Int. J. Anal. Appl. (2023), 21:64 Best Proximity Point and Existence of the Positive Definite Solution for Matrix Equations Satyendra Kumar Jain1, Gopal Meena2,∗, Rashmi Jain2 1Department of Mathematics, St. Aloysius College Jabalpur (M.P.), India 2Department of Applied Mathematics, Jabalpur Engineering College Jabalpur (M.P.), India ∗Corresponding author: gmeena@jecjabalpur.ac.in Abstract. In this research, α−ψ−θ contraction has been defined to find the best proximity point in partially ordered metric spaces. Proper support for the result has been given in the form of a suitable example. The third part is fully devoted to the positive definite solution of matrix equations. 1. Introduction and Preliminaries The concept of the best proximity point was introduced by Basha [5] with the help of the Banach contraction principle. It may be impossible to find a fixed point for two non empty subsets L,M ⊆ W and a mapping S : L → M (for example, when L∩M = φ). However, it is very interesting to find a point x ∈ L, where x and Sx are as close as possible; in other words, find an x ∈ L which minimizes %(x,Sx). Such optimal approximate solutions are called "best proximity points for S." Letter on many Mathematicians [1–3,6,9,10] established best proximity point results. In 2014, idea of θ contraction introduced by Jleli et al. [8] and defined generalization of Banach contraction. In this paper, we define α−ψ−θ contraction and establish the best proximity point in partially ordered metric spaces. Moreover, as a consequence of the result, a fixed point result and the existence of a positive definite solution to matrix equations have been given. In the whole paper, complete metric space and the best proximity point are abbreviated as CMS and BPP, respectively. The subsequent symbols used in our results are: Received: Apr. 5, 2023. 2020 Mathematics Subject Classification. 55M20, 15B48, 54H25. Key words and phrases. best proximity point; matrix equations; positive definite solution. https://doi.org/10.28924/2291-8639-21-2023-64 ISSN: 2291-8639 © 2023 the author(s). https://doi.org/10.28924/2291-8639-21-2023-64 2 Int. J. Anal. Appl. (2023), 21:64 Let (W,%) be a metric space and C,D be non-empty subsets of W. % (C, D) = inf { % (u1,v1) : u1 ∈ C and v1 ∈ D} , C0 = {u1 ∈ C : % (u1,v1) = % (C,D) f or some v1 ∈ D} , D0 = {v1 ∈ D : % (u1,v1) = % (C,D) f or some u1 ∈ C}. In 2012, Samet et al. [13] defined the following contraction: α(x1,y1)%(Tx1,Ty1) ≤ ψ(%(x1,y1)), where ψ : [0,∞) → [0,∞) satisfy the consequent conditions: (1) ψ is non decreasing, (2) ∑∞ m=1ψ m(t) < ∞ f orall t > 0, where ψm is the mth iterate of ψ and ψ(t) < t for any t > 0, T is α- admissible i.e. for all x1,y1 ∈ W, α(x1,y1) ≥ 1 ⇒ α(Tx1,Ty1) ≥ 1, where α : W ×W → [0,∞) is a mapping. Jleli et al. [8] proposed θ contraction in 2014 as follows: Definition 1.1. [8] Let Θ be the set of all functions θ : (0,∞) → (1,∞) satisfy the conditions: θ1. θ is non decreasing, θ2. for every sequence {αn} ⊂ (0,∞), lim n→∞ θ(αn) = 1 ⇔ lim n→∞ αn = 0 +, θ3. there exists s ∈ (0, 1) and L ∈ (0,∞) such that lim α→0 θ(α) − 1 αs = L and prove the following results: Theorem 1.1. [8] Let (V,%) be a CMS and T : V → V be a mapping, if there exists θ ∈ Θ and k ∈ (0, 1) such that for all u,v ∈ V, %(Tu,Tv) 6= 0 ⇒ θ(%(Tu,Tv)) ≤ [θ(%(u,v))]k. (1.1) Then T has a unique fixed point. Also, in 2017, Ahmed et al. [4] used the subsequent weaker condition in place of condition (θ3) : (θ ′ 3) θ is continuous on (0,∞). In this order we denote Ψ the set all functions θ satisfy θ1,θ2,θ ′ 3. In 2017, Piri et al. [11] defined generalized Khan contraction. Int. J. Anal. Appl. (2023), 21:64 3 Theorem 1.2. [11] Let (W,%) be a CMS and A : W → W be a mapping satisfies % (Au, Av) ≤ { k %(u,Au)%(u,Av)+%(v,Av)%(v,Au) max{%(u,Av),%(Au,v)} , if max{%(u,Av),%(Au,v)} 6= 0, 0, if max{%(u,Av),%(Au,v)} = 0, where k ∈ [0, 1) and u,v ∈ W, then A has a unique fixed point. However, the mappings involved in all results were self mappings. Definition 1.2. [14] Let (C,D) be a pair of non-empty subsets of a metric space W with C0 6= φ. Then, the pair (C,D) is said to have the weak P −property if and only if % (u1, v1) = % (C, D) % (u2, v2) = %(C, D) } ⇒ % (u1, u2) ≤ % ( v1, v2) , where u1, u2 ∈ C and v1,v2 ∈ D. Definition 1.3. [13] Let C,D be the subsets of metric space (W,%). A non self mapping A : C → D is said to be α−proximal admissible if α(v1,v2) ≥ 1 % (u1, Av1) = % (C, D) % (u2, Av2) = %(C, D)   ⇒ α (u1, u2) ≥ 1, where u1, u2, v1,v2 ∈ C and α : C ×C → [0,∞) be a function. 2. Main Results Let C,D be two subsets of a partially ordered CMS (V,%,�) and α : C×C → [0,∞) be a function. A mapping T : C → D is said to be α−ψ − θ contraction, if for θ ∈ Ψ, there exists κ ∈ (0, 1) and for every x,y ∈ C with α(x,y) ≥ 1, %(Tx,Ty) > 0, we have α(x,y)θ[%(Tx,Ty)] ≤ [ψ(θ(M(x,y)))]κ, (2.1) where M(x,y) = max{G(x,y),%(x,y)} and G(x,y) = { %(x,Tx)%(x,Ty)+%(y,Ty)%(y,Tx) max{%(x,Ty),%(Tx,y)} , if max{%(x,Ty),%(Tx,y)} 6= 0, 0, if max{%(x,Ty),%(Tx,y)} = 0. Theorem 2.1. Let (V,%,�) be a partially ordered CMS and C, D are closed subsets of V and let T : C → D be a α−ψ −θ contraction satisfies (i) T is α−proximal admissible, (ii) T (C0) ⊆ D0 and the pair (C,D) satisfies week P −property, (iii) T is continuous, (iv) there exists x0,x1 ∈ C0,x0 � x1 with %(x1,Tx0) = %(C,D) such that α(x0,x1) ≥ 1. Then there exists x ∈ V such that %(x,Tx) = %(C,D). 