A refinement of the integral form of Jensen’s inequality

There are a lot of refinements of the discrete Jensen’s inequality, and this problem has been studied by many authors. It is also a natural problem to give analogous results for the classical Jensen’s inequality. In spite of this, few papers have been published dealing with this problem. The purpose of this paper is to give some refinements of the classical Jensen’s inequality. The results give a new approach of this topic. Moreover, new discrete inequalities can be derived, and the integral analogous of discrete inequalities can be obtained. We also have new refinements of the left-hand side of the Hermite-Hadamard inequality.

MSC: 26D07, 26A51.

In view of applications in different parts of mathematics, the classical Jensen’s inequality is especially noteworthy, as well as useful.

Theorem A (see [2])

Letfbe an integrable function on a probability spacetaking values in an interval. Thenlies inI. Ifqis a convex function onIsuch thatisP-integrable, then

The discrete version of the Jensen’s inequality is also particularly important.

Theorem B (see [8])

LetCbe a convex subset of a real vector spaceV, and letbe a convex function. Ifare nonnegative numbers with, and, then

Various attempts have been made by many authors to refine the discrete Jensen’s inequality (2). Basic papers in this direction were [10] and [9]. [6] and [4] contain essential generalizations of some earlier results. Some other types of refinements have been studied in the recent papers [12] and [5]. The treatment of the problem is similar in all the mentioned and other papers: Create an expression satisfying the inequality

where A is a sum whose index set is a subset (mainly a proper subset) of either or for some . For further details, see [5].

It is also a natural problem to give analogous results for the classical Jensen’s inequality (1). In spite of this, few papers have been published dealing with this problem (see [3] and [11]). One important reason is indicated: It is not possible to use the same machinery that is used in the discrete case. To the author’s best knowledge, it does not exist any refinements of (1) in the form

where is an integral over a proper subset of for some .

In this paper, such a refinement of (1) is obtained. The results not only extend and generalize earlier results, but give a new method to refine inequality (1). Moreover, the integral versions of classical discrete inequalities can be obtained. The inspiration for our result comes from the approach applied in [6]. We give such a version which shows the new feature clearly, but the method can be extended.

For the results from integration theory, see [2].

We consider the following conditions:

() Let be a σ-finite measure space such that .

The integrable functions are considered to be measurable.

() Let φ be a positive function on X such that .

In this case, the measure P defined on by

is a probability measure having density φ with respect to μ. An -measurable function is P-integrable if and only if gφ is μ-integrable, and the relationship between the P- and μ-integrals is

() Let be a fixed integer.

The σ-algebra in generated by the projection mappings ()

is denoted by . means the product measure on : This measure is uniquely (μ is σ-finite) specified by

We shall also use the following projection mappings: For define by

For every and for all , the sets

and

are called x-sections of Q () and -sections of Q (), respectively. We note that the sets lie in , while .

() Let such that

and

where are fixed positive numbers.

We stress that the first condition in (4) is necessary. For example, there exists a Borel set in whose image under the first projection map is not a Borel set in (see [7]). The function l is -measurable.

Under the conditions ()-(), we introduce the functions

and ()

Since

ψ is -measurable. Similarly, () is -measurable.

() Suppose ().

Now we formulate the main results.

Theorem 1Assume ()-(). Letbe aP-integrable function taking values in an interval, and letqbe a convex function onIsuch thatisP-integrable. Then

(a)

(b) If () is also satisfied, then

By applying the method used in the proof of the preceding theorem, it is possible to obtain a chain of refinements of the form

but some measurability problems crop up and it is not so easy to construct the expressions (). These difficulties disappear entirely if . In this case, we have the following theorem.

Theorem 2Assume ()-(), and let. Ifis aP-integrable function taking values in an interval, andqis a convex function onIsuch thatisP-integrable, then

where

The following special situations show the force of our results: They extend and generalize some earlier results; new refinements of the discrete Jensen’s inequality can be constructed; the integral version of known discrete inequalities can be derived.

1. Suppose is a probability space, (), , and . Then ()-() are satisfied. Suppose also that is a μ-integrable function taking values in an interval , and q is a convex function on I such that is μ-integrable. In this case, Theorem 1(a) gives Theorem 1.3(a) in [3]:

If () also holds, then Theorem 1.3(b) in [3] comes from Theorem 1(b):

We can see that Theorem 1 is much more general than (5) even if μ is a probability measure. Moreover, Theorem 1(b) makes it possible to obtain a chain of refinements in (5):

2. Let (). The set of the Borel subsets of () is denoted by (). λ means the Lebesgue measure on . Let be a convex function.

The classical Hermite-Hadamard inequality (see [1]) says

We can obtain the following refinement of the left-hand side of the Hermite-Hadamard inequality.

Corollary 3Letbe a Borel set such that

and

where ().

Then

Proof We can apply Theorem 1(a) to the pair of functions , , and . □

If and , then we have from (7) and Theorem 1(b) one of the main results in [13] as a special case:

Another concrete example can be constructed for (7) by using Corollary 5.

3. In the following results, we consider noteworthy proper subsets of .

(a) Let , , and be an integer. The simplex is defined by

Let (), and μ be a finite measure on the trace σ-algebra such that . Suppose is a positive function such that . Fix an integer , and let ().

Choose . Then

once the appropriate identification of with () has been made. Therefore,

where . Thus,

We can see that under the above assumptions ()-() are satisfied, so Theorem 1 can be applied.

Corollary 4Ifis aP-integrable function taking values in an interval, andqis a convex function onIsuch thatisP-integrable, then

Specifically, if , we have

When , this says

and in this case we have the inequality for the cube

Next, we show that inequality (8) extends the following well-known discrete inequality to an integral form. Similar results are quite rare in the literature (see [3]).

