- RIEMANNIAN GEOMETRY 455
(the above calculation holds for any C^2 function¢). That is, if x tJ. Cut (p),
then by (A.17) we have wherever F 2: 0,[¢ (:t -.6.) + 2\7¢. \7] (¢F):::; q;2c2 - (¢F)2 + C¢3/2p + C¢F
:::; 2 1 ( C1^2 - (¢F) 2)
for some constant C1 < oo. Hence
[¢ (:t - .6.) + 2\7¢ · \7] (t¢F):::; ~ (Cf-(¢F)
2
) + ¢2F:::; :t ( Cft^2 - (t¢F)^2 + 2t¢F).
Since limt---+O (t¢F) = 0 and t¢F has compact support, we may apply the
maximum principle to concludeCft^2 - (t¢F)^2 + 2t¢F 2: O
at a maximum point of t¢F on M x [O, £] for any [ E (0, T). In particular,t¢ (x) F (x, t) ::=; C2
on M x [O, T), where C2 depends only on C in (A.15) and A. In particular,
C2
F (x, t) ::=; ton B (p, A/2) x [O, T).
EXERCISE A.11. What happens to the constant C 2 as A---+ oo?In the above we have assumed that at the choice ( x 0 , t 0 ) of maximum
point of t¢F, the function ¢ is C^2. However ¢ is only Lipschitz continuous
since the distance function d (·,p) is only Lipschitz. To solve this problem,
we apply Calabi's trick (see [42]). In particular, suppose for our choice
of maximum point (xo, to) of t¢F, we have that¢ is not smooth at xo, i.e.,
xo E Cut (p). Let
I: [O, d(xo,p)] ---+ M
be a unit speed minimal geodesic joining xo top. Consider q ~ I ( d( xo, p) - c5)
for any small c5 = d(q,p) > 0. We have d(xo, q) + c5 = d(xo,p) and also q is
not in the cut locus of xo since c5 > 0. Consider the function
G (x, t) =;::. trJ (d(x, q) A + c5) F (x, t).
Since d(x, q) + c5 2: d (x,p) and rJ is nonincreasing, we have
G (x, t) ::=; (t¢F) (x, t)at any point (x, t) where F (x, t) 2: 0. Since G (xo, to) =to¢ (xo) F(xo, to)=
maxMx[O,f] (t¢F), we conclude that G (x, t) also achieves its maximum at