HARMONIC ANALYSIS 1. Preliminaries Let Us First Fix Notation

PIOTR HAJLASZ

1. Preliminaries

Let us ﬁrst ﬁx notation that will be used throughout the book. We will be working mostly in the Euclidean spaces and with the Lebesgue measure. The Lebesgue measure of a n measurable set E will be denoted by |E|. The volume of a ball of radius r is |B(0, r)| = ωnr . Then it is well known that the (n − 1)-dimensional measure of the sphere Sn−1(0, r) equals n−1 nωnr . The integral average over a set E of positive measure will be denoted by Z 1 Z f(x) dx = f(x) dx. E |E| E The Lebesgue measure on a surface will be denoted by dσ so the integration in the spherical coordinates will take the form Z Z ∞ Z f(x) dx = f(sθ) dσ(θ) ds. Rn 0 Sn−1 Unless we will explicitly state otherwise, all functions and all function spaces will be complex valued.

n ∞ A generic open set in R will usually be denoted by Ω and C0 (Ω) will stand for a space of smooth functions with compact support in Ω. The space of continuous functions n n vanishing at inﬁnity will be denote by C0(R ), i.e. f ∈ C0(R ) if f is continuous and lim f(x) = 0. |x|→∞

n ∞ n C0(R ) is a Banach space with respect to the supremum norm k · k∞ and C0 (R ) is a n dense subset of C0(R ). We will use the multiindex notation ∂|α| Dαu = , xα = xα1 · ... · xαn α1 αn 1 n ∂x1 . . . ∂xn where α = (α1, . . . , αn), |α| = α1 + ... + αn. Then the product rule takes the form X α! Dα(fg) = DβfDγg, where α! = α ! · ... · α !. β!γ! 1 n β+γ=α

Date: February 27, 2017. 1 2 PIOTR HAJLASZ

The characteristic function of a set E will be denoted by χE. By δa we will denote the Dirac measure centered at a i.e., Z f(x) dδa(x) = f(a). Rn By an increasing function we will mean a non-decreasing function and increasing functions that are actually increasing, will be called strictly increasing. Similar terminology applies to decreasing and strictly decreasing functions. By sgn x we will denote the sign of the number x, i.e. sgn x = 1 for positive x, sgn x = −1 for negative x and sgn 0 = 0. Various constants will usually be denoted by C. By writing C(n, m) we will indicate the the constant C(n, m) depends on n and m only. We adopt the rule that a constant C may change its value within one string of estimates. This will allow us to minimize the use of indices. HARMONIC ANALYSIS 3

2. The Maximal function

2.1. The maximal theorem.

1 n Deﬁnition 2.1. For a locally integrable function f ∈ Lloc(R ) the Hardy-Littlewood maximal function is deﬁned by Z n Mf(x) = sup |f(y)| dy, x ∈ R . r>0 B(x,r)

The operator M is not linear but it is subadditive. We say that an operator T from a space of measurable functions into a space of measurable functions is subadditive if

|T (f1 + f2)(x)| ≤ |T f1(x)| + |T f2(x)| a.e. and |T (kf)(x)| = |k||T f(x)| for k ∈ C. The following integrability result, known also as the maximal theorem, plays a fundamental role in many areas of mathematical analysis.

Theorem 2.2 (Hardy-Littlewood-Wiener). If f ∈ Lp(Rn), 1 ≤ p ≤ ∞, then Mf < ∞ a.e. Moreover

(a) For f ∈ L1(Rn) 5n Z (2.1) |{x : Mf(x) > t}| ≤ |f| for all t > 0. t Rn (b) If f ∈ Lp(Rn), 1 < p ≤ ∞, then Mf ∈ Lp(Rn) and p 1/p kMfk ≤ 2 · 5n/p kfk for 1 < p < ∞, p p − 1 p

kMfk∞ ≤ kfk∞ . Remark 2.3. Note that if f ∈ L1(Rn) is a non-zero function, then Mf 6∈ L1(Rn). Indeed, R if λ = B(0,R) |f| > 0, then for |x| > R, B(0,R) ⊂ B(x, R + |x|) so Z λ Mf(x) ≥ |f| ≥ n , B(x,R+|x|) ωn(R + |x|) Since the function on the right hand side is not integrable on Rn, the statement (b) of the theorem is not true for p = 1. 4 PIOTR HAJLASZ

Remark 2.4. If g ∈ L1(Rn), then the Chebyschev inequality 1 Z (2.2) |{x : |g(x)| > t}| ≤ |g| for t > 0 t Rn 1 is easy to prove. Thus if an (not necessarily linear) operator T is bounded in L , kT fk1 ≤ Ckfk1, it immediately follows from Chebyschev’s inequality that C Z (2.3) |{x : |T f(x)| > t}| ≤ |f(x)| dx. t Rn The maximal function M is not bounded in L1, but it still satisﬁes the weaker estimate (2.3). For this reason estimate (2.1) is called the weak type estimate.

In the proof of Theorem 2.2 we will need the following two results. Theorem 2.5 (Cavalieri’s principle). If µ is a σ-ﬁnite measure on X and Φ : [0, ∞) → [0, ∞) is increasing, absolutely continuous and Φ(0) = 0, then Z Z ∞ Φ(|f|) dµ = Φ0(t)µ({|f| > t}) dt . X 0

Proof. The result follows immediately from the equality Z Z Z |f(x)| Φ(|f(x)|) dµ(x) = Φ0(t) dt dµ(x) X X 0 and the Fubini theorem. Corollary 2.6. If µ is a σ-ﬁnite measure on X and 0 < p < ∞, then Z Z ∞ |f|p dµ = p tp−1µ({|f| > t}) dt . X 0

The next result has many applications that go beyond the maximal theorem. Here and in what follows by 5B we denote the ball concentric with B and with five times the radius. Theorem 2.7 (5r-covering lemma). Let B be a family of balls in a metric space such that sup{diam B : B ∈ B} < ∞. Then there is a subfamily of pairwise disjoint balls B0 ⊂ B such that [ [ B ⊂ 5B. B∈B B∈B0 More precisely, for every B ∈ B there is B0 ∈ B0 such that B ∩ B0 6= ∅ and B ⊂ 5B0. If the metric space is separable, then the family B0 is countable (or finite) and we can 0 ∞ arrange it as a (possibly finite) sequence B = {Bi}i=1 so ∞ [ [ B ⊂ 5Bi . B∈B i=1 Remark 2.8. Here B can be either a family of open balls or closed balls. In both cases the proof is the same. HARMONIC ANALYSIS 5

Proof. Let sup{diam B : B ∈ B} = R < ∞. Divide the balls in the family B according to their diameters: n R R o F = B ∈ B : < diam B ≤ . j 2j 2j−1 S∞ Clearly B = j=1 Fj.

First we select balls as large as possible so we choose balls from the family F1 and we take as many as we can: we deﬁne B1 ⊂ F1 to be a maximal family of pairwise disjoint balls.

Suppose the families B1,..., Bj−1 are already deﬁned.

In the construction of Bj we want to select balls form Fj, but we have to make sure that balls that we choose do not intersects with any of the previously selected balls. Thus we have to select balls from n j−1 o ˆ 0 0 [ Fj = B ∈ Fj : B ∩ B = ∅ for all B ∈ Bi . i=1 ˆ Then we define Bj to be the maximal family of pairwise disjoint balls in Fj 0 S∞ Finally we define B = j=1 Bj. It follows immediately from the definition of Bj that Sj 0 0 Sj every ball B ∈ Fj intersects with a ball in i=1 Bj. If B ∩ B 6= ∅, B ∈ i=1 Bi, then R R diam B ≤ = 2 · < 2 diam B0 2j−1 2j 0 and hence B ⊂ 5B .

1 n Proof of Theorem 2.2. (a) Let f ∈ L (R ) and let Et = {x : Mf(x) > t}. For x ∈ Et, there is rx > 0 such that Z Z −1 |f| > t so |B(x, rx)| < t |f| . B(x,rx) B(x,rx) 1 n Observe that supx∈Et rx < ∞, because f ∈ L (R ). The family of balls {B(x, rx)}x∈Et forms a covering of the set Et so applying the 5r-covering lemma there is a sequence of S∞ pairwise disjoint balls B(xi, rxi ), i = 1, 2,... such that Et ⊂ i=1 B(xi, 5rxi ) and hence ∞ n ∞ Z n Z n X 5 X 5 |Et| ≤ 5 |B(xi, rxi )| ≤ |f| ≤ |f| . t t n i=1 i=1 B(xi,rxi ) R The proof is complete.

p n (b) Let f ∈ L (R ), 1 < p ≤ ∞. Since kMfk∞ ≤ kfk∞ we can assume that 1 < p < ∞. Let f = f1 + f2, where

f1 = fχ{|f|>t/2}, f2 = fχ{|f|≤t/2} be a decomposition of |f| into its ‘lower’ and ‘upper’ parts. It is easy to check that f1 ∈ 1 n ∞ n L (R ) and clearly f2 ∈ L (R ).

The idea now if to apply the weak type estimate from (a) to f1 and the estimate kMfk∞ ≤ kfk∞ to f2. This will give us an estimate for the size of the level set |{Mf > t}| 6 PIOTR HAJLASZ and the result will follow from Corollary 2.6. The same idea is the core of the so called real interpolation method and we will see it again in the proof of the Marcinkiewcz Interpolation Theorem 8.9.

Since |f| ≤ |f1| + t/2 we have Mf ≤ Mf1 + t/2 and hence

{Mf > t} ⊂ {Mf1 > t/2} . Thus 2 · 5n Z (2.4) |Et| = |{Mf > t}| ≤ |f1(x)| dx t Rn 2 · 5n Z = |f(x)| dx . t {|f|>t/2} Cavalieri’s principle gives Z Z ∞ |Mf(x)|p dx = p tp−1|{Mf > t}| dt Rn 0 Z ∞ 2 · 5n Z ≤ p tp−1 |f(x)| dx dt 0 t {|f|>t/2} Z Z 2|f(x)| = 2 · 5np |f(x)| tp−2 dt dx Rn 0 p Z = 2p · 5n |f(x)|p dx p − 1 Rn and the results follows. Remark 2.9. Note that we proved in (2.4) the following inequality 2 · 5n Z (2.5) |{x : Mf(x) > t}| ≤ |f(x)| dx t {|f|>t/2} which is slightly stronger than (2.1). Later we will see that the measure of the set |{Mf > t}| has a similar bound from below, see Proposition 2.33.

Remark 2.10. For a positive measure µ on Rn we deﬁne the maximal function by µ(B(x, r)) Mµ(x) = sup . r>0 |B(x, r)| A minor modiﬁcation of the proof of Theorem 2.2(a) leads to the following result.

Proposition 2.11. If µ is a ﬁnite positive Borel measure on Rn, then 5n |{x : Mµ(x) > t}| ≤ µ( n) for all t > 0. t R Remark 2.12. It is natural to inquire whether in the Euclidean case, there is a version of Theorem 2.7 for families of cubes. The answer is in the positive and it follows immediately from Theorem 2.7 and the fact that cubes with sides parallel to coordinate axes are balls n in R with respect to the metric d∞(x, y) = maxi |xi − yi|. In the following result 5Q will denote a cube concentric with Q and with ﬁve times the diameter. HARMONIC ANALYSIS 7

Corollary 2.13. Let F be a family of open or closed cubes in Rn with sides parallel to coordinate directions such that sup{diam Q : Q ∈ F} < ∞. Then there is a subfamily F 0 ⊂ F of pairwise disjoint cubes such that [ [ Q ⊂ 5Q. Q∈F Q∈F 0 More precisely for every Q ∈ F there is Q0 ∈ F 0 such that Q ∩ Q0 6= ∅ and Q ⊂ 5Q0.

Although we are interested mostly in the Euclidean setting, it was important to formulate Theorem 2.7 in a the setting of metric spaces as we could conclude Corollary 2.13 directly from Theorem 2.7.

In the next two sections we will show applications of the maximal theorem.

2.2. Fractional integration theorem. As a first application of the maximal theorem we will prove a result about integrability of the Riesz potentials. Definition 2.14. For 0 < α < n and n ≥ 2 we define the Riesz potential by n α α 1 Z f(y) π 2 2 Γ (I f)(x) = dy , where γ(α) = 2 . α n−α n−α γ(α) n |x − y| R Γ 2

At this moment the particular value of the constant γ(α) is not important to us. We could even replace this constant by 1.1 Theorem 2.15 (Hardy-Littlewood-Sobolev). Let α > 0, 1 < p < ∞ and αp < n. Then there is a constant C = C(n, p, α) such that p n kIαfkp∗ ≤ Ckfkp for all f ∈ L (R ) , where p∗ = np/(n − αp). 2 p q Remark 2.16. It is not diﬃcult to show directly that if Iα : L → L is bounded, then q = p∗. The idea is to use the change of variables x 7→ tx, t > 0, and see how both sides of the inequality kIαfkq ≤ Ckfkp change. This is so called a scaling argument.

We precede the proof with a technical lemma. Lemma 2.17. If 0 < α < n, and δ > 0, then there is a constant C = C(n, α) such that Z |f(y)| α n n−α dy ≤ Cδ Mf(x) for all x ∈ R . B(x,δ) |x − y|

Proof. For x ∈ Rn and δ > 0 consider the annuli δ δ A(k) = B x, − B x, . 2k 2k+1

1 It will play an important role later: I2f is the convolution with the fundamentals solution of −∆, see Theorem 5.60. Also Riesz potentials will be studied in Section 7 in connection with fractional powers of the Laplace operator. 2A nice exercise. 8 PIOTR HAJLASZ

We have ∞ Z |f(y)| X Z |f(y)| dy = dy |x − y|n−α |x − y|n−α B(x,δ) k=0 A(k) ∞ α−n X δ Z ≤ |f(y)| dy 2k+1 k=0 A(k) ∞ α−n n X δ δ Z ≤ ωn k+1 k |f(y)| dy 2 2 k k=0 B(x,δ/2 ) α−n ∞ ! 1 X 1 ≤ ω δα Mf(x) . n 2 2kα k=0 The proof is complete.

Proof of Theorem 2.15. Fix δ > 0. H¨older’sinequality and integration in polar coordinates yield 0 Z |f(y)| Z dy 1/p n−α dy ≤ kfkp (n−α)p0 Rn\B(x,δ) |x − y| Rn\B(x,δ) |x − y| Z ∞ 1/p0 n−1−(n−α)p0 = kfkp nωn s ds δ α−(n/p) = C(n, p, α)δ kfkp , n−1 because nωn equals the (n − 1)-dimensional measure of the unit sphere S and n − (n − α)p0 < 0. This and the lemma give α α−(n/p) |Iαf(x)| ≤ C δ Mf(x) + δ kfkp . Taking3 Mf(x)−p/n δ = kfkp yields αp αp 1− n |Iαf(x)| ≤ C(Mf(x)) n kfkp which is equivalent to αp ∗ ∗ p p n p |Iαf(x)| ≤ C(Mf(x)) kfkp . Integrating both sides over Rn and applying boundedness of the maximal function in Lp gives the result.

2.3. The Lebesgue diﬀerentiation theorem. We will show now how to prove the Lebesgue diﬀerentiation theorem from the maximal theorem.

n For h ∈ R we deﬁne τhf(x) = f(x + h). p n Lemma 2.18. If f ∈ L (R ), 1 ≤ p < ∞, then kτhf − fkp → 0 as h → 0.

3 α α−(n/p) We apply here a standard trick: we take δ > 0 such that δ Mf(x) = δ kfkp. HARMONIC ANALYSIS 9

This result is obvious if f is a compactly supported smooth function, because in that ∞ p case τhf converges uniformly to f. The general case follows from the density of C0 in L . 1 n R 1 Lemma 2.19. Let f ∈ L (R ) and let fr(x) = B(x,r) f(y) dy. Then fr → f in L as r → 0. In particular, there is a sequence ri → 0 such that Z lim f(y) dy = f(x) a.e. i→∞ B(x,ri)

Proof. We have Z Z Z |fr(x) − f(x)| dx ≤ |f(y) − f(x)| dy dx Rn Rn B(x,r) Z Z = |f(x + y) − f(x)| dy dx Rn B(0,r) Z (2.6) = kτyf − fk1 dy . B(0,r) 1 Since τyf → f in L as y → 0, the right hand side of (2.6) converges to 0 as r → 0. The 1 existence of a sequence ri follows from the fact that from a sequence convergent in L we can extract a subsequence convergent a.e.

The Lebesgue diﬀerentiation theorem gives much more.

1 n Theorem 2.20 (Lebesgue diﬀerentiation theorem). If f ∈ Lloc(R ), then Z lim f(y) dy = f(x) a.e. r→0 B(x,r)

1 n Proof. Since the theorem is local in nature we can assume that f ∈ L (R ). Let fr(x) = R B(x,r) f(y) dy and deﬁne

Ωf(x) = lim sup fr(x) − lim inf fr(x) . r→0 r→0

It suﬃces to prove that Ωf = 0 a.e. Indeed, this property means that fr → g converges a.e. to a measurable function g. Since by Lemma 2.19, fri → f a.e. we get g = f a.e. Observe that Ωf ≤ 2Mf, hence for any ε > 0, Theorem 2.2(a) yields C Z |{x :Ωf(x) > ε}| ≤ |f| . ε Rn 2 Let h be a continuous function such that kf − hk1 < ε . Continuity of h implies Ωh = 0 everywhere and hence Ωf ≤ Ω(f − h) + Ωh = Ω(f − h) , so C Z |{Ωf > ε}| ≤ |{Ω(f − h) > ε}| ≤ |f − h| ≤ Cε . ε Rn Since ε > 0 can be arbitrarily small we conclude Ωf = 0 a.e. 10 PIOTR HAJLASZ

1 n If f ∈ Lloc(R ), then we can deﬁne f at every point by the formula Z (2.7) f(x) := lim sup f(y) dy . r→0 B(x,r) According to the Lebesgue diﬀerentiation theorem this is a representative of f in the class of functions that coincide with f a.e.

1 n n Deﬁnition 2.21. Let f ∈ Lloc(R ). We say that x ∈ R is a Lebesgue point of f if Z lim |f(y) − f(x)| dy = 0 , r→0 B(x,r) where f(x) is deﬁned by (2.7).

1 n Theorem 2.22. If f ∈ Lloc(R ), then the set of points that are not Lebesgue points of f has measure zero.

Proof. For c ∈ Q let Ec be the set of points for which Z (2.8) lim |f(y) − c| dy = |f(x) − c| r→0 B(x,r) does not hold. Clearly |E | = 0 and hence the set E = S E has measure zero too. Thus c c∈Q c for x ∈ Rn \ E and all c ∈ Q, (2.8) is satisﬁed. If x ∈ Rn \ E and f(x) ∈ R, approximating f(x) by rational numbers one can easily check that Z lim |f(y) − f(x)| dy = |f(x) − f(x)| = 0 . r→0 B(x,r) Indeed, if ε > 0 is arbitrary and c ∈ Q is such that |f(x) − c| < ε/3, then there is δ > 0 such that for 0 < r < δ Z Z |f(y) − f(x)| dy ≤ |f(y) − c| dy + |c − f(x)| B(x,r) B(x,r) ε < |f(x) − c| + + |c − f(x)| < ε. 3 The proof is complete. n p Deﬁnition 2.23. We say that x ∈ R is a p-Lebesgue point of f ∈ Lloc, 1 ≤ p < ∞ if Z lim |f(y) − f(x)|p dy = 0 . r→0 B(x,r)

The same method as above leads to the following result that we leave as an exercise.

p n Theorem 2.24. If f ∈ Lloc(R ), 1 ≤ p < ∞, then the set of points that are not p-Lebesgue points of f has measure zero.

Deﬁnition 2.25. Let E ⊂ Rn be a measurable set. We say that x ∈ Rn is a density point of E if |B(x, r) ∩ E| lim = 1 . r→0 |B(x, r)| HARMONIC ANALYSIS 11

Applying the Lebesgue theorem to the characteristic function of the set E, i.e., f = χE we obtain Theorem 2.26. Almost every point of a measurable set E ⊂ Rn is its density point and a.e. point of Rn \ E is not a density point of E.

The Lebesgue theorem shows that for almost all x, the averages of f over balls centered as x converge to f(x). It is natural to inquire whether we can replace balls by other sets like cubes or even balls, but not centered at x. Deﬁnition 2.27. We say that a family F of measurable subsets of Rn is regular at x ∈ Rn if

(a) The sets are bounded and have positive measure. (b) There is a sequence {Si} ⊂ F with |Si| → 0 as i → ∞. (c) There is a constant C > 0 such that for all S ∈ F we have |S| ≥ C|BS|, where BS is the smallest closed ball centered at x that contains S.

Note that if |Si| → 0, then the radius of BSi approaches to zero so the sets Si approach to x. Examples of families regular at x include:

• Family of all balls that contain x. • Family of all cubes centered at x. • Family of all cubes that contain x. 1 n Theorem 2.28. If f ∈ Lloc(R ), x is a Lebesgue point of f, and if F is a family regular at x, then Z lim f(y) dy = f(x) . S∈F |S|→0 S

Proof. We have Z Z Z 1 f(y) dy − f(x) ≤ |f(y) − f(x)| dy ≤ |f(y) − f(x)| dy S S |S| BS |B | Z 1 Z ≤ S |f(y) − f(x)| dy ≤ |f(y) − f(x)| dy → 0 |S| BS C BS as |S| → 0.

Note that if F is a family regular at 0 and if we deﬁne the maximal function associated with F by Z MF f(x) = sup |f(x − y)| dy , S∈F S then a routine calculation shows that

(2.9) MF f(x) ≤ CMf(x) , so MF satisﬁes the claim of Theorem 2.2 (with diﬀerent constants). In particular MF is a bounded operator in Lp, 1 < p ≤ ∞. 12 PIOTR HAJLASZ

Let F be the family of all rectangular boxes in Rn that contain the origin and have sides parallel to the coordinate axes. With the family we can associate the maximal function Z Mff(x) = sup |f(x − y)| dy . S∈F S

Note that the family F is not regular at 0 and hence the boundedness of Mf in Lp, 1 < p ≤ ∞ cannot be concluded from (2.9). However, we have Theorem 2.29 (Zygmund). For 1 < p < ∞ there is a constant C = C(n, p) > 0 such that

(2.10) kMffkp ≤ Ckfkp . p Moreover, if f ∈ Lloc, 1 < p < ∞, then Z (2.11) lim f(x − y) dy = f(x) a.e. diam S→0 S∈F S

Proof. First we will prove how to conclude (2.11) from (2.10). Note that since the family is not regular at 0, (2.11) is not a consequence of Theorem 2.28 (see also Theorem 2.30). However, Z Z 0 ≤ lim sup f(x − y) dy − lim inf f(x − y) dy ≤ 2Mff(x) diam S→0 diam S→0 S S S∈F S∈F and hence (2.11) follows from (2.10) by almost the same argument as the one used to deduce the Lebesgue Diﬀerentiation Theorem from Theorem 2.2. We leave details to the reader. Thus it remains to prove (2.10). In dimension one Z x+b Mff(x) = sup |f(y)| dy a,b>0 x−a is the so called non-centered Hardy-Littlewood maximal function. Clearly Mff ≤ CMf which gives (2.10). For simplicity of notation we will prove the higher dimensional result for n = 2 only. The proof in the case of general n > 1 is similar. We have

Z x2+b2 Z x1+b1 Mff(x1, x2) = sup |f(y1, y2)| dy1 dy2 a ,b >0 1 1 x2−a2 x1−a1 a2,b2>0 Z x2+b2 Z x1+b1 ≤ sup sup |f(y1, y2)| dy1 dy2 . a2,b2>0 x2−a2 a1,b1>0 x1−a1 On the right hand side we have iteration of one dimensional non-centered maximal functions. First we apply the maximal function to variable y1 and evaluate it at x1 and then we apply the maximal function to the variable y2 and evaluate it at x2. It is easy to see now that inequality (2.10) follows from the one dimensional case applied twice and the Fubini theorem. HARMONIC ANALYSIS 13

Surprisingly (2.11) does not hold for p = 1 and hence the maximal function Mff does not satisfy the weak type estimate (2.1).4 Namely one can prove the following result that we leave without a proof. Theorem 2.30 (Saks). Let F be the family of all rectangular boxes in Rn that contain the origin and have sides parallel to the coordinate axes. Then the set of functions f ∈ L1(Rn) such that Z lim sup f(x − y) dy = ∞ for all x ∈ Rn diam S→0 S S∈F 1 n is a dense Gδ subset of L (R ). In particular it is not empty.

2.4. The Calder´on-Zygmund decomposition. The following result will play an important role in what follows. Theorem 2.31 (Calder´on-Zygmund decomposition). Suppose f ∈ L1(Rn), f ≥ 0 and α > 0. Then there is an open set Ω and a closed set F such that

(a) Rn = Ω ∪ F , Ω ∩ F = ∅; (b) f ≤ α a.e. in F ; S∞ (c) Ω can be decomposed into cubes Ω = k=1 Qk with pairwise disjoint interiors such that Z α < f(x) dx ≤ 2nα , k = 1, 2, 3,... Qk

Proof. Decompose Rn into a grid of identical cubes, large enough to have Z f(x) dx ≤ α Q for each cube in the grid. Take a cube Q from the grid and divide it into 2n identical cubes. Let Q0 be one of the cubes from this partition. We have two cases: Z Z f(x) dx > α or f(x) dx ≤ α . Q0 Q0 0 If the fist case holds we include the open cube Q to the family {Qk}. Note that Z Z Z α < f = 2n|Q|−1 f ≤ 2n f ≤ 2nα Q0 Q0 Q so the condition (c) is satisfied. If the second case holds we divide Q0 into 2n identical cubes and proceed as above. We continue this process infinitely many times or until it is S∞ terminated. We apply it to all the cubes of the original grid. Let Ω = k=1 Qk, where the cubes are defined by the first case of the process. It remains to prove that f ≤ α a.e. in the set F = Rn \ Ω. The set F consists of faces of the cubes (this set has measure zero) ˜ ˜ and points x such that there is a sequence of cubes Qi with the property that x ∈ Qi, ˜ R R diam Qi → 0, f ≤ α. According to Theorem 2.28 for a.e. such x we have f → f(x) Q˜i Q˜i and hence f ≤ α a.e. in F . −1 Corollary 2.32. Let f, α and Ω be as in Theorem 2.31. Then |Ω| < α kfk1. 4Such estimate would imply (2.11). 14 PIOTR HAJLASZ

Proof. We have ∞ ∞ Z X X −1 −1 |Ω| = |Qk| ≤ α |f| ≤ α kfk1 . k=1 k=1 Qk The proof is complete.

In Remark 2.9 we obtained an upper bound for the measure of the level set of the maximal function which improves the weak type estimate from Theorem 2.2. We will prove now a similar lower bound.

Proposition 2.33. If f ∈ L1(Rn), then 2−nC(n)−1 Z 2 · 5n Z |f(x)| dx ≤ |{x : Mf(x) > t}| ≤ |f(x)| dx. t {|f|>C(n)t} t {|f|>t/2} n/2 where we can take C(n) = ωnn .

Proof. For the proof of the right inequality see (2.4). In order to prove the left inequality we apply the Calder´on-Zygmnud decomposition to the function |f| and α = s > 0. If S Ω = k Qk is an open set as in Theorem 2.31, then Z s < |f(x)| dx ≤ 2ns for all k. Q k √ Let `k be the edge-length of the cube Qk. Then for any x ∈ Qk, Qk ⊂ Bx = B(x, n`k) so Z Z |Bx| n s < |f(y)| dy ≤ |f(y)| dy ≤ ωnn 2 Mf(x). Qk |Qk| Bx Hence −n Z −1 − n X 2 X |{x : Mf(x) > sω n 2 }| ≥ |Q | ≥ |f(x)| dx n k s k Qk 2−n Z ≥ |f(y)| dy s {|f|>s} where the last inequality follows from the fact that |f| ≤ s for almost all x 6∈ Ω which means S n/2 that the set {|f| > s} is contained in Ω = k Qk. Now it suﬃces to take s = tωnn .

2.5. Integrability of the maximal function. In Remark 2.3 we observed that if f ∈ L1(Rn), then Mf 6∈ L1(Rn). However, we showed that the function cannot be globally integrable. It turns out that, in general, the maximal function need not be even locally integrable. We will actually characterize all functions such that the maximal function is locally integrable. Deﬁnition 2.34. We say that a measurable function f belongs to the Zygmund space L log L if |f| log(e + |f|) ∈ L1.

It is easy to see that for a space with ﬁnite measure we have \ Lp ⊂ L log L ⊂ L1 , p>1 HARMONIC ANALYSIS 15 so the Zygmund space is an intermediate space between all Lp for p > 1 and L1.

Theorem 2.35 (Stein). Suppose that a measurable function f deﬁned in Rn equals zero outside a ball B. Then Mf ∈ L1(B) if and only if f ∈ L log L(B).

Proof. Suppose that f ∈ L log L(B).5 Cavalieri’s principle (Theorem 2.5) and weak type estimates from Proposition 2.33 give Z Z ∞ Mf(x) dx = |{x ∈ B : Mf(x) > t}| dt B 0 Z ∞ ≤ |B| + |{x ∈ B : Mf(x) > t}| dt 1 Z ∞ 2 · 5n Z ≤ |B| + |f(x)| dx dt 1 t {|f|>t/2} Z Z max{2|f(x)|,1} dt = |B| + C(n) |f(x)| dx B 1 t Z = |B| + C(n) |f(x)| max{0, log(2|f(x)|)} dx B Z ≤ |B| + 2C(n) |f(x)| log(e + |f(x)|) dx < ∞. B

To prove the other implication we assume Mf ∈ L1(B) and we want to prove that f ∈ L log L(B). In the proof of the ﬁrst implication we used the right inequality from Proposition 2.33, but now we want to use the left inequality. However, the following calculations seem to disprove integrability of Mf on B. Z Z ∞ Z ∞ C Z Mf(x) dx = |{x ∈ B : Mf(x) > t}| dx ≥ 1 |f(x)| dx dt B 0 0 t {|f(x)|>C2t} Z Z |f(x)|/C2 dt = C1 |f(x)| dx = +∞. B 0 t | {z } = +∞ if f(x) 6= 0. These calculations are erroneous. Although the function f is supported in the ball B only, the maximal function Mf is positive on all of Rn and in the left estimate in Proposition 2.33 we deal with the measure of the set n {x ∈ R : Mf(x) > t} and not of the set {x ∈ B : Mf(x) > t}. We will show not how to overcome this diﬃculty.

1 n Note that Mf ∈ Lloc(R ). Indeed, Mf is integrable in B and locally bounded outside the closed ball B, so we need to verify integrability of Mf in a neighborhood of the boundary of B.6 But it is easy to see that if x is near the boundary and outside the ball,

5The proof of integrability of Mf will be similar to the proof of Lp integrability of the maximal function when f ∈ Lp, p > 1. 6 Integrability on B and local boundedness outside B does not imply local integrability in Rn. The function f(x) = 0 for x ≤ 0 and f(x) = 1/x for x > 0 is integrable on (−∞, 0] and is locally bounded on (0, ∞). 16 PIOTR HAJLASZ we can estimate the value of Mf(x) by (constant times) the value of the maximal function in a point being the reﬂection of x across the boundary. Thus the integrability of Mf near the boundary follows from the integrability of Mf in B. Note also that the set {Mf > 1} is bounded, because Mf(x) decays to zero as x → ∞. Thus local integrability of Mf and boundedness of the set {Mf > 1} implies that the function Mf in integrable in {Mf > 1}. With these remarks we can complete the proof as follows. Z ∞ > Mf(x) dx {Mf>1} Z ∞ ≥ |{Mf > t}| dt 1 Z ∞ 2−nC−1 Z ≥ |f(x)| dx dt 1 t {|f|>Ct} ! Z Z max{|f(x)|/C,1} dt = 2−nC−1 |f(x)| dx B 1 t Z = 2−nC−1 |f(x)| max{0, log(|f(x)|/C)} dx . B This easily implies that f ∈ L log L. Remark 2.36. A more careful analysis leads to the following version of Stein’s theorem. In an open set Ω ⊂ Rn we deﬁne the local maximal function by nZ o MΩf(x) = sup |f| : x ∈ Q ⊂ Ω , Q where the supremum is taken over all cubes Q in Ω that contain x.

n 1 Theorem 2.37 (Stein). Let Q ⊂ R be a cube. Then MQf ∈ L (Q) if and only if f ∈ L log L(Q). Moreover Z Z Z −(n+1) |f| n+2 5 MQf ≤ |f| log e + ≤ 2 MQf , Q Q |f|Q Q where Z |f|Q = |f| . Q

p 2.6. The Minkowski integral inequality. Let Fk ∈ L (Y, ν) and denote the variable by y. If we write F (k, y) instead of Fk(y), then the classical Minkowski inequality k|F1| + ... + |Fm|kp ≤ kF1kp + ... + kFmkp can be rewritten as

!1/p 1/p Z X p X Z (2.12) |F (k, y)| dν(y) ≤ |F (k, y)|p dν(y) . Y k k Y Since the sum over k can be regarded as an integral with respect to the counting measure, the following result is a natural generalization of the Minkowski inequality. HARMONIC ANALYSIS 17

Theorem 2.38 (Minkowski’s integral inequality). If µ and ν are σ-ﬁnite measures on X and Y respectively and if F : X × Y → R is measurable, then for 1 ≤ p < ∞ we have Z Z p 1/p Z Z 1/p |F (x, y)| dµ(x) dν(y) ≤ |F (x, y)|p dν(y) dµ(x) . Y X X Y Remark 2.39. It would be natural to try approximate F by simple functions in x and apply the discrete version of the inequality (2.12). The proof given below uses however, a very diﬀerent and much more elegant argument.

