Almgren’s frequency function and uniquecontinuation

Ali Raad

University of Cambridge

April 2016

Contents

Introduction and Statement of the Theorems 1

How to read this essay 7

1 Notation and Preliminaries 91.1 Lp and Sobolev Spaces . . . . . . . . . . . . . . . . . . . . . . . 91.2 Elliptic Partial Differential Equations Theory . . . . . . . . . . 121.3 Tensor Calculus on Manifolds . . . . . . . . . . . . . . . . . . . 151.4 Hausdorff measure . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2 Proving that the modified Almgren frequency function isnon-decreasing 252.1 Proof of Theorem 0.0.5 . . . . . . . . . . . . . . . . . . . . . . . 252.2 Some results from [AKS62] . . . . . . . . . . . . . . . . . . . . . 362.3 Tying it all up . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3 Proving strong unique continuation 42

4 A non-geometric approach to the Almgren frequency func-tion and some corollaries 46

1

Introduction and Statement ofthe Theorems

The aim of this essay is to prove strong unique continuation for weak solu-tions u ∈ W 1,2

loc (Ω) to a certain elliptic operator L (Ω and L will be definedlater). That is, if u satisfies Lu = 0 and vanishes to infinite order at a pointin Ω, then u is identically 0 in Ω. We will denote by Dr the open ball in Rn

of radius r, and by Br the closed ball of radius r, centred at the origin.

Definition 0.0.1. A function u ∈ L2loc(Ω) is said to vanish to infinite order

at x0 ∈ Ω if for all δ > 0 sufficiently small and natural numbers j ∈ N we havethat

∫

Dδ(x0)

u2 = O(δj). (1)

The aim is thus to prove

Theorem 0.0.2 (Strong Unique Continuation). If a solution u ∈ W 1,2loc (Ω)

to equation (7) defined below vanishes to infinite order at x0 ∈ Ω, then u isidentically 0 in Ω.

To achieve this we are going to make use of the doubling condition:

Theorem 0.0.3 (Doubling Condition). Assume that u ∈ W 1,2loc (Ω) solves

equation (7). Then there exists a constant C depending on n,K, λ and usuch that for every 0 ≤ R < 1

2 we have

2

∫B2R

u2 ≤ C ∫BRu2. (2)

The constants λ and K will be introduced shortly. We will not assume thedoubling condition, and so in order to prove it we need to prove that themodified Almgren frequency function is non-decreasing - see Theorem 0.0.5below.

These results appear in [GL86] and this essay will follow the paper of Garo-falo and Lin closely. However, the proofs in [GL86] are very dense (somederivations are stated without proof) and require a lot of prerequisite knowl-edge of tensor calculus on manifolds, none of which we assume. Hence thisessay will prove the same results but will include a lot of what [GL86] leavesout. We also include a section on tensor calculus in Chapter 1. A novelty inthis essay is a proof that the Almgren frequency function is non-decreasingfor the Laplacian operator (see Theorem 0.0.6 below) without any use of thedifferential geometry in [GL86]. All that follows in this introduction is basedon the setup adopted in [GL86].

• Ω is a smooth, connected and bounded open subset of Rn for n ∈ Nwith n ≥ 3. We will henceforth assume that

B2 = x ∈ Rn such that ∣x∣ ≤ 2 ⊂⊂ Ω. (3)

• L is the operator defined by

L = div(A∇), (4)

where A(x) = (aij(x)) is a symmetric n × n matrix on Ω. L satisfiesthe following conditions:

1. The entries of A are real and Lipschitz continuous. That is, ∃K > 0such that ∀x, y ∈ Ω we have that

∣aij(x) − aij(y)∣ ≤ K∣x − y∣ for all i, j ∈ 1,2, . . . n. (5)

2. L is a strictly elliptic operator. In particular, we assume ∃λ ∈ (0,1)such that for every x ∈ Ω and for all ζ ∈ Rn we have

λ∣ζ ∣2≤

n

∑i,j=1

aij(x)ζiζj ≤1

λ∣ζ ∣

2. (6)

3

• We consider non-trivial weak solutions u ∈W 1,2loc (Ω) to the equation

Lu = div(A(x)∇u(x)) = 0. (7)

When we say that a solution u satisfies (7) we will always mean, unlessotherwise stated, that it does so in the weak, or generalized, sense; i.e.for every v ∈W 1,2

0 (Ω) we have that

∫Ω

⟨A∇u,∇v⟩ = 0. (8)

A special example is when A = Id so that

Lu = ∆u = 0. (9)

This is Laplace’s equation. One can define two important quantities relatedto this. The first is the Dirichlet energy

D(r) ∶= ∫Dr

∣∇u(x)∣2dx. (10)

The second isH(r) ∶= ∫

∂Dru2 dHn−1, (11)

whereHn−1 is the (n−1)-dimensional Hausdorff measure on ∂Dr (see Chapter1, Section 4). Using these two quantities we define the Almgren frequencyfunction, for r ∈ (0,1) and H(r) ≠ 0, by

N(r) ∶= rD(r)

H(r). (12)

In R2 consider the harmonic functions given in polar co-ordinates by uk(r, θ) =bkrk sin(kθ), where k ∈ N and bk ∈ C. They are harmonic because they formthe imaginary parts of the analytic functions fk(z) = bkzk. It is an easycalculation to check that N(r) = k for uk. That is why it was named thefrequency function.

Almgren discovered (see [A79]) that N(r) was a non-decreasing functionof r ∈ (0,1). This is one of the theorems this paper will prove. For ageneral operator of the form (4) however, we want a slight modification of

4

the frequency function. Proving that this modification is also non-decreasingin r ∈ (0,1) is best done by introducing a Lipschitz Riemannian metric tensoron D1,

n

∑i,j=1

gij(x)dxi ⊗ dxj. (13)

We will also assume that under a co-ordinate transformation to polar co-ordinates (r, θ1, . . . , θn−1) this metric tensor takes the form

dr ⊗ dr + r2n−1

∑i,j=1

bij(r, θ)dθi ⊗ dθj. (14)

We assume that the bij’s are real-valued functions on D1 and that

bij(0,0) = δij, for i, j ∈ 1, . . . , n − 1, (15)

where δij is the Kronecker-delta symbol. Additionally we assume that thereis a positive constant Λ such that

∣∂bij(r, θ)

∂r∣ ≤ Λ, for i, j ∈ 1, . . . , n − 1. (16)

On the closed unit ball, B1, we will assume there is given a real-valuedLipschitz function τ satisfying the following two conditions:

1. There are positive constants A and B such that on B1 we have:

A ≤ τ ≤ B. (17)

2. In polar co-ordinates on D1

τ(0,0) = 1, ∣∂τ

∂r∣ ≤ Λ a.e. (18)

It is justified to wonder why we are making these assumptions. It turns out(as we will show in Chapter 2, Section 3) that considering weak solutions (inW 1,2(D1)) to

divM(τ(x)∇Mu(x)) = 0, (19)

with respect to the metric tensor (13), is equivalent to considering solutionsin W 1,2(D1) to

Lu = 0. (20)

5

The notation divMX denotes the intrinsic divergence of a vector field X onD1, and ∇Mu denotes the intrinsic gradient of a function u. Of course, wewill have to define a specific form of the metric tensor that will be compatiblewith (4), (5) and (6), but this will be done in Chapter 2, Section 3. Again,anytime we mention (19) we will always be assuming, unless stated otherwise,that we consider the equation in the weak sense.

Notice how we in (7) look for solutions in W 1,2loc (Ω) but have in (20) switched

to look for solutions in W 1,2(D1). This is because in (3) we assume B2 iscompactly contained in Ω, and so for the sake of simplification we will onlydeal with solutions in W 1,2(D1).

Under the new assumptions we define the general forms of (10) and (11) via:

Definition 0.0.4. Let r ∈ (0,1). Let dVDr and dV∂Dr be the Riemannianvolume elements on Dr and ∂Dr, respectively (as defined in Section 3 ofChapter 1). Define the following:

D(r) ∶= ∫Dr

τ(x)∣∇Mu(x)∣2dVDr , (21)

H(r) ∶= ∫∂Dr

τ(x)u2(x)dV∂Dr . (22)

Now define the generalized Almgren frequency function as

N(r) = rD(r)

H(r), (23)

for H(r) ≠ 0.

We can now state the first theorem that this essay will prove:

Theorem 0.0.5. If u ∈W 1,2(D1) is a non-trivial weak solution of equation(19), then there is a positive constant C depending on n and Λ such that themodified Almgren frequency function

N (r) ∶= exp (Cr)N(r) (24)

is a non-decreasing function of r ∈ (0,1).

6

Finally, we state the last of the main theorems this essay proves:

Theorem 0.0.6. Let u ∈ C2(Ω) be a non-trivial solution of ∆u = 0 in D1.Then N(r), as defined in (12), is non-decreasing in r ∈ (0,1).

7

How to read this essay

The level of this essay is intended for a general graduate student in mathemat-ics who has been exposed to the basics of analysis and differential equationstheory.

Therefore, we do not assume extensive knowledge of Sobolev spaces, tensorcalculus in differential geometry, nor Hausdorff measures. If you have notcovered any of these topics before, it is advised that you read the relevantsections of Chapter 1 before proceeding. If you have covered these thingshowever, you can safely skip Chapter 1, but be aware that starting fromChapter 2 we will always be referring to the results of Chapter 1.

Only once you know the material of Chapter 1 will the Introduction andStatement of the Theorems be useful. It provides a concise display of themain theorems this essay will prove, and will be referred to considerably.

Chapter 1 covers, without proofs, the prerequisite knowledge required to un-derstand the subsequent chapters. Section 1 covers Lp and Sobolev spaces,which are used throughout the essay. Section 2 covers some results from ellip-tic partial differential equations theory from which we can make assumptionsthat will be crucial for the rest of the essay. Section 3 covers the results fromtensor calculus on manifolds required to understand the proofs of Chapter 2.Section 4 covers a very concise account of Hausdorff measures, used mainlyto give the reader an intuition of the type of measure we integrate against.

Chapters 2, 3 and 4 provide the proofs of Theorems 0.0.5, 0.0.2 and 0.0.6,respectively. The results of Chapter 2 are needed for Chapter 3, but Chapter4 can be read in isolation.

8

Chapter 1

Notation and Preliminaries

1.1 Lp and Sobolev Spaces

This essay will consider the operator L defined in (4) and certain solutionsu to Lu = 0. Demanding that these solutions be, for example, infinitelydifferentiable would restrict the type of functions that can solve Lu = 0.To remedy this we want to weaken the classical notion of differentiabilityof functions to allow for more “room” for solutions. A specific space offunctions we will consider closely in this essay is the Sobolev space W 1,2(Ω),where Ω ⊆ Rn. In this section we define this space. To this end we must startwith the definition of a multi-index.

Everything that follows in this section (unless otherwise stated) can be ex-plored in greater detail in sections 5.1, 5.2 and Appendix A of [E02].

