HARMONIC ANALYSIS 1. Maximal Function for a Locally Integrable

Home , Mean of a function, Measurable function

HARMONIC ANALYSIS

PIOTR HAJLASZ

1 n 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.

p n Theorem 1.1 (Hardy-Littlewood-Wiener). If f ∈ L (R ), 1 ≤ p ≤ ∞, then Mf < ∞ a.e. Moreover

1 n (a) For f ∈ L (R ) 5n Z (1.1) |{x : Mf(x) > t}| ≤ |f| for all t > 0. t Rn p n p n (b) If f ∈ L (R ), 1 < p ≤ ∞, then Mf ∈ L (R ) and p 1/p kMfk ≤ 2 · 5n/p kfk for 1 < p < ∞, p p − 1 p

kMfk∞ ≤ kfk∞ .

Date: March 28, 2012. 1 2 PIOTR HAJLASZ

The estimate (1.1) is called weak type estimate.

1 n 1 n Note that if f ∈ L (R ) is a nonzero function, then Mf 6∈ L (R ). Indeed, R if λ = B(0,R) |f| > 0, then for |x| > R we have Z λ Mf(x) ≥ |f| ≥ n , B(x,R+|x|) ωn(R + |x|) n and the function on the right hand side is not integrable on R . Thus the statement (b) of the theorem is not true for p = 1.

1 n If g ∈ L (R ), then the Chebyschev inequality 1 Z |{x : |g(x)| > t}| ≤ |g| for t > 0 t Rn is easy to prove. Hence the inequality at (a) would follow from boundedness of of Mf in L1. Unfortunately Mf is not integrable and (a) is the best what we can get for p = 1. Before we prove the theorem we will show that it implies the Lebesgue diﬀerentiation theorem.

1 n Theorem 1.2 (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 ). R Let fr(x) = B(x,r)f(y) dy and deﬁne

Ωf(x) = lim sup fr(x) − lim inf fr(x) . r→0 r→0 1 It suﬃces to prove that Ωf = 0 a.e. and that fr → f in L . Indeed, the ﬁrst property means that fr converges a.e. to a measurable function g while the second one implies that for a subsequence fri → f a.e. and hence g = f a.e. Observe that Ωf ≤ 2Mf and hence for any ε > 0 Theorem 1.1(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 HARMONIC ANALYSIS 3

Since ε > 0 can be arbitrarily small we conclude Ωf = 0 a.e. We are left 1 with the proof that fr → f in L . 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 (1.2) = kfy − fk1 dy , B(0,r) 1 where fy(x) = f(x + y). Since fy → f in L as y → 0 the right hand side of (1.2) converges to 0 as r → 0. 2

1 n If f ∈ Lloc(R ), then we can deﬁne f at every point by the formula Z (1.3) 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. 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 (1.3). 1 n Theorem 1.3. 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 (1.4) 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 c c∈Q c n n zero. Thus for x ∈ R \ E and all c ∈ Q, (1.4) is satisﬁed. If x ∈ R \ 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) The proof is complete. 2

n We can generalize the above result as follows. We say that x ∈ R is a p 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 the one used above leads to the following result that we leave an an exercise. 4 PIOTR HAJLASZ

p n Theorem 1.4. If f ∈ Lloc(R ), 1 ≤ p < ∞, then the set of points that are not p-Lebesgue points of f has measure zero.

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

Applying the Lebesgue theorem to f = χE we obtain

n Theorem 1.5. Almost every point of a measurable set E ⊂ R is its density n point and a.e. point of R \ E is not a density point of E.

In the Lebesgue theorem we have seen that the averages of f over balls converge to f(x) and it is natural to inquire if we can replace balls by other sets like cubes or even balls, but not centered at x.

n Definition. We say that a family F of measurable subsets of R is regular n at x ∈ R if

(a) The sets are bounded and have positive measure; (b) There is a sequence Sn ∈ F with |Sn| → 0; (c) There is a constant C > 0 such that every S ∈ F is contained in a ball B ⊃ S centered at x such that |S| ≥ C|B|.

n Example. The family of all cubes Q in R such that the distance of Q to x is no more than Cdiam Q is regular.

1 n Theorem 1.6. If f ∈ Lloc(R ), x is a Lebesgue point of f, and F is regular at x, then Z lim f(y) dy = f(x) . S∈F |S|→0 S

Proof. For S ∈ F denote by rS the radius of a ball BS = B(x, rS) such that S ⊂ BS and |S| ≥ C|BS|. Clearly if |S| → 0, then rS → 0. We have Z Z

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

(1.5) MF f(x) ≤ CMf(x) , so MF satisfies the claim of Theorem 1.1 (with different constants). In p particular MF is a bounded operator in L , 1 < p ≤ ∞. In the proof of Theorem 1.1 we will need the following two results. Theorem 1.7 (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. 2 Corollary 1.8. If µ is a σ-ﬁnite measure on X and 0 t}) dt . X 0

The next result has many applications that go beyond the maximal theorem. Theorem 1.9 (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 If the metric space is separable, then the family B0 is countable and we can 0 ∞ arrange it as a sequence B = {Bi}i=1, so ∞ [ [ B ⊂ 5Bi . B∈B i=1

Remark. Here B can be either a family of open balls or closed balls. In both cases proof is the same. 6 PIOTR HAJLASZ

Proof. Let sup{diam B : B ∈ B} = R < ∞. Divide the family B according to the diameter of the balls R R F = {B ∈ B : < diam B ≤ } . j 2j 2j−1 S∞ Clearly B = j=1 Fj. Define B1 ⊂ F1 to be the maximal family of pairwise disjoint balls. Suppose the families B1,..., Bj−1 are already defined. Then we define Bj to be the maximal family of pairwise disjoint balls in

j−1 0 0 [ Fj ∩ {B : B ∩ B = ∅ for all B ∈ Bi} . i=1 0 S∞ Next we deﬁne B = j=1 Bj. Observe that every ball B ∈ Fj intersects Sj Sj with a ball in i=1 Bj. Suppose that B ∩ B1 6= ∅, B1 ∈ i=1 Bi. Then R R diam B ≤ = 2 · ≤ 2 diam B 2j−1 2j 1 and hence B ⊂ 5B1. 2

1 n Proof of Theorem 1.1. (a) Let f ∈ L (R ) and Et = {x : Mf(x) > t}. For x ∈ Et, there is rx > 0 such that Z |f| > t , B(x,rx) so Z −1 |B(x, rx)| < t |f| . 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 pairwise disjoint balls B(xi, rxi ), i = 1, 2,... S∞ 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 ). 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 1 n that f1 ∈ L (R ). Since |f| ≤ |f1| + t/2 we have Mf ≤ Mf1 + t/2 and hence

{Mf > t} ⊂ {Mf1 > t/2} . HARMONIC ANALYSIS 7

Thus 2 · 5n Z (1.6) |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. 2 Note that we proved in (1.6) the following inequality 2 · 5n Z (1.7) |{x : Mf(x) > t}| ≤ |f(x)| dx t {|f|>t/2} which is slightly stronger than (1.1). We will need that inequality later.

n For a positive measure µ on R 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 1.1(a) leads to the following result.

n Proposition 1.10. If µ is a ﬁnite positive Borel measure on R , then 5n |{x : Mµ(x) > t}| ≤ µ( n) for all t > 0. t R

n Let F be the family of all rectangular boxes in R 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 p MF in L , 1 < p ≤ ∞ cannot be concluded from (1.5). However, we have 8 PIOTR HAJLASZ

Theorem 1.11 (Zygmund). For 1 0 such that

(1.8) kMffkp ≤ Ckfkp . p Moreover, if f ∈ Lloc, 1 < p < ∞, then Z (1.9) lim f(x − y) dy = f(x) a.e. diam S→0 S∈F S

Proof. First we will prove how to conclude (1.9) from (1.8). Note that since the family is not regular at 0, (1.9) is not a consequence of Theorem 1.6, see also Theorem 1.12. However Z Z 0 ≤ lim sup f(x − y) dy − lim inf f(x − y) dy ≤ 2Mff(x) diam S→0 S diam S→0 S S∈F S∈F and hence (1.9) follows from (1.8) by almost the same argument that was used to deduce the Lebesgue diﬀerentiation theorem from Theorem 1.1. We leave details to the reader. We are left with the proof of (1.8). For simplicity assume that n = 2. We have

Z x2+b2 Z x1+b1 Mff(x1, x2) = sup |f(y1, y2)| dy1 dy2 a1,b1>0 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 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. These are not exactly the Hardy-Littlewood maximal functions, because we take averages over all intervals that contain x1 and then all the intervals that contain x2, but these maximal functions are bounded by a constant multiplicity of the one dimensional Hardy-Littlewood maximal functions, see (1.5), because the family of all intervals that contain 0 is regular at 0. Thus it is easy to see that inequality (1.8) follows from the one dimensional version of Theorem 1.1 applied twice and the Fubini theorem. 2 Surprisingly (1.9) does not hold for p = 1 and hence the maximal function Mff does not satisfy the weak type estimate (1.1).1 Namely one can prove the following result that we leave without a proof.

n Theorem 1.12 (Saks). Let F be the family of all rectangular boxes in R that contain the origin and have sides parallel to the coordinate axes. Then

1Such estimate would imply (1.9). HARMONIC ANALYSIS 9

1 n the set of functions f ∈ L (R ) such that Z n lim sup f(x − y) dy = ∞ for all x ∈ R diam S→0 S S∈F 1 n is a dense Gδ subset of L (R ). In particular it is not empty.

The proof is based on the category method.

1.1. The Calder´on-Zygmund decomposition. The following result plays a fundamental role in many areas of analysis.

1 n Theorem 1.13 (Calder´on-Zygmund decomposition). Suppose f ∈ L (R ), f ≥ 0 and α > 0. Then there is an open set Ω and a closed set F such that

n (a) R = 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 ≤ 2nα , k = 1, 2, 3,... Qk

n Proof. Decompose R 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 terminated. We apply it to all the cubes of the S∞ original grid. Let Ω = k=1 Qk, where the cubes are defined by the first n case of the process. It remains to prove that f ≤ α a.e. in the set R \ 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 Q˜i with the property that R x ∈ Q˜i, diam Q˜i → 0, f ≤ α. According to Theorem 1.6 for a.e. such x Q˜i R f → f(x) and hence f ≤ α a.e. in F . 2 Q˜i 10 PIOTR HAJLASZ

Corollary 1.14. Let f, α and Ω be as in Theorem 1.13. Then −1 |Ω| ≤ α kfk1 .

Proof. We have ∞ ∞ Z X X −1 −1 |Ω| = |Qk| ≤ α |f| ≤ α kfk1 . k=1 k=1 Qk The proof is complete. 2

1 n 1 n We already observed that if f ∈ L (R ), then Mf 6∈ L (R ), 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. 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 so the Zygmund space is an intermediate space between all Lp for p > 1 and L1. Theorem 1.15 (Stein). Suppose that a measurable function f is supported in a ball B. Then Mf ∈ L1(B) if and only if f ∈ L log L(B).

Proof. Suppose that f ∈ L log L(B). Then Z Z Mf(x) dx ≤ |B| + Mf(x) dx B {Mf≥1} Z ∞ = |B| + |{Mf ≥ 1}| + |{Mf > t}| dx . | {z } 1 ≤|B| The last equality is a consequence of the Cavalieri principle. Applying inequality (1.7) we have ! Z Z ∞ C Z Mf(x) dx ≤ 2|B| + |f(x)| dx dt B 1 t {|f|>t/2} ! Z Z max{2|f(x)|,1} dt = 2|B| + |f(x)| dx B 1 t Z ≤ 2|B| + |f(x)| log(e + 2|f(x)|) dx < ∞ . B This proves the implication from right to left. HARMONIC ANALYSIS 11

To prove the implication from left to right we will show ﬁrst an inequality which is, in some sense, an inverse inequality to (1.7). Namely we will prove that there is a constant C = C(n) > 0 such that

2−n Z (1.10) |{x : Mf(x) > Ct}| ≥ |f(x)| dx . t {|f|>t}

To this end we need to apply the Calder´on-Zygmund decomposition to the S function |f| and α = t. If Ω = k Qk is the Calder´on-Zygmund decomposition, then Z t < |f| ≤ 2nt Qk and hence Mf(x) > Ct for all x ∈ Qk. Thus

∞ X 2−n Z |{x : Mf(x) > Ct}| ≥ |Q | ≥ |f(x)| dx . k t k=1 Ω

Since |f| ≤ t for x 6∈ Ω we have

Z Z |f(x)| dx ≥ |f(x)| dx . Ω {|f|>t}

The last two inequalities combined together prove (1.10). Note that the 1 n inequality (1.10) is satisﬁed by an arbitrary function f ∈ L (R ). Suppose now that f vanished outside a ball B and that Mf ∈ L1(B). Observe that it is not clear if in the inequality (1.10) we can replace the set {x : n R : Mf(x) > Ct} on the left hand side by the set {x ∈ B : Mf(x) > Ct}|. Indeed, the maximal function does not vanish outside B and the proof n involved estimates on R .

1 n Note that Mf ∈ Lloc(R ). Indeed, it 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, but it is easy to see that if x is near the boundary and outside the ball, 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 12 PIOTR HAJLASZ can complete the proof as follows. Z ∞ > Mf(x) dx {Mf>1} Z ∞ ≥ |{Mf > t}| dt 1 Z ∞ C2−n Z ≥ |f(x)| dx dt 1 t {|f|>t/C} ! Z Z max{C|f(x)|,1} dt ≥ C2−n |f(x)| dx B 1 t Z = C2−n |f(x)| log(max{C|f(x)|, 1}) dx . B The proof is complete. 2 A more careful analysis leads to the following version of Stein’s theorem. n In an open set Ω ⊂ R we deﬁne the local maximal function by Z MΩf = sup{ |f| : x ∈ Q ⊂ Ω} , Q where the supremum is taken over all cubes Q in Ω that contain x. n 1 Theorem 1.16 (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

1.2. Fractional integration theorem. As an application of the Hardy- Littlewood-Wiener theorem we will prove a result due to Hardy-Littlewood- Sobolev about integrability of Riesz potentials, called fractional integration theorem. For 0 < α < n and n ≥ 2 we deﬁne the Riesz potentials by 1 Z f(y) (Iαf)(x) = n−α dy , γ(α) Rn |x − y| where n 2 α α π 2 Γ 2 γ(α) = n−α . 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. HARMONIC ANALYSIS 13

Theorem 1.17 (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).

We precede the proof with a technical lemma. Lemma 1.18. 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|

n Proof. For x ∈ R and δ > 0 consider the annuli δ δ A(k) = B x, − B x, . 2k 2k+1 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. 2 Proof of Theorem 1.17. Fix δ > 0. H¨older’sinequality and integration in polar coordinates yield !1/p0 Z |f(y)| Z dy dy ≤ kfk n−α p (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. Thus the lemma gives

α α−(n/p) |Iαf(x)| ≤ C δ Mf(x) + δ kfkp . 14 PIOTR HAJLASZ

Taking Mf(x)−p/n δ = kfkp we obtain 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 . n Integrating both sides over R and applying boundedness of the maximal function in Lp yields the result. 2

2. Fourier transform

2.1. Measures and convolution. Theorem 2.1 (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

Proof. 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

Exercise. Show that the classical Minkowski inequality kf + gkp ≤ kfkp + kgkp follows from the integral Minkowski inequality. HARMONIC ANALYSIS 15

n Recall that the convolution of measurable functions on R is deﬁned by Z (f ∗ g)(x) = f(x − y)g(y) dy . Rn Theorem 2.2. If 1 ≤ p < ∞, then

kf ∗ gkp ≤ kfkpkgk1.

