2.2. THE COVARIANCE FUNCTION AND RIESZ KERNELS 11
(c) Set e =
x
|x|
, x
Rd.
Then
|D2kd−a−b(u,
x)|
|x|b
Rd
dw kd−a(u
|x|w)|D2kd−b(w,
e)|.
(d) For any b ]0, 2[ and each e
Rd
with |e| = 1,
Rd
dw
|D2kd−b(w,
e)| ∞,
and this integral does not depend on e.
(e) For any c R,
¯
D
2kd−a−b(u,
cx, cy)
|c|b
Rd
dw kd−a(u c w)|
¯
D
2kd−b(w,
x, y)|.
Proof. (a) From (2.11), we obtain
Dkd−a−b(u, cx) =
Rd
dz kd−a(u z)Dkd−b(z, cx).
Set z = cw to see that this is equal to
|c|d
Rd
dw kd−a(u c w)Dkd−b(c w, cx)
=
|c|b
Rd
dw kd−a(u c w)Dkd−b(w, x),
which proves (a).
We now check (b). Set
I =
Rd
dw |Dkd−b(w, e)|
and consider the decomposition I = I1 + I2, where I1 (resp. I2) is the integral of
the same expression but over B2(0) (resp. B2(0)c) instead of Rd. Then
I1 2
B3(0)
dw
|w|−d+b
C
3
0
ρb−1dρ
∞,
if b 0. As for I2, we write
I2 =
|w| 2
dw
1
0

d

(
|w
λe|−d+b
)

1
0

|w| 2
dw |w
λe|−d+b−1
C
1
0


1
ρb−2dρ,
which is finite if b 1.
The proof of (c) is analogous to that of (a). It suffices to apply the identity
(2.11) twice, since
D2kd−a−b(u,
x) = Dkd−a−b(u, x) + Dkd−a−b(u, −x), and to use
the change of variables z = |x|w.
Let us now prove (d). Observe that
D2kd−b(ω,
e) = kd−b(w e) 2kd−b(w) + kd−b(w + e)
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