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 (nmdy
JK
hQ \x - x
Q
C\x-xQ\-n-"l(Q)'
= Cl(Q)
-(n/2)
/(G)
\x-x0\
In either case we obtain
\T(aQ)(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.
No w suppos e \x-y\ 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-xQ\-{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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