Lectures in Harmonic Analysis

Transcription

1 Lectures in Harmonic Analysis Joseph Breen Lectures by: Monica Visan Last updated: June 9, 7 Department of Mathematics University of California, Los Angeles

2 Contents Preface 3 Preliminaries 4 Notation and Conventions 4 L p Spaces 4 3 Complex Interpolation 5 4 Some General Principles 7 4 Integrating x α 8 4 Dyadic Sums 8 The Fourier Transform The Fourier Transform on R d The Fourier Transform on T d 6 3 Lorentz Spaces and Real Interpolation 9 3 Lorentz Spaces 9 3 Duality in Lorentz Spaces 3 33 The Marcinkiewicz Interpolation Theorem 6 4 Maximal Functions 9 4 The Hardy-Littlewood Maximal Function 9 4 A p Weights and the Weighted Maximal Inequality 3 43 The Vector-Valued Maximal Inequality 36 5 Sobolev Inequalities 4 5 The Hardy-Littlewood-Sobolev Inequality 4 5 The Sobolev Embedding Theorem The Gagliardo-Nirenberg Inequality 47 6 Fourier Multipliers 49 6 Calderon-Zygmund Convolution Kernels 49 6 The Hilbert Transform The Mikhlin Multiplier Theorem 57 7 Littlewood-Paley Theory 6 7 Littlewood-Paley Projections 6 7 The Littlewood-Paley Square Function Applications to Fractional Derivatives 7 8 Oscillatory Integrals 79 8 Oscillatory Integrals of the First Kind, d = 8 8 The Morse Lemma Oscillatory Integrals of the First Kind, d > 89

3 84 The Fourier Transform of a Surface Measure 9 9 Dispersive Partial Differential Equations 94 9 Dispersion 94 9 The Linear Schrödinger Equation 95 9 The Fundamental Solution 95 9 Dispersive Estimates Strichartz Estimates 96 Restriction Theory 97 The Restriction Conjecture 97 The Tomas-Stein Inequality 3 Restriction Theory and Strichartz Estimates 3 Rearrangement Theory 8 Definitions and Basic Estimates 8 The Riesz Rearrangement Inequality, d = 4 3 The Riesz Rearrangement Inequality, d > 7 4 The Polya-Szego inequality 5 The Energy of the Hydrogen Atom 5 Compactness in L p Spaces 8 The Riesz Compactness Theorem 8 The Rellich-Kondrachov Theorem 3 3 The Strauss Lemma 34 4 The Refined Fatou s Lemma 36 3 The Sharp Gagliardo-Nirenberg Inequality 37 3 The Focusing Cubic NLS 37 4 The Sharp Sobolev Embedding Theorem 38 5 Concentration Compactness in Partial Differential Equations 39 References 39

4 Preface **Eventually I ll probably write a better introduction, this is a placeholder for now** The following text covers material from Monica Visan s Math 47A and Math 47B classes, offered at UCLA in the winter and spring quarters of 7 The content of each chapter is my own adaptation of lecture notes taken throughout the two quarters Chapter contains a selection of background material which was not covered in lectures The rest of the chapters roughly follow the chronological order of lectures given throughout the winter and spring, with some slight rearranging of my own choice For example, basic results about the Fourier transform on the torus are included in Chapter, despite being the subject of lectures in the spring A strictly chronological (and incomplete) version of these notes can be found on my personal web page at mathuclaedu/ josephbreen/ Finally, any mistakes are likely of my own doing and not Professor Visan s Comments and corrections are welcome 3

5 Chapter Preliminaries The material covered in these lecture notes is presented at the level of an upper division graduate course in harmonic analysis As such, we assume that the reader is comfortable with real and complex analysis at a sophisticated level However, for convenience and completeness we recall some of the basic facts and inequalities that are used throughout the text We also establish conventions with notation, and discuss a few select topics that the reader may be unfamiliar with (for example, the Riesz-Thorin interpolation theorem) Any standard textbook in real analysis or harmonic analysis is a suitable reference for this material, for example, [3], [6], and [8] Notation and Conventions If x Cy for some constant C, we say x y If x y and y x, then x y If the implicit constant depends on additional data, this is manifested as a subscript in the inequality sign For example, if A denotes the ball of radius r > in R d and denotes R d -Lebesgue measure, we have A d r d However, the dependencies of implicit constants are usually clear from context, or stated otherwise FINISH L p Spaces ADD Proposition (Young s Convolution inequality) For p, q, r, whenever + r = p + q f g L r (R d ) f L p (R d ) g L q (R d ) Young s convolution inequality may be proved directly, or by using the Riesz-Thorin interpolation theorem (see Section 3) For details on both methods, refer to [3] It is convenient to record a few special cases of Young s inequality that arise frequently Corollary (Young s Convolution inequality, special cases) The following convolution inequalities hold: For any p, f g L f L p g L p In particular, f g L f L g L and f g L f L g L For any p, f g L p f L g L p 4

6 3 Complex Interpolation An important tool in harmonic analysis is interpolation Broadly speaking, interpolation considers the following question: given estimates of some kind on two different spaces, say L p and L p, what can be said about the corresponding estimate on L p for values of p between p and p? There are a number of theorems which answer this kind of question In Chapter 3, we will prove the Marcinkiewicz interpolation theorem using real analytic methods Here, we prove a different interpolation theorem using complex analytic methods Both theorems have their own advantages and disadvantages, and will be used frequently Our treatment of the Riesz-Thorin interpolation theorem follows [8] The main statement is as follows Theorem 3 (Riesz-Thorin interpolation) Suppose that T is a bounded linear map from L p + L p L q + L q and that T f L q M f L p and T f L q M f L p for some p, p, q, q Then where p t = t p + t p and q t = t q T f L q t M t M t f L p t + t q, for any t As noted above, the proof of this theorem relies on complex analysis Specifically, we need the following lemma Lemma 4 (Three Lines lemma) Suppose that Φ(z) is holomorphic in S = { < Re z < } and continuous and bounded on S Let Then for any t M := sup y R Φ(iy) and M := sup Φ( + iy) y R sup Φ(t + iy) M t M t y R Proof We can assume without loss of generality that Φ is nonconstant, otherwise the conclusion is trivial We first prove a special case of the lemma Suppose that M = M = and that sup x Φ(x + iy) as y In this case, φ has a global (on S) supremum of M > Let {z n } be a sequence such that Φ(z n ) M Because sup x Φ(x + iy) as y, the sequence {z n } is bounded, hence there is a point z S such that a subsequence of {z n } converges to z By continuity, Φ(z ) = M Since Φ is nonconstant, by the maximum modulus principle, z is on the boundary of S This means that Φ(z) is globally bounded by Since M = M =, this proves the special case Next, we remove the decay assumption and only suppose that M = M = For ε >, define Φ ε (z) := Φ(z)e ε(z ) First observe that Φ ε (z) if Re z = or Re z = Indeed, note that for y R we have Φ ε (iy) = Φ(iy) e ε((iy) ) e ε(y +) and Φ ε ( + iy) = Φ( + iy) e ε((+iy) ) e ε(+iy y ) e iyε e εy 5