4 Int. J. Anal. Appl. (2023), 21:64 Proof. Let xo ∈ C0, since T (C0) ⊆ D0, there exists an element x1 ∈ C0 such that %(x1,Tx0) = %(C,D) and x0 � x1, by the assumption (iv), α(x0,x1) ≥ 1. Again x1 ∈ C0 and T (C0) ⊆ D0, there exists x2 ∈ C0 such that %(x2,Tx1) = %(C,D) and x1 � x2. By α−proximal admissibility of T, we have α(x1,x2) ≥ 1, continuing this process, we get %(xn+1,Txn) = %(C,D) and α(xn,xn+1) ≥ 1 f or all n ∈ N, (2.2) where x0 � x1 � x2 � x3 · · · � xn � xn+1 � . . . Now, if there exists n0 ∈ N such that xn0 = xn0+1, then we have %(xn0,Txn0) = %(xn0+1,Txn0) = %(C,D). Then xn0 is the best proximity point (BPP)of T. Therefore, we assume that xn 6= xn+1, that is %(xn,xn+1) > 0 for all n ∈ N ∪{0}. By the week P-property of the pair (C,D) and from 2.1, 2.2, we have for all n ∈ N, 1 < θ(%(xn+1,xn)) = θ(%(Txn,Txn−1)) ≤ α(xn,xn−1)θ(%(Txn,Txn−1) ≤ (ψ(θ(M(xn,xn−1))))κ, where M(xn,xn−1) = max{G(xn,xn−1),%(xn,xn−1)} = max{ %(xn−1,Txn−1)%(xn−1,Txn) + %(xn,Txn−1)%(xn,Txn) max{%(xn−1,Txn),%(Txn−1,xn)} ,%(xn,xn−1)} = max{ %(xn−1,xn)%(xn−1,xn+1) %(xn−1,xn+1) ,%(xn,xn−1)} = %(xn−1,xn), so, 1 < θ(%(xn,xn+1)) ≤ (ψ(θ(%(xn,xn−1))))κ ≤ (ψ(θ(%(xn−1,xn−2))))κ 2 ≤ (ψ(θ(%(xn−2,xn−3))))κ 3 Int. J. Anal. Appl. (2023), 21:64 5 ≤ . . . . . . ≤ (ψ(θ(%(x0,x1))))κ n . Taking n →∞ we get θ(%(xn,xn+1)) → 1, therefore, by θ2, we obtain lim n→∞ %(xn,xn+1) = 0. (2.3) Now, we shall show that {xn} is a Cauchy sequence in C. Suppose, on the contrary that, if there exists � > 0, we can find the sequences {pn} and {qn} of natural numbers such that for pn > qn > n, we have %(xpn,xqn ) ≥ �. (2.4) Then, %(xpn−1,xqn ) < � f or all n ∈ N. Thus, by triangular inequality and 2.4, we get � ≤ %(xpn,xqn ) ≤ %(xpn,xpn−1) + %(xpn−1,xqn ) ≤ %(xpn,xpn−1) + �. Taking limit n →∞ and using 2.3, we get lim n→∞ %(xpn,xqn ) = �. (2.5) Again by triangular inequality, we have %(xpn,xqn ) ≤ %(xpn,xpn+1) + %(xpn+1,xqn+1) + %(xqn+1,xqn ) (2.6) and %(xpn+1,xqn+1) ≤ %(xpn+1,xpn ) + %(xpn,xqn ) + %(xqn,xqn+1). (2.7) Taking limit n →∞ and from 2.3, 2.5, we have lim n→∞ %(xpn+1,xqn+1) = �, (2.8) so, equation 2.5 holds. Then by assumption α(xpn,xqn ) ≥ 1, we get 1 ≤ θ(%(xpn+1,xqn+1)) ≤ θ(%(Txpn,Txqn )) ≤ α(xpn,xqn )θ(%(Txpn,Txqn )) ≤ (ψ(θ(M(xpn,xqn )))) κ < θ(M(xpn,xqn )), 6 Int. J. Anal. Appl. (2023), 21:64 by taking limit as n →∞ in above inequality and using [θ ′ 3] in equation 2.3, we get lim n→∞ %(xpn,xqn ) = 0 < �, which is contraction. Therefore, {xn} is a Cauchy sequence. Since {xn} ⊆ C and C is closed in a complete metric space, so we can find x ∈ C, such that xn → x. Now, since T is continuous so, we have Txn → Tx. This implies that %(xn+1,Txn) → %(x,Tx), since the sequence {%(xn+1,Txn)} is a constant sequence with the value %(C,D) . We deduce that %(C,D) = %(x,Tx). So, x is the best