Int. J. Anal. Appl. (2022), 20:40 Some Hermite-Hadamard Type Inequalities via Katugampola Fractional for pq-Convex on the Interval-Valued Coordinates Jen Chieh Lo∗ General Education Center, National Taipei University of Technology, Taipei, Taiwan ∗Corresponding author: jclo@mail.ntut.edu.tw Abstract. In this paper, we established the Hermite-Hadamard inequalities via Katugampola fractional. Meanwhile, interval analysis is a particular case of set-interval analysis. We established the fractional inequalities and these results are an extension of a previous research. 1. Introduction The classical Hermite-Hadamard inequalities such that f ( a + b 2 ) ≤ 1 b−a b∫ a f (t) dt ≤ f (a) + f (b) 2 where f : I →R is a convex function on the closed bounded interval I of R,and a,b ∈ I with a < b. Since then, some improved and generalized results of Hermite-Hadamard inequality on convex function have been explored and study by many authors(e.g. [2], [7], [10–12], [20], [23,24], [27], [29]). On the other hand, interval analysis is a particular case of set-valued analysis which is the study of sets in the spirit of mathematical analysis and general topology. It was introduced as an attempt to handle interval uncertainty that appears in many mathematical or computer models of some deter- ministic real-world phenomena. An old example of interval enclosure is Archimede’s method which is related to the computation of the circumference of circle. In 1966, the first book related to interval analysis was given by Moore who is known as the first user of intervals in computational mathematics. After this book, several scientists started to investigate theory and application of interval arithmetic. Received: Jun. 22, 2022. 2010 Mathematics Subject Classification. 34A40. Key words and phrases. Hermite-Hadamard inequalities; Katugampola fractional; pq-convex; interval-valued. https://doi.org/10.28924/2291-8639-20-2022-40 ISSN: 2291-8639 © 2022 the author(s). https://doi.org/10.28924/2291-8639-20-2022-40 2 Int. J. Anal. Appl. (2022), 20:40 Nowadays, because of its application, interval analysis is a useful tool in various areas related to un- certain data. We can see applications in computer graphics, experimental and computational physics, error analysis, robotics and many others. 2. Interval Calculus A real valued interval X is bounded, closed subset of R and is defined by X = [ X,X ] = { t ∈R : X ≤ t ≤ X } where X,X ∈ R and X ≤ X. The number X and X are called the left and right endpoints of interval X, respectively. When X = X = a, the interval X is said to be degenerate and we use the form X = a = [a,a] . Also we call X positive if X> 0 or negative if X < 0. The set of all closed intervals of R, the sets of all closed positive intervals of R and closed negative intervals of R is denoted by RI,R+I and R − I , respectively. The Pompeiu-Hausdorff distance between the intervals X and Y is defined by d (X,Y ) = d ([ X,X ] , [ Y ,Y ]) = max { |X −Y | , ∣∣X −Y ∣∣} . It is known that (RI,d) is a complete metric space. Now, we give the definitions of basic interval arithmetic operations for the intervals X and Y as follows: X + Y = [ X + Y ,X + Y ] , X −Y = [ X −Y ,X −Y ] , X ·Y = [min S, max S] where S = { XY ,XY ,XY ,XY } , X/Y = [min T, max T ] where T = { X/Y ,X/Y ,X/Y ,X/Y } and 0 /∈ Y. Scalar multiplication of the interval X is defined by λX = λ [ X,X ] =   [ λX,λX ] , λ > 0, 0, λ = 0,[ λX,λX ] , λ < 0, where λ ∈R. The opposits of the interval X is −X := (−1) X = [ −X,−X ] , where λ = −1. The subtraction is given by Int. J. Anal. Appl. (2022), 20:40 3 X −Y = X + (−Y ) = [ X −Y ,X −Y ] . In general, −X is not additive inverse for X, i.e. X −X 6= 0. Use of monotonic