Proof. If p = 1 then we actually have equality by Fubini’s theorem. Thus assume that 1 < p R p < ∞. We will use the fact that the L (Y, ν) norm of the function y 7→ X |F (x, y)| dµ(x) equals to the norm of the corresponding functional on Lq(Y, ν), p−1 + q−1 = 1. Z Z p 1/p |F (x, y)| dµ(x) dν(y) Y X Z Z = sup h(y) |F (x, y)| dµ(x) dν(y) h∈Lq(ν) Y X khkq=1 Z Z = sup h(y)|F (x, y)| dν(y) dµ(x) h∈Lq(ν) X Y khkq=1 Z Z 1/q Z 1/p ≤ sup |h(y)|q dν(y) |F (x, y)|p dν(y) dµ(x) h∈Lq(ν) X Y Y khkq=1 | {z } 1 Z Z 1/p = |F (x, y)|p dν(y) dµ(x) . X Y The proof is complete.

2.7. Convergence of averages. In this section we will show that a large class of averages of f converge to f. Our approach will be based on the maximal function estimates. This will lead to far reaching generalizations of the Lebesgue Diﬀerentiation Theorem 2.20. Note that in the proofs of Theorem 2.20 and its generalizations discussed above (Theorems 2.28 and 2.29), maximal function also plays an important role. The results of this section will be used later in Section 3.3 Since many averages are constructed with the help of convolution, let us recall that the convolution of measurable functions f and g is deﬁned as Z Z (f ∗ g)(x) = f(x − y)g(y) dy = f(y)g(x − y) dy. Rn Rn The equality of the integrals follows from a linear change of variables. This equality actually means that f ∗ g = g ∗ f. One can also easily check that (f ∗ g) ∗ h = f ∗ (g ∗ h).

Theorem 2.40. Let g ∈ L1(Rn). If f ∈ Lp(Rn), 1 ≤ p < ∞, then f ∗ g ∈ Lp(Rn) and

(2.13) kf ∗ gkp ≤ kfkpkgk1. 18 PIOTR HAJLASZ

n n If f ∈ C0(R ), then f ∗ g ∈ C0(R ) and

(2.14) kf ∗ gk∞ ≤ kfk∞kgk1.

Proof. Let f ∈ Lp(Rn), 1 ≤ p < ∞. The inequality can be obtained from the Minkowski Integral Inequality (Theorem 2.38) as follows Z Z p 1/p

kf ∗ gkp = f(x − y)g(y) dy dx Rn Rn Z Z p 1/p ≤ |f(x − y)| |g(y)| dy dx Rn Rn | {z } |{z} dµ dν Z Z 1/p ≤ |f(x − y)|p dx |g(y)| dy Rn Rn = kfkpkgk1 . n The proof in the case of f ∈ C0(R ) is left to the reader as an exercise. Remark 2.41. In the above proof one could use the classical Minkowski inequality instead of the integral one via the following estimate Z Z |f(x − y)| |g(y)| dy = |f(x − y)| |g(y)|1/p|g(y)|1/q dy Rn Rn Z 1/p Z 1/q ≤ |f(x − y)|p|g(y)| dy |g(y)| dy . Rn Rn We leave details to the reader.

Theorem 2.40 is a special case of a more general Young’s inequality. We leave the proof as an exercise. Theorem 2.42 (Young’s inequality). If 1 ≤ p, q, r ≤ ∞ and q−1 = p−1 + r−1 − 1, then

kf ∗ gkq ≤ kfkpkgkr .

For ϕ ∈ L1(Rn) and t > 0 we deﬁne x ϕ (x) = t−nϕ . t t R R Note that n ϕt = n ϕ. R R −1 n −1 Example 2.43. If ϕ = ωn χB(0,1), then ϕt = (ωnt ) χB(0,t) so 1 Z Z f ∗ ϕt(x) = n f(x − y) dy = f(y) dy. ωnt B(0,t) B(x,t) 7 1 1 Thus in that case f ∗ ϕt → f a.e. and in L , provided f ∈ L . Note also that

(2.15) sup(|f| ∗ ϕt)(x) = Mf(x). t>0

7See Lemma 2.19 and Theorem 2.20. HARMONIC ANALYSIS 19

8 Actually this identity plays a crucial role in the proof of the pointwise convergence f ∗ϕt → f a.e. In what follows we will study convergence of f ∗ ϕt for more general functions ϕ. Not surprisingly we will be looking for conditions that will guarantee an estimate similar to (2.15).9

1 n −n p n Theorem 2.44. Let ϕ ∈ L (R ) and let ϕt(x) = t ϕ(x/t) for t > 0. If f ∈ L (R ), n 1 ≤ p < ∞ or f ∈ C0(R ) and p = ∞, then Z (2.16) lim kf ∗ ϕt − afkp = 0, where a = ϕ(x) dx. t→0+ Rn Remark 2.45. The two most interesting cases are when a = 1 and when a = 0.

n n 10 Remark 2.46. If f ∈ C0(R ), then f ∗ ϕt ∈ C0(R ) and (2.16) with p = ∞ means that f ∗ ϕt converges to af uniformly. p Remark 2.47. If a = 1, then (2.16) means that f ∗ ϕt → f in L so the operators f 7→ f ∗ ϕt converge pointwise, i.e. for every f, to the identity operator f 7→ f. For that reason the family of functions {ϕt : t > 0} is often called an approximation of identity (provided a = 1).

Proof. We will prove the result when f ∈ Lp(Rn), 1 ≤ p < ∞, leaving the case of f ∈ n C0(R ) to the reader. We have Z

Under assumptions of Theorem 2.44, if f ∈ Lp(Rn), 1 ≤ p < ∞, there is a sequence ti → 0 such that f ∗ ϕti → af a.e. as i → ∞, but a more challenging question is to ﬁnd conditions that would guarantee f ∗ ϕt → af a.e. as t → 0. That would be a true generalization of the Lebesgue Diﬀerentiation Theorem.

8Along with weak type estimates for the maximal function, see Theorem 2.20. 9See Theorem 2.49. 10See Theorem 2.40. 20 PIOTR HAJLASZ

Deﬁnition 2.48. Let ϕ ∈ L1(Rn). We say that Ψ is a radially decreasing majorant of ϕ if (a) Ψ(x) = η(|x|) for some11 η : [0, ∞) → [0, ∞].12 (b) η is decreasing.13. (c) |ϕ(x)| ≤ Ψ(x) a.e.

Every function ϕ ∈ L1 has the least radially decreasing majorant14

Ψ0(x) = ess sup |ϕ(y)|. |y|≥|x| Thus the existence of an integrable radially decreasing majorant Ψ of ϕ is equivalent to 1 Ψ0 ∈ L . Theorem 2.49. Suppose that ϕ ∈ L1(Rn) has an integrable radially decreasing majorant 1 n Ψ(x) = η(|x|) ∈ L (R ) . 1 n n Then for f ∈ Lloc(R ) and all x ∈ R

(2.17) sup(f ∗ ϕt)(x) ≤ kΨk1Mf(x) . t>0 If in addition f ∈ Lp( n), 1 ≤ p < ∞, and R ϕ(x) dx = a, then R Rn

(2.18) f ∗ ϕt → af a.e. as t → 0.

Proof. In order to prove (2.17) it suﬃces to prove that for a radially decreasing and integrable function Ψ we have

(2.19) |f| ∗ Ψ(x) ≤ kΨk1Mf(x).

Indeed, since Ψt is also radially decreasing and integrable, applying (2.19) to Ψt will give

|f ∗ ϕt(x)| ≤ |f| ∗ Ψt(x) ≤ kΨtk1Mf(x) = kΨk1Mf(x). Thus it remains to prove (2.19). For a positive integer k let

k·2k k·2k X 1 X 1 s (x) = χ k (x) = χ (x) . k 2k {Ψ≥i/2 } 2k Bi i=1 i=1 Since Ψ is radially symmetric the sets n i o B = x : Ψ(x) ≥ i 2k are balls15 centered at zero.

11Functions of this form, i.e. functions constant on spheres Sn−1(0, r) are called radially symmetric. 12It may happen that η(t) = +∞ for all t. 13i.e. non-increasing 14 However, it may happen that Ψ0 = +∞ everywhere. 15Open or closed – ﬁnd an example such that for some t the balls are open while for other values of t they are closed. HARMONIC ANALYSIS 21

The sequence {sk} is increasing and convergent to Ψ almost everywhere. Indeed, for 16 x 6= 0, Ψ(x) < ∞ so Ψ(x) < k0, for some k0 ∈ N. Then for k ≥ k0 ` ` + 1 ≤ Ψ(x) < , for some ` = 0, 1, 2, . . . k · 2k − 1. 2k 2k Since the inequality Ψ ≥ i/2k is satisﬁed for i = 1, 2, . . . , `, the deﬁnition of the function k sk gives that sk(x) = `/2 so 1 Ψ(x) − < s (x) ≤ Ψ(x). 2k k

That means sk(x) → Ψ(x) for all all x 6= 0. Observe also that the sequence {sk} is k k k+1 increasing, sk ≤ sk+1. Indeed, if sk(x) = `/2 , then Ψ(x) ≥ `/2 = (2`)/(2 ) so the deﬁnition of sk+1 gives 1 ` s (x) ≥ · 2` = = s (x). k+1 2k+1 2k k Thus if we can prove

(2.20) |f| ∗ sk(x) ≤ kΨk1Mf(x), inequality (2.19) will follow from the Monotone Convergence Theorem. We have k·2k X |Bi| χB |f| ∗ s (x) = |f| ∗ i (x). k 2k |B | i=1 i Since17 Z k·2k χB X |Bi| |f| ∗ i (x) = |f(x − y)| dy ≤ Mf(x) and = ks k ≤ kΨk , |B | 2k k 1 1 i Bi i=1 inequality (2.20) follows. The proof of (2.17) is complete. It remains now to prove convergence (2.18). If p = 1, then

(2.21) Ωf(x) := lim sup f ∗ ϕt − lim inf f ∗ ϕt ≤ 2kΨk1Mf(x) t→0+ t→0+ and almost the same argument as in the proof of Theorem 2.20 yields that18 Ωf = 0 a.e. 1 so f ∗ ϕt converges a.e. to a measurable function. Since f ∗ ϕt → af in L (Theorem 2.44), (2.18) follows.

The case 1 < p < ∞ uses a similar argument. If f ∈ Lp(Rn), 1 < p < ∞, then Chebyschev’s inequality (2.2) and Theorem 2.2(b) yield Z p p 1 p |{x : Mf(x) > t}| = |{x : |Mf(x)| > t }| ≤ p |Mf(x)| dx t Rn Z C(n, p) p ≤ p |f(x)| dx. t Rn 16Because Ψ ∈ L1. 17 Because the balls Bi are centered at the origin. 18We will actually repeat this argument in the case of 1 < p < ∞. 22 PIOTR HAJLASZ

Hence for ε > 0, inequality (2.21) yields Z C p |{x :Ωf(x) > ε}| ≤ p |f(x)| dx, ε Rn n 19 where the constant C depends on n, p and kΨk1. If h ∈ C0(R ), then Ωh=0. Taking n p p+1 h ∈ C0(R ) such that kf − hkp ≤ ε we conclude that |{x :Ωf(x) > ε}| = |{x : Ω(f − h)(x) > ε}| < Cε so Ωf = 0 a.e. and exactly as in the case of p = 1, (2.18) follows.

19 Because h ∗ ϕt converges uniformly (and hence pointwise) to h. HARMONIC ANALYSIS 23

3. The Fourier transform

3.1. Fourier transform.

Deﬁnition 3.1. For f ∈ L1(Rn) we deﬁne the Fourier transform as Z n ˆ −2πix·ξ X F(f)(x) = f(ξ) = f(x)e dx where x · ξ = xjξj . n R j=1 Theorem 3.2. The Fourier transform has the following properties

∧ 1 n n ˆ (a) : L (R ) → C0(R ) is a bounded linear operator with kfk∞ ≤ kfk1.

(b) (f ∗ g)ˆ = fˆgˆ for f, g ∈ L1(Rn).

(c) If f, g ∈ L1(Rn), then fg,ˆ fgˆ ∈ L1(Rn) and Z Z fˆ(x)g(x) dx = f(x)ˆg(x) dx. Rn Rn (d) If τhf(x) = f(x + h), then ˆ 2πih·ξ 2πih·x ˆ (τhf)ˆ(ξ) = f(ξ)e and (f(x)e )ˆ(ξ) = f(ξ − h). −n (e) If ft(x) = t f(x/t), then ˆ ˆ (ft)ˆ(ξ) = f(tξ) and (f(tx))ˆ(ξ) = (f)t(ξ) . (f) If ρ ∈ O(n) is an orthogonal transformation, then (f(ρ ·))ˆ(ξ) = fˆ(ρξ).

∧ 1 n ∞ n ˆ Proof. (a) Clearly, : L (R ) → L (R ) is a bounded linear mapping with kfk∞ ≤ kfk1. Indeed, Z ˆ −2πix·ξ |f(ξ)| ≤ f(x)e dx = kfk1. Rn The Dominated Convergence Theorem implies that the function fˆ is continuous. It remains to prove that fˆ(ξ) → 0 as |ξ| → ∞. Let ξ 6= 0. Since eπi = −1 we have Z Z fˆ(ξ) = f(x)e−2πix·ξ dx = − f(x)e−2πix·ξeπi dx Rn Rn Z ξ = − f(x) exp − 2πi x − 2 · ξ dx Rn 2|ξ| 24 PIOTR HAJLASZ Z ξ −2πix·ξ = − f x + 2 e dx . Rn 2|ξ| Hence Z ˆ 1 ξ −2πix·ξ f(ξ) = f(x) − f x + 2 e dx 2 Rn 2|ξ| and thus Lemma 2.18 yields 1 |fˆ(ξ)| ≤ f − τ ξ f → 0 as |ξ| → ∞. 2 2|ξ|2 1 (b) Z Z (f ∗ g)ˆ(ξ) = f(x − y)g(y) dy e−2πix·ξ dx Rn Rn Z Z = f(x − y)e−2πi(x−y)·ξ dx g(y)e−2πiy·ξ dy Rn Rn = fˆ(ξ)ˆg(ξ) . (c) fg,ˆ fgˆ ∈ L1, because the functions f,ˆ gˆ are bounded and the equality of the integrals easily follows from the Fubini theorem. (d) Z Z −2πix·ξ −2πi(x−h)·ξ (τhf)ˆ(ξ) = f(x + h)e dx = f(x)e dx Rn Rn Z = e2πih·ξ f(x)e−2πix·ξ dx = e2πih·ξfˆ(ξ) . Rn The second equality follows from a similar argument. (e) Z Z −n x −2πix·ξ −2πi(ty)·ξ (ft)ˆ(ξ) = t f e dx = f(y)e dy Rn t Rn Z = f(y)e−2πiy·(tξ) dy = fˆ(tξ) . Rn The second formula follows from the ﬁrst one if we replace t by t−1. (f)20

Z Z (f(ρ ·))ˆ(ξ) = f(ρx)e−2πix·ξ dx = f(y)e−2πi(ρ−1y)·ξ dy Rn Rn Z = f(y)e−2πiy·(ρξ) dy = fˆ(ρξ) . Rn Equality (ρ−1y) · ξ = y · (ρξ) follows from the fact that the mapping x 7→ ρx is an isometry. 20This property is in some sense obvious: the Fourier transform does not depend on the choice of the coordinate system since it depends on the scalar product x · ξ. Thus the Fourier transform should stay the same if we rotate the coordinate system by ρ. HARMONIC ANALYSIS 25

1 n 1 n Theorem 3.3. Suppose f ∈ L (R ) and xkf(x) ∈ L (R ), where xk is the k-th coordinate ˆ function. Then f is diﬀerentiable with respect to ξk and ∂fˆ (−2πixkf(x))ˆ = (ξ) . ∂ξk

Proof. Let ek be the unit vector along the k-th coordinate. Then the second part of The- orem 3.2(d) gives ∧ fˆ(ξ + he ) − fˆ(ξ) e−2πi(hek)·x − 1 k = f(x) (ξ) → (−2πix f(x))ˆ(ξ) . h h k 1 The convergence follows from the continuity of the Fourier transform in L . 1 n p Definition 3.4. We say that f ∈ Lloc(R ) is L -differentiable with respect to xk if there is g ∈ Lp(Rn) such that Z p f(x + hek) − f(x) − g(x) dx → 0 as h → 0. Rn h p The function g is called the L -partial derivative of f with respect to xk and is denoted by g = ∂f/∂xk. 1 n 1 Theorem 3.5. If f ∈ L (R ) is L -differentiable with respect to xk, then ∧ ∂f ˆ = 2πiξkf(ξ) . ∂xk

Proof. The first part of Theorem 3.2(d) and the L1-differentiability of f give ∧ ∧ ∂f e2πi(hek)·ξ − 1 ∂f f(x + he ) − f(x) − fˆ(ξ) = − k → 0 ∂xk h ∂xk h as h → 0, so ∧ 2πi(hek)·ξ ∂f ˆ e − 1 ˆ (ξ) = lim f(ξ) = 2πiξkf(ξ) . ∂xk h→0 h The proof is complete. Remark 3.6. With each polynomial X α P (x) = aαx |α|≤m of variables x1, . . . , xn we associate a differential operator |α| X α X ∂ P (D) = aαD = aα α . ∂x 1 . . . ∂xαn |α|≤m |α|≤m 1 n Then under suitable assumptions Theorems 3.3 and 3.5 have the following higher order generalizations (3.1) P (D)fˆ(ξ) = (P (−2πix)f(x))ˆ(ξ), (P (D)f)ˆ(ξ) = P (2πiξ)fˆ(ξ).

The next result is our ﬁrst (and actually one of the most important) example of the Fourier transform. 26 PIOTR HAJLASZ

Theorem 3.7. If f(x) = e−π|x|2 , then fˆ(ξ) = e−π|ξ|2 .

Proof. First observe that it suffices to prove the result in dimension 1. Indeed, assuming it for n = 1, the case of general n follows from Fubini’s theorem Z n Z ∞ n 2 Y 2 Y 2 2 ˆ −π|x| −2πix·ξ −πxk −2πixkξk −πξk −π|ξ| f(ξ) = e e dx = e e dxk = e = e . n R k=1 −∞ k=1 In the case n = 1 we will show two different proofs. The first one will use the contour integration while the second one will use differential equations. Thus in the remaining part of the proof we will assume that f(x) = e−πx2 is a function of one variable. Proof by contour integration. The function e−πz2 is holomorphic and hence its integral along the following curve equals zero.

Letting R → ∞ we obtain Z ∞ Z ∞ e−πx2 dx = e−π(x+iξ)2 dx . −∞ −∞ The left hand side equals 1, so Z ∞ 1 = e−πx2 e−2πixξeπξ2 dx . −∞ Hence Z ∞ e−πξ2 = e−πx2 e−2πixξ dx = fˆ(ξ) . −∞ The proof is complete. Proof by diﬀerential equations. We have f 0(x) = −2πxf(x), −i(−2πixf(x)) = f 0(x) so Theorems 3.3 and 3.5 yield ∧ 0 −i − 2πixf(x) (ξ) = fb0(ξ) , (fˆ) (ξ) = −2πξfˆ(ξ). | {z } | {z } (fˆ)0(ξ) 2πiξfˆ(ξ) HARMONIC ANALYSIS 27

Solving this diﬀerential equation for the unknown function ξ 7→ fˆ(ξ) yields Z ∞ fˆ(ξ) = fˆ(0)e−πξ2 = e−πξ2 , because fˆ(0) = e−πx2 dx = 1. −∞

3.2. Measures of finite total variation. Let us first recall basic facts regarding measures of finite total variation.

If µ is a complex (Borel) measure on Rn, then there is a unique positive measure |µ| called the total variation of measure µ and a Borel function h : Rn → C, |h(x)| = 1 for all x ∈ Rn such that Z n µ(E) = h(x) d|µ|(x) for all Borel sets E ⊂ R . E The space B(Rn) of complex measures of ﬁnite total variation kµk := |µ|(Rn) is a Banach space with the norm k · k.

Note that f ∈ L1(Rn) defines a measure of finite total variation by Z µ(E) = f(x) dx. E In that case Z Z |µ|(E) = |f(x)| dx so kµk = |f(x)| dx = kfk1 < ∞. E Rn This proves that L1(Rn) is a closed subspace of B(Rn). In a special case when µ is a real valued measure, called signed measure, we have h : n + − R → {±1} so if we define h = hχ{h=1} and h = hχ{h=−1}, then Z µ±(E) = h±(x) d|µ|(x) E are positive measures such that (3.2) µ = µ+ − µ− and |µ| = µ+ + µ−. The representation (3.2) is called the Hahn decomposition of µ. Note that the measures µ+ and µ− are supported on disjoint sets.

n Theorem 3.8 (Riesz representation theorem). The dual space to C0(R ) is isometri- cally isomorphic to the space of measures of ﬁnite total variation. More precisely, if n ∗ Φ ∈ (C0(R )) , then there is a unique measure µ of ﬁnite total variation such that Z n Φ(f) = f dµ for f ∈ C0(R ). Rn Moreover kΦk = kµk = |µ|(Rn). 28 PIOTR HAJLASZ

This result allows us to deﬁne convolution of measures.

If f, g ∈ L1(Rn), then f ∗ g ∈ L1(Rn) ⊂ B(Rn) and hence it acts as a functional on n C0(R ) by the formula Z Z Z Φ(h) = h(x)(f ∗ g)(x) dx = h(x) f(x − y)g(y) dy dx Rn Rn Rn Z Z = h(x)f(x − y) dx g(y) dy Rn Rn Z Z = h(x + y)f(x)g(y) dx dy . Rn Rn This suggests how to deﬁne convolution of measures.

n If µ1, µ2 ∈ B(R ), then Z Z Φ(h) = h(x + y) dµ1(x) dµ2(y) Rn Rn n n deﬁnes a functional on C0(R ) and hence there is a unique measure µ ∈ B(R ) such that Z Z Z n h(x + y) dµ1(x) dµ2(y) = h(x) dµ(x) for all h ∈ C0(R ). Rn Rn Rn Deﬁnition 3.9. We denote the measure µ by

µ = µ1 ∗ µ2 and call it convolution of measures.

Clearly

µ1 ∗ µ2 = µ2 ∗ µ1 and kµ1 ∗ µ2k ≤ kµ1k kµ2k.

If dµ1 = f dx, dµ2 = g dx, then

d(µ1 ∗ µ2) = (f ∗ g) dx, so the convolution of measures extends the notion of convolution of functions. If dµ1 = f dx and µ ∈ B(Rn), then Z Z Z Z h(x + y) dµ1(x) dµ(y) = h(x + y)f(x) dx dµ(y) Rn Rn Rn Rn Z Z = h(x)f(x − y) dx dµ(y) Rn Rn Z Z = h(x) f(x − y) dµ(y) dx. Rn Rn Thus µ1 ∗ µ can be identiﬁed with a function Z 1 n x 7→ f(x − y) dµ(y) ∈ L (R ), Rn so we can write Z f ∗ µ = f(x − y) dµ(y), kf ∗ µk1 ≤ kfk1 kµk. Rn HARMONIC ANALYSIS 29

Theorem 3.10. If 1 ≤ p < ∞, f ∈ Lp(Rn) and µ ∈ B(Rn), then

kf ∗ µkp ≤ kfkp kµk.

Proof is almost the same as that for Theorem 2.40 as we leave it to the reader.

Definition 3.11. The Fourier transform of a measure of finite total variation µ ∈ B(Rn) is defined by Z µˆ(x) = e−2πix·ξ dµ. Rn Proposition 3.12. If µ ∈ B(Rn), then µˆ is a bounded and continuous function such that kµˆk∞ ≤ kµk.

Proof. Boundedness ofµ ˆ is obvious and continuity follows from the Dominated Conver- gence Theorem. n Proposition 3.13. If µ1, µ2 ∈ B(R ), then µ\1 ∗ µ2 =µ ˆ1 · µˆ2.

We leave the proof as an exercise. Remark 3.14. Convolution of measures and the Fourier transform of measures play a fundamental role in probability.

3.3. Summability methods. The answer to an important question of how to reconstruct a function f from its Fourier transform fˆ is provided by the inversion formula21 Z (3.3) f(x) = fˆ(ξ) e2πix·ξ dξ . Rn It seems that this formula requires both f and fˆ to be integrable. Integrability of f is needed for the existence of fˆ and integrability of fˆ is needed for the integral in (3.3) to make sense. We will actually prove in Corollary 3.26 that if f, fˆ ∈ L1, then (3.3) is true a.e. There is, however, a problem here. Equality (3.3) means that f(x) is the Fourier transform of fˆ ∈ L1 evaluated at −x. Since the Fourier transform of an integrable function fˆ ∈ L1 n n 1 ˆ belongs to C0(R ), it follows that f ∈ C0(R ). That means, if f ∈ L \ C0, then f cannot be integrable so the right hand side of (3.3) does not make much sense. Is there any chance to give meaning to the integral at (3.3), beyond the class of integrable Fourier transforms so that the equality at (3.3) would be true for all f ∈ L1? The answer is in the positive but we need to interpret the integral at (3.3) using suitable summability methods.22

21Compare it with the formula for Fourier series: Z 1 X fˆ(n) = f(x)e−2πixn dx, f(x) = fˆ(n)e2πixn. 0 n∈Z

22An example of a summability method for sequences is given by the so called Ces`aro method: with each sequence {an} we associate the limit of (a1 + ... + an)/n. This extends the notion of limit far beyond the class of convergent sequences. 30 PIOTR HAJLASZ

n If Φ ∈ C0(R ) is such that Φ(0) = 1, then for t > 0 we deﬁne the Φ-mean of a function f as Z MΦ,tf = Mtf = f(x)Φ(tx) dx. Rn 1 n R + If f ∈ L ( ), then MΦ,tf → n f(x) dx as t → 0 , but the limit of Mtf might also exist R R for some non-integrable functions f. R Deﬁnition 3.15. We say that n f is Φ-summable to ` ∈ if lim + MΦ,tf = `. R R t→0

In the case of Φ(x) = e−|x| and Φ(x) = e−|x|2 we obtain the so called the Abel and the Gauss summability methods. More precisely we have Deﬁnition 3.16. For each t > 0 the Abel mean and the Gauss mean of a function f are Z Z −t|x| −t|x|2 At(f) = f(x)e dx and Gt(f) = f(x) e dx . Rn Rn R 23 We say that n f(x) dx is Abel summable (Gauss summable) to ` ∈ if limt→0 At(f) = ` R R (limt→0 Gt(f) = `).

−|x|2 −t2|x|2 Remark 3.17. If Φ(x) = e , then Φ(tx) = e so MΦ,tf = Gt2 (f). This however, does not cause any problems when you compare notions of the Φ-summability and the Gauss summability because limt→0+ MΦ,tf = limt→o Gt2 (f). Remark 3.18. Clearly, in the case of integrable functions, the Abel and the Gauss methods give the usual integral. However, the means At(f) and Gt(f) are ﬁnite whenever f is bounded so for some bounded, but non-integrable f, the integral R f(x) dx might still Rn 1 n n be Abel or Gauss summable to a ﬁnite value. For example if f ∈ L (R ) \ C0(R ), the ˆ 2πix·ξ n the function ξ 7→ f(ξ)e is in C0(R ) but it is not integrable. However, as we will see (Theorems 3.23 and 3.30) for almost all x ∈ n, the integral R fˆ(ξ)e2πix·ξ dξ is both Abel R Rn and Gauss summable to f(x). This gives some meaning of the inversion formula (3.3) for generic f ∈ L1(Rn).

We want to ﬁnd a reasonably large class of Φ-means (including the Abel and the Gauss means) such that Z fˆ(ξ)e2πix·ξ dξ Rn is Φ-summable to f(x) i.e. Z ˆ 2πix·ξ + Mt(x) = f(ξ)e Φ(tξ) dξ → f(x) as t → 0 Rn 1 and we can consider both, convergence Mt → f almost everywhere and in L (with respect to variable x). The main result reads as follows Theorem 3.19. Let Φ ∈ L1( n) be such that ϕ = Φˆ ∈ L1( n) and R ϕ(x) dx = 1. For R R Rn f ∈ L1(Rn) let Z ˆ 2πix·ξ Mt(x) = f(ξ)e Φ(tξ) dξ Rn 23 R a R ∞ −tx Exercise. Prove that if lima→∞ 0 f(x) dx = `, then At = 0 f(x)e dx converges to ` as t → 0. HARMONIC ANALYSIS 31 be the Φ-mean of the integral (3.3). Then 1 (3.4) Mt → f in L as t → 0. If in addition ϕ = Φˆ has an integrable radially decreasing majorant, then

(3.5) Mt → f a.e. as t → 0.

Proof. The lemma below is a link between the Φ-mean and the approximation by convolution discussed in Section 2.7. Lemma 3.20. If f, Φ ∈ L1(Rn) and ϕ = Φˆ, then Z Z ˆ 2πix·ξ f(ξ)e Φ(tξ) dξ = f(x − y)ϕt(−y) dy = f ∗ ϕ˜t(x), where ϕ˜(y) = ϕ(−y). Rn Rn

Proof. For h ∈ Rn deﬁne (h) 2πix·h 1 n g (x) = e Φ(tx) ∈ L (R ) . Recall that24 2πih·x ∧ ∧ (w(x)e ) (ξ) =w ˆ(ξ − h) and (w(tx)) (ξ) = (w ˆ)t(ξ). Hence (h) ∧ ˆ (g ) (ξ) = (Φ)t(ξ − h) = ϕt(ξ − h) so replacing h by x we get (x) ∧ (g ) (ξ) = ϕt(ξ − x). We have Z Z Z fˆ(ξ)e2πix·ξΦ(tξ) dξ = fˆ(ξ)g(x)(ξ) dξ = f(ξ)(g(x))∧(ξ) dξ Rn Rn Rn Z Z = f(ξ)ϕt(ξ − x) dξ = f(x − y)ϕt(−y) dx. Rn Rn We used here Theorem 3.2(c) and in the last equality the change of variables −y = ξ−x.

Now (3.4) follows directly from Theorem 2.44 and (3.5) follows from Theorem 2.49. Remark 3.21. Theorem 3.19 is not easy to apply since given Φ ∈ L1, we have to compute Φˆ and show its integrability. In the case of the Gauss mean Φ(x) = e−|x|2 it can be easily done with the help of Theorem 3.7, but the case of the Abel mean Φ(x) = e−|x| is much harder due to diﬃculty of computing the Fourier transform of Φ(x) = e−|x|.

In Theorems 3.23 and 3.30 we will investigate the case of the the Gauss and the Abel and methods. Theorem 3.22. Let f(x) = e−4π2t|x|2 , t > 0. Then

(a) W (x, t) := fˆ(x) = (4πt)−n/2e−|x|2/(4t). (b) The function W has the following scaling property with respect to t: if ϕ(x) = W (x, 1), then W (x, t) = ϕt1/2 (x).

