Sigma Models with Repulsive Potentials

Home , Sigma model

Jingxian Huang (Ching-Yin Wong)

Department of Mathematics Duke University

Date: Approved:

Mark A. Stern, Supervisor

Hubert Bray

Robert Bryant

Leslie Saper

Dissertation submitted in partial fulﬁllment of the requirements for the degree of Doctor of Philosophy in the Department of Mathematics in the Graduate School of Duke University 2017 Abstract Sigma Models with Repulsive Potentials

Jingxian Huang (Ching-Yin Wong)

Department of Mathematics Duke University

Date: Approved:

Mark A. Stern, Supervisor

Hubert Bray

Robert Bryant

Leslie Saper

An abstract of a dissertation submitted in partial fulﬁllment of the requirements for the degree of Doctor of Philosophy in the Department of Mathematics in the Graduate School of Duke University 2017 Copyright c 2017 by Jingxian Huang (Ching-Yin Wong) All rights reserved except the rights granted by the Creative Commons Attribution-Noncommercial Licence Abstract

Motivated by questions arising in the study of harmonic maps and Yang Mills theory, we study new techniques for producing optimal monotonicity relations for geometric partial diﬀerential equations. We apply these results to sharpen epsilon regularity results. As a sample application, we analyze energy minimizing maps from compact manifolds to the space of Hermitian matrices, where the energy of the map includes the usual kinetic term and a singular potential designed to force the image of the map to lie in a set homotopic to a Grassmannian.

iv To my parents

v Contents

Abstract iv

List of Abbreviations and Symbols viii

Acknowledgements ix

1 Introduction1

1.1 Harmonic Maps...... 1

1.2 Sigma Model Approach...... 3

2 Monotonicity Formulas 10

2.1 Monotonicity Formula for Elliptic Equations...... 12

2.2 Monotonicity Formula for Parabolic Equations...... 19

3 Moser Iteration 30

3.1 Moser Iteration for Elliptic subsolutions...... 30

3.2 Moser-Harnack Iteration for Parabolic Subsolutions...... 36

4 -Regularity 44

4.1 Elliptic -Regularity...... 44

4.2 Parabolic -Regularity...... 49

5 Models with Repulsive Potentials 59

5.1 Sigma Models with Repulsive Potentials...... 59

5.2 Repulsive Potentials...... 60

5.2.1 Estimates of ∇W and HesspW q ...... 61

vi 5.2.2 Short Time Existence of Parabolic Flow...... 62

5.2.3 Induced Equations of Energy Density epfq ...... 63

5.3 Smoothing Potential Wb ...... 64

5.3.1 Improved Bound of sup eb ...... 66 M 5.3.2 -Regularity Results...... 69

5.3.3 Measure of Set with Large ebpfq ...... 71 5.4 Higher Power Potentials...... 74

5.4.1 -Regularity Results...... 77

Lp q 5.4.2 Measure of Set with Large eb f ...... 79 Bibliography 82

Biography 84

vii List of Abbreviations and Symbols

Symbols

Rk k-dimensional Euclidean space

M,N Riemannian manifolds

dimpMq Dimension of M

gij, g˜αβ Metric on Riemannian manifolds

k ˜γ Γij, Γαβ Levi-Civita connection on Riemannian manifolds TM Tangent bundle of M

T ¦M Cotangent bundle of M

Ric Ricci curvature 2-tensor

HesspW q Hessian matrix of some function W

d, d¦ Exterior derivative and its adjoint

dvol Volume form of some Riemannian manifold

∆ Laplace-Beltrami operator

∇X Covariant derivative along vector ﬁeld X

BRpxq Geodesic ball centered at x with radius R

SRpxq Geodesic sphere centered at x with radius R

2 PRpx, tq The set tpy, τq P M||y ¡ x| ¤ R, |τ ¡ t| ¤ R u of Riemannian manifold M.

2 2 TRptq The set M ¢ rt ¡ 4R , t ¡ R s of Riemannian manifold M

viii Acknowledgements

I would like to thank my advisor, Professor Mark A. Stern. I have learned more than I could imagine from him. If I tried to record every single thing he has taught me and said to me, including his “iconic” harsh words, this would become another thesis. Without him, I could never have finished the work as well as this thesis. He always reminded me to work hard and not to get away from difficulties. I still remember when I just came to this program and complained that my bachelor education was not recognized and respected by some of my peers, he told me that “You cannot ask for respect; you have to earn it.” His meticulous and rigorous scholarship has been and will always be the guideline for me, no matter for being a mathematician or for being a moral person. I also want to thank Professor William Pardon. He has always been supportive, no matter when I got harassed, when I had to organize the entire recruitment weekend event as a Graduate Representative, or when I needed to vent about my advisor’s difficult personality. I am grateful to have the chance to work with the staff in the Mathematics Department. They are Ms. Tyffany Kolba, Ms. Monique Brown, Ms. Jenny Hirtz, Ms. Melissa Blakey, Ms. Felina Thorpe-Grissom and Mr. Wayne Williamson. I was never disappointed every time I turned to them. I will always remember their kindness and helpfulness to me during these six years. There were several communities where I have spent significantly long time at

ix Duke. Since spring 2014, I have been working for Duke International House. This is like my second family. I have enjoyed working with Ms. Paige Vinson, Ms. SangHee Jeong, Ms. Lisa Giragosian and other volunteers that I have met during orientations and workshops. Duke International House has been really encouraging and always allowed me to explore any ideas that I purposed so that I can have the freedom to achieve the best I can. I especially thank Ms. Paige Vinson and Ms. Lisa Giragosian for listening to me and helping me out when I had ﬁnancial and academical stress. They have treated me like their family members and I am really thankful that I have been given the opportunity to work in International House and serve for my community as an international student. I have spent two years singing in the Duke Chapel Choir. I have met several interesting people, who have taught me important lessons in life. Particularly, I want to thank Ms. Jenny Vaughn, Dr. Russel Owen, Dr. Carol Weingarten and Mr. Richard Weingarten for staying by my side when I was down and helping me out unconditionally. I had a tough time during my third year when I had to learn how to handle the stress coming from teaching and not being able to focus on what I liked to do. Singing in the choir and spending time cooking and chatting with Jenny in her house really helped me release. As for Russell, he is the person that I can trust whenever I have any confusions and am feeling helpless. His call and his blessings before the date of my defense were so crucial that I would not have picked up my conﬁdence again without talking to him. Not to mention how much I have learned about the culture and tradition of the American society from Jenny and Russel. I got to know Carol when I was randomly assigned to be her roommate during a retreat trip, and this turned out to be an amazing connection with her and her husband Richard. When I needed advice about job searching in industry, they immediately took out time from their busy schedule to meet with me, provided me plenty of advice on revising resume and connected me to their friend. I want to

x express my most sincere appreciation for all their support and kindness. I have worked for Duke Peer Tutoring Program for just a year. However, I have already received a lot of support from Dr. Ben Cooke and Ms. Monique Harris. I was surprised how much my eﬀort and talent was appreciated inside this program as a tutor and a math study group leader. Moreover, I want to thank Ben for connecting me to several his classmates at Duke, which was really helpful for my job search. Finally I want to thank my students. I had struggled for a long time to overcome my shyness of standing in front and speaking publicly, not to mention that I had to teach in a language that I have not fully mastered. I surely have met a group of lovely people who were patient and who could see my worth as a teacher. It was their appreciation and recognition that I decided to bring the best I could as a teacher, even I had to face many unfair dealings outside of the classroom. It has been a pleasure to teach them at Duke.

xi 1

Introduction

1.1 Harmonic Maps

Let M be a compact Riemannian manifold and N a complete Riemannian manifold. Let f : M Ñ N be a smooth map. df is a section of the vector bundle T ¦M bf ¦pTNq over M. Denote the metrics of M and N by gij andg ˜αβ and the respective connections

k rα by Γij and Γβγ. The kinetic energy functional is deﬁned to be » » 1 Epfq |df|2 dvol epfq dvol. (1.1) M 2 M

We are interested in ﬁnding representatives of each homotopy class of M that minimize the above energy. Harmonic maps are the critical points of Epfq. They satisfy the corresponding Euler-Lagrange equation:

Bf β Bf γ ∆f α gijΓrα 0, @α 1, ¤ ¤ ¤ , dimpNq. (1.2) βγ Bxi Bxj

Here, we follow the Einstein summation notation. ∆ is the Laplace-Beltrami operator on M. When the dimension of M is 1, harmonic maps are closed geodesics on N.

1 This Euler-Lagrange equation is a non-linear elliptic equation, where the nonlin- earity arises from the quadratic term

Bf β Bf γ Qαpdf, dfq gijΓrα . (1.3) βγ Bxi Bxj

A typical approach to constructing harmonic maps is to study the solution ft to the corresponding parabolic ﬂow

Bf ∆f Qpdf, dfq 0. (1.4) Bt

This is the gradient ﬂow of Epfq, which decreases Epfq. We want to know

whether ft converges to a harmonic map when t Ñ 8. One of the diﬃculties comes from the quartic term of the induced equation of the energy density

1 epfq |df|2. (1.5) 2

The sign of this quartic term is determined by the Riemannian curvature of the target manifold N.

The following result from Eells and Sampson [6] shows the existence of harmonic maps in each homotopy class of M when the target manifold N has non-positive Riemannian curvature:

Theorem 1 (Eells - Sampson [6]). Let M be compact and N complete. Suppose N has non-positive Riemannian curvature and f : M Ñ N is a continuous map. Let ft

be the solution to (1.4) with ftp0, xq f0pxq. If ft is bounded as t Ñ 8, then ft is

homotopic to a harmonic map f8 with Epf8q ¤ Epfq. Furthermore, if N is compact with non-positive Riemannian curvature, every homotopy class of maps from M to N contains a harmonic map that minimizes the energy Epfq.

2 While negatively curved target manifolds guarantee the existence of harmonic maps, the existence of harmonic maps when f is mapped into a non-negatively curved manifold is more complicated. For example, Schoen and Uhlenbeck [15] proved the following result:

Theorem 2 (Schoen-Uhlenbeck [15]). Consider maps f : Rn Ñ Sk, where Sk is the Euclidean k-sphere. Let " p q 3, k 3 d k t 1 u ¥ min 2 k 1, 6 , k 4

If k ¥ 3 and n ¤ dpkq, there is no non-constant minimizing harmonic map f from Rn to Sk.

Leung ([8]) showed the following

Theorem 3 (Leung [8]). For n ¥ 3, there exists no non-constant stable harmonic maps from any compact Riemannian manifolds to Sn.

Since minimizing harmonic maps may not always exist for a given target manifold, we are interested in understanding when smooth harmonic maps do exist and how singularities develop when smooth minimal harmonic maps do not exist. In the elliptic case, Schoen and Uhlenbeck [14] showed that for any energy minimizing map Ñ } } 8 p q f : M N with f W 1,2 and image f is a compact subset of N, the Hausdorﬀ co-dimension of its singularity set is at least 3.

1.2 Sigma Model Approach » 1 Instead of considering the usual energy functional Epfq |df|2 dvol, which M 2 includes only the kinetic energy, we ﬁrst isometrically embed N in Rk and now allow the image of f to stay in a tubular neighborhood of the target manifold N, i.e.,

3 f : M Ñ Rk N. Imposing a potential energy W pfq, we hope the solution will be pushed to stay on the target manifold. Therefore, we consider » ¢ 1 Epfq |df|2 W pfq dvol. (1.6) M 2

We denote the new energy density as

1 epfq |df|2 W pfq. (1.7) 2

By variational calculus, § » ¢ ¢ § d § Bf Bf § Epfsq§ xdf, d y x∇W pfq, y dvol§ B B ds s 0 M s s s 0 » ¢ § (1.8) Bf § xd¦df ∇W pfq, y dvol§ . B M s s 0

Thus, f is a critical point of E only when f satisﬁes the elliptic equation

d¦df ∇W pfq 0. (1.9)

The induced Laplace equation of the energy density epfq of the critical map f is ¢ 1 ∆epfq ∆ |df|2 W pfq 2

¦ ¦ (1.10) ¡ |∇df|2 xdd df, dfy ¡ Ricpdf, dfq x∇W, d dfy ¡ xd∇W pfq, dfy

2 2 A B ¡ |∇df| ¡ Ricpdf, dfq ¡ |∇W | ¡ 2HesspW qABxdf , df y.

¦ In the last step, we apply the fact that f satisﬁes d df ∇W pfq 0. HesspW qAB is the Hessian matrix of W in the A, B direction. We also examine the corresponding gradient ﬂow. Now assume fpt, xq : M ¢ I Ñ

k R N is a family of functions such that fpt, xq (also denoted as ftpxq) satisﬁes

Bf t d¦df ∇W pf q 0, with fp0, xq f pxq smooth and its image on N. Bt t t 0 (1.11)

4 It can be easily shown that the energy Epftq decreases along the ﬂow » ¢ » § § B §B §2 d p q x ¦ p q ft y ¡ § ft § ¤ E ft d dft ∇W ft , B dvol § B § dx 0. (1.12) dt M t M t

Our focus has turned to the existence and regularity of ftpxq when t Ñ 8 and if so, whether f8pxq lim ftpxq will stay on the target manifold N. tÑ8

The induced heat equation of the energy density epftq of (1.11) is ¢ ¢ ¢ B B 1 ∆ epf q ∆ |df |2 W pf q Bt t Bt 2 t t

2 ¦ 9 ¦ ¡ |∇dft| xdd dft, dfty ¡ Ricpdft, dftq xdft, dfty x∇W pftq, d dfty

x p q 9 y ¡ p q x A By ∇W ft , ft Hess W AB dft , dft

¡ | |2 ¡ | p q|2 ¡ p q ¡ p q x A By ∇dft ∇W ft Ric dft, dft 2Hess W AB dft , dft . (1.13)

Bf Here f9 t and we apply the equation f9 d¦df ∇W pf q 0 in the last t Bt t t t step.

In Chen and Struwe ([1]), W pftq Kχpdistpft,Nqq. Here, χpxq is a smooth function, χpxq x when x RM , the lower bound of the injectivity radius of M, and χpxq is constant when x ¥ RM . distpft,Nq is the distance function between the image of ft and N. Since χpdistpft,Nqq is a bounded function, one can show that

2 both |∇W | and HesspW qAB can be bounded by some constant depending on W , and therefore, we have the inequality ¢ B ∆ epf q ¤ pC epf q C qepf q. (1.14) Bt t 1 t 2 t

Chen and Struwe [1] proved the following result

Theorem 4 (Chen-Struwe [1]). Suppose f : M ¢ R Ñ N is a limit of a sequence tfku with tfku solving the parabolic harmonic equation (1.11) and with uniformly 5 ﬁnite energy

Epfkqptq ¤ E0 8, for t ¡ 0,

2 in the sense that Epfq ¤ E0 almost everywhere and dfk Ñ df weakly in L pQq for

any compact set Q P M ¢ R . Then f solves the parabolic harmonic equation (1.11) in the classical sense and is regular on a dense open set whose complement Σ has locally ﬁnite Hausdorﬀ co-dimension 2 with respect to the parabolic metric.

This thesis began as an exploration of the sigma-model approach to harmonic maps in the context of Grassmannian target spaces. Grassmannians are of fundamental interest due to their role as classifying spaces. They are equipped with a wealth of explicitly computable algebraically defined potential energy functions, and they are positively curved. The latter feature means that we have no effective means for determining which maps are homotopic to stable harmonic maps and which are not. Perhaps in such an explicit example, one might find new criteria guaranteeing the existence or nonexistence of stable harmonic maps. It is convenient to identify the Grassmannian Grpk, Clq of k planes in Cl with the space of hermitian involutions (rather than hermitian projections). With this iden- tification, we note that the connected component of the space of rank l hermitian matrices with nonzero determinant which contains Grpk, Clq is homotopically equivalent to Grpk, Clq. Hence for topological purposes, it may be of interest to replace the target space Grpk, Clq with this connected component in the harmonic mapping problem. In making this replacement, we are led to modify the usual sigma model approach where one chooses a sequence of potential energies which in some limit force a map to lie on the desired target space, with a single repulsive potential W which simply prevents the map from entering a forbidden region - tf : detpfq 0u. In particular, we choose W so that W pfq 8 if and only if detpfq 0, and W is smooth elsewhere.

