A preconditioned new modulus-based matrix splitting method for solving linear complementarity problem of $H_+$-matrices

1.   Introduction

Simon and Volkmann considered in ^[1] the following two equations which are connected with the absolute values of some additive function $\gamma\colon G \to \mathbb{R}$, that is,

for a real function $\eta: G \to \mathbb{R}$ defined on an abelian group $(G, +)$ and both functional equations are satisfied by $\eta(x) = |\gamma(x)|$ where $\gamma(x+y) = \gamma(x)+\gamma(y)$. Moreover, solution of the equation

with supposition about G to be an abelian group was presented in the following theorem as:

Theorem 1. [1, Theorem 2] Let $\eta: G \to \mathbb{R}$, where every element of an abelian group $G$ is divisible by 2 and 3. Then, $\eta$ fulfills Eq $(1.3)$ if and only if $\eta(x) = 0$ or $\eta(x) = e^{|\gamma(x)|}$, $x \in G$, where $\gamma: G \to \mathbb{R}$ is an additive function.

The solutions of Eqs (1.1) and (1.2) presented by Jarczyk et al. ^[2] and are demonstrated as:

Theorem 2. Let $\eta: G \to \mathbb{R}$, where $\eta$ is defined on an abelian group $G$. Then $\eta$ fulfills Eq $(1.1)$ if and only if functional Eq $(1.2)$ holds and also satisfies $\eta(2x) = 2\eta(x)$ for $x \in G$.

Furthermore, the most comprehensive study of the equation

on groups has been presented in ^[3,4]. Volkmann has given the solution of Eq (1.4) with supposition that $\eta$ fulfills the renowned condition called Kannappan condition ^[5], that is defined as, $\eta(xgy) = \eta(xyg)$ for $x, y, g \in G$. Following that, Toborg ^[3] gave the characterization of such mappings exhibited in Eq (1.4) without taking into account the Kannappan condition and abelian group $G$. Their key findings are as follows:

Theorem 3 (For the special case, see ^[3,4] for the general case). Let $\eta: G \to \mathbb{R}$, where $\eta$ is acting on any group $G$. Then, $\eta$ fulfills Eq $(1.4)$ if and only if $\eta(x) = |\gamma(x)|$ for every $x \in G$, where $\gamma \colon G \to \mathbb{R}$ is an additive function.

We suggest the readers consult the articles ^[2,6] and related cited references to get some inclusive results and solutions about the functional Eq (1.4). In addition, some stability results of Eqs (1.2) and (1.4) can be found in ^[7] and ^[6] respectively.

Recently, in ^[8], Eq (1.3) presented in a generalized form as

with the exception of additional suppositions that every element of the abelian group is divisible by 2 and 3. Their main result is demonstrated as:

Theorem 4 (see ^[8]). Let $\eta: G \to \mathbb{R}$, where $G$ is any group. Then a mapping $\eta\colon G \to \mathbb{R}$ fulfills the Eq $(1.5)$ if and only if $\eta\equiv 0$ or there exists a normal subgroup $N_\eta$ such that

and

or $\eta(x) = e^{|\gamma(x)|}$, $x \in G$, where $\gamma\colon G \to \mathbb{R}$ is an additive function.

The main objective of this research article is to determine the solution to the generalized minimum functional equation

With the exception of additional suppositions, we derive some results concerning Eq (1.6) that are appropriate for arbitrary group $G$ rather than abelian group $(G, +)$.

Redheffer and Volkmann ^[9] determined the solution of the Pexider functional equation

for three unknown functions $h$, $f$ and $g$ acting on abelian group $(G, +)$, which is a generalization of Eq (1.1).

We will also examine the general solutions of the generalized Pexider-type functional equation

where real functions $\eta$, $\chi$, and $\psi$ are defined on any group $G$. This Pexider functional Eq (1.8) is a common generalization of two previous Eqs (1.4) and (1.5). Readers can see renowned papers ^[10,11] and associated references cited therein to obtain comprehensive results and discussions concerning the Pexider version of some functional equations.