Definition 1.1.1. Let u ∶ Ω → R be a function, where Ω ⊆ Rn. Let x =

(x1, . . . , xn) ∈ Ω. A multi-index α is a vector of the form α = (α1, . . . , αn),where each αi is a non-negative integer. The order of the multi-index is

∣α∣ =n

∑i=1αi. We define

Dαu(x) ∶=∂ ∣α∣u(x)

∂x1α1 . . . ∂xnαn

. (1.1)

9

We define, for k a non-negative integer,

Dku(x) ∶= Dαu(x) ∶ ∣α∣ = k. (1.2)

We let Du(x) ∶=D1u(x).

Now we define the Lp spaces for 1 ≤ p ≤∞.

Definition 1.1.2. Let Ω ⊆ Rn, and 1 ≤ p <∞. We define

Lp(Ω) ∶=

⎧⎪⎪⎪⎨⎪⎪⎪⎩

u ∶ Ω→ R s.t. ∥u∥Lp(Ω) ∶=⎛

⎝∫Ω

∣u∣p⎞

⎠

1p

<∞

⎫⎪⎪⎪⎬⎪⎪⎪⎭

. (1.3)

For p =∞, we define

L∞(Ω) ∶= u ∶ Ω→ R s.t. ess supΩ

∣u∣ <∞. (1.4)

By essential supremum, ess supΩ

∣u∣, we mean the infimum value of all possible

suprema of u taken over Ω/Z where Z has measure 0. We also define localsummability. For 1 ≤ p ≤∞ define

Lploc(Ω) ∶= u ∶ Ω→ R s.t. u ∈ Lp(V ) whenever V ⊂⊂ Ω. (1.5)

Note that by V ⊂⊂ Ω we mean that V ⊂ V ⊂ Ω and V is compact.

Definition 1.1.3. Let 1 ≤ k <∞. We define the spaces

Ck(Ω) = u ∶ Ω→ R such that Dαu exists and is continuous ∀ ∣α∣ ≤ k,(1.6)

Ck(Ω) = u ∈ Ck(Ω) ∶Dαu is uniformly continuous

on bounded subsets of Ω, ∀ ∣α∣ ≤ k,(1.7)

C∞(Ω) =∞⋂k=1

Ck(Ω), (1.8)

C∞(Ω) =∞⋂k=1

Ck(Ω), (1.9)

CkC(Ω) = u ∈ Ck(Ω) ∶ u vanishes outside a compact subset of Ω. (1.10)

A test function on Ω is a function v ∈ C∞C (Ω).

10

Now we want to go on to discuss norms on such spaces in order to define theHolder space Ck,α(Ω).

Definition 1.1.4. For bounded and continuous u ∶ Ω→ R, we define

∥u∥C(Ω) ∶= supΩ

∣u∣ . (1.11)

We define also

[u]C0,α(Ω) ∶= supx,y∈Ω,x≠y

∣u(x) − u(y)∣

∣x − y∣α . (1.12)

Now define the αth Holder norm as

∥u∥C0,α(Ω) = ∥u∥C(Ω) + [u]C0,α(Ω). (1.13)

We are in a position to define Holder spaces now. The Holder space Ck,α(Ω)

consists of all functions u ∈ Ck(Ω) for which the norm

∥u∥Ck,α(Ω) ∶= ∑∣β∣≤k

∥Dβu∥C(Ω) + ∑∣β∣=k

[Dβu]C0,α(Ω) <∞. (1.14)

Henceforth anytime we mention the space Ck,α(Ω) we implicitly assume0 < α < 1, unless otherwise stated. When we mention Ck,α(Ω) it shouldbe taken to mean approximately the same as Ck,α(Ω) but where the treat-ment is local rather than global. We now move on to study weak derivatives.

Definition 1.1.5. Let u and φ be elements of L1loc(Ω) for Ω ⊆ Rn. Let α be a

multi-index of order k. We say that φ is the αth weak partial derivative of u,and write Dαu = φ, if the follwing equation is satisfied for all test functionsv:

∫Ω

uDαv = (−1)k ∫Ω

φv. (1.15)

To really appreciate the subtlety of such a definition, we need to see an ex-ample.

Example 1.1.6. Let Ω = (−1,1) ⊂ R. Let α = (1). Let u be the functiondefined on (−1,1) by u(x) = ∣x∣. Clearly u fails to be differentiable at x = 0.However I claim that the function φ, defined by being equal to −1 on −1 <

11

x < 0, and 1 on 0 ≤ x < 1, is a weak (first) derivative. Indeed, for all testfunctions v on (−1,1) we have

1

∫−1

u(x)v′(x)dx =

0

∫−1

−xv′(x)dx +

1

∫0

xv′(x)dx

=

0

∫−1

v(x)dx −

1

∫0

v(x)dx

= −

1

∫−1

φ(x)v(x)dx,

(1.16)

where in the second line we used integration by parts and the fact thatv(−1) = v(1) = 0.

Example 1.1.6 highlights that we can now have a notion of derivatives forfunctions which are not differentiable in the classic sense. It also illustrateshow Definition 1.1.5 comes about from the integration by parts formula. In-deed, if u is ∣α∣ times differentiable in the classic sense, (1.15) is equivalentto the classic integration by parts formula.

Definition 1.1.7. Let k be a non-negative integer and let 1 ≤ p ≤ ∞. LetΩ ⊆ Rn. Then the Sobolev space W k,p(Ω) is the space of functions u ∈ L1

loc(Ω)

for which for each multi-index α with ∣α∣ ≤ k, we have that the αth- weakpartial derivative Dαu exists, and belongs to Lp(Ω).

We define W k,p0 (Ω) to be the closure of C∞

C (Ω) in W k,p(Ω).

1.2 Elliptic Partial Differential Equations The-

ory

In this section I include some fundamental results from partial differentialequations theory that we will utilize in Chapter 2. The first one is Theo-rem 8.8 in [GT01]. Note that rather than stating the theorems in the mostgeneral form as they appear in [GT01], I will state them with our specific

12

operator L in mind.

Theorem 1.2.1 (W 2,2-elliptic regularity). Let L be the operator (4) satis-fying (5) and (6). Let Ω ⊂ Rn be smooth and connected. For u ∈ W 1,2(Ω)

satisfying (7) we have that u ∈W 2,2loc (Ω).

The importance of Theorem 1.2.1 is that since we are assuming B2 ⊂⊂ Ω, allthe solutions we will be dealing with will actually belong to W 2,2(D1). Hencewhen we take second (weak) derivatives in all the calculations of Chapters2 and 3, and implicitly assume they belong to L2(D1), we are not breakingany rules. The next is Theorem 8.19 from [GT01], useful for us in Chapter2.

Theorem 1.2.2 (Strong Maximum Principle for Weak Solutions). Let L bethe operator (4) satisfying (5) and (6). Let Ω ⊂ Rn be smooth and connected,and u ∈W 1,2(Ω) satisfying (7). If for some ball b ⊂⊂ Ω we have that

supbu = sup

Ωu ≥ 0, (1.17)

then u is constant on Ω.

The next theorem ensures that solutions in W 1,2(Ω) to equation (7) are au-tomatically in C0,α(b) for every b ⊂⊂ Ω. It is Theorem 8.24 in [GT01].

Theorem 1.2.3 (Interior Holder Estimate). Let L be the operator (4) sat-isfying (5) and (6). Let Ω ⊂ Rn be smooth and connected. Let u ∈ W 1,2(Ω)

satisfying (7). Then for any b ⊂⊂ Ω, we have that u ∈ C0,α(b) for someα = α(n,λ).

The following two theorems, which are Theorem 8.32 and Theorem 8.34 in[GT01], respectively, will ensure, together with Theorem 1.2.3 above, thatthe solutions we are looking for belong to C1,α(B1).

Theorem 1.2.4. Let L be an operator of the type (4) satisfying (5) and(6), with M a bounded domain. Let u ∈ C1,α(M) satisfy (7). Then for anyb ⊂⊂M we have that

∥u∥C1,α(b) ≤ C∥u∥C(M), (1.18)

13

where C depends on n,λ,K and dist(b, ∂M).

Theorem 1.2.5. Let S be a C1,α domain and L an operator of the form (4)satisfying (5) and (6). Let g ∈ L∞(S), f i ∈ C0,α(S) and Φ ∈ C1,α(S). Thenthe generalized Dirichlet problem (in the weak sense)

Lu = g +n

∑i=1

∂f i

∂xiin S, (1.19)

u = Φ on ∂S, (1.20)

is uniquely solvable in C1,α(S).

Finally, we need a lemma before completing our argument. The proof of itis straightforward by basic calculus.

Lemma 1.2.6. Let Ω ⊂ Rn be smooth and connected. Let f ∈ C0,1(Ω). ThenDf ∈ L∞(Ω).

Now consider the following argument. Let the S of Theorem 1.2.5 be D 32.

For our purposes we do not need to know what a C1,α domain is, but the ballbeing smooth certainly satisfies this. Let L be the operator in (4) satisfying(5) and (6). Let g = −div(A∇χ)u and f i = (2uA∇χ)i, where χ is a smoothcut-off function on D 3

2that equals 1 on B1 and 0 near ∂D 3

2. Let Φ = 0. Then

for solutions u ∈ W 1,2loc (Ω) satisfying equation (7) with Ω as in (3), we have

by Theorem 1.2.3 that u ∈ C0,α(B 32), for the α in Theorem 1.2.3.

Using this implies that each f i is C0,α(B 32), because the aij’s and ∇χ are

bounded. Also, note that by Lemma 1.2.6 we have that g ∈ L∞(D 32). Now

14

calculate in the weak sense

L(χu) = div(A∇(χu))

= div(uA∇χ + χA∇u)

= div(uA∇χ) + div(χA∇u)

= div(uA∇χ) + ⟨∇χ,A∇u⟩ + χdiv(A∇u)

= div(uA∇χ) +n

∑i=1

∂χ

∂xi(n

∑j=1

aij∂u

∂xj)

= div(uA∇χ) +n

∑i=1

n

∑j=1

aij∂χ

∂xi

∂u

∂xj

= div(uA∇χ) +n

∑j=1

∂

∂xj(n

∑i=1

aij∂χ

∂xiu) −

n

∑j=1

∂

∂xj(n

∑i=1

aij∂χ

∂xi)u

= div(uA∇χ) + div(uA∇χ) − div(A∇χ)u

= 2div(uA∇χ) − div(A∇χ)u

=n

∑i=1

∂f i

∂xi+ g,

(1.21)

where we have used that div(A∇u) = 0, the fact that A is symmetric, andthe product rule. Theorem 1.2.5 now applies and so the generalized problemL(uχ) = 0 is uniquely solvable in C1,α(B 3

2). Using this in Theorem 1.2.4

implies that uχ ∈ C1,α(B1) and so u ∈ C1,α(B1), as χ is just 1 there.

The conclusion of this entire argument is that we can, in the rest of thisessay, assume that u is continuously differentiable on B1 and so whenever wetake derivatives of u, or derivatives of some form of its integral on B1, weknow that the procedure is justified without further discussion.

1.3 Tensor Calculus on Manifolds

The aim of this section is to quickly summarize the essentials of tensor cal-culus on smooth manifolds in order to understand the proofs in this essay.Most of what I will describe in this section is found in [DP10] and in chapters2 and 3 of [L97], unless otherwise stated. The aim here is to gain an intuitionfor what tensors are and how we can use them.

