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

dλ

d

dλ

(

|w −

λe|−d+b

)

≤

1

0

dλ

|w| 2

dw |w −

λe|−d+b−1

≤ C

1

0

dλ

∞

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)