2.   Analysis of Eq (1.6)

In this research paper, our group $G$ will in general $(G, \cdot)$ not be abelian $(G, +)$, therefore, the group operation will be described multiplicatively as $xy$ for $x, y \in G$. Symbol $e$ will be acknowledged as the neutral element.

Definition 1. Assume that $G$ is any group. A mapping $\eta: G \to \mathbb{R}$ fulfills the Kannappan condition ^[5] if

Remark 1. For every abelian group $G$, a mapping $\eta: G \to \mathbb{R}$ fulfills the Kannappan condition but converse may not be true.

Lemma 1. Suppose that $\eta\colon G \to \mathbb{R}$, where $G$ is an arbitrary group. Let $\eta$ is a strictly positive solution of the functional Eq $(1.6)$, then $\eta(x) = e^{-|\beta(x)|}$, $x \in G$, where $\beta\colon G \to \mathbb{R}$ is an additive function.

Proof. By given assumption, $\eta(x) > 0$ for every $x \in G$. Since $\eta$ satisfies Eq $(1.6)$, as a result, $\frac{1}{\eta}$ also satisfies the functional Eq $(1.3)$, then by well-known theorem from ^[6], we can get that $\eta(x) = e^{-|\beta(x)|}$, $x \in G$, where $\beta\colon G \to \mathbb{R}$ is an additive function.

First, we are going to prove the following important lemma which will be utilized several times during computations especially to prove Theorem 5.

Lemma 2. Let $\eta\colon G \to \mathbb{R}$, where $G$ is an arbitrary group and $\eta$ is a non-zero solution of Eq $(1.6)$, then the following results hold:

(1) $\eta(e) = 1$;

(2) $\eta(x^{-1}) = \eta(x)$;

(3) $\eta(x^{-1}yx) = \eta(y)$;

(4) $\eta$ is central.

Proof. (1). Putting $y = e$ in (1.6), we can obtain that $\eta(x)\eta(e) = \eta(x)$. By given condition, $\eta$ is non-zero, therefore, we obtain $\eta(e) = 1$.

(2). Using $x = e$ in functional Eq (1.6), we can deduce

replacing $y^{-1}$ with $y$ in Eq (2.1) provides that

Eqs (2.1) and (2.2) give that $\eta(y^{-1}) = \eta(y)$. Since $y$ is arbitrary, therefore, we have $\eta(x^{-1}) = \eta(x)$ for any $x \in G$.

(3). From functional Eq (1.6), the proof of property (3) can be obtained from the following simple calculation:

(4). By Lemma 2(3) and replacing $y$ with $xy$, we can see that $\eta(x^{-1}(xy)x) = \eta(xy)$, which gives $\eta(xy) = \eta(yx)$, therefore, $\eta$ is central.

In addition, we concentrate on the main theorem of Section 2 to describe the solutions $\eta$ of Eq (1.6).

Theorem 5. Let $\eta\colon G \to \mathbb{R}$, where $G$ is an arbitrary group. A mapping $\eta$ is a solution of Eq $(1.6)$ if and only if $\eta\equiv 0$ or there exists a normal subgroup $H_\eta$ of $G$ defined as

and fulfills the condition that

or there exists a normal subgroup $H_\eta$ of $G$ fulfills the condition that

or $\eta(x) = e^{-|\beta(x)|}$, $x \in G$, where $\beta\colon G \to \mathbb{R}$ is some additive function.

Proof. The 'if' part obviously demonstrates that every mapping $\eta$ determined in the statement of the theorem is a solution of Eq (1.6). Conversely, suppose that a function $\eta\colon G \to \mathbb{R}$ is a solution of (1.6), then putting $x = y = e$ in Eq (1.6), we get $\eta(e) = \eta(e)\eta(e)$, which gives that either $\eta(e) = 1$ or $\eta(e) = 0$. First, let $\eta(e) = 0$, and then put $y = e$ in (1.6) to get $\eta(x) = 0$ for every $x \in G$. Suppose that $\eta(e) = 1$, then there are the following different cases.