24see Theorem 3.2. 32 PIOTR HAJLASZ

(c) Z W (x, t) dx = 1 for all t > 0. Rn Proof. We can write f(x) = e−4π2t|x|2 = u((4πt)1/2x), where u(x) = e−π|x|2 . ∧ ˆ −π|ξ|2 25 Since (ψ(tx)) (ξ) = (ψ)t(ξ) andu ˆ(ξ) = u(ξ) = e we obtain ˆ −n/2 −|ξ|2/(4t) W (ξ, t) = f(ξ) = (ˆu)(4πt)1/2 (ξ) = u(4πt)1/2 (ξ) = (4πt) e which is (a). In particular

ϕ(ξ) = W (ξ, 1) = u(4π)1/2 (ξ).

Since (ψt1 )t2 (x) = ψt1t2 (x) we get W (ξ, t) = u 1/2 (ξ) = u 1/2 (ξ) = ϕ 1/2 (ξ) (4πt) (4π) t1/2 t R R which is (b). Finally equality n ψt = n ψt , t1, t2 > 0 along with W (ξ, t) = ϕ 1/2 (ξ) R 1 R 2 t yields (c): Z Z 1 Z 2 W (ξ, t) dξ = W ξ, dξ = e−π|ξ| dξ = 1. Rn Rn 4π Rn

As an application we obtain

Theorem 3.23 (The Gauss-Weierstrass summability method). If f ∈ L1(Rn), then Z Z (3.6) fˆ(ξ)e2πix·ξe−4π2t|ξ|2 dξ = f(y)W (x − y, t) dy → f as t → 0+ Rn Rn both in L1(Rn) and almost everywhere.

Proof. First we will prove equality. Let Φ(x) = e−4π2|x|2 . Using notation from Theorem 3.22 ˆ Φ(x) = W (x, 1) = ϕ(x). Since ϕ(x) = ϕ(−x),ϕ ˜t1/2 (x) = ϕt1/2 (x) = W (x, t). Thus Lemma 3.20 yields Z Z fˆ(ξ)e2πix·ξe−4π2t|ξ|2 dξ = fˆ(ξ)e2πix·ξΦ(t1.2ξ) dξ Rn Rn Z = f ∗ ϕ˜t1/2 (x) = W (x − y, t)f(y) dy. Rn Theorem 3.22 shows that the function Φ satisﬁes the assumptions of Theorem 3.19 so both 1 convergences in L and almost everywhere follow.

Remark 3.24. The function W (x, t) = ϕt1/2 (x) satisﬁes the assumptions of Theorem 2.49 with a = 1 so for f ∈ Lp, 1 ≤ p < ∞ Z (3.7) f(y)W (x − y, t) dy → f a.e. and in Lp as t → 0+. Rn 25The function W is deﬁned as a Fourier transform so it will be more convenient for us to prove (a)-(c) with the variable x replaced by ξ. HARMONIC ANALYSIS 33

Remark 3.25. W (x, t) is called the Gauss-Weierstrass kernel. Under suitable assumptions about f, diﬀerentiation under the sign of the integral shows that the function

Z 1 Z |x−y|2 − 4t w(x, t) = W (x − y, t)f(y) dy = n/2 e f(y) dy Rn (4πt) Rn ∂w n+1 satisﬁes the heat equation ∂t = ∆xw in the interior of R+ . As we showed, the function w(x, t) converges to f(x) as t → 0, so w is a solution to ∂w n+1 ∂t = ∆xw on R+ , w(x, 0) = f(x), x ∈ Rn. Corollary 3.26. If both f and fˆ are integrable26, then Z (3.8) f(x) = fˆ(ξ)e2πix·ξ dξ a.e. Rn

Proof. Since fˆ ∈ L1, the left hand side of (3.6) converges to R fˆ(ξ)e2πix·ξ dξ so the result Rn follows from Theorem 3.23.

The above inversion formula motivates the following definitions. Definition 3.27. The inverse Fourier transform is defined by Z fˇ(x) = F −1(f)(x) = f(ξ)e2πix·ξ dξ. Rn 1 n ˆ ˆ n Corollary 3.28. If f1, f2 ∈ L (R ) and f1 = f2 on R , then f1 = f2 a.e.

1 ˆ 1 Proof. Let f = f1 − f2. Then f ∈ L and f = 0 ∈ L so (3.8) yields that f = 0 a.e.

The following result provides another example of a function that satisﬁes the assumptions of Theorem 3.19. Theorem 3.29. Let f(x) = e−2π|x|t, t > 0. Then

(a) t P (x, t) := fˆ(x) = c , n (t2 + |x|2)(n+1)/2 where Γ n+1 c = 2 . n π(n+1)/2 (b) The function P has the following scaling property with respect to t: if ϕ(x) = P (x, 1), then P (x, t) = ϕt(x). (c) Z P (x, t) dx = 1 for all t > 0. Rn By the same arguments as before we obtain.

26The assumption about integrability of fˆ is very strong. As already observed, the equality (3.8) implies n that f equals a.e. to a function in C0(R ). 34 PIOTR HAJLASZ

Theorem 3.30 (The Abel summability method). If f ∈ L1(Rn), then Z Z fˆ(ξ)e2πix·ξe−2π|ξ|t dξ = f(y)P (x − y, t) dy → f as t → 0+ Rn Rn both in L1(Rn) and almost everywhere.

Remark 3.31. The function P (x, t) = ϕt(x) satisfies the assumptions of Theorem 2.49 with a = 1 so for f ∈ Lp, 1 ≤ p < ∞ Z f(y)P (x − y, t) dy → f a.e. and in Lp as t → 0+. Rn Remark 3.32. P (x, t) is called the Poisson kernel. Under suitable assumptions about f, differentiation under the sign of the integral shows that the function Z u(x, t) = P (x − y, t)f(y) dy Rn n+1 satisfies the Laplace equation ∆(x,t)u = 0 in the interior of R+ . As we showed u(x, t) converges to f as t → 0+ so u(x, t) is a solution to the Dirichlet problem in the half-space n+1 ∆(x,t)u = 0 on R+ , u(x, 0) = f(x), x ∈ Rn

The proof of Theorem 3.29 is substantially more difficult than that of Theorem 3.22. Since the formula for the Fourier transform involves the Γ function we need to recall its basic properties. Definition 3.33. For 0 < x < ∞ we define Z ∞ Γ(x) = tx−1e−t dt . 0 Theorem 3.34.

(a) Γ(x + 1) = xΓ(x) for all 0 < x < ∞. (b) Γ(n + 1) =√n! for all n = 0, 1, 2, 3,... (c) Γ(1/2) = π.

Proof. (a) follows from the integration by parts. Since Γ(1) = 1, (b) follows from (a) by induction. The substitution t = s2 gives Z ∞ Z ∞ Z ∞ Γ(x) = tx−1e−t dt = s2(x−1)e−s2 2s ds = 2 s2x−1e−s2 ds 0 0 0 and hence ∞ 1 Z 2 √ Γ = 2 e−s ds = π . 2 0

The formula Γ(x) = Γ(x + 1)/x allows us to deﬁne Γ(x) for all negative non-integer values of x. For example √ 3 Γ(−1/2) Γ(1/2) 4 π Γ − = = = . 2 −3/2 (−3/2)(−1/2) 3 HARMONIC ANALYSIS 35

Deﬁnition 3.35. For x ∈ (−∞, 0) \ {−1, −2, −3,...}, the function Γ(x) is deﬁned by Γ(x + k) Γ(x) = , x(x + 1) · ... · (x + k − 1) where k is a positive integer such that x + k ∈ (0, 1).

Lemma 3.36. √ Z π/2 n+1 n πΓ 2 sin θ dθ = n for n = 1, 2, 3,... 0 nΓ 2

Proof. Denote the left hand side by an and the right hand side by bn. Easy one time integration by parts27 shows that n + 1 a = (n + 1)(a − a ), a = a . n+2 n n+2 n+2 n + 2 n Also elementary properties of the Γ function show that n + 1 b = b n+2 n + 2 n and now it is enough to observe that a1 = 1 = b1 = 1, a2 = π/4 = b2. Lemma 3.37.

(a) The volume of the unit ball in Rn equals 2πn/2 πn/2 (3.9) ω = = . n nΓ(n/2) n Γ 2 + 1 n (b) The (n − 1)-dimensional measure of the unit sphere in R equals nωn

Proof.

It follows from the picture and the Fubini theorem that the volume of the upper half of the ball equals Z 1 1 n−1 ωn = ωn−1r(h) dh . 2 0 The substitution h = 1 − cos θ, dh = sin θ dθ, r(h) = sin θ

27Use sinn+2 Θ = (− cos θ)0 sinn+1 Θ. 36 PIOTR HAJLASZ and Lemma 3.36 give 1 Z π/2 π1/2Γ n+1 ω = ω sinn θ dθ = ω 2 , 2 n n−1 n−1 n 0 nΓ 2 so 1/2 n+1 2π Γ 2 ωn = n ωn−1 . nΓ 2 If 2πn/2 an = n nΓ 2 then a direct computation shows that a1 = 2 = ω1 and that an satisﬁes the same recur- rence relationship as ωn, so ωn = an for all n. The second equality in (3.9) follows from Theorem 3.34(a). (b) follows from the fact that the (n−1)-dimensional measure of a sphere of radius r equals to the derivative with respect to r of the volume of an n-dimensional ball of radius r. 2 We will also need the following result. Lemma 3.38. For β > 0 we have ∞ −u 1 Z e 2 e−β = √ √ e−β /(4u) du . π 0 u

Proof. Applying the theory of residues to the function eiβz/(1 + z2) one can easily prove that Z ∞ −β 2 cos βx e = 2 dx . π 0 1 + x This and an obvious identity Z ∞ 1 −(1+x2)u 2 = e du 1 + x 0 yields Z ∞ −β 2 cos βx e = 2 dx π 0 1 + x ∞ ∞ 2 Z Z 2 = cos βx e−ue−ux du dx π 0 0 ∞ ∞ 2 Z Z 2 = e−u e−ux cos βx dx du π 0 0 ∞ ∞ 2 Z 1 Z 2 = e−u e−ux eiβx dx du π 0 2 −∞ ∞ ∞ (x=−2πy) 2 Z Z 2 2 = e−u π e−4π uy e−2πiβy dy du π 0 −∞ | {z } W (β,u) ∞ r 2 Z 1 π 2 = e−u e−β /(4u) du π 0 2 u HARMONIC ANALYSIS 37

∞ −u 1 Z e 2 = √ √ e−β /(4u) du . π 0 u The proof is complete.

Proof of Theorem 3.29. (a) By a change of variables formula it suﬃces to prove the formula for t = 1. We have ∞ −u Z Z 1 Z e 2 2 e−2π|x|e−2πix·ξ dx = √ √ e−4π |x| /(4u) du e−2πix·ξ dx Rn Rn π 0 u ∞ −u 1 Z e Z 2 2 = √ √ e−4π |x| /(4u)e−2πix·ξ dx du π 0 u Rn | {z } W (ξ,(4u)−1) ∞ −u r n 1 Z e u 2 = √ √ e−u|ξ| du π 0 u π Z ∞ 1 −u(1+|ξ|2) (n−1)/2 = (n+1)/2 e u du π 0 Z ∞ 1 1 −s (n−1)/2 = (n+1)/2 2 (n+1)/2 e s ds . π (1 + |ξ| ) 0 | {z } Γ[(n+1)/2] (b) is obvious. (c) Because of the scaling property (b) it suﬃces to consider t = 1. We have Z Z dx P (x, 1) dx = cn 2 (n+1)/2 Rn Rn (1 + |x| ) Z ∞ Z polar dσ n−1 = cn 2 (n+1)/2 r dr 0 Sn−1(0,1) (1 + r ) Z ∞ rn−1 = cn n ωn 2 (n+1)/2 dr 0 (1 + r ) Z π/2 r=tan θ n−1 = cn n ωn sin θ dθ 0 Γ n+1 2πn/2 π1/2Γ n = 2 n 2 π(n+1)/2 n n−1 n Γ 2 (n − 1)Γ 2 = 1 . The proof is complete.

Corollary 3.26 applied to f(x) = e−4π2t|x|2 and to f(x) = e−2πt|x| yields Corollary 3.39. Z Z W (ξ, t)e2πix·ξ dξ = e−4π2t|x|2 and P (ξ, t)e2πix·ξ dξ = e−2π|x|t Rn Rn for all x ∈ Rn. 38 PIOTR HAJLASZ

The Weierstrass and Poisson kernels have the following semigroup property.

Corollary 3.40. If Wt(x) = W (x, t) and Pt(x) = P (x, t), then for t1, t2 > 0

(Wt1 ∗ Wt2 )(x) = Wt1+t2 (x) and (Pt1 ∗ Pt2 )(x) = Pt1+t2 (x) for all x ∈ Rn.

Proof. It follows from Corollary 3.39 with x replaced by −x that ˆ −4π2t|x|2 ˆ −2π|x|t Wt(x) = e and Pt(x) = e . Hence Theorem 3.2(b) yields ˆ ˆ ˆ (Wt1 ∗ Wt2 )ˆ(x) = Wt1 (x)Wt2 (x) = Wt1+t2 (x) ˆ ˆ ˆ (Pt1 ∗ Pt2 )ˆ(x) = Pt1 (x)Pt2 (x) = Pt1+t2 (x) and the result follows from Corollary 3.28. 2

3.4. The Schwarz class and the Plancherel theorem.

n ∞ n Deﬁnition 3.41. We say that f belongs to the Schwarz class S (R ) = Sn if f ∈ C (R ) and for all multiindices α, β α β sup |x D f(x)| = pα,β(f) < ∞. x∈Rn That means all derivatives of f rapidly decrease to zero as |x| → ∞, faster than the inverse of any polynomial.

∞ n −|x|2 Clearly C0 (R ) ⊂ Sn, but also e ∈ Sn so functions in the Schwarz class need not be compactly supported.

{pα,β} is a countable family of norms in Sn and we can use it to deﬁne a topology in Sn.

Deﬁnition 3.42. We say that a sequence (fk) converges to f in Sn if

lim pα,β(fk − f) = 0 for all multiindices α, β. k→∞

This convergence is metrizable. Indeed, dα,β(f, g) = pα,β(f − g) is a metric and if we 0 0 arrange all these metrics in a sequence d1, d2,..., then ∞ X d0 (f, g) d(f, g) = 2−k k 1 + d0 (f, g) k=1 k deﬁnes a metric in Sn such that fn → f in Sn if and only if fn → f in the metric d.

Proposition 3.43. The space Sn has the following properties.

(a) Sn equipped with the metric d is a complete metric space. ∞ n (b) C0 (R ) is dense in Sn. (c) If ϕ ∈ Sn, then τhϕ → ϕ in Sn as h → 0, where τhϕ(x) = ϕ(x + h). HARMONIC ANALYSIS 39

(d) The mapping α β Sn 3 ϕ 7→ x D ϕ(x) ∈ Sn is continuous. (e) If ϕ ∈ Sn, then ϕ(x + he ) − ϕ(x) ∂ϕ k → (x) as h → 0. h ∂xk

in the topology of Sn. (f) If ϕ, ψ ∈ Sn, then ϕ ∗ ψ ∈ Sn and Dα(ϕ ∗ ψ) = (Dαϕ) ∗ ψ = ϕ ∗ (Dαψ) for any multiindex α.

We leave the proof as an exercise.

Theorem 3.44. The Fourier transform is a continuous, one-to-one mapping of Sn onto Sn such that (a) ∧ ∂f ˆ (ξ) = 2πiξjf(ξ), ∂xj (b) ∂fˆ (−2πixjf)ˆ(ξ) = (ξ) , ∂ξj (c) (f ∗ g)ˆ = fˆg,ˆ (d) Z Z f(x)ˆg(x) dx = fˆ(x)g(x) dx , Rn Rn (e) Z f(x) = fˆ(ξ)e2πix·ξ dξ . Rn

Proof. We already proved (a), (b), (c) and (d). Now we will prove that the Fourier transform is a continuous mapping from Sn into Sn. Formulas (a) and (b) imply that ξαDβfˆ(ξ) = C(Dα(xβf))ˆ(ξ).

Since kgˆk∞ ≤ kgk1 we get ˆ α β ˆ α β pα,β(f) = kξ D f(ξ)k∞ ≤ CkD (x f)k1 . An application of the Leibnitz rule28 implies that Dα(xβf) equals a ﬁnite sum of expressions of the form xβi Dαi f. Since Z βi αi 2 n βi αi 2 −n kx D fk1 = |(1 + |x| ) x D f(x)|(1 + |x| ) dx Rn 28Product rule. 40 PIOTR HAJLASZ

≤ C(n) sup |(1 + |x|2)nxβi Dαi f(x)| < ∞ x∈Rn ˆ it follows that f ∈ Sn. One can also easily deduce that the mapping ˆ : Sn → Sn is continuous. ˆ ˆ If f ∈ Sn than f ∈ Sn and hence both f and f are integrable, so (e) follows from the inversion formula. This formula also shows that the Fourier transform applied four times is an identity on Sn and hence the Fourier transform is a bijection on Sn. 2 Theorem 3.45 (Plancherel). The Fourier transform is an L2 isometry on a dense subset 2 Sn of L ˆ kfk2 = kfk2, f ∈ Sn, and hence it uniquely extends to an isometry of L2 ˆ 2 n kfk2 = kfk2, f ∈ L (R ) . Moreover for f ∈ L2(Rn) Z fˆ(ξ) = lim f(x)e−2πix·ξ dx R→∞ |x|

ˆ¯ ¯ Proof. Given f ∈ Sn let g = f, sog ˆ = f. Indeed, Z ¯ Z gˆ(ξ) = fˆ(x)e−2πix·ξ dx = fˆ(x)e2πix·ξ dx = f¯(x) . Rn Rn Hence Theorem 3.44(d) gives Z Z Z Z ¯ ˆ ˆˆ¯ ˆ kfk2 = ff = fgˆ = fg = ff = kfk2 . Rn Rn Rn Rn 2 2 Thus the Fourier transform is an L isometry on Sn. Since Sn is a dense subset of L it uniquely extends to an isometry of L2. Now for f ∈ L2(Rn) we have 2 1 L L 3 fχB(0,R) −→ f as R → ∞ and hence Z 2 −2πix·ξ L ˆ (fχB(0,R))ˆ(ξ) = f(x)e dx −→ f(ξ) |x|≤R as R → ∞. Similarly

Z 2 fˆ(ξ)e2πix·ξ dξ −→L f(x) as R → ∞. |ξ|

Proposition 3.46. If f, g ∈ L2(Rn), then Z Z fˆ(x)g(x) dx = f(x)ˆg(x) dx . Rn Rn

2 Proof. Approximate f and g in L by functions in Sn, apply Theorem 3.44(d) and pass to the limit.

1 n 2 n Consider the class L (R ) + L (R ) consisting of functions of the form f = f1 + f2, 1 2 f1 ∈ L , f2 ∈ L . Then we define ˆ ˆ ˆ f = f1 + f2. In order to show that the Fourier transform is well defined in the class L1 + L2 we need to show that it does not depend on the particular choice of the representation f = f1 + f2. 1 2 1 2 Indeed, if we also have f = g1 + g2, g1 ∈ L , g2 ∈ L , then f1 − g1 = g2 − f2 ∈ L ∩ L and hence ˆ ˆ f1 − gˆ1 = (f1 − g1)ˆ = (g2 − f2)ˆ =g ˆ2 − f2, ˆ ˆ f1 + f2 =g ˆ1 +g ˆ2 . It is an easy exercise to show that p n 1 n 2 n L (R ) ⊂ L (R ) + L (R ), for 1 ≤ p ≤ 2, and hence the Fourier transform is well defined on Lp(Rn), 1 ≤ p ≤ 2 and p n 2 n n ˆ: L (R ) → L (R ) + C0(R ), for 1 ≤ p ≤ 2. Later we will prove the Hausdorff-Young Inequality which implies that p n p0 n ˆ: L (R ) → L (R ), for 1 ≤ p ≤ 2, where p0 is the Hölderconjugate to p. 42 PIOTR HAJLASZ

4. Miscellaneous results about the Fourier transform

In this chapter we will show some interesting results about the Fourier transform. Some of them, but not all, will be used later.

4.1. The Poisson summation formula.

Theorem 4.1 (Poisson summation formula). If f ∈ C1(R) and C (4.1) |f(x)| + |f 0(x)| ≤ for x ∈ , 1 + x2 R then ∞ ∞ X X f(n) = fˆ(n). n=−∞ n=−∞ Remark 4.2. Here fˆ(n) stand for the Fourier transform of f evaluated at n – do not get confused with the Fourier series. P Note that it follows from the growth condition (4.1) that the series n f(n) converges and that any function f ∈ Sn satisﬁes (4.1).

Proof. It follows from (4.1) that ∞ X (4.2) g(x) = f(x + k) k=−∞ is a periodic function of class C1 with period 1. Therefore g is can be represented as a Fourier series29 ∞ X g(x) = gˆ(n)e2πinx n=−∞ where ∞ ∞ Z 1 X Z 1 X Z k+1 gˆ(n) = g(x)e−2πinx dx = f(x + k)e−2πinx dx = f(x)e−2πinx dx 0 k=−∞ 0 k=−∞ k Z ∞ = f(x)e−2πinx dx = fˆ(n) −∞

29Nowg ˆ(n) are Fourier series coeﬃcients – do not get confused with the Fourier transform! HARMONIC ANALYSIS 43 so ∞ ∞ ∞ X X X f(k) = g(0) = gˆ(n) = fˆ(n). k=−∞ n=−∞ n=−∞

As an application we will prove the Jacobi identity. Deﬁnition 4.3. The Jacobi ‘theta’ function is deﬁned by ∞ X 2 ϑ(t) = e−πn t, t > 0. n=−∞

Clearly ϑ ∈ C∞(R).

− 1 Corollary 4.4 (Jacobi identity). For t > 0 we have ϑ(t) = t 2 ϑ(1/t).

Proof. The function f(x) = t−1/2e−πx2/t, t > 0 can be written as

f(x) = t−1/2u(xt−1/2) where u(x) = e−πx2 . Hence30 ˆ −1/2 −1/2 1/2 −πξ2t f(ξ) = t (ˆu)t−1/2 (ξ) = t ut−1/2 (ξ) = u(ξt ) = e . Thus the Poisson summation formula yields ∞ ∞ X 2 X 2 t−1/2 e−πn /t = e−πn t i.e., t−1/2ϑ(1/t) = ϑ(t). n=−∞ n=−∞ Corollary 4.5. For t > 0 we have ∞ X 1 π 1 + e−2πt = . t2 + n2 t 1 − e−2πt n=−∞ Remark 4.6. It is easy to show by taking the limit as t → 0+ in the above identity that ∞ X 1 π2 = . n2 6 n=1

Proof. In Theorem 3.29 we showed that the Fourier transform of f(x) = e−2π|x|t equals31 t fˆ(x) = π(t2 + x2)

30 ∧ −n We use here two facts: (ϕ(sx)) (ξ) = (ϕ ˆ)s(ξ) = s ϕˆ(ξ/s) with n = 1 andu ˆ(ξ) = u(ξ). 31 We are in dimension n = 1 and c1 = 1. 44 PIOTR HAJLASZ so the Poisson summation formula gives32 ∞ ∞ t X 1 X 1 + e−2πt = e−2π|n|t = π t2 + n2 1 − e−2πt n=−∞ n=−∞ from which the theorem follows.

4.2. The Heisenberg inequality. Celebrated Heisenberg’s uncertainty principle asserts that position and momentum of a particle cannot be measured at the same time and it can be written as an inequality h σ σ ≥ x p 4π where σx and σp are standard deviations of position and momentum and h is the Plank constant. Translating it into our language the above inequality can be formulated as

Theorem 4.7 (Heisenberg’s inequality). For any f ∈ L2(R) and a, b ∈ R, s s Z ∞ Z ∞ kfk2 (x − a)2|f(x)|2 dx (ξ − b)2|fˆ(ξ)|2 dξ ≥ 2 . −∞ −∞ 4π

2 Moreover the equality holds if and only if f(x) = Ce2πibxe−k(x−a) for some C ∈ C and k > 0.

Proof. We will prove the result under the assumption that f ∈ S1. The case of general f ∈ L2 is left to the reader as an exercise. Let f(x) = e2πibxg(x − a). Since fˆ(ξ) =g ˆ(ξ − b)e−2πa(ξ−b) we easily obtain that the Heisenberg inequality is equivalent to Z ∞ 1/2 Z ∞ 1/2 kgk2 (4.3) |x|2|g(x)|2 dx |ξ|2|gˆ(ξ)|2 dξ ≥ 2 . −∞ −∞ 4π Note that (|g|2)0 = (gg)0 = 2 re g0 g so integration by parts gives Z β Z β β Z β 2 0 2 2 0 |g(x)| dx = x |g(x)| dx = x|g(x)| − 2 re xg (x)g(x) dx. α α α α

Letting α → −∞, β → +∞ and using the fact that g ∈ S1 yields Z ∞ Z ∞ |g(x)|2 dx = −2 re xg0(x)g(x) dx −∞ −∞ Z ∞ (4.4) ≤ 2 |x| |g(x)| |g0(x)| dx −∞ Z ∞ 1/2 Z ∞ 1/2 (4.5) ≤ 2 |x|2|g(x)|2 dx |g0(x)|2 dx −∞ −∞

32 In the Poisson summation formula we assumed that f ∈ C1(R) and it satisfies (4.1). However, our function f is not C1. The condition (4.1) was required for the function (4.2) to be represented as a Fourier series. Although the condition (4.1) is not satisfied any longer, it is easy to check that the function defined by (4.2) is periodic and Lipschitz continuous so it is still equal to its Fourier series and the proof carries on. HARMONIC ANALYSIS 45

Z ∞ 1/2 Z ∞ 1/2 = 2 |x|2|g(x)|2 dx |gb0(ξ)|2 dξ −∞ −∞ Z ∞ 1/2 Z ∞ 1/2 = 2 |x|2|g(x)|2 dx |2πiξgˆ(ξ)|2 dξ −∞ −∞ from which (4.3) follows. Finally, let us investigate the case of equality. For the equality in the Schwarz inequality (4.5) we need |g0(x)| to be proportional to |xg(x)|, say |g0(x)| = k|xg(x)|, for some k > 0 so g0(x) = keiϕ(x)xg(x) for some real valued function ϕ(x). But then equality in (4.4) implies that −re e−ϕ(x) = 1 so eiϕ(x) = −1 and we have g0(x) = −kxg(x). Solving this diﬀerential equation yields g(x) = Ce−kx2 and 2πibx −k(x−a)2 hence f(x) = Ce e .

4.3. Eigenfunctions of the Fourier transform. The Fourier transform applied four times is identity on L2(Rn) so if f is an eigenfunction of the Fourier transform, fˆ = λf, then f = |λ|4f so λ ∈ {−1, 1, −i, i}. We know already one eigenfunction with the eigenvalue 1, namely f(x) = e−πx2 and now we will find now all eigenfunctions in dimension n = 1. It is natural to search among functions of the form p(x)e−πx2 , where p(x) is a polynomial.33 Definition 4.8. Hermite functions are defined by the formula n (−1) 2 d n 2 h (x) = eπx e−2πx , for all integers n ≥ 0. n n! dx

−πx2 Note that h0(x) = e .

−πx2 Lemma 4.9. hn(x) = pn(x)e for a polynomial pn(x) of order n of the form (4π)n p (x) = xn + lower order terms. n n!

In particular hn ∈ S1.

This lemma easily follows from the deﬁnition of hn.

0 Lemma 4.10. hn(x) − 2πxhn(x) = −(n + 1)hn+1(x).

Proof. Applying the product rule to n h 2 i h(−1) d n 2 i h (x) = eπx · e−2πx n n! dx

2 2 33Note that evaluation of the Fourier transform of p(x)e−πx is straightforward: since F(e−πx )(ξ) = 2 e−πξ it suﬃces to apply (3.1). 46 PIOTR HAJLASZ gives n h i h (−1) d n+1 2 i h0 (x) = 2πxh (x) + eπx · e−2πx = 2πxh (x) − (n + 1)h (x). m n n! dx n n+1 | {z } (−1)n+1 −(n+1) (n+1)! Lemma 4.11. If f ∈ C∞(R) and n is a positive integer, then d n d n d n−1 (4πx) f(x) − (4πxf(x)) = −4πn f(x) for all x ∈ . dx dx dx R

The result follows from a simple induction.

0 Lemma 4.12. hn(x) + 2πxhn(x) = 4πhn−1(x), where h−1(x) = 0.

Proof. For n = 0 we check the equality directly. For n ≥ 1 Lemma 4.10 gives 0 hn(x) + 2πxhn(x) = 4πxhn(x) − (n + 1)hn+1(x) n n+1 (−1) 2 d n 2 (−1) 2 d n+1 2 = 4πx eπx e−2πx −(n + 1) eπx e−2πx n! dx (n + 1)! dx | {z } | {zn } d n −2πx2 +(−1) /n! ( dx ) (−4πxe ) n (−1) 2 h d n 2 d n 2 i = eπx 4πx e−2πx − (4πxe−2πx ) n! dx dx | {z } d n−1 −2πx2 −4πn( dx ) e by Lemma 4.11 n−1 (−1) 2 d n−1 2 = 4π eπx e−2πx = 4πh (x). (n − 1)! dx n−1 ˆ n ˇ n Lemma 4.13. hn = (−i) hn and hn = i hn. Remark 4.14. The lemma yields ˆ ˆ ˆ ˆ h4n = h4n, h4n+2 = −h4n+2, h4n+3 = ih4n+3, h4n+1 = −ih4n+1 Thus the lemma provides examples of eigenfunctions corresponding to all possible eigen- values 1, −1, i, −i.

Proof. Since hn ∈ S1 we can compute the Fourier transform directly. Lemma 4.10 gives

0 0 ˆ hn − 2πxhn = −(n + 1)hn+1, hcn − 2\πxhn = −(n + 1)hn+1. Since (−2πixf)∧ = (fˆ)0 and fb0 = 2πiξfˆ(ξ) we have ˆ ˆ 0 ˆ (4.6) 2πiξhn(ξ) − i(hn) (ξ) = −(n + 1)hn+1(ξ) ˆ n We will prove the equality hn = (−i) hn by induction. −πx2 ˆ 0 For n = 0, h0 = e so h0 = h0 = (−i) h0. HARMONIC ANALYSIS 47

Suppose now that the equality holds for n and we will prove it for n + 1. The equality (4.6) yields. n n 0 ˆ 2πiξ(−i) hn(ξ) − i((−i) hn(ξ)) = −(n + 1)hn+1(ξ), n+1 n+1 0 ˆ −(−i) 2πξhn(ξ) + (−i) hn(ξ) = −(n + 1)hn+1(ξ), n+1 0 ˆ (−i) hn(ξ) − 2πξhn(ξ) = −(n + 1)hn+1(ξ). | {z } −(n+1)hn+1(ξ) by Lemma 4.10 ˆ n+1 This proves that hn+1 = (−i) hn+1(ξ) which completes the proof of the claim that ˆ n hn = (−i) hn(ξ) fro all n. Finally ˆ ∨ n ∨ nˇ ˇ n hn = (hn) = ((−i) hn) = (−i) hn, hn = i hn.

Let Kf = f 00 − 4π2x2f. 1 Lemma 4.15. Khn = −4π(n + 2 )hn

Proof. Lemmas 4.10 and 4.12 yield 0 hn − 2πxhn = −(n + 1)hn+1, 00 0 0 hn − 2πhn − 2πxhn = −(n + 1)hn+1, 00 0 0 hn = 2πhn + 2πxhn − (n + 1)hn+1. Hence 00 2 2 0 0 2 2 Khn = hn − 4π x hn = 2πhn + 2πxhn − (n + 1)hn+1 − 4π x hn 0 0 = 2πhn − (n + 1)hn+1 + 2πx (hn − 2πxhn) | {z } −(n+1)hn+1 0 = 2πhn − (n + 1)(hn+1 + 2πxhn+1) 1 = −4π n + h . 2 n 34 Lemma 4.16. For real valued functions f, g ∈ S1 we have hKf, gi = hf, Kgi

Proof. The integration by parts yields Z Z Z hKf, gi = (f 00 − 4π2x2f)g = f 00g − 4π2x2fg R R R Z Z Z = fg00 − 4π2x2fg = f(g00 − 4π2x2g) = hf, Kgi. R R

Lemma 4.17. The Hermite functions hn are orthogonal,

hhn, hmi = 0, for n 6= m. 34Here h·, ·i denotes the L2 scalar product. 48 PIOTR HAJLASZ

Proof. We have 1 1 −4π n + hh , h i = hKh , h i = hh , Kh i = −4π m + hh , h i 2 n m n m n m 2 n m and the result easily follows. Lemma 4.18. n 2 (4π) khnk = √ . 2 2n!

