Given a function f defined on a nonempty and convex subset of the d-dimensional Euclidean space, we prove that if f is bounded from below and it satisfies a convexity-type functional inequality with infinite convex combinations, then f has to be convex. We also give alternative proofs of a generalization of some known results on convexity with infinite convex combinations due to Daróczy and Páles (1987) and Pavić (2019) using a probabilistic version of Jensen inequality

Barczy Mátyás

Páles Zsolt

English

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Annales Univ. Sci. Budapest., Sect. Comp. 58 (2025) 47–55A CONVEXITY-TYPE FUNCTIONAL INEQUALITYWITH INFINITE CONVEX COMBINATIONSMátyás Barczy (Szeged, Hungary)Zsolt Páles (Debrecen, Hungary)Communicated by László Szili(Received 8 June 2025; accepted 4 August 2025)Abstract. Given a function f defined on a nonempty and convex subsetof the d-dimensional Euclidean space, we prove that if f is bounded frombelow and it satisfies a convexity-type functional inequality with infiniteconvex combinations, then f has to be convex. We also give alternativeproofs of a generalization of some known results on convexity with infiniteconvex combinations due to Daróczy and Páles (1987) and Pavić (2019)using a probabilistic version of Jensen inequality.1. IntroductionStudying properties of convex functions, in particular inequalities for them,has a long history, and it is still an active area of research, mainly because oftheir importance in many fields of mathematics. For a recent monograph onsome developments in the theory of Jensen-type inequalities (such as Jensen–Steffensen, Jensen–Mercer, Jessen, McShane and Popoviciu inequalities), andfor applications in information theory, see Khan et al. [4]. When convexityof a function comes into play, one usually means convexity with finite convexcombinations. In this paper, we deal with a convexity-type functional inequalitywith infinite convex combinations, which, in the language of probability theory,Key words and phrases: Convex function, Jensen convex function, Jensen-type functionalinequality, infinite convex combination.2010 Mathematics Subject Classification: 26A51, 26B25, 60A05.Zsolt Páles is supported by the K-134191 NKFIH Grant.https://doi.org/10.71352/ac.58.040825First published online: 8 August 202548 M. Barczy and Zs. Pálescan be rephrased as Jensen inequality for discrete random variables. This sub-field of the theory of convex functions seems to be less investigated than thatof convexity for finite convex combinations.Throughout this paper, let N, Z+, R and R+ denote the sets of positiveintegers, non-negative integers, real numbers and non-negative real numbers,respectively. For a real number z ∈ R, its positive part max(z, 0) is denotedby z+. For a vector x ∈ Rd, its Euclidean norm is denoted by ∥x∥. An intervalI ⊆ R will be called nondegenerate if it contains at least two distinct points.Definition 1.1. Let d ∈ N and D ⊆ Rd be a nonempty convex set.(i) Given t ∈ [0, 1], a function f : D → R is called t-convex iff(tx + (1 − t)y) ≤ tf(x) + (1 − t)f(y), x, y ∈ D.If f is 12 -convex, then it is called Jensen convex or midpoint convex aswell. If f is t-convex for all t ∈ (0, 1), then it is said to be convex. Notethat every function f : D → R is automatically 0-convex and 1-convex aswell.