6 In this new problem, issues of the singularity of a minimizing map f become translated into possibly more accessible and explicit issues about singularities of W pfq. In fact, in order to show minimizers are smooth, one simply has to ﬁnd conditions under which the potential energy remains bounded. We have not yet found such conditions. In these early explorations, we have been led ﬁrst to sharpen the -regularity and monotonicity techniques which form the core of the regularity theory of harmonic maps and test them against explicit choices of algebraic potentials. For example we consider the potential

L pp 2L q¡1q Wb : trace f bI

where I is the l ¢ l identity matrix. We would prefer to set b 0, but for many arguments, b ¡ 0 is convenient. Thus we need eﬀective estimates in order to prevent a minimizer from wandering out of the desired component. Of course, we do not expect such a result to hold without additional (as yet undiscovered) hypotheses. So instead, we ﬁrst test our techniques by estimating the size of the bad set. 1 We obtain the following theorems of bounds of eLpfq |df|2 W Lpfq: b 2 b

Theorem 5 (See Theorem 20). Suppose f solves the elliptic equation

¦ Lp q p q d df Wb f 0 with E f ﬁnite. (1.15)

1 b L There exists , with C depending only on M, l, L, such that if there exists 0 2C R ρ, and » L 1 L Φ px0,Rq p1 νpr, xqqe pfq dvol ¤ 0, (1.16) b m¡2 p0 b R BRpx0q

¤ 3 then for any δ 4 ,

1 cpL, l, Mqb L Lp q ¤ sup eb f . (1.17) p q2¡p0 BδRpx0q δR 7 Here, νpr, xq and p0 are deﬁned as in Chapter 2 (2.10) and (2.31).

Theorem 6 (See Theorem 21). Suppose ft solves the parabolic equation

Bf t d¦df W Lpf q 0 with f P Grpk, Clq and Epf q ﬁnite. (1.18) Bt t b t 0 0

Here ft exists and is smooth for some T 8.

1 b L There exists with C depending only on M, l, L, such that if there exists 0 2C ? R with 0 R t0 ¡ t{2 and

» 2 » t0¡R 1 L 2 pp q q p qp q p q ¤ Ψ x0, t0 ,R ν G x0,t0 x, t eb ft ϕ dvol dt 0, (1.19) 2 p ¡ q 0 t0¡4R t0 t M

then for any δ depending only on M,R, we have

1 cpL, l, Mqb L Lp q ¤ sup eb f . (1.20) p q2¡2ν0 PδRpz0q δR

Here z0 px0, t0q, ν0 is the same ratio deﬁned in (2.10) and the theorem holds when

ν0 0.

When L 1, we also derive the absolute bound of ebpfq over the entire manifold

M. Here we denote sup ebpfq eM . M

2m ¡ 2m¡1 ¡ ¡ eM ¤ CpMqEbpfq 2m 2 b m 1 . (1.21)

(See (5.45)) The upper bound of the measure of the bad set deﬁned as " * 1 ΣL x P M|eLpfq ¥ (1.22) b b b is given by the following theorem

8 Theorem 7 (See Theorem 22). Fix a constant b ¡ 0. Let f be the solution to (1.15). ¨ L 2 p ¡ q L Deﬁne the set Σb as in (1.22). The Hausdorﬀ L 1 m 2 -measure of Σb at

1 1 scale b 2 2L (deﬁned in (5.56)) is bounded and independent of b.

9 2

Monotonicity Formulas

When an energy density e satisfies a partial differential inequality of the form ∆e ¤ Ce, Moser iteration allows one to obtain pointwise estimates for e in terms of the integral of e over balls. When studying nonlinear equations, the preceding equality is often replaced by an inequality of the form ∆e ¤ Cep, for some p ¡ 1. In such situations, Moser iteration must be supplemented by -regularity arguments. A crucial component of such arguments are monotonicity relations for rescaled local energies. In this section, we derive such monotonicity relations in the context of reaction diffusion problems. Our derivation is more flexible than the standard method (e.g., [12]); moreover, it clarifies how to find optimal monotonicity relations. We first introduce some basic notation. Denote by |x ¡ y| the geodesic distance between x and y in M. For x0 P M and R ¡ 0 let

BRpx0q tx P M||x ¡ x0| ¤ Ru , (2.1)

and

SRpx0q tx P M||x ¡ x0| Ru . (2.2) 10 For the parabolic case we need additional notation. Let z px, tq P M ¢ R. For a distinguished point z0 px0, t0q and R ¡ 0, let ( 2 PRpz0q z px, tq||x ¡ x0| ¤ R, |t ¡ t0| ¤ R , (2.3)

SRpz0q tz px, tq||x ¡ x0| R, t t0u , (2.4) and ( 2 2 TRpt0q z px, tq|t0 ¡ 4R ¤ t ¤ t ¡ R . (2.5)

Let M be a compact Riemannian manifold. Consider a function f : M Ñ Rk. Let W : Rk Ñ R be a function, which we will initially take to be smooth, but will later allow to be singular. Deﬁne the energy density

1 epfq : |df|2 W pfq. (2.6) 2

1 | |2 p q We will be interested in whether the kinetic term 2 df or the potential term W f dominates the energy density. Therefore we deﬁne the following ratios over geodesic

spheres SRpxq M:

³ | |2 p q fr dσ µpR, xq ³ SR x , (2.7) 1 |df|2 W pfq dσ SRpxq 2 and ³ p q p q W f dσ νpR, xq ³ SR x . (2.8) 1 |df|2 W pfq dσ SRpxq 2

We will see that nontrivial lower bounds for µpR, xq and νpR, xq allow us to sharpen estimates. We will need such lower bounds on balls of small, but not arbitrarily small radius. Hence we will ﬁx a scale ρ ¡ 0 and deﬁne

µpxq inf µpR, xq, νpxq inf νpR, xq. (2.9) R¤ρ R¤ρ

11 When necessary for clarity, we will write instead µρpxq to make the scale explicit. We also set

µ0,S : inf µpxq, ν0,S : inf νpxq. (2.10) xPS xPS

When S is clear from context, we will simply write µ0 and ν0.

It is obvious that 0 ¤ νpxq, ν0 ¤ 1. From [3],

2 lim µpR, xq . (2.11) RÑ0 m

2 Therefore, we have µpxq, µ ¤ . 0 m With these preliminaries, we now turn to the derivation of monotonicity relations.

2.1 Monotonicity Formula for Elliptic Equations

Let RM denote the lower bound of the injectivity radius of M.

Lemma 8. Suppose fpxq : M Ñ Rk is a smooth solution to the diﬀerential equation

d¦df W pfq 0. (2.12)

2ν p q pm ¡ 2qµ p q 0,BR x0 0,BR x0 Fix a scale ρ RM . Let p0 ¡ . Deﬁne 1 µ0,BRpx0q » ¢ ¡ ¡ 1 m p q CR 2 m p0 | |2 p q Φ R, x e R df ¡ W f dvol, (2.13) BRpxq 2 m 2

where m dimpMq, C is a constant which only depends on the geometry of M. Then ΦpR, xq is non-decreasing for any x P M, R ρ.

Proof. Without loss of generality, we assume x 0, the origin of a geodesic co- t jum p q p q ordinate system x j1, and denote BR 0 BR. For a 1-form ω, let e ω denote

12 exterior multiplication on the left by ω, and let e¦pωq denote its adjoint. Let r denote the radial function of the geodesic coordinates. Let us consider » xtd, e¦prdrqudf, dfy vol. (2.14) BR

We compute this quantity in two ways. First, by integration by parts, we have » » xtd, e¦prdrqudf, dfy dvol xdpe¦prdrqdfq, dfy dvol BR BR » » ¦ ¦ ¦ dpe prdrqdf ^ ¦dfq xe prdrqdf, d dfy dvol (2.15) BR BR » » r|e¦pdrqdf|2 dσ xe¦prdrqdf, d¦dfy dvol SR BR

Deﬁne

j ¦ Q epdx qe p∇jdrq. (2.16)

Q is a natural extension of the second fundamental form of the geodesic sphere to an endomorphism of diﬀerential forms. We can compute the action of Q on k-forms as follows:

x epdxiqe¦pdxiq epdrqe¦pdrq k ¡ epdrqe¦pdrq Q epdxjqe¦p∇ j drq ¡ Apxq, j r r r r (2.17)

where Apxq Oprq arises from connection terms and gij ¡ δij.

13 Next, we use (2.17) to compute (2.14),

¦ xt p qu y 2 d, e rdr df, df L pBRq ( j ¦ x p q p q y 2 e dx ∇j, e rdr df, df L pBRq ( ¨ j ¦ j ¦ x p qr p qs p q p q y 2 e dx ∇j, e rdr e dx , e rdr ∇j df, df L pBRq » ¢ B j ¦ 1 2 xepdx qe p∇ prdrqqdf, dfy 2p q r |df| dvol j L BR B BR r 2 » ¢ » » B 1 | |2 x p q ¦p q y x y rB df dvol e dr e dr df, df dvol r Qdf, df dvol BR r 2 BR BR » ¢ » » 1 r ptd, e¦pdrqu ¡ Qq |df|2 dvol |df|2 dvol rxApxqdf, dfy dvol BR 2 BR BR » ¢ » ¢ » 1 2 1 2 2 r pdιr ιrdq |df| dvol ¡ rQ |df| dvol |df| dvol BR 2 BR 2 BR » rxApxqdf, dfy dvol BR » ¢ » » 1 2 1 2 1 2 d rιr |df| dvol ¡ |df| dvol ¡ r |df| Q pdvolq BR 2 BR 2 BR 2 » » |df|2 dvol rxApxqdf, dfy dvol BR BR » ¢ » » 1 1 1 R |df|2 dσ ¡ |df|2 dvol ¡ r |df|2Q pdvolq SR 2 BR 2 BR 2 » » |df|2 dvol rxApxqdf, dfy dvol. BR BR (2.18)

We ﬁrst compute the anticommutator in (2.15) in terms of Q and then integrate. ¢ m ¡ 1 Qpdvolq Hdvol Apxq dvol, (2.19) r

14 where, under an orthonormal frame,

¸m H x∇k pBrq , Bky. (2.20) k

Applying (2.19) on (2.18), we get

¦ x p q y 2 d, e rdr df, df L pBRq » » ¢ » 1 m rApxq R |df|2 dσ ¡ |df|2 dvol p|df|2 rxApxqdf, dfyq dvol SR 2 BR 2 2 BR » ¡ © » m 1 1 ¡ Bpxq |df|2 dvol R |df|2 dσ. BR 2 SR 2 (2.21)

rApxq where Bpxq|df|2 ¡ |df|2 rxApxqdf, dfy Opr2q|df|2. 2 Combining (2.15) and (2.18), we have

» ¡ © » » m 1 1 ¡ Bpxq |df|2 dvol xe¦prdrqdf, d¦dfy dvol R |e¦pdrqdf|2 ¡ |df|2 dσ. BR 2 BR SR 2 (2.22) Since f satisﬁes the elliptic equation

d¦df ∇W pfq 0,

we can replace d¦df by ¡∇W pfq on the right-hand side of (2.22) to get » » xe¦prdrqdf, d¦dfy dvol ¡ xe¦prdrqdf, ∇W pfqy dvol BR BR » » » (2.23) B ¡ © ¡ ¡ ˆp q rB W dvol RW dσ m B x W dvol. BR r SR BR where Bˆpxq Opr2q consists of geometric terms. In the last step we use Stokes’ Theorem.

15 Therefore » ¡ © ¡ © » m 1 1 ¡ Bpxq |df|2 ¡ m Bˆpxq W dvol R |e¦pdrqdf|2 ¡ |df|2 ¡ W dσ. BR 2 SR 2 (2.24) Using the ratios µ and ν deﬁned in (2.7) and (2.8), (2.24) can be rewritten

» ¡ © ¢ » ¢ 1 1 m ¡ 2 2νp1 Bˆpxqq ¡ 2Bpxq |df|2 W dvol R p1 ¡ µq |df|2 W dσ. BR 2 SR 2 (2.25) Let φpRq be a function to be determined later. Multiply (2.25) by φ1pRq and integrate from σ to τ to obtain » ¢ » ¢ 1 1 φpτq pm ¡ 2 2νq |df|2 W dvol ¡ φpσq pm ¡ 2 2νq |df|2 W dvol Bτ 2 Bσ 2 » ¢ » » ¢ 1 τ 1 ¡ φprqpm ¡ 2 2νq |df|2 W dvol φ1pRq Dpxq |df|2 W dvol dR Bτ zBσ 2 σ BR 2 » ¢ 1 φ1prqr p1 ¡ µq |df|2 W dvol, Bτ zBσ 2 (2.26) where Dpxq 2νBˆpxq ¡ 2Bpxq Opr2q. Rearranging the order, we get » ¢ » ¢ 1 1 φpτq pm ¡ 2 2νq |df|2 W dvol ¡ φpσq pm ¡ 2 2νq |df|2 W dvol Bτ 2 Bσ 2 » ¢ 1 rφprq pm ¡ 2 2νq φ1prqrp1 ¡ µqs |df|2 W dvol Bτ zBσ 2 » » ¢ τ 1 φ1pRq Dpxq |df|2 W dvol dR. σ BR 2 (2.27)

16 Now pick φprq to be ¢» ¢» τ ¡ ¡ τ ¡ p ¡ q p q p q 2 m 2ν p q 2 m ¡ 2ν m 2 µ φ τ φ σ exp p ¡ q dr φ σ exp p ¡ q dr . σ r 1 µ σ r r 1 µ (2.28) Then φprq pm ¡ 2 2νq φ1prqrp1 ¡ µq 0. (2.29)

Therefore, » ¢ » ¢ 1 1 φpτq pm ¡ 2 2νq |df|2 W dvol ¡ φpσq pm ¡ 2 2νq |df|2 W dvol Bτ 2 Bσ 2 » » ¢ τ 1 φ1pRq Dpxq |df|2 W dvol dR. σ BR 2 (2.30)

Let

2ν0 pm ¡ 2qµ0 p0 . (2.31) 1 ¡ µ0

When m ¡ 2,

2ν0 pm ¡ 2qµ0 p0 ¥ 0. (2.32) 1 ¡ µ0

We have ¡ © ¡ ¡ τ 2 m p0 φpτq ¤ φpσq . (2.33) σ

17 Since φ1pRq ¤ 0, using

pm ¡ 2 2νq |φ1pRq|R ¡φ1pRqR φpRq, (2.35) 1 ¡ µ

we get » » » » τ τ ¡ 1p q p q ¥ ¡ p q m 2 2ν φ R D x e dvol dR C φ R ¡ e dvol dR σ BR σ BR 1 µ » » τ ¥ ¡ 2C φpRq pm ¡ 2 2νqe dvol dR σ BR » » ¢ τ 1 ¡ 2C φpRq pm ¡ 2 2νq |df|2 W pfq dvol dR. σ BR 2 (2.36)

Deﬁne » ¢ 1 hpRq φpRq pm ¡ 2 2νq |df|2 W pfq dvol. BR 2

We combine (2.30) and (2.36) to get » τ hpτq ¡ hpσq ¥ ¡2C hpRq dR. (2.37) σ

Dividing both sides by τ ¡ σ and taking the limit τ Ñ σ, this becomes

h1pσq ¥ ¡2Chpσq.