functions F (X) = [ F (X) ,F ( X )] . The definitons of operations lead to a number of algebraic properties which allows RI to be quasi- linear space. They can be listed as follows (1)(Associativity of addition) (X + Y ) + Z = X + (Y + Z) for all X,Y,Z ∈RI, (2)(Additivity elemant) X + 0 = 0 + X = X for all X ∈RI, (3)(Commutativity of addition) X + Y = Y + X for all X,Y ∈RI, (4)(Cancellation law) X + Z = Y + Z =⇒ X = Y for all X,Y,Z ∈RI, (5)(Associativity of multiplication) (X ·Y ) ·Z = X · (Y ·Z) for all X,Y,Z ∈RI, (6)(Commutativity of multiplication) X ·Y = Y ·X for all X,Y ∈RI, (7)(Unity element) X · 1 = 1 ·X for all X ∈RI, (8)(Associativity law) λ (µX) = (λµ) X for all X ∈RI, and for all λ,µ ∈R, (9)(First distributiviyu law) λ (X + Y ) = λX + λY for all X,Y ∈RI, and for all λ ∈R, (10)(Second distributiviyu law) (λ + µ) X = λX + µX for all X ∈RI, and for all λ,µ ∈R. But, this law holds in certain cases. If Y ·Z > 0, then X ·Y + Z = X ·Y + X ·Z. What’s more, one of the set property is the inclusion ⊆ that is given by X ⊆ Y ⇐⇒ Y ≤ X and X ≤ Y . Considering together with arthmetic operations and inclusion, one has the following property which is called inclusion isotone of interval operations: Let � be the addition, multiplication, subtraction or division. If X,Y,Z and T areintervals such that X ⊆ Y and Z ⊆ T, then the following relation is valid X �Z ⊆ Y �T. 3. Intgral of Interval-Valued Functions In this section, the notion of integral is mentioned for interval-valued functions. Before the definition of integral, the necessary concepts will be given as the following: A function F is said to be an interval-valued function of t on [a,b] , if it assigns a nonempty interval to each t ∈ [a,b] , 4 Int. J. Anal. Appl. (2022), 20:40 F (t) = [ F (t) ,F (t) ] . A partition of [a,b] is any finite ordered subset P having the form: P : a = t0 < t1 < ... < tn = b. The mesh of a partition P defined by mesh (P ) = max{ti − ti−1 : i = 1, 2, ...,n} . We denoted by P ([a,b]) the set of all partition of [a,b] . Let P (δ, [a,b]) be the set of all P ∈ P ([a,b]) such that mesh (P ) < δ. Choose an arbitrary point ξi in interval [ti−1,ti ] , (i = 1, 2, ...,n) and let us define the sum S (F,P,δ) = n∑ i=1 F (ξi ) [ti − ti−1] , where F : [a,b] →RI. We call S (F,P,δ) a Riemann sum of F corresponding to P ∈ P (δ, [a,b]) . Definition 3.1 A function F : [a,b] →RI is called interval Riemann intrgrable ( (IR)-integrable) on [a,b] , if there exists A ∈RI such that, for each � > 0, there exists δ > 0 such that d (S (F,P,δ) ,A) < � for every Riemann sum S of F corresponding to each P ∈ P (δ, [a,b]) and independent from choice of ξi ∈ [ti−1,ti ] for all 1 ≤ i ≤ n. In this case, A is called the (IR)-integral of F on [a,b] and is denoted by A = (IR) b∫ a F (t) dt. The collection of all functions that are (IR)-integrable on [a,b] will be denoted by IR([a,b]). The following theorem gives relation between (IR)−integrable and Riemann integrable (R)- integrable. Theorem 3.2 Let F : [a,b] → RI be an interval-valued function such that F (t) = [ F (t) ,F (t) ] . F ∈ IR([a,b]) if and only if F (t) ,F (t) ∈ R([a,b]) and (IR) b∫ a F (t) dt =  (R) b∫ a F (t) dt, (R) b∫ a F (t) dt   , where R([a,b]) denoted the all R-integrable functions. Int. J. Anal. Appl. (2022), 20:40 5 It is seen easily that, if F (t) ⊆ G (t) for all t ∈ [a,b] , then (IR) b∫ a F (t) dt ⊆ (IR) b∫ a G (t) dt. Furthermore, if {ti−1,ti} m i=1 is a δ-fine P1 of [a,b] and if { sj−1,sj }n j=1 is a δ-fine P2 of [c,d] , then retangles 4i,j = [ti−1,ti ] × [ sj−1,sj ] are the partition of retangle 4 = [a,b] × [c,d] and the point ( ξi,ηj ) are inside the retangles [ti−1,ti ] × [ sj−1,sj ] . And we denote the set of all δ-fine partition