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DeltaSincov mappings in Banach algebras
Journal of Inequalities and Applications volume 2023, Article number: 147 (2023)
Abstract
We study solutions and approximate solutions of the multiplicative Sincov equation
for mapping T taking values in a commutative Banach algebra.
1 Motivation: Grüsstype inequalities as an approximate multiplicative Sincov equation
Let \([a, b]\subset \mathbb{R}\) be a nondegenerate interval and \(f, g \in L([a,b])\) be Lebesgue integrable functions. Assume that there exist constants \(m_{f}, m_{g}, M_{f}, M_{g} \in \mathbb{R}\) such that
Let us denote the integral mean of a function f by \(I(f)\):
The celebrated Grüss inequality dates as early as 1935 and says that
Nowadays several generalizations and refinements are known. For a comprehensive study of the topic and the list of references, we refer the reader to a monograph by Dragomir [3] and to papers by Otachel [14–16].
A Grüsstype inequality is an inequality that provides an upper bound for the expression of the form
sometimes under additional assumption \(\Vert g\Vert =1\), or
where f, g, h belong to an inner product space (see [3, Theorems 15, 16, 17]). Cases with an inner product replaced by another functional have been studied as well, including discrete versions of the original inequality.
In paper [4] we dealt with functionals that satisfy
with some constant \(c\geq 0\). Research of [4] was motivated by Richard’s inequality, which is an example of Grüsstype inequality. It says that particular solutions of (3) are functionals of the form
In the main result of [4], we proved that every unbounded real or complexvalued solution of (3) is a solution of the Sincov equation
Consequently, S has the representation
with some nevervanishing real or complexvalued mapping φ (see Gronau [8, Theorem]). From the main result of [4] it follows that no generalization of Richard’s inequality, such that the functional S is replaced by an unbounded one, is possible.
The multiplicative Sincov equation is important in the theory of functional equations and has several applications. Recently, Kiss and Schwaiger [9] studied its conditional version and applied their results to the problem of finding a rule of interest compounding if negative interest rates are possible. Besides this, Baczyński et al. [1, 2] investigated fuzzy implications satisfying a version of the Sincov equation. A comprehensive study of this equation with a list of references is due to Gronau [8].
We will conclude this introductory section with an observation that the Sincov equation covers the equation of exponential mappings. Therefore, all our subsequent results generalize the earlier studies concerning exponential mappings.
Proposition 1
Assume that \((X, \cdot )\) is a group, \(\mathcal{A}\) is a commutative ring, \(F\colon X\to \mathcal{A}\) and \(S\colon X\times X\to \mathcal{A}\) are given by
Then F satisfies
if and only if S satisfies (4).
Similarly, approximate solutions of the Sincov equation yield generalizations to approximatively exponential mappings. We will omit the details.
2 DeltaSincov mappings for complexvalued functionals
Assume that X is a nonempty set and S and F are a complexvalued mapping and a nonnegative mapping, respectively, acting on the product \(X \times X\). In this section we study the following functional inequality:
Each solution of (6) is called a deltaSincov mapping with a control function F. Clearly, if \(c\geq 0\) is a fixed constant and one takes as F constant map equal to \(1/2(1+ \sqrt{1+4c})\), then (6) reduces to (3). Therefore, the results of this section extend the findings of [4].
The present study is meant to fall in line with the research initiated by the notion of deltaconvexity by Veselý and Zajíček [17] and then continued by several authors. Dissertation [17] is mainly devoted to the inequality
Continuous solutions F of this inequality are called deltaconvex. One of the theorems of [17] says that every realvalued deltaconvex function can be written as a difference of two convex functionals. This result motivated a study of several related problems. In particular, Ger [6] studied the inequality
using the term deltaexponential map for F. More recently, Olbryś dealt with delta \((s,t)\)convex mappings [10], delta Schurconvex mappings [11, 13], and deltasubadditive and deltasuperadditive mappings [12].
