- Open Access
A new refinement of discrete Jensen’s inequality depending on parameters
© Horváth; licensee Springer. 2013
- Received: 10 June 2013
- Accepted: 23 October 2013
- Published: 22 November 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.
- convex function
- mid-convex function
- discrete Jensen’s inequality
- quasi-arithmetic mean
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 )
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 .
- (b)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.
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.
where means the set in (3).
- (b)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.
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.
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.
is continuous on the interior of .
The restriction of to has a unique continuous (and also convex) extension to .
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 :
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.
We study the monotonicity and convergence of the new means. For this, the next mean is also needed.
We now prove the monotonicity of the means (14) and give limit formulas.
- (c)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.
Before giving the proof of the main result, we introduce some preliminary lemmas. First, a simple inequality is established.
u is strictly increasing on , hence (). □
In the second lemma, the measurability of some functions is studied.
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.
for every .
The proof is now complete. □
The following lemma extends Lemma 14 in .
Proof The lowest common denominator is . Combined with , the result follows. □
We can now prove the first main result.
- I.Since ,
Next, we prove that ().
- III.Finally, we prove that(22)
hence (24) implies (22).
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.
by Theorem C. It follows that the last inequality in (8) is strict (this is true for every ).
are positive in (19), and therefore by Theorem C, it is enough to prove that there exist two different vectors in (25).
and thus , which is a contradiction.
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.
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 .
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).
which gives the result.
We begin the proof of the second part of the result with some limit formulas.
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 .
We have got the second sum in (13).
The proof is complete. □
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.
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