(ii) Given t, s ∈ [0, 1], a function f : D → R is called (t, s)-convex iff(tx + (1 − t)y) ≤ sf(x) + (1 − s)f(y), x, y ∈ D.The following result establishes implications among the convexity notionsintroduced in part (i) of Definition 1.1.Theorem 1.1. Let d ∈ N, D ⊆ Rd be a nonempty convex set and f : D → R.Then the following two assertions hold.(i) If f is t-convex for some t ∈ (0, 1), then it is Jensen convex.(ii) If D is open as well, f is t-convex for some t ∈ (0, 1) and f is boundedfrom above on some nonempty open subset of D, then f is convex andcontinuous.Proof. For the proof of part (i), see Daróczy and Páles [2, Lemma 1] orKuhn [6]. (For completeness, we note that the compactness assumption of thedomain of f in Lemma 1 in Daróczy and Páles [2] is not needed for the validityof their result, so we can indeed apply it.)Next, we prove part (ii). Using part (i), we have that f is a Jensen con-vex function defined on the nonempty open and convex set D, and, by theassumption, it is bounded from above on some nonempty open subset of D.By Theorem 6.2.1 in Kuczma [5], it implies that f is locally bounded at everypoint of D as well. As a consequence, the celebrated Bernstein–Doetsch theo-rem (see Bernstein and Doetsch [1] or Kuczma [5, Theorems 6.4.2 and 7.1.1])yields that f is convex and continuous on D, as desired. ■A convexity-type functional inequality 49Now we recall a result due to Daróczy and Páles [2] about infinite convexcombinations. Let us introduce the setΛ :={(λn)n∈N : λn ≥ λn+1 > 0, n ∈ N, and∞∑n=1λn = 1}.Theorem 1.2. (Daróczy and Páles [2, Theorem 1].) Let (λn)n∈N ∈ Λ, d ∈ N,D ⊆ Rd be a compact, convex set, and f : D → R+. Then the inequalityf( ∞∑i=1λixi)≤∞∑i=1λif(xi)holds for all xi ∈ D, i ∈ N, if and only if f is convex.Pavić [10] recently has proven the following similar result.Theorem 1.3. (Pavić [10, Theorem 2.1].) Let f : [a, b] → R be a convexfunction, where a, b ∈ R with a < b. Let xi ∈ [a, b], i ∈ N, and λi ∈ R+, i ∈ N,be such that∑∞i=1 λi = 1. Thenf(ta + (1 − t)b) = f( ∞∑i=1λixi)≤∞∑i=1λif(xi) ≤ tf(a) + (1 − t)f(b),where t ∈ [0, 1] is such that∑∞i=1 λixi = ta + (1 − t)b.As the main result of the paper, given a function f defined on a nonemptyand convex subset of Rd, we prove that if f is bounded from below and it sat-isfies a convexity-type functional inequality with infinite convex combinations,then f has to be convex, see Theorem 3.2. We also give alternative proofsfor generalizations of some parts of Theorems 1.2 and 1.3 due to Daróczy andPáles [2] and Pavić [10], respectively, see Proposition 3.1. In the proof of The-orem 3.2, results on (s, t)-convexity, while in the proof of Proposition 3.1, aprobabilistic version Jensen inequality (see Section 2) play a crucial role. InRemark 3.1, we give an example which shows that a converse of Theorem 3.2does not hold in general.2. A probabilistic version of the Jensen inequalityIn this section, we recall a probabilistic version of the Jensen inequality(see, e.g., Dudley [3, 10.2.6, page 348]), which will be used for giving alternativeproofs for generalizations of some parts of Theorems 1.2 and 1.3.Proposition 2.1. (Jensen inequality.) Let ξ : Ω → Rd be a random vectorsuch that E(∥ξ∥) < ∞, and let D ⊆ Rd be a Borel measurable, convex set suchthat P(ξ ∈ D) = 1. Then the following statements hold.50 M. Barczy and Zs. Páles(i) We have E(ξ) ∈ D.(ii) For all continuous and convex functions f : D → R, we have thatE(f(ξ)) ∈ (−∞, ∞] and f(E(ξ)) ≤ E(f(ξ)).