By Walter Mead Patterson 3rd (auth.)

**Read or Download Iterative Methods for the Solution of a Linear Operator Equation in Hilbert Space — A Survey, 1st Edition PDF**

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**Extra resources for Iterative Methods for the Solution of a Linear Operator Equation in Hilbert Space — A Survey, 1st Edition**

**Example text**

Proof. We first observe that for n ~ 0 : rn+ I = y - AXn+ I = y - A(xn+tnr n) = rn-tnArn while ro= y . Thus II~n+lll 2= Ilrn-tnArnll 2= Ilrnll 2 -

E n = Ten. I = T2en_ 2 = ... ~ The 0 (Vn} Z llTnll n=O given by (6) converges so that the solution Let so that ZIZ. le011=O . n~ = n~ 58 Subtracting N2Vn+ 1 from both sides of (6): NVn+ I = N2(Vn-Vn+l) + g = N2(Vn-Vn+l) + Nv* or (Vn+l-v*) = N-~2(Vn-Vn+l) from which we obtain (7). We obtain (8) from Lemma (1). I We shall use a fixed relaxation factor be arbitrary and define the sequence [xn} ~ > 0 . Let x0 e iteratively by (9) (D-~S)Xn+ 1 : [(l-w)D+0Q]x n + ~y . Theorem. Let (i0) A be a bounded operator of the form satisfying (2), (a), (b), and (c).

Ax n ~ P M ( A ) y = inf x¢~ to (7). because Thus we can , and 11Ax-yll Noting that PM(A)AXn = APM(A)x n , we obtain PN(A)Xn+l = PN(A)Xn + ~PN(A)y = PN(A)xo + (n+l)~PN(A)y (9) PM(A)Xn+I = PM(A)Xn - ~[APM(A)x n - PM(A)y] • (io) and Assume that (i) is solvable. pM( A)y = y = Ax . 2 and w n E M(A) " 38 w Uslmg Lemma (9), with = w n+l (5) again, n - (YAw n we obtain P~(A)y=O , we get w -* 0 n or M'A)Xn P~(A)x n : P~(A)x o , so that x n = PN(A)Xn + PM(A)Xn ~ PN(A)x0 Now assume that (1) is not solvable then (9) shows that that z , then xn y ~ R(A) [x n] d~verges.