10 2. FUNDAMENTAL SOLUTION AND COVARIANCE FUNCTION
Example 2.5. (a) The basic case is when ϕ 1. In this case, f is termed
a Riesz kernel. We recall that = cd,β Fkd−β [36, Chapter V].
(b) Another possibility is ϕ(x) =
exp(−σ2|x|2/2).
In this case, f(x) is indeed
a covariance function, since f = F(kd−β ψ), where
ψ(ξ) =
(2πσ2)−3/2 exp(−|ξ|2/(2σ2)).
The parameter δ in Assumption 2.4 can be set equal to 1. Condition (1.4) is satisfied
since β ]0, 2]. Indeed,
Rd
(kd−β ψ)(ξ)
|FG(t)(ξ)|2
=
Rd
ψ(η)
Rd
kd−β (ξ) |FG(t)(ξ
η)|2
sup
η∈Rd Rd
kd−β (ξ) |FG(t)(ξ
η)|2,
and the right-hand side is finite when β ]0, 2]: see [9, Lemma 8].
The Riesz potentials Ia associated with the function (x) are defined by
(Iaϕ)(x) =
1
γ(a)
Rd
|x
y|−d+aϕ(y)dy,
for ϕ
S(Rd),
a ]0, d[ and γ(a) =
πd/22aΓ( a
2
)/Γ(
d−a
2
). Riesz potentials can be
interpreted as fractional integrals and have the semigroup property
Ia+b ϕ = Ia(Ibϕ), ϕ
S(Rd),
a + b ]0, d[
(see [36, p.118]). This property implies in particular that
(2.11) |x
y|−d+(a+b)
=
Rd
dz |x
z|−d+a|z

y|−d+b,
provided a + b ]0, d[. This equality can be informally interpreted by saying that
| ·
|−d+(a+b)
is the fractional integral of order a of | ·
|−d+b,
which is natural since
(−∆)a/2(|·|−d+(a+b)) = |·|−d+b, as can be checked by taking Fourier transforms.
We will make heavy use of properties of first and second order increments of
Riesz kernels. For a function f : Rd R, we set
Df(u, x) = f(u + x) f(u),
D2f(u,
x) = f(u x) 2f(u) + f(u + x),
¯
D
2f(u,
x, y) = f(u + x + y) f(u + x) f(u + y) + f(u). (2.12)
Notice that D2f(u, x) =
¯
D 2f(u x, x, x).
Lemma 2.6. Fix u, x, y
Rd,
a + b ]0, d[. The following properties hold.
(a) For any c R,
Dkd−a−b(u, cx) =
|c|b
Rd
dw kd−a(u c w)Dkd−b(w, x).
(b) For any b ]0, 1[ and any vector e
Rd
with |e| = 1,
Rd
dw |Dkd−b(w, e)| ∞.
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