Inequalities Characterizing Linear-Multiplicative Functionals

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 A be an algebra with unit 1 ∈ A over the field of real numbers.A map  : A → R is midpoint convex or Jensen convex, if the inequality  (  +  2 ) ≤  () +  () 2 holds for every ,  ∈ A. If this inequality holds in reverse, then a map  : A → R is midpoint concave or Jensen concave.Further, a map  : A → R is supermultiplicative, if the inequality holds for every ,  ∈ A.
Theorem 1.Let  : A → R 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.
Since  is nonconstant, then  < .Therefore, we can pick some ,  ∈ A 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  ∈ A. Then, Putting these together we get This means that  is odd.
Step 3 ( is additive and multiplicative).Utilizing Step 2 we obtain for all ,  ∈ A. Thus  is midpoint convex and multiplicative.Therefore, by Step 2 we infer that  is additive.
It is clear that if  : A → R is a constant supermultiplicative mapping equal to some  ∈ R, then 0 ≤  ≤ 1.Using this observation we derive an immediate corollary describing Jensen concave and supermultiplicative functions on the real line.
Corollary 3.All functions  : R → R which fulfill the system for all ,  ∈ R are of one of the following forms: with some  ∈ [0, 1].
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showing that if one reverse the inequality sign in one inequality of (15), then the assertion of Corollary 3 remains no longer valid, even if the remaining inequality holds with an equality.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., () = − exp() for  ∈ R).
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.
Proof.The "if " condition is straightforward to check.Therefore, we will prove the "only if " condition.Denote For a fixed  ∈  define a map  : () → R 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 0 <  ≤ 1.Note that the constant  may depend upon  ∈ .Therefore, there exists a function  0 :  → (0, 1] such that ()|  =  0 with every  ∈ ().Let us put  = (0).Note that  and  0 agree on  and moreover (0 2 ) ≥ (0) 2 ≥ (0), so  ≥ 0 and () = 0 for  ∈  0 .Now, fix some  ∈  0 and consider the function  : () → R defined by formula (18) above.By Theorem 1 we get that  is a linear and multiplicative functional of the space ().Therefore, either  ≡ 0 or there exists some (unique) point  0 ∈  (depending upon the choice of  ∈ ) such that Let  ⊆  0 be the set on which the second case holds and denote () =  0 for  ∈ .Note that the case  ≡ 0 is covered in the assertion of our theorem, since () = 0 for  ∈  0 .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  ∈ R the following functions are continuous for every  ∈ R: Observe that the characteristic map of the set  can be written as   = ( 2 ) − ( 1 ).Consequently,   is continuous, because ( 1 ) and ( 2 ) 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.