P of 4 with P1 ×P2, where P1 ∈ P (δ, [a,b]) and P2 ∈ P (δ, [c,d]) . Let 4Ai,j be the area retangle 4i,j, where 1 ≤ i ≤ m, 1 ≤ j ≤ n, choose arbitrary ( ξi,ηj ) and get S (F,P,δ,4) = m∑ i=1 n∑ j=1 F ( ξi,ηj ) 4Ai,j. Definition 3.3 A function F : 4 → RI is called interval double Riemann integrable ( (ID)-integrable) on 4 = [a,b] × [c,d] with the ID-integral I = (ID) ∫∫ 4 F (t,s) dA, if there exists I ∈RI such that, for each � > 0, there exists δ > 0 such that d (S (F,P,δ,4) , I) < � for each P ∈ P (δ,4) . We denote by IR(4) the set of all ID-integrable function on 4, and by R([a,b]), IR([a,b]), the set of all R-integrable and IR-integrable functions on [a,b] ,respectively. Theorem 3.4 Let 4 = [a,b] × [c,d] . If F : 4→RI is ID-integrable on 4, then we have (ID) ∫∫ 4 F (t,s) dA = (IR) ∫ b a (IR) ∫ d c F (s,t) dsdt. In [21], Sadowska obtained the following Hermite-Hadamard inequality for interval-valued functions: Theorem 3.5 Let F : [a,b] → R+ I be an interval-valued function such that F (t) = [ F (t) ,F (t) ] and F ∈ IR([a,b]). Then F (a) + F (b) 2 ⊆ 1 b−a (IR) b∫ a F (t) dt ⊆ F ( a + b 2 ) . 6 Int. J. Anal. Appl. (2022), 20:40 4. Fractional Integrals In [2], Katugampola introduced a new fractional which generalizes the Riemann-Liouville and the Hadamard fractional integrals into a single form as follow. Definition 4.1 Let [a,b] ⊂R be a finite interval. Then, the left- and right-side Katugampola fractional integralsod order α > 0 of f ∈ Xpc (a,b) are defined by ρIαa+f (x) = ρ1−α Γ (α) ∫ x a tρ−1 (xρ − tρ)1−α f (t) dt and ρIαb−f (x) = ρ1−α Γ (α) ∫ b x tρ−1 (tρ −xρ)1−α f (t) dt where a < x < b and ρ > 0, if the integral exists. Theorem 4.2 Let α > 0 and ρ > 0. Then for x > a, 1. lim ρ→1 ρIα a+ f (x) = Jα a+ f (x) , 2. lim ρ→0+ ρIα a+ f (x) = Hα a+ f (x) . Similar results also hold for right-sided operators. Theorem 4.3 Let α > 0 and ρ > 0. Let f : [aρ,bρ] →R be a positive function with 0 ≤ a ≤ b and f ∈ Xpc (a,b) . If f is also a convex function on [a,b] , then the following inequalities hold: f ( aρ + bρ 2 , cρ + dρ 2 ) ≤ ραΓ (α + 1) 2 (bρ −aρ)α [ ρIαa+f (b ρ) +ρ Iαb−f (a ρ) ] ≤ f (aρ) + f (bρ) 2 where the fractional integral are considered for the function f (xρ) and evaluated at a and b, respectively. In [28], Yaldiz established the new definitions and theorem related Katugampola fractional integrals for two variables functions: Definition 4.4 Let f ∈ L1 ([aρ,bρ] × [cρ,dρ]) . The Katugampola fractional integrals ρI α,β a+,c+ f (x,y) , ρI α,β a+,d− f (x,y) ,ρ I α,β b−,c+ f (x,y) and ρIα,β b−,d− f (x,y) of α,β > 0 are defined by ρI α,β a+,c+ f (x,y) : = ρ1−α Γ (α) ρ1−β Γ (β) ∫ x a ∫ y c tρ−1sρ−1 (xρ − tρ)1−α (yρ − sρ)1−β f (t,s) dsdt, x > a,y > c, ρI α,β a+,d− f (x,y) : = ρ1−α Γ (α) ρ1−β Γ (β) ∫ x a ∫ d y tρ−1sρ−1 (xρ − tρ)1−α (sρ −yρ)1−β f (t,s) dsdt, x > a,y < d, Int. J. Anal. Appl. (2022), 20:40 7 ρI α,β b−,c+ f (x,y) : = ρ1−α Γ (α) ρ1−β Γ (β) ∫ b x ∫ y c tρ−1sρ−1 (tρ −xρ)1−α (yρ − sρ)1−β f (t,s) dsdt, x < b,y > c, ρI α,β b−,d− f (x,y) : = ρ1−α Γ (α) ρ1−β Γ (β) ∫ b x ∫ d y tρ−1sρ−1 (tρ −xρ)1−α (sρ −yρ)1−β f (t,s) dsdt, x < b,y < d, with a < x < b and c < y < d with ρ > 0. Similarly, we introduce the following integrals: ρIαa+f ( x, c + d 2 ) := ρ1−α Γ (α) ∫ x a tρ−1 (xρ − tρ)1−α f ( t, c + d 2 ) dt, x > a, ρIαb−f ( x, c + d 2 ) := ρ1−α Γ (α) ∫ b x tρ−1 (tρ −xρ)1−α f ( t, c + d 2 ) dt, x < b, ρI β c+ f ( a + b 2 ,y ) := ρ1−β Γ (β) ∫ y c sρ−1 (yρ − sρ)1−β f ( a + b 2 ,s ) ds, y > c, and ρIαd−f ( a + b 2 ,y ) := ρ1−β Γ (β) ∫ x a sρ−1 (sρ −yρ)1−β f ( a + b 2 ,s ) ds, y < d. Theorem 4.5 Let α,β > 0 and ρ > 0. Let f : 4ρ ⊂R2 →R be a coordinated convex on 4ρ := [aρ,bρ]×[cρ,dρ] in R2 with 0 ≤ a ≤ b, 0 ≤ c ≤ d and f ∈ L1 (4ρ) . Then the following