1547845439-The_Ricci_Flow_-_Techniques_and_Applications_-_Part_I__Chow_

3. STEADY AND EXPANDING BREATHER SOLUTIONS REVISITED 205

(1) the minimum value>.. (g) of 9(g, w) is equal to A.1(g), where >.. 1 (g)

is the lowest eigenvalue of the elliptic operator -4b. + R, and

(2)

(5.50)

wo is the unique positive eigenfunction of

-4b.wo + Rwo = A.1 (g) wo

with L^2 -norm equal to 1.

PROOF. To establish the existence of a minimizer w 0 of (5.46), one takes

a minimizing sequence {wi}: 1 of (5.46) in W^1 ,^2 (M). There then exists a

subsequence { wi} : 1 which converges to w 0 E W^1 ,^2 (M) weakly in W^1 ,^2 (M)

and strongly in L^2 (M) (by the Sobolev embedding theorem). Since

0:::; JM IV (wi - wo)l

2

dμ

=JM 1Vwil

2

dμ +JM 1Vwol

2

dμ- 2 JM (\7wi, Vwo) dμ,

by the weak convergence in W^1 ,^2 , we have limi-+oo JM (\7wi, Vwo) dμ

JM 1Vwol^2 dμ exists, hence

{ 1Vwol^2 dμ:::; li~inf { 1Vwil^2 dμ.

JM i-+oo JM

On the other hand, by the strong convergence of { Wi} : 1 in L^2 , we have

_lim { Rw;dμ = { Rw5dμ,

i->oo JM JM

r w5dμ = _lim r w;dμ = 1.

JM i-tooJM

Hence w 0 is a minimizer of (5.46) in W^1 ,^2 (M) , and w 0 is a weak solution to

the eigenfunction equation (5.48). By standard regularity theory, wo E 000 •

We also have that any minimizer is either nonnegative or nonpositive, since

otherwise ± lwol is a distinct smooth minimizer which agrees with wo on

an open set, contradicting the unique continuation property of solutions to

second-order linear elliptic equations.

We now prove wo is unique up to a sign. Without loss of generality,

we may assume below that wo is nonnegative. Call a minimizer w of g

with JM w^2 dμ = 1 a normalized minimizer. If the nonnegative normalized

minimizer is not unique, then there exist two normalized minimizers wo 2: 0

and w1 2: 0 with JM waw1dμ = 0. Then w2 = awo +bw1 is also a normalized

minimizer for all a, b E JR such that a^2 + b^2 = 1. Indeed, since wo and w 1
satisfy the linear equation (5.50), so does w2 = awo+bw1, and JM w§dμ = 1.
Now it not hard to see that there exist a and b such that w 2 changes

sign. In particular, if there are points x and y such that w1 ( x) = ewo ( x)

and w1 (y) = dwo (y), where c #- d and wo (x) > 0 < wo (y), then by

choosing a and b with a^2 + b^2 = 1 such that a+ be and a + bd have opposite

signs, we have that w2 (x) = (a+ be) wo (x) and w2 (y) = (a+ bd) wo (y)

have opposite signs, which is a contradiction. Hence wo is unique.