Proof. The inequality can be obtained from the Minkowski integral inequality 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 . The proof is complete. 2 The above result is a special case of a more general inequality. Theorem 2.3 (Young’s inequality). If 1 ≤ p, q, r ≤ ∞ and q−1 = p−1 + r−1 − 1, then kf ∗ gkq ≤ kfkpkgkr .

Exercise. Prove it.

n Recall that C0(R ) is the space of continuous functions vanishing at in- n finity, i.e. f ∈ C0(R ) if f is continuous and lim f(x) = 0. |x|→∞ n C0(R ) is a Banach space with respect to the supremum norm k · k∞ ∞ n and C0 (R ) (compactly supported smooth functions) is a dense subset of n C0(R ). n If µ is a signed (Borel) measure on R , then there is a unique Hahn decomposition µ = µ+ − µ−, where µ+ and µ− are positive Borel measures concentrated on disjoint sets. We define the measure |µ| as |µ| = µ+ + µ−. n The number kµk = |µ|(R ) us called total variation of µ. The measures of finite total variation form Banach space with the norm kµk. We denote it n by B(R ). 16 PIOTR HAJLASZ

1 n If f ∈ L (R ), then dµ = f(x) dx is a measure of finite total variation R R 1 n µ(E) = E f(x) dx, |µ|(E) = E |f| dx, kµk = kfk1. Thus L (R ) can be n identified as a closed subspace of B(R ) by the isometry 1 n n L (R ) 3 f 7→ f(x) dx ∈ B(R ). n Theorem 2.4 (Riesz representation theorem). The dual space to C0(R ) is isometrically isomorphic to the space of measures of finite total variation. n ∗ More precisely, if 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 n Moreover kΦk = kµk = |µ|(R ).

1 n 1 n n If f, g ∈ L (R ), then f ∗ g ∈ L (R ) ⊂ B(R ) and hence it acts as a n functional on 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 deﬁnes a functional on C0(R ) and hence there is a unique measure µ ∈ n 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 We denote

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

µ1 ∗ µ2 = (f ∗ g) dx, HARMONIC ANALYSIS 17 so the convolution of measures extends the notion of convolution of func- n tions. If dµ1 = f dx and µ ∈ B(R ), 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 p n n Theorem 2.5. If 1 ≤ p < ∞, f ∈ L (R ) and µ ∈ B(R ), then

kf ∗ µkp ≤ kfkp kµk.

Proof is almost the same as that for Theorem 2.2 as we leave it to the reader.

Exercise. Find δa ∗ δb.

1 n 2.2. Fourier transform. For f ∈ L (R ) we define the Fourier transform as Z fˆ(ξ) = f(x)e−2πix·ξ dx Rn where n X x · ξ = xjξj . j=1 n If µ ∈ B(R ) then we define Z µˆ(ξ) = e−2πix·ξ dµ(x). Rn p n n For f ∈ L (R ) and h ∈ R we define

τhf(x) = f(x + h). We will frequently use the following well known result. Lemma 2.6. For 1 ≤ p < ∞

kf − τhfkp → 0 as |h| → 0. Theorem 2.7. The Fourier transform has the following properties 18 PIOTR HAJLASZ

(a) ∧ 1 n n : L (R ) → C0(R ) is a bounded linear operator with

kfˆk∞ ≤ kfk1. (b) 1 n (f ∗ g)ˆ = fˆgˆ for f, g ∈ L (R ). 1 n 1 n (c) If f, g ∈ L (R ), then fg,ˆ fgˆ ∈ L (R ) and Z Z fˆ(x)g(x) dx = f(x)ˆg(x) dx Rn Rn (d) ˆ 2πih·ξ (τhf)ˆ(ξ) = f(ξ)e (f(x)e2πih·x)ˆ(ξ) = fˆ(ξ − h). −n (e) If fε(x) = ε f(x/ε), then

(fε)ˆ(ξ) = fˆ(εξ), (f(εx))ˆ(ξ) = (fˆ)ε(ξ) . (f) If ρ ∈ O(n) is an orthogonal transformation, then (f(ρ ·))ˆ(ξ) = fˆ(ρξ).

∧ 1 n ∞ n Proof. (a) Clearly : L (R ) → L (R ) is a bounded linear mapping with kfˆk∞ ≤ kfk1. Indeed, Z ˆ −2πix·ξ |f(ξ)| ≤ f(x)e dx = kfk1. Rn The dominate convergence theorem implies that the function fˆis 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| 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 1 |fˆ(ξ)| ≤ f − τ ξ f → 0 as |ξ| → ∞. 2 2|ξ|2 1 HARMONIC ANALYSIS 19

(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 −2πix·ξ (τhf)ˆ(ξ) = f(x + h)e dx Rn Z = f(x)e−2πi(x−h)·ξ dx Rn Z = e2πih·ξ f(x)e−2πix·ξ dx . Rn | {z } fˆ(ξ) The second equality follows from a similar argument. (e) Z −n x −2πix·ξ (fε)ˆ(ξ) = ε f e dx Rn ε Z = f(y)e−2πi(εy)·ξ dy Rn Z = f(y)e−2πiy·(εξ) dy Rn = fˆ(εξ) . The second formula follows from the ﬁrst one if we replace ε by ε−1. (f) Z (f(ρ ·))ˆ(ξ) = f(ρx)e−2πix·ξ dx Rn Z −1 = f(y)e−2πi(ρ y)·ξ dy Rn Z = f(y)e−2πiy·(ρξ) dy Rn = fˆ(ρξ) . Equality (ρ−1y) · ξ = y · (ρξ) follows from the fact that the mapping x 7→ ρx is an isometry. 2 20 PIOTR HAJLASZ

1 n 1 n Theorem 2.8. Suppose f ∈ L (R ) and xkf(x) ∈ L (R ), where xk is the ˆ k-th coordinate 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 Theorem 2.7(d) gives !∧ fˆ(ξ + he ) − fˆ(ξ) e−2πi(hek)·x − 1 k = f(x) (ξ) → (−2πix f(x))ˆ(ξ) . h h k

The convergence follows from the continuity of the Fourier transform in L1. 2

Definition. We say that f is diﬀerentiable in the Lp norm with respect to p n p n xk if f ∈ L (R ) and there is g ∈ L (R ) such that Z p f(x + hek) − f(x) − g(x) dx → 0 as h → 0. Rn h

The function g is called the partial derivative of f (with respect to xk) in p the L norm. We denote it by g = ∂f/∂xk. 1 n Theorem 2.9. If f ∈ L (R ) and ∂f/∂xk is the partial derivative of f in the L1 norm, then ∧ ∂f ˆ = 2πiξkf(ξ) . ∂xk

Proof. The ﬁrst part of Theorem 2.7(d) gives ∧ ∧ ∂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. 2 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 α1 αn . ∂x . . . ∂xn |α|≤m |α|≤m 1 HARMONIC ANALYSIS 21

Then under suitable assumptions Theorems 2.8 and 2.9 have the following higher order generalizations (2.1) P (D)fˆ(ξ) = (P (−2πix)f(x))ˆ(ξ), (P (D)f)ˆ(ξ) = P (2πiξ)fˆ(ξ).

2.3. Summability methods. An important problem is a search for the inversion formula. Namely, given the Fourier transform fˆ, how can we ﬁnd a formula for f? We would like to prove that Z f(x) = fˆ(ξ) e2πix·ξ dξ . Rn As we will see later the formula is true under very restrictive assumptions, but not always. Indeed, in general the right hand side makes no sense as the Fourier transform of an integrable function is not necessarily integrable. To handle this problem we have to use so called summability methods. The two most important summability methods are the Abel and the Gauss (Gauss-Weierstrass) methods. For each ε > 0 the Abel mean of a function f is Z −ε|x| Aε(f) = f(x)e dx . Rn If f is integrable, then Z lim Aε(f) = f(x) dx . ε→0 Rn

However the integral Aε(f) exists also for non integrable functions. For example it exists if f is bounded. If the limit limε→0 Aε(f) = ` exists and is ﬁnite we say that R f is Abel summable to `. Rn R a R ∞ −εx Exercise. Prove that if lima→∞ 0 f(x) dx = `, then Aε = 0 f(x)e dx converges to `. The Gauss mean of f is Z −ε|x|2 Gε(f) = f(x) e dx . Rn R We say that n f(x) dx is Gauss summable to ` if limε→0 Gε(f) = `. R n The two methods can be put is a more general framework. If Φ ∈ C0(R ) and Φ(0) = 1, then we deﬁne the Φ-mean by Z Mε,Φ(f) = MΦ(f) = f(x)Φ(εx) dx . Rn R If limε→0 Mε(f) = `, then we say that n f is Φ-summable to `. R 22 PIOTR HAJLASZ

Our aim is to apply the summability methods to reconstruct f from fˆ. More precisely we want to investigate functions Φ such that Φ-means Z fˆ(ξ)e2πix·ξΦ(εξ) dξ Rn converge in a certain sense to f(x).

1 n n Let f, Φ ∈ L (R ) and ϕ = Φ.ˆ Fix t ∈ R and deﬁne 2πix·t 1 n g(x) = e Φ(εx) ∈ L (R ) . Then Theorem 2.7(d,e) gives

gˆ(ξ) = (Φ)ˆ ε(ξ − t) = ϕε(ξ − t). Hence Theorem 2.7(c) implies Z Z 2πix·t fˆ(x)e Φ(εx) dx = f(x)ϕε(x − t) dx . Rn Rn Replacing x by ξ in the left integral and x by y in the right integral and ﬁnally replacing t by x we have

1 n Theorem 2.10. If f, Φ ∈ L (R ) and ϕ = Φˆ, then Z Z 2πix·ξ fˆ(ξ)e Φ(εξ) dξ = f(y)ϕε(y − x) dy . Rn Rn

To see that the right hand side converges in a certain sense to f(x) and ε → 0 for a reasonable class of functions ϕ we need the next result. 1 n R p n Theorem 2.11. Suppose ϕ ∈ L (R ) with n ϕ(x) dx = 1. If f ∈ L (R ), n R 1 ≤ p < ∞ or f ∈ C0(R ), p = ∞, then

kf ∗ ϕε − fkp → 0 as ε → 0.

R Proof. By a change of variables n ϕε = 1. Hence R Z (f ∗ ϕε)(x) − f(x) = (f(x − y) − f(x))ϕε(y) dy . Rn Therefore the Minkowski integral inequality yields Z Z p 1/p

kf ∗ ϕε − fkp = (f(x − y) − f(x))ϕε(y) dy dx Rn Rn Z Z 1/p ≤ |f(x − y) − f(x)|p dx ε−n|ϕ(y/ε)| dy Rn Rn Z Z 1/p = |f(x − εy) − f(x)|p dx |ϕ(y)| dy Rn Rn Z (2.2) = ω(−εy)|ϕ(y)| dy Rn HARMONIC ANALYSIS 23 where Z 1/p ω(h) = |f(x + h) − f(x)|p dx . Rn Clearly ω(h) ≤ 2kfkp and ω(h) → 0 as h → 0 by Lemma 2.6, so the right hand side of (2.2) converges to 0 by the Lebesgue dominated convergence n theorem. The case f ∈ C0(R ) with p = ∞ follows from a similar argument. 2 1 n R Corollary 2.12. Suppose ϕ ∈ L (R ) and n ϕ(x) dx = 0. Then kf ∗ p n R n ϕεkp → 0 as ε → 0 whenever f ∈ L (R ), 1 ≤ p < ∞ or f ∈ C0(R ), p = ∞.

Proof. Note that Z (f ∗ ϕε)(x) = (f ∗ ϕε)(x) − f(x) · 0 = (f(x − y) − f(x))ϕε(y) dy Rn and the rest follows by the same argument as in the proof of Theorem 2.11. 2 Now we can prove a general result about the inversion formula in terms of the summability methods. Theorem 2.13. If Φ ∈ L1( n) and ϕ = Φˆ ∈ L1( n), R ϕ(x) dx = 1, R R Rn then the Φ-means of the integral R fˆ(ξ)e2πix·ξ dξ converge to f(x) in the Rn L1 norm, i.e. if Z 2πix·ξ Mε(x) = fˆ(ξ)e Φ(εξ) dξ Rn then 1 n Mε → f in L (R ).

Proof. It is a direct consequence of Theorem 2.10 and Theorem 2.11. 2 The existence of functions Φ satisfying the assumptions of Theorem 2.13 follows from the next result. Note that the result also shows a function which is a ﬁxed point of the Fourier transform. Theorem 2.14. 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). (c) Z W (x, t) dx = 1 for all t > 0. Rn In particular, if f(x) = e−π|x|2 , then fˆ(x) = e−π|x|2 , i.e. the function f is a ﬁxed point of the Fourier transform. 24 PIOTR HAJLASZ

Proof. (a) By a simple change of variables it suffices to prove the formula for the Fourier transform for t = (4π)−1. If u(x) = e−πx2 is a function of one variable, then u0 = −2πxu, u0 = −i(−2πixu) and hence (u0)ˆ = −i(−2πixu(x))ˆ(ξ). Applying Theorems 2.8 and 2.9 yields 2πiξuˆ(ξ) = −i(û)0(ξ) (û)0(ξ) = −2πξuˆ(ξ). Solving this differential equation gives

2 uˆ(ξ) =u ˆ(0)e−πξ . Since ∞ Z 2 uˆ(0) = e−πx dx = 1 −∞ we obtainu ˆ(ξ) = e−πξ2 . If f(x) = e−π|x|2 is a function of several variables, then n ∞ Z 2 Z 2 ˆ −π|x| −2πix·ξ Y −πx −2πixkξk f(ξ) = e e dx = e k e dxk n R k=1 −∞ n n Y Y −πξ2 −π|ξ|2 = uˆ(ξk) = e k = e . k=1 k=1 (b) is obvious. (c) It follows from the scaling property (b) that

Z Z Z 2 W (x, t) dx = W (x, (4π)−1) dx = e−π|x| dx = 1 . Rn Rn Rn The proof is complete. 2 The function W (x, t) is called the Gauss-Weierstrass kernel. One can show that the function Z Z 2 1 − |x−y| w(x, t) = W (x − y, t)f(y) dy = e 4t f(y) dy n/2 Rn (4πt) Rn is the solution to the heat equation in the half-space ∂w n+1 ∂t = ∆xw on R+ , n w(x, 0) = f(x), x ∈ R under suitable assumptions about f.

