(0)bψ(1)a)10H(t)f(tψ(0)b+(1t)ψ(1)a)dt, (4.1)

    where

    H(t)={t16if0t12,t56if12<t1.

    Proof. By definition of H and using integrations by parts, we obtain

    (ψ(0)bψ(1)a)10H(t)f(tψ(0)b+(1t)ψ(1)a)dt=(ψ(0)bψ(1)a)120H(t)f(tψ(0)b+(1t)ψ(1)a)dt+(ψ(0)bψ(1)a)112H(t)f(tψ(0)b+(1t)ψ(1)a)dt=[H(t)f(tψ(0)b+(1t)ψ(1)a)]12t=0+[H(t)f(tψ(0)b+(1t)ψ(1)a)]1t=1210f(tψ(0)b+(1t)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)bψ(1)aψ(0)bψ(1)af(x)dx,

    which proves (4.1).

    We have the following Simpson-type inequality.

    Theorem 4.1. Let ψΨ be a decreasing function. If fC1([0,)) and |f| is ψ-convex, then for all a,b0 with a<b, we have

    |16[f(ψ(1)a)+4f(ψ(1)a+ψ(0)b2)+f(ψ(0)b)]1ψ(0)bψ(1)aψ(0)bψ(1)af(x)dx|ψ(0)bψ(1)aψ(0)ψ(1)[(Lψ536ψ(1))ψ(1)|f(a)|+(536ψ(0)Lψ)ψ(0)|f(b)|], (4.2)

    where

    Lψ=120|t16|ψ(t)dt+112|t56|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)bψ(1)aψ(0)bψ(1)af(x)dx|(ψ(0)bψ(1)a)10|H(t)||f(tψ(0)b+(1t)ψ(1)a)|dt. (4.3)

    On the other hand, due to the ψ-convexity of |f|, we have

    |f(tψ(0)b+(1t)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)bψ(1)a)10|H(t)||f(tψ(0)b+(1t)ψ(1)a)|dtψ(0)bψ(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|t16|(ψ(t)ψ(1))dt+112|t56|(ψ(t)ψ(1))dt=ψ(1)(120|t16|dt+112|t56|dt)+120|t16|ψ(t)dt+112|t56|ψ(t)dt.

    Notice that 120|t16|dt=112|t56|dt=572. Hence, we get

    10|H(t)|(ψ(t)ψ(1))dt=536ψ(1)+120|t16|ψ(t)dt+112|t56|ψ(t)dt. (4.5)

    Similarly, we have

    10|H(t)|(ψ(0)ψ(t))dt=120|t16|(ψ(0)ψ(t))dt+112|t56|(ψ(0)ψ(t))dt=ψ(0)(120|t16|dt+112|t56|dt)120|t16|ψ(t)dt112|t56|ψ(t)dt,

    that is,

    10|H(t)|(ψ(0)ψ(t))dt=536ψ(0)120|t16|ψ(t)dt112|t56|ψ(t)dt. (4.6)

    Hence, it follows from (4.4)–(4.6) that

    (ψ(0)bψ(1)a)10|H(t)||f(tψ(0)b+(1t)ψ(1)a)|dtψ(0)bψ(1)aψ(0)ψ(1)[(Lψ536ψ(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)=(m1)tα+1,0t1.

    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)bψ(1)aψ(0)bψ(1)af(x)dx|=|16[f(ma)+4f(ma+b2)+f(b)]1bmabmaf(x)dx|.

    Furthermore, elementary calculations show that

    Lψ=120|t16|ψ(t)dt+112|t56|ψ(t)dt=(m1)(120tα|t16|dt+112tα|t56|dt)+(120|t16|dt+112|t56|dt)=(m1)6α9×2α+5α+2×6α+3α1218(α+1)(α+2)+536:=(m1)ν1+536

    which yields

    (Lψ536ψ(1))ψ(1)=(536ν1)m(1m):=ν2m(1m)

    and

    (536ψ(0)Lψ)ψ(0)=(1m)ν1.

    We also have

    ψ(0)bψ(1)aψ(0)ψ(1)=bma1m.

    Consequently,

    ψ(0)bψ(1)aψ(0)ψ(1)[(Lψ536ψ(1))ψ(1)|f(a)|+(536ψ(0)Lψ)ψ(0)|f(b)|]=(bma)(ν2m|f(a)|+ν1|f(b)|).

    Hence, (4.2) reduces to

    |16[f(ma)+4f(ma+b2)+f(b)]1bmabmaf(x)dx|(bma)(ν2m|f(a)|+ν1|f(b)|)

    and we recover the obtained inequality in [23].

    We introduced the class of ψ-convex functions, where ψC([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 1babaf(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)bψ(1)aψ(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 ψΨ (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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