Some bounds for the derivation of the integral mean of a function defined on a compact disk from the value at the central point and related results are presented. A version of Ostrowski’s inequality for functions defined on the unit disk is also presented

Barnett, Neil S

Cirstea, Florica-Corina

Dragomir, Sever S

English

Victoria University Eprints Repository

SOME INEQUALITIES FOR THE INTEGRAL MEAN OFHO¨LDER CONTINUOUS FUNCTIONS DEFINED ON DISKS INA PLANEN.S. BARNETT, F.C. S¸T. CIˆRSTEA, AND S.S. DRAGOMIRAbstract. Some bounds for the derivation of the integral mean of a functiondefined on a compact disk from the value at the central point and relatedresults are presented. A version of Ostrowski’s inequality for functions definedon the unit disk is also presented.1. IntroductionIn a recent paper [2], S.S. Dragomir proved the following Hermite-Hadamardtype inequality for convex functions on the compact disk D (C,R).Theorem 1. If the mapping f : D (C,R) → R is convex on the disk D (C,R)centered at the point C and having the radius R > 0, then,(1.1) f (C) ≤ 1piR2∫∫D(C,R)f (x, y) dx dy ≤ 12piR∫σ(C,R)f (γ) dl (γ) ,where σ (C,R) is the circle centered at the point C with radius R.The above inequalities are sharp.In fact, the author provided an upper bound for the integral mean,1piR2∫∫D(C,R)f (x, y) dx dy,i.e., the inequality (see [2], inequality (2.4))(1.2)1piR2∫∫D(C,R)f (x, y) dx dy ≤ 23· 12piR∫σ(C,R)f (γ) dl (γ) +13f (C) .Applications for different functionals associated with (1.1) were also provided.For other Hermite-Hadamard type inequalities in several dimensions see [4]-[7]and the monograph on line [3].In this paper we drop the assumption of convexity and compare the integralmean1piR2∫∫D(C,R)f (x, y) dx dywith different quantities including f (C) and 12piR∫σ(C,R)f (γ) dl (γ), under the as-sumption of certain Ho¨lder type conditions for f .Date: October 18, 2001.1991 Mathematics Subject Classification. Primary 26D15; Secondary 26D10.Key words and phrases. Integral Means, Ostrowski’s Inequality, Hermite-Hadamard TypeInequalities.12 N.S. BARNETT, F.C. S¸T. CIˆRSTEA, AND S.S. DRAGOMIR2. Some Inequalities for Ho¨lder Type FunctionsThe following result holds.Theorem 2. Let f : D (C,R)→ R (C = (a, b) ∈ R2) be a function satisfying theHo¨lder type condition:(2.1) |f (a, b)− f (x, y)| ≤ L1 |x− a|α1 + L2 |y − b|α2 ,where L1, L2 > 0 and α1, α2 ∈ (−1,∞), (x, y) ∈ D (C,R) \C, then we have,∣∣∣∣∣f (C)− 1piR2∫∫D(C,R)f (x, y) dx dy∣∣∣∣∣(2.2)≤ 2√pi[Rα1α1 + 2· Γ(α1+12)Γ(α1+22) · L1 + Rα2α2 + 2· Γ(α2+12)Γ(α2+22) · L2] .For α1, α2 > 0, the constant 2√pi is sharp.Proof. Integrating (2.1) on D (C,R), we get∣∣∣∣∣f (C)− 1piR2∫∫D(C,R)f (x, y) dx dy∣∣∣∣∣(2.3)≤ 1piR2∫∫D(C,R)|f (a, b)− f (x, y)| dx dy≤ 