inequalities hold: f ( aρ + bρ 2 , cρ + dρ 2 ) ≤ ρα+βΓ (α + 1) Γ (β + 1) 4 (bρ −aρ)α (dρ −cρ)β × [ ρI α,β a+,c+ f (b,d) +ρ I α,β a+,d− f (b,c) +ρ I α,β b−,c+ f (a,d) +ρ I α,β b−,d− f (a,c) ] ≤ f (aρ,cρ) + f (aρ,dρ) + f (bρ,cρ) + f (bρ,dρ) 4 with a < x < b and c < y < d. Definition 4.6 Let I ⊂ (0,∞) be a real interval and p ∈ R\{0} . A function f : I → R is said to be a p-convex function, if f ( [txp + (1 − t) yp] 1 p ) ≤ tf (x) + (1 − t) f (y) for all x,y ∈ I and t ∈ [0, 1] . If the inequality is reserved, then f is said to be p-concave. In [5], Fang and Shi established the following inequlaity Theorem 4.7 8 Int. J. Anal. Appl. (2022), 20:40 Let f : I →R be a p-convex function and a,b ∈ I with a < b. If f ∈ L [a,b] , then we have f ([ ap + bp 2 ]1 p ) ≤ p bp −ap b∫ a f (x) x1−p dx ≤ f (a) + f (b) 2 . In [27], Toplu et al. established the following inequality Theorem 4.8 Let f : I → R be a p-convex function, p > 0, α > 0 and a,b ∈ I with a < b. If f ∈ L [a,b] , then the following inequality for fractional integrals holds: f ([ ap + bp 2 ]1 p ) ≤ pαΓ (α + 1) bp −ap [ pIαa+f (b) + p Iαb−f (a) ] ≤ f (a) + f (b) 2 . In this paper, we can give a different version of the definition of the pq-convex function as below. Definition 4.9 Let I ⊂ (0,∞) ×(0,∞)be a real interval and p,q ∈ R\{0} . A function f : I → R is said to be a pq-convex function, if f ( [txp + (1 − t) yp] 1 p , [λzq + (1 −λ) wq] 1 q ) ≤ tλf (x,z) + t (1 −λ) f (x,w) + (1 − t) λf (y,z) + (1 − t) (1 −λ) f (y,w) for all (x,z) , (x,w) , (y,z) , (y,w) ∈ I and t,λ ∈ [0, 1] . If the inequality is reserved, then f is said to be pq-concave. We recall the following special functions and inequalities. (1)The Gamma Function: The Gamma Γ function is defined by Γ (z) = Γ (α) = ∫ ∞ 0 e−ttα−1dt for all complex numbers z with Re(z) > 0, respectively. The gamma function is a natural extension of the factorial from integers n to real (and complex) numbers z. (2)The Beta Function: β (x,y) = Γ (x) Γ (y) Γ (x + y) = ∫ 1 0 tx−1 (1 − t)y−1dt, x,y > 0. (3)The Hypergeometric Function: 2F1 (a,b; c,z) = 1 β (b,c −b) ∫ 1 0 tb−1 (1 − t)c−b−1 (1 −zt)−a dt, c > b > 0, |z| < 1. Int. J. Anal. Appl. (2022), 20:40 9 5. Main Result Theorem 5.1 Let f : I × I → R be an interval-valued pq-convex function such that f (t) = [ f (t) , f (t) ] and the order p,q > 0, α,β > 0 and a,b,c,d ∈ I with a < b and c < d. If f ∈ ID([a,b]×[c,d]), then the following inequality for fractional integrals holds: f ([ ap + bp 2 ]1 p , [ cq + dq 2 ]1 q ) ⊇ pαqβΓ (α + 1) Γ (β + 1) (bp −ap)α (dq −cq)β × [ p,qI α,β a+,c+ f (b,d) +p,q I α,β a+,d− f (b,c) +p,q I α,β b−,c+ f (a,d) +p,q I α,β b−,d− f (a,c) ] ⊇ f (a,c) + f (a,d) + f (b,c) + f (b,d) 4 . Proof : Since f is pq-convex function on [a,b] × [c,d] , we have for all (x,z) , (x,w) , (y,z) , (y,w) ∈ [a,b] × [c,d] (with t,λ = 1 2 ) f ([ xp + yp 2 ]1 p , [ zq + wq 2 ]1 q ) ⊇ f (x,z) + f (x,w) + f (y,z) + f (y,w) 4 . By choosing x = [tap + (1 − t) bp] 1 p ,y = [(1 − t) ap + tbp] 1 p ,z = [λcq + (1 −λ) dq] 1 q and w = [(1 −λ) cq + λdq] 1 q , then we get f ([ ap + bp 2 ]1 p , [ cq + dq 2 ]1 q ) ⊇ 1 4 [ f ( [tap + (1 − t) bp] 1 p , [λcq + (1 −λ) dq] 1 q ) + f ( [tap + (1 − t) bp] 1 p , [(1 −λ) cq + λdq] 1 q ) +f ( [(1 − t) ap + tbp] 1 p , [λcq + (1 −λ) dq] 1 q ) + f ( [(1 − t) ap + tbp] 1 p , [(1 −λ) cq + λdq] 1 q )] . Multiplying both sides of the inequality by tα−1λβ−1 and then integrating the resulting inequality with respect to t over [0, 1] and with respect to λ over [0, 1] , then we obtain, 4 αβ f ([ ap + bp 2 ]1 p , [ cq + dq 2 ]1 q ) ⊇ (ID) ∫ 1 0 ∫ 1 0 [ f ( [tap + (1 − t) bp] 1 p , [λcq + (1 −λ) dq] 1 q ) + f ( [tap + (1 − t) bp] 1 p , [(1 −λ) cq + λdq] 1 q ) +f ( [(1 − t) ap + tbp] 1 p , [λcq + (1 −λ) dq] 1 q ) + f ( [(1 − t) ap + tbp] 1 p , [(1 −λ) cq + λdq] 1 q )] dtdλ 10 Int. J. Anal. Appl. (2022), 20:40 ⊇ (ID) ∫ d c ∫ b a ( bp −xp bp −ap )α−1( dq −yq dq −cq )β−1 f (x,y) pxp−1 bp −ap qyq−1 dq −cq dxdy + (ID) ∫ d c ∫ b a ( bp −xp bp −ap )α−1( yq −cq dq −cq )β−1 f (x,y) pxp−1 bp −ap qyq−1 dq −cq dxdy + (ID) ∫ d c ∫ b a ( xp −ap bp −ap )α−1( dq −yq dq −cq )β−1 f (x,y) pxp−1 bp −ap qyq−1 dq −cq dxdy + (ID) ∫ d c ∫ b a ( xp −ap bp −ap )α−1( yq −cq dq −cq )β−1 f (x,y) pxp−1 bp −ap qyq−1 dq −cq dxdy = pq (bp −ap)α (dq −cq)β [ p,qI α,β a+,c+ f (b,d) +p,q I α,β a+,d− f (b,c) +p,q I α,β b−,c+ f (a,d) +p,q I α,β b−,d− f (a,c) ] . Thus we have f ([ ap + bp 2 ]1 p , [ cq + dq 2 ]1 q ) ⊇ pαqβΓ (α + 1) Γ (β + 1) (bp −ap)α (dq −cq)β × [ p,qI α,β a+,c+ f (b,d) +p,q I α,β a+,d− f (b,c) +p,q I α,β b−,c+ f (a,d) +p,q I α,β b−,d− f (a,c) ] which completes the proof of the first inequality. For the second inequalitiy, br using pq-convexity of f , we have f ( [tap + (1 − t) bp] 1 p , [λcq + (1 −λ) dq] 1 q ) ⊇ tλf (a,c) + t (1 −λ) f (a,d) + (1 − t) λf (b,c) + (1 − t) (1 −λ) f (b,d) , f ( [(1 − t) ap + tbp] 1 p , [λcq + (1 −λ) dq] 1 q ) ⊇ (1 − t) λf (a,c) + (1 − t) (1 −λ) f (a,d) + tλf (b,c) + t (1 −λ) f (b,d) , f ( [tap + (1 − t) bp] 1 p , [(1 −λ) cq + λdq] 1 q ) ⊇ t (1 −λ) f (a,c) + tλf (a,d) + (1 − t) (1 −λ) f (b,c) + (1 − t) λf (b,d) , and f ( [(1 − t) ap + tbp] 1 p , [(1 −λ) cq + λdq] 1 q ) ⊇ (1 − t) (1 −λ) f (a,c) + (1 − t) λf (a,d) + t (1 −λ) f (b,c) + tλf (b,d) . By adding these inequalities, then, we have Int. J. Anal. Appl. (2022), 20:40 11 f ( [tap + (1 − t) bp] 1 p , [λcq + (1 −λ) dq] 1 q ) + f ( [(1 − t) ap + tbp] 1 p , [λcq + (1 −λ) dq] 1 q ) +f ( [tap + (1 − t) bp] 1 p , [(1 −λ) cq + λdq] 1 q ) + f ( [(1 − t) ap + tbp] 1 p , [(1 −λ) cq + λdq] 1 q ) ⊇ f (a,c) + f (a,d) + f (b,c) + f (b,d) . Multiplying both sides of the inequality by tα−1λβ−1 ,α > 0,β > 0 and then integrating the resulting inequality with respect to t over [0, 1] and with respect to λ over [0, 1] , then we similarly obtain, pαqβΓ (α + 1) Γ (β + 1) (bp −ap)α (dq −cq)β × [ p,qI α,β a+,c+ f (b,d) +p,q I α,β a+,d− f (b,c) +p,q I α,β b−,c+ f (a,d) +p,q I α,β b−,d− f (a,c) ] ⊇ f (a,c) + f (a,d) + f (b,c) + f (b,d) 4 . Lemma 5.2 Let f : I × I → R be a partial differentiable function with 0 ≤ a < b and 0 ≤ c < d. Then the equality holds. Kf (α,β,a,b,c,d) = (bp −ap) (dq −cq) 4pq   ∫ 1 0 ∫ 1 0 [(1 − t)α − tα] [ (1 −λ)β −λβ ] ∂2 ∂t∂λ f (Mp (a,b,t) ,Mq (c,d,λ)) [tap + (1 − t) bp]1− 1 p [λcq + (1 −λ) dq]1− 1 q dtdλ   = 1 4 {f (a,c) + f (a,d) + f (b,c) + f (b,d) + pαqβΓ (α + 1) Γ (β + 1) (bp −ap)α (dq −cq)β [ p,qI α,β a+,c+ f (b,d) +p,q I α,β a+,d− f (b,c) +p,q I α,β b−,c+ f (a,d) +p,q I α,β b−,d− f (a,c) ] − pαΓ (α + 1) (bp −ap)α [pIαa+f (x,c) + p Iαb−f (x,c) + p Iαa+f (x,d) + p Iαb−f (x,d)] − qβΓ (β + 1) (dq −cq)β [ qI β c+ f (a,y) +q I β d− f (a,y) +q I β c+ f (b,y) +q I β d− f (b,y) ]} where Mp (a,b,t) = [tap + (1 − t) bp] 1 p and Mq (c,d,λ) = [λcq + (1 −λ) dq] 1 q . proof : Let I1 = (bp −ap) (dq −cq) 4pq {∫ 1 0 ∫ 1 0 (1 − t)α (1 −λ)β ∂ 2 ∂t∂λ f (Mp (a,b,t) ,Mq (c,d,λ)) [tap + (1 − t) bp]1− 1 p [λcq + (1 −λ) dq]1− 1 q dtdλ } I2 = (bp −ap) (dq −cq) 4pq {∫ 1 0 ∫ 1 0 (1 − t)α λβ ∂ 2 ∂t∂λ f (Mp (a,b,t) ,Mq (c,d,λ)) [tap + (1 − t) bp]1− 1 p [λcq + (1 −λ) dq]1− 1 q dtdλ } 12 Int. J. Anal. Appl. (2022), 20:40 I3 = (bp −ap) (dq −cq) 4pq {∫ 1 0 ∫ 1 0 tα (1 −λ)β ∂ 2 ∂t∂λ f (Mp (a,b,t) ,Mq (c,d,λ)) [tap + (1 − t) bp]1− 1 p [λcq + (1 −λ) dq]1− 1 q dtdλ } and I4 = (bp −ap) (dq −cq) 4pq {∫ 1 0 ∫ 1 0 tαλβ ∂ 2 ∂t∂λ f (Mp (a,b,t) ,Mq (c,d,λ)) [tap + (1 − t) bp]1− 1 p [λcq + (1 −λ) dq]1− 1 q dtdλ } . By using integrating by part, we have, I1 = 1 4 { f (b,d) − βq (dq −cq) ∫ d c yq−1 (yq −cq)1−β f (b,y) dy − αp (bp −ap) ∫ b a xp−1 (xp −ap)1−α f (x,d) dx + αp (bp −ap) βq (dq −cq) ∫ d c ∫ b a xp−1 (xp −ap)1−α yq−1 (yq −cq)1−β f (x,y) dxdy } and similarly we get I2 = −1 4 { f (b,c) − βq (dq −cq) ∫ d c yq−1 (dq −yq)1−β f (b,y) dy − αp (bp −ap) ∫ b a xp−1 (xp −ap)1−α f (x,c) dx + αp (bp −ap) βq (dq −cq) ∫ d c ∫ b a xp−1 (xp −ap)1−α yq−1 (dq −yq)1−β f (x,y) dxdy } I3 = −1 4 { f (a,d) − βq (dq −cq) ∫ d c yq−1 (yq −cq)1−β f (b,y) dy − αp (bp −ap) ∫ b a xp−1 (bp −xp)1−α f (x,d) dx + αp (bp −ap) βq (dq −cq) ∫ d c ∫ b a xp−1 (bp −xp)1−α yq−1 (yq −cq)1−β f (x,y) dxdy } and I4 = 1 4 { f (a,c) − βq (dq −cq) ∫ d c yq−1 (dq −yq)1−β f (b,y) dy − αp (bp −ap) ∫ b a xp−1 (bp −xp)1−α f (x,d) dx + αp (bp −ap) βq (dq −cq) ∫ d c ∫ b a xp−1 (bp −xp)1−α yq−1 (dq −yq)1−β f (x,y) dxdy } . So that we combine I1 − I2 − I3 + I4, we will get the equality. Theorem 5.3 Let f : I × I →R be an interval-valued pq-convex function such that f (t) = [ f (t) , f (t) ] and the order p,q > 0, α,β > 0 and a,b,c,d ∈ I with a < b and c < d. If f ∈ ID([a,b]×[c,d]) and ∣∣∣ ∂2∂t∂λf ∣∣∣m is pq −convex on [a,b] × [c,d] for m ≥ 1, then the following inequality for fractional integrals holds: Int. J. Anal. Appl. (2022), 20:40 13 ∣∣∣∣14 {f (a,c) + f (a,d) + f (b,c) + f (b,d) + pαqβΓ (α + 1) Γ (β + 1) (bp −ap)α (dq −cq)β [ p,qI α,β a+,c+ f (b,d) +p,q I α,β a+,d− f (b,c) +p,q I α,β b−,c+ f (a,d) +p,q I α,β b−,d− f (a,c) ] − pαΓ (α + 1) (bp −ap)α [pIαa+f (x,c) + p Iαb−f (x,c) + p Iαa+f (x,d) + p Iαb−f (x,d)] − qβΓ (β + 1) (dq −cq)β [ qI β c+ f (a,y) +q I β d− f (a,y) +q I β c+ f (b,y) +q I β d− f (b,y) ]∣∣∣∣ ⊇ (bp −ap) (dq −cq) 4pq M 1− 1 m 1 (α,β) × { M2 (α,β) ∣∣∣∣ ∂2∂t∂λf (a,c) ∣∣∣∣m + M3 (α,β) ∣∣∣∣ ∂2∂t∂λf (a,d) ∣∣∣∣m +M4 (α,β) ∣∣∣∣ ∂2∂t∂λf (b,c) ∣∣∣∣m + M5 (α,β) ∣∣∣∣ ∂2∂t∂λf (b,d) ∣∣∣∣m } 1 m . Proof : From Lemma by using the property of the modulus, the power mean inequality and the pq-convexity of ∣∣∣ ∂2∂t∂λf ∣∣∣m , then we have∣∣∣∣∣∣(bp −ap)(dq −cq)4pq  (ID) ∫ 1 0 ∫ 1 0 [(1− t)α − tα] [ (1−λ)β −λβ ] ∂2 ∂t∂λ f (Mp (a,b,t) ,Mq (c,d,λ)) [tap +(1− t)bp]1− 1 p [λcq +(1−λ)dq]1− 1 q dtdλ   ∣∣∣∣∣∣ ⊇ (bp −ap)(dq −cq) 4pq ×  (ID) ∫ 1 0 ∫ 1 0 ∣∣∣[(1− t)α − tα][(1−λ)β −λβ]∣∣∣ · ∣∣∣ ∂2∂t∂λf (Mp (a,b,t) ,Mq (c,d,λ))∣∣∣ [tap +(1− t)bp]1− 1 p [λcq +(1−λ)dq]1− 1 q dtdλ   ⊇ (bp −ap)(dq −cq) 4pq  (ID) ∫ 1 0 ∫ 1 0 ∣∣∣[(1− t)α − tα][(1−λ)β −λβ]∣∣∣ [tap +(1− t)bp]1− 1 p [λcq +(1−λ)dq]1− 1 q dtdλ   1− 1 m ×  (ID) ∫ 1 0 ∫ 1 0 ∣∣∣[(1− t)α − tα][(1−λ)β −λβ]∣∣∣ · ∣∣∣ ∂2∂t∂λf (Mp (a,b,t) ,Mq (c,d,λ))∣∣∣m [tap +(1− t)bp]1− 1 p [λcq +(1−λ)dq]1− 1 q dtdλ   1 m ⊇ (bp −ap)(dq −cq) 4pq  (ID) ∫ 1 0 ∫ 1 0 ∣∣∣[(1− t)α − tα][(1−λ)β −λβ]∣∣∣ [tap +(1− t)bp]1− 1 p [λcq +(1−λ)dq]1− 1 q dtdλ   1− 1 m ×  (ID) ∫ 1 0 ∫ 1 0 ∣∣∣[(1− t)α − tα][(1−λ)β −λβ]∣∣∣ [tap +(1− t)bp]1− 1 p [λcq +(1−λ)dq]1− 1 q × [ tλ ∣∣∣∣ ∂2∂t∂λf (a,c) ∣∣∣∣m + t (1−λ) ∣∣∣∣ ∂2∂t∂λf (a,d) ∣∣∣∣m +(1− t)λ ∣∣∣∣ ∂2∂t∂λf (b,c) ∣∣∣∣m +(1− t)(1−λ) ∣∣∣∣ ∂2∂t∂λf (b,d) ∣∣∣∣m dtdλ } 1 m = (bp −ap)(dq −cq) 4pq M 1−1 q 1 (α,β)× { M2 (α,β) ∣∣∣∣ ∂2∂t∂λf (a,c) ∣∣∣∣m +M3 (α,β) ∣∣∣∣ ∂2∂t∂λf (a,d) ∣∣∣∣m +M4 (α,β) ∣∣∣∣ ∂2∂t∂λf (b,c) ∣∣∣∣m +M5 (α,β) ∣∣∣∣ ∂2∂t∂λf (b,d) ∣∣∣∣m } 1 m , 14 Int. J. Anal. Appl. (2022), 20:40 where by simple computation, we obtain, M1 (α,β) = ∫ 1 0 ∫ 1 0 [(1 − t)α − tα] [ (1 −λ)β −λβ ] [tap + (1 − t) bp]1− 1 p [λcq + (1 −λ) dq]1− 1 q dtdλ = ∫ 1 0 [(1 − t)α − tα] [tap + (1 − t) bp]1− 1 p dt ∫ 1 0 [ (1 −λ)β −λβ ] [λcq + (1 −λ) dq]1− 1 q dλ = { b1−p α + 1 [ 2F1 ( 1 − 1 p , 1; α + 2, 1 − ap bp ) +2 F1 ( 1 − 1 p ,α + 1; α + 2, 1 − ap bp )]} × { d1−q β + 1 [ 2F1 ( 1 − 1 q , 1; β + 2, 1 − cq dq ) +2 F1 ( 1 − 1 q ,β + 1; β + 2, 1 − cq dq )]} M2 (α,β) = ∫ 1 0 ∫ 1 0 [(1 − t)α − tα] [ (1 −λ)β −λβ ] [tap + (1 − t) bp]1− 1 p [λcq + (1 −λ) dq]1− 1 q tλdtdλ = ∫ 1 0 [(1 − t)α − tα] [tap + (1 − t) bp]1− 1 p tdt ∫ 1 0 [ (1 −λ)β −λβ ] [λcq + (1 −λ) dq]1− 1 q λdλ = { b1−p α + 2 [ 1 α + 12 F1 ( 1 − 1 p , 2; α + 3, 1 − ap bp ) +2 F1 ( 1 − 1 p ,α + 2; α + 3, 1 − ap bp )]} × { d1−q β + 2 [ 1 β + 12 F1 ( 1 − 1 q , 2; β + 3, 1 − cq dq ) +2 F1 ( 1 − 1 q ,β + 2; β + 3, 1 − cq dq )]} M3 (α,β) = ∫ 1 0 ∫ 1 0 [(1 − t)α − tα] [ (1 −λ)β −λβ ] [tap + (1 − t) bp]1− 1 p [λcq + (1 −λ) dq]1− 1 q t (1 −λ) dtdλ = ∫ 1 0 [(1 − t)α − tα] [tap + (1 − t) bp]1− 1 p tdt ∫ 1 0 [ (1 −λ)β −λβ ] [λcq + (1 −λ) dq]1− 1 q (1 −λ) dλ = { b1−p α + 2 [ 1 α + 12 F1 ( 1 − 1 p , 2; α + 3, 1 − ap bp ) +2 F1 ( 1 − 1 p ,α + 2; α + 3, 1 − ap bp )]} × { d1−q β + 1 [ 2F1 ( 1 − 1 q , 1; β + 3, 1 − cq dq ) + 1 β + 12 F1 ( 1 − 1 q ,β + 1; β + 3, 1 − cq dq )]} M4 (α,β) = ∫ 1 0 ∫ 1 0 [(1 − t)α − tα] [ (1 −λ)β −λβ ] [tap + (1 − t) bp]1− 1 p [λcq + (1 −λ) dq]1− 1 q (1 − t) λdtdλ = ∫ 1 0 [(1 − t)α − tα] [tap + (1 − t) bp]1− 1 p (1 − t) dt ∫ 1 0 [ (1 −λ)β −λβ ] [λcq + (1 −λ) dq]1− 1 q λdλ = { b1−p α + 1 [ 2F1 ( 1 − 1 p , 1; α + 3, 1 − ap bp ) + 1 α + 12 F1 ( 1 − 1 p ,α + 1; α + 3, 1 − ap bp )]} × { d1−q β + 2 [ 1 β + 12 F1 ( 1 − 1 q , 2; β + 3, 1 − cq dq ) +2 F1 ( 1 − 1 q ,β + 2; β + 3, 1 − cq dq )]} Int. J. Anal. Appl. (2022), 20:40 15 M5 (α,β) = ∫ 1 0 ∫ 1 0 [ (1 − t)α − tα ][ (1 −λ)β −λβ ] [tap + (1 − t) bp]1− 1 p [λcq + (1 −λ) dq]1− 1 q (1 − t) (1 −λ) dtdλ = ∫ 1 0 [ (1 − t)α − tα ] [tap + (1 − t) bp]1− 1 p (1 − t) dt ∫ 1 0 [ (1 −λ)β −λβ ] [λcq + (1 −λ) dq]1− 1 q (1 −λ) dλ = { b1−p α + 1 [ 2F1 ( 1 − 1 p , 1; α + 3, 1 − ap bp ) + 1 α + 12 F1 ( 1 − 1 p ,α + 1; α + 3, 1 − ap bp )]} × { d1−q β + 1 [ 2F1 ( 1 − 1 q , 1; β + 3, 1 − cq dq ) + 1 β + 12 F1 ( 1 − 1 q ,β + 1; β + 3, 1 − cq dq )]} . Theorem 5.4 Let f : I × I →R be an interval-valued pq-convex function such that f (t) = [ f (t) , f (t) ] and the order p,q > 0, α,β > 0 and a,b,c,d ∈ I with a < b and c < d. If f ∈ ID([a,b]×[c,d]) and ∣∣∣ ∂2∂t∂λf ∣∣∣m is pq −convex on [a,b] × [c,d] for m ≥ 1, then the following inequality for fractional integrals holds:∣∣∣∣14 {f (a,c) + f (a,d) + f (b,c) + f (b,d) + pαqβΓ (α + 1) Γ (β + 1) (bp −ap)α (dq −cq)β [ p,qI α,β a+,c+ f (b,d) +p,q I α,β a+,d− f (b,c) +p,q I α,β b−,c+ f (a,d) +p,q I α,β b−,d− f (a,c) ] − pαΓ (α + 1) (bp −ap)α [pIαa+f (x,c) + p Iαb−f (x,c) + p Iαa+f (x,d) + p Iαb−f (x,d)] − qβΓ (β + 1) (dq −cq)β [ qI β c+ f (a,y) +q I β d− f (a,y) +q I β c+ f (b,y) +q I β d− f (b,y) ]∣∣∣∣ ⊇ (bp −ap) (dq −cq) 4pq { M 1 n 6 (α,β) + M 1 n 7 (α,β) + M 1 n 8 (α,β) + M 1 n 9 (α,β) } × { 1 4 [∣∣∣∣ ∂2∂t∂λf (a,c) ∣∣∣∣m + ∣∣∣∣ ∂2∂t∂λf (a,d) ∣∣∣∣m + ∣∣∣∣ ∂2∂t∂λf (b,c) ∣∣∣∣m + ∣∣∣∣ ∂2∂t∂λf (b,d) ∣∣∣∣m ]} 1 m and 1 m + 1 n = 1. Proof : From Lemma by using the property of the modulus, the Hölder inequality and the pq-convexity of∣∣∣ ∂2∂t∂λf ∣∣∣m , then we have∣∣∣∣∣∣(bp −ap)(dq −cq)4pq  (ID) ∫ 1 0 ∫ 1 0 [(1− t)α − tα] [ (1−λ)β −λβ ] ∂2 ∂t∂λ f (Mp (a,b,t) ,Mq (c,d,λ)) [tap +(1− t)bp]1− 1 p [λcq +(1−λ)dq]1− 1 q dtdλ   ∣∣∣∣∣∣ ⊇ (bp −ap)(dq −cq) 4pq ×   ( (ID) ∫ 1 0 ∫ 1 0 (1− t)αn (1−λ)βn [tap +(1− t)bp]1− 1 p [λcq +(1−λ)dq]1− 1 q dtdλ )1 n × ( (ID) ∫ 1 0 ∫ 1 0 ∣∣∣∣ ∂2∂t∂λf (Mp (a,b,t) ,Mq (c,d,λ)) ∣∣∣∣m dtdλ ) 1 m 16 Int. J. Anal. Appl. (2022), 20:40 + ( (ID) ∫ 1 0 ∫ 1 0 tαn (1 −λ)βn [tap + (1 − t) bp]1− 1 p [λcq + (1 −λ) dq]1− 1 q dtdλ )1 n × ( (ID) ∫ 1 0 ∫ 1 0 ∣∣∣∣ ∂2∂t∂λf (Mp (a,b,t) ,Mq (c,d,λ)) ∣∣∣∣m dtdλ ) 1 m + ( (ID) ∫ 1 0 ∫ 1 0 (1 − t)αn λβn [tap + (1 − t) bp]1− 1 p [λcq + (1 −λ) dq]1− 1 q dtdλ )1 n × ( (ID) ∫ 1 0 ∫ 1 0 ∣∣∣∣ ∂2∂t∂λf (Mp (a,b,t) ,Mq (c,d,λ)) ∣∣∣∣m dtdλ ) 1 m + ( (ID) ∫ 1 0 ∫ 1 0 tαnλβn [tap + (1 − t) bp]1− 1 p [λcq + (1 −λ) dq]1− 1 q dtdλ )1 n × ( (ID) ∫ 1 0 ∫ 1 0 ∣∣∣∣ ∂2∂t∂λf (Mp (a,b,t) ,Mq (c,d,λ)) ∣∣∣∣m dtdλ ) 1 m } ⊇ (bp −ap) (dq −cq) 4pq { M 1 n 6 (α,β) + M 