7 Next, we claim that Φ ε satisfies the decay condition from the previous case Indeed, note that Φ ε (x + iy) = Φ(x + iy) e ε((x+iy) ) = Φ(x + iy) e ε(x y +xyi) Φ(x + iy) e ε(x y ) Because x and because Φ is bounded, as y, Φ ε (x + iy) uniformly in x By the previous case, Φ ε (z) uniformly in S Since this holds for all ε >, letting ε gives the desired result for Φ Finally we assume that M and M are arbitrary positive numbers Define φ(z) := M z φ(z) Note that if Re z =, then M z and if Re z =, then Φ(z) M iy M iy Φ(z) M iy M iy M M iy M M iy The claim then follows by applying the previous case M iy M iy Proof of Theorem 3 By scaling appropriately, we may assume without loss of generality that f L p t = First, consider the case when f is a simple function By duality, T f L q t = sup T f(x) g(x) dx where the supremum is taken over simple functions g with g L q t consider the case p t < and q t > Let ( z γ(z) := p t + z ) ( z and δ(z) := q t p p q + z q Define f z := f γ(z) f f and g z := g δ(z) g g Note that γ(t) = Hence, f t = f Also, if Re z =, then ( ) f z p L p = f p tγ(z) dx = f pt z p t+ p p dx = = Fix such a g First, ) f pt dx = f pt L p t = A similar computation gives f z L p = if Re z = Also, g z L q g z L q = if Re z = Next, define Φ(z) := T f z (x) g z (x) dx = if Re z = and As f and g are simple, we may write f = k a kχ Ek where the sets E k are disjoint and of finite measure, and likewise g = j b jχ Fj Then f z = k a k γ(z) a k a k χ E k and g z = j b j δ(z) b j b j χ F j Also with this notation we have Φ(z) = a k γ(z) b j δ(z) a k b j a k b j j,k T (χ Ek ) χ Fj dx 6

8 From this it is evident that Φ(z) is holomorphic in S = { < Re z < } and continuous and bounded on S Moreover, if Re z =, then Holder s inequality gives Φ(z) T f z (x) g z (x) dx T f z L q g z L q M f L p = M Similarly, if Re z =, Φ(z) M Therefore, by the Three Lines lemma, if Re z = t then Φ(z) M t M t As Φ(t) = T f(x) g(x) dx, this is the desired result The passage from simple functions to general functions is a straightforward limiting argument, and the remaining cases p t = and q t = are handled similarly For more details, consult [8] It is convenient to state a few special cases of Riesz-Thorin interpolation These occur frequently, and also provide more concrete examples of interpolation at work Corollary 5 (Riesz-Thorin interpolation, special cases) Proof Fix p < p Suppose that T is bounded as a linear map from L p L p and also from L p L p Then T is bounded from L p L p for all p p p Suppose that T is bounded as a linear map from L L and also from L L Then T is bounded from L p L p for all p If T is bounded from L p L p and L p L p, then Riesz-Thorin implies that T is bounded from L pt L qt for all t with = t + t p t p p and = t + t q t p p Thus, p t = q t Moreover, as t ranges from to, p t ranges from p to p Therefore, T is bounded from L p L p for all p p p Riesz-Thorin gives boundedness of T : L pt L qt for t with = t + t p t and = t q t + t Thus, p t = t and q t = t so that q t = p t Moreover, p t = t, so as t ranges from to, p t ranges from to The first substantial application of interpolation is found in found in Chapter, where we show that the Fourier transform is well-defined on L p spaces for p As a final remark, the Riesz-Thorin interpolation theorem is an interpolation result based on complex analytic tools, namely, the maximum principle for holomorphic functions In Chapter 3, we prove a different interpolation theorem the Marcinkiewicz interpolation theorem based on real analytic techniques 4 Some General Principles Throughout the rest of this text, there are a number of straightforward principles that we will employ over and over again, often without explicit comment For convenience, we include a short discussion of these recurring principles here 7

9 4 Integrating x α The first concern integration of functions of the form x α In words, integrating x α over a region of R d increases the exponent by d Explicitly, we have the following proposition Proposition 6 Fix R > Then The integral x >R x α dx is finite when α > d, in which case x >R x α dx R α+d The integral x <R x α dx is finite when α d, in which case x <R x α dx R α+d Consequently, x α / L p (R d ) for p <, for any choice of α Proof Fix R > Set r = x Then dx = r d dr Thus, x >R x α dx = r>r r α rd dr = r α+d dr r α+d r=r This limit exists precisely when α + d <, ie, when α > d, in which case x α dx R α+d Similarly, x >R x <R dx r α+d x α This limit exists precisely when α + d, ie, α d, in which case x α dx R α+d x <R R R 4 Dyadic Sums Another computational tool ubiquitous in harmonic analysis is dyadic sums; that is, geometric series with r = The following proposition is easy, but worth recording Proposition 7 Let Z = { n and : n Z } be the set of dyadic numbers Then N Z ; N N N = N N Z ; N N N = N 8

10 Proof Fix a dyadic number N = n Z Then N Z ; N N N = n Z; n n n = Essentially the same computation gives n=n ( ) n ( n = ) N Z ; N N N = N = n = N 9

11 Chapter The Fourier Transform Our study of harmonic analysis naturally begins with the Fourier transform Historically, the Fourier transform was first introduced by Joseph Fourier in his study of the heat equation In this context, Fourier showed how a periodic function could be decomposed into a sum of sines and cosines which represent the frequencies of the function From a modern point of view, the Fourier transform is a transformation which accepts a function and returns a new function, defined via the frequency data of the original function As a bridge between the physical domain and the frequency domain, the Fourier transform is the main tool of use in harmonic analysis The basic theory of the Fourier transform is standard, and there are a wealth of references (for example, [3], [6], and [7]) for the following material The Fourier Transform on R d We begin by discussing the Fourier transform of complex-valued functions on the Euclidean space R d The Fourier transform is defined as an integral transformation of a function; as such, it is natural to first consider integrable functions Definition The Fourier transform of a function f L (R d ) is given by F(f)(ξ) := ˆf(ξ) := (f)ˆ(ξ) := e πix ξ f(x) dx R d There are a number of conventions for placement of constants when defining the Fourier transform For example, another common definition of the Fourier transform is ˆf(ξ) = e ix ξ f(x) dx (π) d R d These choices clearly do not alter the theory in any meaningful way; they are simply a matter of notational preference In this text, we will use either convention when convenient In this section, we will use the former We have two immediate goals One is to understand the basic properties of the Fourier transform and how it interacts with operations such as differentiation, convolution, etc The other goal is to understand which classes of functions the Fourier transform can be reasonably defined on As and aid towards both of these goals, we introduce (or recall) a suitably nice class of functions Here, we use the following multi-index notation: for α = (α,, α d ) N d, define α := α + α d ; x α := x α xα d d ; Dα := α x α xα d d

