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A new refinement of discrete Jensen’s inequality depending on parameters
Journal of Inequalities and Applications volume 2013, Article number: 551 (2013)
In this paper we give a new refinement of discrete Jensen’s inequality, which generalizes a former result. The introduced sequences depend on parameters. The strict monotonicity and the convergence are investigated. We also study the behavior of the sequences when the parameters vary. One of the proofs requires an interesting convergence theorem with probability theoretical background. This result is an extension of a former result, but its proof is simpler. The results are applied to define and study some new quasi-arithmetic means.
We begin with discrete Jensen’s inequality which is a useful tool in different parts of mathematics. The set of nonnegative integers and that of positive integers will be denoted by ℕ and , respectively.
Theorem A (see )
Let C be a convex subset of a real vector space X, and let be a finite subset of C, where is fixed. Let be nonnegative numbers with . If is either a convex or a mid-convex function and in the latter case the numbers () are rational, then
The function is called convex if
and mid-convex or Jensen-convex if
f is strictly convex if strict inequality holds in (2) whenever and .
In recent years, a number of papers have appeared in which the authors constructed different refinements of discrete Jensen’s inequality. See, for example, [2–8] and . These papers also contain a lot of interesting applications. The following parameter-dependent refinement can be found in .
Theorem B Let C be a convex subset of a real vector space X, and let be a finite subset of C, where is fixed. Let be nonnegative numbers with , and let . Suppose that is either a convex or a mid-convex function and in the latter case the numbers () and λ are rational. Introduce the sets
and for define the numbers
Suppose that X is a normed space and f is continuous. For every fixed ,(5)
Our aim is to present a generalization of the previous result.
The sequence depends on one parameter, but we allow the new sequence (see (7)) to depend on n parameters. We give conditions for the strict monotonicity of in some cases.
The proof of limit assertion (5) uses a lemma (see Lemma 15 in ) with a rather difficult probability theoretical proof. Essentially, a fundamental theorem from statistics, which is based on the multidimensional central limit theorem, has been applied. In this work, we obtain a convergence theorem extending (5) which will be deduced from another interesting lemma analogous to Lemma 15 in . The proof of this lemma is simpler than the proof of Lemma 15 in , although it also has probability theoretical background, namely the strong law of large numbers is used.
The behavior of the sequence is studied when the parameters vary. Further refinements of discrete Jensen’s inequality can be derived from these results.
As an application, some new quasi-arithmetic means are constructed, and the monotonicity and convergence of these means are investigated.
2 The main results and some preliminary results
The main results are given by Theorem 1 and Theorem 5. In order to render the main results as transparently as possible, we also give some preparatory lemmas and theorems.
The first result generalizes Theorem B.
Theorem 1 Let C be a convex subset of a real vector space X, and let be a finite subset of C, where is fixed. Let be nonnegative numbers with , and let (). Suppose that is either a convex or a mid-convex function, and in the latter case, the numbers and () are rational.
For , introduce
where means the set in (3).
Assuming X is a normed space and f is continuous, we have(9)
The result contains Theorem B as a special case . Since f is continuous in part (b) of the previous result, f is necessarily convex in this case.
The clue of the proof of Theorem 1(b) is the following decisive assertion, which is deduced from the strong law of large numbers. This result extends Lemma 15 in  with a less complicated proof.
Theorem 2 Let C be a convex subset of a real normed space X, and let be a finite subset of C, where is fixed. Let be positive numbers with , and let (). Suppose is a convex and continuous function. Define, for ,
Then we have
Next, we determine some of the cases of strict inequality in (8).
Theorem 3 Let C be a convex subset of a real vector space X, and let be points of C with at least two different elements, where either or . Let be positive numbers with , and let (). If is strictly convex, then every inequality is strict in (8).
Probably, the result remains true if , but the method applied in the proof of Theorem 3 is getting more and more chaotic. Another treatment of the problem may be successful.
In the proof of Theorem 3, we shall use the following well-known result (see ).