0 Proof. Since by Lemma 4.10 we have (n+1)hn+1 = 2πxhn −hn, integration by parts yields Z Z Z Z 2 0 0 (n + 1)khn+1k2 = 2π xhnhn+1 − hnhn+1 = 2π xhnhn+1 + hnhn+1 R R R R Z 0 2 = hn (hn+1 + 2πxhn+1) = 4πkhnk2. R | {z } 4πhn Replacing n by n − 1 yields 4π kh k2 = kh k2 n 2 n n−1 2 and a simple induction argument yields (4π)n kh k2 = kh k2. n 2 n! 0 2 Now the result follows from a simple observation that Z Z Z 2 −πx2 2 −2πx2 1 −πy2 1 kh0k2 = (e ) dx = e dx = √ e dy = √ . R R 2 R 2

The last two lemmas show that the scaled Hermite functions (4π)n −1/2 en = √ hn 2n! form an orthonormal family in L2(R). ∞ 2 n Theorem 4.19. The fmaily {en}n=0 is an orthonormal basis of L (R ).

2 R Proof. It suﬃces to prove that if f ∈ L (R) and fhn = 0 for all n ≥ 0, then f = 0 a.e. R R Thus suppose that fhn = 0 for all n ≥ 0. Since R 2 (4π) 2 h (x) = xn + ... e−πx , n n! it easily follows that the functions xne−πx2 , n ≥ 0, are linear combinations of the functions h0, h1, . . . , hn. Hence Z f(x)xne−πx2 dx = 0 for all n ≥ 0. R HARMONIC ANALYSIS 49

Since by the Schwarz inequality f(x)e−πx2 ∈ L1 we can compute the Fourier transform directly Z Z ∞ n 2 ∧ 2 2 X (−2πixξ) f(x)e−πx = f(x)e−πx e−2πixξ dx = f(x)e−πx dx n! R R n=0 ∞ n Z X (−2πiξ) 2 = f(x)xne−πx dx = 0. n! n=0 R Vanishing of the Fourier transform of f(x)e−πx2 implies that f(x)e−πx2 = 0 a.e. and hence f(x) = 0 a.e.

∞ 2 2 Since {en}n=0 is an orthonormal basis of L (R) any f ∈ L can be written as ∞ X f = hf, eni en n=0 n This and the fact thate ˆn = (−i) en yield Theorem 4.20 (Wiener’s deﬁnition of the Fourier transform). For f ∈ L2(R) we have ∞ ˆ X n f = hf, eni(−i) en. n=0

2 Let Hk, k = 0, 1, 2, 3 be subspaces of L (R) deﬁned by ∞ Hk = span {e4n+k}n=0.

The subspaces Hk are orthogonal and 2 L (R) = H0 ⊕ H1 ⊕ H2 ⊕ H3.

Clearly, the subspaces Hk are eigenspaces of the Fourier transform: ˆ k f = (−i) f for f ∈ Hk and k = 0, 1, 2, 3.

4.4. The Bochner-Hecke formula. As was pointed out in Section 4.3 computing the Fourier transform of P (x)e−π|x|2 is straightforward due to formula (3.1) and Theorem 3.7, but the computations are often algebraically complicated and it is not always obvious how to write the answer in a compact and nice form. We have see examples of such computations in Section 4.3 when we found eigenfunctions of the Fourier transform and now we will see another example of this type. The results of this section will be used later in Section 6.3

Deﬁnition 4.21. We say that Pk(x) is a homogeneous harmonic polynomial of degree k if

X α Pk(x) = aαx and ∆Pk = 0 . |α|=k

Theorem 4.22 (Bochner-Hecke). If Pk(x) is a homogeneous harmonic polynomial of degree k, then −π|x|2 k −π|ξ|2 F Pk(x)e (ξ) = (−i) Pk(ξ)e . 50 PIOTR HAJLASZ

Proof. Since the polynomial Pk is homogeneous of degree k, X α k Pk(−2πix) = aα(2πix) = (−2πi) Pk(x). |α|=k Hence (3.1) applied to f(x) = e−π|x|2 yields35 Z −π|x|2 −2πix·ξ −π|x|2 Pk(x)e e dx = F Pk(x)e (ξ)(4.7) Rn −k −π|ξ|2 −π|ξ|2 = (−2πi) Pk(D)e = Q(ξ)e for some polynomial Q and it remains to prove that Q(ξ) = Pk(−iξ). 36 Lemma 4.23. If P (x) is a polynomial in Rn, then Z Z −π P (x +iξ )2 −π|x|2 (4.8) P (x)e j j j dx = P (x − iξ)e dx . Rn Rn

Proof. By the same argument involving the same contour integration as in the proof of Theorem 3.7 for any polynomial P of one variable we have Z ∞ Z ∞ (4.9) P (x)e−π(x+iξ)2 dx = P (x − iξ)e−πx2 dx . −∞ −∞ If P is a polynomial in n variables, then (4.8) follows from (4.9) and the Fubini theorem.

Multiplying both sides of (4.7) by eπ|ξ|2 and applying (4.8) we obtain

Z P 2 Z 2 −π (xj +iξj ) −π|x| Q(ξ) = Pk(x)e j dx = Pk(x − iξ)e dx . Rn Rn We can write X α Pk(x − η) = η Pα(x) α and then Z Z −π|x|2 X α −π|x|2 Pk(x − η)e dx = η Pα(x)e dx , n n R α R so clearly the integral is a polynomial is η. With this notation we obtain Z X α −π|x|2 Q(ξ) = (iξ) Pα(x)e dx n α R and hence Z Z X α −π|x|2 −π|x|2 Q(ξ/i) = ξ Pα(x)e dx = Pk(x − ξ)e dx . n n α R R

35 Instead of using (3.1) we could apply the diﬀerential operator Pk(Dξ) to both sides of the equality

Z 2 2 e−π|x| e−2πix·ξ dx = e−π|ξ| n R and diﬀerentiate under the sign of the integral. 36Any polynomial, not necessarily harmonic or homogeneous. HARMONIC ANALYSIS 51

37 Since Pk is a harmonic function, it has the mean value property Z n−1 Pk(sθ − ξ) dσ(θ) = |S |Pk(−ξ) . Sn−1 Thus integration in polar coordinates gives Z ∞ Z n−1 −πs2 Q(ξ/i) = s Pk(sθ − ξ) dσ(θ) e ds 0 Sn−1 Z ∞ n−1 n−1 −πs2 = Pk(−ξ) s |S |e ds 0 Z −π|x|2 = Pk(−ξ) e dx = Pk(−ξ) Rn and hence Q(ξ) = Pk(−iξ). The proof is complete.

Using homogeneity of Pk and the second part of Theorem 3.2(e) one can easily deduce from the Bochner-Hecke formula the folloiwing result.

Corollary 4.24. If Pk is a homogeneous harmonic polynomial of degree k, then for any t > 0 we have −πt|x|2 −k− n k −π|ξ|2/t F Pk(x)e (ξ) = t 2 (−i) Pk(ξ)e .

We leave details as an easy exercise.

4.5. Symmetry of the Fourier transform. If f is radially symmetric i.e., f = f ◦ ρ for all ρ ∈ O(n), then fˆ◦ ρ = f[◦ ρ = fˆ for all ρ ∈ O(n) so fˆ is radially symmetric too. Now we will prove a more delicate result about the symmetry of the Fourier transform. Proposition 4.25. Let f ∈ L1(Rn) be a real valued function of the form x f(x) = j g(|x|). |x| Then there is a continuous function h : [0, ∞) → R, h(0) = 0, h(t) → 0 as t → ∞ such that ξ fˆ(ξ) = i j h(|ξ|), ξ ∈ n. |ξ| R

While the function f is not radially symmetric a more symmetric object is the map x F (x) = g(|x|) ∈ L1( n, n), x ∈ n |x| R R R so f is the jth component of F . We will prove that ξ Fˆ(ξ) = i h(|ξ|), ξ ∈ n |ξ| R from which Proposition 4.25 will readily follow. In the proof we will use the following result which is interesting on its own. It ill be used later one more time in a proof of Theorem 6.2.

37 We integrate here Pk over the sphere centered at −ξ and of radius s. 52 PIOTR HAJLASZ

Lemma 4.26. If m : Rn → Rn is a measurable function that is homogeneous of degree 0, i.e. m(tx) = m(x) for t > 0, and commutes with the orthogonal transformations, i.e. (4.10) m(ρ(x)) = ρ(m(x)) for all x ∈ Rn and ρ ∈ O(n), then there is a constant c such that x (4.11) m(x) = c for all x 6= 0. |x|

n Proof. Let e1, e2, . . . , en be the standard orthogonal basis of R . If [ρjk] is the matrix representation of ρ ∈ O(n), then the condition (4.10) reads as n X (4.12) mj(ρ(x)) = ρjkmk(x), j = 1, 2, . . . , n, k=1 where m(x) = (m1(x), . . . , mn(x)).

Let m1(e1) = c. Consider all ρ ∈ O(n) such that ρ(e1) = e1. This condition means that the ﬁrst column of the matrix [ρjk] equals e1, i.e. ρ11 = 1, ρj1 = 0, for j > 1. Since columns are orthogonal, for k > 1 we have n X 0 = ρj1ρjk = ρ1k . j=1 Thus  1 0 ... 0   0 ρ22 . . . ρ2n  ρ =  . . . .  ,  . . .. .  0 ρn2 . . . ρnn n where [ρjk]j,k=2 is the matrix of an arbitrary orthogonal transformation in the (n − 1)- dimensional subspace orthogonal to e1.

For x = e1 = ρ(e1) = ρ(x) and j ≥ 2 identity (4.12) yields n n X X mj(e1) = ρjkmk(e1) = ρjkmk(e1) , k=1 k=2 and hence  m2(e1)   ρ22 . . . ρ2n   m2(e1)  . . . . .  .  =  . .. .   .  . mn(e1) ρn2 . . . ρnn mn(e1) T That means the vector [m2(e1), . . . , mn(e1)] is ﬁxed under an arbitrary orthogonal transformation of Rn−1, so it must be a zero vector, i.e.

m2(e1) = ... = mn(e1) = 0 .

Now formula (4.12) for any ρ ∈ O(n) and x = e1, takes the form

mj(ρ(e1)) = ρj1m1(e1) = cρj1 . HARMONIC ANALYSIS 53

By homogeneity it suﬃces to prove (4.11) for |x| = 1. Let ρ ∈ O(n) be such that ρ(e1) = x. Then ρj1 = xj, j = 1, 2, . . . , n and hence x m (x) = cρ = cx = c j . j j1 j |x| This completes the proof of the lemma.

Proof of Proposition 4.25. Since the function F is odd Z Z iFˆ(ξ) = iF (x)(cos(2πx · ξ) − i sin(2πx · ξ)) dx = sin(2πx · ξ)F (x) dx Rn Rn ˆ n n−1 n so the mapping ξ 7→ iF (ξ) takes values in R . Fix k > 0 and deﬁne mk : S (0, k) → R ˆ n n by mk(ξ) = iF (ξ) for |ξ| = k. Extend mk to mk : R \{0} → R as a function homogeneous of degree 0, i.e. kξ m (ξ) = i Fˆ for ξ 6= 0. k |ξ| We claim that

(4.13) mk(ρ(ξ)) = ρ(mk(ξ)) for ρ ∈ O(n) and ξ 6= 0. 38 Since mk is homogeneous of degree 0, it suﬃces to check (4.13) for |ξ| = k. We have Z Z x −1 x mk(ρ(ξ)) = sin(2πx · ρ(ξ)) g(|x|) dx = sin(2πρ (x) · ξ) g(|x|) dx Rn |x| Rn |x| Z ρ(x) Z ρ(x) = sin(2πx · ξ) g(|ρ(x)|) dx = sin(2πx · ξ) g(|x|) dx Rn |ρ(x)| Rn |x| Z x = ρ sin(2πx · ξ) g(|x|) dx = ρ(mk(ξ)). Rn |x| According to Lemma 4.26 there is a constant h(k) ∈ R such that mk(ξ) = −h(k)ξ/|ξ|. In particular for |ξ| = k we have ξ iFˆ(ξ) = m (ξ) = − h(|ξ|). k |ξ| ˆ n n Clearly h is continuous, h(0) = 0 and h(t) → 0 as t → ∞, because iF ∈ C0(R , R ).

38First we use equality x·ρ(ξ) = ρ−1(x)·ξ which follows from the fact that the orthogonal transformation ρ−1 preserves the scalar product and then we use the change of variables x = ρ(y) whose absolute value of the Jacobian equals 1. 54 PIOTR HAJLASZ

5. Tempered distributions

5.1. Basic constructions.

0 Deﬁnition 5.1. The space Sn of all continuous linear functionals on Sn is called the 0 space of tempered distributions. The evaluation of u ∈ Sn on ϕ ∈ Sn will usually be 0 denoted by u[ϕ]. The space Sn is equipped with the weak-∗ convergence (usually called 0 weak convergence or just convergence). Namely uk → u in Sn if uk[ψ] → u[ψ] for every ψ ∈ Sn.

Here are examples:

1. If f ∈ Lp(Rn), 1 ≤ p ≤ ∞, then Z 0 Lf [ϕ] = f(x)ϕ(x) dx defines Lf ∈ Sn. Rn 2. If µ is a measure of finite total variation, then Z 0 Lµ[ϕ] = ϕ dµ defines Lµ ∈ Sn. Rn

3. We say that a function f is a tempered Lp function if f(x)(1+|x|2)−k ∈ Lp(Rn) for some nonnegative integer k. If p = ∞ we call f a slowly increasing function. Then Z 0 Lf [ϕ] = f(x)ϕ(x) dµ deﬁnes Lf ∈ Sn for all 1 ≤ p ≤ ∞. Rn Note that slowly increasing functions are exactly measurable functions bounded by polynomials. 4. A tempered measure is a Borel measure µ such that Z (1 + |x|2)−k dµ < ∞ Rn 0 for some integer k ≥ 0. As before Lµ ∈ Sn. α 0 5. L[ϕ] = D ϕ(x0) is a tempered distribution L ∈ Sn. The distributions generated by a function or by a measure will often be denoted by Lf [ϕ] = f[ϕ], Lµ[ϕ] = µ[ϕ]. HARMONIC ANALYSIS 55

0 p Suppose that u ∈ Sn. If there is a tempered L function f such that u[ϕ] = f[ϕ] for ϕ ∈ Sn, then we can identify u with the function f and simply write u = f. The identification is possible, because the function f is uniquely defined (up to a.e. equivalence). This follows from a well known result. n 1 R ∞ Lemma 5.2. If Ω ⊂ R is open and f ∈ Lloc(Ω) satisfies Ω fϕ = 0 for all ϕ ∈ C0 (Ω), then f = 0 a.e.

Note that not every function f ∈ C∞(Rn) deﬁnes a tempered distribution, because it may happen that form some ϕ ∈ Sn the function fϕ is not integrable and hence the integral f[ϕ] = R f(x)ϕ(x) dx does not make sense. Rn

Theorem 5.3. A linear functional L on Sn is a tempered distribution if and only if there is a constant C > 0 and a positive integer m such that X |L(ϕ)| ≤ C pα,β(ϕ) for all ϕ ∈ Sn. |α|,|β|≤m

Proof. If a linear functional L satisﬁes the given estimate, then clearly it is continuous 0 on Sn, so it remains to prove the converse implication. Let L ∈ Sn. We claim that there is a positive integer m such that |L(ϕ)| ≤ 1 for all n X 1 o ϕ ∈ ϕ ∈ Sn : pα,β(ϕ) ≤ := Nm . m |α|,|β|≤m

To the contrary suppose that there is a sequence ϕk ∈ Sn such that |L(ϕk)| > 1 and X 1 (5.1) p (ϕ ) ≤ , k = 1, 2, 3,... α,β k k |α|,|β|≤k

Note that (5.1) implies that ϕk → 0 in Sn, so the inequality |L(ϕk)| > 1 contradicts continuity of L. This proves the claim. Denote X kϕk = pα,β(ϕ) . |α|,|β|≤m

Observe that k · k is a norm. For an arbitrary 0 6= ϕ ∈ Sn,ϕ ˜ = ϕ/(mkϕk) satisﬁes kϕ˜k ≤ 1/m, soϕ ˜ ∈ Nm and hence |L(ϕ)| = mkϕk|L(ϕ ˜)| ≤ mkϕk which proves the theorem. 2

For any function g on Rn we deﬁneg ˜(x) = g(−x). Then it easily follows from the Fubini theorem that for u, ϕ, ψ ∈ Sn Z Z (u ∗ ϕ)(x)ψ(x) dx = u(x)(ϕ ˜ ∗ ψ)(x) dx . Rn Rn Regarding the functions u and u ∗ ϕ as distributions we can rewrite this equality as (u ∗ ϕ)[ψ] = u[ϕ ˜ ∗ ψ] 0 If u ∈ Sn and ϕ ∈ Sn, then ψ 7→ u[ϕ ˜ ∗ ψ] is a tempered distribution so it motivates the following deﬁnition. 56 PIOTR HAJLASZ

0 Deﬁnition 5.4. If u ∈ Sn and ϕ ∈ Sn, then the convolution of u and ϕ is a tempered distribution deﬁned by the formula (u ∗ ϕ)[ψ] := u[ϕ ˜ ∗ ψ] .

The following two results are left as an easy exercise.

0 Proposition 5.5. If u ∈ Sn, ϕ, ψ ∈ Sn, then (u ∗ ϕ) ∗ ψ = u ∗ (ϕ ∗ ψ) .

n Proposition 5.6. If µ ∈ B(R ) and ϕ ∈ Sn, then µ ∗ ϕ = ϕ ∗ µ where µ ∗ ϕ is the 0 convolution of a distribution µ ∈ Sn with a function ϕ ∈ Sn and ϕ ∗ µ is the convolution of a function with a measure.

0 Theorem 5.7. If u ∈ Sn and ϕ ∈ Sn, then the tempered distribution u∗ϕ can be identiﬁed with a slowly increasing function f deﬁned by n f(x) = u[τ−xϕ˜] = u[ϕ(x − ·)] for all x ∈ R . Moreover f ∈ C∞(Rn) and all its derivatives are slowly increasing.

Proof. Note that the equality u[τ−xϕ˜] = u[ϕ(x − ·)] is obvious. First we will prove that ∞ the function f(x) = u[τ−xϕ˜] = u[ϕ(x − ·)] is C and f, as well as all its derivatives, are slowly increasing.

It follows from Theorem 3.43(c) that f is continuous. Observe that for a ﬁxed x ∈ Rn f(x + he ) − f(x) ϕ(x + he − ·) − ϕ(x − ·) k = u k → u[(∂ ϕ)(x − ·)] as h → 0 h h k because it is easy to check using Theorem 3.43(e) that for every x ∈ Rn ϕ(x + he − ·) − ϕ(x − ·) k → (∂ ϕ)(x − ·) as h → 0 h k in the topology of Sn. Thus the partial derivatives

∂kf(x) = u[(∂kϕ)(x − ·)] exist and are continuous by Theorem 3.43(c). Since ∂kϕ ∈ Sn, iterating the above process for any multiindex α we obtain (5.2) Dαf(x) = u[(Dαϕ)(x − ·)] . This implies that f ∈ C∞(Rn). Since u is a tempered distribution, Theorem 5.3 gives the estimate X 0 m (5.3) |f(x)| = |u[τ−xϕ˜]| ≤ C pα,β(τ−xϕ˜) ≤ C (1 + |x|) . |α|,|β|≤m Indeed, α β α β pα,β(τ−xϕ˜) = sup |z (D ϕ˜)(z − x)| = sup |(z + x) (D ϕ˜)(z)| z∈Rn z∈Rn ≤ C1 + |x||α| sup 1 + |z||α| |Dβϕ˜(z)| ≤ C0(1 + |x|)m . z∈Rn HARMONIC ANALYSIS 57

Hence f is slowly increasing, and so it deﬁnes a tempered distribution. Since the derivatives of f satisfy (5.2), which is an expression of the same type as the one in the deﬁnition of f, we conclude that all derivatives Dαf are slowly increasing. It remains to prove that f = u ∗ ϕ in the sense of tempered distributions i.e.,

(5.4) (u ∗ ϕ)[ψ] = f[ψ] for all ψ ∈ Sn. Using f(y) = u[ϕ(y − ·)], (5.4) reads as h Z i Z u ϕ(y − ·)ψ(y) dy = u[ϕ(y − ·)]ψ(y) dy. Rn Rn The idea of proving this equality is to approximate the integral on on the left hand side by Riemann sums, use linearity of u to go under the sign of the sum and to pass to the limit. However, the details of this approximation argument are not obvious. They are elementary and tedious, but not obvious and so in most of the textbooks they are summarized as ‘easy to see’. For the sake of completeness we decided to include details.

∞ First of all observe that is suﬃces to prove (5.4) under the assumption that ϕ, ψ ∈ C0 . Indeed, suppose we know (5.4) for compactly supported functions, but now ϕ, ψ ∈ Sn. 39 ∞ ∞ Let C0 3 ϕk → ϕ in Sn and C0 3 ψ` → ψ in Sn. Observe that fk(x) = u[τ−xϕ˜k] → f(x) pointwise. It follows from the proof of (5.3) that all the functions fk have a common estimate independent of k m |fk(x)| ≤ C(1 + |x|) 0 and hence fk → f in Sn by the Dominated Convergence Theorem. It is also clear from 0 the deﬁnition of the convolution that u ∗ ϕk → u ∗ ϕ in Sn, so

(u ∗ ϕ)[ψ`] = lim (u ∗ ϕk)[ψ`] = lim fk[ψ`] = f[ψ`] . k→∞ k→∞ ∞ The second equality follows from assuming (5.4) for all ϕk, ψ` ∈ C0 . Now letting ` → ∞ yields (5.4). If the support of ψ is contained in a cube Q with integer edge-length we divide Q into −k cubes {Qki}i of edge-length 2 and centes yki. Then the Riemann sums converge in the topology of Sn (as a function of x) to the integral X Z ϕ(yki − x)ψ(yki)|Qki| → ϕ(y − x)ψ(y) dy n i R and hence h Z i X Z u ϕ(y − ·)ψ(y) dy = lim u[ϕ(yki − ·)]ψ(yki)|Qki| = u[ϕ(y − ·)]ψ(y) dy. n k→∞ n R i R Boring (but elementary) details of the convergence of the Riemann sums in the topology of Sn are moved to Appendix 5.7. Note that formula (5.2) implies

0 Theorem 5.8. If u ∈ Sn and ϕ ∈ Sn, then for any multiindex α we have Dα(u ∗ ϕ)(x) = (u ∗ (Dαϕ))(x) . 39See Theorem 3.43(b). 58 PIOTR HAJLASZ

The next result will require the use of the Banach-Steinhaus theorem.

0 Theorem 5.9. If uk → u is Sn and ϕk → ϕ in Sn, then uk[ϕk] → u[ϕ].

Proof. In order to be able to use triangle inequality to prove the convergence we need to 0 know uniform estimates for the sequence uk. Convergence uk → u in Sn means pointwise convergence on a complete metric space Sn so it is natural to use the Banach-Steinhaus Theorem to get uniform estimates for uk. We state it as a lemma.

Lemma 5.10 (Banach-Steinhaus). Let X be a complete metric space and let {fi}i∈I be a family of continuous real-valued functions on X. If the functions in the family are pointwise bounded, i.e.

sup |fi(x)| < ∞ for x ∈ X, i∈I then there is an open set U ⊂ X and a constant M > 0 such that

sup |fi(x)| ≤ M for all x ∈ U. i∈I

It easily follows from the linearity that it suﬃces to consider the case uk → 0 and ϕk → 0. The functions ϕ 7→ |uk[ϕ]| are continuous on the complete metric space (Sn, d). They are pointwise bounded on Sn since |uk[ϕ]| → 0 for every ϕ ∈ Sn. Thus Lemma 5.10 implies that

sup |uk[ϕ]| ≤ M for all ϕ ∈ B(ϕ0, r0). k

Fix ε > 0. There is k1 such that |uk[ϕ0]| < ε for k ≥ k1. Since ϕk → 0 in Sn, then also −1 ε ϕk → 0 and hence −1 −1 d(ϕ0 + ε ϕk, ϕ0) = d(ε ϕk, 0) → 0 , −1 so there is k2 such that for k ≥ k2, ϕ0 + ε ϕk ∈ B(ϕ0, r0). Hence for k ≥ max{k1, k2} we have −1 −1 |uk[ε ϕk]| ≤ |uk[ϕ0 + ε ϕk]| + |uk[ϕ0]| ≤ M + ε , so |uk[ϕk]| ≤ ε(M + ε). 0 Theorem 5.11. If uk → u in Sn, then there is a constant C > 0 and a positive integer m such that X sup |uk[ϕ]| ≤ C pα,β(ϕ) for all ϕ ∈ Sn. k |α|,|β|≤m

0 Proof. Let vk = uk − u. Clearly vk → 0 in Sn. It suﬃces to prove that there is C > 0 and a positive integer m such that X sup |vk[ϕ]| ≤ C pα,β(ϕ) for all ϕ ∈ Sn. k |α|,|β|≤m Indeed, this estimate, the inequality

sup |uk[ϕ]| ≤ |u[ϕ]| + sup |vk[ϕ]| k k and Theorem 5.3 readily yield the result. HARMONIC ANALYSIS 59

As in the proof of Theorem 5.3 it suﬃces to prove that there is a positive integer m such that supk |vk[ϕ]| ≤ 1 for n X 1 o ϕ ∈ Nm := ϕ ∈ Sn : pα,β(ϕ) ≤ . m |α|,|β|≤m

To the contrary suppose that for any positive integer i we can ﬁnd ϕi ∈ Ni such that supk |vk[ϕi]| > 1, i.e.

(5.5) |vki [ϕi]| > 1 for some ki.

Clearly ϕi → 0 in Sn. If the sequence ki is bounded, (5.5) contradicts continuity of functionals vk. If ki is unbounded, then there is a subsequence kij → ∞ and again (5.5) contradicts Theorem 5.9. ∞ n 0 0 Theorem 5.12. C0 (R ) is dense in Sn, i.e. if u ∈ Sn, then there is a sequence wk ∈ ∞ n C0 (R ) such that wk[ψ] → u(ψ) for all ψ ∈ Sn.

0 ∞ n R 40 Proof. Let u ∈ . If ϕ ∈ C ( ), n ϕ = 1, then for every ψ ∈ n,ϕ ˜ε ∗ ψ → ψ in n Sn 0 R R S S so (u ∗ ϕε)[ψ] = u[ϕ ˜ε ∗ ψ] → u[ψ] . ∞ Since uε = u ∗ ϕε ∈ C we obtain a sequence of smooth functions that converge to u in 0 ∞ n Sn. If we take a cut-oﬀ function η ∈ C0 (R ), 0 ≤ η ≤ 1, η(x) = 1 for |x| ≤ 1 and η(x) = 0 for |x| ≥ 2, then one can easily prove that for ψ ∈ Sn

η(x/k)ψ(x) → ψ(x) in Sn as k → ∞. 41 ∞ Using this fact and Theorem 5.9 we obtain that wk(x) = η(x/k)(u∗ϕk−1 ) ∈ C0 converges 0 to u in Sn. Indeed, wk[ψ] = (u ∗ ϕk−1 )[η(·/k)ψ(·)] → u[ψ].

If ϕ, ψ ∈ Sn, then the integration by parts gives Z Z Dαϕ(x)ψ(x) dx = (−1)|α| ϕ(x)Dαψ(x) dx, Dαϕ[ψ] = ϕ[(−1)αψ]. Rn Rn For h ∈ Rn we have Z Z (τhϕ)(x)ψ(x) dx = ϕ(x)(τ−hψ)(x) dx, τhϕ[ψ] = ϕ[τ−hψ]. Rn Rn Moreover Z Z ϕ˜(x)ψ(x) dx = ϕ(x)ψ˜(x) dx, ϕ˜[ψ] = ϕ[ψ˜], n n ZR ZR ϕˆ(x)ψ(x) dx = ϕ(x)ψˆ(x) dx, ϕˆ[ψ] = ϕ[ψˆ], n n ZR ZR ϕˇ(x)ψ(x) dx = ϕ(x)ψˇ(x) dx, ϕˇ[ψ] = ϕ[ψˇ]. Rn Rn 40Why? 41Instead of using Theorem 5.9 which is somewhat not constructive as it is based on the Banach- Steinhaus theorem, we could prove convergence by a brute force estimates. 60 PIOTR HAJLASZ

If η ∈ C∞ is slowly increasing and all derivatives of η are slowly increasing, i.e. every α derivative D η is bounded by a polynomial, then for ψ ∈ Sn, ηψ ∈ Sn and the mapping α ψ 7→ ηψ is continuous in Sn. In particular x ψ(x) ∈ Sn. Moreover Z Z (ηϕ)[ψ] = η(x)ϕ(x)ψ(x) dx = ϕ(x)η(x)ψ(x) dx = ϕ[ηψ]. Rn Rn This motivates the following deﬁnition.

0 Definition 5.13. For u ∈ Sn we define • The distributional partial derivative Dαu is a tempered distribution defined by the formula Dαu[ψ] = u[(−1)|α|Dαψ]. 0 • The translation τhu ∈ Sn is defined by

(τhu)[ψ] = u[τ−hψ]. 0 • The reflectionu ˜ ∈ Sn is defined by u˜[ψ] = u[ψ˜]. 0 0 • The Fourier transform F(u) =u ˆ ∈ Sn and the inverse Fourier transformu ˇ ∈ Sn are uˆ[ψ] = u[ψˆ] andu ˇ[ψ] = u[ψˇ]. • If η ∈ C∞ is slowly increasing and all derivatives of η are slowly increasing, then we define (ηu)[ψ] = u[ηψ]. In particular (xαu)[ψ] = u[xαψ].

0 The formulas preceding the definition show that on the subclass Sn ⊂ Sn the partial derivative, the translation, the reflection, the Fourier transform and the multiplication by a function defined in the distributional sense coincide with those defined in the classical way.

If f ∈ L1(Rn)+L2(Rn), in particular, if f ∈ Lp(Rn), 1 ≤ p ≤ 2, then the classical Fourier transform coincides with the distributional one. That easily follows from Theorem 3.2(c) and Proposition 3.46. The basic properties of the Fourier transform, distributional derivative and convolution 0 in Sn are collected in the next result whose easy proof is left to the reader. 0 42 Theorem 5.14. The Fourier transform in a homeomorphism of Sn onto itself. Moreover 0 for u ∈ Sn and ϕ ∈ Sn we have (a) (ˆu)∨ = u, (b) (u ∗ ϕ)ˆ =ϕ ˆuˆ, (c) Dα(u ∗ ϕ) = Dαu ∗ ϕ = u ∗ Dαϕ, (d) (Dαu)ˆ = (2πix)αuˆ, (e) ((−2πix)αu)ˆ = Dαuˆ.

42 0 With respect to the weak convergence in Sn. HARMONIC ANALYSIS 61

Just for the illustration we will prove the property (d). For any ψ ∈ Sn we have (Dαu)∧[ψ] = u[(−1)|α|Dαψˆ] = u[(−1)|α|((−2πix)αψ)∧] = u[((2πix)αψ)∧] = (2πix)αuˆ[ψ].

0 α Note that in the case (c), u ∗ ϕ ∈ Sn and D (u ∗ ϕ) is understood in the distributional sense. On the other hand u ∗ ϕ ∈ C∞ and Theorem 5.8 shows that the distributional derivative of u ∗ ϕ coincides with the classical one.

Example 5.15. The Dirac measure δa deﬁnes a tempered distribution so Z ˆ ˆ ˆ −2πix·a δa[ψ] = δa[ψ] = ψ(a) = ψ(x)e dx Rn ˆ −2πix·a and hence δa = e .