12 Definition A C -function f : R d C is Schwartz if x α D β f L x (R d ) for every multiindex α, β N d The collection of Schwartz functions, Schwartz space, is denoted S(R d ) In words, a function is Schwartz if it is smooth, and if all of its derivatives decay faster than any polynomial An example of a Schwartz function is e x The set of Schwartz functions forms a Frechet space, ie, a locally convex space (a vector space endowed with a family of seminorms {ρ α } that separates points: if ρ α (f) = for all α, then f = ) which is metrizable and complete In the case of S(R d ), the collection of seminorms is given by {ρ α,β } α,β N d, where ρ α,β (f) := x α D β f L x (R d ) The Frechet space structure of S(R d ) is not our main concern; rather, density in L p spaces (indeed, note that Cc (R d ) S(R d )) and its interaction with the Fourier transform are of more importance Here, we collect a number of basic and important properties of the Fourier transform of Schwartz functions Proposition 3 Let f S(R d ) Then: If g(x) = f(x y), then ĝ(ξ) = e πiy ξ ˆf(ξ) If g(x) = e πix η f(x), then ĝ(ξ) = ˆf(ξ η) 3 If g(x) := f(t x) for T GL(R d ), then ĝ(ξ) = det T ˆf(T t ξ) In particular, if T is a rotation or reflection and f(t x) = f(x) (ie f is rotation or reflection invariant), then ĝ(ξ) = ˆf(ξ) 4 If g(x) := f(x), then ĝ(ξ) = ˆf( ξ) 5 If g(x) := D α f(x), then ĝ(ξ) = (πiξ) α ˆf(ξ) 6 If g(x) := x α f(x), then ĝ(ξ) = ( i π ) α D α ξ ˆf(ξ) 7 If g(x) := (k f)(x) for k L (R d ), then ĝ(ξ) = ˆk(ξ) ˆf(ξ) 8 We have the basic estimate ˆf f L L x ξ Proof Making the change of variables z = x y, we have ĝ(ξ) = e πix ξ f(x y) dx = e πi(z+y) ξ f(z) dz = e πiy e πiz ξ f(z) dz = e πiy ˆf(ξ) Computing directly gives: ĝ(ξ) = e πix ξ e πix η f(x) dx = e πix (ξ η) f(x) dx = ˆf(ξ η) 3 For T GL(R d ), make the change of variables y = T x Then ĝ(ξ) = e πix ξ f(t x) dx = e πi(t y) ξ f(y) det T dy = det T e πiy (T tξ) f(y) dy = det T ˆf(T t ξ)

13 If T is a rotation or reflection (or more generally, T is orthogonal), then T t T = T T t = I and det T = In this case, ĝ(ξ) = det T ˆf(T t ξ) = ˆf(ξ) 4 Computing directly gives: ĝ(ξ) = e πix ξ f(x) dx = e πix ξ f(x) dx = e πix ( ξ) f(x) dx = ˆf( ξ) 5 Integrating by parts, we have: ĝ(ξ) = e πix ξ D α f(x) dx = ( ) α D α e πix ξ f(x) dx = ( ) α ( πiξ) α e πix ξ f(x) dx = (πiξ) α e πix ξ f(x) dx = (πξ) α ˆf(ξ) 6 Note that D α x e πix ξ = ( πix) α e πix ξ, so that x α e πix ξ Thus, integrating by parts again gives: ĝ(ξ) = e πix ξ x α f(x) dx = ( ) i α ˆf(ξ) π 7 Computing, we have ĝ(ξ) = e πix ξ (k f)(x) dx = e πix ξ = e πix ξ k(x y)f(y) dx dy k(x y)f(y) dy dx We make a change of variables z = x y in the inner integral to get ĝ(ξ) = e πi(z+y) ξ k(z)f(y) dz dy = e πiz ξ k(z) dz e πiy ξ f(y) dy = ˆk(ξ) ˆf(ξ) 8 This follows from the triangle inequality for integrals: e πix ξ f(x) dx e πix ξ f(x) dx = f(x) dx From this proposition, it becomes immediately clear why the Fourier transform is such a powerful computational tool For example, properties 5 and 7 above describe how the Fourier transform turns complicated function operations like differentiation and convolution into multiplication This becomes incredibly useful when studying partial differential equations, for instance Also note that properties,,3,4,7, and 8 extend immediately to functions in L (R d ) Proposition 4 If f S(R d ), then ˆf S(R d ) Moreover, if f n f S(R d ), then ˆ f n ˆf S(R d )

14 Proof Suppose that f S(R d ) Note that ξ α D β ˆf(ξ) ξ α xβ f(ξ) D α x β f(ξ) = e πix ξ D α (x β f(x)) dx Because f is Schwartz, D α (x β f(x)) L (R d ) Therefore, ξ α D β L ˆf(ξ) D α (x β f(x)) < ξ L so that ˆf S(R d ) It is clear that {f n } f S(R d ) if and only if { ˆ f n } ˆf S(R d ) by properties 5 and 6 of the previous proposition Corollary 5 (Riemann-Lebesgue Lemma) If f L (R d ), then ˆf is uniformly continuous and vanishes at Proof Let C (R d ) denote the space of continuous functions which vanish at Let {f n } be a sequence of Schwartz functions with f n f in L (R d ) Since f n f f n f L L it follows that ˆ f n ˆf in L (R d ) As ˆf S(R d ), ˆf C (R d ) Since C (R d ) is closed under uniform convergence, we are done Aside from simple properties and estimates of the Fourier transform and its interaction with various function operations, we have not computed any Fourier transforms of actual functions Lemma 6 Let A be a real, symmetric, positive definite d d matrix Then e x Ax e πix ξ dx = π d (det A) e π ξ A ξ Proof Since A is real, symmetric, and positive definite, it is diagonalizable Explicitly, there is an orthogonal matrix O and a diagonal matrix D = diag(λ,, λ d ), λ j >, so that A = O T DO Let y = Ox and η = Oξ Then and Thus, x Ax = x O T DOx = Ox DOx = y Dy = e x Ax e πix ξ dx = x ξ = O T y O T η = y η e d j= λ jy j e πiy η dy = Computing each of these one-variable integrals, we have: e λy πiyη πiη λ(y dy = e λ ) π η λ dy = e π η λ So R R e x Ax e πix ξ dx = d j= d j= R R d λ j yj j= e λ jy j πiy jη j dy j e λy dy = e π η λ λ π e π η j λ j λ j π d = π (λ λ d ) d e j= π η λ j j = π d (det A) e π η D η The result follows from the fact that η D η = Oξ D Oξ = ξ A ξ 3

