It is well-known that the left term of the classical Hermite-Hadamard
inequality is closer to the integral mean value than the right one. We show
that in the multivariate case it is not true. Moreover, we introduce some
related inequality comparing the methods of the approximate integration, which
is optimal. We also present its counterpart of Fejer type.Comment: Submitted to Opuscula Mat

Wasowicz, Szymon

Witkowski, Alfred

English

arXiv

It is well–known that the left term of the classical Hermite–Hadamard ine-quality is closer to the integral mean value than the right one. We show that in the mul-tivariate case it is not true. Moreover, we introduce some related inequality comparing the methods of the approximate integration, which is optimal. We also present its coun-terpart of Fejér type

Szymon Wąsowicz

Alfred Witkowski

CiteSeerX

ON SOME INEQUALITY OF HERMITE–HADAMARD TYPE

It is well-known that the left term of the classical Hermite-Hadamard inequality is closer to the integral mean value than the right one. We show that in the multivariate case it is not true. Moreover, we introduce some related inequality comparing the methods of the approximate integration, which is optimal. We also present its counterpart of Fejér type

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Opuscula Mathematica • Vol. 32 • No. 3 • 2012ON SOME INEQUALITYOF HERMITE–HADAMARD TYPESzymon Wąsowicz and Alfred WitkowskiAbstract. It is well-known that the left term of the classical Hermite–Hadamard inequalityis closer to the integral mean value than the right one. We show that in the multivariate caseit is not true. Moreover, we introduce some related inequality comparing the methods of theapproximate integration, which is optimal. We also present its counterpart of Fejér type.Keywords: convex function, Hermite–Hadamard inequality, Fejér inequality, simplex,approximate integration.Mathematics Subject Classification: 26D15, 26A51, 26B25, 65D30, 65D32.1. INTRODUCTIONThe Hermite–Hadamard inequality is one of the most classical inequalities in thetheory of convex functions. It states that if f : [a, b] → R is a convex function, thenf(a+ b2)61b− ab∫af(x)dx 6f(a) + f(b)2. (1.1)This inequality is present in many textbooks and monographs devoted to convex func-tions and it was also extensively studied by many researchers. Very interesting histo-rical remarks concerning the inequality (1.1) can be found in [5] (see also [9, p. 62]).The following aspect of the inequality (1.1) is also well–known: the inequality on theleft–hand side gives a better estimate of the integral mean value than the inequalityon the right. It means that1b − ab∫af(x)dx− f(a+ b2)6f(a) + f(b)2−1b − ab∫af(x)dx, (1.2)591http://dx.doi.org/10.7494/OpMath.2012.32.3.591592 Szymon Wąsowicz and Alfred Witkowskiwhenever f : [a, b] → R is convex. After slight rearrangement we arrive at1b− ab∫af(x)dx 612[f(a+ b2)+f(a) + f(b)2], (1.3)which is really easy to prove, because it is enough to apply the right–hand sideof (1.1) to the intervals[a, a+b2]and[a+b2 , b]and add the obtained inequalities (cf.e.g. [9, p. 52, Remark 1.9.3]). It is also easy to see by drawing a picture. However,there is also another nice geometrical proof based on the idea of symmetry. Wewould like to quote it here, because