“…By putting 2,nxxyw== in (4) We obtain ()()()(){}2222,,min,,,,,,,,nnnnMAxBwktMSxTwtMAxSxtMBwT wkt≥ Taking limit as and using (5) we get n→∞ ()(){},,min1,1,,,MuBwktMBwukt≥ We have for all (),,MuBukt≥ 0t> Hence (),,MuBut= Thus u Bw= Therefore BwTwu== Since is weak compatible. (,BT We get TBwBTw= that is (10) .BuTu=…”