1 n 7 (α,β) + M 1 n 8 (α,β) + M 1 n 9 (α,β) } × { (ID) ∫ 1 0 ∫ 1 0 tλ ∣∣∣∣ ∂2∂t∂λf (a,c) ∣∣∣∣m + t (1 −λ) ∣∣∣∣ ∂2∂t∂λf (a,d) ∣∣∣∣m + (1 − t) λ ∣∣∣∣ ∂2∂t∂λf (b,c) ∣∣∣∣m + (1 − t) (1 −λ) ∣∣∣∣ ∂2∂t∂λf (b,d) ∣∣∣∣m dtdλ } 1 m = (bp −ap) (dq −cq) 4pq { M 1 n 6 (α,β) + M 1 n 7 (α,β) + M 1 n 8 (α,β) + M 1 n 9 (α,β) } × { 1 4 [∣∣∣∣ ∂2∂t∂λf (a,c) ∣∣∣∣m + ∣∣∣∣ ∂2∂t∂λf (a,d) ∣∣∣∣m + ∣∣∣∣ ∂2∂t∂λf (b,c) ∣∣∣∣m + ∣∣∣∣ ∂2∂t∂λf (b,d) ∣∣∣∣m ]} 1 m , and by calculations of integrals, we obtain, M6 (α,β) = ∫ 1 0 ∫ 1 0 (1 − t)αn (1 −λ)βn [tap + (1 − t) bp]1− 1 p [λcq + (1 −λ) dq]1− 1 q dtdλ = (∫ 1 0 (1 − t)αn [tap + (1 − t) bp]1− 1 p dt )(∫ 1 0 (1 −λ)βn [λcq + (1 −λ) dq]1− 1 q dλ ) = b(1−p)n αp + 1 2F1 ( n− n p , 1; αn + 2, 1 − ap bp ) · d(1−q)n βq + 1 2F1 ( n− n q , 1; βn + 2, 1 − cq dq ) M7 (α,β) = ∫ 1 0 ∫ 1 0 tαn (1 −λ)βn [tap + (1 − t) bp]1− 1 p [λcq + (1 −λ) dq]1− 1 q dtdλ = (∫ 1 0 tαn [tap + (1 − t) bp]1− 1 p dt )(∫ 1 0 (1 −λ)βn [λcq + (1 −λ) dq]1− 1 q dλ ) = b(1−p)n αp + 1 2F1 ( n− n p ,αn + 1; αn + 2, 1 − ap bp ) · d(1−q)n βq + 1 2F1 ( n− n q , 1; βn + 2, 1 − cq dq ) Int. J. Anal. Appl. (2022), 20:40 17 M8 (α,β) = ∫ 1 0 ∫ 1 0 (1 − t)αn λβn [tap + (1 − t) bp]1− 1 p [λcq + (1 −λ) dq]1− 1 q dtdλ = (∫ 1 0 (1 − t)αn [tap + (1 − t) bp]1− 1 p dt )(∫ 1 0 λβn [λcq + (1 −λ) dq]1− 1 q dλ ) = b(1−p)n αp + 1 2F1 ( n− n p ,αn + 1; αn + 2, 1 − ap bp ) · d(1−q)n βq + 1 2F1 ( n− n q ,βn + 1; βn + 2, 1 − cq dq ) M9 (α,β) = ∫ 1 0 ∫ 1 0 tαnλβn [tap + (1 − t) bp]1− 1 p [λcq + (1 −λ) dq]1− 1 q dtdλ = (∫ 1 0 tαn [tap + (1 − t) bp]1− 1 p dt )(∫ 1 0 λβn [λcq + (1 −λ) dq]1− 1 q dλ ) = b(1−p)n αp + 1 2F1 ( n− n p ,αn + 1; αn + 2, 1 − ap bp ) · d(1−q)n βq + 1 2F1 ( n− n q ,βn + 1; βn + 2, 1 − cq dq ) Conclusion In this work, the author established Hermite-Hadamard type inequalities via Katugampola fractional integral. Furthermore, the author extend the ineqalities on interval-valued coordinated. It is an interesting issue, and many researchers work to generalize the Ostrowski’ inequalities, Chebyshev type inequalities and Opial-type inequalities on fuzzy interval-valued set. We hope to establish the general fractional integrals in their future research. Author Contributions: The author contributed has read and agreed to the published version of the manuscript. Acknowledgment: The Author would like to express their sincere to the editor and the anonmous reviewers for their helpful comments and suggestions. Funding: The work was supportes by the Ministry of Science and Technology of Taiwan (MOST110- 2115-M-027-003-MY2). Conflicts of Interest: The author(s) declare that there are no conflicts of interest regarding the publication of this paper. References [1] H.Y. Budak, C.C. Bilişik, A. Kashuri and M.A. Ali, Hermite-Hadamard Type Inequalities For The Interval-Valued Harmonically h-Convex Functions Via Fractional Integrals, Appl. Math. E-Notes, 21 (2021), 12-32. https://www. emis.de/journals/AMEN/2021/AMEN-200121.pdf. https://www.emis.de/journals/AMEN/2021/AMEN-200121.pdf https://www.emis.de/journals/AMEN/2021/AMEN-200121.pdf 18 Int. J. Anal. Appl. 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Appl. 2018 (2018), 302. https://doi.org/10.1186/s13660-018-1896-3. https://dergipark.org.tr/en/download/article-file/351071 https://dergipark.org.tr/en/download/article-file/351071 https://doi.org/10.3390/math8040534 https://doi.org/10.1186/s13662-020-03200-z https://doi.org/10.22190/fumi1901149t https://doi.org/10.22190/fumi1901149t https://doi.org/10.3934/math.2021210 https://doi.org/10.1186/s13662-020-03028-7 https://doi.org/10.1186/s13660-018-1896-3 1. Introduction 2. Interval Calculus 3. Intgral of Interval-Valued Functions 4. Fractional Integrals 5. Main Result Conclusion References