the mentioned above idea will be used in thepaper as the main tool. To this end for a convex function f : [a, b] → R considerg(x) = f(x) + f(a+ b − x), which is also convex and, moreover, symmetric functionwith respect to a+b2 . See Figure 1 below, where the curve symbolizes the graph of g(we silently assume nonnegativity of g for graphical purposes, but our reasoning, afterslight changes, works in the general case). The integralb∫ag(x)dx is not greater thandouble the area of the trapezoid filled in gray. Henceb∫ag(x)dx 6b− a2[g(a) + g(a+ b2)].Taking into account∫ bag(x)dx = 2∫ baf(x)dx (symmetry) and the definition of g, weobtain (1.3).a a+b2bFig. 1The multivariate counterpart of (1.1) can be found in [1,7,8]. In the simplest formit reads as follows.Theorem 1.1. Let T = conv{v0, . . . ,vn } ⊂ Rn be the simplex with the volume |T |.If f : T → R is a convex function, thenf(1n+ 1n∑i=0vi)61|T |∫Tf(x)dx 61n+ 1n∑i=0f(vi). (1.4)The proof given in [1] is straightforward and elementary, but it requires a largeamount of computation. As a byproduct of our main considerations we offer an easierproof, which is also elementary.On some inequality of Hermite–Hadamard type 593Let us adopt the following notation:L(f, T ) =1|T |∫Tf(x)dx− f(1n+ 1n∑i=0vi),R(f, T ) =1n+ 1n∑i=0f(vi)−1|T |∫Tf(x)dx.The natural question arises, whether the inequality similar to (1.2) holds for convexfunctions of multiple variables, i.e. if the inequality L(f, T ) 6 R(f, T ) is true. Thisconjecture turned out to be false in the multidimensional case. To demonstrate this,consider the unit simplex T = conv{(0, 0), (0, 1), (1, 0)}⊂ R2. ThenL(f, T ) = 2∫∫Tf(x, y)dxdy − f(13,13),R(f, T ) =f(0, 0) + f(0, 1) + f(1, 0)3− 2∫∫Tf(x, y)dxdy.Let f be the convex function, whose graph is the surface of the pyramid shown inFigure 2 (with the lower vertex(13 ,13 , 0)). Then L(f, T ) = 23 >13 = R(f, T ). However,there are, of course, convex functions, for which L(f, T ) 6 R(f, T ). Take, for instance,the convex function f : T → R defined byf(x, y) ={1 for (x, y) ∈ {(0, 0), (0, 1), (1, 0)},0 otherwise,for which L(f, T ) = 0 < 1 = R(f, T ).yzx111Fig. 2Let us stay for a while with the two–dimensional case and the unit simplex T .Following the idea of the proof of (1.3) of dividing the interval [a, b] into two parts,divide T into three subsimplices by its barycenter(13 ,13). Applying the right inequalityof (1.4) to each of these subsimplices and summing up the obtained inequalities, aftersome rearrangement we arrive at2∫∫Tf(x, y)dxdy−f(13,13)6 2[f(0, 0) + f(0, 1) + f(1, 0)3− 2∫∫Tf(x, y)dxdy].594 Szymon Wąsowicz and Alfred WitkowskiTherefore for the two-dimensional case we have the inequality L(f, T ) 6 2R(f, T ).The above example of the convex function with the pyramidal graph shows that thisinequality is optimal.In the general case, taking into account the dimension n = 2, we can guess thatL(f, T ) 6 nR(f, T ), where T ⊂ Rn is a (not necessarily unit) simplex and f : T → Ris convex. As a main result we will prove that this conjecture is true. We also provethe similar result for the Hermite–Hadamard–Fejér inequality.2. DEFINITIONS AND BASIC PROPERTIESThe convex hull of n + 1 points v0, . . . ,vn ∈ Rn is called a simplex, if the vectorsv1 − v0, v2 − v0, . . . , vn − v0 are linearly independent. The