Systems of Inequalities Characterizing Ring Homomorphisms

Abstract

Assume that and are arbitrary mappings between two partially ordered rings and . We study a few systems of functional inequalities which characterize ring homomorphisms. For example, we prove that if and satisfy for all and , then and this mapping is a ring homomorphism. Moreover, we find two other systems for which we obtain analogous assertions.

1. Introduction

Let be a compact Hausdorff topological space. By we denote the space of all continuous real valued functions defined on and equipped with the supremum norm. Rădulescu [1] showed that if an operator is super-additive and super-multiplicative simultaneously, that is, it satisfies the systemfor each , then there exist a clopen subset and a continuous function such that where denotes the characteristic function of a given set. In particular, is linear, multiplicative, and continuous. Ercan [2] has shown that Rădulescu’s assumption that is a compact Hausdorff space may be dropped. More results on system (1) and on related questions have been obtained by Dhombres [3], Volkmann [4, 5], J. X. Chen and Z. L. Chen [6], Gusić [7], and the first author [8, 9], among others.

Our purpose is to generalize system (1) to the case of two unknown operators and in various directions. Moreover, in our last result we provide a condition sufficient for the separation of two mappings by a linear and multiplicative operator.

2. Main Results

To the end of the section let and be partially ordered rings and let and be two arbitrary mappings. We will need three crucial assumptions (cf. [3]):)Every nonnegative element of is a square: ()Every square in is nonnegative: () is the only element in whose square is equal to :

We will begin with the following system:assumed for all . Note that (6) with becomes (1).

In our first theorem we assume that (i.e., in ring for every ). We will prove that under assumptions , , and system (6) together with inequality characterizes ring homomorphisms; that is, and is simultaneously additive and multiplicative.

Theorem 1. Assume that conditions , , and hold true. Then and satisfy (6) and if and only if and is a ring homomorphism.

Proof. We will prove the nontrivial implication only. First, put into both inequalities and apply assumption to get and . Since , then .
Next, substitute into the second inequality to obtainUsing we deriveNow, let in the first inequality. We have Consequently, using this we deduceNext, put in the second inequality of (6). We get On the other hand, (10) implies that Join the last two inequalities and use (7) to deriveReplace in (13) by to arrive atThen add (13) and (14) side-by-side to reach Due to the assumptions and the last inequality is equivalent to the equality for every ; that is, is odd. Having this, it is easy to see that (10) implies that . To get the additivity of it is enough to apply the first inequality of the system with replaced by and replaced by and use the oddness of . Similarly, substitution in the second inequality leads to the conclusion that is multiplicative.

Remark 2. The assumption in Theorem 1 is essential and cannot be replaced by the opposite inequality. What is more, it is clear that every even super-additive mapping is nonpositive, whereas every even super-multiplicative mapping is nonnegative; therefore . In particular, both mappings are trivially separated by the zero ring homomorphism (c.f. Theorem 12).

In our next result we will deal with the system:postulated for all . In this case we do not need to assume an inequality between and .

Theorem 3. Assume that conditions , , and hold true and has a unit . Then and satisfy (16) if and only if and is a ring homomorphism.

Proof. Again, one implication is obvious. To prove the nontrivial one put in the first inequality of (16) to getNext, let in the second inequality of (16) and use to arrive atTherefore, thanks to we have whenever . In particular, . On the other hand, from (17) we obtain Consequently, . Having this and using (18) we see that so . Thus, (17) reduces toNext, apply the second inequality of (16) with replaced by . Then, add the result to this inequality side-by-side and then use (21) and the first inequality: Now, apply the above estimate for replaced by and add to the original one side-by-side: In particular, , which in view of and leads to for all ; that is, is odd.
From (21) and from the second inequality of (16) we deduce that Replace by and apply the oddness of to derive the reverse inequality: Therefore, is multiplicative. Using this, from the second inequality of (16) we getSince , then we obtain from this , which is the opposite inequality to (21). Therefore and to finish the proof it is enough to apply Theorem 1.