The function Φ(x) = e−4π2t|x|2 clearly satisﬁes the assumptions of Theo- rem 2.13. Hence we have HARMONIC ANALYSIS 25

Theorem 2.15 (The Gauss-Weierstrass summability method). If f ∈ 1 n L (R ), then

Z 2 2 Z fˆ(ξ)e2πix·ξe−4π t|ξ| dξ = f(y)W (x − y, t) dy → f Rn Rn 1 n + in L (R ) as t → 0 .

Proof. The formula follows from Theorem 2.10 with ε = 1 and the fact that W (y − x, t) = W (x − y, t). The convergence to f follows from Theorem 2.11 and Theorem 2.14(b,c). 2 Corollary 2.16. If both f and fˆ are integrable2, then Z (2.3) f(x) = fˆ(ξ)e2πix·ξ dξ a.e. Rn

R 1 + Proof. Mt(x) = n f(y)W (x − y, t) dy converges to f in L as t → 0 . R Hence Mtk → f a.e. for some sequence tk → 0. On the other hand integrability of fˆ and the Lebesgue dominated convergence theorem yield Z Z 2πix·ξ −4π2t|ξ|2 2πix·ξ + Mt(x) = fˆ(ξ)e e dξ → fˆ(ξ)e dξ as t → 0 Rn Rn n for all x ∈ R . 2 1 n ˆ ˆ n Corollary 2.17. If f1, f2 ∈ L (R ) and f1 = f2 on R , then f1 = f2 a.e.

Proof. Let f = f1 − f2. Then fˆ = 0 and hence the function Mε from Theorem 2.13 equals zero. Since Mε converges to f we conclude that f = 0 a.e. 2 The following result provides another example of a function that satisﬁes the assumptions of Theorem 2.13. Theorem 2.18. 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 2The assumption about integrability of fˆ is very strong. Indeed, the equality (2.3) n implies that f equals a.e. to a function in C0(R ). 26 PIOTR HAJLASZ

The function P (x, t) is called the Poisson kernel. Later we will prove that the function Z u(x, t) = P (x − y, t)f(y) dy Rn is a solution to the Dirichlet problem in the half-space n+1 ∆(x,t)u = 0 on R+ , n u(x, 0) = f(x), x ∈ R under suitable assumptions about f. By the same arguments as before we obtain.

1 n Theorem 2.19 (The Abel summability method). If f ∈ L (R ), then Z Z fˆ(ξ)e2πix·ξe−2π|ξ|t dξ = f(y)P (x − y, t) dy → f Rn Rn 1 n + in L (R ) as t → 0 .

The proof of Theorem 2.18 is substantially more diﬃcult than that of Theorem 2.14. Since the formula for the Fourier transform involves the Γ function we need to recall its basic properties.

Definition. For 0 < x < ∞ we deﬁne Z ∞ Γ(x) = tx−1e−t dt . 0 Theorem 2.20.

(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 2 Z 2 Γ(x) = tx−1e−t dt = s2(x−1)e−s 2s ds = 2 s2x−1e−s ds 0 0 0 and hence ∞ 1 Z 2 √ Γ = 2 e−s ds = π . 2 0 2 Lemma 2.21. √ Z π/2 n+1 n πΓ 2 sin θ dθ = n for n = 1, 2, 3,... 0 nΓ 2 HARMONIC ANALYSIS 27

Proof. Denote the left hand side by an and the right hand side by bn. Easy one time integration by parts 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. 2 Lemma 2.22.

n (a) The volume of the unit ball in R equals 2πn/2 πn/2 (2.4) ωn = = n . nΓ(n/2) 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 and Lemma 2.21 give Z π/2 1/2 n+1 1 n π Γ 2 ωn = ωn−1 sin θ dθ = ωn−1 n , 2 0 nΓ 2 so 1/2 n+1 2π Γ 2 ωn = n ωn−1 . nΓ 2 28 PIOTR HAJLASZ

If 2πn/2 an = n nΓ 2 then a direct computation shows that a1 = 2 = ω1 and that an satisﬁes the same recurrence relationship as ωn, so ωn = an for all n. The second equality in (2.4) follows from Theorem 2.20(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 2.23. For β > 0 we have ∞ −u 1 Z e 2 e−β = √ √ e−β /(4u) du . π 0 u

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 We also need an obvious identity Z ∞ 1 −(1+x2)u 2 = e du . 1 + x 0 We have 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 −∞ ∞ ∞ 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 ∞ −u 1 Z e 2 = √ √ e−β /(4u) du . π 0 u The proof is complete. 2 HARMONIC ANALYSIS 29

Proof of Theorem 2.18. (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 ∞ 1 Z 2 = e−u(1+|ξ| )u(n−1)/2 du (n+1)/2 π 0 1 1 Z ∞ = e−ss(n−1)/2 ds . (n+1)/2 2 (n+1)/2 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 = c n 2 (n+1)/2 Rn Rn (1 + |x| ) ∞ ! polar Z Z dσ = c rn−1 dr n 2 (n+1)/2 0 Sn−1(0,1) (1 + r ) Z ∞ rn−1 = c n ω dr n n 2 (n+1)/2 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 (n+1)/2 n n−1 π n Γ 2 (n − 1)Γ 2 = 1 . The proof is complete. 2

1 n If f, fˆ ∈ L (R ), then Z (2.5) f(x) = fˆ(ξ)e2πix·ξ dξ a.e. Rn by Corollary 2.16. The integral on the right hand side deﬁnes a continuous3 n function of x ∈ R and hence if in addition f is continuous, (2.5) holds everywhere. In particular we can apply the formula to 2 2 f(x) = e−4π t|x| and f(x) = e−2πt|x|

3 n C0(R ). 30 PIOTR HAJLASZ which gives us Corollary 2.24.

Z 2 2 W (ξ, t)e2πix·ξ dξ = e−4π t|x| , Rn Z P (ξ, t)e2πix·ξ dξ = e−2π|x|t Rn n for all x ∈ R .

The Weierstrass and Poisson kernels have the following semigroup property.

Corollary 2.25. For t1, t2 > 0 we have Z W (x − y, t1)W (y, t2) dy = W (x, t1 + t2) , Rn Z P (x − y, t1)P (y, t2) dy = P (x, t1 + t2) Rn n for all x ∈ R . In other words, if Wt(x) = W (x, t) and Pt(x) = P (x, t), then

(Wt1 ∗ Wt2 )(x) = Wt1+t2 (x) ,

(Pt1 ∗ Pt2 )(x) = Pt1+t2 (x) n for all x ∈ R .

Proof. It follows from Corollary 2.24 with x replaced by −x that −4π2t|x|2 −2π|x|t Wˆ t(x) = e , Pˆt(x) = e . Hence Theorem 2.7(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 2.17. 2

p n 1 n R As we know, if f ∈ L (R ), 1 ≤ p < ∞ and ϕ ∈ L (R ), n ϕ = 1, then p R f ∗ ϕε → f in L . In particular Z (2.6) f(y)W (x − y, t) dy → f in Lp as t → 0+. Rn Z (2.7) f(y)P (x − y, t) dy → f in Lp as t → 0+. Rn However, it is also interesting to investigate whether we have a.e. convergence. As we shall see this is true. HARMONIC ANALYSIS 31

1 n Theorem 2.26 (Lebesgue diﬀerentiation theorem). If f ∈ Lloc(R ), then Z 1 Z lim f(y) dy = lim f(y) dy = f(x) r→0 B(x,r) r→0 |B(x, r)| B(x,r) n for a.e. x ∈ R .

We proved this result in Analysis I. Note that the Lebesgue diﬀerentiation theorem can be stated as follows: 1 n −1 if f ∈ Lloc(R ), then (f ∗ ϕε)(x) → f(x) a.e., where ϕ(x) = ωn χB(0,1). The integrals at (2.6) and (2.7) are also convolutions and hence it is an obvious guess that the Lebesgue diﬀerentiation theorem should play central role in the proof of pointwise convergence of these integrals. As a consequence of Theorem 2.26 we have.

1 n Corollary 2.27. If f ∈ Lloc(R ), then Z (2.8) lim |f(y) − f(x)| dy = 0 r→0 B(x,r) n for a.e. x ∈ R .

Exercise. Prove the corollary.4

n Definition. Points x ∈ R for which (2.8) is satisﬁed are called Lebesgue points of f.

1 n Definition. Let ϕ ∈ L (R ). We say that Ψ is a radially decreasing majorant of ϕ if

(a) Ψ(x) = η(|x|) for some5 η : [0, ∞) → [0, ∞]. (b) η is decreasing.6 (c) |ϕ(x)| ≤ Ψ(x) a.e.

1 n Every ϕ ∈ L (R ) has the least radially decreasing majorant. Indeed, if

η0(t) = ess sup |ϕ(y)| |y|≥t then η is decreasing (although it may be equal to inﬁnity on some interval) and the function

Ψ0(x) = η0(|x|) = ess sup |ϕ(y)| |y|≥|x|

4 R Hint: Consider B(x,r) |f(y) − ρ| dy for all rational ρ. 5Functions of this form, i.e. functions constant on spheres Sn−1(0, r) are called radially symmetric. 6i.e. nonincreasing. 32 PIOTR HAJLASZ is the least radially decreasing majorant. Thus the existence of an integrable radially decreasing majorant Ψ of ϕ is equivalent with the condition that 1 Ψ0 ∈ L . Theorem 2.28. Let ϕ ∈ L1( n) be such that R ϕ = a. Suppose that ϕ R Rn has an integrable radially decreasing majorant 1 n Ψ(x) = η(|x|) ∈ L (R ) . p n If f ∈ L (R ), 1 ≤ p ≤ ∞, then

(f ∗ ϕε)(x) → af(x) whenever x is a Lebesgue point of f. In particular the Gauss-Weierstrass (2.6) and the Poisson (2.7) integrals converge to f a.e. as t → 0+.

Proof. We will prove the theorem under the additional assumption that η is absolutely continuous, but the result hods also without this assumption. The integrability of Ψ gives a growth estimate for the function η. Indeed, Z n n n r 2 − 1 n Ψ(x) dx ≥ η(r) ωnr − ωn = ωn n r η(r) . r/2≤|x|≤r 2 2 The left hand side converges to 0 as r → 0 or r → ∞ so does the right hand side (2.9) lim rnη(r) = 0, lim rnη(r) = 0 r→0 r→∞ and hence the right hand side is bounded rnη(r) ≤ M for all r > 0 and some M > 0. Fix a Lebesgue point x of f. Then for every γ > 0 there is δ > 0 such that 1 Z n |f(x − y) − f(x)| dy < γ provided r ≤ δ. r B(x,r) Using polar coordinates we can rewrite this inequality as Z r Z 1 n−1 n s |f(x − sθ) − f(x)| dσ(θ) ds < γ , r 0 Sn−1(0,1) | {z } g(s) i.e. Z r (2.10) G(r) = sn−1g(s) ds < γrn provided r ≤ δ. 0 We have Z

7 We estimate I1 as follows Z δ Z n−1 −n I1 = s |f(x − sθ) − f(x)| dσ(θ) ε η(s/ε) ds 0 Sn−1(0,1) Z δ = sn−1g(s)ε−nη(s/ε) ds 0 Z δ = ε−n G0(s)η(s/ε) ds 0 s δ Z δ −n −n−1 0 = ε G(s)η − ε G(s)η (s/ε) ds ε 0 0 s sn δ Z δ/ε −n −n 0 = G(s)s η − ε G(εt)η (t) dt ε ε 0 0 η0≤0 Z δ/ε ≤ γM − γ tnη0(t) dt 0 η0≤0 Z ∞ ≤ γM − γ tnη0(t) dt 0 ∞ parts & (2.9) Z = γM + nγ tn−1η(t) dt 0 polar −1 = γ(M + ωn kψk1) .

The estimate for the integral I2 is easier Z Z I2 ≤ |f(x − y)|ψε(y) dy + |f(x)| ψε(y) dy = I21 + I22. |y|≥δ |y|≥δ Since Z Z ψε(y) dy = ψ(y) dy → 0 as ε → 0 |y|≥δ |y|≥δ/ε we have I22 → 0 as ε → 0. Z 1/p0 p0 I21 ≤ kfkp ψε (y) dy |y|≥δ Z 1/p0 p0/p = kfkp ψε(y)ψε (y) dy |y|≥δ 1/p Z 1/p0 ≤ kfkp sup ψε(y) ψε(y) dy |y|≥δ |y|≥δ

n 1/p Z 1/p0 −n δ δ ≤ kfkp δ η ψ(y) dy ε ε |y|≥δ/ε → 0 as ε → 0

7If η is decreasing, but not absolutely continuous, then the integration by parts in the estimates below is not allowed. To overcome this diﬃculty one has to use the Stielties integral which allows to integrate by parts functions that are of bounded variation. 34 PIOTR HAJLASZ

−1 by (2.9). Therefore for ε < ε0 the sum I1 + I2 is less than 2γ(M + ωn kψk1) which proves the theorem. 2

2.4. The Schwarz class and the Plancherel theorem. We say that n ∞ n 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 ∞ n −|x|2 than the inverse of any polynomial. Clearly C0 (R ) ⊂ Sn, but also e ∈ Sn so there are functions in the class S that have non compact support. {pα,β} is a countable family of norms in Sn and we can use it to deﬁne a topology in Sn. We say that a sequence (fk) converges to f in Sn if

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

n Recall that (τhf)(x) = f(h + x) for h ∈ R .

Proposition 2.29. 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. (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.

Exercise. Prove the proposition. HARMONIC ANALYSIS 35

Theorem 2.30. The Fourier transform is a continuous, one-to-one mapping of Sn onto Sn such that

(a) ∂f ∧ (ξ) = 2πiξjfˆ(ξ), ∂xj (b) ∂fˆ (−2πixjf)ˆ(ξ) = (ξ) , ∂ξj (c) (f ∗ g)ˆ = fˆg,ˆ (fg)ˆ = fˆ∗ gˆ , (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), first part of (c) and (d). The second part of (c) easily follows from its first part. 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) and hence α β α β |ξ D fˆ(ξ)| ≤ CkD (x f)k1 , ˆ α β pα,β(f) ≤ CkD (x f)k1 . An application of the Leibnitz rule 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 ≤ C(n) sup |(1 + |x|2)nxβi Dαi f(x)| < ∞ x∈Rn it follows that fˆ ∈ Sn. One can also easily deduce that the mapping

ˆ: Sn → Sn is continuous.