15 Corollary 7 The function e π x is an eigenvalue of the Fourier transform with eigenvalue Explicitly, e π x = e π ξ Proof This follows from the previous lemma with A = πi d Indeed, ( e π x e πix ξ dx = π d π d) e π ξ π ξ = e π ξ Above, we computed that the Fourier transform of a Schwartz function is again Schwartz In fact, the Fourier transform is an isometry on Schwartz space and extends to a unitary operator on L The next few results summarize these facts Theorem 8 (Fourier Inversion) If f S(R d ), then ( ˆf)ˆ(x) = f( x) Equivalently, Proof For ε >, let f(x) = ( ˆf)ˆ( x) =: ( ˆf)ˇ(x) =: F ( ˆf)(x) I ε (x) := e πε ξ e πix ξ ˆf(ξ) dξ By the dominated convergence theorem, as ε, I ε (x) e πix ξ ˆf(ξ) dξ Thus, to prove the theorem, it suffices to show that I ε (x) f(x) as ε We have I ε (x) = e πε ξ e πix ξ ˆf(ξ) dξ = e πε ξ e πix ξ e πiy ξ f(y) dy dξ = f(y) e πε ξ e πi(y x) ξ dξ dy Note that e πε ξ e πi(y x) ξ dξ = (e πε ξ )ˆ(y ( d x) = π (πε ) d) = ε d y x π e ε e π (y x) πε (y x) Consequently, I ε (x) = f(y) ε d y x π e ε dy = (f φ ε )(x) where φ ε (x) = ε d φ ( x ε ) with φ(x) = e π x It is a standard result that {φ ε } is an approximation to the identity Thus, as ε, I ε (x) = (f φ ε )(x) f(x) Lemma 9 If f, g S(R d ), then ˆf(ξ)g(ξ) dξ = f(x)ĝ(x) dx In particular, ˆf(ξ)ĝ(ξ) dx = f(x)g(x) dx, so that ˆf = f L L Hence, the Fourier transform is an isometry on S(Rd ) Proof Computing, we have ˆf(ξ)g(ξ) dξ = e πix ξ f(x) dx g(ξ) dξ = = f(x)ĝ(x) dx For the in particular statement, let h = ĝ Then f(x) e πix ξ g(ξ) dξ dx ĥ(x) = ˆĝ = (ĝ)ˆ( x) = g(x) 4

16 Theorem (Plancharel) The Fourier transform extends from an operator on S(R d ) to a unitary operator on L (R d ) Proof Fix f L (R d ) Let {f n } S(R d L ) such that f n f As the Fourier transform is an isometry on S(R d ), we have f ˆ n f ˆ L m = f n f m L This shows that { f ˆ n } is Cauchy in L Let ˆf := lim n fn ˆ, the limit being taken in the L -sense We claim that ˆf is well-defined Let {f n } and {g n } be two sequences of Schwartz functions such that f n, g n f Define L { fk if n = k h n := if n = k g k L Then h n f By the argument above, { h ˆ n } converges in L, which implies by the uniqueness of limits that lim n fn ˆ = lim n gˆ n Next we claim that f L (R d ), then ˆf = f L L, so the Fourier transform is an isometry on L (R d ) Because the norm function is continuous in the L topology, we have ˆf = lim ˆ L f L n = lim f n n n L = lim f n = f n L L Before completing the proof, we remark that in infinite dimensions, an isometry is not necessarily a unitary operator For example, let T : l (N) l (N) be the right-shift operator given by T (a, a, a, ) := (, a, a, ) Then T (a, a, a, ) := (a, a, a 3, ) Clearly T is an isometry, but T T I However, to prove that the isometry F is unitary, it suffices to prove that F is surjective We claim that im F is closed in L From this claim, since S(R d ) im F and S(R d ) is dense in L (R d ), the proof is complete To demonstrate this claim, let g im F Then there is a sequence {f n } of L functions so that f ˆ L n g As the Fourier transform is an isometry on L, this implies that {f n } converges in L Let f := lim n f n Then f ˆ n ˆf = f n f L L so that g = f ˆ n To summarize our progress thus far, we have defined the Fourier transform on L (R d ), investigated its properties on Schwartz functions, and used the class of Schwartz functions to define the Fourier transform on L (R d ) Next, we used Riesz-Thorin interpolation (see Chapter ) to define the Fourier transform on L p (R d ) for p Even more, we will show that the Fourier transform cannot be defined as a bounded operator on L p (R d ) spaces for p > Theorem (Hausdorff-Young) If f S(R d ), then ˆf f L p L p for p, where p + p = Proof We have the estimates ˆf f L L and ˆf L = f L Applying the Riesz- Thorin interpolation theorem with p =, p =, q =, and q =, it follows that the Fourier transform is bounded from L p θ L q θ, where = θ + θ p θ 5 and q θ = θ

17 so that p θ = θ, hence p θ + q θ = Since p θ = θ, as θ [, ] we have p θ [, ] as desired Conversely, we have the following This is one of our first examples of the power of scaling arguments Theorem If ˆf f L q L p for all f S(Rd ) for some p, q, then p and q = p Proof Fix f S(R d ) For λ >, let f λ (x) := f(x/λ) Then ( x ˆf λ (ξ) = e πix ξ f dx = λ λ) d ˆf(λξ) We have ˆf λ L q f λ L p, which implies λ d λ d q ˆf λ d p f L q L p Thus, λ d λ d q λ d p, so that λ d q λ d p and hence λ q λ p for all < λ < Letting λ gives q p, and letting λ gives q p Thus, so q = p Next, we show p It suffices to show p p Let ϕ Cc (R d ) with supp ϕ B(, /) Let ϕ k (x) = e πix λke ϕ(x ke ) Using properties and of the Fourier transform, it follows that ˆϕ k (ξ) = e πiξ ke ˆϕ(ξ λke ) Let f = N k= ϕ k Since the supports of ϕ k are disjoint, it follows that f L p N p Next, we compute, expanding out ˆf as follows: L p N N (ξ)l e πiξ ke ˆϕ(ξ λke )χ B(λke,λ/)(ξ) + e πiξ ke ˆϕ(ξ λke )χ C B(λke,λ/) k= k= p N e πiξ ke ˆϕ(ξ λke )χ B(λke,λ/)(ξ) N p k= L p N e πiξ ke ˆϕ(ξ λke )χ C B(λke,λ/) (ξ) L p k= N ˆϕ(ξ)χ C B(,λ/) (ξ) L p Because ˆϕ is Schwartz, ˆϕ(ξ)χ C B(,λ/) (ξ) L p ( ξ >λ/ ) p dξ λ d p ξ d which as λ Thus, ˆf f L p L p if and only if N p N p for all N, which is true if and only if p p The Fourier Transform on T d The primary focus of this text is harmonic analysis on Euclidean space, and consequently the Fourier transform on R d is the most important tool for our purposes One can also define the Fourier transform on the torus T d := R d /Z d, equivalently, for periodic functions In the interest of studying dispersive partial differential equations on the torus (see Chapter 9), we include a brief discussion of the Fourier transform in this setting 6