We begin the study of (6) with an observation that inequality (6) behaves in a way similar to other inequalities motivated by the notion of deltaconvexity. Proposition 2 below is an analogue to [6, Proposition 1].
Proposition 2
Assume that X is a nonempty set and \(S\colon X \times X \to [0, + \infty ) \) and \(F\colon X \times X \to [0, + \infty )\) satisfy inequality (6). If we denote \(H=S+F\), then
Proof
Fix \(f, g, h \in X\); we have by (6)
□
We will apply an observation from [5] to exclude cases when F attains zero. Indeed, by [5, Proposition 2], if F has a zero, then \(F=0\) on \(X \times X\) and, consequently, S solves Sincov equation (4). Therefore, from now on we will restrict ourselves to the case when F is positive.
In our next result, we describe solutions of (6) that satisfy an additional assumption. We begin with a lemma.
Lemma 1
Assume that X is a nonempty set and \(S\colon X \times X \to \mathbb{C}\) and \(F\colon X \times X \to (0, + \infty )\) satisfy inequality (6). If there exist some \(f, g \in X\) such that the map
is unbounded, then for all \(f, g \in X\) the map (8) is unbounded.
Proof
We will split the proof into a few steps. Let us fix \(f, g \in X\) such that the map (8) is unbounded and let \(h, m \in X\) be arbitrarily fixed.
Step 1. The map
is unbounded.
S and F satisfy (6), thus
It implies immediately the following inequality:
By (9) we have
Clearly, since the lefthand side is unbounded by assumption, then the fraction on the righthand side is unbounded.
Step 2. The map
is unbounded.
From (6) we get
therefore
By Step 1 the first fraction is unbounded, so is the second one.
Step 3. The map
is unbounded.
Using (6) analogously as in Step 1, we have
By Step 2 the lefthand side is unbounded. Thus the fraction on the righthand side is unbounded, which ends the proof. □
The following corollary is straightforward and will be utilized at the end of the proof of the subsequent theorem.
Corollary 1
Under assumptions of Lemma 1, if there exists some \(g \in X\) such that \(S(g,k)= 0\) for each \(k \in X\), then \(S=0\) on \(X\times X\).
Now, we are ready to prove the main result of this section.
Theorem 1
Assume that X is a nonempty set and \(S\colon X \times X \to \mathbb{C}\) and \(F\colon X \times X \to (0, + \infty )\) satisfy inequality (6). If there exist some \(f, g \in X\) such that the map (8) is unbounded, then S solves Sincov equation (4) for all \(f, g, h \in X\).
Proof
Define an auxiliary functional \(\Gamma \colon X \times X \to \mathbb{R}\) as
Using inequality (9), we get
So,
Now, fix arbitrary \(f, g, h, k \in X\). By (6) and (10) we obtain
and
therefore
Using (11), (6), and then (10), we arrive at
Thus, if \(S(h,k)\neq 0\), then
If there were no \(h \in X\) such that \(S(h,k)\neq 0\) for all \(k \in X\), then we would get by Corollary 1 that \(S=0\) on \(X \times X\), contradicting the assumptions. Finally, Lemma 1 implies that the two fractions on the righthand side can be arbitrarily small, while \(f, g, h \in X\) are kept fixed. Thus we get that the lefthand side is equal to zero. □
3 DeltaSincov operators in Banach algebras