If f ∈ Sn than fˆ ∈ Sn and hence both f and fˆ 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 36 PIOTR HAJLASZ

Theorem 2.31 (Plancherel). The Fourier transform is an L2 isometry on 2 a dense subset Sn of L

kfˆk2 = kfk2, f ∈ Sn, and hence it uniquely extends to an isometry of L2 ˆ 2 n kfk2 = kfk2, f ∈ L (R ) . 2 n Moreover for f ∈ L (R ) Z fˆ(ξ) = lim f(x)e−2πix·ξ dx R→∞ |x|

¯ Proof. Given f ∈ Sn let g = fˆ, 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 2.30(d) gives Z Z Z Z ¯ kfk2 = ff¯ = fgˆ = fgˆ = fˆfˆ = kfˆk2 . Rn Rn Rn Rn 2 Thus the Fourier transform is an L isometry on Sn. Since Sn is a dense subset of L2 it uniquely extends to an isometry of L2. Now

2 1 L 2 L 3 fχB(0,R) −→ f for f ∈ L as R → ∞ and hence Z −2πix·ξ L2 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) |ξ|

2 Proof. Approximate f and g in L by functions in Sn, apply Theo- rem 2.30(d) and pass to the limit. 2

1 n 2 n Consider the class L (R ) + L (R ) consisting of functions of the form 1 2 f = f1 + f2, f1 ∈ L , f2 ∈ L . Then we deﬁne

fˆ = fˆ1 + fˆ2. 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 1 the representation f = f1 + f2. Indeed, if we also have f = g1 + g2, g1 ∈ L , 2 1 2 g2 ∈ L , then f1 − g1 = g2 − f2 ∈ L ∩ L and hence

fˆ1 − gˆ2 = (f1 − g1)ˆ = (g2 − f2)ˆ =g ˆ2 − fˆ2,

fˆ1 + fˆ2 =g ˆ1 +g ˆ2 . It is an easy exercise (Problem 8) to show that p n 1 n 2 n L (R ) ⊂ L (R ) + L (R ), for 1 ≤ p ≤ 2, p n and hence the Fourier transform is well deﬁned on L (R ), 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 p n ˆ: L (R ) → L (R ), for 1 ≤ p ≤ 2.

2.5. Tempered distributions.

0 Definition. The space Sn of all continuous linear functionals on Sn is called the space of tempered distributions. Here are examples:

p n 1. If f ∈ L (R ), 1 ≤ p ≤ ∞, then Z Lf (ϕ) = f(x)ϕ(x) dx Rn 0 defines a tempered distribution Lf ∈ Sn. 2. If µ is a measure of finite total variation, then Z Lµ(ϕ) = ϕ dµ Rn 0 defines a tempered distribution Lµ ∈ Sn. 3. We say that a function f is a tempered Lp function if f(x)(1 + |x|2)−k ∈ p n L (R ) for some nonnegative integer k. If p = ∞ we call f a slowly increasing function. Then Z Lf (ϕ) = f(x)ϕ(x) dµ Rn 38 PIOTR HAJLASZ

0 defines a tempered distribution Lf ∈ Sn for all 1 ≤ p ≤ ∞. 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. 5. α L(ϕ) = D ϕ(x0) 0 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]. 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 2.33. If Ω ⊂ R is open and f ∈ Lloc(Ω) satisfies Ω fϕ = 0 for ∞ all ϕ ∈ C0 (Ω), then f = 0 a.e.

∞ n Note that not every function f ∈ C (R ) 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 Rn sense.

Theorem 2.34. A linear functional 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 on Sn, so we are left with the proof of the converse implication. 0 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 ∈ S : p (ϕ) ≤ := N . n α,β m m |α|,|β|≤m

Suppose not. Then there is a sequence ϕk ∈ Sn such that |L(ϕk)| > 1 and X 1 (2.11) p (ϕ ) ≤ , k = 1, 2, 3,... α,β k k |α|,|β|≤k HARMONIC ANALYSIS 39

Note that (2.11) 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

n For any function g on R we deﬁneg ˜(x) = g(−x). Then it easily follows from the Fubini theorem that for u, ϕ, ψ ∈ Sn Z Z (2.12) (u ∗ ϕ)(x)ψ(x) dx = u(x)(ϕ ˜ ∗ ψ)(x) dx . Rn Rn Note that Z ψ 7→ (u ∗ ϕ)(x)ψ(x) dx, ψ ∈ Sn Rn and Z η 7→ u(x)η(x) dx, η ∈ Sn Rn are tempered distributions. We denote them by (u ∗ ϕ)[ψ] and u[η], so we can rewrite (2.12) as (u ∗ ϕ)[ψ] = u[ϕ ˜ ∗ ψ] 0 Note that if u ∈ Sn and ϕ ∈ Sn, then ψ 7→ u[ϕ ˜∗ψ] is a tempered distribution. 0 Definition. 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 result is left as an easy exercise. 0 Proposition 2.35. If u ∈ Sn, ϕ, ψ ∈ Sn, then (u ∗ ϕ) ∗ ψ = u ∗ (ϕ ∗ ψ) . 0 Theorem 2.36. If u ∈ Sn and ϕ ∈ Sn, then the convolution u ∗ ϕ is the n function f whose value at x ∈ R is

f(x) = u[τ−xϕ˜] = u[ϕ(x − ·)] . ∞ n Moreover f ∈ C (R ) and f and all its derivatives are slowly increasing.

∞ Proof. First we will prove that the function f(x) = u[τ−xϕ˜] is C and f as well as all its derivatives are slowly increasing. It follows from Theo- rem 2.29(b) that f is continuous. Observe that ϕ(he − ·) − ϕ(−·) ϕ˜(· − he ) − ϕ˜(·) k = k → −(∂ ϕ˜)(·) = (∂ ϕ)(−·) h h k k 40 PIOTR HAJLASZ

n in the topology of Sn by Theorem 2.29(d), so for every x ∈ R we have ϕ(x + he − ·) − ϕ(x − ·) ϕ(he − ·) − ϕ(−·) k = τ k h x h → τx((∂kϕ)(−·)) = (∂kϕ)(x − ·) in the topology of Sn. Hence f(x + he ) − f(x) ϕ(x + he − ·) − ϕ(x − ·) k = u k → u[(∂ ϕ)(x − ·)] . h h k Thus ∂kf(x) = u[(∂kϕ)(x − ·)] is continuous by Theorem 2.29(b). Since ∂kϕ ∈ Sn, the function on the right hand side is of the same type as the one used to define f, so we can differentiate it again. Iterating this process we obtain (2.13) Dαf(x) = u[(Dαϕ)(x − ·)] . ∞ n This implies f ∈ C (R ). Since u is a tempered distribution, Theorem 2.34 gives the estimate X 0 m (2.14) |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 Hence f is slowly increasing, and so it defines a tempered distribution. Since the derivatives of f satisfy (2.13) 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

(2.15) (u ∗ ϕ)[ψ] = f[ψ] for all ψ ∈ Sn. We have8 Z (u ∗ ϕ)[ψ] = u[ϕ ˜ ∗ ψ] = u ϕ˜(x − y)ψ(y) dy = ♥ . Rn For each x we approximate the last integral by Riemann sums. The Riemann sums as a function of x belong to Sn and it is easy to see that they converge toϕ ˜ ∗ ψ in the topology of Sn. Hence from the linearity and continuity of u we have Z Z ♥ = u[ϕ ˜(· − y)]ψ(y) dy = u[τ−yϕ˜]ψ(y) dy = f[ψ] Rn Rn which completes the proof.

8Valentine’s day is soon. HARMONIC ANALYSIS 41

We used the phrase “easy to see”. This is what most of the textbooks write as an explanation of the convergence of Riemann sums toϕ ˜ ∗ ψ, but it is actually not that easy. The arguments needed in the proof of this fact are elementary, but quite technical and very unpleasant to write down. This is why in the textbooks the authors try to escape from the problem by saying “easy to see”. To make the exposition self-contained we will explain this step with all details, but if we will need a similar argument in the future we will just say “easy to see”. First of all observe that is suﬃces to prove (2.15) under the assumption ∞ that ϕ, ψ ∈ C0 . Indeed, suppose we know (2.15) for compactly supported ∞ ∞ functions, but now ϕ, ψ ∈ Sn. Let C0 3 ϕk → ϕ in Sn and C0 3 ψ` → ψ in Sn. Observe that it follows from the proof of (2.14) that all the functions fk = u[τ−xϕ˜k] have a common estimate independent of k m |fk(x)| ≤ C(1 + |x|) 0 and hence fk → f in Sn. It is also clear from the deﬁnition of the convolution 0 that u ∗ ϕk → u ∗ ϕ in Sn, so

(u ∗ ϕ)[ψ`] = lim (u ∗ ϕk)[ψ`] = lim fk[ψ`] = f[ψ`] . k→∞ k→∞ Now letting ` → ∞ yields (2.15). To approximate the integral R η(y) dy, η ∈ C∞ by Riemann sums, we Rn 0 ﬁx a cube Q = [−N,N]n so large that supp η ⊂ Q and divide the cube into mk −k cubes {Qki}i=1 of 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 actually m Xk Z (2.16) |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 9 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

9 As ﬁnite linear combinations of functions from Sn. 42 PIOTR HAJLASZ

I1 → 0 as k → ∞ by (2.16) and Z I2 ≤ sup sup |ϕ˜(x − yki) − ϕ˜(x − y)| |ψ(y)| dy → 0 as k → ∞, n i y∈Qki R becauseϕ ˜ is uniformly continuous. Since

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) and hence 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(x) = u[τ−xϕ˜] is ∞ smooth, fψ ∈ C0 and hence Z f[ψ] = u[ϕ ˜(· − y)] ψ(y) dy Rn | {z } f(y) m Xk = lim u[ϕ ˜(· − yki)]ψ(yki)|Qki| k→∞ i=1 " m # Xk = lim u ϕ˜(· − yki)ψ(yki)|Qki| k→∞ i=1 = lim u[wk] = u[ϕ ˜ ∗ ψ] . k→∞ This time the proof is really complete. 2 Note that formula (2.13) implies. 0 Theorem 2.37. If u ∈ Sn and ϕ ∈ Sn, then for any multiindex α we have Dα(u ∗ ϕ)(x) = (u ∗ (Dαϕ))(x) . HARMONIC ANALYSIS 43

0 Definition. The space Sn is equipped with the weak-∗ convergence, (called 0 also weak convergence) i.e. we say that uk → u in Sn if

uk(ψ) → u(ψ) for every ψ ∈ Sn.

The following result seems obvious, but the proof requires the use of the Banach-Steinhaus theorem10 (Corollary 9.2 from Functional Analysis). 0 Theorem 2.38. If uk → u is Sn and ϕk → ϕ in Sn, then uk(ϕk) → u(ϕ).

We leave details as an exercise. This result together with Theorem 2.34 easily gives. 0 Theorem 2.39. If Lk → L in Sn, then there is a constant C > 0 and a positive integer m such that X sup |Lk(ϕ)| ≤ C pα,β(ϕ) for all ϕ ∈ Sn. k |α|,|β|≤m

0 ∞ n R Let u ∈ S . If ϕ ∈ C ( ), n ϕ = 1, then for every ψ ∈ Sn,ϕ ˜ε ∗ ψ → ψ n 0 R R in Sn (Why?), so

(u ∗ ϕε)[ψ] = u(ϕ ˜ε ∗ ψ) → u(ψ) . ∞ Since uε = u ∗ ϕε ∈ C we obtain a sequence of smooth functions that 0 ∞ n converge to u in 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 → ∞. Using this fact and Theorem 2.38 one can prove that ∞ wk(x) = η(x/k)(u ∗ ϕk−1 ) ∈ C0 0 converges to u in Sn. This gives ∞ n 0 0 Theorem 2.40. C0 (R ) is dense in Sn, i.e. if u ∈ Sn, then there is a ∞ n sequence wk ∈ C0 (R ) such that

wk[ψ] → u(ψ) for ψ ∈ Sn.

If ϕ, ψ ∈ Sn, then the integration by parts gives Z Z Dαϕ(x)ψ(x) dx = (−1)|α| ϕ(x)Dαψ(x) dx . Rn Rn n For h ∈ R we have Z Z (τhϕ)(x)ψ(x) dx = ϕ(x)(τ−hψ)(x) dx, Rn Rn

10At least I do not know how to avoid the Banach-Steinhaus theorem. 44 PIOTR HAJLASZ

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

0 α Definition. If u ∈ Sn, then the distributional partial derivative D u is a tempered distribution deﬁned by the formula Dα[ψ] = u[(−1)|α|Dαψ]. 0 The translation τhu ∈ Sn is deﬁned by

(τhu)[ψ] = u[τ−hψ]. 0 The reﬂectionu ˜ ∈ Sn is u˜[ψ] = u[ψ˜]. 0 The Fourier transformu ˆ ∈ Sn is uˆ[ψ] = u[ψˆ] . 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 Fourier transform on Sn is often denoted by Fu =u ˆ. 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.

1 n 2 n p n If f ∈ L (R ) + L (R ), in particular, if f ∈ L (R ), 1 ≤ p ≤ 2, then the classical Fourier transform coincides with the distributional one. That easily follows from Theorem 2.7(c) and Proposition 2.32. The basic properties of the Fourier transform, distributional derivative 0 and convolution in Sn are collected in the next result whose easy proof is left to the reader. HARMONIC ANALYSIS 45

0 Theorem 2.41. The Fourier transform in a homeomorphism of Sn onto 11 0 itself. Moreover for u ∈ Sn and ϕ ∈ Sn we have (a) (ˆu)ˆ =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.

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 2.37 shows that the distributional derivative of u ∗ ϕ coincides with the classical one.

0 Let us compare the notion of distributional derivative in Sn with the notion of weak derivative in Sobolev spaces.

n 1 Definition. Let Ω ⊂ R be an open, u, v ∈ Lloc(Ω) and let α be a multiin- α ∞ dex. We say that D u = v in the weak sense if for every ϕ ∈ C0 (Ω) Z Z vϕ = (−1)|α| vDαϕ . Ω Ω Lemma 2.33 implies that the weak derivative Dαu, if exists, is uniquely deﬁned. If u ∈ Cm(Ω), then for |α| ≤ m the the integration by parts gives Z Z α |α| α ∞ D vϕ = (−1) uD ϕ for all ϕ ∈ C0 (Ω), Ω Ω where Dαu is the classical partial derivative, so in this case the weak derivative coincides with the classical one. It is an easy exercise to prove that if f ∈ Lp is diﬀerentiable in Lp and p g = ∂f/∂kk is the partial derivative in the L norm, then actually g is also a weak partial derivative of f.

0 p If u ∈ Sn and there is a tempered L function g such that Z α |α| D u[ψ] = (−1) g(x)ψ(x) dx for ψ ∈ Sn. Rn then Lemma 2.33 shows that g is uniquely deﬁned and we can identify Dαu = g.

n Definition. Let 1 ≤ p ≤ ∞, m a positive integer and Ω ⊂ R an open set. The Sobolev space W m,p(Ω) is the set of functions u ∈ Lp(Ω) such that the weak partial derivatives of order less than or equal to m exist and belong to Lp(Ω). The space is equipped with the norm X α kukm,p = kD ukp. |α|≤m

11 0 With respect to the weak convergence in Sn. 46 PIOTR HAJLASZ

It is an easy exercise to prove Theorem 2.42. W m,p(Ω) is a Banach space.

