#### Abstract

We give a necessary and sufficient mean condition for the quotient of two Jensen functionals and define a new class of mean values where are continuously differentiable convex functions satisfying the relation , . Then we asked for a characterization of such that the inequalities or hold for each positive , where are the harmonic, arithmetic, logarithmic, and identric means, respectively. For a subclass of with , this problem is thoroughly solved.

#### 1. Introduction

It is said that the mean is intermediate relating to the means and if the relation holds for each two positive numbers , .

It is also well known that where are the harmonic, geometric, logarithmic, identric, arithmetic, and Gini mean, respectively.

An easy task is to construct intermediate means related to two given means and with . For instance, for an arbitrary mean , we have that

The problem is more difficult if we have to decide whether the given mean is intermediate or not. For example, the relation holds for each positive and if and only if , where the Stolarsky mean is defined by (cf [1])

Also, holds if and only if , where the Hölder mean of order is defined by

An inverse problem is to find best possible approximation of a given mean by elements of an ordered class of means . A good example for this topic is comparison between the logarithmic mean and the class of Hölder means of order . Namely, since and , it follows from (2) that

Since is monotone increasing in , an improving of the above is given by Carlson [2]:

Finally, Lin showed in [3] that is the best possible approximation of the logarithmic mean by the means from the class .

Numerous similar results have been obtained recently. For example, an approximation of Seiffert’s mean by the class is given in [4, 5].

In this paper we will give best possible approximations for a whole variety of elementary means (2) by the class defined below (see Theorem 5).

Let be twice continuously differentiable (strictly) convex functions on . By definition (cf [6], page 5), if and only if .

It turns out that the expression represents a mean of two positive numbers , ; that is, the relation holds for each , if and only if the relation holds for each .

Let and denote by the set of convex functions satisfying the relation (15). There is a natural question how to improve the bounds in (14); in this sense we come upon the following intermediate mean problem.

*Open Question*. Under what additional conditions on , the inequalities
or, more tightly,
hold for each ?

As an illustration, consider the function defined to be

Since it follows that is a twice continuously differentiable convex function for , .

Moreover, it is evident that .

We will give in the sequel a complete answer to the above question concerning the means defined by

Those means are obviously symmetric and homogeneous of order one.

As a consequence we obtain some new intermediate mean values; for instance, we show that the inequalities hold for arbitrary . Note that

#### 2. Results

We prove firstly the following

Theorem 1. *Let with . The expression represents a mean of arbitrary numbers if and only if the relation (15) holds for .*

*Remark 2. *In the same way, for arbitrary , it can be deduced that the quotient
represents a mean value of numbers , if and only if (15) holds.

A generalization of the above assertion is the next.

Theorem 3. *Let be twice continuously differentiable functions with on and let , be an arbitrary positive weight sequence. Then the quotient of two Jensen functionals
**
represents a mean of an arbitrary set of real numbers if and only if the relation
**
holds for each .*

*Remark 4. *It should be noted that the relation determines in terms of in an easy way. Precisely,
where and and are constants.

Our results concerning the means , are included in the following.

Theorem 5. *For the class of means defined above, the following assertions hold for each .*(1) *The means are monotone increasing in ;*(2) * for each ;*(3) * for ;*(4) * for ;*(5) * there is a number such that for ;*(6) * there is a number such that for ;*(7) * for each ;*(8) * there is no finite such that the inequality holds for each .**The above estimations are best possible.*

#### 3. Proofs

##### 3.1. Proof of Theorem 1

We prove firstly the necessity of the condition (15).

Since is a mean value for arbitrary ; , we have Hence

From the other hand, due to l’Hospital’s rule we obtain Comparing (29) and (30) the desired result follows.

Suppose now that (15) holds and let . Since by the Cauchy mean value theorem there exists such that

But, and, since is strictly increasing, .

Therefore, by (31) we get Finally, integrating (33) over we obtain the assertion from Theorem 1.

##### 3.2. Proof of Theorem 3

We will give a proof of this assertion by induction on .

By Remark 2, it holds for .

Next, it is not difficult to check the identity where

Therefore, by induction hypothesis and Remark 2, we get

The inequality can be proved analogously.

For the proof of necessity, put and proceed as in Theorem 1.

*Remark 6. *It is evident from (15) that if then has to be also convex on . Otherwise, it shouldn't be the case. For example, the conditions of Theorem 3 are satisfied with , . Hence, for an arbitrary sequence of real numbers, we obtain

Because the above inequality does not depend on , a probabilistic interpretation of the above result is contained in the following.

Theorem 7. *For an arbitrary probability law of random variable with support on , one has
*

##### 3.3. Proof of Theorem 5, Part

We will prove a general assertion of this type. Namely, for an arbitrary positive sequence and an associated weight sequence , , denote

For , we have which is equivalent to

Theorem 8. *The sequence * *is monotone increasing in *, .*This assertion follows applying the result from [7, Theorem 2] which states the following.*

Lemma 9. *For , the inequality
**
holds for arbitrary sequences , .*

Putting there , , and , , , we successively obtain

Since , multiplying those inequalities we get the relation (41), that is, the proof of Theorem 8.

The part (1) of Theorem 5 follows for .

A general way to prove the rest of Theorem 5 is to use an easy-checkable identity with .

Since , we get . Also,

Therefore, we have to compare some one-variable inequalities and to check their validness for each .

For example, we will prove that the inequality holds for each positive if and only if .

Since is monotone increasing in , it is enough to prove that

By the above formulae, this is equivalent to the assertion that the inequality holds for each , with

We will prove that the power series expansion of have non-positive coefficients. Thus the relation (48) will be proved.

Since we get

Hence, and, after some calculation, we get

Now, one can easily prove (by induction, e.g.) that is a negative real number for . Therefore , and the proof of the first part is done. For we have

Therefore, for and sufficiently small .

Similarly, we will prove that the inequality holds for each , ; if and only if .

As before, it is enough to consider the expression with

It is not difficult to check the identity

Hence by (48), we get , that is, is monotone increasing for .

Therefore

By monotonicity it follows that for .

For , we have

Hence, for and sufficiently small.

From the other hand,

Examining the function , we find out that it has the only real zero at and is negative for .

*Remark 10. *Since is monotone increasing, we also get
Hence

A calculation gives .

Note also that

Therefore, applying the assertion from the part 1, we get

Finally, we give a detailed proof of the part 7.

We have to prove that for . Since is monotone increasing in , it is sufficient to prove that the inequality holds for each .

Therefore, by the transformation given above, we get and the proof is done.

Further, we have to show that for some positive , whenever .

Indeed, since for and sufficiently small , we get Similarly, Hence, and this expression is positive for and sufficiently small, that is, sufficiently close to .

As for the part 8, applying the above transformation we obtain where , .

Since for , and the last expression is less than one, it follows that the inequality cannot hold whenever is sufficiently large.

The rest of the proof is straightforward.

#### Acknowledgment

The author is indebted to the referees for valuable suggestions.