Assume that \(\mathcal{A}\) is a (complex) commutative Banach algebra. We adopt the convention that Banach algebras and its subsets are denoted by capital calligraphic letters, whereas the capital gothic fonts are reserved for sets consisting of linear multiplicative functionals over Banach algebras and small gothic letters for functionals. Thus the space of all complex homomorphisms on \(\mathcal{A}\) (i.e., nonzero complex mappings on \(\mathcal{A}\) that are linear and multiplicative) is denoted by \(\mathfrak{M}(\mathcal{A})\). \(\mathfrak{M}(\mathcal{A})\) with the weak−^{∗} topology is a compact Hausdorff space. By \(\Phi \colon \mathcal{A}\to C_{\mathbb{C}}(\mathfrak{M}(A))\) we mean the Gelfand transform, i.e.,
The intersection of all maximal ideals on \(\mathcal{A}\) is the (Jacobson) radical of \(\mathcal{A}\), and we denote it by \(\mathrm{rad}(\mathcal{A})\). It is well known that the radical is equal to the kernel of the Gelfand transform. If \(\mathrm{rad}(\mathcal{A}) = \{ 0\}\), i.e., the Gelfand transform is an isomorphism, then algebra \(\mathcal{A}\) is termed semisimple. Further, if the algebra \(\mathcal{A}\) is equipped with an involution \(^{*}\colon \mathcal{A}\to \mathcal{A}\) such that
then \(\mathcal{A}\) is called \(C^{*}\)algebra. The celebrated Gelfand–Naimark theorem says that in the case of \(C^{*}\)algebras, the Gelfand transform is an isometrical isomorphism such that \(\mathfrak{m}(x^{*}) = \overline{\mathfrak{m}(x)}\) for all \(\mathfrak{m}\in \mathfrak{M}\) and \(x \in \mathcal{A}\), or equivalently \(\Phi (x^{*}) = \overline{\Phi (x)}\) for all \(x \in \mathcal{A}\).
We will begin with a description of solutions of the Sincov equation in Banach algebras.
Proposition 3
Assume that X is a nonempty set and \(T\colon X \times X\to \mathcal{A}\) satisfies
Then:

(a)
If \(T(f,g) = 0\) for some \(f, g \in X\), then \(T=0\) on \(X\times X\);

(b)
If \(T(f,g)\) is a zero divisor for some \(f, g \in X\), then \(T(f,g)\) is a zero divisor for all \(f, g \in X\).
Proof
Both assertions follow easily from repetitive use of (12) with the aid of the commutativity of \(\mathcal{A}\). □
Proposition 4
Assume that X is a nonempty set. If \(T\colon X \times X\to \mathcal{A}\) satisfies equation (12), then there exist a (possibly empty) set \(\mathfrak{M}_{1}\subset \mathfrak{M}(\mathcal{A})\) and a mapping \(\varphi \colon \mathfrak{M}(\mathcal{A})\times X \to \mathbb{C} \setminus \{0 \}\) such that \(\Phi (T(\cdot , \cdot ))(\mathfrak{m}) = 0\) for all \(\mathfrak{m}\in \mathfrak{M}(\mathcal{A})\setminus \mathfrak{M}_{1}\) and
Conversely, if \(\mathcal{A}\) is a semisimple Banach algebra, \(\mathfrak{M}_{1}\subset \mathfrak{M}(\mathcal{A})\), \(\varphi \colon \mathfrak{M}(\mathcal{A})\times X \to \mathbb{C} \setminus \{0 \}\) and \(T\colon X \times X\to \mathcal{A}\) is defined by (13) and \(\Phi (T(\cdot , \cdot ) )(\mathfrak{m}) = 0\) for all \(\mathfrak{m}\in \mathfrak{M}(\mathcal{A})\setminus \mathfrak{M}_{1}\), then T satisfies (12).
Proof
To prove the first part, observe that for each \(\mathfrak{m}\in \mathfrak{M}(\mathcal{A})\) the map \(S:= \mathfrak{m}\circ T\) solves (4). Denote
and use Proposition 3 (a) and the representation (5) of scalarvalued solutions recalled in the first section to derive (13).
