Lecture Notes on Introduction to Harmonic Analysis

Home , Dirac measure, Vitali covering lemma

} nrdcint amncAnalysis Harmonic to Introduction nttt fMteais CAS Mathematics, of Institute ETR NOTES LECTURE hncu Hao Chengchun 2016 }

Contents

1 The Fourier Transform and Tempered Distributions ...... 1 1.1 The L1 theory of the Fourier transform...... 1 1.2 The L2 theory and the Plancherel theorem...... 12 1.3 Schwartz spaces...... 13 1.4 The class of tempered distributions...... 17 1.5 *Characterization of operators commuting with translations...... 22

2 Interpolation of Operators ...... 27 2.1 Riesz-Thorin’s and Stein’s interpolation theorems...... 27 2.2 The distribution function and weak Lp spaces...... 33 2.3 The decreasing rearrangement and Lorentz spaces...... 36 2.4 Marcinkiewicz’ interpolation theorem...... 41

3 The Maximal Function and Calderón-Zygmund Decomposition ...... 47 3.1 Two covering lemmas...... 47 3.2 Hardy-Littlewood maximal function...... 49 3.3 Calderón-Zygmund decomposition...... 57

4 Singular Integrals ...... 63 4.1 Harmonic functions and Poisson equation...... 63 4.2 Poisson kernel and Hilbert transform...... 67 4.3 The Calderón-Zygmund theorem...... 77 4.4 Truncated integrals...... 79 4.5 Singular integral operators commuted with dilations...... 82 4.6 The maximal singular integral operator...... 88 4.7 *Vector-valued analogues...... 92

5 Riesz Transforms and Spherical Harmonics...... 95 5.1 The Riesz transforms...... 95 5.2 Spherical harmonics and higher Riesz transforms...... 99 5.3 Equivalence between two classes of transforms...... 106

6 The Littlewood-Paley g-function and Multipliers ...... 111 6.1 The Littlewood-Paley g-function...... 111 6.2 Fourier multipliers on Lp ...... 121

v - vi - Contents

6.3 The partial sums operators...... 128 6.4 The dyadic decomposition...... 130 6.5 The Marcinkiewicz multiplier theorem...... 136

7 Sobolev and Hölder Spaces ...... 141 7.1 Riesz potentials and fractional integrals...... 141 7.2 Bessel potentials...... 145 7.3 Sobolev spaces...... 149 References...... 153

Index ...... 155 Chapter 1 The Fourier Transform and Tempered Distributions

In this chapter, we introduce the Fourier transform and study its more elementary properties, and extend the deﬁnition to the space of tempered distributions. We also give some characteri- zations of operators commuting with translations.

1.1 The L1 theory of the Fourier transform

We begin by introducing some notation that will be used throughout this work. Rn denotes n-dimensional real Euclidean space. We consistently write x = (x1,x2, ,xn), ξ = n n ··· (ξ1,ξ2, ,ξn), for the elements of R . The inner product of x, ξ R is the number ···n ··· n ∈ x ξ = ∑ x jξ j, the norm of x R is the nonnegative number x = √x x. Furthermore, · j=1 ∈ | | · dx = dx dx dx denotes the element of ordinary Lebesgue measure. 1 2 ··· n We will deal with various spaces of functions defined on Rn. The simplest of these p p n are the L = L (R ) spaces, 1 6 p < ∞, of all measurable functions f such that f p = R p 1/p p ∞ n k k ( Rn f (x) dx) < ∞. The number f p is called the L norm of f . The space L (R ) consists | | k k n ∞ n of all essentially bounded functions on R and, for f L (R ), we let f ∞ be the essential n n ∈ k k supremum of f (x) , x R . Often, the space C0(R ) of all continuous functions vanishing at | | ∈ infinity, with the L∞ norm just described, arises more naturally than L∞ = L∞(Rn). Unless otherwise specified, all functions are assumed to be complex valued; it will be assumed, throughout the note, that all functions are (Borel) measurable. In addition to the vector-space operations, L1(Rn) is endowed with a “multiplication” making this space a Banach algebra. This operation, called convolution, is defined in the following way: If both f and g belong to L1(Rn), then their convolution h = f g is the function whose value ∗ at x Rn is ∈ Z h(x) = f (x y)g(y)dy. Rn − One can show by an elementary argument that f (x y)g(y) is a measurable function of the − two variables x and y. It then follows immediately from Fibini’s theorem on the interchange of 1 n the order of integration that h L (R ) and h 1 f 1 g 1. Furthermore, this operation is ∈ k k 6 k k k k commutative and associative. More generally, we have, with the help of Minkowski’s integral R R inequality F(x,y)dy p 6 F(x,y) p dy, the following result: k kLx k kLx Theorem 1.1. If f Lp(Rn), p [1,∞], and g L1(Rn) then h = f g is well defined and ∈ ∈ ∈ ∗ belongs to Lp(Rn). Moreover, h f g . k kp 6 k kpk k1

1 - 2 - 1. The Fourier Transform and Tempered Distributions

Now, we first consider the Fourier1 transform of L1 functions. Definition 1.2. Let ω R 0 be a constant. If f L1(Rn), then its Fourier transform is F f ∈ \{ } ∈ or fˆ : Rn C defined by → Z ωix ξ F f (ξ) = e− · f (x)dx (1.1) Rn for all ξ Rn. ∈ We now continue with some properties of the Fourier transform. Before doing this, we shall introduce some notations. For a measurable function f on Rn, x Rn and a = 0 we define the ∈ 6 translation and dilation of f by τ f (x) = f (x y), (1.2) y − δa f (x) = f (ax). (1.3) Proposition 1.3. Given f ,g L1(Rn), x,y,ξ Rn, α multiindex, a,b C, ε R and ε = 0, we ∈ ∈ ∈ ∈ 6 have (i) Linearity: F (a f + bg) = aF f + bF g. ωiy ξ (ii) Translation: F τy f (ξ) = e− · fˆ(ξ). ωix y (iii) Modulation: F (e · f (x))(ξ) = τy fˆ(ξ). n (iv) Scaling: F δε f (ξ) = ε − δ 1 fˆ(ξ). | | ε− (v) Differentiation: F ∂ α f (ξ) = (ωiξ)α fˆ(ξ), ∂ α fˆ(ξ) = F (( ωix)α f (x))(ξ). − (vi) Convolution: F ( f g)(ξ) = fˆ(ξ)gˆ(ξ). ∗ (vii) Transformation: F ( f A)(ξ) = fˆ(Aξ), where A is an orthogonal matrix and ξ is a ◦ column vector. (viii) Conjugation: fd(x) = fˆ( ξ). − Proof. These results are easy to be verified. We only prove (vii). In fact, Z Z ωix ξ ωiA 1y ξ F ( f A)(ξ) = e− · f (Ax)dx = e− − · f (y)dy ◦ Rn Rn Z Z ωiA y ξ ωiy Aξ = e− > · f (y)dy = e− · f (y)dy = fˆ(Aξ), Rn Rn where we used the change of variables y = Ax and the fact that A 1 = A and detA = 1. − > | | ut Corollary 1.4. The Fourier transform of a radial function is radial.

Proof. Let ξ,η Rn with ξ = η . Then there exists some orthogonal matrix A such that ∈ | | | | Aξ = η. Since f is radial, we have f = f A. Then, it holds ◦ F f (η) = F f (Aξ) = F ( f A)(ξ) = F f (ξ), ◦ by (vii) in Proposition 1.3. ut It is easy to establish the following results: Theorem 1.5 (Uniform continuity). (i) The mapping F is a bounded linear transformation 1 n ∞ n from L (R ) into L (R ). In fact, F f ∞ f 1. k k 6 k k (ii) If f L1(Rn), then F f is uniformly continuous. ∈

1 Jean Baptiste Joseph Fourier (21 March 1768 – 16 May 1830) was a French mathematician and physicist best known for initiating the investigation of Fourier series and their applications to problems of heat transfer and vibrations. The Fourier transform and Fourier’s Law are also named in his honor. Fourier is also generally credited with the discovery of the greenhouse effect. 1.1. The L1 theory of the Fourier transform - 3 -

Proof. (i) is obvious. We now prove (ii). By Z ωix ξ ωix h fˆ(ξ + h) fˆ(ξ) = e− · [e− · 1] f (x)dx, − Rn − we have Z ωix h fˆ(ξ + h) fˆ(ξ) 6 e− · 1 f (x) dx | − | Rn | − || | Z Z ωix h 6 e− · 1 f (x) dx + 2 f (x) dx x 6r | − || | x >r | | Z| | Z | | 6 ω r h f (x) dx + 2 f (x) dx x r | | | || | x >r | | | |6 | | =:I1 + I2, since for any θ > 0 q eiθ 1 = (cosθ 1)2 + sin2 θ = √2 2cosθ = 2 sin(θ/2) θ . | − | − − | | 6 | | Given any ε > 0, we can take r so large that I < ε/2. Then, we fix this r and take h small 2 | | enough such that I1 < ε/2. In other words, for given ε > 0, there exists a sufficiently small δ > 0 such that fˆ(ξ + h) fˆ(ξ) < ε when h δ, where ε is independent of ξ. | − | | | 6 ut Example 1.6. Suppose that a signal consists of a single rectangular pulse of width 1 and height 1. Let’s say that it gets turned on at x = 1 and turned off at x = 1 . The standard name for this − 2 2 “normalized” rectangular pulse is 1 1, if 1 < x < 1 , Π(x) rect(x) := − 2 2 ≡ 0, otherwise. 1 1 x − 2 2 It is also called, variously, the normalized boxcar function, the top hat function, the indicator function, or the characteristic function for the interval ( 1/2,1/2). The Fourier transform of − this signal is 1/2 Z Z 1/2 ωixξ ωixξ ωixξ e− 2 ωξ Πb(ξ) = e− Π(x)dx = e− dx = = sin R 1/2 ωiξ ωξ 2 − 1/2 − − R 1/2 when ξ = 0. When ξ = 0, Πb(0) = 1/2 dx = 1. By l’Hôpital’s rule, 6 − sin ωξ ω cos ωξ lim Πb(ξ) = lim 2 2 = lim 2 2 2 = 1 = Πb(0), ξ 0 ξ 0 ωξ ξ 0 ω → → → so Πb(ξ) is continuous at ξ = 0. There is a standard function called “sinc”2 that is defined by sinξ ωξ sinc(ξ) = ξ . In this notation Πb(ξ) = sinc 2 . Here is the graph of Πb(ξ).

2π 2π ξ − ω ω Remark 1.7. The above definition of the Fourier transform in (1.1) extends immediately to finite Borel measures: if µ is such a measure on Rn, we define F µ by letting

2 The term “sinc” (English pronunciation:["sINk]) is a contraction, ﬁrst introduced by Phillip M. Woodward in 1953, of the function’s full Latin name, the sinus cardinalis (cardinal sine). - 4 - 1. The Fourier Transform and Tempered Distributions Z ωix ξ F µ(ξ) = e− · dµ(x). Rn Theorem 1.5 is valid for this Fourier transform if we replace the L1 norm by the total variation of µ. The following theorem plays a central role in Fourier Analysis. It takes its name from the fact that it holds even for functions that are integrable according to the deﬁnition of Lebesgue. We prove it for functions that are absolutely integrable in the Riemann sense.3 Theorem 1.8 (Riemann-Lebesgue lemma). If f L1(Rn) then F f 0 as ξ ∞; thus, in ∈ n → | | → view of the last result, we can conclude that F f C0(R ). ∈ The Riemann-Lebesgue lemma states that the integral of a function like the left is small. The integral will approach zero as the number of oscillations increases.

Proof. First, for n = 1, suppose that f (x) = χ(a,b)(x), the characteristic function of an interval. Then Z b ωiaξ ωibξ ωixξ e− e− fˆ(ξ) = e− dx = − 0, as ξ ∞. a ωiξ → | | → Similarly, the result holds when f is the characteristic function of the n-dimensional rectangle n I = x R : a1 x1 b1, ,an xn bn since we can calculate F f explicitly as an { ∈ 6 6 ··· 6 6 } iterated integral. The same is therefore true for a ﬁnite linear combination of such characteristic functions (i.e., simple functions). Since all such simple functions are dense in L1, the result for a general f L1(Rn) follows easily by approximating f in the L1 norm by such a simple function ∈ g, then f = g+( f g), where F f F g is uniformly small by Theorem 1.5, while F g(ξ) 0 − − → as ξ ∞. | | → ut Theorem 1.8 gives a necessary condition for a function to be a Fourier transform. However, that belonging to C0 is not a sufﬁcient condition for being the Fourier transform of an integrable function. See the following example. Example 1.9. Suppose, for simplicity, that n = 1. Let  1  , ξ > e,  lnξ g(ξ) =  ξ  , 0 ξ e, e 6 6 g(ξ) = g( ξ), ξ < 0. − − It is clear that g(ξ) is uniformly continuous on R and g(ξ) 0 as ξ ∞. → | | → Assume that there exists an f L1(R) such that fˆ(ξ) = g(ξ), i.e., ∈ Z ∞ ωixξ g(ξ) = e− f (x)dx. ∞ − Since g(ξ) is an odd function, we have Z ∞ Z ∞ Z ∞ ωixξ g(ξ) = e− f (x)dx = i sin(ωxξ) f (x)dx = sin(ωxξ)F(x)dx, ∞ − ∞ 0 − − g(ξ) where F(x) = i[ f ( x) f (x)] L1(R). Integrating over (0,N) yields − − ∈ ξ

3 Let us very briefly recall what this means. A bounded function f on a finite interval [a,b] is integrable if it can be approximated by Riemann sums from above and below in such a way that the difference of the integrals of these sums can be made as small as we wish. This definition is then extended to unbounded functions and infinite intervals by taking limits; these cases are often called improper integrals. If I is any interval and f is a function on I such that the (possibly improper) integral R f (x) dx has a finite value, then f is I | | said to be absolutely integrable on I. 1.1. The L1 theory of the Fourier transform - 5 -

Z N g(ξ) Z ∞ Z N sin(ωxξ) Z ∞ Z ωxN sint dξ = F(x) dξ dx = F(x) dt dx. 0 ξ 0 0 ξ 0 0 t Noticing that Z N sint π lim dt = , N ∞ 0 t 2 → and by Lebesgue dominated convergence theorem,we get that the integral of r.h.s. is convergent as N ∞. That is, → Z N g(ξ) π Z ∞ lim dξ = F(x)dx < ∞, N ∞ 0 ξ 2 0 → R ∞ g(ξ) R e g(ξ) which yields e ξ dξ < ∞ since 0 ξ dξ = 1. However, Z N g(ξ) Z N dξ lim dξ = lim = ∞. N ∞ e ξ N ∞ e ξ lnξ → → This contradiction indicates that the assumption was invalid.

We now turn to the problem of inverting the Fourier transform. That is, we shall consider the question: Given the Fourier transform fˆ of an integrable function f , how do we obtain f back again from fˆ ? The reader, who is familiar with the elementary theory of Fourier series and integrals, would expect f (x) to be equal to the integral Z ωix ξ C e · fˆ(ξ)dξ. (1.4) Rn Unfortunately, fˆ need not be integrable (for example, let n = 1 and f be the characteristic function of a finite interval). In order to get around this difficulty, we shall use certain summability methods for integrals. We first introduce the Abel method of summability, whose analog for series is very well-known. For each ε > 0, we define the Abel mean Aε = Aε ( f ) to be the integral Z ε x Aε ( f ) = Aε = e− | | f (x)dx. (1.5) Rn 1 n R It is clear that if f L (R ) then lim Aε ( f ) = n f (x)dx. On the other hand, these Abel ∈ ε 0 R means are well-defined even when f is not→ integrable (e.g., if we only assume that f is bounded, then Aε ( f ) is defined for all ε > 0). Moreover, their limit Z ε x lim Aε ( f ) = lim e− | | f (x)dx (1.6) ε 0 ε 0 n → → R may exist even when f is not integrable. A classical example of such a case is obtained by letting f (x) = sinc(x) when n = 1. Whenever the limit in (1.6) exists and is finite we say that R Rn f dx is Abel summable to this limit. A somewhat similar method of summability is Gauss summability. This method is defined by the Gauss (sometimes called Gauss-Weierstrass) means Z ε x 2 Gε ( f ) = e− | | f (x)dx. (1.7) Rn R We say that Rn f dx is Gauss summable (to l) if Z ε x 2 lim Gε ( f ) = lim e− | | f (x)dx (1.6’) ε 0 ε 0 n → → R exists and equals the number `. We see that both (1.6) and (1.6’) can be put in the form Z Mε,Φ ( f ) = Mε ( f ) = Φ(εx) f (x)dx, (1.8) Rn - 6 - 1. The Fourier Transform and Tempered Distributions R where Φ C0 and Φ(0) = 1. Then n f (x)dx is summable to ` if limε 0 Mε ( f ) = `. We shall ∈ R → call Mε ( f ) the Φ means of this integral. ε x 2 ε x We shall need the Fourier transforms of the functions e− | | and e− | |. The first one is easy to calculate. Theorem 1.10. For all a > 0, we have n ξ 2 a ωx 2 ω − n/2 | | F e− | | (ξ) = | | (4πa)− e− 4a . (1.9) 2π Proof. The integral in question is Z ωix ξ a ωx 2 e− · e− | | dx. Rn Notice that this factors as a product of one variable integrals. Thus it is sufficient to prove the R πx2 case n = 1. For this we use the formula for the integral of a Gaussian: R e− dx = 1. It follows that Z ∞ Z ∞ 2 ωixξ aω2x2 a(ωx+iξ/(2a))2 ξ e− e− dx = e− e− 4a dx ∞ ∞ − − 2 Z ∞+iξ/(2a) 1 ξ ax2 = ω − e− 4a e− dx | | ∞+iξ/(2a) − 2 Z ∞ 1 ξ p πy2 = ω − e− 4a π/a e− dy | | ∞ − 1 2 ω − 1/2 ξ = | | (4πa)− e− 4a , 2π where we used contour integration at the next to last one. ut The second one is somewhat harder to obtain: Theorem 1.11. For all a > 0, we have n a ωx ω − cna Γ ((n + 1)/2) F (e− | |) = | | , cn = . (1.10) 2π (a2 + ξ 2)(n+1)/2 π(n+1)/2 | | Proof. By a change of variables, i.e., Z Z a ωx ωix ξ a ωx n ix ξ/a x F (e− | |) = e− · e− | |dx = (a ω )− e− · e−| |dx, Rn | | Rn we see that it suffices to show this result when a = 1. In order to show this, we need to express the decaying exponential as a superposition of Gaussians, i.e., Z ∞ η γ 1 e− γ2/4η e− = e− dη, γ > 0. (1.11) √π 0 √η Then, using (1.9) to establish the third equality, Z Z Z ∞ η ix t x ix t 1 e− x 2/4η e− · e−| |dx = e− · e−| | dη dx Rn Rn √π 0 √η Z ∞ η Z 1 e− ix t x 2/4η = e− · e−| | dx dη √π 0 √η Rn Z ∞ η 1 e− n/2 η t 2 = (4πη) e− | | dη √π 0 √η Z ∞ n (n 1)/2 η(1+ t 2) n 1 =2 π − e− | | η −2 dη 0 1.1. The L1 theory of the Fourier transform - 7 -

n+1 Z ∞ n+1 n (n 1)/2 2 2 ζ 1 =2 π − 1 + t − e− ζ 2 − dζ | | 0 n (n 1)/2 n + 1 1 =2 π − Γ . 2 (1 + t 2)(n+1)/2 | | Thus, n n n a ωx (a ω )− (2π) cn ω − cna F (e− | |) = | | = | | . (1 + ξ/a 2)(n+1)/2 2π (a2 + ξ 2)(n+1)/2 | | | | Consequently, the theorem will be established once we show (1.11). In fact, by changes of variables, we have Z ∞ η 1 γ e− γ2/4η e e− dη √π 0 √η Z ∞ 2√γ γ(σ 1 )2 2 = e− − 2σ dσ (by η = γσ ) √π 0 ∞ 2√γ Z 1 2 1 1 γ(σ 2σ ) = e− − 2 dσ (by σ ) √π 0 2σ 7→ 2σ ∞ √γ Z 1 2 1 γ(σ 2σ ) = e− − 1 + 2 dσ (by averaging the last two formula) √π 0 2σ Z ∞ √γ γu2 1 = e− du (by u = σ ) √π ∞ − 2σ − Z πx2 =1, (by e− dx = 1) R which yields the desired identity (1.11). n n ut ω a ωx 2 ω a ωx We shall denote the Fourier transform of |2π| e− | | and |2π| e− | |, a > 0, by W and P, respectively. That is, ξ 2 n/2 | | cna W(ξ,a) = (4πa)− e− 4a , P(ξ,a) = . (1.12) (a2 + ξ 2)(n+1)/2 | | The ﬁrst of these two functions is called the Weierstrass (or Gauss-Weierstrass) kernel while the second is called the Poisson kernel. Theorem 1.12 (The multiplication formula). If f ,g L1(Rn), then Z Z ∈ fˆ(ξ)g(ξ)dξ = f (x)gˆ(x)dx. Rn Rn Proof. Using Fubini’s theorem to interchange the order of the integration on R2n, we obtain the identity. ut 1 n n Theorem 1.13. If f and Φ belong to L (R ), ϕ = Φˆ and ϕε (x) = ε− ϕ(x/ε), then Z Z ωix ξ e · Φ(εξ) fˆ(ξ)dξ = ϕε (y x) f (y)dy Rn Rn − for all ε > 0. In particular, n Z Z ω ωix ξ ε ωξ | | e · e− | | fˆ(ξ)dξ = P(y x,ε) f (y)dy, 2π Rn Rn − and n Z Z ω ωix ξ ε ωξ 2 | | e · e− | | fˆ(ξ)dξ = W(y x,ε) f (y)dy. 2π Rn Rn − - 8 - 1. The Fourier Transform and Tempered Distributions

ωix ξ Proof. From (iii) and (iv) in Proposition 1.3, it implies (F e Φ(εξ))(y) = ϕε (y x). The · − first result holds immediately with the help of Theorem 1.12. The last two follow from (1.9), (1.10) and (1.12). ut R Lemma 1.14. (i) n W(x,ε)dx = 1 for all ε > 0. R R (ii) Rn P(x,ε)dx = 1 for all ε > 0. Proof. By a change of variable, we first note that Z Z x 2 Z n/2 | | W(x,ε)dx = (4πε)− e− 4ε dx = W(x,1)dx, Rn Rn Rn and Z Z c ε Z P(x,ε)dx = n dx = P(x,1)dx. n n (ε2 + x 2)(n+1)/2 n R R | | R Thus, it suffices to prove the lemma when ε = 1. For the first one, we use a change of variables R πx2 and the formula for the integral of a Gaussian: R e− dx = 1 to get Z Z x 2 Z n/2 | | n/2 π y 2 n n/2 W(x,1)dx = (4π)− e− 4 dx = (4π)− e− | | 2 π dy = 1. Rn Rn Rn For the second one, we have Z Z 1 P(x,1)dx = c dx. n 2 (n+1)/2 Rn Rn (1 + x ) n 1 n| | Letting r = x , x0 = x/r (when x = 0), S − = x R : x = 1 , dx0 the element of surface area n 1 | | 4 6 { ∈ | | } on S − whose surface area is denoted by ωn 1 and, finally, putting r = tanθ, we have Z Z ∞−Z 1 1 n 1 dx = dx0r − dr n (1 + x 2)(n+1)/2 0 Sn 1 (1 + r2)(n+1)/2 R | | − Z ∞ n 1 r − =ωn 1 dr − 0 (1 + r2)(n+1)/2 Z π/2 n 1 =ωn 1 sin − θdθ. − 0 n 1 But ωn 1 sin − θ is clearly the surface area of the sphere of radius xn+1 − n Sn sinθ obtained by intersecting S with the hyperplane x1 = cosθ. n

Thus, the area of the upper half of S is obtained by summing these θ (n 1) dimensional areas as θ ranges from 0 to π/2, that is, O θ sin 1 cos θ x1 − n 1 Z π/2 S − n 1 ωn ωn 1 sin − θdθ = , − 0 2 which is the desired result by noting that 1/c = ω /2. n n ut 1 n R n Theorem 1.15. Suppose ϕ L (R ) with Rn ϕ(x)dx = 1 and let ϕε (x) = ε− ϕ(x/ε) for ε > 0. p n ∈ n ∞ n If f L (R ), 1 p < ∞, or f C0(R ) L (R ), then for 1 p ∞ ∈ 6 ∈ ⊂ 6 6 f ϕε f 0, as ε 0. k ∗ − kp → → In particular, the Poisson integral of f : Z u(x,ε) = P(x y,ε) f (y)dy Rn − and the Gauss-Weierstrass integral of f : Z s(x,ε) = W(x y,ε) f (y)dy Rn −

4 n/2 ωn 1 = 2π /Γ (n/2). − 1.1. The L1 theory of the Fourier transform - 9 - converge to f in the Lp norm as ε 0. → Proof. By a change of variables, we have Z Z Z n ϕε (y)dy = ε− ϕ(y/ε)dy = ϕ(y)dy = 1. Rn Rn Rn Hence, Z ( f ϕε )(x) f (x) = [ f (x y) f (x)]ϕε (y)dy. ∗ − Rn − − Therefore, by Minkowski’s inequality for integrals and a change of variables, we get Z n f ϕε f p 6 f (x y) f (x) pε− ϕ(y/ε) dy k ∗ − k Rn k − − k | | Z = f (x εy) f (x) p ϕ(y) dy. Rn k − − k | | p n We point out that if f L (R ), 1 6 p < ∞, and denote f (x t) f (x) p = ∆ f (t), then 5 ∈ n ∞ n k −∞ − k ∆ f (t) 0, as t 0. In fact, if f1 D(R ) := C (R ) of all C functions with compact → → ∈ 0 support, the assertion in that case is an immediate consequence of the uniform convergence f1(x t) f1(x), as t 0. In general, for any σ > 0, we can write f = f1 + f2, such that f1 − → → n p n is as described and f2 p σ, since D(R ) is dense in L (R ) for 1 p < ∞. Then, ∆ f (t) k k 6 6 6 ∆ (t)+∆ (t), with ∆ (t) 0 as t 0, and ∆ (t) 2σ. This shows that ∆ (t) 0 as t 0 f1 f2 f1 → → f2 6 f → → for general f Lp(Rn), 1 6 p < ∞. ∈ n n For the case p = ∞ and f C0(R ), the same argument gives us the result since D(R ) is n ∈ dense in C0(R ) (cf. [Rud87, p.70, Proof of Theorem 3.17]). Thus, by the Lebesgue dominated convergence theorem (due to ϕ L1 and the fact ∈ ∆ f (εy) ϕ(y) 6 2 f p ϕ(y) ) and the fact ∆ f (εy) 0 as ε 0, we have | | k k | | Z → Z → lim f ϕε f p 6 lim ∆ f (εy) ϕ(y) dy = lim ∆ f (εy) ϕ(y) dy = 0. ε 0 k ∗ − k ε 0 n | | n ε 0 | | → → R R → This completes the proof. ut With the same argument, we have 1 n R Corollary 1.16. Let 1 6 p 6 ∞. Suppose ϕ L (R ) and Rn ϕ(x)dx = 0, then f ϕε p 0 p n ∈ n ∞ n k ∗ k → as ε 0 whenever f L (R ), 1 p < ∞, or f C0(R ) L (R ). → ∈ 6 ∈ ⊂ Proof. Once we observe that Z ( f ϕε )(x) =( f ϕε )(x) f (x) 0 = ( f ϕε )(x) f (x) ϕε (y)dy ∗ ∗ − · ∗ − Rn Z = [ f (x y) f (x)]ϕε (y)dy, Rn − − the rest of the argument is precisely that used in the last proof. ut In particular, we also have 1 n R n Corollary 1.17. Suppose ϕ L (R ) with n ϕ(x)dx = 1 and let ϕε (x) = ε− ϕ(x/ε) for ε > 0. ∈ R Let f (x) L∞(Rn) be continuous at 0 . Then, ∈ {Z } lim f (x)ϕε (x)dx = f (0). ε 0 n → R R R Proof. Since n f (x)ϕε (x)dx f (0) = n ( f (x) f (0))ϕε (x)dx, then we may assume without R − R − loss of generality that f (0) = 0. Since f is continuous at 0 , then for any η > 0, there exists a { } δ > 0 such that

5 This statement is the continuity of the mapping t f (x t) of Rn to Lp(Rn). → − - 10 - 1. The Fourier Transform and Tempered Distributions η f (x) < , | | ϕ 1 R k k whenever x < δ. Noticing that n ϕ(x)dx 6 ϕ 1, we have | | | R | k k Z η Z Z f (x)ϕε (x)dx 6 ϕε (x) dx + f ∞ ϕε (x) dx Rn ϕ 1 x <δ | | k k x >δ | | k ηk | | Z | | 6 ϕ 1 + f ∞ ϕ(y) dy ϕ 1 k k k k y δ/ε | | k k | |> =η + f ∞Iε . k k But Iε 0 as ε 0. This proves the result. → → ut From Theorems 1.13 and 1.15, we obtain the following solution to the Fourier inversion problem: ˆ R Theorem 1.18. If both Φ and its Fourier transform ϕ = Φ are integrable and Rn ϕ(x)dx = 1, n R ωix ξ 1 then the Φ means of the integral ( ω /2π) n e fˆ(ξ)dξ converges to f (x) in the L norm. | | R · In particular, the Abel and Gauss means of this integral converge to f (x) in the L1 norm. We have singled out the Gauss-Weierstrass and the Abel methods of summability. The former is probably the simplest and is connected with the solution of the heat equation; the latter is intimately connected with harmonic functions and provides us with very powerful tools in Fourier analysis. n ω R ωix ξ ε ωξ 2 | | ˆ 1 Since s(x,ε) = 2π Rn e · e− | | f (ξ)dξ converges in L to f (x) as ε > 0 tends to 0, we can ﬁnd a sequence ε 0 such that s(x,ε ) f (x) for a.e. x. If we further assume that k → k → fˆ L1(Rn), the Lebesgue dominated convergence theorem gives us the following pointwise ∈ equality: Theorem 1.19 (Fourier inversion theorem). If both f and fˆ are integrable, then n Z ω ωix ξ f (x) = | | e · fˆ(ξ)dξ, 2π Rn for almost every x. Remark 1.20. We know from Theorem 1.5 that fˆ is continuous. If fˆ is integrable, the inte- R ωix ξ ˆ gral n e fˆ(ξ)dξ also deﬁnes a continuous function (in fact, it equals fˆ( x)). Thus, by R · − changing f on a set of measure 0, we can obtain equality in Theorem 1.19 for all x. It is clear from Theorem 1.18 that if fˆ(ξ) = 0 for all ξ then f (x) = 0 for almost every x. Ap- plying this to f = f f , we obtain the following uniqueness result for the Fourier transform: 1 − 2 1 n n Corollary 1.21 (Uniqueness). If f1 and f2 belong to L (R ) and fˆ1(ξ) = fˆ2(ξ) for ξ R , n ∈ then f1(x) = f2(x) for almost every x R . ∈ We will denote the inverse operation to the Fourier transform by F 1 or ˇ. If f L1, then we − · ∈ have n Z ω ωix ξ fˇ(x) = | | e · f (ξ)dξ. (1.13) 2π Rn We give a very useful result. Theorem 1.22. Suppose f L1(Rn) and fˆ 0. If f is continuous at 0, then ∈ > ω n Z f (0) = | | fˆ(ξ)dξ. 2π Rn Moreover, we have fˆ L1(Rn) and ∈ 1.1. The L1 theory of the Fourier transform - 11 - n Z ω ωix ξ f (x) = | | e · fˆ(ξ)dξ, 2π Rn for almost every x.

Proof. By Theorem 1.13, we have n Z Z ω ε ωξ | | e− | | fˆ(ξ)dξ = P(y,ε) f (y)dy. 2π Rn Rn From Lemma 1.14, we get, for any δ > 0,

Z Z P(y,ε) f (y)dy f (0) = P(y,ε)[ f (y) f (0)]dy Rn − Rn −

Z Z 6 P(y,ε)[ f (y) f (0)]dy + P(y,ε)[ f (y) f (0)]dy y <δ − y δ − | | | |> =I1 + I2. Since f is continuous at 0, for any given σ > 0, we can choose δ small enough such that f (y) f (0) σ when y < δ. Thus, I σ by Lemma 1.14. For the second term, we have, | − | 6 | | 1 6 by a change of variables, that Z I2 6 f 1 sup P(y,ε) + f (0) P(y,ε)dy k k y δ | | y >δ | |> | | c ε Z = f n + f (0) P(y,1)dy 0, 1 2 2 (n+1)/2 k k (ε + δ ) | | y >δ/ε → n | | ω R ε ωξ as ε 0. Thus, | | n e fˆ(ξ)dξ f (0) as ε 0. On the other hand, by Lebesgue → 2π R − | | → → dominated convergence theorem, we obtain n Z n Z ω ω ε ωξ | | fˆ(ξ)dξ = | | lim e− | | fˆ(ξ)dξ = f (0), 2π n 2π ε 0 n R → R which implies fˆ L1(Rn) due to fˆ 0. Therefore, from Theorem 1.19, it follows the desired ∈ > result. ut An immediate consequence is R ωix ξ ε ωx 2 Corollary 1.23. i) Rn e · W(ξ,ε)dξ = e− | | . R ωix ξ ε ωx ii) Rn e · P(ξ,ε)dξ = e− | |. Proof. Noticing that n n ω ε ωx 2 ω ε ωx W(ξ,ε) = F | | e− | | , and P(ξ,ε) = F | | e− | | , 2π 2π we have the desired results by Theorem 1.22. ut We also have the semigroup properties of the Weierstrass and Poisson kernels.

Corollary 1.24. If α1 and α2 are positive real numbers, then R i) W(ξ,α1 + α2) = n W(ξ η,α1)W(η,α2)dη. RR − ii) P(ξ,α + α ) = n P(ξ η,α )P(η,α )dη. 1 2 R − 1 2 Proof. It follows, from Corollary 1.23, that n ω (α +α ) ωx 2 W(ξ,α + α ) = | | (F e− 1 2 | | )(ξ) 1 2 2π n ω α ωx 2 α ωx 2 = | | F (e− 1| | e− 2| | )(ξ) 2π - 12 - 1. The Fourier Transform and Tempered Distributions n ω 2 Z α1 ωx ωix η = | | F e− | | e · W(η,α2)dη (ξ) 2π Rn n ω Z 2 Z ωix ξ α1 ωx ωix η = | | e− · e− | | e · W(η,α2)dηdx 2π Rn Rn n Z Z ω 2 ωix (ξ η) α1 ωx = e− · − | | e− | | dx W(η,α2)dη Rn Rn 2π Z = W(ξ η,α1)W(η,α2)dη. Rn − A similar argument can give the other equality. ut Finally, we give an example of the semigroup about the heat equation. Example 1.25. Consider the Cauchy problem to the heat equation n ut ∆u = 0, u(0) = u0(x), t > 0, x R . − ∈ Taking the Fourier transform, we have uˆ + ωξ 2uˆ = 0, uˆ(0) = uˆ (ξ). t | | 0 Thus, it follows, from Theorem 1.10, that 1 ωξ 2t 1 ωξ 2t n/2 x 2/4t u =F − e−| | F u = (F − e−| | ) u = (4πt)− e−| | u 0 ∗ 0 ∗ 0 =W(x,t) u =: H(t)u . ∗ 0 0 Then, we obtain H(t +t )u =W(x,t +t ) u = W(x,t ) W(x,t ) u 1 2 0 1 2 ∗ 0 1 ∗ 2 ∗ 0 =W(x,t ) (W(x,t ) u ) = W(x,t ) H(t )u 1 ∗ 2 ∗ 0 1 ∗ 2 0 =H(t1)H(t2)u0, i.e., H(t1 +t2) = H(t1)H(t2).

1.2 The L2 theory and the Plancherel theorem

The integral defining the Fourier transform is not defined in the Lebesgue sense for the general function in L2(Rn); nevertheless, the Fourier transform has a natural definition on this space and a particularly elegant theory. If, in addition to being integrable, we assume f to be square-integrable then fˆ will also be square-integrable. In fact, we have the following basic result: n/2 1 n 2 n ω − Theorem 1.26 (Plancherel theorem). If f L (R ) L (R ), then fˆ 2 = | | f 2. ∈ ∩ k k 2π k k Proof. Let g(x) = f ( x). Then, by Theorem 1.1, h = f g L1(Rn) and, by Proposition 1.3, − ∗ ∈ hˆ = fˆgˆ. Butg ˆ = fˆ, thus hˆ = fˆ 2 > 0. Applying Theorem 1.22, we have hˆ L1(Rn) and h(0) = ω n R | | ∈ | | ˆ 2π Rn h(ξ)dξ. Thus, we get n Z Z ω − fˆ(ξ) 2dξ = hˆ(ξ)dξ = | | h(0) Rn | | Rn 2π n ω − Z = | | f (x)g(0 x)dx 2π Rn − 1.3. Schwartz spaces - 13 -

n n ω − Z ω − Z = | | f (x) f (x)dx = | | f (x) 2dx, 2π Rn 2π Rn | | which completes the proof. ut Since L1 L2 is dense in L2, there exists a unique bounded extension, F , of this operator to ∩ all of L2. F will be called the Fourier transform on L2; we shall also use the notation fˆ = F f whenever f L2(Rn). ∈ A linear operator on L2(Rn) that is an isometry and maps onto L2(Rn) is called a unitary ω n/2 operator. It is an immediate consequence of Theorem 1.26 that |2π| F is an isometry. ω n/2 Moreover, we have the additional property that |2π| F is onto: n/2 ω 2 n Theorem 1.27. |2π| F is a unitary operator on L (R ).

n/2 ω 2 n Proof. Since |2π| F is an isometry, its range is a closed subspace of L (R ). If this sub- 2 n R ˆ 2 space were not all of L (R ), we could ﬁnd a function g such that Rn f gdx = 0 for all f L 2 R R ∈ and g = 0. Theorem 1.12 obviously extends to L ; consequently, n f gdxˆ = n fˆ gdx = 0 k k2 6 R R for all f L2. But this implies thatg ˆ(x) = 0 for almost every x, contradicting the fact that ∈ ω n/2 gˆ = | | − g = 0. k k2 2π k k2 6 ut Theorem 1.27 is a major part of the basic theorem in the L2 theory of the Fourier transform: 1 Theorem 1.28. The inverse of the Fourier transform, F − , can be obtained by letting n 1 ω (F − f )(x) = | | (F f )( x) 2π − for all f L2(Rn). ∈ We can also extend the deﬁnition of the Fourier transform to other spaces, such as Schwartz space, tempered distributions and so on.

1.3 Schwartz spaces

Distributions (generalized functions) aroused mostly due to Paul Dirac and his delta function δ. The Dirac delta gives a description of a point of unit mass (placed at the origin). The mass density function is such that if its integrated on a set not containing the origin it vanishes, but if the set does contain the origin it is 1. No function (in the traditional sense) can have this property because we know that the value of a function at a particular point does not change the value of the integral. In mathematical analysis, distributions are objects which generalize functions and probability distributions. They extend the concept of derivative to all integrable functions and beyond, and are used to formulate generalized solutions of partial differential equations. They are important in physics and engineering where many non-continuous problems naturally lead to differential equations whose solutions are distributions, such as the Dirac delta distribution. “Generalized functions” were introduced by Sergei Sobolev in 1935. They were indepen- dently introduced in late 1940s by Laurent Schwartz, who developed a comprehensive theory of distributions. - 14 - 1. The Fourier Transform and Tempered Distributions

The basic idea in the theory of distributions is to consider them as linear functionals on some space of “regular” functions — the so-called “testing functions”. The space of testing functions is assumed to be well-behaved with respect to the operations (differentiation, Fourier transform, convolution, translation, etc.) we have been studying, and this is then reflected in the properties of distributions. We are naturally led to the definition of such a space of testing functions by the following considerations. Suppose we want these operations to be defined on a function space, S , and to preserve it. Then, it would certainly have to consist of functions that are indefinitely differentiable; this, in view of part (v) in Proposition 1.3, indicates that each function in S , after being multiplied by a polynomial, must still be in S . We therefore make the following definition: Definition 1.29. The Schwartz space S (Rn) of rapidly decaying functions is defined as n ∞ n α β n S (R ) = ϕ C (R ) : ϕ α,β := sup x (∂ ϕ)(x) < ∞, α,β N0 , (1.14) ∈ | | x Rn | | ∀ ∈ ∈ where N0 = N 0 . ∪ { } m If ϕ S , then ϕ(x) Cm(1 + x )− for any m N0. The second part of next example ∈ | | 6 | | ∈ shows that the converse is not true. ε x 2 ε x Example 1.30. ϕ(x) = e− | | , ε > 0, belongs to S ; on the other hand, ϕ(x) = e− | | fails to be differential at the origin and, therefore, does not belong to S . ε(1+ x 2)γ Example 1.31. ϕ(x) = e− | | belongs to S for any ε,γ > 0. Example 1.32. S contains the space D(Rn). But it is not immediately clear that D is nonempty. To find a function in D, consider the function e 1/t, t > 0, f (t) = − 0, t 6 0. Then, f C∞, is bounded and so are all its derivatives. Let ϕ(t) = f (1+t) f (1 t), then ϕ(t) = 2/(1 t2∈) 1 − e− − if t < 1, is zero otherwise. It clearly belongs to D = D(R ). We can easily obtain | | n-dimensional variants from ϕ. For examples, n n (i) For x R , define ψ(x) = ϕ(x1)ϕ(x2) ϕ(xn), then ψ D(R ); ∈ n 2/(1 x 2)··· ∈ n (ii) For x R , define ψ(x) = e− −| | for x < 1 and 0 otherwise, then ψ D(R ); ∈ | | ∈ (iii) If η C∞ and ψ is the function in (ii), then ψ(εx)η(x) defines a function in D(Rn); ∈ moreover, e2ψ(εx)η(x) η(x) as ε 1. → → Example 1.33. We observe that the order of multiplication by powers of x , ,x and dif- 1 ··· n ferentiation, in (1.14), could have been reversed. That is, ϕ S if and only if ϕ C∞ and β α ∈ ∈ sup n ∂ (x ϕ(x)) < ∞ for all multi-indices α and β of nonnegative integers. This shows x R | | that∈ if P is a polynomial in n variables and ϕ S then P(x)ϕ(x) and P(∂)ϕ(x) are again in S , ∈ where P(∂) is the associated differential operator (i.e., we replace xα by ∂ α in P(x)). Example 1.34. Sometimes S (Rn) is called the space of rapidly decaying functions. But observe x2 iex that the function ϕ(x) = e− e is not in S (R). Hence, rapid decay of the value of the function alone does not assure the membership in S (R). n p n n Theorem 1.35. The spaces C0(R ) and L (R ), 1 6 p 6 ∞, contain S (R ). Moreover, both n p n S and D are dense in C0(R ) and L (R ) for 1 6 p < ∞. Proof. S C L∞ is obvious by (1.14). The Lp norm of ϕ S is bounded by a finite linear ⊂ 0 ⊂ ∈ combination of L∞ norms of terms of the form xα ϕ(x). In fact, by (1.14), we have 1.3. Schwartz spaces - 15 -

Z 1/p ϕ(x) pdx Rn | | Z 1/p Z 1/p p p 6 ϕ(x) dx + ϕ(x) dx x 1 | | x >1 | | | |6 | | Z 1/p Z 1/p 2n 2np 6 ϕ ∞ dx + x ϕ(x) ∞ x − dx k k x 1 k| | | |k x >1 | | | |6 | | 1/p 1/p ωn 1 ωn 1 2n = − ϕ ∞ + − x ϕ n k k (2p 1)n | | | | ∞ − <∞. For the proof of the density, we only need to prove the case of D since D S . We will ⊂ use the fact that the set of ﬁnite linear combinations of characteristic functions of bounded measurable sets in Rn is dense in Lp(Rn), 1 6 p < ∞. This is a well-known fact from functional analysis. Now, let E Rn be a bounded measurable set and let ε > 0. Then, there exists a closed set ⊂ F and an open set Q such that F E Q and m(Q F) < ε p (or only m(Q) < ε p if there is ⊂ ⊂ \ no closed set F E). Here m is the Lebesgue measure in Rn. Next, let ϕ be a function from D ⊂ such that suppϕ Q, ϕ F 1 and 0 6 ϕ 6 1. Then, ⊂ | ≡Z Z p p m p ϕ χE p = ϕ(x) χE(x) dx 6 dx = (Q F) < ε k − k Rn | − | Q F \ or \ ϕ χ < ε, k − Ekp n p n where χE denotes the characteristic function of E. Thus, we may conclude that D(R ) = L (R ) with respect to Lp measure for 1 6 p < ∞. For the case of C , we leave it to the interested reader. 0 ut Remark 1.36. The density is not valid for p = ∞. Indeed, for a nonzero constant function f n ≡ c0 = 0 and for any function ϕ D(R ), we have 6 ∈ f ϕ ∞ c > 0. k − k > | 0| Hence we cannot approximate any function from L∞(Rn) by functions from D(Rn). This example also indicates that S is not dense in L∞ since lim ϕ(x) = 0 for all ϕ S . x ∞ | | ∈ | |→ From part (v) in Proposition 1.3, we immediately have Theorem 1.37. If ϕ S , then ϕˆ S . ∈ ∈ If ϕ,ψ S , then Theorem 1.37 implies that ϕˆ,ψˆ S . Therefore, ϕˆψˆ S . By part (vi) ∈ ∈ ∈ in Proposition 1.3, i.e., F (ϕ ψ) = ϕˆψˆ , an application of the inverse Fourier transform shows ∗ that Theorem 1.38. If ϕ,ψ S , then ϕ ψ S . ∈ ∗ ∈ n The space S (R ) is not a normed space because ϕ α,β is only a semi-norm for multi-indices | | α and β, i.e., the condition ϕ = 0 if and only if ϕ = 0 | |α,β fails to hold, for example, for constant function ϕ. But the space (S ,ρ) is a metric space if the metric ρ is deﬁned by - 16 - 1. The Fourier Transform and Tempered Distributions

α β ϕ ψ α,β ρ(ϕ,ψ) = 2−| |−| | | − | . ∑ 1 + ϕ ψ α,β Nn α,β ∈ 0 | − | Theorem 1.39 (Completeness). The space (S ,ρ) is a complete metric space, i.e., every Cauchy sequence converges.

γ ∞ n 2−| |σ Proof. Let ϕk k=1 S be a Cauchy sequence. For any σ > 0 and any γ N0, let ε = 1+2σ , { } ⊂ ∈ ∞ then there exists N0(ε) N such that ρ(ϕk,ϕm) < ε when k,m N0(ε) since ϕk is a ∈ > { }k=1 Cauchy sequence. Thus, we have ϕ ϕ γ σ | k − m|0, < , 1 + ϕ ϕ 1 + σ | k − m|0,γ and then γ sup ∂ (ϕk ϕm) < σ x K | − | n ∈ ∞ for any compact set K R . It means that ϕk k=1 is a Cauchy sequence in the Banach space γ ⊂ { γ } C| |(K). Hence, there exists a function ϕ C| |(K) such that ∈ γ lim ϕk = ϕ, in C| |(K). k ∞ → Thus, we can conclude that ϕ C∞(Rn). It only remains to prove that ϕ S . It is clear that ∈ ∈ for any α,β Nn ∈ 0 α β α β α β sup x ∂ ϕ 6sup x ∂ (ϕk ϕ) + sup x ∂ ϕk x K | | x K | − | x K | | ∈ ∈ ∈ β α β 6Cα (K)sup ∂ (ϕk ϕ) + sup x ∂ ϕk . x K | − | x K | | ∈ ∈ Taking k ∞, we obtain → α β sup x ∂ ϕ 6 limsup ϕk α,β < ∞. x K | | k ∞ | | ∈ → The last inequality is valid since ϕ ∞ is a Cauchy sequence, so that ϕ is bounded. The { k}k=1 | k|α,β last inequality doesn’t depend on K either. Thus, ϕ < ∞ and then ϕ S . | |α,β ∈ ut Moreover, some easily established properties of S and its topology, are the following: Proposition 1.40. i) The mapping ϕ(x) xα ∂ β ϕ(x) is continuous. 7→ ii) If ϕ S , then limh 0 τhϕ = ϕ. ∈ → n iii) Suppose ϕ S and h = (0, ,hi, ,0) lies on the i-th coordinate axis of R , then the ∈ ··· ··· difference quotient [ϕ τ ϕ]/h tends to ∂ϕ/∂x as h 0. − h i i | | → iv) The Fourier transform is a homeomorphism of S onto itself. v) S is separable. Finally, we describe and prove a fundamental result of Fourier analysis that is known as the uncertainty principle. In fact this theorem was "discovered" by W. Heisenberg in the context of quantum mechanics. Expressed colloquially, the uncertainty principle says that it is not possible to know both the position and the momentum of a particle at the same time. Expressed more precisely, the uncertainty principle says that the position and the momentum cannot be simultaneously localized. In the context of harmonic analysis, the uncertainty principle implies that one cannot at the same time localize the value of a function and its Fourier transform. The exact statement is as follows. 1.4. The class of tempered distributions - 17 -

Theorem 1.41 (The Heisenberg uncertainty principle). Suppose ψ is a function in S (R). Then 1/2 2 ω − ψ 2 xψ 2 ξψˆ 2 > | | k k , k k k k 2π 2 ω | | Bx2 and equality holds if and only if ψ(x) = Ae− where B > 0 and A R. ∈ Moreover, we have 1/2 2 ω − ψ 2 (x x0)ψ 2 (ξ ξ0)ψˆ 2 > | | k k k − k k − k 2π 2 ω | | for every x0, ξ0 R. ∈ ωixξ Proof. The last inequality actually follows from the first by replacing ψ(x) by e− 0 ψ(x+x0) ωix (ξ+ξ ) (whose Fourier transform is e 0 0 ψˆ (ξ + ξ0) by parts (ii) and (iii) in Proposition 1.3) and changing variables. To prove the first inequality, we argue as follows. Since ψ S , we know that ψ and ψ are rapidly decreasing. Thus, an integration by parts ∈ 0 gives Z ∞ Z ∞ 2 2 d 2 ψ 2 = ψ(x) dx = x ψ(x) dx k k ∞ | | − ∞ dx| | −Z ∞ − = xψ0(x)ψ(x) + xψ0(x)ψ(x) dx. − ∞ − The last identity follows because ψ 2 = ψψ. Therefore, Z| ∞ | 2 ψ 2 6 2 x ψ(x) ψ0(x) dx 6 2 xψ 2 ψ0 2, k k ∞ | || || | k k k k − where we have used the Cauchy-Schwarz inequality. By part (v) in Proposition 1.3, we have F (ψ0)(ξ) = ωiξψˆ (ξ). It follows, from the Plancherel theorem, that ω 1/2 ω 1/2 ψ0 = | | F (ψ0) = | | ω ξψˆ . k k2 2π k k2 2π | |k k2 Thus, we conclude the proof of the inequality in the theorem. If equality holds, then we must also have equality where we applied the Cauchy-Schwarz inequality, and as a result, we find that ψ0(x) = βxψ(x) for some constant β. The solutions to 2 this equation are ψ(x) = Aeβx /2, where A is a constant. Since we want ψ to be a Schwartz function, we must take β = 2B < 0. − ut

1.4 The class of tempered distributions

The collection S 0 of all continuous linear functionals on S is called the space of tempered distributions. That is Deﬁnition 1.42. The functional T : S C is a tempered distribution if → i) T is linear, i.e., T,αϕ + βψ = α T,ϕ + β T,ψ for all α,β C and ϕ,ψ S . h i h i h i ∈ ∈ ii) T is continuous on S , i.e., there exist n0 N0 and a constant c0 > 0 such that ∈ T,ϕ 6 c0 ∑ ϕ α,β |h i| α , β n | | | | | |6 0 for any ϕ S . ∈ In addition, for Tk,T S 0, the convergence Tk T in S 0 means that Tk,ϕ T,ϕ in C ∈ → h i → h i for all ϕ S . ∈ - 18 - 1. The Fourier Transform and Tempered Distributions

Remark 1.43. Since D S , the space of tempered distributions S is more narrow than the ⊂ 0 space of distributions D , i.e., S D . Another more narrow distribution space E which 0 0 ⊂ 0 0 consists of continuous linear functionals on the (widest test function) space E := C∞(Rn). In short, D S E implies that ⊂ ⊂ E 0 S 0 D0. ⊂ ⊂ p n Example 1.44. Let f L (R ), 1 6 p 6 ∞, and define T = Tf by letting ∈ Z T,ϕ = Tf ,ϕ = f (x)ϕ(x)dx h i h i Rn for ϕ S . It is clear that T is a linear functional on S . To show that it is continuous, therefore, ∈ f it suffices to show that it is continuous at the origin. Then, suppose ϕ 0 in S as k ∞. From k → → the proof of Theorem 1.35, we have seen that for any q 1, ϕ is dominated by a finite linear > k kkq combination of L∞ norms of terms of the form xα ϕ (x). That is, ϕ is dominated by a finite k k kkq linear combination of semi-norms ϕ . Thus, ϕ 0 as k ∞. Choosing q = p , i.e., | k|α,0 k kkq → → 0 1/p + 1/q = 1, Hölder’s inequality shows that T,ϕ f ϕ 0 as k ∞. Thus, |h ki| 6 k kpk kkp0 → → T S . ∈ 0 Example 1.45. We consider the case n = 1. Let f (x) = ∑m a xk be a polynomial, then f S k=0 k ∈ 0 since

Z m k Tf ,ϕ = ∑ akx ϕ(x)dx |h i| R k=0 m Z 1 ε 1+ε k 6 ∑ ak (1 + x )− − (1 + x ) x ϕ(x) dx k=0 | | R | | | | | | | | m Z 1 ε 6C ∑ ak ϕ k+1+ε,0 (1 + x )− − dx, k=0 | || | R | | so that the condition ii) of the definition is satisfied for ε = 1 and n0 = m + 2. n n Example 1.46. Fix x0 R and a multi-index β N . By the continuity of the semi-norm α,β ∈ ∈ 0 |·| in S , we have that T,ϕ = ∂ β ϕ(x ), for ϕ S , defines a tempered distribution. A special h i 0 ∈ case is the Dirac δ-function: T ,ϕ = ϕ(0). h δ i The tempered distributions of Examples 1.44-1.46 are called functions or measures. We shall write, in these cases, f and δ instead of Tf and Tδ . These functions and measures may be considered as embedded in S 0. If we put on S 0 the weakest topology such that the linear functionals T T,ϕ (ϕ S ) are continuous, it is easy to see that the spaces Lp(Rn), 1 → h i ∈ 6 p 6 ∞, are continuously embedded in S 0. The same is true for the space of all finite Borel measures on Rn, i.e., B(Rn). There exists a simple and important characterization of tempered distributions: Theorem 1.47. A linear functional T on S is a tempered distribution if and only if there exists a constant C > 0 and integers ` and m such that

T,ϕ 6 C ∑ ϕ α,β |h i| α `, β m | | | |6 | |6 for all ϕ S . ∈ Proof. It is clear that the existence of C, `, m implies the continuity of T. Suppose T is continuous. It follows from the definition of the metric that a basis for the neighborhoods of the origin in S is the collection of sets Nε,`,m = ϕ : ∑ α `, β m ϕ α,β < ε , { | |6 | |6 | | } where ε > 0 and ` and m are integers, because ϕ ϕ as k ∞ if and only if ϕ ϕ 0 k → → | k − |α,β → 1.4. The class of tempered distributions - 19 - for all (α,β) in the topology induced by this system of neighborhoods and their translates. Thus, there exists such a set N satisfying T,ϕ 1 whenever ϕ N . ε,`,m |h i| 6 ∈ ε,`,m Let ϕ = ∑ α `, β m ϕ α,β for all ϕ S . If σ (0,ε), then ψ = σϕ/ ϕ Nε,`,m if k k | |6 | |6 | | ∈ ∈ k k ∈ ϕ = 0. From the linearity of T, we obtain 6 σ T,ϕ = T,ψ 1. ϕ |h i| |h i| 6 k k But this is the desired inequality with C = 1/σ. ut n Example 1.48. Let T S 0 and ϕ D(R ) with ϕ(0) = 1. Then the product ϕ(x/k)T is well- ∈ ∈ defined in S 0 by ϕ(x/k)T,ψ := T,ϕ(x/k)ψ , h i h i for all ψ S . If we consider the sequence T := ϕ(x/k)T , then ∈ k T ,ψ T,ϕ(x/k)ψ T,ψ h k i ≡ h i → h i as k ∞ since ϕ(x/k)ψ ψ in S . Thus, T T in S as k ∞. Moreover, T has compact → → k → 0 → k support as a tempered distribution in view of the compactness of ϕk = ϕ(x/k). Now we are ready to prove more serious and more useful fact. ∞ Theorem 1.49. Let T S 0, then there exists a sequence Tk k=0 S such that ∈ Z { } ⊂ Tk,ϕ = Tk(x)ϕ(x)dx T,ϕ , as k ∞, h i Rn → h i → where ϕ S . In short, S is dense in S with respect to the topology on S . ∈ 0 0 Proof. If h and g are integrable functions and ϕ S , then it follows, from Fubini’s theorem, ∈ that Z Z Z Z h g,ϕ = ϕ(x) h(x y)g(y)dydx = g(y) h(x y)ϕ(x)dxdy h ∗ i Rn Rn − Rn Rn − Z Z = g(y) Rh(y x)ϕ(x)dxdy = g,Rh ϕ , Rn Rn − h ∗ i where Rh(x) := h( x) is the reflection of h. − n R n Let now ψ D(R ) with n ψ(x)dx = 1 and ψ( x) = ψ(x). Let ζ D(R ) with ζ(0) = 1. ∈ R − ∈ Denote ψ (x) := knψ(kx). For any T S , denote T := ψ T˜ , where T˜ = ζ(x/k)T. From k ∈ 0 k k ∗ k k above considerations, we know that ψ T˜ ,ϕ = T˜ ,Rψ ϕ . h k ∗ k i h k k ∗ i Let us prove that these Tk meet the requirements of the theorem. In fact, we have T ,ϕ ψ T˜ ,ϕ = T˜ ,Rψ ϕ = ζ(x/k)T,ψ ϕ h k i ≡h k ∗ k i h k k ∗ i h k ∗ i = T,ζ(x/k)(ψ ϕ) T,ϕ , as k ∞, h k ∗ i → h i → by the fact ψ ϕ ϕ in S as k ∞ in view of Theorem 1.15, and the fact ζ(x/k) 1 k ∗ → → → pointwise as k ∞ since ζ(0) = 1 and ζ(x/k)ϕ ϕ in S as k ∞. Finally, since ψk, ζ n → n n → → ∈ D(R ), it follows that Tk D(R ) S (R ). ∈ ⊂ ut Definition 1.50. Let L : S S be a linear continuous mapping. Then, the dual/conjugate → mapping L : S S is defined by 0 0 → 0 L0T,ϕ := T,Lϕ , T S 0, ϕ S . h i h i ∈ ∈ Clearly, L0 is also a linear continuous mapping. Corollary 1.51. Any linear continuous mapping (or operator) L : S S admits a linear → continuous extension L˜ : S S . 0 → 0 - 20 - 1. The Fourier Transform and Tempered Distributions

Proof. If T S , then by Theorem 1.49, there exists a sequence T ∞ S such that T T ∈ 0 { k}k=0 ⊂ k → in S as k ∞. Hence, 0 → LT ,ϕ = T ,L0ϕ T,L0ϕ := LT˜ ,ϕ , as k ∞, h k i h k i → h i h i → for any ϕ S . ∈ ut Now, we can list the properties of tempered distributions about the multiplication, differentiation, translation, dilation and Fourier transform. Theorem 1.52. The following linear continuous operators from S into S admit unique linear continuous extensions as maps from S into S : For T S and ϕ S , 0 0 ∈ 0 ∈ i) ψT,ϕ := T,ψϕ , ψ S . h α i h i α ∈ α n ii) ∂ T,ϕ := T,( 1)| |∂ ϕ , α N0. h i h − i n ∈ iii) τhT,ϕ := T,τ hϕ , h R . h i h − ni ∈ iv) δλ T,ϕ := T, λ − δ /λ ϕ , 0 = λ R. h i h | | 1 i 6 ∈ v) F T,ϕ := T,F ϕ . h i h i Proof. See the previous deﬁnition, Theorem 1.49 and its corollary. ut 1 1 1 1 Remark 1.53. Since F − F T,ϕ = F T,F − ϕ = T,FF − ϕ = T,ϕ , we get F − F = 1 h i h i h i h i FF − = I in S 0. Example 1.54. Since for any ϕ S , ∈ Z F 1,ϕ = 1,F ϕ = (F ϕ)(ξ)dξ h i h i Rn n n Z ω − ω ωi0 ξ = | | | | e · (F ϕ)(ξ)dξ 2π 2π Rn n n ω − 1 ω − = | | F − F ϕ(0) = | | ϕ(0) 2π 2π n ω − = | | δ,ϕ , 2π h i we have n ω − 1ˆ = | | δ, in S 0. 2π ω n Moreover, δˇ = | | 1. 2π · Example 1.55. For ϕ S , we have ∈ Z ωix 0 δˆ,ϕ = δ,F ϕ = ϕˆ(0) = e− · ϕ(x)dx = 1,ϕ . h i h i Rn h i Thus, δˆ = 1 in S 0. Example 1.56. Since α α α α α ∂dδ,ϕ = ∂ δ,ϕˆ = ( 1)| | δ,∂ ϕˆ = δ,F [(ωiξ) ϕ] h i h i − h i h i = δˆ,(ωiξ)α ϕ = (ωiξ)α ,ϕ , h i h i we have ∂dα δ = (ωiξ)α .

