The Hodge Theorem and the Bochner Technique: a Vanishingly Short Proof

THE HODGE THEOREM AND THE BOCHNER TECHNIQUE: A VANISHINGLY SHORT PROOF

HADRIAN QUAN

Abstract. This paper serves as a relatively self-contained introduction to the Hodge Decomposition theorem, and related techniques in Differential geometry. Assuming minimal exposure to analysis and modern geometry, we motivate and prove some results in the classical theory of an Elliptic Partial Differential Op- erator. This includes constructing the W k,p Sobolev spaces, and proving some related norm inequalities. Armed with the Hodge theorem, we will be able to prove several “Vanishing theorems” of geometric interest. The pacing and tone are ideally written to an undergraduate audience. A number of necessary definitions and results in geometry and analysis are stated or proven in the appendix.

Contents 1. A review of Differential forms and de Rham Cohomology 1 2. Preliminaries for Hodge Theory 5 3. An Interlude: The Hodge Theorem as a Least Norm Solution 8 4. Weak Solutions and the Hodge Decomposition Theorem 8 5. Solving Poissons Equation: Laplace and Dirac 11 6. Distributions and The Schwartz Space of Test Functions 17 7. Density, Completeness, and Difference Quotients of W k,p(Ω): The Sobolev Spaces 19 8. Inequalities on Sobolev Norms: Nirenberg and Morrey 22 9. Partial Differential Operators, Weak Existence, and Elliptic Regularity 27 10. The Bochner Technique 30 11. Appendix 32 11.1. A Rapid Introduction to Smooth Manifolds 32 11.2. Riemannian Geometry 33 11.3. Measure Theory 33 11.4. Essentials of Functional Analysis 39 11.5. Orthogonal Decomposition in a Hilbert Space 41 12. References 42

1. A review of Diﬀerential forms and de Rham Cohomology

Consider the fundamental theorem of calculus, which states that for f ∈ C1(R) we have Z b df dx = f(b) − f(a). a dx The student familiar with vector calculus will recall its analogs for 1-dimensional curves C, 2-dimensional surfaces S, and 3-dimensional regions V : Z Z Z ∇φ · ~tdl = φ(p2) − φ(p1) ∇ × F~ · ~ndS = F~ · ~tdl C S ∂S Z Z ∇ · F~ dV = F~ · ~ndS V ∂V where C has endpoints p1, p2 and ~t, ~n are outer normal vectors for a given choice of orientation. In every case we have an equality of two integrals; exchanging the integral of a derivative over a region, for the integral over the boundary of a region. We can generalize these statements, but each of them also involved a more subtle

Date: May 2015. 1 2 HADRIAN QUAN phenomenon, namely the ability to integrate. To define the integral we shall introduce differential forms, some of the main players in our program. One reason for this need of differential forms is that even integrals over Rn are not invariant under diffeomorphisms (smooth, bijective maps with smooth inverses). Diffeomorphism is the natural form of equivalence for two smooth manifolds. In the context of calculus, such diffeomorphisms are referred to as ‘changes of coordinates.’ Recall from vector calculus, the change of variables formula states that if f : U ⊂ Rn → Rn is a local diffeomorphism then Z Z h(x)dx1 . . . dxn = h ◦ f(y) |det(Df)| dy1 . . . dyn, f(U) U where |det(Df)| denotes the Jacobian of the map f. Heuristically, if we want to integrate over a manifold we need a mathematical object which transforms similarly to the determinant of a linear map, and resembles the Lebesgue measure dx1 . . . dxn (for more on measure, consult the appendix). Recall first that the dual vector space V ∗ to a vector space V is the space of linear functionals θ : V → R. These are linear maps which assign a real value to every element of V . One way of defining this space is by defining them with respect to the basis for V . Explicitly, if {v1, . . . vn} is a basis for V , then we can uniquely define a basis {θ1, . . . , θn} for V ∗ by requiring ( i 1 i = j θ (vj) = δij = 0 i 6= j and elements of V ∗ can be written with respect to this basis. If we have (x1, . . . , xn) a choice of local coordinates at a point p ∈ M, this generates a basis ∂ ∂ ,..., ∂x1 ∂xn for the tangent space TpM at p. Then this basis for the vector space TpM uniquely determines a basis {dx1, . . . , dxn} ∗ for the dual vector space Tp M. For n-dimensional M, there are differential forms of degree from 0 up to n. A degree-k differential form (generally referred to as a k-form) can be defined as iteratively follows. The 0-forms are merely functions of n variables, f(x) = f(x1, . . . , xn). The 1-forms are all expressible as 1 n α = f1dx + ... + fndx where the fi coefficients are 0-forms. The 2-forms are of the form 1 n β = α1 ∧ dx + ... + αn ∧ dx where the ai coefficients are 1-forms. The wedge product “∧” is a multiplication operation between two forms which is associative and distributive, and defined uniquely by i j i j j i i i fidx ∧ fjdx = (fifj)dx ∧ dx = −(fifj)dx ∧ dx dx ∧ dx = 0 for all i, j and all functions fi, fj. Continuing as above, we can say in general that k-forms are expressions of the type 1 n ω = ψ1 ∧ dx + ... + ψn ∧ dx where the ψi are (k − 1)-forms.

What are differential forms intuitively? If we examine 1-forms, our general definition shows they are linear combinations of linear functionals, with coefficients in C∞(M, R). So at each point p ∈ M the form αp(·): TpM → R defines a linear functional on TpM as follows: 1 n αp(v) = f1(p)dx (ν) + ... + fn(p)dx (ν) ∀ν ∈ TpM. So 1-forms act as smoothly varying linear functionals on each tangent space. The space of degree p-differential forms is denoted ΛpM, and forms a vector space. Similarly, degree k differential forms are smoothly varying linear functionals which take elements of TpM × ... × TpM (k-tuple) as their arguments. As an example, consider the differential 1-form defined on R2 \{0} xdy − ydx ω = . x2 + y2 THE HODGE THEOREM AND THE BOCHNER TECHNIQUE: A VANISHINGLY SHORT PROOF 3

If we ﬁx the point (1, 1) ∈ R2 \ 0, it simplies to the linear functional 1 ω (·) = (−dx + dy): T 2 → , (1,1) 2 (1,1)R R which takes in tangent vectors based at the point (1, 1), and returns a scalar.

Given the above discussion we can now generalize our definition of the integral as follows. Since every n 1 n n differential n−form ω on R can be expressed as ωp = f(p)dx ∧ ... ∧ dx , we define the integral of f on R as follows Z Z ω = f(x)dx1 . . . dxn, n n R R where the right hand side is the regular Lebesgue integral of f on Rn. If we want the integral to be finite we typically require that f vanish outside a compact set. This is not any major restriction as any smooth function can be modified in such a way as to remain smooth. The reason for such a definition of integral is that by definition, our differential n−form transforms appropriately as to preserve the change of variables formula. To extend this definition to a manifold, we define similarly to the above, and use the change of variables formula. 1 n If an n−form ω on a subset W ⊂ M can be represented in its chart (W, φ) as ωp = f(p)dx ∧ ... ∧ dx , we define Z Z ω = f ◦ φ(x)|det(Dφ)|dx1 . . . dxn, W φ(W ) where dx1 . . . dxn is the n−dimensional Lebesgue measure on Rn. This definition makes sense as U ⊂ M, and φ(W ) ⊂ Rn is an integrable subset. It is a small exercise to show that this definition is independent of the choice of chart, however we shall exclude this for the sake of brevity. Let us return to the land of vector calculus, and resurface to take a gulp of air before diving back into the depths of differential forms. We have three linear differential operators familiar to students, namely the gradient ∇, curl ∇×, and divergence ∇·. The attentive calculus student may recall that we have the following properties for these operators: for all smooth functions f and all vector fields W that (1.1) ∇ × (∇f) = 0 ∇ · (∇ × W ) = 0. We may recast these statements in a more linear algebraic fashion as follows Im(∇) ⊆ Ker(∇×) Im(∇×) ⊆ Ker(∇·), the image of the gradient operator is contained in the kernel of the curl operator, and the image of the curl operator is contained in the the kernel of the divergence operator. In physics terminology, if a vector field X : R3 → R3 is the gradient of a function (i.e. X = ∇f) then X is called a conservative field. A vector field Y : R3 → R3 with zero curl is called irrotational. So (1.1) implies that every conservative field is irrotational, since ∇ × X = ∇ × (∇f) = 0. This has the interesting physical interpretation that physical fields of force (which tend to be conservative vector fields) such as the gravitational force or electromagnetic force cannot have any localized rotations in their flows. The phenomena we have observed generalize quite nicely in the spaces of forms. For every differential operator we have described from vector calculus, there is an analog defined on the spaces ΛpM.

The exterior derivative, d :ΛpM → Λp+1M is a linear operator which is deeply related to the spaces of differential forms. It may be defined in an inductive fashion, similarly to differential forms. For a differential p-form ω ∈ ΛpM, we say the form is closed if dω = 0, i.e. its exterior derivative vanishes. A closely related property is that exactness. A p-form α ∈ ΛpM is exact if there exists β ∈ Λp−1M such that dβ = α. We say these are properties are related as one implies the other. Since d2 ≡ 0, then if α is exact it is also closed. This follows from dα = d(dβ) = 0. What is interesting about these definitions is that both conditions are equivalent to the forms α or ω satisfying a certain differential equation. Given a set of coordinates (x1, . . . , xn) on our manifold M, we can write some differential p-forms ω in local coordinates as

ω = fdxi1 ∧ ... ∧ dxip , 4 HADRIAN QUAN

where the set I = {i1, . . . , ip} is taken to be some ordered subset of {1, . . . , n}. In these coordinates its exterior derivative equals n X ∂f dω = df ∧ dxi1 ∧ ... ∧ dxip = dxi1 ∧ ... ∧ dxip , ∂xk k=1 where we interpret the differential of the function f as the exterior derivative of the zero-form f. To say ∂f this object equals zero is equivalent to requiring every term ∂xk vanishes. An arbitrary differential p-form is a linear combination of terms of the form ω = fdxi1 ∧ ... ∧ dxip , so we may expect that closedness and exactness conditions to be equivalent to solving linear first order partial differential equations. In fact, they mirror the conditions required for a vector field to be conservative, or irrotational. Above we observed that divergence, curl, and gradient operators in R3 obeyed certain relationships, namely ∇ × (∇f) = 0 ∇ · (∇ × W ) = 0. We can see that he statement curl◦grad ≡ 0 says little more than that d2 ≡ 0, given the right perspective. To make this observation explicit, we shall show the proceeding diagram commutes. Denote the space of vector 3 3 fields on R by X(R ). Then there exists linear isomorphisms φ1, φ2, φ3 such that the following diagram commutes. ∇ ∇× ∇· C∞(R3) X(R3) X(R3) C∞(R3)

id Ψ1 Ψ2 Ψ3 d d d M Λ1(R3) Λ2(R3) Λ3(R3) When we say this diagram commutes, we mean that we can travel along the arrows in any direction, and the result will not change. Recall the differential 1-form we observed earlier, which was defined on R2 \{0} and given by xdy − ydx ω = . x2 + y2 We can see this is closed as −y x dω = d ∧ dx + d ∧ dy x2 + y2 x2 + y2 ∂ y ∂ x = − dy ∧ dx + dx ∧ dy ∂y x2 + y2 ∂x x2 + y2 (x2 + y2) − 2y2 (x2 + y2) − 2x2 = − (−1)dx ∧ dy + dx ∧ dy (x2 + y2)2 (x2 + y2)2 2(x2 + y2) − 2x2 − 2y2 = dx ∧ dy = 0, (x2 + y2)2 where we used the property of the wedge product that dx ∧ dy = −dy ∧ dx. Since we have seen already that Exact form =⇒ Closed form, it is a natural question to ask if the reverse implication is true. Sadly not. In fact we already have a xdy−ydx counterexample: to see that the form ω = x2+y2 is not exact, we note the following property: a differential form α is exact only if Z α = 0 γ for all simple closed curves γ (this is if and only if, but we shall only need necessity). To see this, let α = dβ then we may use Stokes theorem to show Z Z Z α = dβ = β = 0, γ γ ∂γ=∅ since a closed curve has no boundary, so ∂γ = ∅. Now that we have this criterion for exactness, lets xdy−ydx 1 examine our form ω = x2+y2 . If we take γ = S , the unit circle, define a parametrization of this curve: THE HODGE THEOREM AND THE BOCHNER TECHNIQUE: A VANISHINGLY SHORT PROOF 5

R R 2π 1∗ 1∗ C(t) = (cos(t), sin(t)), over the region t ∈ [0, 2π]. The line integral of a curve S1 ω = 0 S ω, where S ω is the pullback of ω onto [0, 2π]. Calculating this integral we obtain Z Z 2π (cost)(costdt) − (sint)(−sintdt) Z 2π ω = = dt = 2π 6= 0, 2 2 S1 0 cos (t) + sin (t) 0 xdy−ydx hence ω = x2+y2 is not exact. So we have Exact =6 ⇒ Closed. This result should not rattle us too severely. Any good mathematician should be ready and willing to take an obstruction and turn it into an invariant. In the above example, the 1-form was not exact, in that xdy−ydx there did exist any function θ such that dθ = ω = x2+y2 . This is an issue more with the function θ then the form ω. We defined ω on R2 \{0}. In fact, there is no way to extend this function θ (and hence the form ω) to a smooth single-valued function on all of R2. The space R2 has a “hole” at the origin, which is the source of the 1-form’s problems. (The Poincare Lemma states that every closed form is locally exact). In other words, global properties of the space influence the analytic properties of the forms, and in this regard the difference between closedness and exactness becomes a measure of topological behavior.

To solidify this analogy, we can place an equivalence relation on closed p-forms. If we have two closed p p−1 forms α1, α2 ∈ Λ M we say they are cohomologous if they diﬀer by an exact form, i.e. ∃β ∈ Λ M such that

α1 ∼ α2 ⇐⇒ α1 − α2 = dβ. The equivalence relation of being cohomologous requires that β is a globally defined, and the equivalence classes with respect to this relation are referred to as cohomology classes. The set of cohomology classes [ω] differential forms of degree p is referred to as the pth de Rham Cohomology Group (or space), and is p denoted by HdR(M). p Proposition 1.1. For every fixed p, the pth de Rham Cohomology group of M, HdR(M) forms a vector space over the field R, with respect to the operations of addition and scaling of representatives of the equivalence relation. Proof. First, we shall prove these linear operations are well-defined with respect to the choice of representative. If α, β are two closed forms then α + β is a closed form since the exterior derivative d is linear. If β is cohomologous to β0, then β0 = β + dγ. So replacing one form with the other will keep the sum in the same cohomology class, as α + β0 = α + (β + dγ) = (α + β) + dγ =⇒ α + β0 ∼ α + β. p So addition is well-defined on the cohomology classes. The proof for multiplication is similar. So HdR(M) is a vector space with the operations [α1] + [α2] := [α1 + α2] and k[α] := [kα], for k ∈ R. p Since these are vector spaces over R we typically denote them by HdR(M, R). They can be defined similarly p as vector spaces over C, in which case we denote them HdR(M, C). The terminology of cohomology group may be confusing, since these objects are vector spaces. The naming originates in the classical cohomology theories where the analogous objects were in fact abelian groups. Our final definition is that of the Betti numbers. Definition 1.2. Betti Numbers p The integer bp(M) = HdR(M) is called the p−th betti number. For every p, this integer is a topological invariant.

2. Preliminaries for Hodge Theory We begin with some preliminary definitions, and hint at the results we plan to show. Unless otherwise stated, all manifolds are taken to compact, connected, and oriented. Our first player is a purely linear- algebraic object, which can be defined with respect to the Exterior Algebra of any Vector Space. For our purposes we are considering the space of differential p−forms. Note that if a statement or proposition is made with respect to ΛpM, then unless otherwise specified it holds for all ΛpM for all 1 ≤ p ≤ n. 6 HADRIAN QUAN

Definition 2.1. The Hodge Star The Hodge Star ∗ :ΛpM → Λn−pM is a linear operator which satisfies ∗∗ = (−1)p(n−p). It can be defined ∗ with respect to any ordered basis {e1, ..., en} of the cotangent space Tp M (in particular any reordering of that basis) that

∗(1) = e1 ∧ ... ∧ en ∗ (e1 ∧ ... ∧ en) = 1.

Additionally, for any ordered subset {i1, . . . , ip} ⊆ {1, . . . , n} we have

∗(ei1 ∧ ... ∧ eip ) = ±ej1 ∧ . . . ejn−p

where {j1, . . . jn−p} is the complement of {i1, . . . , ip} in {1, . . . , n}. In other words

{1, . . . , n}\{i1, . . . , ip} = {j1, . . . jn−p}. The sign is determined such that

(2.1) ei1 ∧ ... ∧ eip ∧ ∗(ei1 ∧ ... ∧ eip ) = ei1 ∧ ... ∧ eip ∧ (±ej1 ∧ . . . ejn−p ) = e1 ∧ ... ∧ en after reordering the left side.