Example 5.16. Let µ be a measure on R deﬁned by µ(E) = #(E ∩ Z), where #A stands for the cardinality of a set A. In other words µ is the counting measure on Z trivially extended to R. Clearly, µ is a tempered measure so it deﬁnes a tempered distribution. 0 Moreoverµ ˆ = µ in S1. Indeed, this is a simple consequence of the Poisson summation formula Theorem 4.1 that ∞ ∞ X X µˆ[ϕ] = µ[ϕ ˆ] = ϕˆ(n) = ϕ(n) = µ[ϕ]. n=−∞ n=−∞

5.2. Tempered distributions as derivatives of functions. Using the notion of distri- 0 butional derivative we can provide a new class of examples of a distributions in Sn. If fα, |α| ≤ m are slowly increasing functions and aα ∈ C for |α| ≤ m, then

X α 0 (5.6) u = aαD fα ∈ Sn . |α|≤m

α where the derivative D fα is understood in the distributional sense.

0 The aim of this section is to prove a surprising result that every distribution in Sn can be represented in the form (5.6), see Theorem 5.21. Proposition 5.17. Let N be a positive integer, then the operators43

N N 0 0 (I − ∆) : Sn → Sn and (I − ∆) : Sn → Sn are homeomorphisms. The inverse operator in both cases is given by the operator (I − ∆)−N := F −1(1 + 4π2|ξ|2)−N F.

0 N Remark 5.18. That means for every v ∈ Sn (v ∈ Sn) the equation (I − ∆) u = v has 0 a unique solution u ∈ Sn (u ∈ Sn) given by u = (I − ∆)N v = F −1(1 + 4π2|ξ|2)−N F(v).

43I − ∆ stands for the identity minus the Laplace operator. 62 PIOTR HAJLASZ

N Proof. Suppose that u, v ∈ Sn satisfy (I − ∆) u = v. Taking the Fourier transform yields (1 + 4π2|ξ|2)N F(u) = F(v) and hence F(u) = (1 + 4π2|ξ|2)−N F(v).

Since F(u) ∈ Sn, the right hand side is also in Sn so we can take the inverse Fourier transform and we obtain (5.7) u = F −1(1 + 4π2|ξ|2)−N F(v).

Thus if a solution u ∈ Sn exists, it is unique and given by (5.7). To see that the formula 2 2 −N indeed, deﬁnes a solution u ∈ Sn observe that the function (1 + 4π |ξ| ) and all its 2 2 −N derivatives are slowly increasing so if f ∈ Sn, then (1 + 4π |ξ| ) f ∈ Sn. Hence for any v ∈ Sn, the right hand side of (5.7) deﬁnes a function in Sn and backward computations shows that (I − ∆)N u = v.

0 N Similarly, if u, v ∈ Sn, then (I − ∆) u = v if and only if the Fourier transforms are equal i.e., (1 + 4π2|ξ|2)N F(u) = F(v) which is equivalent to F(u) = (1 + 4π2|ξ|2)−N F(v), u = F −1(1 + 4π2|ξ|2)−N F(v). Note that we were allowed to multiply both sides by (1 + 4π2|ξ|2)−N because this function and all its derivatives are slowly increasing.

The operator (I −∆)N maps functions of class Ck to functions of class Ck−2N so it lowers regularity of functions. That means the inverse operator (I − ∆)−N increases regularity. Hence we should expect that it also increases regularity of distributions. Since every tempered distribution has finite order in a sense that it can be estimated by a finite sum as in 0 Theorem 5.3, one could expect that for u ∈ Sn and sufficiently large N, the distribution (I − ∆)−N u is actually a regular function. As we will see this intuition is correct. Remark 5.19. The operator (I −∆)−N is called a Bessel potential and one can prove that it can be represented as an integral operator. In Section ?? we will study Bessel potentials in detail. We will find an explicit integral formula for the Bessel potentials and we will show how use them to characterize Sobolev spaces. k,1 Definition 5.20. By Cloc we will denote the class of k times continuously differentiable 0,1 function whose derivatives of order k are locally Lipschitz continuous. In particular Cloc denotes the class of locally Lipschitz continuous functions. 0 44 Theorem 5.21. Suppose u ∈ Sn satisfies the estimate X |u(ϕ)| ≤ C pα,β(ϕ) . |α|,|β|≤m If k is a nonnegative integer and N > (n + m + k)/2, then the tempered distribution v = (I − ∆)−N u has the following properties:

44See Theorem 5.3. HARMONIC ANALYSIS 63

(a) If k = 0, then v is a slowly increasing function. k−1,1 (b) If k ≥ 1, then v is a slowly increasing function of the class Cloc . Also all derivatives of v or order |α| ≤ k are slowly increasing.

Proof. We will need

Lemma 5.22. If P (x) is a polynomial in Rn of degree p and N > (n + p)/2, then all derivatives of P (x) f(x) = (1 + |x|2)N belong to L1(Rn).

Proof. Since 2N − p > n, f ∈ L1(Rn). We have ∂f Q(x) = 2 2N , deg Q = 2N + p − 1. ∂xi (1 + |x| ) The function on the right hand side is of the same form as f. Since n + (2N + p − 1) 2N > 2 1 we conclude that ∂f/∂xi ∈ L . The integrability of higher order derivatives follows by induction. 2 We will prove that the distributional derivatives of v of orders |γ| ≤ k are slowly increasing functions. We have (−1)|γ|Dγv[ψ] = uF(1 + 4π2|ξ|2)−N F −1(Dγψ) . Hence X |Dγv[ψ]| ≤ C sup xαDβ F(1 + 4π2|ξ|2)−N F −1(Dγψ) x∈ n |α|,|β|≤m R β+γ X α ξ −1 = C Cα,β,γ sup F D 2 2 N F (ψ) x∈ n (1 + 4π |ξ| ) |α|,|β|≤m R β+γ 0 X α x −1 ≤ C D F (ψ) . (1 + 4π2|x|2)N |α|,|β|≤m 1 Note that xβ+γ Dα F −1(ψ) (1 + 4π2|x|2)N α! xβ+γ X αi βi −1 = D 2 2 N D (F (ψ)) . α1!βi! (1 + 4π |x| ) αi+βi=α Since deg xβ+γ ≤ m + k and N > (n + m + k)/2, Lemma 5.22 gives xβ+γ Dαi ∈ L1( n) , (1 + 4π2|x|2)N R 64 PIOTR HAJLASZ so

γ X −1 βi |D v[ψ]| ≤ C kF (x ψ)k∞ i

X βi ≤ C kx ψk1 i 0 2 m/2 ≤ C k(1 + |x| ) ψ(x)k1 . This proves that for |γ| ≤ k the functional ψ 7→ Dγv[ψ] 1 2 m/2 ∞ is bounded on L ((1 + |x| ) dx). Thus there are functions gγ ∈ L such that Z γ 2 m/2 D v[ψ] = ψ(x)gγ(x)(1 + |x| ) dx , Rn i.e. γ 2 m/2 D v = gγ(x)(1 + |x| ) 2 m/2 in the distributional sense. In particular, if γ = 0, v(x) = g0(x)(1 + |x| ) is slowly increasing which proves (a). The part (b) follows from the following result whose proof is postponed to Section ??, see Theorem ?? Proposition 5.23. If a function u is slowly increasing and its distributional derivatives k−1,1 of order less than or equal k, k ≥ 1, are slowly increasing functions, then u ∈ Cloc .

5.3. Tempered distributions with compact support. We will investigate now properties of the Fourier transform of tempered distributions with compact support.

n 0 ∞ Lemma 5.24. Let Ω ⊂ R be open and u ∈ Sn. If u[ϕ] = 0 for all ϕ ∈ C0 (Ω), then u[ϕ] = 0 for all ϕ ∈ Sn such that supp ϕ ⊂ Ω.

∞ n Proof. Let ϕ ∈ Sn, supp ϕ ⊂ Ω and let η be a cut-oﬀ function, i.e. η ∈ C0 (R ), 0 ≤ η ≤ 1, η(x) = 1 for |x| ≤ 1 and η(x) = 0 for |x| ≥ 2. One can easily prove that η(x/R)ϕ(x) → ϕ(x) ∞ in Sn as R → ∞. Since η(x/R)ϕ(x) ∈ C0 (Ω), the lemma follows. 0 Deﬁnition 5.25. Let u ∈ Sn. The support of u (supp u) is the intersection of all closed sets E ⊂ Rn such that ∞ n ϕ ∈ C0 (R \ E) ⇒ u[ϕ] = 0. 0 ∞ n Proposition 5.26. If u ∈ Sn and ϕ ∈ C0 (R \ supp u), then u[ϕ] = 0.

Proof. Indeed, complement of supp u is the union of all open sets Ui such that if ϕ ∈ ∞ ∞ S C0 (Ui), then u[ϕ] = 0, and we need to prove that if ϕ ∈ C0 ( i Ui), then still u[ϕ] = 0. Since the support of ϕ is compact we can select a ﬁnite subfamily of open sets Ui covering the support and the result follows from a simple argument involving partition of unity.

Thus the support of u is the smallest closed set such that the distribution vanishes on ∞ C0 functions supported outside that set. ∞ n The lemma shows that we can replace ϕ ∈ C0 (R \E) in the above deﬁnition by ϕ ∈ Sn with supp ϕ ⊂ Rn \ E. HARMONIC ANALYSIS 65

Before we state the next result we need some facts about analytic and holomorphic functions in several variables.

Deﬁnition 5.27. We say that a function f :Ω → C deﬁned in an open set Ω ⊂ Rn is R-analytic, if in a neighborhood on any point x0 ∈ Ω it can be expanded as a convergent power series

X α (5.8) f(x) = aα(x − x0) , aα ∈ C, α i.e. if in a neighborhood of any point f equals to its Taylor series.

We say that a function f :Ω → C deﬁned in an open set Ω ⊂ Cn is C-analytic if in a neighborhood of any point z0 ∈ Ω it can be expanded as a convergent power series

X α f(z) = aα(z − z0) . α

Rn has a natural embedding into Cn, just like R into C. n n R 3 x = (x1, . . . , xn) 7→ (x1 + i · 0, . . . , xn + i · 0) = x + i · 0 ∈ C . It is easy to see that an R-analytic function f in Ω ⊂ Rn extends to a C-analytic function f˜ in an open set Ω˜ ⊂ Cn,Ω ⊂ Ω.˜ Namely, if f satisﬁes (5.8), we set ˜ X α f(z) = aα(z − z0) , z0 = x0 + i · 0. α On the other hand, if f˜ is C-analytic in Ω˜ ⊂ Cn, then the restriction f of f˜ to Ω = Ω˜ ∩ Rn is R-analytic. For example for any ξ ∈ Rn, f(x) = ex·ξ is R-analytic and f˜(z) = ez·ξ is its C-analytic extension.

Deﬁnition 5.28. We say that a continuous function f :Ω → C deﬁned in an open set Ω ⊂ Cn is holomorphic if ∂f = 0 for i = 1, 2, . . . , n. ∂zi

It is easy to see that C-analytic function are holomorphic, but the converse implication is also true. Lemma 5.29 (Cauchy). If f is holomorphic in45 n 1 1 n D (w, r) = D (w1, r1) × ... × D (wn, rn) ⊂ C and continuous in the closure Dn(x, r), then Z Z 1 f(ξ) dξ1 . . . dξn (5.9) f(z) = n ... 1 1 (2πi) ∂D (w1,r1) ∂D (wn,rn) (ξ1 − z1) ... (ξn − zn) for all z ∈ Dn(w, r).

45Product of one dimensional discs. 66 PIOTR HAJLASZ

Proof. The function f is holomorphic in each variable separately, so (5.9) follows from one dimensional Cauchy formulas and the Fubini theorem. 2 Just like in the case of holomorphic functions of one variable one can prove that the integral on the right hand side of (5.9) can be expanded as a power series and hence deﬁnes a C-analytic function. We proved Theorem 5.30. A function f is holomorphic in Ω ⊂ Cn if and only if it is C-analytic.

Now we can state an important result about compactly supported distributions.

0 ∞ Theorem 5.31. If u ∈ Sn has compact support, then uˆ is a slowly increasing C function and all derivatives of uˆ are slowly increasing. Moreover uˆ is R-analytic on Rn and has a holomorphic extension to Cn.

∞ n Proof. Let η ∈ C0 (R ) be such that η(x) = 1 in a neighborhood of supp u. Then u = ηu 0 in Sn and hence for ψ ∈ Sn we have Z uˆ[ψ] = (ηu)ˆ[ψ] = u η(·)ψ(x)e−2πix·(·) dx Rn Z = u η(·)e−2πix·(·) ψ(x) dx . Rn We could pass with u under the sign of the integral, because of an argument with approximation of the integral by Riemann sums.46 Note that the function −2πix·(·) F (x1, . . . , xn) = u η(·)e is C∞ smooth and α α −2πix·(·) D F (x1, . . . , xn) = u (−2πi(·)) η(·)e . Indeed, we could diﬀerentiate under the sign of u because the corresponding diﬀerence quotients converge in the topology of Sn. It also easily follows from Theorem 5.3 that F and all its derivatives are slowly increasing. Thus we may identifyu ˆ with F , sou ˆ ∈ C∞.

Moreover F has a holomorphic extension to Cn by the formula −2πiz·(·) F (z1, . . . , zn) = u η(·)e so in particular F (x1, . . . , xn) is R-analytic. 2 0 Remark 5.32. Note that if u ∈ Sn has compact support, then we can reasonably deﬁne u[e−2πix·(·)] by the formula ue−2πix·(·) := uη(·)e−2πix·(·) , ∞ n where η ∈ C0 (R ) is such that η = 1 in a neighborhood of supp u. Note that this construction does not depend on the choice of η. In the same way we can deﬁne u[f] for any f ∈ C∞(Rn) 46Compare with the proof of Theorem 5.7. HARMONIC ANALYSIS 67

0 Theorem 5.33. If u ∈ Sn and supp u = {x0}, then there is an integer m ≥ 0 and complex numbers aα, |α| ≤ m such that X α u = aαD δx0 . |α|≤m

Proof. Without loss of generality we may assume that x0 = 0. According to Theorem 5.3 the distribution u satisﬁes the estimate X |u[ϕ]| ≤ C pα,β(ϕ) . |α|,|β|≤m

First we will prove that for ϕ ∈ Sn we have (5.10) Dαϕ[0] = 0 for |α| ≤ m =⇒ u[ϕ] = 0 . Indeed, it follows from Taylor’s formula that ϕ(x) = O(|x|m+1) as x → 0 and hence also (5.11) Dβϕ(x) = O(|x|m+1−|β|) as x → 0 for all |β| ≤ m. ∞ n Let η ∈ C0 (R ) be a cut-oﬀ function, i.e. 0 ≤ η ≤ 1, η(x) = 1 for |x| ≤ 1, η(x) = 0 for |x| ≥ 2 and deﬁne ηε(x) = η(x/ε). The estimate (5.11) easily implies that

pα,β(ηεϕ) → 0 as ε → 0 47 for all |α|, |β| ≤ m. Note that ϕ − ηεϕ = 0 in a neighborhood of 0, so u[ϕ − ηεϕ] = 0 and hence X |u[ϕ]| ≤ |u[ϕ − ηεϕ]| + |u[ηεϕ]| ≤ 0 + pα,β(ηεϕ) → 0 as ε → 0. |α|,|β|≤m This completes the proof of (5.10).

Let now ψ ∈ Sn be arbitrary and let X Dαψ(0) h(x) = ψ(x) − xα . α! |α|≤m Clearly (5.12) Dαh(0) = 0 for |α| ≤ m. We have   X Dαψ(0) ψ(x) = η(x) xα + η(x)h(x) + (1 − η(x))ψ(x) .  α!  |α|≤m Since (1−η)ψ vanishes in a neighborhood of 0 we have u[(1−η)ψ] = 0. The equality (5.12) ∞ n implies that ϕ = ηh ∈ C0 (R ) satisﬁes the assumptions of (5.10), so u[ηh] = 0. Hence h X Dαψ(0) i u[ψ] = u η(x) xα α! |α|≤m

47Check it! 68 PIOTR HAJLASZ

X η(x)xα = (−1)|α|u (−1)|α|Dαψ(0) . α! |α|≤m | α{z } | {z } D δ0[ψ] aα The proof is complete. 2

0 2πix·ξ0 Corollary 5.34. Let u ∈ Sn. If suppu ˆ = {ξ0}, then u = P (x)e for some polynomial P (x) in Rn. In particular if suppu ˆ = {0}, then u is a polynomial.

We leave the proof as an exercise.

0 Corollary 5.35. If u ∈ Sn satisﬁes ∆u = 0, then u is a polynomial.

Proof. We have 2 2 0 −4π |ξ| uˆ = (∆u)ˆ = 0 in Sn. This implies that suppu ˆ = {0}, so u is a polynomial by Corollary 5.34. 2

5.4. Homogeneous distributions. A very important class of distributions is provided by so called homogeneous distributions that we will study now.

Deﬁnition 5.36. A function f ∈ C(R \{0}) is homogeneous of degree a ∈ R if for all x 6= 0 and all t > 0 we have f(tx) = taf(x). More generally a measurable function on Rn is homogeneous of degree a ∈ R if a n f(tx) = t f(x) for almost all (x, t) ∈ R × (0, ∞).

If f ∈ C(Rn \{0}) is homogeneous of degree a > −n, then it defines a tempered distribution. Indeed, a x a |f(x)| = |x| f ≤ C|x| |x| and |x|a is integrable in a neighborhood of the origin (because a > −n). If f is a measurable homogeneous function of degree a we can assume (by changing f on a set of measure zero) that f is continuous on almost all open rays going out of 0. Then on the remaining rays (which form a set of measure zero) we can make f homogeneous. This is to say that we can find a representative of f (in class of functions that are equal a.e.) such that f(tx) = taf(x) for all x 6= 0 and t > 0. If a measurable function that is not a.e. equal to zero is homogeneous of degree a ≤ −n, then the integration in the spherical coordinate system shows that it is not integrable in any neighborhood of 0 and hence it cannot define a tempered distribution.

1 n Lemma 5.37. Suppose that f ∈ Lloc(R ) deﬁnes a tempered distribution. Then the func- 48 tion f is homogeneous of degree a ∈ R if and only if a (5.13) f[ϕt] = t f[ϕ] for all ϕ ∈ Sn and all t > 0. Remark 5.38. If f is not equal to zero a.e., then necessarily a > −n.

48 −n As always ϕt(x) = t ϕ(x/t). HARMONIC ANALYSIS 69

Proof. For ϕ ∈ Sn we have Z Z Z a a f[ϕt] = f(x)ϕt(x) dx = f(tx)ϕ(x) dx and t f[ϕ] = t f(x)ϕ(x) dx. Rn Rn Rn Hence condition (5.13) is equivalent to the equality Z Z a (5.14) f(tx)ϕ(x) dx = t f(x)ϕ(x) dx for all ϕ ∈ Sn and all t > 0 Rn Rn which, in turn, is equivalent to the fact that for all t > 0, equality f(tx) = taf(x) holds a.e.

The above result motivates the following deﬁnition.

0 49 Deﬁnition 5.39. We say that a distribution u ∈ Sn is homogeneous of degree a ∈ R if a u[ϕt] = t u[ϕ] for all ϕ ∈ Sn and all t > 0.

0 0 Proposition 5.40. u ∈ Sn is homogeneous of degree a ∈ R if and only if uˆ ∈ Sn is homogeneous of degree −n − a.

Proof. Let ϕ ∈ Sn. Then −n −1 −n −1 −n ϕbt(ξ) =ϕ ˆ(tξ) = t (t ) ϕˆ(ξ/t ) = t (ϕ ˆ)t−1 (ξ) and hence −n −n−a −n−a uˆ[ϕt] = u[ϕbt] = t u[(ϕ ˆ)t−1 ] = t u[ϕ ˆ] = t uˆ[ϕ]. 0 If u ∈ Sn is homogeneous of degree a, the above equality shows thatu ˆ is homogeneous of degree −n − a. Now ifu ˆ is homogeneous of degree −n − a, then its Fourier transform i.e., u˜ is homogeneous of degree −n − (−n − a) = a and hence also u is homogeneous of degree a. Example 5.41. u(x) = |x|a, a ≥ 0, is homogeneous of degree a and it deﬁnes a tempered distribution. However,u ˆ is homogeneous of degree −n − a ≤ −n and hence it cannot be represented as a function.

The notion of a homogeneous distribution will play a signiﬁcant role in the proof of the next result. Theorem 5.42. For 0 < a < n, n n−a a− 2 −a π Γ 2 a−n F |x| (ξ) = a |ξ| . Γ 2 Remark 5.43. |x|−a is homogeneous of degree −a > −n and |ξ|a−n is homogeneous of degree a − n > −n so both |x|−a and |ξ|a−n are tempered distributions represented as functions.

49Now a can be any real number and we do not impose restriction a > −n that was necessary in the case of functions. 70 PIOTR HAJLASZ

Proof. First we will prove the theorem under the assumption that n/2 < a < n. In that case −a −a −a 1 2 |x| = |x| χ{|x|≤1} + |x| χ{|x|>1} ∈ L + L . −a 2 Thus the Fourier transform of |x| is a function in C0 + L . The Fourier transform commutes with orthogonal transformations f[◦ ρ = fˆ◦ ρ, for ρ ∈ O(n) and f ∈ L1 + L2. If f is radially symmetric i.e., f = f ◦ ρ for all ρ ∈ O(n), then fˆ ◦ ρ = f[◦ ρ = fˆ for all ρ ∈ O(n) so fˆ is radially symmetric too. Thus the Fourier transform of |x|−a ∈ L1 + L2 is 2 a radially symmetric function in C0 + L , homogeneous of degree a − n as a distribution. Hence Lemma 5.37 yields −a a−n F |x| (ξ) = ca,n|ξ| −π|x|2 and it remains to compute the coeﬃcient ca,n. Employing the fact that ϕ(x) = e is a ﬁxed point of the Fourier transform we have a−n −a −a −a ca,n|x| [ϕ] = F(|x| )[ϕ] = |x| [ϕ ˆ] = |x| [ϕ], i.e., Z Z −π|x|2 a−n −π|x|2 −a (5.15) ca,n e |x| dx = e |x| dx. Rn Rn The integrals in this identity are easy to compute. Lemma 5.44. For γ > −n we have Z n+γ −π|x|2 γ Γ 2 e |x| dx = γ n . n 2 R π Γ 2

Proof. For γ > −n we have50 Z Z ∞ −π|x|2 γ −πs2 γ+n−1 e |x| dx = nωn e s ds Rn 0 Z ∞ t=πs2 nωn n+γ −t 2 −1 = n+γ e t dt 2π 2 0 nω n + γ Γ n+γ = n Γ = 2 . n+γ γ n 2 2 2 2π π Γ 2

Applying the lemma to both sides of (5.15) yields a n−a n−a Γ 2 Γ 2 a− n Γ 2 c = so c = π 2 . a,n a−n n − a n a,n a 2 π 2 Γ Γ π Γ 2 2 2 This completes the proof when n/2 < a < n.

50 n Recall that the volume of the unit sphere in R equals nωn so the ﬁrst equality is just integration in 2πn/2 the spherical coordinates. In the last equality we use ωn = nΓ(n/2) , see (3.9). HARMONIC ANALYSIS 71

Suppose now that 0 < a < n/2. Since n/2 < n − a < n we have n a 2 −a −(n−a) π Γ 2 −a F |x| (ξ) = n−a |ξ| Γ 2 and applying the Fourier transform to this equality yields n a 2 −a a−n π Γ 2 −a | − x| = n−a F |ξ| (x). Γ 2 Replacing x by ξ and ξ by x yields the result when 0 < a < n/2.

We are left with the case of a = n/2. Take a sequence n/2 < ak < n, ak → a = n/2. Since the result is true for ak we have n a − n−ak Z π k 2 Γ Z |x|−ak ϕˆ(x) dx = 2 |x|ak−nϕ(x) dx. ak Rn Γ 2 Rn Passing to the limit ak → a = n/2 yields the result for a = n/2. ak−n a−n −ak −a 0 Remark 5.45. Clearly |ξ| → |ξ| and |x| → |x| in Sn. The last convergence −ak −a 0 and the continuity of the Fourier transform yields that F(|x| ) → F(|x| ) in Sn. Hence the result for a = n/2 follows by passing to the limit in the equality n a − n−ak π k 2 Γ F |x|−ak (ξ) = 2 |ξ|ak−n in 0 . ak Sn Γ 2 Remark 5.46. In the case a = n/2 we obtain a particularly simple formula Z Z − n − n − n − n F(|x| 2 ) = |x| 2 , i.e., |x| 2 ϕˆ(x) dx = |x| 2 ϕ(x) dx Rn Rn for all ϕ ∈ Sn. This formula can be regarded as a version of the Poisson summation formula.51

5.5. The principal value. 0 α Proposition 5.47. If u ∈ Sn is homogeneous of degree a, then D u is homogeneous of degree a − |α|.

We leave the proof as an (easy) exercise.

In particular if f ∈ C∞(Rn \{0}) is homogeneous of degree 1 − n, then it defines a distribution, but its first order distributional partial derivatives are homogeneous of degree −n and hence they cannot be represented as functions. It is however, still possible to represent these distributional derivatives as improper integrals, the so called principal value of the integral. Definition 5.48. If for any ε > 0, a function f is integrable in Rn \B(0, ε), then we define the principal value of the integral as Z Z p.v. f(x) dx = lim f(x) dx ε→0 Rn |x|≥ε provided the limit exists.

51cf. Theorem 4.1. 72 PIOTR HAJLASZ

If a function f is slowly increasing, but has singularity at 0, then we can try to define a distribution p.v. f by Z Z (p.v. f)[ϕ] = p.v. f(x)ϕ(x) dx = lim f(x)ϕ(x) dx for ϕ ∈ Sn. ε→0 Rn |x|≥ε In many interesting cases p.v. f defines a tempered distribution, but not always. Theorem 5.49. Let Ω(x/|x|) K (x) = , x 6= 0, Ω |x|n where Ω ∈ L1(Sn−1) satisfies Z Ω(θ) dσ(θ) = 0. Sn−1 Then 0 p.v.KΩ ∈ Sn and |(p.v.KΩ)[φ]| ≤ C(n)kΩkL1(Sn−1) k∇ϕk∞ + k|x|ϕ(x)k∞ for all ϕ ∈ Sn.

0 Remark 5.50. Distributions p.v.KΩ ∈ Sn described in the above result will be called distributions of the p.v. type or simply p.v. distributions.

Remark 5.51. Note that the function KΩ is homogeneous of degree −n so the function KΩ does not define a tempered distribution. R Remark 5.52. It is not difficult to verify that if Sn−1 Ω(θ) dσ(θ) 6= 0, then p.v.KΩ does not define a tempered distribution because p.v.KΩ[ϕ] will diverge for some functions ϕ ∈ Sn.

Proof. For ϕ ∈ Sn we have Z Z Ω(x/|x|) Ω(x/|x|) (p.v.KΩ)[ϕ] = lim n (ϕ(x) − ϕ(0)) + n ϕ(x) dx ε→0 ε≤|x|≤1 |x| |x|≥1 |x| Z |Ω(x/|x|)| Z |Ω(x/|x|)| ≤ k∇ϕk∞ n−1 dx + kϕ(y)|y|k∞ n+1 dx |x|≤1 |x| |x|≥1 |x| and the result follows upon integration in spherical coordinates. We could subtract ϕ(0) R Ω(x/|x|) in the ﬁrst integral because ε≤|x|≤1 |x|n dx = 0. In the last step used the following estimates Z 1 Z 1 d |ϕ(x) − ϕ(0)| = ϕ(tx) dt = ∇ϕ(tx) · x dt ≤ |x| k∇ϕk∞ 0 dt 0 and |ϕ(x)| |x| kϕ(y)|y|k |ϕ(x)| = ≤ ∞ . |x| |x| HARMONIC ANALYSIS 73

The next result illustrates a simple situation when the principal value appears in a n+1 natural way. Note that the function xj/|x| satisﬁes the assumptions of Theorem 5.49. However, the proof of Proposition 5.53 will be straightforward and we will not use Theo- rem 5.49.

1−n 0 Proposition 5.53. For n ≥ 2, the distributional partial derivatives of |x| ∈ Sn satisfy ∂ 1−n xj 0 |x| = (1 − n) p.v. n+1 ∈ Sn, ∂xj |x| i.e., Z ∂ 1−n xj (5.16) |x| [ϕ] = (1 − n) lim n+1 ϕ(x) dx. for ϕ ∈ Sn. ∂xj ε→0 |x|≥ε |x|

Before we prove the proposition let us recall the integration by parts formula for functions deﬁned in a domain in Rn. If Ω ⊂ Rn is a bounded domain with piecewise C1 boundary and f, g ∈ C1(Ω), then Z Z (5.17) (∇f(x)g(x) + f(x)∇g(x)) dx = fg~νdσ , Ω ∂Ω where ~ν = (ν1, . . . , νn) is the unit outer normal vector to ∂Ω. Comparing jth components on both sides of (5.17) we have Z ∂f Z ∂g Z (5.18) (x) g(x) dx = − f(x) (x) dx + fg νj dσ . Ω ∂xj Ω ∂xj ∂Ω

Proof of Proposition 5.53. Let A(ε, R) = {x : ε ≤ |x| ≤ R}. We have ∂ Z ∂ϕ |x|1−n [ϕ] = − |x|1−n dx ∂xj Rn ∂xj Z ∂ϕ = lim lim − |x|1−n dx ε→0 R→∞ ε≤|x|≤R ∂xj Z Z ∂ 1−n 1−n = lim lim ϕ(x) |x| dx − ϕ(x)|x| νj dσ ε→0 R→∞ ε≤|x|≤R ∂xj ∂A(ε,R)

| {z n+1} (1−n)xj /|x| Z Z xj 1−n = lim (1 − n) ϕ(x) n+1 − ϕ(x)|x| νj dσ ε→0 |x|≥ε |x| |x|=ε Indeed, we could pass to the limit as R → ∞, because the part of the integral over ∂A(ε, R) 1−n corresponding to {|x| = R} converges to zero. Since the integral of the function |x| νj over the sphere |x| = ε equals zero we have Z Z 1−n 1−n ϕ(x)|x| νj dσ = (ϕ(x) − ϕ(0))|x| νj dσ ≤ Cε → 0 as ε → 0, |x|=ε |x|=ε because 1−n 1−n 2−n ϕ(x) − ϕ(0) |x| νj ≤ Cε · ε = ε n−1 and the surface area of the sphere {|x| = ε} equals Cε . 74 PIOTR HAJLASZ

n+1 Remark 5.54. We proved that p.v. (xj/|x| ) deﬁnes a tempered distribution somewhat 1−n 0 indirectly by showing that for ϕ ∈ Sn equality (5.16) is satisﬁed. Since ∂|x| /∂xj ∈ Sn, n+1 0 equality (5.16) implies that p.v. (xj/|x| ) ∈ Sn. Remark 5.55. When n = 1 the counterpart of Theorem 5.53 is that d x 1 (log |x|) = p.v. = p.v. in S 0 . dx |x|2 x n We leave the proof as an exercise.

We plan to generalize Proposition 5.53 to a class of more general functions. Let’s start with the following elementary result.

Theorem 5.56. Suppose that K ∈ C1(Rn \{0}) is such that both K and |∇K| have polynomial growth52 for |x| ≥ 1 and there are constants C, α > 0 such that C |K(x)| ≤ for 0 < |x| < 1, |x|n−1−α C |∇K(x)| ≤ for 0 < |x| < 1. |x|n−α 0 Then K ∈ Sn and the distributional partial derivatives ∂K/∂xj, 1 ≤ j ≤ n, coincide with pointwise derivatives, i.e. for ϕ ∈ Sn ∂K Z ∂ϕ Z ∂K [ϕ] := − K(x) (x) dx = (x) ϕ(x) dx . ∂xj Rn ∂xj Rn ∂xj

Proof. Let A(ε) = {x : ε ≤ |x| ≤ ε−1}. From (5.18) we have ∂K Z ∂ϕ [ϕ] = lim − K(x) (x) dx ∂xj ε→0 ε≤|x|≤ε−1 ∂xj Z ∂K Z = lim (x) ϕ(x) dx − K(x)ϕ(x) νj dσ(x) . ε→0 ε≤|x|≤ε−1 ∂xj ∂A(ε) Since Z ∂K Z ∂K lim (x) ϕ(x) dx = (x) ϕ(x) dx ε→0 ε≤|x|≤ε−1 ∂xj Rn ∂xj it remains to show that Z lim K(x)ϕ(x) νj dσ = 0 . ε→0 ∂A(ε) The integral on the outer sphere |x| = ε−1 converges to 0 since K has polynomial growth and ϕ rapidly converges to 0 as |x| → ∞ and on the inner sphere |x| = ε we have Z C n−1 α K(x)ϕ(x) νj dσ ≤ n−1−α ε = Cε → 0 as ε → 0. |x|=ε ε The proof is complete.