proximity point. If we take C = D = V and α(x,y) = 1, we obtain the subsequent result: Corollary 2.1. Let (V,%,�) be a complete metric space and T : V → V be a mapping satisfying θ[%(Tx,Ty)] ≤ [ψ(θ(M(x,y)))]κ, where M(x,y) = max{G(x,y),%(x,y)} and G(x,y) = { %(x,Tx)%(x,Ty)+%(y,Ty)%(y,Tx) max{%(x,Ty),%(Tx,y)} , if max{%(x,Ty),%(Tx,y)} 6= 0, 0, if max{%(x,Ty),%(Tx,y)} = 0, for all x,y ∈ V with θ ∈ Θ and κ ∈ (0, 1), suppose that (i) T is continuous, (ii) there exist x0 ∈ V such that x0 � Tx0. Then T has a unique fixed point. Proof. By the Theorem 2.1, {xn} is a Cauchy sequence. Since {xn} ⊆ V and V is a complete metric space, so we can find x ∈ V such that xn → x. Now, we shall show that x is a fixed point of T. %(Tx,x) = lim n→∞ %(Txn,x) = lim n→∞ %(xn+1,x) = 0. Then x is a fixed point of T. Uniqueness: let, if possible there are two fixed points x1 and x2 such that x1 6= x2. Since x1 and x2 Int. J. Anal. Appl. (2023), 21:64 7 are fixed points, so Tx1 = x1 and Tx2 = x2. θ[%(Tx1,Tx2)] ≤ [ψ(θ(M(x1,x2)))]κ θ[%(x1,x2)] ≤ [ψ(θ(max{G(x1,x2),%(x1,x2)}))]κ ≤ [ψ(θ(max{ %(x1,Tx1)%(x1,Tx2) + %(x2,Tx2)%(x2,Tx1) max{%(x1,Tx2),%(Tx1,x2)} ,%(x1,x2)}))]κ ≤ [ψ(θ(max{ %(x1,x1)%(x1,x2) + %(x2,x2)%(x2,x1) max{%(x1,x2),%(x1,x2)} ,%(x1,x2)}))]κ ≤ [ψ(θ(%(x1,x2)))]κ ≤ [(θ(%(x1,x2)))]κ, which is contradiction, so x1 = x2. Therefore, T has a unique fixed point. Note. In this result, if ψ(t) = t and M(x,y) = %(x,y), then we get theorem 1.1. Example 2.1. Let W = { 0, 1, 2, 3} with the usual order ≤, be a partially ordered set and let % : W ×W → R be given as %(0, 0) = %(1, 1) = %((2, 2) = %(3, 3) = 0,%(0, 1) = %(1, 0) = 2, %(0, 2) = %(2, 0) = 3 2 ,%(0, 3) = %(3, 0) = 5 2 ,%(2, 3) = %(1, 3) = 5 2 ,%(1, 2) = 3. Consider C = {0, 1},D = {2, 3} and T : C → D defined by T (0) = 2,T (1) = 3. So, %(C,D) = %(0, 2) = 3 2 . Also, C0 = {0} and D0 = {2}. Clearly T (C0) ⊆ D0 and % (u1, v1) = % (C, D) = 3 2 % (u2, v2) = %(C, D) = 3 2 } ⇒ % (u1, u2) ≤ % ( v1, v2) , where u1, u2 ∈ C and v1,v2 ∈ D. Then, we have u1 = 0,v1 = 2 and u2 = 0,v2 = 2. In this case, %(0, 0) = 0 = %(2, 2), that is, the pair (C,D) has the weak P property. Taking θ(u) = u + 1 and ψ(u) = 999 1000 u for all u ≥ 0 and define α : W ×W → [0,∞) as follows,{ α(u,v) = 1, if (u,v) ∈{(0, 0), (0, 1), (1, 1)}, α(u,v) = 0, if not. Let u1,v1,u1 and u2 in C such that  α(u1,u2) ≥ 1, % (v1, Tu1) = % (C, D) = 3 2 , % (v2, Tu2) = %(C, D) = 3 2 . Then we have u1 = v1 = u1 = u2 = 0. So, α(v1,v2) ≥ 1, 8 Int. J. Anal. Appl. (2023), 21:64 that is, T is α− proximal admissible. By the symmetry of % and α, it suffices to study the cases (u = 0,v = 1) and (u = v = 0). If (u = 0,v = 1),u ≤ v, α(0, 1)θ(%(T 0,T 1)) = θ( 11 10 ) = 7 2 , M(u,v) = max{ %(u,Tv)%(u,Tv) + %(v,Tv)%(v,Tu) max{%(u,Tv),%(Tu,v)} ,%(u,v)} = max{ %(0,T 0)%(0,T 1) + %(1,T 1)%(1,T 0) max{%(0,T 1),%(T 0, 1)} ,%(0, 1)} = max{ %(0, 2)%(0, 3) + %(1, 3)%(1, 2) max{%(0, 3),%(2, 1)} ,%(0, 1)} = max{ 3 2 × 5 2 + 5 2 × 3 max{5 2 , 3)} , 2} = 15 4 . So, [ψ(θ(M(0, 1)]κ = ( 999 1000 × 19 4 )κ. Therefore, for κ = .805, we have 7 2 = ( 999 1000 × 19 4 )κ. If (u = 0,v = 0), then α(0, 0)θ(%(T 0,T 0)) = 1, M(u,v) = max{ %(u,Tv)%(u,Tv) + %(v,Tv)%(v,Tu) max{%(u,Tv),%(Tu,v)} ,%(u,v)} = max{ %(0,T 0)%(0,T 0) + %(0,T 0)%(1,T 0) max{%(0,T 0),%(T 0, 0)} ,%(0, 1)} = max{ 3 2 × 3 2 + 3 2 × 3 2 max{3 2 , 3 2 )} , 0} = 3. So, [ψ(θ(M(0, 0)]κ = ( 999 1000 × 4)κ. Therefore, for κ = .005, we have α(u,v)θ[%(Tu,Tv)] ≤ [ψ(θ(M(u,v)))]κ Hence, all the conditions of the theorem 2.1 are fulfilled. So T has a Best proximity point and it is u = 0. Int. J. Anal. Appl. (2023), 21:64 9 3. Application to Matrix Equations In this part, we will use the subsequent symbols: C(m) represents the collection of m ×m complex matrices, H(m) ⊂ C(m) represents the collection of the m×m hermitian matrices, ℘(m) ⊂ H(m) represents the collection of m×m positive definite matrices, H1(m) ⊂ H(m) is the set of positive semi definite matrices of m×m. In addition, U1,V1 ∈ C(m). So, if U1 ∈ ℘(m) this means that U1 � 0 and U1 � 0, means U1 ∈ H(m) . Moreover, U1 � V1(U1 � V1) is replaced by U1 − V1 � 0(U1 − V1 � 0). The spectral norm of the matrix B is denoted by the notation ||.||, i.e., ||B|| = √ λ+(B∗B), where λ+(B∗B) is the largest eigenvalue of B∗B and B∗ is the traconjugate of B. We write ||B||Y = m∑ j=1 Sj(B), where Sj(B) is the singular value of B ∈ C(m). For a given G ∈ ℘(m), we denoted the modified norm by ||B||Y,G = ||G 1 2BG 1 2 ||Y . The set H(m) equipped with the metric induced by ||.|| is CMS. Furthermore, H(m) is Poset with partial order �, where U1 � V1 ⇔ V1 � U1. In this Part, we use %(U1,V1) = ||V1 −U1||Y,G = tr(G 1 2 (V1 −U1)G 1 2 ). We assume that the subsequent nonlinear matrix equation is U = G ± n∑ j=1 B∗j τ(U)Bj. (3.1) Where G ∈ ℘(m), Bj, j = 1, 2, . . .n, are arbitrary m×m matrices and τ : H(m) → H(m) is continuous mapping, which maps ℘(m) into ℘(m). Consider τ is order preserving, that is , if C,D ∈ H(m) ⇒ τ(C) � τ(D), where C � D. Lemma 3.1. [12] Let C � 0 and D � 0 be m×m matrices. Then 0 ≤ tr(CD) ≤ ||C||.tr(D). Theorem 3.1. Let T : H(m) → H(m) be continuous (order preserving) mapping, which maps ℘(m) into ℘(m) and G ∈ ℘(m). Consider that (i) for all U � V and M > 1, %(τ(U),τ(V )) ≤ %(T (U),T (V ))(θ(tr(M(U,V )))) 1 2 M 1 2θ(tr(T (U) −T (V ))) , 10 Int. J. Anal. Appl. (2023), 21:64 where M(U,V ) = max{G(U,V ),%(U,V )} and G(U,V ) = { %(U,TU)%(U,TV )+%(V,TV )%(V,TU) max{%(U,TV ),%(TU,V )} , if max{%(U,TV ),%(TU,V )} 6= 0, 0, if