Now, we shall show that the convolution can be defined on the class S 0. We first recall a notation we have used: If g is any function on Rn, we define its reflection, Rg, by letting Rg(x) = g( x). A direct application of Fubini’s theorem shows that if u, ϕ and ψ are all in S , − then 1.4. The class of tempered distributions - 21 - Z Z (u ϕ)(x)ψ(x)dx = u(x)(Rϕ ψ)(x)dx. Rn ∗ Rn ∗ R R The mappings ψ n (u ϕ)(x)ψ(x)dx and θ n u(x)θ(x)dx are linear functionals on S . 7→ R ∗ 7→ R If we denote these functionals by u ϕ and u, the last equality can be written in the form: ∗ u ϕ,ψ = u,Rϕ ψ . (1.15) h ∗ i h ∗ i If u S and ϕ, ψ S , the right side of (1.15) is well-defined since Rϕ ψ S . Furthermore, ∈ 0 ∈ ∗ ∈ the mapping ψ u,Rϕ ψ , being the composition of two continuous functions, is continu- 7→ h ∗ i ous. Thus, we can define the convolution of the distribution u with the testing function ϕ, u ϕ, ∗ by means of equality (1.15). It is easy to show that this convolution is associative in the sense that (u ϕ) ψ = u (ϕ ψ) ∗ ∗ ∗ ∗ whenever u S and ϕ, ψ S . The following result is a characterization of the convolution ∈ 0 ∈ we have just described.

Theorem 1.57. If u S 0 and ϕ S , then the convolution u ϕ is the function f , whose value n ∈ ∈ ∗ at x R is f (x) = u,τxRϕ , where τx denotes the translation by x operator. Moreover, f ∈ h i belongs to the class C∞ and it, as well as all its derivatives, are slowly increasing.

Proof. We ﬁrst show that f is C∞ slowly increasing. Let h = (0, ,h , ,0), then by part iii) ··· j ··· in Proposition 1.40, τx+hRϕ(y) τxRϕ(y) ∂Rϕ − τx (y), h j → − ∂y j as h 0, in the topology of S . Thus, since u is continuous, we have | | → f (x + h) f (x) τx+hRϕ(y) τxRϕ(y) ∂Rϕ − = u, − u, τx (y) h j h h j i → h − ∂y j i as h j 0. This, together with ii) in Proposition 1.40, shows that f has continuous ﬁrst partial → β derivatives. Since ∂Rϕ/∂y j S , we can iterate this argument and show that ∂ f exists and ∈ n β β β is continuous for all multi-index β N . We observe that ∂ f (x) = u,( 1)| |τx∂ Rϕ . Con- ∈ 0 h − i sequently, since ∂ β Rϕ S , if f were slowly increasing, then the same would hold for all the ∈ derivatives of f . In fact, that f is slowly increasing is an easy consequence of Theorem 1.47: There exist C > 0 and integers ` and m such that

f (x) = u,τxRϕ 6 C ∑ τxRϕ α,β . | | |h i| α `, β m | | | |6 | |6 α β α β But τ Rϕ = sup n y ∂ Rϕ(y x) = sup n (y+x) ∂ Rϕ(y) and the latter is clearly | x |α,β y R | − | y R | | bounded by a polynomial∈ in x. ∈ R In order to show that u ϕ is the function f , we must show that u ϕ,ψ = n f (x)ψ(x)dx. ∗ h ∗ i R But, Z u ϕ,ψ = u,Rϕ ψ = u, Rϕ(y x)ψ(x)dx h ∗ i h ∗ i h Rn − i Z = u, τxRϕ(y)ψ(x)dx h Rn i Z Z = u,τxRϕ(y) ψ(x)dx = f (x)ψ(x)dx, Rn h i Rn R since u is continuous and linear and the fact that the integral Rn τxRϕ(y)ψ(x)dx converges in S , which is the desired equality. ut - 22 - 1. The Fourier Transform and Tempered Distributions

1.5 *Characterization of operators commuting with translations

Having set down these facts of distribution theory, we shall now apply them to the study of the basic class of linear operators that occur in Fourier analysis: the class of operators that commute with translations. Deﬁnition 1.58. A vector space X of measurable functions on Rn is called closed under trans- n lations if for f X we have τy f X for all y R . Let X and Y be vector spaces of measurable ∈ ∈ ∈ functions on Rn that are closed under translations. Let also T be an operator from X to Y. We say that T commutes with translations or is translation invariant if

T(τy f ) = τy(T f ) for all f X and all y Rn. ∈ ∈ It is automatic to see that convolution operators commute with translations. One of the main goals of this section is to prove the converse, i.e., every bounded linear operator that commutes with translations is of convolution type. We have the following: Theorem 1.59. Let 1 6 p,q 6 ∞. Suppose T is a bounded linear operator from Lp(Rn) into Lq(Rn) that commutes with translations. Then there exists a unique tempered distribution u such that T f = u f , f S . ∗ ∀ ∈ The theorem will be a consequence of the following lemma. Lemma 1.60. Let 1 p ∞. If f Lp(Rn) has derivatives in the Lp norm of all orders n+1, 6 6 ∈ 6 then f equals almost everywhere a continuous function g satisfying α g(0) 6 C ∑ ∂ f p, | | α n+1 k k | |6 where C depends only on the dimension n and the exponent p.

n Proof. Let ξ R . Then there exists a Cn0 such that ∈ 2 (n+1)/2 n+1 α (1 + ξ ) 6 (1 + ξ1 + + ξn ) 6 Cn0 ∑ ξ . | | | | ··· | | α n+1 | | | |6 Let us ﬁrst suppose p = 1, we shall show fˆ L1. By part (v) in Proposition 1.3 and part (i) ∈ in Theorem 1.5, we have ˆ 2 (n+1)/2 α ˆ f (ξ) 6Cn0 (1 + ξ )− ∑ ξ f (ξ) | | | | α n+1 | || | | |6 2 (n+1)/2 α α =Cn0 (1 + ξ )− ∑ ω −| | F (∂ f )(ξ) | | α n+1 | | | | | |6 2 (n+1)/2 α 6C00(1 + ξ )− ∑ ∂ f 1. | | α n+1 k k | |6 2 (n+1)/2 n 1 n Since (1 + ξ )− deﬁnes an integrable function on R , it follows that fˆ L (R ) and, | | R 2 (n+1)/2 ∈ letting C = C N (1 + ξ ) dξ, we get 000 00 R | | − ˆ α f 1 6 C000 ∑ ∂ f 1. k k α n+1 k k | |6 Thus, by Theorem 1.19, f equals almost everywhere a continuous function g and by Theorem 1.5, 1.5. *Characterization of operators commuting with translations - 23 -

ω n g(0) f | | fˆ C ∂ α f . 6 ∞ 6 π 1 6 ∑ 1 | | k k 2 k k α n+1 k k | |6 Suppose now that p > 1. Choose ϕ D(Rn) such that ϕ(x) = 1 if x 1 and ϕ(x) = 0 if ∈ | | 6 x > 2. Then, it is clear that f ϕ L1(Rn). Thus, f ϕ equals almost everywhere a continuous | | ∈ function h such that α h(0) 6 C ∑ ∂ ( f ϕ) 1. | | α n+1 k k | |6 α α! µ ν By Leibniz’ rule for differentiation, we have ∂ ( f ϕ) = ∑µ+ν=α µ!ν! ∂ f ∂ ϕ, and then Z α α! µ ν ∂ ( f ϕ) 1 6 ∑ ∂ f ∂ ϕ dx k k x 2 µ!ν!| || | | |6 µ+ν=α Z ν µ 6 ∑ C sup ∂ ϕ(x) ∂ f (x) dx | | x 2 | | µ+ν=α x 62 | |6 Z| | µ µ 6A ∑ ∂ f (x) dx 6 AB ∑ ∂ f p, µ α x 62 | | µ α k k | |6| | | | | |6| | ν where A ∂ ϕ ∞, ν α , and B depends only on p and n. Thus, we can find a constant K > k k | | 6 | | such that α h(0) 6 K ∑ ∂ f p. | | α n+1 k k | |6 Since ϕ(x) = 1 if x 1, we see that f is equal almost everywhere to a continuous function | | 6 g in the sphere of radius 1 centered at 0, moreover, α g(0) = h(0) 6 K ∑ ∂ f p. | | | | α n+1 k k | |6 But, by choosing ϕ appropriately, the argument clearly shows that f equals almost everywhere a continuous function on any sphere centered at 0. This proves the lemma. ut Now, we turn to the proof of the previous theorem. Proof of Theorem 1.59. We first prove that β β n ∂ T f = T∂ f , f S (R ). (1.16) ∀ ∈ In fact, if h = (0, ,h , ,0) lies on the j-th coordinate axis, we have ··· j ··· τ (T f ) T f T(τ f ) T f τ f f h − = h − = T h − , h j h j h j τ f f since T is linear and commuting with translations. By part iii) in Proposition 1.40, h − h j → ∂ f in S as h 0 and also in Lp norm due to the density of S in Lp. Since T is bounded − ∂x j | | → p q τh(T f ) T f ∂T f q operator from L to L , it follows that − in L as h 0. By induction, we h j → − ∂x j | | → get (1.16). By Lemma 1.60, T f equals almost everywhere a continuous function g f satisfying β β g f (0) 6C ∑ ∂ (T f ) q = C ∑ T(∂ f ) q | | β n+1 k k β n+1 k k | |6 | |6 β 6 T C ∑ ∂ f p. k k β n+1 k k | |6 From the proof of Theorem 1.35, we know that the Lp norm of f S is bounded by a finite ∈ linear combination of L∞ norms of terms of the form xα f (x). Thus, there exists a m N such α β ∈ that g f (0) 6 C ∑ α m, β n+1 x ∂ f ∞ = C ∑ α m, β n+1 f α,β . Then, by Theorem 1.47, | | | |6 | |6 k k | |6 | |6 | | - 24 - 1. The Fourier Transform and Tempered Distributions the mapping f g (0) is a continuous linear functional on S , denoted by u . We claim that 7→ f 1 u = Ru is the linear functional we are seeking. Indeed, if f S , using Theorem 1.57, we 1 ∈ obtain

(u f )(x) = u,τxR f = u,R(τ x f ) = Ru,τ x f = u1,τ x f ∗ h i h − i h − i h − i =(T(τ x f ))(0) = (τ xT f )(0) = T f (x). − − We note that it follows from this construction that u is unique. The theorem is therefore proved. ut Combining this result with Theorem 1.57, we obtain the fact that T f , for f S , is almost ∈ everywhere equal to a C∞ function which, together with all its derivatives, is slowly increasing. Now, we give a characterization of operators commuting with translations in L1(Rn). Theorem 1.61. Let T be a bounded linear operator mapping L1(Rn) to itself. Then a necessary and sufficient condition that T commutes with translations is that there exists a measure µ in B(Rn) such that T f = µ f , for all f L1(Rn). One has then T = µ . ∗ ∈ k k k k Proof. We first prove the sufficiency. Suppose that T f = µ f for a measure µ B(Rn) and 1 n ∗ ∈ all f L (R ). Since B S 0, by Theorem 1.57, we have ∈ ⊂ τh(T f )(x) =(T f )(x h) = µ,τx hR f = µ(y), f ( y x + h) − h − i h − − i = µ,τ Rτ f = µ τ f = Tτ f , h x h i ∗ h h i.e., τ T = Tτ . On the other hand, we have T f = µ f µ f which implies h h k k1 k ∗ k1 6 k kk k1 T = µ . k k k k Now, we prove the necessariness. Suppose that T commutes with translations and T f 1 6 1 n k k T f 1 for all f L (R ). Then, by Theorem 1.59, there exists a unique tempered distribution k kk k ∈ µ such that T f = µ f for all f S . The remainder is to prove µ B(Rn). ∗ 1 ∈ ∈ We consider the family of L functions µε = µ W( ,ε) = TW( ,ε), ε > 0. Then by assump- ∗ · · tion and Lemma 1.14, we get

µε T W( ,ε) = T . k k1 6 k kk · k1 k k 1 1 n That is, the family µε is uniformly bounded in the L norm. Let us consider L (R ) as embed- { } n n n ded in the Banach space B(R ). B(R ) can be identiﬁed with the dual of C0(R ) by making R each ν B corresponding to the linear functional assigning to ϕ C the value n ϕ(x)dν(x). ∈ ∈ 0 R Thus, the unit sphere of B is compact in the weak* topology. In particular, we can ﬁnd a ν B ∈ and a null sequence εk such that µεk ν as k ∞ in this topology. That is, for each ϕ C0, { } Z → → Z ∈ lim ϕ(x)µε (x)dx = ϕ(x)dν(x). (1.17) k ∞ n k n → R R We now claim that ν, consider as a distribution, equals µ. R Therefore, we must show that µ,ψ = Rn ψ(x)dν(x) for all ψ S . Let ψε = W( ,ε) n h α i α ∈ · ∗ ψ. Then, for all α N0, we have ∂ ψε = W( ,ε) ∂ ψ. It follows from Theorem 1.15 that α ∈ α · ∗ ∂ ψε (x) converges to ∂ ψ(x) uniformly in x. Thus, ψε ψ in S as ε 0 and this implies → → that µ,ψε µ,ψ . But, since W( ,ε) = RW( ,ε), h i → h i · · Z µ,ψε = µ,W( ,ε) ψ = µ W( ,ε),ψ = µε (x)ψ(x)dx. h i h · ∗ i h ∗ · i Rn Thus, putting ε = εk, letting k ∞ and applying (1.17) with ϕ = ψ, we obtain the desired R → equality µ,ψ = n ψ(x)dν(x). Hence, µ B. This completes the proof. h i R ∈ ut For L2, we can also give a very simple characterization of these operators. 1.5. *Characterization of operators commuting with translations - 25 -

Theorem 1.62. Let T be a bounded linear transformation mapping L2(Rn) to itself. Then a necessary and sufficient condition that T commutes with translation is that there exists an m ∞ n 2 n ∈ L (R ) such that T f = u f with uˆ = m, for all f L (R ). One has then T = m ∞. ∗ ∈ k k k k Proof. If v S and ψ S , we define their product, vψ, to be the element of S such that ∈ 0 ∈ 0 vψ,ϕ = v,ψϕ for all ϕ S . With the product of a distribution with a testing function so h i h i ∈ defined we first observe that whenever u S and ϕ S , then ∈ 0 ∈ F (u ϕ) = uˆϕˆ. (1.18) ∗ To see this, we must show that F (u ϕ),ψ = uˆϕˆ,ψ for all ψ S . It follows immediately, h ∗ i h i ∈ from (1.15), part (vi) in Proposition 1.3 and the Fourier inversion formula, that 1 F (u ϕ),ψ = u ϕ,ψˆ = u,Rϕ ψˆ = uˆ,F − (Rϕ ψˆ ) h ∗ i h ∗ i h ∗ i h ∗ i ω n = uˆ, | | (F (Rϕ ψˆ ))( ξ) 2π ∗ − ω n = uˆ, | | (F (Rϕ))( ξ)(F ψˆ )( ξ) = uˆ,ϕˆ(ξ)ψ(ξ) 2π − − h i = uˆϕˆ,ψ . h i Thus, (1.18) is established. Now, we prove the necessariness. Suppose that T commutes with translations and T f 2 6 2 n k k T f 2 for all f L (R ). Then, by Theorem 1.59, there exists a unique tempered distribution k kk k ∈ u such that T f = u f for all f S . The remainder is to proveu ˆ L∞(Rn). ω ∗ ∈ n/∈2 | | x 2 ω Let ϕ = e 2 , then, we have ϕ S and ϕˆ = | | − ϕ by Theorem 1.10 with 0 − | | 0 ∈ 0 2π 0 a = 1/2 ω . Thus, Tϕ = u ϕ L2 and therefore Φ := F (u ϕ ) = uˆϕˆ L2 by (1.18) and | | 0 ∗ 0 ∈ 0 ∗ 0 0 ∈ ω n/2 ω 2 | 2 | ξ the Plancherel theorem. Let m(ξ) = |2π| e | | Φ0(ξ) = Φ0(ξ)/ϕˆ0(ξ). We claim that F (u ϕ) = mϕˆ (1.19) ∗ for all ϕ S . By (1.18), it suffices to show that uˆϕˆ,ψ = mϕˆ,ψ for all ψ D since D is ∈ h n/2 i h i ∈ ω ω ξ 2 dense in S . But, if ψ D, then (ψ/ϕˆ )(ξ) = | | ψ(ξ)e | 2 | D; thus, ∈ 0 2π | | ∈ uˆϕˆ,ψ = uˆ,ϕψˆ = uˆ,ϕˆϕˆ ψ/ϕˆ = uˆϕˆ ,ϕψˆ /ϕˆ h i h i h 0 0i h 0 0i n/2 Z ω ω | | ξ 2 = Φ0(ξ)ϕˆ(ξ) | | ψ(ξ)e 2 | | dξ Rn 2π Z = m(ξ)ϕˆ(ξ)ψ(ξ)dξ = mϕˆ,ψ . Rn h i It follows immediately thatu ˆ = m: We have just shown that uˆ,ϕψˆ = mϕˆ,ψ = m,ϕψˆ h i h i h i for all ϕ S and ψ D. Selecting ϕ such that ϕˆ(ξ) = 1 for ξ suppψ, this shows that ∈ ∈ ∈ uˆ,ψ = m,ψ for all ψ D. Thus,u ˆ = m. h i h i ∈ Due to n/2 ω − mϕˆ = F (u ϕ) = | | u ϕ k k2 k ∗ k2 2π k ∗ k2 n/2 ω − 6 | | T ϕ 2 = T ϕˆ 2 2π k kk k k kk k for all ϕ S , it follows that ∈ - 26 - 1. The Fourier Transform and Tempered Distributions Z 2 2 2 T m ϕˆ dξ > 0, Rn k k − | | | | for all ϕ S . This implies that T 2 m 2 0 for almost all x Rn. Hence, m L∞(Rn) and ∈ k k −| | > ∈ ∈ m ∞ T . k k 6 k k Finally, we can show the sufficiency easily. Ifu ˆ = m L∞(Rn), the Plancherel theorem and ∈ (1.18) immediately imply that ω n/2 T f 2 = u f 2 = | | m fˆ 2 6 m ∞ f 2 k k k ∗ k 2π k k k k k k which yields T 6 m ∞. k k k∞ k Thus, if m = uˆ L , then T = m ∞. ∈ k k k k ut For further results, one can see [SW71, p.30] and [Gra04, p.137-140]. Chapter 2 Interpolation of Operators

2.1 Riesz-Thorin’s and Stein’s interpolation theorems

We ﬁrst present a notion that is central to complex analysis, that is, the holomorphic or analytic function. Let Ω be an open set in C and f a complex-valued function on Ω. The function f is holomorphic at the point z Ω if the quotient 0 ∈ f (z + h) f (z ) 0 − 0 (2.1) h converges to a limit when h 0. Here h C and h = 0 with z0 + h Ω, so that the quotient → ∈ 6 ∈ is well deﬁned. The limit of the quotient, when it exists, is denoted by f 0(z0), and is called the derivative of f at z0: f (z0 + h) f (z0) f 0(z0) = lim − . h 0 h → It should be emphasized that in the above limit, h is a complex number that may approach 0 from any directions. The function f is said to be holomorphic on Ω if f is holomorphic at every point of Ω. If C is a closed subset of C, we say that f is holomorphic on C if f is holomorphic in some open set containing C. Finally, if f is holomorphic in all of C we say that f is entire. Every holomorphic function is analytic, in the sense that it has a power series expansion near every point, and for this reason we also use the term analytic as a synonym for holomorphic. For more details, one can see [SS03, pp.8-10].

Example 2.1. The function f (z) = z is holomorphic on any open set in C, and f 0(z) = 1. Example 2.2. The function 1/z is holomorphic on any open set in C that does not contain the origin, and f (z) = 1/z2. 0 − Example 2.3. The function f (z) = z¯ is not holomorphic. Indeed, we have f (z + h) f (z ) h¯ 0 − 0 = h h which has no limit as h 0, as one can see by ﬁrst taking h real and then h purely imaginary. → The next result pertains to the size of a holomorphic function. Theorem 2.4 (Maximum modulus principle). Suppose that Ω is a region with compact closure Ω¯ . If f is holomorphic on Ω and continuous on Ω¯ , then sup f (z) 6 sup f (z) . z Ω | | z Ω¯ Ω | | ∈ ∈ \

27 - 28 - 2. Interpolation of Operators

Proof. See [SS03, p.92]. ut For convenience, let S = z C : 0 ℜz 1 be the closed strip, S◦ = z C : 0 < ℜz < 1 { ∈ 6 6 } { ∈ } be the open strip, and ∂S = z C : ℜz 0,1 . { ∈ ∈ { }} Theorem 2.5 (Phragmen-Lindelöf theorem/Maximum principle). Assume that f (z) is analytic on S◦ and bounded and continuous on S. Then sup f (z) 6 max sup f (it) , sup f (1 + it) . z S | | t R | | t R | | ∈ ∈ ∈ Proof. Assume that f (z) 0 as ℑz ∞. Consider the mapping h : S C deﬁned by → | | → → eiπz i h(z) = − , z S. (2.2) eiπz + i ∈ Then h is a bijective mapping from S onto U = z C : z 1 1 , that is analytic in S◦ { ∈ | | 6 } \ {± } and maps ∂S onto z = 1 1 . Therefore, g(z) := f (h 1(z)) is bounded and continuous on {| | }\{± } − U and analytic in the interior U◦. Moreover, because of lim ℑz ∞ f (z) = 0, limz 1 g(z) = 0 | |→ →± and we can extend g to a continuous function on z C : z 1 . Hence, by the maximum { ∈ | | 6 } modulus principle (Theorem 2.4), we have g(z) 6 max g(ω) = max sup f (it) , sup f (1 + it) , | | ω =1 | | t R | | t R | | | | ∈ ∈ which implies the statement in this case. Next, if f is a general function as in the assumption, then we consider δ(z z )2 f (z) = e − 0 f (z), δ > 0, z S◦. δ,z0 0 ∈ 2 2 2 δ(z z0) δ(x y ) Since e − e − with z z0 = x + iy, 1 x 1 and y R, we have fδ,z (z) 0 | | 6 − − 6 6 ∈ 0 → as ℑz ∞. Therefore | | →

f (z0) = fδ,z0 (z0) 6 max sup fδ,z0 (it) , sup fδ,z0 (1 + it) | | | | t R | | t R | | ∈ ∈ δ 6e max sup f (it) , sup f (1 + it) . t R | | t R | | ∈ ∈ Passing to the limit δ 0, we obtain the desired result since z S is arbitrary. → 0 ∈ ut As a corollary we obtain the following three lines theorem, which is the basis for the proof of the Riesz-Thorin interpolation theorem and the complex interpolation method.

Theorem 2.6 (Hadamard three lines theorem). Assume that f (z) is analytic on S◦ and bounded and continuous on S. Then 1 θ θ − sup f (θ + it) 6 sup f (it) sup f (1 + it) , t R | | t R | | t R | | ∈ ∈ ∈ for every θ [0,1]. ∈ Proof. Denote

A0 := sup f (it) , A1 := sup f (1 + it) . t R | | t R | | ∈ ∈ Let λ R and deﬁne ∈ λz Fλ (z) = e f (z). Then by Theorem 2.5, it follows that F (z) max(A ,eλ A ). | λ | 6 0 1 2.1. Riesz-Thorin’s and Stein’s interpolation theorems - 29 -

Hence, λθ λ f (θ + it) e− max(A ,e A ) | | 6 0 1 A0 λ for all t R. Choosing λ = ln such that e A1 = A0, we complete the proof. ∈ A1 ut In order to state the Riesz-Thorin theorem in a general version, we will state and prove it in measurable spaces instead of Rn only. Let (X, µ) be a measure space, µ always being a positive measure. We adopt the usual convention that two functions are considered equal if they agree except on a set of µ-measure zero. Then we denote by Lp(X,dµ) (or simply Lp(dµ), Lp(X) or even Lp) the Lebesgue-space of (all equivalence classes of) scalar-valued µ-measurable functions f on X, such that Z 1/p p f p = f (x) dµ k k X | | is ﬁnite. Here we have 1 6 p < ∞. In the limiting case, p = ∞, Lp consists of all µ-measurable and bounded functions. Then we write

f ∞ = sup f (x) . k k X | | In this section, scalars are supposed to be complex numbers. Let T be a linear mapping from Lp = Lp(X,dµ) to Lq(Y,dν). This means that T(α f +βg) = αT( f ) + βT(g). We shall write T : Lp Lq → if in addition T is bounded, i.e., if T f q A = sup k k f =0 f p 6 k k is finite. The number A is called the norm of the mapping T. It will also be necessary to treat operators T defined on several Lp spaces simultaneously. p p Definition 2.7. We define L 1 + L 2 to be the space of all functions f , such that f = f1 + f2, with f Lp1 and f Lp2 . 1 ∈ 2 ∈ Suppose now p1 < p2. Then we observe that Lp Lp1 + Lp2 , p [p , p ]. ⊂ ∀ ∈ 1 2 In fact, let f Lp and let γ be a fixed positive constant. Set ∈ f (x), f (x) > γ, f (x) = | | 1 0, f (x) γ, | | 6 and f2(x) = f (x) f1(x). Then Z− Z Z p p p p p p p f (x) 1 dx = f (x) f (x) 1− dx γ 1− f (x) dx, | 1 | | 1 | | 1 | 6 | | since p1 p 6 0. Similarly, − Z Z Z p p p p p p p f (x) 2 dx = f (x) f (x) 2− dx γ 2− f (x) dx, | 2 | | 2 | | 2 | 6 | | so f Lp1 and f Lp2 , with f = f + f . 1 ∈ 2 ∈ 1 2 Now, we have the following well-known theorem. Theorem 2.8 (The Riesz-Thorin interpolation theorem). Let T be a linear operator with domain p p (L 0 + L 1 )(X,dµ), p0, p1,q0,q1 [1,∞]. Assume that ∈ p T f q A f p , if f L 0 (X,dµ), k kL 0 (Y,dν) 6 0k kL 0 (X,dµ) ∈ and - 30 - 2. Interpolation of Operators

p T f q A f p , if f L 1 (X,dµ), k kL 1 (Y,dν) 6 1k kL 1 (X,dµ) ∈ for some p = p and q = q . Suppose that for a certain 0 < θ < 1 0 6 1 0 6 1 1 1 θ θ 1 1 θ θ = − + , = − + . (2.3) p p0 p1 q q0 q1 Then p T f q A f p , if f L (X,dµ), k kL (Y,dν) 6 θ k kL (X,dµ) ∈ with 1 θ θ Aθ 6 A0− A1 . (2.4)

Remark 2.9. 1) (2.4) means that Aθ is logarithmically convex, 1 i.e., lnAθ is convex. q (1, 1) ( 1 , 1 ) 2) The geometrical meaning of (2.3) is that the points (1/p,1/q) p0 q0 are the points on the line segment between (1/p0,1/q0) and (1/p ,1/q ). 1 1 1 1 ( p , q ) 3) The original proof of this theorem, published in 1926 by Mar- cel Riesz, was a long and difﬁcult calculation. Riesz’ student G. ( 1 , 1 ) p1 q1

Olof Thorin subsequently discovered a far more elegant proof O 1 and published it in 1939, which contains the idea behind the com- p plex interpolation method. Proof. Denote Z h,g = h(y)g(y)dν(y) h i Y and 1/q = 1 1/q. Then we have, by Hölder inequality, 0 − h q = sup h,g , and Aθ = sup T f ,g . k k g =1 |h i| f p= g =1 |h i| k kq0 k k k kq0 p Noticing that Cc(X) is dense in L (X, µ) for 1 6 p < ∞, we can assume that f and g are bounded with compact supports since p, q < ∞.1 Thus, we have f (x) M < ∞ for 0 | | 6 all x X, and supp f = x X : f (x) = 0 is compact, i.e., µ(supp f ) < ∞ which implies ∈ { ∈ 6 } R f (x) `dµ(x) = R f (x) `dµ(x) M`µ(supp f ) < ∞ for any ` > 0. So g does. X | | supp f | | 6 For 0 6 ℜz 6 1, we put 1 1 z z 1 1 z z = − + , = − + , p(z) p0 p1 q0(z) q0 q10 and p f (x) η(z) =η(x,z) = f (x) p(z) , x X; | | f (x) ∈ | | q0 g(y) ζ(z) =ζ(y,z) = g(y) q0(z) , y Y. | | g(y) ∈ | | Now, we prove η(z), η (z) Lp j for j = 0,1. Indeed, we have 0 ∈

p 1 z z 1 ℜz ℜz ℑz ℑz p( − + ) p( − + )+ip( ) η(z) = f (x) p(z) = f (x) p0 p1 = f (x) p0 p1 p1 − p0 | | | | | | | | 1 ℜz ℜz p p( − + ) = f (x) p0 p1 = f (x) p(ℜz) . | | | | Thus,

1 1 1/q0 Otherwise, it will be p0 = p1 = ∞ if p = ∞, or θ = 1/q− 1/q > 1 if q0 = ∞. 1− 0 2.1. Riesz-Thorin’s and Stein’s interpolation theorems - 31 -

Z Z pp j p j p j p(ℜz) η(z) p j = η(x,z) dµ(x) = f (x) dµ(x) < ∞. k k X | | X | | We have p p 0 f (x) η0(z) = f (x) p(z) ln f (x) | | p(z) f (x) | | | | 1 1 p f (x) =p f (x) p(z) ln f (x) . p1 − p0 | | f (x) | | α | | On one hand, we have lim f (x) 0+ f (x) ln f (x) = 0 for any α > 0, that is, ε > 0, δ > 0 | |→ | | | | ∀ ∃ s.t. f (x) α ln f (x) < ε if f (x) < δ. On the other hand, if f (x) > δ, then we have || | | || | | | | f (x) α ln f (x) Mα ln f (x) Mα max( lnM , lnδ ) < ∞. Thus, f (x) α ln f (x) C. || | | || 6 | | || 6 | | | | || | | || 6 Hence,

1 1 p α p(z) α η0(z) =p f (x) − f (x) ln f (x) | | p1 − p0 | | | | | | || p α p α C f (x) p(z) − = C f (x) p(ℜz) − , 6 | | | | which yields Z p p j ( p(ℜz) α)p j η0(z) p j 6 C f (x) − dµ(x) < ∞. k k X | | Therefore, η(z), η (z) Lp j for j = 0,1. So ζ(z), ζ (z) Lq0j for j = 0,1 in the same way. It 0 ∈ 0 ∈ follows that Tη(z) Lq j , and (Tη) (z) Lq j with 0 < ℜz < 1, for j = 0,1. This implies the ∈ 0 ∈ existence of F(z) = Tη(z),ζ(z) , 0 ℜz 1. h i 6 6 Since dF(z) d d Z = Tη(z),ζ(z) = (Tη)(y,z)ζ(y,z)dν(y) dz dzh i dz Y Z Z = (Tη)z(y,z)ζ(y,z)dν(y) + (Tη)(y,z)ζz(y,z)dν(y) Y Y = (Tη)0(z),ζ(z) + Tη(z),ζ 0(z) < ∞, h i h i F(z) is analytic on the open strip 0 < ℜz < 1. Moreover it is easy to see that F(z) is bounded and continuous on the closed strip 0 6 ℜz 6 1. Next, we note that for j = 0,1 p p η( j + it) = f j = 1. k kp j k kp Similarly, we also have ζ( j + it) = 1 for j = 0,1. Thus, for j = 0,1 k kq0j

F( j + it) = Tη( j + it),ζ( j + it) 6 Tη( j + it) q j ζ( j + it) q | | |h i| k k k k 0j

6A j η( j + it) p j ζ( j + it) q = A j. k k k k 0j Using Hadamard three line theorem, reproduced as Theorem 2.6, we get the conclusion 1 θ θ F(θ + it) A − A , t R. | | 6 0 1 ∀ ∈ 1 θ θ Taking t = 0, we have F(θ) 6 A0− A1 . We also note that η(θ) = f and ζ(θ) = g, thus | | 1 θ θ 1 θ θ F(θ) = T f ,g . That is, T f ,g A − A . Therefore, A A − A . h i |h i| 6 0 1 θ 6 0 1 ut Now, we shall give two rather simple applications of the Riesz-Thorin interpolation theorem.

Theorem 2.10 (Hausdorff-Young inequality). Let 1 6 p 6 2 and 1/p + 1/p0 = 1. Then the Fourier transform deﬁned as in (1.1) satisﬁes - 32 - 2. Interpolation of Operators

n/p ω − 0 F f p 6 | | f p. k k 0 2π k k

1 ∞ Proof. It follows by interpolation between the L -L result F f ∞ f (cf. Theorem 1.5) k k 6 k k1 ω n/2 and Plancherel’s theorem F f = | | − f (cf. Theorem 1.26). k k2 2π k k2 ut Theorem 2.11 (Young’s inequality for convolutions). If f Lp(Rn) and g Lq(Rn), 1 ∈ ∈ 6 p,q,r ∞ and 1 = 1 + 1 1, then 6 r p q − f g f g . k ∗ kr 6 k kpk kq Proof. We ﬁx f Lp, p [1,∞] and then will apply the Riesz-Thorin interpolation theorem to ∈ ∈ the mapping g f g. Our endpoints are Hölder’s inequality which gives 7→ ∗ f g(x) f g | ∗ | 6 k kpk kp0 and thus g f g maps Lp0 (Rn) to L∞(Rn) and the simpler version of Young’s inequality 7→ ∗ (proved by Minkowski’s inequality) which tells us that if g L1, then ∈ f g f g . k ∗ kp 6 k kpk k1 Thus g f g also maps L1 to Lp. Thus, this map also takes Lq to Lr where 7→ ∗ 1 1 θ θ 1 1 θ θ = − + , and = − + . q 1 p0 r p ∞ Eliminating θ, we have 1 = 1 + 1 1. r p q − The condition q > 1 is equivalent with θ > 0 and r > 1 is equivalent with the condition θ 6 1. Thus, we obtain the stated inequality for precisely the exponents p, q and r in the hypothesis. ut Remark 2.12. The sharp form of Young’s inequality for convolutions can be found in [Bec75, Theorem 3], we just state it as follows. Under the assumption of Theorem 2.11, we have f g (A A A )n f g , k ∗ kr 6 p q r0 k kpk kq 1/m 1/m 1/2 where A = (m /m 0 ) for m (1,∞), A = A∞ = 1 and primes always denote dual m 0 ∈ 1 exponents, 1/m + 1/m0 = 1. The Riesz-Thorin interpolation theorem can be extended to the case where the interpolated operators allowed to vary. In particular, if a family of operators depends analytically on a parameter z, then the proof of this theorem can be adapted to work in this setting. We now describe the setup for this theorem. Suppose that for every z in the closed strip S there is an associated linear operator Tz deﬁned on the space of simple functions on X and taking values in the space of measurable functions on Y such that Z Tz( f )g dν < ∞ (2.5) Y | | whenever f and g are simple functions on X and Y, respectively. The family T is said to be { z}z analytic if the function Z z Tz( f )gdν (2.6) → Y is analytic in the open strip S◦ and continuous on its closure S. Finally, the analytic family is of admissible growth if there is a constant 0 < a < π and a constant Cf ,g such that Z a ℑz e− | | ln Tz( f )gdν 6 Cf ,g < ∞ (2.7) Y for all z S. The extension of the Riesz-Thorin interpolation theorem is now stated. ∈ 2.2. The distribution function and weak Lp spaces - 33 -

Theorem 2.13 (Stein interpolation theorem). Let Tz be an analytic family of linear operators of admissible growth. Let 1 6 p0, p1,q0,q1 6 ∞ and suppose that M0 and M1 are real-valued functions such that b t supe− | | lnMj(t) < ∞ (2.8) t R ∈ for j = 0,1 and some 0 < b < π. Let 0 < θ < 1 satisfy 1 1 θ θ 1 1 θ θ = − + , and = − + . (2.9) p p0 p1 q q0 q1 Suppose that T ( f ) M (t) f , T ( f ) M (t) f (2.10) k it kq0 6 0 k kp0 k 1+it kq1 6 1 k kp1 for all simple functions f on X. Then T ( f ) M(θ) f , when 0 < θ < 1 (2.11) k θ kq 6 k kp for all simple functions f on X, where sinπθ Z lnM (t) lnM (t) M(θ) = exp 0 + 1 dt . 2 R coshπt cosπθ coshπt + cosπθ − p q By density, Tθ has a unique extension as a bounded operator from L (X, µ) into L (Y,ν) for all p and q as in (2.9). The proof of the Stein interpolation theorem can be obtained from that of the Riesz-Thorin theorem simply “by adding a single letter of the alphabet”. Indeed, the way the Riesz-Thorin theorem is proven is to study an expression of the form F(z) = Tη(z),ζ(z) , h i the Stein interpolation theorem proceeds by instead studying the expression F(z) = T η(z),ζ(z) . h z i One can then repeat the proof of the Riesz-Thorin theorem more or less verbatim to obtain the Stein interpolation theorem. Of course, the explicit expression of M(θ) need an extension of the three lines theorem. For the detailed proof, one can see [SW71, p. 205-209] or [Gra04, p.38-42].

2.2 The distribution function and weak Lp spaces

We shall now be interested in giving a concise expression for the relative size of a function. Thus we give the following concept. Definition 2.14. Let f (x) be a measurable function on Rn. Then the function f : [0,∞) [0,∞] ∗ 7→ defined by f (α) = m( x : f (x) > α ) ∗ { | | } is called to be the distribution function of f . The distribution function f provides information about the size of f but not about the be- ∗ havior of f itself near any given point. For instance, a function on Rn and each of its translates have the same distribution function. In particular, the decrease of f (α) as α grows describes the relative largeness of the function; ∗ this is the main concern locally. The increase of f (α) as α tends to zero describes the relative ∗ smallness of the function “at infinity”; this is its importance globally, and is of no interest if, for example, the function is supported on a bounded set. - 34 - 2. Interpolation of Operators

Now, we give some properties of distribution functions. Proposition 2.15. For the distribution function, we have following fundamental properties. (i) f (α) is decreasing and continuous on the right. ∗ (ii) If f (x) 6 g(x) , then f (α) 6 g (α). | | | | ∗ ∗ (iii) If f (x) 6 liminfk ∞ fk(x) for a.e. x, then f (α) 6 liminfk ∞( fk) (α) for any α > 0. | | → | | ∗ → ∗ (iv) If f (x) 6 g(x) + h(x) , then f (α1 + α2) 6 g (α1) + h (α2) for any α1,α2 > 0. | | | | | | ∗ ∗ ∗ (v) ( f g) (α1α2) 6 f (α1) + g (α2) for any α1,α2 > 0. ∗ ∗ ∗ p R p (vi) For any p (0,∞) and α > 0, it holds f (α) 6 α− x: f (x) >α f (x) dx. p ∈ p ∗ { | | p} | | (vii) If f L , p [1,∞), then limα +∞ α f (α) = 0 = limα 0 α f (α). R∈∞ p 1 ∈ → ∗ p → ∗ (viii) If α − f (α)dα < ∞, p [1,∞), then α f (α) 0 as α +∞ and α 0, re- 0 ∗ ∈ ∗ → → → spectively.

Proof. For simplicity, denote E (α) = x : f (x) > α for α > 0. f { | | } (i) Let αk is a decreasing positive sequence which tends to α, then we have E f (α) = ∞ { } k= E f (αk). Since E f (αk) is a increasing sequence of sets, it follows limk ∞ f (αk) = f (α). ∪ 1 { } → ∗ ∗ This implies the continuity of f (α) on the right. ∗ (iii) Let E = x : f (x) > α and Ek = x : fk(x) > α , k N. By the assumption and the { | | } { | | } ∈ deﬁnition of inferior limit, i.e.,

f (x) 6 liminf fk(x) = sup inf fk(x) , | | k ∞ | | ` N k>` | | → ∈ for x E, there exists an integer M such that for all k > M, fk(x) > α. Thus, E S∞ ∈T∞ | | ⊂ M=1 k=M Ek, and for any ` > 1, ∞ ! \ m m m m Ek 6 inf (Ek) 6 sup inf (Ek) = liminf (Ek). k ` k ` k ∞ k=` > ` > → Since T∞ E ∞ is an increasing sequence of sets, we obtain { k=M k}M=1 ∞ ∞ ! ∞ ! m [ \ \ f (α) = (E) 6 m Ek = lim m Ek 6 liminf( fk) (α). ∗ M ∞ k ∞ ∗ M=1 k=M → k=M → (v) Noticing that x : f (x)g(x) > α α x : f (x) > α x : g(x) > α , we have the { | | 1 2} ⊂ { | | 1}∪{ | | 2} desired result. m R R f (x) p (vi) f (α) = ( x : f (x) > α ) = x: f (x) >α dx 6 x: f (x) >α ( | α | ) dx p R ∗ {p | | } { | | } { | | } = α− x: f (x) >α f (x) dx. { | | } | | p R p R p m (vii) From (vi), it follows α f (α) 6 x f (x) >α f (x) dx 6 n f (x) dx. Thus, ( x : ∗ : | | R | | { f (x) > α ) 0 as α +∞ and { | | } | | } → → Z lim f (x) pdx = 0. α +∞ x: f (x) >α | | → { | | } Hence, α p f (α) 0 as α +∞ since α p f (α) > 0. ∗ → → ∗ For any 0 < α < β, we have, by noticing that 1 6 p < ∞, that p p p lim α f (α) = lim α ( f (α) f (β)) = lim α m( x : α < f (x) 6 β ) α 0 ∗ α 0 ∗ − ∗ α 0 { | | } → Z→ → p 6 f (x) dx. x: f (x) β | | { | |6 } By the arbitrariness of β, it follows α p f (α) 0 as α 0. R α p p ∗p → → (viii) Since α/ (t )0dt = α (α/2) and f (α) 6 f (t) for t 6 α, we have 2 − ∗ ∗ Z α p p p 1 f (α)α (1 2− ) 6 p t − f (t)dt ∗ − α/2 ∗ 2.2. The distribution function and weak Lp spaces - 35 - which implies the desired result. For other ones, they are easy to verify. ut From this proposition, we can prove the following equivalent norm of Lp spaces. Theorem 2.16 (The equivalent norm of Lp). Let f (x) be a measurable function in Rn, then R ∞ p 1 1/p i) f p = p α − f (α)dα , if 1 6 p < ∞, k k 0 ∗ ii) f ∞ = inf α : f (α) = 0 . k k { ∗ } Proof. In order to prove i), we first prove the following conclusion: If f (x) is finite and f (α) < ∗ ∞ for any α > 0, then Z Z ∞ f (x) pdx = α pd f (α). (2.12) Rn | | − 0 ∗ Indeed, the r.h.s. of the equality is well-defined from the conditions. For the integral in the l.h.s., we can split it into Lebesgue integral summation. Let 0 < ε < 2ε < < kε < and n ··· ··· E j = x R : ( j 1)ε < f (x) jε , j = 1,2, , { ∈ − | | 6 } ··· then, m(E j) = f (( j 1)ε) f ( jε), and ∗ ∗ Z − −∞ ∞ p pm p f (x) dx = lim ∑ ( jε) (E j) = lim ∑ ( jε) [ f ( jε) f (( j 1)ε)] n | | ε 0 − ε 0 ∗ − ∗ − R → j=1 → j=1 Z ∞ = α pd f (α). − 0 ∗ Now we return to prove i). If the values of both sides are infinite, then it is clearly true. If one of the integral is finite, then it is clear that f (α) < +∞ and f (x) is finite almost everywhere. ∗ Thus (2.12) is valid. p n R ∞ p 1 If either f L (R ) or α − f (α)dα < ∞ for 1 6 p < ∞ , then we always have ∈ 0 ∗ α p f (α) 0 as α +∞ and α 0 from the property (vii) and (viii) in Proposition 2.15. ∗ → → → Therefore, integrating by part, we have Z ∞ Z ∞ Z ∞ p p 1 p +∞ p 1 α d f (α) =p α − f (α)dα α f (α) 0 = p α − f (α)dα. − 0 ∗ 0 ∗ − ∗ | 0 ∗ Thus, i) is true. For ii), we have inf α : f (α) = 0 =inf α : m( x : f (x) > α ) = 0 { ∗ } { { | | } } ∞ =inf α : f (x) 6 α, a.e. = esssupx Rn f (x) = f L . { | | } ∈ | | k k We complete the proofs. ut Notice that the same argument yields the more general fact that for any increasing continuously differentiable function ϕ on [0,∞) with ϕ(0) = 0 we have Z Z ∞ ϕ( f )dµ = ϕ0(α) f (α)dα. (2.13) X | | 0 ∗ Using the distribution function f , we now introduce the weak Lp-spaces denoted by Lp. ∗ ∗ p Definition 2.17. The space L , 1 6 p < ∞, consists of all f such that ∗ 1/p f Lp = supα f (α) < ∞. k k ∗ α ∗ In the limiting case p = ∞, we put L∞ = L∞. ∗ By the part (iv) in Proposition 2.15 and the triangle inequality of Lp norms, we have 1/p f + g Lp 6 2 ( f Lp + g Lp ). k k ∗ k k ∗ k k ∗ - 36 - 2. Interpolation of Operators

Thus, one can verify that Lp is a quasi-normed vector space. The weak Lp spaces are larger than ∗ the usual Lp spaces. We have the following: p p p Theorem 2.18. For any 1 6 p < ∞, and any f L , we have f Lp 6 f p, hence L L . ∈ k k ∗ k k ⊂ ∗ Proof. From the part (vi) in Proposition 2.15, we have Z 1/p 1/p p α f (α) 6 f (x) dx ∗ x: f (x) >α | | { | | } which yields the desired result. ut p p n/p The inclusion L L is strict for 1 6 p < ∞. For example, let h(x) = x − . Obviously, h ⊂ ∗ p | | is not in Lp(Rn) but h is in L (Rn) and we may check easily that ∗ 1/p m n/p 1/p h Lp =supαh (α) = supα( ( x : x − > α )) k k ∗ α ∗ α { | | } p/n 1/p p 1/p =supα(m( x : x < α− )) = supα(α− Vn) α { | | } α 1/p =Vn , n/2 n where Vn = π /Γ (1 + n/2) is the volume of the unit ball in R and Γ -function Γ (z) = R ∞ z 1 t 0 t − e− dt for ℜz > 0. It is not immediate from their deﬁnition that the weak Lp spaces are complete with respect to the quasi-norm p . For the completeness, we will state it later as a special case of Lorentz k · kL spaces. ∗

2.3 The decreasing rearrangement and Lorentz spaces

The spaces Lp are special cases of the more general Lorentz spaces Lp,q. In their definition, ∗ we use yet another concept, i.e., the decreasing rearrangement of functions. Definition 2.19. If f is a measurable function on Rn, the decreasing rearrangement of f is the function f : [0,∞) [0,∞] defined by ∗ 7→ f ∗(t) = inf α > 0 : f (α) 6 t , { ∗ } where we use the convention that inf∅ = ∞. Now, we first give some examples of distribution function and decreasing rearrangement. The first example establish some important relations between a simple function, its distribution function and decreasing rearrangement. Example 2.20 (Decreasing rearrangement of a simple function). Let f be a simple function of the following form k

f (x) = ∑ a jχA j (x) j=1 where a1 > a2 > > ak > 0, A j = x R : f (x) = a j and χA is the characteristic function of ··· { ∈ } the set A (see Figure (a)). Then k k f (α) = m( x : f (x) > α ) = m( x : a jχA (x) > α ) = b jχB (α), ∗ ∑ j ∑ j { | | } { j=1 } j=1 2.3. The decreasing rearrangement and Lorentz spaces - 37 - where b = ∑ j m(A ), B = [a ,a ) for j = 1,2, ,k and a = 0 which shows that the j i=1 i j j+1 j ··· k+1 distribution function of a simple function is a simple function (see Figure (b)). We can also ﬁnd the decreasing rearrangement k

f ∗(t) =inf α > 0 : f (α) 6 t = inf α > 0 : b jχB j (α) 6 t ∗ ∑ { } { j=1 } k

= ∑ a jχ[b j 1,b j)(t) j=1 − which is also a simple function (see Figure (c)).

