Abstract
The paper deals with convex combinations, convex functions, and Jensen’s functionals. The main idea of this work is to present the given convex combination by using two other convex combinations with minimal number of points. For example, as regards the presentation of the planar combination, we use two trinomial combinations. Generalizations to higher dimensions are also considered.
1. Introduction
Let be a real vector space. A set is affine if it contains the lines passing through all pairs of its points (all binomial affine combinations in , i.e., the combinations of points and coefficients of the sum ). A function is affine if it satisfies the equality for all binomial affine combinations in .
A set is convex if it contains the line segments connecting all pairs of its points (all binomial convex combinations in , i.e., the combinations of points and nonnegative coefficients of the sum ). A function is convex if it satisfies the inequality for all binomial convex combinations in .
Using the mathematical induction, it can be proved that every affine function satisfies the equality for all affine combinations in and that every convex function satisfies the Jensen inequality for all convex combinations in .
For an affine or a convex combination the point itself is called the combination center, and it is important to mathematical inequalities. Recognizing the importance of the combination center, the authors (see [1]) have recently considered inequalities on simplexes and their cones.
A general overview of convex sets, convex functions, and its applications can be found in [2]. In working with means and their inequalities, we can rely on the book in [3]. Many details of the branch of mathematical inequalities are written in [4].
2. Convex Combinations of the Line
The section shows the importance of convex combination centers in deriving inequalities. The main result is Theorem 2.
If are different numbers, say , then every number can be uniquely presented as the affine combination The above binomial combination is convex if and only if the number belongs to the interval . Given the function , let be the function of the line passing through the points and of the graph of . Using the affinity of , we get the equation If the function is convex, then, using the definition of convexity, we obtain the inequality and the reverse inequality
By the end of this section we will use an interval with the nonempty interior .
The following lemma represents a systematised version of [5, Proposition 2] and deals with two convex combinations having the same center. Assigning the convex function to such convex combinations, we obtain the following Jensen type inequality.
Lemma 1. Let be points such that . Let be a convex combination with points , and let be a convex combination with points .
If the above convex combinations have the same center
then every convex function satisfies the inequality
If is concave, then the reverse inequality is valid in (8).
Proof. Assume is convex. If , the right-hand side follows from the series of inequalities derived applying the inequality in (5) to and the inequality in (6) to . If , we use any support line instead of the chord line .
Lemma 1 is the generalization of Jensen’s inequality: applying this lemma to the convex combination center equality with the assumption , we come to the Jensen inequality
So, the discrete form of the famous Jensen inequality (discrete form in [6] and integral form in [7]) can be derived applying the convexity definition and the affinity of the chord or support line. The different forms of Jensen’s inequality can be seen in [8].
Theorem 2. Let be points such that . Let be a convex combination of the points with the center .
Then there exist two binomial convex combinations and so that
and consequently, every convex function satisfies the inequality
If is concave, then the reverse inequality is valid in (13).
Proof. We use the formula in (3) to calculate the coefficients and that satisfy and also for and . Now we need to apply Lemma 1 to both sides of the obtained equality in (12).
The graphical representation of the equality in (12) and the inequality in (13) is shown in Figure 1.
Binomial convex combinations are included into the definition of convexity. The following corollary demonstrates how the binomial combinations may be assigned to each convex combination.
Corollary 3. Let be a convex combination in . Then there exist two binomial convex combinations and with points from the set so that and consequently, every convex function satisfies the inequality
Proof. Put and suppose . Take and . If , we take . If , we calculate the coefficients and by the formula in (3) to get .
If is equal to some , then we take and . Otherwise, it must be for some pair, in which case we take and . Calculating and by (3), we also get .
It remains to apply Lemma 1 to both sides of the obtained equality in (14).
Respecting the Jensen inequality, the formula in (15) can be expressed in the extended form:
Let us show how the intermediate member in (13) can be transformed into integral. Let be a bounded set with the length , and let be an integrable (in the sense of Riemann or Lebesgue) function on . Given the positive integer , we employ a partition where each of pairwise disjoint subsets contracts to the point as approaches infinity. Take one point for every and then compose the convex combination of the points with the coefficients ; that is, Letting to infinity, the sequence approaches the point
Using the integral method with convex combinations, we obtain the mixed discrete-integral form of Theorem 2 as follows.
Corollary 4. Let be points such that . Let the barycenter of the set belongs to ; that is,
Then there exist two binomial convex combinations and so that
and consequently, every convex function satisfies the inequality
If is concave, then the reverse inequality is valid in (22).