We solve h to get e2Cτ hpτq ¥ e2Cσhpσq, which is » » e2Cτ φpτq pm ¡ 2 2νqe dvol ¥ e2Cσφpσq pm ¡ 2 2νqe dvol. (2.38) Bτ Bσ

18 Combining (2.30), (2.33), and (2.38), we get » ¢ » ¢ ¡ ¡ 1 ¡ ¡ 1 e2Cτ τ 2 m p0 pm¡2 2νq |df|2 W dvol ¥ e2Cσσ2 m p0 pm¡2 2νq |df|2 W dvol. Bτ 2 Bσ 2 (2.39) That is, » ¢ » ¢ ¡ ¡ 1 m ¡ ¡ 1 m 2Cτ 2 m p0 | |2 ¥ 2Cσ 2 m p0 | |2 e τ df ¡ W dvol e σ df ¡ W dvol. Bτ 2 m 2 Bσ 2 m 2 (2.40) When M is a Euclidean space, C 0 since Dprq 0.

2.2 Monotonicity Formula for Parabolic Equations

k Lemma 9. Suppose fpt, xq ftpxq : M ¢ I Ñ R is a regular solution to the diﬀerential equation Bf t d¦df W pf q 0, (2.41) Bt t t

where I is a subset of the interval when ftpxq exists in short time. Let t0 P I, and

2 2 0 R R0 RM such that rt0 ¡ 4R , t0 ¡ R s I.

Deﬁne » 1 2 φpR, x0, t0q Gpx ,t qpx, tqepfqϕ pxq dvol, (2.42) 2¡2ν0 0 0 R M and ? » ¡ 2 » ¡ 2 » t0 R p ¡ q t0 R φ t0 t 1 2 p q p qp q p q p q Ψ R, x0, t0 dt ν G x0,t0 x, t e f ϕ x dvol dt, 2 ¡ 2 p ¡ q 0 t0¡4R t0 t t0¡4R t0 t M (2.43)

where ν0 is the same ratio deﬁned in (2.10), ¢ 2 1 |x ¡ x0| Gp qpx, tq exp ¡ (2.44) x0,t0 m p ¡ q p4πpt0 ¡ tqq 2 4π t0 t 19 is the Euclidean backward Gaussian, and ϕpxq ϕp|x¡x0|q is a compactly supported

1 4 radial function on BR px0q such that ϕpRq 1 for R RM {2 and |ϕ pRq| ¤ . M RM Then we have the following monotonicity formulas for φpRq and ΨpRq: ¨ p q ¤ p 2 ¡ 2q p q ˆ p p qqp 2 ¡ 2q φ R exp c R0 R φ R0 CE f0 x R0 R , (2.45)

and p q ¤ p p 2 ¡ 2qq p q ˆ p p qqp 2 ¡ 2q Ψ R exp c R0 R Ψ R0 CE f0 x R0 R , (2.46) ? ˆ for any 0 R R0 t0 ¡ t{2. Here constants c, C only depend on the geometry of M, and m dimpMq. E is the energy of ft over M: » ¢ 1 2 Epftq |dft| W pftq dvol. (2.47) M 2

It is clear that when νpR, xq 0, which implies ν0 0, the above results (2.45) and (2.46) reduce to results of Struwe [1, Lemma 4.2].

Proof. Without loss of generality, we assume x0 0. Let ϕprq be the radial cut-oﬀ function in BRp0q for some R RM . Consider » xtd, e¦pdY qudf, ϕ2dfy dvol, (2.48) M where Y is another radial function. Our goal is to ﬁnd a Y so that (2.48) produces new monotonicity result. We

20 calculate this quantity in two ways. On one hand, » xtd, e¦pdY qudf, ϕ2dfy dvol M » ( j ¦ 2 x epdx q∇j, e pdY q df, ϕ dfy dvol M » ( ¨ j ¦ j ¦ 2 x epdx qr∇j, e pdY qs epdx q, e pdY q ∇j df, ϕ dfy dvol M » » j ¦ 2 2 xepdx q∇j pYre pdrqq df, ϕ dfy dvol Yrx∇rpdfq, ϕ dfy dvol M M » » 2 j ¦ 2 j ¦ ∇jpYrqϕ xepdx qe pdrqdf, dfy dvol Yrϕ xepdx qe p∇jdrqdf, dfy dvol M M » ¢ B 1 | |2 2 Yr B df ϕ dvol M r 2 » » » ¢ B 1 | ¦p q |2 2 x y 2 | |2 2 Yrr e dr df ϕ dvol Yr Qdf, df ϕ dvol Yr B df ϕ dvol M M M r 2

I II III. (2.49)

For the second term II, we use

1 ¡ epdrqe¦pdrq Q epdxjqe¦p∇ drq Apxq, (2.50) j r where Apxq Oprq, to replace Q: » ¢ 1 2 ¦ 2 2 II Yr p|df | ¡ |e pdrqdf| q xApxqdf, dfy ϕ dvol. (2.51) M r

Integration by parts in the third term III, we have » ¢ » ¢ » B 1 1 2 | |2 x 2 | |2 y ¡ | |2 1 III Yrϕ B df dvol dY, d ϕ df dx Yr df ϕϕ dvol M r 2 M 2 M » ¢ » ¦ 1 2 2 2 1 d dY |df| ϕ dvol ¡ Yr|df| ϕϕ dvol. M 2 M (2.52) 21 Therefore, » xtd, e¦pdY qudf, ϕ2dfy dvol M I II III » » ¢ ¦ 2 2 1 2 ¦ 2 2 Yrr|e pdrqdf| ϕ dvol Yr p|df| ¡ |e pdrqdf| q xApxqdf, dfy ϕ dvol M M r » ¢ » ¦ 1 2 2 2 1 d dY |df| ϕ dvol ¡ Yr|df| ϕϕ dvol. M 2 M (2.53)

On the other hand, we may apply integration by parts at the very beginning and

¦ 2 get another expression about xtd, e pdY qudf, ϕ dfyL2pMq, » xtd, e¦pdY qudf, ϕ2dfy dvol M » ¨ xe¦pdY qdf, d¦ ϕ2df y dvol (2.54) M » » xe¦pdY qdf, d¦dfyϕ2 dvol ¡ xe¦pdY qdf, 2ϕϕ1prqe¦pdrqdfy dvol. M M

22 For the ﬁrst term, we replace d¦df using (2.41) to get » xe¦pdY qdf, d¦dfyϕ2 dvol M » ¦ 2 ¡ Yrxe pdrqdf, ft ∇W yϕ dvol M » » 2 ¦ 2 ¡ xdf, epdY qftyϕ dvol ¡ Yre pdrqdW ϕ dvol M M » » » » 2 2 2 1 ¡ xdf, dpY ftqyϕ dvol xdf, Y dpftqyϕ dvol ¡ xdY, dpϕ W qy dvol YrW 2ϕϕ dvol M M M M » » » 2 2 2 ¡ xdf, dpY ftqyϕ dvol ¡ xdf, Y dpftqyϕ dvol 2 xdf, Y pdftqyϕ dvol M M M » » ¦ 2 1 ¡ d dY W ϕ dx YrW 2ϕϕ dvol M M » ¢ » » d 1 2 2 2 2 ¡ Y |df| ϕ dvol ¡ 2 xdf, rd, Y sftyϕ dvol xdf, dpY ftϕ qy dvol M dt 2 M M » » » 1 ¦ 2 1 ¡ Y xfr, fty2ϕϕ dvol ¡ d dY W ϕ dvol YrW 2ϕϕ dvol M M M » ¢ » » d 1 2 2 2 ¦ 2 ¡ Y |df| W ϕ dvol ¡ 2 Yrxfr, ftyϕ dvol Y xd df, ftyϕ dvol M dt 2 M M » » » » 1 2 ¦ 2 1 ¡ Y xfr, fty2ϕϕ dvol Y x∇W, ftyϕ dvol ¡ d dY W ϕ dx YrW 2ϕϕ dvol M M M M » ¢ » ¢ » d 1 2 2 1 2 2 ¦ 2 ¡ Y |df| W ϕ dvol Yt |df| W ϕ dvol ¡ d dY W ϕ dvol dt M 2 M 2 M » » » » 2 2 2 1 1 ¡ Y |ft| ϕ dvol ¡ 2 Yrxfr, ftyϕ dvol ¡ Y xfr, fty2ϕϕ dvol YrW 2ϕϕ dvol. M M M M (2.55)

In the last step, we use the equation (2.41) again.

23 Therefore, » xtd, e¦pdY qudf, ϕ2dfy dvol M » ¢ » ¢ » d 1 2 2 1 2 2 ¦ 2 ¡ Y |df| W ϕ dvol Yt |df| W ϕ dvol ¡ d dY W ϕ dvol dt M 2 M 2 M » » » 2 2 2 1 ¡ Y |ft| ϕ dvol ¡ 2 Yrxfr, ftyϕ dvol ¡ Y xfr, fty2ϕϕ dvol M M M » » 1 2 1 YrW 2ϕϕ dvol ¡ Yr|fr| 2ϕϕ dvol. M M (2.56)

Equating (2.53) and (2.56), we get » ¢ » ¢ » d 1 2 2 ¦ 1 2 2 2 2 0 Y |df| W ϕ dvol pd dY ¡ Ytq |df| W ϕ dvol Y |ft| ϕ dvol dt M 2 M 2 M » » ¢ 2 Yr 2 2 2 Yrxfr, ftyϕ dvol Yrr ¡ |fr| ϕ dvol M M r » ¢ » » Yr 2 2 1 2 1 |df| YrxApxqdf, dfy ϕ dvol Y xfr, fty2ϕϕ dvol Yr|fr| 2ϕϕ dvol M r M M » ¢ 1 2 1 ¡ Yr |df| W 2ϕϕ dvol M 2 » ¢ » ¢ ¢ d 1 2 2 ¦ 2Yr 1 2 2 Y |df| W ϕ dvol d dY ¡ Yt |df| W ϕ dvol dt M 2 M r 2 » § § » ¢ » § §2 2 § Yr § 2 Yr Yr 2 2 Yr 2 Y §ft fr§ ϕ dvol Yrr ¡ ¡ |fr| ϕ dvol ¡ 2W ϕ dvol M Y M r Y M r » » ¢ Yr 1 1 2 1 Y xfr, ft fry2ϕϕ dvol ¡ Yr |df| W 2ϕϕ dvol M Y M 2 » 2 YrxApxqdf, dfyϕ dvol. M (2.57)

24 Recall νpR, xq deﬁned in (2.8). The above equality can be rewritten as » ¢ d 1 0 Y |df|2 W ϕ2 dvol dt M 2 » ¢ ¢ ¦ 2Yr 1 2 2 d dY ¡ Yt p1 ¡ νq |df| W ϕ dvol M r 2 » § § » ¢ § §2 2 § Yr § 2 Yr Yr 2 2 Y §ft fr§ ϕ dvol Yrr ¡ ¡ |fr| ϕ dvol (2.58) M Y M r Y » » ¢ Yr 1 1 2 1 Y xfr, ft fry2ϕϕ dvol ¡ Yr |df| W 2ϕϕ dvol M Y M 2 » 2 YrxApxqdf, dfyϕ dvol. M

To simplify the equation, we choose Y so that

Y Y 2 Y ¡ r ¡ r 0. (2.59) rr r Y

One family of solutions is given by ¨ Y pr, tq gptq exp hptqr2 , (2.60) for any positive function gptq and arbitrary function hptq. 2Y With this Y , we now simplify the expression d¦dY ¡Y r p1¡νpr, xqq as much t r as possible. n ¡ 1 Notice d¦dY ¡Y ¡ Y Bprq with Bprq Opr2qY . Since Bprq will lead rr r r m ¡ 1 2Y to a lower order term, we focus on eliminating ¡Y ¡ Y r p1 ¡ νpr, xqq. rr r r r A detailed computation shows

m ¡ 1 2Y ¡ Y ¡ Y r p1 ¡ νpr, xqq ¡4h2r2Y ¡ pm ¡ 2 2νpr, xqq2hY. (2.61) rr r r r

25 Since m ¡ 2 and νpr, xq is small, we require hptq ¤ 0 so that we may eliminate m ¡ 1 2Y ¡Y ¡ Y r p1 ¡ νpr, xqq. Therefore, we pick rr r r r ¨ Y pr, tq gptq exp hptqr2 (2.62) with gptq ¡ 0 and hptq ¤ 0. We remove the dependence of the r and x in νpr, xq to simplify our computations.

By the deﬁnition of ν0, we then get » ¢ ¢ ¡ m 1 2Yr 1 2 2 ¡Yrr ¡ Yr p1 ¡ νpr, xqq |df| W ϕ dvol M r r 2 » ¢ ¢ (2.63) ¡ m 1 2Yr 1 2 2 ¥ ¡Yrr ¡ Yr p1 ¡ ν0q |df| W ϕ dvol. M r r 2

(2.58) becomes inequality (Yr ¤ 0): » ¢ d 1 0 ¥ Y |df|2 W ϕ2 dvol dt M 2 » ¢ ¢ ¡ m 1 2Yr 1 2 2 ¡Yt ¡ Yrr ¡ Yr p1 ¡ ν0q |df| W ϕ dvol M r r 2 » § § » » § §2 § Yr § 2 2 2 Y §ft fr§ ϕ dvol BprqW ϕ dvol YrxApxqdf, dfyϕ dvol M Y M M » » ¢ Yr 1 1 2 1 Y xfr, ft fry2ϕϕ dvol ¡ Yr |df| W 2ϕϕ dvol. M Y M 2 (2.64)

Imposing the equation on Y

m ¡ 1 2Y ¡ Y ¡ Y ¡ Y r p1 ¡ ν q 0, (2.65) t rr r r r 0

we get ¢ g1ptq ¡4h2ptqr2 ¡ h1ptqr2 ¡ 2h ¡ pm ¡ 3 2ν q2h ¡ Y 0. (2.66) 0 gptq 26 Here Y ¡ 0 since gptq ¡ 0. h and g only depend on t. We conclude that

g1ptq 4h2ptq h1ptq 0, pm ¡ 2 2ν q2hptq 0. (2.67) 0 gptq

From the ﬁrst equation, we get

1 hptq , (2.68) 4pt ¡ Cq

for any constant C ¡ t. We then solve for gptq by plugging this h into the second equation:

1¡ m ¡ν gptq D |t ¡ C| 2 0 , (2.69)

for any D ¡ 0.

Here we pick D 1, C t0 ¡ t, then we get ¢ m 2 1¡ ¡ν0 r Y px, tq pt0 ¡ tq 2 exp ¡ . (2.70) 4pt0 ¡ tq

We can simplify (2.64) into » ¢ » § § » § §2 d 1 2 2 § Yr § 2 2 0 ¥ Y |df| W ϕ dvol Y §ft fr§ ϕ dvol YrxApxqdf, dfyϕ dvol dt M 2 M Y M » » » ¢ 2 Yr 1 1 2 1 BprqW ϕ dx Y xfr, ft fry2ϕϕ dvol ¡ Yr |df| W 2ϕϕ dvol M M Y M 2 » ¢ » § § § §2 d 1 2 2 § Yr § 2 Y |df| W ϕ dvol Y §ft fr§ ϕ dvol dt M 2 M Y

I II III IV. (2.71)

2 Since R ¤ pt0 ¡ tq{4, we can easily show that §» ¢ § » ¢ § r2 1 § 1 | | § | |2 2 § ¤ | |2 2 I § p ¡ qY df W ϕ dvol§ c Y df W ϕ dvol, M 2 t0 t 2 M 2 (2.72)

27 and » ¢ 1 2 |II| ¤ C |df| W dvol ¤ CEpf0q. (2.73) M 2

2 2 For III, since R ¤ t0 ¡ t ¤ 4R with R RM , we have » § § » § §2 1 § Yr § 2 2 1 2 |III| ¤ Y §ft fr§ ϕ dvol C Y |fr| |ϕ | dvol 2 M Y M » § § » § §2 1 § Yr § 2 2 1 ¤ | | r { s (2.74) Y §ft fr§ ϕ dvol c Y fr 2 χ RM 2,RM dvol 2 M Y M RM » § § » § §2 1 § Yr § 2 2 ¤ Y §ft fr§ ϕ dvol c Y eϕ dvol CEpf0pxqq, 2 M Y M

z where χrRM {2,RM s is the indicator of the region BRM BRM {2.