points vi are calledvertices of the simplex. Its barycentric coordinates will be denoted by (ξ0, ξ1 . . . , ξn).The pointb =(1n+ 1, . . . ,1n+ 1)=v0 + · · ·+ vnn+ 1is called the barycenter of T .We denote by Sn the group of permutations of n elements. Any σ ∈ Sn+1 generatesan affine mapping σ : T → T byσ(ξ0, . . . , ξn) = (ξσ(0), . . . , ξσ(n)).For a function f : T → R and σ ∈ Sn+1 we define the function fσ byfσ(x) = f(σ(x)). (2.1)Lemma 2.1. If f : T → R is integrable, then∫Tf(x)dx =∫Tfσ(x)dx.Proof. Let σ(x) = Σx+b, where Σ is a linear mapping. Since T = σ(T ) and |σ(T )| =| detΣ||T |, we conclude that | detΣ| = 1 (see e.g. [4, Th. 5.4.8]). Changing variablesin integrals we have∫Tf(x)dx =∫σ(T )f(x)dx =∫Tf(σ(x))| detΣ| dx =∫Tfσ(x)dx.Lemma 2.2. If f : T → R is convex, then so is fσ.Proof. For 0 < t < 1 and x,y ∈ T , we havefσ(tx+ (1 − t)y)= f(σ(tx+ (1− t)y))= f(tσ(x) + (1− t)σ(y))≤≤ tf(σ(x))+ (1− t)f(σ(y))= tfσ(x) + (1 − t)fσ(y).On some inequality of Hermite–Hadamard type 5953. MAIN RESULTOur main result is the following refinement of the classical Hermite–Hadamard in-equality.Theorem 3.1. Let T ⊂ Rn be simplex with vertices v0, . . . ,vn and let f : T → R bea convex function. Then0 6 L(f, T ) 6 nR(f, T ). (3.1)The constant n in this inequality cannot be improved.Proof. The inequality L(f, T ) > 0 follows trivially by (1.4).Denote by b the barycenter of T and let M = 1n+1(f(v0)+ · · ·+ f(vn)). Let σ bea cyclic permutation of order n + 1 and C be the subgroup of Sn+1 generated by σ.DefineF (x) =1n+ 1∑σ∈Cfσ(x). (3.2)ThenF (b) = f(b). (3.3)By (2.1), we getF(σ(x))= F (x). (3.4)The formula (3.2) gives usF (vi) =f(v0) + · · ·+ f(vn)n+ 1= M, i = 0, . . . , n, (3.5)while Lemma 2.1 (together with (3.2)) yields∫TF (x)dx =∫Tf(x)dx. (3.6)By Lemma 2.2, the function F is convex. Of course,1n+ 1∑σ∈Cσ(x) = b, x ∈ T,whence for any x = ξ0v0 + · · ·+ ξnvn ∈ T we haveF (b) = F(1n+ 1∑σ∈Cσ(x))61n+ 1∑σ∈CF(σ(x))= F (x) (3.7)andF (x) = F (ξ0v0 + · · ·+ ξnvn) 6 ξ0F (v0) + · · ·+ ξnF (vn) = M. (3.8)596 Szymon Wąsowicz and Alfred WitkowskiConsider the functionG(x) = (n+ 1) min06j6n{ξj}F (b) +(1− (n+ 1) min06j6n{ξj})M.Note that G(b) = F (b) and G ≡ M on the boundary of T (since at least one of thebarycentric coordinates vanishes here). Our goal is to show that for all x ∈ T theinequality F (x) 6 G(x) holds. To this end fix x = ξ0v0+ · · ·+ξnvn ∈ T and suppose,that ξi = minj{ξj}. Then, taking into account that b = 1n+1∑j vj and∑j ξj = 1,we can writex =∑jξjvj = (n+ 1)ξib+∑j 6=i(ξj − ξi)vj .Clearly all the coefficients are nonnegative and sum up to 1, so the convexity and (3.5)yieldF (x) 6 (n+ 1)ξiF (b) +∑j 6=i(ξj − ξi)F (vj) == (n+ 1)ξiF (b) +(1− (n+ 1)ξi)M = G(x).(3.9)We claim that the function G is convex. To check this fix i ∈ {0, . . . , n}. Since thebarycentric coordinate ξi of x = ξ0v0 + · · · + ξnvn ∈ T is constant on hyperplanesparallel to the face of T opposite to the vertex vi, the function hi(x) = min(1n+1 , ξi)is concave. Thus the mapping x = ξ0v0 + · · ·+ ξnvn 7→ mini hi(x) = minj{ξj} is alsoconcave. But G(x) = M −(M − F (b))minj{ξj}, hence G is convex (remember thatF (b) 6 M). Moreover, the above argument shows that the graph of G coincides withthe lateral the surface of the pyramid in Rn+1 with base (T,M) and apex(b, F (b))(in the case n = 2 see