The summarizing Jensen’s functional of a function for the given convex combination in is defined with If the conditions of Theorem 2 hold, we get the functional inequality
The integrating Jensen’s functional of an integrable function for the given subset with the length is defined with If the conditions of Corollary 4 are satisfied, we get
The inequalities in (24) and (26) offer the local bounds of Jensen’s functionals. The global bounds of Jensen’s summarizing functional were investigated in [9]. Some new Jensen type inequalities were obtained in [10].
3. Convex Combinations of the Plane
This section contains the main results, Theorem 6, and its consequences.
We assume that is the real vector space treating its points as the vectors with the standard coordinate addition and the scalar multiplication .
If , , and are the planar points that do not belong to one line, respectively, the convex hull is a real triangle, and then every point can be presented by the unique affine combination where The above trinomial combination is convex if and only if the point belongs to the triangle .
Given the triangle with vertices , , and , the convex cone with the vertex at is the set spanned by the vectors and (similarly and ; all three cones can be viewed in Figure 2); that is,
Given the function , let be the function of the plane passing through the points , , and of the graph of . Due to the affinity of , it follows For a convex function , using the convexity definition, we get the inequality and using the affinity of , the reverse inequality
By the end of the section we will use a planar convex set with the nonempty interior . The area of a planar set will be denoted with .
Lemma 5. Let be points, , and be the cone union. Let be a convex combination of the points , and let be a convex combination of the points .
If the above convex combinations have the same center
then every convex function satisfies the inequality
Proof. If the set is a real triangle, we can apply the proof of Lemma 1 using instead of respecting the plane inequalities in (31)-(32). If , then we rely on the proof of Lemma 1 with the chord line . If , we use any support line at the point .
Theorem 6. Let be points such that
with , and let be the cone union. Let be a convex combination of the points with the center .
Then there exist two trinomial convex combinations and so that
and consequently, every convex function satisfies the inequality
Proof. First we calculate the coefficients by the formula in (28) to get the equality in (36) and then apply Lemma 5 to its both sides.
The graphical representation of the equality in (36) can be seen in Figure 2.
Applying the integral method with convex combinations, we obtain the form as follows.
Corollary 7. Let be points such that
with , and let be the cone union. Let the barycenter of the set belong to ; that is,
Then there exist two trinomial convex combinations and so that
and consequently, every convex function satisfies the inequality
If is concave, then the reverse inequality is valid in (41).
If the conditions of Theorem 6 are valid, then using the summarizing Jensen functional of for the given convex combination , we get the functional inequality If the conditions of Corollary 7 are satisfied, then applying another integrating Jensen functional of for the given subset with the area , we have the inequality
4. Generalization
The aim of the section is to generalize Theorem 6 and Corollary 7 to more dimensions.
If are points such that the vectors are linearly independent, then the convex hull is called the -simplex with the vertices . All the simplex vertices can not belong to the same hyperplane in . Any point can be presented by the unique affine combination If we use the point coordinates, then the coefficients can be determined by the generalized coefficient formula in (28). The combination in (47) is convex if and only if the point belongs to the -simplex .
Given the -simplex with vertices , let be the convex cone with the vertex at spanned by the vectors (similarly ); that is,
Given the function , let be the function of the hyperplane (in ) passing through the points of the graph of . Applying the affinity of to the combination in (47), it follows If we use the convex function , then we get the inequality and the reverse inequality which can be proved in the same way as the inequality in (32).
By the end of the section we will use a convex set with the nonempty interior . The volume of a set will be denoted with .
The generalization of Theorem 6 applied to -simplexes is as follows.
Theorem 8. Let ; be points such that
with , and let be the cone union. Let be a convex combination of the points with the center .
Then there exist two -membered convex combinations and so that
and consequently every convex function satisfies the inequality
Using the integrals, we get the form as follows.
Corollary 9. Let be points such that
with , and let be the cone union. Let the barycenter of the set belong to ; that is,
Then there exist two -membered convex combinations and so that
and consequently every convex function satisfies the inequality
If is concave, then the reverse inequality is valid in (58).
If the conditions of Theorem 8 are valid, then using the summarizing Jensen functional of for the given convex combination , we get the functional inequality If the conditions of Corollary 9 are satisfied, then applying another integrating Jensen functional of for the given subset with the volume , we have the inequality
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgments
The work of the first and third authors has been fully supported by Mechanical Engineering Faculty in Slavonski Brod. The work of the second author has been supported by the Natural Science Foundation of Fujian province of China under Grant no. 2012J01014. The authors are grateful to Velimir Pavić (graphic designer at Školska knjiga Zagreb) who has graphically edited Figures 1 and 2.