» ¢ ¨ 1 1 ¡m{2¡ν0 2 2 |IV| ¤C |rϕ |ϕpt0 ¡ tq exp ¡r {4pt0 ¡ tq |df| W dvol M 2 » (2.75) 2 ¤c Y eϕ dvol CEpf0pxqq. M

Here r is only deﬁned when r is close to the boundary of M since that’s the only place where ϕ1 is supported. Hence » ¢ » § § » § § § §2 § § d 1 2 2 § Yr § 2 1 § Yr § 2 0 ¥ Y |df| W ϕ dvol Y §ft fr§ ϕ dvol ¡ Y §ft fr§ ϕ dvol dt M 2 M Y 2 M Y

¡ CEpf0pxqq » ¢ » ¢ d 1 2 2 1 2 2 ¥ Y |df| W ϕ dvol ¡ c Y |df| W ϕ dvol ¡ CEpf0pxqq. dt M 2 M 2 (2.76)

Denote » ¢ » ¢ 1 ¡ 1 p q | |2 2 p ¡ q1 ν0 p q | |2 2 φ t Y df W ϕ dvol t0 t Gz0 x, t df W ϕ dvol. M 2 M 2 (2.77)

28 Integrate (2.76) in the t direction between rt1, t2s, where 0 t1 t2 t0, we have ˜ φpt2q ¤ exp pcpt2 ¡ t1qq φpt1q CEpfp0, xqq pexp pcpt2 ¡ t1qq ¡ 1q (2.78) ˆ exp pcpt2 ¡ t1qq φpt1q CEpfp0, xqqpt2 ¡ t1q.

¡ 2 ¡ 2 Substituting t2 t0 R , and t1 t0 R0 gives (2.45).

Deﬁne ¢ » ¡ 2 » ¡ 2 » t0 R 1 t0 R 1 1 p q p q | |2 2 Ψ R φ t dt ν Gz0 df W ϕ dvol dt. 2 p ¡ q 2 p ¡ q 0 t0¡4R t0 t t0¡4R M t0 t 2 (2.79) We have d ¡2R ¡8R 2 ¨ ΨpRq φpt ¡ R2q ¡ φpt ¡ 4R2q φpt ¡ 4R2q ¡ φpt ¡ R2q dR R2 0 4R2 0 R 0 0 2 pφpt q ¡ φpt qq . R 1 2 (2.80)

Now we turn to prove (2.46). We have » » » t2 t2 t2 p q p q¡ p q 1p q ¤ p q p ¡ q φ t p ¡ q φ t2 φ t1 φ t dt cφ t CE0 dt CE0 t2 t1 c ¡ t0 t dt. t1 t1 t1 t0 t (2.81)

So » d 2CE c t2 φptq ΨpRq ¥ ¡ 0 pt ¡ t q ¡ pt ¡ tq dt dR R 2 1 R t ¡ t 0 t1 0 (2.82) ˜ ¥ ¡ CE0R ¡ cR˜ ΨpRq.

Integrating from R to R (assuming 0 R R R ), we have 0 0 ¡ M © c˜ p 2¡ 2q c˜ p 2¡ 2q 2 R0 R ˆ R0 R ΨpRq ¤ΨpR0qe CEpfp0, xqq e 2 ¡ 1 (2.83) ¤ p p 2 ¡ 2qq p q ˆ p p qqp 2 ¡ 2q exp c R0 R Ψ R0 CE f 0, x R0 R .

29 3

Moser Iteration

In this chapter we recall the Moser iteration arguments that allow us to derive L8 bounds for smooth functions e satisfying either an elliptic inequality

∆e ¤ Ce, or a parabolic inequality d p ∆qe ¤ Ce. dt

The contents of this section are not original, but will be used in the next section to obtain cleaner and more eﬀective -regularity arguments than we have found in the literature (see [13] and [1]).

3.1 Moser Iteration for Elliptic subsolutions

Lemma 10. Let M be a compact manifold. Let e be a positive function from M Ñ R1 satisfying

∆e ¤ C0e, (3.1) where ∆ is the Laplace-Beltrami operator on M.

30 Then

¢ m » r 16 2 }e} 8p p qq ¤ CpMq C0 e dvol, (3.2) L BδR x0 p ¡ q2 R δR BRpx0q

r for 0 ¤ δ 1 and any geodesic ball BRpx0q P M where R RM . Here CpMq is a constant only depends on M. Particularly, when δ 0, (3.2) implies a single point estimate of epfq:

¢ m » 2 p qp q ¤ ˜p q 16 e f x0 C M C0 2 e dvol. (3.3) R BRpx0q

If sup epfq exists and is achieved at x0, then M

¢ m » 2 p q ¤ ˜p q 16 sup e f C M C0 2 e dvol, (3.4) M R BRpx0q

for any R RM .

Proof. Without loss of generality, we may assume x0 0 and denote BRp0q BR. p q p q ¤ 3 We define the radial cut-off function η s such that η s 1 for s 2 , and 1 ηpsq 0 for s ¥ 2, with |η psq| ¤ 4. Define radial function ηkpxq ηkp|x|q ¢ ¢ ¢ |x| 1 η 1 2k¡1 ¡ 1 . We can show that η pxq 1 in B where r R 1 , R k rk k 2k 2k 1 and ηpxq is supported in B , |dη | ¤ , and dη is supported in the set rk¡1 k R k z Brk¡1 Brk .

31 pk¡1 Multiply (3.1) by ηke on both sides, and integrate by parts to get » ¡ ¡ | pk |2 ¥x 2 2pk 1y x p 2 2pk 1qy C0 ηke ∆e, ηke L2 de, d ηke L2 BR » ¡ ¡ ¡ xp pk 1q p pk 1qy 2 2pk 2| |2 2 ηke de, ed ηke L2 ηke de BR » »

pk pk pk¡1 pk¡1 xdpηke q, dpηke qydvol ¡ xdpηke qe, dpηke qeydvol BR BR ¡ 2 } p pk q}2 ¡ p pk 1q d ηke 2p q d ηke e 2 . L BR L pBRq (3.5)

Therefore, ¡ 2 } p pk q}2 ¤ } pk }2 p pk 1q d ηke 2p q C0 ηke 2p q d ηke e 2 L BR L BR L pBRq ¡ 2 } pk }2 pk p ¡ q pk 1 C0 ηke 2p q e dηk pk 1 ηke de 2 L BR L pBRq p ¡ 1 2 pk 2 k pk pk C }η e } 2 η dpe q e dη 0 k L pBRq k k p 2 k L pBRq p ¡ 1 1 2 (3.6) pk 2 k pk pk C }η e } 2 dpη e q e dη 0 k L pBRq k k p p 2 k k L pBRq ¢ p ¡ 1 2 pk 2 k pk 2 C }η e } 2 }dpη e q} 2 0 k L pBRq k L pBRq pk p ¡ 1 1 k pk pk pk 2 2 xdpη e q, e dη y 2p q }e dη } 2 . 2 k k L BR 2 k L pBRq pk pk

We apply Cauchy-Schwarz to the cross term to get,

2p ¡ 1 p ¡ 1 1 k pk 2 pk 2 k pk 2 pk 2 }dpη e q} 2 ¤ C }η e } 2 }dpη e q} 2 }e dη } 2 . 2 k L pBRq 0 k L pBRq 2 k L pBRq 2 k L pBRq pk pk pk (3.7) Therefore,

1 1 pk 2 pk 2 pk 2 }dpη e q} 2 ¤ C }η e } 2 }e dη } 2 . (3.8) k L pBRq 0 k L pBRq 2 k L pBRq pk pk 32 That is,

pk 2 pk 2 pk 2 }dpη e q} 2 ¤C p }η e } 2 }e dη } 2 k L pBRq 0 k k L pBRq k L pBRq (3.9) 4k 1 pk 2 pk 2 ¤C0pk }ηke } 2p q }e χk¡1} 2p q , L BR r2 L BR

z where χk¡1 is the indicator function of Brk¡1 Brk . Since M is compact, from the Sobolev inequality [7, Theorem 2.6] we have

pk pk }dpη e q} 2 ¥ CpMq }η e } 2m , (3.10) k L pBRq k m¡2 L pBRq where CpMq is a constant depending only M.

m Setting p1 1 and pk 1 pk m¡2 , we have » } }2pk 2pk ¥ } pk }2 C0pk e 2pk p q C0pk e dvol C0pk ηke L2pB q L Brk¡1 R Brk¡1

2k 2 pk 2 pk 2 ¥ }dpηke q} 2 ¡ }e χk¡1} 2p q L r2 L BR 4k 1 pk 2 pk 2 ¥CpMq }ηke } 2m ¡ }e χk¡1} 2p q L m¡2 r2 L BR (3.11) £» m¡2 m k 1 m 4 2 2pk m¡2 pk CpMq e dvol ¡ }e χk¡1} 2p q r2 L BR Brk

£» 2pk 2pk 1 4k 1 2pk 1 pk 2 CpMq e dvol ¡ }e χk¡1} 2p q . r2 L BR Brk

Rearranging, we obtain ¢ 4 k } }2pk ¥ p q } }2pk C0pk 4 e 2p C M e 2p . (3.12) 2 L k pBr q L k 1 pB q r k¡1 rk

That is,

¢ 1 C p 4 pk 0 k k } }2 ¥ } }2 4 e L2pk pB q e L2pk 1 pB q . (3.13) CpMq CpMqr2 rk¡1 rk 33 Iterating, we get

8 ¢ 1 ¹ p C0pk k 4 k 2 2 4 }e} 2p q ¥ }e} 8p q . (3.14) CpMq CpMqr2 L B2R L BR k1

¢ ¡ m k 1 Since p , p ¤ 4k¡1. The product factor on the left-hand side of k m ¡ 2 k (3.14) can be bounded by

¢ ¢ 1 ¹8 ¹8 k C p 16 4 16 pk 0 k 4k¡1 ¤ C CpMq r2 CpMq 0 r2 k1 k1

¢ m ¢ m m2 1 2 16 2 4 4 C (3.15) CpMq 0 r2

¢ m 16 2 CrpMq C . 0 r2

Hence,

¢ m » 2 2 16 2 }e} 8 ¤ C˜pMq C e dvol. (3.16) L pBRq 0 2 R B2R

} } 2 } } 2 To pass from a bound in terms of e L pB2Rq to one in terms of e L pBRq, we ¨ 1¡δ ﬁrst consider rk R δ 2k . This changes the bound of the derivative of ηk to 2 ¤ 2k |dη | ¤ . Then (3.16) will turn into k Rp1 ¡ δq

¢ m » 16 2 sup e2 ¤ C˜pMq C e2 dvol. (3.17) 0 p ¡ q2 xPBδRpx0q R δR BRpx0q

To show that sup e can be bounded by its L1 norm, we use the following lemma from [2, Lemma 4.1]

Lemma 11. Let ϕptq be a bounded nonnegative function deﬁned on the interval rT0,T1s, where 0 ¤ T0 T1. Suppose that for any T0 ¤ t s ¤ T1, ϕ satisﬁes

A ϕptq ¤ θϕpsq B, (3.18) ps ¡ tqα 34 where θ, A, B and α are nonnegative constants, θ 1. Then A ϕpρq ¤ C B , @T ¤ ρ R ¤ T , (3.19) pR ¡ ρqα 0 1 where C depends only on α, θ.

We apply this lemma by setting t δR, s R, and ϕptq sup e. By Cauchy- Btpx0q Schwarz and (3.17), we get

ϕptq sup e BδRpx0q

£ m d» 16C˜ 4 ¤ C C˜ e2 dvol 0 p ¡ q2 R δR BRpx0q

£ m d» b 16C˜ 4 ¤ sup e C C˜ e dvol 0 p ¡ q2 (3.20) BRpx0q R δR BRpx0q

£ m » 1 16C˜ 2 ¤ sup e 2 C˜ e dvol 0 p ¡ q2 2 BRpx0q R δR BRpx0q » 1 A ϕpsq e dvol, p ¡ qm 2 s t BRpx0q

˜ 2 ˜ where A can be taken as a constant slightly larger than C0pR ¡ δRq 16C since pR ¡ δRq2 is bounded. Applying the lemma, we have

» A ϕptq ¤ C e dvol . (3.21) p ¡ qm s t BRpx0q

That is,

¢ m » C pMq 2 sup e ¤ C pMqC 2 e dvol. (3.22) 1 0 p ¡ q2 BδRpx0q R δR BRpx0q

35 In particular, we have the single point estimate at the center x0 (a special case when δ 0):

¢ m » p q 2 p qp q ¤ p q C2 M p q e f x0 C1 M C0 2 e f dvol. (3.23) R BRpx0q

If sup epfq epx1q, then from the Moser iteration, we can easily get the result M

¢ m » p q 2 p q ¤ p q C2 M p q sup e f C1 M 2 e f dvol, (3.24) M R BRpxq

for x P M such that epfqpxq sup epfq and any R RM with BRpxq M. M

3.2 Moser-Harnack Iteration for Parabolic Subsolutions

Lemma 12. Let z0 px0, t0q P M ¢ I with M compact. Suppose a positive function e : M Ñ R1 satisﬁes ¢ B ∆ e ¤ C e in P pz q, (3.25) Bt 0 R 0

for some C0 ¥ 0 where it may depend on e, and PRpz0q is deﬁned to be the cylinder ( 2 PRpz0q px, tq P M ¢ I ||x ¡ x0| ¤ R, |t ¡ t0| ¤ R (3.26)

for R RM .

Then

¢ m 2 » C pMq 2 sup e ¤ C pMqC 2 e dvol dt, (3.27) 1 0 p ¡ q2 PδRpz0q R δR PRpz0q

where C1pMq and C2pMq only depend on M.

36 In particular, when δ 0, we have

¢ m 2 » p q 2 p qp q ¤ p q C2 M e f z0 C1 M C0 2 e dvol dt. (3.28) R PRpz0q

If sup epfq exists, then M¢I

¢ m 2 » 2 C2pMq sup epfq ¤ C1pMqC0 e dvol dt, (3.29) ¢ 2 M I R PRpz0q

for any R RM .

