It turns out that some meaning can be assigned to K ∗f for some functions K which are
not in L^{1}. This involves assuming a certain form for K and exploiting cancellation. The
resulting theory of singular integrals is very useful. To illustrate, an application will be
given to the Helmholtz decomposition of vector fields in the next section. Like Mihlin’s
theorem, the theory presented here rests on Theorem 31.3.3, restated here for
convenience.
Theorem 31.4.1Let ρ ∈ L^{2}
(ℝn )
∩ L^{∞}
(ℝn)
and suppose
∫
||F−1ρ(x − y)− F− 1ρ (x )||dx ≤ C .
|x|≥2|y| 1
Then for each p ∈
(1,∞ )
, there exists a constant, A_{p}, depending only on
p,n,||ρ||∞ ,
and C_{1}such that for all ϕ ∈G,
||||F− 1ρ∗ϕ|||| ≤ Ap ||ϕ||.
p p
Lemma 31.4.2Suppose
K ∈ L2 (ℝn ), ||F K||∞ ≤ B < ∞, (31.4.39)
(31.4.39)
and
∫
|K (x − y)− K (x)|dx ≤ B.
|x|>2|y|
Then for all p > 1, there exists a constant, A
(p,n,B)
, depending only on the indicatedquantities such that
||K ∗f||p ≤ A(p,n,B )||f||p
for all f ∈G.
Proof:Let FK = ρ so F^{−1}ρ = K. Then from 31.4.39ρ ∈ L^{2}
∫
|x|≥2|y||K ε(x − y )− Kε(x)|dx =
∫ |K ε(x − y )− Kε(x)|dx+
|x|≥2|y|,|x−y|> ε,|x|<ε
∫
+|x|≥2|y|,|x−y|<ε,|x|≥ ε |K ε(x− y )− Kε(x)|dx+ (31.4.46)
∫
|K ε(x − y )− Kε(x)|dx+
|x|≥2|y|,|x−y|> ε,|x|>ε
+ ∫ |K (x − y)− K (x)|dx.
|x|≥2|y|,|x−y|<ε,|x|<ε ε ε
(31.4.46)
Now consider the terms in the above expression. The last integral in 31.4.46 equals 0
from the definition of K_{ε}. The third integral on the right is no larger than B by the
definition of K_{ε} and 31.4.42. Consider the second integral on the right. This integral is no
larger than
∫
B |x|−ndx.
|x|≥2|y|,|x|≥ε,|x−y|<ε
Now
|x |
≤
|y|
+ ε ≤
|x|
∕2 + ε and so
|x |
< 2ε. Thus this is no larger than
∫ ∫ ∫
−n 2ε n−1 1-
B |x| dx = B Sn−1 ε ρ ρndρdσ ≤ BC (n)ln2 = C(n)B.
ε≤ |x|≤2ε
It remains to estimate the first integral on the right in 31.4.46. This integral is bounded
by
∫
B |x− y|−ndx
|x|≥2|y|,|x−y|> ε,|x|<ε
In the integral above,
|x|
< ε and so
|x − y|
−
|y|
< ε. Therefore,
|x − y|
< ε +
|y|
< ε +
|x|
∕2 < ε + ε∕2 =
(3∕2)
ε. Hence ε ≤
|x− y |
≤
(3∕2)
|x − y|
.
Therefore, the above integral is no larger than
∫ (3∕2)ε ∫ ∫ (3∕2)ε
B |z|−ndz = B ρ− 1dρdσ = BC (n)ln(3∕2).
ε Sn−1 ε