f(x) f (α) f ∗(t) ∗

a1 a1

a2 a2 a3 b5 a3 b4 a4 a4 b3 a5 a5 b2

A3 A4 A1 A5 A2 x a5 a4 a3 a2 a1 α b1 b2 b3 b4b5 t (a) (b) (c)

Example 2.21. Let f : [0,∞) [0,∞) be 7→ 1 (x 1)2, 0 x 2, f (x) = − − 6 6 0, x > 2. It is clear that f (α) = 0 for α > 1 since f (x) 6 1. For α [0,1], we have ∗ | | ∈ f (α) =m( x [0,∞) : 1 (x 1)2 > α ) ∗ { ∈ − − } =m( x [0,∞) : 1 √1 α < x < 1 + √1 α ) = 2√1 α. { ∈ − − − } − That is, 2√1 α, 0 α 1, f (α) = − 6 6 ∗ 0, α > 1. The decreasing rearrangement f ∗(t) = 0 for t > 2 since f (α) 6 2 for any α > 0. For t 6 2, we ∗ have

f ∗(t) =inf α 0 : 2√1 α t { > − 6 } =inf α 0 : α 1 t2/4 = 1 t2/4. { > > − } − Thus, 1 t2/4, 0 6 t 6 2, f ∗(t) = − 0, t > 2. - 38 - 2. Interpolation of Operators

f f f ∗ ∗ 2 2 2

1 1 1

1 2 x 1 2 α 1 2 t (a) (b) (c)

Observe that the integral over f , f and f ∗ are all the same, i.e., ∗ Z ∞ Z 2 Z 1 Z 2 f (x)dx = [1 (x 1)2]dx = 2√1 αdα = (1 t2/4)dt = 4/3. 0 0 − − 0 − 0 − Example 2.22. We define an extended function f : [0,∞) [0,∞] as  7→ 0, x = 0,   1 ln( 1 x ), 0 < x < 1,  − f (x) = ∞, 1 6 x 6 2,  1 ln( x 2 ), 2 < x < 3,  − 0, x > 3. Even if f is infinite over some interval the distribution function and the decreasing rearrangement are still defined and can be calculated, for any α > 0 1 f (α) =m( x [1,2] : ∞ > α x (0,1) : ln( ) > α ∗ { ∈ } ∪ { ∈ 1 x } 1 − x (2,3) : ln( ) > α ) ∪ { ∈ x 2 } α − α =1 + m((1 e− ,1)) + m((2,e− + 2)) α − =1 + 2e− , and  ∞, 0 6 t 6 1,  2 f ∗(t) = ln(t 1 ), 1 < t < 3,  − 0, t > 3.

f f ∗ f ∗ 5 5 5 4 4 4 3 3 3 2 2 2 1 1 1

1 2 3 x 1 2 3 α 1 2 3 t (a) (b) (c)

Example 2.23. Consider the function f (x) = x for all x [0,∞). Then f (α) = m( x [0,∞) : ∈ ∗ { ∈ x > α ) = ∞ for all α 0, which implies that f (t) = inf α 0 : ∞ t = ∞ for all t 0. } > ∗ { > 6 } > 2.3. The decreasing rearrangement and Lorentz spaces - 39 -

x Example 2.24. Consider f (x) = 1+x for x > 0. It is clear that f (α) = 0 for α > 1 since f (x) < 1. For α [0,1), ∗ | | ∈ we have x f (α) =m( x [0,∞) : > α ) ∗ { ∈ 1 + x } f α 1 ∗ =m( x [0,∞) : x > ) = ∞. α { ∈ 1 } f That is, − ∞, 0 6 α < 1, f (t) = 1 2 ∗ 0, α > 1. Thus, f ∗(t) = inf α > 0 : f (α) 6 t = 1. { ∗ } n Proposition 2.25. The decreasing rearrangement f ∗ of the measurable function f on R has the following properties: (i) f ∗(t) is a non-negative and non-increasing function on [0,∞). (ii) f ∗(t) is right continuous on [0,∞). (iii) (k f )∗ = k f ∗ for k C. | | ∈ (iv) f g a.e. implies that f g . | | 6 | | ∗ 6 ∗ (v) ( f + g)∗(t1 +t2) 6 f ∗(t1) + g∗(t2). (vi) ( f g)∗(t1 +t2) 6 f ∗(t1)g∗(t2). (vii) f 6 liminfk ∞ fk a.e. implies that f ∗ 6 liminfk ∞ fk∗. | | → | | → (viii) f f a.e. implies that f f . | k| ↑ | | k∗ ↑ ∗ (ix) f ∗( f (α)) 6 α whenever f (α) < ∞. ∗ m ∗ m (x) f ( f ∗(t)) = ( f > f ∗(t) ) 6 t 6 ( f > f ∗(t) ) if f ∗(t) < ∞. ∗ {| | } {| | } (xi) f ∗(t) > α if and only if f (α) > t. ∗ (xii) f ∗ is equimeasurable with f , that is, ( f ∗) (α) = f (α) for any α > 0. ∗ ∗ (xiii) ( f p) (t) = ( f (t))p for 1 p < ∞. | | ∗ ∗ 6 (xiv) f = f for 1 p < ∞. k ∗kp k kp 6 (xv) f ∞ = f ∗(0). k k s s (xvi) supt>0 t f ∗(t) = supα>0 α( f (α)) for 0 < s < ∞. ∗ Proof. (v) Assume that f ∗(t1)+g∗(t2) < ∞, otherwise, there is nothing to prove. Then for α1 = f ∗(t1) and α2 = g∗(t2), by (x), we have f (α1) 6 t1 and g (α2) 6 t2. From (iv) in Proposition ∗ ∗ 2.15, it holds

( f + g) (α1 + α2) 6 f (α1) + g (α2) 6 t1 +t2. ∗ ∗ ∗ Using the deﬁnition of the decreasing rearrangement, we have

( f + g)∗(t1 +t2) = inf α : ( f + g) (α) 6 t1 +t2 6 α1 + α2 = f ∗(t1) + g∗(t2). { ∗ } (vi) Similar to (v), by (v) in Proposition 2.15, it holds that ( f g) (α1α2) 6 f (α1)+g (α2) 6 ∗ ∗ ∗ t1 +t2. Then, we have ( f g)∗(t1 +t2) = inf α : ( f g) (α) 6 t1 +t2 6 α1α2 = f ∗(t1)g∗(t2). { ∗ } (xi) If f (α) > t, then by the decreasing of f , we have α < inf β : f (β) 6 t = f ∗(t). ∗ ∗ { ∗ } Conversely, if f ∗(t) > α, i.e., inf β : f (β) 6 t > α, we get f (α) > t by the decreasing of f { ∗ } ∗ ∗ again. (xii) By the definition and (xi), we have m m ( f ∗) (α) = ( t > 0 : f ∗(t) > α ) = ( t > 0 : f (α) > t ) = f (α). ∗ { } { ∗ } ∗ (xiii) For α [0,∞), we have ∈ - 40 - 2. Interpolation of Operators p m p ( f )∗(t) =inf α > 0 : ( x : f (x) > α ) 6 t | | { p { | | } } p =inf σ 0 : m( x : f (x) > σ ) t = ( f ∗(t)) , { > { | | } 6 } where σ = α1/p. (xiv) From Theorem 2.16, we have Z ∞ Z ∞ p p p 1 f ∗(t) p = f ∗(t) dt = p α − ( f ∗) (α)dα k k 0 | | 0 ∗ Z ∞ p 1 p =p α − f (α)dα = f p. 0 ∗ k k We remain the proofs of others to interested readers. ut Having disposed of the basic properties of the decreasing rearrangement of functions, we proceed with the definition of the Lorentz spaces. Definition 2.26. Given f a measurable function on Rn and 1 6 p,q 6 ∞, define  1 Z ∞ q  1 q dt  t p f ∗(t) , q < ∞, 0 t f Lp,q = k k  1  supt p f ∗(t), q = ∞.  t>0 p,q n The set of all f with f Lp,q < ∞ is denoted by L (R ) and is called the Lorentz space with k k indices p and q. As in Lp and in weak Lp, two functions in Lp,q will be considered equal if they are equal almost everywhere. Observe that the previous definition implies that Lp,∞ = Lp in view of (xvi) , ∗ in Proposition 2.25 and Lp p = Lp in view of (xiv) in Proposition 2.25 for 1 6 p < ∞. By (i) and (xv) in Proposition 2.25, we have f ∞,∞ = sup f (t) = f (0) = f ∞ which implies that k kL t>0 ∗ ∗ k k L∞,∞ = L∞ = L∞. Thus, we have ∗ Theorem 2.27. Let 1 6 p 6 ∞. Then it holds, with equality of norms, that Lp,p = Lp, Lp,∞ = Lp. ∗ , Remark 2.28. For the Lorentz space Lp q, the case when p = ∞ and 1 6 q < ∞ is not of any n ∞,q interest. The reason is that f L∞,q < ∞ implies that f = 0 a.e. on R . In fact, assume that L k k is a non-trivial space, there exists a nonzero function f L∞,q on a nonzero measurable set, that ∈ is, there exists a constant c > 0 and a set E of positive measure such that f (x) > c for all x E. | | ∈ Then, by (iv) in Proposition 2.25, we have Z ∞ Z ∞ Z m(E) q q dt q dt q dt f L∞,q = ( f ∗(t)) > [( f χE)∗(t)] > c = ∞, k k 0 t 0 t 0 t m n since ( f χE)∗(t) = 0 for t > (E). Hence, we have a contradiction. Thus, f = 0 a.e. on R . The next result shows that for any fixed p, the Lorentz spaces Lp,q increase as the exponent q increases.

Theorem 2.29. Let 1 6 p 6 ∞ and 1 6 q < r 6 ∞. Then, there exists some constant Cp,q,r such that f p,r C f p,q , (2.14) k kL 6 p,q,rk kL where C = (q/p)1/q 1/r. In other words, Lp,q Lp,r. p,q,r − ⊂ Proof. We may assume p < ∞ since the case p = ∞ is trivial. Since f ∗ is non-creasing, we have Z t 1/q Z t 1/q 1/p q q/p 1 q 1/p q ds t f ∗(t) = s − ds f ∗(t) = [s f ∗(t)] p 0 p 0 s 2.4. Marcinkiewicz’ interpolation theorem - 41 -

Z t 1/q 1/q q 1/p q ds q 6 [s f ∗(s)] 6 f Lp,q . p 0 s p k k Hence, taking the supremum over all t > 0, we obtain q 1/q f p,∞ f p,q . (2.15) k kL 6 p k kL This establishes (2.14) in the case r = ∞. Finally, when r < ∞, we have by (2.15) Z ∞ 1/r 1/p r q+q dt f Lp,r = [t f ∗(t)] − k k 0 t 1 q Z ∞ q r 1/p (r q)/r 1/p q dt · 6sup[t f ∗(t)] − [t f ∗(t)] t>0 0 t r q − (r q)/r q/r q rq = f p−,∞ f p,q f p,q . k kL k kL 6 p k kL This completes the proof. ut p,q In general, L is a quasi-normed space, since the functional p,q satisﬁes the conditions k · kL of normed spaces except the triangle inequality. In fact, by (v) in Proposition 2.25, it holds 1/p+1/q+1 f + g p,q 2 ( f p,q + g p,q ). (2.16) k kL 6 k kL k kL However, is this space complete with respect to its quasi-norm? The next theorem answers this question. , Theorem 2.30. Let 1 6 p,q 6 ∞. Then the spaces Lp q(Rn) are complete with respect to their quasi-norms and they are therefore quasi-Banach spaces.

Proof. See [Gra04, p. 50, Theorem 1.4.11]. ut For the duals of Lorentz spaces, we have

Theorem 2.31. Let 1 < p,q < ∞, 1/p + 1/p0 = 1 and 1/q + 1/q0 = 1. Then we have 1,1 1 ∞ 1,q p,q p ,q (L )0 = (L )0 = L , (L )0 = 0 , (L )0 = L 0 0 . { } Proof. See [Gra04, p. 52-55, Theorem 1.4.17]. ut For more results, one can see [Gra04, Kri02].

2.4 Marcinkiewicz’ interpolation theorem

We first introduce the definition of quasi-linear operators. Definition 2.32. An operator T mapping functions on a measure space into functions on another measure space is called quasi-linear if T( f +g) is defined whenever T f and Tg are defined and if T(λ f )(x) κ λ T f (x) and T( f + g)(x) K( T f (x) + Tg(x) ) for a.e. x, where κ and | | 6 | || | | | 6 | | | | K is a positive constant independent of f and g. The idea we have used, in Definition 2.7, of splitting f into two parts according to their respective size, is the main idea of the proof of the theorem that follows. There, we will also use two easily proved inequalities, which are well-known results of Hardy’s (see [HLP88, p. 245–246]): - 42 - 2. Interpolation of Operators

Lemma 2.33 (Hardy inequalities). If q > 1, r > 0 and g is a measurable, non-negative function on (0,∞), then Z ∞ Z t q 1/q Z ∞ 1/q r dt q q r dy g(y)dy t− 6 (yg(y)) y− , (2.17) 0 0 t r 0 y Z ∞ Z ∞ q 1/q Z ∞ 1/q r dt q q r dy g(y)dy t 6 (yg(y)) y . (2.18) 0 t t r 0 y Proof. To prove (2.17), we use Jensen’s inequality2 with the convex function ϕ(x) = xq on (0,∞). Then Z t q 1 Z t q Z t q g(y)dy = g(y)y1 r/qyr/q 1dy yr/q 1dy R t r/q 1 − − − 0 0 y − dy 0 0 Z t q 1 Z t q r/q 1 − 1 r/q r/q 1 6 y − dy g(y)y − y − dy 0 0 q 1 Z t q r/q − q r/q 1 r = t (yg(y)) y − − dy. r 0 By integrating both sides over (0,∞) and use the Fubini theorem, we get that Z ∞ Z t q r 1 g(y)dy t− − dt 0 0 q 1 Z ∞ Z t q − 1 r/q q r/q 1 r 6 t− − (yg(y)) y − − dy dt r 0 0 q 1 Z ∞ Z ∞ q − q r/q 1 r 1 r/q = (yg(y)) y − − t− − dt dy r 0 y q Z ∞ q q 1 r = (yg(y)) y− − dy, r 0 which yields (2.17) immediately. To prove (2.18), we denote f (x) = g(1/x)/x2. Then by taking t = 1/s and y = 1/x, and then applying (2.17) and changing variable again by x = 1/y, we obtain Z ∞ Z ∞ q 1/q Z ∞ Z ∞ q 1/q r 1 r 1 g(y)dy t − dt = g(y)dy s− − ds 0 t 0 1/s Z ∞ Z s q 1/q 2 r 1 = g(1/x)/x dx s− − ds 0 0 Z ∞ Z s q 1/q r 1 = f (x)dx s− − ds 0 0 Z ∞ 1/q Z ∞ 1/q q q r 1 q q r 1 6 (x f (x)) x− − dx = (g(1/x)/x) x− − dx r 0 r 0 Z ∞ 1/q q q r 1 = (g(y)y) y − dy . r 0 Thus, we complete the proofs. ut

2 Jensen’s inequality: If f is any real-valued measurable function on a set Ω and ϕ is convex over the range of f , then

1 Z 1 Z ϕ f (x)g(x)dx 6 ϕ( f (x))g(x)dx, G Ω G Ω R where g(x) > 0 satisﬁes G = Ω g(x)dx > 0. 2.4. Marcinkiewicz’ interpolation theorem - 43 -

Now, we give the Marcinkiewicz3 interpolation theorem4 and its proof due to Hunt and Weiss in [HW64].

Theorem 2.34 (Marcinkiewicz interpolation theorem). Assume that 1 6 pi 6 qi 6 ∞, p0 < p1, q = q and T is a quasi-linear mapping, deﬁned on Lp0 +Lp1 , which is simultaneously of weak 0 6 1 types (p0,q0) and (p1,q1), i.e.,

T f q0,∞ A f , T f q1,∞ A f . (2.19) k kL 6 0k kp0 k kL 6 1k kp1 If 0 < θ < 1, and 1 1 θ θ 1 1 θ θ = − + , = − + , p p0 p1 q q0 q1 then T is of type (p,q), namely T f A f , f Lp. k kq 6 k kp ∈ Here A = A(Ai, pi,qi,θ), but it does not otherwise depend on either T or f .

2 Proof. Let σ be the slope of the line segment in R joining (1/p0,1/q0) with (1/p1,1/q1). Since (1/p,1/q) lies on this segment, we can denote the slope of this segment by 1/q 1/q 1/q 1/q σ = 0 − = − 1 , 1/p 1/p 1/p 1/p 0 − − 1 which may be positive or negative, but is not either 0 or ∞ since q = q and p 0, we split an arbitrary function f Lp as follows: t ∈ f = f + ft where f (x), f (x) > f (tσ ), f t(x) = | | ∗ 0, otherwise, and f = f f t. t − Then we can verify that f (y), 0 y tσ , ( f t) (y) 6 ∗ 6 6 ∗ = 0, y > tσ , σ σ (2.20) f ∗(t ), 0 6 y 6 t , ( ft)∗(y) 6 σ f ∗(y), y > t . t t In fact, by (iv) in Proposition 2.25, f 6 f implies ( f )∗(y) 6 f ∗(y) for all y > 0. More- t | | | | t t σ over, by the deﬁnition of f and (x) in Proposition 2.25, we have ( f ) (α) 6 ( f ) ( f ∗(t )) = σ σ σ t ∗ ∗ f ( f ∗(t )) 6 t for any α > 0. Thus, for y > t , we get ( f )∗(y) = 0. Similarly, by (iv) in ∗ Proposition 2.25, we have ( ft)∗(y) 6 f ∗(y) for any y > 0 since ft 6 f . On the other hand, σ | | | | for y > 0, we have ( ft)∗(y) 6 ( ft)∗(0) = ft ∞ 6 f ∗(t ) with the help of the non-increasing of k k σ ( ft)∗(y) and (xv) in Proposition 2.25. Thus, ( ft)∗(y) 6 min( f ∗(y), f ∗(t )) for any y > 0 which implies (2.20). Suppose p < ∞. Notice that p q, because p q . By T f K( T f t + T f ), Theorems 1 6 i 6 i | | 6 | | | t| 2.27 and 2.29,(2.19), and then by a change of variables and Hardy’s inequalities (2.17) and (2.18), we get

3 Józef Marcinkiewicz (1910–1940) was a Polish mathematician. He was a student of Antoni Zygmund; and later worked with Juliusz Schauder, and Stefan Kaczmarz. 4 The theorem was ﬁrst announced by Marcinkiewicz (1939), who showed this result to Antoni Zygmund shortly before he died in World War II. The theorem was almost forgotten by Zygmund, and was absent from his original works on the theory of singular integral operators. Later Zygmund (1956) realized that Marcinkiewicz’s result could greatly simplify his work, at which time he published his former student’s theorem together with a generalization of his own. - 44 - 2. Interpolation of Operators

t t T f K( T f + T f ) = K( T f q,q + T f q,q ) k kq 6 k kq k tkq k kL k tkL 1/p 1/q t 6K(p/q) − ( T f Lq,p + T ft Lq,p ) (k k k k 1/p 1/q Z ∞ p 1/p p − h 1/q t i dt =K t (T f )∗(t) q 0 t ) Z ∞ p 1/p h 1/q i dt + t (T ft)∗(t) 0 t 1/p 1/q ( 1/p − Z ∞ h ip p 1/q 1/q0 t dt 6K A0 t − f p0 q 0 k k t ) Z ∞ h ip 1/p 1/q 1/q1 dt + A1 t − ft p1 0 k k t  1/p 1/q " 1 1/p #p !1/p −  Z ∞ − 0 p 1/q 1/q0 1 t dt 6K A0 t − f Lp0,1 q  0 p0 k k t  " 1 1/p #p !1/p Z ∞ − 1  1/q 1/q1 1 dt +A1 t − ft Lp1,1 0 p1 k k t 

1/p 1/q( 1 1/p0 p − 1 − =K A0 q p0

Z ∞ Z tσ p 1/p 1/q 1/q 1/p dy dt t − 0 y 0 f ∗(y) · 0 0 y t 1 1/p p 1/p − 1 Z ∞ Z ∞ 1 1/q 1/q1 1/p1 dy dt + A1 t − y f ∗(y) p1 0 tσ y t 1 1/p σ p 1/p) − 1 Z ∞ Z t 1 1/q 1/q1 1/p1 σ dy dt + A1 t − y f ∗(t ) p1 0 0 y t

1/p 1/q ( 1 1/p0 p − 1 1 − =K σ − p A0 q | | p0 Z ∞ Z s p 1/p p(1/p 1/p) 1/p dy ds s− 0− y 0 f ∗(y) · 0 0 y s 1 1/p p 1/p − 1 Z ∞ Z ∞ 1 p(1/p 1/p1) 1/p1 dy ds + A1 s − y f ∗(y) p1 0 s y s 1 1/p p 1/p) − 1 Z ∞ Z s 1 p(1/p 1/p1) 1/p1 dy ds + A1 s − y f ∗(s) p1 0 0 y s ( 1/p 1/q 1 1/p0 Z ∞ p 1/p p − 1 1 − 1 1/p dy 6K σ − p A0 y f ∗(y) q | | p (1/p 1/p) 0 y 0 0 − 1 1/p1 Z ∞ p 1/p 1 − 1 1/p dy + A1 y f ∗(y) p (1/p 1/p ) 0 y 1 − 1 2.4. Marcinkiewicz’ interpolation theorem - 45 -

1 1/p 1/p − 1 Z ∞ 1 1 p/p1 1/p1 p ds + A1 s − (p1s f ∗(s)) p1 0 s  1 1/p 1 1/p  1 − 0 1 − 1 1/p 1/q A0 A1  p − 1/p p0 p1 1/p1 =K σ − 1 1 + 1 1 + A1 p1 f p q | |  k k  p0 − p p − p1  =A f . k kp σ In case p = ∞ the proof is the same except for the use of the estimate f ∞ f (t ), we 1 k tk 6 ∗ can get  1 1/p  1 − 0 1/p 1/q A0  p − 1/p  p0  A = K σ − 1 1 + A1 . q | |    p0 − p  Thus, we complete the proof. ut From the proof given above it is easy to see that the theorem can be extended to the following situation: The underlying measure space Rn of the Lpi (Rn) can be replaced by a general measurable space (and the measurable space occurring in the domain of T need not be the same as the one entering in the range of T). A less superﬁcial generalization of the theorem can be given in terms of the notation of Lorentz spaces, which unify and generalize the usual Lp spaces and the weak-type spaces. For a discussion of this more general form of the Marcinkiewicz interpolation theorem see [SW71, Chapter V] and [BL76, Chapter 5]. As an application of this powerful tool, we present a generalization of the Hausdorff-Young inequality due to Paley. The main difference between the theorems being that Payley introduced a weight function into his inequality and resorted to the theorem of Marcinkiewicz. In what follows, we consider the measure space (Rn, µ) where µ denotes the Lebesgue measure. Let w be a weihgt function on Rn, i.e., a positive and measurable function on Rn. Then we denote by Lp(w) the Lp-space with respect to wdx. The norm on Lp(w) is Z 1/p p f Lp(w) = f (x) w(x)dx . k k Rn | | With this notation we have the following theorem. Theorem 2.35 (Hardy-Littlewood-Paley theorem on Rn). Assume that 1 6 p 6 2. Then

F f Lp( ξ n(2 p)) 6 Cp f p. k k | |− − k k Proof. We considering the mapping (T f )(ξ) = ξ n fˆ(ξ). By Plancherel theorem, we have | | T f 2 2n T f 2 2n = fˆ C f , L ( ξ − ) 6 L ( ξ − ) 2 6 2 k k ∗ | | k k | | k k k k which implies that T is of weak type (2,2). We now work towards showing that T is of weak type (1,1). Thus, the Marcinkiewicz interpolation theorem implies the theorem. n Now, consider the set Eα = ξ : ξ fˆ(ξ) > α . For simplicity, we let ν denote the measure 2n { | | } n ξ dξ and assume that f = 1. Then, fˆ(ξ) 1. For ξ Eα , we therefore have α ξ . | |− k k1 | | 6 ∈ 6 | | Consequently, Z Z 2n 2n 1 (T f ) (α) = ν(Eα ) = ξ − dξ ξ − dξ Cα− . 6 n 6 ∗ Eα | | ξ α | | | | > Thus, we proves that

α (T f ) (α) 6 C f 1, · ∗ k k - 46 - 2. Interpolation of Operators which implies T is of weak type (1,1). Therefore, we complete the proof. ut Chapter 3 The Maximal Function and Calderón-Zygmund Decomposition

3.1 Two covering lemmas

Lemma 3.1 (Finite version of Vitali covering lemma). Suppose B = B1,B2, ,BN is a ﬁnite n { ··· } collection of open balls in R . Then, there exists a disjoint sub-collection B j ,B j , ,B j of B 1 2 ··· k such that N ! k [ n m m B` 6 3 ∑ (B ji ). `=1 i=1 Proof. The argument we give is constructive and relies on the following simple observation: Suppose B and B0 are a pair of balls that intersect, with the radius of B0 being not greater than that of B. Then B0 is contained in the ball B˜ that is concentric with B but with 3 times its radius. (See Fig 3.1.)

As a first step, we pick a ball B j1 in B with maximal (i.e., largest) radius, and then delete from B the ball B j1 as well as any balls that intersect B j1 . ˜ B0 Thus all the balls that are deleted are contained in the ball B j1 concentric B with j1 , but with 3 times its radius. B The remaining balls yield a new collection B0, for which we repeat the B˜ procedure. We pick B j2 and any ball that intersects B j2 . Continuing this way, we find, after at most N steps, a collection of disjoint balls B j , B j , , B j . 1 2 ··· k Fig. 3.1 The balls B and Figure 1: The balls B and B˜ Finally, to prove that this disjoint collection of balls satisfies the inequal- B˜ ity in the lemma, we use the observation made at the beginning of the proof. ˜ Let B ji denote the ball concentric with B ji , but with 3 times its radius. Since any ball B in B must intersect a ball B ji and have equal or smaller radius than B ji , we must have ˜ B B j =∅B B ji , thus ∪ ∩ i 6 ⊂ N ! k ! k k [ [ ˜ m ˜ n m m B` 6 m B ji 6 ∑ (B ji ) = 3 ∑ (B ji ). `=1 i=1 i=1 i=1 In the last step, we have used the fact that in Rn a dilation of a set by δ > 0 results in the multiplication by δ n of the Lebesgue measure of this set. ut For the infinite version of Vitali covering lemma, one can see the textbook [Ste70, the lemma on p.9]. The decomposition of a given set into a disjoint union of cubes (or balls) is a fundamental tool in the theory described in this chapter. By cubes we mean closed cubes; by disjoint we mean that their interiors are disjoint. We have in mind the idea first introduced by Whitney and formulated as follows.

47 - 48 - 3. The Maximal Function and Calderón-Zygmund Decomposition

Theorem 3.2 (Whitney covering lemma). Let F be a non-empty closed set in Rn and Ω be its complement. Then there exists a collection of cubes F = Q whose sides are parallel to the { k} axes, such that S∞ c (i) k=1 Qk = Ω = F , (ii) Q◦ Q◦ = ∅ if j = k, where Q◦ denotes the interior of Q, j ∩ k 6 (iii) there exist two constants c1,c2 > 0 independent of F (In fact we may take c1 = 1 and c2 = 4.), such that c1 diam(Qk) 6 dist(Qk,F) 6 c2 diam(Qk). Proof. Consider the lattice of points in Rn whose coordinates are integral. This lattice deter- mines a mesh M0, which is a collection of cubes: namely all cubes of unit length, whose vertices are points of the above lattice. The mesh M0 leads to a two-way infinite chain of such ∞ k meshes Mk ∞, with Mk = 2− M0. { }− n Thus each cube in the mesh Mk gives rise to 262 Chapter 4 Calderon-Zygmund´ Decomposition cubes in the mesh Mk+1 by bisecting the sides. The k cubes in the mesh Mk each have sides of length 2− M 1 k − and are thus of diameter √n2− . M0 Mk Mk M1 In addition to the meshes Mk, we consider the layers O 1 2 3 Q Ωk, defined by F n k k+1o Ωk = x : c2− < dist(x,F) 6 c2− , Ωk+1 Ωk where c is a positive constant which we shall fix mo- Fig. 4.1 Meshes and layers: M0 with dashed (green) lines; M1 with dotted lines; M 1 with solid (blue) lines S∞ Fig. 3.2 Meshes and layers: M0 with dashed− (green) mentarily. Obviously, Ω = k= ∞ Ωk. lines; M1 with dotted lines; M 1 with solid (blue) − diam (Q) 6 dist (Q, F) 6 4 diam− (Q), Q F0. (4.1) Now we make an initial choice of cubes, and denote lines ∈ k Let us prove (4.1) first. Suppose Q M ; then diam (Q) = √n2− . Since Q F , there exists an ∈ k ∈ 0 x Q Ω . Thus dist (Q, F) 6 dist (x, F) 6 c2 k+1, and dist (Q, F) > dist (x, F) diam (Q) > the resulting collection by F0. Our choice is made as∈ ∩ k − − c2 k √n2 k. If we choose c = 2 √n we get (4.1). − − − Then by (4.1) the cubes Q F are disjoint from F and clearly cover Ω. Therefore, (i) is also follows. We consider the cubes of the mesh M , (each such cube∈ 0 is of size approximately k proved. k 2 ), and include a cube of this mesh in F if it intersectsNotice that theΩ collection, (theF0 has points all our required of properties, the latter except that arethe cubes all in it are not − 0 necessarily disjoint.k To finish the proof of the theorem, we need to refine our choice leading to F0, k eliminating those cubes which were really unnecessary. approximately at a distance 2− from F). Namely, We require the following simple observation. Suppose Q1 and Q2 are two cubes (taken respectively from the mesh M and M ). Then if Q and Q are not disjoint, one of the two must be contained [ k1 k2 1 2 in the other. (In particular, Q Q , if k > k .) F0 = Q Mk : Q Ωk = . 1 ⊂ 2 1 2 Start now with∅ any cube Q F , and consider the maximal cube in F which contains it. In { ∈ ∩ 6 } ∈ 0 0 view of the inequality (4.1), for any cube Q0 F which contains Q F , we have diam (Q0) 6 k ∈ 0 ∈ 0 dist (Q0, F) 6 dist(Q, F) 6 4 diam (Q). Moreover, any two cubes Q0 and Q00 which contain Q have obviously a non-trivial intersection. Thus by the observation made above each cube Q F has We then have ∈ 0 [ a unique maximal cube in F0 which contains it. By the same taken these maximal cubes are also Ωdisjoint. We let F denote the collection of maximal cubes of F0. Then obviously Q = . (i) Q F Q = Ω, ∈ (ii) The cubes of F are disjoint, S Q F 0 (iii) diam (Q) 6 dist (Q, F) 6 4 diam (Q), Q F . ∈ ∈ Therefore, we complete the proof. For appropriate choice of c, we claim that ut 4.2 Calderon-Zygmund´ Fundamental Lemma diam(Q) 6 dist(Q,F) 6 4diam(Q), Q F0. (3.1) Now, we give an important∈ theorem in harmonic analysis. Let us prove (3.1) first. Suppose Q M ; then diam(Q) = √n2 k. Since Q F , there exists an ∈ k − ∈ 0 x Q Ω . Thus dist(Q,F) dist(x,F) c2 k+1, and dist(Q,F) dist(x,F) diam(Q) > ∈ ∩ k 6 6 − > − c2 k √n2 k. If we choose c = 2√n we get (3.1). − − − Then by (3.1) the cubes Q F are disjoint from F and clearly cover Ω. Therefore, (i) is ∈ 0 also proved. Notice that the collection F0 has all our required properties, except that the cubes in it are not necessarily disjoint. To finish the proof of the theorem, we need to refine our choice leading to F0, eliminating those cubes which were really unnecessary. We require the following simple observation. Suppose Q1 and Q2 are two cubes (taken respectively from the mesh Mk1 and Mk2 ). Then if Q1 and Q2 are not disjoint, one of the two must be contained in the other. (In particular, Q Q , if k k .) 1 ⊂ 2 1 > 2 3.2. Hardy-Littlewood maximal function - 49 -

Start now with any cube Q F , and consider the maximal cube in F which contains ∈ 0 0 it. In view of the inequality (3.1), for any cube Q F which contains Q F , we have 0 ∈ 0 ∈ 0 diam(Q0) 6 dist(Q0,F) 6 dist(Q,F) 6 4diam(Q). Moreover, any two cubes Q0 and Q00 which contain Q have obviously a non-trivial intersection. Thus by the observation made above each cube Q F has a unique maximal cube in F which contains it. By the same taken these ∈ 0 0 maximal cubes are also disjoint. We let F denote the collection of maximal cubes of F0. Then obviously S (i) Q F Q = Ω, (ii) The∈ cubes of F are disjoint, (iii) diam(Q) dist(Q,F) 4diam(Q), Q F . 6 6 ∈ Therefore, we complete the proof. ut

3.2 Hardy-Littlewood maximal function

Maximal functions appear in many forms in harmonic analysis. One of the most important of these is the Hardy-Littlewood maximal function. They play an important role in understanding, for example, the differentiability properties of functions, singular integrals and partial differential equations. They often provide a deeper and more simplified approach to understanding problems in these areas than other methods. First, we consider the differentiation of the integral for one-dimensional functions. If f is given on [a,b] and integrable on that interval, we let Z x F(x) = f (y)dy, x [a,b]. a ∈ To deal with F0(x), we recall the definition of the derivative as the limit of the quotient F(x+h) F(x) h− when h tends to 0, i.e., F(x + h) F(x) F0(x) = lim − . h 0 h → We note that this quotient takes the form (say in the case h > 0) 1 Z x+h 1 Z f (y)dy = f (y)dy, h x I I | | where we use the notation I = (x,x + h) and I for the length of this interval. | | At this point, we pause to observe that the above expression in the “average” value of f over I, and that in the limit as I 0, we might expect that these averages tend to f (x). Reformulating | | → the question slightly, we may ask whether 1 Z lim f (y)dy = f (x) I 0 I I | x|→I ∈ | | holds for suitable points x. In higher dimensions we can pose a similar question, where the averages of f are taken over appropriate sets that generalize the intervals in one dimension. In particular, we can take the sets involved as the ball B(x,r) of radius r, centered at x, and denote its measure by m(B(x,r)). It follows 1 Z lim f (y)dy = f (x), for a.e. x? (3.2) r 0 m(B(x,r)) B(x,r) → Let us first consider a simple case, when f is continuous at x, the limit does converge to f (x). Indeed, given ε > 0, there exists a δ > 0 such that f (x) f (y) < ε whenever x y < δ. Since | − | | − | - 50 - 3. The Maximal Function and Calderón-Zygmund Decomposition 1 Z 1 Z f (x) f (y)dy = ( f (x) f (y))dy, − m(B(x,r)) B(x,r) m(B(x,r)) B(x,r) − we find that whenever B(x,r) is a ball of radius r < δ, then

1 Z 1 Z f (x) f (y)dy 6 f (x) f (y) dy < ε, − m(B(x,r)) B(x,r) m(B(x,r)) B(x,r) | − | as desired. In general, for this “averaging problem” (3.2), we shall have an affirmative answer. In order to study the limit (3.2), we consider its quantitative analogue, where “limr 0” is replaced by → “supr>0”, this is the (centered) maximal function. Since the properties of this maximal function are expressed in term of relative size and do not involve any cancelation of positive and negative values, we replace f by f . | | Definition 3.3. If f is locally integrable1 on Rn, we define its maximal function M f : Rn [0,∞] → by Z 1 n M f (x) = sup m f (y) dy, x R . (3.3) r>0 (B(x,r)) B(x,r) | | ∈ Moreover, M is also called as the Hardy-Littlewood maximal operator. The maximal function that we consider arose first in the one-dimensional situation treated by Hardy and Littlewood.2 It is to be noticed that nothing excludes the possibility that M f (x) is infinite for any given x. It is immediate from the definition that Theorem 3.4. If f L∞(Rn), then M f L∞(Rn) and ∈ ∈ M f ∞ f ∞. k k 6 k k By the previous statements, if f is continuous at x, then we have 1 Z f (x) = lim f (y) dy | | r 0 m(B(x,r)) B(x,r) | | → 1 Z 6sup m f (y) dy = M f (x). r>0 (B(x,r)) B(x,r) | | Thus, we have proved Theorem 3.5. If f C(Rn), then ∈ f (x) M f (x) | | 6 for all x Rn. ∈ Sometimes, we will define the maximal function with cubes in place of balls. If Q(x,r) is the cubes [x r,x + r]n, define − Z 1 n M0 f (x) = sup n f (y) dy, x R . (3.4) r>0 (2r) Q(x,r) | | ∈ When n = 1, M and M0 coincide. If n > 1, then there exist constants cn and Cn, depending only on n, such that cnM0 f (x) 6 M f (x) 6 CnM0 f (x). (3.5)

1 n A measurable function f on R is called to be locally integrable, if for every ball B the function f (x)χB(x) is integrable. We shall 1 n denote by Lloc(R ) the space of all locally integrable functions. Loosely speaking, the behavior at inﬁnity does not affact the local x 1/2 n integrability of a function. For example, the functions e| | and x − are both locally integrable, but not integrable on R . | | 2 The Hardy-Littlewood maximal operator appears in many places but some of its most notable uses are in the proofs of the Lebesgue differentiation theorem and Fatou’s theorem and in the theory of singular integral operators. 3.2. Hardy-Littlewood maximal function - 51 -

Thus, the two operators M and M0 are essentially interchangeable, and we will use whichever is more appropriate, depending on the circumstances. In addition, we can deﬁne a more general maximal function 1 Z M00 f (x) = sup m f (y) dy, (3.6) Q x (Q) Q | | 3 where the supremum is taken over all cubes containing x. Again, M00 is pointwise equivalent to M. One sometimes distinguishes between M0 and M00 by referring to the former as the centered and the latter as the non-centered maximal operator. Alternatively, we could deﬁne the non- centered maximal function with balls instead of cubes: Z ˜ 1 M f (x) = sup m f (y) dy B x (B) B | | 3 at each x Rn. Here, the supremum is taken over balls B in Rn which contain the point x and ∈ m(B) denotes the measure of B (in this case a multiple of the radius of the ball raised to the power n).

Example 3.6. Let f : R R, f (x) = χ(0,1)(x). Then →  1 , x > 1,  2x M f (x) =M0 f (x) = 1, 0 6 x 6 1,  1  2(1 x) , x < 0, −  1 , > ,  x x 1 M˜ f (x) =M00 f (x) = 1, 0 6 x 6 1,  1 1 x , x < 0. − In fact, for x > 1, we get 1 Z x+h M f (x) = M0 f (x) =sup χ(0,1)(y)dy h>0 2h x h − ! 1 x + h 1 1 =max sup − , sup = , x h>0 2h x h 0 2h 2x − − 6 1 Z x+h2 M˜ f (x) = M00 f (x) = sup χ(0,1)(y)dy h1,h2>0 h1 + h2 x h1 − ! 1 x + h 1 1 =max sup − 1 , sup = . 00 2h x h − 2h 1 x + h =max sup , sup − , 00 h1 + h2 x h1 1 2 − - 52 - 3. The Maximal Function and Calderón-Zygmund Decomposition

h + h x + h =max sup 1 2 , sup 2 , 00 x+h>1 − ! x + h2 1 M˜ f (x) = M00 f (x) =max sup , sup h1,h2>0,00,x+h2>1 h1 + h2 1 = . 1 x 1 − 1 Observe that f L (R), but M f ,M0 f ,M00 f ,M˜ f / L (R). ∈ ∈ Remark 3.7. (i) M f is defined at every point x Rn and if f = g a.e., then M f (x) = Mg(x) at ∈ every x Rn. ∈ (ii) It may be well that M f = ∞ for every x Rn. For example, let n = 1 and f (x) = x2. ∈ (iii) There are several definitions in the literature which are often equivalent. Next, we state some immediate properties of the maximal function. The proofs are left to interested readers. Proposition 3.8. Let f ,g L1 (Rn). Then ∈ loc (i) Positivity: M f (x) 0 for all x Rn. > ∈ (ii) Sub-linearity: M( f + g)(x) 6 M f (x) + Mg(x). (iii) Homogeneity: M(α f )(x) = α M f (x), α R. | | ∈ (iv) Translation invariance: M(τ f ) = (τ M f )(x) = M f (x y). y y − With the Vitali covering lemma, we can state and prove the main results for the maximal function. Theorem 3.9 (The maximal function theorem). Let f be a given function defined on Rn. (i) If f Lp(Rn), p [1,∞], then the function M f is finite almost everywhere. ∈ ∈ (ii) If f L1(Rn), then for every α > 0, M is of weak type (1,1), i.e., ∈ 3n m( x : M f (x) > α ) f . { } 6 α k k1 (iii) If f Lp(Rn), p (1,∞], then M f Lp(Rn) and ∈ ∈ ∈ M f A f , k kp 6 pk kp n where A = 3 p/(p 1) + 1 for p (1,∞) and A∞ = 1. p − ∈ Proof. We first prove the second one, i.e., (ii). Denote

Eα = x : M f (x) > α , { } then from the deﬁnitions of M f and the supremum, for each x Eα and 0 < ε < M f (x) α, ∈ − there exists a r > 0 such that 1 Z f (y) dy > M f (x) ε > α. m(B(x,r)) B(x,r) | | − 3.2. Hardy-Littlewood maximal function - 53 -

We denote that ball B(x,r) by Bx that contains x. Therefore, for each Bx, we have 1 Z m(Bx) < f (y) dy. (3.7) α Bx | | 3 Fix a compact subset K of Eα . Since K is covered by x Eα Bx, by Heine-Borel theorem, we ∪ ∈ may select a finite subcover of K, say K SN B . Lemma 3.1 guarantees the existence of a ⊂ `=1 ` sub-collection B , , B of disjoint balls with j1 ··· jk N k m [ n m ( B`) 6 3 ∑ (B ji ). (3.8) `=1 i=1 Since the balls B , , B are disjoint and satisfy (3.7) as well as (3.8), we find that j1 ··· jk N k n k Z m m [ n m 3 (K) 6 ( B`) 6 3 ∑ (B ji ) 6 ∑ f (y) dy α B | | `=1 i=1 i=1 ji 3n Z 3n Z = f (y) dy 6 f (y) dy. α Sk α n i=1 B ji | | R | | Since this inequality is true for all compact subsets K of Eα , the proof of the weak type inequality (ii) for the maximal operator is complete. The above proof also gives the proof of (i) for the case when p = 1. For the case p = ∞, by Theorem 3.4, (i) and (iii) is true with A∞ = 1. Now, by using the Marcinkiewicz interpolation theorem between L1 L1,∞ and L∞ L∞, → → we can obtain simultaneously (i) and (iii) for the case p (1,∞). ∈ ut Now, we make some clarifying comments. Remark 3.10. (1) The weak type estimate (ii) is the best possible for the distribution function of M f , where f is an arbitrary function in L1(Rn). Indeed, we replace f (y) dy in the definition of (3.3) by a Dirac measure dµ whose total | | R measure of one is concentrated at the origin. The integral B(x,r) dµ = 1 only if the ball B(x,r) contains the origin; otherwise, it will be zeros. Thus, 1 n 1 M(dµ)(x) = sup m = (Vn x )− , r>0, 0 B(x,r) (B(x,r)) | | ∈ i.e., it reaches the supremum when r = x . Hence, the distribution function of M(dµ) is | | n 1 (M(dµ)) (α) =m( x : M(dµ)(x) > α ) = m( x : (Vn x )− > α ) ∗ { | | } | | n 1 1/n =m( x : V x < α− ) = m(B(0,(V α)− )) n| | n 1 =Vn(Vnα)− = 1/α. But we can always find a sequence f (x) of positive integrable functions, whose L1 norm { m } is each 1, and which converges weakly to the measure dµ. So we cannot expect an estimate essentially stronger than the estimate (ii) in Theorem 3.9, since, in the limit, a similar stronger version would have to hold for M(dµ)(x). (2) It is useful, for certain applications, to observe that 1 A = O , as p 1. p p 1 → −

3 The Heine-Borel theorem reads as follows: A set K Rn is closed and bounded if and only if K is a compact set (i.e., every open ⊂ cover of K has a ﬁnite subcover). In words, any covering of a compact set by a collection of open sets contains a ﬁnite sub-covering. For the proof, one can see the wiki: http://en.wikipedia.org/wiki/Heine%E2%80%93Borel_theorem. - 54 - 3. The Maximal Function and Calderón-Zygmund Decomposition

In contrast with the case p > 1, when p = 1 the mapping f M f is not bounded on L1(Rn). 7→ So the proof of the weak bound (ii) for M f requires a less elementary arguments of geometric measure theory, like the Vitali covering lemma. In fact, we have Theorem 3.11. If f L1(Rn) is not identically zero, then M f is never integrable on the whole ∈ of Rn, i.e., M f / L1(Rn). ∈ Proof. We can choose an N large enough such that Z 1 f (x) dx > f 1. B(0,N) | | 2k k Then, we take an x Rn such that x N. Let r = 2( x + N), we have ∈ | | > | | 1 Z 1 Z M f (x) > f (y) dy = n f (y) dy m(B(x,r)) B(x,r) | | V (2( x + N)) B(x,r) | | n | | 1 Z 1 > n f (y) dy > n f 1 V (2( x + N)) B(0,N) | | 2V (2( x + N)) k k n | | n | | 1 > f 1. 2V (4 x )n k k n | | It follows that for sufﬁciently large x , we have | | n n 1 M f (x) c x − , c = (2V 4 )− f . > | | n k k1 This implies that M f / L1(Rn). ∈ ut Moreover, even if we limit our consideration to any bounded subset of Rn, then the integrability of M f holds only if stronger conditions than the integrability of f are required. In fact, we have + Theorem 3.12. Let E be a bounded subset of Rn. If f ln f L1(Rn) and supp f E, then Z Z | | ∈ ⊂ + M f (x)dx 6 2m(E) +C f (x) ln f (x) dx, E E | | | | where ln+ t = max(lnt,0).

Proof. By Theorem 2.16, it follows that Z Z ∞ M f (x)dx =2 m( x E : M f (x) > 2α )dα E 0 { ∈ } Z 1 Z ∞ =2 + m( x E : M f (x) > 2α )dα 0 1 { ∈ } Z ∞ 62m(E) + 2 m( x E : M f (x) > 2α )dα. 1 { ∈ } Decompose f as f1 + f2, where f1 = f χ x: f (x) >α and f2 = f f1. Then, by Theorem 3.4, it { | | } − follows that

M f (x) M f ∞ f ∞ α, 2 6 k 2k 6 k 2k 6 which yields x E : M f (x) > 2α x E : M f (x) > α . { ∈ } ⊂ { ∈ 1 } Hence, by Theorem 3.9, we have Z ∞ Z ∞ m m ( x E : M f (x) > 2α )dα 6 ( x E : M f1(x) > α )dα 1 { ∈ } 1 { ∈ } Z ∞ Z Z Z max(1, f (x) ) 1 | | dα 6C f (x) dxdα 6 C f (x) dx 1 α x E: f (x) >α | | E | | 1 α { ∈ | | } 3.2. Hardy-Littlewood maximal function - 55 - Z =C f (x) ln+ f (x) dx. E | | | | This completes the proof. ut As a corollary of Theorem 3.9, we have the differentiability almost everywhere of the integral, expressed in (3.2). Theorem 3.13 (Lebesgue differentiation theorem). If f Lp(Rn), p [1,∞], or more generally ∈ ∈ if f is locally integrable (i.e., f L1 (Rn)), then ∈ loc 1 Z lim f (y)dy = f (x), for a.e. x. (3.9) r 0 m(B(x,r)) B(x,r) → Proof. We first consider the case p = 1. It suffices to show that for each α > 0, the set 1 Z Eα = x : limsup m f (y)dy f (x) > 2α r 0 (B(x,r)) B(x,r) − → S∞ has measure zero, because this assertion then guarantees that the set E = k=1 E1/k has measure zero, and the limit in (3.9) holds at all points of Ec. Fix α, since the continuous functions of compact support are dense in L1(Rn), for each ε > 0 we may select a continuous function g of compact support with f g < ε. As we remarked k − k1 earlier, the continuity of g implies that 1 Z lim g(y)dy = g(x), for all x. r 0 m(B(x,r)) B(x,r) → Since we may write the difference 1 R f (y)dy f (x) as m(B(x,r)) B(x,r) − 1 Z ( f (y) g(y))dy m(B(x,r)) B(x,r) − 1 Z + g(y)dy g(x) + g(x) f (x), m(B(x,r)) B(x,r) − − we find that

1 Z limsup m f (y)dy f (x) 6 M( f g)(x) + g(x) f (x) . r 0 (B(x,r)) B(x,r) − − | − | → Consequently, if

Fα = x : M( f g)(x) > α and Gα = x : f (x) g(x) > α , { − } { | − | } then Eα Fα Gα , because if u and u are positive, then u + u > 2α only if u > α for at ⊂ ∪ 1 2 1 2 i least one ui. On the one hand, Tchebychev’s inequality4 yields 1 m(Gα ) f g , 6 α k − k1 and on the other hand, the weak type estimate for the maximal function gives 3n m(Fα ) f g . 6 α k − k1 Since the function g was selected so that f g < ε, we get k − k1 3n 1 3n + 1 m(E ) ε + ε = ε. α 6 α α α

4 Tchebychev inequality (also spelled as Chebyshev’s inequality): Suppose f > 0, and f is integrable. If α > 0 and Eα = x Rn : f (x) > α , then { ∈ } Z m 1 (Eα ) 6 f dx. α Rn - 56 - 3. The Maximal Function and Calderón-Zygmund Decomposition

Since ε is arbitrary, we must have m(Eα ) = 0, and the proof for p = 1 is completed. Indeed, the limit in the theorem is taken over balls that shrink to the point x, so the behavior of f far from x is irrelevant. Thus, we expect the result to remain valid if we simply assume integrability of f on every ball. Clearly, the conclusion holds under the weaker assumption that f is locally integrable. For the remained cases p (1,∞], we have by Hölder inequality, for any ball B, Z ∈ 1/p ( ) p m( ) 0 . f x dx 6 f L (B) 1 Lp0 (B) 6 B f p B | | k k k k k k Thus, f L1 (Rn) and then the conclusion is valid for p (1,∞]. Therefore, we complete the ∈ loc ∈ proof of the theorem. ut By the Lebesgue differentiation theorem, we have 1 n Theorem 3.14. Let f Lloc(R ). Then ∈ n f (x) M f (x), a.e. x R . | | 6 ∈ Combining with the maximal function theorem (i.e., Theorem 3.9), we get Corollary 3.15. If f Lp(Rn), p (1,∞], then we have ∈ ∈ f M f A f . k kp 6 k kp 6 pk kp As an application, we prove the (Gagliardo-Nirenberg-) Sobolev inequality by using the maximal function theorem for the case 1 < p < n. We note that the inequality also holds for the case p = 1 and one can see [Eva98, p.263-264] for the proof. Theorem 3.16 ((Gagliardo-Nirenberg-) Sobolev inequality). Let p (1,n) and its Sobolev con- n ∈ jugate p∗ = np/(n p). Then for f D(R ), we have − ∈ f C ∇ f , k kp∗ 6 k kp where C depends only on n and p.