Proof. Without loss of generality, we may assume z0 px0, t0q p0, 0q. We denote

PRp0q PR and so on. First we assume ¢ B ∆ epfq ¤ C epfq in P pz q. (3.30) Bt 0 2R 0

2pn¡1 2 2 Multiply the equation by e ηngn on both sides, then integrate over P2R. With

37 integration by parts, we have »

2pn 2 2 C0 e ηngn dvol dt P2R » » Be ¡ ¡ ¥ x , e2pn 1η2g2y dvol dt x∆e, e2pn 1η2g2y dvol dt B n n n n P2R t P2R » » ¨ Be ¡ ¡ x , e2pn 1η2g2y dvol dt xde, d e2pn 1η2 g2y dvol dt B n n n n P2R t P2R » » 1 d ¨ 2p ¡ 1 2pn 2 2 n | p pn q|2 2 2 e ηngn dvol dt 2 d e ηngn dvol dt (3.31) 2pn P2R dt pn P2R » 1 pn 2x p pn q p 2qy e gn d e , d ηn dvol dt pn P2R » » 1 d ¨¨ 1 d ¨ 2pn 2 2 ¡ 2pn 2 2 e ηngn dvol dt e ηn gn dvol dt 2pn P2R dt 2pn P2R dt » » 2p ¡ 1 1 n | p pn q|2 2 2 pn 2x p pn q p 2qy 2 d e ηngn dvol dt e gn d e , d ηn dvol dt. pn P2R pn P2R

pn We bound the last two terms by }dpe η g q} 2 , and pick p 1, p n n L pP2Rq 1 k ¢ m 2 p ¡ ¥ 1 m k 1 »

τ ¢ r¡ 2 q The preceding inequality still holds if we replace P2R by P2R : B2R 4R , τ P p¡ 2 s 2 ¡ 2 for any τ 4R , 0 . Since gn 0 when t 4R , integration by parts in the t direction yields » » 2 2 2pn | p pn q|2 ηngne dvol 2 d e ηngn dvol dt B2R¢tτu P2R » ¢ (3.34) d ¨ ¤ 2pn 2 2 2 2 2 | p q|2 e 2C0pnηngn ηn gn 2gn d ηn dvol dt. P2R dt

1 1 By Young’s inequality, for any a, b ¥ 0, and 1, p q

ap bq ¥ ab. (3.35) p q » m 2 m 2 ap 2pn 2 2 Pick p and q and view e ηngn dvol dt , 2 m B2R¢tτu p » q pn 2 b 2 |d pe ηngnq| dvol dt , we have P2R q

¢ » 1 ¢ » 1 m 2 p 2m q e2pn η2g2 dvol |d pepn η g q|2 dvol dt n n n n 2 B2R¢tτu m 2 P2R » ¢ (3.36) d ¨ ¤ 2pn 2 2 2 2 2 | p q|2 e 2C0pnηngn ηn gn 2gn d ηn dvol dt. P2R dt

39 This works for any τ P p¡4r2, 0s, therefore

£ 1 ¢ » p » 1 m 2 2m q 2pn 2 2 | p pn q|2 sup e ηngn dx d e ηngn dvol dt 2 τPp¡4R ,0s 2 B2R¢tτu m 2 P2R » ¢ d ¨ ¤ 2pn 2 2 2 2 2 | p q|2 e 2C0pnηngn ηn gn 2gn d ηn dvol dt. P2R dt (3.37)

By the Sobolev Lemma, for I an interval in R1,

£ 2 » » m » 2 m 2 2 2 u m dvol dt ¤ CpMq sup u dvol ¢ |du| dvol dt (3.38) P Bρ¢I τ I Bρ¢ttu Bρ¢I

pn By (3.38) with u e ηngn, we have

¢» m

m 2 m 2 r pn 2 CpMq pe ηngnq m dvol dt P2R

£ m 2 m 2 » m » pn 2 pn 2 ¤ sup pe ηngnq ¢ |dpe ηngnq| dvol dt (3.39) 2 τPp¡4R ,0s B2R¢tτu P2R » ¢ d ¨ ¤ 2pn 2 2 2 2 2 | p q|2 e 2C0pnηngn ηn gn 2gn d ηn dvol dt. P2R dt

2n 2n 1 Since |dpg q| ¤ , |dpη q| ¤ , and g , η are cutoﬀ functions both com- n R2 n R n n

pactly supported by PRp1 1 q, we can use the information to restrict the domain 2n¡1 of the integrals on both sides

¤ m » m 2 ¢ » n 1 m 2 4 r ¥ 2pn 2pn CpMq e m dvol dt ¤ 2C p e dvol dt. 0 n R2 P 1 P 1 Rp1 n q Rp1 q 2 2n¡1 (3.40) That is ¢ 4n rp q } }2pn ¤ } }2pn C M e 2pn 1 p q C0pn e L2pn pP q . (3.41) L Prn R2 rn¡1 40 Hence,

£ 1 n pn } }2 ¤ C0pn 4 } }2 e L2pn 1 pP q e L2pn pP q . (3.42) rn CrpMq CrpMqR2 rn¡1

Iterate until pn Ñ 8, we have £ ¢ 1 ¹8 n pn 2 2 1 2 }e} 8p q ¤ C0 }e} 2p q L PR rp q R2 L P2R n1 C M

£ ¢ m 2 2 4 1 2 (3.43) C0 }e} 2p q CrpMq R2 L P2R

¢ m 2 2 C2pMq 2 C1pMqC0 }e} 2p q , R2 L P2R

for C1pMq,C2pMq being just the Sobolev constants that only depend on M.

As for the general case when (3.25) is true in PRpz0q, we have £ 2 ¢ m 2 » C pMq 2 sup e ¤ C pMqC 2 e2 dvol dt. (3.44) 1 0 p ¡ q2 zPPδRpz0q R δR PRpz0q

We then apply Lemma 11 in order to get the L8 norm bounded by the L1 norm of epfq in a larger cylinder. We use the same trick as in the elliptic case. From Cauchy-Schwarz, we have d ¢ m 2 » C pMq 4 sup epfq ¤ C pMqC 2 e2 dvol dt 1 0 p ¡ q2 PδRpz0q R δR PRpz0q d ¢ m 2 c » p q 4 ¤ p q C2 M p q p q C1 M C0 2 sup e f e f dvol dt pR ¡ δRq P p q Rpz0q PR z0

¢ m 2 » p q 2 ¤1 p q p q C2 M p q sup e f C1 M C0 2 e f dvol dt. 2 P pR ¡ δRq p q Rpz0q PR z0 (3.45) 41 Now apply the lemma by setting t δR, s R, and

ϕptq sup epfq, Ptpz0q we have

¢ m 2 » 1 C pMq 2 ϕptq ¤ ϕpsq C pMqC 2 epfq dvol dt. 1 0 p ¡ q2 2 R δR PRpz0q

Since C1pMqC0 is bounded, we get

1 A ϕptq ¤ ϕpsq B, 2 ps ¡ tqm 2 where » » ˜ ˜ A C2pMq epfq dvol dt, B C1pMqC0 epfq dvol dt. PRpz0q PRpz0q

From the lemma, we get

ϕptq sup epfq PδRpz0q A ¤C B pR ¡ δRqm 2 (3.46)

£ m 2 » C˜ pMq 2 ¤ C˜ pMqC 2 epfq dvol dt, 1 0 p ¡ q2 R δR PRpz0q

for any 0 ¤ δ 1.

In particular, we have

£ m 2 » ˜ p q 2 p qp q ¤ ˜ p q C2 M p q e f z0 C1 M C0 2 e f dvol dt, (3.47) R PRpz0q

42 for any R where (3.25) holds.

Suppose sup epfq exists and is achieved at z0, we also have M¢I

£ m 2 » ˜ p q 2 ˜ C2 M sup epfq ¤ C1pMqC0 epfq dvol dt, (3.48) ¢ 2 M I R PRpz0q for any R.

43 4

-Regularity

In this chapter, we apply Moser iteration and the monotonicity formulas derived in Chapter3 to derive new -regularity results. We obtain better bounds for sup e under

suitable hypotheses on the ratios µ0, ν0, and p0. This proof is derived directly from Moser iteration without introducing any rescaling of the function e. Therefore, we

can see clearly how 0 depends on the constants playing a role in the Moser iteration.

4.1 Elliptic -Regularity

1 We obtain estimates for positive functions e : BRpx0q Ñ R which satisfy a monotonicity relation and the elliptic inequality

∆e ¤ Ce2. (4.1)

Theorem 13. Suppose e : M Ñ Rk is positive and satisﬁes (4.1). Suppose there exists C2, p0 ¥ 0, and a smooth function ν with 0 ¤ ν ¤ 1 such that » 1 ΦpR, xq : p1 νpr, xqqe dvol (4.2) m¡2 p0 R BRpx0q

44 is monotonically increasing in R. There exists a constant 0 ¡ 0, such that if

ΦpR, x0q 0, (4.3)

3 then for any δ , 4 C sup e ¤ , (4.4) p q2¡p0 xPBδRpx0q δR where C is an absolute constant only depending on M and the monotonicity constant. Moreover, we have » sup epfq ¤ CR¡m p1 νpr, xqqepfq dvol. (4.5) B 1 px0q BRpx0q 2 R

Proof. From the monotonicity formula, which applies to all x P BRpx0q, we have » » ¡ ¡ ¡ ¡ ρ2 m p0 p1 νpr, xqqepfq dvol ¤ Cτ 2 m p0 p1 νpr, xqqepfq dvol, (4.6) Bρpxq Bτ pxq

for any x P BRpx0q and 0 ρ τ R ¡ |x ¡ x0|, where C1 only depends on m and M. 3 Now we set R δR R. Consider 0 ρ R ¡ |x ¡ x | R ¡ |x ¡ x |. By 1 4 1 0 0 p q p q monotonicity and the fact that BR¡|x¡x0| x BR x0 the above inequality implies,

45 P p q for any x BR1 x0 , that » ¡ ¡ ρ2 m p0 p1 νpr, xqqepuq dvol Bρpxq »

2¡m¡p0 ¤C1 pR1 ¡ |x ¡ x0|q p1 νpr, xqqepfq dvol B pxq R1¡|x¡x0| » ¡ ¡ ¤ p ¡ | ¡ |q2 m p0 p p qq p q C2 R x x0 1 ν r, x e f dvol (4.7) B pxq R¡|x¡x0|

¢ ¡ ¡ » 1 2 m p0 ¤C2 R1 p1 νpr, xqqepfq dvol 3 BRpx0q ¢ » 2¡m¡p0 1 ¡ ¡ C R2 m p0 p1 νpr, xqqepfq dvol. 4 BRpx0q

¡ | ¡ | 4 ¡ | ¡ | ¥ 1 The last step holds because R x x0 3 R1 x x0 3 R1.

Now consider the following function

2¡p0 hpσq pR1 ¡ σq sup epfq. (4.8) Bσpx0q

p q p q P p q Let σ0 be the point where h σ0 max h σ . Suppose at x1 Bσ0 x0 we have 0¤σ R1

p qp q p q p q e f x1 Bσ0 x0 e f : e0. (4.9)

1 Let ρ pR ¡ σ q. Since B px q B px q, we have 0 2 1 0 ρ0 1 ρ0 σ0 0

sup epfq ¤ sup epfq. (4.10) p q p q Bρ0 x1 Bρ0 σ0 x0

From the deﬁnition of hpσq, we get

2 2 pB1 ¡ pρ0 σ0qq sup epfq ¤ pB1 ¡ σ0q sup epfq. (4.11) p q p q Bρ0 σ0 x0 Bσ0 x0

46 This implies

sup epfq ¤ sup epfq ¤ 4 sup epfq 4e0. (4.12) p q p q p q Bρ0 x1 Bρ0 σ0 x0 Bσ0 x0

p q Now consider the elliptic equation in Bρ0 x1 . We have £ p q ¤ 2p q ¤ p q p q ¤ p q p q ∆e f Ce f C sup e f e f 4Ce0e f in Bρ0 x1 . (4.13) p q Bρ0 x1

By the single point Moser iteration result, we have

¢ m » p q 2 p q ¤ p q C2 M p q e0 e x1 C1 M 4Ce0 2 e f dvol. (4.14) ρ p q 0 Bρ0 x1

Rearranging, we get » ¨ ¡ m 1 2 p0 ¤ p q 2 p q 2 p q ρ0 e0 C1 M ρ0e0 C2 M m¡2 p e f dvol 0 p q ρ0 Bρ0 x1 » (4.15) ¨ ¡ m 1 ¤ p q 2 p0 p q 2 p p qq p q C1 M ρ0 e0 C2 M m¡2 p 1 ν r, x e f dvol. 0 p q ρ0 Bρ0 x1

This is » m 1 p q ¤ p p q p q p qq 2 p p qq p q h σ0 C1 M h σ0 C2 M m¡2 p 1 ν r, x e f dvol. (4.16) 0 p q ρ0 Bρ0 x1

P p q 3 Since x1 BR1 x0 and ρ0 σ0 R1 4 R, we can apply the previous result that » » 1 1 p1 νpr, xqqepfq dvol ¤ C p1 νpr, xqqepfq dvol, m¡2 p0 m¡2 p p q R 0 p q ρ0 Bρ0 x1 BR x0 (4.17) and therefore, » m 1 hpσ0q ¤ pC1pMqhpσ0q C2pMqq 2 p1 νpr, xqqepfq dvol. (4.18) m¡2 p0 R BRpx0q 47 3 p q Suppose there exists R1 4 R such that h σ0 1. Then » m{2 1 1 hpσ0q ¤ pC1 C2q p1 νpr, xqqepfq dvol. (4.19) m¡2 p0 R BRpx0q

¤ 1 We can pick 0 m{2 such that 2pC1 C2q » m{2 1 m{2 pC1 C2q p1 νpr, xqqepfq dvol ¤ pC1 C2q 0 1, (4.20) m¡2 p0 R BRpx0q

for any R such that ΨpR, x0q ¤ 0. This leads to a contradiction. Therefore,

hpσ0q ¤ 1. That is,

2¡p0 hpσq pR1 ¡ σq sup epfq ¤ hpσ0q ¤ 1. (4.21) Bσpx0q

3 We can see that the above proof will hold when we replace R1 4 R with R1 δR 3 with any δ ¤ . 4

1 1 Finally, picking σ 2 R1 2 δR, we get

p q ¤ C sup e f 1 ¡ . (4.22) p q p q2 p0 B 1 x0 2 δR 2 δR

1 with C a constant depends on C0. (We can denote new δ as 2 δ to get the desired result.)

As for (4.5), we go back to the original estimate (4.18) and use the fact that hpσ0q ¤ 1, » p m 1 p q ¤ p p q 0 p qq 2 p p qq p q h σ0 C1 M ρ0 C2 M m¡2 p 1 ν r, x e f dvol 0 p q ρ0 Bρ0 x1 » (4.23) 1 ¤C p1 νpr, xqqepfq dvol. m¡2 p0 R BRpx0q 48 Using the deﬁnition of hpσ0q again, we have » 1 2¡p0 pR1 ¡ σq sup p1 νpr, xqqepfq ¤ hpσ0q ¤ C epfq dvol. (4.24) m¡2 p0 Bσpx0q R BRpx0q

3 1 Here we pick δ to be 4 , and σ 2 R R1 δR, we have » sup epfq ¤ CR¡m p1 νpR, xqqepfq dvol. (4.25) B 1 px0q BRpx0q 2 R

4.2 Parabolic -Regularity

In this section, we consider a positive function e : M Ñ R1 that satisﬁes the following parabolic inequality Be ∆e ¤ Ce2 in P pz q (4.26) Bt R 0 for some R RM . Here z0 px0, t0q. The regions PRpz0q and TRpt0q are the same as the ones deﬁned in Chapter 2.

Theorem 14. Suppose e : M ¢ I Ñ Rk is a positive function that satisﬁes (4.26).

Suppose there exists ν0 ¥ 0 such that

» 2 » t0¡R 1 2 pp q q p qp q p qp q Ψ x0, t0 ,R : ν G x0,t0 x, t e f x, t ϕ dvol dt (4.27) 2 p ¡ q 0 t0¡4R t0 t M

is monotonically increasing in R. There exists 0 such that, if there exists R ¤ ? t0 ¡ t{2 with

» 2 » t0¡R 1 2 pp q q p qp q p q ¤ Ψ x0, t0 ,R ν G x0,t0 x, t e x, t ϕ dvol dt 0, (4.28) 2 p ¡ q 0 t0¡4R t0 t M

49 then for any δ that depends only on M,Epf0pxqq,R, and C an absolute constant, we have C sup e ¤ . (4.29) 2¡2ν0 PδRpx0,t0q R

Here Gpx0,t0q is the backward Gaussian ¢ 2 1 |x ¡ x0| Gp qpx, tq exp ¡ , (4.30) x0,t0 m p ¡ q p4πpt0 ¡ tqq 2 4 t0 t

p q p| ¡ |q p q and ϕ x ϕ x x0 is a cut-oﬀ function supported in BRM x0 . (Sometimes we p q p q denote Gpx0,t0q x, t G x, t or just Gpx0,t0q, Gz0 .)

