Ngày tải lên :
23/07/2014, 13:20
... and B Suppose that AAu = Au By (1), we have d(Au, AAu) (max{d(Bu, BAu), d(Bu, AAu), d(AAu, BAu)}) < max{d(Au, AAu), d(Au, AAu), 0} = d(Au, AAu) This is a contradiction Hence Au = AAu = BAu and ... follows that d(Au, AAu) (max{d(Bu, BAu), d(Bu, AAu)}) = (d(Au, AAu)) < d(Au, AAu) This is a contradiction Therefore Au = AAu = BAu and Au is a common fixed point of A and B Assume that there exist ... Au = Bu The weak compatibility of A and B implies that ABu = BAu, and then AAu = ABu = BAu = BBu Let us show that Au is a common fixed point of A and B Suppose that AAu = Au In view of (1),...