Proof. Since f D(Rn), we have ∈ Z ∞ ∂ f (x) = f (x + rz)dr, − 0 ∂r where z Sn 1. Integrating this over the whole unit sphere surface Sn 1 yields ∈ − − Z Z Z ∞ ∂ ωn 1 f (x) = f (x)dσ(z) = f (x + rz)drdσ(z) n 1 n 1 − S − − S − 0 ∂r Z Z ∞ = ∇ f (x + rz) zdrdσ(z) n 1 − S − 0 · Z ∞ Z = ∇ f (x + rz) zdσ(z)dr. n 1 − 0 S − · Changing variables y = x + rz, dσ(z) = r (n 1)dσ(y), z = (y x)/ y x and r = y x , we − − − | − | | − | get Z ∞ Z y x ωn 1 f (x) = ∇ f (y) − n dσ(y)dr − − 0 ∂B(x,r) · y x | − | Z y x = ∇ f (y) − dy, − n · y x n R | − | which implies that 1 Z ∇ f (y) f (x) 6 | n |1 dy. | | ωn 1 Rn y x − − | − | 3.3. Calderón-Zygmund decomposition - 57 - R R R We split this integral into two parts as Rn = B(x,r) + Rn B(x,r). For the ﬁrst part, we have \ 1 Z ∇ f (y) | n |1 dy ωn 1 B(x,r) x y − − | − | 1 ∞ Z ∇ f (y) = ∑ | n |1 dy ωn 1 B(x,2 kr) B(x,2 k 1r) x y − − k=0 − \ − − | − | 1 ∞ Z ∇ f (y) 6 ∑ | k 1 n| 1 dy ωn 1 B(x,2 kr) B(x,2 k 1r) (2− − r) − − k=0 − \ − − ∞ k Z 2− r n 1 ∇ f (y) 2 − | | dy 6 ∑ k k k n 1 k=0 nVn2− r B(x,2− r) (2− r) − ∞ Z 1 k+n 1 1 2− − r ∇ f (y) dy 6 ∑ m k k n k=0 (B(x,2− r)) B(x,2− r) | | n 1 ∞ n 2 − k 2 6 rM(∇ f )(x) ∑ 2− = rM(∇ f )(x). n k=0 n For the second part, by Hölder inequality, we get for 1 < p < n Z ∇ f (y) | n |1 dy Rn B(x,r) x y − \ | − | Z 1/p Z 1/p0 p (1 n)p 6 ∇ f (y) dy x y − 0 dy Rn B(x,r) | | Rn B(x,r) | − | \ \ Z ∞ 1/p0 (1 n)p0 n 1 6 ωn 1 ρ − ρ − dρ ∇ f p − r k k 1/p0 (p 1)ωn 1 1 n/p = − − r − ∇ f . n p k kp − (p 1)/n p/n (p 1) n ∇ f p Choosing r = − − k k satisfying (p 1)/n 1/n p M(∇ f )(x) (n p) − ωn 12 − − 1/p0 n n (p 1)ωn 1 1 n/p 2 rM(∇ f )(x) = − − r − ∇ f p, ωn 1 n p k k − − then we get p/n 1 p/n f (x) C ∇ f (M(∇ f )(x)) − . | | 6 k kp Thus, by part (iii) in Theorem 3.9, we obtain for 1 < p < n p/n 1 p/n ∇ ∇ − f p∗ 6C f p M( f ) p (1 p/n) k k k k k k ∗ − p/n 1 p/n =C ∇ f M(∇ f ) − C ∇ f . k kp k kp 6 k kp This completes the proof. ut

3.3 Calderón-Zygmund decomposition

Applying Lebesgue differentiation theorem, we give a decomposition of Rn, called Calderón- Zygmund decomposition, which is extremely useful in harmonic analysis. Theorem 3.17 (Calderón-Zygmund decomposition of Rn). Let f L1(Rn) and α > 0. Then ∈ there exists a decomposition of Rn such that - 58 - 3. The Maximal Function and Calderón-Zygmund Decomposition

(i) Rn = F Ω,F Ω = ∅. ∪ ∩ (ii) f (x) 6 α for a.e. x F. | | ∈ S (iii) Ω is the union of cubes, Ω = k Qk, whose interiors are disjoint and edges parallel to the coordinate axes, and such that for each Qk Z 1 n α < m f (x) dx 6 2 α. (3.10) (Qk) Qk | | (0) Proof. We decompose Rn into a mesh of equal cubes Q (k = 1,2, ), whose interiors are k ··· disjoint and edges parallel to the coordinate axes, and whose common diameter is so large that 1 Z α ( ) f (x) dx 6 , (3.11) m (0) Q 0 | | (Qk ) k since f L1. ∈ Split each Q(0) into 2n congruent cubes. These we denote by Q(1), k = 1,2, . There are two k k ··· possibilities: 1 Z 1 Z α α either ( ) f (x) dx 6 , or ( ) f (x) dx > . m (1) Q 1 | | m (1) Q 1 | | (Qk ) k (Qk ) k In the ﬁrst case, we split Q(1) again into 2n congruent cubes to get Q(2) (k = 1,2, ). In the k k ··· second case, we have Z Z 1 1 n α < f (x) dx 6 f (x) dx 6 2 α (1) (1) n (0) (0) m(Q ) Qk | | 2 m(Q ) Q˜ | | k − k˜ k in view of (3.11) where Q(1) is split from Q(0), and then we take Q(1) as one of the cubes Q . k k˜ k k S∞ ( j) A repetition of this argument shows that if x / Ω =: Qk then x Q ( j = 0,1,2, ) ∈ k=1 ∈ k j ··· for which 1 Z m(Q( j)) 0 as j ∞, and f (x) dx α ( j = 0,1, ). k j ( j) ( j) 6 → → m(Q ) Qk | | ··· k j j Thus f (x) α a.e. x F = Ω c by a variation of the Lebesgue differentiation theorem. Thus, | | 6 ∈ we complete the proof. ut We now state an immediate corollary.

Corollary 3.18. Suppose f , α,F, Ω and Qk have the same meaning as in Theorem 3.17. Then there exists two constants A and B (depending only on the dimension n), such that (i) and (ii) of Theorem 3.17 hold and A (a) m(Ω) f , 6 α k k1 1 Z (b) m f dx 6 Bα. (Qk) Qk | | Proof. In fact, by (3.10) we can take B = 2n, and also because of (3.10) Z m m 1 1 (Ω) = ∑ (Qk) < f (x) dx 6 f 1. k α Ω | | α k k This proves the corollary with A = 1 and B = 2n. ut It is possible however to give another proof of this corollary without using Theorem 3.17 from which it was deduced, but by using the maximal function theorem (Theorem 3.9) and also the theorem about the decomposition of an arbitrary open set as a union of disjoint cubes. This 3.3. Calderón-Zygmund decomposition - 59 - more indirect method of proof has the advantage of clarifying the roles of the sets F and Ω into which Rn was divided. Another proof of the corollary. We know that in F, f (x) α, but this fact does not determine | | 6 F. The set F is however determined, in effect, by the fact that the maximal function satisﬁes M f (x) 6 α on it. So we choose F = x : M f (x) 6 α and Ω = Eα = x : M f (x) > α . Then { n } { } by Theorem 3.9, part (b) we know that m(Ω) 3 f . Thus, we can take A = 3n. 6 α k k1 Since by deﬁnition F is closed, we can choose cubes Qk according to Theorem 3.2, such that S Ω = k Qk, and whose diameters are approximately proportional to their distances from F. Let Qk then be one of these cubes, and pk a point of F such that

dist(F,Qk) = dist(pk,Qk).

Let Bk be the smallest ball whose center is pk and which contains the interior of Qk. Let us set m(Bk) γk = . m(Qk) We have, because p x : M f (x) α , that k ∈ { 6 } 1 Z 1 Z α > M f (pk) > m f (x) dx > m f (x) dx. (Bk) Bk | | γk (Qk) Qk | | Thus, we can take a upper bound of γk as the value of B. The elementary geometry and the inequality (iii) of Theorem 3.2 then show that

radius(Bk) 6dist(pk,Qk) + diam(Qk) = dist(F,Qk) + diam(Qk) 6(c2 + 1)diam(Qk), and so m n n n (Bk) =Vn(radius(Bk)) 6 Vn(c2 + 1) (diam(Qk)) n n/2 =Vn(c2 + 1) n m(Qk), m n n n/2 since (Qk) = (diam(Qk)/√n) . Thus, γk 6 Vn(c2 + 1) n for all k. Thus, we complete the proof with A = 3n and B = V (c + 1)nnn/2. n 2 ut Remark 3.19. (1) Notice that this second proof of the lemma also rewarded us with an unex- pected beneﬁt: the cubes Qk are at a distance from F comparable to their diameters. (2) Theorem 3.17 may be used to give another proof of the fundamental inequality for the maximal function in part (ii) of Theorem 3.9. (See [Ste70, §5.1, p.22–23] for more details.) The Calderón-Zygmund decomposition is a key step in the real-variable analysis of singular integrals. The idea behind this decomposition is that it is often useful to split an arbitrary integrable function into its “small” and “large” parts, and then use different techniques to analyze each part. The scheme is roughly as follows. Given a function f and an altitude α, we write f = g + b, where g is called the good function of the decomposition since it is both integrable and bounded; hence the letter g. The function b is called the bad function since it contains the singular part of f (hence the letter b), but it is carefully chosen to have mean value zero. To obtain the decomposition f = g + b, one might be tempted to “cut” f at the height α; however, this is not what works. Instead, one bases the decomposition on the set where the maximal function of f has height α. Indeed, the Calderón-Zygmund decomposition on Rn may be used to deduce the Calderón- Zygmund decomposition on functions. The later is a very important tool in harmonic analysis. - 60 - 3. The Maximal Function and Calderón-Zygmund Decomposition

Theorem 3.20 (Calderón-Zygmund decomposition for functions). Let f L1(Rn) and α > 0. ∈ Then there exist functions g and b on Rn such that f = g + b and n (i) g f and g ∞ 2 α. k k1 6 k k1 k k 6 b = b , where each b is supported in a dyadic cube Q satisfying R b (x)dx = and (ii) ∑ j j j j Q j j 0 b 2n+1αm(Q ). Furthermore, the cubes Q and Q have disjoint interiors when j = k. k jk1 6 j j k 6 (iii) ∑ m(Q ) α 1 f . j j 6 − k k1 Proof. Applying Corollary 3.18 (with A = 1 and B = 2n), we have 1) Rn = F Ω, F Ω = ∅; ∪ ∩ 2) f (x) 6 α, a.e. x F; | |S∞ ∈ 3) Ω = j=1 Q j, with the interiors of the Q j mutually disjoint; 1 R 1 R n 4) m(Ω) 6 α− n f (x) dx, and α < f (x) dx 6 2 α. R | | m(Q j) Q j | | Now deﬁne 1 Z b = f f dx χ , j m Q j − (Q j) Q j b = ∑ b and g = f b. Consequently, j j − Z Z Z m 1 b j dx 6 f (x) dx + (Q j) m f (x)dx Q j | | Q j | | (Q j) Q j Z n+1 m 62 f (x) dx 6 2 α (Q j), Q j | | n+1 m which proves b j 1 6 2 α (Q j). k k n Next, we need to obtain the estimates on g. Write R = jQ j F, where F is the closed set ∪ 1 ∪R obtained by Corollary 3.18. Since b = 0 on F and f b j = m f (x)dx, we have − (Q j) Q j  f , on F,  g = 1 Z (3.12)  m f (x)dx, on Q j. (Q j) Q j 1 R n On the cube Q j, g is equal to the constant f (x)dx, and this is bounded by 2 α by m(Q j) Q j n 4). Then by 2), we can get g ∞ 2 α. Finally, it follows from (3.12) that g f . This k k 6 k k1 6 k k1 completes the proof. ut As an application of Calderón-Zygmund decomposition and Marcinkiewicz interpolation theorem, we now prove the Fefferman-Stein inequality on the Hardy-Littlewood maximal operator M.

Theorem 3.21 (Fefferman-Stein inequality). For p (1,∞), there exists a constant C = C , ∈ n p such that, for any measurable function on Rn, ϕ(x) > 0 and f , we have the inequality Z Z p p (M f (x)) ϕ(x)dx 6 C f (x) Mϕ(x)dx. (3.13) Rn Rn | | Proof. Except when Mϕ(x) = ∞ a.e., in which case (3.13) holds trivially, Mϕ is the density of a positive measure µ. Thus, we may assume that Mϕ(x) < ∞ a.e. x Rn and Mϕ(x) > 0. If we ∈ denote dµ(x) = Mϕ(x)dx and dν(x) = ϕ(x)dx, then by the Marcinkiewicz interpolation theorem in order to get (3.13), it sufﬁces to prove that M is both of type (L∞(µ),L∞(ν)) and of weak type (L1(µ),L1(ν)). ∞ ∞ Let us ﬁrst show that M is of type (L (µ),L (ν)). In fact, if f ∞ α, then k kL (µ) 6 3.3. Calderón-Zygmund decomposition - 61 - Z n Mϕ(x)dx = µ( x R : f (x) > α ) = 0. x Rn: f (x) >α { ∈ | | } { ∈ | | } Since Mϕ(x) > 0 for any x Rn, we have m( x Rn : f (x) > α ) = 0, equivalently, n ∈ n{ ∈ | | } f (x) α a.e. x R . Thus, M f (x) α a.e. x R and this follows M f ∞(ν) α. Therefore, | | 6 ∈ 6 ∈ k kL 6 M f ∞ f ∞ . k kL (ν) 6 k kL (µ) Before proving that M is also of weak type (L1(µ),L1(ν)), we give the following lemma. 1 n Lemma 3.22. Let f L (R ) and α > 0. If the sequence Qk of cubes is chosen from the ∈ { } Calderón-Zygmund decomposition of Rn for f and α > 0, then n n [ x R : M0 f (x) > 7 α Q∗, { ∈ } ⊂ k k where Qk∗ = 2Qk. Then we have m n n n m ( x R : M0 f (x) > 7 α ) 6 2 ∑ (Qk). { ∈ } k S Proof. Suppose that x / k Qk∗. Then there are two cases for any cube Q with the center x. If n S ∈ Q F := R Qk, then ⊂ \ k 1 Z f (x) dx 6 α. m(Q) Q | | If Q Qk = ∅ for some k, then it is easy to check that Qk 3Q, and ∩ 6 [ ⊂ Qk : Qk Q = ∅ 3Q. { ∩ 6 } ⊂ k Hence, we have Z Z Z f (x) dx 6 f (x) dx + ∑ f (x) dx Q | | Q F | | Q | | Qk Q=∅ k ∩ ∩ 6 m n m 6α (Q) + ∑ 2 α (Qk) Qk Q=∅ ∩ 6 n 6αm(Q) + 2 αm(3Q) n 67 αm(Q). n S Thus we know that M0 f (x) 6 7 α for any x / k Qk∗, and it yields that ∈ ! m n n m [ n m ( x R : M0 f (x) > 7 α ) Q∗ = 2 (Qk). { ∈ } 6 k ∑ k k We complete the proof of the lemma. ut Let us return to the proof of weak type (L1(µ),L1(ν)). We need to prove that there exists a constant C such that for any α > 0 and f L1(µ) Z ∈ n ϕ(x)dx =ν( x R : M f (x) > α ) x Rn:M f (x)>α { ∈ } { ∈ } (3.14) C Z 6 f (x) Mϕ(x)dx. α Rn | | 1 n 1 n We may assume that f L (R ). In fact, if we take fk = f χB(0,k), then fk L (R ), 0 6 ∈n | | ∈ fk(x) 6 fk+1(x) for x R and k = 1,2, . Moreover, limk ∞ fk(x) = f (x) . ∈ ··· → | | By the pointwise equivalence of M and M0, there exists cn > 0 such that M f (x) 6 cnM0 f (x) n n n for all x R . Applying the Calderón-Zygmund decomposition on R for f and α0 = α/(cn7 ), ∈ we get a sequence Qk of cubes satisfying { } Z 1 n α0 < m f (x) dx 6 2 α0. (Qk) Qk | | - 62 - 3. The Maximal Function and Calderón-Zygmund Decomposition

By Lemma 3.22 and the pointwise equivalence of M and M00, we have that Z ϕ(x)dx x Rn:M f (x)>α Z{ ∈ } ϕ(x)dx 6 n n x R :M0 f (x)>7 α0 Z{ ∈ } Z ϕ(x)dx ϕ(x)dx 6 S 6 ∑ k Qk∗ k Qk∗ 1 Z 1 Z ϕ(x)dx f (y) dy 6∑ m( ) α k Qk Qk∗ 0 Qk | | c 7n Z 2n Z = n f (y) ϕ(x)dx dy α ∑ m( ) k Qk | | Qk∗ Qk∗ n Z cn14 6 ∑ f (y) M00ϕ(y)dy α k Qk | | C Z 6 f (y) Mϕ(y)dy. α Rn | | Thus, M is of weak type (L1(µ),L1(ν)), and the Fefferman-Stein inequality can be obtained by applying the Marcinkiewicz interpolation theorem. ut Chapter 4 Singular Integrals

4.1 Harmonic functions and Poisson equation

Among the most important of all PDEs are undoubtedly Laplace equation ∆u = 0 (4.1) and Poisson equation ∆u = f . (4.2) − In both (4.1) and (4.2), x Ω and the unknown is u : Ω¯ R, u = u(x), where Ω Rn is a ∈ → ⊂ given open set. In (4.2), the function f : Ω R is also given. Remember that the Laplacian of → ∆ = n ∂ 2 u is u ∑k=1 xk u. Definition 4.1. A C2 function u satisfying (4.1) is called a harmonic function. Now, we derive a fundamental solution of Laplace’s equation. One good strategy for investi- gating any PDEs is first to identify some explicit solutions and then, provided the PDE is linear, to assemble more complicated solutions out of the specific ones previously noted. Furthermore, in looking for explicit solutions it is often wise to restrict attention to classes of functions with certain symmetry properties. Since Laplace equation is invariant under rotations, it consequently seems advisable to search first for radial solutions, that is, functions of r = x . Let us therefore | | attempt to find a solution u of Laplace equation (4.1) in Ω = Rn, having the form u(x) = v(r), where r = x and v is to be selected (if possible) so that ∆u = 0 holds. First note for k = 1, ,n | | ··· that ∂r x = k , x = 0. ∂xk r 6 We thus have 2 2 xk 2 xk 1 xk ∂ u = v0(r) , ∂ u = v00(r) + v0(r) xk r xk r2 r − r3 for k = 1, ,n, and so ··· n 1 ∆u = v00(r) + − v0(r). r Hence ∆u = 0 if and only if n 1 v00 + − v0 = 0. (4.3) r If v = 0, we deduce 0 6

63 - 64 - 4. Singular Integrals

v00 1 n (lnv0)0 = = − , v0 r a and hence v0(r) = rn 1 for some constant a. Consequently, if r > 0, we have −  blnr + c, n = 2, v(r) = b  n 2 + c, n > 3, r − where b and c are constants. These considerations motivate the following Definition 4.2. The function  1  ln x , n = 2,  − 2π | | Φ( ) = x : 1 1 (4.4)  , n > 3,  n(n 2)V x n 2 − n | | − defined for x Rn, x = 0, is the fundamental solution of Laplace equation. ∈ 6 The reason for the particular choices of the constants in (4.4) will be apparent in a moment. We will sometimes slightly abuse notation and write Φ(x) = Φ( x ) to emphasize that the | | fundamental solution is radial. Observe also that we have the estimates C C ∇Φ(x) , ∆Φ(x) , (x = 0) (4.5) | | 6 x n 1 | | 6 x n 6 | | − | | for some constant C > 0. By construction, the function x Φ(x) is harmonic for x = 0. If we shift the origin to a new 7→ 6 point y, the PDE (4.1) is unchanged; and so x Φ(x y) is also harmonic as a function of x 7→ − for x = y. Let us now take f : Rn R and note that the mapping x Φ(x y) f (y) (x = y) is 6 → 7→ − 6 harmonic for each point y Rn, and thus so is the sum of finitely many such expression built ∈ for different points y. This reasoning might suggest that the convolution  1 Z Z  (ln x y ) f (y)dy, n = 2,  − 2π R2 | − | u(x) = Φ(x y) f (y)dy = Z (4.6) n − 1 f (y) R  dy, n > 3  n(n 2)V n x y n 2 − n R | − | − would solve Laplace equation (4.1). However, this is wrong: we cannot just compute Z ∆u(x) = ∆xΦ(x y) f (y)dy = 0. (4.7) Rn − Indeed, as intimated by estimate (4.5), ∆Φ(x y) is not summable near the singularity at y = x, − and so the differentiation under the integral sign above is unjustified (and incorrect). We must proceed more carefully in calculating ∆u. Let us for simplicity now assume f C2(Rn), that is, f is twice continuously differentiable, ∈ c with compact support. Theorem 4.3 (Solving Poisson equation). Let f C2(Rn), define u by (4.6). Then u C2(Rn) ∈ c ∈ and ∆u = f in Rn. − We consequently see that (4.6) provides us with a formula for a solution of Poisson’s equation (4.2) in Rn. Proof. Step 1: To show u C2(Rn). We have ∈ Z Z u(x) = Φ(x y) f (y)dy = Φ(y) f (x y)dy, Rn − Rn − 4.1. Harmonic functions and Poisson equation - 65 - hence u(x + he ) u(x) Z f (x + he y) f (x y) k − = Φ(y) k − − − dy, h Rn h where h = 0 and e = (0, ,1, ,0), the 1 in the kth-slot. But 6 k ··· ··· f (x + he y) f (x y) ∂ f k − − − (x y) h → ∂xk − uniformly on Rn as h 0, and thus → ∂u Z ∂ f (x) = Φ(y) (x y)dy, k = 1, ,n. ∂xk Rn ∂xk − ··· Similarly, ∂ 2u Z ∂ 2 f (x) = Φ(y) (x y)dy, k, j = 1, ,n. (4.8) ∂xk∂x j Rn ∂xk∂x j − ··· As the expression on the r.h.s. of (4.8) is continuous in the variable x, we see that u C2(Rn). ∈ Step 2: To prove the second part. Since Φ blows up at 0, we will need for subsequent calcu- lations to isolate this singularity inside a small ball. So fix ε > 0. Then Z Z ∆u(x) = Φ(y)∆x f (x y)dy + Φ(y)∆x f (x y)dy =: Iε + Jε . (4.9) B(0,ε) − Rn B(0,ε) − \ Now ( Z Cε2(1 + lnε ), n = 2, ∆ Φ( ) | | Iε 6 C f ∞ y dy 6 2 (4.10) | | k k B(0,ε) | | Cε , n > 3, since Z Z ε Z ε 2 ε ln y dy = 2π r lnrdr = π r lnr 0 rdr B(0,ε) | | || − 0 − | − 0 = π(ε2 lnε ε2/2) − −π =πε2 lnε + ε2, | | 2 for ε (0,1] and n = 2 by an integration by parts. ∈ An integration by parts yields Z Jε = Φ(y)∆x f (x y)dy Rn B(0,ε) − \ Z ∂ f Z (4.11) = Φ(y) (x y)dσ(y) ∇Φ(y) ∇y f (x y)dy ∂B(0,ε) ∂ν − − Rn B(0,ε) · − \ =:Kε + Lε , where ν denotes the inward pointing unit normal along ∂B(0,ε). We readily check Z Z n 1 Kε 6 ∇ f ∞ Φ(y) dσ(y) 6 C Φ(ε) dσ(y) = C Φ(ε) ε − | | k k ∂B(0,ε) | | | | ∂B(0,ε) | | ( (4.12) Cε lnε , n = 2, 6 | | Cε, n > 3, since Φ(y) = Φ( y ) = Φ(ε) on ∂B(0,ε) = y Rn : y = ε . | | { ∈ | | } We continue by integrating by parts once again in the term Lε , to discover Z ∂Φ Z Lε = (y) f (x y)dσ(y) + ∆Φ(y) f (x y)dy − ∂B(0,ε) ∂ν − Rn B(0,ε) − \ Z ∂Φ = (y) f (x y)dσ(y), − ∂B(0,ε) ∂ν − - 66 - 4. Singular Integrals

1 y y y since Φ is harmonic away from the origin. Now, ∇Φ(y) = n for y = 0 and ν = − = − nVn y 6 y − ε ∂Φ 1 | | n | 1| on ∂B(0,ε). Consequently, ∂ν (y) = ν ∇Φ(y) = n 1 on ∂B(0,ε). Since nVnε − is the · nVnε − surface area of the sphere ∂B(0,ε), we have 1 Z Lε = n 1 f (x y)dσ(y) − nVnε ∂B(0,ε) − − (4.13) 1 Z = f (y)dσ(y) f (x) as ε 0. − m(∂B(x,ε)) ∂B(x,ε) → − → by Lebesgue differentiation theorem. Combining now (4.9)-(4.13) and letting ε 0, we ﬁnd that ∆u(x) = f (x), as asserted. → − ut Remark 4.4. (i) We sometimes write n ∆Φ = δ0 in R , − n where δ0 denotes the Dirac measure on R giving unit mass to the point 0. Adopting this notation, we may formally compute Z Z n ∆u(x) = ∆xΦ(x y) f (y)dy = δx f (y)dy = f (x), x R , − Rn − − Rn ∈ in accordance with Theorem 4.3. This corrects the erroneous calculation (4.7). (ii) Theorem 4.3 is in fact valid under far less stringent smoothness requirements for f : see [GT01]. Consider now an open set Ω Rn and suppose u is a harmonic function within Ω. We next ⊂ derive the important mean-value formulas, which declare that u(x) equals both the average of u over the sphere ∂B(x,r) and the average of u over the entire ball B(x,r), provided B(x,r) Ω. ⊂ Theorem 4.5 (Mean-value formula for harmonic functions). If u C2(Ω) is harmonic, then for ∈ each ball B(x,r) Ω, ⊂ 1 Z 1 Z u(x) = u(y)dσ(y) = u(y)dy. m(∂B(x,r)) ∂B(x,r) m(B(x,r)) B(x,r) Proof. Denote 1 Z 1 Z f (r) = u(y)dσ(y) = u(x + rz)dσ(z). m(∂B(x,r)) ∂B(x,r) ωn 1 Sn 1 − − Obviously, 1 Z n 1 Z ∂u f 0(r) = ∑ ∂x j u(x + rz)z jdσ(z) = (x + rz)dσ(z), ωn 1 Sn 1 ωn 1 Sn 1 ∂ν − − j=1 − − ∂ where ∂ν denotes the differentiation w.r.t. the outward normal. Thus, by changes of variable 1 Z ∂u f 0(r) = (y)dσ(y). ωn 1 ∂B(x,r) ∂ν − By Stokes theorem, we get 1 Z f 0(r) = ∆u(y)dy = 0. ωn 1 B(x,r) − Thus f (r) = const. Since limr 0 f (r) = u(x), hence, f (r) = u(x). → Next, observe that our employing polar coordinates gives Z Z r Z Z r u(y)dy = u(y)dσ(y) ds = m(∂B(x,s))u(x)ds B(x,r) 0 ∂B(x,s) 0 Z r n 1 n =u(x) nVns − ds = Vnr u(x). 0 4.2. Poisson kernel and Hilbert transform - 67 -

This completes the proof. ut Theorem 4.6 (Converse to mean-value property). If u C2(Ω) satisﬁes ∈ 1 Z u(x) = u(y)dσ(y) m(∂B(x,r)) ∂B(x,r) for each ball B(x,r) Ω, then u is harmonic. ⊂ Proof. If ∆u 0, then there exists some ball B(x,r) Ω such that, say, ∆u > 0 within B(x,r). 6≡ ⊂ But then for f as above, 1 Z 0 = f 0(r) = n 1 ∆u(y)dy > 0, r − ωn 1 B(x,r) − is a contradiction. ut

4.2 Poisson kernel and Hilbert transform

We shall now introduce a notation that will be indispensable in much of our further work. Indeed, we have shown some properties of Poisson kernel in Chapter1. The setting for the application of this theory will be as follows. We shall think of Rn as the boundary hyperplane of the (n + 1) dimensional upper-half space Rn+1. In coordinate notation, n+ n R 1 = (x,y) : x R ,y > 0 . + { ∈ } We shall consider the Poisson integral of a function f given on Rn. This Poisson integral is n+1 effectively the solution to the Dirichlet Problem for R+ : find a harmonic function u(x,y) on n+1 n R+ , whose boundary values on R (in the appropriate sense) are f (x), that is ( n+1 ∆x,yu(x,y) = 0, (x,y) R+ , n ∈ (4.14) u(x,0) = f , x R . ∈ The formal solution of this problem can be given neatly in the context of the L2 theory. In fact, let f L2(Rn), and consider ∈ n Z ω ωiξ x ωξ y u(x,y) = | | e · e−| | fˆ(ξ)dξ, y > 0. (4.15) 2π Rn 2 n ωξ y This integral converges absolutely (cf. Theorem 1.15), because fˆ L (R ), and e−| | is ∈ rapidly decreasing in ξ for y > 0. For the same reason, the integral above may be differentiated | | w.r.t. x and y any number of times by carrying out the operation under the sign of integration. This gives ∂ 2u n ∂ 2u ∆x,yu = 2 + ∑ 2 = 0, ∂y k=1 ∂xk ωiξ x ωξ y because the factor e · e−| | satisfies this property for each fixed ξ. Thus, u(x,y) is a har- n+1 monic function on R+ . By Theorem 1.15, we get that u(x,y) f (x) in L2(Rn) norm, as y 0. That is, u(x,y) satis- → → fies the boundary condition and so u(x,y) structured above is a solution for the above Dirichlet problem. This solution of the problem can also be written without explicit use of the Fourier transform. For this purpose, we define the Poisson kernel Py(x) := P(x,y) by - 68 - 4. Singular Integrals n Z ω ωiξ x ωξ y 1 ωξ y Py(x) = | | e · e−| | dξ = (F − e−| | )(x), y > 0. (4.16) 2π Rn Then the function u(x,y) obtained above can be written as a convolution Z u(x,y) = Py(z) f (x z)dz, (4.17) Rn − as the same as in Theorem 1.15. We shall say that u is the Poisson integral of f . For convenience, we recall (1.12) and (1.10) as follows. Proposition 4.7. The Poisson kernel has the following explicit expression: cny Γ ((n + 1)/2) Py(x) = n+1 , cn = n+1 . (4.18) ( x 2 + y2) 2 π 2 | | Remark 4.8. We list the properties of the Poisson kernel that are now more or less evident: (i) Py(x) > 0 for y > 0. R ωξ y (ii) Rn Py(x)dx = Pby(0) = 1, y > 0; more generally, Pby(ξ) = e−| | by Lemma 1.14 and Corollary 1.23, respectively. n (iii) Py(x) is homogeneous of degree n: Py(x) = y− P1(x/y), y > 0. − p n (iv) Py(x) is a decreasing function of x , and Py(x) L (R ), 1 p ∞. Indeed, by changes | | ∈ 6 6 of variables, we have for 1 6 p < ∞ Z p p p y Py(x) p =cn dx k k n ( x 2 + y2)(n+1)/2 R | | x=yz Z 1 ==cpy n(p 1) dz n − − 2 p(n+1)/2 Rn (1 + z ) Z ∞| | z=rz0 p n(p 1) 1 n 1 ==cn y− − ωn 1 r − dr − 0 (1 + r2)p(n+1)/2 Z 1 Z ∞ p n(p 1) n 1 p(n+1) 6cn y− − ωn 1 dr + r − − dr − 0 1 p n(p 1) 1 6cn y− − ωn 1 1 + . − p(n + 1) n n − For p = ∞, it is clear that P (x) ∞ c y . k y k 6 n − (v) Suppose f Lp(Rn), 1 6 p 6 ∞, then its Poisson integral u, given by (4.17), is harmonic n+1 ∈ n+1 in R+ . This is a simple consequence of the fact that Py(x) is harmonic in R+ ; the latter is immediately derived from (4.16). (vi) We have the “semi-group property” P P = P if y ,y > 0 in view of Corollary y1 ∗ y2 y1+y2 1 2 1.24. The boundary behavior of Poisson integrals is already described to a significant extent by the following theorem. Theorem 4.9. Suppose f Lp(Rn), 1 p ∞, and let u(x,y) be its Poisson integral. Then ∈ 6 6 (a) sup u(x,y) M f (x), where M f is the maximal function. y>0 | | 6 (b) limy 0 u(x,y) = f (x), for almost every x. → (c) If p < ∞, u(x,y) converges to f (x) in Lp(Rn) norm, as y 0. → The theorem will now be proved in a more general setting, valid for a large class of approximations to the identity. n n Let ϕ be an integrable function on R , and set ϕε (x) = ε− ϕ(x/ε), ε > 0. 4.2. Poisson kernel and Hilbert transform - 69 -

Theorem 4.10. Suppose that the least decreasing radial majorant of ϕ is integrable; i.e., let R ψ(x) = sup y > x ϕ(y) , and we suppose Rn ψ(x)dx = A < ∞. Then with the same A, | | | | | | p n (a) supε>0 ( f ϕε )(x) 6 AM f (x), f L (R ), 1 6 p 6 ∞. | ∗ R | ∈ (b) If in addition n ϕ(x)dx = 1, then limε 0( f ϕε )(x) = f (x) almost everywhere. R → ∗ (c) If p < ∞, then f ϕε f 0, as ε 0. k ∗ − kp → → Proof. For the part (c), we have shown in Theorem 1.15. Next, we prove assertion (a). We have already considered a special case of (a) in Chapter3, 1 with ϕ(x) = m(B) χB. The point of the theorem is to reduce matters to this fundamental special case. With a slight abuse of notation, let us write ψ(r) = ψ(x), if x = r; it should cause | | no confusion since ψ(x) is anyway radial. Now observe that ψ(r) is decreasing and then R ψ(x)dx ψ(r)R dx = cψ(r)rn. Therefore the assumption ψ L1 proves that r/26 x 6r > r/26 x 6r ∈ rnψ(r| )| 0 as r 0 or r ∞|.| To prove (a), we need to show that → → → ( f ψε )(x) AM f (x), (4.19) ∗ 6 p n R where f 0, f L (R ), ε > 0 and A = n ψ(x)dx. > ∈ R Since (4.19) is clearly translation invariant w.r.t f and also dilation invariant w.r.t. ψ, it suf- ﬁces to show that ( f ψ)(0) AM f (0). (4.20) ∗ 6 In proving (4.20), we may clearly assume that M f (0) < ∞. Let us write λ(r) = R R n 1 f (rx )dσ(x ), and Λ(r) = f (x)dx, so S − 0 0 x 6r Z r Z | | Z r n 1 n 1 n 1 Λ(r) = f (tx0)dσ(x0)t − dt = λ(t)t − dt, i.e., Λ 0(r) = λ(r)r − . n 1 0 S − 0 We have Z Z ∞ Z n 1 ( f ψ)(0) = f (x)ψ(x)dx = r − f (rx0)ψ(r)dσ(x0)dr n n 1 ∗ R 0 S − Z ∞ Z N n 1 n 1 = r − λ(r)ψ(r)dr = lim λ(r)ψ(r)r − dr 0 ε 0 ε N→∞ → Z N Z N N = lim Λ 0(r)ψ(r)dr = lim [Λ(r)ψ(r)]ε Λ(r)dψ(r) . ε 0 ε ε 0 − ε N→∞ N→∞ → → Λ( ) = R ( ) n ( ) nψ( ) ∞ Since r x 6r f x dx 6 Vnr M f 0 , and the fact r r 0 as r 0 or r , we have | | n→ → → 0 6 lim Λ(N)ψ(N) 6 VnM f (0) lim N ψ(N) = 0, N ∞ N ∞ → → which implies limN ∞ Λ(N)ψ(N) = 0 and similarly limε 0 Λ(ε)ψ(ε) = 0. Thus, by integra- → → tion by parts, we have Z ∞ Z ∞ n ( f ψ)(0) = Λ(r)d( ψ(r)) 6 VnM f (0) r d( ψ(r)) ∗ 0 − 0 − Z ∞ Z n 1 =nVnM f (0) ψ(r)r − dr = M f (0) ψ(x)dx, 0 Rn since ψ(r) is decreasing which implies ψ0(r) 6 0, and nVn = ωn 1. This proves (4.20) and then − (4.19). Finally, we prove (b) in a familiar way as follows. First, we can verify that if f C 0, then 1 ∈ c ( f1 ϕε )(x) f1(x) uniformly as ε 0 (cf. Theorem 1.15). Next we can deal with the case ∗ p n → → f L (R ), 1 p < ∞, by writing f = f1 + f2 with f1 as described and with f2 p small. ∈ 6 k k The argument then follows closely that given in the proof of Corollary 3.13 (the Lebesgue - 70 - 4. Singular Integrals differentiation theorem). Thus we get that limε 0 fε (x) exists almost everywhere and equals → f (x). To deal with the remaining case, that of bounded f , we ﬁx any ball B, and set ourselves the task of showing that

lim( f ϕε )(x) = f (x), for almost every x B. ε 0 ∗ ∈ → c Let B1 be any other ball which strictly contains B, and let δ = dist(B,B ) be the distance 1 f (x), x B1, from B to the complement of B1. Let f1(x) = ∈ ; f (x) = f1(x) + f2(x). Then, 0, x / B1, 1 n ∈ f1 L (R ), and so the appropriate conclusion holds for it. However, for x B, ∈ ∈ Z Z ( f2 ϕε )(x) = f2(x y)ϕε (y)dy 6 f2(x y) ϕε (y) dy | ∗ | Rn − y >δ>0 | − || | Z | | 6 f ∞ ϕ(y) dy 0, as ε 0. k k y δ/ε | | → → | |> Thus, we complete the proof. ut Proof of Theorem 4.9. Theorem 4.10 then applies directly to prove Theorem 4.9, because of properties (i)–(iv) of the Poisson kernel in the case ϕ(x) = ψ(x) = P (x). 1 ut There are also some variants of the result of Theorem 4.10, which apply equally well to Poisson integrals. The first is an easy adaptation of the argument already given, and is stated without proof. n Corollary 4.11. Suppose f is continuous and bounded on R . Then ( f ϕε )(x) f (x) uni- ∗ → formly on compact subsets of Rn. The second variant is somewhat more difficult. It is the analogue for finite Borel measures in place of integrable functions, and is outlined in further result of [Ste70, §4.1, p.77–78]. Now, we give the definition of harmonic conjugate functions as follows. Definition 4.12. The harmonic conjugate to a given function u(x,y) is a function v(x,y) such that f (x,y) = u(x,y) + iv(x,y) is analytic, i.e., satisfies the Cauchy-Riemann equations u = v , u = v , x y y − x where u ∂u/∂x, u ∂u/∂y. It is given by x ≡ y ≡ Z (x,y) v(x,y) = uxdy uydx +C, (x0,y0) − along any path connecting (x0,y0) and (x,y) in the domain, where C is a constant of integration. Given a function f in S (R), its harmonic extension to the upper half-plane is given by u(x,y) = Py f (x), where Py is the Poisson kernel. We can also write, in view of (4.15), ∗ Z ω ωiξ x ωξ y u(z) =u(x,y) = | | e · e−| | fˆ(ξ)dξ 2π R Z ∞ Z 0 ω ωiξ x ω ξy ωiξ x ω ξy =| | e · e−| | fˆ(ξ)dξ + e · e| | fˆ(ξ)dξ 2π 0 ∞ − Z ∞ Z 0 ω ωiξ (x+isgn(ω)y) ωiξ (x isgn(ω)y) =| | e · fˆ(ξ)dξ + e · − fˆ(ξ)dξ , 2π 0 ∞ − 4.2. Poisson kernel and Hilbert transform - 71 - where z = x + iy. If we now define Z ∞ ω h ωiξ (x+isgn(ω)y) isgn(ω)v(z) = | | e · fˆ(ξ)dξ 2π 0 Z 0 ωiξ (x isgn(ω)y) i e · − fˆ(ξ)dξ , − ∞ − 2 then v is also harmonic in R+ and both u and v are real if f is. Furthermore, u + iv is analytic since it satisfies the Cauchy-Riemann equations u = v = ωiξu(z) and u = v = ωiξv(z), x y y − x − so v is the harmonic conjugate of u. Clearly, v can also be written as, by Theorem 1.12, Proposition 1.3 and Theorem 1.28, Z ω ωiξ x ωξ y v(z) =| | isgn(ω)sgn(ξ)e · e−| | fˆ(ξ)dξ 2π R − Z ω ωiξ x ωξ y =| | isgn(ω)Fξ [sgn(ξ)e · e−| | ](η) f (η)dη 2π R − Z ω ωξ y =| | isgn(ω)Fξ [sgn(ξ)e−| | ](η x) f (η)dη 2π R − − Z 1 ωξ y = isgn(ω)Fξ− [sgn(ξ)e−| | ](x η) f (η)dη, R − − which is equivalent to v(x,y) = Q f (x), (4.21) y ∗ where ωξ y Qˆ (ξ) = isgn(ω)sgn(ξ)e−| | . (4.22) y − Now we invert the Fourier transform, we get, by a change of variables and integration by parts, Z ω ωix ξ ωξ y Qy(x) = isgn(ω)| | e · sgn(ξ)e−| | dξ − 2π R Z ∞ Z 0 ω ωix ξ ω ξy ωix ξ ω ξy = isgn(ω)| | e · e−| | dξ e · e| | dξ − 2π 0 − ∞ − Z ∞ Z ∞ ω ωix ξ ω ξy ωix ξ ω ξy = isgn(ω)| | e · e−| | dξ e− · e−| | dξ − 2π 0 − 0 Z ∞ ω ξy ω ωix ξ ωix ξ ∂ξ e−| | = isgn(ω)| | e · e− · dξ − 2π 0 − ω y −|∞ | 1 h ωix ξ ωix ξ ω ξy =isgn(ω) e · e− · e−| | 2πy − 0 Z ∞ ωix ξ ωix ξ ω ξy i ωix e · + e− · e−| | dξ − 0 Z ∞ ω x ωix ξ ωix ξ ω ξy =| | e · + e− · e−| | dξ 2πy 0 Z ω x ωix ξ ωξ y x ω ωξ y =| | e− · e−| | dξ = F | |e−| | 2πy R y 2π x x c y c x = P (x) = 1 = 1 , y y y y2 + x2 y2 + x2 where c1 = Γ (1)/π = 1/π. That is, 1 x Q (x) = . y π y2 + x2 - 72 - 4. Singular Integrals

One can immediately verify that Q(x,y) = Qy(x) is a harmonic function in the upper half-plane and the conjugate of the Poisson kernel Py(x) = P(x,y). More precisely, they satisfy Cauchy- Riemann equations 1 2xy 1 x2 y2 ∂ P = ∂ Q = , ∂ P = ∂ Q = − . x y −π (y2 + x2)2 y − x π (y2 + x2)2 In Theorem 4.9, we studied the limit of u(x,t) as y 0 using the fact that P is an approxi- → { y} mation of the identity. We would like to do the same for v(x,y), but we immediately run into an obstacle: Q is not an approximation of the identity and, in fact, Q is not integrable for any { y} y y > 0. Formally, 1 lim Qy(x) = , y 0 πx → this is not even locally integrable, so we cannot define its convolution with smooth functions. We define a tempered distribution called the principal value of 1/x, abbreviated p.v.1/x, by 1 Z φ(x) p.v. ,φ = lim dx, φ S . x ε 0 x >ε x ∈ → | | To see that this expression defines a tempered distribution, we rewrite it as 1 Z φ(x) φ(0) Z φ(x) p.v. ,φ = − dx + dx, x x <1 x x 1 x | | | |> this holds since the integral of 1/x on ε < x < 1 is zero. It is now immediate that | | 1 p.v. ,φ C( φ 0 ∞ + xφ ∞). x 6 k k k k 1 1 Proposition 4.13. In S 0(R), we have lim Qy(x) = p.v. . y 0 π x → 1 Proof. For each ε > 0, the functions ψε (x) = x− χ x >ε are bounded and define tempered dis- | | tributions. It follows at once from the definition that in S 0, 1 lim ψε (x) = p.v. . ε 0 x → Therefore, it will suffice to prove that in S 0 1 lim Qy ψy = 0. y 0 − π → Fix φ S , then by a change of variables, we have ∈ Z xφ(x) Z φ(x) πQy ψy,φ = 2 2 dx dx h − i R y + x − x y x | |> Z xφ(x) Z x 1 = 2 2 dx + 2 2 φ(x)dx x Z xφ(yx) Z φ(yx) = 2 dx 2 dx. x <1 1 + x − x 1 x(1 + x ) | | | |> If we take the limit as y 0 and apply the dominated convergence theorem, we get two integrals → of odd functions on symmetric domains. Hence, the limit equals 0. ut As a consequence of this proposition, we get that 1 Z f (x t) lim Qy f (x) = lim − dt, y 0 ∗ π ε 0 t >ε t → → | | and by the continuity of the Fourier transform on S 0 and by (4.22), we get 4.2. Poisson kernel and Hilbert transform - 73 -

1 1 F p.v. (ξ) = isgn(ω)sgn(ξ). π x − Given a function f S , we can deﬁne its Hilbert transform by any one of the following ∈ equivalent expressions:

H f = lim Qy f , y 0 ∗ → 1 1 H f = p.v. f , π x ∗ 1 H f =F − ( isgn(ω)sgn(ξ) fˆ(ξ)). − The third expression also allows us to define the Hilbert transform of functions in L2(R), which satisfies, with the help of Theorem 1.26, ω 1/2 ω 1/2 H f = | | F (H f ) = | | fˆ = f , (4.23) k k2 2π k k2 2π k k2 k k2 that is, H is an isometry on L2(R). Moreover, H satisfies 2 1 2 H f = H(H f ) =F − (( isgn(ω)sgn(ξ)) fˆ(ξ)) = f , (4.24) − − By Theorem 1.28, we have Z Z 1 H f ,g = H f gdx = F − ( isgn(ω)sgn(ξ) fˆ(ξ)) gdx h i R · R − · Z = isgn(ω)sgn(ξ) fˆ(ξ) gˇ(ξ)dξ R − · Z = f (x) F [ isgn(ω)sgn(ξ)gˇ(ξ)](x)dx R · − Z ω = f (x) F [ isgn(ω)sgn(ξ)| |gˆ( ξ)](x)dx R · − 2π − Z 1 = f (x) F − [isgn(ω)sgn(η)gˆ(η)](x)dx R · Z = f Hgdx = f , Hg , (4.25) − R · h − i namely, the dual/conjugate operator of H is H = H. Similarly, the adjoint operator H of H 0 − ∗ is uniquely defined via the identity Z Z ( f ,Hg) = f Hgdx = H f gdx¯ = ( H f ,g) =: (H∗ f ,g), R · − R − that is, H = H. ∗ − Note that for given x R, H f (x) is defined for all integrable functions f on R that satisfy a ∈ Hölder condition near the point x, that is, f (x) f (y) C x y εx | − | 6 x| − | for some C > 0 and ε > 0 whenever y x < δ . Indeed, suppose that this is the case, then x x | − | x 1 Z f (y) 1 Z f (y) Qy f (x) = dy + dy ∗ π ε< x y <δ x y π x y δ x y | − | x − | − |> x − 1 Z f (y) f (x) 1 Z f (y) = − dy + dy. π ε< x y <δ x y π x y δ x y | − | x − | − |> x − Both integrals converge absolutely, and hence the limit of Q f (x) exists as ε 0. Therefore, y ∗ → the Hilbert transform of a piecewise smooth integrable function is well defined at all points of Hölder-Lipschitz continuity of the function. On the other hand, observe that Q f is well y ∗ - 74 - 4. Singular Integrals defined for all f Lp, 1 p < ∞, as it follows from the Hölder inequality, since 1/x is in Lp0 ∈ 6 on the set x ε. | | > Example 4.14. Consider the characteristic function χ[a,b] of an interval [a,b]. It is a simple calculation to show that 1 x a H(χ )(x) = ln | − |. (4.26) [a,b] π x b | − | Let us verify this identity. By the definition, we have 1 Z χ[a,b](x y) 1 Z 1 H(χ[a,b])(x) = lim − dy = lim dy. π ε 0 π ε 0 y >ε y >ε y x b| |y x a y → | | → − 6 6 − Thus, we only need to consider three cases: x b > 0, x a < 0 and x b < 0 < x a. For the − − − − first two cases, we have Z x a 1 − 1 1 x a H(χ[a,b])(x) = dy = ln | − |. π x b y π x b − | − | For the third case we get (without loss of generality, we can assume ε < min( x a , x b )) Z ε Z x a | − | | − | 1 − 1 − 1 H(χ[a,b])(x) = lim dy + dy π ε 0 x b y ε y → − 1 x a ε = lim ln | − | + ln π ε 0 ε x b → | − | 1 x a = ln | − |, π x b | − | where it is crucial to observe how the cancellation of the odd kernel 1/x is manifested. Note 1 that H(χ[a,b])(x) blows up logarithmically for x near the points a and b and decays like x− as x ∞. See the following graph with a = 1 and b = 3: → ±

The following is a graph of the function H(χ[ 10,0] [1,2] [4,7]): − ∪ ∪

It is obvious, for the dilation operator δε with ε > 0, by changes of variables (εy y), that → 4.2. Poisson kernel and Hilbert transform - 75 -

1 Z f (εx εy) (Hδε ) f (x) = lim − dy σ 0 π y σ y → | |> Z f (εx y) = lim − dy = (δε H) f (x), σ 0 y εσ y → | |> so Hδε = δε H; and it is equally obvious that Hδε = δε H, if ε < 0. − These simple considerations of dilation “invariance” and the obvious translation invariance in fact characterize the Hilbert transform. Proposition 4.15 (Characterization of Hilbert transform). Suppose T is a bounded linear operator on L2(R) which satisﬁes the following properties: (a) T commutes with translations; (b) T commutes with positive dilations; (c) T anticommutes with the reﬂection f (x) f ( x). → − Then, T is a constant multiple of the Hilbert transform.

Proof. Since T commutes with translations and maps L2(R) to itself, according to Theorem 1.62, there is a bounded function m(ξ) such that Tc f (ξ) = m(ξ) fˆ(ξ). The assumptions (b) and 2 (c) may be written as Tδε f = sgn(ε)δε T f for all f L (R). By part (iv) in Proposition 1.3, ∈ we have 1 F (Tδε f )(ξ) =m(ξ)F (δε f )(ξ) = m(ξ) ε − fˆ(ξ/ε), | | 1 1 sgn(ε)F (δε T f )(ξ) =sgn(ε) ε − Tc f (ξ/ε) = sgn(ε) ε − m(ξ/ε) fˆ(ξ/ε), | | | | which means m(εξ) = sgn(ε)m(ξ), if ε = 0. This shows that m(ξ) = csgn(ξ), and the propo- 6 sition is proved. ut The next theorem shows that the Hilbert transform, now deﬁned for functions in S or L2, can be extended to functions in Lp, 1 6 p < ∞. Theorem 4.16. For f S (R), the following assertions are true: ∈ (i) (Kolmogorov) H is of weak type (1,1): m C ( x R : H f (x) > α ) f 1. { ∈ | | } 6 α k k (ii) (M. Riesz) H is of type (p, p), 1 0. From the Calderón-Zygmund decomposition of f at height α (Theorem 3.20), there exist two functions g and b such that f = g + b and (1) g f and g ∞ 2α. k k1 6 k k1 k k 6 b = b b I R b (x)dx = (2) ∑ j j, where each j is supported in a dyadic interval j satisfying Ij j 0 and b 4αm(I ). Furthermore, the intervals I and I have disjoint interiors when j = k. k jk1 6 j j k 6 (3) ∑ m(I ) α 1 f . j j 6 − k k1 Let 2I be the interval with the same center as I and twice the length, and let Ω = I and j j ∪ j j Ω = 2I . Then m(Ω ) 2m(Ω) 2α 1 f . ∗ ∪ j j ∗ 6 6 − k k1 Since H f = Hg + Hb, from parts (iv) and (vi) of Proposition 2.15,(4.23) and (1), we have (H f ) (α) 6 (Hg) (α/2) + (Hb) (α/2) ∗ Z ∗ ∗ 2 2 m m 6(α/2)− Hg(x) dx + (Ω ∗) + ( x / Ω ∗ : Hb(x) > α/2 ) R | | { ∈ | | } Z Z 4 2 1 1 6 2 g(x) dx + 2α− f 1 + 2α− Hb(x) dx α R | | k k R Ω | | \ ∗ - 76 - 4. Singular Integrals 8 Z 2 2 Z 6 g(x) dx + f 1 + ∑ Hb j(x) dx α R | | α k k α R Ω | | \ ∗ j 8 2 2 Z 6 f 1 + f 1 + ∑ Hb j(x) dx. α k k α k k α R 2Ij | | j \ For x / 2I , we have ∈ j 1 Z b j(y) 1 Z b j(y) Hb j(x) = p.v. dy = dy, π I x y π I x y j − j − since suppb I and x y m(I )/2 for y I . Denote the center of I by c , then, since b j ⊂ j | − | > j ∈ j j j j is mean zero, we have Z Z Z 1 b j(y) Hb j(x) dx = dy dx R 2Ij | | R 2Ij π Ij x y \ \ − 1 Z Z 1 1 = b j(y) dy dx π R 2Ij Ij x y − x c j \ − − 1 Z Z y c j 6 b j(y) | − | dx dy π Ij | | R 2Ij x y x c j \ | − || − | 1 Z Z m(Ij) 6 b j(y) 2 dx dy. π Ij | | R 2Ij x c j \ | − | The last inequality follows from the fact that y c < m(I )/2 and x y > x c /2. Since | − j| j | − | | − j| x c > m(I ), the inner integral equals | − j| j Z ∞ 1 1 m(I ) dr = m(I ) = . 2 j 2 2 j m 2 m(Ij) r (Ij) Thus, by (2) and (3), Z 10 4 10 4 m (H f ) (α) 6 f 1 + ∑ b j(y) dy 6 f 1 + ∑4α (Ij) ∗ α k k απ j Ij | | α k k απ j 10 16 1 10 + 16/π f + f = f . 6 α k k1 π α k k1 α k k1 (ii) Since H is of weak type (1,1) and of type (2,2), by the Marcinkiewicz interpolation theorem, we have the strong (p, p) inequality for 1 2, we apply the dual estimate with the help of (4.25) and the result for p < 2:

H f p = sup H f ,g = sup f ,Hg k k g 61 |h i| g 61 |h i| k kp0 k kp0

6 f p sup Hg p0 6 Cp0 f p. k k g 61 k k k k k kp0 This completes the proof. ut Remark 4.17. i) Recall from the proof of the Marcinkiewicz interpolation theorem that the coefﬁcient  10 + 16/π (1/2)1/2  + + 21/2, 1 2.  1/2 1/p − So the constant Cp tends to inﬁnity as p tends to 1 or ∞. More precisely, 1 C = O(p) as p ∞, and C = O((p 1)− ) as p 1. p → p − → 4.3. The Calderón-Zygmund theorem - 77 -

ii) The strong (p, p) inequality is false if p = 1 or p = ∞, this can easily be seen from the 1 x a previous example Hχ[a,b] = π ln |x−b| which is neither integrable nor bounded. See the following ﬁgure. | − |

The integral

Hχ[1,2]

iii) By using the inequalities in Theorem 4.16, we can extend the Hilbert transform to functions in Lp, 1 p < ∞. If f L1 and f is a sequence of functions in S that converges to f 6 ∈ { n} in L1, then by the weak (1,1) inequality the sequence H f is a Cauchy sequence in measure: { n} for any ε > 0, m lim ( x R : (H fn H fm)(x) > ε ) = 0. m,n ∞ → { ∈ | − | } Therefore, it converges in measure to a measurable function which we deﬁne to be the Hilbert transform of f . If f Lp, 1 < p < ∞, and f is a sequence of functions in S that converges to f in Lp, by ∈ { n} the strong (p, p) inequality, H f is a Cauchy sequence in Lp, so it converges to a function in { n} Lp which we call the Hilbert transform of f . In either case, a subsequence of H f , depending on f , converges pointwise almost every- { n} where to H f as deﬁned.