Suppose that there exists $z_\circ \in G$ such that $\eta(z_\circ) \leq 0$. Putting $x = y$ in (1.6), we have $\min \{\, \eta(x^2), \eta(e)\, \} = \eta(x)^2$, which gives that $\eta(x)^2 \leq 1$, so $-1 \leq \eta(x) \leq 1$ but $\eta(z_\circ) \leq 0$, therefore, $-1 \leq \eta(z_\circ) \leq 0$. Let $-1 < \eta(z_\circ) < 0$, then we can compute

which implies that $\eta(z_\circ^2) = \eta(z_\circ)^2$. Moreover,

which gives a contradiction, consequently, either $\eta(z_\circ) = 0$ or $\eta(z_\circ) = -1$. Additionally, it is not possible that $\eta(x) = 0$ and $\eta(y) = -1$ for some $x, y \in G$. Since $\eta(e) = 1$, therefore, either $\eta(x) \in \{\, 0, 1\, \}$ or $\eta(x) \in \{\, -1, 1\, \}$. Moreover, define $H_\eta = \{\, x \in G \mid \eta(x) = 1 \, \}$.

It is obvious that $e \in H_\eta$ for the reason that $\eta(e) = 1$. Suppose that $h \in H_\eta$; then from Lemma 2(2) we obtain $\eta(h^{-1}) = \eta(h) = 1$; therefore, $h^{-1} \in H_\eta$. Let $h_1, h_2 \in H_\eta$; then, $\eta(h_1) = \eta(h_2) = 1$, and we can deduce from Eq (1.6) that

By (2.5) and (2.6) we can get that $\eta(h_1h_2) = \eta(h_1)\eta(h_2) = 1$, therefore, we have $h_1h_2 \in H_\eta$. Consequently, $H_\eta$ is a subgroup of $G$. Assume that $h \in H_\eta$; then Lemma 2(3) yields that $\eta(x^{-1}hx) = \eta(h)$ for any $x \in G$ and $h \in H_\eta$; accordingly, $H_\eta$ is a normal subgroup of $G$.

First, suppose that $\eta(x) \in \{\, 0, 1\, \}$ and $x, y \in G\setminus H_\eta$; therefore, $\eta(x) = \eta(y) = 0$, then, by functional Eq (1.6), we have $\min \{\, \eta(xy), \eta(xy^{-1})\, \} = \eta(x)\eta(y) = 0$. In a consequence, we can determine that $xy^{-1} \in H_\eta \vee xy \in H_\eta$ for any $x, y \in G\setminus H_\eta$.

In addition, considering the second case, let $\eta(x) \in \{\, -1, 1\, \}$ and let $x, y \in G\setminus H_\eta$; thus, $\eta(x)\neq 1$ and $\eta(y)\neq 1$; then $\eta(x) = \eta(y) = -1$. Consequently, Eq (1.6) gives that $\min \{\, \eta(xy), \eta(xy^{-1})\, \} = \eta(x)\eta(y) = 1$. In either case, we can conclude that $\eta(xy) = 1$ and $\eta(xy^{-1}) = 1$, which infers that $xy^{-1} \in H_\eta \wedge xy \in H_\eta$ for all $x, y \in G\setminus H_\eta$.

Furthermore, let $\eta(x) > 0$ for all $x \in G$, then from Lemma 1, we can conclude that $\eta(x) = e^{-|\beta(x)|}$, $x \in G$.

Corollary 1. Let $\eta \colon G \to \mathbb{R}$, where $G$ is an arbitrary group. Assume that $\eta$ is a non-zero solution of the functional Eq $(1.6)$; then the commutator subgroup $G^\prime$ is a normal subgroup of $H_\eta$.

