96 29. COMPACT 2-DIMENSIONAL ANCIENT SOLUTIONSLEMMA 29.29. There exists a constant c 1 > 0 such that for all t E (-oo, -1],
(29.85) c1^1 itl :::; Pt :::; C1 itl ·
Hence diam(g(t)):::; Cltl.
PROOF. STEP l. There exists a constant C < oo such that for each t ,
(29.86) c-^1 pt :::; Pt:::; Cpi.
Since Lg(t) ht):::; C by (29.66), we h ave(29.87) diam(g(t)):::; Pt+ Pt+ C.
Since B^9 x, ~) (pt) c S~Jt) and by Lemma 29.23, there exists a constant c > 0
such that for all t ,c < A ±ht)
- A =Fht)
Area 9 (t)(B!~\Pt +Pt+ C) - B!~)(pi))
< l l
Areag(t) (Bx,^9 ~) (p'i))< (Pt+ Pt+ C)^2 - (/[')^2
(pi)2
by the relative Bishop- Gromov volume comparison theorem. The above inequality
says that
v'C+Ipi :::; Pt + Pt + C.Hence Pi :::; C Pt for some constant C < oo.
STEP 2. Bounds for Pt. By (29.79) and (29.82), we have Pt :::; ~~ ltl.
By (29.87) and the relative Bishop- Gromov volume comparison theorem, we
haveArea g(t )(Bg(t)(p'f) xi t - Bg(xi t)(p'f t - 1))
Areag(t)(Bxg~)(Pt +Pt+ C) - Bx^9 ~\p't))
< l .,- Area g(t) (Bg(txi )(p'f)-Bgt xi (t)(p'f-l)) t
< (Pt+ Pt+ C)^2 - (pi)^2
(/[')2 - (/! - 1 )2
:::; Cpt
by (29.86).
From the geometry of t he neck we obtain
(29.88) Area 9 (t)(B^9 xt ~)(pt) - B^9 xt ~)(Pt - 1)):::; C.Hence
(29.89)where the last inequality follows from (29.72). This completes the proof of the
lemma. O