1547845440-The_Ricci_Flow_-_Techniques_and_Applications_-_Part_III__Chow_

498 J. SOLUTIONS TO SELECTED EXERCISES

Hence, by (J.2) and (J.4), for all 1::::; k::::; m + 1 the distance of Xk to xo at
time tk has the following upper bound:
1

dg(tk) (xk, xo) ::::; dg(tk_ 1 ) (xk-1, xo) +Al Rm 1-2 (xk-1, tk-l)

' 1

::::; dg(ti) (xi, xo) +Al Rm 1-2 (x1, t1)

1

.+ ·· · +AIRml-2(xk-r,tk-1).

(J.5) :Sc:+A ( 1+^1 1 )^1

2

+···+

2

k_ 2 IRmJ-2(x1,t1).

1

Since I Rm 1-2 (x1, ti) < c:, we have

dg(tk)(xk,xo) < (2A+ l)c:

for 1::::; k::::; m + 1. In particular, Xm+i E Bg(trn+i) (xo; (2A + l)c:).

By (J.3), if the point (x, f) in the claim cannot be taken to be (xki tk)

for any k EN, then we have

lim I Rm l(xk, tk) = oo,

k--too

which contradicts the fact that the metrics g (t) are continuous in time so

that the space-time set UtE[O,c2] Bg(t) (xo, (2A + l)c:) x {t} is compact and

IRml has a uniform upper bound on this set. Hence there exists £ E N such

that (x, f) can be taken to be (xe, te).

SOLUTION TO EXERCISE 22.18. By (22.93) and the co-area formula, we

have

F (t) =Vol {y E ]Rn: h (y) 2: t} = r= ( r IV'hl-l d(J") ds,

Jt J{h=s}

where d(J" is the area form of {y E ]Rn : h (y) = s}. Hence

• ddF (s) = r IV'hl-l d(J" for a.e. s > 0.
  S J{h=s}
  Applying (22.95) to this and again applying the co-area formula yields

1

00

d>. -d (s) F (s) ds = 1= >. (s) J IV'hl-^1 d(J"ds

O S O {h=s}

= r >. (h) dμJR.n.

j {yEJR.n:h(y)>O}

SOLUTION TO EXERCISE 25. 7. With Ti ~To ( 1 + ?:=~) we have

('y - 1)

Ti - Ti+l = 2i. To,

2 2 ('y-1) 2

Ti - Ti+l 2: 2 i-1 To,