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Refinements of Hermite-Hadamard inequality on simplices Flavia-Corina Mitroi and Cătălin Irinel Spiridon 2 1 0 Some refinements of the Hermite-Hadamard inequality are obtained in the case of continuous 2 convexfunctionsdefinedonsimplices. n a AMS2010 Subject Classification: 26A51. J Key words: Hermite-Hadamardinequality,convex function,simplex. 1 ] A 1 Introduction C . h The aim of this paper is to provide a refinement of the Hermite-Hadamard t a inequality on simplices. m Suppose that K is a metrizable compact convex subset of a locally convex [ Hausdorff space E. Given a Borel probability measure µ on K, one can prove 7 the existence of a unique point b ∈ K (called the barycenter of µ) such that µ v 3 4 x′(b ) = x′(x)dµ(x) µ 0 ZK 5 . for all continuous linear functionals x′ on E. The main feature of barycenter 5 0 is the inequality 1 f(b )≤ f(x)dµ(x), 1 µ : ZK v valid for every continuous convex function f : K → R. It was noted by several i X authors that this inequality is actually equivalent to the Jensen inequality. r The following theorem due to G. Choquet complements this inequality and a relates the geometry of K to a given mass distribution. THEOREM 1. (The general form of Hermite-Hadamard inequality). Let µ be a Borel probability measure on a metrizable compact convex subset K of a locally convex Hausdorff space. Then there exists a Borel probability measure ν on K which has the same barycenter as µ, is zero outside ExtK, and verifies the double inequality (1) f(b ) ≤ f(x)dµ(x) ≤ f(x)dν(x) µ ZK ZExtK 1 for all continuous convex functions f :K → R. Here ExtK denotes the set of all extreme points of K. The details can be found in [7], pp. 192-194. See also [6]. Intheparticularcaseofsimplicesonecantakeadvantageofthebarycentric coordinates. Suppose that ∆ ⊂ Rn is an n-dimensional simplex of vertices P ,...,P . 1 n+1 InthiscaseExt∆ = {P ,...,P }andeachx ∈ ∆canberepresenteduniquely 1 n+1 as a convex combination of vertices, n+1 (2) λ (x)P = x, k k k=1 X where the coefficients λ (x) are nonnegative numbers (depending on x) and k n+1 (3) λ (x) = 1. k k=1 X Each function λ : x → λ (x) is an affine function on ∆. This can be easily k k seen by considering the linear system consisting of the equations (2) and (3). The coefficients λ (x) can be computed in terms of Lebesgue volumes (see k [1], [5]). We denote by ∆ (x) the subsimplex obtained when the vertex P is j j replaced by x ∈ ∆. Then one can prove that Vol(∆ (x)) k (4) λ (x) = k Vol(∆) for all k = 1,...,n + 1 (the geometric interpretation is very intuitive). Here Vol(∆)= dx. ∆ Intheparticularcasewheredµ(x)= dx/Vol(∆)wehaveλ b = R k dx/Vol(∆) 1 for every k = 1,...,n+1. n+1 (cid:0) (cid:1) The above discussion leads to the following form of Theorem 1 in the case of simplices: COROLLARY 1. Let ∆ ⊂ Rn be an n-dimensional simplex of vertices P ,...,P and µ be a Borel probability measure on ∆ with barycenter b . 1 n+1 µ Then for every continuous convex function f : ∆ → R, n+1 f(b ) ≤ f(x)dµ(x)≤ λ (b )f (P ). µ k µ k Z∆ k=1 X Proof. In fact, n+1 n+1 f(x)dµ(x)= f λ (x)P dµ(x) ≤ λ (x) f(P )dµ(x) k k k k Z∆ Z∆ k=1 ! Z∆k=1 X X n+1 n+1 = f(P ) λ (x) dµ(x) = λ (b )f (P ). k k k µ k k=1 Z∆ k=1 X X 2 On the other hand n+1 n+1 λ (b )f(P )= fd λ (b )δ k µ k k µ Pk k=1 ZExt∆ k=1 ! X X and n+1λ (b )δ is the only Borel probability measure ν concentrated at k=1 k µ Pk the vertices of ∆ which verifies the inequality P f(x)dµ(x) ≤ f(x)dν(x) Z∆ ZExt∆ for every continuous convex functions f : ∆ → R. Indeed ν must be of the form ν = n+1α δ , with b = n+1α P = b and the uniqueness of k=1 k Pk ν k=1 k k µ barycentric coordinates yields the equalities P P α = λ (b ) for k = 1,...,n+1. k k µ It is worth noticing that the Hermite–Hadamard inequality is not just a consequence of convexity, it actually characterizes it. See [9]. The aim of the present paper is to improve the result of Corollary 1, by providing better bounds for the arithmetic mean of convex functions defined on simplices. 