15

In what follows, and for the rest of this essay, we will use Einstein summationconvention. For example, a summation on indexed coefficients of the form

n

∑β=1

n

∑γ=1

AαβBβγCγδ, (1.22)

will be denoted byAαβBβγCγδ. (1.23)

In general, we assume that a summation takes place over indices which appearat least twice in the expression, and no summation is assumed over indicesthat appear once.

Assume for now that we are in Rn and that we have a (not necessarilyorthonormal) system of basis vectors E1, E2, . . . , En. Let

E =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

E1

E2

⋮

En

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

(1.24)

Imagine we transform our co-ordinate system linearly:

F = AE, (1.25)

where A is some invertible matrix. A straight forward calculation using linearalgebra would show that if X is a vector in Rn with components expressedin terms of the first co-ordinate system, and V is the corresponding vectorin the new co-ordinate system, then

V = (A−1)T X. (1.26)

If we had an orthonormal co-ordinate system then (A−1)T = A and so thevectors would vary in “conformity” with the change of co-ordinates. In gen-eral this will not be the case, and so normal vectors vary “contrary” to thebasis transformation.

However, for certain types of vectors, the transformation is always in “con-formity”. Gradient vectors are such an example. Let f ∶ Rn → R be a

16

bounded smooth function. Let the co-ordinate system in Rn be denoted byx = (x1, x2, . . . , xn). With the basis transform described earlier, we havethat

vµ = Bµγxγ, (1.27)

where for ease of notation B = (A−1)T from equation (1.26), and v representsa vector in terms of the new co-ordinates. Let

wα =∂f

∂xα, (1.28)

and let

zα =∂f

∂vα. (1.29)

Another simple calculation using the chain rule shows that

z = (B−1)T w. (1.30)

But (B−1)T = A and therefore we see that gradient vectors vary in “con-formity” with the change of co-ordinates. This is why in literature normalvectors are also sometimes called contravariant vectors and gradient vectorsare called covariant vectors. When given a co-ordinate chart on a mani-fold, the components of contravariant vectors are usually upper-indexed: xγ,whilst those for covariant vectors are lower-indexed: wα. This makes summa-tion using the Einstein convention very convenient. On a simple vector spacehowever, like Rn, one usually denotes the normal vectors by X = (X1, . . . ,Xn)

and the vectors of the covariant (dual) space by ω = (ω1, . . . , ωn). This alsomakes Einstein summation more convenient.

It is easily checked that the usual inner product between a contravariantand covariant vector, ωβXβ, remains unchanged after a basis transformation.However the same is not true for the inner product between two contravariantvectors, XβYβ. But we can modify this to some gµγXµYγ under which takingan inner product becomes invariant under the basis transformations. Theobject gµγ is a covariant 2 -tensor as it “acts” on two contravariant vectorsand yields a real number. We can also have an l-contravariant tensor whichacts on l covariant vectors and yields a real number.

With this intuition at hand we are now ready to rigorously define tensors onfinite dimensional real vector spaces. The notation in the upcoming defini-tions, including what is upper and lower indexed, is what we will adopt in

17

the remainder of this essay.

Definition 1.3.1. Let V be a real vector space. Let V ⋆ denote the dual ofV , i.e. the space of covariant vectors. A k-covariant, l-contravariant tensor,or a tensor of type (

kl), is a multilinear map

g ∶ V ⋆ × . . . × V ⋆ × V × . . . × V → R, (1.31)

where the V ⋆’s appears l times and the V ’s appear k times.

The space of all tensors of type (kl) is denoted by T kl (V ).

The tensor product of two tensors F ∈ T kl (V ) and G ∈ T rs (V ) is F ⊗ G ∈

T k+rl+s (V ), defined by:

F ⊗G(ω1, . . . , ωl+s,X1, . . . ,Xk+r) =

F (ω1, . . . , ωl,X1, . . . ,Xk)G(ωl+1, . . . , ωl+s,Xk+1, . . . ,Xk+r).(1.32)

If e1, . . . , en is a basis set for V , we define the dual basis for V ⋆ byφ1, . . . , φn via φi(ej) = δij.

The basis elements for the space T kl (V ) are defined by the tensors of the form

ej1 ⊗ . . .⊗ ejl ⊗ φi1 ⊗ . . .⊗ φik , (1.33)

where j1, . . . , jl, i1, . . . , ik ∈ 1, . . . , n. These are defined to act on basis ele-ments of V and V ⋆ by:

ej1⊗ . . .⊗ejl⊗φi1⊗ . . .⊗φik(φs1 , . . . , φsl , er1 , . . . , erk) = δs1j1 . . . δsljlδi1r1 . . . δikrk .

(1.34)We denote by Λk(V ) the space of k-forms on V . This is the space of k-covectors that change sign whenever two arguments are interchanged. Wedefine the wedge product on 1-forms ω1, . . . , ωk by setting

ω1 ∧ . . . ∧ ωk(X1, . . . ,Xk) = det(ωi(Xj)). (1.35)

With the above definition at hand any tensor F ∈ T kl (V ) can be written interms of the basis elements by

F = F j1...jli1...ik

ej1 ⊗ . . .⊗ ejl ⊗ φi1 ⊗ . . .⊗ φik . (1.36)

18

We want to use our knowledge now to define the Riemannian metric tensor.In general, we would like to have tensors defined on manifolds with some givenco-ordinate chart, and not just finite dimensional vector spaces. Knowledgeof what exactly a manifold is (and all related information about local co-ordinate charts, partitions of unity, functions on manifolds etc.) will not befundamental to this essay, and so the precise definitions are omitted. If youwould still like to find out, a good source is Chapter 1 of [L03]. For now, thinkof them as “nice” structures in Rn with some specified co-ordinate system.Note that because we are limited by our wish not to involve an extensivediscussion on manifolds, what follows is just meant for intuition and is notentirely rigorous.

For a manifold M , one can think of the space of tangent vectors at a pointp of the manifold, call it TpM . This is a vector space, and so it would makesense to define a (

kl)-tensor at p ∈M to be an element of T kl (TpM).

We will also need to have the intuition of a bundle of (kl) tensors on M ,

written asT kl (M) = ⊍

p∈MT kl (TpM). (1.37)

It is important to note that the tensor bundle is not just a disjoint union asin (1.37); there is a smooth structure that needs to go along with this, butsince this section is meant mainly for intuition, we will omit the details ofthis. Similarly, we can think of a bundle of k-forms on M , written as

Λk(M) = ⊍p∈M

Λk(TpM). (1.38)

And finally, we will let TM denote the tangent bundle of a smooth manifoldM , which can be thought of as a disjoint union (with respect to p ∈ M) ofTpM (where again there is a smooth structure that needs to go with this!)

If xi are local co-ordinates on a subset U ⊂ M , and p ∈ U , then the set ∂∂x1 , . . . ,

∂∂xn forms a basis for TpM , and the basis of the covariant space

(the dual space) is dx1, . . . ,dxn.

A local frame e1, . . . , en for TM are n smooth vector fields defined onsome open set U ⊂ M , such that e1∣p , . . . , en∣p forms a basis for TpM ateach p ∈ M . The dual coframe, φ1, . . . , φn, are smooth 1-forms satisfyingφi(ej) = δij.

19

A smooth section of some tensor bundle T kl (M) is what is known as a tensor

field on M . The space of (kl) tensor fields is denoted by T kl (M). For a co-

ordinate frame ∂∂xi

and dual coframe dxi, a (kl)- tensor field F evaluated

at a point p ∈M has co-ordinate expression

F ∣p = Fj1...jli1...ik

(p)∂

∂xj1⊗ . . .⊗

∂

∂xjl⊗ dxi1 ⊗ . . .⊗ dxik , (1.39)

where we use F ∣p to emphasize that this is where the tensor field is evaluated,in order not to confuse it with the vectors or covectors to which it is applied.We now define the Riemannian metric, adopting definition 9.1.1 of [A]:

Definition 1.3.2. A Riemannian metric on a smooth manifold M is anelement of T 2(M) in the sense that it is a smoothly chosen inner product oneach tangent space TpM , that satisfies the following:

• Symmetry: g(Xp, Yp) = g(Yp,Xp) for all Xp, Yp ∈ TpM , and

• Positive semi-definiteness: g(Xp,Xp) ≥ 0 for all Xp ∈ TpM , and equalityholds if and only if Xp = 0.

By smoothly chosen we mean that for smooth vector fields X and Y , the mapp→ g(Xp, Yp) is smooth.

For this essay we will need the notion of a Lipschitz metric tensor. This issimply a Riemannian metric such that when written as

gij dxi ⊗ dxj, (1.40)

for a co-ordinate frame ∂∂x1 , . . . ,

∂∂xn and dual coframe dx1, . . . ,dxn, then

gij = g(∂∂xi, ∂∂xj

) is Lipschitz for all i, j ∈ 1, . . . , n.

I now state five essential derivations that will be very important for the mainproofs of this paper. I will give quick remarks on how one can prove somethem. All can be found in Chapter 3 of [L97], except Lemma 1.3.8, which Ihave extracted from page 387 of [H64].

Lemma 1.3.3 (Intrinsic Gradient). Let M be a smooth manifold on whichwe have a Riemannian metric g. Let f ∶M → R be a smooth map. Then

∇Mf = gij∂f

∂xi∂

∂xj, (1.41)

20

where ∇Mf denotes the intrinsic gradient of f on the manifold, gij(x) is theinverse of gij(x), and ∂

∂xi is our co-ordinate frame basis.

Remark 1.3.4. Using the definition of a Riemannian metric notice thatwe can create covariant vectors from contravariant ones and vice-versa. Forinstance, we can define ω(Y ) = g(X,Y ) for X,Y ∈ TM . Note that ω iscovariant as it acts on contravariant vectors. In particular, expanding X inco-ordinates as X =X i ∂

∂xi, we get

ω(⋅) = g(X, ⋅) = g(X i ∂

∂xi, ⋅) =X igij dxj ∶=Xj dxj. (1.42)

This construction of creating a covariant vector is called lowering an index.In a similar fashion one can raise an index and create contravariant vectorsfrom covariant ones. After having raised an index we usually denote ω by ω#.The intrinsic gradient of f can then be obtained via ∇Mf = df#. Unpackingthe definitions yields Lemma 1.3.3.

Another important point to discuss is the fact that the lemma assumes fis smooth. The types of functions u we will be dealing with however are inW 1,2

loc (Ω) and we would still want to apply the lemma. In order to do so wecan use a sequence of mollifiers on u which make it smooth (see [E02], Ap-pendix C, Theorem 6), then insert these smooth functions in (1.41) and passto a limit. Since Theorem 1.2.4 ensured us that on B1 the functions u we areconsidering will be C1,α, apply part (iii) of of Theorem 6 in [E02] and we getthat Lemma 1.3.3 also applies for u ∈W 1,2(D1), which is what we care about.

Now we turn to inner products of tensors, which we will use in Chapter 2.