Proof. Let R1 δR for δ 1{2. For any σ R1, consider the following situation. Suppose

e0 sup epfq. Pσpz0q

We are interested in the point z1 px1, t1q P Pσpz0q such that

epfqpz1q e0.

p q p q Then inside some cylinder Pρ z1 PR1 z0 , we have ¢ £ B ∆ epfq ¤ C sup epfq e. Bt p q PR1¡σ z0

From the single point Moser iteration, we have

£ £ m 2 » p q 2 p qp q p q ¤ p q p q C2 M p q e f z1 sup e f C1 M sup e f 2 e f dvol. Pσpz0q Pρpz1q ρ Pρpz1q

Rearranging the above inequality so that it attains a form in which we may apply the monotonicity formula, we get £ pm 2q{2 » 1 2¡2ν0 2¡p0 p0 ρ sup epfq ¤ C1pMqρ p sup eqρ C2pMq epfq dvol dt. m 2ν0 Pσpz0q Pρpz1q ρ Pρpz1q 50 To relate sup epfq back to sup epfq, we deﬁne the function Pρpz1q Pσpz0q

2¡2ν0 hpσq pR1 ¡ σq sup epfq, (4.31) Pσpz0q

1 p ¡ q and pick ρ 2 R1 σ .

Since epfq is regular in PRpz0q, we are able to ﬁnd a point σ0 such that

hpσ0q max hpσq. 0¤σ R1

p q P p q There exists z1 x1, t1 Pσ0 z0 such that

epfqpz1q sup epfq. p q Pσ0 z0

1 Let ρ pR ¡ σ q. We will argue that 0 2 1 0

sup epfq ¤ 4 sup epfq. p q p q Pρ0 z1 Pσ0 z0

1 It is easy to check that ρ σ pR σ q R δR R and P pz q 0 0 2 1 0 1 ρ0 1 p q Pρ0 σ0 z0 . Similar argument as in the elliptic case, we have

sup epfq ¤ sup ¤ 4 sup 4epfqpz1q. (4.32) p q p q p q Pρ0 z1 Pρ0 σ0 z0 Pσ0 z0

p q Now we base our estimate inside the cylinder Pρ0 z1 since z1 is the point that p q p q e f achieves its max within Pσ0 z0 . From the single point Moser iteration, we get

£ £ m 2 » p q 2 p qp q ¤ p q p q C2 M p q e f z1 C1 M sup e f 2 e f dvol dt p q ρ p q Pρ0 z1 0 Pρ0 z1 (4.33) £ £ m 2 » p q 2 ¤ p q p q C2 M p q 4C1 M sup e f 2 e f dvol dt. p q ρ p q Pσ0 z0 0 Pρ0 z1

51 Do the same rearranging as before, we get the following,

¡ 2 2ν0 p qp q ρ0 e f z1

1 ¡ p ¡ q2 2ν0 p q 2¡2ν R1 σ0 sup e f 2 0 p q Pσ0 z0 £ pm 2q{2 » ¡ 1 ¤ p qp ¡ q2 2ν0 p q p q p q C1 M R1 σ0 sup e f C2 M m 2ν e f dvol dt. p q ρ 0 p q Pσ0 z0 0 Pρ0 z1 (4.34)

That is, » p q ¤ p p q p q p qqpm 2q{2 1 p q h σ0 C1 M h σ0 C2 M m 2ν e f dvol dt. ρ 0 p q 0 Pρ0 z1

Here, as long as we can show that » 1 p q m 2ν e f dvol dt ρ 0 p q 0 Pρ0 z1

can be bounded by 0 using the monotonicity formula, we could argue that hpσ0q

cannot be arbitrarily large, provided σ0 R1 δR R, where 0 gets achieved.

Therefore, we will have control for all sup epfq, @σ R1 from hpσ0q. Pσpz0q We present the following lemma and will prove it after the proof of this theorem.

1 P p q Lemma 15. Let R1 δR for some δ 2 , and R RM . Let ρ, σ 0,R1 with

ρ σ R1. Then for any z1 px1, t1q P Pσpz0q, the monotonicity formula implies » 1 epuq dvol dt m 2ν0 ρ Pρpz1q ¨ ¡2 ¤C1pRqΨppx0, t0q,Rq C2REpf0pxqq C3 exp p2 ¡ m ¡ 2ν0q ln R ¡ c2δ Epf0pxqq. (4.35)

Here C1pRq is a constant depends on R and M, C3, c2 are absolute constants.

Epf0pxqq is the initial energy. 52 With the lemma, we can proceed to argue for bounds of hpσ0q as follows. Assume

Ψppx0, t0q,R0q ¤ 0. We then pick δ small so that

¡2 C3 exp pp2 ¡ m ¡ 2ν0q ln R ¡ c2δ q Epf0pxqq ¤ 0, for some small R RM . We have » p q ¤ p p q qpm 2q{2 1 p q h σ0 C1h σ0 C2 m 2ν e f dvol dt ρ 0 p q 0 Pρ0 z1

pm 2q{2 (4.36) ¤ pC1hpσ0q C2q pc10 c2REpf0pxqqq

pm 2q{2 ¤ pC1hpσ0q C2q 0.

Suppose there exists R1 δR, such that for any σ R1, hpσ0q max 1. 0¤σ R1 By the above inequality, we have

pm 2q{2 1 ¤ pC1 C2q 0. (4.37)

pm 2q{2 Therefore, we can choose 0 so that pC1 C2q 0 1. That said, with

R1 δR δR0, and for all σ R1, we always have

hpσq ¤ hpσ0q ¤ 1. (4.38)

1 1 Since hpσ0q max hpσq, we can pick σ R1 δR, and we have 0¤σ R1 2 2

2¡p0 pR1 ¡ σq sup epfq ¤ hpσ0q ¤ 1, (4.39) Pσpz0q

which is,

p q ¤ C sup e f 1 ¡ , (4.40) p q p q2 p0 P 1 z0 2 δR 2 δR

where C is an absolute constant. Picking new δ to be half of the original δ, we will get the desired estimate.

53 Proof of Lemma 15. We want to show that » 1 epfq dvol dt m 2ν0 ρ Pρpz1q

can be controlled by some Ψppx0, t0q,Rq and eventually bounded by 0.

First, we need an estimate between the backward Gaussian G and ρ¡m. Since

Gaussians blow up at t t0, we cannot derive the inequality directly from z1 px1, t1q. To get a bound on the Gaussian, we move the center a little above so that

2 2 it does not lie in the region TRpt0q M ¢ rt0 ¡ 4R , t0 ¡ R s, where we later apply the monotonicity formula.

Let R1 δR and σ, ρ P p0,R1q such that ρ σ R1. Let z1 px1, t0q P Pσpz0q.

For any point in z P Pρpz1q, we have the following estimate ¢ | ¡ |2 2 ¡m{2 x x1 p 2q p p ¡ qq ¡ G x1,t1 2ρ 4π t1 2ρ t exp 2 4pt1 2ρ ¡ tq ¢ ¨¡ { ρ2 ¥ p 2 ¡ p ¡ 2qq m 2 ¡ (4.41) 4π t1 2ρ t1 ρ exp 2 2 4pt1 2ρ ¡ pt1 ρ qq

Cρ¡m.

Therefore, we have » »

¡m¡2ν0 ¡2ν0 p q ¤ 2 p q p q ρ e f dvol dt C1ρ Gpx1,t1 2ρ q x, t e f dvol dt. (4.42) Pρpz1q Pρpz1q

2 2 It is obvious that Pρpz1q Tρpt1 2ρ q. Suppose ϕ is the cut-oﬀ function that is used in the proof of the monotonicity formula that is compactly supported in

54 p q BRM x0 , we have »

¡2ν0 2 p q p q ρ Gpx1,t1 2ρ q x, t e f dvol dt Pρpz1q »

¡2ν0 2 ¤ 2 p q p q ρ Gpx1,t1 2ρ q x, t e f ϕ dvol dt p 2q Tρ t1 2ρ (4.43)

» 2 » t1 ρ 1 2 ¤ p 2qp q p q C 2 ν G x1,t1 2ρ x, t e f ϕ dvol dt 2 p ¡ q 0 t1¡3ρ t1 2ρ t M

2 CΨppx1, t1 2ρ q, ρq.

By the monotonicity formula, we have ? 2 2 Ψppx1, t1 2ρ q, ρq ¤ C1pRqΨppx1, t1 2ρ q, 3R{2q C2REpf0pxqq. (4.44)

¤ 1 2 ¡ With the fact that R1 δR 2 R and ρ, σ R1, we can check that t1 2ρ 2 2 2 3R ¡ t0 ¡ 4R . We then split the stripe TR{2pt1 2ρ q into 3 parts ? 2 Ψppx1, t1 2ρ q, 3R{2q £ » 2 » 2 2 » t0¡R t1 2ρ ¡3R {4 1 2 ¤ p 2qp q p q 2 ν G x1,t1 2ρ x, t e f ϕ dvol dt 2 2 p ¡ q 0 t0¡4R t0¡R t1 2ρ t M » 1 2 ¤C Gpx ,t 2ρ2qepfqϕ dvol dt p ¡ qν0 1 1 TRpt0q t0 t

» 2 2 » t1 2ρ ¡3R {4 1 2 p 2q p q 2 ν G x1,t1 2ρ e f ϕ dvol dt 2 p ¡ q 0 t0¡R t1 2ρ t M

I II. (4.45)

2 2 ¤ p q2 2 2{ For II, since σ ρ σ ρ R1 R 4, we apply the monotonicity formula

55 again and get

» 2 2 » t1 2ρ ¡3R {4 1 2 p 2q p q 2 ν G x1,t1 2ρ e f ϕ dvol dt 2 p ¡ q 0 t0¡R t1 2ρ t M

» 2 2 2 t0 σ 2ρ ¡3R {4 1 2 ¤ p 2q p q C ν G x1,t1 2ρ e f ϕ dvol dt 2 p ¡ q 0 t0¡R t0 t (4.46) » 2 t0¡R {4 1 2 ¤ p 2q p q C ν G x1,t1 2ρ e f ϕ dvol dt 2 p ¡ q 0 t0¡R t0 t

» 2 t0¡R 1 2 ¤ p 2q p q C ν G x1,t1 2ρ e f ϕ dvol dt. 2 p ¡ q 0 t0¡4R t0 t

So far we have shown » ¡ ¡ ρ m 2ν0 epfq dvol dt Pρpz1q » » (4.47) 1 2 ¤C1pRq Gpx ,t 2ρ2qpx, tqepfqϕ dvol dt C2pRqEpf0pxqq. p ¡ qν0 1 1 TRpt0q t0 t M

2 p q Now we need to get a bound for Gpx1,t1 2ρ q x, t from Gx0,t0 . Here we consider two regions: |x ¡ x1| ¥ R{δ and |x ¡ x1| R{δ.

Case 1: when |x ¡ x1| ¥ R{δ.

p q P p q ¡ ¥ ¡ 2 ¥ ¡ 2 ¥ ¡ 1 2 ¥ ¡ 1 p ¡ q For x, t TR t0 , t0 t R and t1 t0 2ρ σ 2 R 2 t0 t .

2 m{2 m{2 Therefore, pt1 2ρ ¡ tq ¥ cpt0 ¡ tq . Hence, ¢ ¨¡ m |x ¡ x |2 2 2 1 p 2qp q p ¡ q ¡ G x1,t1 2ρ x, t 4π t1 2ρ t exp 2 4pt1 2ρ ¡ tq ¢ 2 ¡ m pR{δq (4.48) ¤c p4πpt ¡ tqq 2 exp ¡ 0 4R2 σ2 2ρ2 ¨ ¡m ¡2 ¤c1R exp ¡c2δ .

Case 2: when |x ¡ x1| R{δ. We use the same argument that

2 2 2 2 2 2 t1 2ρ ¡ t ¤ t0 ¡ t σ 2ρ ¤ t0 ¡ t 2δ R ¤ p1 2δ q pt0 ¡ tq, and the triangle 56 inequality to get ¢ ¢ 2 2 2 ¡ m |x ¡ x0| |x ¡ x0| |x ¡ x1| p 2qp q ¤ p p ¡ qq 2 ¡ ¢ ¡ G x1,t1 2ρ x, t c 4π t0 t exp exp 2 4pt0 ¡ tq 4pt0 ¡ tq 4pt1 2ρ ¡ tq ¢ 2 2 |x ¡ x0| |x ¡ x1| p q ¡ cG x0,t0 exp 2 4pt0 ¡ tq 4pt1 2ρ ¡ tq ¢ | ¡ |2 1 2 x x1 ¤ p q | ¡ | ¡ cG x0,t0 exp x x0 2 4pt0 ¡ tq 1 2δ ¢ 2 1 2 |x ¡ x1| ¤ p q p| ¡ | | ¡ |q ¡ cG x0,t0 exp x x1 x1 x0 2 4pt0 ¡ tq 1 2δ ¢ 2 1 2δ 2 2 p q | ¡ | | ¡ || ¡ | | ¡ | cG x0,t0 exp 2 x x1 2 x x1 x x0 x x0 4pt0 ¡ tq 1 2δ £ ¢ ¢ 1 R 2 R ¤ 2 2 cGpx0,t0q exp 2δ 2 σ σ 4pt0 ¡ tq δ δ ¢ 1 ¤ 2 cGpx0,t0q exp c2R 4pt0 ¡ tq ¤ p { q cGpx0,t0q exp c2 4 cG˜ px0,t0q. (4.49)

Therefore, we have for any px, tq P TRpt0q, ¨ ¡m ¡2 2 p q ¤ ¡ Gpx1,t1 2ρ q x, t cGpx0,t0q c1R exp c2δ ¨ ¡ ¡ ¤ p q 2 2ν0 p ¡ ¡ q ¡ 2 cGpx0,t0q x, t c1R exp 2 m 2ν0 ln R c2δ . (4.50)

With the estimate of the Gaussian, we have for any z1 P Pσpz0q, and

57 σ ρ R1 δR R{2, » ¡ ¡ ρ m 2ν0 epfq dvoldt Pρpz1q » » 1 ¨¨ ¡2 2ν0 ¡2 ¤C1pRq cGpx ,t q c1R exp p2 ¡ m ¡ 2ν0q ln R ¡ c2δ p ¡ q2ν0 0 0 TRpt0q t0 t M 2 ¢ epfqϕ dvol dt C2pRqEpf0pxqq ¨ ¡2 ¤C1pRqΨppx0, t0q,Rq C3pRq exp p2 ¡ m ¡ 2ν0q ln R ¡ c2δ C2pRq Epf0pxqq ¨ ¡2 C1pRqΨppx0, t0q,Rq C3pRq exp p2 ¡ m ¡ 2ν0q ln R ¡ c2δ Epf0pxqq C2REpf0pxqq. (4.51)

¡1{2 ¡2 To get the second term small, we may pick δ cpm, ν0q| ln R| so that c2δ is

much larger than p2 ¡ m ¡ 2ν0q ln R when R 1, and δ can be chosen independently if R ¥ 1.

58 5

Models with Repulsive Potentials

5.1 Sigma Models with Repulsive Potentials

The sigma model approach from [1] illustrates an idea to replace the nonlinear constraints from the target manifold with potential energy constraints. This approach allows us to replace a nonlinear problem that may not have solutions with a sequence of problems each of which has a solution. Then the problem becomes showing this sequence of solutions converges to a solution of the original problem. The repulsive potential is motivated by the desire to keep the homotopy class of the projection of f onto the Grassmannian ﬁxed. We are particularly interested in the existence of harmonic maps from a compact Riemannian manifold M to complex Grassmannian manifolds. It is well-known that Grassmannian manifolds have positive sectional curvature. So the approach from [6] cannot be applied. However, Grassmannian manifolds with their canonical metrics are Kahler, and holomorphic maps between Kahler manifolds are harmonic. Hence it is easy to show there exist many harmonic maps to the Grassmannian. The diﬃculty is producing them without knowing holomorphicity in advance.

59 The rank k complex Grassmannian Grpk, Clq can be deﬁned as the set of Hermi- tian matrices in Rl¢l that have k p 1q eigenvalues and the remaining eigenvalues are 0. An equivalent representation is: ( Grpk, Clq u P Rl¢l|u is hermitian and k eigenvalues of u are 1, the rest are -1 . (5.1)

l Consider the classifying map fF : M Ñ Grpk, C q of a vector bundle F over M, and consider the energy » ¢ 1 2 EpfF q |dfF | W pfF q dvol. M 2

To guarantee that the projection of energy minimizers to the Grassmannian re- main in the same homotopy class, we have introduced a potential term W to our

energy. W pfF q is chosen so that W pfF q Ñ 8 if one of the eigenvalues of fF Ñ 0,

and W is smooth in the complement of the set tdetpfF q 0u.

