CALDERON'S FORMULA AND A DECOMPOSITION OF L 2 (R") 15 \(TaQ)(x)\ = | ^ [K(x -y)~ K(x - x Q ))aQ(y) dy C \y-*Q\a si"' n—a ///^va+rt—(n/2) i ^^-AQT (nm dy JK hQ \x - x Q C\x-xQ\-n-"l(Q)' = Cl(Q) -(n/2) /(G) \x-x 0 \ In either case we obtain \T(a Q )(x)\Cl(Q)-{nmll + l ^j^- - l , with C = C(n,a) an d w e have (1.11 ) fo r m Q = C~ Ta Q . Finally, w e sho w tha t (1.12 ) i s satisfied . I f \x - y\ l(Q), the n (1.11 ) gives us \(TaQ)(x) - (Ta Q )(y)\ \(Ta Q )(x)\ + \(Ta Q )(y)\ C\Q\ -1/2 -1/2 1 + /(G) 1 + \y-xc C\Q\ '" su p 2 Ml*-y| 1 + \x - z -x r KQ) W) -(/?+a) -{n+a)y C\Q\ -1/2 \x-y\ /(G) sup 2 Ml*-l 1 + \X - Z - X r KQ) -(n+a) Now suppose \x-y\ /((?) . I f x £ Ty/nQ, the n |x-y | /(g) \\x-z\ which gives (1.12) fo r thi s case. 1{Q). when z e 3Q . Thus , we can apply (1.17 ) to obtai n \(TaQ)(y) - (Ta )(x)\ = I / [K(y - z) - K(x - z)]a Q (z) d C f \x-y\ a \x-z\-{n+a)\Q\-l'2dz. J 30 But x £ ly/nQ an d z e 3Q impl y \x - z\ C\x - x Q \. Hence , th e las t expression doe s not excee d '\x-y\- . /(G ) . a '\x-xQ\'i /(G) , n —n—a C\Q\*/2\x-y\a\x-x Q \-{n+a) = C\Q\- l/2 which gives us (1.12) with e = a (sinc e I*-*,) ! 3/(G) , hence \x-z-x Q \ \x-xQ\ fo r \z\\x-y\W)).
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