18 Definition 3 For a C -function f : T d C, we define its Fourier transform ˆf : Z d C by ˆf(k) = e πik x f(x) dx T d Then ˆf(k) = f, e k where e k (x) := e πik x The characters e k are orthonormal, since e k, e l = e πi(k l) x = δ kl T d We quickly establish some familiar properties of the Fourier transform Proposition 4 (Bessel s Inequality) For f C (T d ), f, e k f L (T d ) k Z d Proof Let S Z d be a finite set Then f f, e k e k = f f, e k e k, f f, e l e l k S L (T d ) k S l S = f L (T d ) f, e k e k, f + f, e k e k, f, e l e l k S k S l S = f L (T d ) f, e k + f, e k k S where the last equality follows from orthogonality of the e k s Simplifying and rearranging gives f, e k f L (R d ) k S for every finite set S, hence the desired inequality This fact shows that if f C (T d ), then k Z d f, e k e k L (T d ) Unsurprisingly, these two objects can be identified Proposition 5 (Fourier inversion) If f C (T d ), then f = k Z d f, e k e k = k Z d ˆf(k)e πik x Proof Suppose for the sake of contradiction that f k Z d f, e k e k Then by Stone- Weierstrass, there is a trigonometric polynomial g such that f k Z d f, e k e k, g For any character e l, f k Zd f, e k e k, e l = f, e l f, e l = This is a contradiction 7

19 The following result is unique to the periodic case Lemma 6 (Poisson summation) Let ϕ S(R) Then n Z ϕ(x + n) = n Z for all x R In particular, n Z ϕ(n) = n Z ˆϕ(n) ˆϕ(n)e πinx Proof Define F (x) = n Z ϕ(x + n) and F (x) = n Z ˆϕ(n)eπinx Note that both F and F are -periodic in x Also, since ϕ is Schwartz, each sum converges absolutely in n and converges uniformly in x on compact sets Therefore, F and F are both continuous functions on T As such, to give the desired equality it suffices to prove that F and F have the same Fourier coefficients The definition of F immediately gives ˆF (k) = ˆϕ(k) for all k Z Next, we compute: ˆF (k) = n Z n+ ϕ(x + n)e πikx dx = n Z = ϕ(y)e πiky dy n Z n = ϕ(y)e πiky dy R = ˆϕ(k) n+ The in particular statement follows by considering x = n ϕ(y)e πik(y n) dy 8

20 Chapter 3 Lorentz Spaces and Real Interpolation In this chapter, we introduce a new class of functions space, namely, Lorentz spaces These spaces measure the size of functions in a more refined manner than the classical L p spaces After developing some of the basic theory of Lorentz spaces, we then state and prove the Marcinkiewicz interpolation theorem in its most general setting Some references for Lorentz spaces are [5] and [6] 3 Lorentz Spaces Before defining Lorentz spaces in their full generality, we first consider a simpler and more familiar class of functions Definition 3 For p and f : R d C measurable, define { } f L p := sup λ x R d p : f(x) > λ (3) weak λ> The weak-l p space, written L p weak (Rd ), is the family of measurable functions f for which f L p is finite weak The quantity defined in (3) contains an asterisk as a superscript because is not a norm However, it is a quasinorm Recall that a quasinorm satisfies the same properties as a norm, except that the triangle inequality is replaced by the quasi triangle inequality: f + g C ( f + g ) for some universal constant C > To clarify the definition of the weak-l p (quasi)norm, consider a function f L p (R d ) Using the fundamental theorem of calculus, we can write f(x) f p L p (R d ) R = f(x) p dx = pλ p dλ dx d By Tonelli s theorem, f p L p (R d ) = pλ p dx dλ = { x R d : f(x) >λ } ( { = p λ x R d So f L p (R d ) = p p λ { x R d R d } ) p p dλ : f(x) > λ λ pλ p { x R d : f(x) > λ } dλ : f(x) > λ } p Lp ((, ), dλ λ ) Using the convention p =, we can then write f L p weak (Rd ) = p { } λ x R d p : f(x) > λ L ((, ), dλ λ ) In this way, weak-l p spaces are a clear generalization of L p spaces 9

21 Example 3 Let f(x) = x d p Then f L p weak (Rd ), but f / L p (R d ) To see this, first note that R d f(x) p dx = R d x d dx r d rd dr r dr This latter integral does not converge, hence the L p -norm of f is not finite, so f / L p (R d ) On the other hand, { x R d } p : f(x) > λ { = x R d : x d p > λ } p { = x R d : x < ( ) p d λ The set being measured is a ball of radius (/λ) p/d The volume of a such a ball scales according to ( (/λ) p/d) d = (/λ) p Taking pth roots as above then gives { x R d } p : f(x) > λ λ } p Hence, so that f L p weak (Rd ) f L p weak (Rd ) sup λ λ> λ = < Lorentz spaces are an even further generalization of L p and weak-l p spaces Definition 33 For p and q, the Lorentz space L p,q (R d ) is the space of measurable functions f : R d C for which the quantity f L p,q (R d ) := p { } q λ x R d p : f(x) > λ (3) Lq ((, ), dλ λ ) is finite By our previous computation, L p (R d ) = L p,p (R d ), and L p weak (Rd ) = L p, (R d ) Also, as with the weak-l p norm, the L p,q -norm defined in (3) is not actually a norm Lemma 34 The quantity L p,q (R d ) is a quasinorm Proof First, note that if f L p,q (R d ) =, then { } λ x R d p : f(x) > λ = Lq ((, ), dλ λ ) which implies that, for almost all λ >, { x R d : f(x) > λ } = It follows that f(x) = almost everywhere Thus, f = Next, let a C Then af L p,q (R d ) = p { } q λ x R d p : af(x) > λ Lq ((, ), dλ λ ) { = a p λ q a x R d : f(x) > λ } p a Lq ((, ), dλ λ ) By making the change of variables η = λ/a, we have af L p,q (R d ) = a f L p,q (R d )

22 It remains to prove that quasi-triangle inequality To do this, we invoke the fact that for < α <, the map x x α is concave It is a standard fact that concave functions are subadditive, so that (x + y) α x α + y α With this in mind, we compute: f + g L p,q (R d ) = p { } q λ x R d p : f(x) + g(x) > λ Lq ((, ), dλ λ ) Observe that { } x R d : f(x) + g(x) > λ { x R d : f(x) > λ } { x R d : g(x) > λ } Thus, ({ f + g L p,q (R d ) p q λ x : f(x) > λ } { + x : g(x) > λ }) p Lq ((, ), dλ λ ) By subbaditivity of the concave map x x p, we have ( { f + g L p,q (R d ) p q λ x : f(x) > λ } p + { x : g(x) > λ } ) p Lq ((, ), dλ λ ) Distributing the λ, factoring out a, and appliyng the usual L q -Minkowski inequality then yields: { f + g L p,q (R d ) p λ q x : f(x) > λ } { p λ + x : g(x) > λ } p Lq ((, ), dλ λ ) { p q λ x : f(x) > λ } p Lq ((, ), dλ λ ) { λ + x : g(x) > λ } p Lq ((, ), dλ λ ) ( ) = f L p,q (R d ) + g L p,q (R d ) For < p < and q, we will show that the quasinorm L p,q (R d ) is in fact equivalent to a norm For p =, q, the quasinorm is not equivalent to a norm However, in this case, there does exist a metric that generates the same topology Thus, in either of these cases, L p,q (R d ) is a complete metric space As another quick remark, we note that if g f, then for any λ >, { } { } x R d : g(x) > λ x R d : f(x) > λ This implies that g L p,q (R d ) f L p,q (R d ) It is natural to ask whether different Lorentz spaces have any simple embedding properties The next result gives a partial answer to this question Proposition 35 Let p <, q, and f L p,q (R d ) Write f = m Z f m, where f m (x) := f(x) χ { x R d : m f(x) m+ } Then f L p,q (R d ) p,q f m L p (R d ) l q m (Z) In particular, L p,q (R d ) L p,q (R d ) whenever q q