Proof. Since $\eta$ is a non-zero, then by the main theorem, we can derive the following cases:

Case 1. According to the main theorem, there exists a normal subgroup $H_\eta$ defined as $\eta(x) = 1$ for every $x\in H_\eta$ and also satisfies the condition (2.4); consequently, by Lemma 2, we can compute that

which indicates that $\eta([x, y]) = 1$.

Case 2. There exists a normal subgroup $H_\eta$ which satisfies the condition (2.3), that is $xy^{-1} \in H_\eta \vee xy \in H_\eta$ for all $x, y \in G\setminus H_\eta$; accordingly, applying Lemma 2 and Case 1, we can deduce that $\eta([x, y]) = 1$.

Case 3. Assume that $\eta(x) > 0$ for any $x \in G$; consequently by Theorem 5, we have $\eta(x) = e^{-|\beta(x)|}$ for any $x \in G$, where $\beta\colon G \to \mathbb{R}$ is an additive function, thus, $\eta([x, y]) = 1$ for the reason that $\beta([x, y]) = 0$ for any $x, y \in G$.

Hence, in either case, the required proof is completed.

Corollary 2. Any solution $\eta\colon G \to \mathbb{R}$ of Eq (1.6) on any group $G$ fulfills the Kannappan condition.

Proof. The proof relies on the following cases:

Case 1. Assume that $\eta\equiv 0$ on group $G$, then it is obvious that $\eta$ fulfills the Kannappan condition.

Case 2. Let $\eta(x)\leq 0$ for all $x \in G$. Then from Theorem 5 and Corollary 1, there exists normal subgroup $H_\eta$ such that $G^\prime \subseteq H_\eta$, consequently, $\eta(xyg) = 1$ if and only if $xyg \in H_\eta$ if and only if $[y^{-1}, x^{-1}]xyg = xgy \in H_\eta$ if and only if $\eta(xgy) = 1$. It is sufficient to prove the Kannappan condition because $\eta$ only takes the values 1, 0, and -1.

Case 3. Suppose that $\eta(x) > 0$, $x \in G$, then $\eta(x) = e^{-|\beta(x)|}$, therefore $\eta(xyg) = \eta(xgy)$ for any $x, g, y \in G$ because $\beta$ is an additive function.

Corollary 3. If $\eta$ is a strictly positive solution of $(1.6)$, then $\max \{\, \eta(xy^{-1}), \eta(xy) \, \} \in (0, 1]$.

Theorem 6. Let $\eta\colon G \to \mathbb{R}$ and $\eta$ is a non-zero solution of $(1.6)$, then:

(1) Assume that $g \in G$ and $\eta(gx^{-1}) = \eta(gx)$ for some elements $x \in G$ with the restriction that $\eta(x^2)\neq 1$. Then $\eta(g^2) = 1$.

(2) Suppose that $G_\eta = \{\, g \in G \mid \, \eta(g^2) = 1 \, \}$, then $G_\eta$ is a normal subgroup of $G$.

(3) If $\eta$ is strictly positive, then $G_\eta = H_\eta$.

Proof. Assume that $x, y \in G$, then by Eq (1.6) and Corollary 2, we have

(1). By given condition $\eta(gx) = \eta(gx^{-1})$ for some $x \in G$ and by Eq (2.7), we can see that either $\eta(x^2) = 1$ or $\eta(g^2) = 1$, therefore, given condition $\eta(x^2) \neq 1$ implies that $\eta(g^2) = 1$.

(2). Since $\eta(e) = 1$, therefore $e \in G_\eta$. Let $g \in G_\eta$; then $\eta(g^2) = 1$. Moreover, $\eta(x^{-1}) = \eta(x)$ gives that $\eta(g^{-2}) = \eta(g^2) = 1$, therefore $g^{-1} \in G_\eta$. Let $g, y \in G_\eta$; then, $\eta(y^2) = 1$ and $\eta(g^2) = 1$, therefore, a simple calculation yields

So, inequalities (2.8) and (2.9) implies that $\eta(y^2g^2) = \eta(y^2)\eta(g^2) = 1$, which yields that $yg \in G_\eta$, consequently, $G_\eta$ is a subgroup of $G$. Additionally, Lemma 2 provides that $\eta(xg) = \eta(gx)$ for every $x \in G$ and $g \in G_\eta$. As a result, $G_\eta$ is a normal subgroup of a group $G$.