Lemma 1.3.5 (Tensor Inner Product). Let xi be any local co-ordinatesystem on a manifold M , and g the Riemannian metric. Let F,G ∈ T kl (M)

be expressed as in (1.36). Then the tensor inner product is given by

⟨F,G⟩ = gi1r1 . . . gikrkgj1s1 . . . gjlslFj1...jli1...ik

Gs1...slr1...rk

. (1.43)

Lemma 1.3.6. Let M be an oriented n-dimensional manifold, with a Rie-mannian metric g. Then there exists a unique n - form dV such that for anyoriented orthonormal basis (e1, . . . , en) in TpM , we have that

dV (e1, . . . , en) = 1. (1.44)

21

In particular, if ei is any oriented local frame with dual coframe φi, then

dV =√

det (gij)φ1 ∧ . . . ∧ φn. (1.45)

Remark 1.3.7. The dV in Lemma 1.3.6 is sometimes called the Riemannianvolume element. It is used to define the integral of a function over manifolds.Indeed, if f is a smooth and compactly supported function on an orientedmanifold M with Riemannian metric g, then note that f dV is a compactlysupported n-form, and so we define the integral of f over M to be

∫M

f dV. (1.46)

The volume of M is defined to be

∫M

1 dV. (1.47)

Lemma 1.3.8 (Intrinsic Divergence). Let X = X i ∂∂xi

be a vector field onM . Then the intrinsic divergence of X on M is given by the function on Mwhich locally is defined by:

divMX =1

√∣det (gij)∣

∂

∂xi(√

∣det (gij)∣Xi). (1.48)

Remark 1.3.9. Let M be a smooth oriented manifold with Riemannianmetric g and volume element dV (as described in Lemma 1.3.6). Let X be avector field on M , and t be a k-form. Consider the (k − 1)-form iXt definedby

iXt(V1, . . . , Vk−1) = t(X,V1, . . . , Vk−1). (1.49)

Then actually the definition of the intrinsic divergence operator divM ∶ T (M)→

C∞(M) is given by:d(iX dV ) = (divMX)dV, (1.50)

where T (M) denotes the space of smooth sections of TM (the space ofsmooth vector fields). For details, see Chapter 4 of [C13].

Lemma 1.3.10. Let ψ ∶M → R be a smooth function, where M is a smoothmanifold. Let X ∈ T (M). Then

divM(ψX) = ψdivMX + ⟨∇Mψ,X⟩. (1.51)

22

Remark 1.3.11. Exactly as in Remark 1.3.4, the lemma can be shown tohold for functions in W 1,2

loc (Ω).

Lemma 1.3.12. Let M be a bounded and compact oriented manifold. Let X ∈

T (M). Let n be the outward unit normal to ∂M and let dV and dV denotethe volume elements (as per Lemma 1.3.6) with respect to the Riemannianmetric on M and ∂M , respectively. Then

∫M

divMX dV = ∫

∂M

⟨X, n⟩dV . (1.52)

1.4 Hausdorff measure

We begin with a definition of the Hausdorff measure, taken from [F85].

Definition 1.4.1. Let β be the Borel σ-algebra on Rn. For any set Ω ∈ β wedefine its diameter as

diam(Ω) = sup∣x − y∣ ∶ x, y ∈ Ω ∈ [0,∞], diam(∅) = 0. (1.53)

We say that Uii∈N is a δ - cover of Ω if Ω ⊆∞⋃i=1Ui and 0 < diam(Ui) ≤ δ for

every i ∈ N, with Ui ∈ β. For k ∈ N, define

Hkδ(Ω) = inf ∞∑i=1

(diam(Ui))k ∶ Uii∈N is a δ − cover for Ω . (1.54)

We then define the k-dimensional Hausdorff measure on Ω as

Hk(Ω) = limδ→0Hkδ(Ω). (1.55)

Remark 1.4.2. 1. We used Borel σ-algebras on Rn to define the Haus-dorff measure, but of course this could be done for general metricspaces. We will not need this however.

2. For open subsets of Rn, the n-dimensional Hausdorff measure can bethought of as the classic “volume” of the set, and the (n−1)-dimensionalHausdorff measure on a surface in Rn can be thought of as “surfacearea”. When used in integration, dHn will denote the classic “volumeelement” measure, and dHn−1 the “area element” measure. That is allthe intuition we will need for this essay.

23

3. Note that we can also write Hk(Ω) = limδ→0Hkδ(Ω) = sup

δ>0Hkδ(Ω). This

is because as δ becomes smaller the number of δ-covers become morelimited and so the infimum, as in Definition 1.4.1, always increases.

24

Chapter 2

Proving that the modifiedAlmgren frequency function isnon-decreasing

In the first section of this chapter, I prove Theorem 0.0.5, using the lemmasin Chapter 1. The proof given in [GL86] is very dense and omits many non-trivial mathematical derivations. Therefore, I will provide the proofs of theseand state them as lemmas.

The second section will discuss some important derivations presented in[AKS62], which will be essential in relating Theorem 0.0.5 to our weak solu-tions u ∈W 1,2

loc (Ω) of equation (7). The relationship will be established in thethird section.

2.1 Proof of Theorem 0.0.5

Lemma 2.1.1. Theorem 0.0.5 will be proved if one can show that there is aconstant C = C(n,Λ) satisfying

C(n,Λ) ≥H ′(r)

H(r)−

1

r−D′(r)

D(r), (2.1)

where D and H are as in (21) and (22), respectively.

25

Proof. N (r) is non-decreasing if N ′(r) ≥ 0. Now, by the quotient rule, forN(r) as in (23),

N ′(r) =D(r)H(r) + rD′(r)H(r) − rD(r)H ′(r)

(H(r))2

=D(r)

H(r)+rD′(r)

H(r)−rD(r)H ′(r)

(H(r))2

=N(r)

r+N(r)D′(r)

D(r)−N(r)H ′(r)

H(r)

= N(r)(1

r+D′(r)

D(r)−H ′(r)

H(r))

(2.2)

and so

N ′(r) ≥ 0 ⇐⇒ C exp (Cr)N(r) + exp (Cr)N ′(r) ≥ 0

⇐⇒ C +1

r+D′(r)

D(r)−H ′(r)

H(r)≥ 0,

(2.3)

where we obtain the second equivalence of (2.3) by dividing by the positivequantity exp(Cr)N(r), and using (2.2). The claim follows.

Lemma 2.1.1 informs us that we need to calculate H ′(r) and D′(r). The fol-lowing lemmas serve this purpose. We start however with some definitions.

Definition 2.1.2. Let gij(x) denote the inverse of the matrix gij(x) pre-sented in equation (13). Let

g(x) ∶= ∣det(gij(x))∣. (2.4)

Whenever we apply a coordinate transformation into polar co-ordinates, wewill abuse notation and write g(r, θ) and gij(r, θ) for the tensors in the newco-ordinates. We will also let

b(r, θ) ∶= ∣det(bij(r, θ))∣, (2.5)

where the bij’s were introduced in (14).

26

Lemma 2.1.3. The quantity H(r) as in equation (22) can be written as

H(r) = rn−1∫

∂D1

τ(r, θ)u2(r, θ)√b(r, θ)dθ (2.6)

when we transform our co-ordinates into polar co-ordinates (r, θ).

Proof. We have that any point on ∂Dr can be represented in our polar co-ordinates by (r, θ) = (r, θ1, . . . , θn−1), where r is fixed as we are on the surface.This is an oriented choice of co-ordinates on ∂Dr, and so by equation (1.45)

of Lemma 1.3.6, we have that dV∂Dr =√

det(gij(r, θ))dθ. Therefore

H(r) = ∫∂Dr

τu2 dV∂Dr = ∫∂D1

τ(r, θ)u2(r, θ)√

det(gij(r, θ))dθ. (2.7)

Now√

det(gij(r, θ)) =√g(r, θ) by definition. Notice that by equation (14) we

obtain:

(gij(r, θ)) =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 0 . . . 00 r2b11(r, θ) . . . r2b1,n−1(r, θ)0 r2b21(r, θ) . . . r2b2,n−1(r, θ)⋮ ⋮ ⋱ ⋮

0 r2bn−1,1(r, θ) . . . r2bn−1,n−1(r, θ)

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

. (2.8)

Thus, it is clear that g(r, θ) = r2(n−1)∣det((bij(r, θ)))∣ = r2(n−1)b(r, θ), and thus√

det(gij(r, θ)) = rn−1√b(r, θ), from which the result follows.

Now we calculate H ′(r). Note that in what follows we might sometimes omitthe arguments that the functions take in order to simplify notation.

Lemma 2.1.4. The derivative of H is given by

H ′(r) = (n − 1

r+O(1))H(r) + 2 ∫

∂Dr

τuuρ dV∂Dr , (2.9)

where O(1) is some function of r and θ bounded in absolute value by someconstant C = C(n,Λ), almost everywhere, and uρ = ⟨∇Mu,

xρ ⟩ is derivative of

u in the radial direction (normal to ∂Dr).

27

Proof. Using Lemma 2.1.3 with the product and chain rule we obtain

H ′(r) =d

dr

⎛⎜⎝rn−1

∫

∂D1

τu2√bdθ

⎞⎟⎠

= (n − 1)rn−2∫

∂D1

τu2√bdθ + rn−1

∫

∂Dr

∂

∂ρ(τu2

√b)dρdθ

=n − 1

rH(r) + rn−1

∫

∂Dr

∂

∂ρ(τu2

√b)dρdθ.

(2.10)

Thus

H ′(r) =n − 1

rH(r) + rn−1

∫

∂Dr

∂

∂ρ(τ√b)u2 dρdθ + rn−1

∫

∂Dr

τ√b∂

∂ρ(u2)dρdθ

=n − 1

rH(r) + ∫

∂Dr

1√b

∂

∂ρ(τ√b)u2 dV∂Dr + 2 ∫

∂Dr

τuuρ dV∂Dr ,

(2.11)

where we have used that dV∂Dr = rn−1

√bdr dθ from the proof of Lemma 2.1.3.

Now notice that

1√b

∂

∂ρ(τ√b)u2 = τρu

2 +1

2bbρτu

2 = (τρτ+bρ2b

) τu2. (2.12)

By (16) and (18), we conclude that τρ and bρ are bounded in absolute value bysome constant depending on Λ and n, (almost everywhere for τρ), where then appears because taking a determinant gives us a dimensional dependence.We deduce that b is also automatically bounded (because if on a boundeddomain a function has bounded derivative then it must itself be bounded).We also have that τ is bounded from below and above due to (17). Thus theintegral of (2.12) on ∂Dr yields O(1)H(r) where O(1) is a function of (r, θ),bounded in absolute value almost everywhere by some constant C = C(n,Λ).

Putting all this together we get

H ′(r) = (n − 1

r+O(1))H(r) + 2 ∫

∂Dr

τuuρ dV∂Dr , (2.13)

as desired.

28

Lemma 2.1.5. The quantity ∫∂Dr

τuuρ dV∂Dr obtained in (2.13) equals D(r).

