Journal of Function Spaces

Volume 2015 (2015), Article ID 945758, 3 pages

http://dx.doi.org/10.1155/2015/945758

## Inequalities Characterizing Linear-Multiplicative Functionals

Institute of Mathematics, University of Silesia, Bankowa 14, 40-007 Katowice, Poland

Received 14 October 2014; Accepted 13 January 2015

Academic Editor: Bruce A. Watson

Copyright © 2015 Włodzimierz Fechner. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

We prove, in an elementary way, that if a nonconstant real-valued mapping defined on a real algebra with a unit satisfies certain inequalities, then it is a linear and multiplicative functional. Moreover, we determine all Jensen concave and supermultiplicative operators , where and are compact Hausdorff spaces.

Let be an algebra with unit over the field of real numbers. A map is* midpoint convex* or* Jensen convex*, if the inequality
holds for every , . If this inequality holds in reverse, then a map is* midpoint concave* or* Jensen concave*. Further, a map is* supermultiplicative*, if the inequality
holds for every , .

First, we show that a nonconstant mapping is midpoint concave and supermultiplicative if and only if it is a linear and multiplicative functional. In the proof we follow some ideas of J. X. Chen and Z. L. Chen [1], Dhombres [2], Ercan [3], Gusic [4], Radulescu [5], and Volkmann [6, 7].

Theorem 1. *Let be a nonconstant mapping. Then is midpoint concave and supermultiplicative if and only if it is a linear and multiplicative functional.*

*Proof. *The “if” condition is straightforward; thus we will prove the “only if” condition. We will divide the proof into four steps.

*Step 1* (). Suppose for a contradiction that . Note that for every we have
Thus, in particular, and, consequently, we deduce that for all . Next, for each one can note the following estimate:
Thus, we arrive at
Let us define
Therefore, we can write
Moreover, since , then .

Since is nonconstant, then . Therefore, we can pick some , such that

Utilizing the fact that is midpoint concave, we obtain the inequality From the previous estimates we get a contradiction.

*Step 2* ( is an odd mapping). Let . Then,
Putting these together we get
This means that is odd.

*Step 3* ( is additive and multiplicative). Utilizing Step 2 we obtain
for all , . Thus is midpoint convex and multiplicative. Therefore, by Step 2 we infer that is additive.

*Step 4* ( is -homogeneous). Observe that . Therefore, either or . In the first case we get that for every . Since is nonconstant, we get a contradiction. Therefore .

Define a map by . Clearly, is additive and multiplicative and . Therefore for all . Now, fix arbitrary and . We have

*Remark 2. *Theorem 1 can be easily generalized to even more general spaces. An inspection of the proof shows that if we assume that is an arbitrary ring (not necessarily unital), then every nonconstant midpoint concave and superadditive mapping is additive and multiplicative.

It is clear that if is a constant supermultiplicative mapping equal to some , then . Using this observation we derive an immediate corollary describing Jensen concave and supermultiplicative functions on the real line.

Corollary 3. *All functions which fulfill the system
**
for all , are of one of the following forms: *(i)* for all ,*(ii)* for all ,**with some .*

We will provide two easy examples showing that if one reverse the inequality sign in one inequality of (15), then the assertion of Corollary 3 remains no longer valid, even if the remaining inequality holds with an equality.

*Example 4. *The absolute value of a real number satisfies the second inequality of (15) (with equality) and the reverse of the first one.

Every constant mapping equal to a negative number satisfies the converse to the second inequality of (15) and is both Jensen convex and Jensen concave. More generally, every real mapping with nonpositive values is submultiplicative. Therefore, to show that there are nonconstant solutions one can take any nonpositive convex mapping (e.g., for ).

In our next result we apply Theorem 1 to determine all Jensen concave and supermultiplicative operators , where and are compact Hausdorff spaces. Symbols and denote the spaces of all real continuous functions defined on or , respectively. These spaces are furnished with standard algebraic operations, pointwise partial order, and the supremum norm.

Theorem 5. *Assume that and are compact Hausdorff spaces and let be an arbitrary mapping. Then is a Jensen concave and supermultiplicative operator if and only if there exist a clopen (i.e., closed and open) set , function , and continuous function such that is continuous on and
**
for every .*

*Proof. *The “if” condition is straightforward to check. Therefore, we will prove the “only if” condition.

Denote
For a fixed define a map by the formula
It is clear that is a Jensen concave and supermultiplicative functional. Therefore, we can utilize Theorem 1. If , then is constant. If is constant, say for all , then we have . Note that the constant may depend upon . Therefore, there exists a function such that with every . Let us put . Note that and agree on and moreover , so and for .

Now, fix some and consider the function defined by formula (18) above. By Theorem 1 we get that is a linear and multiplicative functional of the space . Therefore, either or there exists some (unique) point (depending upon the choice of ) such that
Let be the set on which the second case holds and denote for . Note that the case is covered in the assertion of our theorem, since for . Therefore, formula (16) is fully proved.

We will show that is a clopen set. Let be any constant mapping equal to some . Since is a continuous mapping, according to formula (16), for every the following functions are continuous for every :
Observe that the characteristic map of the set can be written as . Consequently, is continuous, because and are continuous. From this we get that is a clopen set.

To finish the proof we need to show that is continuous on . Note that is not involved in the assertion of the theorem outside the set . And the continuity of on follows from the equality on and from the Urysohn’s lemma, which implies that elements of separate points of .

If we assume additionally that the space is connected, then the form of operators in question can be specified further.

Corollary 6. *Under assumptions of Theorem 5, if additionally the space is connected, then is a Jensen concave and supermultiplicative operator if and only if either there exists a continuous function such that
**
or there exists a continuous function such that
*

*Conflict of Interests*

*The author declares that there is no conflict of interests regarding the publication of this paper.*

*References*

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