2 Main results We start by extending the following well known inequality concerning the continuous convex functions defined on intervals: 1 b 1 f (a)+f(b) a+b f(x)dx ≤ +f . b−a 2 2 2 Za (cid:18) (cid:18) (cid:19)(cid:19) See [7], p. 52. THEOREM 2. Let ∆ ⊂ Rn be an n-dimensional simplex of vertices P ,...,P ,endowedwiththenormalizedLebesguemeasuredx/Vol(∆).Then 1 n+1 for every continuous convex function f : ∆ → R and every point P ∈ ∆ we have n+1 1 1 (5) f(x)dx ≤ (1−λ (P))f (P )+f(P) k k Vol(∆) n+1 Z∆ k=1 ! X In particular, n+1 1 1 n (6) f(x)dx ≤ f (P )+f b . k dx/Vol(∆) Vol(∆) n+1 n+1 Z∆ k=1 ! X (cid:0) (cid:1) 3 Proof. Consider the barycentric representation P = λ (P)P ∈ ∆ . k k k According to Corollary 1, P 1 1 f(x)dx ≤ f (P )+f(P) , k λ (P)Vol(∆) n+1  i Z∆i(P) k6=i X   where ∆ (P) denotes the simplex obtained from ∆ by replacing the vertex P i i by P. Notice that (4) yields λ (P)Vol(∆) = Vol(∆ (P)). Multiplying both i i sides by λ (P) and summing up over i we obtain the inequality (5). i In the particular case when P = b all coefficients λ (P) equal dx/Vol(∆) k 1/(n+1). REMARK 1. Using [1, Theorem 1] instead of Corollary 1 of the present paper, one may improve the statement of Theorem 2 by the cancellation of the continuity condition. Such statement is proved independently, by a different approach, in [12]. An extension of Theorem 2 is as follows: THEOREM 3. Let ∆ ⊂ Rn be an n-dimensional simplex of vertices P ,...,P endowed with the Lebesgue measure and let ∆′ ⊆ ∆ a subsimplex 1 n+1 of vertices P′,...,P′ which has the same barycenter as ∆. Then for every 1 n+1 continuous convex function f : ∆ → R, and every index j ∈ {1,...,n+1}, we have the estimates 1 1 (7) f(x)dx ≤ λ P′ f (P )+f P′ Vol(∆) n+1  i k i j  Z∆ k6=j i XX (cid:0) (cid:1) (cid:0) (cid:1) 1   ≤ f (P ) i n+1 i X and 1 1 (8) f(x)dx ≥ λ P′ f P +P′ Vol(∆) i j n+1 k j Z∆ i k6=i X (cid:0) (cid:1) X    1 ≥ f P i n+1 ! i X Proof. We will prove here only the inequalities (7), the proof of (8) being similar. By Corollary 1, for each index i, 1 1 f(x)dx ≤ f(P )+f P′ , λi Pj′ Vol(∆)Z∆i(Pj′) n+1 Xk6=i k (cid:0) j(cid:1) (cid:16) (cid:17)   4 where∆ P′ denotes thesimplexobtainedfrom ∆by replacingthevertexP i j i by P′. No(cid:16)tice(cid:17)that (4) yields λ P′ Vol(∆)= Vol(∆ P′ ). By multiplying j i j i j both sides by λ P′ and summ(cid:16)ing(cid:17)over i we obtain (cid:16) (cid:17) i j (cid:16) (cid:17) 1 1 f(x)dx ≤ λ P′ f (P )+f P′ Vol(∆) n+1 i j k j  Z∆ i k6=i X (cid:0) (cid:1)X (cid:0) (cid:1)   1 (9) = λ P′ f(P )−f(P ) +f P′ n+1 i j k i j ! ! i k X (cid:0) (cid:1) X (cid:0) (cid:1) 1 = 1−λ P′ f (P )+f P′ . n+1 i j i j ! i X(cid:0) (cid:0) (cid:1)(cid:1) (cid:0) (cid:1) Furthermore, since ∆′ and ∆ have the same barycenter, we have 1−λ P′ = λ P′ i j i k k6=j (cid:0) (cid:1) X (cid:0) (cid:1) for all i= 1,...,n+1. Then 1 1 f (x)dx ≤ λ P′ f(P )+f P′ . Vol(∆) n+1 i k i j  Z∆ k6=j i XX (cid:0) (cid:1) (cid:0) (cid:1)   The fact that ∆′ and ∆ have the same barycenter and the convexity of the function f yield f(P ) = λ P′ f(P ) j i k i j k i X XX (cid:0) (cid:1) = λ P′ f(P )+ λ P′ f(P ) i k i i j i k6=j i i XX (cid:0) (cid:1) X (cid:0) (cid:1) ≥ λ P′ f(P )+f λ P′ P i k i i j i ! k6=j i i XX (cid:0) (cid:1) X (cid:0) (cid:1) = λ P′ f(P )+f P′ . i k i j k6=j i XX (cid:0) (cid:1) (cid:0) (cid:1) This concludes the proof (7). When ∆′ as a singleton, the inequality (7) coincides with the inequality (6). Notice that if we omit the barycenter condition then the inequality (9) translates into (5). 5 An immediate consequence of Theorem 2 is the following result due to Farissi [3]: COROLLARY 2. Assume that f : [a,b] → R is a convex function and λ ∈ [0,1]. Then 1 b (1−λ)f(a)+λf(b)+f (λa+(1−λ)b) f(x)dx ≤ b−a 2 Za f (a)+f (b) ≤ 2 and 1 b a+(1−λ)a+λb b+(1−λ)a+λb f(x)dx ≥λf +(1−λ)f b−a 2 2 Za (cid:18) (cid:19) (cid:18) (cid:19) a+b ≥f . 2 (cid:18) (cid:19) Proof. Apply Theorem 2 for n = 1 and ∆′ the subinterval of endpoints (1−λ)a+λb and λa+(1−λ)b. Anotherrefinement oftheHermite-Hadamard inequality inthecaseofsim- plices is as follows. THEOREM 4. Let ∆ ⊂ Rn be an n-dimensional simplex of vertices P ,...,P endowed with the Lebesgue measure and let ∆′ ⊆ ∆ be a subsim- 1 n+1 plexwhosebarycenterwithrespecttothenormalizedLebesguemeasure dx Vol(∆′) is P = λ (P)P . Then for every continuous convex function f : ∆ → R, k k k 1 P (10) f(P)≤ f(x)dx≤ λ (P)f (P ). Vol(∆′)Z∆′ j j j X Proof. Let P′, k = 1,...,n+1 be the vertices of ∆′. By Corollary 1, k 1 1 f (P) ≤ f(x)dx ≤ f P′ , Vol(∆′) Z∆′ n+1 k k X (cid:0) (cid:1) The barycentric representation of each of the points P′ ∈ ∆ gives us k λ (P′)P =P′ j j k j k . λ P′ = 1 ( Pk k j (cid:16) (cid:17) Since f is a convex function, P 1 1 f P′ = f λ P′ P n+1 k n+1  j k j k k j X (cid:0) (cid:1) X X (cid:0) (cid:1)   1 ≤ λ P′ f(P ) = λ (P)f(P ) n+1 j k j j j ! j k j X X (cid:0) (cid:1) X 6 and the assertion of Theorem 2 is now clear. Asacorollary ofTheorem 2wegetthefollowing resultdueVasićandLack- ović [10], and Lupaş [2] (cf. J. E. Pečarić et al. [8]). COROLLARY3. Letpandq betwopositivenumbersanda ≤ a ≤ b ≤ b . 1 1 Then the inequalities pa+qb 1 A+y pf(a)+qf(b) f ≤ f(x)dx ≤ p+q 2y p+q (cid:18) (cid:19) ZA−y hold for A = pa+qb, y > 0 and all continuous convex functions f : [a ,b ] → R p+q 1 1 if and only if b−a y ≤ min{p,q}. p+q The right-hand side of the inequality stated in Theorem 2 can be improved as follows. THEOREM 5. Suppose that ∆ is an n-dimensional simplex of vertices P ,...,P and let P ∈ ∆. Then for every subsimplex ∆′ ⊂ ∆ such that 1 n+1 P = jλj(P)Pj = bdx/Vol(∆′) we have P 1 1 f(x)dx ≤ n λ (P)f (P )+f (P) Vol(∆′) Z∆′ n+1  j j j  X   for every continuous convex function f : ∆ → R. Proof. Let P′, k = 1,...,n + 1 be the vertices of ∆′. We denote by ∆′ k i the subsimplex obtained by replacing the vertex P′ by the barycenter P of the i normalized measure dx/Vol(∆′) on ∆′. According to Corollary 1, 1 1 1 f(x)dx≤ f P′ + f (P) Vol(∆′i) Z∆′i n+1 k6=i k n+1 X (cid:0) (cid:1) 1 1 = f λ P′ P + f(P) n+1  j k j n+1 k6=i j X X (cid:0) (cid:1)   1 1 ≤ λ P′ f(P )+ f (P) n+1 j k  j n+1 j k6=i X X (cid:0) (cid:1)  1  1 = λ (P)− λ P′ f(P )+ f (P) j n+1 j i j n+1 j (cid:18) (cid:19) X (cid:0) (cid:1) 7 for each index i = 1,...,n+1. Summing up over i we obtain n+1 f(x)dx Vol(∆′)Z∆′ 1 ≤ (n+1) λ (P)f (P )− λ P′ f (P )+f (P) j j n+1 j i j j i j X X X (cid:0) (cid:1) = n λ (P)f(P )+f (P). j j j X and the proof of the theorem is completed. Of course, the last theorem yields an improvement of Corollary 2. This was first noticed in [8, pp. 146]. We end this paper with an alternative proof of some particular case of a result established by A. Guessab and G. Schmeisser [4, Theorem 2.4]. Before we start we would like to turn the reader’s attention to the paper [12] by S. Wąsowicz, wheretheresultwearegoingtopresentwas obtainedbysomemore general approach (see [11, Theorems 2 and Corollary 4]). THEOREM 6. Let ∆ ⊂ Rn be an