4.3 The Calderón-Zygmund theorem

From this section on, we are going to consider singular integrals whose kernels have the same essential properties as the kernel of the Hilbert transform. We can generalize Theorem 4.16 to get the following result. Theorem 4.18 (Calderón-Zygmund Theorem). Let K be a tempered distribution in Rn which coincides with a locally integrable function on Rn 0 and satisﬁes \{ } Kb(ξ) B, (4.27) | | 6 Z n K(x y) K(x) dx 6 B, y R . (4.28) x 2 y | − − | ∈ | |> | | Then we have the strong (p, p) estimate for 1 < p < ∞ K f C f , (4.29) k ∗ kp 6 pk kp and the weak (1,1) estimate C (K f ) (α) 6 f 1. (4.30) ∗ ∗ α k k - 78 - 4. Singular Integrals

We will show that these inequalities are true for f S , but they can be extended to arbi- ∈ trary f Lp as we did for the Hilbert transform. Condition (4.28) is usually referred to as the ∈ Hörmander condition; in practice it is often deduced from another stronger condition called the gradient condition (i.e., (4.31) as below). Proposition 4.19. The Hörmander condition (4.28) holds if for every x = 0 6 C ∇K(x) . (4.31) | | 6 x n+1 | | Proof. By the integral mean value theorem and (4.31), we have Z Z Z 1 K(x y) K(x) dx 6 ∇K(x θy) y dθdx x 2 y | − − | x 2 y 0 | − || | | |> | | | |> | | Z 1 Z C y Z 1 Z C y 6 | |n+1 dxdθ 6 | |n+1 dxdθ 0 x >2 y x θy 0 x >2 y ( x /2) | | | | | −Z ∞ | | | | | | | n+1 1 n+1 1 n 62 C y ωn 1 2 dr = 2 C y ωn 1 = 2 Cωn 1. | | − 2 y r | | − 2 y − | | | | This completes the proof. ut Proof of Theorem 4.18. Since the proof is (essentially) a repetition of the proof of Theorem 4.16, we will omit the details. Let f S and T f = K f . From (4.27), it follows that ∈ ∗ ω n/2 ω n/2 T f 2 = | | Tc f 2 = | | Kb fˆ 2 k k 2π k k 2π k k ω n/2 ω n/2 (4.32) 6 | | Kb ∞ fˆ 2 6 B | | fˆ 2 2π k k k k 2π k k =B f , k k2 by the Plancherel theorem (Theorem 1.26) and part (vi) in Proposition 1.3. It will suffice to prove that T is of weak type (1,1) since the strong (p, p) inequality, 1 2 it follows from the duality since the conjugate operator T 0 has kernel K0(x) = K( x) which also satisfies (4.27) and (4.28). In fact, Z − Z Z T f ,ϕ = T f (x)ϕ(x)dx = K(x y) f (y)dyϕ(x)dx h i Rn Rn Rn − Z Z Z Z = K( (y x))ϕ(x)dx f (y)dy = (K0 ϕ)(y) f (y)dy Rn Rn − − Rn Rn ∗ = f ,T 0ϕ . h i To show that f is of weak type (1,1), fix α > 0 and form the Calderón-Zygmund decomposition of f at height α. Then as in Theorem 4.16, we can write f = g + b, where n (i) g f and g ∞ 2 α. k k1 6 k k1 k k 6 b = b b Q R b (x)dx = (ii) ∑ j j, where each j is supported in a dyadic cube j satisfying Q j j 0 and b 2n+1αm(Q ). Furthermore, the cubes Q and Q have disjoint interiors when j = k. k jk1 6 j j k 6 (iii) ∑ m(Q ) α 1 f . j j 6 − k k1 The argument now proceeds as before, and the proof reduces to showing that Z Z Tb j(x) dx C b j(x) dx, (4.33) n 6 R Q∗ | | Q j | | \ j where Q∗j is the cube with the same center as Q j and whose sides are 2√n times longer. Denote their common center by c j. Inequality (4.33) follows from the Hörmander condition (4.28): 4.4. Truncated integrals - 79 - since each b j has zero average, if x / Q∗j Z ∈ Z Tb j(x) = K(x y)b j(y)dy = [K(x y) K(x c j)]b j(y)dy; Q j − Q j − − − hence, Z Z Z ! Tb j(x) dx K(x y) K(x c j) dx b j(y) dy. n 6 n R Q∗ | | Q j R Q∗ | − − − | | | \ j \ j However, by changing variables x c = x and y c = y , and the fact that x c 2 y c − j 0 − j 0 | − j| > | − j| for all x / Q∗j and y Q j as an obvious geometric consideration shows, and (4.28), we get ∈ Z ∈ Z K(x y) K(x c j) dx K(x0 y0) K(x0) dx0 B. n 6 6 R Q∗ | − − − | x 2 y | − − | \ j | 0|> | 0| This completes the proof. ut

4.4 Truncated integrals

There is still an element which may be considered unsatisfactory in our formulation, and this is because of the following related points: 1) The L2 boundedness of the operator has been assumed via the hypothesis that Kb L∞ and ∈ not obtained as a consequence of some condition on the kernel K; 2) An extraneous condition such as K L2 subsists in the hypothesis; and for this reason our ∈ results do not directly treat the “principal-value” singular integrals, those which exist because of the cancelation of positive and negative values. However, from what we have done, it is now a relatively simple matter to obtain a theorem which covers the cases of interest. 1 n Deﬁnition 4.20. Suppose that K(x) Lloc(R 0 ) and satisﬁes the following conditions: ∈ \{ }n K(x) B x − , x = 0, | | 6 | | ∀ 6 Z (4.34) K(x y) K(x) dx 6 B, y = 0, x 2 y | − − | ∀ 6 | |> | | and Z K(x)dx = 0, 0 < R1 < R2 < ∞. (4.35) R < x 0 and f Lp(Rn), ∈ 1 Then the following conclusions hold. (i) We have

Tε f A f (4.37) k kp 6 pk kp where Ap is independent of f and ε. p n p (ii) For any f L (R ), limε 0 Tε ( f ) exists in the sense of L norm. That is, there exists an ∈ → operator T such that Z T f (x) = p.v. K(y) f (x y)dy. Rn − p n (iii) T f p Ap f p for f L (R ). k k 6 k k ∈ - 80 - 4. Singular Integrals

Remark 4.22. 1) The linear operator T defined by (ii) of Theorem 4.21 is called the Calderón- Zygmund singular integral operator. Tε is also called the truncated operator of T. 2) The cancelation property alluded to is contained in condition (4.35). This hypothesis, together with (4.34), allows us to prove the L2 boundedness and from this the Lp convergence of the truncated integrals (4.37). 3) We should point out that the kernel K(x) = 1 , x R1, clearly satisfies the hypotheses of πx ∈ Theorem 4.21. Therefore, we have the existence of the Hilbert transform in the sense that if f Lp(R), 1 exists in the Lp norm and the resulting operator is bounded in Lp, as has shown in Theorem 4.16. For L2 boundedness, we have the following lemma. Lemma 4.23. Suppose K satisfies the conditions (4.34) and (4.35) of the above theorem with bound B. Let K(x), x > ε, Kε (x) = | | 0, x < ε. | | Then, we have the estimate sup Kbε (ξ) 6 CB, ε > 0, (4.38) ξ | | where C depends only on the dimension n.

Proof. First, we prove the inequality (4.38) for the special case ε = 1. Since Kˆ1(0) = 0, thus we can assume ξ = 0 and have 6 Z ωix ξ Kb1(ξ) = lim e− · K1(x)dx R ∞ x 6R Z→ | | Z ωix ξ ωix ξ = e− · K1(x)dx + lim e− · K1(x)dx x <2π/( ω ξ ) R ∞ 2π/( ω ξ )< x R | | | || | → | || | | |6 =:I1 + I2. R By the condition (4.35), 1< x <2π/( ω ξ ) K(x)dx = 0 which implies | | Z| || | K1(x)dx = 0. x <2π/( ω ξ ) | | | || | R ωix ξ R ωix ξ Thus, x <2π/( ω ξ ) e− · K1(x)dx = x <2π/( ω ξ )[e− · 1]K1(x)dx. Hence, from the fact θ | | | || | | | | || | − ei 1 6 θ (see Section 1.1) and the ﬁrst condition in (4.34), we get | − | | | Z Z n+1 I1 6 ω x ξ K1(x) dx 6 ω B ξ x − dx | | x <2π/( ω ξ ) | || || || | | | | | x <2π/( ω ξ ) | | | | | || | | | | || | Z 2π/( ω ξ ) | || | =ωn 1B ω ξ dr = 2πωn 1B. − | || | 0 − To estimate I , choose z = z(ξ) such that e ωiξ z = 1. This choice can be realized if z = 2 − · − πξ/(ω ξ 2), with z = π/( ω ξ ). Since, by changing variables x + z = y, we get | |Z | | | || |Z Z ωix ξ ωi(x+z) ξ ωiy ξ e− · K1(x)dx = e− · K1(x)dx = e− · K1(y z)dy Rn − Rn − Rn − Z ωix ξ = e− · K1(x z)dx, − Rn − 4.4. Truncated integrals - 81 -

R ωix ξ 1 R ωix ξ which implies n e K (x)dx = n e [K (x) K (x z)]dx, then we have R − · 1 2 R − · 1 − 1 − Z Z ωix ξ I2 = lim e− · K1(x)dx R ∞ x R − x 2π/( ω ξ ) → 6 6 Z| | | | | || | Z 1 ωix ξ ωix ξ = lim e− · [K1(x) K1(x z)]dx e− · K1(x)dx 2 R ∞ x R − − − x 2π/( ω ξ ) → 6 6 Z| | | | | || | 1 ωix ξ = lim e− · [K1(x) K1(x z)]dx 2 R ∞ 2π/( ω ξ ) x R − − → 6 6 Z | || | | | Z 1 ωix ξ 1 ωix ξ e− · K1(x)dx e− · K1(x z)dx. − 2 x 2π/( ω ξ ) − 2 x 2π/( ω ξ ) − | |6 | || | | |6 | || | The last two integrals are equal to, in view of the integration by parts, Z Z 1 ωix ξ 1 ωi(y+z) ξ e− · K1(x)dx e− · K1(y)dy − 2 x 62π/( ω ξ ) − 2 y+z 62π/( ω ξ ) Z| | | || | Z| | | || | 1 ωix ξ 1 ωix ξ = e− · K1(x)dx + e− · K1(x)dx − 2 x 62π/( ω ξ ) 2 x+z 62π/( ω ξ ) Z| | | || | Z | | | || | 1 ωix ξ 1 ωix ξ = e− · K1(x)dx + e− · K1(x)dx. − x 62π/( ω ξ ) x+z 62π/( ω ξ ) 2 x|+|z >2π/|( ω|| ξ| ) 2 | x >| 2π/( ω| ξ|| )| | | | || | | | | || | For the first integral, we have 2π/( ω ξ ) x x + z z > | || | > | | > | | − | | 2π/( ω ξ ) π/( ω ξ ) = π/( ω ξ ), and for the second one, | || | − | || | | || | 2π/( ω ξ ) < x 6 x + z + z 6 3π/( ω ξ ). These two integrals | || | | | | | | | | || | 2π are taken over a region contained in the spherical shell, π/( ω ξ ) < O ω ξ x 1 | || | | || | | | x 6 3π/( ω ξ ) (see the figure), and is bounded by Bωn 1 ln3 | | | || | 2 − since K (x) B x n. By z = π/( ω ξ ) and the condition (4.34), | 1 | 6 | |− | | | || | the first integral of I2 is majorized by 1 Z K1(x z) K1(x) dx 2 x 2π/( ω ξ ) | − − | | |> | || | 1 Z 1 = K1(x z) K1(x) dx 6 B. 2 x 2 z | − − | 2 | |> | | Thus, we have obtained 1 1 Kb1(ξ) 6 2πωn 1B + B + Bωn 1 ln3 6 CB, | | − 2 2 − where C depends only on n. We finish the proof for K1. To pass to the case of general Kε , we use a simple observation (dilation argument) whose significance carries over to the whole theory presented in this chapter. Let δε be the dilation by the factor ε > 0, i.e., (δε f )(x) = f (εx). Thus if T is a convolution operator Z T f (x) = ϕ f (x) = ϕ(x y) f (y)dy, ∗ Rn − then Z 1 δ 1 Tδ f (x) = ϕ(ε− x y) f (εy)dy ε− ε Rn − Z n 1 =ε− ϕ(ε− (x z)) f (z)dz = ϕε f , Rn − ∗ n 1 where ϕ (x) = ε ϕ(ε x). In our case, if T corresponds to the kernel K(x), then δ 1 Tδ ε − − ε− ε n 1 corresponds to the kernel ε− K(ε− x). Notice that if K satisfies the assumptions of our theorem, n 1 then ε− K(ε− x) also satisfies these assumptions with the same bounds. (A similar remark - 82 - 4. Singular Integrals holds for the assumptions of all the theorems in this chapter.) Now, with our K given, let K0 = n ε K(εx). Then K0 satisfies the conditions of our lemma with the same bound B, and so if we denote K0(x), x > 1, K0 (x) = | | 1 0, x < 1, | | n 1 then we know that Kb10 (ξ) 6 CB. The Fourier transform of ε− K10 (ε− x) is Kb10 (εξ) which | | n 1 is again bounded by CB; however ε− K10 (ε− x) = Kε (x), therefore the lemma is completely proved. ut We can now prove Theorem 4.21.

Proof of Theorem 4.21. Since K satisﬁes the conditions (4.34) and (4.35), then Kε (x) satisﬁes the same conditions with bounds not greater than CB. By Lemma 4.23 and Theorem 4.18, we p have that the L boundedness of the operators Kε ε>0, are uniformly bounded. { } p 1 n Next, we prove that Tε f1 ε>0 is a Cauchy sequence in L provided f1 C (R ). In fact, we { } ∈ c have Z Z Tε f1(x) Tη f1(x) = K(y) f1(x y)dy K(y) f1(x y)dy − y >ε − − y >η − | | Z | | =sgn(η ε) K(y)[ f1(x y) f1(x)]dy, − min(ε,η) y max(ε,η) − − 6| |6 because of the cancelation condition (4.35). For p (1,∞), we get, by the mean value theorem ∈ with some θ [0,1], Minkowski’s inequality and (4.34), that ∈ Z Tε f1 Tη f1 p 6 K(y) ∇ f1(x θy) y dy k − k min(ε,η) y max(ε,η) | || − || | 6| |6 p Z 6 K(y) ∇ f1(x θy) p y dy min(ε,η)6 y 6max(ε,η) | |k − k | | Z | | 6C K(y) y dy min(ε,η)6 y 6max(ε,η) | || | Z | | n+1 6CB y − dy min(ε,η) y max(ε,η) | | 6| |6 Z max(ε,η) =CBωn 1 dr − min(ε,η)

=CBωn 1 η ε − | − | p which tends to 0 as ε,η 0. Thus, we obtain Tε f converges in L as ε 0 by the completeness → 1 → of Lp. Finally, an arbitrary f Lp can be written as f = f + f where f is of the type described ∈ 1 2 1 above and f2 p is small. We apply the basic inequality (4.37) for f2 to get Tε f2 p 6 C f2 p, k k p k k k k then we see that limε 0 Tε f exists in L norm; that the limiting operator T also satisﬁes the → inequality (4.37) is then obvious. Thus, we complete the proof of the theorem. ut

4.5 Singular integral operators commuted with dilations

In this section, we shall consider those operators which not only commute with translations but also with dilations. Among these we shall study the class of singular integral operators, falling under the scope of Theorem 4.21. 4.5. Singular integral operators commuted with dilations - 83 -

If T corresponds to the kernel K(x), then as we have already pointed out, δ 1 Tδ cor- ε− ε n 1 responds to the kernel ε K(ε x). So if δ 1 Tδ = T we are back to the requirement − − ε− ε K(x) = ε nK(ε 1x), i.e., K(εx) = ε nK(x), ε > 0; that is K is homogeneous of degree n. − − − − Put another way Ω(x) K(x) = , (4.39) x n | | with Ω homogeneous of degree 0, i.e., Ω(εx) = Ω(x), ε > 0. This condition on Ω is equivalent with the fact that it is constant on rays emanating from the origin; in particular, Ω is completely n 1 determined by its restriction to the unit sphere S − . Let us try to reinterpret the conditions of Theorem 4.21 in terms of Ω. 1) By (4.34), Ω(x) must be bounded and consequently integrable on Sn 1; and another con- − R Ω(x y) Ω(x) dition x 2 y x −y n x n dx 6 C which is not easily restated precisely in terms of Ω. How- | |> | | − ever, what is evident| − | is that| | it requires a certain continuity of Ω. Here we shall content ourselves in treating the case where Ω satisfies the following “Dini-type” condition suggested by (4.34): Z 1 w(η) if w(η) := sup Ω(x) Ω(x0) , then dη < ∞. (4.40) x x η | − | 0 η | − 0|6 x = x =1 | | | 0| Of course, any Ω which is of class C1, or even merely Lipschitz continuous, satisfies the condition (4.40). 2) The cancelation condition (4.35) is then the same as the condition Z Ω(x)dσ(x) = 0 (4.41) n 1 S − n 1 where dσ(x) is the induced Euclidean measure on S − . In fact, this equation implies that Z Z R2 Z Ω(rx0) n 1 K(x)dx = n dσ(x0)r − dr R < x | | (a) Then there exists a bound Ap (independent of f and ε) such that Tε f A f . k kp 6 pk kp p (b) limε 0 Tε f = T f exists in L norm, and → T f A f . k kp 6 pk kp (c) If f L2(Rn), then the Fourier transforms of f and T f are related by Tc f (ξ) = m(ξ) fˆ(ξ), ∈ where m is a homogeneous function of degree 0. Explicitly, Z πi m(ξ) = sgn(ω)sgn(ξ x) + ln(1/ ξ x ) Ω(x)dσ(x), ξ = 1. (4.42) n 1 S − − 2 · | · | | | Proof. The conclusions (a) and (b) are immediately consequences of Theorem 4.21, once we Ω(x) have shown that any K(x) of the form x n satisfies | | - 84 - 4. Singular Integrals Z K(x y) K(x) dx 6 B, (4.43) x 2 y | − − | | |> | | if Ω is as in condition (4.40). Indeed, Ω(x y) Ω(x) 1 1 K(x y) K(x) = − − + Ω(x) . − − x y n x y n − x n | − | | − | | | The second group of terms is bounded since Ω is bounded and

Z 1 1 Z x n x y n n n dx = | | − | n− n| dx x 2 y x y − x x 2 y x y x | |> | | | − | | | | |> | | | − | | | n 1 n 1 j j Z x x y ∑ − x x y || | − | − || j=0 | | − − | − | = n n dx x 2 y x y x | |> | | | − | | | Z n 1 − j 1 j n 6 y ∑ x − − x y − dx x 2 y | | | | | − | | |> | | j=0 Z n 1 − j 1 j n 6 y ∑ x − − ( x /2) − dx (since x y > x y > x /2) x 2 y | | | | | | | − | | | − | | | | | |> | | j=0 Z n 1 Z − n j n 1 n n 1 = y ∑ 2 − x − − dx = 2(2 1) y x − − dx x 2 y | | | | − | | x 2 y | | | |> | | j=0 | |> | | n 1 n =2(2 1) y ωn 1 = (2 1)ωn 1. − | | − 2 y − − | | To estimate the ﬁrst group of terms, we notice that if x 2 y , the distance PQ between | | > | | | | the projections of x y and x on the unit sphere as in the following picture. −

x x x P : x |x| y 1 y Q: x−y 1 | − | y P P x y x y − θ Q − θ Q O O

y Case 1: x x y , sin θ | | Case 2: x x y , y y | | ≥ | − | x sin θ x| |y |x| ≤ | | | | ≤ | − | ≤ | − | ≤ | |

π θ sinθ sin −2 π By the sine theorem, we have PQ = OP where OP = 1. Since y 6 x /2, we have θ 6 2 |q | | | | | | | | | θ 1+cosθ √ and so cosθ > 0. Thus, cos 2 = 2 > 1/ 2. Then, we have

x y x sinθ sinθ y y − = PQ = = 6 √2| | 6 2| | x y − x | | sin( π θ ) cos θ x x | − | | | 2 − 2 2 | | | | y since sinθ 6 |x| for both cases. So the integral corresponding to the ﬁrst group of terms is dominated by | | Z Z Z ∞ n y dx n dz n dr 2 w 2| | n = 2 w(2/ z ) n = 2 ωn 1 w(2/r) x 2 y x x z 2 | | z − 2 r | |> | | | | | | | |> | | Z 1 n w(η)dη =2 ωn 1 < ∞ − 0 η in view of changes of variables x = y z and the Dini-type condition (4.40). | | 4.5. Singular integral operators commuted with dilations - 85 -

Now, we prove (c). Since T is a bounded linear operator on L2 which commutes with translations, we know, by Theorem 1.62 and Proposition 1.3, that T can be realized in terms of a multiplier m such that Tc f (ξ) = m(ξ) fˆ(ξ). For such operators, the fact that they commute with dilations is equivalent with the property that the multiplier is homogeneous of degree 0. For our particular operators we have not only the existence of m but also an explicit expression of the multiplier in terms of the kernel. This formula is deduced as follows. Since K(x) is not integrable, we first consider its truncated function. Let 0 < ε < η < ∞, and  Ω( )  x n , ε 6 x 6 η, Kε,η (x) = x | | 0|, | otherwise. 1 n 2 n Clearly, Kε,η L (R ). If f L (R ) then K\ε,η f (ξ) = Kdε,η (ξ) fˆ(ξ). ∈ ∈ ∗ We shall prove two facts about Kdε,η (ξ). (i) sup Kdε,η (ξ) A, with A independent of ε and η; ξ | | 6 (ii) if ξ = 0, lim ε 0 Kdε,η (ξ) = m(ξ), see (4.42). η→∞ 6 → For this purpose, it is convenient to introduce polar coordinates. Let x = rx0, r = x , x0 = n 1 n 1 | | x/ x S − , and ξ = Rξ 0, R = ξ , ξ 0 = ξ/ ξ S − . Then we have | | ∈ Z | | | | ∈ Z ωix ξ ωix ξ Ω(x) Kdε,η (ξ) = e− · Kε,η (x)dx = e− · n dx Rn ε x η x 6| |6 Z Z η | | ωiRrx ξ n n 1 = Ω(x0) e− 0· 0 r− r − dr dσ(x0) n 1 S − ε Z Z η ωiRrx ξ dr = Ω(x0) e− 0· 0 dσ(x0). n 1 S − ε r Since Z Ω(x0)dσ(x0) = 0, n 1 S − we can introduce the factor cos( ω Rr) (which does not depend on x ) in the integral defining | | 0 Kdε,η (ξ). We shall also need the auxiliary integral Z η ωiRrx0 ξ 0 dr Iε,η (ξ,x0) = [e− · cos( ω Rr)] , R > 0. ε − | | r Thus, it follows Z Kdε,η (ξ) = Iε,η (ξ,x0)Ω(x0)dσ(x0). n 1 S − Now, we first consider Iε,η (ξ,x0). For its imaginary part, we have, by changing variable ωRr(x0 ξ 0) = t, that · Z η sinωRr(x0 ξ 0) ℑIε,η (ξ,x0) = · dr − ε r Z ω Rη(x ξ ) | | 0· 0 sint = sgn(ω)sgn(x0 ξ 0) dt, − · ω Rε(x ξ ) t | | 0· 0 converges to Z ∞ sint π sgn(ω)sgn(x0 ξ 0) dt = sgn(ω)sgn(x0 ξ 0), − · 0 t − 2 · as ε 0 and η ∞. → → For its real part, since cosr is an even function, we have Z η dr ℜIε,η (ξ,x0) = [cos( ω Rr x0 ξ 0 ) cos( ω Rr)] . ε | | | · | − | | r - 86 - 4. Singular Integrals

If x ξ = 1, then ℜIε η (ξ,x ) = 0. Now we assume 0 < ε < 1 < η. For the case x ξ = 1, 0 · 0 ± , 0 0 · 0 6 ± we get the absolute value of its real part

Z 1 ω ω dr ℜIε,η (ξ,x0) 6 2sin | |Rr( x0 ξ 0 + 1)sin | |Rr( x0 ξ 0 1) ε − 2 | · | 2 | · | − r

Z η dr Z η dr + cos ω Rr x0 ξ 0 cos ω Rr 1 | | | · | r − 1 | | r 2 Z 1 ω 2 2 6| | R (1 x0 ξ 0 ) rdr 2 − | · | ε

Z ω Rη ξ 0 x0 cost Z ω Rη cost + | | | · | dt | | dt ω R ξ x t − ω R t | | | 0· 0| | | 2 ω 2 6| | R + I1. 4 If η ξ x > 1, then we have | 0 · 0| Z ω R Z ω Rη | | cost | | cost I1 = dt dt ω R ξ x t − ω Rη ξ x t | | | 0· 0| | | | 0· 0| Z ω R Z ω Rη | | dt | | dt 6 + ω R ξ x t ω Rη ξ x t | | | 0· 0| | | | 0· 0| 2ln(1/ ξ 0 x0 ). 6 | · | If 0 < η ξ x 1, then | 0 · 0| 6 Z ω R/ ξ x | | | 0· 0| dt I1 6 6 2ln(1/ ξ 0 x0 ). ω R ξ x t | · | | | | 0· 0| Thus, 2 ω 2 ℜIε,η (ξ,x0) 6 | | R + 2ln(1/ ξ 0 x0 ), 4 | · | and so the real part converges as ε 0 and η ∞. By the fundamental theorem of calculus, → → we can write Z η cos(λr) cos(µr) Z η Z λ Z λ Z η − dr = sin(tr)dtdr = sin(tr)drdt ε r − ε µ − µ ε Z λ Z η ∂ cos(tr) Z λ cos(tη) cos(tε) = r drdt = − dt µ ε t µ t Z λη cos(s) Z λ cos(tε) sins λη Z λη sins Z λ cos(tε) = ds dt = + ds dt µη s − µ t s µη µη s2 − µ t Z λ 1 0 dt = ln(λ/µ) = ln(µ/λ), as η ∞, ε 0. → − µ t − → → Take λ = ω R x ξ , and µ = ω R. So | | | 0 · 0| | | Z ∞ dr lim ℜ(Iε,η (ξ,x0)) = [cos ω Rr(x0 ξ 0) cos ω Rr] = ln(1/ x0 ξ 0 ). ε 0 | | · − | | r | · | η→∞ 0 → By the properties of Iε,η just proved, we have Z 2 π ω 2 Kdε,η (ξ) + | | R + 2ln(1/ ξ 0 x0 ) Ω(x0) dσ(x0) 6 n 1 | | S − 2 4 | · | | | 2 Z π ω 2 C( + | | R )ωn 1 + 2C ln(1/ ξ 0 x0 )dσ(x0). 6 n 1 2 4 − S − | · | 4.5. Singular integral operators commuted with dilations - 87 -

0 R For n = 1, we have S = 1,1 and then n 1 ln(1/ ξ 0 x0 )dσ(x0) = 2ln1 = 0. For n > 2, we {− } S − | · | can pick an orthogonal matrix A such that Ae1 = ξ 0, and so by changes of variables and using the notationy ¯ = (y2,y3,...,yn), Z Z ln(1/ ξ 0 x0 )dσ(x0) = ln(1/ Ae1 x0 )dσ(x0) n 1 n 1 S − | · | S − | · | Z 1 Z 1 A− x0=y = ln(1/ e1 A− x0 )dσ(x0) ==== ln(1/ e1 y )dσ(y) n 1 n 1 S − | · | S − | · | Z Z 1 Z dy1 = ln(1/ y1 )dσ(y) = ln(1/ y1 ) dσ(y¯)q Sn 1 | | | | y2Sn 2 − 1 √1 1 − 1 y2 − − − 1 z¯=y¯/√1 y2 Z 1 Z − 1 2 (n 3)/2 ======ln(1/ y1 ) (1 y1) − dσ(z¯)dy1 1 | | Sn 2 − − − Z 1 2 (n 3)/2 =ωn 2 ln(1/ y1 )(1 y1) − dy1 − 1 | | − − Z 1 2 (n 3)/2 =2ωn 2 ln(1/ y1 )(1 y1) − dy1 − 0 | | − Z π/2 y1=cosθ n 2 ====2ωn 2 ln(1/cosθ)(sinθ) − dθ = 2ωn 2I2. − 0 − For n > 3, we have, by integration by parts, Z π/2 Z π/2 I2 6 ln(1/cosθ)sinθdθ = sinθdθ = 1. 0 0 π For n = 2, we have, by the formula R /2 ln(cosθ)dθ = π ln2 (see [GR, 4.225.3, p.531]), 0 − 2 Z π/2 Z π/2 π I2 = ln(1/cosθ)dθ = ln(cosθ)dθ = ln2. 0 − 0 2 R n 1 Hence, n 1 ln(1/ ξ 0 x0 )dσ(x0) 6 C for any ξ 0 S − . S − | · | ∈ Thus, we have proved the uniform boundedness of Kdε,η (ξ), i.e., (i). In view of the limit of Iε η (ξ,x ) as ε 0, η ∞ just proved, and the dominated convergence theorem, we get , 0 → → lim Kdε,η (ξ) = m(ξ), ε 0 η→∞ → if ξ = 0, that is (ii). 6 2 n 2 By the Plancherel theorem, if f L (R ), Kε,η f converges in L norm as ε 0 and η ∞, ∈ ∗ → → and the Fourier transform of this limit is m(ξ) fˆ(ξ). R However, if we keep ε fixed and let η ∞, then clearly Kε,η (y) f (x y)dy converges R → − everywhere to K(y) f (x y)dy, which is Tε f . y >ε − Letting now ε| | 0, we obtain the conclusion (c) and our theorem is completely proved. → ut n 1 Remark 4.25. 1) In the theorem, the condition that Ω is mean zero on S − is necessary and cannot be neglected. Since in the estimate Z Ω(y) Z Z Ω(y) n f (x y)dy = + n f (x y)dy, Rn y − y 1 y >1 y − | | | |6 | | | | the main difficulty lies in the first integral. For instance, if we assume Ω(x) 1, f is a nonzero ≡ constant, then this integral is divergent. 2) From the formula of the symbol m(ξ), it is homogeneous of degree 0 in view of the mean zero property of Ω. - 88 - 4. Singular Integrals

3) The proof of part (c) holds under very general conditions on Ω. Write Ω = Ωe +Ωo where Ω is the even part of Ω, Ω (x) = Ω ( x), and Ω (x) is the odd part, Ω ( x) = Ω (x). Then, e e e − o o − − o because of the uniform boundedness of the sine integral, i.e., ℑIε,η (ξ,x0), we required only R n 1 Ωo(x0) dσ(x0) < ∞, i.e., the integrability of the odd part. For the even part, the proof S − | | requires the uniform boundedness of Z Ωe(x0) ln(1/ ξ 0 x0 )dσ(x0). n 1 S − | | | · | This observation is suggestive of certain generalizations of Theorem 4.21, see [Ste70, §6.5, p.49–50].

4.6 The maximal singular integral operator

Theorem 4.24 guaranteed the existence of the singular integral transformation Z Ω(y) lim n f (x y)dy (4.44) ε 0 y ε y − → | |> | | in the sense of convergence in the Lp norm. The natural counterpart of this result is that of convergence almost everywhere. For the questions involving almost everywhere convergence, it is best to consider also the corresponding maximal function. Theorem 4.26. Suppose that Ω satisﬁes the conditions of the previous theorem. For f Lp(Rn), ∈ 1 6 p < ∞, consider Z Ω(y) Tε f (x) = n f (x y)dy, ε > 0. y ε y − | |> | | (The integral converges absolutely for every x.) (a) limε 0 Tε f (x) exists for almost every x. → 1 n (b) Let T ∗ f (x) = sup Tε f (x) . If f L (R ), then the mapping f T ∗ f is of weak type ε>0 | | ∈ → (1,1). (c) If 1 < p < ∞, then T f A f . k ∗ kp 6 pk kp Proof. The argument for the theorem presents itself in three stages. The ﬁrst one is the proof of inequality (c) which can be obtained as a relatively easy conse- p quence of the L norm existence of limε 0 Tε , already proved, and certain general properties of → “approximations to the identity”. p Let T f (x) = limε 0 Tε f (x), where the limit is taken in the L norm. Its existence is guar- → anteed by Theorem 4.24. We shall prove this part by showing the following Cotlar inequality

T ∗ f (x) 6 M(T f )(x) +CM f (x). Let ϕ be a smooth non-negative function on Rn, which is supported in the unit ball, has integral equal to one, and which is also radial and decreasing in x . Consider | | ( Ω(x) x n , x > ε, Kε (x) = | | | | 0, x < ε. | | This leads us to another function Φ deﬁned by

Φ = ϕ K K1, (4.45) R ∗ − where ϕ K = limε 0 ϕ Kε = limε 0 x y ε K(x y)ϕ(y)dy. ∗ → ∗ → | − |> − 4.6. The maximal singular integral operator - 89 -

We shall need to prove that the smallest decreasing radial majorant of Φ is integrable (so as to apply Theorem 4.10). In fact, if x < 1, then | | Z Z Φ = ϕ K = K(y)ϕ(x y)dy = K(y)(ϕ(x y) ϕ(x))dy | | | ∗ | Rn − Rn − − Z Z ϕ(x y) ϕ(x) 6 K(y) ϕ(x y) ϕ(x) dy 6 C | − n− |dy 6 C, Rn | || − − | Rn y R | | since (4.41) implies Rn K(y)dy = 0 and by the smoothness of ϕ. If 1 x 2, then Φ = ϕ K K is again bounded by the same reason and K is bounded in 6 | | 6 ∗ − this case. Finally if x > 2, | | Z Z Φ(x) = K(x y)ϕ(y)dy K(x) = [K(x y) K(x)]ϕ(y)dy. Rn − − y 1 − − | |6 Similar to (4.43), we can get the bound for y 6 1 Z | |Z K(x y) K(x) dx 6 K(x y) K(x) dx 6 C. x 2 | − − | x 2 y | − − | | |> | |> | | Thus we obtain Z Z Φ(x) dx 6C ϕ(y)dy 6 C. x 2 | | y 1 | |> | |6 Therefore, we have proved that Φ L1(Rn) from three cases discussed above. ∈ From (4.45), it follows, because the singular integral operator ϕ ϕ K commutes with → ∗ dilations, that n ϕε K Kε = Φε , with Φε (x) = ε− Φ(x/ε). (4.46) ∗ − Now, we claim that for any f Lp(Rn), 1 < p < ∞, ∈ (ϕε K) f (x) = T f ϕε (x), (4.47) ∗ ∗ ∗ where the identity holds for every x. In fact, we notice ﬁrst that

(ϕε K ) f (x) = T f ϕε (x), for every δ > 0 (4.48) ∗ δ ∗ δ ∗ because both sides of (4.48) are equal for each x to the absolutely convergent double integral R R q n n ( ) ( )ϕ ( ) ϕ ( ) < < ∞ / + z R y >δ K y f z y ε x z dydz. Moreover, ε L R , with 1 q and 1 p ∈ | | − − q ∈ p 1/q = 1, so ϕε K ϕε K in L norm, and T f T f in L norm, as δ 0, by Theorem ∗ δ → ∗ δ → → 4.24. This proves (4.47), and so by (4.46)

Tε f = Kε f = ϕε K f Φε f = T f ϕε f Φε . ∗ ∗ ∗ − ∗ ∗ − ∗ Passing to the supremum over ε and applying Theorem 4.10, part (a), Theorem 3.9 for maximal funtions and Theorem 4.24, we get T ∗ f p 6 sup T f ϕε p + sup f Φε p k k k ε>0 | ∗ |k k ε>0 | ∗ |k C M(T f ) +C M f C T f +C f C f . 6 k kp k kp 6 k kp k kp 6 k kp Thus, we have proved (c). The second and most difﬁcult stage of the proof is the conclusion (b). Here the argument proceeds in the main as in the proof of the weak type (1,1) result for singular integrals in Theorem 4.18. We review it with deliberate brevity so as to avoid a repetition of details already examined. 6.4 The Maximal Singular Integral Operator 99

(i) If x < Q∗, then x c > 2 y c for all y Q , as an obvious geometric consideration shows. j | − j| | − j| ∈ j (ii) Suppose x Rn Q and assume that for some y Q , x y = ε. Then the closed ball centered ∈ \ ∗j ∈ j | − | at x, of radius γ ε, contains Q , i.e. B(x, r) Q , if r = γ ε. n j ⊃ j n (iii) Under the same hypotheses as (ii), we have that x y > γ ε, for every y Q . | − | n0 ∈ j Here γn and γn0 depend only on the dimension n, and not the particular cube Q j. - 90 - 4. Singular Integrals

For a given α > 0, we split f = g+b as in the proof of Theo- rem 4.18. We also consider for each cube Q j its mate Q∗j, which B(x, r) γ has the same center c j but whose side length is expanded 2√n n ε times. The following geometric remarks concerning these cubes γ ε are nearly obvious (The first one has given in the proof of The- n0 x Q j orem 4.18). y ε Q∗j Rn (i) If x / Q∗, then x c j > 2 y c j for all y Q j, as an jQ∗j ∈ j | − | | − | ∈ \ ∪ obvious geometric consideration shows. n (ii) Suppose x R Q∗ and assume that for some y Q j, ∈ \ j ∈ x y = ε. Then the closed ball centered at x, of radius γ ε, Observation for (ii) and (iii) | − | n Fig. 4.1 Observation for (ii) and (iii) contains Q , i.e., B(x,r) Q , if r = γ ε. j ⊃ j n (iii) Under the same hypotheses as (ii), we have thatWithx thesey observations, and following the development in the proof of Theorem 6.1, we shall > n prove| that− if|x R jQ∗, γ ε, for every y Q . ∈ \ ∪ j n0 ∈ j Here γ and γ depend only on the dimension n, and not the particular cube Q . n n0 sup Tεb(x) 6 j K(x y) K(x c j) b(y) dy ε>0 | | Q j | − − − || | With these observations, and following the development in the proof of TheoremXj Z 4.18, we (6.23) n 1 shall prove that if x R jQ∗, + C sup b(y) dy, j m(B(x, r)) | | ∈ \ ∪ Z r>0 ZB(x,r) sup Tε b(x) K(x y) K(x Ωc(x)j) b(y) dy 6∑ with K(x) = x n . ε>0 | | j Q j | − − −| | || | The addition of the maximal function to the r.h.s(4.49) of (6.23) is the main new element of the proof. ToZ prove (6.23), fix x Rn Q , and ε > 0. Now the cubes Q fall into three classes: 1 ∈ \ ∪ j ∗j j +C sup 1) for all y bQ(jy, )x dyy, < ε; r>0 m(B(x,r)) B(x,r)∈| | |− | 2) for all y Q j, x y > ε; Ω ∈ | − | (x) 3) there is a y Q j, such that x y = ε. with K(x) = n . ∈ | − | x We now examine The addition| | of the maximal function to the r.h.s of (4.49) is the main new element of the T b(x) = K (x y)b(y)dy. (6.24) proof. ε ε − n j ZQ j To prove (4.49), fix x R jQ∗, and ε > 0. Now the cubes Q j fall into three classes:X ∈ \ ∪ j Case 1). Kε(x y) = 0 if x y < ε, and so the integral over the cube Q j in (6.24) is zero. 1) for all y Q j, x y < ε; − | − | ∈ | − | Case 2). Kε(x y) = K(x y), if x y > ε, and therefore this integral over Q j equals 2) for all y Q , x y > ε; − − | − | ∈ j | − | 3) there is a y Q j, such that x y = ε. K(x y)b(y)dy = [K(x y) K(x c j)]b(y)dy. ∈ | − | Q − Q − − − We now examine Z j Z j Z This term is majorized in absolute value by Tε b(x) = ∑ Kε (x y)b(y)dy. (4.50) j Q j − Case 1). Kε (x y) = 0 if x y < ε, and so the integral over the cube Q in (4.50) is zero. − | − | j Case 2). Kε (x y) = K(x y), if x y > ε, and therefore this integral over Q j equals − Z − | − |Z K(x y)b(y)dy = [K(x y) K(x c j)]b(y)dy. Q j − Q j − − − This term is majorized in absolute value by Z K(x y) K(x c j) b(y) dy, Q j | − − − || | which expression appears in the r.h.s. of (4.49). Case 3). We write simply

Z Z Kε (x y)b(y)dy 6 Kε (x y) b(y) dy Q j − Q j | − || | Z = Kε (x y) b(y) dy, Q B(x,r) | − || | j∩ by (ii), with r = γnε. However, by (iii) and the fact that Ω is bounded, we have 4.6. The maximal singular integral operator - 91 -

Ω(x y) C Kε (x y) = − . | − | x y n 6 (γ ε)n | − | n0 Thus, in this case,

Z C Z Kε (x y)b(y)dy 6 b(y) dy. Q − m(B(x,r)) Q B(x,r) | | j j∩ If we add over all cubes Q j, we ﬁnally obtain, for r = γnε, Z Tε b(x) 6∑ K(x y) K(x c j) b(y) dy | | j Q j | − − − || | C Z + b(y) dy. m(B(x,r)) B(x,r) | | Taking the supremum over ε gives (4.49). This inequality can be written in the form

T ∗b(x) Σ +CMb(x), x F∗, | | 6 ∈ and so m n ( x R jQ∗ : T ∗b(x) > α/2 ) { ∈ \ ∪ j | | } m n m n 6 ( x R jQ∗j : Σ > α/4 ) + ( x R jQ∗j : CMb(x) > α/4 ). { ∈ \ ∪ } { ∈ \ ∪ R } The ﬁrst term in the r.h.s. is similar to (4.33), and we can get n Σ(x)dx C b which R jQ∗ 6 1 \∪ j k k m n 4C implies ( x R jQ∗ : Σ > α/4 ) b 1. { ∈ \ ∪ j } 6 α k k For the second one, by Theorem 3.9, i.e., the weak type estimate for the maximal function n C M, we get m( x R jQ∗ : CMb(x) > α/4 ) b 1. { ∈ \ ∪ j } 6 α k k The weak type (1,1) property of T ∗ then follows as in the proof of the same property for T, in Theorem 4.18 for more details. The ﬁnal stage of the proof, the passage from the inequalities of T ∗ to the existence of the limits almost everywhere, follows the familiar pattern described in the proof of the Lebesgue differential theorem (i.e., Theorem3.13). More precisely, for any f Lp(Rn), 1 6 p < ∞, let ∈

Λ f (x) = limsupTε f (x) liminfTε f (x) . ε 0 − ε 0 → → Clearly, Λ f (x) 2T f (x). Now write f = f + f where f C1, and f δ. 6 ∗ 1 2 1 ∈ c k 2kp 6 We have already proved in the proof of Theorem 4.21 that Tε f converges uniformly as ε 0, 1 → so Λ f (x) 0. By (4.37), we have Λ f 2A f 2A δ if 1 < p < ∞. This shows Λ f = 1 ≡ k 2kp 6 pk 2kp 6 p 2 0, almost everywhere, thus by Λ f (x) 6 Λ f1(x) +Λ f2(x), we have Λ f = 0 almost everywhere. So limε 0 Tε f exists almost everywhere if 1 α ) f , { } 6 α k 2k1 6 α and so again Λ f (x) = 0 almost everywhere, which implies that limε 0 Tε f (x) exists almost → everywhere. ut - 92 - 4. Singular Integrals

4.7 *Vector-valued analogues

It is interesting to point out that the results of this chapter, where our functions were assumes to take real or complex values, can be extended to the case of functions taking their values in a Hilbert space. We present this generalization because it can be put to good use in several problems. An indication of this usefulness is given in the Littlewood-Paley theory. We begin by reviewing quickly certain aspects of integration theory in this context. Let H be a separable Hilbert space. Then a function f (x), from Rn to H is measurable if the scalar valued functions ( f (x),ϕ) are measurable, where ( , ) denotes the inner product of · · H , and ϕ denotes an arbitrary vector of H . If f (x) is such a measurable function, then f (x) is also measurable (as a function with | | non-negative values), where denotes the norm of H . | · | Thus, Lp(Rn,H ) is defined as the equivalent classes of measurable functions f (x) from Rn R p 1/p to H , with the property that the norm f = ( n f (x) dx) is finite, when p < ∞; when k kp R | | p = ∞ there is a similar definition, except f ∞ = esssup f (x) . k k | | Next, let H1 and H2 be two separable Hilbert spaces, and let L(H1,H2) denote the Banach space of bounded linear operators from H1 to H2, with the usual operator norm. n We say that a function f (x), from R to L(H1,H2) is measurable if f (x)ϕ is an H2-valued measurable function for every ϕ H1. In this case f (x) is also measurable and we can de- p n ∈ | | fine the space L (R ,L(H1,H2)), as before; here again denotes the norm, this time in | · | L(H1,H2). The usual facts about convolution hold in this setting. For example, suppose K(x) q n p n R ∈ L (R ,L(H1,H2)) and f (x) L (R ,H1), then g(x) = n K(x y) f (y)dy converges in the ∈ R − norm of H2 for almost every x, and Z Z g(x) 6 K(x y) f (y) dy 6 K(x y) f (y) dy. | | Rn | − | Rn | − || | Also g K f , if 1/r = 1/p + 1/q 1, with 1 r ∞. k kr 6 k kqk kp − 6 6 Suppose that f (x) L1(Rn,H ). Then we can define its Fourier transform fˆ(ξ) = R ωix ξ ∈ ∞ n 1 n 2 n Rn e− · f (x)dx which is an element of L (R ,H ). If f L (R ,H ) L (R ,H ), then n/2 ∈ ∩ 2 n ω − fˆ(ξ) L (R ,H ) with fˆ 2 = | | f 2. The Fourier transform can then be extended ∈ k k 2π k k by continuity to a unitary mapping of the Hilbert space L2(Rn,H ) to itself, up to a constant multiplication. These facts can be obtained easily from the scalar-valued case by introducing an arbitrary orthonormal basis in H . Now suppose that H1 and H2 are two given Hilbert spaces. Assume that f (x) takes values in H1, and K(x) takes values in L(H1,H2). Then Z T f (x) = K(y) f (x y)dy, Rn − whenever defined, takes values in H2. Theorem 4.27. The results in this chapter, in particular Theorem 4.18, Proposition 4.19, Theo- rems 4.21, 4.24 and 4.26 are valid in the more general context where f takes its value in H1,K takes its values in L(H1,H2) and T f and Tε f take their value in H2, and where throughout the absolute value is replaced by the appropriate norm in H ,L(H ,H ) or H respectively. | · | 1 1 2 2 This theorem is not a corollary of the scalar-valued case treated in any obvious way. However, its proof consists of nothing but a identical repetition of the arguments given for the scalar- 4.7. *Vector-valued analogues - 93 - valued case, if we take into account the remarks made in the above paragraphs. So, we leave the proof to the interested reader.

Remark 4.28. 1) The final bounds obtained do not depend on the Hilbert spaces H1 or H2, but only on B, p, and n, as in the scalar-valued case. 2) Most of the argument goes through in the even greater generality of Banach space-valued functions, appropriately defined. The Hilbert space structure is used only in the L2 theory when applying the variant of Plancherel’s formula. The Hilbert space structure also enters in the following corollary. Corollary 4.29. With the same assumptions as in Theorem 4.27, if in addition 2 n T f 2 = c f 2, c > 0, f L (R ,H1), k k k k ∈ p n then f p A0 T f p, if f L (R ,H1), if 1 < p < ∞. k k 6 pk k ∈ 2 n Proof. We remark that the L (R ,H j) are Hilbert spaces. In fact, let ( , ) j denote the inner · · 2 n product of H j, j = 1,2, and let , j denote the corresponding inner product in L (R ,H j); h· ·i that is Z f ,g j = ( f (x),g(x)) jdx. h i Rn 2 n Now T is a bounded linear transformation from the Hilbert space L (R ,H1) to the Hilbert 2 n space L (R ,H2), and so by the general theory of inner products there exists a unique adjoint 2 n 2 n transformation T˜, from L (R ,H2) to L (R ,H1), which satisfies the characterizing property 2 n T f1, f2 2 = f1,T˜ f2 1, with f j L (R ,H j). h i h i ∈ But our assumption is equivalent with the identity (see the theory of Hilbert spaces, e.g. [Din07, Chapter 6]) 2 2 n T f ,Tg 2 = c f ,g 1, for all f ,g L (R ,H1). h i h i ∈ Thus using the definition of the adjoint, TT˜ f ,g = c2 f ,g , and so the assumption can be h i1 h i1 restated as 2 2 n TT˜ f = c f , f L (R ,H1). (4.51) ∈ T˜ is again an operator of the same kind as T but it takes function with values in H2 to functions with values in H , and its kernel K˜(x) = K ( x), where denotes the adjoint of an element in 1 ∗ − ∗ L(H1,H2). This is obvious on the formal level since Z Z T f1, f2 2 = (K(x y) f1(y), f2(x))2dydx h i Rn Rn − Z Z = ( f1(y),K∗( (y x)) f2(x))1dxdy = f1,T˜ f2 1. Rn Rn − − h i The rigorous justification of this identity is achieved by a simple limiting argument. We will not tire the reader with the routine details. This being said we have only to add the remark that K ( x) satisfies the same conditions ∗ − as K(x), and so we have, for it, similar conclusions as for K (with the same bounds). Thus by (4.51), c2 f = TT˜ f A T f . k kp k kp 6 pk kp This proves the corollary with A = A /c2. 0p p ut - 94 - 4. Singular Integrals

Remark 4.30. This corollary applies in particular to the singular integrals commuted with dilations, then the condition required is that the multiplier m(ξ) have constant absolute value. 1 This is the case, for example, when T is the Hilbert transform, K(x) = πx , and m(ξ) = isgn(ω)sgn(ξ). − Chapter 5 Riesz Transforms and Spherical Harmonics

5.1 The Riesz transforms

We look for the operators in Rn which have the analogous structural characterization as the Hilbert transform. We begin by making a few remarks about the interaction of rotations with the n-dimensional Fourier transform. We shall need the following elementary observation. Let ρ denote any rotation about the origin in Rn. Denote also by ρ its induced action on functions, ρ( f )(x) = f (ρx). Then Z Z ωix ξ ωiρ 1y ξ (F ρ) f (ξ) = e− · f (ρx)dx = e− − · f (y)dy Rn Rn Z ωiy ρξ = e− · f (y)dy = F f (ρξ) = ρF f (ξ), Rn that is, F ρ = ρF . n Let `(x) = (`1(x),`2(x),...,`n(x)) be an n-tuple of functions defined on R . For any rotation ρ about the origin, write ρ = (ρ jk) for its matrix realization. Suppose that ` transforms like a vector. Symbolically this can be written as `(ρx) = ρ(`(x)), or more explicitly ` j(ρx) = ∑ρ jk`k(x), for every rotation ρ. (5.1) k Lemma 5.1. Suppose ` is homogeneous of degree 0, i.e., `(εx) = `(x), for ε > 0. If ` transforms x according to (5.1) then `(x) = c x for some constant c; that is | | x j ` (x) = c . (5.2) j x | | Proof. It suffices to consider x Sn 1 due to the homogeneousness of degree 0 for `. Now, ∈ − let e1, e2, ..., en denote the usual unit vectors along the axes. Set c = `1(e1). We can see that ` (e ) = 0, if j = 1. j 1 6 In fact, we take a rotation arbitrarily such that e1 fixed under the acting of ρ, i.e., ρe1 = e1. 1 1 Thus, we also have e1 = ρ− ρe1 = ρ− e1 = ρ>e1. From ρe1 = ρ>e1 = e1, we get ρ11 = 1 1 1 0 1 0 − 1 0 and ρ1k = ρ j1 = 0 for k = 1 and j = 1. So ρ = . Because = 1 and 6 6 0 A 0 A 0 A− 1 1 n 1 ρ− = ρ>, we obtain A− = A> and detA = 1, i.e., A is a rotation in R − . On the other hand, by (5.1), we get ` (e ) = ∑n ρ ` (e ) for j = 2,...,n. That is, the n 1 dimensional vector j 1 k=2 jk k 1 −

95 - 96 - 5. Riesz Transforms and Spherical Harmonics

(` (e ),` (e ), ,` (e )) is left fixed by all the rotations on this n 1 dimensional vector 2 1 3 1 ··· n 1 − space. Thus, we have to take ` (e ) = ` (e ) = = ` (e ) = 0. 2 1 3 1 ··· n 1 Inserting again in (5.1) gives ` j(ρe1) = ρ j1`1(e1) = cρ j1. If we take a rotation such that ρe = x, then we have ρ = x , so ` (x) = cx ,( x = 1), which proves the lemma. 1 j1 j j j | | ut We now define the n Riesz transforms. For f Lp(Rn), 1 p < ∞, we set ∈ 6 Z y j R j f (x) = lim cn n+1 f (x y)dy, j = 1,...,n, (5.3) ε 0 y ε y − → | |> (n+1)/|2 | c = Γ ((n+1)/2) /c = π Sn with n π(n+1)/2 where 1 n Γ ((n+1)/2) is half the surface area of the unit sphere of n+1 Ω j(x) x j R . Thus, R j is defined by the kernel Kj(x) = x n , and Ω j(x) = cn x . Next, we derive the multipliers which correspond| | to the Riesz transforms,| | and which in fact justify their definition. Denote

Ω(x) = (Ω1(x),Ω2(x),...,Ωn(x)), and m(ξ) = (m1(ξ),m2(ξ),...,mn(ξ)). Let us recall the formula (4.42), i.e., Z m(ξ) = Φ(ξ x)Ω(x)dσ(x), ξ = 1, (5.4) n 1 S − · | | with Φ(t) = πi sgn(ω)sgn(t) + ln 1/t . For any rotation ρ, since Ω commutes with any ro- − 2 | | tations, i.e., Ω(ρx) = ρ(Ω(x)), we have, by changes of variables, Z Z ρ(m(ξ)) = Φ(ξ x)ρ(Ω(x))dσ(x) = Φ(ξ x)Ω(ρx)dσ(x) Sn 1 · Sn 1 · Z − Z − 1 = Φ(ξ ρ− y)Ω(y)dσ(y) = Φ(ρξ y)Ω(y)dσ(y) n 1 n 1 S − · S − · =m(ρξ).