Since W pfF q is smooth as long as all eigenvalues of fF are away from 0, singularities of fF can only arise when some eigenvalues of fF approach to 0. This translates questions about singularities of fF into questions about singularities of W pfF q.

Because W pfF q is pushing fF away from the set tdetpfF q 0u, we call W a “repulsive potential”.

5.2 Repulsive Potentials

Ñ l¢l p Clq t ul Let f : M R Gr k, . Let λi i1 be the eigenvalues of f. Consider the energy functional » ¢ 1 Epfq |df|2 W pfq dvol, M 2 with repulsive potential

1 1 ¸l 1 1 W : Rl¢l Ñ R1, with W pfq tracepf ¡2q : |f ¡1|2. (5.2) 2 2 λ2 2 i i 60 Here | ¤ | is the Hilbert-Schmidt norm of matrices.

When the eigenvalues of f are bounded, we have bounded potential W pfq and f lies in a set that is homotopy equivalent to the Grassmannian Grpk, Clq. This energy functional E has a nice scaling property, if we replace M with a Euclidean space. Assuming

f λf,˜ x λ2y, we have » ¢ ˜ 1 ˜ 2 1 ˜¡2 Epfpyqq |dyf| |f | dvoly M 2 2 » ¢ 1 1 1 | |2 | ¡2| (5.3) 2m¡2 dxf f dvolx λ M 2 2 1 Epfpxqq. λ2m¡2

5.2.1 Estimates of ∇W and HesspW q

For this speciﬁc potential W , we calculate some of its related quantities, which will be used later. The directional derivative of W along the B direction is § d § WBpfq W pf sBq§ ds s0 § d 1 ¨ § trace pf sBq¡2 § ds 2 s0 ¢ § (5.4) 1 d § trace pf sBq¡2 § 2 ds s0 1 ¨ ¨ trace ¡f ¡2Bf ¡1 ¡ f ¡1Bf ¡2 ¡trace f ¡3B . 2

Therefore ∇W pfq ¡f ¡3, and |∇W |2 |f ¡3|2. (5.5)

61 Under this W , the static elliptic equation and the corresponding parabolic gradient equation are d¦df ¡ f ¡3 0, (5.6)

and Bf t d¦df ¡ f ¡3 0, fp0, xq f pxq P Grpk, Clq. (5.7) Bt t t 0

5.2.2 Short Time Existence of Parabolic Flow

We assume f0, the initial data of (5.7), is smooth. We can convert (5.7) into the integral equation » t p q ¡td¦d ps¡tqd¦d ¡3 f t, x e f0 e fs ds. (5.8) 0

Consider the subset ! ) P 1p q| } ¡ } ¤ Z f C M f f0 C1pMq α , (5.9)

for a ﬁxed α 1. We have (from [17, p.272, (1.3) and p.273, (1.11)])

¦ e¡td d : C1pMq Ñ C1pMq is a strongly continous semigroup for t ¥ 0, (5.10)

and ¦ e¡td d ¤ Ct¡1{2, for t P p0, 1q. (5.11) LpC0pMq,C1pMqq

Furthermore, the functional

F : f Ñ f ¡3 (5.12)

is deﬁned from C1pMq to C0pMq when f is invertible. The derivative of F is

¡1 ¡3 ¡2 ¡2 ¡3 ¡1 DFf pgq ¡f gf ¡ f gf ¡ f gf . (5.13)

62 We have the estimate } p q} ¤ ¡1 ¡3 ¡2 ¡2 ¡3 ¡1 DFf g C0pMq f gf C0 f gf C0 f gf C0 (5.14) ¤ p q } } ¤ p q } } C f g C0 C f g C1 ,

where Cpfq depends on the C0 norm of f ¡1, f ¡2, and f ¡3. } ¡ } ¤ When f f0 C1 α, we can show ¨¡ ¡1 ¤ ¡1p ¡ q 1 ¡1 ¤ p ¡1 q f 0 f0 f f0 I f0 0 C f0 0 (5.15) C C0 C C ¡1 } ¡2} } ¡3} as long as f0 C0 α 1. We can show similar results for f C0 and f C0 . Therefore, for f P Z, } p q} ¤ p q } } DFf g C0 C f0 g C1 . (5.16)

Since » » 1 1 p q ¡ p q d p p ¡ qq p ¡ q F f F f0 F f0 t f f0 dt DFf0 tpf¡f0q f f0 dt, (5.17) 0 dt 0 we have » 1 } p q ¡ p q} ¤ p ¡ q } ¡ } ¤ p q } ¡ } F f F f0 C0 DFf0 t f f0 LpC1,C0q f f0 C1 dt C f0 f f0 C1 , 0 (5.18) which implies that F : C1pMq Ñ C0pMq is locally Lipschitz.

From [17, Chapter 15, Proposition 1.1A], ft exists in short time.

5.2.3 Induced Equations of Energy Density epfq

A B As for the Hessian term HesspW qABxdf , df y xd∇W pfq, dfy, we have xd∇W pfq, dfy xdp¡f ¡3q, dfy

xf ¡3df f ¡1, dfy xf ¡2df f ¡2, dfy xf ¡1df f ¡3, dfy (5.19)

|f ¡2df|2 2xf ¡2df, f ¡1df f ¡1y. 63 When we study the induced Laplace or heat equation of the energy density, we have ¢ 1 1 ∆epfq ∆ |df|2 |f ¡1|2 2 2 (5.20) ¡ |∇df|2 ¡ |f ¡3|2 ¡ Ricpdf, dfq ¡ 2|f ¡2df|2 ¡ 4xf ¡2df, f ¡1df f ¡1y

2 2 2 3 ¤|Ric|8|df| CW |df| ¤ C1epfq C2e pfq,

and ¢ ¢ ¢ B B 1 1 ∆ epf q ∆ |df |2 |f ¡1|2 Bt t Bt 2 t 2 t

¡ | |2 ¡ | ¡3|2 ¡ p q ¡ | ¡2 |2 ¡ x ¡2 ¡1 ¡1y ∇dft ft Ric dft, dft 2 ft dft 4 ft dft, ft dft ft

3 ¤C1epftq C2e pftq. (5.21)

Since the right-hand sides of the inequality in the last steps of both cases involve e3pfq, we cannot simply apply the previous -regularity to get the bound of the energy density e.

5.3 Smoothing Potential Wb

Since the unboundedness of W gives undesired bounds for the Laplace or heat equation of the energy density e, we start with a bounded version of repulsive potential by deﬁning

1 ¨ 1 ¸l 1 W pfq trace pf 2 bIq¡1 , (5.22) b 2 2 λ2 b i i

where I is the l ¢ l identity matrix. For b ¡ 0, we then analyze the bound of epfq in terms of b using -regularity. Later, we will show how the approximate co-dimension of the singularity set depends on b.

64 Since l Cplq Cplq W pfq ¤ , |HesspW q| ¤ W pfq ¤ , (5.23) b 2b b b b b2 we have the estimate of the Laplace and heat equation of the energy density e ¢ b 1 1 ¨ ∆e pfq ∆ |df|2 trace pf 2 bIq¡1 b 2 2 ¢ (5.24) p q p q 2 C l 2 C l ¤|Ric|8|df| W |df| ¤ e C e , b b b 2 b

and ¢ C plq ∆e pfq ¤ C 2 e . (5.25) b 1 b2 b

Similarly, ¢ ¢ B Cplq ∆ e pf q ¤ e C e , (5.26) Bt b t b 2 b and ¢ ¢ B C plq ∆ e pf q ¤ C 2 e . (5.27) Bt b t 1 b2 b

Here C1 only depends on M, C2plq depends on l, the size of f. ¢ B Since ∆e or ∆ e is bounded by e , we immediately get the following b Bt b b

result on the regularity of eb:

Theorem 16 (Elliptic). Fix a constant b ¡ 0. Let f : M Ñ Rl¢l Grpk, Clq satisfy the following equation

¦ d df Wbpfq 0, (5.28) » 1 2 1 2 8 with Ebpfq |df| Wbpfq dv bounded. Then ebpfq |df| Wbpfq P C pMq. M 2 2

l¢l l Theorem 17 (Parabolic). Fix a constant b ¡ 0. Let ft : M ¢ I Ñ R Grpk, C q satisfy the equation Bf t d¦df W pf q 0, (5.29) Bt t b t 65 l 8 with fp0, xq f0pxq P Grpk, C q and smooth. Then ebpftq P C pMq. »

Proof. Here we only prove the elliptic case. Since Ebpfq ebpfq dvol is bounded, M 1 2 1 ebpfq P L pMq. That is, |df| ,Wbpfq P L pMq. Recall the inequality that

C ∆e pfq ¤ e pfq. b b2 b

From Moser iteration, we have

¢ m{2 » p qp q ¤ C C2 p q eb f x 2 2 e f dvol, (5.30) b R BRpxq for any x P M. Since b is ﬁxed, we can take R Ñ 8 and get »

ebpfqpxq ¤ Cpbq ebpfq dvol CpbqEbpfq, (5.31) M

8 8 which is bounded. Therefore, ebpfq P L and ∆ebpfq P L . Here we get ebpfq P

1,α 8 C pMq for some α ¡ 0. ebpfq P C can be easily derived by bootstraping.

8 Since ebpfq P C , we consider sup ebpfq and can get the following bound:

¢ { » » Cplq m 2 p q ¤ p q ¤ ¡m p q ¡m p q sup eb f 2 eb f dvol Cb eb f dvol Cb Eb f . (5.32) M b M M

5.3.1 Improved Bound of sup eb M

In the previous section, we have obtained two bounds for ¡HesspWbqAB. They are either

¡2 ¡ HesspWbqAB ¤ cplqb ebpfq, (5.33) or

¡1 2 ¡ HesspWbqAB ¤ cplqb pebpfqq . (5.34) 66 We can get an intermediate bound of ¡HesspWbqAB between these two

¡1¡q 2¡q ¡ HesspWbqAB ¤ cplqb pebpfqq , (5.35) for 0 ¤ q ¤ 1.

Using this intermediate bound, we can get ﬁner control of sup ebpfq. In the M following, we denote sup ebpfq eM . We illustrate the idea using the elliptic version. M We have in M,

2 2 A B ∆ebpfq ¡ |∇df| ¡ |∇Wb| ¡ Ricpdf, dfq ¡ 2HesspWbqABxdf , df y (5.36) ¡1¡q 2¡q ¤cplqb pebpfqq .

8 We have shown that for a ﬁxed b ¡ 0, ebpfq P C pMq, and therefore eM exists. From (5.36), we get ¨ p q ¤ p q ¡1¡q 1¡q p q ∆eb f c l b eM eb f . (5.37)

Moser iteration gives

¢ ¡ m » C pmqcplqe1 q C pmq 2 ¤ 1 M 2 p q eM 1 q 2 eb f dvol. (5.38) b R BRpx0q

Here epx0q eM and R ¤ 1. In the following, for notational simplicity, we will assume RM ¥ 1.

2 ¡ 1 q Multiplying R 1¡q b 1¡q on both sides and factoring out 1¡q inside the parentheses

67 on the right-hand side of (5.38), we get

¤ £ m £ 2 2 1¡q 2 2 » 1¡q 1¡q 1¡q R eM ¤ ¥ R eM R 1 p q 1 q C1 1 q C2 1 q m eb f dvol 1¡q 1¡q 1¡q R b b b BRpx0q

¤ m £ ¡ 2 1 q 2 2q » 1¡q p0 1¡q ¥ R eM R 1 p q C1 1 q C2 1 q eb f dvol m¡2 p0 1¡q 1¡q R b b BRpx0q

¤ m £ ¡ 2 1 q 2 2q » 1¡q p0 1¡q ¤ ¥ R eM R p q C1 1 q C2 1 q eb f dvol (5.39) ¡ ¡ b 1 q b 1 q B1px0q

¤ m £ ¡ 2 1 q 2 2q » 1¡q p0 1¡q ¤ ¥ R eM R p q C1 1 q C2 1 q eb f dvol b 1¡q b 1¡q M

¤ m £ ¡ 2 1 q 2 2q 1¡q p0 1¡q ¥ R eM R p q C1 1 q C2 1 q Eb f . b 1¡q b 1¡q

Here, we use the monotonicity formula and the fact that R ¤ 1. Suppose we can pick R so that

2 1¡q R eM 1 q 1. (5.40) b 1¡q

This means

¡ ¨ 1 q 1 q ¡m 2 (5.41) ¤ CEbpfqb b 2

¡mp1¡qq{2 m 1 q¡ m¡1 CEbpfq b 2 2 .

m ¡ 1 m 1 m ¡ 1 When q ¡ , we can guarantee the power q ¡ ¡ 0, which m 1 2 2 ensures R is small when b 1.

68 m ¡ 1 1 m ¡ 1 Thus we pick q 1 ¡ ¡ . We have m m m 1

p0 2pm¡1q p0 2pm¡1q m R R 1 ¤ pC C q 2 E pfq Cpmq E pfq. (5.42) 1 2 b2m¡1 b b2m¡1 b

This gives

¢ 1 2m¡1 p ¡ q b p0 2 m 1 R ¥ . (5.43) CpmqEbpfq

Recall we pick R so that (5.40) holds. We get

¢ 1 q ¡ 2m 1¡q 2m 1 p q p q p0 2pm¡1q 2m p2m¡1qpp ¡2q b b 2m¡1 C m Eb f 0 ¤ ¤ p q p q p0 2m¡2 p0 2m¡2 eM 2 2m b 2m¡1 C m Eb f b . R 1¡q R b (5.44)

In the special case when p0 0, we have

2m ¡ 2m¡1 ¡ ¡ sup ebpfq eM ¤ CpmqEbpfq 2m 2 b m 1 , (5.45) M

¡m which gives a better bound than eM ¤ Cpmqb Ebpfq.

If (5.40) does not hold for any R ¤ RM , we have

b2m¡1 p q ¤ ¤ 2m¡1 sup eb f eM 2m b . (5.46) M RM

5.3.2 -Regularity Results

Now we apply the -regularity and get the bound of epfq in both cases in terms of the ﬁxed b:

Theorem 18 (Elliptic). Suppose f : M Ñ Rl¢l is a smooth solution to (5.28). There b P exists 0 2C for C depend only on M and l, such that if there exists some x0 M and R RM with BRpx0q M such that, » 1 Φpx0,Rq p1 νpr, xqqebpfq dvol ¤ 0, (5.47) m¡2 p0 R BRpx0q 69 then we have the estimate ¢ p q 1 2 c l b sup epfq sup |df| Wbpfq ¤ , (5.48) p q2¡p0 BδRpx0q BδRpx0q 2 δR

with p0 as deﬁned before, and δ ¤ 3{4.