Example 2.2. On R4 with the Euclidean coordinates (x, y, z, w), we have {dx, dy, dz, dw} as the coordinate 1-forms, and dx ∧ dy ∧ dz ∧ dw as the top degree form. From the deﬁnition above, we compute ∗dx = dy ∧ dz ∧ dw ∗ dy = −dx ∧ dz ∧ dw ∗ dz = dx ∧ dy ∧ dw ∗ dw = −dx ∧ dy ∧ dz

∗(dx ∧ dy) = dz ∧ dw ∗ (dx ∧ dz) = −dy ∧ dw to demonstrate a few examples. The ﬁrst are so, since dx ∧ ∗dx = dx ∧ (dy ∧ dz ∧ dw) dy ∧ ∗dy = dy ∧ (−dx ∧ dz ∧ dw) = (−1)2dx ∧ dy ∧ dz ∧ dw, which show that ∗ satisﬁes the critical property (2.1). The equalities of the remaining examples are true by similar calculations. Remark 2.3. We can prove relatively quickly the identity ∗∗ = (−1)p(n−p). Consider some reordering of the

standard basis for which we take the ﬁrst p terms and wedge them, i.e. ei1 ∧ ... ∧ eip , then the deﬁnition implies that

ei1 ∧ ... ∧ eip ∧ ∗(ei1 ∧ ... ∧ eip ) = e1 ∧ ... ∧ en as this held for any reordering. But then observe that

ei1 ∧ ... ∧ eip ∧ ∗(ei1 ∧ ... ∧ eip ) = ∗(ei1 ∧ ... ∧ eip ) ∧ ∗ ∗ (ei1 ∧ ... ∧ eip ), | {z } | {z } (i) (ii) and if we interchange (i) and (ii), we are in a place to evaluate the effect of ∗∗. Since (i) ∈ Λn−pM and (ii) ∈ ΛpM, if we interchange them we pick up a (−1)n−p for every term in (ii). Since there are p of them we have p(n−p) ei1 ∧ ... ∧ eip ∧ ∗(ei1 ∧ ... ∧ eip ) = (−1) ∗ ∗(ei1 ∧ ... ∧ eip ) ∧ ∗(ei1 ∧ ... ∧ eip ) hence ∗∗ = (−1)p(n−p). With this operator in place, we can define an inner-product on ΛpM as follows Z hα, βi = α ∧ ∗β for α, β ∈ ΛpM, M and denote the corresponding norm by ||α||. This defines an inner product as α ∧ ∗β is a top degree form and can be integrated over M. This integral is guaranteed to exist since M is compact. Definition 2.4. The δ-operator Define the operator δ :ΛpM → Λp−1M as follows, δ = (−1)n(p+1)+1 ∗ d∗. Note that on 0−forms, i.e. functions, this is simply the zero linear operator. Proposition 2.5. We can show that the operator δ is the formal adjoint of the exterior derivative d. Additionally, it satisfies δ2 ≡ 0. THE HODGE THEOREM AND THE BOCHNER TECHNIQUE: A VANISHINGLY SHORT PROOF 7

Proof. We shall show that hdα, βi = hα, δβi with respect to the inner product deﬁned above. By orthogo- nality of the ΛpM, we need only consider the case when α is a (p − 1) form and β is a p−form. In this case observe that since d ∗ β = (−1)p ∗ δβ for all β ∈ ΛpM, then we have

d(α ∧ ∗β) = dα ∧ ∗β + (−1)p−1α ∧ d ∗ β = dα ∧ ∗β + (−1)2p−1α ∧ ∗δβ = dα ∧ ∗β − α ∧ ∗δβ.

Hence, Z Z Z 0 = α ∧ ∗β = d(α ∧ ∗β) = dα ∧ ∗β − α ∧ ∗δβ = hdα, βi − hα, δβi ∂M=∅ M M where this is zero from an application of Stokes Theorem since the manifold is compact. Hence we have hdα, βi = hα, δβi as claimed. To see that δ2 ≡ 0, note that for any α ∈ Λp−2M, β ∈ ΛpM we have

hα, δ2βi = hd2α, βi = h0, βi = 0

2 and since hα, ·i is positive deﬁnite, we must have δ β = 0. p was arbitrary, so we have our result.

Since we now have the linear operators d and δ we can use them to deﬁne the Hodge Laplacian ∆ : ΛpM → ΛpM, by ∆ = dδ + δd. This map is a symmetric, linear operator on forms. Linearity is clear, as it is a linear combination of linear operators. To see that it is symmetric with respect to our inner product on forms, note that for α, β ∈ ΛpM,

h∆α, βi = h(dδ + δd)α, βi = hdδα, βi + hδdα, βi = hδα, δβi + hdα, dβi = hα, dδβi + hα, δdβi = hα, ∆βi.

And as a corollary of the above calculation, we also have the following.

Corollary 2.6. ∆α = 0 if and only if dα = 0 and δα = 0.

Proof. From above we see that

h∆α, αi = hdα, dαi + hδα, δαi = ||dα|| + ||δα||

So if ∆α = 0, then dα = 0 and δα = 0. Conversely, from the deﬁnition of ∆, if dα and δα are both zero, then ∆α is clearly zero.

And as a corollary to the corollary,

Corollary 2.7. The only Harmonic functions (∆f = 0) on compact, connected, oriented manifolds are constant functions.

Proof. Since ∆f = 0 implies df = 0, and the exterior derivative on functions is just the diﬀerential. So df = 0 implies f is a constant function.

However, the Hodge laplacian allows us to deﬁne harmonicity for forms as well as functions.

Deﬁnition 2.8. Harmonic Forms We denote Hp by the space of Harmonic p−forms, i.e.

Hp = {α ∈ ΛpM|∆α = 0} = Ker(∆).

A worthwhile observation is that despite notational similarity this space is distinct (a priori) from p HdR(M, R), the pth de Rham cohomology group. However, corollary 2.4 shows that a form is closed if p p and only if it is Harmonic, hence the inclusion HdR(M, R) ⊆ H . We cannot immediately say the reverse inclusion is true, as a given Harmonic form may be closed but not exact. In fact, for a compact manifold M this is always true, and this fact will be a corollary of our results. 8 HADRIAN QUAN

3. An Interlude: The Hodge Theorem as a Least Norm Solution There is a pleasing interpretation as of the Hodge theorem as a criterion for selecting a single differential form from a multitude. Among all possible differential forms α in a fixed cohomology class [ω], there a ‘best’ choice, in some regard. The Hodge star ∗ has allowed us to define an inner product on ΛpM, Z hα, βi = α ∧ ∗β for α, β ∈ ΛpM, M so we should be able to select a differential form of minimum norm, (provided we can prove some type of p p uniqueness result). Fix [ω] ∈ HdR(M), and select α ∈ Λ M such that Z ||α||2 = inf η ∧ ∗η. η∈[ω] M Since α = η + dτ, we would readily be able to select α if τ belonged to a finite dimensional vector space. However, this may not be the case. Say we assume that we can find a solution to this minimization problem exists. Let τ ∈ Lp−1(Ω) then ||α + dτ||2 ≥ ||α||2 =⇒ 2hα, dτi + 2||dτ||2 ≥ 0 hence hα, dτi = 0. By adjointness, we have δα = 0, since hδα, τi = 0 for all τ ∈ Λp−1(Ω). From our corollary in the previous section, δα = 0 implies ∆α = 0, i.e. α is Harmonic! So the assumption that α is a minimizer of this norm implies that it solves our PDE.

4. Weak Solutions and the Hodge Decomposition Theorem The real power of the Hodge Theorem is it gives an orthogonal decomposition of the de Rham Cohomology groups, with respect to the ∆ operator. The dimensions of the de Rham Cohomology groups are geometric invariants. By combining simple geometric arguments with this control over the de Rham cohomology groups we gain much insight into the geometry and topology of our manifold. The statement of the Hodge Decomposition Theorem can be written simply as ΛpM = Ker(∆) ⊕ Im(∆), for M a compact, oriented, Riemannian manifold. This result is highly surprising, as we have decomposed a (possibly infinite dimensional) vector space in terms of the kernel and image of an operator. Another way this can be interpreted is that we have an orthogonal decomposition ΛpM = Ker(∆) ⊕ (Ker(∆))⊥, and that (Ker(∆))⊥ = Im(∆). To prove this statement is difficult. To begin with, some subspaces of infinite dimensional spaces do not have orthogonal complements! The natural finite-dimensional proofs of decomposition theorems (such as the Rank-Nullity Theorem for L : V n → V n) are quite constructive, but typically rely on our ability to extend the basis of Im(L) or Ker(L) to a basis of the entire vector space. Consider for example the following proof of Rank-Nullity: dim(Im(L)) + dim(Ker(L)) = dim(V n).

Proof. Suppose {u1,..., um} forms a basis of Ker(L). We can extend this to form a basis of V :

{u1,..., um, w1,..., wn}. Since the dimension of Ker(L) is m and the dimension of V is m+n, it suﬃces to show that the dim(Im(L)) = n. Let us see that {Lw1,...,Lwn} is a basis of Im(L). Let v be an arbitrary vector in V . There exist unique scalars such that: v = a1u1 + ··· + amum + b1w1 + ··· + bnwn

⇒ Lv = a1Lu1 + ··· + amLum + b1Lw1 + ··· + bnLw ⇒ Lv = b1Lw1 + ··· + bnLwn ∵ Lui = 0 Thus, {Lw1,...,Lwn} spans Im(L). We only now need to show that this list is not redundant; that is, that {Lw1,...,Lwn} are linearly independent. We can do this by showing that a linear combination of these vectors is zero if and only if the coeﬃcient on each vector is zero. Let:

c1Lw1 + ··· + cnLwn = 0 ⇔ L{c1w1 + ··· + cnwn} = 0

∴ c1w1 + ··· + cnwn ∈ ker(L) THE HODGE THEOREM AND THE BOCHNER TECHNIQUE: A VANISHINGLY SHORT PROOF 9

Then, since ui span Ker(L), there exists a set of scalars di such that:

c1w1 + ··· + cnwn = d1u1 + ··· + dmum

But, since {u1,..., um, w1,..., wn} form a basis of V , all ci, di must be zero. Therefore, {Lw1,...,Lwn} is linearly independent and indeed a basis of Im(L). This proves that the dim(Im(L)) = n, as desired. In an infinite dimensional vector space this approach may break down, as some basis elements of the image or kernel of an operator may ‘diverge.’ The issue of orthogonal decompositions in a topological vector space is discussed somewhat in the appendix, under the section “Orthogonal Decomposition in a Hilbert Space.” Now that we have ΛpM = Ker(∆) ⊕ (Ker(∆))⊥, (provided dimKer(∆) < ∞, which we have not yet shown) it remains to be proven that (Ker(∆))⊥ = Im(∆). One of these inclusions is almost immediate: Im(∆) ⊆ (Ker(∆))⊥. To see this, we note that if ∆ω = α (i.e. α ∈ Im(∆)) and β ∈ Ker(∆) then 0 > hα, βi = h∆ω, βi = hω,∆βi = 0 so α ∈ (Ker(∆))⊥. To prove the reverse inclusion requires us to prove existence of a solution to ‘Poissons Equation’ for differential forms: find ω such that ∆ω = α, provided α ∈ (Ker(∆))⊥. From the discussion and definitions of differential forms in section 1, we see that ∆ω = α really does represent a partial differential equation. How might we solve this PDE? The Hodge Laplacian is a linear operator, so a natural place to start looking is by considering the analogous problem in linear algebra. In finite dimensions, denote by L : V n → V n a linear operator. Given some prescribed v ∈ V n we seek to prove existence of u ∈ V n such that Lu = v. If V n is also equipped with an inner product h·, ·i, this question becomes equivalent to finding u ∈ V n such that hLu, wi = hv, wi ∀w ∈ V n ⇐⇒ hu, L∗wi = hv, wi ∀w ∈ V n. ∗ ∗ where L denotes the conjugate-transpose map to L, i.e. Lij = Lji. The advantage of this definition is that it allows us to say u solves Lu = v without explicitly applying L to u. To consider now the case at hand, we have a linear operator ∆ : ΛpM → ΛpM on a possibly infinite dimensional vector space. The associated problem we seek to solve: ∆ω = α, also admits the following a weak solution Z Z Z Z ∆ω ∧ ∗β = α ∧ ∗β ∀β ∈ ΛpM ⇐⇒ ω ∧ ∗∆β = α ∧ ∗β ∀β ∈ ΛpM M M M M

Again, we are seeking to show in this context that Suppose that ω ∈ ΛpM solves ∆ω = α, then h∆ω, βi = hα, βi for all β ∈ ΛpM. which implies hω, ∆∗βi = hα, βi for all β ∈ ΛpM, since hω, ∆∗βi = h∆ω, βi. This may seem silly but it is actually a very powerful reformulation, since we can now do the following: the solution ω defines a bounded linear functional ` :ΛpM → R by the formula `(β) = hω, βi. And this functional should satisfy `(∆∗β) = hω, ∆∗βi = h∆ω, βi = hα, βi, for all β ∈ ΛpM to ensure it acts similarly to our original PDE. Such a linear functional is called a weak solution to the equation ∆ω = α. We formalize this discussion in the following definitions. Definition 4.1. Let α ∈ ΛpM.A strong solution is ω ∈ ΛpM such that ∆ω = α.A weak solution of ∆ω = α is a bounded linear functional ` :ΛpM → R such that `(∆β) = hα, βi ∀β ∈ ΛpM. 10 HADRIAN QUAN

From our discussion above, it is clear that every solution of the equation ∆ω = α determines an accompa- nying weak solution `. The major technical obstacle in Hodge’s original proof of the Hodge Decomposition theorem is to prove the converse: every weak solution `(∆∗β) = hα, βi determines a unique solution ω ∈ ΛpM of the equation ∆ω = α. So the linear functional given by the weak solution demonstrates existence of a This will be our first major theorem Theorem 4.2. Regularity of ∆ Let α ∈ ΛpM and let ` be a weak solution of the equation ∆ω = α. Then there exists ω ∈ ΛpM such that `(β) = hω, βi for all β ∈ ΛpM. And hence determines a solution of ∆ω = α. Theorem 4.3. ∆ is a compact operator p Let {αn} be a sequence in Λ M such that ||αn|| ≤ c and ||∆αn|| ≤ c for all n and for some c > 0. Then p there exists a subsequence {αnk } which is Cauchy in Λ M. The previous two results shall be instrumental to proving our main theorems. For now, let’s assume them. Proving these analytic results shall be the major technical obstacles. Theorem 4.4. Hodge Decomposition Theorem Let M be a compact, orientable Riemannian manifold. For all 0 ≤ p ≤ n we have that Hp is finite dimensional, and we have the following orthogonal direct sum decomposition ΛpM = Im(∆) ⊕ Ker(∆) = ∆(ΛpM) ⊕ Hp = dδ(ΛpM) ⊕ δd(ΛpM) ⊕ Hp = d(Λp−1M) ⊕ δ(Λp+1M) ⊕ Hp hence the equation ∆ω = α has a solution ω ∈ ΛpM if and only if the p−form α is orthogonal to Hp. Proof. In this proof, Theorem 3.2 is what ensures that Hp is finite dimensional, and Theorem 3.1 gives existence of a solutions to ‘Poissons equation,’ which will help us prove that (Hp)⊥ ⊆ ∆(ΛpM). Assume for the sake of contradiction that Hp were not finite dimensional. This would imply there was an p infinite sequence of orthonormal vectors {αk} in H , via the Gram-Schimdt process. However, since these are all normalized vectors in Hp, there exists c > 0 such that

||αk|| = 1 < c ||∆αk|| = ||0|| < c, p so Theorem 3.2 implies there exists a Cauchy subsequence {αnk } of elements in H . However, these vectors are orthogonal, and hence all have distance from each other equal to √1 , contradicting that this is a Cauchy 2 subsequence. So Hp must be ﬁnite-dimensional. p p Let {h1, . . . , hr} be a basis for H . Then for arbitrary α ∈ Λ M, we may write it as r X α = β + hα, hkihk k=1 where β ∈ (Hp)⊥. This gives us an orthogonal decomposition ΛpM = (Hp)⊥ ⊕ Hp. To conclude our proof we need only show that ∆(ΛpM) = (Hp)⊥. We have immediately that ∆(ΛpM) ⊂ (Hp)⊥: if α ∈ ΛpM and γ ∈ Hp then h∆α, γi = hα, ∆γi = hα, 0i = 0, so α ∈ (Hp)⊥. Now we wish to show the reverse inclusion. To do this, we need the following inequality: ∃c > 0 such that ||β|| ≤ c||∆β|| ∀β ∈ (Hp)⊥. ∞ p ⊥ Suppose otherwise, then ∃{βj}j=1 ⊆ (H ) with ||βj|| = 1 and ||∆βj|| → 0. By theorem 4.3, a subsequence p p of βj is cauchy. So limj→∞hβj, ωi exists for each ω ∈ Λ M. Then we can deﬁne a linear function on Λ M by `(ω) = lim hβj, ωi. j→∞ Each ` is bounded, via the Cauchy-Schwartz inequality. And

`(∆ω) = lim hβj, ∆ωi = lim h∆βj, ωi = 0, j→∞ j→∞ THE HODGE THEOREM AND THE BOCHNER TECHNIQUE: A VANISHINGLY SHORT PROOF 11 hence ` is a weak solution of ∆β = 0. By theorem 4.2, there exists β ∈ ΛpM such that `(ω) = hβ, ωi. Then p ⊥ p βj → β. Additionally, since ||βj|| = 1, βj ∈ (H ) this implies ||β|| = 1 and β ∈ H . But ∆β = 0, so β ∈ Hp, and Hp ∩ (Hp)⊥ = {0}! A contradiction. So our inequality holds. Now, let α ∈ (Hp)⊥. Define a function on Im(∆) by `(∆φ) = hαφi ∀φ ∈ ΛpM. ` is well-defined since ∆ is injective. Additionally, ` is a bounded linear functional on Im(∆). To see this, p p ⊥ let φ ∈ Λ M and define ψ := φ − πHp (φ) ∈ (H ) . Then from our above inequality, we have |`(∆φ)| = |`(∆ψ)| = |hα, ψi| ≤ ||α|| · ||ψ|| ≤ c||α|| · ||∆ψ|| = c||α|| · ||∆φ||. Since ` is a bounded linear functional, the Hahn-Banach theorem (see appendix, under “Fundamentals of Functional Analysis”) says that ` extends uniquely to a bounded linear functional on all ΛpM. Thus ` defines a weak solution of ∆ω = α. Hence we have shown (Hp)⊥ = Im(∆). Having assumed theorems 4.2 and 4.3, the Hodge decomposition theorem has been proven. Theorem 4.5. For all p, every de Rham Cohomology Class in ΛpM contains a unique harmonic representative. p Proof. Let α ∈ [η] for some arbitrary [η] ∈ HdR(M, R). From the Hodge decomposition, we may write α as α = ∆ω + β = dδω + δdω + β, where β ∈ Hp, and for some ω ∈ ΛpM. Since α is closed, by applying d to both sides we have 0 : 0 0 = dα =ddδω + dδdω + dβ> where dβ = 0 since ∆β = 0. So dδdω = 0. If dδdω = 0 then we can see that 0 = h0, dωi = hdδdω, dωi = hδdω, δdωi = ||δdω||2, hence we must have that δdω = 0 identically. So our original decomposition of α becomes α = dδω + 0 + β =⇒ α − β = dδω, thus α and β differ only by an exact form d(δω). So they are cohomologous. We have shown that an arbitrary cohomology class contains a harmonic representative. To complete our proof, we need only show that any two harmonic forms in the same cohomology class p are equal, and hence unique. Let α1, α2 ∈ H be in the same cohomology class, i.e. α1 − α2 = dτ. Their difference is also harmonic, so we have 2 ||dτ|| = hdτ, dτi = hα1 − α2, dτi = hδ(α1 − α2), τi = h0, τi = 0 so dτ = 0, thus α1 = α2. In the next section, we will attempt to solve Poisson’s equation. Since a smooth, compact manifold is geometric space which is locally euclidean and not too ‘large’, it is similar to an open bounded domain in Rn (at least locally). Our hope is that if we can solve Poisson’s equation in as much generality as possible on such a domain in Rn, this may be sufficient to solve it for differential forms defined on the charts on our n m n manifold. In particular, if we can prove our results for arbitrary functions f : R → R , where m = q , this will allow us to define every element of ΛqM as images of such functions in an open chart of M. To this end, we shall first try and solve the equation formally, trying to employ any heuristics which serve our purpose. If this gives us some insight into the structure of the equation, we shall follow this intuition, and fill in the mathematical details as we go along.