To compare Sobolev spaces with tempered distributions note that ele- m,p n 0 ments of W (R ) belong to Sn and the distributional derivatives of orders 0 less than or equal m in the sense of Sn coincide with the weak derivatives in the sense used to define the Sobolev space. Note also that if f ∈ Lp has partial derivatives Dαf, |α| ≤ m in the Lp norm, then f ∈ W m,p and the Lp derivatives coincide with the weak derivatives. Using the notion of distributional derivative we can provide a new class 0 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 (2.17) u = aαD (Lfα ) ∈ Sn . |α|≤m 0 Surprisingly, every distribution in Sn can be represented in the form (2.17). We will prove it now, but the proof will require some preparations. First we will show how to solve, for each positive integer N, the partial differential equation (2.18) (I − ∆)N v = u , 0 where u ∈ Sn is a given tempered distribution. A direct application of (2.1) shows that if ψ ∈ Sn, then F(I − ∆)N ψ(ξ) = (1 + 4π2|ξ|2)N (Fψ)(ξ) and hence also (2.19) F −1(I − ∆)N ψ(ξ) = (1 + 4π2|ξ|2)N (F −1ψ)(ξ) . Therefore the operator (I − ∆)N can be represented as (2.20) (I − ∆)N = F −1(1 + 4π2|ξ|2)N F Observe that the function (1 + 4π2|ξ|2)−N and all its derivatives are slowly increasing, so we can multiply tempered distributions by that function. For 0 u ∈ Sn we define −1 2 2 −N 0 v = F (1 + 4π |ξ| ) (Fu) ∈ Sn .

We claim that v solves (2.18). Indeed for ψ ∈ Sn we have h i (I − ∆)N v[ψ] = u F(1 + 4π2|ξ|2)−N F −1(I − ∆)N ψ = u[ψ] by (2.19). HARMONIC ANALYSIS 47

In view of (2.20) it is natural to deﬁne (2.21) (I − ∆)−N = F −1(1 + 4π2|ξ|2)−N F , so v = (I − ∆)−N u is a solution to (2.18). The operator (I − ∆)−N is called a Bessel potential and one can prove that it can be represented as an integral operator.12

0 We proved that every distribution u ∈ Sn can be represented as (2.22) u = (I − ∆)N v , −N 0 where v = (I − ∆) u ∈ Sn. Now we will show that if N is sufficiently large, then v is actually a slowly increasing locally Lipschitz function, so (2.22) gives a representation of the form (2.17). We will also show that for k n any positive integer k we can find N so large that v ∈ C (R ). 0 Theorem 2.43. 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:

(a) If k = 0, then v is a slowly increasing function. (b) If k = 1, then v is a locally Lipschitz continuous slowly increasing function. (c) If k > 1, then v is a slowly increasing function of the class Ck−1.

Proof. We will need Lemma 2.44. If P (x) is a polynomial of degree p and N > (n + p)/2, then all derivatives of P (x) f(x) = (1 + |x|2)N 1 n belong to L (R ).

1 n Proof. Since 2N − p > n, f ∈ L (R ). We have ∂f Q(x) = 2 2N , deg Q = 2N + p − 1. ∂xi (1 + |x| )

12Later we will carefully study Bessel potentials. We will ﬁnd an explicit integral formula for the Bessel potential and we will show how to characterize Sobolev spaces in terms of Bessel potentials. 48 PIOTR HAJLASZ

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|x|2)−N F −1(Dγψ) . Hence X |Dγv[ψ]| ≤ C sup xαDβ F(1 + 4π2|x|2)−N F −1(Dγψ) x∈ n |α|,|β|≤m R β+γ X α x −1 = C Cα,β,γ sup F D 2 2 N F (ψ) x∈ n (1 + 4π |x| ) |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 2.44 gives xβ+γ Dαi ∈ L1( n) , (1 + 4π2|x|2)N R 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→ (−1)|γ|Dγv[ψ] 1 2 m/2 ∞ is bounded on L ((1 + |x| ) dx). Thus there are functions gγ ∈ L such that Z |γ| γ 2 m/2 (−1) D v[ψ] = ψ(x)gγ(x)(1 + |x| ) dx , Rn HARMONIC ANALYSIS 49 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). Moreover the distributional derivatives n of orders |γ| ≤ k are bounded on every bounded subset of R , so v belongs k,∞ n to the Sobolev space W on every bounded subset of R and the result follows from the following result that we state without proof. n Lemma 2.45. If Ω ⊂ R is a bounded domain with smooth boundary, then the Sobolev space W 1,∞(Ω) is the same as the space Lip (Ω) of Lipschitz functions on Ω. More precisely every Lipschitz function on Ω belongs to W 1,∞(Ω) and every function in W 1,∞(Ω) equals a.e. to a Lipschitz function. If k > 1, then W k,∞(Ω) ⊂ Ck−1(Ω) in the sense that every function in W k,∞ equals a.e. to a function in Ck−1(Ω).

The proof is complete. 2 We will investigate now properties of the Fourier transform of tempered distributions with compact support. n 0 Lemma 2.46. 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.

Proof. Let ϕ ∈ Sn, supp ϕ ⊂ Ω. Let η be a cut-oﬀ function, i.e. η ∈ ∞ n C0 (R ), 0 ≤ η ≤ 1, η(x) = 1 for |x| ≤ 1 and η(x) = 0 for |x| ≥ 2, then 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. 2 0 Definition. Let u ∈ Sn. The support of u (supp u) is the intersection of all n closed sets E ⊂ R such that ∞ n ϕ ∈ C0 (R \ E) ⇒ u(ϕ) = 0. 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 n deﬁnition by ϕ ∈ Sn with supp ϕ ⊂ R \ E. Before we state the next result we need some facts about analytic and holomorphic functions in several variables.

Definition. We say that a function f :Ω → C deﬁned in an open set n Ω ⊂ R is R-analytic, if in a neighborhood on any point x0 ∈ Ω it can be expanded as a convergent power series X α (2.23) f(x) = aα(x − x0) , aα ∈ C, α i.e. if in a neighborhood of any point f equals to its Taylor series. 50 PIOTR HAJLASZ

n We say that a function f :Ω → C deﬁned in an open set Ω ⊂ C 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

n n R has a natural embedding into C , just like R into C. n n R 3 x = (x1, . . . , xn) 7→ (x1 + i · 0, . . . , xn + i · 0) = x + i · 0 ∈ C . n It is easy to see that an R-analytic function f in Ω ⊂ R extends to a C- analytic function f˜ in an open set Ω˜ ⊂ C,Ω ⊂ Ω.˜ Namely, if f satisﬁes (2.23), we set X α f˜(z) = aα(z − z0) . α n On the other hand, if f˜ is C-analytic in Ω˜ ⊂ C , then the restriction f of f˜ n to Ω = Ω˜ ∩ R is R-analytic. n x·ξ z·ξ For example for any ξ ∈ R , f(x) = e is R-analytic and f˜(x) = e is its C-analytic extension. Definition. We say that a continuous function f :Ω → C deﬁned in an n open set Ω ⊂ C 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 2.47 (Cauchy). If f is holomorphic in13 n 1 1 D (w, r) = D (w1, r1) × ... × D (wn, rn) and continuous in the closure Dn(x, r), then 1 Z Z f(ξ) dξ . . . dξ (2.24) f(z) = ... 1 n n 1 1 (2πi) ∂D (w1,r1) ∂D (wn,rn) (ξ1 − z1) ... (ξn − zn) for all z ∈ Dn(w, r).

Proof. The function f is holomorphic in each variable separately, so (2.24) 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 (2.24) can be expanded as a power series and hence deﬁnes a C-analytic function. We proved. n Theorem 2.48. A function f is holomorphic in Ω ⊂ C if and only if it is C-analytic.

13Product of one dimensional discs. HARMONIC ANALYSIS 51

Now we can state an important result about compactly supported distributions. 0 Theorem 2.49. 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 n n R-analytic on R and has a holomorphic extension to C .

∞ n Proof. Let η ∈ C0 (R ) be such that η(x) = 1 in a neighborhood of supp u. 0 Then u = ηu in Sn and hence for ψ ∈ Sn we have Z uˆ[ψ] = (ηu)ˆ[ψ] = u η(·)ψ(x)e−2πix·(·) dx Rn Z h i = 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.14 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 2.34 that F and all its derivatives are slowly increasing. Thus we may identifyu ˆ with F , sou ˆ ∈ C∞.

n Moreover F has a holomorphic extension to C by the formula −2πiz·(·) F (z1, . . . , zn) = u η(·)e so in particular F (x1, . . . , xn) is R-analytic. 2 0 Remark. 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 since the right hand side does not depend on the choice of η ∈ C0 (R ) such that η = 1 in a neighborhood of supp u. 0 Theorem 2.50. If u ∈ Sn and supp u = {x0}, then there is an integer m and complex numbers aα, |α| ≤ m such that X α u = aαD δx0 . |α|≤m

14Compare with the proof of Theorem 2.36. 52 PIOTR HAJLASZ

Proof. Without loss of generality we may assume that x0 = 0. According to Theorem 2.34 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 (2.25) 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 (2.26) 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 (2.26) easily implies that

pα,β(ηεϕ) → 0 as ε → 0 15 for all |α|, |β| ≤ m. Note that ϕ − ηεϕ = 0 in a neighborhood of 0, so u(ϕ − ηεϕ) = 0. 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 (2.25).

Let now ψ ∈ Sn be arbitrary and let X Dαψ(0) h(x) = ψ(x) − xα . α! |α|≤m Clearly (2.27) 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 ∞ n equality (2.27) implies that ϕ = ηh ∈ C0 (R ) satisﬁes the assumptions of

15Check it! HARMONIC ANALYSIS 53

(2.25), so u(ηh) = 0. Hence h X Dαψ(0) i u(ψ) = u η(x) xα α! |α|≤m X η(x)xα = (−1)|α|u (−1)|α|Dαψ(0) . α! | {z } |α|≤m α | {z } D δ0(ψ) aα The proof is complete. 2 0 Corollary 2.51. Let u ∈ Sn. If suppu ˆ = {ξ0}, then u is a ﬁnite linear combination of functions (−2πix)αe2πix·ξ0 . In particular if suppu ˆ = {0}, then u is a polynomial.

We leave the proof as an exercise. 0 Corollary 2.52. 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 2.51. 2

3. Interpolation of operators

If 1 ≤ p < q ≤ ∞ and T is a linear operator such that kT fkp ≤ Akfkp and also kT fkq ≤ Bkfkq, then it turns out that for any p < r < q there is a constant C such that kT fkr ≤ Ckfkr. To be more precise we should clarify on what space T is defined. For example we can assume that T is defined on Lp + Lq or just on the class of all simple functions. The result seems very natural, but the proof is surprisingly difficult. Results of this type are called interpolation theorems. More generally interpolation theorems assume that an operator is bounded between some spaces and as a conclusion they give a larger class of spaces between which the operator is also bounded. The result stated above can be proved by the real variable methods, but such methods require a lot of estimates and hence the result does not give the sharp estimate for the norm of the operator T : Lr → Lr. To get the sharp estimate we need to use holomorphic functions. First we will prove an interpolation result (the Riesz-Thorin theorem) using holomorphic functions and this method is known as complex interpolation. Later we will prove another, more general result (the Marcinkiewicz theorem) using real methods. The Marcinkiewicz theorem is more general, but the estimate for the norm is not sharp. However, the Marcinkiewicz theorem being more general applies to the situations where the Riesz-Thorin theorem does not apply. 54 PIOTR HAJLASZ

3.1. Complex interpolation. In the following result we assume that all Lr spaces consist of complex valued functions. Theorem 3.1 (Riesz-Thorin). Let (X, µ) and (Y, ν) be two measure spaces. Let T be a linear operator deﬁned on the set of all simple functions on X and taking values in the set of measurable functions on Y . Let 1 ≤ p0, p1, q0, q1 ≤ ∞ and assume that

kT fkq0 ≤ M0kfkp0

kT fkq1 ≤ M1kfkp1 for all simple functions on X. Then for all 0 < θ < 1 we have 1−θ θ kT fkq ≤ M0 M1 kfkp for all simple functions f on X, where 1 1 − θ θ 1 1 − θ θ = + and = + . p p0 p1 q q0 q1 By density, T has a unique extension as a bounded operator from Lp(µ) to Lq(ν).

Before we prove the theorem we will show some applications. If p0 = q0 < p1 = q1, then 1/p = 1/q is a convex combination of 1/p0 and 1/p1, so p can p p be any number between p0 and p1, and hence T : L → L is bounded for any p0 < p < p1 as stated at the beginning of this section. Although the Young inequality, Theorem 2.3 has an elementary proof it can also be concluded from the Riesz-Thorin theorem. Proof of Thorem 2.3. Let g ∈ Lr, 1 ≤ r ≤ ∞. The operator T f = f ∗ g satisﬁes the estimate

kT fkr ≤ kgkrkfk1 (Theorem 2.2) and kT fk∞ ≤ kgkrkfkr0 (H¨older’sinequality). 0 Thus taking q0 = r, p0 = 1 and q1 = ∞, p1 = r we obtain from the Riesz-Thorin theorem 1−θ θ kf ∗ gkq = kT fkq ≤ kgkr kgkrkfkp = kgkrkfkp , where 1 1 − θ θ 1 1 − θ θ = + , = + , θ ∈ (0, 1) , p 1 r0 q r ∞ i.e. r r p = , q = , θ ∈ (0, 1) , r − θ 1 − θ so p ranges from 1 to r0, while q ranges from r to ∞ and q−1 = p−1 +r−1 −1. 2 p n ˆ 2 n n As we know, if f ∈ L (R ), 1 ≤ p ≤ 2, then f ∈ L (R ) + C0(R ), but we also have HARMONIC ANALYSIS 55

p n Theorem 3.2 (Hausdorﬀ-Young). If f ∈ L (R ), 1 ≤ p ≤ 2, then ˆ kfkp0 ≤ kfkp .

Proof. Since

kfˆk∞ ≤ kfk1, kfˆk2 ≤ kfk2 , then the Riesz-Thorin theorem gives 1−θ θ kfˆkq ≤ 1 1 kfkp = kfkp , where 1 1 − θ θ 1 1 − θ θ = + , = + , θ ∈ (0, 1) , q ∞ 2 p 1 2 i.e. 2 2 q = , p = , θ ∈ (0, 1) , θ 2 − θ so p−1 + q−1 = 1 and p is any exponent between 1 and 2. 2 Remark. As we know (Problem 26), if p > 2, then there is a function f ∈ Lp ˆ 0 such that the distributional Fourier transform f ∈ Sn is not a function. Proof of Theorem 3.1. Let m X iαk f = ake χAk k=1 be a simple function, where ak > 0, αk ∈ R and Ak are pairwise disjoint subsets in X of ﬁnite measure. We want to estimate Z

kT fkq = sup (T f)(x)g(x) dν(x) , Y where the supremum is over all simple functions n X iβ j 0 g = bje χBj with kgkq ≤ 1 , j=1 where bj > 0, βj ∈ R and Bj are pairwise disjoint subsets of Y of ﬁnite measure.