Figure 2). Therefore,∫TG(x)dx = |T |M −1n+ 1|T |(M − F (b))= |T |(nn+ 1M +1n+ 1F (b)). (3.10)Then integrating (3.9) over T we obtain∫TF (x)dx 6 |T |(nn+ 1M +1n+ 1F (b)),which, taking into account (3.3)–(3.6), can be written asn+ 1|T |∫Tf(x)dx 6 nf(v0) + · · ·+ f(vn)n+ 1+ f(b),or, equivalently L(f, T ) 6 nR(f, T ). Clearly, by (3.10) we get L(G, T ) = nR(G, T ),so the inequality (3.1) is optimal.On some inequality of Hermite–Hadamard type 597Remark 3.2. As we could see above, the inequality (3.1) is sharp. For that reasonin the multivariate case the left inequality of (1.4) does not (in general) estimatethe integral mean value better than the right one. The counterexample could be, ofcourse, the example given at the end of the above proof.Remark 3.3. Observe that by integrating the inequalities (3.7) and (3.8) andtaking into account the properties (3.4)–(3.5) we obtain an easy proof of theHermite–Hadamard inequality (1.4). In the authors’ opinion this proof is easier thanthe proof given in [1].4. FEJÉR VERSIONThe classical Fejér version of the Hermite–Hadamard inequality (cf. [3], see also [2,5,10]) states that if f : [a, b] → R is convex and g : [a, b] → R is nonnegative, integrableand symmetric with respect to the midpoint of [a, b], thenf(a+ b2)b∫ag(x)dx 6b∫af(x)g(x)dx 6f(a) + f(b)2b∫ag(x)dx.In this section we reprove analogous result for simplices and provide some estimatesfor the upper and lower bounds. We start with the multivariate version of the Fejér’sinequality. It was proved, for instance, in [7]. We provide an alternative proof.Theorem 4.1. Let T be a simplex with vertices v0, . . . ,vn, f : T → R be a convexfunction and g : T → R be a nonnegative integrable function such that g(σ(x))= g(x)holds for certain cyclic permutation σ ∈ Sn+1. Thenf(1n+ 1n∑i=0vi)∫Tg(x)dx 6∫Tf(x)g(x)dx 6(1n+ 1n∑i=0f(vi))∫Tg(x)dx.Proof. The proof goes exactly along the same lines as that of Theorem 3.1 with twoslight differences: we multiply inequalities (3.7) and (3.8) by g(x) before integrating,and we use the identity∫Tfσ(x)g(x)dx =∫Tfσ(x)gσ(x)dx =∫Tf(x)g(x)dx.598 Szymon Wąsowicz and Alfred WitkowskiTheorem 4.2. Under hypotheses of Theorem 4.1, ifL(f, T ; g) =∫Tf(x)g(x)dx− f(1n+ 1n∑i=0vi)∫Tg(x)dx,R(f, T ; g) =(1n+ 1n∑i=0f(vi))∫Tg(x)dx−∫Tf(x)g(x)dx,∆(f, T ) =1n+ 1n∑i=0fvi)− f(1n+ 1n∑i=0vi),then0 6 ∆(f, T )∫Tg(x) · (n+ 1)minj{ξj} dx 6 R(f, t; g), (4.1)0 6 L(f, t; g) 6 ∆(f, T )∫Tg(x) ·(1− (n+ 1)minj{ξj})dx. (4.2)Proof. Let α =∫Tg(x) · (n + 1)minj{ξj} dx. Multiplying (3.9) by g and integratingwe obtain∫Tf(x)g(x)dx 6 f(b)α+M∫Tg(x)dx−Mαwhich immediately yields (4.1). Subtracting f(b)∫Tg(x)dx from both sides we obtain(4.2) (the inequality L(f, t; g) > 0 follows trivially by Theorem 4.1).Remark 4.3. The full counterpart of Theorem 3.1 cannot be obtained. Namely, theinequality L(f, T ; g) 6 nR(f, T ; g) does not hold for the arbitrarily chosen positivesymmetric weight function g and a convex function f . A counterexample is quite easyto show. It is enough to take n = 1, T = [−1, 1], f(x) = g(x) = x2. In these settingswe haveL(f, T ; g) =1∫−1x4 dx =25, R(f, T ; g) =1∫−1x2 dx−1∫−1x4 dx =415.Another example could be given for any simplex T ⊂ Rn. We will show that forthe arbitrarily chosen constant N > 0 it is possible to find a convex function F :T → R and