Abstract :
Suppose X = Lp (or lp), 1 < p ≤ 2. Let T: X → X be a Lipschitzian and strongly accretive map with constant k ∈ (0, 1) and Lipschitz constant L satisfying p(p − 1)L2 < p − 2(p − 1)(1 − k). Define S: X → X by Sx = ƒ − Tx + x. For arbitrary x0 ∈ X, the sequence {xn}∞n=1 is defined by xn+1 = (1 - αn)xn + αnSyn,yn = (1 - βn)xn + βnSxn, n ≥ 0, where {αn}∞n=1, {βn}∞n=1 are two real sequences satisfying (i) 0 ≤ αn ≤ 2−1{p − 2(p − 1)(1 − k − kβn + L2βn) − p(2 − p)L2[1 − wβn + (w + L2 − 1)β2n]}{p − 2(p − 1)(1 − k − kβn + L2βn) + (p − 1)2L2[1 − wβn + (w + L2 − 1)β2n] − 1}−1 for each n,(ii) 0 ≤ βn ≤ min{w(w + L2 − 1)−1, w[4(p − 1)(L2 - k) + (p − 1)2L2w]−1} for each n,(iii) Σn αn = ∞, where w = p − 2(p − 1)(1 − k) − p(2 − p)L2. Then {x}∞n=1 converges strongly to the unique solution of Tx = ƒ. Moreover, if αn = 2−1{p − 2(p − 1)(1 − k − kβ + L2β) − p(2 − p)L2[1 − wβ + (w + L2 − 1)β2]}{p − 2(p − 1)(1 − k − kβ + L2β) + (p − 1)2L2[1 − wβ + (w + L2 − 1)β2] − 1}−1 and βn = β for each n and some 0 ≤ β ≤ min {w(w + L2 − 1)−1, w[4(p − 1)(L2 − k) + (p − 1)2L2w]−1} then ||xn+1 − q|| ≤ pn/2||x1 − q||, where q denotes the solution of Tx = ƒ and ρ = [1 − 4−1{p − 2(p − 1)(1 − k − kβ + L2 β) − p(2 − p) × L2[1 − wβ + (w + L2 − 1)β2]}2{p − 2(p − 1)(1 − k − kβ + L2β) + (p − 1)2L2[1 − wβ + (w + L2 − 1)β2] − 1}−1] ∈ (0, 1). A related result details with the iterative approximation of Lipschitz strongly pseudocontractive maps in X.