(iii) Supposing, in addition, that ξ is discrete as well, for all convex functionsf : D → R, we have that E(f(ξ)) ∈ (−∞, ∞] and f(E(ξ)) ≤ E(f(ξ)).Proof. We plan to apply Dudley [3, 10.2.6, page 348] to derive the assertions.First note that, for all continuous functions f : D → R, we have that f(ξ)is a random variable. Further, if ξ is discrete as well, then, for all functionsf : D → R, we have that f(ξ) is a random variable. Indeed, if {xi : i ∈ N}denotes the range of ξ, then the range of f(ξ) is {f(xi) : i ∈ N} is also countable,and, for each j ∈ N, we get that{f(ξ) = f(xj)} =⋃{i∈N:f(xi)=f(xj)}{ξ = xi},which is an event, since {ξ = xi} is an event for each i ∈ N and a σ-algebra isclosed under countable union. Consequently, if ξ(ω) ∈ D for all ω ∈ Ω, then(i)–(iii) directly follow from Dudley [3, 10.2.6, page 348].In the more general case when P(ξ ∈ D) = 1 holds, let us introduce themapping ξ̃ : Ω → Rd,ξ̃(ω) :=(ξ1{ξ∈D} + d01{ξ ̸∈D})(ω) ={ξ(ω) if ω ∈ Ω is such that ξ(ω) ∈ D,d0 otherwise,where d0 is arbitrarily fixed element of D. Then ξ̃(ω) ∈ D for all ω ∈ Ω, and,since P(ξ ̸∈ D) = 0, we haveE(∥ξ̃∥) = E(∥ξ̃∥1{ξ∈D})+ E(∥ξ̃∥1{ξ ̸∈D})== E(∥ξ∥1{ξ∈D})+ E(∥d0∥1{ξ ̸∈D})== E(∥ξ∥) − E(∥ξ∥1{ξ ̸∈D})+ ∥d0∥P(ξ ̸∈ D) = E(∥ξ∥) < ∞.Using again Dudley [3, 10.2.6, page 348], we obtain that E(ξ̃) ∈ D, E(f(ξ̃))∈∈ (−∞, ∞] and f(E(ξ̃))≤ E(f(ξ̃)). Since P(ξ ̸∈ D) = 0, we get thatE(ξ̃) = E(ξ̃1{ξ∈D})+ E(ξ̃1{ξ ̸∈D})= E(ξ1{ξ∈D})+ E(d01{ξ ̸∈D})== E(ξ) − E(ξ1{ξ ̸∈D})+ d0P(ξ ̸∈ D) = E(ξ),and therefore E(ξ) ∈ D holds as well. Similarly, one can check that E(f(ξ̃))== E(f(ξ)), and consequently (i)–(iii) also follow in the case P(ξ ∈ D) = 1. ■Note that if d > 1 and D ⊆ Rd is a convex set, then D may be not Borelmeasurable. Therefore, the assumption about the Borel measurability of D inthe above proposition is indispensable. Namely, the union of an open ball inRd with any non Borel measurable subset of its boundary provides an exampleof a non Borel measurable convex set of Rd, cf. Lang [8].A convexity-type functional inequality 513. Main resultsFirst, we give alternative proofs for a generalized form of the ‘only if’ partof Theorem 1.2 and the first inequality of Theorem 1.3 using a probabilisticversion of Jensen inequality (see part (iii) of Proposition 2.1).Proposition 3.1. Let λi ∈ R+, i ∈ N, be such that∑∞i=1 λi = 1. Let d ∈ N,D ⊆ Rd be a convex set, and let xi ∈ D, i ∈ N, be such that∑∞i=1 λi∥xi∥ < ∞.Then, for any convex function f : D → R, we have thatf( ∞∑i=1λixi)≤∞∑i=1λif(xi).Proof. Let ξ be a discrete random variable with range {xi : i ∈ N} anddistribution P(ξ = xi) = λi, i ∈ N. Since λi ∈ R+, i ∈ N, and∑∞i=1 λi = 1,the distribution of ξ is indeed well-defined. Then, by the assumption, we havethatE(∥ξ∥) =∞∑i=1λi∥xi∥ < ∞.Consequently, by part (i) of Proposition 2.1, we get E(ξ) ∈ D, i.e.,∞∑i=1λixi =∞∑i=1xiP(ξ = xi) = E(ξ) ∈ D.Furthermore, by part (iii) of Proposition 2.1, for any convex function f : D →→ R, we have that f(E(ξ)) ≤ E(f(ξ)), i.e.,f( ∞∑i=1λixi)= f(E(ξ)) ≤ E(f(ξ)) =∞∑i=1f(xi)P(ξ = xi) =∞∑i=1λif(xi).This completes the proof of the statement. ■Next, we present a generalization of the ‘if part‘ of Theorem 1.2.Theorem 3.2. Let (λn)n∈N, (µn)n∈N ∈ Λ, let d ∈ N, D ⊆ Rd be nonemptyand convex, and let f : D → R be bounded from below. If the inequalityf( ∞∑i=1λixi)≤∞∑i=1µif(xi)(3.1)holds for all bounded sequences xi ∈ D, i ∈ N, then f is convex. Here the righthand side of (3.1) is either convergent, or