Thus, m commutes with rotations and so m satisﬁes (5.1). However, the m j are each homoge- ξ j neous of degree 0, so Lemma 5.1 shows that m j(ξ) = c ξ , with Z | | c =m1(e1) = Φ(e1 x)Ω1(x)dσ(x) n 1 S − · Z πi = [ sgn(ω)sgn(x1) + ln 1/x1 ]cnx1dσ(x) n 1 S − − 2 | | πi Z = sgn(ω) cn x1 dσ(x) (the 2nd is 0 since it is odd w.r.t. x1) n 1 − 2 S − | | πi Γ ((n + 1)/2) 2π(n 1)/2 = sgn(ω) − = sgn(ω)i. − 2 π(n+1)/2 Γ ((n + 1)/2) − R (n 1)/2 Here we have used the fact n 1 x1 dσ(x) = 2π − /Γ ((n + 1)/2). Therefore, we obtain S − | | ξ j Rdj f (ξ) = sgn(ω)i fˆ(ξ), j = 1,...,n. (5.5) − ξ | | This identity and Plancherel’s theorem also imply the following “unitary” character of the Riesz transforms n 2 2 ∑ R j f 2 = f 2. j=1 k k k k

By m(ρξ) = ρ(m(ξ)) proved above, we have m j(ρξ) = ∑k ρ jkmk(ξ) for any rotation ρ and then m j(ρξ) fˆ(ξ) = ∑k ρ jkmk(ξ) fˆ(ξ). Taking the inverse Fourier transform, it follows 1 ˆ 1 ˆ F − m j(ρξ) f (ξ) =F − ∑ρ jkmk(ξ) f (ξ) k 5.1. The Riesz transforms - 97 - 1 ˆ =∑ρ jkF − mk(ξ) f (ξ) = ∑ρ jkRk f . k k But by changes of variables, we have 1 F − m j(ρξ) fˆ(ξ) n Z ω ωix ξ = | | e · m j(ρξ) fˆ(ξ)dξ 2π Rn n Z ω ωiρx η 1 = | | e · m j(η) fˆ(ρ− η)dη 2π Rn 1 1 1 1 =(F − (m j(ξ) fˆ(ρ− ξ)))(ρx) = ρF − (m j(ξ) fˆ(ρ− ξ))(x) 1 =ρR jρ− f , since the Fourier transform commutes with rotations. Therefore, it reaches 1 ρR jρ− f = ∑ρ jkRk f , (5.6) k which is the statement that under rotations in Rn, the Riesz operators transform in the same manner as the components of a vector. We have the following characterization of Riesz transforms. 2 n Proposition 5.2. Let T = (T1,T2,...,Tn) be an n-tuple of bounded linear transforms on L (R ). Suppose n (a) Each Tj commutes with translations of R ; n (b) Each Tj commutes with dilations of R ; n 1 (c) For every rotation ρ = (ρ jk) of R , ρTjρ− f = ∑k ρ jkTk f . Then the Tj is a constant multiple of the Riesz transforms, i.e., there exists a constant c such that Tj = cR j, j = 1,...,n. Proof. All the elements of the proof have already been discussed. We bring them together. 2 n (i) Since the Tj is bounded linear on L (R ) and commutes with translations, by Theorem 1.62 they can be each realized by bounded multipliers m j, i.e., F (Tj f ) = m j fˆ. (ii) Since the Tj commutes with dilations, i.e., Tjδε f = δε Tj f , in view of Proposition 1.3, n n we see that F T δ f = m (ξ)F δ f = m (ξ)ε δ 1 fˆ(ξ) = m (ξ)ε fˆ(ξ/ε) and F δ T f = j ε j ε j − ε− j − ε j n n n ε δ 1 F T f = ε δ 1 (m fˆ) = ε m (ξ/ε) fˆ(ξ/ε), which imply m (ξ) = m (ξ/ε) or − ε− j − ε− j − j j j equivalently m j(εξ) = m j(ξ), ε > 0; that is, each m j is homogeneous of degree 0. (iii) Finally, assumption (c) has a consequence by taking the Fourier transform, i.e., the relation (5.1), and so by Lemma 5.1, we can obtain the desired conclusion. ut One of the important applications of the Riesz transforms is that they can be used to mediate between various combinations of partial derivatives of a function. 2 n n ∂ 2 f Proposition 5.3. Suppose f Cc (R ). Let ∆ f = ∑ j=1 2 . Then we have the a priori bound ∈ ∂x j

∂ 2 f 6 Ap ∆ f p, 1 < p < ∞. (5.7) ∂x j∂xk p k k

Proof. Since F (∂x j f )(ξ) = ωiξ jF f (ξ), we have 2 ∂ f 2 F (ξ) = ω ξ jξkF f (ξ) ∂x j∂xk − iξ j iξ = sgn(ω) sgn(ω) k ( ω2 ξ 2)F f (ξ) − − ξ − ξ − | | | | | | - 98 - 5. Riesz Transforms and Spherical Harmonics

= F R R ∆ f . − j k ∂ 2 f p Thus, = R jRk∆ f . By the L boundedness of the Riesz transforms, we have the desired ∂x j∂xk − result. ut 1 2 Proposition 5.4. Suppose f Cc (R ). Then we have the a priori bound ∈ ∂ f ∂ f ∂ f ∂ f + 6 Ap + i , 1 < p < ∞. ∂x1 p ∂x2 p ∂x1 ∂x2 p Proof. The proof is similar to the previous one. Indeed, we have 2 2 iξ j iξ j ξ + ξ F ∂ f =ωiξ F f (ξ) = ω ξ F f (ξ) = ω 1 2 F f (ξ) x j j ξ | | ξ ξ | | | | | | iξ j (ξ iξ )(ξ + iξ ) =ω 1 − 2 1 2 F f (ξ) ξ ξ | | | | sgn(ω)iξ j sgn(ω)i(ξ iξ ) = − − 1 − 2 F (∂ f + i∂ f ) − ξ ξ x1 x2 | | | | = F R (R iR )(∂ f + i∂ f ). − j 1 − 2 x1 x2 That is, ∂ f = R (R iR )(∂ f + i∂ f ). Also by the Lp boundedness of the Riesz trans- x j − j 1 − 2 x1 x2 forms, we can obtain the result. ut We shall now tie together the Riesz transforms and the theory of harmonic functions, more particularly Poisson integrals. Since we are interested here mainly in the formal aspects we shall restrict ourselves to the L2 case. For Lp case, one can see the further results in [Ste70, §4.3 and §4.4, p.78]. 2 n Theorem 5.5. Let f and f1, ..., fn all belong to L (R ), and let their respective Poisson integrals be u (x,y) = P f , u (x,y) = P f , ..., u (x,y) = P f . Then a necessary and sufﬁcient 0 y ∗ 1 y ∗ 1 n y ∗ n condition of

f j = R j( f ), j = 1,...,n, (5.8) is that the following generalized Cauchy-Riemann equations hold:  n ∂u j  = 0,  ∑ ∂x j=0 j (5.9)  ∂u j ∂uk  = , j = k, with x0 = y.  ∂xk ∂x j 6 Remark 5.6. At least locally, the system (5.9) is equivalent with the existence of a harmonic ∂g function g of the n + 1 variables, such that u j = , j = 0,1,2,...,n. ∂x j iξ j ˆ Proof. Suppose f j = R j f , then bf j(ξ) = sgn(ω) ξ f (ξ), and so by (4.15) − | | n Z ω iξ j ωiξ x ωξ y u j(x,y) = sgn(ω) | | fˆ(ξ) e · e−| | dξ, j = 1,...,n, − 2π n ξ R | | and n Z ω ωiξ x ωξ y u0(x,y) = | | fˆ(ξ)e · e−| | dξ. 2π Rn The equation (5.9) can then be immediately veriﬁed by differentiation under the integral sign, which is justiﬁed by the rapid convergence of the integrals in question. 5.2. Spherical harmonics and higher Riesz transforms - 99 -

n ω R ωiξ x ωξ y | | Conversely, let u j(x,y) = 2π Rn bf j(ξ)e · e−| | dξ, j = 0,1,...,n with f0 = f . Then ∂u ∂u the fact that ∂u0 = j = j , j = 1,...,n, and Fourier inversion theorem, show that ∂x j ∂x0 ∂y ωξ y ωξ y ωiξ j fb (ξ)e−| | = ωξ bf j(ξ)e−| | , 0 −| | iξ j therefore bf j(ξ) = sgn(ω) ξ fb0(ξ), and so − | | f j = R j f0 = R j f , j = 1,...,n.

5.2 Spherical harmonics and higher Riesz transforms

We return to the consideration of special transforms of the form Z Ω(y) T f (x) = lim n f (x y)dy, (5.10) ε 0 y ε y − → | |> | | n 1 where Ω is homogeneous of degree 0 and its integral over S − vanishes. y j We have already considered the example, i.e., the case of Riesz transforms, Ω j(y) = c y , j = 1,...,n. For n = 1, Ω(y) = csgny, and this is the only possible case, i.e., the Hilbert transform.| | To study the matter further for n > 1, we recall the expression Z m(ξ) = Λ(y ξ)Ω(y)dσ(y), ξ = 1 n 1 S − · | | where m is the multiplier arising from the transform (5.10). We have already remarked that the mapping Ω m commutes with rotations. We shall n 1 → 2 n 1 therefore consider the functions on the sphere S − (more particularly the space L (S − )) from the point of view of its decomposition under the action of rotations. As is well known, this decomposition is in terms of the spherical harmonics, and it is with a brief review of their properties that we begin. We fix our attention, as always, on Rn, and we shall consider polynomials in Rn which are also harmonic. α Definition 5.7. Denote α = (α ,...,α ), α = ∑n α and xα = x 1 xαn . Let P denote the 1 n | | j=1 j 1 ··· n k linear space of all homogeneous polynomials of degree k, i.e., α P := P(x) = aα x : α = k . k ∑ | | α Each such polynomial corresponds its dual object, the differential operator P(∂x) = ∑aα ∂x , α α1 αn ¯ where ∂x = ∂x1 ∂xn . On Pk, we define a positive inner product P,Q = P(∂x)Q. Note that ··· α α h i two distinct monomials x and x 0 in Pk are orthogonal w.r.t. it, since there exists at least one αi αi0 2 i such that α α , then ∂ x = 0. P,P = ∑ aα α! where α! = (α !) (α !). i > i0 xi i h i | | 1 ··· n Definition 5.8. We define Hk to be the linear space of homogeneous polynomials of degree k which are harmonic: the solid spherical harmonics of degree k. That is, H := P(x) P : ∆P(x) = 0 . k { ∈ k } n 1 It will be convenient to restrict these polynomials to S − , and there to define the standard inner product, Z (P,Q) = P(x)Q(x)dσ(x). n 1 S − - 100 - 5. Riesz Transforms and Spherical Harmonics

n 1 For a function f on S − , we define the spherical Laplacean∆S by ∆ f (x) = ∆ f (x/ x ), S | | where f (x/ x ) is the degree zero homogeneous extension of the function f to Rn 0 , and ∆ | | \{ } is the Laplacian of the Euclidean space.1 Proposition 5.9. We have the following properties. (1) The finite dimensional spaces H ∞ are mutually orthogonal. { k}k=0 (2) Every homogeneous polynomial P P can be written in the form P = P + x 2P , where ∈ k 1 | | 2 P1 Hk and P2 Pk 2. ∈ ∈ − 2 (3) Let Hk denote the linear space of restrictions of Hk to the unit sphere. The elements of Hk are the surface spherical harmonics of degree k, i.e., H = P(x) H : x = 1 . k { ∈ k | | } 2 n 1 ∞ 2 Then L (S − ) = ∑k=0 Hk. Here the L space is taken w.r.t. usual measure, and the infinite direct sum is taken in the sense of Hilbert space theory. That is, if f L2(Sn 1), then f has the ∈ − development ∞ f (x) = ∑ Yk(x), Yk Hk, (5.11) k=0 ∈ 2 n 1 where the convergence is in the L (S − ) norm, and Z Z 2 2 f (x) dσ(x) = Yk(x) dσ(x). n 1 ∑ n 1 S − | | k S − | | (4) If Y (x) H , then ∆ Y (x) = k(k + n 2)Y (x). k ∈ k S k − − k (5) Suppose f has the development (5.11). Then f (after correction on a set of measure zero, n 1 ∞ n 1 if necessary) is indefinitely differentiable on S − (i.e., f C (S − )) if and only if Z ∈ 2 N Yk(x) dσ(x) = O(k− ), as k ∞, for each fixed N. (5.12) n 1 S − | | → α Proof. (1) If P Pk, i.e., P(x) = ∑aα x with α = k, then n∈ n | | n α α j 1 1 − αn α ∑ x j∂x j P = ∑ x j ∑aα α jx1 x j xn = ∑ α j ∑aα x = kP. j=1 j=1 ··· ··· j=1 n 1 ∂P ∂ On S − , it follows kP = ∂ν where ∂ν denotes differentiation w.r.t. the outward normal vector. Thus, for P H , and Q H , then by Green’s theorem ∈ k ∈ j Z Z ∂P ∂Q¯ (k j) PQd¯ σ(x) = Q¯ P dσ(x) n 1 n 1 − S − S − ∂ν − ∂ν Z = [Q¯∆P P∆Q¯]dx = 0, x 1 − | |6 where ∆ is the Laplacean on Rn. 2 2 (2) Indeed, let x Pk 2 be the subspace of Pk of all polynomials of the form x P2 where | | − | | P2 Pk 2. Then its orthogonal complement w.r.t. , is exactly Hk. In fact, P1 is in this or- ∈ − h· ·i thogonal complement if and only if x 2P ,P = 0 for all P . But x 2P ,P = (P (∂ )∆)P = h| | 2 1i 2 h| | 2 1i 2 x 1

1 This is implied by the well-known formula for the Euclidean Laplacian in spherical polar coordinates: 1 n ∂ n 1 ∂ f 2 ∆ f = r − r − + r− ∆ f . ∂r ∂r S

2 Sometimes, in order to emphasize the distribution between Hk and Hk, the members of Hk are referred to as the surface spherical harmonics. 5.2. Spherical harmonics and higher Riesz transforms - 101 -

2 P2,∆P1 , so ∆P1 = 0 and thus Pk = Hk x Pk 2, which proves the conclusion. In addition, h i ⊕| | − we have for P Pk ∈ k 2 x P0(x), k even, P(x) = Pk(x) + x Pk 2(x) + + | |k 1 | | − ··· x P (x), k odd, | | − 1 where P H by noticing that P = H for j = 0,1. j ∈ j j j (3) In fact, by the further result in (2), if x = 1, then we have | | P0(x), k even, P(x) = Pk(x) + Pk 2(x) + . + − ··· P1(x), k odd, with Pj H j. That is, the restriction of any polynomial on the unit sphere is a ﬁnite linear ∈ 2 n 1 combination of spherical harmonics. Since the restriction of polynomials is dense in L (S − ) in the norm (see [SW71, Corollary 2.3, p.141]) by the Weierstrass approximation theorem,3 the conclusion is then established. (4) In fact, for x = 1, we have | | k k k k ∆ Y (x) =∆( x − Y (x)) = x − ∆Y + ∆( x − )Y + 2∇( x − ) ∇Y S k | | k | | k | | k | | · k 2 k 2 2 k 2 =(k + (2 n)k) x − − Y 2k x − − Y − | | k − | | k k 2 = k(k + n 2) x − Yk = k(k + n 2)Yk, n − − | | − − since ∑ j=1 x j∂x jYk = kYk for Yk Pk. ∈ ∞ 0 0 (5) To prove this, we write (5.11) as f (x) = ∑k=0 akYk (x), where the Yk are normalized such R 0 2 N/2 that n 1 Y (x) dσ(x) = 1. Our assertion is then equivalent with ak = O(k− ), as k ∞. If S − | k | → f is of class C2, then an application of Green’s theorem shows that Z Z 0 0 ∆S fY dσ = f ∆SY dσ. n 1 k n 1 k S − S − Thus, if f C∞, then by (4) ∈Z Z Z ∞ r 0 r 0 r 0 0 ∆ fY dσ = f ∆ Y dσ = [ k(k + n 2)] a jY Y dσ n 1 S k n 1 S k n 1 ∑ j k S − S − − − S − j=0 Z r 0 2 r =[ k(k + n 2)] ak Y dσ = ak[ k(k + n 2)] . n 1 k − − S − | | − − 2r So ak = O(k− ) for every r and therefore (5.12) holds. To prove the converse, from (5.12), we have for any r N ∞ ∞ ∈ r 2 r r r r ∆S f 2 =(∆S f ,∆S f ) = ( ∑ ∆SYj(x), ∑ ∆SYk(x)) k k j=0 k=0 ∞ ∞ r r =( ∑ [ j( j + n 2)] Yj(x), ∑ [ k(k + n 2)] Yk(x)) j=0 − − k=0 − − ∞ 2r = ∑ [ k(k + n 2)] (Yk(x),Yk(x)) k=0 − − ∞ 2r N = ∑ [ k(k + n 2)] O(k− ) 6 C, k=0 − − if we take N large enough. Thus, f C∞(Sn 1). ∈ − ut Theorem 5.10 (Hecke’s identity). It holds

3 n 1 n 1 If g is continuous on S − , we can approximate it uniformly by polynomials restricted to S − . - 102 - 5. Riesz Transforms and Spherical Harmonics

n/2 ω ω | | x 2 ω − k | | ξ 2 n F (Pk(x)e− 2 | | ) = | | ( isgn(ω)) Pk(ξ)e− 2 | | , Pk Hk(R ). (5.13) 2π − ∀ ∈ Proof. That is to prove n/2 Z ω ω ωix ξ | | x 2 ω − k | | ξ 2 Pk(x)e− · − 2 | | dx = | | ( isgn(ω)) Pk(ξ)e− 2 | | . (5.14) Rn 2π − Applying the differential operator Pk(∂ξ ) to both sides of the identity (cf. Theorem 1.10) n/2 Z ω ω ωix ξ | | x 2 ω − | | ξ 2 e− · − 2 | | dx = | | e− 2 | | , Rn 2π we obtain n/2 Z ω ω k ωix ξ | | x 2 ω − | | ξ 2 ( ωi) Pk(x)e− · − 2 | | dx = | | Q(ξ)e− 2 | | . − Rn 2π n Since Pk(x) is polynomial, it is obvious analytic continuation Pk(z) to all of C . Thus, by a change of variable n/2 Z ω ω k ω ωix ξ | | x 2+ | | ξ 2 Q(ξ) =( ωi) | | Pk(x)e− · − 2 | | 2 | | dx − 2π Rn n/2 Z ω k ω | | (x+isgn(ω)ξ)2 =( ωi) | | Pk(x)e− 2 dx − 2π Rn n/2 Z ω k ω | | y 2 =( ωi) | | Pk(y isgn(ω)ξ)e− 2 | | dy. − 2π Rn − So, n/2 Z ω k ω | | y 2 Q(isgn(ω)ξ) =( ωi) | | Pk(y + ξ)e− 2 | | dy − 2π Rn n/2 Z ∞ ω Z k ω n 1 | | r2 =( ωi) | | r − e− 2 Pk(ξ + ry0)dσ(y0)dr. n 1 − 2π 0 S − Since Pk is harmonic, it satisﬁes the mean value property, i.e., Theorem 4.5, thus Z Z Pk(ξ + ry0)dσ(y0) = ωn 1Pk(ξ) = Pk(ξ) dσ(y0). n 1 n 1 S − − S − Hence n/2 Z ∞ ω Z k ω n 1 | | r2 Q(isgn(ω)ξ) =( ωi) | | Pk(ξ) r − e− 2 dσ(y0)dr n 1 − 2π 0 S − n/2 Z ω k ω | | x 2 k =( ωi) | | Pk(ξ) e− 2 | | dx = ( ωi) Pk(ξ). − 2π Rn − Thus, Q(ξ) = ( ωi)kP ( isgn(ω)ξ) = ( ωi)k( isgn(ω))kP (ξ), which proves the theorem. − k − − − k ut The theorem implies the following generalization of itself, whose interest is that it links the various components of the decomposition of L2(Rn), for different n. If f is a radial function, we write f = f (r), where r = x . | | n 2 n Corollary 5.11. Let Pk(x) Hk(R ). Suppose that f is radial and Pk(x) f (r) L (R ). Then the ∈ ∈ Fourier transform of Pk(x) f (r) is also of the form Pk(x)g(r), with g a radial function. Moreover, the induced transform f g, T f = g, depends essentially only on n + 2k. More precisely, we → n,k have Bochner’s relation 5.2. Spherical harmonics and higher Riesz transforms - 103 -

ω k T = | | ( isgn(ω))kT . (5.15) n,k 2π − n+2k,0 Proof. Consider the Hilbert space of radial functions Z ∞ 2 2 2k+n 1 R = f (r) : f = f (r) r − dr < ∞ , k k 0 | | with the indicated norm. Fix now Pk(x), and assume that Pk is normalized, i.e., Z 2 Pk(x) dσ(x) = 1. n 1 S − | | Our goal is to show that ω k (T f )(r) = | | ( isgn(ω))k(T f )(r), (5.16) n,k 2π − n+2k,0 for each f R. ∈ ω | | r2 First, if f (r) = e− 2 , then (5.16) is an immediate consequence of Theorem 5.10, i.e., n/2 ω ω | | r2 ω − k | | R2 (T e− 2 )(R) = | | ( isgn(ω)) e− 2 n,k 2π − k ω ω k | | r2 = | | ( isgn(ω)) (T e− 2 )(R), 2π − n+2k,0 k ω ω k | | r2 which implies T f = | | ( isgn(ω)) T f for f = e 2 . n,k 2π − n+2k,0 − ω | | εr2 Next, we consider e− 2 for a fixed ε > 0. By the homogeneity of Pk and the interplay of n dilations with the Fourier transform (cf. Proposition 1.3), i.e., F δ = ε δ 1 F , and Hecke’s ε − ε− identity, we get ω ω | | ε x 2 k/2 1/2 | | ε x 2 F (Pk(x)e− 2 | | ) = ε− F (Pk(ε x)e− 2 | | ) ω k/2 n/2 | | x 2 =ε− − δ 1/2 F (P (x)e− 2 | | ) ε− k n/2 ω k/2 n/2 ω − k | | ξ 2 =ε− − | | ( isgn(ω)) δ 1/2 (Pk(ξ)e− 2 | | ) 2π − ε− n/2 ω ω − k k/2 n/2 1/2 | | ξ 2/ε = | | ( isgn(ω)) ε− − P (ε− ξ)e− 2 | | 2π − k n/2 ω ω − k k n/2 | | ξ 2/ε = | | ( isgn(ω)) ε− − P (ξ)e− 2 | | . 2π − k ω n/2 ω | | εr2 ω k k n/2 | | r2/ε This shows that T e 2 = | | − ( isgn(ω)) ε e 2 , and so n,k − 2π − − − − k n/2 ω ω | | εr2 ω − − 0 0 (n+2k)/2 | | r2/ε T e− 2 = | | ( isgn(ω)) ε− − e− 2 n+2k,0 2π − k n/2 ω ω − − k n/2 | | r2/ε = | | ε− − e− 2 . 2π ω k ω | | εr2 ω k | | εr2 Thus, T e 2 = | | ( isgn(ω)) T e 2 for ε > 0. n,k − 2π − n+2k,0 − ω | | εr2 To finish the proof, it suffices to see that the linear combination of e 2 ∞ is dense { − }0<ε< in R. Suppose the contrary, then there exists a (almost everywhere) non-zero g R, such that ω ∈ | | εr2 g is orthogonal to every e− 2 in the sense of R, i.e., - 104 - 5. Riesz Transforms and Spherical Harmonics

Z ∞ ω | | εr2 2k+n 1 e− 2 g(r)r − dr = 0, (5.17) 0 2 for all ε > 0. Let ψ(s) = R s e r g(r)rn+2k 1dr for s 0. Then, putting ε = 2(m+1)/ ω , where 0 − − > | | m is a positive integer, and by integration by parts, we have Z ∞ Z ∞ mr2 mr2 0 = e− ψ0(r)dr = 2m e− ψ(r)rdr. 0 0 r2 By the change of variable z = e− , this equality is equivalent to Z 1 m 1 p 0 = z − ψ( ln1/z)dz, m = 1,2,.... 0 Since the polynomials are uniformly dense in the space of continuous functions on the closed p interval [0,1], this can only be the case when ψ( ln1/z) = 0 for all z in [0,1]. Thus, ψ0(r) = 2 e r g(r)rn+2k 1 = 0 for almost every r (0,∞), contradicting the hypothesis that g(r) is not − − ∈ equal to 0 almost everywhere. k ω k Since the operators T and | | ( isgn(ω)) T are bounded and agree on the dense n,k 2π − n+2k,0 subspace, they must be equal. Thus, we have shown the desired result. ut We come now to what has been our main goal in our discussion of spherical harmonics.

Theorem 5.12. Let Pk(x) Hk, k > 1. Then the multiplier corresponding to the transform Pk(x) ∈ (5.10) with the kernel x k+n is | | P (ξ) Γ (k/2) γ k , with γ = πn/2( isgn(ω))k . k ξ k k − Γ (k/2 + n/2) | | Remark 5.13. 1) If k > 1, then Pk(x) is orthogonal to the constants on the sphere, and so its mean value over any sphere centered at the origin is zero. 2) The statement of the theorem can be interpreted as P (x) P (ξ) F k = γ k . (5.18) x k+n k ξ k | | | | 3) As such it will be derived from the following closely related fact, Pk(x) Pk(ξ) F k+n α = γk,α k+α , (5.19) x − ξ α | | | | n/2 ω − k Γ (k/2+α/2) where γk,α = π | 2 | ( isgn(ω)) Γ (k/2+n/2 α/2) . − − Lemma 5.14. The identity (5.19) holds in the sense that Z Z Pk(x) Pk(ξ) ϕˆ(x)dx = γk,α ϕ(ξ)dξ, ϕ S . (5.20) n x k+n α n ξ k+α ∀ ∈ R | | − R | | It is valid for all non-negative integer k and for 0 < α < n. Remark 5.15. For the complex number α with ℜα (0,n), the lemma and (5.19) are also valid, ∈ see [SW71, Theorem 4.1, p.160-163]. Proof. From the proof of Corollary 5.11, we have already known that n/2 ω ω | | ε x 2 ω − k k n/2 | | ξ 2/ε F (P (x)e− 2 | | ) = | | ( isgn(ω)) ε− − P (ξ)e− 2 | | , k 2π − k so we have by the multiplication formula, Z ω Z ω | | ε x 2 | | ε x 2 Pk(x)e− 2 | | ϕˆ(x)dx = F (Pk(x)e− 2 | | )(ξ)ϕ(ξ)dξ Rn Rn 5.2. Spherical harmonics and higher Riesz transforms - 105 -

n/2 Z ω ω − k k n/2 | | ξ 2/ε = | | ( isgn(ω)) ε− − Pk(ξ)e− 2 | | ϕ(ξ)dξ, 2π − Rn for ε > 0. We now integrate both sides of the above w.r.t. ε, after having multiplied the equation by a β 1 suitable power of ε,(ε − , β = (k + n α)/2, to be precise). That is Z ∞ Z ω − β 1 | | ε x 2 ε − Pk(x)e− 2 | | ϕˆ(x)dxdε 0 Rn n/2 (5.21) Z ∞ Z ω ω − k β 1 k n/2 | | ξ 2/ε = | | ( isgn(ω)) ε − ε− − Pk(ξ)e− 2 | | ϕ(ξ)dξdε. 2π − 0 Rn By changing the order of the double integral and a change of variable, we get Z Z ∞ ω β 1 | | ε x 2 l.h.s. of (5.21) = Pk(x)ϕˆ(x) ε − e− 2 | | dεdx Rn 0 2 Z β Z ∞ t= ω ε x /2 ω 2 − β 1 t ====| | | | Pk(x)ϕˆ(x) | | x t − e− dtdx Rn 2 | | 0 β Z ω − 2β = | | Γ (β) Pk(x)ϕˆ(x) x − dx. 2 Rn | | Similarly, n/2 Z ω − k r.h.s. of (5.21) = | | ( isgn(ω)) Pk(ξ)ϕ(ξ) 2π − Rn Z ∞ ω (k/2+α/2+1) | | ξ 2/ε ε− e− 2 | | dεdξ 0 ω 2 n/2 (k+α)/2 t= | | ξ /ε ω − Z ω − 2 | | k 2 ==== | | ( isgn(ω)) Pk(ξ)ϕ(ξ) | | ξ 2π − Rn 2 | | Z ∞ k/2+α/2 1 t t − e− dtdξ 0 n/2 (k+α)/2 ω − ω − = | | ( isgn(ω))k | | Γ (k/2 + α/2) 2π − 2 Z (k+α) Pk(ξ)ϕ(ξ) ξ − dξ. Rn | | Thus, we get (k+n α)/2 Z ω − − (k+n α) | | Γ ((k + n α)/2) Pk(x)ϕˆ(x) x − − dx 2 − Rn | | n/2 (k+α)/2 ω − ω − = | | ( isgn(ω))k | | Γ (k/2 + α/2) 2π − 2 Z (k+α) Pk(ξ)ϕ(ξ) ξ − dξ · Rn | | which leads to (5.20). Observe that when 0 < α < n and ϕ S , then double integrals in the above converge abso- ∈ lutely. Thus the formal argument just given establishes the lemma. ut Proof of Theorem 5.12. By the assumption that k > 1, we have that the integral of Pk over any sphere centered at the origin is zero. Thus for ϕ S , we get Z Z ∈ Pk(x) Pk(x) k+n α ϕˆ(x)dx = k+n α [ϕˆ(x) ϕˆ(0)]dx Rn x − x 1 x − − | | | |6 | | - 106 - 5. Riesz Transforms and Spherical Harmonics Z Pk(x) + k+n α ϕˆ(x)dx. x >1 x − | | | | R Pk(x) ϕˆ( ) α Obviously, the second term tends to x >1 x k+n x dx as 0 by the dominated convergence | | | | → theorem. As in the proof of part (c) of Theorem 4.26, Pk(x) [ϕˆ(x) ϕˆ(0)] is locally integrable, x k+n − thus we have, by the dominated convergence theorem,| the| limit of the ﬁrst term in the r.h.s. of the above Z Z Pk(x) Pk(x) lim k+n α [ϕˆ(x) ϕˆ(0)]dx = k+n [ϕˆ(x) ϕˆ(0)]dx α 0 x 1 x − x 1 x − → 6 − 6 Z | | | | Z | | | | Pk(x) Pk(x) = k+n ϕˆ(x)dx = lim k+n ϕˆ(x)dx. x 1 x ε 0 ε x 1 x | |6 | | → 6| |6 | | Thus, we obtain Z Z Pk(x) Pk(x) lim k+n α ϕˆ(x)dx = lim k+n ϕˆ(x)dx. (5.22) α 0+ Rn x − ε 0 x ε x → | | → | |> | | Similarly, Z Z Pk(ξ) Pk(ξ) lim k+α ϕ(ξ)dξ = lim k ϕ(ξ)dξ. α 0+ Rn ξ ε 0 ξ ε ξ → | | → | |> | | Thus, by Lemma 5.11, we complete the proof with γk = limα 0 γk,α . → ut For ﬁxed k 1, the linear space of operators in (5.10), where Ω(y) = Pk(y) and P H , > y k k ∈ k form a natural generalization of the Riesz transforms; the latter arise in the| | special case k = 1. Those for k > 1, we call the higher Riesz transforms, with k as the degree of the higher Riesz transforms, they can also be characterized by their invariance properties (see [Ste70, §4.8, p.79]).

5.3 Equivalence between two classes of transforms

We now consider two classes of transforms, defined on L2(Rn). The first class consists of all transforms of the form Z Ω(y) T f = c f + lim n f (x y)dy, (5.23) · ε 0 y ε y − → | |> ∞ n 1 | | where c is a constant, Ω C (S − ) is a homogeneous function of degree 0, and the integral R ∈ n 1 Ω(x)dσ(x) = 0. The second class is given by those transforms T for which S − F (T f )(ξ) = m(ξ) fˆ(ξ) (5.24) where the multiplier m C∞(Sn 1) is homogeneous of degree 0. ∈ − Theorem 5.16. The two classes of transforms, defined by (5.23) and (5.24) respectively, are identical.

Proof. First, support that T is of the form (5.23). Then by Theorem 4.24, T is of the form (5.24) with m homogeneous of degree 0 and Z πi m(ξ) = c + sgn(ω)sgn(ξ x) + ln(1/ ξ x ) Ω(x)dσ(x), ξ = 1. (5.25) n 1 S − − 2 · | · | | | Now, we need to show m C∞(Sn 1). Write the spherical harmonic developments ∈ − 5.3. Equivalence between two classes of transforms - 107 -

∞ ∞ N N ˜ ˜ Ω(x) = ∑ Yk(x), m(x) = ∑ Yk(x), ΩN(x) = ∑ Yk(x), mN(x) = ∑ Yk(x), (5.26) k=1 k=0 k=1 k=0 where Yk,Y˜k Hk in view of part (3) in Proposition 5.9. k starts from 1 in the development of R ∈ Ω, since n 1 Ω(x)dx = 0 implies that Ω(x) is orthogonal to constants, and H contains only S − 0 constants. Then, by Theorem 5.12, if Ω = ΩN, then m(x) = mN(x), with Y˜k(x) = γkYk(x), k > 1. R h πi 1 i But mM(x) mN(x) = n 1 sgn(ω)sgn(y x) + ln [ΩM(y) ΩN(y)]dσ(y). Moreover − S − − 2 · y x − by Hölder’s inequality, | · |

sup mM(x) mN(x) x Sn 1 | − | ∈ − 2 !1/2 Z πi sup sgn(ω)sgn(y x) + ln(1/ y x ) dσ(y) 6 n 1 x S − − 2 · | · | Z 1/2 2 ΩM(y) ΩN(y) dσ(y) 0, (5.27) n 1 × S − | − | → as M, N ∞, since4 for n = 1, S0 = 1,1 , → {− } 2 Z πi π2 sgn(ω)sgn(y x) + ln(1/ y x ) dσ(y) = , S0 − 2 · | · | 2 and for n 2, we can pick a orthogonal matrix A satisfying Ae = x and detA = 1 for x = 1, > 1 | | and then by a change of variable, 2 Z πi sup sgn(ω)sgn(y x) + ln(1/ y x ) dσ(y) n 1 x S − − 2 · | · | Z π2 =sup + (ln(1/ y x ))2 dσ(y) n 1 x S − 4 | · | 2 Z π 2 = ωn 1 + sup (ln y Ae1 ) dσ(y) n 1 4 − x S − | · | 2 Z π 1 2 = ωn 1 + sup (ln A− y e1 ) dσ(y) n 1 4 − x S − | · | 1 2 Z z=A− y π 2 ==== ωn 1 + (ln z1 ) dσ(z) < ∞. n 1 4 − S − | | Here, we have used the boundedness of the integral in the r.h.s., i.e., (with the notationz ¯ = (z2,...,zn), cf. [Gra04, p.A-20,p.267]) Z Z 1 Z 2 2 dz1 (ln z1 ) dσ(z) = (ln z1 ) dσ(z¯)q Sn 1 | | | | z2Sn 2 − 1 √1 1 − 1 z2 − − − 1 y=z¯/√1 z2 Z 1 Z − 1 2 2 (n 3)/2 ======(ln z1 ) (1 z1) − dσ(y)dz1 1 | | Sn 2 − − − Z 1 2 2 (n 3)/2 =ωn 2 (ln z1 ) (1 z1) − dz1 − 1 | | − Z−π z1=cosθ 2 n 2 ====ωn 2 (ln cosθ ) (sinθ) − dθ = ωn 2I1. − 0 | | −

4 There the argument is similar with some part of the proof of Theorem 4.24. - 108 - 5. Riesz Transforms and Spherical Harmonics

If n > 3, then, by integration by parts, Z π Z π Z π 2 I1 6 (ln cosθ ) sinθdθ = 2 ln cosθ sinθdθ = 2 sinθdθ = 4. 0 | | − 0 | | 0 R π/2 2 π 2 2 If n = 2, then, by the formula 0 (ln(cosθ)) dθ = 2 [(ln2) +π /12], cf. [GR, 4.225.8, p.531], we get Z π Z π/2 2 2 2 2 I1 = (ln cosθ ) dθ = 2 (ln(cosθ)) dθ = π[(ln2) + π /12]. 0 | | 0 Thus, (5.27) shows that ∞ m(x) = c + ∑ γkYk(x). k=1 Since Ω C∞, we have, in view of part (5) of Proposition 5.9, that ∈ Z 2 N Yk(x) dσ(x) = O(k− ) n 1 S − | | as k ∞ for every fixed N. However, by the explicit form of γ , we see that γ k n/2, so m(x) → k k ∼ − is also indefinitely differentiable on the unit sphere, i.e., m C∞(Sn 1). ∈ − Conversely, suppose m(x) C∞(Sn 1) and let its spherical harmonic development be as in ∈ − (5.26). Set c = Y˜ , and Y (x) = 1 Y˜ (x). Then Ω(x), given by (5.26), has mean value zero in 0 k γk k the sphere, and is again indefinitely differentiable there. But as we have just seen the multiplier corresponding to this transform is m; so the theorem is proved. ut As an application of this theorem and a final illustration of the singular integral transforms we shall give the generalization of the estimates for partial derivatives given in 5.1. n Let P(x) Pk(R ). We shall say that P is elliptic if P(x) vanishes only at the origin. For ∈ any polynomial P, we consider also its corresponding differential polynomial. Thus, if P(x) = α ∂ ∂ α ∑aα x , we write P( ∂x ) = ∑aα ( ∂x ) as in the previous definition. ∂ α Corollary 5.17. Suppose P is a homogeneous elliptic polynomial of degree k. Let ( ∂x ) be any k differential monomial of degree k. Assume f Cc , then we have the a priori estimate α ∈ ∂ ∂ f 6 Ap P f , 1 < p < ∞. (5.28) ∂x p ∂x p α ∂ ∂ Proof. From the Fourier transform of ∂x f and P ∂x f , Z ∂ ωix ξ ∂ k F P f (ξ) = e− · P f (x)dx = (ωi) P(ξ) fˆ(ξ), ∂x Rn ∂x and ∂ α F f (ξ) = (ωi)kξ α fˆ(ξ), ∂x we have the following relation ∂ α ∂ P(ξ)F f (ξ) = ξ α F P f (ξ). ∂x ∂x ξ α Since P(ξ) is non-vanishing except at the origin, P(ξ) is homogenous of degree 0 and is indefinitely differentiable on the unit sphere. Thus ∂ α ∂ f = T P f , ∂x ∂x 5.3. Equivalence between two classes of transforms - 109 - where T is one of the transforms of the type given by (5.24). By Theorem 5.16, T is also given by (5.23) and hence by the result of Theorem 4.24, we get the estimate (5.28). ut

Chapter 6 The Littlewood-Paley g-function and Multipliers

In harmonic analysis, Littlewood-Paley theory is a term used to describe a theoretical frame- work used to extend certain results about L2 functions to Lp functions for 1 < p < ∞. It is typically used as a substitute for orthogonality arguments which only apply to Lp functions when p = 2. One implementation involves studying a function by decomposing it in terms of functions with localized frequencies, and using the Littlewood-Paley g-function to compare it with its Poisson integral. The 1-variable case was originated by J. E. Littlewood and R. Paley (1931, 1937, 1938) and developed further by Zygmund and Marcinkiewicz in the 1930s using complex function theory (Zygmund 2002 [1935], chapters XIV, XV). E. M. Stein later extended the theory to higher dimensions using real variable techniques.

6.1 The Littlewood-Paley g-function

The g-function is a nonlinear operator which allows one to give a useful characterization of the Lp norm of a function on Rn in terms of the behavior of its Poisson integral. This characterization will be used not only in this chapter, but also in the succeeding chapter dealing with function spaces. Let f Lp(Rn) and write u(x,y) for its Poisson integral ∈ n Z Z ω ωiξ x ωξ y u(x,y) = | | e · e−| | fˆ(ξ)dξ = Py(t) f (x t)dt 2π Rn Rn − n+1 as defined in (4.15) and (4.17). Let ∆ denote the Laplace operator in R+ , that is ∆ = ∂ 2 n ∂ 2 2 ∂u 2 2 ∂ 2 + ∑ j=1 2 ; ∇ is the corresponding gradient, ∇u(x,y) = ∂y + ∇xu(x,y) , where y ∂x j | | | | | | 2 n ∂u 2 ∇xu(x,y) = ∑ . | | j=1 | ∂x j | Definition 6.1. With the above notations, we define the Littlewood-Paley g-function g( f )(x), by Z ∞ 1/2 g( f )(x) = ∇u(x,y) 2ydy . (6.1) 0 | | We can also define two partial g-functions, one dealing with the y differentiation and the other with the x differentiations, 1/2 Z ∞ 2 ! Z ∞ 1/2 ∂u 2 g1( f )(x) = (x,y) ydy , gx( f )(x) = ∇xu(x,y) ydy . (6.2) 0 ∂y 0 | | 2 2 2 Obviously, g = g1 + gx.

111 - 112 - 6. The Littlewood-Paley g-function and Multipliers

The basic result for g is the following. Theorem 6.2. Suppose f Lp(Rn), 1 < p < ∞. Then g( f )(x) Lp(Rn), and ∈ ∈ A0 f g( f ) A f . (6.3) pk kp 6 k kp 6 pk kp Proof. Step 1: We ﬁrst consider the simple case p = 2. For f L2(Rn), we have Z Z ∞ Z ∞ Z ∈ 2 2 2 g( f ) 2 = ∇u(x,y) ydydx = y ∇u(x,y) dxdy. k k Rn 0 | | 0 Rn | | In view of the identity n Z ω ωiξ x ωξ y u(x,y) = | | e · e−| | fˆ(ξ)dξ, 2π Rn we have n Z ∂u ω ωiξ x ωξ y = | | ωξ fˆ(ξ)e · e−| | dξ, ∂y 2π Rn −| | and n Z ∂u ω ωiξ x ωξ y = | | ωiξ j fˆ(ξ)e · e−| | dξ. ∂x j 2π Rn Thus, by Plancherel’s formula, " 2 n 2# Z Z ∂u ∂u ∇u(x,y) 2dx = + dx n n ∑ R | | R ∂y j=1 ∂x j 2 n 2 ∂u ∂u = + ∑ ∂y 2 ∂x j 2 Lx j=1 Lx " n # 1 ˆ ωξ y 2 1 ˆ ωξ y 2 = F − ( ωξ f (ξ)e−| | ) 2 + ∑ F − (ωiξ j f (ξ)e−| | ) 2 k −| | k j=1 k k n " n # ω ˆ ωξ y 2 ˆ ωξ y 2 = | | ωξ f (ξ)e−| | 2 + ∑ ωiξ j f (ξ)e−| | 2 2π k − | | k j=1 k k n ω 2 ωξ y 2 =2 | | ω ξ fˆ(ξ)e−| | 2π k| | k2 Z n ω 2 2 2 2 ωξ y = 2 | | ω ξ fˆ(ξ) e− | | dξ, Rn 2π | | | | and so Z ∞ Z n 2 ω 2 2 ˆ 2 2 ωξ y g( f ) 2 = y 2 | | ω ξ f (ξ) e− | | dξdy k k 0 Rn 2π | | | | Z n Z ∞ ω 2 2 2 2 ωξ y = 2 | | ω ξ fˆ(ξ) ye− | | dydξ Rn 2π | | | | 0 Z n n ω 2 2 ˆ 2 1 1 ω ˆ 2 = 2 | | ω ξ f (ξ) dξ = | | f 2 n 2π | | | | 4ω2 ξ 2 2 2π k k R | | 1 = f 2. 2k ||2 Hence, 1/2 g( f ) = 2− f . (6.4) k k2 k k2 We have also obtained g ( f ) = g ( f ) = 1 f . k 1 k2 k x k2 2 k k2 6.1. The Littlewood-Paley g-function - 113 -

Step 2: We consider the case p = 2 and prove g( f ) A f . We define the Hilbert 6 k kp 6 pk kp spaces H1 and H2 which are to be consider now. H1 is the one-dimensional Hilbert space of 0 2 complex numbers. To define H2, we define first H2 as the L space on (0,∞) with measure ydy, i.e., Z ∞ 0 2 2 H2 = f : f = f (y) ydy < ∞ . | | 0 | | 0 Let H2 be the direct sum of n + 1 copies of H2 ; so the elements of H2 can be represented 0 as (n + 1) component vectors whose entries belong to H2 . Since H1 is the same as the complex numbers, then L(H1,H2) is of course identifiable with H2. Now let ε > 0, and keep it temporarily fixed. Define ∂Py+ε (x) ∂Py+ε (x) ∂Py+ε (x) Kε (x) = , , , . ∂y ∂x1 ··· ∂xn Notice that for each fixed x, Kε (x) H . This is the same as saying that ∈ 2 Z ∞ 2 Z ∞ 2 ∂Py+ε (x) ∂Py+ε (x) ydy < ∞ and ydy < ∞, for j = 1,...,n. 0 ∂y 0 ∂x j

cny ∂Py ∂Py A In fact, since Py(x) = , we have that both and are bounded by . ( x 2+y2)(n+1)/2 ∂y ∂x j ( x 2+y2)(n+1)/2 | | | | So the norm in H2 of Kε (x), Z ∞ 2 2 ydy Kε (x) 6A (n + 1) 2 2 n+1 | | 0 ( x + (y + ε) ) Z ∞ | | 2 dy 6A (n + 1) 2n+1 6 Cε , 0 (y + ε) and in another way Z ∞ 2 2 2 ydy A (n + 1) 2 2 n 2n Kε (x) 6 A (n + 1) + = ( x + ε )− 6 C x − . | | ε ( x 2 + y2)n 1 2n | | | | | | Thus, 1 n Kε (x) L (R ). (6.5) | | ∈ loc Similarly, 2 Z ∞ Z ∞ ∂Kε (x) ydy ydy 2n 2 6 C 2 2 n+2 6 C 2 2 n+2 6 C x − − . ∂x 0 ( x + y ) ε ( x + y ) | | j | | | | Therefore, Kε satisﬁes the gradient condition, i.e.,

∂Kε (x) (n+1) 6 C x − , (6.6) ∂x j | | with C independent of ε. Now we consider the operator Tε deﬁned by Z Tε f (x) = Kε (t) f (x t)dt. Rn − The function f is complex-valued (take their value in H1), but Tε f (x) takes its value in H2. Observe that 1 1 Z ∞ 2 Z ∞ 2 2 2 Tε f (x) = ∇u(x,y + ε) ydy 6 ∇u(x,y) ydy 6 g( f )(x). (6.7) | | 0 | | ε | | 1/2 2 n Hence, Tε f (x) 2 6 2− f 2, if f L (R ), by (6.4). Therefore, k k k k ∈ 1/2 Kˆε (x) 2− . (6.8) | | 6 - 114 - 6. The Littlewood-Paley g-function and Multipliers

Because of (6.5), (6.6) and (6.8), by Theorem 4.27 (cf. Theorem 4.18), we get Tε f k kp 6 A f , 1 < p < ∞ with A independent of ε. By (6.7), for each x, Tε f (x) increases to g( f )(x), pk kp p | | as ε 0, so we obtain ﬁnally → g( f ) A f , 1 < p < ∞. (6.9) k kp 6 pk kp Step 3: To derive the converse inequalities,

A0 f g( f ) , 1 < p < ∞. (6.10) pk kp 6 k kp 1 2 n In the ﬁrst step, we have shown that g1( f ) 2 = f 2 for f L (R ). Let u1,u2 are the k k 2 k k ∈ Poisson integrals of f , f L2, respectively. Then we have g ( f + f ) 2 = 1 f + f 2, i.e., 1 2 ∈ k 1 1 2 k2 4 k 1 2k2 R R ∞ ∂(u1+u2) 2 1 R 2 n ydydx = n f + f dx. It leads to the identity R 0 | ∂y | 4 R | 1 2| Z Z ∞ Z ∂u1 ∂u2 4 (x,y) (x,y)ydydx = f1(x) f2(x)dx. Rn 0 ∂y ∂y Rn This identity, in turn, leads to the inequality, by Hölder’s inequality and the deﬁnition of g1,

1 Z Z f1(x) f2(x)dx 6 g1( f1)(x)g1( f2)(x)dx. 4 Rn Rn p n p0 n Suppose now in addition that f1 L (R ) and f2 L (R ) with f2 p 1 and 1/p+1/p0 = ∈ ∈ k k 0 6 1. Then by Hölder inequality and the result (6.9).

Z f1(x) f2(x)dx 6 4 g1( f1) p g1( f2) p0 6 4Ap0 g1( f1) p. (6.11) Rn k k k k k k Now we take the supremum in (6.11) as f ranges over all function in L2 Lp0 , with f 2 ∩ k 2kp0 6 1. Then, we obtain the desired result (6.10), with A0p = 1/4Ap0 , but where f is restricted to be in L2 Lp. The passage to the general case is provided by an easy limiting argument. Let f be ∩ m a sequence of functions in L2 Lp, which converges in Lp norm to f . Notice that g( f )(x) ∩ | m − g( fn)(x) = ∇um L2(0,∞;ydy) ∇un L2(0,∞;ydy) 6 ∇um ∇un L2(0,∞;ydy) = g( fm fn)(x) by | k k − k k k − kp − the triangle inequality. Thus, g( fm) is a Cauchy sequence in L and so converges to g( f ) in p { } L , and we obtain the inequality (6.10) for f as a result of the corresponding inequalities for fm. ut We have incidentally also proved the following, which we state as a corollary. 2 n p n p n Corollary 6.3. Suppose f L (R ), and g1( f ) L (R ), 1 < p < ∞. Then f L (R ), and ∈ ∈ ∈ A f g ( f ) . 0pk kp 6 k 1 kp Remark 6.4. There are some very simple variants of the above that should be pointed out: (i) The results hold also with g ( f ) instead of g( f ). The direct inequality g ( f ) A f x k x kp 6 pk kp is of course a consequence of the one for g. The converse inequality is then proved in the same way as that for g1. (ii) For any integer k > 1, deﬁne 1/2 Z ∞ k 2 ! ∂ u 2k 1 gk( f )(x) = k (x,y) y − dy . 0 ∂y p Then the L inequalities hold for gk as well. both (i) and (ii) are stated more systematically in [Ste70, Chapter IV, §7.2, p.112-113]. (iii) For later purpose, it will be useful to note that for each x, gk( f )(x) > Akg1( f )(x) where the bound Ak depends only on k. It is easily veriﬁed from the Poisson integral formula that if f Lp(Rn), 1 p ∞, then ∈ 6 6 6.1. The Littlewood-Paley g-function - 115 -

∂ ku(x,y) 0 for each x, as y ∞. ∂yk → → Thus, k Z ∞ k+1 ∂ u(x,y) ∂ u(x,s) k ds k = k+ s k . ∂y − y ∂s 1 s By Schwarz’s inequality, therefore, k 2 Z ∞ k+1 2 !Z ∞ ∂ u(x,y) ∂ u(x,s) 2k 2k k 6 k+ s ds s− ds . ∂y y ∂s 1 y Hence, by Hardy’s inequality (2.18) (on p.42, with q = r = 1 there), we have Z ∞ k 2 2 ∂ u 2k 1 (gk( f )(x)) = k (x,y) y − dy 0 ∂y Z ∞ Z ∞ k+1 2 !Z ∞ ∂ u 2k 2k 2k 1 6 k+1 (x,s) s ds s− ds y − dy 0 y ∂s y Z ∞ Z ∞ k+1 2 ! 1 ∂ u 2k = k+1 (x,s) s ds dy 2k 1 0 y ∂s − Z ∞ k+1 2 1 ∂ u 2k+1 6 k+1 (x,s) s ds 2k 1 0 ∂s − Z ∞ k+1 2 1 ∂ u 2(k+1) 1 = k+1 (x,s) s − ds 2k 1 0 ∂s − 1 = (g ( f )(x))2. 2k 1 k+1 Thus, the assertion is proved by− the induction on k. The proof that was given for the Lp inequalities for the g-function did not, in any essential way, depend on the theory of harmonic functions, despite the fact that this function was defined in terms of the Poisson integral. In effect, all that was really used is the fact that the Poisson kernels are suitable approximations to the identity. There is, however, another approach, which can be carried out without recourse to the theory of singular integrals, but which leans heavily on characteristic properties of harmonic functions. We present it here (more precisely, we present that part which deals with 1 < p 6 2, for the inequality (6.9)), because its ideas can be adapted to other situations where the methods of Chapter4 are not applicable. Everything will be based on the following three observations. Lemma 6.5. Suppose u is harmonic and strictly positive. Then p p 2 2 ∆u = p(p 1)u − ∇u . (6.12) − | | Proof. The proof is straightforward. Indeed, p p 1 2 p p 2 2 p 1 2 ∂ u = pu − ∂ u, ∂ u = p(p 1)u − (∂ u) + pu − ∂ u, x j x j x j − x j x j which implies by summation p p 2 2 p 1 p 2 2 ∆u = p(p 1)u − ∇u + pu − ∆u = p(p 1)u − ∇u , − | | − | | since ∆u = 0. ut + + Lemma 6.6. Suppose F(x,y) C(Rn 1) C2(Rn 1), and suitably small at infinity. Then ∈ + ∩ + - 116 - 6. The Littlewood-Paley g-function and Multipliers Z Z y∆F(x,y)dxdy = F(x,0)dx. (6.13) n+1 n R+ R Proof. We use Green’s theorem Z Z ∂v ∂u (u∆v v∆u)dxdy = u v dσ D − ∂D ∂N − ∂N n+1 n+1 where D = Br R , with Br the ball of radius r in R centered at the origin, N is the ∩ + outward normal vector. We take v = F, and u = y. Then, we will obtain our result (6.13) if Z Z y∆F(x,y)dxdy y∆F(x,y)dxdy, n+1 D → R+ and Z ∂F ∂y y F dσ 0, ∂D0 ∂N − ∂N → as r ∞. Here ∂D is the spherical part of the boundary of D. This will certainly be the case, if → 0 for example ∆F 0, and F O(( x +y) n ε ) and ∇F = O(( x +y) n 1 ε ), as x +y ∞, > | | 6 | | − − | | | | − − − | | → for some ε > 0. ut Lemma 6.7. If u(x,y) is the Poisson integral of f , then sup u(x,y) 6 M f (x). (6.14) y>0 | | Proof. This is the same as the part (a) of Theorem 4.9. It can be proved with a similar argument as in the proof of part (a) for Theorem 4.10. ut Now we use these lemmas to give another proof for the inequality g( f ) A f , 1 < k kp 6 pk kp p 6 2. n Another proof of g( f ) p 6 Ap f p, 1 0, since Py > 0 for any x R and y > 0; and the majorizations u(x,y) = O(( x + y)− ) ∈ | | and ∇u = O(( x +y) n 1), as x +y ∞ are valid. We have, by Lemma 6.5, Lemma 6.7 and | | | | − − | | → the hypothesis 1 < p 6 2, Z ∞ Z ∞ 2 2 1 2 p p (g( f )(x)) = y ∇u(x,y) dy = yu − ∆u dy 0 | | p(p 1) 0 − 2 p Z ∞ [M f (x)] − p 6 y∆u dy. p(p 1) 0 − We can write this as (2 p)/2 1/2 g( f )(x) 6 Cp(M f (x)) − (I(x)) , (6.15) R ∞ p where I(x) = 0 y∆u dy. However, by Lemma 6.6, Z Z Z I(x)dx = y∆updydx = up(x,0)dx = f p. (6.16) n n+1 n p R R+ R k k This immediately gives the desired result for p = 2. Next, suppose 1 < p < 2. By (6.15), Hölder’s inequality, Theorem 3.9 and (6.16), we have, for 0 6 f D(Rn), ∈Z Z p p p(2 p)/2 p/2 (g( f )(x)) dx 6 Cp (M f (x)) − (I(x)) dx Rn Rn Z 1/r0 Z 1/r p p p/r0 p/r p 6Cp (M f (x)) dx I(x)dx 6 C0p f p f p = C0p f p, Rn Rn k k k k k k 6.1. The Littlewood-Paley g-function - 117 - where r = 2/p (1,2) and 1/r + 1/r0 = 1, then r0 = 2/(2 p). ∈ − n Thus, g( f ) p Ap f p, 1 < p 2, whenever 0 f D(R ). k k 6 k k 6 6 ∈ For general f Lp(Rn) (which we assume for simplicity to be real-valued), write f = f + ∈ − f − as its decomposition into positive and negative part; then we need only approximate in norm + n f and f −, each by a sequences of positive functions in D(R ). We omit the routine details that are needed to complete the proof. ut Unfortunately, the elegant argument just given is not valid for p > 2. There is, however, a more intricate variant of the same idea which does work for the case p > 2, but we do not intend to reproduce it here. We shall, however, use the ideas above to obtain a significant generalization of the inequality for the g-functions. Definition 6.8. Define the positive function Z ∞ Z λn 2 y 2 1 n (gλ∗ ( f )(x)) = ∇u(x t,y) y − dtdy. (6.17) 0 n t + y | − | R | | Before going any further, we shall make a few comments that will help to clarify the meaning of the complicated expression (6.17).