23 Proof By the preceding remark, note that m χ { x : m f m+ } f m m Z L p,q (R d ) m Z L p,q (R d ) m+ χ { x : m f m+ } m Z L p,q (R d ) Therefore, it suffices to prove the proposition for a function of the form f(x) = m Z m χ Em where {E m } is a pairwise disjoint collection of sets In this case, we need to show that f L p,q (R d ) p,q m E m p We compute: ( ) q f L p,q (R d ) = p = p m Z λ q { x : f(x) > λ } q p m l q m (Z) dλ λ m+ λ q { x : f(x) > λ } q p dλ λ Next, observe that for λ [ m+, m ), { x : f(x) > λ } = n m E M As the E m s are disjoint, we then have q ( ) q m f L p,q (R d ) = p λ q p E m dλ m Z m+ λ n m This removes the dependence on λ in the set inside the integral Because it follows that Thus, p m m+ λ q dλ λ = p q (mq (m+)q ) = p q ( q ) mq ( ) q f L p,q (R d ) p,q mq m Z f L p,q (R d ) p,q m E n n m E n n m p q p l q m (Z) = m E m n m p q l q m(z) m p,q ( E m ) p l q m (Z) This gives half of the p,q relation that we need To get the other inequality, we compute as follows First, invoking concavity of fractional powers as before, f L p,q (R d ) p,q m p E n n m m E n p n m l q m (Z) l q m (Z) Next, making the change of variables n = m + k, = m E m+k p = k m+k E m+k p k m+k E m+k p k l q k m (Z) l q k m (Z) = k m E m p k = m E m p l q m (Z) l q m (Z) l q m (Z)

24 so that f L p,q (R d ) p,q m E m p Therefore, l q m (Z) f L p,q (R d ) p,q m E m p l q m (Z) as desired The in particular statement follows from the fact that if q q, then l q (Z) l q (Z) As noted in the proof of the above proposition, the monotonicty of the quasinorm allows for the following reduction in many computations involving Lorentz spaces If f L p,q (R d ) is real-valued and nonnegative, then we can assume without loss of generality that f = m Z m χ Em, where the sets E m are pairwise disjoint In this case, f L p,q (R d ) = m E m p l q m (Z) We can further decompose a general function into its real and imaginary parts, and then into their corresponding positive and negative parts, to apply this reduction 3 Duality in Lorentz Spaces One of the most important principles in L p space theory is that of duality The analogous fact for Lorentz spaces is given by the following proposition Proposition 36 Let < p < and q, and let p and q denote the respective Holder conjugates Then f L p,q (R d ) p,q sup f(x)g(x) dx (33) R d g L p,q (R d ) Proof As the quasinorm is positively homogeneous, we can scale the function f appropriately and assume without loss of generality that f L p,q (R d ) = Also, as remarked previously, we can assume without loss of generality that f and g are real-valued and nonnegative, hence we can take f = n Z n χ Fn for disjoint sets F n and g = m Z m χ Em for disjoint sets E m In this case, we have ( q = f L p,q (R )) = n d F n p q = nq F n q p l q n(z) n Z We decompose the above sum as follows Let Z denote the set of dyadic numbers Then nq F n q p = nq N q p = N q p nq N q p q n N Z N Z N Z n Z n: F n N n: F n N n: F n N The latter equivalence is a straightforward exercise in dealing with dyadic sums, together with the fact that the l q -norm is bounded by the ell -norm Thus, N Z q n N p n: F n N Similarly, since g L p,q, the same computation gives (R d ) q m E m p M Z m: E m M 3

25 With these computations and our reductions in mind, we demonstrate the equivalence in (33) We first show that the right hand side is bounded by the left f(x)g(x) dx = n m χ Fn Em (x) dx = n m F n E m N,M Z n: F n N n,m Z N,M Z n: F n N m: E m M n m: E m M n,m Z m min{n, M} By our above computations, we eventually want to introduce the quantities N p and E m p We force them in the sum as follows: { } n N p m E m N M p min, N,M Z n: F n N m: E m M N p M p N p M p = { ( ) ( ) n F n p m E m p N p min } M p, M N Next, we use Holder s inequality on the above sum, viewing the n-sum as living in l q N and the m-sum as living in l q M This yields N,M Z n: F n N n F n p min N,M Z q m: E m M { ( ) N M M E m p p, ( M N q min ) } q p { ( ) N M p, ( M N ) } q p In the first term, freeze N and sum over M Because p > and hence p <, we have { ( ) ( ) N p } M p, = ( ) M p ( ) N p + + M N N M M N M>N M Z min Similarly, in the second term we freeze M and sum over N This gives f(x)g(x) dx N Z n: F n N q n F n p This gives one inequality Next, we consider the converse inequality n n χ Fn with f L p,q (R d ) Let g = n q M Z m: E m M M E m p q q Again we can take f to be of the form ( ) n F n q p Fn p χ Fn = n(q ) F n q p χ Fn We make two claims, from which the desired inequality clearly follows: f(x)g(x) dx ; g L p,q (R d ) n 4

26 The first claim follows from f(x)g(x) dx = n(q ) n F n q p χ Fn = n n = n F n p q l q n(z) nq F n q p To see the second claim, write where S n = g N Z Nχ n Sn Fn { } n : n(q ) F n q p N Note that the S n s are disjoint sets Then ( ) q ( ) q g L p,q N Z p N q F (R d ) n n S n N Z N q ( N Z N q n S n n S n N ) q (N n(q )) p p q p p q p q p p nq q p where in the last equivalence we have used the fact about dyadic sums and l q norms from above and the fact that q = q q It remains to compute and simplify all of the Holder exponents Doing this gives ( ) q g L p,q ) ( n(q ) F (R d ) n q p q q p p q nq p q p n Z F n q p nq n Z This proves both claims, the the converse inequality, and hence completes the proof The right hand side of (33) defines a norm which is equivalent to the quasinorm Lorentz quasinorm With respect to this norm, L p,q (R d ) for < p <, q, is a Banach space The proof of completeness is the same as the proof of completeness in L p spaces The dual space in this case is L p,q (R d ) As remarked above, if p = and q, there is no norm which is equivalent to the quasinorm The following example demonstrates this explicitly Example 37 Consider the case p =, q =, d = saw in a previous example that x n = L, (R) x n N n= x n N L, (R) We claim that N n= N log N We have N x n n= x n L, (R) L, (R) = sup λ λ> { x R d : L weak N n= Let f(x) = N n= x n We (R) This implies that } x n > λ 5

