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7.1 Pseudohomogeneous Kernels 391

for any |β|≥ 0. Now, choose |β| = m. Then, with the previous remark we
have
β
β!a β (x) = a m (x, D) (•− x) ψ(|•−x|)

β
− p.f. k κ (x, x − y)(y − x) ψ(|y − x|)dy
Ω

β i(x−y)·ξ
= (2π) −n/2 (y − x) e ψ(|y − x|)a m (x, ξ)dydξ
n
IR IR n

β
− p.f. k κ (x, x − y)(y − x) ψ(|y − x|)dy .
IR n
We want to show a β (x) = 0 under assumption (7.1.76). First we note that
Lemma 7.1.9 implies the parity conditions
k κ+j (x, x − y)=(−1) m−j+1 k κ+j (x, x − y) (7.1.79)
for the homogeneous term of the asymptotic kernel expansion. A straightfor-
ward computation shows for j = 0 by substituting z = y − x and replacing ξ
by −ξ, that

β iz·ξ
a β (x) = 1 (2π) −n/2 z e a m (x, −ξ)ψ(|z|)dzdξ
β!
n
IR IR n

β
− p.f. k κ (x, −z)z ψ(|z|)dz .
IR n
Now, use the parity conditions for a m as well as for k κ after transforming z
to −z and obtain
a β (x)= −a β (x) , i.e. a β (x) = 0 for all |β| = m.
Substituting this into (7.1.78) and choosing |β| = m − 1, we obtain

β
β!a β (x)= a 1 (x, D) (•− x) ψ(|•−x|)

β
− p.f. k κ+1 (x, x − y)(y − x) ψ(|•−x|)dy
Ω

β i(x−y)·ξ
= (2π) −n/2 (y − x) e ψ(|•−x|)a 1 (x, ξ)dydξ
n
IR IR n

β
− p.f. k κ+1 (x, x − y)(y − x) ψ(|•−x|)dy
Ω
with the application of our theorem to the particular operator a 0 (x, D). Ap-
plying again the parity conditions (7.1.76) and (7.1.79), we find in the same
manner as for |β| = m,now