max{%(U,TV ),%(TU,V )} = 0, (ii) 0 < ∑n j=1B ∗ j τ(G)Bj ≤ G, hold. Then 3.1 has a positive definite solution Ū ∈ ℘(m). Proof. Define T : H(m) → H(m) by T (U) = G ± n∑ j=1 B∗j τ(U)Bj, (3.2) and ψ(v) = v/M, then solution of 3.1 is a fixed point of T. Let U,V ∈ H(m) with U � V, then T (U) � T (V ) . %(T (U),T (V )) = ||T (V ) −T (U)||Y,G = tr(G 1 2 (T (V ) −T (U))G 1 2 ) = tr( n∑ j=1 B∗j G 1 2 (τ(V ) −τ(U))G 1 2Bj) = n∑ j=1 tr(B∗j G 1 2 (τ(V ) −τ(U))G 1 2Bj) = n∑ j=1 tr(B∗j GBj(τ(V ) −τ(U))) = n∑ j=1 tr(B∗j GBjG 1 2G −1 2 (τ(V ) −τ(U))G 1 2G −1 2 ) = n∑ j=1 tr(G −1 2 B∗j GBjG −1 2 G 1 2 (τ(V ) −τ(U))G 1 2 ) = tr( n∑ j=1 G −1 2 B∗j GBjG −1 2 )(G 1 2 (τ(V ) −τ(U))G 1 2 ), by lemma 3.1, we get %(T (U),T (V )) = || n∑ j=1 G −1 2 B∗j GBjG −1 2 ||.tr(G 1 2 (τ(V ) −τ(U))G 1 2 ) = || n∑ j=1 G −1 2 B∗j GBjG −1 2 ||.||τ(V ) −τ(U)||Y,G %(T (U),T (V )) = || n∑ j=1 G −1 2 B∗j GBjG −1 2 ||.%(τ(V ),τ(U)). Int. J. Anal. Appl. (2023), 21:64 11 So, by condition (i) and (ii), we get %(T (U),T (V )) ≤ %(T (U),T (V ))(θ(tr(M(U,V )))) 1 2 M 1 2θ(tr(T (U) −T (V ))) θ(tr(T (U) −T (V ))) ≤ (θ(tr(M(U,V )))) 1 2 M 1 2 θ(tr(T (U) −T (V ))) ≤ ( (θ(tr(M(U,V )))) M 1 2 ) 1 2 θ(tr(T (U) −T (V ))) ≤ (ψ(θ(M(U,V )))) 1 2 . Hence, by corollary 2.1, T has a fixed point. Therefore, matrix equation 3.1 has a unique solution Ū ∈ ℘(m). Numerical Experiment: Example 3.1. Consider the matrix equation U = G + 2∑ j=1 B∗j τ(U)Bj, (3.3) where G,B1 and B2 are given by G =   4 2 1 2 4 2 1 2 4   , B1 =   0.0241 0.047 0.047 0.047 0.0241 0.0241 0.047 0.0241 0.0241   , B2 =   0.58 0.0671 0.58 0.0671 0.58 0.0671 0.58 0.0671 0.58   . Define θ(u) = u + 1 and T (u) = u 9 . Then conditions (i) and (ii) of Theorem 3.1 are satisfied for M = 2. By using the iteration Un+1 = G + 2∑ j=1 B∗j UnBj with U0 =   0 0 0 0 0 0 0 0 0   . 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Cheng, A note on ’A best proximity point theorem for Geraghty-contractions’, Fixed Point Theory Appl. 2013 (2013), 99. https://doi.org/10.1186/1687-1812-2013-99. https://doi.org/10.1007/s10957-011-9810-x https://doi.org/10.1007/s11590-011-0379-y https://doi.org/10.1007/s13398-012-0074-6 https://doi.org/10.1007/s13398-012-0074-6 https://doi.org/10.22436/jnsa.010.05.07 https://doi.org/10.1080/01630563.2010.485713 https://doi.org/10.1007/s10898-011-9774-2 https://doi.org/10.1007/s10898-011-9774-2 https://doi.org/10.1186/1029-242x-2014-38 https://doi.org/10.1007/s11590-012-0529-x https://doi.org/10.1007/s11590-013-0709-3 https://doi.org/10.22436/jnsa.010.09.02 https://doi.org/10.22436/jnsa.010.09.02 https://doi.org/10.1016/j.na.2011.10.014 https://doi.org/10.1186/1687-1812-2013-99 1. Introduction and Preliminaries 2. Main Results 3. Application to Matrix Equations References