Consequently, we can write

H ′(r) = (n − 1

r+O(1))H(r) + 2D(r). (2.14)

Proof. We first use Lemma 1.3.3 to write

∇M(u2) = gij∂(u2)

∂xi∂

∂xj= 2u∇Mu, (2.15)

and then we use Lemma 1.3.10 to write

divM(τ∇M(u2)) = divM(2uτ∇Mu) = 2udivM(τ∇Mu) + ⟨2∇Mu, τ∇Mu⟩

= 2τ ∣∇Mu∣2,

(2.16)

where we have applied divM(τ∇Mu) = 0 by equation (19). Thus

∫Dr

divM(τ∇M(u2))dVDr = 2∫Dr

τ ∣∇Mu∣2dVDr = 2D(r). (2.17)

On the other hand, Lemma 1.3.12 tells us that

∫Dr

divM(τ∇M(u2))dVDr = ∫Dr

divM(2uτ∇Mu)dVDr

= 2 ∫∂Dr

⟨uτ∇Mu,x

ρ⟩dV∂Dr

= 2 ∫∂Dr

τuuρ dV∂Dr .

(2.18)

The claim follows.

Now the aim is to find an expression for D′(r). To this end we start withthe following setup:

Fix an r and h between 0 and 12 . Define a map ρ ∶ Rn → R≥0 by ρ(x) = ∣x∣.

Also define, for t ∈ R+, a map wt ∶ Rn → R+ by

wt(x) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

t, if ρ(x) ≤ r

t r+h−ρ(x)h +ρ(x)−rh , if r ≤ ρ(x) ≤ r + h

1, if ρ(x) ≥ r + h

(2.19)

29

Now for 0 < t < 1 + hr+h , define another map lt ∶ Rn → Rn by

lt(x) = wt(x)x. (2.20)

For u ∈W 1,2(D1) we define ut(x) = u(l−1t (x)). Also set

I[ut] = ∫D1

τ ∣∇M(ut)∣2dVD1 . (2.21)

Lemma 2.1.6. The map lt is bi-Lipschitz, and consequently ut ∈W 1,2(D1).

Proof. We will show that ∃K ≥ 1 such that

1

K∣x1 − x2∣ ≤ ∣wt(x1)x1 −wt(x2)x2∣ ≤K ∣x1 − x2∣, (2.22)

for all x1, x2 ∈ Rn. Now by (2.19) we see that for all x1, x2 with ρ(x1), ρ(x2) ≥

r + h the claim is clear for all K ≥ 1. The other cases must be analysedindividually. I will do the case ρ(x1) ≤ r and ρ(x2) ≥ r + h. The other casesfollow a similar logic. We want to show that ∃K ≥ 1 such that

1

K∣x1 − x2∣ ≤ ∣tx1 − x2∣ ≤K ∣x1 − x2∣, (2.23)

for all x1, x2 ∈ Rn, with ρ(x1) ≤ r and ρ(x2) ≥ r + h. First note that

∣x1 − x2∣ = ∣x1 − tx1 + tx1 − x2∣

≤ ∣tx1 − x2∣ + ∣1 − t∣∣x1∣

≤ ∣tx1 − x2∣ + ∣1 − t∣r

≤ (1 +∣1 − t∣r

r + h − rt) ∣tx1 − x2∣,

(2.24)

where in order to get the last inequality we first note that since x1 ∈ Br

and x2 ∈ BCr+h we can deduce that ∣tx1 − x2∣ ≥ r + h − rt. Then note that

r +h− rt > r +h− r(1+ hr+h) =

h2

r+h > 0, so that ∣tx1−x2∣r+h−rt ≥ 1. This gives us a firstbound. For the second we get that

∣tx1 − x2∣ = ∣tx1 + tx2 − tx2 − x2∣

≤ t∣x1 − x2∣ + ∣t − 1∣∣x2∣

= t∣x1 − x2∣ + ∣t − 1∣∣x2 − x1 + x1∣

≤ (t + ∣t − 1∣)∣x1 − x2∣ + r∣t − 1∣

≤ (t + ∣t − 1∣ +r∣t − 1∣

h) ∣x1 − x2∣,

(2.25)

30

where in the last inequality we used the fact that ∣x1 − x2∣ ≥ h. Putting allthis together and choosing a maximal constant proves this special case. Thegeneral case is similar, and it follows easily that ut ∈W 1,2(D1).

Lemma 2.1.7. For u ∈W 1,2(D1) solving divM(τ∇Mu) = 0 in D1, it followsthat

d

dtI[ut]∣

t=1

= 0. (2.26)

Proof. This is an important concept, and so I want to discuss the originsof this result and how it can be generalized. Assume for now that we arelooking for a solution u ∈ C2(Ω) to the problem ∆u = 0 (we are in the usualCartesian co-ordinates). Assume that Ω is as in (3). Define the functional

E(u) = ∫Ω

∣∇u∣2. (2.27)

It is simply checked that if ∆u = 0, then

d

dtE(u + tψ)∣

t=0

= 0 (2.28)

for all ψ ∈ C∞C (Ω) (i.e. u is a critical point of E), and for all w ∈ C2(Ω) with

w∣∂Ω = u∣∂Ω, we have thatE(u) ≤ E(w). (2.29)

This is known as Dirichlet’s Principle (see [H97], Theorem 1); harmonicfunctions minimize the Dirichlet energy. However, for our purposes, we wanta similar statement for solutions in W 1,2

loc (Ω). Theorem 6 in [H97] tells us thatfor w ∈ W 1,2(Ω), there exists a unique u ∈ W 1,2(Ω) with the same trace asw on the boundary such that this u minimizes (2.27) amongst all functionsthat have the same trace as w on ∂Ω.

And indeed, we may define a functional (now with respect to the Riemannianmetric (13)) via

I(u) = ∫D1

τ ∣∇Mu∣2dVD1 . (2.30)

From the same ideas as above, we can show that I has a minimizer amongstfunctions in W 1,2(Ω) with fixed boundary data, and by Theorem 1.2.2, it

31

can be shown that if u satisfies equation (19), then u is the unique minimizeramongst all functions with the same boundary data.

The radial deformation introduced in (2.19) serves the same purpose. Noticethat for t = 1, w1(x) = 1 and so l1(x) = x and so u1(x) = u(x). Thus amongstall perturbations near t = 1 the minimizer of I is at t = 1, and that is whyequation (2.26) follows. Notice that the radial deformation introduced fixesthe boundary data, because when x ∈ ∂D1, ρ(x) ≥ r+h and so wt(x) = 1 andut(x) = u(x).

Now

I[ut] = ∫D1

τ ∣∇M(ut)∣2dVD1

= ∫Drt

τ ∣∇M(ut)∣2dVDrt + ∫

Dr+h/Drt

τ ∣∇M(ut)∣2dVDr+h/Drt + ∫

D1/Dr+h

τ ∣∇M(ut)∣2dVD1/Dr+h

∶= I1 + I2 + I3.

(2.31)

We now compute dI1dt

∣t=1, dI2

dt∣t=1, dI3

dt∣t=1

.

Lemma 2.1.8. We have that

dI1

dt∣t=1

= (n − 2 +O(r))D(r), (2.32)

where O(r) denotes a function of the polar co-ordinates r and θ which isbounded in absolute value almost everywhere by a constant Cr, where C issome constant depending on Λ and n.

Proof. First we want to calculate ∣∇M(ut)∣2, when x ∈ Drt. Now ut(x) =

u(l−1t (x)) for x ∈ Drt. So we are looking for y such that lt(y) = wt(y)y = x ∈

Drt. Note if ρ(y) ≤ r then wt(y) = t and so ty = x ∈Drt is satisfied. Thereforey = x

t . Thus

ut(x) = u(x

t) . (2.33)

Now by Lemma 1.3.3 we have that

∇M(ut(x)) = gij∂ut

∂xi∂

∂xj∶= F j ∂

∂xj. (2.34)

32

Now use Lemma 1.3.5 for this tensor of type (01), to get

∣∇M(ut)∣2= ⟨∇M(ut),∇M(ut)⟩

= gmnFmF n

= gmngim∂u

t

∂xigkn

∂ut

∂xk

= δin∂ut

∂xigkn

∂ut

∂xk

= gkn∂ut

∂xk∂ut

∂xn.

(2.35)

Now of course since our metric tensor is given in polar co-ordinates (ρ, θ), sat-isfying the form (13), we have that g11(ρ, θ) = 1 and gij(ρ, θ) = ρ−2bi−1,j−1(ρ, θ)for i, j ∈ 2, . . . , n, and where (bij)−1 = (bij). To show the latter statementinvert the matrix in (2.8). Thus

∫Drt

τ ∣∇M(ut)∣2dVDrt =

rt

∫0

∫

Sn−1

τ(ρ, θ)gkn(ρ, θ)∂

∂xku(

ρ

t, θ)

∂

∂xnu(

ρ

t, θ)dVDrt

(2.36)where x1 = ρ and xi = θi−1 for i = 2,3, . . . , n. Now by Lemma 1.3.6, we havethat dVDrt =

√g(ρ, θ)dρdθ. Write s = ρ

t . Putting this together and using thechain rule on the partial derivatives in (2.36), we obtain the expression

∫Drt

τ ∣∇M(ut)∣2dVDrt =

r

∫0

∫

Sn−1

τ(st, θ)u2s(s, θ)

√g(st, θ)

tdsdθ+

r

∫0

∫

Sn−1

tτ(st, θ)uθi(s, θ)uθj(s, θ)bij(st, θ)

√g(st, θ)dsdθ.

(2.37)

The next step is to estimate√g(st,θ)t and tbij(st, θ)

√g(st, θ). For this purpose

let

√b(s, θ) = 1 + ε(s, θ), (1 + ε(s, θ))bij(s, θ) = δij + εij(s, θ), (2.38)

where ε(0,0) = 0 and εij(0,0) = 0. Seeing that b(0,0) = 1, we have that εrepresents an error function of how much the square root of the determinant

33

of (bij) differs from 1. Likewise, note that (1 + ε)bij = (1 + ε)(δij + γij) =

δij + εδij + γij + εγij, and so we have simply chosen to call εij = εδij + γij + εγij,where (γij) measures the error of (bij) being the identity matrix.

Now use (2.38) together with the fact that√g(st, θ) = (st)n−1

√b(st, θ) from

the proof of Lemma 2.1.3 to conclude that

√g(st, θ) = (st)n−1(1 + ε(st, θ)), (2.39)

and the fact that going from bij(st, θ) to bij(st, θ) picks up the factor (st)−2

to conclude that

bij(st, θ)√g(st, θ) = (st)n−3(δij + εij(st, θ)). (2.40)

We now want to plug (2.39) and (2.40) into (2.37), differentiate with respectto t and evaluate the expression at t = 1. When we differentiate on ε and εijwe will use the chain rule to get factors of s out of the expressions and thendifferentiate with respect to the radial variable ρ. It follows from (2.38) thatερ = (

√b)ρ. Using (16) we get that ∣ερ∣ is bounded by a constant depending

on Λ (coming from the bij’s) and n (coming from the fact that taking adeterminant will give us a dimensional dependence). We clearly see that thesame thing applies to εij. Using this fact in (2.37) we get equation (2.32), asdesired.

Lemma 2.1.9. We have that

dI2

dt∣t=1

= 2r ∫∂Dr

τu2ρ dV∂Dr − r ∫

∂Dr

τ ∣∇Mu∣2dV∂Dr (2.41)

Proof. The proof is very similar to the case above, but now x ∈ Dr+h/Drt

and at the end of the calculation you take the limit as h → 0+, to get theboundary integral.