(3). As $\eta(x) > 0$ for every $x \in G$, then the proof can be seen easily from Lemma 1 and Theorem 6 (2).

Definition 2. Suppose that group $G$ is abelian. Then a mapping $\eta: G \to \mathbb{R}$ is called a discrete norm if $\eta(x) > \gamma$, where $\gamma > 0$ and $x \in G\setminus \{e\}$. Then $(G, \eta, e)$ is said to be a discretely normed abelian group ^[12].

Theorem 7. Assume that $(G, \eta, e)$ is a discretely normed abelian group. A mapping $\eta: G \to \mathbb{R}$ is a solution of $(1.6)$ if and only if $\eta(x) = e^{-|\beta(x)|}$, $x \in G\setminus \{e\}$, where $\beta: G \to \mathbb{R}$ is some additive function.

Proof. Since $(G, \eta, e)$ is a discretely normed, then there exists a mapping $\eta: G \to \mathbb{R}$ such that $\eta(x) > \gamma$, where $\gamma > 0$ and $x \in G\setminus \{e\}$. Setting $\eta(x) = \log \eta(x)$, and using Lemma 1, we get

if and only if $\eta(x) = e^{-|\beta(x)|}$, $x \in G\setminus \{e\}$, where $\beta: G \to \mathbb{R}$ is some additive function.

Corollary 4. For free abelian group $G$, a mapping $\eta$ is a solution of Eq $(1.6)$ if and only if $\eta(x) = e^{-|\beta(x)|}$, $x \in G\setminus \{e\}$, where $\beta: G \to \mathbb{R}$ is some additive function.

3.   Generalized Pexider-type functional Eq (1.8)

Theorem 8. Let $\eta\colon G \to \mathbb{R}$ fulfills the Kannappan condition, where $G$ is an arbitrary group. Then $\eta, \chi, \psi$ are solutions of the functional Eq $(1.8)$ if and only if

or

where $\xi\colon G \to \mathbb{R}$ is a solution of Eq $(1.4)$;

or

where $\xi\colon G \to \mathbb{R}$ is a solution of Eq $(1.5)$;

or

where $\xi\colon G \to \mathbb{R}$ is a solution of Eq $(1.6)$.

Proof. The 'if' part of the theorem can easily be seen that every function $\eta$, $\chi$, and $\xi$ presented in the statement is a solution of Eq (1.8). Conversely, suppose that $\eta, \chi, \psi$ are solutions of Eq (1.8), then we have the following cases:

(1). $\eta$ is constant.

Assuming that $\eta(x) = \lambda_1$ for $x\in G$ and $\lambda_1 \in \mathbb{R}$, we may deduce from Eq (1.8) that $\chi(x) = 1-\lambda_1^{-1}\psi(x)$ when $\psi$ is an arbitrary function, which is required result described in the statement.

(2). $\eta$ is not constant.

Setting $y = e$ in Eq (1.8) gives that

Using Eq (3.1) in (1.8), we conclude that

Setting $x = e$, we can obtain

We are going to show that $\chi(e) = 1$, but on the contrary, assume that $\chi(e)\neq 1$. Setting

If $y \in H^\prime$ then $y^{-1} \in H$. Also, if $y \in H$, then from Eq (3.3), we have

which implies that $\eta(y) = \eta(e)$ for all $y \in H$. Moreover, $H^\prime \neq \emptyset$ because $\eta$ is not constant. Assume that $y^\prime \in H^\prime$ then $\eta(y^\prime) < \eta(y^{\prime^{-1}}) = \eta(e)$, which implies that

Writing $y^\prime$ instead of $y$ in Eq (3.3) and using (3.4) we can get that

which implies that $(\eta(y^\prime)-\eta(e))\chi(e) = 0$, so $\chi(e) = 0$. Setting $x = y^\prime$ and $y = y^{\prime^{-1}}$ in (3.2) we have

which is a contradiction, thus, we have $\chi(e) = 1$. Moreover, from Eq (3.3), we can see that

writting $y^{-1}$ instead of $y$ in (3.5) we have

from (3.5) and (3.6) we can get that $\eta(y^{-1}) = \eta(y)$.