5. Solving Poissons Equation: Laplace and Dirac We wish to solve Poisson’s Equation ∆u = f, in full generality. This is a second order, linear, inhomogeneous, partial diﬀerential equation. That it is a second order PDE says it demands the full force of our mathematical methods. That it is a linear equation says we may stand some chance of developing a general theory to solve it. Hopefully we can model our approach from the theory of solving linear equations in ﬁnite dimensions. 12 HADRIAN QUAN

In analogy with solving algebraic equations, sometimes solutions of simple equations (say with rational coefficients) can only be found by taking detours through the real or complex number system. As an example, the equation p(x) = x5 + 20x3 + 20x2 + 30x + 10 = 0 has a highly non-trivial root, given by x = 21/5 − 22/5 + 23/5 − 24/5. If we care only about existence of such a solution, then this question can quickly be answered using calculus. Since this polynomial is of odd-order, p(x) → ±∞ as x → ±∞. As a polynomial, continuity is ensured, so the intermediate value theorem implies it attains every value in its range. In particular, there exists a point x0 such that p(x0) = 0. So there exists at least one real solution of this equation. Such limiting operations require we consider Q as a subset of R or C. In both cases, we ‘completed’ Q by considering it as a subset of a larger number system (here as a topological space). Ideally, a similar scheme may be implemented to solve this partial differential equation. Our solution to Poisson’s equation will hopefully be at least C2 (why shouldn’t it be? Its definition involves two derivatives), but to solve this equation we may need to ‘complete’ C2 appropriately, and perform the majority of our analysis in this larger space of functions. This may be hard to do, as functions are inherently more complicated objects than numbers. In particular, we must be careful how we specify a topology on our collection of functions. As an example, there are several distinct ways to state ∞ “The sequence {fn}n=1 converges to f.” At the undergraduate level, a student may have already experienced two separate notions of function convergence: Definition 5.1. Pointwise vs. Uniform Convergence ∞ (1) The sequence {fn}n=1 converges pointwisely to f on A if for every x0 ∈ A, and for every > 0, there exists an index N such that

|fn(x0) − f(x0)| ≤ ∀n ≥ N.

∞ (2) The sequence {fn}n=1 converges uniformly to f on A if for every > 0, there exists an index N such that

|fn − f| ≤ ∀n ≥ N, ∀x ∈ A. Intuitively, the first definition has an explicit dependence on the point in the set A. The sequence converges at every point, but the rate of convergence may differ. Hence, to find the index N after which the sequence is −close to its limit, we may need to consider what point we evaluated on. Conversely, the uniform convergence ensures that for a given , a single N guarantees at most an ’s worth of distance for every x ∈ A.

Before we begin our function space odyssey, let’s deﬁne some convenient notation for derivatives, which we shall use throughout this paper.

Deﬁnition 5.2. Derivatives, Multi-Indices, and the Laplacian on Rn n A multi-index α = (α1, . . . , αn) ∈ Z is an n−tuple of integers. We deﬁne the following operations on multi-indices: 2 2 1/2 |α| := (α1 + ... + αn) , and [α] = α1 + ... + αn, if αi ≥ 0 We shall use the following notation for derivative operators:

[α] ∂f α ∂ Dx f := D f := (f) j α1 αn ∂xj ∂x1 ··· ∂xn And the Laplacian operator ∆ on Rn is given by n X ∂2f ∆f = ∂x2 i=1 i THE HODGE THEOREM AND THE BOCHNER TECHNIQUE: A VANISHINGLY SHORT PROOF 13

To solve Poisson’s equation, one approach may start with solving Laplace’s equation: ∆Φ = 0, and then convolving this particular solution with our inhomogeneous term f, to obtain Z (5.1) u(x) = f(y)Φ(x − y)dy.

Definition 5.3. Convolution The convolution of two integrable functions f, g is denoted f ∗ g and is defined to be Z f ∗ g(x) = f(x − y)g(y)dy. n R Note that the operation of convolution is commutative, as may be observed by the substitution of z = x − y. (Here the asterisk denotes convolution, not the Hodge star. We shall do our best to be unambiguous as to which operation we are using.) Intuitively, we are exploiting linearity of the differential operator ∆. By solving Laplace’s equation, we have solved a homogeneous version of Laplace’s equation. Then, convolution uses integration to sum together the inhomogeneous parts (i.e. f) to build up the solution u. We haven’t yet decided what properties to impose on f to ensure a solution to Poisson’s equation in this manner. Let alone that this scheme will actually work. Let’s be bold and try this approach anyway. To solve Laplace’s equation, one approach is to assume a spherical solution of the form v(r) = u(|x|), and then solve Laplace’s equation in this new coordinate system. We can compute how the Laplacian changes in this new coordinate system as follows, 2 2 ∂ 0 xi ∂ 00 xi 0 |x| − xi v(r) = v (r) , 2 v(r) = v (r) + v (r) 2 , ∂xi |x| ∂xi |x| |x| from the chain rule. So n 2 n 2 X ∂ X x |x| − xi ∆v = v = v00(r) i + v0(r) ∂x2 |x| |x|2 i=1 i i=1 |x| n|x| − |x| n − 1 = v00 + v0 = v00 + v0 |x| |x|2 |x| n − 1 = v00 + v0 . r Setting this equal to zero, and rearranging gives us v00 1 − n 1 − n = =⇒ (ln(v0))0 = . v0 r r We may integrate both sides with respect to r, and exponentiate to obtain C ln(v0) = (1 − n) ln(r) + C =⇒ v0 = . rn−1 So we have  C ln(r) + D n = 2 v(r) = C + D n ≥ 3. rn−2 We have not yet made choices for the constants C and D. Additionally, since we made the substitution r = |x|, this implies our general solution can be written as follows Definition 5.4. Fundamental solution of Laplace’s Equation The function  1 − ln |x| n = 2  2π Φ(x) = 1 1 n ≥ 2  n−2 n(n − 2)ωn |x|

(where ωn denotes the volume of the unit-ball) is the Fundamental Solution of Laplace’s equation, deﬁned for all Rn \{0}. On this set it satisﬁes ∆Φ = 0. 14 HADRIAN QUAN

Why do we say that this function is defined only on the punctured space Rn \{0}? Recall that we obtained this solution by assuming the solution of Laplace’s equation was a radial function of the variable r = |x|. However no such function can be continuous at the origin, let alone differentiable, so it makes little sense to consider the Laplacian of such a function at the origin. 2 n We make two further claims: (1) For a function f ∈ C0 (R ) (twice-differentiable functions which vanish 1 outside a compact set) we can interchange integration and the Laplace operator ∆ ,. And (2) ∆Φ(x) = δ0(x), the dirac delta function, on all of Rn. We use these two claims to show the following: The equation defined by (5.1) is a solution to Poisson’s equation, since Z Z ∆u(x) = ∆ f(y)Φ(x − y)dy = ∆x(f(y)Φ(x − y))dy n n R R Z = f(y)∆xΦ(x − y)dy n ZR = f(y)δx(y)dy n R = f(x) Physically, we can imagine the delta function as a ‘point mass,’ concentrating the value of the integral of f around the value of x. Mathematically, this idea is meaningless. Why? Well... the dirac delta isn’t a function. With some knowledge of measure theory, the reader will have little trouble verifying that there exists no function whose integral over all Rn is 1, and vanishes at every point except for the origin. The integral of a function should not be altered by the value of a function at a single point, i.e. a set of measure zero. 2 So let’s say what we mean by this equality. We shall define δ0(x) below shortly. ∞ Definition 5.5. The space C0 n ∞ Let Ω ⊂ R be an open, and bounded set. The space C0 (Ω) is the set of smooth functions defined on Ω which have compact support. In other words the collection of functions f ∈ C∞(Ω) such that the set supp(f) = {x ∈ Ω: f(x) 6= 0} is a compact set. Smoothness is a convenient analytic property, and the condition of compact support ensures we can comfortably integrate. We only care about this distinction is because C∞(Ω) 6⊂ Lp(Ω). For example 1 ∞ 1 1 x ∈ C (0, 1) but x ∈/ L (0, 1). Definition 5.6. Mollifiers and the dirac delta function The dirac delta function is a limit of a sequence of mollifiers:

δ0(x) = lim ϕn(x), n→∞ Where the molliﬁers are smooth functions which satisfy

(1) Positivity: ϕn(x) ≥ 0, for all n R (2) Unit norm: n ϕn(x)dx = 1, for all n R (3) Compact Support: supp(ϕn) is a compact set for all n (4) Concentration around zero: For all δ > 0 small, Z lim ϕn(x)dx = 0, ∀n. n→∞ |x|≥δ ∞ In particular, every molliﬁer is an element of C0 (Ω). As an example, we can consider the function ( − 1 Me 1−|x|2 |x| ≤ 1 η(x) := 0 |x| > 1

1when this operation is justiﬁed is discussed in the appendix, under the title diﬀerentiating under the integral. It is contained on the section reviewing measure theory 2More about Measure can be found in the appendix. For the readers interest, the author recommends the highly accessible text of Royden-Fitzpatrick. THE HODGE THEOREM AND THE BOCHNER TECHNIQUE: A VANISHINGLY SHORT PROOF 15

which is smooth, and has support contained in the unit ball B1(0), a compact set. The constant M is so chosen such that η has unit integral over all of Rn. Now if we define 1 x (5.2) η (x) := η , n this defines a sequence of Mollifiers as we let → 0. One inspiring property of Mollifiers is that they play quite nicely with convolutions. 1 n Proposition 5.7. Let f ∈ L (R ), and let K be a compact set on which f is continuous. Let {ϕn} be a sequence of Mollifiers. Then {ϕn ∗ f} converges to f uniformly on A.

Proof. Denote fn = ϕn ∗ f, and let x ∈ K. Since, from the second property of unit norm for Molliﬁers, we have Z Z f(x) = f(x) ϕn(y)dy = f(x)ϕn(y)dy, n n R R then Z fn(x) − f(x) = [f(x − y) − f(x)]ϕn(y)dy. n R Using this, we can estimate this quantity as follows Z |fn(x) − f(x)| ≤ |f(x − y) − f(x)|ϕn(y)dy n ZR Z = |f(x − y) − f(x)|ϕn(y)dy + |f(x − y) − f(x)|ϕn(y)dy n R \Bδ (x) Bδ (x) | {z } | {z } (1) (2) Since f is uniformly continuous on K, the second integral may be made small for suﬃciently small |x| < δ, as ! 1 Z Z : (2) := |f(x − y) − f(x)|ϕn(y)dy ≤ max |f(x − y) − f(x)| ϕn(y)dy ≤ /2. x∈K Bδ (x) Bδ (x)

On the other hand, ϕn is compactly supported, hence the integral in (1) vanishes outside a compact set. On that compact set, f is bounded, say by M. Additionally, for n sufficiently large we may make R n ϕn(y)dy ≤ /2M, since ϕn(x) concentrates around zero. So R \B(0) Z Z (1) := |f(x − y) − f(x)|ϕn(y)dy ≤ M ϕn(y)dy ≤ /2, n n R \Bδ (x) R \B(0) thus |fn(x) − f(x)| ≤ as claimed. The previous result may seem impractical due to its hypotheses: why should a function f in L1(Rn) necessarily be continuous on a compact set? The following result, due to Nikolai Lusin, states informally that a measurable function is a continuous function on ‘almost all’ of its domain. Theorem 5.8. Lusin’s Theorem Let (Rn, Σ, µ) be a measure space with µ the standard Lebesgue measure, and let f : Rn → R be a Lebesgue- measurable function. For every Ω ⊂ Rn of finite measure, and for all > 0, there exists K ⊂ Ω closed such that µ(Ω \ K) < and f|K is continuous. Furthermore, if the set Ω is bounded, we can choose K to be compact. One further property of Mollifiers is stated in the following.

Proposition 5.9. Let {ϕn} be a sequence of molliﬁers. Then, for any f which is integrable, the support of ϕn ∗ f is contained in supp(f) + supp(ϕn). Proof. Since Z ϕn ∗ f(x) = f(x − y)ϕn(y)dy, n R since ϕn is a Molliﬁer, it has compact support, so this integral restricts to supp(ϕn). If ϕn ∗ f(x) 6= 0, then we must have x − y ∈ supp(f) for some y ∈ supp(ϕn). Hence we have

x ∈ supp(f) + supp(ϕn) 16 HADRIAN QUAN

as claimed. Proposition 5.10. C∞ Urysohn’s Lemma ∞ Given K ⊂⊂ U, there exists f ∈ C0 (Ω) such that supp(f) ⊂ U, 0 ≤ f ≤ 1, and f|K = 1.

Proof. Let ϕn(x) be a Mollifier whose support is contained in the ball B/3(0), and define = dist(K,U c) = inf inf {d(x, y)}. x∈U c y∈K Since K T U c 6= ∅, then > 0. Now we define the following sets V (t) = {x ∈ Ω: dist(x, K) ≤ t},

i.e. sets of points which are at least some distance from K. Deﬁne f := ϕn ∗ 1V (/3), where 1A(x) denotes the indicator function of a set A. This function will have the properties required by our proposition, ﬁrst seen by its support:

supp(f) = supp(ϕn) + supp(1V (/3)) ⊂ V (2/3) ⊂ U,

which is contained where we need it to be. Now, for any x ∈ K, we have that 1V (/3)(x) = 1, so Z Z f(x) = 1V (/3)(y)ϕn(x − y)dy = ϕn(x − y)dy = 1, n n R R so f|K = 1. Additionally, as an integral of two non-negative functions, the function f is non-negative. Thus 0 ≤ f ≤ 1 as claimed. The reason for the naming of the previous proposition is that a similar but more classical result regarding continuous (instead of smooth) functions is attributed to Pavel Urysohn. His result on the existence of a continuous function which separates a compact set and an open set is true in general for any locally compact, Hausdorff space (or T 1 space). 3 2 ∞ p Proposition 5.11. The space C0 (Ω) is dense in L (Ω) ∞ Proof. We shall first show that C0 (Ω) is dense in the space of simple functions, which we shall show are themselves dense in Lp(Ω), so this result at least seems plausible. Let A ⊆ Ω be a fixed subset of finite measure. The Lebesgue measure is regular, hence we know there exists a compact set K and an open set U such that K ⊆ A ⊆ U, µ(U \ K) ≤ . ∞ ∞ Now, we can use the previous proposition (C Urysohn’s Lemma) to select a function f ∈ C0 (Ω) with supp(f) ⊂ U, 0 ≤ f ≤ 1, and f|K = 1. Then we have Z Z p p p |1A − f| dx = |1A − f| dx ≤ µ(U \ K) · max |1A − f| ≤ · (1), Ω U\K U\K 1 1 p ∞ this demonstrates || A − f||Lp ≤ , i.e. that A can be approximated in L by elements of C0 . By linearity, we can extend this result to approximate simple functions in the Lp−norm. But as we know, simple functions are dense in the space Lp(Ω) (The proof of simple function approximation is quite powerful, and very constructive. However, the proof itself is rather technical and not immediately enlightening, so we ∞ p have left it to the appendix). So C0 (Ω) is dense in L (Ω) as claimed. But back to solving Poisson’s equation. Using Green’s second formula: Z Z ∂Φ ∂f [f∆Φ − Φ∆f]dx = f − Φ dS, Ω ∂Ω ∂ν ∂ν + − and the fact that ∂(Ω \ B(0)) = ∂Ω ∪ B(0) we can compute the following. Additionally, away from x = 0, we have ∆Φ = 0. So, Z Z ∂Φ ∂f [f∆Φ − Φ∆f]dx = f − Φ dS =⇒ Ω\B(0) ∂(Ω\B(0)) ∂ν ∂ν

Z Z Z ∂Φ ∂f Z ∂Φ ∂f f∆Φdx = Φ∆fdx − f − Φ dS + f − Φ dS Ω\B(0) Ω\B(0) ∂Ω ∂ν ∂ν ∂B(0) ∂ν ∂ν THE HODGE THEOREM AND THE BOCHNER TECHNIQUE: A VANISHINGLY SHORT PROOF 17

6. Distributions and The Schwartz Space of Test Functions

Let Ω ⊂ Rn be an open, and bounded set. Functions with these properties are quite tractable, having as many derivatives as needed, and vanishing outside of a compact set. Additionally, we saw in previous sections that they interact favorably with Molliﬁers. Ideally, we can look for solutions to our PDE in the ∞ space C0 (Ω). The question remains how we can identify elements in this space, and relate them to our ∞ PDE. In essence, we seek an appropriate topology on C0 (Ω). If we denote for each compact K ⊂ Ω the following space, ∞ DK (Ω) := {f ∈ C0 (Ω) : supp(f) ⊆ K}, ∞ ∞ then we can place a vector space topology on C0 (Ω) such that each Dk is a closed subspace. Let {Ki}i=1 be S a sequence of compact sets such that each Ki is contained in the interior of Ki+1, and Ω = Ki. Proceeding in this way, we deﬁne [ D(Ω) := DK (Ω), K⊂Ω compact

to be the space of Test Functions. Then we can deﬁne a separating family of semi-norms P = {pN } by α pN (f) = max {|D f(x)| : |α| ≤ N}. x∈KN This induces a vector topology on D(Ω) by deﬁning a prebasis of sets of the form: 1 U = f ∈ C∞(Ω) : p (f) ≤ . N N N

Finite intersections of the UN sets create a basis for the topology we place on D(Ω). Being induced by a separating family of semi-norms this gives D(Ω) the structure of a locally convex topological vector space. ∞ The mapping φx : f 7→ f(x) is a continuous linear functional on C (Ω). Observe that we have \ DK (Ω) = Ker(φx) x∈Kc

so each DK (Ω) is an intersection of closed sets, hence closed in this topology. This set equality can be seen c by noting that the functions in the kernel of the evaluation maps φx are precisely those which vanish on K , i.e. those functions have supp(f) ⊆ K. ∞ Something to note is the spaces D(Ω) and C0 (Ω) are equivalent as sets. The notation D(Ω) is mostly to ∞ emphasize the topology we have placed on C0 (Ω). In later sections we may use either notation.