52i.e., |K(x)| + |∇K(x)| ≤ C(1 + |x|)m for some C > 0, m ≥ 1 and all |x| ≥ 1. HARMONIC ANALYSIS 75

An interesting problem is the case α = 0, i.e. when K and ∇K satisfy the estimates C C (5.19) |K(x)| ≤ , |∇K(x)| ≤ for x 6= 0. |x|n−1 |x|n One such situation was described in Proposition 5.53.

Here we make an additional assumption about K. We assume that K ∈ C1(Rn \{0}) is homogenelus of degree 1 − n, i.e. K(x/|x|) K(x) = for x 6= 0. |x|n−1 0 Since K is bounded in {|x| ≥ 1} and integrable in {|x| ≤ 1} we have K ∈ Sn and there is no need to interpret K through the principal value of the integral. The first estimate at (5.19) is satisfied. To see that the second estimate if satisfied too we observe that ∇K is homogeneous of degree −n. Indeed, for 1 ≤ j ≤ n and t > 0 ∂K ∂ ∂ (tx)t = (K(tx)) = t1−n K(x) ∂xj ∂xj ∂xj and hence (∇K)(tx) = t−n∇K(x) . Thus (∇K)(x/|x|) ∇K(x) = , x 6= 0 |x|n from which the second estimate at (5.19) follows. Observe that ∇K(x) is not integrable at any neighborhood of 0, but we may try to consider the principal value of ∇K(x), i.e. the principal value of each of the partial derivatives ∂K/∂xj. Theorem 5.57. Suppose that K ∈ C1(Rn \{0}) is homogeneous of degree 1 − n. Then ∇K(x) is homogeneous of degree −n. Moreover Z (5.20) ∇K(θ) dσ(θ) = 0 Sn−1 and 0 p.v. ∇K(x) ∈ Sn is a well defined tempered distribution, i.e. for each 1 ≤ j ≤ n

∂K 0 p.v. (x) ∈ Sn . ∂xj Finally the distributional gradient ∇K satisﬁes

(5.21) ∇K = cδ0 + p.v. ∇K(x) , |{z} | {z } dist. pointwise where Z x c = K(x) dσ(x) . Sn−1 |x| 76 PIOTR HAJLASZ

In other words for ϕ ∈ Sn and 1 ≤ j ≤ n we have ∂K Z ∂ϕ Z ∂K [ϕ] := − K(x) (x) dx = cjϕ(0) + lim (x)ϕ(x) dx , ε→0 ∂xj Rn ∂xj |x|≥ε ∂xj where Z xj cj = K(x) dσ(x) . Sn−1 |x| Remark 5.58. Note that Proposition 5.53 follows from this result: since K(x) = |x|1−n, R −n the constant c = Sn−1 |x| x dσ(x) = 0. 0 Remark 5.59. The fact that p.v. ∇K ∈ Sn is a straightforward consequence of (5.20) and Theorem 5.49. However, our argument will be direct without referring to Theorem 5.49.

Proof. We already checked that ∇K(x) is homogeneous of degree −n. For r > 1 let A(1, r) = {x : 1 ≤ |x| ≤ r}. From the integration by parts formula (5.17) we have Z Z ∇K(x) dx = K(x) ~ν(x) dσ(x) 1≤|x|≤r ∂A(1,r) Z x Z x = − K(x) dσ(x) + K(x) dσ(x) = 0 . |x|=1 |x| |x|=r |x| Indeed, the last two integrals are equal by a simple change of variables and homogeneity of K. Thus the integral on the left hand side equals 0 independently of r. Hence its derivative with respect to r is also equal zero. d Z Z 0 = ∇K(x) dx = ∇K(θ) dσ(θ) . + dr r=1 1≤|x|≤r |x|=1 0 This proves (5.20). As we will see (5.20) implies that p.v. ∇K(x) ∈ Sn is a well deﬁned tempered distribution as well as that the distributional gradient ∇K satisﬁes (5.21). Let ϕ ∈ Sn. We have Z ∇K[ϕ] := − K(x)∇ϕ(x) dx Rn Z = lim lim − K(x)∇ϕ(x) dx ε→0 R→∞ ε≤|x|≤R Z Z = lim lim ∇K(x) ϕ(x) dx − K(x)ϕ(x) ~ν(x) dσ(x) ε→0 R→∞ ε≤|x|≤R ∂A(ε,R) Z Z x = lim ∇K(x) ϕ(x) dx + K(x)ϕ(x) dσ(x) . ε→0 |x|≥ε |x|=ε |x| It remains to prove that Z x Z x lim K(x)ϕ(x) dσ(x) = ϕ(0) K(x) dσ(x) . ε→0 |x|=ε |x| |x|=1 |x| Let Z x Z x c = K(x) dσ(x) = K(x) dσ(x) . |x|=1 |x| |x|=ε |x| HARMONIC ANALYSIS 77

The last equality follows from a simple change of variables and homogeneity of K. We have Z x K(x)ϕ(x) dσ(x)(5.22) |x|=ε |x| Z x = cϕ(0) + K(x)ϕ(x) − ϕ(0) dσ(x) |x|=ε |x| → cϕ(0) as ε → 0. Indeed, for |x| = ε

x 1−n 2−n K(x) ϕ(x) − ϕ(0) ≤ Cε ε = Cε . |x| n−1 Since the surface area of the sphere {|x| = ε} is nωnε , the integral on the right hand side of (5.22) converges to 0 as ε → 0.

5.6. Fundamental solution of the Laplace equation. A straightforward application of Theorem 5.57 gives a well known formula for the fundamental solution to the Laplace equation. Theorem 5.60. For n ≥ 2 we have53

∆Φ = δ0 , where  1 log |x| if n = 2,  2π Φ(x) = 1 1  − n−2 if n ≥ 3. n(n−2)ωn |x|

Proof. We will prove the theorem for n ≥ 3, but a similar argument works for n = 2. According to Theorem 5.5654

1 x 0 (5.23) ∇Φ(x) = n in Sn. nωn |x| Note that the function Φ(x) is harmonic in Rn \{0} and hence n 1 X ∂ xj 0 = ∆Φ(x) = div ∇Φ(x) = for x 6= 0. nω ∂x |x|n n j=1 j Now Theorem 5.57 gives a formula for the distributional Laplacian n 1 X ∂ xj ∆Φ = div ∇Φ = cδ + p.v. = cδ , 0 nω ∂x |x|n 0 n j=1 j | {z } 0 where n Z Z X 1 xj xj 1 c = dσ(x) = dσ(x) = 1 . nω |x|n |x| nω j=1 |x|=1 n n |x|=1 The proof is complete. 53 Note that when n ≥ 3, then Φ ∗ f = −I2f, where I2 is the Riesz potential, see Deﬁnition 2.14. 54Formula (5.23) is also true for n = 2. 78 PIOTR HAJLASZ

55 ∞ n For ϕ ∈ Sn let u(x) = (Φ ∗ ϕ)(x). Then u ∈ C (R ) and ∆u(x) = ∆(Φ ∗ ϕ)(x) = (∆Φ) ∗ ϕ (x) = (δ0 ∗ ϕ)(x) = ϕ(x) . Hence convolution with the fundamental solution of the Laplace operator provides an explicit solution of the Poisson equation ∆u = ϕ . This explains the importance of the fundamental solution in partial diﬀerential equations. Observe that the above calculation gives also ϕ(x) = ∆(Φ ∗ ϕ)(x) n X ∂2 = (Φ ∗ ϕ)(x) ∂x2 j=1 j n X ∂Φ ∂ϕ = ∗ (x) ∂x ∂x j=1 j j Z = ∇Φ(x − y) · ∇ϕ(y) dy Rn 1 Z (x − y) · ∇ϕ(y) = n dy nωn Rn |x − y| for every x ∈ Rn. In the last equality we employed (5.23). Thus we proved

Theorem 5.61. For ϕ ∈ Sn, n ≥ 2 we have Z 1 (x − y) · ∇ϕ(y) n ϕ(x) = n dy for all x ∈ R . nωn Rn |x − y|

From this theorem we can conclude a similar result for higher order derivatives.

Theorem 5.62. For ϕ ∈ Sn, n ≥ 2 and m ≥ 1 we have Z α α m X D ϕ(y) (x − y) n ϕ(x) = n dy for all x ∈ R . nωn n α! |x − y| R |α|=m

Proof. Fix x ∈ Rn and deﬁne X (x − y)β ψ(y) = Dβϕ(y) . β! |β|≤m−1 Then ψ(x) = ϕ(x) and56 ∂ψ X (x − y)β (y) = Dβ+δj ϕ(y) ∂yj β! |β|=m−1

55 0 As a convolution of Φ ∈ Sn with ϕ ∈ Sn. 56 We compute ∂ψ/∂yj using the Leibniz rule and observe that the lower order terms cancel out. HARMONIC ANALYSIS 79

57 where δj = (0,..., 1,..., 0). Hence Theorem 5.60 applied to ψ yields ϕ(x) = ψ(x) Z n 1 X ∂ψ xj − yj = (y) n dy nω n ∂y |x − y| n R j=1 j 1 n Z (x − y)β x − y X X β+δj j j = D ϕ(y) n dy nωn n β! |x − y| j=1 |β|=m−1 R m X Z Dαϕ(y) (x − y)α = n dy , nωn n α! |x − y| |α|=m R because for α with |α| = m X 1 m = . β! α! j,β: β+δj =α The proof is complete.

5.7. Appendix. We will present here details of the last step in the proof of Theorem 5.7. 0 ∞ n Namely we will show that if u ∈ Sn and ϕ, ψ ∈ C0 (R ), then h Z i Z u ϕ(y − ·)ψ(y) dy = u[ϕ(y − ·)ψ(y)] dy. Rn Rn To approximate the integral R η(y) dy, η ∈ C∞ by Riemann sums, we ﬁx a cube Rn 0 n mk Q = [−N,N] so large that supp η ⊂ Q and divide the cube into cubes {Qki}i=1 of −k sidelength 2 . Denote the centers of these cubes by yki. Clearly m Xk Z η(yki)|Qki| → η(y) dy as k → ∞, n i=1 R and m Xk Z (5.24) |η(y) − η(yki)| dy → 0 as k → ∞. i=1 Qki ∞ n ∞ n n Suppose ϕ, ψ ∈ C0 (R ), soϕ ˜ ∗ ψ ∈ C0 (R ). Hence for every x ∈ R m Xk Z wk(x) = ϕ˜(x − yki)ψ(yki)|Qki| → ϕ˜(x − y)ψ(y) dy as k → ∞. n i=1 R 58 The functions wk(x) belong to Sn. We will prove that Z wk(x) → ϕ˜(x − y)ψ(y) dy = (ϕ ˜ ∗ ψ)(x) as k → ∞ Rn in the topology of Sn. First let us show that

sup |wk(x) − (ϕ ˜ ∗ ψ)(x)| → 0 as k → ∞. x∈Rn

571 on jth coordinate, 0 otherwise. 58 As ﬁnite linear combinations of functions from Sn. 80 PIOTR HAJLASZ

mk β X β D wk(x) = D ϕ˜(x − yki)ψ(yki)|Qki| , i=1 exactly the same argument as above shows that β β sup |D wk(x) − (D ϕ˜ ∗ ψ)(x) | → 0 as k → ∞. x∈Rn | {z } Dβ (ϕ ˜∗ψ)(x) Therefore pα,β(wk − ϕ˜ ∗ ψ) → 0 as k → ∞, because |x|α is bounded as the functions have compact support.

This proves that wk → ϕ˜ ∗ ψ in Sn. ∞ Since the function f(y) = u[ϕ ˜(· − y)] is smooth, u[ϕ ˜(· − y)]ψ(y) ∈ C0 and hence m h Z i Xk u ϕ(y − ·)ψ(y) dy = u[ϕ ˜ ∗ ψ] = lim u[wk] = lim u[ϕ ˜(· − yki)]ψ(yki)|Qki| n k→∞ k→∞ R i=1 Z Z = u[ϕ ˜(· − y)]ψ(y) dy = u[ϕ(y − ·)]ψ(y) dy. Rn Rn HARMONIC ANALYSIS 81

6. Convolution with principal value distributions

In Theorem 5.49 we proved that if Ω ∈ L1(Sn−1) satisﬁes Z Ω(θ) dσ(θ) = 0, Sn−1 then Ω(x/|x|) W = p.v.K , where K (x) = Ω Ω Ω |x|n deﬁnes a tempered distribution. In this chapter we will study convolution operators of the form Z Ω(y/|y|) TΩϕ(x) = WΩ ∗ ϕ(x) = lim n ϕ(x − y) dy. ε→0 |y|≥ε |y| ∨ Since TΩϕ = (ϕ ˆWdΩ) for ϕ ∈ Sn, the main objective will be to compute the Fourier transform WdΩ and conclude properties of TΩ from those of WdΩ. We will focus on the setting of tempered distributions and L2 spaces only.

Operators TΩ are called singular integrals and the main objective is to study their boundedness in Lp spaces for 1 < p < ∞. We will return to this problem in Chapters ?? and ??. First we will consider the Riesz transforms which are amongst the most important singular integrals, but have a very simple structure. Finally we will consider WΩ in its general form.

6.1. Riesz transforms: L2 theory. In Proposition 5.53 we proved that for n ≥ 2 ∂ 1−n xj 0 |x| = (1 − n) p.v. n+1 ∈ Sn. ∂xj |x| In this section we will investigate the Riesz transforms which are convolutions with the distributions n+1 xj Γ 2 Wj = cn p.v. n+1 , cn = n+1 , n ≥ 2. |x| π 2 The choice of the coeﬃcient cn will be justiﬁed in Theorem 6.2; with this choice the Fourier transform of Wj has a particularly simple form. 59 Note that when n = 1, Wj becomes 1 x 1 1 W = p.v. = p.v. π |x|2 π x

59cf. Remark 5.55. 82 PIOTR HAJLASZ and convolution with W is known as the Hilbert transform. Thus the Riesz transforms are higher dimensional generalizations of the Hilbert transform. The Hilbert transform will be carefully studied in Section 10. The reader may choose to read Section 10.1 now. This section contains elementary properties of the Hilbert transform and it ﬁts well into the content of Chapter 6.

Deﬁnition 6.1. For 1 ≤ j ≤ n the Riesz transform Rj of a function ϕ ∈ Sn is deﬁned by

(Rjϕ)(x) = (Wj ∗ ϕ)(x) Z yj = cn p.v. n+1 ϕ(x − y) dy Rn |y| Z xj − yj = cn lim n+1 ϕ(y) dy . ε→0 |x−y|≥ε |x − y| Theorem 6.2. ξj Wcj(ξ) = −i . |ξ|

Proof. Applying Proposition 5.53 and then Theorem 5.42 we have ∧ ∧ xj 1 ∂ 1−n p.v. n+1 = |x| |x| 1 − n ∂xj 2πiξ = j |x|1−n∧ 1 − n Γ n−(n−1) 2πiξj n−1− n 2 −1 = π 2 |ξ| 1 − n n−1 Γ 2 n+1 π 2 −iξ = j . n+1 |ξ| Γ 2 We used here equalities n − 1 n − 1 n + 1 Γ(1/2) = π1/2 and Γ = Γ . 2 2 2

The proof is complete.

Corollary 6.3. For ϕ ∈ Sn and 1 ≤ j ≤ n iξ ∨ (6.1) (R ϕ)(x) = − j ϕˆ(ξ) (x) . j |ξ| ∞ Rjϕ ∈ C is slowly increasing and all its derivatives are slowly increasing. Moreover

kRjϕk2 ≤ kϕk2 for ϕ ∈ Sn.

ξj Remark 6.4. One has to be careful here. Although ϕ ∈ Sn, in general −i |ξ| ϕˆ 6∈ Sn so the inverse Fourier transform in (6.1) has to be understood in the distributional or in the L2 sense. HARMONIC ANALYSIS 83

Proof. Equality (6.1) follows from ξ (R ϕ)∧ = (W ∗ ϕ)∧ =ϕ ˆWˆ = −i j ϕ.ˆ j j j |ξ|

Since Rjϕ is a convolution with a tempered distribution, it is smooth, slowly increasing and all its derivatives are slowly increasing, see Theorem 5.7. Finally the L2 estimate follows from the fact that the Fourier transform is an isometry on L2 and the multiplication by 2 the function −iξj/|ξ| is bounded in L with norm 1 (since the supremum of the function 60 is 1).

The above result allows us to extend the Riesz transforms to bounded operators in L2.

Definition 6.5. The Riesz transforms are bounded operators on L2(Rn), n ≥ 2, 2 n kRjfk2 ≤ kfk2 for f ∈ L (R ) and j = 1, 2, . . . , n, defined by the formula iξ ∨ (R f)(x) = − j fˆ(ξ) (x) for f ∈ L2( n). j |ξ| R Remark 6.6. The Riesz transform on L2 was defined as an extension of a bounded operator 2 n+1 Rj defined on a subspace Sn ⊂ L by the formula Rjϕ = cn(p.v. xj/|x| ) ∗ ϕ. An interesting question is whether the L2 extension can also be defined by the same p.v. integral formula and if yes, whether the limit in the definition of the p.v. integral exists in the L2 sense or in the a.e. sense. These highly non-trivial questions will be answered in Section ??. Corollary 6.7. The Riesz transforms satisfy n X 2 2 n Rj = −I on L (R ). j=1

Proof. For f ∈ L2 we have n n ∧ 2 X X iξj R2f (ξ) = − fˆ(ξ) = −fˆ(ξ) j |ξ| j=1 j=1 and the result follows upon taking the inverse Fourier transform.

The next general and elementary result applies to Riesz transforms.

Theorem 6.8. If m ∈ L∞(Rn), then the operator ˆ ∨ 2 Tmf = (mf) , f ∈ L is bounded in L2, kTmfk2 ≤ kmk∞kfk2. 0 Moreover, there is a tempered distribution u ∈ Sn such that

Tmϕ = u ∗ ϕ for all ϕ ∈ Sn.

60cf. Theorem 6.8 and estimates (6.3). 84 PIOTR HAJLASZ

Hence Tmϕ is smooth, slowly increasing, all its derivatives are slowly increasing and α α (6.2) D (Tmϕ) = Tm(D ϕ) for all multiindices α. ∞ If m1, m2 ∈ L , then Tm1 ◦ Tm2 = Tm2 ◦ Tm1 = Tm1m2 .

2 Proof. Boundedness of Tm on L is an obvious consequence of the Plancherel theorem ˆ ˆ (6.3) kTmfk2 = kTdmfk2 = kmfk2 ≤ kmk∞kfk2 = kmk∞kfk2. 0 ∞ To prove existence of u ∈ Sn, observe that m ∈ L is a tempered distribution and 0 ∧ ∧ u =m ˇ ∈ Sn satisﬁesu ˆ = m. Thus (u ∗ ϕ) =ϕm ˆ = (Tmϕ) so Tmϕ = u ∗ ϕ.

Smoothness of Tmϕ = u ∗ ϕ along with equality (6.2) follow from Theorems 5.7 and 5.8.

Finally, the composition formula Tm1 ◦ Tm2 = Tm2 ◦ Tm1 = Tm1m2 follows directly from the deﬁnition of Tm.

An amazing property of the Riesz transforms is that they allow us to compute mixed partial derivatives ∂j∂ku if we only know ∆u. More precisely we have

Proposition 6.9. If ϕ ∈ Sn, then for 1 ≤ j, k ≤ n we have ∂2ϕ = −RjRk∆ϕ(x) = −∆(RjRkϕ)(x),. ∂xj∂xk

Proof. For ϕ ∈ Sn we have ∧ (∂j∂kϕ) (ξ) = (2πiξj)(2πiξk)ϕ ˆ(ξ) −iξ −iξ = − j k (−4π2|ξ|2)ϕ ˆ(ξ) |ξ| |ξ| −iξj −iξk = − ∆dϕ(ξ) |ξ| |ξ| ∧ = (−RjRk∆ϕ) (ξ). Thus taking the inverse Fourier transform in L2 yields ∂2ϕ = −RjRk∆ϕ(x). ∂xj∂xk

The composition of the Riesz transforms RjRk is understood as composition of bounded 2 0 operators in L and according to Theorem 6.8 there is u ∈ Sn such that −iξ −iξ ∨ R R ϕ = j k ϕˆ = u ∗ ϕ. j k |ξ| |ξ| ∞ so RjRkϕ ∈ C is slowly increasing and all its derivatives are slowly increasing. Also ∂2ϕ = −RjRk∆ϕ(x) = −u ∗ ∆ϕ = −∆(u ∗ ϕ) = −∆(RjRkϕ). ∂xj∂xk The proof is complete. HARMONIC ANALYSIS 85

Remark 6.10. One can prove that if j 6= k, then RjRk can be written as a convolution 2 with a p.v. distribution. However, Rj cannot be written as a convolution with a p.v. distribution. Fortunately, it is still a convolution with a tempered distribution, but not of a p.v. type.61

As a direct consequence of Proposition 6.9 and of boundedness of the Riesz transforms in L2 we obtain Corollary 6.11.

∂ϕ ≤ k∆ϕk2 for ϕ ∈ Sn. ∂xj∂xk 2 Remark 6.12. The same estimate can be proved directly using integration by parts. Later we will see that a similar estimate is also true for the Lp norm, 1 < p < ∞. But this is a much harder result and a simple integration by parts is not enough.

0 Remark 6.13. It is quite convincing that the proof of Proposition 6.9 applied to u ∈ Sn 2 such that ∆u = f ∈ L , gives ∂j∂ku = −RjRkf. However this is not true. For example 0 2 if u = xy, then, as a slowly increasing function, u ∈ S2. Clearly ∆u = 0 ∈ L , but ∂x∂yu = 1 6= 0 = −RxRy0. The problem is of a very delicate nature and it is important to understand it well. Remember thatu ˆ is a tempered distribution, not a function so in the equality −iξ −iξ (2πiξ )(2πiξ )û(ξ) = − j k [(−4π2|ξ|2)û(ξ)] j k |ξ| |ξ| the left hand side is well defined since we are allowed to multiply distributions by 2 2 (2πiξj)(2πiξk). However, the right hand side is not well defined: (−4π |ξ| )û(ξ) is a distribution and we are not allowed to multiply this distribution by a non-smooth function. It is actually still confusing. After all, (−4π2|ξ|2)û = ∆cu = fˆ ∈ L2 so why are we not allowed to multiply it by a bounded function? We are, but if we do, the above equality is no longer true, because we switch from the distributional context to the L2 context. In fact, the distributionu ˆ may have a non-trivial part that is supported at 0 such that the multiplication by (2πiξj)(2πiξk) does not ‘kill’ this part. However, multiplication by −4π2|ξ|2 does ‘kill’ it so it may happen that −iξ −iξ (2πiξ )(2πiξ )û(ξ) = − j k fˆ+ v j k |ξ| |ξ| where v is a distribution supported at 0. This will be carefully explained in the next result.

0 Theorem 6.14. If u ∈ Sn satisﬁes 2 n (6.4) ∆u = f ∈ L (R ) , then for any 1 ≤ j, k ≤ n ∂2u = −RjRkf + Pjk , ∂xj∂xk where Pjk is a harmonic polynomial.

61c.f. Problems ?? and ?? and Example 6.41. 86 PIOTR HAJLASZ

Proof. We will only prove that Pjk are polynomials and we leave the proof that they are harmonic to the reader. According to Corollary 5.34 it suffices to prove that the tempered distribution ∂2u ∧ + RjRkf ∂xj∂xk has support contained in {0}. To this end we need to show that if ϕ ∈ Sn, supp ϕ ⊂ Rn \{0}, say ϕ(x) = 0 for |x| ≤ r, then 2 ∧ ∂ u ∧ ∧ + RjRkf [ϕ] = 0, i.e., (∂j∂ku) [ϕ] = (−RjRkf) [ϕ] . ∂xj∂xk Let η ∈ C∞(Rn) be such that η(x) = 0 for |x| ≤ r/2 and η(x) = 1 for |x| ≥ r. The function η and its derivatives are slowly increasing and ηϕ = ϕ. Hence ∧ (∂j∂ku) [ϕ] = (2πiξi)(2πiξj)û [ηϕ] = η(ξ)(2πiξi)(2πiξj)û [ϕ] = ♥. Observe that −iξ −iξ η(ξ)(2πiξ )(2πiξ ) = − η(ξ) j k (−4π2|ξ|2) i j |ξ| |ξ| | {z } λ(ξ) and λ ∈ C∞ and its derivatives are slowly increasing since η vanishes in a neighborhood of ξ = 0. Thus62 ♥ = − λ(ξ)(−4π2|ξ|2)û[ϕ] = (−λ(ξ)fˆ(ξ))[ϕ] −iξ −iξ = −η(ξ) j k fˆ(ξ) [ϕ](6.5) |ξ| |ξ| −iξ −iξ = − j k fˆ(ξ) [ηϕ] |ξ| |ξ| −iξ −iξ = − j k fˆ(ξ) [ϕ] |ξ| |ξ| ∧ = (−RjRkf) [ϕ] . The proof is complete.

6.2. The Beurling-Ahlfors transform. We proved that ∂2ϕ = −RjRk∆ϕ = −∆(RjRkϕ) for ϕ ∈ Sn ∂xj∂xk

62 2 2 ˆ 0 Taking the Fourier transform of (6.4) yields −4π |ξ| uˆ = f in Sn. We should understand this equality 0 2 2 2 ˆ 2 in Sn and not in L , because despite the fact that −4π |ξ| uˆ = f ∈ L , it may happen that the distribution uˆ is not a function. If λ ∈ C∞(Rn) and all its derivatives are slowly increasing, then −λ(ξ)(−4π2|ξ|2)û = ˆ 0 ˆ 2 ˆ 2 −λ(ξ)f in Sn. Now −λf ∈ L and f ∈ L are functions. Functions that are equal a.e. define the same distribution. Hence equality (6.5) is true because the functions that define distributions on the left hand side and on the right hand side of (6.5) are equal for all ξ 6= 0 so they are equal a.e. HARMONIC ANALYSIS 87 so the Riesz transforms, in some sense, can change the direction of partial derivatives. Now we will find an operator S such that ∂ϕ ∂ϕ S = for ϕ ∈ S ( 2) = S ( ). ∂z¯ ∂z R C We will be using complex notation for points in R2 = C

z = x + iy and ξ = ξ1 + iξ2, and also for the Lebesgue measure dz¯ ∧ dz dz = dz + idy, dz¯ = dx − idy so = dx ∧ dy = dx dy. 2i Deﬁnition 6.15. The operator 2 2 S = (iR1 + R2) deﬁned on L (C). is called the Beurling-Ahlfors transform.

The next result collects basic properties of the Beurling-Ahlfors transform. Theorem 6.16. The Beurling-Ahlfors transform has the following properties

(1) S : L2(C) → L2(C) is a surjective isometry, 2 kSfk2 = kfk2 for f ∈ L (C). (2) The inverse operator is S−1f = Sf.¯ (3) For f ∈ L2(C) we have ξ¯ ∨ ξ ∨ Sf(ξ) = fˆ(ξ) ,S−1f(ξ) = fˆ(ξ) . ξ ξ¯ (4) ∂ϕ ∂ ∂ϕ S = (Sϕ) = for ϕ ∈ S ( 2) = S ( ). ∂z¯ ∂z¯ ∂z R C

Proof. Since ξ¯ ξ¯ 2 ξ − iξ 2 iξ iξ 2 = = 1 2 = i − 1 + − 2 ξ |ξ| |ξ| |ξ| |ξ| we conclude that ξ¯ ∨ Sf = (iR + R )2f = fˆ for f ∈ L2( ). 1 2 ξ C Clearly ξ ξ¯ ξ ∨ · = 1 so S−1f = fˆ ξ¯ ξ ξ¯ which proves (3). The fact that S is an isometry easily follows from the Plancherel theorem ¯ ξ ˆ ˆ kSfk2 = kSfck2 = f = kfk2 = kfk2. ξ 2 88 PIOTR HAJLASZ

Since S has the inverse operator, it maps L2(C) onto L2(C). This establishes (1). A simple computation based on the fact that F(f¯(ξ)) = F(f)(−ξ) and F −1(f¯(ξ)) = F −1(f)(−ξ) yields ξ ∨ Sf¯ = fˆ = S−1f ξ¯ which is (2). Equality ∂ϕ ∂ S = (Sϕ) for ϕ ∈ S ( 2) = S ( ) ∂z¯ ∂z¯ R C follows from Theorem 6.8, see (6.2). Recall that ∂ 1 ∂ ∂ ∂ 1 ∂ ∂ = + i and = − i ∂z¯ 2 ∂x ∂y ∂z 2 ∂x ∂y so ∂ϕ∧ 1 = (2πiξ + i(2πiξ ))ϕ ˆ = πiξϕˆ ∂z¯ 2 1 2 and ∂ϕ∧ 1 = (2πiξ − i(2πiξ ))ϕ ˆ = πiξ¯ϕ.ˆ ∂z 2 1 2 Therefore ∂ϕ∧ ξ¯ ∂ϕ∧ S = πiξϕˆ = πiξ¯ϕˆ = . ∂z¯ ξ ∂z This completes the proof of (4). Corollary 6.17. For ϕ ∈ S (C) we have

∂ϕ ∂ϕ = . ∂z 2 ∂z¯ 2

Proof. This is an immediate consequence of parts (4) and (1) of Theorem 6.16.