Lemma 2.1.10. We have that dI3dt

∣t=1

= 0.

Proof. For x ∈ D1/Dr+h we have that ρ(x) ≥ r + h and so l−1t (x) = x, and so

ut(x) = u(x). Hence I3 is independent of t and the result follows.

34

Putting Lemmas 2.1.7, 2.1.8, 2.1.9, and 2.1.10 together we get that

(n − 2 +O(r))D(r) + 2r ∫∂Dr

τu2ρ dV∂Dr − r ∫

∂Dr

τ ∣∇Mu∣2dV∂Dr = 0, (2.42)

where we recognize:

∫

∂Dr

τ ∣∇Mu∣2dV∂Dr =D

′(r). (2.43)

Thus we have

rD′(r) − (n − 2 +O(r))D(r) = 2r ∫∂Dr

τu2ρ dV∂Dr . (2.44)

From this we prove Theorem 0.0.5:

Proof of Theorem 0.0.5. We have that

1

r+D′(r)

D(r)−H ′(r)

H(r)=

1

r+n − 2 +O(r)

r+

2

D(r) ∫∂Dr

τu2ρ dV∂Dr −

H ′(r)

H(r)

=n − 1

r+O(r)

r+

2

D(r) ∫∂Dr

τu2ρ dV∂Dr

−n − 1

r−O(1) − 2

∫∂Dr

τuuρ dV∂Dr

H(r)

= O(1) + 2

∫∂Dr

τu2ρ dV∂Dr

∫∂Dr

τuuρ dV∂Dr− 2

∫∂Dr

τuuρ dV∂Dr

∫∂Dr

τu2 dV∂Dr,

(2.45)

where in the first equality we used (2.44), in the second (2.13), and in the last(22) and Lemma 2.1.5. Now since u and its (weak) derivative are in L2(D1)

and τ is bounded, we can use the Cauchy-Schwarz inequality to concludethat

⟨√τu,

√τuρ⟩

2 ≤ ∥√τu∥

22∥√τuρ∥

22. (2.46)

35

This implies that

2

∫∂Dr

τu2ρ dV∂Dr

∫∂Dr

τuuρ dV∂Dr− 2

∫∂Dr

τuuρ dV∂Dr

∫∂Dr

τu2 dV∂Dr≥ 0, (2.47)

so that1

r+D′(r)

D(r)−H ′(r)

H(r)≥ O(1) ≥ −C(n,Λ). (2.48)

By Lemma 2.1.1, Theorem 0.0.5 is now proved.

2.2 Some results from [AKS62]

Now I explore some of the results presented in sections 2 and 3 of [AKS62].This section is independent of the work we have done thus far, but everythingwill be tied up neatly in section 3 of this chapter. I commence with theassumptions made in order for Theorem 2 in [AKS62] to hold.

• For some r0 > 0, the manifold D2r0 is endowed with a Lipschitz metrictensor Gij(x).

• There are positive constants k1 and k2 such that for all ζ ∈ Rn, and allx ∈D2r0 , we have

k1∣ζ ∣2≤ Gij(x)ζ

iζj ≤ k2∣ζ ∣2. (2.49)

• There is a positive constant k such that for all ζ ∈ Rn and x ∈ D2r0 wehave that

∣∂Gij

∂eζ iζj∣ ≤ k∣ζ ∣

2a.e. (2.50)

Here ∂∂e is the radial derivative in D2r0 .

We then define a function r ∶D2r0 → R via

r(x) ∶=√Gij(0)xixj, (2.51)

and a new metric tensor on D2r0

Gij(x) ∶= Gij(x) (Gkl(x)

∂r

∂xk∂r

∂xl) , (2.52)

36

where of course Gij(x) denotes the inverse of Gij(x). In the proof of Theorem2 in [AKS62] the authors show that for

k1 ∶=k2

1

k2

, k2 ∶=k2

2

k1

, (2.53)

we have that for all ζ ∈ Rn and x ∈D2r0

k1∣ζ ∣2≤ Gij(x)ζ

iζj ≤ k2∣ζ ∣2. (2.54)

Also, almost everywhere on D2r0 and for all ζ ∈ Rn, we have that

∣∂Gij

∂eζ iζj∣ ≤ k∣ζ ∣

2, (2.55)

where

k =6kk2

2

k21

√k2

k1

. (2.56)

With this at hand, section 3 of [AKS62] shows that in the metric Gij thefunction r is actually the geodesic distance from 0 to x. To show this considerthe following system of ODE’s:

dxi

dσ= Gij

∂r

∂xj, i = 1,2, . . . n. (2.57)

Here Gij(x) denotes the inverse of Gij(x). Consider the ellipsoid

S ∶= r(x) < q ∶= r0

√k1. (2.58)

It can be checked that S is contained in Dr0 . Let Σ denote the boundaryof S. Let t be a point on Σ and consider (2.57) together with an initialcondition x(q) = t. The ODE can now be solved for a unique solution x(σ).Now notice that

dr(x(σ))

dσ=∂r

∂xidxi

dσ

= Gij∂r

∂xi∂r

∂xj

= Gij(x) (Gkl(x)∂r

∂xk∂r

∂xl)

−1 ∂r

∂xi∂r

∂xj

= 1,

(2.59)

37

where in the second line we used equation (2.57) and in the third line equation(2.52). We conclude that r(x(σ)) = σ + c, but recall that q = r(t) = r(x(q))so c must be 0.

Notice that at the origin, the right hand side of (2.57) fails to be continuousas ∂r

∂xjfails to be continuous there due to (2.51). Thus we only consider our

solutions x(σ) ∈ (0, q].

Now our solution r(x(σ)) = σ means that for points t on Σ we get a systemof simple arcs joining 0 to t, mutually disjoint and filling out S.

The next step is to introduce polar co-ordinates (t1, . . . , tn−1, r) on Σ, wherer = r(x). Note that by the chain rule

Gij dxi dxj = Gij (∂xi

∂tαdtα +

∂xi

∂rdr)(

∂xj

∂tβdtβ +

∂xj

∂rdr) , (2.60)

where α and β run from 1 to n − 1. Expanding (2.60), and using (2.57) and(2.59) we get that the terms involving the partial derivative with respect tor satisfy

Gij∂xi

∂r

∂xj

∂r= GijGik

∂r

∂xkGjl

∂r

∂xl= Gkl

∂r

∂xk∂r

∂xl= 1, (2.61)

whilst those that have one partial derivative with respect to r and anotherwith respect to tα satisfy

Gij∂xi

∂tα∂xj

∂r= Gij

∂xi

∂tαGjl

∂r

∂xl=∂xl

∂tα∂r

∂xl=∂r

∂tα= 0. (2.62)

Putting all this together we can conclude that our metric tensor takes theform

Gij dxi dxj = (dr)2 + r2Bαβ dtα dtβ, (2.63)

where

Bαβ =1

r2Gij

∂xi

∂tα∂xj

∂tβ. (2.64)

From these derivations it can thereafter be shown that the polar co-ordinateswe have chosen are geodesic relative to the metric Gij, and hence r is thegeodesic distance from 0 to x. Furthermore, it can be deduced by consideringa point t ∈ Σ and a contravariant vector T tangential to Σ, that as a functionof r, Bαβ satisfies

∣∂Bαβ

∂r∣ ≤ ω, (2.65)

38

with

ω =30kk5

2

√k2

k71

. (2.66)

2.3 Tying it all up

Now we want to use the discussion of section 2.2 in order to provide a theoremlike 0.0.5, but for weak solutions u ∈W 1,2

loc (Ω) of equation (7); which was ouroriginal equation of interest. We substitute in the following:

Let r0 =12 , where r0 is as defined in section 2.2 above, so that on D1 ⊂ Rn we

define a Lipschitz Riemannian metric by

Gij(x)dxi ⊗ dxj = aij(x)(detA)1n−2 dxi ⊗ dxj, (2.67)

where aij(x) denote entries of A(x)−1, the inverse of the matrix defined for(4). Now recall that (6) told us that A had lowest eigenvalue λ and highest 1

λ

on Ω. By writing A = P −1DP where D is the matrix of eigenvalues of A it iseasy to check that (detA)

1n−2A−1 has lowest eigenvalue a constant depending

on λ and n (the n coming from taking a determinant) and a highest eigenvaluea bigger constant also depending on λ and n. Translating this into equation(2.49), we have that the k1 and k2 appearing depend on λ and n.

Also, note that the Lipschitz condition in (5) implies that partial derivativesof aij are bounded in absolute value by K, almost everywhere. Translatingthis into the partial derivative defined in (2.50), we notice that we need to

take partial derivatives of a−1ij (detA)

1n−2 . To do this we will use the chain

rule and product rule, and exactly because the aij’s are bounded in absolutevalue from below (due to (6)), the final outcome will be bounded in absolutevalue almost everywhere by a constant that depends on n (coming fromthe determinant), λ (coming from the aij bounds), and K (coming from thepartial derivative that will act on aij). Hence it follows that the k in (2.50)depends on n,λ and K.

We set (Gij(x)) = (Gij(x))−1, and we define the function r ∶D1 → R via

r(x) =√Gij(0)xixj. (2.68)

With this at hand we can introduce a new Riemannian metric

Gij(x)dxi ⊗ dxj, (2.69)

39

where Gij(x) ∶= Gij(x) (Gkl(x) ∂r∂xk

∂r∂xl

), just as in (2.52).

Lemma 2.3.1. Lu = 0 in D1 if and only if divM(∇Mu) = 0 in the Rieman-nian metric defined by equation (2.67).

Proof. The following equivalences should be considered in the weak sense,where G = ∣det((Gij))∣:

divM(∇Mu) = 0 ⇐⇒1

√G

∂

∂xk(√GGik ∂u

∂xi) = 0

⇐⇒∂

∂xk(√GGik ∂u

∂xi) = 0

⇐⇒∂

∂xk(aik

∂u

∂xi) = 0

⇐⇒ div(A∇u) = 0

⇐⇒ Lu = 0,

(2.70)

where in the first equivalence we used Lemmas 1.3.3 and 1.3.8, in the fourthequivalence we used that A is symmetric, and in the third equivalence wecomputed:

√GGik =

√det (Gij)G

ik

=

√

det ((aij)(detA)1n−2 )((aik)(detA)

1n−2 )−1

=

√

det(A−1)(detA)nn−2aik(detA)

−1n−2

= (detA)1n−2aik(detA)

−1n−2

= aik.