A first encounter with topologies defined in terms of bases can be non-intuitive. Lets recast our familiar notion of convergence on R in terms of its bases. To show a sequence of reals {xk} is Cauchy in the topology on R, we show ∀ > 0, that |xj − xk| ≤ for i, j sufficiently large. In other words, that for a fixed B(0) (a fixed basis element for this topology), the difference of any two elements sufficiently far along in the sequence is an element of B(0). Hence, to understand convergence in this manner we need only understand the structure of the bases. In the next definition, we summarize a convenient convergence criterion for this space, and prove it is equivalent to the topology we have defined on D(Ω). ∞ Definition 6.1. Convergence in the Space of Test Functions, D(Ω) = C0 (Ω) ∞ ∞ ∞ A sequence {φn}n=1 ⊂ C0 (Ω) is said to converge in the sense of D(Ω) to φ ∈ C0 (Ω) if the following are satisfied: (1) ∃K ⊂⊂ Ω such that supp(φn − φ) ⊂ K α α (2) lim D φn = D φ uniformly on K for each α n→∞ where K ⊂⊂ Ω is defined to mean K ⊆ Ω and K ⊆ Ω is compact. Condition (2) implies the map Dα : D(Ω) → D(Ω) is continuous. Further, the space D(Ω) is complete in this topology. Proof. The topology on D(Ω) is defined by the prebasis of sets of the form: 1 U = f ∈ C∞(Ω) : p (f) ≤ . N N N

Finite intersections of the UN sets create a basis for the topology we place on D(Ω). So two elements f, g are in the same open set only if they are both contained in some basis set. Since the basis sets are ﬁnite 18 HADRIAN QUAN

intersections of the UN , and finite intersections of compact sets remain compact, there exists some KN 0 compact such that supp(f − g) ⊂ K0. 0 This proves condition (1). Additionally, if f − g is in a finite intersection of UN sets, select the largest N amongst this finite collection of sets to which f − g belongs. The definition of UN 0 implies

α 0 1 0 0 f − g ∈ UN ⇐⇒ pN (f − g) = max {|D (f(x) − g(x))| : |α| ≤ N } ≤ 0 . x∈KN0 N α α 0 0 In other words, |D f(x) − D g(x)| ≤ 1/N on KN 0 , if |α| ≤ N . ∞ The structure of the open sets on C0 (Ω) implies the following: If {fi} is Cauchy in this topology, then for a fixed N, we have fi − fj ∈ UN for i, j sufficiently large. In other words 1 |Dαf (x) − Dαf (x)| ≤ ∀|α| ≤ N, i j N α α for all x ∈ KN . So D fi converges (uniformly on compact subsets of Ω) to some gα ∈ D(Ω). So D is a α continuous map. In particular, fi → g0 ∈ D(Ω). Further g0 ∈ D(Ω), and D g0 = gα. Thus fi → g0 in the topology on D(Ω). So D(Ω) is complete, as claimed. Now, let Now that we have formed this topology on D(Ω), we can consider what topology has been formed over (D(Ω))∗: the space of continuous linear functionals on D(Ω). The space (D(Ω))∗ is called the (Schwartz) Space of Distributions. Definition 6.2. Convergence in the Schwartz Space of Distributions The topology which we have defined on (D(Ω))∗ has the following notion of sequential convergence: a ∗ ∗ sequence {φn} ⊂ (D(Ω)) converges to φ ∈ (D(Ω)) if and only if

φn(f) → φ(f) ∀f ∈ D(Ω). 3

Since D(Ω) is locally convex, we ‘know’ mathematically that its dual space (D(Ω))∗ is non-trivial. How- ever, we may take reassurance in the fact that we always have a large class of elements of (D(Ω))∗ which we can construct. Let u be locally integrable on Ω, i.e. u ∈ L1(A) for all measurable sets A ⊂⊂ Ω. This collection of 1 1 ∗ functions is denoted by u ∈ Lloc(Ω). For every u ∈ Lloc(Ω) there corresponds a distribution Tu ∈ (D(Ω)) deﬁned by Z Tu(f) = u(x)f(x)dx f ∈ D(Ω) Ω ∗ We shall prove that Tu ∈ (D(Ω)) shortly. 1 ∗ Proposition 6.3. For u ∈ Lloc(Ω), the functional Tu is linear, and continuous in the topology on (D(Ω)) . ∗ p 1 Hence Tu ∈ (D(Ω)) . Further, L (Ω) ⊂ Lloc(Ω) for all 1 ≤ p < ∞. Proof. Linearity of this functional is inherited from linearity of the integral. To prove it is continuous in the ∗ topology on (D(Ω)) , we must show that for all convergent sequences fn → f ∈ D(Ω) we have

Tu(fn) → Tu(f). ∞ This agrees with our familiar notion of continuity, as it says Tu takes convergent sequences {fn}n=1 in D(Ω) ∞ to convergent sequences {Tu(fn)}n=1 in R. To prove this note Z |Tu(fn) − Tu(f)| ≤ sup |fn(x) − f(x)| |u|dx → 0 x∈K K since fn converges to f in D(Ω). p 1 To show L (Ω) ⊂ Lloc(Ω), we note ﬁrst that this inclusion is immediate for p = 1. This is because for A ⊂ Ω measurable, we have Z Z |f|dx ≤ |f|dx < ∞ A Ω

3The topology on distributions is given by the weak-star topology as the dual of D(Ω), and hence is also a locally-convex topological vector space. THE HODGE THEOREM AND THE BOCHNER TECHNIQUE: A VANISHINGLY SHORT PROOF 19

1 1 1 since f ∈ L (Ω). A was arbitrary, so f ∈ Lloc(Ω). Additionally, we have for K , where Z 1/p Z 1/p p 1/p ||1K ||Lp = |1K | dx = dx = µ(K) < ∞. Ω K p 1 1 Hence, for any f ∈ L (Ω), if we have p + q = 1, then Hölder’sinequality implies Z Z Z 1/p Z 1/q p q 1/q |f|dx = |f1K |dx ≤ |f| dx |1K | dx = ||f||Lp µ(K) < ∞ K Ω Ω Ω 1 hence f ∈ Lloc(Ω), as claimed. (The statement of Hölder’s inequality may be found in the appendix.) With everything we have discussed, we can now formulate a notion of a distributional derivative. We begin by recalling the multivariate integration by parts formula. Say f, g ∈ C1(Ω), and let ν = (ν1, . . . , νn) denote the unit outer normal vector. Then we have the classical divergence theorem: Z Z Z g(x)∇ · f(x) + f(x)∇ · g(x)dx = ∇ · (f(x)g(x))dx = f(x)g(x)νdS. Ω Ω ∂Ω By considering the above equation component by component, we can rephrase this equivalently as Z ∂f Z Z ∂g g dx = fgνidS − f dx, Ω ∂xi ∂Ω Ω ∂xi which gives us a way of integrating by parts. Now, if u ∈ C1(Ω) and f ∈ D(Ω), since f is zero outside some compact set contained in Ω (thus identically zero on ∂Ω) we have Z ∂ Z ∂ u fdx = − u f dx. Ω ∂xi Ω ∂xi Similarly, integrating by parts |α| times gives Z Z (Dαu)fdx = (−1)|α| u(Dαf)dx, Ω Ω provided that u ∈ C|α|(Ω). This is highly suggestive, as it allows us to consider ‘generalized’ derivatives of u by converting them into actual derivatives of f. In other words, if some function were to exist which was integrable, and replaced Dαu on the left hand side of the above equation, it could play a similar role to Dαu; even if u were not actually differentiable! This all motivates the following definition. Definition 6.4. Weak Derivatives 1 Say u ∈ Lloc(Ω). We say that v is a weak (or distributional) α-th derivative of u if we have Z Z v(x)f(x)dx = (−1)|α| u(x)Dαf(x)dx ∀f ∈ D(Ω), Ω Ω α 1 ∗ which we denote by D u = v (weakly). Bearing in mind that u ∈ Lloc(Ω) defines a distribution Tu ∈ (D(Ω)) as shown above, we denote the derivative of this distribution by α |α| α D Tu(f) = (−1) Tu(D f) f ∈ D(Ω).

Since f ∈ D(Ω), it is infinitely differentiable. So Tu has infinitely many weak (or distributional) derivatives as defined in this manner. For a distribution T ∈ (D(Ω))∗ which is not defined by integration against a locally integrable function, we choose to keep this definition DαT (f) = (−1)|α|T (Dαf). Now we can finally define the much-sought after Sobolev Spaces.

7. Density, Completeness, and Diﬀerence Quotients of W k,p(Ω): The Sobolev Spaces

In this section, and throughout, we shall continue to assume that Ω ⊂ Rn is open and bounded. Deﬁnition 7.1. The Sobolev Spaces W k,p(Ω) The k, p−Sobolev space, denoted k,p p α p W0 (Ω) := {f ∈ L (Ω) : D f ∈ L (Ω), ∀|α| ≤ k}, is the space of all functions f which are in Lp(Ω), and have weak derivatives up to order k which are also in Lp(Ω). The integer k is sometimes referred to as the order of the Sobolev space. 20 HADRIAN QUAN

For all 1 ≤ p < ∞, this space may be given the norm  1/p  1/p Z X α p X α p ||f||W k,p :=  ||D f||Lp  =  |D f| dx . |α|≤k |α|≤k Ω We claim that for this norm makes W k,p a Banach space, and is in fact a Hilbert space for p = 2. We shall prove this shortly. Our proof that W k,p is a complete space relies heavily on the completeness of Lp(Ω). For a proof that Lp(Ω) is complete, the reader may consult the appendix. Theorem 7.2. W k,p(Ω) is Complete The space W k,p(Ω) for 1 ≤ p ≤ ∞ is complete in its norm (or its inner product for p = 2). Hence W k,p(Ω) is a Banach space, and is a Hilbert space for p = 2. k,p α Proof. Let {ui} be a Cauchy sequence in W (Ω). Then, by definition, the sequence {D ui} is Cauchy in Lp(Ω) for all 0 ≤ |α| ≤ k. From the inclusion W k,p(Ω) ⊂ Lp(Ω), and the fact that Lp(Ω) is complete, there p exist functions u, uα ∈ L (Ω) for all 1 ≤ |α| ≤ k such that α ||u − ui||Lp → 0, ||uα − D ui||Lp → 0 1 ≤ |α| ≤ k. p p What remains to be shown is that the elements uα ∈ L (Ω) correspond to the weak derivatives of u ∈ L (Ω). k,p p 1 If so, then u ∈ W (Ω), proving completeness. To prove this, we note that L (Ω) ⊂ Lloc(Ω), hence ui, u ∗ α also define a distributions Tui ,Tu ∈ (D(Ω)) as in the above proposition. We claim first that TD ui → Tuα ∗ α p in (D(Ω)) . This is because D ui → uα in L , so Z α α α α 1 0 p |TD ui (f) − Tuα (f)| ≤ |D ui − uα||f|dx = ||(D ui − uα)f||L ≤ ||f||Lp ||D ui − uα||L → 0 Ω where 1/p + 1/p0 = 1. The second inequality follows from HöldersInequality. Having shown this, we see that |α| α |α| α Tu (f) = lim TDαu (f) = (−1) lim Tu (D f) = (−1) Tu(D f) ∀|α| ≤ k α i→∞ i i→∞ i α p p k,p i.e. (weakly) D u = uα ∈ L (Ω). So u and its first k weak derivatives are in L (Ω). Hence u ∈ W (Ω). Further, since lim ||ui − u||W k,p = 0, i→∞ k,p we have proven that W (Ω) is complete. Something to note is that if we choose p = 2, then W k,p(Ω) is a Hilbert space with the inner product  1/2 Z X α α hg, fiW k,2 =  D g · D fdx . |α|≤k Ω Definition 7.3. Related Function Spaces k,p ∞ A closely related space is that of W0 (Ω) which is defined to be the closure of C0 (Ω) under the Sobolev- norm || · ||W k,p . A different construction of the Sobolev spaces occurs by considering the W k,p−norm on the space ∞ S := {u ∈ C (Ω) : ||u||W k,p < ∞}. Then we can define W k,p(Ω) to be the closure of space S, i.e. S constructing W k,p(Ω) in the usual way as the equivalence classes of Cauchy sequences from elements in S. Now that we know W k,p(Ω) is complete, lets examine it further. We built this function space as an ∞ extension of C0 (Ω), in hopes of solving PDE. Luckily, they relate in quite nice ways. ∞ k,p Theorem 7.4. C0 (Ω) is dense in W0 (Ω) ∞ p Proof. We have already shown C0 (Ω) is dense in L (Ω), so this result seems plausible. We actually prove k,p this by mollifying an element of W0 (Ω); this makes it smooth, and then letting that sequence converge to k,p the original element. If we can show this convergence occurs in the topology defined by the W0 (Ω). k,p Let u ∈ W0 (Ω), and let {φn} be a sequence of Mollifiers. Define Z un(x) := u ∗ φn(x) = u(y)φn(x − y)dy. Ω THE HODGE THEOREM AND THE BOCHNER TECHNIQUE: A VANISHINGLY SHORT PROOF 21

Since φn is both smooth, and compactly supported, we satisfy the conditions to differentiate under the integral, so Z Z α α |α| α D un(x) = u(y)Dx φn(x − y)dy = (−1) u(y)Dy φn(x − y)dy Ω Ω α where Dy denotes differentiation with respect to the y variable. Since for each fixed x ∈ Ω the function ∞ φn(x − y) ∈ C0 (Ω). So from the definition of weak derivative we have Z Z α |α| α u(y)Dy φn(x − y)dy = (−1) Dy φn(x − y)dy, Ω Ω k,p provided |α| ≤ k, as u ∈ W0 (Ω). Hence Z α α |α|+|α| α α D (φn ∗ u)(x) = D un(x) = (−1) Dy u(y)φn(x − y)dy = (φn ∗ D u)(x), Ω for all |α| ≤ k. Next, we note Z α α α α D un(x) − D u(x) = (D u(x − y) − D u(y))φn(y)dy, Ω so we have

α α α α ||D un − D u||Lp = ||(D u(x − y) − D u(x))φn(y)||Lp α α ≤ ||D u(x − y) − D u(x)||Lp ||φn||L1

∞ 1 using Young’s inequality for convolutions. Additionally, φn ∈ C0 (Ω) ⊆ L (Ω) (i.e. that ||φn||L1 ≤ C) so we have shown that for |α| ≤ k that

α α α α ||D un − D u||Lp ≤ C||D u(x − y) − D u(x)||Lp → 0

as |y| → 0. But since φn is a sequence of Molliﬁers, it concentrates around the origin, so as n → ∞, we have |y| → 0. Thus α α ||D un − D u||Lp → 0 as n → ∞, for all |α| ≤ k. If we have shown holds for |α| ≤ k, then the deﬁnition of norm on W k,p(Ω) implies

||un − u||W k,p → 0 as n → ∞,

∞ k,p as claimed. So C0 (Ω) is dense in W0 (Ω). Something to note is we cannot extend this density result immediately to functions which are not compactly supported, such as W k,p(Ω). Luckily, by mollification, we can extend such a local approximation to a nearly global approximation of smooth functions. Our smooth approximations will hold in a subset of Ω which is bounded away from the boundary ∂Ω, such that these approximations hold in a region which limits to all Ω. Definition 7.5. Difference Quotients We denote the h−translate of u ∈ W k,p(Ω) to be

