- THE ALMOST EVERYWHERE VANISHING OF R 00
PROOF. We compute that\7^3 (v
22) =\7^2 (v\7v) =v\7^3 v+2\7v®\7^2 v+\7^2 v®\7v,
so that, by (29.34) and Lemma 29.10, we have on (-oo, -1],(29.39) llv'V3
vllea :S ~ ll\73
(v2
) Ilea+ 311 ~Ilea llvv'V2
vllea :SC.79Hence, from \7 (v\7^2 v) = v\7^3 v + \7v ® \7^2 v E LP for p ~ 1, we have (29.38) on
(-oo, -1]. D
The above estimates shall be combined in the following sections with various
monotonicity formulas to strengthen our understanding of the ancient solution g(t)
and especially its backward limits.5. The almost everywhere vanishing of R=
Let (S^2 , g (t)), t E (-oo, 0), be an ancient solution to the Ricci fl.ow on a
maximal time interval. By the trace Harnack estimate ~~ ~ 0, the backward limit
(29.40) R= (x) ~ lim R (x, t)
t-+- =
exists and is a bounded nonnegative upper semicontinuous measurable function.
However, R= might not be continuous, which can be seen by the King-Rosenau
example.
Adopting the notation of §3 of this chapter, we have thatR~~(t) ~max RKR(x, t) = 4 (a+ jJ) (t)
x(attained at the north and south poles) and R~l;,(t) ~ minx RKR(x, t) = 4a(t)
(attained on the equator, which is invariant under the Z2 reflectional symmetry).
Let
R~R (x) ~ lim RKR (x , t).
t-+- =
We a lso observe that R~R (N) = R~R (S) = 4μ = limt-+-= R~~x(t), whereas
R~R(x) = 0 for x E S^2 - {N,S}.
In this section we give two proofs that R= = 0 a .e. on 52. The first proof
uses the evolution of the area form, whereas the second proof relies on an energy
monotonicity formula.
5.1. Proof that R= = 0 a.e. using area evolution.
Define Ee~ {x E 52 : R=(x) > c}, which is a measurable set for each c ~ 0.
Let m 9 denote the Riemannian measure with respect to g. For any i E .N and
t E (-oo, -1] we have
- dt d m 9 (tJ(E1;i) =^1 R 9 (t)dA 9 (t) ~ -;,m^1 9 ctJ(E 1 ;i) ,
E11•
where we used R (x, t) ~ R=(x). This implies that
m lt+ll/'
9 (t)(E1;i) ~ e 'm 9 c-1J(E1;J
for t E ( -oo, -1 J. On the other hand,