Similarly as in the case of the Riesz transforms, the Beurling-Ahlfors transform can be represented as a convolution with a principal value distribution. In order to ﬁnd a formula we need to investigate a connection between the Beurling-Ahlfors transform and the fundamental solution to the Laplace equation. Recall that63 1 1 Φ(z) = log |z| = log |z|2 2π 4π satisﬁes ∂2 ∂2 ∂2 ∆Φ = + Φ = 4 Φ = δ . ∂x2 ∂y2 ∂z∂z¯ 0 Hence 1 ∂ ∂ϕ 1 ∂ ∂ 1 ∂2 (6.6) (4πΦ) ∗ = (4πΦ) ∗ ϕ = (4πΦ) ∗ ϕ = ϕ. π ∂z ∂z¯ π ∂z¯ ∂z π ∂z∂z¯

63Theorem 5.60. HARMONIC ANALYSIS 89

An argument similar to that used in the proof of Theorem 5.56 allows us to diﬀerentiate Φ once in the pointwise sense without necessity of taking the principal value distribution so64 ∂ ∂ 1 ∂ z¯ 1 (6.7) (4πΦ(z)) = log |z|2 = (zz¯) = = for z 6= 0 ∂z ∂z |z|2 ∂z |z|2 z and hence ∂ 1 ∂ ∂ = (4Φ) = ∆Φ = δ . ∂z¯ πz ∂z¯ ∂z 0 We proved 1 Proposition 6.18. The function G(z) = πz is a fundamental solution of the operator ∂/∂z¯, i.e. ∂ 1 = δ0 in S ( ). ∂z¯ πz C

Also (6.7) allows us to rewrite (6.6) as65 1 Z ϕ (ξ) 1 ∂ Z ϕ(ξ) (6.8) z¯ dξ¯∧ dξ = dξ¯∧ ξ = ϕ(z). 2πi C z − ξ 2πi ∂z¯ C z − ξ This is a representation formula similar to that in Theorem 5.61. It allows us to reconstruct the function ϕ from its derivative ϕz¯. In paricular it follows immediately that the only holomorphic function in S (C) is the zero function. Deﬁnition 6.19. The Cauchy transform of a function f is deﬁned by 1 Z f(ξ) Cf(z) = (G ∗ f)(z) = dξ¯∧ ξ. 2πi C z − ξ Thus (6.8) reads as Corollary 6.20. ∂ ∂ϕ (Cϕ) = C = ϕ for ϕ ∈ S ( ). ∂z¯ ∂z¯ C

The next result allows us to identify the Beurling-Ahlfors transform with the convolution with a principal value distribution. A diﬀerent proof will be presented in Section 6.3, see Example 6.27. Theorem 6.21. For ϕ ∈ S (C) we have (6.9) Z ∂ ∂ϕ 1 ϕ(ξ) ¯ 1 1 S(ϕ) = (Cϕ) = C = − lim 2 dξ ∧ dξ = − p.v. 2 ∗ ϕ. ∂z ∂z 2πi ε→0 |ξ−z|≥ε (z − ξ) π z

Proof. First we will prove that for ϕ ∈ S (C) we have Z ∂ ∂ϕ 1 ϕ(ξ) ¯ 1 1 (6.10) (Cϕ) = C = − lim 2 dξ ∧ dξ = − p.v. 2 ∗ ϕ. ∂z ∂z 2πi ε→0 |ξ−z|≥ε (z − ξ) π z

64 ∂ ∂f ∂g If f is holomorphic and g is smooth in the real sense, then ∂z (f ◦ g)(z) = ∂z (g(z)) ∂z (z). 65 1 ¯ Recall that dξ1 dξ2 = 2i dξ ∧ dξ. 90 PIOTR HAJLASZ

We could actually conclude this equality from Theorem 5.57 but we prefer to provide a direct proof. Note that ∂ϕ ∂ (6.11) C = (Cϕ), for ϕ ∈ S ( ) ∂z ∂z C is a consequence of a general rule how we diﬀerentiate convolution of a tempered distribution with ϕ ∈ S (C). It is easy to check that in the complex notation Green’s theorem can be rewritten as follows Lemma 6.22 (Green’s theorem). If Ω ⊂ C is a bounded domain with the piecewise C1 boundary and f, g ∈ C1(Ω), then Z ∂f ∂g Z + dz¯ ∧ dz = g dz − f dz.¯ Ω ∂z ∂z¯ ∂Ω Here Z Z Z f dz¯ = f (dx − idy) = f¯ dz. ∂Ω ∂Ω ∂Ω In particular Z ∂ Z (fg) dz¯ ∧ dz = − fg dz¯ Ω ∂z ∂Ω i.e. we have the following version of the integration by parts formula Z ∂f Z ∂g Z (6.12) g dz¯ ∧ dz = − f dz¯ ∧ dz − fg dz.¯ Ω ∂z Ω ∂z ∂Ω Applying (6.12) to our situation yields66 ∂ϕ Z ∂ϕ(ξ)/∂ξ 2πi C = dξ¯∧ dξ ∂z C z − ξ Z ∂ϕ(ξ)/∂ξ = lim dξ¯∧ dξ ε→0 |ξ−z|≥ε z − ξ Z ∂ 1 Z ϕ(ξ) = lim − ϕ(ξ) dξ¯∧ dξ + lim dξ¯ = ♥. ε→0 |ξ−z|≥ε ∂ξ z − ξ ε→0 |ξ−z|=ε z − ξ The sign ‘+’ in the last limit is opposite to the sign ‘-’ in the boundary integral in (6.12). This is because the positive orientation of the boundary |ξ − z| = ε of the exterior domain |ξ − z| ≥ ε is clockwise. Note that Z ¯ Z ¯ Z Z dξ dξ dξ −2 = − = − ¯ = −ε ξ dξ = 0 |ξ−z|=ε z − ξ |ξ|=ε ξ |ξ|=ε ξ |ξ|=ε and hence Z ϕ(ξ) Z ϕ(ξ) − ϕ(z) lim dξ¯ = lim dξ¯ = 0. ε→0 |ξ−z|=ε z − ξ ε→0 |ξ−z|=ε z − ξ

66In the integration by parts we actually should consider the domain ε ≤ |ξ − z| ≤ R and let ε → 0 and R → ∞ since (6.12) applies to bounded domains. We simply skipped one step and passed to the limit as R → ∞, ‘in mind’, cf. the proof of Proposition 5.53. HARMONIC ANALYSIS 91

Since ∂ 1 1 = ∂ξ z − ξ (z − ξ)2 we obtain Z ϕ(ξ) ¯ ♥ = − lim 2 dξ ∧ dξ. ε→0 |ξ−z|≥ε (z − ξ) This completes the proof of (6.10). Note that (6.10) was a quite straightforward consequence of Green’s formula. It remains ∂ 67 to prove that S(ϕ) = ∂z (Cϕ) and this is not obvious. Let us start with an interesting remark. Remark 6.23. The space

Sz¯ = {ϕz¯ : ϕ ∈ S (C)} 2 2 is dense in L (C). For if not, we would ﬁnd 0 6= f ∈ L which is orthogonal to Sz¯, i.e., Z ¯ ϕz¯f dz¯ ∧ dz = 0 for all ϕ ∈ S (C). C ¯ 0 ¯ ¯ 0 That means fz¯ = 0 in S (C) and hence ∆f = 4fzz¯ = 0 in S (C). According to Corol- lary 5.35, f¯ is a polynomial. Since f¯ ∈ L2, we conclude that f = 0 which is a contradiction. Observe also that ∂ S(ϕ ) = ϕ and (Cϕ ) = ϕ . z¯ z ∂z z¯ z 2 This shows that equality (6.9) holds on the subspace Sz¯ ⊂ S (C) which is dense in L . 2 ∂ Since the operator S is an isometry on L , we conclude that ϕ 7→ ∂z C(ϕ) is an isometry on 2 ∂ a dense subset Sz¯ of L . We could conclude the theorem if we could prove that ϕ 7→ ∂z C(ϕ) is bounded in the L2 norm on S (C). After all, boundedness of this operator follows from Theorem 6.21, but we cannot use it before we prove Theorem 6.21. Vicious circle! In our proof given below we will follow the idea described in this remark by constructing a suitable 2 L approximation of ϕ ∈ S (C) by functions of the class Sz¯(C).

68 Let’s return to the proof. Let ϕ ∈ S (C) and let ψ = Cϕ. As we know ψz¯ = ϕ. However, 69 in general ψ 6∈ S (C). Since the Cauchy transform is bounded by the Riesz potential I1 and ϕ ∈ Lp for all p, the Fractional Integration Theorem 2.15 implies that ψ ∈ Lp for all 2 < p < ∞, but in general ψ 6∈ L2.70

67 The function ξ/ξ¯ ∈ L∞ defines a tempered distribution. If u = (ξ/ξ¯ )∨ ∈ S 0(C), then S(ϕ) = u ∗ ϕ (c.f. Theorem 6.8) and it remains to prove that u is the p.v. distribution given by (6.10). However, finding ¯ 2 the Fourier transform of ξ/ξ is not easy, see Example 6.27. On the other hand since S = (iR1 + R2) is a composition of convolutions with p.v. distributions, S should be easy to represent as a convolution with 2 a p.v. distribution. However, finding a formula for the composition is not easy either. For example Rj cannot be represented as a composition with a p.v. distribution (see Problem ?? and Example 6.41.) We will discuss this topic in detail in Section 6.6. 68Corollary 6.20. 69cf. Problem ?? 70cf. Problem ?? 92 PIOTR HAJLASZ

Let η ∈ C∞(C), η(z) = 1 for |z| ≤ 1 and η(z) = 0 for |z| ≥ 2 be a standard cut- oﬀ function. Let ψR(z) = ψ(z)η(z/R). Since the function has compact support we have R R S(ψz¯ ) = ψz . We claim that R 2 (6.13) ψz¯ → ϕ in L as R → ∞. The chain rule gives z z 1 z z 1 ψR = ψ η + ψ(z)η = ϕ(z)η + ψ(z)η . z¯ z¯ R z¯ R R R z¯ R R Clearly z ϕ(z)η → ϕ in L2. R

The function ηz¯(z/R) is bounded an supported in B(0, 2R). Hence for any 2 < p < ∞ we have71 Z 1/2 Z 1/p z 1 2 p ψ(z)ηz¯ ≤ C |ψ| ≤ C |ψ| R R 2 B(0,2R) B(0,2R) −2/p ≤ CR kψkp → 0 as R → ∞. This proves (6.13) and thus72 R 2 (6.14) S(ψz¯ ) → S(ϕ) in L . On the other hand by the arguing in the same way as in the proof of (6.13) we have z z 1 ∂ (6.15) S(ψR) = ψR = ψ (z)η + ψ(z)η → ψ = (Cϕ) in L2 as R → ∞. z¯ z z R z R R z ∂z ∂ Comparing (6.14) and (6.15) yields S(ϕ) = ∂z (Cϕ). The proof is complete.

6.3. Higher Riesz transforms. Deﬁnition 6.24. Higher Riesz transforms of degree k are deﬁned as a convolution with P (x) (6.16) p.v. k , |x|n+k where Pk(x) is a homogeneous harmonic polynomial of degree k ≥ 1.

If k = 1 and P1(x) = cnxj, we obtain the Riesz transform Rj.

If Pk is a homogeneous harmonic polynomial of degree k ≥ 1 we can write P (x) P (x)|x|−k P (x/|x|) Ω(x/|x|) k = k = k = |x|n+k |x|n |x|n |x|n and in order to prove that (6.16) deﬁnes a tempered distribution we need to prove that Z Z (6.17) Ω(θ) dσ(θ) = Pk(θ) dσ(θ) = 0 . Sn−1 Sn−1

71In the ﬁrst inequality we use the fact that the area of the disc B(0, 2R) equals CR2. We also use here H¨older’sinequality and the fact that ψ ∈ Lp. 72S is bounded in L2. HARMONIC ANALYSIS 93

If ~ν is an outward normal vector to the unit sphere, then

∂Pk d k−1 = Pk(tx) = kt Pk(x) = kPk(x) ∂~ν dt t=1 t=1 and hence Green’s formula73 yields Z Z Z ∂Pk k Pk(θ) dσ(θ) = (θ) dσ(θ) = ∆Pk dx = 0 . Sn−1 Sn−1 ∂~ν Bn Thus according to Theorem 5.49 Z Pk(x) WΩ[ϕ] = p.v. n+k ϕ(x) dx Rn |x| is a well deﬁned tempered distribution. Our aim is to prove the following result.

Theorem 6.25. If Pk is a homogeneous harmonic polynomial of degree k ≥ 1, then P (x) ∧ P (ξ) p.v. k (ξ) = γ k , |x|n+k k |ξ|k where k k n/2 Γ 2 γk = (−i) π k+n . Γ 2 Example 6.26. This result immediately implies Theorem 6.2. Example 6.27. We will show now that the formula for the Beurling-Ahlfors transform 1 1 (6.18) S(ϕ) = − p.v. ∗ ϕ for ϕ ∈ S ( ) π z2 C originally proved in Theorem 6.21 is a straightforward consequence of Theorem 6.25. 74 ¯2 2 2 Let n = 2 and let P2(ξ) = ξ = ξ1 − 2iξ1ξ2 − ξ2 . Clearly P2 is a homogeneous harmonic polynomial of degree 2. Since γ2 = −π we obtain 1 ∧ P (z)∧ P (ξ) ξ¯ p.v. (ξ) = p.v. 2 (ξ) = −π 2 = −π z2 |z|4 |ξ|2 ξ so 1 1 ∧ ξ¯ − p.v. ∗ ϕ = ϕˆ = S[(ϕ) π z2 ξ which proves (6.18).

We will deduce Theorem 6.25 from the following result.

Theorem 6.28. If Pk is a homogeneous harmonic polynomial of degree k and 0 < α < n, then P (x) ∧ P (ξ) k (ξ) = γ k , |x|k+n−α k,α |ξ|k+α

73R R ∂g ∂f Ω(f∆g − g∆f) dx = ∂Ω(f ∂~ν − g ∂~ν ) dσ where ~ν is the outward normal. 74 We use complex notation in R2 = C. 94 PIOTR HAJLASZ where k+α n Γ k 2 −α 2 γk,α = (−i) π k+n−α . Γ 2 Remark 6.29. Note that the function

Pk(x) |x|k+n−α is a tempered L1 function, so it deﬁnes a tempered distribution without necessity of taking the principal value of the integral.

75 Proof. For any t > 0 and ϕ ∈ Sn, Corollary 4.24 gives Z Z −πt|x|2 k −π|x|2/t −k− n Pk(x)e ϕˆ(x) dx = (−i) Pk(x)e t 2 ϕ(x) dx . Rn Rn Now we multiply both sides by k + n − α tβ−1, where β = > 0 2 and integrate with respect to 0 < t < ∞. Since Z ∞ e−πt|x|2 tβ−1 dt = (π|x|2)−βΓ(β) 0 the integral on the left hand side will be equal to Z k+n−α ∧ Γ(β) Pk(x) Γ 2 Pk(·) (6.19) β 2β ϕˆ(x) dx = k+n−α k+n−α [ϕ] . π Rn |x| π 2 | · | Similarly Z ∞ Z ∞ −π|x|2/t −k− n β−1 −π|x|2/t − k+α −1 e t 2 t dt = e t 2 dt 0 0 2 Z ∞ s=π|x| /t 2 − k+α −s k+α −1 = (π|x| ) 2 e s 2 ds 0 2 − k+α k + α = (π|x| ) 2 Γ . 2 Thus the integral on the right hand side equals k+α Z k Γ 2 Pk(x) (6.20) (−i) k+α k+α ϕ(x) dx . π 2 Rn |x| Since integrals at (6.19) and (6.20) are equal one to another, the theorem follows.

Proof of Theorem 6.25. For ϕ ∈ Sn we take the identity Z Z Pk(x) Pk(x) k+n−α ϕˆ(x) dx = γk,α k+α ϕ(x) dx Rn |x| Rn |x|

75R R ˆ n f(x)ϕ ˆ(x) dx = n f(x)ϕ(x) dx. R R HARMONIC ANALYSIS 95 and let α → 0. The right hand side converges to k Z k n Γ 2 Pk(x) (−i) π 2 ϕ(x) dx . k+n k n |x| Γ 2 R −(k+n−α) To compute the limit on the left hand side we observe that the integral of Pk(x)|x| on the unit ball equals zero, see (6.17), and hence Z Pk(x) k+n−α ϕˆ(x) dx Rn |x| Z Z Pk(x) Pk(x) = k+n−α (ϕ ˆ(x) − ϕˆ(0)) dx + k+n−α ϕˆ(x) dx |x|≤1 |x| |x|≥1 |x| Z Z α→0+ Pk(x) Pk(x) −→ k+n (ϕ ˆ(x) − ϕˆ(0)) dx + k+n ϕˆ(x) dx |x|≤1 |x| |x|≥1 |x| Z Z Pk(x) Pk(x) = lim k+n (ϕ ˆ(x) − ϕˆ(0)) dx + k+n ϕˆ(x) dx ε→0 ε≤|x|≤1 |x| |x|≥1 |x| Z Pk(x) = lim k+n ϕˆ(x) dx ε→0 |x|≥ε |x| P (x) ∧ = p.v. k [ϕ] . |x|k+n Comparing the above limits yields the result.

6.4. Another proof of Theorem 6.2. Two proof presented above76 are somewhat in- direct since they were based on other results: Theorem 5.42 and Theorem 6.25. In this section we will present a diﬀerent, more straightforward proof. A one dimensional version of this argument will be presented again in Section 10 in the proof of Theorem 10.2.

For ϕ ∈ Sn we have Z ξj Wcj[ϕ] = Wj[ϕ ˆ] = lim cn ϕˆ(ξ) n+1 dξ ε→0 |ξ|≥ε |ξ| Z Z −2πix·ξ ξj = lim cn ϕ(x)e dx dξ ε→0 n+1 ε≤|ξ|≤ε−1 Rn |ξ| Z Z −2πix·ξ ξj = lim cn ϕ(x) e dξ dx = ♥ . ε→0 n+1 Rn ε≤|ξ|≤ε−1 |ξ| | {z } I Expressing the integral I is the spherical coordinates yields Z ε−1 Z n−1 −2πix·(sθ) sθj I = s e n+1 dσ(θ) ds ε Sn−1 s Z ε−1 Z ds = cos(2πsx · θ) − i sin(2πsx · θ) θj dσ(θ) ε Sn−1 s

76The second proof was given in Remark 6.26. 96 PIOTR HAJLASZ

−1 Z ε Z ds = −i sin(2πsx · θ)θj dσ(θ) ε Sn−1 s −1 ! Z Z ε sin(2πs|x · θ|) = −i ds sgn (x · θ)θj dσ(θ) Sn−1 ε s −1 ! Z Z 2π|x·θ|ε sin t = −i dt sgn (x · θ)θj dσ(θ) . Sn−1 2π|x·θ|ε t Thus Z π Z ♥ = ϕ(x) −icn sgn (x · θ)θj dσ(θ) dx . Rn 2 Sn−1 Indeed, we could pass to the limit under the sign of the integral using the Dominated Convergence Theorem and we employed the equality77 Z ∞ sin t π dt = . 0 t 2 Thus it remains to prove that Z π xj (6.21) cn sgn (x · θ)θj dσ(θ) = . 2 Sn−1 |x| It is obvious that the function m : Rn → Rn deﬁned as Z m(x) = sgn (x · θ)θ dσ(θ) Sn−1 is homogeneous of degree 0. We will use Lemma 4.26 so we need to check that m commutes with orthogonal transformations. It does because Z m(ρ(x)) = sgn (ρ(x) · θ)θ dσ(θ) Sn−1 Z = sgn (x · ρ−1(θ))θ dσ(θ) Sn−1 Z = sgn (x · θ)ρ(θ) dσ(θ)(6.22) Sn−1 Z (6.23) = ρ sgn (x · θ)θ dσ(θ) Sn−1 = ρ(m(x)) . Note that (6.22) follows from the fact that ρ induces a volume preserving change of variables on Sn−1, while (6.23) is a direct consequence of linearity of ρ. Thus the Lemma 4.26 yields Z x sgn (x · θ)θ dσ(θ) = c Sn−1 |x| and hence looking at the jth component we have Z xj (6.24) sgn (x · θ)θj dσ(θ) = c . Sn−1 |x|

77Recall that this is understood as an improper integral which is consistent with our setting since we pass to the limit as ε → 0. HARMONIC ANALYSIS 97

Thus it remains to prove that

π −1 2π(n−1)/2 c = c = = 2ω . n 2 n+1 n−1 Γ 2

Taking x = ej in (6.24) we have Z |θj| dσ(θ) = c . Sn−1

n−1 n−1 The unit ball B in coordinates perpendicular to xj split the sphere S into two half n−1 spheres S± . Thus Z c = 2 θj dσ(θ) . n−1 S+

Recall that if M ⊂ Rn is a graph of a C1 function f :Ω → R,Ω ⊂ Rn−1, then for a measurable function g on M we have Z Z g dσ = g(x, f(x))p1 + |∇f(x)|2 dx . M Ω

n−1 In our situation we parametrize S+ as a graph of the function

f(x) = p1 − |x|2 , x ∈ Bn−1 . Form the picture 98 PIOTR HAJLASZ

we conclude Z Z Z p 2 θj dσ(θ) = (1 − h) dσ(θ) = (1 − h) 1 + tan α dx n−1 n−1 n−1 S+ S+ B Z = dx = ωn−1 , Bn−1 because p 1 1 1 + tan2 α = = cos α 1 − h and the result follows. 2 The proof of the next result is based on an argument similar to that used to prove (6.24); we leave details to the reader as an exercise. Lemma 6.30. Let K be a function of one variable, then for n ≥ 2 we have Z Z 1 2 n−3 K(x · θ) dσ(θ) = (n − 1)ωn−1 K(s|x|)(1 − s ) 2 ds Sn−1 −1 for all x ∈ Rn \{0}.

6.5. The general case. In this section we will consider operators TΩϕ = WΩ ∗ ϕ in the 1 n−1 R most general form where Ω ∈ L (S ), Sn−1 Ω(θ) dσ(θ) = 0, and Ω(x/|x|) W = p.v.K , where K (x) = . Ω Ω Ω |x|n The following lemma will play an important role. 1 n−1 Lemma 6.31. If F is a function homogeneous of degree zero and F Sn−1 ∈ L (S ), then F is a tempered L1 function Z |F (x)|(1 + |x|n+1)−1 dx < ∞. Rn 0 Hence F ∈ Sn. Moreover n+1 |F [ϕ]| ≤ C(n)kF kL1(Sn−1) sup (1 + |x| )|ϕ(x)|. x∈Rn

Proof. We have Z Z ∞ Z |F [ϕ]| ≤ |F (x)ϕ(x)| dx = sn−1 |F (θ)ϕ(sθ)| dσ(θ) ds Rn 0 Sn−1 Z Z ∞ = |F (θ)| sn−1|ϕ(sθ)| ds dσ(θ) = ♥ Sn−1 0 Z ∞ Z ∞ n−1 n−1 s n+1 s |ϕ(sθ)| ds ≤ n+1 sup(1 + |x| )|ϕ(x)| ds 0 0 1 + s |x|=s ≤ C(n) sup (1 + |x|n+1)|ϕ(x)|. x∈Rn Hence n+1 ♥ ≤ C(n)kF kL1(Sn−1) sup(1 + |ξ| )|ϕ(ξ)|. ξ∈Rn HARMONIC ANALYSIS 99

In particular taking78 ϕ(x) = (1 + |x|n+1)−1 yields Z n+1 −1 |F (x)|(1 + |x| ) ≤ CkF kL1(Sn−1) < ∞ Rn 1 so F is a tempered L function.

The main result of this section generalizes Theorem 6.2. Theorem 6.32. Let n ≥ 2 and Ω ∈ L1(Sn−1) be such that Z Ω(θ) dσ(θ) = 0 . Sn−1 1 Then the Fourier transform of the distribution WΩ is a tempered L function, homogeneous of degree zero such that Z (6.25) kWdΩkL1(Sn−1) ≤ CkΩkL1(Sn−1) and WdΩ(ξ) dσ(ξ) = 0. Sn−1

Therefore WdΩ is a tempered distribution satisfying n+1 (6.26) |WdΩ[ϕ]| ≤ CkΩkL1(Sn−1) sup(1 + |ξ| )|ϕ(ξ)| . ξ∈Rn Moreover Z 1 iπ WdΩ(ξ) = Ω(θ) log − sgn (ξ · θ) dσ(θ)(6.27) Sn−1 |ξ · θ| 2 Z 1 iπ 0 = Ω(θ) log 0 − sgn (ξ · θ) dσ(θ) , Sn−1 |ξ · θ| 2 where ξ0 = ξ/|ξ|. Finally, if Ω is odd, then Z iπ 0 (6.28) WdΩ(ξ) = − Ω(θ) sgn (ξ · θ) dσ(θ) is odd 2 Sn−1 and if Ω is even, then Z 1 (6.29) WdΩ(ξ) = Ω(θ) log 0 dσ(θ) is even. Sn−1 |ξ · θ|

We will precede the proof with some technical lemmas.

n−1 n−1 0 1 Lemma 6.33. For θ ∈ S deﬁne a function on S by fθ(ξ ) = log |ξ0·θ| . Then fθ ∈ Lp(Sn−1) for all 1 ≤ p < ∞ and

sup kfθkLp(Sn−1) = C(n, p) < ∞. θ∈Sn−1

78 n+1 −1 ϕ(x) = (1 + |x| ) does not belong to Sn, but the above estimates work for this function too. 100 PIOTR HAJLASZ

Proof. According to Lemma 6.3079 1 Z 1 Z 1 n−3 p 0 p 2 2 (6.30) log 0 dσ(ξ ) = (n − 1)ωn−1 log (1 − s ) dx < ∞ Sn−1 |ξ · θ| −1 |s| The last integral has three singularities80

p 1 n−3 n−3 log at s = 0, (1 + s) 2 at s = −1, (1 − s) 2 at s = 1 |s| and all these singularities are integrable. Lemma 6.34. If h : [0, ∞) → R is continuous, bounded and the improper integral Z ∞ h(s) ds 1 s converges, then for µ > λ > 0 and N > ε > 0 we have Z N h(λs) − h(µs) µ (6.31) ds ≤ 2khk∞ log . ε s λ Moreover Z N h(λs) − h(µs) µ (6.32) lim ds = h(0) log . N→∞ s λ ε→0 ε

Proof. We have Z N h(λs) − h(µs) Z λN h(s) Z µN h(s) ds = ds − ds ε s λε s µε s Z µε h(s) Z µN h(s) = ds − ds . λε s λN s Estimating the absolute value of the last two integrals gives (6.31). Since Z µε h(s) µ ds → h(0) log as ε → 0 λε s λ and Z µN h(s) ds → 0 as N → ∞ λN s (6.32) follows. 2 Corollary 6.35. For a 6= 0 we have Z N e−isa − cos s 1 π lim ds = log − i sgn a N→∞ s |a| 2 ε→0 ε and Z N −isa e − cos s 1 ds ≤ 2 log + C, ε s |a| where C is some positive constant independent of a, ε and N.

79 0 1 Since |ξ · θ| ≤ 1, log |ξ0·θ| ≥ 0 and we do not have to take the absolute value of the log term. 80The last two are singularities only when n = 2. HARMONIC ANALYSIS 101

Proof. We have Z N e−isa − cos s Z N cos(sa) − cos s Z N sin(sa) ds = ds − i ds ε s ε s ε s Z N cos(s|a|) − cos s Z N sin(s|a|) = − i sgn (a) ds ε s ε s Z N cos(s|a|) − cos s Z N|a| sin s = − i sgn (a) ds ε s ε|a| s and the result follows from Lemma 6.34.

Proof of Theorem 6.32. The structure of the proof is as follows

(1) We will prove that the two integrals at (6.27) are equal. (2) Deﬁne Z 1 iπ 0 F (ξ) = Ω(θ) log 0 − sgn (ξ · θ) dσ(θ) , Sn−1 |ξ · θ| 2 Clearly F is homogeneous of degree zero. We will prove that Z kF kL1(Sn−1) ≤ CkΩkL1(Sn−1), F (ξ) dσ(ξ) = 0. Sn−1 1 0 Hence according to Lemma 6.31, F is a tempered L function so F ∈ Sn and n+1 |F [ϕ]| ≤ CkΩkL1(Sn−1) sup(1 + |ξ| )|ϕ(ξ)| . ξ∈Rn 1 0 (3) Since ξ 7→ log |ξ0·θ| is even and ξ 7→ sgn (ξ · θ) is odd, it follows that if Ω is odd or even, then Z Z iπ 0 1 F (ξ) = − Ω(θ) sgn (ξ · θ) dσ(θ) or F (ξ) = Ω(θ) log 0 dσ(θ) 2 Sn−1 Sn−1 |ξ · θ| are odd and even respectively. (4) Finally we will prove that WdΩ(ξ) = F (ξ). First observe that the equality of the two integrals in (6.27) follows from 1 1 1 Z 1 log = log + log 0 and Ω(θ) log dσ(θ) = 0 . |ξ · θ| |ξ| |ξ · θ| Sn−1 |ξ| This proves (1). Let Z 1 iπ 0 F (ξ) = Ω(θ) log 0 − sgn (ξ · θ) dσ(θ) Sn−1 |ξ · θ| 2 be the function deﬁned by the integral in the second line of (6.27). It is clear that F (ξ) is homogeneous of degree zero. We will show that it is integrable on the unit sphere and

kF kL1(Sn−1) ≤ CkΩkL1(Sn−1). 102 PIOTR HAJLASZ

Note that Z iπ ξ 7→ Ω(θ) sgn (ξ0 · θ) dσ(θ) Sn−1 2 π is a bounded by 2 kΩkL1(Sn−1), so this component of F (ξ) does not cause any troubles and hence we only need to estimate Z 1 G(ξ) = Ω(θ) log 0 dσ(θ) . Sn−1 |ξ · θ| Lemma 6.33 with p = 1 yields Z Z 1 0 1 n−1 kGkL (S ) ≤ Ω(θ) log 0 dσ(θ) dσ(ξ ) Sn−1 Sn−1 |ξ · θ| Z Z 1 0 1 n−1 ≤ |Ω(θ)| log 0 dσ(ξ ) dσ(θ) = C(n)kΩkL (S ). Sn−1 Sn−1 |ξ · θ| R It easily follow now from the Fubini theorem that Sn−1 F (ξ) dσ(ξ) = 0. That completes the proof of (2). Since F ∈ L1(Sn−1), the step (3) is obvious.

To prove (4) we need to show that WdΩ = F in the sense of distributions i.e., Z Z 1 iπ 0 WdΩ[ϕ] = ϕ(ξ) Ω(θ) log 0 − sgn (ξ · θ) dσ(θ) dξ. Rn Sn−1 |ξ · θ| 2 We have

WdΩ[ϕ] = WΩ[ϕ ˆ] Z Ω(x/|x|) = lim ϕˆ(x) dx ε→0 |x|n N→∞ ε≤|x|≤N Z Z Ω(x/|x|) −2πix·ξ = lim ϕ(ξ) n e dx dξ ε→0 n |x| N→∞ R ε≤|x|≤N Z Z Z N ds = lim ϕ(ξ) Ω(θ) e−2πisθ·ξ dσ(θ) dξ ε→0 n n−1 s N→∞ R S ε Z Z Z N ds = lim ϕ(ξ) Ω(θ) e−2πisθ·ξ − cos(2πs|ξ|) dσ(θ) dξ ε→0 n n−1 s N→∞ R S ε = ♦ . The last equality follows from the fact that the integral of Ω over the sphere vanishes. We have Z N ds Z 2π|ξ|N e−isθ·ξ0 − cos s e−2πisθ·ξ − cos(2πs|ξ|) = ds ε s 2π|ξ|ε s 1 π −→ log − i sgn (θ · ξ0) |θ · ξ0| 2 by Corollary 6.35. Also Corollary 6.35 and integrability of G easily imply that Z N −2πisθ·ξ ds ϕ(ξ)Ω(θ) e − cos(2πs|ξ|) ε s HARMONIC ANALYSIS 103 1 ≤ C|ϕ(ξ)| |Ω(θ)| 1 + log ∈ L1( n × Sn−1) |ξ0 · θ| R so the Dominated Convergence Theorem gives Z Z 1 π 0 ♦ = ϕ(ξ) Ω(θ) log 0 − i sgn (θ · ξ ) dσ(θ) dξ . Rn Sn−1 |θ · ξ | 2 The proof is complete. 2 If Ω is an odd function, then Z iπ 0 WdΩ(ξ) = − Ω(θ) sgn (ξ · θ) dσ(θ) . 2 Sn−1

n−1 In particular the Fourier transform WdΩ is bounded. More generally any function Ω on S can be decomposed into its even and odd parts 1 1 Ω (θ) = (Ω(θ) + Ω(−θ)), Ω (θ) = (Ω(θ) − Ω(−θ)) . e 2 o 2 Corollary 6.36. Let Ω ∈ L1(Sn−1) be such that Z Ω(θ) dσ(θ) = 0 . Sn−1 1 n−1 q n−1 If Ωo ∈ L (S ) and Ωe ∈ L (S ) for some q > 1, then the Fourier transform of WΩ is a bounded function. In particular the operator TΩϕ = WΩ ∗ ϕ for ϕ ∈ Sn extends to a bounded operator in L2.

Proof. Lemma 6.33 implies that log(1/|ξ0 · θ|) belongs to Lq0 (Sn−1) and hence Z 1 q n−1 Ωe(θ) log 0 dσ(θ) ≤ CkΩekL (S ) , Sn−1 |ξ · θ| with a constant C independent of ξ. Now the formulas (6.28) and (6.29) yield Z Z iπ 0 1 WdΩ(ξ) = − Ωo(θ) sgn (ξ · θ) dσ(θ) + Ωe(θ) log 0 dσ(θ) 2 Sn−1 Sn−1 |ξ · θ| from which we have

kWdΩk∞ ≤ C(kΩokL1(Sn−1) + kΩekLq(Sn−1)) .

∨ Since TΩϕ = (ϕ ˆWdΩ) and multiplication by a bounded function constitutes a bounded operator in L2 it follows that

kTΩϕk2 ≤ kWdΩk∞kϕk2 for ϕ ∈ Sn.