In what follows we will study one more system:postulated for all . It turns out that in general (27) is not equivalent to (16), but some arguments which worked for (16) can be utilized for (27). Again, no inequality between and will be assumed.

Lemma 4. Assume that conditions and hold true. If and satisfy (27), then (i) for ,(ii) for .

Proof. Since the second inequalities of (16) and (27) are identical, then from the proof of Theorem 3 we derive (i). In particular, . Next, put in the first inequality of (27) to get

Theorem 5. Assume that conditions , , and hold true. Then and satisfy (27) and if and only if and is a ring homomorphism.

Proof. We will justify the nontrivial implication. By repeating the respective calculations of the proof of Theorem 3 which involved the second inequality only together with and , we get that is odd and then, following this proof further, that is multiplicative and also .

An easy example shows that, even in the case the target space is the real line, the assumption cannot be dropped.

Example 6. For arbitrary functions and given by and for satisfy (27). Therefore, assumption in Theorem 5 is essential. In fact, both functions are subadditive.
Further, if is constant and equal to some , then for every function such that for all the pair solves (27). Similarly, if is constant and equal to some , then for every mapping such that for all the pair solves (27). Nonconstant solutions can be provided in a similar fashion by giving some small “freedom” for both functions.

In what follows, we will state some observations for real solutions of (21), which are not covered by the previous theorem, that is, such that .

Proposition 7. Assume that condition holds true, and the pair solves (27). If , then and for every .

Proof. Put in both inequalities of (27) to get and for all . Therefore, we obtainClearly . Now, suppose that for some . We claim that . Let . From (29) we deduce the estimate which justifies our claim. Therefore, there exists a sequence such that for . From this and from the second inequality of (27) we derive that as , which leads to a contradiction.
If for some , then .

In view of Proposition 7, the next one is self-evident.

Proposition 8. Assume that condition holds true, , the pair solves (27), and . Denote and . Then In particular,

Next, we will show that if contains a unit, then and attain positive values only.

Proposition 9. Assume that condition holds true, contains a unit , , and the pair solves (27). If , then and for every .

Proof. Put into the second inequality of (27) to getTherefore, since , then and since , then also for all .
Now, suppose that for some . Then from (34) we have so . Consequently, and thus . Finally, (34) implies that therefore , which is a contradiction.

Example 6 shows that it can happen that both sets , defined in Proposition 8 are empty, even if mappings , are regular. In what follows we will show that under some additional assumptions if one of the sets is nonempty, then both are equal to ; that is, and are constant and equal to and , respectively.

Corollary 10. Assume that condition holds true, contains a unit , , the pair solves (27), and the sets , are defined as in Proposition 8. If and contains an invertible element, then .

Proof. Let be an invertible element. Apply the second inequality of (27) for to get for all . Join this with the first inequality to obtainNow, suppose that is nonconstant. Define . Clearly, . Take such that . Put and apply (38) with to get a contradiction. Thus on . Equivalently, and, as a consequence, we get

Corollary 11. Assume that condition holds true, contains a unit , , the pair solves (27), and the sets , are defined as in Proposition 8. If and contains an invertible element, then .

Proof. We have , so by Proposition 8 also . Consequently and we can apply Corollary 10.

Now, let be a nonempty set and let denote the space of all bounded real valued functions defined on and equipped with the supremum norm. We consider with an order relation defined, as usual, coordinatewise; that is,

We will terminate the paper with an application of Theorem 1 to a separation problem. We will give a sufficient condition for the separation of two operators by operator which is additive and multiplicative simultaneously. Note that if , then the conditions are satisfied.

Theorem 12. Let satisfy the inequalityfor all , . Then there exists an additive and multiplicative operator such that

Proof. Let us define the operator by the formula Directly from the definition and from condition (42) we infer that and consequently, attains values in . We shall prove that operator is super-additive and super-multiplicative. Take arbitrary and arbitrary such that By the definition of operator , there exist , such that and there exist , such that Obviously, where for all , and , .
Therefore, we get On the other hand, We can choose and such that for every the values and are arbitrarily close to and , respectively. Therefore, we get for all To finish the proof it remains to apply Theorem 1.

Competing Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.