“…B1) for ∈ (0, 1), there exists ( ) ∈ ( , 1) such that A( x + ( − 1)e, −1 + ( −1 − 1)e) ≥ ( )A(x, ) + ( ( ) − 1)e, ∀x, ∈ P h,e ; (B2) for ∈ (0, 1), there exists ( ) ∈ ( , 1) such that B( x + ( − 1)e) ≥ ( )Bx + ( ( ) − 1)e, x ∈ P h,e ; (B3) there exists a constant > 0 such that A(x, ) ≥ (Bx + e) − e.Then, operator Equation(31) has a unique solution x * in P h,e ;(2) for any x 0 , y 0 ∈ P h,e , making the sequencesx n = A(x n−1 , n−1 ) + Bx n−1 + e, n = A( n−1 , x n−1 ) + B n−1 + e, n = 1, 2, … , one has x n → x * , y n → x * as n → ∞.Proof. On account of conditions (B1) and (B2), for ∈ (0, 1), we haveA( −1 x + ( −1 − 1)e, + ( − 1)e) ≤ −1 ( )A(x, ) + ( −1 ( ) − 1)e, x, ∈ P h,e , B( −1 x + ( −1 − 1)e) ≤ −1 ( )Bx + ( −1 ( ) − 1)e, x ∈ P h,e .From theorem 3.1 in Sang and Ren,34 A ∶P h,e → P h,e is gotten; then, we prove that B ∶ P h,e → P h,e . Because Bh ∈ P h,e , there existr 1 , r 2 > 0 such that r 1 h + (r 1 − 1)e ≤ Bh ≤ r 2 h + (r 2 − 1)e.For x ∈ P h,e , we can choose c ∈ (0, 1) such that ch + (c − 1)e ≤ x ≤ c −1 h + (c −1 − 1)e; we thus obtain Bx ≤ B(c −1 h + (c −1 − 1)e) ≤ −1 (c)Bh + ( −1 (c) − 1)e ≤ −1 (c)(r 2 h + (r 2 − 1)e) + ( −1 (c) − 1)e = ( −1 (c) − 1)r 2 h + ( −1 (c)r 2 − 1)e, and Bx ≥ B(ch + (c − 1)e) ≥ (c)Bh + ( (c) − 1)e ≥ (c)(r 1 h + (r 1 − 1)e) + ( (c) − 1)e = ( (c) − 1)r 1 h + ( (c)r 1 − 1)e, that is, Bx ∈ P h,e , so B ∶ P h,e → P h,e .…”