τh(u)(x) = u(x + h)

for some fixed, non-zero h ∈ Rn. And further we define the difference quotient of u to be τ (u) − u uh = h ∈ W k,p(Ω) |h|

k,p Something to note is that ||τh(u)||W k,p = ||u||W k,p , so the translation map is an isometry on W (Ω). α h 1 α α k,p h k,p Additionally, since D u = |h| [D u(x + h) − D u(x)], if u ∈ W then u ∈ W for all h. We can go even further by noting that: Lemma 7.6. If u ∈ W k+1,p then the diﬀerence quotients uh are uniformly bounded in the W k,p-norm. In fact, for all K ⊆ Ω compact, for all 0 < |h| < dist(K, Ω) we have that

h ||u ||W k,p ≤ ||u||W k+1,p . 22 HADRIAN QUAN

Proof. The reason we are considering 0 < |h| < dist(K, Ω) is so we can consider functions which are ∞ k,p smooth on each compact K. Since C0 (Ω) is dense in W0 (Ω), it will suffice to prove this bound for every ∞ element of C0 (Ω), and exploit that this bound will pass through to the limit (since limits preserve non-strict inequalities). If our proof holds on each compact set, we can build a global solution in that way. ∞ Now, if f ∈ C0 (Ω) its difference quotient can be represented as Z h h 1 ∂f f (x) = (x + tei)dt h 0 ∂xi since it is continuousl differentiable. So 1/p Z h Z h p ! h 1 ∂f 1 ∂f |f (x)| = (x + tei) dt ≤ (x + tei) dx , h 0 ∂xi h 0 ∂xi The second inequality holds for a general integrable function. Hence  1/pp Z Z Z h p ! h p h p 1 ∂f ||f ||Lp = |f | dx ≤  (x + tei) dt  dx Ω Ω h 0 ∂xi Z Z h p Z h Z p 1 ∂f 1 ∂f = (x + tei) dtdx = (x + tei) dxdt h Ω 0 ∂xi h 0 Ω ∂xi Z h 1 ∂f ∂f = τtei dt = h 0 ∂xi Lp ∂xi Lp where we can interchange the order of integration due to the Fubini-Tonelli theorem, and use that the p translation map τtei is an isometry on L . ∞ h p p We have shown that any f ∈ C0 (Ω) satisfies ||f ||L ≤ ||∂xi f||L . By density, for an arbitrary u ∈ k,p ∞ W0 (Ω) there exists a sequence {fn} in C0 (Ω) converging to u, so h p p ||fn ||L ≤ ||∂xi fn||L   y y h p p ||u ||L ≤ ||∂xi u||L as these non-strict inequalities are preserved in the limit, (note that ∂xi u is representing the weak derivative k,p of u). If we have shown this inequality holds for every u ∈ W0 (Ω), in particular we have proven that α h α p p ||D u ||L ≤ ||D (∂xi u)||L ∀|α| ≤ k, ∀h, ∀xi α h β =⇒ ||D u ||Lp ≤ ||D u||Lp ∀|α| ≤ k, ∀h, such that |β| = |α| + 1 h n hence ||u ||W k,p ≤ ||u||W k+1,p , for all non-zero h ∈ R as claimed. In essence, we have traded a power of k to gain control on the size of the derivative. The converse is also true: if the difference quotients of a function are uniformly bounded, then the function belongs to a Sobolev space of at least one higher order. Lemma 7.7. Let u ∈ W k,p(Ω), and K ⊆ Ω compact. If ∃C > 0 such that for all K ⊆ Ω compact, we have h ||u ||W k,p ≤ C for all 0 < |h| < dist(K, ∂Ω). Then u ∈ W k+1,p(Ω). This gives a converse to our first lemma. However, we can prove an even stronger converse, which will be one of the main results of our paper: The Sobolev Embedding Theorem. To prove this, we shall first show a fundamental inequality originally due to Sobolev.

8. Inequalities on Sobolev Norms: Nirenberg and Morrey In lieu of Sobolev’s original proof, we shall follow a proof due to Louis Nirenberg 4, which is striking in that it relies on little more than H¨older’s Inequality, induction, and the fundamental theorem of calculus. This proof is remarkable in how elementary its methods are. However, this does not mean the proof is easy.

4At the time of this writing Professor Nirenberg had recently been awarded the Abel prize for his lifetime acheivement in mathematics. THE HODGE THEOREM AND THE BOCHNER TECHNIQUE: A VANISHINGLY SHORT PROOF 23

Theorem 8.1. Gagliardo-Nirenberg-Sobolev Inequality 1,p For u ∈ W0 (Ω) for all 1 ≤ p < n, there exists a constant C (depending only on p and n) such that

||u|| np ≤ C||∇u||Lp , L n−p where the constant C depends only on p and n, the dimension of the domain. In particular this says that 1,p q np [||u|| np ≤ ||u||W 1,p , i.e. that the Sobolev space W (Ω) ,→ L (Ω) embeds continuously, for all q < . L n−p 0 n−p Proof. We can prove this result ﬁrst for p = 1. In its proof, we note crucially that we use a slight gen- 1 1 eralization of H¨older’sInequality (the original inequality is in the appendix). If p + q = 1 implies that ||fg||L1 ≤ ||f||Lp ||g||Lq , then the following can be proven through induction: Let

1 1 + ... + = 1, p1 pm then

1 1 Z Z p1 Z pm g1 ··· fmdx ≤ g1dx ··· gmdx . A A A

∞ Now to our proof. By density, if we can prove this for bound for arbitrary f ∈ C0 (Ω), then we will have 1,1 ∞ proven it for W0 (Ω). Note that if f ∈ C0 (Ω) then we have

Z xi Z xi i i f(x) = Dxi f(x1, . . . , xi−1, yi, xi+1, . . . , xn)dy := Dxi fdy . −∞ −∞

[Here we have deﬁned (for simplicity) integration w.r.t dyi to denote integrating the function g in its ith component variable i.e. the integral of the function g(x1, . . . , xi−1, yi, xi+1, . . . , xn) w.r.t yi.] So

Z xi Z ∞ i i |f| ≤ |Dxi f| dy ≤ |Dxi f|dy −∞ −∞

n 1 Z ∞ n−1 n Y n−1 i =⇒ |f| ≤ |Dxi f|dy . i=1 −∞

Something important to note is that the integral of any quantity w.r.t to yi is independent of the integral w.r.t xi. Our goal is to successively integrate this inequality with respect to x1, x2, . . . , xn and use H¨olders inequality at each step. So

n 1 Z ∞ Z ∞ Z ∞ n−1 n Y n−1 1 i 1 |f| dx ≤ |Dxi f|dy dx −∞ −∞ i=1 −∞   1 n 1 Z ∞ Z ∞ n−1 Y Z ∞ n−1  =  |D f|dy1 |D f|dyi  dx1  x1 xi  −∞  −∞ i=2 −∞  | {z } | {z } independent of x1 independent of xi 1 n 1 Z ∞ n−1 Z ∞ Z ∞ n−1 1 Y i 1 = |Dx1 f|dy |Dxi f|dy dx −∞ −∞ i=2 −∞ | {z } Apply generalized H¨olderw.r.t x1 1 n 1 Z ∞ n−1 Z ∞ Z ∞ n−1 1 Y i 1 ≤ |Dx1 f|dy |Dxi f|dy dx −∞ i=2 −∞ −∞ in the second underbrace, we have used the Generalized H¨olderinequality, by excluding the variable x1. This 1 R Qn 1 Qn R n−1 1 n−1 is done by noting i=2 |gi|dx ≤ i=2 |gi| dx . Integrating this new inequality with respect to 24 HADRIAN QUAN

x2 yields " 1 n 1 # Z ∞ Z ∞ Z ∞ Z ∞ n−1 Z ∞ Z ∞ n−1 n Y n−1 1 2 1 i 1 2 |f| dx dx ≤ |Dx1 f|dy |Dxi f|dy dx dx −∞ −∞ −∞ −∞ i=2 −∞ −∞   1 1 n 1 Z ∞ Z ∞ n−1 Z ∞ Z ∞ n−1 Y Z ∞ Z ∞ n−1  =  |D f|dy1 |D f|dy2dx1 |D f|dyidx1  dx2  x1 x2 xi  −∞  −∞ −∞ −∞ i=3 −∞ −∞  | {z } independent of xi  

1 1 n 1 Z ∞ Z ∞ n−1 Z ∞ Z ∞ n−1 Z ∞ Z ∞ n−1  2 1  1 Y i 1  2 = |Dx2 f|dy dx  |Dx1 f|dy |Dxi f|dy dx  dx   −∞ −∞ −∞  −∞ i=3 −∞ −∞  | {z } generalized H¨older w.r.t. x2 1 1 n 1 Z ∞ Z ∞ n−1 Z ∞ Z ∞ n−1 Z ∞ Z ∞ Z ∞ n−1 2 1 1 2 Y i 1 2 ≤ |Dx2 f|dy dx |Dx1 f|dy dx |Dxi f|dy dx dx . −∞ −∞ −∞ −∞ i=3 −∞ −∞ −∞ n At the jth step of this inductive inequality development, we have integrated |f| n−1 w.r.t to the ﬁrst j variables to obtain Z ∞ Z ∞ n 1 j ··· |f| n−1 dx ··· dx −∞ −∞ j 1 n 1 Z ∞ Z ∞ n−1 Z ∞ Z ∞ n−1 Y k 1 j−1 j+1 j Y i 1 j ≤ ··· |Dxk f|dy dx ··· dx dx ··· dx ··· |Dxi f|dy dx ··· dx . k=1 −∞ −∞ i=j −∞ −∞ Hence this scheme allows us to obtain n 1 n 1 Z Z n−1 Z n−1 n Y Y n−1 k 1 j−1 j+1 n |f| dx ≤ |Dxk f|dy dx ··· dx dx ··· dx = |Dxk f|dx n n R k=1 R k=1 Ω

n 1 n !1/n n Z n−1 Z n−1 Y Y =⇒ ||f|| n ≤ |Dx f|dx =⇒ ||f|| n ≤ |Dx f|dx L n−1 k L n−1 k k=1 Ω k=1 Ω ∞ Since f ∈ C0 (Ω), we can restrict the domain of integration to Ω. Now we can apply the Arithmetic- Geometric Mean Inequality: a + ··· + a (a ··· a )1/m ≤ 1 m a ∈ +, 1 m m i R to obtain n !1/n n Y Z 1 X Z 1 Z ||f|| n ≤ |D f|dx ≤ √ |D f|dx = √ |∇f|dx ≤ C||∇f|| 1 , L n−1 xk n xk n L k=1 Ω k=1 Ω Ω ∞ 1,1 so ||f|| n ≤ C||∇f|| 1 for all f ∈ C (Ω) as claimed! By density, this inequality extends to all of W (Ω) L n−1 L 0 We’re almost done. We’ve shown the Gagliardo-Nirenberg-Sobolev Inequality in the special case of p = 1. In particular, this implies that W 1,1(Ω) ,→ L1(Ω) is a continuous embedding. To prove our inequality for the general case 1 ≤ p < n, ﬁrst note that for f α our inequality implies

n−1 Z n α αn 1 α ||f || n = |f| n−1 dx ≤ √ ||∇(f )|| 1 L n−1 L Ω n Z Z 1 α α α−1 α α−1 = √ |∇(f )|dx = √ |f| |∇f|dx ≤ √ ||f || p ||∇f|| p , p−1 L n Ω n Ω n L from H¨oldersinequality, since p − 1 1 + = 1. p p So α α α−1 n p p ||f|| α ≤ √ ||f|| (α−1) ||∇f||L L n−1 n L p−1 THE HODGE THEOREM AND THE BOCHNER TECHNIQUE: A VANISHINGLY SHORT PROOF 25

i.e. a clever choice of constant α will complete our proof. Since we are clever, we choose α such that αn p p(n − 1) = (α − 1) ⇐⇒ α = , n − 1 p − 1 n − p This implies p(n−1) np n αn p = n−p = = (α − 1) . n − p n − 1 n − 1 p − 1 Hence α =⇒ ||f|| np ≤ √ ||∇f||Lp , L n−p n as claimed. So we have completed the proof of our inequality. k,p Corollary 8.2. If kp < n then there exists a constant C > 0 such that for all u ∈ W0 (Ω) we have,

||u|| np ≤ C||u||W k,p . L n−kp np k,p n−kp In particular, we have that the Sobolev space W0 (Ω) ,→ L (Ω) Proof. First, we observe that if u ∈ W k,p(Ω) then Dαu ∈ W k−1,p(Ω) for all |α| ≤ k − 1. By the Gagliardo- np Nirenberg-Sobolev Inequality, this implies Dαu ∈ L n−p (Ω), again for all |α| ≤ k − 1. np k−1, np Now, if Dαu ∈ L n−p (Ω), for all |α| ≤ k − 1, then Dαu ∈ W n−p (Ω), for all |α| ≤ k − 1. We have lost one order of weak differentiability, and shown that each derivative is bounded in a slightly higher Lq norm. By iterating the above argument k−times, and using that 1 1 k = − , np p n n−kp we obtain our continuous inclusion as stated. Since we went to the trouble of proving this difficult result, let’s reward ourselves. What results does this inequality imply? To some extent we have rigorously proven our previous intuition: weak derivatives are a form of ‘currency’ in which we can exchange weak derivatives of a function for control on the growth of the higher Lp norms of our function. Like good analysts we can, and we will, push this analogy to its breaking point. Having control of the Lp norms gives some pleasant properties of the functions in consideration. However, this is not quite as nice knowing that our functions are continuous (or differentiable?!). In what follows, by considering when kp > n, we can prove a class of norm-estimates due to Morrey. Our approach will follow a proof of Sobolev’s: Theorem 8.3. Morrey-Sobolev Inequality ∞ n For u ∈ C0 (R ), if p > n ≥ 2, there exists C > 0, depending on p and n such that 1 − 1 n p p n sup |u| ≤ µ[supp(u)] ||Dxi u||L (R ) Which we can use to prove: Theorem 8.4. Sobolev’s Embedding Theorem n k,p r k,p If k − p > r then W (Ω) ⊆ C (Ω). Hence, a function belonging W (Ω) for sufficiently large k in fact has r continuous partial derivatives. Proof. We have excluded the previous two proof for brevity, but recommend the comprehensive text of Adams, Sobolev Spaces. Now that we have been able to rigorously prove several results about the structure of W k,p(Ω), and k,p W0 (Ω) we may feel we are ready to use these enlarged function spaces to tackle our PDE, however there is still one major issue we have yet to address. In our attempt to solve the boundary value problem ( ∆u = f x ∈ Ω u = 0 x ∈ ∂Ω our scheme is to use the Sobolev spaces W k,p(Ω) to seek solutions u which reside in the space Lp(Ω). As we have discussed previously, and in the appendix, the Lebesgue integral ‘ignores’ sets of measure zero. Usually this is an advantage, as measure theory allows us to work with pathological functions, provided such strangeness occurs on a ‘small’ set. However, since two functions which differ on a measure zero set have 26 HADRIAN QUAN

the same integral (and thus the same Lp norm), functions which are in Lp(Ω) are equivalent up to a set of measure zero. Here’s our problem: the boundary of an open set is a set of measure zero. But if ∂Ω is a set of measure zero, the phrase

u|∂Ω = 0 isn’t meaningful, as we could modify u in any manner on the boundary ∂Ω and u would remain the same function when considered as an element of Lp(Ω). This problem can also be resolved, but again we must exclude its development. The theory developed to deal with this boundary problem involves Sobolev Traces, linear operators which extend our solution in a unique way to the boundary. The author again recommends the read consult Sobolev Spaces, by Adams. Before we leave the land of Sobolev spaces, and begin examining our PDE once more, we require one further result. This will be crucial in proving the second major analytic result we require. Theorem 8.5. Rellich-Kondrachkov Lemma 1,p q np The embedding W (Ω) ,→ L (Ω) is compact for all q < n−p , i.e. the image of a bounded set has compact ∞ 1,p closure. In particular, for a sequence {ui}i=1 ⊆ W (Ω) such that ||ui||k ≤ 1, then there exists a subsequence ∞ q {uin }n=1 which converges in L (Ω). Proof. Structurally, this theorem acts as a ‘selection theorem.’ Given mild assumptions on the a sequence of functions, it serves as a criterion for ‘selecting’ a convergent subsequence. As such, it strongly resembles another classical selection criterion: the Arzela-Ascoli theorem. The Arzela-Ascoli gives criterion on when a sequence of continuous functions has a uniformly convergent subsequence, (this result is of primary importance in the theory of PDEs, and is proven in the appendix). Our goal is to try and reduce this result to the point where we may cite Arzela-Ascoli to conclude our np proof. Luckily, we can prove this strictly for the case q = 1, then since the embedding W 1,p(Ω) ,→ L n−p (Ω) is continuous (from the Gagliardo-Nirenberg-Sobolev Inequality) we can continue using the trick of Hölder inequality interpolation. 1,p 1,p Let {uk} ⊆ W (Ω) be a sequence which is bounded in the W (Ω) norm, i.e. ||uk||W 1,p ≤ α. From ∞ ∞ density of C0 (Ω), there exists a sequence {vk} ⊆ C0 (Ω) such that 1 ||u − v || 1,p ≤ . k k W k 0 Then we have ||vk||W 1,p ≤ 1 + α for all k. If we can show that the vks have a subsequence which converges in 1 1 L (Ω), then {uk} has a subsequence which converges in L (Ω). This follows from the Gagliardo-Nirenberg- Sobolev inequality: C ||u − v || 1 ≤ C||u − v || 1,p ≤ . k k L k k W k ∞ So we can now restrict our attention to the {vk} ⊆ C0 (Ω) sequence. To this end, we shall employ mollifica- ∞ ∞ tion, as has been useful in other analytic proofs involving elements of C0 (Ω). Denote by vk, ∈ C0 (A) the function 1 Z x − y Z vk,(x) := η ∗ vk(x) = n η vk(y)dy = η(z)vk(x − z)dz Ω Ω n where η is the function defined by (5.2), and A = {x ∈ R : dist(x, Ω) < 1}. Since vk is smooth, we may reexpress the difference of it as Z 1 d Z 1 vk,(x − z) − vk(x) = vk(x − tz)dt = − Dvk(x − tz) · zdt 0 dt 0 Z Z Z 1

