- CLASSIFYING THE BACKWARD POINTWISE LIMIT 103
Now let z = w - w. Then lieucZ = 0 in Bz 0 (r). Since z+ is subha rmonic on
B z 0 (r), by the mean value inequ ality we h ave thatOn the other hand, since z+ :S: iil+ + lwl, we have
llz+llL1(B, 0 (r)) :S: llw+llL1(Bo(lzol+r)) + llwllL1(B, 0 (r))
:S: 7r ln iloo( O)(lzol + r)^2 + C(lzol + r, Umin(-1)) + C(r, c)
~C'by (29.99) and (29.105). Hence llz+ llu-'°(B, 0 (3r/4)) :S: .,;-;2 C'.
This implies that(29.106) (^0) < -uA eucW - - R ge w < R ge z++w _ < C ie w in Bz 0 (3r/4),
(^16) C' 1- 1
where C1 = e;;::-:r sup 8 2x(-oo,-l] R. By (29.104), we have that e w E Lq(Bz 0 (r)).
Now we may apply a standard interior elliptic estimate to (29.106) to obtain (see
Theorem 4.1 in [144])
llw+llL^00 (B, 0 (r/2)) :S: Cllw+llL1(B, 0 (3r/4)) + ClllieucWllLq(B, 0 (3r/4)) :S: C.
That is, u :S: ec in B z 0 (r/2). D
9.3. A nontrivial v 00 must vanish at no more than two zeros.
With the above lemma we can prove the following result. Note that by (29.47)
we have that v 00 has exactly two zeros for the King- Rosenau solution.
LEMMA 29.35 (Nontrivial v 00 have at most two zeros). For any ancient solutionon 52 , eith er v 00 = 0 or v 00 has at most two zeros.
PROOF. Suppose that v 00 is zero at distinct points N1, N2, and N3. We shallthen show that v 00 = 0. By applying a conformal diffeomorphism of 52 , we may
choose the north and south poles to satisfy v 00 (N) > 0 and v 00 (S) > 0. Note that
the points N, S, N 1 , N 2 , N 3 are distinct. Let O': 52 - {N}---+ JR^2 be stereographic
projection, so that O'(S) = 0 E JR^2 satisfies u 00 (0) < oo. Let zk = O'(Nk) for
k = 1,2,3. Let r > 0 be such that the balls Bzk(r) are disjoint. Let c E (0 , 47r/3).
Consider any sequence of times ti ---+ -oo. Since limi---+oo v(zk, ti) = 0, by Lemma
29.33 we have that for i sufficiently large
for each k = 1, 2, 3. Since the Bzk (r) are disjoint, we conclude that
3
r Rg(t;)dμg(t;) ;::::::: L r R9(t;)u(ti)dμeuc > 87r,
Js 2 k=l Js.k(r)which contradicts the Gauss-Bonnet formula. D