(2.71)

Now I proved in the section above that in intrinsic geodesic polar co-ordinateswith pole at (0,0), we can write the metric (2.69) as

dr ⊗ dr + r2bij(r, θ)dθi ⊗ dθj, (2.72)

where according to (2.65) and (2.66) the bij’s satisfy (16), where the constantΛ in (16) depends on ω = ω(k, k1, k2) but we have already established thatk, k1 and k2 depend on n,K and λ. Finally, by (2.72) we can write

divM(∇Mu) = 0 ⇐⇒ divM(τ∇Mu) = 0, (2.73)

40

where we use M to denote the manifold endowed with the Riemannian metricdefined in (2.67), and M to denote the one endowed with the metric definedby (2.72), and where τ is Lipschitz on D1 satisfying conditions (17) and (18)where the constants in these conditions all depend on Λ but which in turndepends on n,K and λ. Putting Theorem 0.0.5 together with Lemma 2.3.1and condition (2.73), we obtain the following theorem:

Theorem 2.3.2. Let Ω be a smooth, bounded and connected subset of Rn

compactly containing B2, where n ∈ N, n ≥ 3. Consider an elliptic operatoracting on u via Lu = div(A∇u), where A is an n×n matrix satisfying condi-tions (5) and (6). If Lu = 0 weakly, then there is a constant C = C(n,λ,K)such that N (r), as defined in (24), but this time with respect to the Rieman-nian metric (2.69), is a non-decreasing function of r ∈ (0,1).

41

Chapter 3

Proving strong uniquecontinuation

The aim of this section is to provide the proof for Theorem 0.0.2. To doso we must first prove Theorem 0.0.3, as we will need the doubling condi-tion in our proof. The proofs are based on those in [GL86], but here I provethe steps that the paper leaves out. We commence with an important remark.

Remark 3.0.1. If we go back to the proof of Theorem 0.0.5, we notice thatwe could have shown that the modified Almgren frequency function was non-decreasing in r ∈ (0,1) by using balls centred at Dr(x0) for any x0 ∈ Ω aslong as B2r(x0) ⊂⊂ Ω, so that we do not hit the boundary. The choice x0 = 0was just for convenience. Hence it follows that the constant C = C(n,λ,K)in Theorem 2.3.2 works relative to any such ball Dr(x0).

The proof below for the doubling condition will therefore without loss ofgenerality be given for DR(0) for 0 < R < 1

2 (the case R = 0 is trivially true),but note that the same conclusion will hold for any ball DR(x0), 0 < R < 1

2

as long as B2R(x0) ⊂⊂ Ω.

Proof of Theorem 0.0.3. Assume the conditions stated in Theorem 0.0.3. Sincediv(A∇u) = 0 applies in Ω, we have that Theorem 2.3.2 applies. I.e. the func-tion N (r) given is non-decreasing in r ∈ (0,1).

42

Let H be defined as in (22), where τ is as in (2.73). Notice that

d

dr(log

H(r)

rn−1) = (

H ′(r)rn−1 − (n − 1)H(r)rn−2

r2(n−1) )rn−1

H(r)

=H ′(r)

H(r)−n − 1

r.

(3.1)

Now by (2.14) we can write

H ′(r)

H(r)−n − 1

r= O(1) + 2

D(r)

H(r)= O(1) +

2N (r) exp(−Cr)

r, (3.2)

where in the last equality we used the definition of N (r). Note that the Cthat appears depends on n,K and λ according to Theorem 2.3.2. Now for0 < R < 1

2 , integrating (3.1) from R to 2R yields

logH(2R)

(2R)n−1− log

H(R)

Rn−1= log(

H(2R)

H(R)2n−1) =

2R

∫R

(O(1) +2N (r) exp(−Cr)

r)dr

≤ C ′R + 2N (1)

2R

∫R

1

rdr

= C ′R + 2N (1) log(2),

(3.3)

where C ′ is a constant that depends on n and Λ, but as discussed in Chap-ter 2, Λ depends on n,λ and K. In the second line we used the fact thatN (r) is non-decreasing in r ∈ (0,1) to conclude that N (r) ≤ N (1), and thatexp(−Cr) is decreasing on (0,1) to conclude exp(−Cr) ≤ 1. Exponentiatingnow yields

H(2R)

H(R)2n−1≤ exp(C ′R) exp(2N (1) log(2))

= exp(C ′R)(4N (1))

=D′,

(3.4)

where D′ is a constant depending on n,K, λ and u. The u dependence comesfrom the fact that N (1) depends on u. Thus

H(2R) ≤ 2n−1D′H(R) Ô⇒

∫

∂D2R

τu2 dV∂D2R≤ 2n−1D′

∫

∂DR

τu2 dV∂DR .(3.5)

43

Integrating in R now gives

∫D2R

τu2 dVD2R≤ 2n−1D′

∫DR

τu2 dVDR = C(n,λ,K, u)∫DR

τu2 dVDR . (3.6)

Divide by the positive and bounded τ to complete the proof.

Now we prove strong unique continuation, but before that we make anotherimportant remark:

Remark 3.0.2. 1. Assume u vanishes to infinite order at x0 ∈ Ω. Defini-tion 0.0.1 then applies and so for a sufficiently small ball Dδ(x0) andfor all natural numbers j we have that ∫

Dδ(x0)u2 = O(δj).

2. Assume that we show that u ≡ 0 on Dδ(x0).

3. Pick a point y ∈Dδ(x0) near the boundary of the ball. Enclose it withthe ball Dγ(y) such that Dγ(y) ⊂Dδ(x0). Then u ≡ 0 on Dγ(y) also.

4. Recall from Remark 3.0.1 that the conclusion of the doubling conditionholds with respect to any ball as long as the radius is less that 1

2 andthat twice the ball is compactly contained in Ω. I.e. the radius isindependent of the solution function u. In particular it holds for Dγ(y)as long as γ < 1

2 and D2γ(y) ⊂⊂ Ω.

5. By the doubling condition, we have that ∫D2γ(y)

u2 ≤ C(n,λ,K, u) ∫Dγ(y)

u2 =

C ×0 = 0. Hence even though the constant C of the doubling conditiondepends on u, it is multiplied by 0 and so the u dependence is wipedout. And hence u is identically 0 on D2γ(y).

6. Now take a union Dδ(x0)∪D2γ(y), on which u is identically 0. We picka point y1 ∈Dδ(x0) ∪D2γ(y) that is near the boundary and repeat thesame thing, until we have “patched up” Ω.

Proof of Theorem 0.0.2. Assume without loss of generality that u, whichsolves div(A∇u) = 0 weakly, vanishes to infinite order at 0 ∈ Ω. I.e. forδ > 0 small enough, we have that ∫

Dδ

u2 = O(δN) for every N ∈ N. Notice that

this will also hold for all δ0 ≤ δ. We will show that u ≡ 0 on Dδ.

44

Let β ∈ R, to be chosen. Using the doubling condition we get that

∫Dδ

u2 dVDδ ≤ Ck∫

Dδ2−k

u2 dVDδ2−k

= Ck∣Dδ2−k ∣β 1

∣Dδ2−k ∣β ∫Dδ2−k

u2 dVDδ2−k

= Ckωβn(δn2−kn)β

1

∣Dδ2−k ∣β ∫Dδ2−k

u2 dVDδ2−k

,

(3.7)

where ωn denotes the volume of the unit ball in Rn. Now choose β such thatC2−nβ = 1, and note that we obtain

∫Dδ

u2 dVDδ ≤ (ωnδn)β

1

∣Dδ2−k ∣β ∫Dδ2−k

u2 dVDδ2−k

. (3.8)

Since u vanishes of infinite order at x = 0, taking a limit as k →∞ yields

∫Dδ

u2 dVDδ ≤ limk→∞

(ωnδn)β

1

∣Dδ2−k ∣β ∫Dδ2−k

u2 dVDδ2−k

= 0, (3.9)

so that u ≡ 0 on Dδ. This completes the proof.

45

Chapter 4

A non-geometric approach tothe Almgren frequencyfunction and some corollaries

In the rest of this chapter we assume that our operator L = ∆ and thatLu = 0 (no longer in the weak sense) for solutions u belonging to C2(Ω),where Ω satisfies condition (3). For non-trivial solutions u we set the Almgrenfrequency function to be

N(r) ∶= rD(r)

H(r), (4.1)

whereD(r) ∶= ∫

Dr

∣∇u∣2

(4.2)

andH(r) ∶= ∫

∂Dr

u2, (4.3)

where H(r) ≠ 0. We prove something very similar to Theorem 0.0.5, usingthe ideas from [HL], section 2.2. Before that however, we need a couple oflemmas.

46

Lemma 4.0.1. Let uρ = ⟨∇u, xρ ⟩ denote the normal derivative of u on ∂Dr.Then

H ′(r) =n − 1

rH(r) + 2 ∫

∂Dr

uuρ. (4.4)

Proof. We have

H(r) = ∫∂Dr

u2(x)dHn−1(x)

= rn−1∫

∣y∣=1

u2(ry)dHn−1(y),(4.5)

and therefore, by the product rule,

H ′(r) = (n − 1)rn−2∫

∣y∣=1

u2(ry)dHn−1(y) + 2 ∫∂Dr

uuρ

=n − 1

rH(r) + 2 ∫

∂Dr

uuρ,(4.6)

as desired.

Lemma 4.0.2. Let D(r) be as in (4.2). Then

D′(r) =n − 2

rD(r) + 2 ∫

∂Dr

u2ρ. (4.7)

Proof. First recall that

D′(r) = ∫∂Dr

∣∇u∣2dHn−1. (4.8)

Now if we denote (in Rn) x = (x1, . . . , xn), note that for x ∈ ∂Dr, we have

1

r

n

∑i=1

x2i

r∣∇u∣

2= ∣∇u∣

2. (4.9)

So (4.8) can be written as

D′(r) =1

r ∫∂Dr

n

∑i=1

x2i

r∣∇u∣

2dHn−1 =

1

r

n

∑i=1∫

∂Dr

xi∣∇u∣2xir

dHn−1. (4.10)

47

We have written it in such a form in order to apply integration by parts:

∫Dr

∂

∂xi(xi∣∇u∣

2)dHn = ∫

∂Dr

xi∣∇u∣2xir

dHn−1. (4.11)

Using the product rule we compute:

∫Dr

∂

∂xi(xi∣∇u∣

2)dHn = ∫

Dr

∣∇u∣2dHn + ∫

Dr

xi∂

∂xi(n

∑k=1

(∂u

∂xk)

2

)dHn

= ∫Dr

∣∇u∣2dHn + ∫

Dr

2xi (n

∑k=1

∂u

∂xk

∂2u

∂xi∂xk)dHn

= ∫Dr

∣∇u∣2dHn + 2

n

∑k=1∫Dr

xi∂u

∂xk

∂2u

∂xk∂xidHn.

(4.12)

Note that in the very last term of (4.12) we used Clairaut’s theorem (sym-metry of second derivatives theorem) to switch the order of differentiation,valid because u ∈ C2(Ω).

Now denote n = xρ on ∂Dr, so that ni =

xiρ . Using integration by parts on the

second term of (4.12) we obtain:

∫Dr

xi∂u

∂xk

∂2u

∂xk∂xidHn = ∫

∂Dr

xi∂u

∂xk

∂u

∂xink dHn−1 − ∫

Dr

(δik∂u

∂xk+ xi

∂2u

∂x2k

)∂u

∂xidHn

= r ∫∂Dr

ni∂u

∂xi

∂u

∂xknk dHn−1 − ∫

Dr

(δik∂u

∂xk+ xi

∂2u

∂x2k

)∂u

∂xidHn.