Theorem C (see [9])

LetIbe an interval in, and letbe a convex function. If, then for each

Let ( is an integer), and let μ be the measure on the trace σ-algebra defined by , where is the unit mass at m (). Suppose (), is a fixed integer, and ().

Some easy combinatorial considerations yield that for every and

where is the largest natural number that does not exceed x. Therefore,

Now, if I is an interval in , is a convex function, and defined by

then (9) follows immediately from (8).

(b) Let be an integer, , and . The open ball of radius r centered at the point z is denoted by .

Consider the measure space (). Suppose is a positive function such that . Fix an integer , and let (). Choose

Then for all and

Consequently,

According to this, for all ,

It is not hard to check that ()-() are satisfied in this situation, and thus Theorem 1 says:

Corollary 5Ifis aP-integrable function taking values in an interval, andqis a convex function onIsuch thatisP-integrable, then

By applying this result (, and ), we can have a special case of the refinement of the left-hand side of the Hermite-Hadamard inequality in (7).

4. We turn now to the case where X is a countable set.

() Consider the measure space , where either for some positive integer n or , denotes the power set of X, and is a positive number for all .

() Let be a sequence of positive numbers for which .

() Let be a fixed integer.

We define the functions (, ) on by

Then means the number of occurrences of v in . If , we introduce the following sums:

and

Every sum is either a nonnegative integer or ∞.

() Let such that for all , and

where ().

Since for all , for all . By the definition of the measure μ,

In this case, the function ψ has the form

Now Theorem 1(a) can be formulated in the following way.

Corollary 6Assume ()-(), and letbe a sequence taking values in an intervalsuch that. Ifqis a convex function onIsuch that, then

Assume ()-(), and suppose μ is the counting measure on , that is, for all . For a set , let denote the number of elements of A. Then ,

and

We note explicitly this particular case of Corollary 6.

Corollary 7Assume ()-(), whereμis the counting measure on, and letbe a sequence taking values in an intervalsuch that. Ifqis a convex function onIsuch that, then

Corollary 6 corresponds to Theorem 2 in [4], but in [4] only finite sets are considered. If and (), then Theorem 1(a) in [6] contains Corollary 7, but Corollary 6 makes sense in a lot of other cases (for example, for countably infinite sets).

Next, some examples are given.

The first example deals with a relatively flexile case.

Example 8 (a) Assume ()-(), and let () such that () and . Define . Then (4) holds and

where (), and means the characteristic function of (). We can see that () is satisfied and

The condition () is also true, since

Moreover, for ,

It follows that Theorem 1 can be applied.

(b) We consider the special case of (a), when the sets () are pairwise disjoint (a special partition of X). Let the function τ be defined on X by

Then

and

The second example corresponds to Corollary 6.

Example 9 Let , let be a sequence of positive numbers for which , and let be a sequence taking values in an interval such that . Define

An easy calculation shows that (≥2) for all , where denotes the greatest integer that does not exceed . If q is a convex function on I such that , then by Corollary 7

The final example illustrates the case .

Example 10 Consider the measure space . The function , is the density of the standard normal distribution on , and thus . Let

Then ()-() are satisfied. Let be a Borel measurable function taking values in an interval such that fφ is integrable, and let q be a convex function on I such that is integrable. By Theorem 1(a),

We first establish a result which will be fundamental to our treatment.

Lemma 11Assume ()-(), and letbe aP-integrable function. Then

Proof The functions

are obviously -measurable on S.

Suppose first that the function f is nonnegative. By (), , and hence the theorem of Fubini implies that

It follows from (10) that

If , then let

where

The sets () are pairwise disjoint and measurable. Moreover, by (4),

These establishments with (11) imply that

Choose . It is clear that if and , and hence

Therefore, (12) gives

Having disposed of the nonnegativity of the function f, we have from the first part of the proof that

and, therefore, the functions

are -integrable over S. By using this, the result follows by an argument entirely similar to that for the nonnegative case.

The proof is complete. □

Remark 12 Under the conditions of Lemma 11, we have

(a) The functions

are -integrable over S.

(b) The measure defined on by

is a probability measure.

Lemma 13Assume ()-(). Letbe aP-integrable function taking values in an interval, and letqbe a convex function onIsuch thatisP-integrable.

(a) The function

is-integrable overS.

(b) The functions

are-integrable overS.

Proof (a) It is easy to check that for fixed

are positive numbers with

This gives immediately that for every

and, therefore, by Theorem B,

Since the function q is convex on I, it is lower semicontinuous on I and, therefore, the function g is -measurable.

Choose an interior point a of I. The convexity of q on I implies that

where means the right-hand derivative of q at a. It follows from this and from (13) that

Now we can apply Remark 12(a), by the P-integrability of the functions , f and .

(b) Fix i from the set . We can prove as in (a) by using the -measurability of and the estimates

The proof is complete. □

Now we are able to prove the main results.

Proof of Theorem 1 (a) By Lemma 11,

Since is a probability measure on , it follows from the previous part, Theorem A, Lemma 11, and Lemma 13(a) that

Now (a) has been proven.

(b) By using the convexity of q, an easy manipulation leads to

Consequently, by applying (), Lemma 13(b), and the theorem of Fubini, we have

The proof is complete. □

Proof of Theorem 2 Apply Theorem 1 with and (). Then the conditions () (by using ) and () are satisfied,

and () has the form

Therefore,

and

The proof is complete. □

The author declares that they have no competing interests.

Supported by the Hungarian National Foundations for Scientific Research Grant No. K101217.

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