- ALMOST K;-SOLUTIONS 133
Note that^9
(20.16)
Bgk(t*k) (Y*k' 1
1
0 Ak
12
R;
1
l
2
(Yk' tk)) c Bgk(t*k) ( xok, ~~~).
Since Re 2:: 0, (20.16), and dgk(t*k) (Y*k' Xok) < !, we have by the Bishop-
Gromov volume comparison theorem that^10
Volgk(t.k) Bgk(t*k) (Y*b fo-Ak/2 R;1/2(Y*k' t*k))
(
lOAk^1 1/2 Rgk -1/2( Y*k' t*k) )n
Volgk(tk) Bgk(tk) (Y*k' !)
- (!t
Volgk(t.k) Bgk(t*k) (xok, i). (-
3
1)n
2:: (i) n
2:: 3-nvolgk(t*k) Bgk(t.k) (xok, 1)
2:: 3-nwo,
where we used (20.14) with t = t*k to obtain the last inequality.
On the other hand, we can apply Proposition 20.4 to gk (t) ~ gk (t + t*k),
the geodesic ball Bgk(t*k) (Y*k' loAk^12 R;k
1
!
2
(Y*k' t*k)), and the time interval
[-DkR;k^112 (Y*k' t*k), OJ, to obtain that for every c > 0 there exists A ( c) < oo
(^10) Since B 9 k(t.k) (xok, !~6) C B 9 k(t.k) (xok, 1) C B9k(o) (xok, 1), which is compactly
contained in Mk, this enables us to apply the volume comparison theorem.