To prove the converse implication, observe that since Φ is an isometry, then T is well defined by (13). Further, for each \(\mathfrak{m}\in \mathfrak{M}_{1}\) and each \(f, g, h \in X\), we have
or equivalently
Since \(\mathfrak{m}\in \mathfrak{M}_{1}\) was taken arbitrarily and \(\mathfrak{m}\circ T = (\Phi \circ T)(\mathfrak{m})=0\) for all \(\mathfrak{m}\in \mathfrak{M}(\mathcal{A})\setminus \mathfrak{M}_{1}\) by assumption, then
Finally, utilizing the fact that \(\mathcal{A}\) is a semisimple Banach algebra, we arrive at
□
The next example shows that the assumption that \(\mathcal{A}\) is semisimple cannot be dropped.
Example 1
Let \(X= (0, + \infty )\), \(\mathcal{A}\) is a subalgebra of \(2 \times 2\) complex matrices of the form (\begin{array}{cc}x& y\\ 0& z\end{array}) for \(x, y, z \in \mathbb{C}\) and \(T\colon X \times X\to \mathcal{A}\) is given by
There are precisely two distinct linearmultiplicative functionals on \(\mathcal{A}\), namely
Therefore,
formula (13) is satisfied with \(\varphi = 1\) (for both \(\mathfrak{m}_{1}\) and \(\mathfrak{m}_{2}\)) and T fails to satisfy equation (12).
In the proof of the next theorem, we follow some ideas of the article by Ger and Šemrl [7].
Assume that X is a nonempty set and \(T\colon X \times X\to \mathcal{A}\) and \(F\colon X \times X\to (0, +\infty )\) are arbitrary mappings. We will study the following functional inequality:
We do not impose any additional assumptions upon T. Therefore, since we can embed \(\mathcal{A}\) into algebra with a unit without loss of generality, we may assume that \(\mathcal{A}\) has a unit.
Theorem 2
Let X be a nonempty set and let \(\mathcal{A}\) be a semisimple commutative Banach algebra. Assume that \(T \colon X \times X \to \mathcal{A}\) and \(F \colon X \times X \to (0, + \infty )\) satisfy inequality (14) together with the condition:
Then T solves Sincov equation (12).
Proof
Let \(\mathfrak{m}\in \mathfrak{M}(\mathcal{A})\) be fixed arbitrarily. Clearly, \(\Vert \mathfrak{m}\Vert = 1\), therefore for every \(f,g,h \in X\) we have
Define \(S \colon X \times X \to \mathbb{C}\) as \(S := \mathfrak{m}\circ T\). Note that S satisfies inequality (6). Therefore by Theorem 1 we infer that S satisfies equation (4). Consequently, equality (13) holds true with \(\varphi (\mathfrak{m}, \cdot ) = \varphi \) given by (5) and for every \(\mathfrak{m}\in \mathfrak{M}(\mathcal{A})\). To finish the proof, apply the second part of Proposition 4. □
One can ask whether there is an analogue of the second theorem of Ger and Šemrl, i.e., [7, Theorem 3.2]. The following example shows that a full analogue is not true. Namely, we will show that if \(\mathcal{A}\) is a \(C^{*}\)algebra, then the set of all \(\mathfrak{m}\in \mathfrak{M}(\mathcal{A})\) for which \((\Phi \circ T)(\cdot )(\mathfrak{m})\) is bounded does not need to be closed. Consequently, it is not possible to decompose algebra \(\mathcal{A}\) as a direct sum of two closed ideals such that, if Q and R are the corresponding projections, then \(Q\circ T\) solves the Sincov equation and \(R\circ T\) is norm bounded.
Example 2
Let
\(\mathcal{A}:= C_{\mathbb{C}}([0,1])\) and \(T \colon X \times X \to \mathcal{A}\) and \(F \colon X \times X \to (0, + \infty )\) be given by \(F= (1+\sqrt{5})/2\) (a constant map) and
One can check that inequality (14) is satisfied by T and F. Further, since each \(\mathfrak{m}\in \mathfrak{M}(\mathcal{A})\) is of the form \(\mathfrak{m}(f) = f(x)\) for some \(x \in [0,1]\), then for fixed \(f, g \in X\) such that \(g(x) \neq 1\) for all \(x \in [0,1]\), we have
We see that the supremum is finite if and only if \(x \in (0,1]\). Therefore, the set of all \(\mathfrak{m}\in \mathfrak{M}(\mathcal{A})\), for which \((\Phi \circ T)(X)(\mathfrak{m})\) is bounded, is not closed.