=⇒ |vk,(x) − vk(x)| = η(z)|vk,(x − z) − vk(x)|dz = η(z) Dvk(x − tz) · zdt dz. Ω B1(0) 0

since η(z) is supported in B1(0). Hence Z Z Z 1 Z ||vk, − vk||L1 = |vk,(x) − vk(x)|dx = η(z) |Dvk(x − tz) · z| dxdzdt Ω B1(0) 0 Ω Z ≤ |Dvk|dx = ||Dvk||L1 ≤ ||Dvk||Lp = C Ω THE HODGE THEOREM AND THE BOCHNER TECHNIQUE: A VANISHINGLY SHORT PROOF 27

∞ Something to note is this function is supported in A. So for a fixed > 0, our sequence {vk,}k=1 satisfies the following bounds: 1 Z x − y M |vk,| ≤ n |vk(y)|η dy ≤ n sup η(x) Ω x∈B1(0) since vk is smooth (thus bounded on a compact set) and η is supported on B1(0). For our bounds on the gradient, we note Z 1 x − y M |Dvk,| ≤ n+1 |vk| Dη dy ≤ n+1 sup |Dη(x)|. Ω x∈B1(0) For both inequalities, for each > 0, our bounds do not depend on k. This implies that the sequence ∞ 1,p {vk,}k=1 is uniformly bounded and equicontinuous in the W (Ω) norm. By the Arzela-Ascoli theorem, this implies there exists a subsequence {vkl,} ⊆ {vk,} which converges uniformly on Ω to an element in W 1,p(Ω). np By the Gagliardo-Nirenberg-Sobolev inequality, if q < n−p then γ γ np ||u||Lq ≤ C||Du|| =⇒ ||u||Lq ≥ C||Du|| np L n−p L n−p for any γ < 1. Now, using Hölder’sinequality interpolation, we have θ 1−θ ||vk, − vk||Lq ≤ ||vk, − vk||L1 ||vk, − vk|| np L n−p 1−θ 2−a ≤ M||Dvk, − Dvk||Lp ≤ CM → 0 q so vk, is arbitrarily close to vk in the L −norm. Combining this with our convergent subsequence {vkl,} in 1,p q W (Ω), we have shown we have a convergent subsequence in L , as claimed.

9. Partial Differential Operators, Weak Existence, and Elliptic Regularity Definition 9.1. Linear Differential Operators A linear partial differential operator L of order k, defined on Rm−valued functions consists of: an m × m matrix L = (Lij)i,j=m in which, X α α Lij = aijD , |α|≤k n α ∞ where α = (α1, . . . , αn) ∈ Z is a multi-index, and aij ∈ C (Ω). To avoid trivial cases, we require at least α one αij 6≡ 0 for at least one pair i, j and α satisfying |α| = k. A particularly important class of linear differential operators are Elliptic Operators which have a very tractable analytic structure. Definition 9.2. Elliptic Operators A linear partial differential operator L of order k defined on a domain Ω ⊆ Rn is called Uniformly Elliptic if, ∀x ∈ Ω and every non-zero ξ ∈ Rn we have n X α α 2k L2k(ξ) = aij(x)ξ > C|ξ| , i,j=1 |α|=2k

α α1 αn where ξ = ξ1 ··· ξn , and C > 0. In other words, if we consider the matrix L2k(ξ) obtained by considering the components of the highest order derivatives, and substituting ξα for Dα; L is an elliptic operator if L2k(ξ) is a positive-deﬁnite matrix, for every x ∈ Ω.

The condition that the matrix L2k(ξ) is positive-definite implies a great deal about the associated partial differential operator. In fact, this analytic property will be crucial to the final steps in our proof. A simple (and perhaps the most important) example of an elliptic operator is the Laplacian on Rn: n X ∂2f ∆f = . ∂x2 i=1 i α This is an operator in which the smooth coefficient functions aij satisfy ( α 1 i = j a = δij = , ij 0 i 6= j 28 HADRIAN QUAN

α and aij ≡ 0 for |α| < 2. To see that this operator is uniformly elliptic, we note that the associated matrix is the identity operator: ∆2(ξ) = δij = I. This is a positive-definite matrix, with no dependence on x, hence ∆ is a uniformly elliptic operator. Going further, we claim that in any open chart, the Hodge Laplacian satisfies the uniform ellipticity. Unfortunately, there was not time to include a proof of this result, so we shall have to assume it for this current draft of our paper. Solutions of elliptic equations may be found in a fairly systematic manner. Why is this result plausible? If we think of a heuristic property of the Fourier transform, Z ∞ fˆ(ξ) = f(x)e−2πx·ξdx −∞ it converts differential equations into algebraic equations. Even further the Fourier transform can be inverted. Informally, we can think of PDEs defined by an elliptic linear partial differential operator as having an associated system of linear equations. The ellipticity condition implies that the matrix corresponding to this system of linear equations is positive-definite, hence invertible! Ideally we would have a situation analogous to Elliptic PDE −−−−−−−−−−−→Fourier Transform Linear System

−−−−→Invert Solution to Linear System −−−−−−−−−→Inverse Fourier Solution to PDE, but obviously this approach doesn’t work perfectly. If it did, we would have just used this scheme to solve our PDE. k,2 Instead, we shall exploit the structure of W0 (Ω), the Hilbert spaces which we have so painstakinly constructed. The following variation on the Riesz Representation theorem is particularly convenient for proving existence of weak solution to Elliptic PDE. Theorem 9.3. Lax-Milgram Theorem Let H be a Hilbert space, and B : H × H → R a bilinear form which satisfies: (1) Boundedness, |B[x, y]| ≤ c||x|| · ||y|| (2) Coercivity, γ||x||2 ≤ B[x, x]. Then there exists a unique T ∈ L (H,H) with T −1 ∈ L (H,H), which satisfy 1 ||T −1|| ≤ , ||T || ≤ c, hx, yi = B(T x, y) γ Proof. For each fixed y ∈ H, the mapping x 7→ B[x, y] is a bounded linear functional. Hence the Riesz representation theorem asserts the existence of an element z ∈ H such that B[x, y] = hz, yi, ∀x ∈ H. If so, we shall say T x = z whenever the above holds for all x ∈ H. In this representation, we say B[x, y] = hT x, yi, ∀x, y ∈ H. We claim that T is a bounded linear operator. Linearity is not difficult to verify, so we need only check ||T x||2 = hT x, T xi = B[x, T x] ≤ c||x|| · ||T x||, thus ||T x|| ≤ c||x||, for every x ∈ H. So T is bounded as claimed. To show that Im(T ) is closed, we note γ||x||2 ≤ B[x, x] = hT x, xi ≤ ||T x|| · ||x||, thus γ||x|| ≤ ||T x||. This implies the norm of the image of T is bounded away from zero, hence T is an injective linear map. It also gives that Im(T ) is a closed subspace. To prove surjectivity, we assume for the sake of contradiction that Im(T ) ⊂ H. Then there exists non-zero x ∈ H not in Im(T ). Since Im(T ) is closed, our orthogonal decomposition of H implies x ∈ Im(T )⊥. But γ||x||2 ≤ B[x, x] = hT x, xi = 0, a contradiction. Thus we have surjectivity. This additionally implies the inequalities we have on the operators. The final statement: “hx, yi = B(T x, y)” is a consequence of the Riesz Representation Theorem, whose conditions we have satisfied. THE HODGE THEOREM AND THE BOCHNER TECHNIQUE: A VANISHINGLY SHORT PROOF 29

Using Lax-Milgram, we can prove that there always exists a weak solution of ∆ω = α. We shall define a 1,2 1,2 bilinear form L : W0 (Ω) × W0 (Ω) → R defined by  n  Z X L[u, v] =  ai,j(x)Dxi uDxj v + h(uv) dx, M i,j=1 such that the coefficients ai,j(x) satisfy the uniform ellipticity condition. Since the Hodge Laplacian repre- sents a uniformly ellipticity condition in any open chart U ⊆ M, we can create an associated bilinear form defined as above, using the coefficients of ∆ in this chart U. This bilinear form will be denoted by L∆[·, ·]. Theorem 9.4. Existence of a Weak Solution to ∆ω = α For the bilinear form L∆, there exists C1,C2 > 0 such that it satisfies

(1) |L∆[u, v]| ≤ C1||u|| 1,2 · ||v|| 1,2 W0 W0 2 (2) C2||u|| 1,2 ≤ L∆[u, u] W0 1,2 ∗ 1,2 hence we can ﬁnd for every α ∈ (W0 (Ω)) , there exists u ∈ W0 (Ω) such that

hα, vi = L∆[u, v], i.e. u is a weak solution to the equation ∆u = α.

Proof. The first inequality on L∆ follows from boundedness of the coefficients of ∆, and that our bilinear form requires only first weak derivatives. The second inequality is a consequence of the uniform ellipticity requirement:   Z Z n 2 2 X θ||Du||L2 = θ |Du| dx ≤  ai,j(x)Dxi uDxj u dx M M i,j=1 Z 2 2 ≤ L∆[u, u] − h u dx = L∆[u, u] − h||u||L2 M 2 2 2 hence we have C2||u|| 1,2 = θ||Du||L2 + h||u||L2 ≤ L∆[u, u], as claimed. Having satisfied the conditions of W0 the Lax-Milgram theorem, we can cite it to ensure existence of a weak solution. We now have existence of a weak solution, which is incredible! We’re nearly done with our work. All that remains is to prove are the analytic results we began with: Theorem 9.5. Regularity of ∆ n 1 Let Ω ⊆ R be open and bounded, and L∆ be given as in the above theorem, with coefficients in ai,j ∈ C (Ω) 2 1,2 2,2 and f ∈ L (Ω). If u ∈ W0 (Ω) is a weak solution of ∆u = f then u ∈ W (A) for any A ⊂⊂ Ω. Moreover, we have 2 2 2 ||u||W 2,2(A) ≤ C(||f||L2 + ||u||L2 ) Proof. The idea of the proof is to exploit the uniform ellipticity condition to obtain a uniform bound on the difference quotients of our element u. If so, this will demonstrate that u actually has one additionally weak derivative, in any compactly contained subset A of Ω. By iterating the above result, we can prove that a weak solution of ∆u = f is actually C∞(Ω). This proves one of the two results we originally needed to complete our proof of the Hodge theorem. Theorem 9.6. ∆ is a Compact Operator p Let {αn} be a sequence in Λ M such that ||αn|| ≤ c and ||∆αn|| ≤ c for all n and for some c > 0. Then p there exists a subsequence {αnk } which is Cauchy in Λ M. Proof. We give a sketch of the proof, which depends crucially on the Rellich-Kondrachkov Lemma. The idea goes as follows: if we can prove such a Cauchy subsequence exists in any open chart on M, by using a partition of unity we can stitch together a global collection of differential forms which are cauchy. In particular, since M is compact we can select a finite collection of open charts, and create our sequence of ‘global forms.’ ∞ As usual, we rely on density of Cc (Ω), and on this set || · ||∞ ≡ || · ||L2 . Using the Rellich-Kondrachkov Lemma, to prove existence of such a Cauchy sequence, it suffices to prove that this sequence is bounded in the || · || 1,2 −norm. Using our estimates on this norms similarly to our proof of weak existence, we can show W0 this without too much difficulty. 30 HADRIAN QUAN

Given our assumption that ∆ is uniformly elliptic, we have completed our proof of the Hodge theorem. Armed with this result, lets examine some geometric consequences.

10. The Bochner Technique The Bochner technique relies on a formula in differential geometry, which has its analogues in vector calculus. If u : Rn → R, since derivatives commute on Rn we can prove that 1 ∆|∇u|2 = |Hess |2 + h∇u, ∇∆ui. 2 u This can act as an integration by parts formula, allowing us to express the Hessian of a smooth function in terms of more easily computed derivative terms. To see this, note that n n 1 1 X ∂ 1 X ∂ ∂ ∆|∇u|2 = h∇u, ∇ui = 2 ∇u, ∇u 2 2 ∂x2 2 ∂x ∂x i=1 i i=1 i i n X ∂ ∂ ∂ = ∇u, ∇u + ∇u, ∇u ∂x2 ∂x ∂x i=1 i i i n  n  X ∂ X ∂ ∂ = ∇ u , ∇u + Tr ∇u, ∇u ∂x2  ∂x ∂x  i=1 i i,j=1 i j t = h∇∆u, ∇ui + Tr HessuHessu 2 = h∇∆u, ∇ui + |Hessu| , as claimed. In the third equality, we used that derivatives commute. This formula holds for any real-valued 3 n 1 2 2 u ∈ C (R ). In particular, if u is a Harmonic function, then 2 ∆|∇u| = |Hessu| . This has the added advantage that this formula, which should be third order in its number of derivatives, reduces to a term of second order. On a Riemannian manifold, we have a different situation. For starters, all of the above definitions relied on a choice of Euclidean coordinates. Since on Rn, in Euclidean coordinates we have that derivatives commute, this formula changes slightly, if u ∈ C3(M) then 1 ∆|∇u|2 = |Hess |2 + h∇u, ∇∆ui + Ric(∇u, ∇u). 2 u The third term, the Ricci tensor, appears as a consequence of this commutation failure. We shall explore this shortly. On a Riemannian manifold, the objects of interest (gradient, divergence, laplacian, etc.) are defined in a coordinate invariant manner (e.g. their definitions reflect how they transform under different coordinates). But they general depend on the choice of Riemannian metric. We denote by ∆gu = 0 the Laplace-Beltrami operator (not the Hodge Laplacian as we have used for most of these notes). The relevant objects in the proceeding calculation are defined on a Riemannian manifold by

h∇f, Xi = DX f, Hessf (X,Y ) = h∇X ∇f, Y i, ∆gf = tr(Hessf ) = Div(∇f)

One convenience is the existence of normal coordinates: {Xi} a basis for TpM which is orthonormal and locally euclidean, up to second order. In other words

hXi,Xji = δij, ∇Xi Xj = 0, ∀i, j. In such a coordinate frame, many calculations are greatly simplified. Since definitions are coordinate invariant, it suffices to prove a formula such as ours in a convenient frame. Lemma 10.1. The Bochner-Weitzenbock Formula If M n is a complete, Riemannian manifold, then for all f ∈ C3(M) we have 1 ∆|∇u|2 = |Hess |2 + h∇u, ∇∆ui + Ric(∇u, ∇u). 2 u 1 2 2 In particular if u is Harmonic, then 2 ∆|∇u| = |Hessu| + Ric(∇u, ∇u). THE HODGE THEOREM AND THE BOCHNER TECHNIQUE: A VANISHINGLY SHORT PROOF 31

Proof. Let {Xi} be a choice of normal coordinates, as described above (i.e. hXi,Xji = δij and ∇Xi Xj = 0). In this coordinate frame, the laplacian reduces to

n X ∆f = DXi DXi f, i=1 as in the Euclidean case. The proof of our formula is a long, and uninspired calculation, using the deﬁnition of these objects.

n n 1 1 X 1 X ∆|∇u|2 = D D h∇u, ∇ui = 2D h∇ ∇u, ∇ui 2 2 Xi Xi 2 Xi Xi i=1 i=1 n n X X = DXi Hessu(Xi, ∇u) = DXi Hessu(∇u, Xi) (since Hessu is symmetric) i=1 i=1 n n X X : 0 = DXi h∇∇u∇u, Xii = h∇Xi ∇∇u∇u, Xii + h∇∇u∇u,∇Xi Xii i=1 i=1 n X = h∇Xi ∇∇u∇u, Xii. i=1 The next term we can rewrite in terms of the Riemann Curvature Tensor. By definition for any three vector fields X,Y,Z the curvature tensor is defined to be

RX,Y Z := ∇X ∇Y Z − ∇Y ∇X Z − ∇[X,Y ]Z. So n 1 X ∆ |∇u|2 = h∇ ∇ ∇u, X i 2 g Xi ∇u i i=1 n n n X X X = hRXi,∇u∇u, Xii + h∇∇u∇Xi ∇u, Xii + h∇[Xi,∇u]∇u, Xii . i=1 i=1 i=1 | {z } | {z } | {z } =:Ric(∇u,∇u) (1) (2) To complete our calculation, we now must simplify terms (1) and (2).

n n X X (1) = h∇∇u∇Xi ∇u, Xii = D∇uh∇Xi ∇u, Xii − h∇Xi ∇u, ∇∇uXii i=1 i=1 n X j : 0 = D∇u h∇Xi ∇u, Xii − h∇Xi ∇u, a ∇Xj Xii i=1

= D∇uDiv(∇u) = D∇u(∆gu) = h∇u, ∇u∆gui, and n n X X (2) := h∇[Xi,∇u]∇u, Xii = hHessu([Xi, ∇u],Xi) i=1 i=1 n X = Hessu(∇Xi ∇u − ∇∇uXi,Xi) i=1 n n 0 X X j : = Hessu(∇Xi ∇u, Xi) − Hessu(a ∇Xj Xi,Xi) i=1 i=1 n n X X = Hessu(∇Xi ∇u, Xi) = h∇Xi ∇u, ∇Xi ∇ui i=1 i=1 2 = |Hessu|

Collecting all terms, we have completed the proof of our formula. 32 HADRIAN QUAN

If u : M → R is Harmonic, i.e. ∆gu = 0 (note that this is with respect to the Laplace-Beltrami operator, not the Hodge Laplacian as we have used for most of these notes), then we have the following formula due to Salomon Bochner: 1 ∆ |∇u|2 = |∇2u|2 + Ric(∇u, ∇u) g 2 where ∇ is the gradient, and ∇2 is the second covariant derivative. 1 If αX is a harmonic 1-form on (M, g), then we shall consider f = 2 g(αX , αX ). If X is dual to αX i.e. αX (v) = g(X, v) then 1 1 1 f = g(α , α ) = g(X,X) = α (X). 2 X X 2 2 X Additionally we claim that divX = −δαX , for any dual X, αX . Theorem 10.2. Bochner’s Vanishing Theorem If M is compact, oriented and has Ric ≤ 0 then every harmonic 1-form θ is parallel (i.e. ∇θ ≡ 0).