First, gλ∗ ( f )(x) will turn out to be a pointwise majorant of g( f )(x). To understand this situation better we have to introduce still another quantity, which is roughly midway between g and gλ∗ . It is defined as follows. n+1 Definition 6.9. Let Γ be a fixed proper cone in R+ with vertex at the origin and which contains (0,1) in its interior. The exact form of Γ will not really matter, but for the sake of definiteness let us choose for Γ the up circular cone: n+ Γ = (t,y) R 1 : t < y,y > 0 . ∈ + | | For any x Rn, let Γ (x) be the cone Γ translated such that its vertex is at x. Now define the ∈ positive Luzin’s S-function S( f )(x) by Z Z 2 2 1 n 2 1 n [S( f )(x)] = ∇u(t,y) y − dydt = ∇u(x t,y) y − dydt. (6.18) Γ (x) | | Γ | − | y We assert, as we shall momentarily prove, that Γ Γ(x) Proposition 6.10.

(0, 1) g( f )(x) 6 CS( f )(x) 6 Cλ gλ∗ ( f )(x). (6.19) π 4 What interpretation can we put on the inequalities relating O x t these three quantities? A hint is afforded by considering three Fig. 6.1 Γ and Γ (x) for n = 1 Figure 1: Γ and Γ(x) for n = 1 corresponding approaches to the boundary for harmonic functions. (a) With u(x,y) the Poisson integral of f (x), the simplest approach to the boundary point x Rn is obtained by letting y 0, (with x ﬁxed). This is the perpendicular approach, and for ∈ → it the appropriate limit exists almost everywhere, as we already know. (b) Wider scope is obtained by allowing the variable point (t,y) to approach (x,0) through any cone Γ (x), (where vertex is x). This is the non-tangential approach which will be so important for us later. As the reader may have already realized, the relation of the S-function to the g-function is in some sense analogous to the relation between the non-tangential and the perpendicular approaches; we should add that the S-function is of decisive signiﬁcance in its own right, but we shall not pursue that matter now. - 118 - 6. The Littlewood-Paley g-function and Multipliers

(c) Finally, the widest scope is obtained by allowing the variable point (t,y) to approach (x,0) in an arbitrary manner, i.e., the unrestricted approach. The function gλ∗ has the analogous role: it takes into account the unrestricted approach for Poisson integrals.

Notice that gλ∗ (x) depends on λ. For each x, the smaller λ the greater gλ∗ (x), and this behavior p is such that that L boundedness of gλ∗ depends critically on the correct relation between p and λ. This last point is probably the main interest in gλ∗ , and is what makes its study more difﬁcult than g or S. After these various heuristic and imprecise indications, let us return to ﬁrm ground. The only thing for us to prove here is the assertion (6.19).

Proof of Proposition 6.10. The inequality S( f )(x) 6 Cλ gλ∗ ( f )(x) is obvious, since the integral (6.17) majorizes that part of the integral taken only over Γ , and y λn 1 > t + y 2λn | | since t < y there. The non-trivial part of the assertion is: | | g( f )(x) 6 CS( f )(x). y

It sufﬁces to prove this inequality for x = 0. Let us denote by By the B Γ n+1 y ball in R+ centered at (0,y) and tangent to the boundary of the cone (0, y) Γ ; the radius of By is then proportional to y. Now the partial derivatives ∂u and ∂u are, like u, harmonic functions. Thus, by the mean ∂y ∂xk O t value theorem of harmonic functions (i.e., Theorem 4.5 by noticing Fig. 6.2 FigureΓ and 1:By Γ and By (0,y) is the center of By), ∂u(0,y) 1 Z ∂u(x,s) = dxds ∂y m(By) By ∂s n+1 where m(By) is the n + 1 dimensional measure of By, i.e., m(By) = cy for an appropriate constant c. By Schwarz’s inequality 2 2 ∂u(0,y) 1 Z ∂u(x,s) Z 6 2 dxds dxds ∂y (m(By)) By ∂s By 2 1 Z ∂u(x,s) = dxds. m(By) By ∂s If we integrate this inequality, we obtain Z ∞ 2 Z ∞ Z 2 ! ∂u(0,y) 1 n ∂u(x,s) y dy 6 c− y− dxds dy. 0 ∂y 0 By ∂s However, (x,s) B clearly implies that c s y c s, for two positive constants c and c . ∈ y 1 6 6 2 1 2 Thus, apart from a multiplicative factor by changing the order of the double integrals, the last integral is majorized by

Z Z c2s 2 Z 2 n ∂u(x,s) ∂u(x,s) 1 n y− dy dxds 6 c0 s − dxds. Γ c1s ∂s Γ ∂s This is another way of saying that, Z ∞ 2 Z 2 ∂u(0,y) ∂u(x,y) 1 n y dy 6 c00 y − dxdy. 0 ∂y Γ ∂y The same is true for the derivatives ∂u , j = 1,...,n, and adding the corresponding estimates ∂x j proves our assertion. ut 6.1. The Littlewood-Paley g-function - 119 -

p We are now in a position to state the L estimates concerning gλ∗ . Theorem 6.11. Let λ > 1 be a parameter. Suppose f Lp(Rn). Then n ∈ (a) For every x R , g( f )(x) Cλ g∗ ( f )(x). ∈ 6 λ (b) If 1 2/λ, then

g∗ ( f ) A f . (6.20) k λ kp 6 p,λ k kp Proof. The part (a) has already been proved in Proposition 6.10. Now, we prove (b). For the case p > 2, only the assumption λ > 1 is relevant since 2/λ < 2 6 p. Let ψ denote a positive function on Rn, we claim that Z Z 2 2 (gλ∗ ( f )(x)) ψ(x)dx 6 Aλ (g( f )(x)) (Mψ)(x)dx. (6.21) Rn Rn The l.h.s. of (6.21) equals Z ∞ Z Z ψ(x) y ∇u(t,y) 2 yλny ndx dtdy, λn − 0 t Rn | | x Rn ( t x + y) ∈ ∈ | − | so to prove (6.21), we must show that Z ψ(x) yλny ndx A Mψ(t). sup λn − 6 λ (6.22) y>0 x Rn ( t x + y) ∈ | − | However, we know by Theorem 4.10, that sup(ψ ϕε )(t) 6 AMψ(t) ε>0 ∗ n λn for appropriate ϕ, with ϕε (x) = ε ϕ(x/ε). Here, we have in fact ϕ(x) = (1 + x ) , ε = y, − | | − and so with λ > 1 the hypotheses of that theorem are satisﬁed. This proves (6.22) and thus also (6.21). The case p = 2 follows immediately from (6.21) by inserting in this inequality the function 2 ψ = 1 (or by the deﬁnitions of gλ∗ ( f ) and g( f ) directly), and using the L result for g. Suppose now p > 2; let us set 1/q + 2/p = 1, and take the supremum of the l.h.s. of (6.21) q n 2 over all ψ 0, such that ψ L (R ) and ψ q 1. Then, it gives g∗ ( f ) ; Hölder’s inequality > ∈ k k 6 k λ kp yields an estimate for the right side: A g( f ) 2 Mψ . λ k kpk kq However, by the inequalities for the g-function, g( f ) A f ; and by the theorem of k kp 6 0pk kp the maximal function Mψ A ψ A , since q > 1, if p < ∞. If we substitute these in k kq 6 qk kq 6 q00 the above, we get the result:

g∗ ( f ) A f , 2 p < ∞, λ > 1. k λ kp 6 p,λ k kp 6 The inequalities for p < 2 will be proved by an adaptation of the reasoning used for g. Lem- mas 6.5 and 6.6 will be equally applicable in the present situation, but we need more general version of Lemma 6.7, in order to majorize the unrestricted approach to the boundary of a Poisson integral. It is at this stage where results which depend critically on the Lp class ﬁrst make their appear- ance. Matters will depend on a variant of the maximal function which we deﬁne as follows. Let µ > 1, and write Mµ f (x) for Z 1/µ 1 µ Mµ f (x) = sup f (y) dy . (6.23) r>0 m(B(x,r)) B(x,r) | | µ 1/µ Then M f (x) = M f (x), and Mµ f (x) = ((M f )(x)) . From the theorem of the maximal 1 | | function, it immediately follows that, for p > µ, - 120 - 6. The Littlewood-Paley g-function and Multipliers

µ 1/µ µ 1/µ µ 1/µ Mµ f = ((M f )(x)) = ((M f )(x)) f = f . (6.24) k kp k | | kp k | | kp/µ 6 k| | kp/µ k kp This inequality fails for p 6 µ, as in the special case µ = 1. The substitute for Lemma 6.7 is as follows. Lemma 6.12. Let f Lp(Rn), p µ 1; if u(x,y) is the Poisson integral of f , then ∈ > > t n u(x t,y) A 1 + | | M f (x), (6.25) | − | 6 y and more generally t n/µ u(x t,y) Aµ 1 + | | Mµ f (x). (6.26) | − | 6 y We shall now complete the proof of the inequality (6.20) for the case 1 2/λ. 2 p Let us observe that we can always ﬁnd a µ [1, p) such that if we set λ 0 = λ −µ , then one 2 ∈p − still has λ > 1. In fact, if µ = p, then λ − > 1 since λ > 2/p; this inequality can then be 0 − µ maintained by a small variation of µ. With this choice of µ, we have by Lemma 6.12 y n/µ u(x t,y) Aµ Mµ f (x). (6.27) | − | y + t 6 | | We now proceed the argument with which we treated the function g. 2 (gλ∗ ( f )(x)) Z λn 1 1 n y 2 p p = y − u − (x t,y) ∆u (x t,y) dtdy p(p 1) n+1 y + t − | − | − R+ | | 1 2 p 2 p 6 Aµ− (Mµ f (x)) − I∗(x), (6.28) p(p 1) − where Z λ 0n 1 n y p I∗(x) = y − ∆u (x t,y)dtdy. n+1 y + t − R+ | | It is clear that Z Z Z λ 0n 1 n y p I∗(x)dx = y − ∆u (t,y)dxdtdy n n+1 n R R+ Rx y + t x Z | − | =C y∆up(t,y)dtdy. λ 0 n+1 R+ The last step follows from the fact that if λ 0 > 1 Z λ 0n Z λ 0n n y n y y− dx =y− dx n y + t x n y + x R | − | R | | λ 0n x=yz Z 1 == dz n 1 + z R | | = < ∞. Cλ 0 So, by Lemma 6.6 Z Z ( ) = p( , ) = p. I∗ x dx Cλ 0 u t 0 dt Cλ 0 f p (6.29) Rn Rn k k Therefore, by (6.28), Hölder’s inequality, (6.24) and (6.29), 1 p/2 1/2 1 p/2 1/2 g∗ ( f ) C Mµ f (x) − (I∗(x)) C Mµ f − I∗ C f . k λ kp 6 k kp 6 k kp k k1 6 k kp 6.2. Fourier multipliers on Lp - 121 -

That is the desired result. ut Finally, we prove Lemma 6.12. Proof of Lemma 6.12. One notices that (6.25) is unchanged by the dilation (x,t,y) → (δx,δt,δy), it is then clear that it sufﬁces to prove (6.25) with y = 1. Setting y = 1 in the Poisson kernel, we have P (x) = c (1 + x 2) (n+1)/2, and u(x t,1) = 1 n | | − − f (x) P (x t), for each t. Theorem 4.10 shows that u(x t,1) A M f (x), where A = ∗ 1 − | − | 6 t t R Q (x)dx, and Q (x) is the smallest decreasing radial majorant of P (x t), i.e., t t 1 − 1 Q (x) = c sup . t n 2 (n+1)/2 x x (1 + x0 t ) | 0|>| | | − | 2 (n+1)/2 For Qt(x), we have the easy estimates, Qt(x) 6 cn for x 6 2t and Qt(x) 6 A0(1+ x )− , | | n | | for x > 2 t , from which it is obvious that At 6 A(1 + t ) and hence (6.25) is proved. | | | | R R | | Since u(x t,y) = n P (s) f (x t s)ds, and n P (s)ds = 1, by Hölder inequality, we have − R y − − R y 1/µ 1/µ0 u(x t,y) P f µ P − 6k y k k y kµ0 Z 1/µ µ 1/µ 6 Py(s) f (x t s) ds = U (x t,y), Rn | − − | − where U is the Poisson integral of f µ . Apply (6.25) to U, it gives | | u(x t,y) A1/µ (1 + t /y)n/µ (M( f µ )(x))1/µ | − | 6 | | | | n/µ =Aµ (1 + t /y) Mµ f (x), | | and the Lemma is established. ut

6.2 Fourier multipliers on Lp

In this section, we introduce briefly the Fourier multipliers on Lp, and we prove two (or three) main multiplier theorems. In the study of PDEs, we often investigate the estimates of semigroups. For example, we consider the linear heat equation u ∆u = 0, u(0) = u . t − 0 1 t ωξ 2 It is clear that u = F − e− | | F u0 =: H(t)u0 is the solution of the above heat equation. The natural question is: Is H(t) a bounded semigroup from Lp to Lp? In other word, is the following inequality true? 1 t ωξ 2 F − e− | | F u u , for 1 p ∞. k 0kp . k 0kp 6 6 Of course, we have known that this estimate is true. From this example, we can give a general concept. Definition 6.13. Let ρ S . ρ is called a Fourier multiplier on Lp if the convolution (F 1ρ) ∈ 0 − ∗ f Lp for all f S , and if ∈ ∈ 1 ρ Mp = sup (F − ρ) f p k k f =1 k ∗ k k kp is finite. The linear space of all such ρ is denoted by Mp. Since S is dense in Lp (1 p < ∞), the mapping from S to Lp: f (F 1ρ) f can be 6 → − ∗ extended to a mapping from Lp to Lp with the same norm. We write (F 1ρ) f also for the − ∗ values of the extended mapping. - 122 - 6. The Littlewood-Paley g-function and Multipliers

For p = ∞ (as well as for p = 2) we can characterize Mp. Considering the map: 1 f (F − ρ) f for f S , → ∗ ∈ we have 1 ρ M∞ F − ρ f (0) C f ∞, f S . (6.30) ∈ ⇔ | ∗ | 6 k k ∈ Indeed, if ρ M∞, we have ∈ 1 1 F − ρ f ∞ F − ρ f (0) 6 k ∗ k f ∞ 6 C f ∞. | ∗ | f ∞ k k k k 1 k k On the other hand, if F − ρ f (0) 6 C f ∞, we can get 1 | ∗ | 1 k k 1 F − ρ f ∞ = sup F − ρ f (x) = sup [(F − ρ) ( f (x + ))](0) k ∗ k x Rn | ∗ | x Rn | ∗ · | ∈ ∈ C f (x + ) ∞ = C f ∞, 6 k · k k k which yields ρ M∞ 6 C, i.e., ρ M∞. k k ∈1 n But (6.30) also means that F − ρ is a bounded measure on R . Thus M∞ is equal to the space of all Fourier transforms of bounded measures. Moreover, ρ M∞ is equal to the total mass 1 k k of F − ρ. In view of the inequality above and the Hahn-Banach theorem, we may extend the mapping f F 1ρ f from S to L∞ to a mapping from L∞ to L∞ without increasing its norm. → − ∗ We also write the extended mapping as f F 1ρ f for f L∞. → − ∗ ∈ Theorem 6.14. Let 1 6 p 6 ∞ and 1/p + 1/p0 = 1, then we have

Mp = Mp0 (equal norms). (6.31) Moreover, 1 M1 = ρ S 0 : F − ρ is a bounded measure ∈ Z 1 1 (6.32) ρ M1 =total mass of F − ρ = F − ρ(x) dx k k Rn | | and ∞ M2 = L (equal norm). (6.33) For the norms (1 6 p0, p1 6 ∞) 1 θ θ ρ Mp 6 ρ M− ρ M , ρ Mp0 Mp1 (6.34) k k k k p0 k k p1 ∀ ∈ ∩ if 1/p = (1 θ)/p +θ/p (0 θ 1). In particular, the norm p decreases with p in the − 0 1 6 6 k·kM interval 1 6 p 6 2, and M M M M , (1 p q 2). (6.35) 1 ⊂ p ⊂ q ⊂ 2 6 6 6 p p Proof. Let f L , g L 0 and ρ Mp. Then, we have ∈ ∈ 1 ∈ 1 ρ M = sup (F − ρ) g p = sup (F − ρ) g(x), f ( x) p0 0 k k g =1 k ∗ k f p= g =1 |h ∗ − i| k kp0 k k k kp0 1 1 = sup (F − ρ) g f (0) = sup (F − ρ) f g(0) f p= g =1 | ∗ ∗ | f p= g =1 | ∗ ∗ | k k k kp0 k k k kp0 Z 1 = sup ((F − ρ) f )(y)g( y)dy n f p= g =1 | R ∗ − | k k k kp0 1 = sup (F − ρ) f p = ρ Mp . f =1 k ∗ k k k k kp The assertion (6.32) has already been established because of M1 = M∞. The Plancherel theorem immediately gives (6.33). In fact, 6.2. Fourier multipliers on Lp - 123 -

n/2 1 ω ˆ ρ M2 = sup (F − ρ) f 2 = sup | | ρ f 2 6 ρ ∞. k k f =1 k ∗ k f =1 2π k k k k k k2 k k2 On the other hand, for any ε > 0, we can choose a non-zero measurable set E such that ρ(ξ) > 2 | | ρ ∞ ε for ξ E. Then choose a function f L such that suppF f E, we can obtain k k − ∈ ∈ ⊂ ρ M2 > ρ ∞ ε. k k k k − 1 Invoking the Riesz-Thorin theorem, (6.34) follows, since the mapping f (F − ρ) f maps p p p p → ∗ L 0 L 0 with norm ρ M and L 1 L 1 with norm ρ M . → k k p0 → k k p1 Since 1/q = (1 θ)/p + θ/p for some θ and p q 2 p , by using (6.34) with p = p, − 0 6 6 6 0 0 p1 = p0, we see that ρ ρ , k kMq 6 k kMp from which (6.35) follows. ut Proposition 6.15. Let 1 6 p 6 ∞. Then Mp is a Banach algebra under pointwise multiplication.

Proof. It is clear that Mp is a norm. Note also that Mp is complete. Indeed, let ρk is a k · k ∞ ∞ { }∞ Cauchy sequence in Mp. So does it in L because of Mp L . Thus, it is convergent in L and ∞ 1 ⊂ 1 we denote the limit by ρ. From L S 0, we have F − ρkF f F − ρF f for any f S in ⊂ 1 → ∈ sense of the strong topology on S 0. On the other hand, F − ρkF f is also a Cauchy sequence in Lp S , and converges to a function g Lp. By the uniqueness of limit in S , we know that ⊂ 0 ∈ 0 g = F 1ρF f . Thus, ρ ρ 0 as k ∞. Therefore, M is a Banach space. − k k − kMp → → p Let ρ1 Mp and ρ2 Mp. For any f S , we have ∈ 1 ∈ 1 ∈ 1 1 (F − ρ ρ ) f = (F − ρ ) (F − ρ ) f ρ (F − ρ ) f k 1 2 ∗ kp k 1 ∗ 2 ∗ kp 6 k 1kMp k 2 ∗ kp ρ ρ f , 6k 1kMp k 2kMp k kp which implies ρ ρ M and 1 2 ∈ p ρ ρ ρ ρ . k 1 2kMp 6 k 1kMp k 2kMp Thus, M is a Banach algebra. p ut n In order to clarify the next theorem we write Mp = Mp(R ) for Fourier multipliers which n n are functions on R . The next theorem says that Mp(R ) is isometrically invariant under afﬁne transforms of Rn. n m 1 m Theorem 6.16. Let a : R R be a surjective afﬁne transform with n m, and ρ Mp(R ). → > ∈ Then

ρ(a( )) n = ρ m . k · kMp(R ) k kMp(R ) If m = n, the mapping a∗ is bijective. In particular, we have

ρ(c ) n = ρ( ) n , c = 0, (6.36) k · kMp(R ) k · kMp(R ) ∀ 6 ρ( x, ) n = ρ( ) , x = 0, (6.37) k h ·i kMp(R ) k · kMp(R) ∀ 6 where x,ξ = ∑n x ξ . h i i=1 i i Proof. It sufﬁces to consider the case that a : Rn Rm is a linear transform. Make the coordinate → transform ηi = ai(ξ), 1 6 i 6 m; η j = ξ j, m + 1 6 j 6 n, (6.38)

1 An afﬁne transform of Rn is a map F : Rn Rn of the form F(p) = Ap + q for all p Rn, where A is a linear transform of Rn and → ∈ q Rn. ∈ - 124 - 6. The Littlewood-Paley g-function and Multipliers which can be written as η = A 1ξ or ξ = Aη where detA = 0. Let A be the transposed matrix − 6 > of A. It is easy to see, for any f S (Rn), that ∈ n Z 1 ω ωixξ F − ρ(a(ξ))F f (x) = | | e ρ(a(ξ)) fˆ(ξ)dξ 2π Rn n Z ω ωix Aη = detA | | e · ρ(η1, ,ηm) fˆ(Aη)dη | | 2π Rn ··· ω n Z ωiA>x η = detA | | e · ρ(η1, ,ηm) fˆ(Aη)dη | | 2π Rn ··· 1 = detA (F − ρ(η , ,η ) fˆ(Aη))(A>x) | | 1 ··· m h 1 1 i = F − ρ(η , ,η ) F f ((A>)− ) (η) (A>x). 1 ··· m · m n It follows from ρ Mp(R ) that for any f S (R ) ∈1 ∈ F − ρ(a(ξ))F f k kp 1/p 1 1 = detA − F − ρ(η , ,η ) F f ((A>)− ) (η) | | k 1 ··· m · kp 1/p 1 1 = ρ(η , ,η ) (( ) ) p n m detA − Fη−1, ,ηm 1 m f A> − L (R ) | | ··· ··· ∗ k · k − Lp(Rm) ρ m f . 6k kMp(R )k kp Thus, we have

ρ(a( )) n ρ m . (6.39) k · kMp(R ) 6 k kMp(R ) Taking f ((A ) 1 ) = f (x , ,x ) f (x , ,x ), one can conclude that the inverse inequal- > − · 1 1 ··· m 2 m+1 ··· n ity (6.39) also holds. ut Now we give a simple but very useful theorem for Fourier multipliers. Theorem 6.17 (Bernstein multiplier theorem). Assume that k > n/2 is an integer, and that α 2 n n ∂ ρ L (R ), j = 1, ,n and 0 α k. Then we have ρ Mp(R ), 1 p ∞, and x j ∈ ··· 6 6 ∈ 6 6 n/2k n ! 1 n/2k k ρ ρ − ∂ ρ Mp . 2 ∑ x j 2 . k k k k j=1 k k

Proof. Let t > 0 and J(x) = ∑n x k. By the Cauchy-Schwartz inequality and the Plancherel j=1 | j| theorem, we obtain Z Z n 1ρ 1 1ρ n/2 k ∂ k ρ F − (x) dx = J(x)− J(x) F − (x) dx . t − ∑ x j 2. x >t | | x >t | | k k | | | | j=1 Similarly, we have Z 1 n/2 F − ρ(x) dx . t ρ 2. x t | | k k | |6 Choosing t such that ρ = t k ∑n ∂ k ρ , we infer, with the help of Theorem 6.14, that k k2 − j=1 k x j k2 n/2k Z n ! 1 1 n/2k k ρ Mp ρ M = F − ρ(x) dx ρ − ∂ ρ 2 . 6 1 n . 2 ∑ x j k k k k R | | k k j=1 k k This completes the proof. ut The ﬁrst application of the theory of the functions g and gλ∗ will be in the study of multipliers. Our main tool when proving theorems for the Sobolev and Besov spaces, deﬁned in the 6.2. Fourier multipliers on Lp - 125 - following chapters, is the following theorem. Note that 1 n/2 ∈ α \{ } ∂ is an integer. Assume also that for every differential monomial ∂ξ , α = (α1,α2,...,αn), with α = α1 + α2 + ... + αn, we have Mihlin’s condition | | α ∂ α ρ(ξ) A ξ −| |, whenever α k. (6.40) ∂ξ 6 | | | | 6 Then ρ M , 1 < p < ∞, and ∈ p ρ C A. k kMp 6 p,n The proof of the theorem leads to a generalization of its statement which we formulate as a corollary. Corollary 6.19 (Hörmander multiplier theorem). The assumption (6.40) can be replaced by the weaker assumptions, i.e., Hörmander’s condition ρ(ξ) A, | | 6 Z α 2 (6.41) 2 α n ∂ sup R | |− ρ(ξ) dξ 6A, α 6 k. 0

g1(F)(x) 6 Aλ gλ∗ ( f )(x), where λ = 2k/n. (6.42) Thus in view of the lemma, the g-functions and their variants are the characterizing expressions which deal at once with all the multipliers considered. On the other hand, the fact that the relation (6.42) is pointwise shows that to a large extent the mapping Tρ is “semi-local”. Proof of Theorem 6.18 and Corollary 6.19. The conclusion is deduced from the lemma as follows. Our assumption on k is such that λ = 2k/n > 1. Thus, Theorem 6.11 shows us that 2 p g∗ ( f )(x) A f , 2 p < ∞, if f L L . k λ kp 6 λ,pk kp 6 ∈ ∩ However, by Corollary 6.3, A F g (F)(x) , therefore by Lemma 6.20, 0pk kp 6 k 1 kp 2 p Tρ f = F A g∗ ( f )(x) A f , if 2 p < ∞ and f L L . k kp k kp 6 λ k λ kp 6 pk kp 6 ∈ ∩ That is, ρ M , 2 p < ∞. By duality, i.e., (6.31) of Theorem 6.14, we have also ρ M , ∈ p 6 ∈ p 1 < p 2, which gives the assertion of the theorem. 6 ut Now we shall prove Lemma 6.20. Proof of Lemma 6.20. Let u(x,y) denote the Poisson integral of f , and U(x,y) the Poisson integral of F. Then withˆdenoting the Fourier transform w.r.t. the x variable, we have ωξ y ωξ y ωξ y uˆ(ξ,y) = e−| | fˆ(ξ), and Ub(ξ,y) = e−| | Fb(ξ) = e−| | ρ(ξ) fˆ(ξ). n ω R ωix ξ ωξ y ωξ y | | Deﬁne M(x,y) = 2π Rn e · e−| | ρ(ξ)dξ. Then clearly Mb(ξ,y) = e−| | ρ(ξ), and so - 126 - 6. The Littlewood-Paley g-function and Multipliers

Ub(ξ,y1 + y2) = Mb(ξ,y1)uˆ(ξ,y2), y = y1 + y2, y1,y2 > 0. This can be written as Z U(x,y1 + y2) = M(t,y1)u(x t,y2)dt. Rn − We differentiate this relation k times w.r.t. y1 and once w.r.t. y2, and set y1 = y2 = y/2. This gives us the identity Z U(k+1)(x,y) = M(k)(t,y/2)u(1)(x t,y/2)dt. (6.43) Rn − Here the superscripts denote the differentiation w.r.t. y. Next, we translates the assumptions (6.40) (or (6.41)) on ρ in terms of M(x,y). The result is (k) n k M (t,y) 6A0y− − , (6.44) Z | | 2k (k) 2 n t M (t,y) dt 6A0y− . (6.45) Rn | | | | In fact, by the deﬁnition of M and the condition ρ(ξ) 6 A, it follows that n Z | | (k) ω k k ωξ y M (x,y) 6 | | ω ξ e−| | ρ(ξ)dξ | | 2π | | Rn | | n Z ∞ ω k k ω ry n 1 n k 6Aωn 1 | | ω r e−| | r − dr = A0y− − , − 2π | | 0 which is (6.44). To prove (6.45), let us show more particularly that Z α (k) 2 n x M (x,y) dx 6 A0y− , Rn | | where α = k. | | By Plancherel’s theorem n/2 α α (k) ω k ∂ k ωξ y x M (x,y) 2 = | | ω ( ξ ρ(ξ)e−| | ) . (6.46) k k 2π | | ∂ξ | | 2 So we need to evaluate, by using Leibniz’ rule, α β γ ∂ k ωξ y ∂ k ∂ ωξ y ( ξ ρ(ξ)e−| | ) = ∑ Cβ,γ ( ξ ρ(ξ)) e−| | . (6.47) ∂ξ | | β+γ=α ∂ξ | | ∂ξ Case I: (6.40) = (6.45). By the hypothesis (6.40) and Leibniz’ rule again, we have ⇒ β ∂ k k β ( ξ ρ(ξ)) 6 A0 ξ −| |, with β 6 k. ∂ξ | | | | | | Thus, α ∂ k ωξ y ( ξ ρ(ξ)e−| | ) ∂ξ | | k β γ ωξ y r r ωξ y C ξ −| |y| |e−| | C ξ y e−| | . 6 ∑ | | 6 ∑ | | β + γ =k 06r6k | | | | Since for r > 0 Z Z ∞ 2r 2r 2 ωξ y 2r 2r 2 ω Ry n 1 y ξ e− | | dξ =Cy R e− | | R − dR Rn | | 0 Z ∞ n 2r 2 ω z n 1 n =Cy− z e− | | z − dz 6 Cy− , 0 we get α (k) 2 n x M (x,y) A0y− , α = k, k k2 6 | | 6.2. Fourier multipliers on Lp - 127 - which proves the assertion (6.45). Case II: (6.41) = (6.45). From (6.46) and (6.47), we have, by Leibniz’ rule again and ⇒ (6.41), xα M(k)(x,y) k k2  2 2 1/2 Z β 0 β 00 ∂ k ∂ 2 ωξ y 2 γ C  ξ ρ(ξ) e− | | y | |dξ 6 ∑ n β + β + γ =k R ∂ξ | | ∂ξ | 0| | 00| | |  2 1/2 Z β 00 γ 2(k β ) ∂ 2 ωξ y 6C ∑ y| |  ∑ ξ −| 0| ρ(ξ) e− | | dξ jy ξ j+|1y | ∂ξ β + β + γ =k j Z 2 6 62 | 0| | 00| | | ∈ | | j+1 k β γ ω 2 jy2 6C ∑ ∑ (2 y) −| 0|y| |e−| | β + β + γ =k j Z | 0| | 00| | | ∈  2 1/2 Z β 00 j β +n/2 j 2 β n ∂ (2 y)−| 00| (2 y) | 00|− ρ(ξ) dξ · 2 jy ξ 2 j+1y ∂ξ 6| |6 j γ +n/2 γ ω 2 jy2 n/2 j 2 r ω 2 jy2 6CA ∑ ∑ (2 y)| | y| |e−| | = CAy− ∑ ∑ (2 y ) e−| | γ k j Z 06r6k j Z | |6 ∈ ∈ n/2 6Cy− , which yields (6.45). Now, we return to the identity (6.43), and for each y divide the range of integration into two parts, t y/2 and t > y/2. In the ﬁrst range, use the estimate (6.44) on M(k) and in the second | | 6 | | range, use the estimate (6.45). This together with Schwarz’ inequality gives immediately Z (k+1) 2 n 2k (1) 2 U (x,y) 6Cy− − u (x t,y/2) dt | | t y/2 | − | | |6 Z (1) 2 n u (x t,y/2) dt +Cy− | − 2k | t >y/2 t | | | | =:I1(y) + I2(y). Now Z ∞ 2 Z ∞ 2 (k+1) 2 2k+1 2k+1 (gk+1(F)(x)) = U (x,y) y dy 6 ∑ Ij(y)y dy. 0 | | j=1 0 However, by a change of variable y/2 y, Z ∞ Z →∞ Z 2k+1 (1) 2 n+1 I1(y)y dy 6C u (x t,y/2) y− dtdy 0 0 t 6y/2 | − | Z | | 2 n+1 2 6C ∇u(x t,y) y− dtdy = C(S( f )(x)) Γ | − | 2 6Cλ (gλ∗ ( f )(x)) . Similarly, with nλ = 2k, Z ∞ Z ∞ Z 2k+1 n+2k+1 2k 2 I2(y)y dy 6C y− t − ∇u(x t,y) dtdy 0 0 t >y | | | − | | | 2 6C(gλ∗ ( f )(x)) . - 128 - 6. The Littlewood-Paley g-function and Multipliers

This shows that gk+1(F)(x) 6 Cλ gλ∗ ( f )(x). However by Remark 6.4 (iii) of g-functions after Corollary 6.3, we know that g1(F)(x) 6 Ckgk+1(F)(x). Thus, the proof of the lemma is con- cluded. ut

6.3 The partial sums operators

We shall now develop the second main tool in the Littlewood-Paley theory, (the ﬁrst being the usage of the functions g and g∗). Let ρ denote an arbitrary rectangle in Rn. By rectangle we shall mean, in the rest of this chapter, a possibly inﬁnite rectangle with sides parallel to the axes, i.e., the Cartesian product of n intervals.

Deﬁnition 6.21. For each rectangle ρ denote by Sρ the partial sum operator, that is the multiplier operator with m = χρ = characteristic function of the rectangle ρ. So 2 n p n F (Sρ ( f )) = χρ fˆ, f L (R ) L (R ). (6.48) ∈ ∩ For this operator, we immediately have the following theorem. Theorem 6.22. 2 p Sρ ( f ) A f , f L L , k kp 6 pk kp ∈ ∩ if 1 < p < ∞. The constant Ap does not depend on the rectangle ρ. However, we shall need a more extended version of the theorem which arises when we replace complex-valued functions by functions taking their value in a Hilbert space. Let H be the sequence Hilbert space, ∞ 2 1/2 H = (c j) j=1 : (∑ c j ) = c < ∞ . { j | | | | } Then we can represent a function f Lp(Rn,H ), as sequences ∈ f (x) = ( f (x), , f (x), ), 1 ··· j ··· ∞ 2 1/2 where each f j is complex-valued and f (x) = (∑ j=1 f j(x) ) . Let ℜ be a sequence of rect- ∞ | | | | 2 n angle, ℜ = ρ j j=1. Then we can deﬁne the operator Sℜ, mapping L (R ,H ) to itself, by the rule

S ( f ) = (Sρ ( f ), ,Sρ ( f ), ), where f = ( f , , f , ). (6.49) ℜ 1 1 ··· j j ··· 1 ··· j ··· We ﬁrst give a lemma, which will be used in the proof of the theorem or its generalization. Recall the Hilbert transform f H( f ), which corresponds to the multiplier isgn(ω)sgn(ξ) → − in one dimension. 2 n p n Lemma 6.23. Let f (x) = ( f1(x), , f j(x), ) L (R ,H ) L (R ,H ). Denote He f (x) = ··· ··· ∈ ∩ (H f (x), ,H f (x), ). Then 1 ··· j ··· He f p Ap f p, 1 < p < ∞, k k 6 k k where Ap is the same constant as in the scalar case, i.e., when H is one-dimensional.

Proof. We use the vector-valued version of the Hilbert transform, as is described more generally in Sec. 4.7. Let the Hilbert spaces H1 and H2 be both identical with H . Take in R, K(x) = I · 1/πx, where I is the identity mapping on H . Then the kernel K(x) satisﬁes all the assumptions of Theorem 4.27 and Theorem 4.24. Moreover, 6.3. The partial sums operators - 129 - Z lim K(y) f (x y)dy = He f (x), ε 0 y ε − → | |> and so our lemma is proved. ut The generalization of Theorem 6.22 is then as follows. Theorem 6.24. Let f L2(Rn,H ) Lp(Rn,H ). Then ∈ ∩ S ( f ) A f , 1 < p < ∞, (6.50) k ℜ kp 6 pk kp where Ap does not depend on the family ℜ of rectangles.

Proof. The theorem will be proved in four steps, the first two of which already contain the essence of the matter. Step 1: n = 1, and the rectangles ρ1, ρ2, , ρ j, are the semi-infinite intervals ( ∞,0). 1 ··· ···1 1 sgn(ξ) − It is clear that S( ∞,0) f = F − χ( ∞,0)F f = F − − 2 F f , so − − I isgn(ω)H S( ∞,0) = − , (6.51) − 2 where I is the identity, and S( ∞,0) is the partial sum operator corresponding to the interval ( ∞,0). − − Now if all the rectangles are the intervals ( ∞,0), then by (6.51), − I isgn(ω)He S = − ℜ 2 and so by Lemma 6.23, we have the desired result. Step 2: n = 1, and the rectangles are the intervals ( ∞,a1), ( ∞,a2), , ( ∞,a j), . ωix a − − ··· − ··· Notice that F ( f (x)e− · ) = fˆ(ξ + a), therefore ωix a F (H(e− · f (x))) = isgn(ω)sgn(ξ) fˆ(ξ + a), − ωix a ωix a and hence F (e · H(e− · f (x))) = isgn(ω)sgn(ξ a) fˆ(ξ). From this, we see that − ωix−a ωix a f j isgn(ω)e · j H(e− · j f j) (S( ∞,a ) f j)(x) = − . (6.52) − j 2 ωix a If we now write symbolically e− · f for ωix a ωix a (e− · 1 f , ,e− · j f , ) 1 ··· j ··· with f = ( f , , f , ), then (6.52) may be written as 1 ··· j ··· ωix a ωix a f isgn(ω)e · He(e− · f ) S f = − , (6.53) ℜ 2 and so the result again follows in this case by Lemma 6.23. Step 3: General n, but the rectangles ρ j are the half-spaces x1 < a j, i.e., ρ j = x : x1 < a j . (1) 2 n { } Let S( ∞,a ) denote the operator defined on L (R ), which acts only on the x1 variable, by the − j action given by S( ∞,a ). We claim that − j (1) Sρ j = S( ∞,a ). (6.54) − j This identity is obvious for L2 functions of the product form

f 0(x ) f 00(x , ,x ), 1 2 ··· n since their linear span is dense in L2, the identity (6.54) is established. We now use the Lp inequality, which is the result of the previous step for each fixed x , x , , 2 3 ··· x . We raise this inequality to the pth power and integrate w.r.t. x , , x . This gives the desired n 2 ··· n result for the present case. Notice that the result holds as well if the half-space x : x < a ∞ , { 1 j} j=1 - 130 - 6. The Littlewood-Paley g-function and Multipliers is replaced by the half-space x : x > a ∞ , or if the role of the x axis is taken by the x axis, { 1 j} j=1 1 2 etc. Step 4: Observe that every general finite rectangle of the type considered is the intersection of 2n half-spaces, each half-space having its boundary hyperplane perpendicular to one of the axes of Rn. Thus a 2n-fold application of the result of the third step proves the theorem, where the family ℜ is made up of finite rectangles. Since the bounds obtained do not depend on the family ℜ, we can pass to the general case where ℜ contains possibly infinite rectangles by an obvious limiting argument. ut We state here the continuous analogue of Theorem 6.24. Let (Γ ,dγ) be a σ-finite measure space,2 and consider the Hilbert space H of square integrable functions on Γ , i.e., H = L2(Γ ,dγ). The elements p n f L (R ,H ) ∈ n are the complex-valued functions f (x,γ) = fγ (x) on R Γ , which are jointly measuable, and R R 2 p/2 1/p × for which ( n ( Γ f (x,γ) dγ) dx) = f p < ∞, if p < ∞. Let ℜ = ργ γ Γ , and suppose R | | k k { } ∈ that the mapping γ ργ is a measurable function from Γ to rectangles; that is, the numerical- → valued functions which assign to each γ the components of the vertices of ργ are all measurable. 2 n Suppose f L (R ,H ). Then we define F = Sℜ f by the rule ∈ F(x,γ) = Sργ ( fγ )(x), ( fγ (x) = f (x,γ)). Theorem 6.25. S f A f , 1 < p < ∞, (6.55) k ℜ kp 6 pk kp 2 n p n for f L (R ,H ) L (R ,H ), where the bound Ap does not depend on the measure space ∈ ∩ (Γ ,dγ), or on the function γ ργ . → Proof. The proof of this theorem is an exact repetition of the argument given for Theorem 6.24. The reader may also obtain it from Theorem 6.24 by a limiting argument. ut

6.4 The dyadic decomposition

We shall now consider a decomposition of Rn into rectangles. First, in the case of R, we decompose it as the union of the “disjoint” intervals (i.e., whose interiors are disjoint) [2k,2k+1], ∞ < k < ∞, and [ 2k+1, 2k], ∞ < k < ∞. This double − − − − collection of intervals, one collection for the positive half-line, the other for the negative half- line, will be the dyadic decomposition of R.3 Having obtained this decomposition of R, we take the corresponding product decomposition for Rn. Thus we write Rn as the union of “disjoint” rectangles, which rectangles are products of the intervals which occur for the dyadic decomposition of each of the axes. This is the dyadic decomposition of Rn. The family of resulting rectangles will be denoted by ∆. We recall the partial sum operator Sρ , deﬁned in (6.48) for each rectangle. Now in an obvious sense, (e.g. L2 convergence)

2 If µ is measure on a ring R, a set E is said to have σ-finite measure if there exists a sequence E of sets in R such that E { n} ⊂ ∞ E , and µ(E ) < ∞, n = 1,2, . If the measure of every set E in R is σ-finite, the measure µ is called σ-finite on R. ∪n=1 n n ··· 3 Strictly speaking, the origin is left out; but for the sake of simplicity of terminology, we still refer to it as the decomposition1 of R.

Fig. 6.3 The dyadic decomposition Figure 1: The dyadic decomposition 6.4. The dyadic decomposition - 131 -

∑ Sρ = Identity. ρ ∆ ∈ 2 Also in the L case, the different blocks, Sρ f , ρ ∆, behave ∈ as if they were independent; they are of course mutually orthogonal. To put the matter precisely: The L2 norm of f can be given 2 exactly in terms of the L norms of Sρ f , i.e., 2 2 ∑ Sρ f 2 = f 2, (6.56) ρ ∆ k k k k ∈ (and this is true for any decomposition of Rn). For the general Lp case not as much can be hoped for, but the following important theorem can nevertheless be established. Theorem 6.26 (Littlewood-Paley square function theorem). Suppose f Lp(Rn), 1 < p < ∞. ∈ Then 2 1/2 ( ∑ Sρ f (x) ) p f p. k ρ ∆ | | k ∼ k k ∈ The Rademacher functions provide a very useful device in p the study of L norms in terms of quadratic expressions. r1(t) These functions, r (t), r (t), , r (t), are deﬁned on the 0 1 ··· m ··· interval (0,1) as follows: 1 1 t 1, 0 t 1/2, 2 r (t) = 6 6 0 1, 1/2 < t < 1, r0(t) − r0 is extended outside the unit interval by periodicity, i.e., Fig. 6.4 r0(t) and r1(t) m Figure 1: r (t) and r (t) r0(t + 1) = r0(t). In general, rm(t) = r0(2 t). The sequences 0 1 of Rademacher functions are orthonormal (and in fact mutually independent) over [0,1]. In fact, for m < k, the integral Z 1 Z 1 Z 2m m k m k m rm(t)rk(t)dt = r0(2 t)r0(2 t)dt = 2− r0(s)r0(2 − s)ds 0 0 0 Z 1 Z 1/2 Z 1 k m k m k m = r0(s)r0(2 − s)ds = r0(2 − s)ds r0(2 − s)ds 0 0 − 1/2 " k m 1 k m # Z 2 − − Z 2 − m k =2 − r0(t)dt r0(t)dt k m 1 0 − 2 − − Z 1 Z 1 1 =2− r0(t)dt r0(t)dt = 0, 0 − 0 R 1 2 so, they are orthogonal. It is clear that they are normal since 0 (rm(t)) dt = 1. For our purposes, their importance arises from the following fact. Suppose ∑∞ a 2 < ∞ and set F(t) = ∑∞ a r (t). Then for every 1 < p < ∞, F(t) m=0 | m| m=0 m m ∈ Lp[0,1] and ∞ 2 1/2 Ap F p 6 F 2 = ( ∑ am ) 6 Bp F p, (6.57) k k k k m=0 | | k k for two positive constants Ap and Bp. Thus, for functions which can be expanded in terms of the Rademacher functions, all the Lp norms, 1 < p < ∞, are comparable. We shall also need the n-dimensional form of (6.57). We consider the unit cube Q Rn, ⊂ Q = t = (t ,t , ,t ) : 0 t 1 . Let m be an n-tuple of non-negative integers m = { 1 2 ··· n 6 j 6 } - 132 - 6. The Littlewood-Paley g-function and Multipliers

(m ,m , ,m ). Deﬁne r (t) = r (t )r (t ) r (t ). Write F(t) = ∑a r (t). With 1 2 ··· n m m1 1 m2 2 ··· mn n m m Z 1/p p F p = F(t) dt , k k Q | | we also have (6.57), whenever ∑ a 2 < ∞. That is | m| 2 Lemma 6.27. Suppose ∑ am < ∞. Then it holds | | / ∞ !1 2 2 F p F 2 = ∑ am , 1 < p < ∞. (6.58) k k ∼ k k m=0 | | Proof. We split the proof into four steps. Step 1: Let µ, a , a , , a , be real numbers. Then because the Rademacher functions are 0 1 ··· N mutually independent variables, we have, in view of their deﬁnition, Z 1 Z 1 Z 2m Z 1 µa r (t) µa r (2mt) m µa r (s) µa r (s) e m m dt = e m 0 dt = 2− e m 0 ds = e m 0 ds 0 0 0 0 1 µam µam =2− (e + e− ) = cosh µam. and for m < k Z 1 Z 1 m k eµamrm(t)eµakrk(t)dt = eµamr0(2 t)eµakr0(2 t)dt 0 0 Z 2m Z 1 m µa r (s) µa r (2k ms) µa r (s) µa r (2k ms) =2− e m 0 e k 0 − ds = e m 0 e k 0 − ds 0 0 Z 1/2 Z 1 µa µa r (2k ms) µa µa r (2k ms) = e m e k 0 − ds + e− m e k 0 − ds 0 1/2 " k m 1 k m # Z 2 − − Z 2 − m k µa µa r (t) µa µa r (t) =2 − e m e k 0 dt + e− m e k 0 dt k m 1 0 2 − − Z 1 Z 1 Z 1 1 µa µa µa r (t) µa r (t) µa r (t) =2− (e m + e− m ) e k 0 dt = e m m dt e k k dt. 0 0 0 Thus, by induction, we can verify Z 1 N Z 1 µ N ( ) µ ( ) e ∑m=0 amrm t dt = ∏ e amrm t dt. 0 m=0 0 x2 ∞ x2k If we now make use of this simple inequality coshx 6 e (since coshx = ∑k=0 (2k)! 6 2k 2 ∑∞ x = ex for x < ∞ by Taylor expansion), we obtain k=0 k! | | Z 1 N N µF(t) µ2a2 µ2 ∑N a2 e dt = ∏ cosh µam 6 ∏ e m = e m=0 m , 0 m=0 m=0 N with F(t) = ∑m=0 amrm(t). N 2 µ F µF Step 2: Let us make the normalizing assumption that ∑n=0 am = 1. Then, since e | | 6 e + µF e− , we have Z 1 µ F(t) µ2 e | |dt 6 2e . 0 Recall the distribution function F (α) = m t [0,1] : F(t) > α . If we take µ = α/2 in the ∗ { ∈ | | } above inequality, we have Z 2 Z 2 α2 2 α α F(t) α α F (α) = dt 6 e− 2 e 2 | |dt 6 e− 2 2e 22 6 2e− 4 . ∗ F(t) >α F(t) >α | | | | 6.4. The dyadic decomposition - 133 -

R ∞ b ax2 √ b+1 From Theorem 2.16, the above and the formula 0 x e− dx = Γ ((b + 1)/2)/2 a , it follows immediately that Z ∞ 1/p p 1 F p = p α − F (α)dα k k 0 ∗

Z ∞ 2 1/p p 1 α 1/p 6 2p α − e− 4 dα = 2(pΓ (p/2)) , 0 for 1 6 p < ∞, and so in general / ∞ !1 2 2 F p 6 Ap ∑ am , 1 6 p < ∞. (6.59) k k m=0 | | Step 3: We shall now extend the last inequality to several variables. The case of two variables is entirely of the inductive procedure used in the proof of the general case. We can also limit ourselves to the situation when p > 2, since for the case p < 2 the desired inequality is a simple consequence of Hölder’s inequality. (Indeed, for p < 2 and some q > 2, we have F Lp(0,1) 6 F Lq(0,1) 1 qp/(q p) 6 F Lq(0,1) by Hölder’s inequality.) k k k k k kL − (0,1) k k We have N N N

F(t1,t2) = ∑ ∑ am1m2 rm1 (t1)rm2 (t2) = ∑ Fm1 (t2)rm1 (t1). m1=0 m2=0 m1=0 By(6.59), it follows !p/2 Z 1 F(t ,t ) pdt Ap F (t ) 2 . | 1 2 | 1 6 p ∑| m1 2 | 0 m1 Integrating this w.r.t. t2, and using Minkowski’s inequlaity with p/2 > 1, we have !p/2 p/2 !p/2 Z 1 2 2 2 ∑ Fm1 (t2) dt2 = ∑ Fm1 (t2) 6 ∑ Fm1 (t2) p/2 0 | | | | k| | k m1 m1 p/2 m1 !p/2 = F (t ) 2 . ∑k m1 2 kp m1

However, Fm1 (t2) = ∑m2 am1m2 rm2 (t2), and therefore the case already proved shows that F (t ) 2 A2 a2 . k m1 2 kp 6 p ∑ m1m2 m2 Inserting this in the above gives !p/2 Z 1 Z 1 F(t ,t ) pdt dt Ap a2 , | 1 2 | 1 2 6 p ∑∑ m1m2 0 0 m1 m2 which leads to the desired inequality F A F , 2 p < ∞. k kp 6 pk k2 6 Step 4: The converse inequality F B F , p > 1 k k2 6 pk kp is a simple consequence of the direct inequality. In fact, for any p > 1, (here we may assume p < 2) by Hölder inequality 1/2 1/2 F 2 6 F p F . k k k k k kp0 We already know that F p 6 A0 F 2, p0 > 2. We therefore get k k 0 p0 k k - 134 - 6. The Littlewood-Paley g-function and Multipliers

2 F 2 6 (A0 ) F p, k k p0 k k which is the required converse inequality. ut Now, let us return to the proof of the Littlewood-Paley square function theorem. Proof of Theorem 6.26. It will be presented in ﬁve steps. Step 1: We show here that it sufﬁces to prove the inequality

1/2 2 ∑ Sρ f (x) 6 Ap f p, 1 < p < ∞, (6.60) ρ ∆ | | k k ∈ p for f L2(Rn) Lp(Rn). To see this sufﬁciency, let g L2(Rn) Lp0 (Rn), and consider the ∈ ∩ ∈ ∩ identity Z Z Sρ f Sρ gdx = f gdx¯ ∑ n n ρ ∆ R R ∈ which follows from (6.56) by polarization. By Schwarz’s inequality and then Hölder’s inequality 1 1 Z Z ! 2 ! 2 2 2 f gdx¯ 6 ∑ Sρ f ∑ Sρ g dx Rn Rn ρ | | ρ | | 1 1 ! 2 ! 2 2 2 Sρ f Sρ g . 6 ∑| | ∑| | ρ ρ p p0 Taking the supremum over all such g with the additional restriction that g 1, gives f k kp0 6 k kp for the l.h.s. of the above inequality. The r.h.s. is majorized by

21/2 Ap0 ∑ Sρ f , | | p since we assume (6.60) for all p. Thus, we have also

!1/2 2 Bp f p 6 ∑ Sρ f . (6.61) k k ρ | | p To dispose of the additional assumption that f L2, for f Lp take f L2 Lp such that ∈ ∈ j ∈ ∩ f f 0; use the inequality (6.60) and (6.61) for f and f f ; after a simple limiting k j − kp → j j − j0 argument, we get (6.60) and (6.61) for f as well. Step 2: Here we shall prove the inequality (6.60) for n = 1. We shall need first to introduce a little more notations. We let ∆1 be the family of dyadic intervals in R, we can enumerate them as I0,I1, , Im, (the order is here im- ··· ··· material). For each I ∆ , we consider the partial sum operator S , and a modification ∈ 1 I of it that we now define. Let ϕ C1 be a fixed function with the following properties: ∈ 1, 1 6 ξ 6 2, ϕ(ξ) ϕ(ξ) = 1 0, ξ 6 1/2, or ξ > 4. Suppose I is any dyadic interval, and assume that it is k k+1 1 2 3 4 ξ of the form [2 ,2 ]. Define S˜I by k F (S˜I f )(ξ) = ϕ(2− ξ) fˆ(ξ) = ϕI(ξ) fˆ(ξ). (6.62) Fig. 6.5 ϕ(ξ) Figure 1: ϕ(ξ) That is, S˜I, like SI, is a multiplier transform where the multiplier is equal to one on the interval I; but unlike SI, the multiplier of S˜I is smooth. 6.4. The dyadic decomposition - 135 -

A similar deﬁnition is made for S˜ when I = [ 2k+1, 2k]. We observe that I − − SIS˜I = SI, (6.63) since SI has as multiplier the characteristic function of I. Now for each t [0,1], consider the multiplier transform ∈ ∞ ˜ ˜ Tt = ∑ rm(t)SIm . m=0 That is, for each t, T˜t is the multiplier transform whose multiplier is mt(ξ), with ∞

mt(ξ) = ∑ rm(t)ϕIm (ξ). (6.64) m=0

By the deﬁnition of ϕIm , it is clear that for any ξ at most three terms in the sum (6.64) can be non-zero. Moreover, we also see easily that

dmt B m (ξ) B, (ξ) , (6.65) | t | 6 dξ 6 ξ | | where B is independent of t. Thus, by the Mihlin multiplier theorem (Theorem 6.18) T˜ f A f , for f L2 Lp, (6.66) k t kp 6 pk kp ∈ ∩ and with Ap independent of t. From this, it follows obviously that Z 1 1/p p T˜t f pdt 6 Ap f p. 0 k k k k However, by Lemma 6.27 about the Rademacher functions, Z 1 Z 1 Z ˜ p ˜ p Tt f pdt = ∑rm(t)(SIm f )(x) dxdt 0 k k 0 R1 Z p/2 ˜ 2 >A0p ∑ SIm f (x) dx. R1 m | | Thus, we have

1/2 S˜ ( f ) 2 B f . (6.67) ∑| Im | 6 pk kp m p Now using (6.63), applying the general theorem about partial sums, Theorem 6.24, with ˜ ˜ ˜ ℜ = ∆1 here and (6.67), we get, for F = (SI0 f ,SI1 f , ,SIm f , ), ··· ··· 1/2 1/2 S ( f ) 2 = S S˜ ( f ) 2 = S F ∑| Im | ∑| Im Im | k ∆1 kp m p m p

1/2 A F = A S˜ ( f ) 2 A B f = C f , (6.68) 6 pk kp p ∑| Im | 6 p pk kp pk kp m p which is the one-dimensional case of the inequality (6.60), and this is what we had set out to prove. Step 3: We are still in the one-dimensional case, and we write Tt for the operator

Tt = ∑rm(t)SIm . m Our claim is that Tt f p 6 Ap f p, 1 < p < ∞, (6.69) k kLt,x k k with A independent of t, and f L2 Lp. p ∈ ∩ - 136 - 6. The Littlewood-Paley g-function and Multipliers

N N N Write Tt = ∑m=0 rm(t)SIm , and it suffices to show that (6.69) holds, with Tt in place of Tt 2 p (and Ap independent of N and t). Since each SIm is a bounded operator on L and L , we have that T N f L2 Lp and so we can apply (6.61) to it, which has already been proved in the case t ∈ ∩ n = 1. So 1/2 N ! N 2 B T f p S f C f , p t Lt,x 6 ∑ Im 6 p p k k m=0 | | k k p by using (6.68). Letting N ∞, we get (6.69). → (1) Step 4: We now turn to the n-dimensional case and define Tt1 , as the operator Tt1 acting only on the x1 variable. Then, by the inequality (6.69), we get Z 1 Z Z (1) p p p Tt1 f (x1,x2, ,xn) dx1dt1 6 Ap f (x1, ,xn) dx1, (6.70) 0 R1 | ··· | R1 | ··· | 2 1 p 1 for almost every fixed x2,x3, ,xn, since x1 f (x1,x2, ,xn) L (R ) L (R ) for almost ···2 n p n → ··· ∈ ∩ every fixed x2, ,xn, if f L (R ) L (R ). If we integrate (6.70) w.r.t. x2, ,xn, we obtain ··· ∈ ∩ ··· (1) p 2 p Tt1 f L 6 Ap f p, f L L , (6.71) k k t1,x k k ∈ ∩ with Ap independent of t1. The same inequality of course holds with x1 replaced by x2, or x3, etc. Step 5: We first describe the additional notation we shall need. With ∆ representing the col- n lection of dyadic rectangles in R , we write any ρ ∆, as ρ = Im Im Imn where ∈ 1 × 2 × ··· × I ,I , ,I , represents the arbitrary enumeration of the dyadic intervals used above. Thus 0 1 ··· m ··· if m = (m1,m2, ,mn), with each m j > 0, we write ρm = Im1 Im2 Imn . ··· (1) × × ··· × We now apply the operator Tt1 for the x1 variable, and successively its analogues for x2,x3, etc. The result is n Tt f p 6 Ap f p. (6.72) k kLt,x k k Here

Tt = ∑ rm(t)Sρm ρ ∆ m∈ with r (t) = r (t ) r (t ) as described in the previous. The inequality holds uniformly for m m1 1 ··· mn n each (t ,t , ,t ) in the unit cube Q. 1 2 ··· n We raise this inequality to the pth power and integrate it w.r.t. t, making use of the properties of the Rademacher functions, i.e., Lemma 6.27. We then get, as in the analogous proof of (6.67), that

!1/2 2 Sρ f A f , ∑ | m | 6 pk kp ρm ∆ ∈ p if f L2(Rn) Lp(Rn). This together with the ﬁrst step concludes the proof of Theorem 6.26. ∈ ∩ ut

6.5 The Marcinkiewicz multiplier theorem

We now present another multiplier theorem which is one of the most important results of the whole theory. For the sake of clarity, we state ﬁrst the one-dimensional case. 6.5. The Marcinkiewicz multiplier theorem - 137 -

Theorem 6.28. Let m be a bounded function on R1, which is of bounded variation on every finite interval not containing the origin. Suppose (a) m(ξ) B, ∞ < ξ < ∞, | | 6 − (b) R m(ξ) dξ B, for every dyadic interval I. I | | 6 Then m M , 1 < p < ∞; and more precisely, if f L2 Lp, ∈ p ∈ ∩ T f A f , k m kp 6 pk kp where Ap depends only on B and p. To present general theorem, we consider R as divided into its two half-lines, R2 as divided into its four quadrants, and generally Rn as divided into its 2n “octants”. Thus, the first octants in Rn will be the open “rectangle” of those ξ all of whose coordinates are strictly positive. We shall assume that m(ξ) is defined on each such octant and is there continuous together with its partial derivatives up to and including order n. Thus m may be left undefined on the set of points where one or more coordinate variables vanishes. For every k 6 n, we regard Rk embedded in Rn in the following obvious way: Rk is the subspace of all points of the form (ξ ,ξ , ,ξ ,0, ,0). 1 2 ··· k ··· Theorem 6.29 (Marcinkiewicz’ multiplier theorem). Let m be a bounded function on Rn that is Cn in all 2n “octant”. Suppose also (a) m(ξ) B, | | 6 (b) for each 0 < k 6 n,

Z ∂ km sup dξ1 dξk 6 B ξ , ,ξ ρ ∂ξ1∂ξ2 ∂ξk ··· k+1 ··· n ··· as ρ ranges over dyadic rectangles of Rk. (If k = n, the “sup” sign is omitted.) (c) The condition analogous to (b) is valid for every one of the n! permutations of the variables ξ ,ξ , ,ξ . 1 2 ··· n Then m M , 1 < p < ∞; and more precisely, if f L2 Lp, T f A f , where A ∈ p ∈ ∩ k m kp 6 pk kp p depends only on B, p and n.

