Systems of Inequalities Characterizing Ring Homomorphisms

Copyright © 2016 W. Fechner and A. Olbryś. 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. Assume that T : P → R and U : P → R are arbitrary mappings between two partially ordered rings P and R. We study a few systems of functional inequalities which characterize ring homomorphisms. For example, we prove that if T and U satisfy T(f+g) ≥ T(f)+T(g), U(f⋅g) ≥ U(f) ⋅U(g), for allf, g ∈ P andT ≥ U, thenU = T and this mapping is a ring homomorphism. Moreover, we find two other systems for which we obtain analogous assertions.


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.Rȃdulescu [1] showed that if an operator  : () → () is super-additive and super-multiplicative simultaneously, that is, it satisfies the system  ( + ) ≥  () +  () ,  ( ⋅ ) ≥  () ⋅  () , for 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 Rȃdulescu'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.

Main Results
To the end of the section let (P, ≤) and (R, ≤) be partially ordered rings and let  : P → R and  : P → R be two arbitrary mappings.We will need three crucial assumptions (cf.[3]): (1) Every nonnegative element of P is a square: (2) Every square in R is nonnegative: (3) 0 is the only element in R whose square is equal to 0: We will begin with the following system: assumed for all ,  ∈ P. Note that (6) with  =  becomes (1). 2

Journal of Function Spaces
In our first theorem we assume that  ≤  (i.e., () ≤ () in ring R for every  ∈ P).We will prove that under assumptions (1), (2), and (3) system (6) together with inequality  ≤  characterizes ring homomorphisms; that is,  =  and  is simultaneously additive and multiplicative.
Next, substitute  =  into the second inequality to obtain Using (1) we derive Now, let  = − in the first inequality.We have Consequently, using this we deduce Next, 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 derive ∈ P. ( Replace in (13)  by − to arrive at Then add (13) and ( 14) side-by-side to reach Due to the assumptions (2) and (3) the last inequality is equivalent to the equality () + (−) = 0 for every  ∈ P; 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  ≤ 0 ≤ .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 ,  ∈ P. In this case we do not need to assume an inequality between  and .
Theorem 3. Assume that conditions (1), (2), and (3) hold true and P has a unit 1.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  = 0 in the first inequality of ( 16) to get Next, let  =  in the second inequality of (16) and use (2) to arrive at Therefore, thanks to (1) we have () ≥ 0 whenever  ≥ 0. In particular, (0) ≥ 0. On the other hand, from (17) we obtain Consequently, (0) = 0. Having this and using (18) we see that so (0) = 0. Thus, (17) reduces to Next, 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, [() + (−)] 2 ≤ 0, which in view of (2) and (3) leads to () = −() for all  ∈ P; that is,  is odd.
From (21) and from the second inequality of (16) we deduce that  ()  () ≤  ( ⋅ ) ≤  ( ⋅ ) , , ∈ P. (24) Replace  by − and apply the oddness of  to derive the reverse inequality: Therefore,  is multiplicative.Using this, from the second inequality of (16) we get Since 1 ∈ P, 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 ,  ∈ P. 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.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 (0) = 0, 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 (0) = 0 cannot be dropped.
Further, if  is constant and equal to some  < 1/4, then for every function  : P → R such that 2 ≤ () ≤ √ for all  ∈ P the pair (, ) solves (27).Similarly, if  is constant and equal to some  < 1/2, then for every mapping  : P → R such that  2 ≤ () ≤ (1/2) for all  ∈ P 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 (0) ̸ = 0.
In view of Proposition 7, the next one is self-evident.
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 P; that is,  and  are constant and equal to 1/2 and 1/4, respectively.
Proof.Let  ∈  be an invertible element.Apply the second inequality of (27) for  =  to get (1/2)() ≤ () for all  ∈ P. Join this with the first inequality to obtain Now, suppose that  is nonconstant.Define  = inf (P).
Consequently  ⊆  and we can apply Corollary 10.Now, let  be a nonempty set and let B() denote the space of all bounded real valued functions defined on  and equipped with the supremum norm.We consider B() with an order relation defined, as usual, coordinatewise; that is,  ≤  ⇐⇒  () ≤  () ,  ∈ , , ∈ B () .(41) 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 ,  : B() → B() by operator Φ : B() → B() which is additive and multiplicative simultaneously.Note that if P = R = B(), then the conditions (1)-(3) are satisfied.