1 Proof. The energy density epfq |df|2 W pfq satisﬁes the inequality in the ball 2 b p q Bρ0 x1 , where ρ0, x1 are deﬁned as in the previous -regularity proof. £ £ Cplq ∆ebpfq ¤ sup eb C2 eb, (5.49) b p q Bρ0 x1 with

sup ebpfq ¤ 4 sup ebpfq. p q p q Bρ0 x1 Bσ0 x0

Let ¢ 1 2 e0 sup ebpfq sup |df| Wbpfq ebpx1q. (5.50) BδRpx0q BδRpx0q 2

Then we have ¢ Cplq C˜plq ∆epfq ¤ e C e ¤ e epfq. (5.51) b 0 2 b 0

We can apply Lemma8 and Theorem 13 to get ¢ » ¢ ¡ Cplq ¡ 1 Cplq ¡ 2 p0 ¤ 2 p0 p q ¤ 2 p0 ρ e0 ρ e0 C2 m¡2 p eb f dvol ρ e0 C2 0. b 0 p q b ρ0 Bρ0 x1 (5.52)

¡ b Suppose Cplqρ2 p0 e {b 1. Picking where 0 0 0 2CpM, lq

m{2 CpM, lq Cplqp1 C2pMqq

70 can lead to a contradiction. Therefore,

b cplqb sup ebpfq ¤ , (5.53) p qp q2¡p0 p q2¡p0 BδRpx0q C l δR δR

C˜plq 1 with C and cplq . b C˜plq

We have the similar result for the parabolic case

l¢l Theorem 19 (Parabolic). Suppose ft : M ¢ r¡T, 0s Ñ R is a regular solution to parabolic equation (5.29). b There exists , where C only depends on M, and l, such that if for some 0 2C z0 px0, t0q P M ¢ r¡T, 0s, R RM , we have

» ¡ 2 » t0 R 1 p q p q p q 2 ¤ Ψ z0,R ν Gz0 x, t eb f ϕ dvol dt 0, (5.54) 2 p ¡ q 0 t0¡4R t0 t M

with PRpz0q P M ¢ r¡T, 0s, then we have the estimate ¢ p q 1 2 c l b sup epfq sup |dft| Wbpftq ¤ ¡ , (5.55) p q2 2ν0 PδRpz0q PδRpz0q 2 δR with δ only depending on M, inftR, 1u.

Proof is similar to the elliptic case.

5.3.3 Measure of Set with Large ebpfq

Definition 5.3.1. We define the Hausdorff d-measure at scale b to be # + 8 § 8 ¸ § ¤ dp q d § p q ¤ Hb S inf ri Bri xi S, ri b . (5.56) i1 i1

71 In this part we discuss the estimate of the Hausdorﬀ measure at scale b of the

8 set where ebpfq is large. In the previous section, we have seen that ebpfq is C with b ¡ 0 ﬁxed. In order to analyze the eﬀect when b Ñ 0, we are interested in the size of the set " * 1 Σ x P M|epfqpxq ¥ . (5.57) b b

Since ebpfq is at least continuous and Σb is a closed set in a compact manifold

M,Σb is compact.

Here we only discuss the case when p0pxq 0 and state the elliptic case. We have similar results with the parabolic metrics.

Theorem 20 (Elliptic). Fix a constant b ¡ 0. Let f be the solution to (5.28) with

finite energy Epfq. Define the set Σb as in (5.57). The Hausdorff pm¡1q-measure

of Σb at scale b is bounded and is independent of b

m¡1p q ¤ p q Hb Σb CE f . (5.58)

b Proof. From -regularity, there exists , for C only depends on m and M, 0 2C such that, if there exists R, » 1 p p qq p q ¤ b m¡2 1 ν r, x e f dvol 0 , (5.59) R BRpx0q 2C then p q p q ¤ 4c l b sup e f 2 . (5.60) BR{2px0q R

Here cplq is an absolute constant depending on l, the size of f.

For x0 P Σb, this implies

» a 1 p p qq p q ¡ b @ ¡ p q m¡2 1 ν r, x e f dvol R 4c l b. (5.61) R BRpx0q 2C 72 ( a p q p p q Let Brj xj j be a cover of Σb with rj 2 4c l b. Also assume b is small

enough so that rj RM {3. By Vitali’s covering lemma, we have ﬁnite sub-cover, still ( ¤J p q J P p q p q denoted as Brj xj j1 such that xj Σb, Brj xj are disjoint and B3rj xj Σb. j1 Inside each ball, we have » 1 p q ¡ b m¡2 e f dvol . (5.62) r p q 2C j Brj xj

That is » p p qq p q ¡ m¡2 2C 1 v r, x e f dvol brj . (5.63) p q Brj xj

Summing over all j, we get

¸J ¸J m¡1 m¡2 p qm¡1 3 brj 3rj j j » p m¡1q p p qq p q 2C 3 1 ν r, x eb f dvol p q Brj xj (5.64) » »

¤C p1 νpr, xqqebpfq dx ¤ 2C ebpfq dvol M M ¤CEpfq.

Since we can pick the initial map f0 such that Epf0q is bounded and f minimizes the energy functional E, Epfq is ﬁnite. We get

¸J m¡1p q ¤ p qm¡1 p q Hb Σb 3rj CE f , (5.65) j

which is ﬁnite.

Heuristically, Theorem 20 implies that as b Ñ 0, the dimension of singularity set of f will be at most m¡1.

73 5.4 Higher Power Potentials

We can see from Wbpfq that the smaller b is, the larger ebpfq can be. However, since

0 depends on b, 0 Ñ 0 when b Ñ 0. The -regularity results cannot be directly applied when b Ñ 0. We consider changing powers of f in the potential and deﬁne

1 ¨ W Lpfq trace pf 2L bIq¡1 ,L ¥ 1. (5.66) b 2L

We hope to get tighter estimates for the expected co-dimension of the limiting singular set. By changing powers in the potential, we can get arbitrarily close to 2, suggesting that the earlier co-dimension 1 result is an artifact of an interaction of the speciﬁc potential and our techniques.

L We consider the elliptic and parabolic equation for f with this potential Wb

¦ Lp q Lp q d df Wb f 0, with Eb f ﬁnite, (5.67)

and

Bf t d¦df W Lpfq 0, with f P Grpk, Clq and Epf q ﬁnite. (5.68) Bt b 0 0

It is obvious that cplq W Lpfq ¤ . (5.69) b Lb

Detailed computations give

¨ ¨ ¨ L ¡ p 2L q¡2 2L¡1 L ¡ 2L ¡2 2L¡1 Wb B trace f bI f B , ∇Wb f bI f . (5.70) t ul We write f in a local orthonormal basis ei i1

¸l b ¦ f λiei ei . (5.71) i

x L y The Hessian term of d∇Wb , df can be computed as follows x L y d∇Wb , df

¸ λ2L¡1 ¡ xdp i q, dλ y pλ2L bq2 i i i

¡ ¸ λ2L¡1 λ2L 1 ¡ xp i ¡ j qpλ ¡ λ q|xde , e y|2 pλ2L bq2 pλ2L bq2 i j i j i,j i j

¸ λ2L¡2 λ4L¡2 ¡ p2L ¡ 1q i |dλ |2 4L i |dλ |2 pλ2L bq2 i pλ2L bq3 i i i i (5.74) £ ¡ ¸ λ2L¡1 λ2L 1 ¡ pλ ¡ λ q¡1 i ¡ j pλ ¡ λ q2|xde , e y|2 i j pλ2L bq2 pλ2L bq2 i j i j i,j i j

¸ p2L 1qλ2L ¡ p2L ¡ 1qb λ2L¡2 i |dλ |2 i pλ2L bq3 i i i £ ¡ ¸ λ2L¡1 λ2L 1 ¡ pλ ¡ λ q¡1 i ¡ j pλ ¡ λ q2|xde , e y|2. i j pλ2L bq2 pλ2L bq2 i j i j i,j i j

p q 2L ¡ p ¡ q 1 2L¡2 2L 1 λi 2L 1 b 2 Observe that when |λ | ¡ b 2L , the ﬁrst term λ |dλ | i i p 2L q3 i λi b

1 is positive. When |λi b 2L |, we have § § § p q 2L ¡ p ¡ q § § 2L¡2 2L 1 λi 2L 1 b§ 1¡ 1 ¡2 λ ¤ cpLqb L b . (5.75) § i p 2L q3 § λi b

75 For the second term in (5.74), denote

λ2L¡1 gpλq . pλ2L bq2

Then £ ¡ λ2L¡1 λ2L 1 pλ ¡ λ q¡1 i ¡ j i j p 2L q2 p 2L q2 λi b λj b

p q ¡ p q (5.76) g λi g λj λi ¡ λj

1 ¤C|g pλq|, for some λ between λi and λj.

Since p2L 1qλ2L ¡ p2L ¡ 1qb g1pλq λ2L¡2 , pλ2L bq3

1 we have shown previously that when |λ| b 2L ,

1 ¡1¡ 1 |g pλq| ¤ cpLqb L

When |λ| ¥ b1{2L, we have

1 ¡2L¡2 ¡1¡ 1 |g pλq| ¤ cpLqλ ¤ cpLqb L .

Therefore, in general we always have § § L ¡1¡ 1 2 §x y§ ¤ p q L | | d∇Wb , df c l, L b df . (5.77)

cpl, Lq Since W L ¤ , we can conclude that eLpfq P C8pMq. We have two versions b b b ¡ x L y to control the term 2 d∇Wb , df inside the induced elliptic or parabolic equation Lp q Lp q of eb f , which will give us diﬀerent controls of eb f . We illustrate the idea using the elliptic equation.

76 When using Lp q ¡ | |2 ¡ | L|2 ¡ p q ¡ x L y ∆eb f ∇df ∇Wb Ric df, df 2 d∇Wb , df (5.78) ¡1¡ 1 L ¤ p q L c l, L b eb ,

we get the upper bound of sup e by using Moser iteration directly » ¡p1 1 q m ¡p1 1 q m sup epfq ¤ cpl, Lqb L 2 epfq dvol cpl, Lqb L 2 Epfq. (5.79) M M

5.4.1 -Regularity Results

1 1 When we want to estimate the Hausdorﬀ measure at scale b 2 2L of the set " * 1 ΣL x P M|eLpfq ¥ , (5.80) b b b

L we need a ﬁner estimate of eb Lp q ¡ | |2 ¡ | L|2 ¡ p q ¡ x L y ∆eb f ∇df ∇Wb Ric df, df 2 d∇Wb , df (5.81) ¡ 1 L 2 ¤ p q L p q c l, L b eb .

L Since the monotonicity formula (Lemma8) still holds for eb , we derive the - regularity results

Theorem 21 (Elliptic). Suppose f solves the elliptic equation (5.67) with ﬁnite Lp q energy Eb f .

1 b L There exists with C depending only on M, l, L, such that if there exists 0 2C R ρ, and » L 1 L Φ px0,Rq p1 νpr, xqqe pfq dvol ¤ 0, (5.82) b m¡2 p0 b R BRpx0q

¤ 3 then for any δ 4 ,

1 cpL, l, Mqb L Lp q ¤ sup eb f . (5.83) p q2¡p0 BδRpx0q δR

Here, νpr, xq and p0 are deﬁned as before. 77 Proof. The key to the proof lies in the elliptic inequality

L ¡ 1 L 2 p q ¤ p q L p q ∆eb f c l, L b eb . (5.84)

We apply Moser iteration to get

¢ m » C pM, l, Lq C pMq 2 L ¤ 1 2 Lp q sup eb 1 2 eb f dvol. (5.85) L p ¡ q BδRpx0q b R δR BRpx0q

This implies that the function hpσq satisﬁes

¢ { » C pM, l, Lqhpσ q m 2 1 p q ¤ 1 0 p q Lp q h σ0 1 C2 M e f dvol. (5.86) m¡2 p0 b L b R BRpx0q

1 b L We can pick ¤ , where CpM, l, Lq is determined by C pM, l, Lq and 0 2CpM, l, Lq 1

C2pMq. Then a contradiction argument implies

hpσq ¤ hpσ0q ¤ 1.

That is,

1 cpM, l, Lqb L Lp q ¤ sup eb f . (5.87) p q2¡p0 BδRpx0q δR

Theorem 22 (Parabolic). Suppose ft solves the parabolic equation (5.68) with ﬁnite Lp q initial energy Eb f0 .

1 b L There exists , with C depending only on M, l, L, such that if there exists 0 2C ? R with 0 R t0 ¡ t{2 and

» 2 » t0¡R 1 L 2 pp q q p qp q p q ¤ Ψ x0, t0 ,R ν G x0,t0 x, t eb ft ϕ dvol dt 0, (5.88) 2 p ¡ q 0 t0¡4R t0 t M

78 then for any δ depending only on M,R, we have

1 cpL, l, Mqb L Lp q ¤ sup eb f . (5.89) p q2¡2ν0 PδRpz0q δR

Here z0 px0, t0q, ν0 is the same ratio deﬁned in (2.10) and the theorem holds when

ν0 0.

We omit the proof here since it is an easy analog of the elliptic case.

Lp q 5.4.2 Measure of Set with Large eb f

In this part we assume p0 0. We consider the set " * 1 ΣL x P M|eLpfq ¥ , (5.90) b b b

Lp q L L Since eb f is continuous and Σb is a closed set in a compact manifold M,Σb is compact.

Theorem 23 (Elliptic). Fix a constant b ¡ 0. Let f be the solution to (5.67). ¨ L 2 p ¡ q L Deﬁne the set Σb as in (5.90). The Hausdorﬀ L 1 m 2 -measure of Σb at

1 1 scale b 2 2L is bounded and independent of b.

1 b L Proof. From -regularity, there exists , for C only depending on M, l, L, such 0 2C that, if there exists R,

» 1 1 b L p p qq Lp q ¤ m¡2 1 ν r, x eb f dvol 0 , (5.91) R BRpx0q 2C then

1 4cpM, l, Lqb L Lp q ¤ sup eb f 2 . (5.92) BR{2px0q R

79 For x0 P Σb, this implies

» 1 a 1 b L 1 p p qq p q ¡ @ ¡ 1 L m¡2 1 ν r, x e f dvol R 4cb (5.93) R BRpx0q 2C

( b 1 p q L p 1 L Let Brj xj j be a cover of Σb with rj 2 4cb . By Vitali’s covering ( p q J P L lemma, we have a ﬁnite sub-cover, still denoted as Brj xj j1 such that xj Σb ,

¤J p q p q L Brj xj are disjoint and B3rj xj Σb . Inside each ball, we have j1

» 1 1 b L Lp q ¡ m¡2 eb f dvol . (5.94) r p q 2C j Brj xj

That is » » L L 1 m¡2 p p qq p q ¡ p q ¡ L 2C 1 v r, x eb f dvol 2C eb f dvol b rj . (5.95) p q p q Brj xj Brj xj

Summing over all j, we get

¸J ¸J 2 pm¡2q 1 m¡2 2 pm¡2q L 1 L p q L 1 3 b rj 3rj j j » 2 pm¡2q p L 1 q p p qq Lp q 2C 3 1 ν r, x eb f dvol p q Brj xj (5.96) » » ¤ p p qq Lp q ¤ p q C 1 ν r, x eb f dx 2C e f dvol M M ¤ Lp q CEb f0 .

Since we can pick f0 such that Epf0q is bounded and f minimizes the energy functional E, we get

¸J p 2 p ¡ q 2 L 1 m 2 pm¡2q p Lq ¤ p q L 1 Lp q Hb Σb 3rj CEb f0 . (5.97) j 80 2 1 L p ¡ q L 2L which is ﬁnite. Therefore, the Hausdorﬀ L 1 m 2 -measure of Σb at scale b is bounded.

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83 Biography

Wong, Ching-Yin (Lizzy), born on January 27, 1988 in Guangzhou (Canton), China, is a Ph.D. candidate from the Department of Mathematics at Duke University. She received her bachelor of science in mathematics and bachelor of economics from Peking University, China in 2010. Her research focuses on deriving analytic techniques in solving partial diﬀerential equations, especially the ones that arise from geometric problems. During her time at Duke, Lizzy has given talks about her research at Duke as well as in other regional conferences. She was awarded the Outstanding Graduate Presentation Award in November 2016 for her excellent presentation on The Geometry of Tension in the UNCG Mathematics and Statistics Regional Con- ference. As a graduate instructor, Lizzy has taught more than ten mathematics courses and has received overwhelmingly positive evaluations from Duke students. Besides research and academic activities, Lizzy has been an active member in several communities at Duke. She has been volunteering for Duke International House for over three years, providing consulting services, organizing orientations, as well as synthesizing local information as technical blogs which have served more than 3000 international students. Lizzy sang in Duke Chapel Choir for two years, having performed Messiah and other choral pieces to the community. She also volunteered to serve in the Sunday worship services. Apart from volunteering, Lizzy also worked for Duke Intramurals Program and Duke Peer Tutoring Program from where she both has received positive feedback and recognition as an outstanding employee.