a nonnegative, integrable and symmetric function g : T → R such thatL(F, T ; g) > NR(F, T ; g). We only sketch the construction.Let us build the convex function F similarly as in the proof of Theorem 3.1, withthe additional conditions F (b) = 0, F (vi) = 1, i = 0, . . . , n. Then for x = ξ0v0 +· · ·+ ξnvn ∈ T we have G(x) = 1− (n+ 1)minj{ξj}. Next, for a ∈ (0, 1) take g(x) =On some inequality of Hermite–Hadamard type 599αmax{G(x)−a, 0}with a factor α, that∫Tg(x)dx = 1. Then∫TF (x)g(x)dx −−−→a→11,so L(F, T ; g) −−−→a→11 and R(F, T ; g) −−−→a→10.If n = 1 and T = [−1, 1], then the above construction leads toF (x) = G(x) = |x|, g(x) =max{|x| − a, 0}(1− a)2and we arrive at1∫−1F (x)g(x)dx =a+ 23−−−→a→11,1∫−1g(x)dx = 1.Remark 4.4. After submission of this paper it came to our attention that similarresult were obtained by Mitroi and Spiridon, see the paper [6].REFERENCES[1] M. Bessenyei, The Hermite–Hadamard inequality on simplices, Amer. Math. Monthly115 (2008), 339–345.[2] S.S. Dragomir, C.E.M. Pearce, Selected topics on Hermite–Hadamard in-equalities and applications, RGMIA Monographs, Victoria University, 2002.http://www.rgmia.org/monographs/hermite_hadamard.html.[3] L. Fejér, Über die Fourierreihen II, Math. Naturwiss. Anz. Ungar. Akad. Wiss.24 (1906), 369–390.[4] S. Łojasiewicz, An introduction to the theory of real functions, John Wiley & Sons,Chichester, 1988.[5] D.S. Mitrinović, I.B. Lacković, Hermite and convexity, Aequationes Math. 28 (1985),229–232.[6] F.–C. Mitroi, C.I. Spiridon, Refinements of Hermite–Hadamard inequality on simplices,arXiv:1105.5043v6 [math.CA], to appear in Math. Rep. (Bucur.) 15(65) 2013.http://arxiv.org/PS_cache/arxiv/pdf/1105/1105.5043v6.pdf.[7] E. Neuman, Inequalities involving multivariate convex functions II, Proc. Amer. Math.Soc. 109 (1990), 965–974.[8] E. Neuman, J. Pečarić, Inequalities involving multivariate convex functions, J. Math.Anal. Appl. 137 (1989), 541–549.[9] C. P. Niculescu, L.E. Persson, Convex functions and their applications. A contemporaryapproach, Springer, New York, 2006.[10] J.E. Pečarić, F. Proschan, Y.L. Tong, Convex functions, partial orderings, and statisticalapplications, Mathematics in Science and Engineering, vol. 187, Academic Press Inc.,Boston, MA, 1992.600 Szymon Wąsowicz and Alfred WitkowskiSzymon Wąsowiczswasowicz@ath.euUniversity of Bielsko-BiałaDepartment of Mathematics and Computer ScienceWillowa 2, 43–309 Bielsko-Biała, PolandAlfred Witkowskialfred.witkowski@utp.edu.plUniversity of Technology and Life SciencesInstitute of Mathematics and Physicsal. prof. Kaliskiego 7, 85-796 Bydgoszcz, PolandReceived: September 29, 2011.Accepted: December 30, 2011.

Opuscula Mathematica

On some inequality of Hermite-Hadamard type

It is well-known that the left term of the classical Hermite-Hadamard inequality is closer to the integral mean value than the right one. We show that in the multivariate case it is not true. Moreover, we introduce some related inequality comparing the methods of the approximate integration, which is optimal. We also present its counterpart of Fej&eacute;r type

On some inequality of Hermite-Hadamard type

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On some inequality of Hermite-Hadamard type

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