主要研究求解增生算子零点问题的一类算法:xn+1=αnu+(1-αn)((1-λ)xn+λJrnxn),其u是固定向量, λ∈(0,1),{rn}和{αn}是实数列, Jrn表示增生算子A的预解式.其中(rn)收敛是保证算法收敛的一个充分条件,该文主要证明了此条件可减弱为limn |1-rn+1/rn|=0.
This paper deals with the problem:find x so that 0 ∈ Ax,where A is an accretive operator.One algorithm solving this problem has the following scheme:xn+1=αnu + (1-αn)((1-λ)xn + λJrnxn),where u is a fixed element,λ ∈ (0,1),{rn} and {αn} are real sequences,and Jrn denotes the resolvent of A.The algorithm is known to converge provided that (rn) is a convergent sequence.In this paper we show that such a condition can be relaxed as limn→∞ |1-rn+1/rn| =0.