For z ∈ C let m n X P (z) iαk X Q(z) iβj fz = ak e χAk , gz = bj e χBj , k=1 j=1 where p p q0 q0 P (z) = (1 − z) + z, Q(z) = 0 (1 − z) + 0 z . p0 p1 q0 q1 56 PIOTR HAJLASZ

P (z) Q(z) P (z) and Q(z) are linear functions of z and hence the functions ak , bj are entire since ak, bj > 0. Observe that P (θ) = Q(θ) = 1, so fθ = f, gθ = g. Deﬁne Z F (z) = (T fz)(x)gz(x) dν(x) . Y In particular Z F (θ) = (T f)g dν . Y R Thus we extended the integral Y (T f)g dν to a family of integrals depending on the parameter z ∈ C whose values form an entire function of z. Indeed, from the linearity of T we have m n Z X X P (z) Q(z) iαk iβj F (z) = ak bj e e T (χAk ) dν . k=1 j=1 Bj The integrals in the last formula are ﬁnite, so F (z) is an entire function. Let us now restrict z to be in the closed strip

S = {z ∈ C : 0 ≤ re z ≤ 1} . It easily follows from the formula that deﬁnes F (z) that |F (z)| is bounded on S. Now we will obtain a more precise estimate for |F (z)| on the boundary on that strip. If re z = 0, then m m p0 X P (z) p0 X p p kfzkp0 = |ak | µ(Ak) = akµ(Ak) = kfkp , k=1 k=1

P (z) re P (z) p/p0 because the sets Ak are pairwise disjoint and |ak | = ak = ak . Similarly 0 0 q0 q kgzk 0 = kgk 0 . q0 q Now H¨older’sinequality and the assumptions about T give

0 0 p/p0 q /q0 |F (z)| ≤ k(T f )k kg k 0 ≤ M kf k kg k 0 = M kfk kgk 0 . z q0 z q0 0 z p0 z q0 0 p q If rez = 1, then the same argument gives

0 0 p1 p q1 q kfzkp = kfkp, kgzk 0 = kgk 0 1 q1 q and hence 0 0 p/p1 q /q1 |F (z)| ≤ M1kfkp kgkq0 . The entire function F (z) is bounded on S and we can use arguments from the theory of holomorphic functions to estimate |F (z)| inside the strip by the estimates on the boundary. We need HARMONIC ANALYSIS 57

Lemma 3.3 (The Hadamard three lines lemma). Let F be a holomorphic function in the open strip S = {z ∈ C : 0 < re z < 1} , continuous and bounded on the closure S. If |F (z)| ≤ B0 when re z = 0 and |F (z)| ≤ B1 when re z = 1, then 1−re z re z |F (z)| ≤ B0 B1 for z ∈ S.

Before we prove the lemma we show how to use it to complete the proof of the Riesz-Thorin theorem. If re z = θ, the estimate for F (z) on the boundary of S and the three lines lemma give 0 0 1−θ 0 0 θ p/p0 q /q0 p/p1 q /q1 |F (z)| ≤ M0kfkp kgkq0 M1kfkp kgkq0 1−θ θ = M0 M1 kfkpkgkq0 . In particular if z = θ, then Z 1−θ θ (T f)g dν = |F (θ)| ≤ M0 M1 kfkpkgkq0 Y and the result follows upon taking supremum over all simple functions g with kgkq0 ≤ 1. Proof of Lemma 3.3. Deﬁne the function 1−z z −1 G(z) = F (z)(B0 B1 ) . Note that |G(z)| ≤ 1 on the boundary of S, i.e. if re z = 0 or re z = 1. It suﬃces to prove that |G(z)| ≤ 1 for all z ∈ S.

(z2−1)/n Consider auxiliary functions Gn(z) = G(z)e that will help us estimate G(z).

Thus Gn(x + iy) converges uniformly to 0 in 0 ≤ x ≤ 1 as |y| → ∞. Let z ∈ S. By the uniform convergence to 0 there is y0 > |im z| such that

|Gn(x ± iy0)| ≤ 1 for all x ∈ [0, 1].

Since |G| is bounded by 1 on the boundary of the strip, also |Gn| is bounded by 1 on that boundary. Thus |Gn| is bounded by 1 on the boundary of the rectangle [0, 1] × [−y0, y0], so |Gn| ≤ 1 in the entire rectangle by the maximum principle. In particular |Gn(z)| ≤ 1. Letting n → ∞ we conclude that |G(z)| ≤ 1. 2 This completes the proof of the Riesz-Thorin theorem. 2 58 PIOTR HAJLASZ

3.2. Real methods. If f ∈ Lp(µ), 0 0 p p t µ({x : |f(x)| > t}) ≤ kfkp . Hence 1/p sup t µ({x : |f(x)| > t}) ≤ kfkp . t>0 This suggests the following deﬁnition. Definition. Let (X, µ) be a measure space and 0 t}) < ∞ . t>0 ∞,∞ ∞ We also identify L = L with kfk∞,∞ = kfk∞. The previous argument gives p p,∞ Proposition 3.4. For 0 < p ≤ ∞, L (µ) ⊂ L (µ) and kfkp ≤ kfkp,∞.

The inclusion Lp ⊂ Lp,∞ is strict for 0 < p < ∞. For example |x|−1 ∈ 1,∞ −1 1 L (R), but |x| 6∈ L (R). p,∞ n p n Exercise. For any 0 < p < ∞ ﬁnd f ∈ L (R ) \ L (R ). Note that kkfkp,∞ = |k|kfkp,∞ for k ∈ C but kf + gkp,∞ ≤ Cp(kfkp,∞ + kgkp,∞) 1/p p,∞ where Cp = max(2, 2 ), so L is not a normed space, but a quasi-normed linear space when 0 < p < ∞.

p,∞ We say that fn → f in L if kf − fnkp,∞ → 0. Definition. 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. We say that a subadditive operator T : Lp(µ) → Lq(ν) is of strong type (p, q) if it is bounded, i.e. kT fkq ≤ Ckfkp for some C > 0 and all f ∈ Lp(µ) and it is of weak type (p, q) if

kT fkq,∞ ≤ Ckfkp for some C > 0 and all f ∈ Lp. If q = ∞, then weak type (p, ∞) is the same as the strong type (p, ∞), because L∞,∞ = L∞.

16 p p p The Marcinkiewicz space is also called the weak L and denoted by weak-L or Lw. HARMONIC ANALYSIS 59

Equivalently T is of weak type (p, q) if for all t > 0 Ckfk q µ({x : |T f(x)| > t}) ≤ p . t It follows immediately from Proposition 3.4 that operators of strong type are also of weak type. It is important to observe that the Lp norm of a function f can be com- puted if we know measures of the level sets {x : |f(x)| > t}. Namely we have Theorem 3.5 (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. 2 Corollary 3.6. If µ is a σ-ﬁnite measure on X and 0 t}) dt . X 0

While the space Lp,∞ is larger than Lp the diﬀerence is not big, because we have Theorem 3.7 (Kolmogorov). If µ(X) < ∞, then Lp,∞(µ) ⊂ Lq(µ) for all 0 < q < p. Moreover q 1/p kfk ≤ 21/q µ(X)1/q−1/pkfk . q p − q p,∞

Proof. Fix t0 > 0. Using Corollary 3.6 and the estimates µ({|f| > t}) ≤ p −p µ(X) for 0 < t ≤ t0 and µ({|f| > t}) ≤ kfkp,∞t for t > t0 we get Z Z t0 Z ∞ q q−1 p q−p−1 |f| dµ ≤ q t µ(X) dt + kfkp,∞ t dt X 0 t0 q = tqµ(X) + tq−pkfkp . 0 p − q 0 p,∞ 17 1/p −1/p Then the result follows by choosing t0 = (q/(p − q)) µ(X) kfkp,∞. 2

17This is a general trick. The right hand side is a sum of two expressions depending on t0 and the inequality is true for any t0, so we minimize the right hand side over t0. Equivalently, we choose t0 such that both summands on the right hand side are equal to each other. 60 PIOTR HAJLASZ

Theorem 3.8 (Marcinkiewicz). Let (X, µ) and (Y, ν) be two measure spaces with σ-ﬁnite measures. Let 0 < p0 ≤ q0 ≤ ∞, 0 < p1 ≤ q1 ≤ ∞ and q0 6= q1. Let T be a subadditive operator deﬁned on the space Lp0 (µ) + Lp1 (µ) taking values into the space of measurable functions on Y . Assume that

p0 kT fkq0,∞ ≤ M0kfkp0 for f ∈ L (µ).

p1 kT fkq1,∞ ≤ M1kfkp1 for f ∈ L (µ). Then for any 0 < θ < 1 there is a constant M depending on M0, M1, p0, p1, q0, q1 and θ only such that p kT fkq ≤ Mkfkp for f ∈ L (µ) , where 1 1 − θ θ 1 1 − θ θ = + , = + . p p0 p1 q q0 q1

Although the Marcinkiewicz theorem seems very similar to the Riesz- Thorin one, the two results are very different. Here the Lr spaces can consist of real valued functions, but in the Riesz-Thorin theorem they must be complex valued, because of the use of holomorphic functions. The price we pay for this is that the constant M is not as good as the one in the Riesz- Thorin theorem. Moreover we assume now that p0 ≤ q0 and p1 ≤ q1 and there was no such requirement in the previous interpolation result. On the other hand we allow exponents to be just greater than 0 and not greater than or equal to 1 and the operator needs only to be subadditive. However, the main difference which makes the Marcinkiewicz theorem so powerful is that we require the operator to be of weak type, while in the Riesz-Thorin theorem the operator had to be of strong type. Later we will see that in many situations it is possible to verify the weak type, so we can apply the Maricnkiewicz theorem, while in such situations the Riesz-Thorin theorem is useless. We will prove only a special case of the Marcinkiewicz theorem, the case which is the most important in its applications. Theorem 3.9 (Marcinkiewicz). Let (X, µ) and (Y, ν) be two measure spaces with σ-finite measures. Let T be a subadditive operator defined on Lp0 (µ) + p L 1 (µ), where 0 < p0 < p1 ≤ ∞, and taking values into measurable functions on Y . Assume that there are constants M0,M1 > 0 such that

p0 kT fkp0,∞ ≤ M0kfkp0 for f ∈ L (µ)

p1 kT fkp1,∞ ≤ M1kfkp1 for f ∈ L (µ). Then for any p0 < p < p1 p kT fkp ≤ Mkfkp for f ∈ L (µ), where 1/p p p 1−θ θ M = 2 + M0 M1 p − p0 p1 − p HARMONIC ANALYSIS 61 if 1 1 − θ θ = + , 0 < θ < 1 . p p0 p1

p Proof. Let f ∈ L (µ) and t > 0. We decompose f as f = f0 + f1, where

f0 = fχ{|f|>ct} f1 = fχ{|f|≤ct} . p The constant c will be chosen later. It is easy to see that f0 ∈ L 0 (µ) and p f1 ∈ L 1 (µ). We have

|T f(x)| ≤ |T f0(x)| + |T f1(x)| , so {|T f| > t} ⊂ {|T f0| > t/2} ∪ {|T f1| > t/2} and hence

ν({|T f| > t}) ≤ ν({|T f0| > t/2}) + ν({|T f1| > t/2}) . We will split the proof into two cases.

−1 Case 1: p1 = ∞. Choose c = (2M1) . We have t kT f k ≤ M kf k ≤ M ct = , 1 ∞ 1 1 ∞ 1 2 so ν({|T f1| > t/2}) = 0 .

By the weak (p0, p0) type we have 2M p0 ν({|T f | > t/2}) ≤ 0 kf k 0 t 0 p0 and hence Z ∞ p p−1 kT fkp = p t ν {|T f| > t} dt 0 Z ∞ p0 p−p0−1 p0 ≤ p(2M0) t kf0kp0 dt 0 Z ∞ Z p0 p−p0−1 p0 = p(2M0) t |f(x)| dµ(x) dt 0 {x:|f(x)|>ct} Z Z |f(x)|/c p0 p0 p−p0−1 = p(2M0) |f(x)| t dt dµ(x) X 0 p Z |f(x)|p−p0 p0 p0 = (2M0) |f(x)| dµ(x) p − p0 X c

p p p0 p−p0 p = 2 M0 M1 kfkp . p − p0 If 1 1 − θ θ = + , 0 < θ < 1 , p p0 ∞ 62 PIOTR HAJLASZ then the above estimate reads as 1/p p 1−θ θ kT fkp ≤ 2 M0 M1 kfkp . p − p0

Case 2: p1 < ∞. Now we have two inequalities arising from the weak type estimates 2M p0 ν({|T f | > t/2}) ≤ 0 kf k , 0 t 0 p0 2M p1 ν({|T f | > t/2}) ≤ 1 kf k , 1 t 1 p1 By an argument similar to the one used in Case 1 we have Z ∞ p p−p0−1 p0 p0 kT fkp ≤ p t (2M0) kf0kp0 dt 0 Z ∞ p−p1−1 p1 p1 + p t (2M1) kf1kp1 dt 0 p p p0 p0−p p1 p1−p p = (2M0) c + (2M1) c kfkp . p − p0 p − p1 If we choose now c in a way that

p0 p0 p1 p1 (2M0) c = (2M1) c the desired estimate will follow. 2

3.3. The Hardy-Littlewood maximal function. As an immediate application of the Marcinkiewicz theorem we will prove the Hardy-Littlewood 1 n maximal theorem. 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.

p n Theorem 3.10. If f ∈ L (R ), 1 ≤ p ≤ ∞, then Mf < ∞ a.e. Moreover

1 n (a) The operator M is of weak type (1, 1), i.e. for f ∈ L (R ) 5n Z (3.1) |{x : Mf(x) > t}| ≤ |f| for all t > 0. t Rn p n p n (b) If f ∈ L (R ), 1 < p ≤ ∞, then Mf ∈ L (R ) and p 1/p kMfk ≤ 2 · 5n/p kfk . p p − 1 p HARMONIC ANALYSIS 63

Proof. It immediately follows from the deﬁnition of the maximal function that kMfk∞ ≤ kfk∞ . Once we prove that the operator is on weak type (1, 1), i.e.18 n kMfk1,∞ ≤ 5 kfk1 , the boundedness of M on Lp, i.e. p 1/p kMfk ≤ 2 · 5n/p kfk p p − 1 p will follow from the Marcinkiewicz theorem. Thus we are left with the proof of the inequality (3.1). To this end we need an important covering lemma. Theorem 3.11 (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 If the metric space is separable, then the family B0 is countable and we can 0 ∞ arrange it as a sequence B = {Bi}i=1, so ∞ [ [ B ⊂ 5Bi . B∈B i=1

Remark. Here B can be either a family of open balls or closed balls. In both cases proof is the same. Proof. Let sup{diam B : B ∈ B} = R < ∞. Divide the family B according to the diameter of the balls R R F = {B ∈ B : < diam B ≤ } . j 2j 2j−1 S∞ Clearly B = j=1 Fj. Define B1 ⊂ F1 to be the maximal family of pairwise disjoint balls. Suppose the families B1,..., Bj−1 are already defined. Then we define Bj to be the maximal family of pairwise disjoint balls in j−1 0 0 [ Fj ∩ {B : B ∩ B = ∅ for all B ∈ Bi} . i=1 0 S∞ Next we define B = j=1 Bj. Observe that every ball B ∈ Fj intersects Sj Sj with a ball in i=1 Bj. Suppose that B ∩ B1 6= ∅, B1 ∈ i=1 Bi. Then R R diam B ≤ = 2 · ≤ 2 diam B 2j−1 2j 1 and hence B ⊂ 5B1. 2

18This inequality is equivalent to (3.1). 64 PIOTR HAJLASZ

1 n Let f ∈ L (R ) and Et = {x : Mf(x) > t}. For x ∈ Et, there is rx > 0 such that Z |f| > t , B(x,rx) so Z −1 |B(x, rx)| < t |f| . B(x,rx) 1 n Observe that supx∈Et rx < ∞, because f ∈ L (R ). The family of balls

