1547845440-The_Ricci_Flow_-_Techniques_and_Applications_-_Part_III__Chow_

J. SOLUTIONS TO SELECTED EXERCISES 499

T/i in (25.30) now satisfies 0:::; TJ~ :::; ('Y!~)r 6 , and i/Ji in (25.31) now satisfies

ai/J. 2i 2i+i

0:::; ari :::; (! - 1) r@ and IY'i/lil_g :::; (! - 1) ro.

Instead of (25.33), we now have

{)nl.. 4i+2r-2

n/,._'f'_i '!-'i or + l\i'n/,.12 '!-'i g -< £· i ...!... -;-- "( - Q 1

Corresponding to (25.27) now define (instead of (25.34))

Mi~ ec(i+VKri) ((1+"!:-1)2 cn+I 4i+2 + 1) (2an4i+2r-2) ~.

"{Tl2/nB~( vO g g XQ, Ti ·) 2i-l 0 "(-1 0 0

Analogous to (25.35) and (25.36), we have

llvllL=(Qro) S 6 ilvllL2(Q 1 r 0 )'

where

6 ~ e^0 (1+-yVKro)i (Vol_g B_g (xo, ro))-~

oo (( i+3) ~) Hn~

2

ri

x !! 1+6 0 +i ~ _ 1 ( 260 4H^2 r 02 ) n

Again assuming 60 +14^4 2::: 1, we have 6 :::; C (!), where C (!) is given by

(25.40).

SOLUTION TO EXERCISE G.3. It is clear that f*dy is nonnegative. If

x1,x2,x3 EX, then

(fdy) (x1, xs) = dy (f (x1), f (xs))
:::; dy (f (x1), f (x2)) + dy (f (x2), f (xs))
= (fdy) (x1, x2) + (fdy) (x2, xs),
so the triangle inequality holds. If f is injective, then (fdy) (x1, x2) =
dy (f (x1), f (x2)) > 0 provided x1 # x2.
EXAMPLE OF A SOLUTION TO EXERCISE G.24. First consider the subset
of JR^2 defined by the zigzagging set
00
y -:-{ (0, O)} U LJ ( [ (2k~l)!' c2l)1] x {O})
k=O

U ( {O} x Q [ {'k~')" { 2 k~l)!]) U Q

0

{ (;!i,1;;;') : t E [O, 11}

Now define a subset of JR^3 by

X = {(r,0,z): (r,z) E Y and e E [0,27r)},