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By F. Brackx, Richard Delanghe, F. Sommen

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1 44 A decomposition of the Laplace and Laplace-Beltrami operator First we set up the framework in which a function theory will be developed· on the one hand we have the Euclidean space Rm+ 1 with m ~ 1; points in Rm+ 1 are denoted alternatively by x = (x 0 ,x 1, ••• ,xm) = (x 0 ,x) x where clearly = (x 1, ••• ,xm) lays in the hyperplane x0 = 0 which is identified with Rm. Frequent use will be made of the notation R~+ 1 which means Rm+1 = Rm+1,{0}. 0 On the other hand we have the Clifford algebra A, already introduced in Section 1, the vector-subspace A1 of which has dimension n.

AI~ <1e ,(f,f)>. 0 <1 iAI~ eo ,(f,f)> ~ 0 and <1 eo ,(f,f)> = 0 if and for all A E A and f E L 2 ,(r)(H;A;~),<1e 0 ' So all conditions for L 2 (r)(H;A;~) to be a unitary right 1nner product A-module are satisfied. Since L 2 (H;A;~) IT L 2 (H;~) we have that L2 ( )(H;A;~) is complete; in other words AEPN • r L 2 ,(r)(H;A;~) is a right Hilbert A-module. 39 Notes to Chapter 1 Ever since the nineteenth century, mathematicians have developed different algebras adapted to meet the need of physicists.

8 Proposition Let E(r) I H(r) be~a closed submodule of H(r)" exists f E H(r)'{O} such that f E E (r)" Then there ~ Take g E H(r)'E(r)' call g0 the unique element in E(r) for which c5 = llg 0 -gll = inf llh-gll and put f = g0 -g. Then f 1 0 hEE(r) and for all A E Rand hE E(r)' g0 + Ah E E(r)" Hence, using a classical result, we find that = 0. eo Now call for hE H(r) fixed, ~ = (f,h). 9 Let E c H(r) and denote by spA(E) the right A-linear envelope of E. 11 = {0}. 13 If E(r) is a closed submodule of H(r)' then the foregoing theorem tells us that any f E H(r) admits a unique decomposition f = g + h with g E E(r) ~ and h E E(x).

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