1547845447-The_Ricci_Flow_-_Techniques_and_Applications_-_Part_IV__Chow_

3. 2 -DIMENSIONAL ANCIENT SOLUTIONS WITH FINITE WIDTH 61

We conclude that since 2 ~ < 2,

1 (

o lw(x)I ) d ( ) 4n

2

exp μIE x < 2!,; r^21 lb.w(y)I d μIE ( ) y

B(r) llLiwllL'(B(r)) - 4n - 0 B(r) llb.wllu(B(r))

16n^2 r^2

<--.

4n - o

D

REMARK 28.51. It would be interesting to further investigate the geometric

aspects of the Brezis- Merle result. See Struwe [395] for work in this direction.

Using the above lemma we can now prove the following.

LEMMA 28 .52 (Upper bound for u). For each t E (-oo, 0) there exist constants

so(t) and C2(t) such that

(28.62) u(s, e, t):::; C 2 (t) in [s 0 (t), oo) x 51 ,

where C2(t) is uniformly bounded on compact time intervals.

PROOF. R ecall that Licyl ln u = -Ru < 0. On the other hand, since 4n >

fR2 Rdμ 9 = f Rxs 1 Rudμcyl by the Cohn-Vossen inequality, near infinity Ru is small

in the following sense. For each c; > 0 and t < 0 there exists st:(t) E (0, oo) such

that

f

00

( RudBds < c:.

Js,(t) }51

Lets E [st:(t) +n, oo) and B E 51. Let Brr = B(;,

1

11 ) (n), so that (Brr, 9cyl) is isometric

to a Euclidean 2-ball. Let w be the unique solution to

-LicylW =Ru in Brr,

w=O on8Brr·

By Lemma 28.50 with o = 1 and since f 8n IRuldμcyl < c:, we have

(28.63)

1

• 11 · 1 l67r 4
  et: w dμ cy l < - -- 4 1 < 135.
  Bn 7r -

By J ensen 's inequality we have JBn c^11 w1n-^3 dμcyl ::::: ln(f Bn et:-'lwln-^3 dμcyl), so

that

(28.64)

Now let z ln u - w. Then Licy1z = 0 in Brr. Since z+ max{z, O} is
subharmonic on Brr, by the mean value inequality we h ave that

llz+llL^00 (Bn;2) :::'.: A tB ) llz+llu(Bn) =

4

rea rr ;2 n^3 llz+llu(Bn)>

where Brr; 2 = B(;,^111 ) (n/2). On the other hand , since z+ :::'.: (lnu)+ + lwl, we h ave

llz+llL'(Bn) :::'.: ll(ln u) +llL'(Bn) + llwllL1(Bn) :::'.: C

by (28.60) and (28.64). Hence llz+llL=(Bn; 2 ) :::'.: ~- This implies that

(28.65)