326 7. THE REDUCED DISTANCE
Hence b..L(q 7 , r) ?:: 0. We have
d+h(-)<l. L(q7,r+s)-L(q7,r)
-d T _ imsup
T s-+O+ S
1
:::; im. sup L(q7,r+s)-L(q^7 ,r)8L( - -;::;- q )
7 , T.
s-+0+ S UT
Then using (7.84), which holds for the barrier function L, we have
at A
OT (q 7 , r) :::; 2n - b..L(q7, r) :::; 2n.
We have proved that d;'Th (r) :::; o when Lis not C^2 at (q 7 , r). Hence we have
proved that d; 7 h(r):::; 0 for all r E (0, T). By the monotonicity principle for
Lipschitz functions stated in §3 (Lemma 3.1) of [179], h(T) is nonincreasing.
(ii) This follows from (i) and
_lim h(r) = !im min L (q, r)
T-+0+ 'T-+0 qEM
:S !imL(p,r) = (d 9 (o) (p,p))
2
'T-+0
= 0.
D
6.3. The reduced distance function£. To get even better equations
than those in Lemma 7.45, we introduce the reduced distance function£.
DEFINITION 7.49. The reduced distance f is defined by
(7.87)
1 1 -
f(x,T) ~ r;;:.L(x,T) = -
4
L(x,T).
2yT T
Let 7 E (0, T) and let'/': [O, r] ----t M be a minimal £-geodesic from p to
q and let K = K ('!', r) be defined as in (7.75). By (7.79), (7.80), and (7.81),
we have at (q, r),
(7.88)
(7.89)
(7.90)
8f = -^1 -K - ~ R
or 273/^2 r + '
2 1 g
IV'fl = -R--K 73/2 +-7'
1 n
!:lf < - ---K 273/2 +- 27 -R.
From these equations (which involve the trace Harnack integral K), (7.86)
and Lemma 7.48, we easily deduce the following which do not involve K.