11. Lemma. Let ve T,(M). (1) If f, ge F(M) are equal on a neighbor-
hood of p, then Uf) = v(g). (2) If he F(M) is constant on a neighborhood of p,
then u(h)
0.
三
Proof.(1) By linearity it suffices to show that if f = 0 on a neighbor-
hood W of p, then u(f) = 0. Let g be a bump function at p with support in W;
then fg = 0onall of M. But v(0)= u(0 + 0) = u(0) + u(0) implies v(0) = 0. Thus
0= (fg) = v(S)g(p) + f(p)v{g) = u(f),
since f(p)
(2) By(1) we can assume that h has constant value c on all of M. If 1 is the
constant function of value 1, then
0 and g(p)
1.
ニ
(1) = (1.1) = (1)1 + lu(1) = 2v(1).
Hence v(1) = 0, and v(h) = u(c· 1) = cu{1) = 0.