Since $\eta(x^{-1}) = \eta(x)$ for every $x \in G$, then from Eq (3.2) and Kannappan condition we have

which infers that $\chi(x^{-1})-\chi(x) = 0$ because $\eta$ is not constant. Moreover, when $\eta$ is not constant then $\eta(x^{-1}) = \eta(x)$ and $\chi(x^{-1}) = \chi(x)$ for every $x \in G$, consequently, by Eq (3.1) we can get that $\psi(x^{-1}) = \psi(x)$. Also, by Eq (3.2) and Kannappan condition we can see that

Suppose that $\eta(y^\prime)\neq \eta(e)$ for $y^\prime \in G$, then we can obtain that

Moreover, assume that $\beta: = \frac{\chi(y^\prime)-1}{\eta(y^\prime)-\eta(e)}$, then $\chi(x)-1 = \beta(\eta(x)-\eta(e))$, so we can write as $\chi_1(x) = \beta\eta_1(x)$, where

Also $\eta_1(e) = 0$. By functional Eq (3.2) and definition of $\eta_1$, we have

According to the different values of $\beta$, we can discuss the following three different cases.

Case 1. $\beta = 0$.

By Eq (3.7), we see that $\chi(x) = 1$, $x \in G$. Furthermore, by functional Eq (3.9), we have

for all $x, y \in G$ and also $\eta_1$ satisfies the functional Eq (1.4), then from well-known theorem of Toborg ^[3], there exists some additive function $g\colon G \to \mathbb{R}$ such that $\eta_1(x) = |g(x)|$ for all $x\in G$, then from Eqs (3.1), (3.7) and (3.8), we can deduce

where $\lambda_1 = \eta(e)$ and $\xi\colon G \to \mathbb{R}$ is a solution of Eq (1.4) such that $\xi(x) = |g(x)|$.

Case 2. $\beta > 0$.

Let $\eta_2: = \beta \eta_1(x)$ for all $x \in G$, then multiplying functional Eq (3.9) by $\beta$, we conclude that

then by setting $\xi(x): = \eta_2(x)+1$ for $x \in G$, we get

It is clear that $\xi\colon G \to \mathbb{R}$ satisfies Eq (1.5), then from Eq (3.8) we get

which gives that $\eta(x) = \lambda_2\xi(x)+\lambda_1$ where $\lambda_2 = \beta^{-1}, \lambda_1 = \eta(e)-\beta^{-1}$.

Also, from Eqs (3.1), (3.7) and (3.8), we can see that

where $\xi\colon G \to \mathbb{R}$ is a solution of (1.5).

Case 3. $\beta < 0$.

Assume that $\eta_2: = -\beta \eta_1(x)$ for every $x \in G$, then multiplying functional Eq (3.9) by $-\beta$, we have

for any $x, y \in G$, then by setting $\xi_1(x): = \eta_2(x)-1$ for $x \in G$, we have

then by setting $\xi(x): = -\xi_1(x)$, $x \in G$, we can see that $\xi\colon G \to \mathbb{R}$ satisfies the Eq (1.6), then from Eqs (3.1), (3.7) and (3.8), we have

which completes the proof.

Acknowledgments

The authors are grateful to the referees and the editors for valuable comments and suggestions, which have improved the original manuscript greatly. This work was supported by the National Natural Science Foundation of P. R. China (Grant number 11971493 and 12071491).

Conflict of interest

All authors declare no conflict of interest in this paper.