Theorem C Let C be a convex subset of a real vector space X, and let be points of C with at least two different elements, where is a fixed integer. Let be positive numbers with . If is strictly convex, then the inequality is strict in (1).
The following notations and observations may help to understand the next results, and they will be used in their proofs. We motivate proceeding in this direction as follows: consider Theorem 1, and let . It is easy to see that only the restriction of f to the convex hull of has significance.
Let C be a convex subset of a real vector space X, and let be a finite subset of C, where is a fixed integer. Suppose that is a convex function. Choose the integers and , where and (). Define the convex set
and the function on by
Then is well defined since the set
is the convex hull of , and hence it is a subset of C. A consequence of the convexity of f is that is also convex. Further properties of the function are studied in the following lemma.
Lemma 4 Let C be a convex subset of a real vector space X, and let be a finite subset of C, where is a fixed integer. Suppose that is a convex function. Choose the integers and , where and (). Then:
is continuous on the interior of .
The restriction of to has a unique continuous (and also convex) extension to .
There are some trivial cases of Theorem 1: if for an , then
and therefore () (thus ) will be supposed in the next main result, in which the parameter varies.
Theorem 5 Let C be a convex subset of a real vector space X, and let be a finite subset of C, where is a fixed integer. Let be nonnegative numbers such that () and , and let (). Suppose that is a convex function. Choose . Then, for every fixed (, ) and for each fixed :
(b1) In the case ,
(b2) In the case (<1), there exists an integer such that m elements of the sequence belong to the open interval , and members are 0. We can assume that ( and they are different from l) and the others are 0. By using the function (see (10) and Lemma 4(c)), we have
By (8), the limits in (11) and (13) also refine discrete Jensen’s inequality. It should be mentioned that in Theorem 1 the topology on X is needed in showing (b), but Theorem 5 is true without any topology on X.
Definition 6 Let be an interval, (), be nonnegative numbers such that , and let (). Let be continuous and strictly monotone functions. We define the quasi-arithmetic means with respect to (7) by
We study the monotonicity and convergence of the new means. For this, the next mean is also needed.
Definition 7 Let be an interval, (), be nonnegative numbers such that . For a continuous and strictly monotone function , we introduce the following mean:
We now prove the monotonicity of the means (14) and give limit formulas.
Proposition 8 Let be an interval, (), be nonnegative numbers such that , and let (). Let be continuous and strictly monotone functions. Then:
if either is convex and ψ is increasing or is concave and ψ is decreasing.
if either is convex and ψ is decreasing or is concave and ψ is increasing.
Moreover, in both cases
Proof Theorem 1(a) can be applied to the function if it is convex ( if it is concave) and to the n-tuples , then upon taking , we get (a) and (b). (c) Comes from Theorem 1(b). □
As an illustration, we consider the following special case.
Example 9 If , and (), then by Proposition 8(b), we have the following sharpened version of the weighted arithmetic mean - geometric mean inequality: for every (), () and ,
4 Proofs and some auxiliary results
Before giving the proof of the main result, we introduce some preliminary lemmas. First, a simple inequality is established.
Lemma 10 If , then
Proof We shall show that
The left inequality is well known, and the right inequality is equivalent to
u is strictly increasing on , hence (). □
In the second lemma, the measurability of some functions is studied.
Lemma 11 Let be a measurable space, and let X be a real normed space. ℬ denotes the σ-algebra of Borel sets in X. If , is measurable, and is fixed, then the function
Proof Since X is a normed space, the mapping , is continuous. Now the result follows from the fact that function (16) can be written as . □
After these preparations, we can now prove Theorem 2.
Proof of Theorem 2 We can obviously suppose that and .
Then () and , that is, we have a discrete distribution.
Let be a probability space, and let
be identically distributed and independent random variables such that
Define the random variables
where means the characteristic function of the set . The joint distribution of the vector random variable is a multinomial distribution
for every .