1piR2[L1∫∫D(C,R)|x− a|α1 dx dy + L2∫∫D(C,R)|y − b|α2 dx dy].DenoteA1 :=∫∫D(C,R)|x− a|α1 dx dy and A2 :=∫∫D(C,R)|y − b|α2 dx dy.If we use the change of coordinates x = a+ r cos θ, r ∈ [0, R] , θ ∈ [0, 2pi] ,y = b+ r sin θ,then∂ (x, y)∂ (r, θ)= randA1 =∫ R0∫ 2pi0rα1 |cos θ|α1 · r drdθ =∫ R0rα1+1dr ·∫ 2pi0|cos θ|α1 dθ.Since ∫ R0rα1+1dr =Rα1+2α1 + 2and ∫ 2pi0|cos θ|α1 dθ = 4 ·∫ pi20(cos θ)α1 dθ = 4 ·√pi2· Γ(α1+12)Γ(α1+22) ,thenA1 =2√piα1 + 2Rα1+2 · Γ(α1+12)Γ(α1+22) .HO¨LDER CONTINUOUS FUNCTIONS 3Also,A2 = 4 · Rα2+2α2 + 2·∫ pi20(sin θ)α2 dθand since ∫ pi20(sin θ)α2 dθ =√pi2· Γ(α2+12)Γ(α2+22)we deduceA2 =2√piα2 + 2Rα2+2 · Γ(α2+12)Γ(α2+22) .Now using (2.3) we deduce the desired inequality (2.2).To prove the sharpness of the constant 2√pi, we observe that if we choose (forα1, α2 > 0)f : D (C,R)→ R, f (x, y) = L1 |x− a|α1 + L2 |y − b|α2 ,then ∣∣∣∣∣f (C)− 1piR2∫∫D(C,R)f (x, y) dx dy∣∣∣∣∣=1piR2∫∫D(C,R)|f (a, b)− f (x, y)| dx dy=1piR2[L1∫∫D(C,R)|x− a|α1 dx dy + L2∫∫D(C,R)|y − b|α2 dx dy].The theorem is now completely proved.The following corollary for Lipschitzian functions holds.Corollary 1. If f : D (C,R)→ R satisfies the Lipschitzian condition(2.4) |f (a, b)− f (x, y)| ≤M1 |x− a|+M2 |y − b| , (x, y) ∈ D (C,R) ,where M1,M2 > 0, then we have the inequality,(2.5)∣∣∣∣∣f (C)− 1piR2∫∫D(C,R)f (x, y) dx dy∣∣∣∣∣ ≤ 43piR (M1 +M2) .The constant in (2.5), 43pi , is sharp.The proof follows by the above theorem with α1 = α2 = 1.In practical applications, the following corollary may be useful.Corollary 2. If f : D (C,R) → R has continuous partial derivatives on D (C,R)and ∥∥∥∥∂f∂x∥∥∥∥D(C,R),∞: = sup(x,y)∈D(C,R)∣∣∣∣∂f (x, y)∂x∣∣∣∣ <∞,∥∥∥∥∂f∂y∥∥∥∥D(C,R),∞: = sup(x,y)∈D(C,R)∣∣∣∣∂f (x, y)∂y∣∣∣∣ <∞;4 N.S. BARNETT, F.C. S¸T. CIˆRSTEA, AND S.S. DRAGOMIRthen ∣∣∣∣∣f (C)− 1piR2∫∫D(C,R)f (x, y) dx dy∣∣∣∣∣(2.6)≤ 43piR[∥∥∥∥∂f∂x∥∥∥∥D(C,R),∞+∥∥∥∥∂f∂y∥∥∥∥D(C,R),∞].The constant 43pi is sharp.The following theorem also holds.Theorem 3. Assume that the function f : D (C,R)→ R satisfies the condition|f (r cos θ + a, r sin θ + b)− f (R cos θ + a,R sin θ + b)|(2.7)≤ L1 (R− r)α1 |cos θ|α2 + L2 (R− r)α3 |sin θ|α4for all r ∈ [0, R], θ ∈ [0, 2pi], where αi ∈ (−1,∞)(i = 1, 4), then,∣∣∣∣∣ 1piR2∫∫D(C,R)f (x, y) dx dy − 12piR∫σ(C,R)f (γ) dl (γ)∣∣∣∣∣(2.8)≤ 2√pi[Rα1(α1 + 1) (α1 + 2)· Γ(α2+12)Γ(α2+22) · L1+Rα3(α3 + 1) (α3 + 2)· Γ(α4+12)Γ(α4+22) · L2] .Proof. If we multiply the condition (2.7) by r > 0, we get,|f (r cos θ + a, r sin θ + b) r − f (R cos θ + a,R sin θ + b) r|(2.9)≤ L1r (R− r)α1 |cos θ|α2 + L2r (R− r)α3 |sin θ|α4 .Integrating (2.9) over [0, R]× [0, 2pi] and using the change of variable x = r cos θ + a, r ∈ [0, R] , θ ∈ [0, 2pi] ,y = r sin θ + b,for which we have∂ (x, y)∂ (r, θ)= r,we obtain ∣∣∣∣∣∫ R0∫ 2pi0f (r cos θ + a, r sin θ + b) r drdθ(2.10)− R22∫ 2pi0f (R cos θ + a,R sin θ + b) dθ∣∣∣∣≤ L1∫ R0r (R− r)α1 dr ·∫ 2pi0|cos θ|α2 dθ+L2∫ R0r (R− r)α3 dr ·∫ 2pi0|sin θ|α4 dθ.HO¨LDER CONTINUOUS FUNCTIONS 5Since ∫ R0∫ 2pi0f (r cos θ + a, r sin θ + b) r drdθ =∫∫D(C,R)f (x, y) dx dy,∫ 2pi0f (R cos θ + a,R sin θ + b) dθ =∫σ(C,R)f (γ) dl (γ)(=∫ 2pi0f (x (θ) , y (θ))[(x˙ (θ))2 + (y˙ (θ))2] 12dθ),∫ R0r (R− r)α1 dr = Rα1+2(α1 + 1) (α1 + 2),∫ R0r (R− r)α3 dr = Rα3+2(α3 + 1) (α3 + 2),∫ 2pi0|cos θ|α2 dθ = 4∫ pi20[cos θ]α2 dθ = 2√piΓ(α2+12)Γ(α2+22) ,∫ 2pi0|sin θ|α4 dθ = 4∫ pi20[sin θ]α4 dθ = 2√piΓ(α4+12)Γ(α4+22) ,then, by (2.10), on dividing by piR2, we deduce the desired result (2.8).The following corollary for Lipschitzian functions holds.Corollary 3. Assume that the function f : D (C,R) → R is Lipschitzian onD (C,R) with the constants K1,K2 > 0, i.e.,(2.11) |f (x, y)− f (u, v)| ≤ K1 |x− u|+K2 |y − v| for any (x, y) ∈ D (C,R) ,then, ∣∣∣∣∣ 1piR2∫∫D(C,R)f (x, y) dx dy − 12piR∫σ(C,R)f (γ) dl (γ)∣∣∣∣∣(2.12)≤ 2R3pi(K1 +K2) .A more practical result is embodied in the following corollary.Corollary 4. Assume that f : D (C,R)→ R has partial derivatives continuous onD (C,R), then, ∣∣∣∣∣ 1piR2∫∫D(C,R)f (x, y) dx dy − 12piR∫σ(C,R)f (γ) dl (γ)∣∣∣∣∣(2.13)≤ 2R3pi[∥∥∥∥∂f∂x∥∥∥∥D(C,R),∞+∥∥∥∥∂f∂y∥∥∥∥D(C,R),∞].It is also quite natural to compare the value of the function in C with the integralmean taken on the boundary.Theorem 4. Assume that the function f : D (C,R)→ R satisfies the condition:|f (a, b)− f (R cos θ + a,R sin θ + b)|(2.14)≤ L1Rα1 |cos θ|α2 + L2Rα3 |sin θ|α4 , θ ∈ [0, 2pi] ,6 N.S. BARNETT, F.C. S¸T. CIˆRSTEA, AND S.S. DRAGOMIRwhere L1, L2 > 0 and α1, α2, α3, α4 ∈ (−1,∞), then we have,∣∣∣∣∣f (C)− 12piR∫σ(C,R)f (γ) dl (γ)∣∣∣∣∣(2.15)≤ 1√pi[L1Rα1 · Γ(α2+12)Γ(α2+22) + L2Rα3 · Γ (α4+12 )Γ (α4+22 )].The constant 1√piis sharp for α1 = α2 and α3 = α4.Proof. Integrating the condition (2.14) and θ ∈ [0, 2pi], we have∣∣∣∣2pif (a, b)− ∫ 2pi0f (R cos θ + a,R sin θ + b) dθ∣∣∣∣≤ L1Rα1∫ 2pi0|cos θ|α2 dθ + L2Rα3∫ 2pi0|sin θ|α4 dθ= L1Rα1 · 2√pi · Γ(α2+12)Γ(α2+22) + L2Rα3 · 2√pi · Γ (α4+12 )Γ (α4+22 )and dividing by 2pi, we deduce (2.15).The equality in (2.15) holds for f (x, y) = L1 |x− a|α1 + L2 |y − b|α3 .Corollary 5. If the function f : D (C,R)→ R satisfies the Lipschitzian condition(2.4), then we have the inequality:(2.16)∣∣∣∣∣f (C)− 12piR∫σ(C,R)f (γ) dl (γ)∣∣∣∣∣ ≤ 2Rpi [M1 +M2] .The constant 2pi is sharp.Corollary 6. If f : D (C,R)→ R has continuous partial derivatives on D (C,R),then ∣∣∣∣∣f (C)− 12piR∫σ(C,R)f (γ) dl (γ)∣∣∣∣∣(2.17)≤ 2Rpi[∥∥∥∥∂f∂x∥∥∥∥D(C,R),∞+∥∥∥∥∂f∂y∥∥∥∥D(C,R),∞].The constant 2pi is sharp.3. An Ostrowski Type Inequality on the DiskUsing the notation(3.1) pFq a1, . . . , ap ; xb1, . . . , bq := ∞∑n=0(a1)n · · · (ap)n(b1)n · · · (bq)n· xnn!,where(a)0 := 1, (a)n := a (a+ 1) · · · (a+ n− 1) , n ≥ 1, n ∈ Nwe have Euler’s representation formula(3.2) 2F1[a, bc; y]=Γ (c)Γ (b) Γ (c− b)∫ 10(1− yt)−a tb−1 (1− t)c−b−1 dt,HO¨LDER CONTINUOUS FUNCTIONS 7where |y| < 1, Re c > Re b > 0 (see for example [1]).The following technical lemma holds.Lemma 1. For x ∈ [−1, 1] and α > −1, consider the integral(3.3) I (α, x) :=∫ 1−1|x− u|α√1− u2 du.We then have the representation,(3.4) I (α, x) =√2pi2· Γ (α+ 1)Γ(α+ 52) {(x+ 1)α+ 32 2F1 [ − 12 , 32α+ 52 ; x+ 12]+ (1− x)α+ 32 2F1[ − 12 , 32α+ 52;1− x2]}, if x ∈ (−1, 1)2α+2B(32, α+32), if x ∈ {−1, 1}Proof. Let x ∈ (−1, 1). First, observe thatI (α, x) =∫ x−1(x− u)α (1− u) 12 (u+ 1) 12 du+∫ 1x(u− x)α (1− u) 12 (1 + u) 12 du.Applying the change of variable,u = (1− λ) (−1) + λx, λ ∈ [0, 1]in the first integral, we obtain,I1 (α, x) =∫ x−1(x− u)α (1− u) 12 (u+ 1) 12 du= (x+ 1)α+32∫ 10(1− λ)α λ 12 [2− (1 + x)λ] 12 dλ=√2 (x+ 1)α+32∫ 10(1− λ)α+ 52− 32−1 λ 32−1(1−(x+ 12)λ)−(− 12 )dλ.Using (3.2) for a = − 12 , b = 32 , c = α + 52 , y = x+12 , we observe that |y| < 1,Re c > Re b > 0 and then∫ 10(1− λ)α+ 52− 32−1 λ 32−1(1−(x+ 12)λ)−(− 12 )dλ=Γ(32)Γ (α+ 1)Γ(α+ 52) 2F1 [ − 12 , 32α+ 52 ; x+ 12]giving(3.5) I1 (α, x) =√2pi2· Γ (α+ 1)Γ(α+ 52) (x+ 1)α+ 32 2F1 [ − 12 , 32α+ 52 ; x+ 12].If we consider the change of variableu = (1− λ)x+ λ, λ ∈ [0, 1]8 N.S. BARNETT, F.C. S¸T. CIˆRSTEA, AND S.S. DRAGOMIRin the second integral, we obtain:I2 (α, x) =∫ 1x(u− x)α (1− u) 12 (1 + u) 12 du= (1− x)α+ 32∫ 10λα (1− λ) 12 [1 + x+ λ (1− x)] 12 dλ= (1− x)α+ 32∫ 10(1− µ)α µ 12 [2− (1− x)µ] 12 dµ=√2 (1− x)α+ 32∫ 10(1− µ)α µ 12[1−(1− x2)µ] 12dµ=√2 (1− x)α+ 32∫ 10(1− µ)α+ 52− 32−1 µ 32−1(1−(1− x2)µ)−(− 12 )dµ.Using (3.2) for a = −12 , b =32 , c = α +52 , y =1−x2 , we observe that |y| < 1,Re c > Re b > 0 and thus∫ 10(1− µ)α+ 52− 32−1 µ 32−1(1−(1− x2)µ)−(− 12 )dµ=Γ(32)Γ (α+ 1)Γ(α+ 52) 2F1 [ − 12 , 32α+ 52 ; 1− x2]givingI2 (α, x) =√2pi2· Γ (α+ 1)Γ(α+ 52) (1− x)α+ 32 2F1 [ − 12 , 32α+ 52 ; 1− x2].For x = −1 or x = 1 the identity (3.4) is obvious and we omit the details.We may now state the following Ostrowski type inequality for functions of twoindependent variables defined on the compact unity disk.Theorem 5. Assume that the function f : D (0, 1) ⊂ R2 → R where D (0, 1) ={(x, y) ∈ R2|x2 + y2 ≤ 1} satisfy the condition(3.6) |f (x, y)− f (u, v)| ≤ L1 |x− u|α1 + L2 |y − v|α2 , (α1, α2 > −1)for any (x, y), (u, v) ∈ D (0, 1), (u, v) 6= (x, y). We have the inequality,(3.7)∣∣∣∣∣f (x, y)− 1pi∫∫D(0,1)f (u, v) du dv∣∣∣∣∣ ≤ 2pi [L1I (α1, x) + L2I (α2, y)]for any (x, y) ∈ D (0, 1), where the functions I (·, ·) are defined by (3.3).HO¨LDER CONTINUOUS FUNCTIONS 9Proof. By (3.6) we have, for (x, y) ∈ D (0, 1),∣∣∣∣∣f (x, y)− 1pi∫∫D(0,1)f (u, v) du dv∣∣∣∣∣≤ 1pi∫∫D(0,1)|f (x, y)− f (u, v)| du dv≤ 1pi∫∫D(0,1)[L1 |x− u|α1 + L2 |y − v|α2 ] du dv=2pi[L1I (α1, x) + L2I (α2, y)]and the inequality (3.7) is proved.Corollary 7. Assume that the function f : D (0, 1) ⊂ R2 → R satisfy the Lips-chitzian condition,(3.8) |f(x, y)− f(u, v)| ≤M1 |x− u|+M2 |y − v|for any (x, y), (u, v) ∈ D(0, 1), then we have the inequality,∣∣∣∣∣f (x, y)− 1pi∫∫D(0,1)f (u, v) du dv∣∣∣∣∣(3.9)≤ 2pi[M1(x arcsinx+13√1− x2 (2 + x2))+M2(y arcsin y +13√1− y2 (2 + y2))].for any (x, y) ∈ D (0, 1).Remark 1. Finally, we observe that if f has bounded partial derivatives on D(0, 1),then the inequality (3.9) becomes∣∣∣∣∣f (x, y)− 1pi∫∫D(0,1)f (u, v) du dv∣∣∣∣∣(3.10)≤ 2pi[∥∥∥∥∂f∂x∥∥∥∥D(0,1),∞(x arcsinx+13√1− x2 (2 + x2))+∥∥∥∥∂f∂y∥∥∥∥D(0,1),∞(y arcsin y +13√1− y2 (2 + y2))].for any (x, y) ∈ D (0, 1).References[1] W.N. Bailey, Generalized Hypergeometric Series, Stechert-Hafner Publ. Co., New York, 1964.[2] S.S. Dragomir, On Hadamard’s inequality on a disk, J. Ineq. Pure & Appl. Math., 1 (1)(2000), Article 2, http://jipam.vu.edu.au[3] S.S. Dragomir and C.E.M. Pearce, Selected Topics on Hermite-Hadamard Inequal-ities and Applications, RGMIA Monographs, Victoria University, 2000 (ON LINE:http://rgmia.vu.edu.au/monographs)[4] B. Gavrea, On Hadamard’s inequality for convex mappings defined on a convex domain inthe space, J. Ineq. Pure & Appl. Math., 1 (1) (2000), Article 9, http://jipam.vu.edu.au[5] E. Neuman, Inequalities involving generalized symmetric means, J. Math. Anal. Appl., 120(1986), 315-320.10 N.S. BARNETT, F.C. S¸T. CIˆRSTEA, AND S.S. DRAGOMIR[6] E. Neuman, Inequalities involving multivariate convex functions II, Proc. Amer. Math. Soc.,109 (1990), 965-974.[7] E. Neuman and J. Pecˇaric´, Inequalities involving multivariate convex functions, J. Math.Anal. Appl., 137 (1989), 514-549.School of Communications and Informatics, Victoria University of Technology, POBox 14428, Melbourne City MC, 8001, Victoria, Australia.E-mail address: neil@matilda.vu.edu.auURL: http://sci.vu.edu.au/scistaff/neilb.htmlE-mail address: florica@sci.vu.edu.auE-mail address: sever@matilda.vu.edu.auURL: http://rgmia.vu.edu.au/SSDragomirWeb.html

cari´ c, Inequalities involving multivariate convex functions,

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On Hadamard’s inequality for convex mappings deﬁned on a convex domain in the space,

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Some Inequalities for the Integral Mean of Hölder Continuous Functions Defined on Disks in a Plane

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Some Inequalities for the Integral Mean of Hölder Continuous Functions Defined on Disks in a Plane

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