“…Now, let's consider case (ii). Without loss of generality, we assume that k n increases monotonically with n. Then, by using(25) and the inequality z h,δ k+1 − |z h,δk | = 2(G + A * h A h ) −1 A * h (y δ − A h |z h,δ k |), we deduce together with the positivity ofx † that z h,δ k+1 − x † 2 − z h,δ k − x † 2 ≤ z h,δ k+1 − x † 2 − |z h,δ k | − x † 2 = 2(|z h,δ k | − x † , z h,δ k+1 − |z h,δ k |) + z h,δ k+1 − |z h,δ k | 2 = 4 |z h,δ k | − x † , (G + A * h A h ) −1 A * h (y δ − A h |z h,δ k |) +4 (G + A * h A h ) −1 A * h (y δ − A h |z h,δ k |) 2 = 4 A h (|z h,δ k | − x † ), (G + A h A * h ) −1 (y δ − A h |z h,δ k |) +4 (G + A * h A h ) −1 A * h (y δ − A h |z h,δ k |) 2 = 4 (G + A h A * h ) −1 (y δ − A h |z h,δ k |), y δ − A h x † −4 (G + A h A * h ) −1 (y δ − A h |z h,δ k |),G(G + A h A * h ) −1 (y δ − A h |z h,δ k |) ≤ 4 (G + A h A * h ) −1 (y δ − A h |z h,δ k |) y δ − A h x † −4µ (G + A h A * h ) −1 (y δ − A h |z h,δ k |) 2, which implies together with the inequalityy δ − A h x † ≤ δ + hC † that, (29) z h,δ k+1 − x † 2 ≤ z h,δ k − x † 2 + 4 (G + A h A * h ) −1 (y δ − A h |z h,δ k |) Y • δ + hC † − µ (G + A h A * h ) −1 (y δ − A h |z h,δ k |) Y ,…”