Lemma 11 and the continuity of f imply that the function ,
is measurable, and therefore it is also a random variable. By applying Lemma 4 with (), f is bounded on the convex hull of the set . It follows that is also bounded, and thus () is P-integrable. By (17),
It is a consequence of Lebesgue’s convergence theorem that
By the strong law of large numbers,
Therefore, by Lemma 10,
and this leads to
Consequently, we get from (18) that
The proof is now complete. □
The following lemma extends Lemma 14 in .
Lemma 12 Let and be fixed. If we set
Proof The lowest common denominator is . Combined with , the result follows. □
We can now prove the first main result.
Proof of Theorem 1 (a) The proof is divided into three steps.
Next, we prove that ().
It is not hard to see that for every ,
According to Theorem A, this yields that
Bringing in Lemma 12, we find that the right-hand side of (21) can be written in the form
which is just .
Finally, we prove that(22)
It follows from Theorem A that
The multinomial theorem shows that
hence (24) implies (22).
(b) It is enough to confirm that for ,
This is evident if . If , then we have it from Theorem 2.
The proof is complete. □
Proof of Theorem 3 It is enough to prove that there is strict inequality for a fixed () in (20) and in (23), respectively.
The treatment of inequality (23) is pretty immediate. Under the conditions of the theorem,
by Theorem C. It follows that the last inequality in (8) is strict (this is true for every ).
It remains to study (20). Let be fixed, and let
Under the hypotheses, the coefficients of the vectors
are positive in (19), and therefore by Theorem C, it is enough to prove that there exist two different vectors in (25).
We begin with the case . Then , and (25) consists of two vectors, namely
Elementary considerations show that implies that
and thus , which is a contradiction.
Now, we continue with the case . Then the following three vectors belong to (25):
Suppose . A simple but troublesome calculation confirms that
Therefore, recalling that
we have a homogeneous system of two linear equations with solutions and . A few easy calculations yield that the determinant of the matrix of this system is
() implies that
and hence (26) is negative (it is only important that different from 0). From this we get
that is, , which is also a contradiction.
The proof is complete. □
Proof of Lemma 4 (a) It is well known (see ) that a convex function on an open convex set in is continuous.
(b) is bounded above since discrete Jensen’s inequality shows
Let be fixed. Since is convex, there is a linear functional L on (a support of f at ) such that
and therefore the continuity of L on the compact set yields that is bounded below on . It is easy to check that inequality (27) holds for every boundary point of , too.
(c) Because is a polytope, this follows from (b) (see ).
The proof is complete. □
Proof of Theorem 5 (a) We can obviously suppose that .
An elementary calculation gives
If , then for every ,
and therefore, by (28)
which we wanted.
In the rest of the proof, . From this, the existence of an integer such that m elements of the sequence belong to the open interval and members are 0 follows. We can assume that and .
Consider the function defined in (10).
Fix . Then
which shows that
is an interior point of . In light of the continuity of on (see Lemma 4), this implies that
The limit relations (28) and (29) entail
which gives the result.
We begin the proof of the second part of the result with some limit formulas.
It is obvious that
It is also easy to calculate that for each ,
(b1) The expression
does not depend on for each , from which we have (11) by (30) and (31).
(b2) If , then (32) is independent of , hence we can have the coefficient in (12) and the first member in the sum (13) from (30) and (31).
Suppose such that .
By using (31), it remains to study the existence of the limit of (32) as . Under the conditions on the zero elements of the sequence , we have to consider
is an interior point of the domain of for every , Lemma 4(c) implies that the limit (33) exists and it is . According to this and (31)
We have got the second sum in (13).
The proof is complete. □
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The author thanks the anonymous referees for their careful reading and valuable comments. This work was supported by Hungarian National Foundations for Scientific Research Grant No. K101217.
The author declares that they have no competing interests.
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Horváth, L. A new refinement of discrete Jensen’s inequality depending on parameters. J Inequal Appl 2013, 551 (2013). https://doi.org/10.1186/1029-242X-2013-551
- convex function
- mid-convex function
- discrete Jensen’s inequality
- quasi-arithmetic mean