Proof. It will be best to prove Theorem 6.29 in the case n = 2. This case is already completely typical of the general situation, and in doing only it we can avoid some notational complications. 2 2 p 2 Let f L (R ) L (R ), and write F = Tm f , that is F (F(x)) = m(ξ) fˆ(ξ). ∈ ∩ Let ∆ denote the dyadic rectangles, and for each ρ ∆, write fρ = Sρ f , Fρ = Sρ F, thus ∈ Fρ = Tm fρ . In view of Theorem 6.26, it suffices to show that 1/2 1/2 2 2 ∑ Fρ 6 Cp ∑ fρ . (6.73) ρ ∆ | | p ρ ∆ | | p ∈ ∈ The rectangles in ∆ come in four sets, those in the first, the second, the third, and fourth quadrants, respectively. In estimating the l.h.s. of (6.73), consider the rectangles of each quadrant separately, and assume from now on that our rectangles belong to the first quadrant. We will express Fρ in terms of an integral involving fρ and the partial sum operators. That this is possible is the essential idea of the proof. Fix ρ and assume ρ = (ξ ,ξ ) : 2k ξ 2k+1,2l ξ 2l+1 . Then, for (ξ ,ξ ) ρ, it { 1 2 6 1 6 6 2 6 } 1 2 ∈ is easy to verify the identity Z ξ Z ξ 2 Z ξ 2 1 ∂ m(t1,t2) 1 ∂ l m(ξ1,ξ2) = dt1dt2 + m(t1,2 )dt1 2l 2k ∂t1∂t2 2k ∂t1 - 138 - 6. The Littlewood-Paley g-function and Multipliers

Z ξ 2 ∂ k k l + m(2 ,t2)dt2 + m(2 ,2 ). 2l ∂t2 Now let S denote the multiplier transform corresponding to the rectangle (ξ ,ξ ) : 2k+1 > t { 1 2 ξ > t , 2l+1 > ξ > t . Similarly, let S(1) denote the multiplier corresponding to the interval 1 1 2 2} t1 2k+1 > ξ > t , similarly for S(2). Thus in fact, S = S(1) S(2). Multiplying both sides of the above 1 1 t2 t t1 · t2 equation by the function χρ fˆ and taking inverse Fourier transforms yields, by changing the (1) (1) order of integrals in view of Fubini’s theorem and the fact that Sρ Tm f = Fρ , and St1 Sρ = St1 , (2) (2) St2 Sρ = St2 , StSρ = St, we have 1 Fρ =TmSρ f = F − mχρ fˆ n Z Z ξ Z ξ 2 ω ωix ξ h 2 1 ∂ m(t1,t2) i = | | e · dt1dt2χρ (ξ) fˆ(ξ) dξ 2π R2 2l 2k ∂t1∂t2 n Z Z ξ ω ωix ξ h 1 ∂ l i + | | e · m(t1,2 )dt1χρ (ξ) fˆ(ξ) dξ 2π R2 2k ∂t1 n Z Z ξ ω ωix ξ h 2 ∂ k i + | | e · m(2 ,t2)dt2χρ (ξ) fˆ(ξ) dξ 2π R2 2l ∂t2 1 k l + F − m(2 ,2 )χρ (ξ) fˆ(ξ) n Z Z 2l+Z1 2k+1 2 ω ωix ξ ∂ m(t1,t2) = e χ k (t )χ l (t )dt dt | | · [2 ,ξ1] 1 [2 ,ξ2] 2 1 2 2π R2 2l 2k ∂t1∂t2 χρ (ξ) fˆ(ξ)dξ · n Z Z 2k+1 ω ωix ξ ∂ l + e m(t , )χ k (t )dt χ (ξ) fˆ(ξ)dξ | | · 1 2 [2 ,ξ1] 1 1 ρ 2π R2 2k ∂t1 n Z Z 2l+1 ω ωix ξ ∂ k + e m( ,t )χ l (t )dt χ (ξ) fˆ(ξ)dξ | | · 2 2 [2 ,ξ2] 2 2 ρ 2π R2 2l ∂t2 k l + m(2 ,2 ) fρ n Z 2l+Z1 2k+Z1 ω ωix ξ = e χ k+1 (ξ )χ l+1 (ξ )χ (ξ) fˆ(ξ)dξ | | · [t1,2 ] 1 [t2,2 ] 2 ρ 2π 2l 2k R2 2 ∂ m(t1,t2) dt1dt2 · ∂t1∂t2 n Z 2k+1 Z ω ωix ξ ∂ l + e χ k+1 (ξ )χ (ξ) fˆ(ξ)dξ m(t , )dt | | · [t1,2 ] 1 ρ 1 2 1 2π 2k R2 ∂t1 n Z 2l+1 Z ω ωix ξ ∂ k + e χ l+1 (ξ )χ (ξ) fˆ(ξ)dξ m( ,t )dt | | · [t2,2 ] 2 ρ 2 2 2 2π 2l R2 ∂t2 k l + m(2 ,2 ) fρ Z 2 Z 2k+1 ∂ m(t1,t2) (1) ∂ l = St fρ dt1dt2 + St1 fρ m(t1,2 )dt1 ρ ∂t1∂t2 2k ∂t1 Z 2l+1 (2) ∂ k k l + St2 fρ m(2 ,t2)dt2 + m(2 ,2 ) fρ . 2l ∂t2 We apply the Cauchy-Schwarz inequality in the ﬁrst three terms of the above w.r.t. the measures ∂ ∂ m(t ,t ) dt dt , ∂ m(t ,2l) dt , ∂ m(2k,t ) dt , respectively, and we use the assump- | t1 t2 1 2 | 1 2 | t1 1 | 1 | t2 2 | 2 tions of the theorem to deduce 6.5. The Marcinkiewicz multiplier theorem - 139 -

Z 2 Z 2 2 2 ∂ m ∂ m Fρ . St fρ dt1dt2 dt1dt2 | | ρ | | ∂t1∂t2 ρ ∂t1∂t2 Z 2k+1 Z 2k+1 (1) 2 ∂ l ∂ l + St1 fρ m(t1,2 ) dt1 m(t1,2 ) dt1 2k | | ∂t1 2k ∂t1 Z 2l+1 Z 2l+1 (2) 2 ∂ k ∂ k + St2 fρ m(2 ,t2) dt2 m(2 ,t2) dt2 2l | | ∂t2 2l ∂t2 k l 2 2 + m(2 ,2 ) fρ | | | | Z 2 Z l 2 ∂ m (1) 2 ∂m(t1,2 ) 6B0 St fρ dt1dt2 + St1 fρ dt1 ρ | | ∂t1∂t2 I1 | | ∂t1 Z k (2) 2 ∂m(2 ,t2) 2 + St2 fρ dt2 + fρ I2 | | ∂t2 | | =ℑ1 + ℑ2 + ℑ3 + ℑ4 , with ρ = I I . ρ ρ ρ ρ 1 × 2 2 1/2 To estimate (∑ Fρ ) , we estimate separately the contributions of each of the four terms k ρ | | kp on the r.h.s. of the above inequality by the use of Theorem 6.25. To apply that theorem in the 2 1 ∂ m(t1,t2) case of ℑρ we take for Γ the ﬁrst quadrant, and dγ = dt1dt2, the functions γ ργ are | ∂t1∂t2 | → constant on the dyadic rectangles. Since for every rectangle, Z Z 2 ∂ m(t1,t2) dγ = dt1dt2 6 B, ρ ρ ∂t1∂t2 then

!1/2 !1/2 1 2 ∑ ℑρ 6 Cp ∑ fρ . ρ | | ρ | | p p Similarly, for ℑ2 , ℑ3 and ℑ4 , which concludes the proof. ρ ρ ρ ut

Chapter 7 Sobolev and Hölder Spaces

7.1 Riesz potentials and fractional integrals

Let f be a sufficiently smooth function which is small at infinity, then the Fourier transform of its Laplacean ∆ f is F ( ∆ f )(ξ) = ω2 ξ 2 fˆ(ξ). (7.1) − | | From this, we replace the exponent 2 in ξ 2 by a general exponent s, and thus to define (at | | least formally) the fractional power of the Laplacean by s/2 1 s ( ∆) f = F − (( ω ξ ) fˆ(ξ)). (7.2) − | || | Of special significance will be the negative powers s in the range n < s < 0. In general, with − a slight change of notation, we can define Definition 7.1. Let s > 0. The Riesz potential of order s is the operator s/2 I = ( ∆)− . (7.3) s − For 0 < s < n, Is is actually given in the form Z 1 n+s Is f (x) = x y − f (y)dy, (7.4) γ(s) Rn | − | with πn/22sΓ (s/2) γ(s) = . Γ ((n s)/2) − The formal manipulations have a precise meaning. Lemma 7.2. Let 0 < s < n. (a) The Fourier transform of the function x n+s is the function γ(s)( ω ξ ) s, in the sense | |− | || | − that Z Z n+s s x − ϕ(x)dx = γ(s)( ω ξ )− ϕˆ(ξ)dξ, (7.5) Rn | | Rn | || | whenever ϕ S . ∈ s (b) The identity F (Is f ) = ( ω ξ )− fˆ(ξ) holds in the sense that Z | || | Z s Is f (x)g(x)dx = fˆ(ξ)( ω ξ )− gˆ(ξ)dξ Rn Rn | || | whenever f ,g S . ∈ Proof. Part (a) is merely a restatement of Lemma 5.14 since γ(s) = ω sγ . | | 0,s Part (b) follows immediately from part (a) by writing

141 - 142 - 7. Sobolev and Hölder Spaces

Z Z 1 n+s s \ Is f (x) = f (x y) y − dy = ( ω ξ )− f (x )dξ γ(s) Rn − | | Rn | || | − · Z Z s ωiξ x s ωiξ x = ( ω ξ )− fˆ(ξ)e · dξ = ( ω ξ )− fˆ(ξ)e− · dξ, Rn | || | Rn | || | so Z Z Z s ωiξ x Is f (x)g(x)dx = ( ω ξ )− fˆ(ξ)e− · dξg(x)dx Rn Rn Rn | || | Z s = ( ω ξ )− fˆ(ξ)gˆ(ξ)dξ. Rn | || | This completes the proof. ut Now, we state two further identities which can be obtained from Lemma 7.2 and which reflect essential properties of the potentials Is. I (I f ) = I f , f S , s,t > 0, s +t < n. (7.6) s t s+t ∈ ∆(Is f ) = Is(∆ f ) = Is 2 f , f S , n > 3, 2 6 s 6 n. (7.7) − − ∈ The deduction of these two identities have no real difficulties, and these are best left to the interested reader to work out. A simple consequence of (7.6) is the n-dimensional variant of the beta function,1 Z n+s n+t γ(s)γ(t) n+(s+t) x y − y − dy = x − (7.8) Rn | − | | | γ(s +t) | | with s,t > 0 and s+t < n. Indeed, for any ϕ S , we have, by the definition of Riesz potentials ∈ and (7.6), that ZZ n+s n+t x y − y − dyϕ(z x)dx Rn Rn | − | | | − Z × Z n+t n+s = y − x y − ϕ(z y (x y))dxdy Rn| | Rn | − | − − − Z n+t = y − γ(s)Isϕ(z y)dy = γ(s)γ(t)It(Isϕ)(z) = γ(s)γ(t)Is+tϕ(z) Rn | | − Z γ(s)γ(t) n+(s+t) = x − ϕ(z x)dx. γ(s +t) Rn | | − By the arbitrariness of ϕ, we have the desired result. We have considered the Riesz potentials formally and the operation for Schwartz functions. But since the Riesz potentials are integral operators, it is natural to inquire about their actions on the spaces Lp(Rn). For this reason, we formulate the following problem. Given s (0,n), for what pairs p and p n q n ∈ q, is the operator f Is f bounded from L (R ) to L (R )? That is, when do we have the → inequality I f A f ? (7.9) k s kq 6 k kp There is a simple necessary condition, which is merely a reflection of the homogeneity of the kernel (γ(s)) 1 y n+s. In fact, we have − | |− Proposition 7.3. If the inequality (7.9) holds for all f S and a finite constant A, then 1/q = ∈ 1/p s/n. −

1 The beta function, also called the Euler integral of the ﬁrst kind, is a special function deﬁned by B(x,y) = R 1 tx 1(1 t)y 1dt for 0 − − − ℜx > 0 and ℜy > 0. It has the relation with Γ -function: B(x,y) = Γ (x)Γ (y)/Γ (x + y). 7.1. Riesz potentials and fractional integrals - 143 -

Proof. Let us consider the dilation operator δε , deﬁned by δε f (x) = f (εx) for ε > 0. Then clearly, for ε > 0 Z 1 1 n+s (δ 1 I δ f )(x) = ε− x y − f (εy)dy ε− s ε γ(s) Rn | − | Z z=εy n 1 1 n+s ==ε− ε− (x z) − f (z)dz γ(s) Rn | − | s =ε− Is f (x). (7.10) Also n/p n/q δε f p = ε− f p, δ 1 Is f q = ε Is f q. (7.11) k k k k k ε− k k k Thus, by (7.9) s s+n/q Is f q =ε δ 1 Isδε f q = ε Isδε f q k k k ε− k k k s+n/q s+n/q n/p Aε δε f = Aε − f . 6 k kp k kp If I f = 0, then the above inequality implies k s kq 6 1/q = 1/p s/n. (7.12) − If f = 0 is non-negative, then I f > 0 everywhere and hence I f > 0, and we can conclude 6 s k s kq the desired relations. ut Next, we observe that the inequality must fail at the endpoints p = 1 (then q = n/(n s)) and − q = ∞ (then p = n/s). Let us consider the case p = 1. It is not hard to see that the presumed inequality

Is f n/(n s) 6 A f 1, (7.13) k k − k k cannot hold. In fact, we can choose a nice positive function ϕ L1 with R ϕ = 1 and a compact n ∈+ support. Then, with ϕε (x) = ε− ϕ(x/ε), we have that as ε 0 , 1 n→+s I (ϕε )(x) (γ(s))− x − . s → | | 2 If Isϕε n/(n s) 6 A ϕε 1 = A were valid uniformly as ε, then Fatou’s lemma will imply that k k − k k Z n x − dx < ∞, Rn | | and this is a contradiction. The second atypical case occurs when q = ∞. Again the inequality of the type (7.9) cannot hold, and one immediate reason is that this case is dual to the case p = 1 just considered. The failure at q = ∞ may also be seen directly as follows. Let f (x) = x s(ln1/ x ) (1+ε)s/n, for | |− | | − x 1/2, and f (x) = 0, for x > 1/2, where ε is positive but small. Then f Ln/s(Rn), since | | 6 | | ∈ f n/s = R x n(ln1/ x ) 1 ε dx < ∞. However, I f is essentially unbounded near the k kn/s x 61/2 | |− | | − − s origin since| | Z 1 n (1+ε)s/n Is f (0) = x − (ln1/ x )− dx = ∞, γ(s) x 1/2 | | | | | |6 as long as (1 + ε)s/n 6 1. After these observations, we can formulate the following Hardy-Littlewood-Sobolev theorem of fractional integration. The result was ﬁrst considered in one dimension on the circle by Hardy and Littlewood. The n-dimensional result was considered by Sobolev.

2 Fatou’s lamma: If f is a sequence of nonnegative measurable functions, then { k} Z Z liminf fkdµ 6 liminf fkdµ. k ∞ k ∞ → → - 144 - 7. Sobolev and Hölder Spaces

Theorem 7.4 (Hardy-Littlewood-Sobolev theorem of fractional integrations). Let 0 < s < n, 1 6 p < q < ∞, 1/q = 1/p s/n. p n − (a) If f L (R ), then the integral (7.4), deﬁning Is f , converges absolutely for almost every ∈ x. (b) If, in addition, p > 1, then Is f q 6 Ap,q f p. 1 n k k k k 1 q (c) If f L (R ), then m x : Is f (x) > α (Aα− f 1) , for all α > 0. That is, the map- ∈ { | | } 6 k k ping f I f is of weak type (1,q), with 1/q = 1 s/n. → s − Proof. We ﬁrst prove parts (a) and (b). Let us write Z Z n+s n+s γ(s)Is f (x) = x y − f (y)dy + x y − f (y)dy B(x,δ) | − | Rn B(x,δ) | − | \ =:Lδ (x) + Hδ (x). j ( j+1) Divide the ball B(x,δ) into the shells E j := B(x,2− δ) B(x,2− δ), j = 0,1,2,..., thus \ ∞ Z ∞ Z n+s n+s Lδ (x) 6 ∑ x y − f (y)dy 6 ∑ x y − f (y) dy | | j=0 E j | − | j=0 E j | − | | | ∞ Z ( j+1) n+s 6 ∑ (2− δ)− f (y) dy j=0 E j | | ∞ Z ( j+1) n+s (2− δ)− f (y) dy 6 ∑ j j=0 B(x,2− δ) | | ∞ (2 ( j+1)δ) n+sm(B(x,2 jδ)) Z = − − − f (y) dy ∑ j j j=0 m(B(x,2− δ)) B(x,2− δ) | | ∞ (2 ( j+1)δ) n+sV (2 jδ)n Z = − − n − f (y) dy ∑ j j j=0 m(B(x,2− δ)) B(x,2− δ) | | ∞ s n s n s s j Vnδ 2 6Vnδ 2 − 2− M f (x) = M f (x). ∑ 2s 1 j=0 − Now, we derive an estimate for Hδ (x). By Hölder’s inequality and the condition 1/p > s/n (i.e., q < ∞), we obtain Z 1/p0 ( n+s)p0 Hδ (x) 6 f p x y − dy | | k k Rn B(x,δ) | − | \ Z Z ∞ 1/p0 ( n+s)p0 n 1 = f p r − r − drdσ n 1 k k S − δ Z ∞ 1/p0 1/p0 ( n+s)p0+n 1 =ωn 1 f p r − − dr − k k δ 1/p0 ωn 1 n/p (n s) s n/p = − δ 0− − f = C(n,s, p)δ − f . (n s)p n k kp k kp − 0 − By the above two inequalities, we have s s n/p γ(s)I f (x) C(n,s)δ M f (x) +C(n,s, p)δ − f =: F(δ). | s | 6 k kp Choose δ = C(n,s, p)[ f /M f ]p/n, such that the two terms of the r.h.s. of the above are equal, k kp i.e., the minimizer of F(δ), to get 1 ps/n ps/n γ(s)I f (x) C(M f ) − f . | s | 6 k kp 7.2. Bessel potentials - 145 -

Therefore, by part (i) of Theorem 3.9 for maximal functions, i.e., M f is ﬁnite almost everywhere if f Lp (1 p ∞), it follows that I f (x) is ﬁnite almost everywhere, which proves ∈ 6 6 | s | part (a) of the theorem. By part (iii) of Theorem 3.9, we know M f A f (1 < p ∞), thus k kp 6 pk kp 6 1 ps/n ps/n I f C M f − f = C f . k s kq 6 k kp k kp k kp This gives the proof of part (b). n+s n+s Finally, we prove (c). Since we also have Hδ (x) 6 f 1δ − , taking α = f 1δ − , i.e., 1/(n s) | | k k k k δ = ( f 1/α) − , by part (ii) of Theorem 3.9, we get k k 1 m x : I f (x) > 2(γ(s))− α { | s | } m x : L (x) > α + m x : H (x) > α 6 { | δ | } { | δ | } m x : Cδ sM f (x) > α + 0 6 { | | } C n/(n s) q 6 s f 1 = C[ f 1/α] − = C[ f 1/α] . δ − α k k k k k k This completes the proof of part (c). ut

7.2 Bessel potentials

While the behavior of the kernel (γ(s)) 1 x n+s as x 0 is well suited for their smoothing − | |− | | → properties, their decay as x ∞ gets worse as s increases. | | → We can slightly adjust the Riesz potentials such that we maintain their essential behavior near zero but achieve exponential decay at infinity. The simplest way to achieve this is by replacing the “nonnegative” operator ∆ by the “strictly positive” operator I ∆, where I = identity. − − Here the terms nonnegative and strictly positive, as one may have surmised, refer to the Fourier transforms of these expressions. Definition 7.5. Let s > 0. The Bessel potential of order s is the operator s/2 J = (I ∆)− s − whose action on functions is given by 1 Js f = F − GcsF f = Gs f , ∗ where 1 2 2 s/2 G (x) = F − ((1 + ω ξ )− )(x). s | | Now we give some properties of Gs(x) and show why this adjustment yields exponential decay for Gs at infinity. Proposition 7.6. Let s > 0. 2 x s n 1 R ∞ t | | dt (a) G (x) = e e 4t t −2 . s (4π)n/2Γ (s/2) 0 − − t n 1 n R (b) Gs(x) > 0, x R ; and Gs(x) L (R ), precisely, n Gs(x)dx = 1. ∀ ∈ ∈ R (c) There exist two constants 0 < C(s,n),c(s,n) < ∞ such that x /2 G (x) C(s,n)e−| | , when x 2, s 6 | | > and such that 1 Gs(x) 6 6 c(s,n), when x 6 2, c(s,n) Hs(x) | | where Hs is a function that satisfies - 146 - 7. Sobolev and Hölder Spaces

 s n s n+2 x − + 1 + O( x − ), 0 < s < n, | | 2 | 2| Hs(x) = ln x + 1 + O( x ), s = n, | | | | 1 + O( x s n), s > n, | | − as x 0. | | → p n (d) Gs(x) L 0 (R ) for any 1 p ∞ and s > n/p. ∈ 6 6 Proof. (a) For A,s > 0, we have the Γ -function identity Z ∞ s/2 1 tA s/2 dt A− = e− t , Γ (s/2) 0 t which we use to obtain Z ∞ 2 2 s/2 1 t t ωξ 2 s/2 dt (1 + ω ξ )− = e− e− | | t . | | Γ (s/2) 0 t Note that the above integral converges at both ends (as ξ 0, or ∞). Now take the inverse | | → Fourier transform in ξ and use Theorem 1.10 to obtain Z ∞ 1 1 t t ωξ 2 s/2 dt Gs(x) = Fξ− e− e− | | t Γ (s/2) 0 t Z ∞ 1 t 1 t ωξ 2 s/2 dt = e− Fξ− e− | | t Γ (s/2) 0 t Z ∞ x 2 s n 1 t | | dt = e− e− 4t t −2 . (4π)n/2Γ (s/2) 0 t 3 R 1 n (b) We have easily Rn Gs(x)dx = F Gs(0) = 1. Thus, Gs L (R ). 2 ∈ 2 x 1 x (c) First, we suppose x 2. Then t + | | t + and also t + | | x . This implies that | | > 4t > t 4t > | | x 2 t 1 x t | | 6 | |, − − 4t −2 − 2t − 2 from which it follows that when x 2 | | > Z ∞ t 1 s n x x 1 − dt | | | | Gs(x) 6 e− 2 e− 2t t 2 e− 2 6 C(s,n)e− 2 , (4π)n/2Γ (s/2) 0 t s n /2 2| − | Γ ( s n /2) 4 where C(s,n) = | − | for s = n, and C(s,n) = for s = n since (4π)n/2Γ (s/2) 6 (4π)n/2Γ (s/2) Z ∞ Z 1 Z ∞ Z ∞ t 1 dt 1 dt t y dy 1/2 e− 2 e− 2t 6 e− 2t + e− 2 dt = e− + 2e− 0 t 0 t 1 1/2 y Z ∞ y 62 e− dy + 2 6 4. 1/2 1 2 3 Next, suppose that x 6 2. Write Gs(x) = Gs (x) + Gs (x) + Gs (x), where | | 2 Z x x 2 s n 1 1 | | t | | − dt Gs (x) = e− e− 4t t 2 , (4π)n/2Γ (s/2) 0 t

3 R π x 2/t n/2 Or use (a) to show it. From part (a), we know Gs(x) > 0. Since Rn e− | | dx = t , by Fubini’s theorem, we have

∞ 2 Z Z 1 Z x s n dt G (x)dx = e t e |4|t t −2 dx s n/2 − − Rn Rn (4π) Γ (s/2) 0 t ∞ 2 1 Z Z x s n dt = e t e |4|t dxt −2 n/2 − − (4π) Γ (s/2) 0 Rn t Z ∞ 1 t n/2 s n dt = e− (4πt) t −2 (4π)n/2Γ (s/2) 0 t Z ∞ 1 t s 1 = e− t 2 − dt = 1. Γ (s/2) 0 7.2. Bessel potentials - 147 -

Z 4 x 2 s n 2 1 t | | − dt Gs (x) = e− e− 4t t 2 , (4π)n/2Γ (s/2) x 2 t | | Z ∞ x 2 s n 3 1 t | | − dt Gs (x) = e− e− 4t t 2 . (4π)n/2Γ (s/2) 4 t 2 Since t x 2 16 in G1, we have e t x = 1 + O(t x 2) as x 0; thus after changing variables, | | 6 s − | | | | | | → we can write Z 1 2 1 s n 1 s n 1 t x − dt Gs (x) = x − e− | | e− 4t t 2 | | (4π)n/2Γ (s/2) 0 t Z 1 s n+2 Z 1 s n 1 1 s n dt O( x − ) 1 s n = x − e− 4t t −2 + | | e− 4t t −2 dt | | (4π)n/2Γ (s/2) 0 t (4π)n/2Γ (s/2) 0 n s 2 s n Z ∞ n s 4 s n+2 Z ∞ 2 − − x − s n dy 2 − − O( x − ) s n dy y −2 y −2 = | | e− y + | | e− y 2 (4π)n/2Γ (s/2) 1/4 y (4π)n/2Γ (s/2) 1/4 y 1 s n s n+2 =c x − + O( x − ), as x 0. s,n| | | | | | → 2 x 2 x 1 / t | | Since 0 | | and 0 t 4 in G2, we have e 17 4 e 4t 1, thus as x 0, we 6 4t 6 4 6 6 s − 6 − − 6 | | → obtain s n s n+1  x − 2 | | − , < , Z 4  n s n s s n 2 (s n)/2 dt  − 2− − G (x) t − = 2ln , s = n, s 2 x ∼ x t s n+| 1| | |  2 − s n , s > n. − x 2 1/4 | | 3 3 Finally, we have e− 6 e− 4t 6 1 in Gs , which yields that Gs (x) is bounded above and j below by fixed positive constants. Combining the estimates for Gs(x), we obtain the desired conclusion. (d) For p = 1 and so p = ∞, by part (c), we have G (x) ∞ C for s > n. 0 k s k 6 Next, we assume that 1 2, p0 p x /2 | | that Gs 6 Ce− 0| | , and then the integration over this range x > 2 is clearly finite. R p0 | | On the range x 6 2, it is clear that x 2 Gs (x)dx 6 C for s > n. For the case s = n and | | | |6 R p0 n = 1, we also have x 2 Gs (x)dx 6 C by noticing that 6 | |6 Z q Z 2 q 2 2 n 1 ln dx = C ln r − dr 6 C x 2 x 0 r | |6 | | ε R 2 q for any q > 0 since limr 0 r ln(2/r) = 0. For the case s = n = 1, we have x 2(ln x ) dx = → | |6 R 2 q R 1 q R 1 p 1 | | 2 0 (ln2/r) dr = 4 0 (ln1/r) dr = 4Γ (q+1) for q > 0 by the formula 0 (ln1/x) − dx = Γ (p) R 2 for ℜp > 0. For the final case s < n, we have r(s n)p0 rn 1dr C if (s n)p + n > 0, i.e., 0 − − 6 − 0 s > n/p. Thus, we obtain G (x) C for any 1 p ∞ and s > n/p, which implies the desired k s kp0 6 6 6 result. ut We also have a result analogues to that of Riesz potentials for the operator Js. r n Theorem 7.7. (a) For all 0 < s < ∞, the operator Js maps L (R ) into itself with norm 1 for all 1 6 r 6 ∞. (b) Let 0 < s < n and 1 < p < q < ∞ satisfy 1/q = 1/p s/n. Then there exists a constant p n − Cn,s,p < ∞ such that for all f L (R ), we have ∈ J f C f . k s kq 6 n,s,pk kp - 148 - 7. Sobolev and Hölder Spaces

1 n 1 q (c) If f L (R ), then m x : Js f (x) > α (Cn,sα− f 1) , for all α > 0. That is, the ∈ { | | } 6 k k mapping f J f is of weak type (1,q), with 1/q = 1 s/n. → s − Proof. By Young’s inequality, we have J f = G f G f = f . This proves k s kr k s ∗ kr 6 k sk1k kr k kr the result (a). In the special case 0 < s < n, we have, from the above proposition, that the kernel Gs of Js satisfies x n+s, x 2, G (x) | |− | | 6 s ∼ e x /2, x 2. −| | | | > Then, we can write Z Z n+s y /2 Js f (x) 6Cn,s f (x y) y − dy + f (x y) e−| | dy y 2 | − || | y 2 | − | | |6 | |> Z y /2 6Cn,s Is( f )(x) + f (x y) e−| | dy . | | Rn | − | We now use that the function e y /2 Lr for all 1 r ∞, Young’s inequality and Theorem −| | ∈ 6 6 7.4 to complete the proofs of (b) and (c). ut The affinity between the two potentials is given precisely in the following lemma. Lemma 7.8. Let s > 0. n (i) There exists a finite measure µs on R such that its Fourier transform µbs is given by ωξ s µs(ξ) = | | . b (1 + ωξ 2)s/2 | | n (ii) There exist a pair of finite measures νs and λs on R such that (1 + ωξ 2)s/2 = ν (ξ) + ωξ sλb (ξ). | | bs | | s Remark 7.9. 1) The first part states in effect that the following formal quotient operator is bounded on every Lp(Rn), 1 6 p 6 ∞, ( ∆)s/2 − , s > 0. (7.14) (I ∆)s/2 − 2) The second part states also to what extent the same thing is true of the operator inverse to (7.14). Proof. To prove (i), we use the Taylor expansion ∞ s/2 m (1 t) = 1 + ∑ Am,st , t < 1, (7.15) − m=1 | | s ( s 1) ( s m+1) ( s )(1 s ) (m s 1) where A = ( 1)mCm = ( 1)m 2 2 − ··· 2 − = − 2 − 2 · − 2 − . All the A are of m,s − s/2 − m! m! m,s same sign for m > s + 1, so ∑ A < ∞, since (1 t)s/2 remains bounded as t 1, if s 0. 2 | m,s| − → > Let t = (1 + ωξ 2) 1. Then | | − 2 s/2 ∞ ωξ 2 m | | = 1 + A , (1 + ωξ )− . (7.16) 1 + ωξ 2 ∑ m s | | | | m=1 R ωix ξ 2 m However, G (x) 0 and n G (x)e dx = (1 + ωξ ) . 2m > R 2m − · | | − We noticed already that R G (x)dx = 1 and so G = 1. 2m k 2mk1 Thus from the convergence of ∑ A , it follows that if µ is defined by | m,s| s ∞ ! µs = δ0 + ∑ Am,sG2m(x) dx (7.17) m=1 7.3. Sobolev spaces - 149 - with δ0 the Dirac measure at the origin, then µs represents a finite measure. Moreover, by (7.16), ωξ s µs(ξ) = | | . (7.18) b (1 + ωξ 2)s/2 | | 1 n For (ii), we now invoke the n-dimensional version of Wiener’s theorem, to wit: If Φ1 L (R ) 1 n ∈ 1 and Φc1(ξ) + 1 is nowhere zero, then there exists a Φ2 L (R ) such that (Φc1(ξ) + 1)− = ∈ Φc2(ξ) + 1. For our purposes, we then write ∞ Φ1(x) = ∑ Am,sG2m(x) + Gs(x). m=1 Then, by (7.18), we see that ωξ s + 1 Φc1(ξ) + 1 = | | , (1 + ωξ 2)s/2 | | which vanishes nowhere. Thus, for an appropriate Φ L1, by Wiener’s theorem, we have 2 ∈ (1 + ωξ 2)s/2 = (1 + ωξ s)[Φc(ξ) + 1], | | | | 2 and so we obtain the desired conclusion with ν = λ = δ + Φ (x)dx. s s 0 2 ut

7.3 Sobolev spaces

We start by weakening the notation of partial derivatives by the theory of distributions. The appropriate definition is stated in terms of the space D(Rn). Let ∂ α be a differential monomial, whose total order is α . Suppose we are given two locally | | integrable functions on Rn, f and g. Then we say that ∂ α f = g (in the weak sense), if Z Z α α f (x)∂ ϕ(x)dx = ( 1)| | g(x)ϕ(x)dx, ϕ D. (7.19) Rn − Rn ∀ ∈ Integration by parts shows us that this is indeed the relation that we would expect if f had continuous partial derivatives up to order α , and ∂ α f = g had the usual meaning. | | Of course, it is not true that every locally integrable function has partial derivatives in this i/ x n sense: consider, for example, f (x) = c | | . However, when the partial derivatives exist, they are determined almost everywhere by the defining relation (7.19). In this section, we study a quantitative way of measuring smoothness of functions. Sobolev spaces serve exactly this purpose. They measure the smoothness of a given function in terms of the integrability of its derivatives. We begin with the classical definition of Sobolev spaces. , Definition 7.10. Let k be a nonnegative integer and let 1 6 p 6 ∞. The Sobolev space W k p(Rn) is defined as the space of functions f in Lp(Rn) all of whose distributional derivatives ∂ α f are also in Lp(Rn) for all multi-indices α that satisfies α k. This space is normed by the | | 6 expression α f W k,p = ∑ ∂ f p, (7.20) k k α k k k | |6 where ∂ (0,...,0) f = f . The index k indicates the “degree” of smoothness of a given function in W k,p. As k increases, the functions become smoother. Equivalently, these spaces form a decreasing sequence Lp W 1,p W 2,p ⊃ ⊃ ⊃ ··· - 150 - 7. Sobolev and Hölder Spaces meaning that each W k+1,p(Rn) is a subspace of W k,p(Rn) in view of the Sobolev norms. k,p n We next observe that the space W (R ) is complete. Indeed, if fm is a Cauchy sequence k,p α p { } in W , then for each α, ∂ fm is a Cauchy sequence in L , α 6 k. By the completeness of p {(α) } (α) α p| | L , there exist functions f such that f = limm ∂ fm in L , then clearly Z Z Z α α α (α) ( 1)| | fm∂ ϕdx = ∂ fmϕdx f ϕdx, − Rn Rn → Rn for each ϕ D. Since the first expression converges to ∈ Z α α ( 1)| | f ∂ ϕdx, − Rn α (α) k,p n it follows that the distributional derivative ∂ f is f . This implies that f j f in W (R ) → and proves the completeness of this space. First, we generalize Riesz and Bessel potentials to any s R by s 1 s n ∈ I f =F − ωξ F f , f S 0(R ), 0 / supp fˆ, | | ∈ ∈ s 1 2 s/2 n J f =F − (1 + ωξ ) F f , f S 0(R ). | | ∈ s s It is clear that I− = Is and J− = Js for s > 0 are exactly Riesz and Bessel potentials, respectively. we also note that Js Jt = Js+t for any s,t R from the definition. · ∈ Next, we shall extend the spaces W k,p(Rn) to the case where the number k is real. Definition 7.11. Let s R and 1 6 p 6 ∞. We write ∈ s s f ˙ s = I f p, f Hs = J f p. k kHp k k k k p k k s n Then, the homogeneous Sobolev space H˙ p(R ) is defined by s n n n 1 n o H˙ (R ) = f S 0(R ) : fˆ L (R ), and f ˙ s < ∞ , (7.21) p ∈ ∈ loc k kHp s n The nonhomogeneous Sobolev space Hp(R ) is defined by s n n n o H (R ) = f S 0(R ) : f Hs < ∞ . (7.22) p ∈ k k p ˙ s n ˙ s n s n s n If p = 2, we denote H2(R ) by H (R ) and H2(R ) by H (R ) for simplicity. s n It is clear that the space Hp(R ) is a normed linear space with the above norm. Moreover, it is complete and therefore Banach space. To prove the completeness, let fm be a Cauchy s p p { } sequence in Hp. Then, by the completeness of L , there exists a g L such that s s ∈ f J− g s = J f g 0, as m ∞. k m − kHp k m − kp → → Clearly, J sg S and thus Hs is complete. − ∈ 0 p We give some elementary results about Sobolev spaces. Theorem 7.12. Let s R and 1 6 p 6 ∞, then we have ∈s (a) S is dense in Hp, 1 6 p < ∞. s+ε s (b) Hp Hp, ε > 0. s ⊂∞ ∀ (c) Hp L , s > n/p. ⊂ ∀ s n s 1 n (d) Suppose 1 ∈ p ∈ p ∂ f s 1 n j, H − (R ). Moreover, the two norms are equivalent: ∂x j ∈ p n ∂ f s f Hp f Hs 1 + ∑ . k k ∼ k k p− ∂x s 1 j=1 j Hp− (e) Hk(Rn) = W k,p(Rn), 1 < p < ∞, k N. p ∀ ∈ 7.3. Sobolev spaces - 151 -

Proof. (a) Take f Hs, i.e., J f Lp. Since S is dense in Lp (1 p < ∞), there exists a g S ∈ p s ∈ 6 ∈ such that s s f J− g s = J f g k − kHp k − kp s s is smaller than any given positive number. Since J− g S , therefore S is dense in Hp. s+ε ∈ p p (b) Suppose that f H . By part (a) in Theorem 7.7, we see that Jε maps L into L with ∈ p norm 1 for ε > 0. Form this, we get the result since s ε s+ε s+ε s+ε f Hs = J f p = J− J f p = Jε J f p 6 J f p = f s+ε . k k p k k k k k k k k k kHp (c) By Young’s inequality, the deﬁnition of the kernel Gs(x) and part (d) of Proposition 7.6, we get for s > 0 1 2 s/2 2 s/2 f ∞ = F − (1 + ωξ )− (1 + ωξ ) F f ∞ k k k | | | | k 1 2 s/2 s = F − (1 + ωξ )− J f ∞ k | | ∗ k 1 2 s/2 s F − (1 + ωξ )− J f 6k | | kp0 k kp = G (x) f s C f s . k s kp0 k kHp 6 k kHp (d) From the Mihlin multiplier theorem, we can get (ωξ )(1+ ωξ 2) 1/2 M for 1 < p < ∞ j | | − ∈ p (or use part (i) of Lemma 7.8 and properties of Riesz transforms), and thus

∂ f 1 2 (s 1)/2 = F − (1 + ωξ ) − (ωiξ j)F f p ∂x s 1 k | | k j Hp− 1 2 1/2 2 s/2 = F − (1 + ωξ )− (ωξ )(1 + ωξ ) F f k | | j | | kp 1 2 1/2 s s = F − (1 + ωξ )− (ωξ ) J f C J f = C f s . k | | j ∗ kp 6 k kp k kHp Combining with f s 1 6 f Hs , we get k kHp− k k p n ∂ f s f Hs 1 + ∑ 6 C f Hp . k k p− ∂x s 1 k k j=1 j Hp− Now, we prove the converse inequality. We use the Mihlin multiplier theorem once more and an auxiliary function χ on R, inﬁnitely differentiable, non-negative and with χ(x) = 1 for x > 2 and χ(x) = 0 for x < 1. We obtain | | | | n 2 1/2 1 1 (1 + ωξ ) (1 + ∑ χ(ξ j) ξ j )− Mp, χ(ξ j) ξ j ξ j− Mp, 1 < p < ∞. | | j=1 | | ∈ | | ∈ Thus, s 1 2 1/2 s 1 f s = J f = F − (1 + ωξ ) F J − f k kHp k kp k | | kp n 1 s 1 6C F − (1 + ∑ χ(ξ j) ξ j )F J − f p k j=1 | | k n 1 1 s 1 ∂ f C f s 1 +C F − χ(ξ ) ξ ξ − F J − 6 Hp− ∑ j j j p k k j=1 k | | ∂x j k n ∂ f 6C f Hs 1 + ∑ . k k p− ∂x s 1 j=1 j Hp− Thus, we have obtained the desired result. (e) It is obvious that W 0,p = H0 = Lp for k = 0. However, from part (d), if k 1, then f Hk p > ∈ p ∂ f k 1 k,p k if and only if f and H − , j = 1,...,n. Thus, we can extends the identity of W = H ∂x j ∈ p p from k = 0 to k = 1,2,.... ut - 152 - 7. Sobolev and Hölder Spaces

We continue with the Sobolev embedding theorem.

Theorem 7.13 (Sobolev embedding theorem). Let 1 < p 6 p1 < ∞ and s,s1 R. Assume that n n ∈ s = s1 . Then the following conclusions hold − p − p1 Hs Hs1 , H˙ s H˙ s1 . p ⊂ p1 p ⊂ p1 Proof. It is trivial for the case p = p1 since we also have s = s1 in this case. Now, we assume 1 1 s s1 that p < p1. Since = − , by part (b) of Theorem 7.7, we get p1 p − n s s s s s s s 1 1 s f H 1 = J f p1 = J − J f p1 = Js s1 J f p1 6 C J f p = C f Hp . k k p1 k k k k k − k k k k k Similarly, we can show the homogeneous case. Therefore, we complete the proof. ut σ s s σ Theorem 7.14. Let s,σ R and 1 p ∞. Then J is an isomorphism between H and H − . ∈ 6 6 p p Proof. It is clear from the deﬁnition. ut Corollary 7.15. Let s R and 1 6 p < ∞. Then ∈ s s (H )0 = H− . p p0 Proof. It follows from the above theorem and the fact that (Lp) = Lp0 , if 1 p < ∞. 0 6 ut Next we give the connection between the homogeneous and the nonhomogeneous spaces, whose proof will be postponed to next chapter. n Theorem 7.16. Suppose that f S 0(R ) and 0 / supp fˆ . Then ∈s s ∈ f H˙ f H , s R, 1 p ∞. ∈ p ⇔ ∈ p ∀ ∈ 6 6 Moreover, for 1 6 p 6 ∞, we have Hs =Lp H˙ s, s > 0, p ∩ p ∀ Hs =Lp + H˙ s, s < 0, p p ∀ 0 p 0 Hp =L = H˙ p. References - 153 -

References

Abe11. Helmut Abels. Short lecture notes: Interpolation theory and function spaces. May 2011. Bec75. William Beckner. Inequalities in Fourier analysis on Rn. Proc. Nat. Acad. Sci. U.S.A., 72:638–641, 1975. BL76. Jöran Bergh and Jörgen Löfström. Interpolation spaces. An introduction. Springer-Verlag, Berlin, 1976. Grundlehren der Mathematischen Wissenschaften, No. 223. Din07. Guanggui Ding. New talk of functional analysis. Science press, Beijing, 2007. Duo01. Javier Duoandikoetxea. Fourier analysis, volume 29 of Graduate Studies in Mathematics. American Mathematical Society, Providence, RI, 2001. Translated and revised from the 1995 Spanish original by David Cruz-Uribe. Eva98. Lawrence C. Evans. Partial differential equations, volume 19 of Graduate Studies in Mathematics. American Mathemat- ical Society, Providence, RI, 1998. GR. I. S. Gradshteyn and I. M. Ryzhik. Table of integrals, series, and products. Seventh edition. Gra04. Loukas Grafakos. Classical and modern Fourier analysis. Pearson Education, Inc., Upper Saddle River, NJ, 2004. GT01. David Gilbarg and Neil S. Trudinger. Elliptic partial differential equations of second order. Classics in Mathematics. Springer-Verlag, Berlin, 2001. Reprint of the 1998 edition. HLP88. G. H. Hardy, J. E. Littlewood, and G. Pólya. Inequalities. Cambridge Mathematical Library. Cambridge University Press, Cambridge, 1988. Reprint of the 1952 edition. HW64. Richard A. Hunt and Guido Weiss. The Marcinkiewicz interpolation theorem. Proc. Amer. Math. Soc., 15:996–998, 1964. Kri02. Erik Kristiansson. Decreasing rearrangement and Lorentz L(p,q) spaces. Master’s thesis, Luleå University of Technology, 2002. http://bioinformatics.zool.gu.se/~erikkr/MasterThesis.pdf. Mia04. Changxing Miao. Harmonic analysis and its application in PDEs, volume 89 of Fundamental series in mordon Mathe- matics. Science Press, Beijing, 2nd edition, 2004. Rud87. Walter Rudin. Real and complex analysis. McGraw-Hill Book Co., New York, third edition, 1987. SS03. Elias M. Stein and Rami Shakarchi. Complex analysis. Princeton Lectures in Analysis, II. Princeton University Press, Princeton, NJ, 2003. Ste70. Elias M. Stein. Singular integrals and differentiability properties of functions. Princeton Mathematical Series, No. 30. Princeton University Press, Princeton, N.J., 1970. Ste93. Elias M. Stein. Harmonic analysis: real-variable methods, orthogonality, and oscillatory integrals, volume 43 of Prince- ton Mathematical Series. Princeton University Press, Princeton, NJ, 1993. With the assistance of Timothy S. Murphy, Monographs in Harmonic Analysis, III. SW71. Elias M. Stein and Guido Weiss. Introduction to Fourier analysis on Euclidean spaces. Princeton University Press, Princeton, N.J., 1971. Princeton Mathematical Series, No. 32. WHHG11. Baoxiang Wang, Zhaohui Huo, Chengchun Hao, and Zihua Guo. Harmonic analysis method for nonlinear evolution equations, volume I. World Scientiﬁc Publishing Co. Pte. Ltd., 2011.

Index

n Vn: the volume of the unit ball in R , 36 Gauss-Weierstrass kernel,7 Hk, 100 gradient condition, 78 Lp: weak Lp spaces, 35 Green theorem, 114 ∗ C: complex number ﬁeld,2 n C0(R ),1 Hörmander condition, 78 Γ -function, 36 Hadamard three lines theorem, 28 Hk, 99 Hardy inequality, 42 N0 = N 0 , 14 Hardy-Littlewood maximal function, 50 ∪ { } Pk, 99 Hardy-Littlewood maximal operator, 50 R: real number ﬁeld,2 Hardy-Littlewood-Paley theorem on Rn, 45 n/2 n 1 ωn 1 = 2π /Γ (n/2): surface area of the unit sphere S − ,8 Hardy-Littlewood-Sobolev theorem of fractional integrations, − gλ∗ -function, 115 142 harmonic conjugate, 70 Abel method of summability,5 harmonic function, 63, 67 analytic, 27 Hausdorff-Young inequality, 31 heat equation, 12 Bessel potential, 143 Hecke’s identity, 101 beta function: n-dimensional variant, 140 Heine-Borel theorem, 53 Bochner’s relation, 102 Heisenberg uncertainty principle, 16 higher Riesz transforms, 106 Calderón-Zygmund decomposition Hilbert transform, 73 for functions, 60 Hilbert transform of Rn, 57 Characterization, 75 Calderón-Zygmund kernel, 79 holomorphic, 27 s n Calderón-Zygmund singular integral operator, 80 homogeneous Sobolev spaces H˙ p(R ), 148 Calderón-Zygmund Theorem, 77 Cauchy-Riemann equations, 70 Jensen’s inequality, 42 generalized ..., 98 convolution,1 Laplace equation, 63 Cotlar inequality, 88 Lebesgue differentiation theorem, 55 Littlewood-Paley g-function, 109 decreasing rearrangement, 36 Littlewood-Paley square function theorem, 129 dilation,2 locally integrable, 50 dilation argument, 81 Lorentz space, 40 Dini-type condition, 83 Luzin’s S-function, 115 distribution function, 33 dyadic decomposition of Rn, 128 Marcinkiewicz interpolation theorem, 43 maximal function, 50 elliptic homogeneous polynomial of degree k, 108 maximal function theorem, 52 entire function, 27 maximum modulus principle, 27 maximum principle, 28 Fatou’s lemma, 141 Mean-value formula for harmonic functions, 66 Fefferman-Stein inequality, 60 Minkowski integral inequality,1 Fourier inversion theorem, 10 multiplication formula,7 Fourier transform,2 multiplier theorem Bernstein’s multiplier theorem, 122 Gagliardo-Nirenberg-Sobolev inequality, 56 Hörmander’s multiplier theorem, 123 Gauss summability,5 Marcinkiewicz’ multiplier theorem, 135 Gauss-Weierstrass integral,8 Mihlin’s multiplier theorem, 123

155 - 156 - Index

s n k,p n nonhomogeneous Sobolev spaces Hp(R ), 148 Sobolev space W (R ), 147 solid spherical harmonics of degree k, 99 partial g-functions, 109 Stein interpolation theorem, 33 partial sum operator, 126 Phragmen-Lindelöf theorem, 28 Tchebychev inequality, 55 Plancherel theorem, 12 tempered distribution, 17 Poisson equation, 63 The equivalent norm of Lp, 35 Poisson integral,8, 67, 68 translation,2 Poisson kernel,7, 67 translation invariant, 22 principal value of 1/x, 72 truncated operator, 80 quasi-linear mapping, 41 unitary operator, 13 Rademacher functions, 129 Riemann-Lebesgue lemma,4 Vitali covering lemma, 47 Riesz potentials, 139 Riesz transform, 96 Weierstrass kernel,7 Riesz-Thorin interpolation theorem, 29 Whitney covering lemma, 48 Wiener’s theorem, 147 Schwartz space, 14 Sobolev embedding theorem, 150 Young’s inequality for convolutions, 32