Proof. If αX is a harmonic 1-form, and its dual vector ﬁeld is X, then from our representation formula we have 1 : 0 ∆ |X|2 = |∇X|2 + D divX + Ric(X,X) = |∇X|2 + Ric(X,X), g 2 X since divX = 0 if αX is harmonic. Using Stokes theorem, and Ric ≥ 0 we have Z Z 1 2 2 0 = ∆g |X| dV ol = |∇X| + Ric(X,X)dV ol M 2 M Z ≥ |∇X|2dV ol ≥ 0 M hence we have |∇X| = 0.

Theorem 10.3. If (M, g) is compact, orientable, and satisﬁes Ric ≥ 0, then b1(M) ≤ n with equality holding if and only if (M, g) is a ﬂat torus.

1 Proof. The Hodge theorem tells us b1(M) = dim(H (M)), the space of Harmonic 1-forms. If every Harmonic 1-form is parallel, the evaluation map 1 ∗ H (M) → Tp M

ω 7→ ωp ∗ n 1 is an injective map. Since Tp M is linearly isomorphic to R , this injection implies, dim(H (M)) ≤ n.

11. Appendix 11.1. A Rapid Introduction to Smooth Manifolds. One of the most salient properties of a smooth manifold is it’s locally euclidean structure. Intuitively, a smooth manifold may be thought of as a geometric space that resembles Rn, at least locally. Deﬁnition 11.1. Atlas of Charts n For every open set U ⊆ M, there exists a homeomorphism φU : U → φ(U) ⊆ R . Such a pair of open n set and mapping into R ,(φU ,U) is referred to as a Chart. Two charts (Uα, φα) and (Uβ, φβ) are called compatible if we have that the maps −1 φβ ◦ φα : φα(Uα ∩ Uβ) → φβ(Uα ∩ Uβ), −1 φα ◦ φβ : φβ(Uα ∩ Uβ) → φα(Uα ∩ Uβ), n n n are smooth as functions from R to R . This deﬁnition is sensical since φα(Uα ∩Uβ) ⊂ R and φβ(Uα ∩Uβ) ⊂ Rn, so these transition maps between the open sets are just functions on Rn. A collection of such compatible charts, {(φα,Uα)}α∈A with open sets forming an open cover of M (M ⊆ S 5 α∈A Uα) is referred to as an Atlas .

5Much of the nomenclature in differential geometry has a delightful cartographic influence. Mapmaking puns and references abound, in part owing to the history of differential geometry. Some of the earliest occurrences of this branch of mathematics can be found in Carl Gauss’s geodesic survey of Hanover. THE HODGE THEOREM AND THE BOCHNER TECHNIQUE: A VANISHINGLY SHORT PROOF 33

Definition 11.2. Definition of smooth function on a Manifold A function defined on a smooth manifold M, f : M → Rk is smooth if the composite function −1 n k f ◦ φα : φα(Uα) ⊂ R → R is smooth for every chart {(Uα, φα)}α∈A in some smooth Atlas. Similarly, if F : M m → N n is a map between manifolds, then we require that the composite function

−1 m n ψβ ◦ F ◦ φα : φα(Uα) ⊂ R → ψβ(Vβ) ⊂ R m n is smooth as a map from R to R , for every pair of coordinate charts {(Uα, φα)}α∈A and {(Vβ, ψβ)}β∈B on M and N respectively. The following definition and theorem are technical results in the study topological manifolds. Unfortu- nately we shall not furnish a proof of the theorem on the existence of a partition of unity. Definition 11.3. Partition of Unity Subordinate to an Open Cover Let {Uα}α∈A be an open cover of a topological space M.A Partition of Unity, subordinate to the Open Cover {Uα}α∈A is a collection of functions {fα}α∈A satisfying the following (1)0 ≤ f ≤ 1 (2) supp(fα) ⊂ Uα, ∀α (3) Local finiteness: ∀x ∈ M, there is a neighbourhood of x where all but a finite number of the functions of fα are zero P (4) fα ≡ 1 α∈A Theorem 11.4. Existence of a Partition of Unity If M is a paracompact, Hausdorff, topological space (in particular, if M is a smooth manifold) then for any open cover of M there exists a partition of unity subordinate to the open cover.

11.2. Riemannian Geometry. A layperson might say with some confidence, if asked, that geometry was all about shapes. After some questioning they might add that angles, lengths, and area also play some role. Given an Atlas of charts on a smooth manifold, we can express the topology and smooth structure of a manifold, but this gives us no way to express lengths and angles of tangent vectors. Conveniently, an inner product can encode all of that information. Inconveniently, the tangent vectors based from point to point on a manifold lie in distinct tangent spaces. We can resolve this with an additional gadget on a smooth manifold: a riemannian metric. Definition 11.5. Riemannian Manifold A riemannian manifold (M, g) consists of a smooth manifold M equipped with a riemannian metric g. On each tangent space TpM, g assigns of a Euclidean inner product gp to TpM. Further we require that it varies smoothly: for any two smooth vector fields X,Y we require that gp(Xp,Yp) is a smooth function of p.

11.3. Measure Theory. Deﬁnition 11.6. σ−Algebra and Measurable Spaces A σ−algebra of sets is a collection of sets Σ which is closed under the operations of countable unions, countable intersections, and complements. A measurable space (X, Σ) is a set X with a σ−algebra Σ of Measurable Sets. Deﬁnition 11.7. Measure and Measure Spaces A measure µ on a set X is a non-negative set function µ : X → [0, ∞], satisfying the following properties: (1) Empty set: µ(∅) = 0 (2) Monotonicity: If A ⊆ B then µ(A) ≤ µ(B) ∞ S P∞ (3) Countable Sub-Additivity: µ Ak ≤ k=1 µ(Ak) k=1 34 HADRIAN QUAN

A measure space (X, Σ, µ) is a measurable space (X, Σ) together with a measure µ, such that the inequality in property (3) becomes an equality when restricted to the measurable sets Σ. In other words, if the Ak ∈ Σ then ∞ ! ∞ [ X µ Ak = µ(Ak). k=1 k=1 Definition 11.8. Sets of Measure Zero Let (X, Σ, µ) be a measure space. A set A ⊆ X is said to be of measure zero if µ(A) = 0. A statement is said to hold almost everywhere (denoted a.e.) if such a statement is true, except possibly when ignoring a set of measure zero. For example f = g a.e., if there exists A ⊂ X such that f = g on X \ A, and µ(A) = 0. Definition 11.9. Measurable Functions Let (X, Σ, µ) be a measure space. A measurable function f : X → R has the property that for every α ∈ R, the set f −1((−∞, α]) ⊆ X is a measurable. Definition 11.10. Indicator function of a set Let A be a set. The function Indicator function of A, 1A, is defined to be ( 1 x ∈ A 1A = 0 x∈ / 0 as it indicates if a point is in a set or not. These functions are sometimes also referred to as the characteristic function of A. This function is measurable if and only if the set A is measurable. Definition 11.11. Simple functions A Simple Function ψ(x) is a measurable function which can be represented as a finite sum of the form r X 1 ψ(x) = ak · Ak (x), k=1 n for ak ∈ R and sets Ak of finite measure. Simple functions are finite sums of step functions, but weighted to different heights. Hence the ’area’ under the function is appropriately easy to define: it is a finite sum of areas of ‘generalized rectangles.’ The height of the is the value ak, the length of the base is the measure of the base set. Definition 11.12. Lebesgue Integral of Simple Functions Let (X, Σ, µ) be a measure space. The integral of a simple function ψ(x), as defined above is r Z X ψdx = ak · µ(Ak) X k=1 The next theorem demonstrates that simple functions are appropriately ‘dense’ in the space of measurable functions. Theorem 11.13. Simple Approximation Theorem Let f be a measurable function with respect to an underlying measure space (X, Σ, µ). Then there exists ∞ a sequence of simple functions {ψi(x)}i=1 such that: (1) ∀i, ∀x ∈ X, we have |ψi(x)| ≤ |ψi+1(x)| (2) ∀x ∈ X, we have limi→∞ ψi(x) = f(x) (3) If f is bounded, the convergence is uniform. In particular, this implies that a function is measurable if and only if it is a pointwise limit of simple functions. Hence, simple functions are dense in the space Lp(X), as every element of Lp(X) is measurable.

Proof. For all k, n ∈ N such that −n2n + 1 ≤ k ≤ n2n THE HODGE THEOREM AND THE BOCHNER TECHNIQUE: A VANISHINGLY SHORT PROOF 35

deﬁne the following families of sets: k − 1 k A := x ∈ X : ≤ f(x) ≤ , n,k 2n 2n

Bn := {x ∈ X : f(x) ≤ −n},Cn := {x ∈ X : f(x) ≥ n}. In essence we have, for each n, partitioned the range of f into three pieces: those above, below, and including the interval (−n, n). We further subdivide the range along the interval (−n, n) for each k. So, for each n we have that ! [ X = Bn ∪ An,k ∪ Cn. k Define n2n 0 X k − 1 X k ψ (x) := n1 (x) − n1 (x) + 1 (x) + 1 (x). n Cn Bn 2n An,k 2n An,k k=1 k=−n2n+1 Oh boy... The first thing to prove is |ψn(x)| ≤ |ψn+1(x)|. Heuristically, if we refine the partition of the range, this increases the lower bounds of the function over the subintervals of the range. Furthermore, 1 |ψ (x) − f(x)| ≤ ∀x ∈ (B ∪ C )c n 2n n n c and in the limit n → ∞, the set (Bn ∪ Cn) = ∅. This, and the above inequality, implies our pointwise convergence ψn → f. Definition 11.14. Lebesgue Integral of a Measurable Function If f is a non-negative, measurable function with respect to an underlying measure space (X, Σ, µ), we define its Lebesgue Integral as Z Z

fdx = sup φdx φ is a simple function, 0 ≤ φ ≤ f X X With the definition of the integral given, we can now consider some function spaces of crucial analytic interest. Since integration defines a norm || · || in a natural manner, we can consider function spaces which have finite norm. Definition 11.15. The Lp spaces Let (X, Σ, µ) be a measure space. The L1(X) space of functions is Z 1 L (X) := f : X → R, measurable : ||f||L1 = |f|dx < ∞ . X For 1 < p < ∞, the Lp(X) space of functions is defined by ( 1 ) Z p p p L (X) := f : X → R, measurable : ||f||Lp = |f| dx < ∞ X And for p = ∞, the space L∞(X) is the space of essentially bounded functions, i.e. ∞ L (X) := f : X → R, measurable : ||f||L∞ = sup |f| < ∞, except possibly on a set of measure zero X R R One word of caution: if f, g are both measurable and f = g a.e., then X fdx = X gdx, as the Lebesgue integral of a function ‘ignores’ measure zero sets. Since the Lp norms are defined via integration, if two functions are equal a.e., they have the same Lp norm. So the Lp spaces are only well-defined if we consider an element of the space as an equivalence class of measurable functions which are equal a.e. Some further results about the Lp(Ω) norms are summarized below. Theorem 11.16. Hölder’sInequality 1 1 1 Let (X, Σ, µ) be a complete measure space. If for 1 ≤ p, q ≤ ∞ we have p + q = s , then

||fg||Ls ≤ ||f||Lp ||g||Lq Theorem 11.17. Interpolation Inequality Let (X, Σ, µ) be a complete measure space. If for 1 ≤ p, q ≤ ∞, then for all 0 ≤ θ ≤ 1 satisfying 1 θ 1−θ r = p + q we have θ 1−θ ||fg||Lr ≤ ||f||Lp ||g||Lq 36 HADRIAN QUAN

Proof. Using H¨older’sInequality we obtain Z Z r r θr (1−θ)r ||f||Lr = |fg| dx = |f| |g| dx Ω Ω θr (1−θ)r Z p Z q p (1−θ)r q (1−θ)r θr θr (1−θ)r θr ≤ |f| dx |g| dx = ||f||Lp ||g||Lq Ω Ω Theorem 11.18. Young’s Inequality Let (X, Σ, µ) be a complete measure space. If 1 ≤ p, q, r ≤ ∞ satisfy 1/p + 1/q = 1 + 1/r, then

||f ∗ g||Lr ≤ ||f||Lp ||g||Lq where ∗ denotes the convolution of two functions. Theorem 11.19. Lp norms are ordered Let (X, Σ, µ) be a complete measure space. If X has finite measure (i.e. µ(X) < ∞), then ||f||Lp ≤ ||f||Lq for all p < q. Proof. We already have 1 1 Z p p p p 1 |f| dx ≤ µ(X) sup |f| = µ(X) p ||f||L∞ , X X q 1 q so we have ||f||Lp ≤ ||f||L∞ . Now, if p < q, then we have p > 1. Choose r such that r = 1 − p . Then, since r ||1X ||Lr = µ(X) for all r, we can use Hölder’sinequality to derive p p r 1 q 1 ||f|| p = ||f · X ||L1 ≤ ||f || || X ||Lr = µ(X) ||f||Lq , L L p 1 − 1 =⇒ ||f||Lp ≤ µ(X) p q ||f||Lq , so ||f||Lp ≤ ||f||Lq as claimed. The following three theorems are of crucial importance. All are ‘interchange criterion’ of a sort: they give conditions on when it is allowable to interchange two limiting operations. When can a limit be interchanged with an integral? This is a crucial step in many analytic proofs. Theorem 11.20. Monotone Convergence Theorem Let (X, Σ, µ) be a complete measure space (meaning all subsets of measure zero sets are measurable sets). ∞ Say {fn}n=1 is a monotone sequence of non-negative functions, i.e. |fn(x)| ≤ |fn+1(x)|, and that this sequence converges pointwisely a.e. to f. Then ||fn − f||L1 → 0, i.e. Z Z lim fndx = fdx n→∞ Ω Ω so we have interchanged integration and limit. Proof. Since the sequence is non-negative, then Z fndx → α ∈ [0, +∞]. Ω R R Since fn ≤ f for all n, we have α ≤ Ω fdx. We claim further that α ≥ Ω fdx. Let 0 < c < 1 be arbitrary, and consider any simple function φ such that 0 ≤ φ ≤ f. Now, we define the family of sets

Ωn,c := {x ∈ Ω: cφ(x) ≤ fn(x)}, and note that ∞ [ Ω = Ωn. n=1 This family of sets is monotone increasing: Ωn ⊆ Ωn+1 ⊆ ..., and Z Z Z Z α ≥ fdx ≥ fn(x)dx ≥ cφ(x)dx = c φ(x)dx. Ω Ω Ωn Ωn From continuity of measure, Z Z α ≥ lim c φdx = c φdx, n→∞ Ωn Ω THE HODGE THEOREM AND THE BOCHNER TECHNIQUE: A VANISHINGLY SHORT PROOF 37 since limits preserve non-strict inequalities. We have proven this result for every c ∈ (0, 1), so it follows that Z α ≥ φdx. Ω But our choice of simple function φ satisfying 0 ≤ φ ≤ f was arbitrary. So Z Z

α ≥ sup φdx φ is a simple function, 0 ≤ φ ≤ f = fdx, Ω Ω hence we have proven our claim. Theorem 11.21. Fatou’s Lemma ∞ Let {fn}n=1 be a sequence of non-negative, measurable functions. Then Z Z lim inf fndx ≤ lim inf fndx. Ω Ω

Proof. If we set gk = infn≥k fn, then lim inf fn = limk→∞ gk, and Z Z gkdx ≤ fkdx ∀k Ω Ω ∞ from the deﬁnition of inﬁmum. The sequence {gk}k=1 is monotone increasing, as it converges to lim inf fn. Hence, from the Monotone Convergence Theorem, we have Z Z Z lim inf fndx = lim gkdx ≤ lim inf fdx, Ω k→∞ Ω Ω as claimed. Theorem 11.22. Lebesgue’s Dominated Convergence Theorem Let (X, Σ, µ) be a complete measure space. Let {fn} be a sequence of measurable functions, which converge 1 pointwisely a.e. to f. Assume additionally that ∀n we have |fn(x)| ≤ g(x) for some g ∈ L (Ω). Then ||fn − f||L1 → 0, i.e. Z Z lim fndx = fdx, n→∞ Ω Ω so we have interchanged integration and limit. (Colloquially, the condition that ∃g such that |fn(x)| ≤ g(x), ∀n is referred to as the sequence {fn} being dominated by g.)

