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7.2 Coordinate Changes and Pseudohomogeneous Kernels  405

                           Theorem 7.2.6. (Kieser [156, Theorem 2.2.12])
                                       m
                              For A ∈L (Ω) with m ∈ IN 0 let the Schwartz kernel of A be k(x, x−y) ∼
                                       c

                               k κ+j (x, x − y) and let k κ+j satisfy the parity conditions (7.1.74) with
                           j≥0
                                              σ j = j − m +1 for 0 ≤ j ≤ m.            (7.2.31)
                           Then the finite part integral is invariant under change of coordinates; namely


                               p.f.  k(x, x − y)u(y)dy =p.f.  k(x ,x − y ) u(y )J(y )dy    (7.2.32)






                                   Ω                       Ω
                           where k is given by (7.2.21).

                           Remark 7.2.1: This class of operators includes all of the boundary inte-
                           gral operators generated by the reduction to the boundary of regular elliptic
                           boundary value problems based on Green’s formula. The proof of this the-
                           orem is delicate and will be presented after we establish some preliminary
                           results.
                              Let P denote the class of all polynomials in ε of the form
                                                        j                    j
                                       ℘(ω, ε)=   a j (ω)ε  with a j (−ω)=(−1) a j (ω)
                                               j≥0
                                       n
                           where ω ∈ IR with |ω| =1.
                           Lemma 7.2.7. If ℘ 1 ,℘ 2 ∈P then ℘ 1 + ℘ 2 ∈P and ℘ 1 ℘ 2 ∈P.

                                                    j                  k
                           Proof: Let ℘ 1 =    a j (ω)ε and ℘ 2 =  b k (ω)ε . Then
                                           j≥0                k≥0

                                          ℘ 1 + ℘ 2 =  c   (ω)ε  with c   = a   + b   .
                                                     ≥0

                           Clearly,




                                c   (−ω)= a   (−ω)+ b   (−ω)=(−1) a   (ω) − b   (ω) =(−1) c   (ω) .
                           Similarly, by the use of the Cauchy product,

                                     ℘ 1 ℘ 2 =  c   (ω)ε  where c   (ω)=  a j (ω)b k (ω) .
                                             ≥0                      j+k=
                           Hence,


                              c   (−ω)=     a j (−ω)b k (−ω)=(−1)    a j (ω)b k (ω)=(−1) c   (ω) .
                                      j+k=                      j+k=
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