27 Fix x [, N] Note that, for n x, x n < n, so that x n > n For n < x, we can consider each finite number and rearrange them so that their sum is comparable to n<x n Thus, for x [, N] we have f(x) N log N Choose λ = C log N for some appropriate { constant C, we then have x R d n= n : } N n= x n > C log N [, N] Therefore, f L, (R) C log N { x R d : N n= } x n > C log N CN log N as claimed Now, suppose that the quasinorm was equivalent to a norm Then by the triangle inequality, N N N log N x n x n N, which is a contradiction n= L, (R) n= L, (R) 33 The Marcinkiewicz Interpolation Theorem Finally, we arrive at the Marcinkiewicz interpolation theorem Before stating and proving the theorem, we recall the notion of sublinearity and introduce some language that will be used throughout the remainder of this text Definition 38 A mapping T on a class of measurable functions is sublinear if T (cf) c T (f) and T (f + g) T (f) + T (g) for all c C and f, g in the support of T Any linear operator is obviously sublinear A more interesting class of examples, and perhaps the primary motivation for defining sublinearity, is the following Example 39 Given a family of linear mappings {T t } t I, the mapping defined by T (f)(x) := T t (f)(x) L q t is sublinear When q =, operators of this form are called maximal functions When q =, the term square function is used Definition 3 Let p, q A mapping of functions T is of (strong) type (p, q) if T f L q (R d ) T f L p (R d ) That is, T is of type (p, q) if it is bounded as a mapping from L p (R d ) L q (R d ) Similarly, for q <, T is of weak type (p, q) if T f L q, (R d ) T f L p (R d ), and is of restricted weak type (p, q) if T χ F L q, (R d ) T F p for all finite measure sets F Note that type (p, q) implies weak type (p, q), which implies restricted weak type (p, q) An alternative characterization of restricted weak type operators is given by the following proposition Proposition 3 A mapping T is of restricted weak type (p, q) if and only if T χ F χ E dx E p F q for all finite measure sets E, F 6

28 Proof First, suppose that T is of restricted type (p, q) Then T χ F L q, (R d ) T F p By the duality of Lorentz quasinorms, this implies T χ F χ E dx T χ F L q, (R d ) χ E L q, (R d ) Note that Hence, ( ) q χ E = q L q, (R d ) λ q { x : χ E (x) > λ } dλ λ E T χ F χ E dx E p F q Conversely, suppose that the above inequality holds for all finite measure sets E and F We again invoke duality to write T χ F L q, (R d ) sup T χ F g dx sup T χ F g dx g L q, g L q, Consider g := m m E m for an appropriate disjoint collection of sets E m Then T χ F g dx m m F p E q F p g L q, (R d ) F p as desired The formulation of the Marcinkiewicz interpolation theorem that we state below is due to Hunt (see [5]) The classical statement of the theorem is given as Corollary 34 below Theorem 3 (Marcinkiewicz interpolation) Fix p, p, q, q such that p p and q q Let T be a sublinear operator of restricted weak type (p, q ) and of restricted weak type (p, q ) Then for any r and < θ <, T f L q θ,r f L p θ,r where p θ = θ p + θ p and q θ = θ q + θ q Before proving the theorem, we make a few remarks First, if p θ q θ, taking r = q θ gives T f L q θ f L p θ,q θ f L p θ so that T is of strong type (p θ, q θ ) The requirement that p θ q θ is essential for the strong type conclusion, as evidenced by the following example Example 33 Define T (f)(x) := f(x) Then T is bounded as an operator from L p (R) x L p p+, (R) for any p, but is not bounded as an operator from L p (R) L p p+ (R) To prove this, we invoke Holder s inequality in Lorentz spaces: for p, p, p < and q, q, q satisfying p = p + p and q = q + q, we have fg Lp,q f L p,q g L p,q By this inequality, we have: T f L p+ p f, L (R) p, (R) x f L, L (R) p (R) 7

29 ( ) On the other hand, consider the function f(x) = x p log x + p+ p x We first claim that f L p (R) Indeed, we can write ( f p L p (R) = x log x + x ( ( = x log x + x ) p+ dx ) p+ ( dx + x log x + x ) p+ dx ) Consider the second integral Since p, we have x ( log x + x ) p+ dx = < log ( x log x + ) x u du dx x(log x) dx By symmetry, the singularity at behaves similarly Hence, f L p (R) However, it is not difficult to show that the quantity is not finite Proof T f p p+ L p+ p We now prove Theorem 3 = x p+ Corollary 34 (Marcinkiewicz interpolation) ( log x + ) dx x 8

30 Chapter 4 Maximal Functions The notion of averaging is one which is ubiquitous in analysis For example, convolution can be viewed as a sort of averaging against a given function Convolving a function with a smooth function then smooths the original function, a fact which is immensely useful In this chapter, we study averaging in more depth via maximal functions We begin by recalling the classical Hardy-Littlewood maximal inequality, and spend the rest of the chapter developing generalizations 4 The Hardy-Littlewood Maximal Function The reader may already be familiar with the Hardy-Littlewood maximal function, defined as M(f)(x) := sup f(y) dy (4) r> B r (x) B r(x) The quantity B r(x) B r(x) f(y) dy represents the average value of the function on a ball centered at x Thus, the maximal function measures the largest possible average values for f on various balls The following theorem is standard Theorem 4 (Hardy-Littlewood maximal inequality) Let M(f) denote the Hardy-Littlewood maximal function as in (4) Then For f L p (R d ), p, M(f) is finite almost everywhere The operator M is of weak type (, ), and of strong type (p, p) for < p Later in this chapter, we will prove a generalization of Theorem 4, and so we defer a proof until then Unraveling statement of Theorem 4, we see that M being of weak type (, ) means and hence Mf L, (R d ) f L (R d ) { x : (Mf)(x) > λ } C λ f L (R d ) for all λ > This is the usual formulation of the Hardy-Littlewood maximal inequality We remark that the p > condition in statement is necessary, because M is not of strong type (, ) To see this, let φ Cc (R d ) with supp φ B / () Fix x > If r < x /, then B r(x) φ(y) dy = If r > x + /, then φ(y) dy = B r (x) B r(x) φ(y) dy B r (x) B x +/ (x) 9 φ(y) dy B x +/ (x) B x +/ (x)

31 Thus, (M φ)(x) = sup φ(y) dy x / r x +/ B r (x) B r(x) x d But x d is not in L (R d ) We close this section with a classic application of the Hardy-Littlewood maximal inequality Theorem 4 (Lebesgue differentiation) Let f L loc (Rd ) Then for almost every x R d Proof ADD IN lim r f(y) dy = f(x) B r (x) B r(x) More details on the classical Hardy-Littlewood maximal function can be found in [9] 4 A p Weights and the Weighted Maximal Inequality The main theorem of this section, which is a generalization of Theorem 4, is the following Theorem 43 (Weighted maximal inequality) Let ω : R d [, ) be locally integrable Associate to ω the measure defined by ω(e) := E ω(y) dy Then and for < p Explicitly, and M : L (M(ω) dx) L, (ω dx) M : L p (M(ω) dx) L p (ω dx) ω ({ x : (Mω)(x) > λ }) λ R d f(x) (Mω)(x) dx (Mf)(x) p ω(x) dx f(x) p (Mω)(x) dx R d R d Note that if ω, then M(ω), and we recover the classical Hardy-Littlewood maximal inequality in Theorem 4 Before proving the weighted maximal inequality, we establish some preliminary facts on A p weights Definition 44 A function ω satisfies the A condition, written ω A, if M(ω) ω almost everywhere Note that if ω A, then Theorem 43 gives M : L (ω dx) L, (ω dx) and M : L p (ω dx) L p (ω dx) for < p Lemma 45 The following are equivalent: ω A ; ω(b) 3 B B ω(x) for almost all x B and all balls B; B f(y) dy ω(b) B f(y)ω(y) dx for all f and all balls B 3

