“…IYlf(-a,Y;t) dyIff(')= A f'"', we have that 11 f(*);X(')Il 5 11 All 11 f(");X(")I( , hence + l + aI (1 yg ds)dy 2 11 f(");X(")II (1 -ll All , -1-awhere 11A11 is the norm of the bounded operator A, as defined in(10). Therefore, if llA\l 5 1 and z > 0, we obtain from (16):(1 g ( 1 2 ZII f II , )IR(z,TA)gll 5 &gll ; (19) ((exp(tT,)fII 5 llfll , vf E x, i.e.…”