n-dimensional simplex of vertices P ,...,P endowed with the Lebesgue measure and let M ,...,M be points 1 n+1 1 m in ∆ such that b is a convex combination β M of them. Then for dx/Vol(∆) j j j every continuous convex function f :∆ → R, the following inequalities hold: P m n+1 1 f b ≤ β f (M )≤ f(P ). dx/Vol(∆) j j i n+1 j=1 i=1 (cid:0) (cid:1) X X Proof. The first inequality follows from Jensen’s inequality. In order to establish the second inequality, since we have 1 b = β λ (M )P = β λ (M ) P = P , dx/Vol(∆) j i j i j i j i i   n+1 ! j i i j i X X X X X   we infer that β λ (M ) = 1 for every i and j j i j n+1 P β f (M )= β f λ (M )P ≤ β λ (M ) f (P ) j j j i j i j i j i   ! j j i i j X X X X X 1   = f(P ). i n+1 i X This completes the proof. Acknowledgement. The first author has been supported by CNCSIS Grant 420/2008. The authors are indebted to Szymon Wąsowicz for pointing out to them the references [11] and [4]. 8 References [1] M. Bessenyei, The Hermite–Hadamard Inequality on Simplices, American Mathematical Monthly, volume 115, April 2008, pages 339–345. [2] A.Lupaş, A generalization of Hadamard inequalities for convex functions, Univ.Beograd.Publ.Elektrotehn.Fak.Ser.Mat.Fiz.(1976),no.544-576, 115-121. [3] A. El Farissi, Simple proof and refinement of Hermite-Hadamard inequality, J. Math. Inequal., 4 (2010), No. 3, 365–369. [4] A. Guessab and G. Schmeisser, Convexity results and sharp error estimates in approximate multivariate integration, Math. Comp., 73 (2004), pp. 1365–1384. [5] E. Neuman, J. E. Pečarić, Inequalities involving multivariate convex functions, J. Math. Anal. Appl. 137 (1989), no. 2, 541-549. [6] C. P. Niculescu, The Hermite-Hadamard inequality for convex functions of a vector variable, Math. Inequal. Appl., 4 (2002), 619–623. [7] C. P. Niculescu, L.-E. Persson, Convex Functions and their Applications. A Contemporary Approach, CMSBooksinMathematicsvol.23,Springer- Verlag, New York, 2006. [8] J. E. Pečarić, F. Proschan, Y. L. 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Cuza 13, Craiova, RO-200585 Romania [email protected], [email protected] 9 Refinements of Hermite-Hadamard inequality on simplices Flavia-Corina Mitroi and Cătălin Irinel Spiridon 2 1 0 Some refinements of the Hermite-Hadamard inequality are obtained in the case of continuous 2 convexfunctionsdefinedonsimplices. n a J AMS2010 Subject Classification: 26A51. 1 ] Key words: Hermite-Hadamardinequality,convex function,simplex. A C . 1.INTRODUCTION h t a The aim of this paper is to provide a refinement of the Hermite-Hadamard m inequality on simplices. [ Suppose that K is a metrizable compact convex subset of a locally convex 7 Hausdorff space E. Given a Borel probability measure µ on K, one can prove v the existence of a unique point b ∈ K (called the barycenter of µ) such that 3 µ 4 0 x′(b ) = x′(x)dµ(x) µ 5 ZK . 5 for all continuous linear functionals x′ on E. The main feature of barycenter 0 1 is the inequality 1 f(b )≤ f(x)dµ(x), : µ v ZK Xi valid for every continuous convex function f : K → R. It was noted by several r authors that this inequality is actually equivalent to the Jensen inequality. a The following theorem due to G. Choquet complements this inequality and relates the geometry of K to a given mass distribution. THEOREM 1. (The general form of Hermite-Hadamard inequality). Let µ be a Borel probability measure on a metrizable compact convex subset K of a locally convex Hausdorff space. Then there exists a Borel probability measure ν on K which has the same barycenter as µ, is zero outside ExtK, and verifies the double inequality (1) f(b ) ≤ f(x)dµ(x) ≤ f(x)dν(x) µ ZK ZExtK 1