else, it diverges to the extended realnumber +∞.52 M. Barczy and Zs. PálesThe inequality (3.1) can be reformulated asf(E(ξ)) ≤ E(f(η)),where (Ω, A,P) is a probability space, ξ : Ω → Rd and η : Ω → Rd arerandom vectors with common ranges {xi : i ∈ N} and with possibly differentdistributions P(ξ = xi) = λi, i ∈ N, and P(η = xi) = µi, i ∈ N, respectively.Proof of Theorem 3.2. First, we check that the left hand side of (3.1)is well-defined for all bounded sequences xi ∈ D, i ∈ N, that is,∑∞i=1 λixibelongs to D. Indeed, one can interpret the sum∑∞i=1 λixi as the expectationof a discrete random vector ξ with range {xi : i ∈ N} and with distributionP(ξ = xi) = λi, i ∈ N. Since (xi)i∈N is bounded, there exists K > 0 such that∥xi∥ ≤ K, i ∈ N, yielding that∞∑i=1∥λixi∥ =∞∑i=1λi∥xi∥ ≤ K∞∑i=1λi = K < ∞.As a consequence, we have that ξ is integrable, and, by part (i) of Proposi-tion 2.1, we get that∑∞i=1 λixi = E(ξ) ∈ D.Next, we show that the right hand side of (3.1) is either convergent, orelse, it diverges to the extended real number +∞. Assume that B is a lowerbound for f . Then f −B ≥ 0, and hence the series∑∞i=1 µi(f(xi)−B) is eitherconvergent, or else, it diverges to the extended real number +∞. Therefore,observing that∞∑i=1µif(xi) =∞∑i=1µi(f(xi) − B) +∞∑i=1(µiB) =∞∑i=1µi(f(xi) − B) + B,we can conclude that the series∑∞i=1 µif(xi) is also either convergent, or else,it diverges to the extended real number +∞.Now, suppose that (3.1) holds for all bounded sequences xi ∈ D, i ∈ N.Let x, y ∈ D be fixed. By choosing x1 := x and xi := y, i ∈ N \ {1}, in (3.1),we have thatf(λ1x + (1 − λ1)y) = f(λ1x +( ∞∑i=2λi)y)≤≤ µ1f(x) +( ∞∑i=2µi)f(y) == µ1f(x) + (1 − µ1)f(y).Since λi > 0 and µi > 0 for each i ∈ N and∑∞i=1 λi =∑∞i=1 µi = 1, wehave that λ1, µ1 ∈ (0, 1). Consequently, we get that f is (λ1, µ1)-convex onD. By Fact on page 135 in Matkowski and Pycia [9], we get that f is JensenA convexity-type functional inequality 53convex on D. Here, for historical fidelity, we mention that this result is due toKuhn [7] and, then, using the idea in the proof of Lemma 1 in Daróczy andPáles [2], later Matkowski and Pycia [9] gave an elementary proof of the resultof Kuhn [7]. This together with Theorem 5.3.5 in Kuczma [5] imply that f isq-convex for all q ∈ Q∩ [0, 1]. (For completeness, we note that the openness ofD in Theorem 5.3.5 in Kuczma [5] is assumed, but, in fact, it is not needed forthe validity of the result.)By Lemma 2 in Daróczy and Páles [2], for any t ∈ [0, 1], there exist qi ∈Q ∩ [0, 1], i ∈ N, such that t =∑∞i=1 λiqi, and hence1 − t =∞∑i=1λi −∞∑i=1λiqi =∞∑i=1λi(1 − qi).Therefore, for any x, y ∈ D and t ∈ [0, 1], we get thatf(tx + (1 − t)y) = f(( ∞∑i=1λiqi)x +( ∞∑i=1λi(1 − qi))y)== f( ∞∑i=1λi(qix + (1 − qi)y)).Using (3.1) for the bounded sequence qix + (1 − qi)y ∈ D, i ∈ N, and that f isqi-convex for each i ∈ N, we have thatf(tx + (1 − t)y) ≤∞∑i=1µif(qix + (1 − qi)y) ≤≤∞∑i=1µi(qif(x) + (1 − qi)f(y)) ==( ∞∑i=1µiqi)f(x) +( ∞∑i=1µi(1 − qi))f(y) ≤≤( ∞∑i=1µiqi)(f(x))+ +( ∞∑i=1µi(1 − qi))(f(y))+ ≤≤( ∞∑i=1µi)(f(x))+ +( ∞∑i=1µi)(f(y))+ == (f(x))+ + (f(y))+.This proves that f is bounded from above on the segment [x, y] := {rx ++(1 − r)y : r ∈ [0, 1]} connecting x and y. All in all, f : D → R is a Jensenconvex function defined on a non-empty and convex set D ⊆ Rd such that, forall x, y ∈ D, it is bounded from above on the segment [x, y]. To finish the proof,for any x, y ∈ D with x ̸= y, let us apply the Bernstein–Doetsch theorem (see54 M. Barczy and Zs. PálesBernstein and Doetsch [1] or Kuczma [5, Theorem 6.4.2]) to the convex andopen set (x, y) := {rx + (1 − r)y : r ∈ (0, 1)} and the Jensen convex functionfx,y : (x, y) → R, fx,y(v) := f(v), v ∈ (x, y), which is bounded from above.Then, for all x, y ∈ D with x ̸= y, we have that fx,y is continuous and convexon (x, y). Since fx,y is a restriction of f onto (x, y), it readily yields that f isconvex on D, as desired. ■In the next remark, we provide an example, which shows that a converseof Theorem 3.2 does not hold in general.Remark 3.1. Let (λn)n∈N ∈ Λ and (µn)n∈N ∈ Λ be such that (eλ1 − 1)//(e − 1) > µ1. Let d := 1, D := R and f : R → R, f(x) := ex, x ∈ R. Then fis convex, and, with x1 := 1 and xi := 0, i ∈ N \ {1}, we get thatf( ∞∑i=1λixi)= f(λ1) = eλ1 ,and∞∑i=1µif(xi) = µ1f(1) + (1 − µ1)f(0) = µ1e + 1 − µ1 = 1 + µ1(e − 1).Since (eλ1 − 1)/(e − 1) > µ1, we have that the inequality (3.1) does not hold.References[1] Bernstein, F. and G. Doetsch, Zur Theorie der konvexen Funktionen,Math. Ann., 76(4) (1915), 514–526.https://doi.org/10.1007/BF01458222[2] Daróczy, Z. and Zs. Páles, Convexity with given infinite weight se-quences, Stochastica, 11(1) (1987), 5–12.[3] Dudley, R.M., Real Analysis and Probability, Cambridge Studies in Ad-vanced Mathematics 74, Cambridge University Press, Cambridge, 2002.https://doi.org/10.1017/CBO9780511755347[4] Khan, M.A., K.A. Khan, D. Pečarić and J. Pečarić, Some newimprovements of Jensen’s inequality—Jensen’s type inequalities in infor-mation theory, Monographs in Inequalities 19, ELEMENT, Zagreb, 2020.[5] Kuczma, M., An Introduction to the Theory of Functional Equationsand Inequalities, Cauchy’s equation and Jensen’s inequality, BirkhäuserVerlag, Basel, 2nd edition, 2009.https://doi.org/10.1007/978-3-7643-8749-5[6] Kuhn, N., A note on t-convex functions, in: W. Walter (Ed.) General In-equalities, 4 (Oberwolfach, 1983), International Series of Numerical Math-ematics 71, Birkhäuser Verlag, Basel, 1984, 269–276.https://doi.org/10.1007/978-3-0348-6259-2_25A convexity-type functional inequality 55[7] Kuhn, N., On the structure of (s, t)-convex functions, in: W. Walter(Ed.) General Inequalities, 5 (Oberwolfach, 1986), International Series ofNumerical Mathematics 80, Birkhäuser Verlag, Basel, 1987, 161–174.https://doi.org/10.1007/978-3-0348-7192-1_13[8] Lang, R., A note on the measurability of convex sets. Arch. Math. (Basel),47(1) (1986), 90–92.https://doi.org/10.1007/BF01202504[9] Matkowski, J. and M. Pycia, On (α, a)-convex functions, Arch. Math.(Basel), 64(2) (1995), 132–138.https://doi.org/10.1007/BF01196632[10] Pavić, Z., Inequalities with infinite convex combinations, Turkish J. Ineq.,3(2) (2019), 53–61.https://www.tjinequality.com/articles/03-02-006.pdfMátyás Barczyhttps://orcid.org/0000-0003-3119-7953HUN-REN–SZTE Analysis and Applications Research GroupBolyai InstituteUniversity of SzegedAradi vértanúk tere 1H–6720 SzegedHungarybarczy@math.u-szeged.huZsolt Páleshttps://orcid.org/0000-0003-2382-6035Institute of MathematicsUniversity of DebrecenEgyetem tér 1H–4032 DebrecenHungarypales@science.unideb.hu

A convexity-type functional inequality with infinite convex combinations

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A convexity-type functional inequality with infinite convex combinations

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