32 Proof We first show () () Fix a ball B of radius r and let x B Then ω(x) (Mω)(x) = sup ω(y) dy ω(y) dy r> B r (x) B r (x) B r(x) B r (x) Because B r (x) B (drawing a picture makes this clearer) and because B r (x) d B, ω(x) ω(y) dy = ω(b ) B B B as desired The implication () () follows immediately from the definition of the maximal function Next, we show () (3) Fix a ball B and f Then ω(b) B f(y)ω(y) dy ω(b) B f(y) ω(b) B dy = B B f(y) dy Finally (3) () Fix a ball B and let x B be a Lebesgue point of B (ie, a point for which the conclusion of the Lebesgue differentiation theorem holds) Choose r << so that B r (x) B Let f = χ Br(x) Then by (3), B B χ Br(x)(y) dy ω(b) Since x is a Lebesgue point, as r ω(b) B B χ Br(x)(y)ω(y) dx B r(x) B ω(y) dy ω(x) B r (x) B r(x) ω(y) dy ω(b) B r(x) Definition 46 A function ω : R d [, ) satisfies the A p condition, written ω A p, if sup B ω(b) B ( ) p ω(y) p p p dy, B B where the supremum is taken over all open balls in R d We note for convenience that the above condition is equivalent to ( ) p ω(b) sup B B p ω(y) p p p dy B Also, if < p < q <, then A p A q This follows from Holder s inequality The importance of the class of A p weights is the following theorem, which characterizes when the maximal function is a bounded operator on a weighted L p space Theorem 47 Fix < p < Then M : L p (ω dx) L p (ω dx) if and only if ω A p Proof Here, we only explicitly prove one direction ( ) of this statement For the other direction, we will provide an outline of the argument and leave the details as an exercise for the reader Suppose that M : L p (ω dx) L p (ω dx) Fix a ball B and let f = (ω + ε) p p χ B For x B, (Mf)(x) = sup r> B B(x, r) B B(x,r) (ω(y) + ε) p p dy =: λ (ω(y) + ε) p p χ B (y) dy (ω(y) + ε) p p dy B B 3

33 Here our definition of λ includes any implicit constants from the above inequality Then ({ ω(b) ω x : (Mf)(x) > λ }) λ p f(y) p ω(y) dy where the final inequality follows from the fact that M : L p (M(ω) dx) L p (ω dx), and hence M : L p (M(ω) dx) L p, (ω dx) Plugging in our definition of λ gives Thus, ω(b) B p ( B p ( = B p ( = B p ( Letting ε gives ω A p B B B B ω(b) B p (ω(y) + ε) p p (ω(y) + ε) p p (ω(y) + ε) p p ) p dy (ω(y) + ε) p ω(y) dy B ) p dy (ω(y) + ε) p + dy B ) p dy (ω(y) + ε) p p dy) (p ) ( (ω(y) + ε) p p dy B (ω(y) + ε) p p dy B ) p p Now, we sketch the proof of the converse direction Suppose that ω A p Define the following weighted maximal function: (M ω f)(x) := sup f(y) ω(y) dy r> ω(b(x, r)) B(x,r) It can be shown (via techniques from this section) that M ω : L (ω dx) L, (ω dx), and moreover (Mf) p M ω (f p ) Fix f L p (ω dx) Then ( Mf L p, (ω dx)) p = sup λ> Since (Mf) p (x) M ω ( f p )(x), λ p ω ({ x : (Mf)(x) > λ }) = sup λ p ω ({ x : (Mf) p (x) > λ p }) λ> = sup η ω ({ x : (Mf) p (x) > η }) η> ω ({ x : (Mf) p (x) > η }) ω ({ x : M ω ( f p )(x) > η }) Thus, ( p Mf L p, (ω dx) sup η ω ({ x : M ω ( f p )(x) > η }) = M ω ( f p ) L, (ω dx) η> Since M ω : L (ω dx) L, (ω dx), M ω ( f p ) L, (ω dx) f p L (ω dx) = f(x) p ω(x) dx = f p L p (ω dx) Therefore, Mf L p, (ω dx f L p (ω dx) so that M : L p (ω dx) L p, (ω dx) 3

34 Theorem 48 Fix p, and let dµ be a nonnegative Borel measure If M : L p (dµ) L p, (dµ), then dµ = ω dx for some ω A p Proof If we can prove that dµ is absolutely continuous, then in light of the previous proof, we are done Decompose dµ = ω(x) dx + dν where dν is the singular part of dµ, ie, there exists a compact set K with K = but ν(k) > Define U n = { x : d(x, K) < /n } and f n = χ Un\K As U n \ K U n+ \ K, and (U n \ K) =, it follows that f n pointwise We claim that dµ is finite on compact sets To see this, pick a measurable set E with < µ(e) < ; this possible since µ is nontrivial We may assume that E is compact by inner regularity Then (Mχ E )(x) = sup χ E (y) dy r> B(x, r) B(x,r) Let r = d(x, E) + diam E Then (Mχ E )(x) E r d Since E is compact, if we restrict x to a compact set then d(x, E) is bounded below and above Thus, Mχ E E,F uniformly for x F with F compact So suppose for the sake of contradiction that there is a compact set F with µ(f ) = Then = µ(f ) µ ({ x : Mχ E (x) E,F }) E,F dµ = µ(e) < which is a contradiction With this claim proven, next, note that f n p dµ by the dominated convergence theorem Fix x K Then since B(x, /n) U n and since K =, (Mf)(x) = sup r> B(x, r) B(x,r) = χ B(x, /n) B(x,/n) (y) dy K C = R d K C χ B(x,/n) (y) dy = χ Un\K(y) dy E B(x, /n) B(x,/n) So x { x : (Mf n )(x) > / } Thus, K { x : (Mf n )(x) > / } So < ν(k) < µ(k) µ ({ x : (Mf n )(x) > / }) f n p dµ χ Un\K(y) dy which is a contradiction Thus, dµ = ω(x) dx, and by the previous theorem, ω A p Lemma 49 We have ω A p if and only if ( ) p f(y) dy f(y) p ω(y) dy B B ω(b) B uniformly for all f and all balls B Proof First, suppose that ω A p Then sup B ω(b) B p ( B ) p w(y) p p p 33