Master Dissertation

However, for x2 = ∅ we can have X1 = ∅. From Equation (5.3) it follows
that
˜T (X1)T (Y1) = 0. (5.19)
P 0 2
As a consequence, in (5.18) only terms with X2 = ∅ and hence Y2 = Q
remain. That is, remembering that T0(∅) = 1 = ˜ T0(∅),
A ′
n1+1(Y, x) = ˜T (X1)T (Y1) T (Q, x),
P2
where P2 is set set of all partitions such that P = X1 ∪ Y1 and X1 = ∅.
Including ∅ would give 0 according to (5.19), hence
A ′
n1+1(Y, x) = ˜T (X1)T (Y1) − ˜
T (∅)T (P ) T (Q, x)
as wanted.
P 0 2
Fixme: investigate the meaning of this.
= (0 − T (P ))T (Q, x) = −T (P )T (Q, x),
Corollary 5.4 (Support property 1). The supports of A and R are
respectively advanced and retarded, that is
and
Proof. Equation (5.9) gives us
An1+1(Y, x) = A ′ n1+1 + Tn(Y, x)
supp An1+1(Y, x) ⊂ Γ − n1+1 (x)
supp Rn1+1(Y, x) ⊂ Γ + n1+1 (x)
= −Tn1−n2 (P )Tn2+1(Q, x) + Tn1+1(P ∪ Q, x)
= −Tn1−n2 (P )Tn2+1(Q, x) + Tn1−n2 (P )Tn2+1(Q, x) = 0.
2. follows the same way from (5.10).
The corollary explains why A and R are called advanced and retarded
distributions. An+1(X, x) vanishes if there exists a point x ′ ∈ X with
x ′ > x, i.e if P exists. On the other hand Rn+1(X, x) vanishes if there
exists a point x ′ ∈ X with x ′ < x.
Theorem 5.5 (Support property 2). If n ≥ 3 then
supp Dn(x1, . . . , xn−1, xn) ⊆ Γ − n (xn) ∪ Γ + n (xn). (5.20)
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