(4.13)

Now we would like to sum (4.13) over k. Note thatn

∑k=1

∂u∂xk

nk = uρ,n

∑k=1δik

∂u∂xk

∂u∂xi

=

( ∂u∂xi)2, and

n

∑k=1

xi∂2u∂x2k

∂u∂xi

= xi(∆u)∂u∂xi

= 0. Putting this together, summing

(4.13) gives

n

∑k=1∫Dr

xi∂u

∂xk

∂2u

∂xk∂xidHn = r ∫

∂Dr

ni∂u

∂xiuρ dHn−1 − ∫

Dr

(∂u

∂xi)

2

dHn. (4.14)

48

Hence (4.12) becomes

∫Dr

∂

∂xi(xi∣∇u∣

2)dHn = ∫

Dr

∣∇u∣2dHn+2r ∫

∂Dr

ni∂u

∂xiuρ dHn−1−2∫

Dr

(∂u

∂xi)

2

dHn.

(4.15)Because of (4.10) we now need to also sum (4.15) over i, yielding

n

∑i=1∫Dr

∂

∂xi(xi∣∇u∣

2)dHn = n∫

Dr

∣∇u∣2dHn + 2r ∫

∂Dr

u2ρ dHn−1 − 2∫

Dr

∣∇u∣2dHn

= (n − 2)∫Dr

∣∇u∣2dHn + 2r ∫

∂Dr

u2ρ dHn−1

= (n − 2)D(r) + 2r ∫∂Dr

u2ρ dHn−1.

(4.16)

And now finally using (4.16) in (4.10) we obtain

D′(r) =n − 2

rD(r) + 2 ∫

∂Dr

u2ρ dHn−1. (4.17)

Lemma 4.0.3. The quantity D(r) can also be written as

D(r) = ∫∂Dr

uuρ. (4.18)

Proof. Note that by the product rule we have that

∆u2 =n

∑i=1

∂

∂xi(2u

∂u

∂xi) = 2∣∇u∣

2+ 2u∆u = 2∣∇u∣

2. (4.19)

And hence

D(r) = ∫Dr

∣∇u∣2=

1

2 ∫Dr

∆u2 =1

2 ∫Dr

n

∑i=1

∂

∂xi(∂(u2)

∂xi) = ∫

∂Dr

uuρ. (4.20)

49

Proof of Theorem 0.0.6. We want to show N ′(r) ≥ 0. Using (4.1) and thequotient rule we get

N ′(r) =(D(r) + rD′(r))H(r) − rD(r)H ′(r)

H(r)2

=D(r)

H(r)+rD′(r)

H(r)−rD(r)

H(r)

H ′(r)

H(r)

=N(r)

r+N(r)

D′(r)

D(r)−N(r)

H ′(r)

H(r)

= N(r)(1

r+D′(r)

D(r)−H ′(r)

H(r)) .

(4.21)

As N(r) is non-negative, all we need to show is that 1r +

D′(r)D(r) −

H′(r)H(r) ≥ 0. By

Lemmas 4.0.1, 4.0.2 and 4.0.3, we have

1

r+D′(r)

D(r)−H ′(r)

H(r)=

1

r+n − 2

r+ 2

∫∂Dr

u2ρ

∫∂Dr

uuρ−n − 1

r− 2

∫∂Dr

uuρ

∫∂Dr

u2

= 2

⎛⎜⎜⎝

∫∂Dr

u2ρ

∫∂Dr

uuρ−

∫∂Dr

uuρ

∫∂Dr

u2

⎞⎟⎟⎠

.

(4.22)

If we show that ( ∫∂Dr

uuρ)

2

≤ ( ∫∂Dr

u2)( ∫∂Dr

u2ρ) then we’re done. And indeed,

the Cauchy-Schwarz inequality tells us that ∣⟨u,uρ⟩∣2≤ ∥u∥

22∥uρ∥

22 on ∂Dr,

which is the inequality desired.

We now discuss some corollaries of Theorem 0.0.6. We begin by a corollarythat states that the surface integral of u2 on a large sphere is controlled bythe surface integral on a smaller sphere.

Corollary 4.0.4. Let u ∈ C2(Ω) be a non-trivial solution to ∆u = 0. Thenfor any 0 < R1 ≤ R2 ≤ 1 we have that

H(R2)

H(R1)≤ (

R2

R1

)

n−1+2N(R2). (4.23)

50

Proof. The claim is obvious if 0 < R1 = R2. So assume R1 < R2. Lemmas4.0.1 and 4.0.3 imply that

H ′(r) =n − 1

rH(r) + 2D(r), (4.24)

and sod

drlog(H(r)) =

H ′(r)

H(r)=n − 1

r+

2

rN(r). (4.25)

Now integrate (4.25) with respect to r between R1 and R2, to obtain

logH(R2) − logH(R1) = log(H(R2)

H(R1)) = (n − 1) log (

R2

R1

) + 2

R2

∫R1

N(r)

rdr

= log((R2

R1

)

n−1

) + 2

R2

∫R1

N(r)

rdr

(4.26)

Now exponentiate in (4.26), and use Theorem 0.0.6 to get

H(R2)

H(R1)= (

R2

R1

)

n−1

exp⎛⎜⎝

2

R2

∫R1

N(r)

rdr

⎞⎟⎠

≤ (R2

R1

)

n−1

exp(2N(R2) log (R2

R1

))

= (R2

R1

)

n−1

exp(log((R2

R1

)

2N(R2)))

= (R2

R1

)

n−1

(R2

R1

)

2N(R2)

= (R2

R1

)

n−1+2N(R2),

(4.27)

which is the result we desire.

The importance of this corollary is also that it allows us to easily prove avery general form of the doubling condition (cf. Theorem 0.0.3) for harmonicfunctions on D1. We state this result as the next corollary.

51

Corollary 4.0.5. Let u ∈ C2(Ω) be a non-trivial solution to ∆u = 0. Thenfor any R ∈ (0, 1

2) and any S ∈ [1,2] we have that

∫DSR

u2 ≤ S2N(1)+n−1∫DR

u2. (4.28)

In particular, the case S = 2 gives the doubling condition (cf. Theorem 0.0.3)for harmonic functions.

Proof. Substitute R1 = R and R2 = SR in Corollary 4.0.4. We obtain

H(SR)

H(R)≤ Sn−1+2N(SR) ≤ Sn−1+2N(1), (4.29)

because the frequency function is non-decreasing. Using the definition of Hand re-arranging, this implies

∫

∂DSR

u2 ≤ Sn−1+2N(1)∫

∂DR

u2. (4.30)

Now integrate (4.30) from 0 to R to get the volume integral (4.28) as required.

We now prove that the volume integral of u2 on D1 is can be bounded frombelow and above by terms that relate to its surface integral on ∂D1.

Corollary 4.0.6. Let u ∈ C2(Ω) be a non-trivial solution to ∆u = 0. Then

1

2N(1) + n ∫∂D1

u2 ≤ ∫D1

u2 ≤1

n ∫∂D1

u2. (4.31)

Proof. Let’s start with the upper bound. Let R ∈ (0,1). We will substituteR2 = 1 and R1 = R in the first line of equation (4.27) to write

H(1)

H(R)=

1

Rn−1exp

⎛

⎝2

1

∫R

N(r)

rdr

⎞

⎠≥

1

Rn−1, (4.32)

52

as 21

∫R

N(r)r dr ≥ 0. Now notice that

∫D1

u2 =

1

∫0

∫

∂DR

u2 =

1

∫0

H(R)dR ≤H(1)

1

∫0

Rn−1 dR =H(1)

n, (4.33)

which is the upper bound required, by the definition of H.

As for the lower bound we have by equation (4.32) and Theorem 0.0.6 that

H(1)

H(R)=

1

Rn−1exp

⎛

⎝2

1

∫R

N(r)

rdr

⎞

⎠≤

1

Rn−1exp(2N(1) log (

1

R)) =

1

R2N(1)+n−1.

(4.34)Thus

1

∫0

H(R)dR ≥H(1)

1

∫0

R2N(1)+n−1 dR =H(1)

2N(1) + n, (4.35)

which is the lower bound desired, by the definition of H.

Notice that equality can indeed happen in (4.31), just take u = 1. Hence thisis in fact an optimal inequality.

Our final corollary will tell us that the limit of the frequency function as rapproaches 0 from above is actually the order of vanishing of u at 0. Firstwe need to define however the order of vanishing of a function. We adoptthe definition found in [Z14].

Definition 4.0.7. Let u ∶ Ω → R be a function, where Ω ⊆ Rn. It’s order ofvanishing at x0 ∈ Ω is k if

Dαu(x0) = 0, (4.36)

for all multi-indices α with ∣α∣ < k.

Now that we have the definition, we prove the following:

Corollary 4.0.8. Let u ∈ C2(Ω) be a non-trivial solution to ∆u = 0. Let kbe the order of vanishing of u at 0. Then

limr→0+

N(r) = k. (4.37)

53

Proof. First, by Theorem 0.0.2 we know that u does not vanish to infiniteorder at 0 (as otherwise u would be identically 0). Also, lim

r→0+N(r) exists as

N(r) is non-increasing as r → 0+, and bounded below by 0.

Now if u is harmonic then u is also real analytic (for the proof, see Theorem1.28 in [ABR01]). Hence, we can express u by a Taylor series. Now sinceDαu(0) = 0 for all ∣α∣ < k, we have that the term of smallest degree in theTaylor expansion of u has degree k. So we can write

u(x) = P (x) +R(x), (4.38)

where P is a homogenous polynomial of degree k, and R is the remainderpolynomial where the term of smallest degree has degree at least k + 1. Now∆P = 0, because if not, then there must be some terms in ∆R which cancelaway the the non-zero terms of ∆P (as ∆u = 0), which means there is atleast one term in R of degree k, which is a contradiction to what we haveassumed. Hence both ∆P and ∆R equal 0. Now

N(r) =

r ∫Dr

∣∇(P +R)∣2

∫∂Dr

(P +R)2=

r (∫Dr

∣∇P ∣2+ 2 ∫

Dr

∣∇P ∣∣∇R∣ + ∫Dr

∣∇R∣2)

∫∂Dr

P 2 + 2 ∫∂Dr

PR + ∫∂Dr

R2. (4.39)

Notice that if we factor out the term involving the ∇P from the numeratorof (4.39), and the term involving the P in the denominator, and then take alimit as r → 0+, all the terms remaining that involve an R or ∇R will vanish,because the term of smallest degree in R is of higher degree than the degreeof P . Hence we obtain

limr→0+

N(r) = limr→0+

r ∫Dr

∣∇P ∣2

∫∂Dr

P 2. (4.40)

Now since P is a homogenous harmonic polynomial of degree k, it formsthe real or imaginary part of an analytic function of the form that allowsus to write in polar co-ordinates P (x) = rkχ(θ), where χ is the restrictionof P to the surface of the (n − 1)-dimensional hypersphere. Recall that

54

D(r) = ∫Dr

∣∇P ∣2= ∫∂Dr

PPρ, by Lemma 4.0.3. Clearly Pρ = krk−1χ(θ), so that

r ∫Dr

∣∇P ∣2

∫∂Dr

P 2=

r ∫∂Dr

kr2k−1χ2(θ)

∫∂Dr

r2kχ2(θ)= k, (4.41)

and the result follows.

55

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