Proof. If |fn| ≤ g for all n, then |f| ≤ g, as pointwise limits preserve non-strict inequalities. Thus Z Z |f|dx ≤ gdx < ∞ Ω Ω so f ∈ Lp(Ω). If f is non-negative, then 2g − |fn − f| ≥ 0 so by applying Fatou’s lemma we obtain Z Z lim inf(2g − |fn − f|)dx ≤ lim inf 2g − |fn − f|dx. Ω Ω

Since fn converges pointwisely to f, we have Z Z Z Z 2gdx = lim inf(2g − |fn − f|)dx ≤ lim inf 2gdx − |fn − f|dx , Ω Ω Ω Ω which simpliﬁes further when we note that − sup(A) = inf(−A) for an arbitrary set A, (the reader may enjoy proving this fact). So Z Z Z 2gdx ≤ 2g − lim sup |fn − f|dx Ω Ω Ω Z =⇒ lim sup |fn − f|dx ≤ 0, =⇒ lim ||fn − f||L1 = 0 Ω n→0 Next, we can use Dominated convergence to prove a further ‘interchange criterion.’ This is a result which I have always hoped was more explicitly discussed in an introductory analysis course: diﬀerentiation under the integral sign. 38 HADRIAN QUAN

Theorem 11.23. Differentiation Under the Integral Let Ω ⊂ Rn and f : Rn × Ω → R satisfies the following: (1) x 7→ f(x, y) is differentiable for all y ∈ Ω

(2) ∂f ≤ g(y) for some g ∈ L1(Ω) ∂xi R (3) Ω |f(x, y)|dy < ∞ Then we have that ∂ Z Z ∂f f(x, y)dy = (x, y)dy ∂xi Ω Ω ∂xi n Proof. Let {~ei}i=1 denote the standard Euclidean basis. First we consider the left hand side. From the definition of partial differentiation we have ∂ Z 1 Z Z f(x, y)dy = lim f(x + h~ei, y)dy − f(x, y)dy ∂xi Ω h→0 h Ω Ω Z f(x + h~e , y) − f(x, y) = lim i dy, h→0 Ω h hence our result would be done if we could interchange integration and the limit. From condition (3), we know that f(x, y) ∈ L1(Ω) for all x ∈ Ω. Additionally from condition (1), our difference quotient f(x + h~e , y) − f(x, y)∞ 1 i , by setting h = h n=1 n forms a sequence of measurable functions which converge pointwisely to ∂f . Finally, condition (2) plus the ∂xi mean value theorem imply that our difference quotients are uniformly bounded by g(y) ∈ L1(Ω). This is because for every h > 0, ∃z ∈ Bh(x) such that

f(x + h~ei, y) − f(x, y) ∂f = (z, y) ≤ g(y), h ∂xi from the mean-value theorem. This result is applicable since f is diﬀerentiable for every x ∈ Ω. From everything we have shown Lebesgue’s Dominated Convergence Theorem allows us to conclude ∂ Z Z f(x + h~e , y) − f(x, y) Z f(x + h~e , y) − f(x, y) Z ∂f f(x, y)dy = lim i dy = lim i dy = dy, ∂xi Ω h→0 Ω h Ω h→0 h Ω ∂xi as claimed. Theorem 11.24. Riesz-Fischer Theorem The space Lp(Ω) is complete in its norm for 1 ≤ p ≤ ∞. Proof. We begin by making the following claim: A normed space Y is complete ⇐⇒ Every Absolutely Convergent series of elements is a convergent series ∞ in the norm of Y i.e. if {fn}n=1 ⊆ Y satisﬁes

∞ k X X ||fn||Y < ∞ =⇒ ∃f ∈ Y : f − fn → 0. n=1 n=1 Y (The motivation behind the name absolutely convergent series comes from the case of a series of elements in R: there the absolute value function acts as the norm on R. Here a collection of elements in a normed space is absolutely convergent if the numerical series of norms of those elements is convergent.) To prove this claim, we ﬁrst assume Y is complete in its norm. Let {fn} ⊆ Y be a sequence of elements P in Y whose norms are absolutely convergent, i.e. ||fn||Y < ∞. The partial sums of this series are a convergent sequence of real numbers, hence Cauchy. If we consider the partial sums of the elements of Y : Pk which we denote Sk := n=1 fn ∈ Y , we have j j

X X ||Sj − Sk||Y = fn ≤ ||fn||Y ≤ n=k+1 Y n=k+1 hence the Sk are Cauchy in Y . Since Y is complete in its norm, they are convergent as claimed. Conversely, assume every absolutely convergent series is a convergent series in the norm of Y . We shall show Y is complete in its norm. Let {fn} ⊆ Y be Cauchy in the norm of Y . Then there exists a subsequence THE HODGE THEOREM AND THE BOCHNER TECHNIQUE: A VANISHINGLY SHORT PROOF 39

1 {fnk } such that ||fnk − fnk+1 ||Y ≤ 2k . We can deﬁne a new sequence uk = fnk+1 − fnk , and observe that this sequence is absolutely convergent: ∞ ∞ X X 1 ||v || ≤ = 1. k Y 2k k=1 k=1 P So there exists f ∈ Y such that vk = f. However ∞ r X X f = vk = lim (fn − fn ) = lim fn . r→∞ k+1 k r→∞ r+1 k=1 k=1

So this subsequence converges to f ∈ Y . However, since limits are unique in Hausdorff spaces, if {fn} is a Cauchy sequence, and a subsequence converges to f, then {fn} converges to f. Hence Y is complete. Now that we have proven our claim, we can use this lemma as an easier criterion to prove that the space Lp(Ω) is complete: given an absolutely convergent series of elements, we shall show it is convergent in Lp(Ω). p P∞ Say {fn} ⊆ L (Ω) forms an absolutely convergent series: n=1 ||fn||Y < ∞. Define ∞ k !p X X g(x) := |fn(x)|, φk(x) := |fn(x)| . n=1 n=1 p Since |φk(x)| ≤ |φk+1(x)|, and limits to g (x) we can use the Monotone convergence theorem to conclude 1/p 1/p k !p ! k !p ! Z X Z X ||g||Lp = lim |fn(x)| dx = lim |fn(x)| dx k→∞ k→∞ Ω n=1 Ω n=1 k k X X = lim |fn(x)| ≤ lim ||fn||Lp < ∞, k→∞ k→∞ k=1 Lp n=1 where we use that the triangle inequality for the norm (Minkowski’s inequality), and that this series was absolutely convergent. Hence g ∈ Lp(Ω). Now we can define ( g(x) g(x) < ∞ f(x) = , 0 g(x) = ∞ which is also in Lp(Ω) since g < ∞ a.e. (being an element of Lp(Ω)). Now |f| ≤ |g|, so the definition of g implies fn converges pointwisely a.e. to f. Additionally p ∞ X p f − fn ≤ 4|g| , n=1 so we may use Lebesgue’s Dominated Convergence Theorem to conclude

k X f − fn → 0. n=1 Lp Thus we have shown that every absolutely convergent series is convergent in the Lp norm. Hence Lp(Ω) is complete. 11.4. Essentials of Functional Analysis. Theorem 11.25. Arzela-Ascoli Theorem Let F ⊂ C0(M,N) be a collection of uniformly equicontinuous functions from M to N, where M is a compact metric space, and N is a complete metric space. Then for all sequences {fn} ⊂ F , there is a subsequence {fnk } that converges uniformly. Proof. The assumption that the closure of the union of the images is totally bounded is quite natural. In the original proof of the Arzela-Ascoli theorem (∼ 1884) for functions on R, assuming the functions are uniformly bounded implies the closure of the union of their images is a compact set. S Since the closure of f∈F f(M) is a closed subset of a complete space it is complete. Additionally we have assumed this set is totally bounded, hence it is compact. Let {fn} ⊂ F be a sequence of functions in F and let x1 ∈ M. Since {fn(x1)} is a sequence of points in Im(F ) a compact space, it is sequentially compact and hence has a convergent subsequence {f1,k(x1)} ⊂ {fn(x1)}. 40 HADRIAN QUAN

By the same argument, ∀x2 ∈ M there exists a subsequence {f2,k(x)} ⊂ {f1,k(x)} ⊂ {fn(x)} such that {f2,k(xi)} converges at both x1 and x2. This is because {f1,k(x2)} is a sequence in Im(F ) so it has a convergent sequence {f2,k(x2)}. But {f2,k(x1)} ⊂ {f1,k(x1)} is a subsequence of a convergent sequence so it converges to the same point. Additionally since M is a compact metric space, it has a countable dense subset. Now given a countable dense-in-M set [x1, x2, ..., xn, ...] ⊂ M we have a diagonal argument by considering the sequences where

f1,1(x) f1,2(x) f1,3(x) ...

f2,1(x) f2,2(x) f2,3(x) ...

f3,1(x) f3,2(x) f3,3(x) ......

the first row is a subsequence which converges to x1, and the second row is a subsequence of the first row which converges to x1, x2, the third row is a subsequence of the second row which converges x1, x2, x3 etc. And for ∀l ∈ we have that {f (x)}∞ converges at [x , x , ..., x ]. Since we have that {f (x)}∞ ⊂ {f (x)}∞ N l,k k=1 1 2 l l,l l=l0 k,n n=1 are subsequences for all l0 ≥ k, by construction this diagonal subsequence converges at every point in [x1, x2, ..., xn, ...]. We claim that {fl,l(x)} converges uniformly on M. Continuity gives us that for all > 0, ∃δ > 0 such that d(x, x0)M ≤ δ implies d(fn,n(x), fn,n(x0))N ≤ /3 for all x, x0 ∈ M and for all n. Fix δ. Density of 0 0 0 the set says there exists a subset {x1, ..., xk} ⊂ [x1, x2, ..., xn, ...] such that for all x ∈ M there exists xi in 0 the subset such that d(x, xi) ≤ δ. Convergence of the subsequence implies there exists m ∈ N such that 0 0 0 0 0 d(fk,k(xi), fl,l(xi))N ≤ /3 for all k, l ≥ m and for all xi ∈ {x1, ..., xk}. Our set-up gives us that 0 0 0 0 d(fk,k(x), fl,l(x))N ≤ d(fk,k(x), fk,k(xi))N + d(fk,k(xi), fl,l(xi))N + d(fl,l(xi), fl,l(x))N ≤ + + = 3 3 3 for all k, l ≥ m and for all x ∈ M. Hence this sequence is Cauchy. Since Im(F ) is complete, this sequence is uniformly convergent as claimed. So we win. Theorem 11.26. The Hahn-Banach Theorem Let X be a linear space over F, a complete ordered field, and p : X → F a semi-norm (has homogeneity and triangle inequality). Let Y ⊆ X be a subspace of X, with a linear functional ` : Y → F defined over it, and which is dominated by p: `(y) ≤ p(y) ∀y ∈ Y.

Proof. Suppose Y is not equal to X, then ∃x ∈ X not in Y . Write Z = F · z + Y = {az + y|y ∈ Y, a ∈ F}. We wish to extend ` from Y to Z such that ` is still dominated by p on Z, i.e. `(y) + a`(z) = `(y + az) ≤ p(y + az). This already holds for a = 0. Due to homogeneity, we need only prove this for a = ±1 `(y) + `(z) ≤ p(y + z) `(y0) − `(z) ≤ p(y0 − z) which is true if and only if `(y0) − p(y0 − z) ≤ `(z) ≤ p(y + z) − `(z), but such an `(z) exists if and only if for all y, y0 `(y0) − p(y0 − z) ≤ p(y + z) − `(y) ⇐⇒ `(y0) + `(y) = `(y0 + y) ≤ p(y + z) + p(y0 − z), which we aim to show. However y + y0 ∈ Y , and since ` is dominated by p on Y we have `(y0 + y) ≤ p(y0 + y) = p(y0 + y + z − z) ≤ p(y + z) + p(y0 − z), as needed. So it is possible to extend ` from Y to Z in a bounded fashion. Consider all extensions of ` from Y to any linear space Z where ` is dominated by p. We can put a partial order on this collection if we say 0 0 0 0 (Z, `) ≤ (Z , ` ) if Z ⊆ Z and ` |Z = `. Now, let {Zν , `ν } be a totally ordered collection of extensions `. Then we can define ` on the union S Z = Zν by defining it to be `ν on Zν . This satisfies that ` is still dominated by p on Z, and (Zν , `ν ) ≤ (Z, `) for each ν. So every totally ordered collection of extensions has an upper bound. By Zorn’s lemma, there exists a maximal extension, but this maximal extension must be the entire space X. THE HODGE THEOREM AND THE BOCHNER TECHNIQUE: A VANISHINGLY SHORT PROOF 41

11.5. Orthogonal Decomposition in a Hilbert Space. The issue of decomposing a topological vector space in terms of a subspace is non-trivial, and forms the motivating problem of this entire paper. Luckily, in a linear space with enough structure, this goal may be achieved. Even further, such a decomposition allows us to find weak solutions of linear equations. This final statement is made rigorous via the Riesz Representation Theorem. Definition 11.27. Hilbert spaces and Inner products H is a Hilbert space if it is a topologically complete, inner product space. Given a subspace W ⊆ H, we define its ortho-complement W ⊥ by W ⊥ := {x ∈ H : hx, wi = 0, ∀w ∈ W }. Additionally we use the following notation to denote an element is orthogonal to a set: x ⊥ U if hx, ui = 0 for all u ∈ U. With these definitions at the ready, we prove a technical result which we shall need for our decomposition, namely that we can always find an element of minimum norm in a closed subspace. Proposition 11.28. Let H be a Hilbert space and U a closed, convex subset. Let x ∈ U. Then there exists a unique u ∈ U such that ||u|| = min ||z|| z∈U

Proof. Let c = inf ||z||, and consider a sequence {zn} ⊆ U such that z∈U 1 c2 ≤ ||z ||2 ≤ c2 + . n n zn+zm We claim such a sequence is Cauchy. Observe ﬁrst that 2 ∈ U, since U is convex. Hence ||(zn +zm)/2|| ≤ c. So, since 2 2 zn + zm 2 4c ≤ 4 = ||zn + zm|| 2 we have 2 2 2 ||zn − zm|| = ||zn|| − 2hzn, zmi + ||zm|| 2 2 = 2||zn|| + 2||zm|| − (hzn, zni + hzm, zmi + 2hzn, zmi) 2 2 = 2||zn|| + 2||zm|| − hzn + zm, zn + zmi 2 2 2 = 2||zn|| + 2||zm|| − ||zn + zm|| 1 1 1 1 ≤ 2 c2 + + 2 c2 + − 4c2 = 2 + n m n m

Thus {zn} is Cauchy. Since U is a closed subset of a complete space, this sequence converges to an element of U: zn → u ∈ U. Additionally, we see that

2 2 2 1 ||u|| = lim ||zn|| ≤ lim c + = c n→∞ n→∞ n so u is the element we desired. To prove uniqueness, assume we have two distinct elements u1, u2 ∈ U deﬁned as above. If each satisﬁes this minimum requirement, and u1 6= u2, then as before

u1 + u2 u1 + u2 ||u1|| ≤ , ||u2|| ≤ . 2 2 Thus

u1 + u2 ||u1|| + ||u2|| ≤ 2 = ||u1 + u2|| ≤ ||u1|| + ||u2||, 2 which implies ||u1 + u2|| = ||u1|| + ||u2||. The triangle inequality only gives equality when one vector is a scalar multiple of another. Hence u1 = cu2 for some c ∈ R. However, this would contradict that both elements have minimum norm! Hence this minimizer is unique. Theorem 11.29. Orthogonal Decomposition of Hilbert Spaces Let H be a Hilbert space. If W ⊆ H is a closed subspace, then we have H = W ⊕ W ⊥ 42 HADRIAN QUAN

Proof. Clearly, W ∩ W ⊥ = {0}. So we need only show that for all x ∈ X, there exists a ∈ W, b ∈ W ⊥ such that x = a + b. Consider translates x + W , since translation is a homeomorphism, this space is a closed convex set. So from our previous result, we know that we can select w⊥, as the unique element in x + W with minimum norm. We claim that hw, w⊥i = 0 for all w ∈ W . To see this, note ||w⊥||2 ≤ ||w + w⊥||2 ∀w ∈ W since 0 ≤ 2hw⊥, wi + ||w||2 = hw + w⊥, w + w⊥i − ||w⊥||2. ⊥ ⊥ Now, assume for the sake of contradiction w is not perpendicular to W , then ∃w0 ∈ W such that hw , w0i = a 2 6= 0. If so, set x = −aw0 ∈ W . Then ⊥ 2 ⊥ 2 2 2 2 2 0 ≤ 2hw , xi + ||x|| = −2ahw , w0i + a||w0|| = −a + a ||w0|| 2 2 2 =⇒ a ≤ a||w0|| which is impossible for > 0 suﬃciently small. Hence hw, w⊥i = 0 for all w ∈ W . So w⊥ ∈ x + W , and ⊥ w ⊥ W , hence we have our orthogonal decomposition. Theorem 11.30. The Riesz Representation Theorem Let H be a Hilbert space. Then for each φ ∈ H∗, there exists a unique h ∈ H such that φ(x) = hx, hi, ∀x ∈ H In other words, every linear functional can be used represent an element in H, via the pairing with the inner product on H. Proof. Let φ ∈ H∗ and set W = Ker(φ). If φ ≡ 0, our result follows trivially, so assume W = Ker(φ) ⊂ H. From our result on orthogonal decompositions, we know there exists a subspace W ⊥ ⊆ H such that H = Ker(φ) ⊕ W ⊥. ⊥ ⊥ Since W is non-trivial, we may choose non-zero x0 ∈ W and deﬁne a linear functional by ψ(x) = hx, x0i. We clearly have that Ker(ψ) = Ker(φ), so λψ(x) = φ(x), ∀x ∈ H.

Our desired h we can deﬁne by h = λx0, and we have our representation: φ(x) = hx, hi ∀x ∈ H.

To prove uniqueness, we note that if there exists h1, h2 such that

hx, h1i = φ(x) = hx, h2i

=⇒ hx, h1 − h2i = 0 ∀x ∈ H hence h1 − h2 = 0. 12. References 1-Adams, D. Sobolev Spaces.

2-Petersen, P. Riemannian Geometry.

3-Lee, J. Introduction to Smooth Manifolds.

4-Royden, H. Fitzpatrick, P. Real Analysis.

5-Warner, F. Foundations of Diﬀerentiable Manifolds and Lie Groups.