36 NONLINEAR REPRESENTATION AND SPACES
and that
dv
f is a continuous function vanishing at oo.
Let (p e C^°(R3), (p(x) = 1 for 1/2 |x| 2, cp(x) = 0 for |x| 1/3, p(x) = 0 for
\x\ 3 and (p(x) 0, x e
M3.
We define ga(x) = f(ax). Since
a^{duf){ax)
=
duga(x)
and dvga{x) = d"(gap)(x) for 1/2 \x\ 2, we obtain from (2.58)
sup \a^(d»f)(ax)\
H A ^ H L -
(2.59)
l/2|x|2
£ H*W)HLP
\(3\n
Cn £ II^NaX^V)!!^
where (n \u\)p 3 and where we have used the Leibniz rule in the last inequalities.
A change of coordinates gives
\\(d^9a)(d0^)\\LP
=
a\M-^[j\{d^f){x){d^^){x/a)\^dx)L'P.
Since the integrand vanishes outside the set {x|a/3 \x\ 3a}, it follows that
a3/p\\(d^ga)(d^^)\\LP C
n
,
p
||^VllLoo|||x|^l^/||
L P
C^p £ \\Mad^f\\LP
Ml0i|
c;
p
£ II^^/IILP-
\a\\(3\n
This inequality and (2.59) show that for 1/2 |y| 2
a3/p+M|(^/)Mi a,p £
IIM«^/IILP,
|a||/3|n
which with a = \x\ 0,y = x/\x\, together with (2.58) gives estimate (2.57). This proves
the lemma.
Remark 2.11. Lemma 2.10 is still true if p oo, n \u\. In this case the proof is trivial.
Theorem 2.12. If (/,0) e M{, 1/2 p 1, then
(l + |x|)
3
/
2
-^|/(x)|C||(/,0)||
M f
(2.60a)
and if (/, / ) e
| + 2
,1/2 p 1, then
(1 + | X | ) 5 / 2 + I H - P ( | ^ ^
/ ( X )
|
+
|^/
( X
)| ) C
M
| | ( / , / ) | |
M H + 2
. (2.606)
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