- BACKWARD LIMIT OF ANCIENT A;-SOLUTION IS A SHRINKER 409
PROOF. First we shall prove that there exists o > 0 such that for any
T > 0,
(8.44)
for all (q, f) E B 9 ( 7 ) ( qn ~) x [A-^1 T, AT]. From (8.43), since our an-
cient solution has Rm~ 0 (in particular, R ~ 0)
1
\1 ve (q,f) I - ::; {343_.
g(T) v 4f
For any q E B g( 7 ) ( qn J E-^1 T) , let / ( s) be a minimal normal geodesic from
q 7 to q with respect to the metric g ( T). Since £ is locally Lipschitz, we have
I I
rdg(-r)(q,q(T)) I I
yf£(q,T)-yf£(qnT) ::; Jo \lyf£(t(s) ,T) g(T) ds
< f3.~= !3.
-V4r V4c
Since £(qn T) ::; ~'by the above estimate, we have
(8.45) R(q,T) <; (!%+ff)' for q E Bg(r) (qr,~).
From Lemma 7.65, we have for q EM
A-2 < _£(q,f) £ (q, T) 1 "f -TE [A-1 T,T, ]
£(q,f) £ (q, T) ::; A^2 1 "f -T E [ T, A T ].
(Note that the inequalities in the above two lines have the same form.)
Hence for q E B 9 ( 7 ) ( q 7 , J c^1 T) and f E [ A-^1 T, AT J ,
R(q,T) <; A
2
(!%+ff)
2
Now (8.44) is proved by choosing
o-
1
~ 3A
2
(!%+ff)
2
Now f R (q,f) ::; 5-^1 follows directly from (8.43) and (8.44); the lemma
is proved. D
For any sequence Ti --+ oo, consider the sequence of solutions