3.4. Theorem 2.28 revisited. The following result is related to Theo- rem 2.28

1 n Theorem 3.12. Suppose that ϕ ∈ L (R ) 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

(3.2) sup(f ∗ ϕε)(x) ≤ kΨk1Mf(x) . ε>0

Proof. The proof is similar to that of Theorem 2.28 and also now we will prove the result under the additional assumption that η is absolutely continuous. As in the proof of Theorem 2.28 we conclude from the integrability of Ψ that (3.3) lim rnη(r) = 0, lim rnη(r) = 0 . r→0 r→∞ Since both sides of the inequality (3.2) commute with translations we can assume that x = 0. We can also assume that Mf(0) < ∞ as otherwise the inequality is obvious. We have Z |(f ∗ ϕε)(0)| ≤ |f(y)|Ψε(y) dy Rn Z ∞ Z = sn−1 |f(sθ)| dσ(θ) ε−nη(s/ε) ds = ♥ 0 Sn−1 HARMONIC ANALYSIS 65

Let Z g(s) = |f(sθ)| dσ(θ) Sn−1 and Z r Z G(r) = sn−1g(s) ds = |f(y)| dy . 0 B(0,r) Clearly Z n n G(r) = ωnr |f(y)| dy ≤ ωnr Mf(0) . B(0,r) We have Z ∞ ♥ = G0(s)ε−nη(s/ε) ds 0 R Z R (3.4) = lim G(s)ε−nη(s/ε) − lim G(s)ε−n−1η0(s/ε) ds = ♦ R→∞ r R→∞ r→0 r→0 r Since −n n G(s)ε η(s/ε) ≤ ωnMf(0)(s/ε) η(s/ε) , and the right hand converges to 0 as s → 0 or s → ∞ by (3.3), the ﬁst limit at (3.4) equals 0 and hence Z R ♥ = − lim G(s)ε−n−1η0(s/ε) ds R→∞ r→0 r Z R/ε = − lim G(sε)ε−nη0(s) ds R→∞ r→0 r/ε Z R/ε n 0 ≤ − lim ωnMf(0) s η (s) ds R→∞ r→0 r/ε Z ∞ parts & (3.3) n−1 = nωnMf(0) s η(s) ds 0 = Mf(0)kΨk1 .

The last equality follows from the fact that nωn equals to the (n − 1) dimensional measure of the sphere Sn−1(0, 1) and the integration in spherical coordinates. 2

4. Translation invariant operators

p n q n Definition. We say that a bounded operator T : L (R ) → L (R ), 1 ≤ p, q ≤ ∞ commutes with translations if p n n T (τhf) = τhT (f) for f ∈ L (R ) and h ∈ R . p,q n p,q n The set of all such operators is denoted by M (R ). Clearly M (R ) is a normed linear space as a subspace of all bounded operators B(Lp,Lq). 66 PIOTR HAJLASZ

p,q n Theorem 4.1. Suppose T ∈ M (R ), 1 ≤ p, q ≤ ∞. Then there is a 0 tempered distribution u ∈ Sn such that

T ϕ = u ∗ ϕ for ϕ ∈ Sn.

Proof. We will need two lemmas. p,q n Lemma 4.2. If T ∈ M (R ), 1 ≤ p, q ≤ ∞ and ϕ ∈ Sn, then for all multiindices α, the distributional derivatives Dα(T ϕ) belong to Lq and Dα(T ϕ) = T (Dαϕ) . n+1,q n Lemma 4.3 (Sobolev). If f ∈ W (R ), 1 ≤ q ≤ ∞, then f equals a.e. to a continuous function (still denoted by f) such that

|f(0)| ≤ C(n, q)kfkn+1,q .

Assuming for a moment the two lemmas we will show how to prove the p,q n theorem. Let T ∈ M (R ), 1 ≤ p, q ≤ ∞ and ϕ ∈ Sn. According to m,q n n+1,q n Lemma 4.2, T ϕ ∈ W (R ) for all m. In particular T ϕ ∈ W (R ), so T ϕ is a continuous function and the Sobolev lemma gives X β |(T ϕ)(0)| ≤ C(n, q) kD (T ϕ)kq |β|≤n+1 X β = C(n, q) kT (D ϕ)kq |β|≤n+1 X β ≤ C(n, q)kT kMp,q kD ϕkp . |β|≤n+1 We have Z 1/p β (n+1)p −(n+1)p β p kD ϕkp = (1 + |x|) (1 + |x|) |D ϕ(x)| dx Rn Z 1/p ≤ sup (1 + |x|)n+1|Dβϕ(x)| (1 + |x|)−(n+1)p dx x∈Rn Rn | {z } <∞ X ≤ C pα,β(ϕ) , |α|≤n+1 so X |(T ϕ)(0)| ≤ C pα,β(ϕ) . |α|,|β|≤n+1 Thus v(ϕ) = (T ϕ)(0) 0 deﬁnes a tempered distribution v ∈ Sn. Let u =v ˜. We have

(u ∗ ϕ)(x) = (˜v ∗ ϕ)(x) =v ˜[τ−xϕ˜] = v[τxϕ] = T (τxϕ) (0) = τx(T ϕ) (0) = (T ϕ)(x) . HARMONIC ANALYSIS 67

We are left with the proofs of Lemmas 4.2 and 4.3.

q Proof of Lemma 4.2. Let ϕ, ψ ∈ Sn. T ϕ ∈ L , so the distributional deriv- 0 ative ∂j(T ϕ) ∈ Sn is well deﬁned. Since ϕ(x) − ϕ(x − he ) j → ∂ ϕ(x) as h → 0 h j in the topology of Sn we have Z ∂j(T ϕ)[ψ] = − (T ϕ)(x)∂jψ(x) dx Rn Z ψ(x + he ) − ψ(x) = − lim (T ϕ)(x) j dx h→0 n R h Z T ϕ(x) − (T ϕ)(x − he ) = lim j ψ(x) dx h→0 n R h Z T ϕ(x) − (τ−he T )ϕ(x) = lim j ψ(x) dx h→0 n R h Z T ϕ(x) − T (τ−he ϕ)(x) = lim j ψ(x) dx h→0 n R h Z ϕ(·) − ϕ(· − he ) = lim T j (x) ψ(x) dx h→0 n R h Z = T (∂jϕ)(x)ψ(x) dx . Rn Thus

∂j(T ϕ) = T (∂jϕ) and by induction we have α α q n D (T ϕ) = T (D ϕ) ∈ L (R ) . The proof is complete. 2

∞ n 19 Proof of Lemma 4.3. Let η ∈ C0 (R ) be a cut-oﬀ function and let 1 ηR(x) = η(x/R). Then ηRf ∈ L . Since the weak derivatives satisfy the Leibniz rule α! α X βi γi (4.1) D (ηRf) = D ηRD f βi!γi! βi+γi=α

βi ∞ n+1,1 n and D ηR ∈ C0 we conclude that ηRf ∈ W (R ). The elementary inequality X 1 ≤ C(n)(1 + |x|)−(n+1) |(−2πix)α| |α|≤n+1

190 ≤ η ≤ 1, η(x) = 1 if |x| ≤ 1, η(x) = 0 if |x| ≥ 2. 68 PIOTR HAJLASZ gives −(n+1) X α |(ηRf)ˆ(x)| ≤ C(n)(1 + |x|) (−2πix) (ηRf)ˆ(x) |α|≤n+1 −(n+1) X α ≤ C(n)(1 + |x|) (D ηRf)ˆ ∞ |α|≤n+1 −(n+1) X α ≤ C(n)(1 + |x|) kD (ηRf)k1 |α|≤n+1 −(n+1) X α ≤ C(n, R)(1 + |x|) kD fkq , |α|≤n+1 where in the last inequality we applied (4.1) and H¨older’sinequality. Inte- grating both sides with respect to x we have

k(ηRf)ˆk1 ≤ C(n, R)kfkn+1,q . 1 1 Since ηRf ∈ L and (ηRf)ˆ ∈ L we conclude that ηRf ∈ C0. Since f(x) = n (ηRf)(x) for |x| ≤ R continuity of f on R follows. Moreover the inversion formula gives Z

|f(0)| = |(ηRf)(0)| = (ηRf)ˆ ≤ C(n, R)kfkn+1,q . Rn This completes the proof of Lemma 4.3 and hence that of Theorem 4.1. 2 If 1 ≤ p, q, r ≤ ∞ and q−1 = p−1 + r−1 − 1, then according to Young’s inequality (Theorem 2.3),

kf ∗ gkq ≤ kfkpkgkr , so T f = f ∗g deﬁnes an operator in Mp,q. Observe that q ≥ p and if r ranges from r = 1 to r = p0 = p/(p − 1), then q can be any exponent p ≤ q ≤ ∞. The fact that q has to be greater or equal to p is a very general one. p,q n Theorem 4.4 (H¨ormander). If 1 ≤ q < p < ∞, then M (R ) = {0}.

Proof. We will need the following result which is interesting on its own. p n Lemma 4.5. If f ∈ L (R ), 1 ≤ p < ∞, then 1/p lim kτhf + fkp = 2 kfkp . |h|→∞

∞ n 20 Proof. Let ϕ ∈ C0 (R ) be such that kf − ϕkp < ε. Then

kτhf + fkp − kτhϕ + ϕkp ≤ kτh(f − ϕ) + (f − ϕ)kp < 2ε .

If |h| is large enough, then the supports of ϕ and τhϕ are disjoint and hence 1/p kτhϕ + ϕkp = 2 kϕkp .

20 kxk − kyk ≤ kx − yk. HARMONIC ANALYSIS 69

Thus 1/p lim sup kτhf + fkp − 2 kϕkp ≤ 2ε . |h|→∞ Since ε > 0 can be arbitrary the lemma follows. 2

p n Now we can complete the proof of the theorem. For f ∈ L (R ) we have

kτh(T (f)) + T (f)kq = kT (τhf + f)kq ≤ kT kMp,q kτhf + fkp . Letting |h| → ∞ the lemma yields 1/q 1/p 2 kT (f)kq ≤ kT kMp,q 2 kfkp , so 1/p−1/q kT kMp,q ≤ kT kMp,q 2 which implies T = 0. 2 The same argument shows that the only translation invariant operator n q n T : C0(R ) → L (R ), 1 ≤ q < ∞ is the zero operator. p,q n Theorem 4.6. Let 1 ≤ p ≤ q ≤ ∞, p < ∞, q > 1 and T ∈ M (R ). Then 0 0 T : Lp ∩ Lq → Lp is a bounded operator, so it uniquely extends to T ∈ Mq0,p0 . Moreover

kT kMq0,p0 = kT kMp,q . In other words, we have isometric identiﬁcation p,q n q0,p0 n M (R ) = M (R ) .

Proof. Let g ∈ Lq0 . Then Z Lp 3 f 7→ T (f)g Rn is a bounded linear functional on Lp, so there is a unique function T ∗g ∈ Lp0 such that Z Z T (f)g = fT ∗g . Rn Rn In other words T ∗ : Lq0 → Lp0 is the adjoint operator. It is easy to see that T ∗ is translation invariant, so T ∗ ∈ Mq0,p0 and thus ∗ kT kMq0,p0 = kT kMp,q as the norm of the adjoint operator equals to the norm of the operator. Indeed, Z ∗ ∗ ∗ kT kMq0,p0 = sup kT gkp0 = sup sup fT g n kgkq0 ≤1 kgkq0 ≤1 kfkp≤1 R Z = sup sup T (f)g = kT kMp,q . n kfkp≤1 kgkq0 ≤1 R 70 PIOTR HAJLASZ

0 Let u ∈ Sn be such that

T ϕ = u ∗ ϕ for ϕ ∈ Sn.

For ϕ, ψ ∈ Sn we have (T ∗ϕ)[ψ] = (T ψ)[ϕ] = (u ∗ ψ)[ϕ] = u[ψ˜ ∗ ϕ] =u ˜[ψ ∗ ϕ˜] =u ˜[ϕ ˜ ∗ ψ] = (˜u ∗ ϕ)[ψ] , i.e. ∗ T ϕ =u ˜ ∗ ϕ for ϕ ∈ Sn. Since T ∗ϕ =u ˜ ∗ ϕ = (u ∗ ϕ˜)˜= (T ϕ˜)˜ we see that T ∗ ∈ Mq0,p0 implies that T ∈ Mq0,p0 and ∗ kT kMq0,p0 = kT kMq0,p0 = kT kMp,q . Thus Mp,q ⊂ Mq0,p0 isometrically, but the same argument applied to Mq0,p0 in place of Mp,q gives the opposite inclusion, so Mp,q = Mq0,p0 isometrically. 2 1,1 n Theorem 4.7. T ∈ M (R ) if and only if Z 1 n (T f)(x) = (f ∗ µ)(x) = f(x − y) dµ(y), f ∈ L (R ) Rn n for some complex-valued measure of ﬁnite total variation µ ∈ B(R ). More- over kT kM1,1 = kµk .

n Proof. If µ ∈ B(R ), then T f = f ∗ µ is a translation invariant operator and

(4.2) kT kM1,1 ≤ kµk 1,1 0 by Theorem 2.5. Suppose now that T ∈ M . Then there is u ∈ Sn such that T ψ = u ∗ ψ for ψ ∈ Sn. R −n Let ϕ ∈ Sn, ϕ ≥ 0, n ϕ = 1 and ϕε(x) = ε ϕ(x/ε). It follows from R Problem 14 that for ψ ∈ Sn,ϕ ˜ε ∗ ψ → ψ in the topology of Sn as ε → 0. 1 Since kϕεk1 = 1, the family {ϕε}ε is bounded in L , so the family 1 n n {u ∗ ϕε}ε = {T ϕε}ε ⊂ L (R ) ⊂ B(R ) is also bounded. According to the Riesz representation theorem (Theo- n n rem 2.4), the space B(R ) is dual to the separable Banach space C0(R ), and hence it follows from the separable case of the Banach-Alaoglu theorem21 that there is a weakly-∗ convergent subsequence ∗ n u ∗ ϕεk * µ ∈ B(R ) as k → ∞,

21Theorem 14.6 in Functional Analysis. HARMONIC ANALYSIS 71

n i.e. for every g ∈ C0(R ) Z Z

(4.3) (u ∗ ϕεk )(x)g(x) dx → g(x) dµ(x) . Rn Rn We claim that u = µ. Indeed, for g = ψ ∈ Sn (4.3) means that

(u ∗ ϕεk )[ψ] = u[ϕ ˜εk ∗ ψ] → µ[ψ] as k → ∞.

Sinceϕ ˜εk ∗ ψ → ψ in Sn, continuity of u gives

u[ψ] = µ[ψ] for ψ ∈ Sn, n i.e. u = µ. For g ∈ C0(R ), (4.3) yields Z g dµ ≤ kgk sup kT ϕ k ≤ kgk kT k 1,1 . ∞ εk 1 ∞ M Rn k n Now taking the supremum over g ∈ C0(R ), kgk∞ ≤ 1 and applying Theo- rem 2.4 we obtain kµk ≤ kT kM1,1 which together with (4.2) yields kT kM1,1 = kµk. 2 2,2 n ∞ n Theorem 4.8. T ∈ M (R ) if and only if there is a function m ∈ L (R ) such that22 −1 2 n T f = F (m(Ff)) for f ∈ L (R ). Moreover kT kM2,2 = kmk∞ .