In our next result, we will show that it is possible to get a partial analogue to [7, Theorem 3.2].
Proposition 5
Let X be a nonempty set and let \(\mathcal{A}\) be a \(C^{*}\)algebra. Assume that \(T \colon X \times X \to \mathcal{A}\) and \(F \colon X \times X \to (0, + \infty )\) satisfy inequality (14). Then there exist: a Banach \(C^{*}\)algebra \(\mathcal{B}\), a normed \(C^{*}\)algebra \(\mathcal{C}\), and ^{∗}homomorphisms \(\Lambda _{1}\colon \mathcal{A}\to \mathcal{B}\) and \(\Lambda _{2}\colon \mathcal{A}\to \mathcal{C}\) such that the map \(\Lambda _{1} \circ T\) solves Sincov equation (12). For each \(f, g \in X\), the set
is norm bounded in \(\mathcal{C}\) and \((\Lambda _{1},\Lambda _{2})\colon \mathcal{A}\to \mathcal{B}\oplus \mathcal{C}\) is an isometrical ^{∗}homomorphism.
Proof
Define \(S \colon \mathfrak{M}(\mathcal{A})\times X \times X \to \mathbb{C}\) as
Using the basic properties of the Gelfand transform, we obtain
Define
By the Gelfand–Naimark theorem, the Gelfand transform is an isometry. Hence \(\mathfrak{M}_{s}\) is a closed subset of \(\mathfrak{M}(\mathcal{A})\) (and thus compact).
Define \(\mathcal{B}:=C_{\mathbb{C}}(\mathfrak{M}_{s})\) and \(\Lambda _{1}\colon \mathcal{A}\to \mathcal{B}\) via
Above we identified the algebra \(\mathcal{A}\) with its image by the Gelfand transform. Directly from the definition mapping \(\Lambda _{1} \circ T\colon X \times X \to \mathcal{B}\) solves Sincov equation (12).
Further, put \(\mathcal{C}:= L^{\infty}(\mathfrak{M}(\mathcal{A})\setminus \mathfrak{M}_{s})\cap C_{\mathbb{C}}(\mathfrak{M}(\mathcal{A}) \setminus \mathfrak{M}_{s})\) and define \(\Lambda _{2}\colon \mathcal{A}\to \mathcal{C}\) as
Lemma 1 implies the boundedness in \(\mathcal{C}\) of the sets \(\{ \Lambda _{2} (\frac{ T(g,k)}{F(f,k) } ) : k \in X \}\) for all \(f, g \in X\). Finally, if the direct sum \(\mathcal{B}\oplus \mathcal{C}\) is equipped with the norm
then it is clear that the map \((\Lambda _{1},\Lambda _{2})\colon \mathcal{A}\to \mathcal{B}\oplus \mathcal{C}\) is an isometry. □
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Acknowledgements
This article has been completed while one of the authors, Aleksandra Świątczak, was the Doctoral Candidate in the Interdisciplinary Doctoral School at the Lodz University of Technology, Poland.
Funding
The work of Aleksandra Świątczak is financed under the program “FU^{2}N  Fund for Improving Skills of Young Scientists” supporting scientific excellence of the Lodz University of Technology  grant no 5/2023.
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Fechner, W., Świątczak, A. DeltaSincov mappings in Banach algebras. J Inequal Appl 2023, 147 (2023). https://doi.org/10.1186/s13660023030518
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DOI: https://doi.org/10.1186/s13660023030518