It is natural to expect that if Ω in Theorem 6.32 is smooth, then WdΩ is smooth away from the origin. As we will see this it true. The next result is quite surprising since it gives a complete characterization of such Fourier transforms WdΩ. 104 PIOTR HAJLASZ

Theorem 6.37. If Ω ∈ C∞(Sn−1) satisﬁes Z (6.33) Ω(θ) dσ(θ) = 0, Sn−1 then ∞ n (6.34) m = WdΩ ∈ C (R \{0}) is homogeneous of degree zero and Z (6.35) m(ξ) dσ(ξ) = 0. Sn−1 Conversely if m ∈ C∞(Rn \{0}) is homogeneous of degree zero and it satisﬁes (6.35), then ∞ n−1 there is Ω ∈ C (S ) satisfying (6.33) such that m = WdΩ. Lemma 6.38. If Ω ∈ C∞(Sn−1) and Z 1 2 n−3 (6.36) |K(s)|(1 − s ) 2 ds < ∞ −1 then m ∈ C∞(Sn−1) where Z m(ξ) = Ω(θ)K(ξ · θ) dσ(θ). Sn−1 Remark 6.39. We do not assume here that the integral of Ω vanishes. Remark 6.40. According to Lemma 6.30 inequality (6.36) implies that for every ξ ∈ Sn−1 Z Z 1 2 n−3 |K(ξ · θ)| dσ(θ) = (n − 1)ωn−1 |K(s)|(1 − s ) 2 ds < ∞. Sn−1 −1

Proof. Observe that it suﬃces to prove that m is smooth in a neighborhood of e1. Indeed, n−1 if ξ0 ∈ S and ρ(e1) = ξ0 for some ρ ∈ O(n), then Z m˜ (ξ) := (m ◦ ρ)(ξ) = Ω(θ)K(ρ(ξ) · θ) dσ(θ) Sn−1 Z Z = Ω(θ)K(ξ · ρ−1(θ)) dσ(θ) = Ω(ρ(θ))K(ξ · θ) dσ(θ) Sn−1 Sn−1 ∞ n−1 and Ω ◦ ρ ∈ C (S ). Thus smoothness ofm ˜ is a neighborhood of e1 will imply smooth- −1 −1 n−1 n−1 ness of m in a neighborhood of ξ0, because m =m ˜ ◦ ρ and ρ : S → S is a diﬀeomorphism that maps a neighborhood of ξ0 onto a neighborhood of e1, wherem ˜ is smooth.

Thus we will prove that m is smooth in a neighborhood of e1.

If ξ is in the right hemisphere generated by e1, we can represent it in a local coordinate system v  u n u X 2 n−1 ξ = q(ξ2, . . . , ξn) = t1 − ξj , ξ2, . . . , ξn , (ξ2, . . . , ξn) ∈ B (0, 1). j=2

n−1 It suﬃces to prove that (m ◦ q)(ξ2, . . . , ξn) is smooth in B (0, 1). HARMONIC ANALYSIS 105

Given ξ = q(ξ2, . . . , ξn) let

ρ(ξ2, . . . , ξn) ∈ SO(n), ρ(ξ2, . . . , ξn)(ξ) = e1 be a rotation that rotates ξ to e1 in the plane span {e1, ξ} and ﬁxes the orthogonal complement of that plane. Such a rotation is unique and it is easy to see that components of the matrix ρ(ξ2, . . . , ξn) smoothly depend in (ξ2, . . . , ξn). By an argument similar to the one used at the beginning of the proof Z Z −1 (m ◦ q)(ξ2, . . . , ξn) = Ω(θ)K(ξ · θ) dσ(θ) = (Ω ◦ ρ (ξ2, . . . , ξn))(θ)K(θ1) dσ(θ). Sn−1 Sn−1

Now it is easy to see that we can diﬀerentiate with respect to ξ2, . . . , ξn under the sign of 81 the integral inﬁnitely many times.

Proof of Theorem 6.37. Let Ω ∈ C∞(Sn−1) satisfy (6.33). For ξ 6= 0, Z m(ξ) = WdΩ(ξ) = Ω(θ)K(ξ · θ) dσ(θ) Sn−1 where 1 iπ K(s) = log − sgn (s) |s| 2 and we already proved in (6.30) that Z 1 2 n−3 |K(s)|(1 − s ) 2 ds < ∞. −1 Thus m ∈ C∞(Sn−1) by Lemma 6.38 and since m is homogeneous of degree zero, m ∈ C∞(Rn \{0}). The fact that the integral of m over the sphere vanishes was already proved in Theorem 6.32, see (6.25).

Suppose now that m ∈ C∞(Rn \{0}) is homogeneous of degree zero and it satisﬁes (6.35). Since m is a tempered distribution,m ˆ is a tempered distribution too and n ∧ ∂ m n n (ξ) = (2πiξj) mˆ (ξ). ∂xj n n The function ∂ m/∂xj is homogeneous of degree −n. Since ! ∂nm ∂ ∂n−1m n = n−1 ∂xj ∂xj ∂xj it is the derivative of a function that is homogeneous of degree 1 − n, Theorem 5.57 yield Z ∂nm(θ) ∂nm ∂nm n dσ(θ) = 0 and n = cδ0 + p.v. n . Sn−1 ∂xj ∂xj ∂xj

81Perhaps it will be easier to see it if we rewrite the integral in a way that it does not involve integration of functions on manifolds (sphere). Let F ∈ C∞(Rn) be a smooth extension of Ω from Sn−1 to Rn. Then Z −1 (m ◦ q)(ξ2, . . . , ξn) = F (ρ (ξ2, . . . , ξn)(x))dµ(x) n R n−1 where µ is a measure concentrated on the sphere S . Namely µ is the extension of K(θ1)dσ(θ) from n−1 n R S to R by zero, i.e., µ(E) = Sn−1∩E K(θ1) dσ(θ). 106 PIOTR HAJLASZ

Taking the Fourier transform yields n ∧ n n ∂ m (2πi) ξj mˆ (ξ) = c + p.v. n (ξ). ∂xj Note that the function on the right hand side is of class C∞(Rn \{0}), homogeneous of degree zero (by the ﬁrst part of the proof). The above equality yields n n ∧ X X ∂nm (6.37) (2πi)n (ξ2n)m ˆ (ξ) = ξn c + p.v. (ξ) . j j ∂xn j=1 j=1 j We claim that the distributionm ˆ coincides in Rn \{0} with the function that is smooth in Rn \{0}, homogeneous of degree −n ∧ Pn n ∂nm ξj c + p.v. n (ξ) j=1 ∂xj (6.38)m ˆ (ξ) = in n \{0}. n Pn 2n R (2πi) j=1(ξj ) n That means, if ϕ ∈ Sn with supp ϕ ⊂ R \{0}, then ∧ Pn n ∂nm ξj c + p.v. n (ξ) Z j=1 ∂xj (6.39)m ˆ [ϕ] = ϕ(x) dx. n Pn 2n Rn (2πi) j=1(ξj ) Pn 2n To see this observe that since ϕ vanishes in a neighborhood of 0, the function j=1 ξj is positive on the support of ϕ so ϕ ψ = ∈ n Pn 2n Sn (2πi) j=1(ξj ) is in the Schwarz class.82 Now it is easy to see that if we evaluate both sides of the distributional equality (6.37) on ψ we obtain equality (6.39).

We proved that in Rn \{0},m ˆ coincides with a smooth function, homogeneous of degree −n which is given by the formula on the right hand side of (6.38). We still do not know what happens at 0. There is a possibility thatm ˆ has a part supported at 0. Let Ω(˜ ξ) =m ˆ (ξ) for ξ ∈ Sn−1. Thus Ω(˜ ξ/|ξ|) mˆ (ξ) = for ξ 6= 0. |ξ|n We claim that Z (6.40) Ω(˜ θ) dσ(θ) = 0, Sn−1

Indeed, let ϕ ∈ Sn be a radial function supported in the annulus 1 ≤ |x| ≤ 2 and positive in the interior of the annulus. Then the integration in the spherial coordinates gives Z ˜ Z Ω(x/|x|) ˜ (6.41)m ˆ [ϕ] = n ϕ(x) dx = C Ω(θ) dσ(θ), where C > 0. Rn |x| Sn−1 82Why? HARMONIC ANALYSIS 107

On the other hand the Fourier transform commutes with orthogonal transformations and henceϕ ˆ(ξ) is also a radial functionϕ ˆ(ξ) = f(|ξ|) so Z Z Z ∞ (6.42)m ˆ [ϕ] = m[ϕ ˆ] = m(ξ)ϕ ˆ(ξ) dξ = m(θ)dσ(θ) sn−1f(s) ds = 0. Rn Sn−1 0 Now (6.40) follows from the equality between (6.41) and (6.42).

We proved thatm ˆ (x) coincides with Ω(˜ x/|x|)/|x|n in Rn \{0}. Hence the distribution Ω(˜ x/|x|) (6.43)m ˆ − p.v. |x|n is supported at the origin. According to Corollary 5.34 its Fourier transform is a polynomial. However, the Fourier transform of (6.43) is a bounded function so the polynomial must be constant. The constant must be zero because m and (p.v. Ω(˜ x/|x|)/|x|n)∧ have the zero integrals over Sn−1. Hence Ω(˜ x/|x|) Ω(x/|x|)∧ mˆ = p.v. so m(ξ) = p.v. (ξ) |x|n |x|n ˜ where Ω(θ) = Ω(−θ).

6.6. Algebra of singular integrals. In this section we will investigate compositions of singular integrals Z Ω(y/|y|) Z TΩf(x) = WΩ ∗ f(x) = lim n f(x − y) dy where Ω(θ) dσ(θ) = 0. ε→0 |y|≥ε |y| Sn−1 We will assume that Ω ∈ C∞(Sn−1). In general we cannot expect that the composition of singular integrals is a singular integral. This can be seen in the following example Example 6.41. There is a function Ω ∈ C∞(Sn−1) with the vanishing integral such that that the square of a Riesz transform satisﬁes83 1 (6.44) R2 = − I + T j n Ω Indeed, 1 ∧ 1 ξ2 R2 + I [ϕ] = − j ϕˆ(ξ) = m(ξ)ϕ ˆ(ξ) j n n |ξ|2 and Z Z Z n ! 1 2 1 X 2 m(ξ) dσ(ξ) = − ξj dσ(ξ) = 1 − ξk dσ(ξ) = 0 n−1 n−1 n n n−1 S S S k=1 so the existence of Ω satisfying (6.44) follows from Theorem 6.37.

However, the class of operators of the form aI + TΩ is closed under compositions.

83I stands for the identity. 108 PIOTR HAJLASZ

Theorem 6.42. If m ∈ C∞(Rn \{0}) is homogeneous of degree zero, then there is a ∈ C and Ω ∈ C∞(Sn−1) with vanishing integral such that ∨ T ϕ := (mϕˆ) = aϕ + TΩϕ for ϕ ∈ Sn. R Proof. Let a ∈ C be such that Sn−1 (m(θ) − a) dσ(θ) = 0. Then by Theorem 6.37 there is Ω such that T f − af = TΩ. Corollary 6.43. The class A of operators of the form

aI + TΩ where a ∈ C and Ω ∈ C∞(Sn−1) has vanishing integral coincides with the class of operators of the form T ϕ = (mϕˆ)∨, where m ∈ C∞(Rn \{0}) is homogeneous of degree zero. Hence A is a commutative algebra. An operator in the class A is invertible if and only if m = a + WdΩ does not vanish at any point of Sn−1. HARMONIC ANALYSIS 109

7. Riesz potentials and fractional powers of the Laplacian

84 Recall that for 0 < α < n the Riesz potantial Iα is deﬁned by 1 Z f(y) (Iαf)(x) = (Uα ∗ f)(x) = n−α dy , γ(α) Rn |x − y| where n α n n−α 2 α α− 2 π 2 Γ 2 −α π Γ 2 α−n γ(α) = n−α so Uα = (2π) α |x| . Γ 2 Γ 2 Clearly, Iαϕ is smooth, slowly increasing, and all its derivatives are slowly increasing when 85 ϕ ∈ Sn. A relationship between Riesz potentials and Riesz transforms is explained in the next result

Proposition 7.1. If n ≥ 2 and ϕ ∈ Sn, then for 1 ≤ j ≤ n ∂ (I1ϕ)(x) = −Rjϕ(x) . ∂xj

Proof. This result easily follows from Proposition 5.53 n−1 ∂ ∂U2 Γ 2 ∂ 1−n (I1ϕ) = ∗ ϕ = n+1 |x| ∗ ϕ ∂xj ∂xj 2π 2 ∂xj n−1 Γ 2 xj = (1 − n) n+1 p.v. n+1 ∗ ϕ = −Rjϕ. 2π 2 |x|

−α −α 2 2 −α/2 ∧ According to Theorem 5.42, Ucα(ξ) = (2π) |ξ| = (4π |ξ| ) . Since (Iαϕ) =ϕ ˆUcα we get

86 Theorem 7.2. For 0 < α < n and ϕ ∈ Sn, ∧ 2 2 −α/2 (7.1) (Iαϕ) (ξ) = (4π |ξ| ) ϕˆ(ξ),

84c.f. Deﬁnition 2.14. 85 However, in general Iαϕ 6∈ Sn, see Remark 7.5. 86 The Fourier transform in (7.1) is understood as a Fourier transform of a tempered distribution Uα ∗ϕ. On the other hand (4π2|ξ|2)−α/2ϕˆ ∈ L1 so the inverse Fourier transform in (7.2) is just a regular inverse Fourier transform of an integrable function. 110 PIOTR HAJLASZ i.e.,

2 2 −α/2 ∨ (7.2) Iαϕ = (4π |ξ| ) ϕˆ(ξ) .

Recall that when n ≥ 3 the fundamental solution of the Laplace operator is87 1 1 Φ(x) = − n−2 = −U2(x). n(n − 2)ωn |x| That means

(7.3) −∆(I2ϕ) = −∆(U2 ∗ ϕ) = U2 ∗ ((−∆)ϕ) = I2((−∆)ϕ) = ∆(Φ ∗ ϕ) = δ0 ∗ ϕ = ϕ.

Hence in some sense the Riesz potential I2 is the inverse of −∆ so it is reasonable to use notation −1 2 2 −1 ∨ (−∆) ϕ = I2ϕ = (4π |ξ| ) ϕˆ(ξ) . With this notation (7.3) reads as (−∆)(−∆)−1ϕ = (−∆)−1(−∆)ϕ = ϕ.

Also for ϕ ∈ Sn and a positive integer k we have −∆ϕ = 4π2|ξ|2ϕˆ(ξ)∨ so (−∆)kϕ = (4π2|ξ|2)kϕˆ(ξ)∨. This suggests how to deﬁne fractional powers of the Laplace operator, including powers with negative exponents

Deﬁnition 7.3. For α > −n and ϕ ∈ Sn we deﬁne (−∆)α/2ϕ = (4π2|ξ|2)α/2ϕˆ(ξ)∨.

Thus 2 2 −α/2 ∨ −α/2 Iαϕ = (4π |ξ| ) ϕˆ(ξ) = (−∆) ϕ for 0 < α < n.

0 α/2 Proposition 7.4. For any α > −n there is u ∈ Sn such that (−∆) ϕ = u ∗ ϕ for α/2 ϕ ∈ Sn. In particular (−∆) ϕ is smooth, slowly increasing, and all its derivatives are slowly increasing.

α/2 Proof. If 0 > α > −n, then (−∆) ϕ = I−αϕ = U−α ∗ ϕ so it remains to consider the case α ≥ 0. However, our argument will work for all α > −n. Let88

2 2 α/2 0 v(ξ) =v ˜(ξ) = (4π |ξ| ) ∈ Sn and u =v ˆ sou ˆ =v ˜ = v. Then (u ∗ ϕ)∧ =ϕ ˆuˆ =ϕv ˆ = (4π2|ξ|2)α/2ϕˆ(ξ) = (−∆)α/2ϕ∧(ξ).

87Theorem 5.60. 88Since α > −n, the singularity of v at ξ = 0 is integrable and hence the function v deﬁnes a tempered distribution. HARMONIC ANALYSIS 111

Remark 7.5. Although (−∆)α/2ϕ is always smooth, it is not always in the Schwarz class α/2 Sn. Namely for every α 6∈ {0, 2, 4, 6,...}, α > −n, there is ϕ ∈ Sn such that (−∆) ϕ 6∈ Sn. Suppose to the contrary that for some α 6∈ {0, 2, 4, 6,...}, α > −n, and all ϕ ∈ Sn we α/2 have (−∆) ϕ ∈ Sn. Then the Fourier transform of this function also belongs to Sn i.e., 2 2 α/2 (7.4) (4π |ξ| ) ϕˆ(ξ) ∈ Sn. However |ξ|α is not C∞ smooth at ξ = 0 so ifϕ ˆ(0) 6= 0, the function (7.4) is not C∞ smooth at ξ = 0 and hence it cannot belong to Sn. On the other hand if α = 2k ∈ {0, 2, 4, 6,...}, then α/2 k (−∆) = (−∆) : Sn → Sn since the Laplace operator maps Sn to Sn and so does the k-fold composition of (−∆).

Proposition 7.6. If α > −n, then for ϕ ∈ Sn we have α/2 α/2 α+2 (−∆)(−∆) ϕ = (−∆) (−∆)ϕ = (−∆) 2 ϕ. Remark 7.7. We need to understand this result as follows. The function (−∆)α/2ϕ is smooth so we can apply the classical Laplace operator −∆ to it. Also since (−∆)α/2ϕ is slowly increasing and all its derivatives are slowly increasing, the classical Laplace operator coincides with the distributional one (since there is no problem with the integration by parts). This is how we understand the left hand side. In the middle term (−∆)ϕ ∈ Sn so we can apply (−∆)α/2 to it. And ﬁnally the the term on the right hand side is well deﬁned because ϕ ∈ Sn.

α/2 0 Proof. According to Proposition 7.4, (−∆) ϕ = u ∗ ϕ for some u ∈ Sn. Hence (−∆)((−∆)α/2ϕ) = −∆(u ∗ ϕ) = u ∗ (− ∆ϕ ) = (−∆)α/2(−∆ϕ) |{z} in Sn ∨ ∨ 2 2 α/2 2 2 α/2 2 2 α+2 = (4π |ξ| ) (−∆dϕ) = (4π |ξ| ) (4π |ξ| )ϕ ˆ = (−∆) 2 ϕ. Remark 7.8. Formally we can extend the above result to composition of operators (−∆)α/2(−∆)β/2 as follows ∧ (−∆)α/2(−∆)β/2ϕ) = (4π2|ξ|2)α/2(4π2|ξ|2)β/2ϕˆ(ξ) ∧ 2 2 α+β α+β = (4π |ξ| ) 2 ϕˆ(ξ) = (−∆) 2 ϕ so formally α/2 β/2 α+β (7.5) (−∆) (−∆) = (−∆) 2 on Sn provided α, β > −n and α + β > −n.

β/2 The problem is that in general if ϕ ∈ Sn, then (−∆) ϕ 6∈ Sn so in the composition (−∆)α/2(−∆)β/2ϕ α/2 we apply the operator (−∆) to a function which is not in Sn and Deﬁnition 7.3 requires α/2 the function to which we apply (−∆) to be in the space Sn. Well, the formula that 112 PIOTR HAJLASZ

α/2 deﬁnes (−∆) actually applies to a much larger class of function than just Sn and with this extension in mind we can justify the above composition formula. There is however, a 89 α/2 β/2 price we have to pay for it. If ϕ ∈ Sn, then (−∆) ϕ = uα ∗ ϕ and (−∆) ϕ = uβ ∗ ϕ 0 for some uα, uβ ∈ Sn and we would like to write α/2 β/2 (7.6) (−∆) (−∆) ϕ = uα ∗ (uβ ∗ ϕ) β/2 and this does not make any sense if uβ ∗ ϕ = (−∆) ϕ 6∈ Sn. An example where this creates a problem will be seen in the next result. A formal proof, similar to the arguments used above will be very short and elegant, but the actual rigorous proof will be quite long and boring. Theorem 7.9. If α, β > 0, α + β < n, then

(7.7) Iα(Iβϕ) = Iα+βϕ for ϕ ∈ Sn. Remark 7.10. Formally we can write this equality as

−α/2 −β/2 − α+β (7.8) (−∆) (−∆) ϕ = (−∆) 2 ϕ when α, β > 0, α + β < n and ϕ ∈ Sn. This looks like a special case of the situation considered in Remark 7.8 so, does it complete the proof of Theorem 7.9? Not really, because we run into the problem described in (7.6). The equality can be written as

Uα ∗ (Uβ ∗ ϕ) = Uα+β ∗ ϕ

If we would know that Iβϕ = Uβ ∗ ϕ ∈ Sn we could apply the Fourier transform and easily prove the identity, but in general, without this information, we have no reason to claim ∧ that (Uα ∗ (Uβ ∗ ϕ)) = U\β ∗ ϕ Ucα.

Proof.

90 Lemma 7.11. If α, β > 0, α + β < n, then there is a constant C0 = C0(α, β, n) such that Z dy C0 n−α n−β = n−(α+β) . Rn |x − y| |y| |x|

Proof. First observe that the integral on the left hand side is finite whenever x 6= 0. To see this let R = |x| and divide the integral over Rn into four integrals over mutually disjoint regions Z Z Z Z ... dy, ... dy, ... dy, ... dy. R R R R n B(0, 2 ) B(x, 2 ) B(0,2R)\(B(0, 2 )∪B(x, 2 )) R \B(0,2R) It is very easy to see that the first three integrals are finite. Indeed, in the first integral we have a singularity |y|n−β at y = 0 only and this singularity is integrable. In the second integral we have a singularity |x − y|n−α at y = x only which is integrable again and in the third integral we do not have singularities at all. It remains to prove finiteness of the last integral.

89Proposition 7.4. 90 In Corollary 7.12 we will ﬁnd the exact value of C0. HARMONIC ANALYSIS 113

If |y| ≥ 2R, then |x − y| ≥ |y| − |x| ≥ |y| − |y|/2 = |y|/2 so Z Z dy n−α dy n−α n−β ≤ 2 2n−(α+β) < ∞, Rn\B(0,2R) |x − y| |y| Rn\B(0,2R) |y| because 2n − (α + β) > n. Note that the integral from the lemma is invariant under rotations and hence Z dy n−β n−α |y| = f(|x|). Rn |x − y|

For x 6= 0 let x0 = x/|x| and t = |x| so x = tx0. A simple change of variables gives Z dy Z dy n−α n−β = n−α n−β Rn |x − y| |y| Rn |tx0 − y| |y| Z α+β−n dy α+β−n = t n−α n−β = |x| f(1). Rn |x0 − y| |y|

We have 1 Z 1 Z ϕ(z) Iα(Iβϕ) = n−α n−β dz dy γ(α)γ(β) Rn |x − y| Rn |y − z| 1 Z Z dy = n−α n−β ϕ(z) dz γ(α)γ(β) Rn Rn |x − y| |y − z| 1 Z Z dy = n−α n−β ϕ(z) dz γ(α)γ(β) Rn Rn |(x − z) − y| |y| Z C0 ϕ(z) = n−(α+β) dz γ(α)γ(β) Rn |x − z| C γ(α + β) = 0 I ϕ(x). γ(α)γ(β) α+β We could use the Fubini theorem here because the function that we integrated over Rn ×Rn was integrable. To see the integrability, repeat the above computations with ϕ replaced by |ϕ|. It remains to show that γ(α)γ(β) C = . 0 γ(α + β)

To prove this it suﬃces to verify that Iα(Iβϕ) = Iα+βϕ for just one non-zero function ϕ. To this end let ϕ ∈ Sn be such thatϕ ˆ = 0 in a neighborhood of 0. Then ∨ 2 2 −β/2 2 2 −α/2 2 2 −β/2 ∨ Iα(Iβϕ) = Iα (4π |ξ| ) ϕˆ = (4π |ξ| ) (4π |ξ| ) ϕˆ = Iα+βϕ. | {z } ∈Sn The proof is complete.

As a corollary we obtain 114 PIOTR HAJLASZ

Corollary 7.12. If α, β > 0, α + β < n, then Z dy γ(α)γ(β) 1 n−α n−β = n−(α+β) . Rn |x − y| |y| γ(α + β) |x|

In the next result we will prove an interesting formula for the fractional Laplacian (−∆)α/2 when 0 < α < 2.

Theorem 7.13. For 0 < α < 2 and ϕ ∈ Sn we have Z α/2 1 ϕ(y) − ϕ(x) (−∆) ϕ = lim n+α dy ε→0 γ(−α) |x−y|≥ε |y − x| 1 Z ϕ(x + y) + ϕ(x − y) − 2ϕ(x) = n+α dy, 2γ(−α) Rn |y| where91 n α 2 −α π 2 Γ − 2 γ(−α) = n+α . Γ 2

Proof. First we will prove equality of the integrals 1 Z ϕ(x + y) + ϕ(x − y) − 2ϕ(x) Z ϕ(y) − ϕ(x) (7.9) dy = lim dy. n+α ε→0 n+α 2 Rn |y| |x−y|≥ε |y − x| It easily follows from Taylor’s formula that |ϕ(x + y) + ϕ(x − y) − 2ϕ(x)| kD2ϕk ≤ C ∞ , |y|n+α |y|n+α−2 where 1/2 n 2 2! 2 X ∂ ϕ kD ϕk∞ = sup (x) . x∈ n ∂xj∂xk R j,k=1 Since n + α − 2 < n, both sides of this inequality are integrable over the ball {|y| ≤ 1}. They are also integrable in {|y| > 1} because of rapid decay of ϕ. Thus the first integral at (7.9) is well defined and finite. For the second integral we have Z ϕ(y) − ϕ(x) Z ϕ(x + y) − ϕ(x) Z ϕ(x − y) − ϕ(x) n+α dy = n+α dy = n+α dy |x−y|≥ε |y − x| |y|≥ε |y| |y|≥ε |y| so Z ϕ(y) − ϕ(x) 1 Z ϕ(x + y) + ϕ(x − y) − 2ϕ(x) n+α dy = n+α dy |x−y|≥ε |y − x| 2 |y|≥ε |y| and hence passing to the limit as ε → 0+ yields (7.9). Observe that ϕ(x + y) + ϕ(x − y) − 2ϕ(x) (7.10) (x, y) 7→ ∈ L1( n × n). |y|n+α R R

91 α 2 α α We use here Γ evaluated at a negative number. Actually, Γ(− 2 ) = − α Γ(1 − 2 ), where 1 − 2 > 0, see Deﬁnition 3.35. Note also that γ(−α) is given by the same formula as the constant in the deﬁnition of the Riesz potential. HARMONIC ANALYSIS 115

Indeed, for (x, y) ∈ Rn × B(0, 1) we have 2 −2n ϕ(x + y) + ϕ(x − y) − 2ϕ(x) kD ϕkL∞(B(x,1)) (1 + |x|) 1 n ≤ C ≤ C ∈ L (R × B(0, 1)) |y|n+α |y|n+α−2 |y|n+α−2 and Z Z |ϕ(x + y) + ϕ(x − y) − 2ϕ(x)| Z dy n+α dx dy ≤ 3kϕk1 n+α < ∞. |y|≥1 Rn |y| |y|≥1 |y| This allows us to compute the Fourier transform of the integrable function Z ϕ(x + y) + ϕ(x − y) − 2ϕ(x) (7.11) f(x) = n+α dy Rn |y| by changing the order of integration. We have Z Z ˆ −2πix·ξ ϕ(x + y) + ϕ(x − y) − 2ϕ(x) f(ξ) = e n+α dy dx Rn Rn |y| Z Z 1 −2πix·ξ = n+α e (ϕ(x + y) + ϕ(x − y) − 2ϕ(x)) dx dy Rn |y| Rn Z e2πiy·ξ + e−2πiy·ξ − 2 =ϕ ˆ(ξ) n+α dy Rn |y| Z 1 − cos(2πy · ξ) = −2ϕ ˆ(ξ) n+α dy. Rn |y|

Let ρ ∈ O(n) be such that ρ(ξ) = |ξ|e1. Then the change of variablesy ˜ = ρ(y) gives Z 1 − cos(2πy · ξ) Z 1 − cos(2πρ−1(y) · ξ) Z 1 − cos(2πy · ρ(ξ)) n+α dy = −1 n+α dy = n+α dy Rn |y| Rn |ρ (y)| Rn |y| Z 1 − cos(2π|ξ|y1) = n+α dy Rn |y| Z α 1 − cos y1 = (2π|ξ|) n+α dy Rn |y| where in the last equality we used another change of variablesy ˜ = 2π|ξ|y. Actually it is n+α a consequence of the above calculation that the function (1 − cos y1)/|y| is integrable n 2 on R , but one can see it more directly by using the estimate |1 − cos y1| ≤ C|y| in a neighborhood of the origin. Let Z 1 − cos y1 C(α) = n+α dy. Rn |y| We proved that fˆ(ξ) = −2ϕ ˆ(ξ)(2π|ξ|)αC(α) = −2C(α)(−∆)α/2ϕ∧(ξ). Hence Z α/2 1 1 ϕ(x + y) + ϕ(x − y) − 2ϕ(x) (−∆) ϕ(x) = − f(x) = − n+α dy 2C(α) 2C(α) Rn |y| 1 Z ϕ(x + y) − ϕ(x) = − lim n+α dy. C(α) ε→0 |y|≥ε |y| 116 PIOTR HAJLASZ

It remains to prove that n α 2 −α π 2 Γ − 2 (7.12) C(α) = − n+α . Γ 2 Let ϕ(x) = e−π|x|2 . Sinceϕ ˆ = ϕ we have 1 Z ϕ(0 + y) − ϕ(0) 1 Z e−π|y|2 − 1 (7.13) (−∆)α/2ϕ(0) = − lim dy = − dy. ε→0 n+α n+α C(α) |y|≥ε |y| C(α) Rn |y| Both sides of this equality are easy to compute. Recall that according to Lemma 5.44 for γ > −n we have Z Z ∞ n+γ −π|x|2 γ −πs2 n+γ−1 Γ 2 e |x| dx = nωn e s ds = γ n . n 2 R 0 π Γ 2 Since for g ∈ L1,g ˇ(0) = R g(x) dx, we obtain Rn Z (7.14) (−∆)α/2ϕ(0) = (4π2|ξ|2)α/2ϕˆ(ξ)∨(0) = (2π|ξ|)αe−π|ξ|2 dξ Rn n+α α n+α α 2 α Γ 2 2 π Γ 2 = (2π) α n = n . 2 π Γ 2 Γ 2 On the other hand Z −π|y|2 Z ∞ −πs2 Z ∞ −α 0 e − 1 n−1 e − 1 −πs2 s n+α dy = nωn s n+α ds = nωn (e − 1) ds Rn |y| 0 s 0 −α ∞ ∞ nω Z 2 nω Z 2 = n (e−πs − 1)0s−α ds = −2π n e−πs s1−α ds α 0 α 0 ∞ n+γ 2π Z 2 2π Γ γ=2=−α−n − nω e−πs sn+γ−1 ds = − 2 n γ n α α 2 0 π Γ 2 α α+n Γ − 2 2 = π n . Γ 2 This identity, (7.13) and (7.14) yield α n+α α α 2 2 π Γ 2 1 α+n Γ − 2 = − π 2 n C(α) n Γ 2 Γ 2 which easily implies (7.12). Corollary 7.14. For 0 < α < 2 n −α α Z 1 − cos y π 2 2 Γ − 1 dy = − 2 . n+α n+α n |y| R Γ 2 α/2 1 n Corollary 7.15. If 0 < α < 2 and ϕ ∈ Sn, then (−∆) ϕ ∈ L (R ).

Proof. Z α/2 1 ϕ(x + y) + ϕ(x − y) − 2ϕ(x) 1 1 (−∆) ϕ = n+α dy = f(x) ∈ L 2γ(−α) Rn |y| 2γ(−α) since we proved in (7.10) that the function f deﬁned in (7.11) is integrable. HARMONIC ANALYSIS 117

It follows from Proposition 7.6 that is 0 < α < 2 and k ≥ 0 is an integer, then α +k α k k α k (−∆) 2 ϕ = (−∆) 2 (−∆) ϕ = (−1) (−∆) 2 (∆ ϕ) and hence Theorem 7.13 can be generalized to the case of higher powers as follows

Corollary 7.16. If 0 < α < 2, k ≥ 0 is an integer and ϕ ∈ Sn, then k k k α (−1) Z ∆ ϕ(y) − ∆ ϕ(x) 2 +k (−∆) ϕ = lim n+α dy ε→0 γ(−α) |x−y|≥ε |y − x| (−1)k Z ∆kϕ(x + y) + ∆kϕ(x − y) − 2∆kϕ(x) = n+α dy, 2γ(−α) Rn |y| where n α 2 −α π 2 Γ − 2 γ(−α) = n+α . Γ 2