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/2x2
£ 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 {xa/3 \x\ 3a}, it follows that
a3/p\\(d^ga)(d^^)\\LP C
n
,
p
^VllLoox^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/3n
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 M£
 + 2
,1/2 p 1, then
(1 +  X  ) 5 / 2 + I H  P (  ^ ^
/ ( X )

+
^/
( X
) ) C
M
  ( / , / )  
M H + 2
. (2.606)