- INTEGRAL CURVES TO GRADIENTS OF CONCAVE FUNCTIONS 453
PROOF. (1) This is trivial.
(2) This follows from Lemma H.55 directly.
(3) This follows from
f (r (x, Pk) (sk,i)) = f bi (sk,i)) = sk,i
for all k and i.
(4) Given so E [a, T] and E > 0, by Lemma H.54 there is a neighborhood
U of r 00 (so) and S& > 0 such that for ally EC n U and O" E (0, s&) we have
(H.43)
s
d e ( y, YO' ) 2: IV! (r oo (so))I - s.
Here we chose S& small enough such that r 00 ([so, s 0 + s&]) c U. By the
uniform convergence of r (x, Pk) there is a ko such that for k 2: k 0 we have
IP kl < S& and r (x, Pk) ([so, so+ S&]) cu.
For any O" E (0, s&) and k 2: ko, we define
imin(so, k) ~ min{i: so :S sk,i} and imax(so, O", k) ~ max{i: so+ O" 2: Sk,i}·
Then it follows from (H.43) that for any i = imin (so, k)+l, ... , imax (so, O", k),
we have
de (r (x, Pk) (sk,(i-1)), (r (x, Pk) (sk,i)))
Hence
=de (r(x,Pk) (sk,(i-1)), (r(x,Pk) (sk,(i-l)))sk,i-sk,(i-^1 J)
< Sk,i - Sk,(i-1)
- Iv J (r oo (so)) I -s ·
imax (so ,O',k)
< I:: de (r(x,Pk) (sk,(i-1)), (r(x,Pk) (sk,i)))
i=imin,(so,k)+l
< sk,imax(so,O',k) - sk,imin(so,k).
- IV f (r oo (so))I - s
Now the lemma follows from taking the limit as k -+ oo of the above in-
equality. D
The following says that r 00 is an integral curve for \7 f /IV fl^2.
LEMMA H.58 (Existence of integral curve for \7 f /IV' Jl^2 ). Let C, x, a,
and f be as in Proposition H.53. Suppose TE (a, !sup)· Let Pk= {sk,i}~ 0
be a sequence of partitions with limk---+oo IPkl = 0 and such that the broken
geodesics r (x, Pk) converge as k-+ oo to a path r oo : [a, T] -+ C uniformly
on [a, T]. Then r 00 is the integral curve for \7 f / I \7 f I 2 emanating from x.