Proof. If m ∈ L∞, then if follows from the Plancherel theorem that the operator T f = F −1(m(Ff)) is bounded on L2. Indeed, −1 kF (m(Ff))k2 = km(Ff)k2 ≤ kmk∞kFfk2 = kmk∞kfk2 . It is easy to see that the operator T is translation invariant, so T ∈ M2,2 and the above estimate yields

(4.4) kT kM2,2 ≤ kmk∞ . 2,2 0 Now let T ∈ M and let u ∈ Sn be such that T ϕ = u ∗ ϕ for ϕ ∈ Sn. −π|x|2 Let ϕ0(x) = e . Recall thatϕ ˆ0 = ϕ0 (Theorem 2.14). We have 2 n ϕ0uˆ =ϕ ˆ0uˆ = (u ∗ ϕ0)ˆ = (T ϕ0)ˆ ∈ L (R ) . Hence (ϕ0uˆ)(x) 2 m(x) = ∈ Lloc . ϕ0(x) π|x|2 0 Note that the multiplication by 1/ϕ0(x) = e is not allowed in Sn, because eπ|x|2 is not slowly increasing, so we cannot conclude that m =u ˆ (at least

22Compare with the operators deﬁned by the formulas (2.20) and (2.21). According to the theorem the operator (I − ∆)−N is bounded on L2, while (I − ∆)N is not. 72 PIOTR HAJLASZ

2 not now). However, m(x) is a well defined function in Lloc and, of course, we want to prove that m =u ˆ, but we have to be very cautious and check that every step in our proof is justifiable. First we will prove that ∞ n (4.5)u ˆ[ψ] = m[ψ] for ψ ∈ C0 (R ). ∞ n Observe that for ψ ∈ C0 (R ) both sides of this equality are well defined with the right hand side understood as Z m[ψ] = m(x)ψ(x) dx . Rn We have Z Z (ϕ0uˆ)(x) −1 m[ψ] = ψ(x) dx = (ϕ0uˆ)(x) ϕ0 (x)ψ(x) dx n ϕ0(x) n R R | {z∞ } C0 −1 −1 = (ϕ0uˆ)[ϕ0 ψ] =u ˆ[ϕ0ϕ0 ψ] =u ˆ[ψ] ∞ which proves (4.5). We could do this calculation only for ψ ∈ C0 and not for ψ ∈ Sn, because we do not know if mψ is integrable and also, because −1 ϕ0 ψ does not necessarily belong to Sn for ψ ∈ Sn. ∞ 2 0 If ϕ ∈ C0 , then ϕm ∈ L ⊂ Sn and (4.5) easily implies that 0 ϕuˆ = ϕm in Sn. Thus kϕmk2 = kϕuˆk2 = k u ∗ ϕˇ ˆk2

= ku ∗ ϕˇk2 = kT (ϕ ˇ)k2

≤ kT kM2,2 kϕˇk2 = kT kM2,2 kϕk2 and hence Z 2 2 2 kT kM2,2 − |m(x)| |ϕ(x)| ≥ 0 Rn ∞ n ∞ for all ϕ ∈ C0 (R ). This, however, easily implies that m ∈ L with

(4.6) kmk∞ ≤ kT kM2,2 . Since m ∈ L∞, equality (4.5) yields 0 uˆ = m in Sn.

Finally for ϕ ∈ Sn we have (T ϕ)ˆ = (u ∗ ϕ)ˆ =ϕ ˆuˆ = mϕˆ , so T ϕ = F −1(m(Fϕ)) and by density T f = F −1(m(Ff)) for f ∈ L2. HARMONIC ANALYSIS 73

Moreover inequalities (4.4) and (4.6) give

kT kM2,2 = kmk∞ . The proof is complete. 2 p,p n Theorem 4.9. If T ∈ M (R ), 1 ≤ p < ∞, then there is a bounded ∞ n function m ∈ L (R ) such that −1 T ϕ = F (m(Fϕ)) for ϕ ∈ Sn. In other words we can identify Mp,p with a subspace of M2,2.

Proof. Let 1 < p < ∞ and T ∈ Mp,p. Then T ∈ Mp0,p0 by Theorem 4.6. Since 2 is between p and p0 it follows from the interpolation theorem23 that T ∈ M2,2. If p = 1, then T ϕ = ϕ ∗ µ = F −1(m(Fϕ)), where m =µ ˆ ∈ L∞. 2

n Definition. Given 1 ≤ p < ∞ we deﬁne Mp(R ) to be the space of bounded ∞ n functions m ∈ L (R ) such that the operator −1 Tmϕ = F (m(Fϕ)), ϕ ∈ Sn p n is bounded on L . The norm of m in Mp(R ) is deﬁned by

p,p kmkMp = kTmkM . n p p Elements of the space Mp(R ) are called L multipliers or L Fourier multipliers. The function m is also called the symbol of the operator Tm. It follows from Theorem 4.8 that n ∞ n M2(R ) = L (R ), kmkM2 = kmk∞ n and Theorem 4.7 shows that M1(R ) consists of functions that are Fourier transforms of measures of ﬁnite bounded variation.

If 1 < q ≤ 2 and m =µ ˆ ∈ M1, then by Theorem 2.5 and Theorem 4.7 we have

kTmϕkq = kϕ ∗ µkq ≤ kµkkfkq = kmkM1 kfkq , so M1 ⊂ Mq and kmkMq ≤ kmkM1 .

If 1 < q ≤ p ≤ 2 and m ∈ Mq, then according to Theorem 4.6 q q q0 q0 Tm : L → L ,Tm : L → L , kTmkMq,q = kTmkMq0,q0 . Since q ≤ p ≤ q0, the Riesz-Thorin theorem gives p p Tm : L → L , kTmkMp,p ≤ kTmkMq,q .

Hence Mq ⊂ Mp, kmkMp ≤ kmkMq . Thus for 1 ≤ q ≤ p ≤ 2 we have ∞ M1 ⊂ Mq ⊂ Mp ⊂ M2 = L ,

(4.7) kmk∞ ≤ kmkMp ≤ kmkMq ≤ kmkM1 .

23Riesz-Thorin or Marcinkiewicz. 74 PIOTR HAJLASZ

Finally, if 1 < p ≤ 2, then Theorem 2.7 implies that Mp = Mp0 isometrically.

2πiξ·h p n Example. The function m(ξ) = e is an L multiplicator for all h ∈ R and the corresponding operator is Tmf(x) = f(x + h). n Theorem 4.10. If 1 ≤ p < ∞, then Mp(R ) is a commutative Banach algebra with respect to a pointwise multiplication.

Proof. Since Tam1+bm2 = aTm1 + bTm2 it follows that Mp is a linear space and

p,p p p kam1 + bm2kMp = kTam1+bm2 kM ≤ |a|km1kM + |b|km2kM shows that k · kMp is a norm. Since Tm1m2 = Tm1 Tm2 we have

p,p km1m2kMp = kTm1m2 kM ≤ km1kMp km2kMp , so Mp is a complex commutative algebra with the unit element m ≡ 1, kmkMp = 1 that corresponds to the identity mapping.

Thus it remains to prove that Mp is complete with respect to the norm. 0 If 2 < p < ∞, then Mp is isometric to Mp0 , 1 < p < 2, so we can assume that 1 ≤ p ≤ 2. Let {mk} be a Cauchy sequence in Mp. By (4.7)

kmj − mlk∞ ≤ kmj − mkkMp ∞ ∞ and hence {mj} is a Cauchy sequence in L . Thus it converges in the L ∞ norm to a bounded function m ∈ L . We have to prove that m ∈ Mp and mj → m in Mp.

Fix ϕ ∈ Sn. The dominated convergence theorem yields Z Z 2πix·ξ 2πix·ξ (Tmk ϕ)(x) = ϕˆ(ξ)mk(ξ)e dξ → ϕˆ(ξ)m(ξ)e dξ = (Tmϕ)(x) . Rn Rn p,p Given ε > 0 let N be such that kTmj − Tmk kM < ε for j, k ≥ N. For j ≥ N Fatou’s lemma implies Z Z p p |(Tmj − Tm)ϕ| dx = lim |(Tmj − Tmk )ϕ| dx n n k→∞ R R Z p ≤ lim inf |(Tmj − Tmk )ϕ| dx k→∞ n R p p ≤ ε kϕkp , p,p p,p p,p i.e. kTmj − TmkM ≤ ε. Thus Tm ∈ M and Tmj → Tm in M , i.e. m ∈ Mp and mj → m in Mp. 2 The following result easily follows from the properties of the Fourier transform. n n Theorem 4.11. Let m ∈ Mp(R ), 1 ≤ p < ∞, x ∈ R and h > 0. Then

(a) kτxmkMp = kmkMp . HARMONIC ANALYSIS 75

h h (b) kδ mkMp = kmkMp , where (δ m)(x) = m(hx) is a dilation.

(e) km(ρ·)kMp = kmkMp , when ρ ∈ O(n) is an orthogonal transformation.

The following result is often useful.

n 1 n Proposition 4.12. Let m ∈ Mp(R ), 1 ≤ p < ∞ and ψ ∈ L (R ). Then n m ∗ ψ ∈ Mp(R ) and

km ∗ ψkMp ≤ kψk1kmkMp .

We leave the proof as an exercise.

∞ n R −n In particular if ψ ∈ C0 (R ), n ψ dx = 1 and ψε = ε ψ(x/ε), then ∞ R mε = m ∗ ψε satisﬁes mε ∈ C , mε ∈ Mp, kmεkMp ≤ kmkMp . Hence for every ϕ ∈ Sn

(Tmε ϕ)(x) → (Tmϕ)(x) as ε → 0

n for every x ∈ R . The next result is a kind of Fubini theorem for Fourier multipliers.

n+m Theorem 4.13. Suppose that m(ξ, η) ∈ Mp(R ), 1 < p < ∞. Then for n n almost every ξ ∈ R , the function η 7→ m(ξ, η) is in Mp(R ) and

m n+m km(ξ, ·)kMp(R ) ≤ kmkMp(R ) .

∞ n+m Proof. Since m ∈ L (R ) it follows from the Fubini theorem that for n ∞ m a.e. ξ ∈ R , m(ξ, ·) ∈ L (R ) and

(4.8) km(ξ, ·)k∞ ≤ kmk∞ .

Fix ϕ1, ψ1 ∈ Sn and ϕ2, ψ2 ∈ Sm. For ξ such that (4.8) is satisﬁed we deﬁne Z Z ∨ M(ξ) = (m(ξ, ·)ϕ ˆ2(·)) (η)ψ2(η) dη = m(ξ, η)ϕ ˆ2(η)ψˇ2(η) dη . Rm Rm

∞ n Observe that M ∈ L (R ) with

kMk∞ ≤ kmk∞kϕˆ2ψˇ1k1 . 76 PIOTR HAJLASZ

Thus M deﬁnes a multiplier (at least an L2 multiplier) and we have Z Z ∨ ˇ (M(·)ϕ ˆ1(·)) (ξ)ψ1(ξ) dξ = M(ξ)ϕ ˆ1(ξ)ψ1(ξ) dξ Rn Rn ZZ ˇ ˇ = m(ξ, η)ϕ ˆ1(ξ)ϕ ˆ2(η)ψ1(ξ)ψ2(η) dη dξ Rn+m ZZ ∧ ∨ = m(ξ, η)(ϕ1ϕ2) (ξ, η)(ψ1ψ2) (ξ, η) dξ dη Rn+m ZZ ∧∨ = m(ϕ1ϕ2) (ψ1ψ2) dξ dη Rn+m n+m 0 0 ≤ kmkMp(R )kϕ1kpkϕ2kpkψ1kp kψ2kp .

Taking the supremum over ψ1 ∈ Sn with kψ1kp0 ≤ 1 it follows that ∨ (Mϕˆ ) ≤ kmk n+m kϕ k kψ k 0 kϕ k , 1 p Mp(R ) 2 p 2 p 1 p n i.e. M ∈ Mp(R ) with

n n+m 0 kMkMp(R ) ≤ kmkMp(R )kϕ2kpkψ2kp .

n Since kMk∞ ≤ kMkMp(R ) we conclude that Z ∨ m(ξ, ·)ϕ ˆ (·) (η)ψ (η) dη ≤ kmk n+m kϕ k kψ k 0 2 2 Mp(R ) 2 p 2 p Rm and taking the supremum over kψ2kp0 ≤ 1 yields the result. 2 The above theory of Fourier multipliers is very beautiful, but there is one problem: we do not have good examples. Indeed, we characterized all L2 multipliers as bounded functions, but we do not know what bounded functions define Lp multipliers for p 6= 2. The only examples of Lp multipliers that we know so far come from translation invariant operators on L1. Indeed, every such an operator is a convolution with a measure and it is also bounded in Lp for all p. Thus all functions that are Fourier transforms of measures of finite total variation define Lp multipliers, but on the other hand there is no need to use the theory of Fourier multipliers to deal with convolutions of measures and there is a huge gap between the space of Fourier transforms of measures and all bounded functions. In the following sections we will construct more and more Lp multipliers, but as we shall see it is always a very difficult task.

5. The Hilbert transform

The function f(x) = sin x/x is not integrable on [0, ∞), but we deﬁne its integral as an improper one Z ∞ sin x Z R sin x π (5.1) dx = lim dx = . 0 x R→∞ 0 x 2 HARMONIC ANALYSIS 77

A similar problem appears if we want to define the integral Z ∞ ϕ(x) dx, ϕ ∈ S(R) . −∞ x Since the function 1/x is not integrable in any neighborhood of 0, in the case in which ϕ(0) 6= 0, the integral diverges. This suggests that we should define this integral as a kind of an improper integral known as the principal value of the integral 1 Z ∞ ϕ(x) Z ϕ(x) (5.2) p.v. [ϕ] = p.v. dx = lim dx . x −∞ x ε→0 |x|≥ε x The limit exists when ϕ ∈ S(R) and it actually defines a tempered distribution.

Theorem 5.1. If ϕ ∈ S(R), then the limit at (5.2) exists and deﬁnes a 0 tempered distribution p.v. 1/x ∈ S (R) such that 1 0 p.v. [ϕ] ≤ 2(kϕ k∞ + kxϕk∞) . x

Proof. Note that Z ϕ(0) dx = 0 , ε≤|x|≤1 x so we have Z ϕ(x) Z ϕ(x) − ϕ(0) Z ϕ(x) dx = dx + dx . |x|≥ε x ε≤|x|≤1 x |x|≥1 x For the ﬁrst integral on the right hand side we have Z ϕ(x) − ϕ(0) Z Z 1 dx = ϕ0(tx) dt dx ε≤|x|≤1 x ε≤|x|≤1 0 Z Z 1 → ϕ0(tx) dt dx as ε → 0 |x|≤1 0 and hence Z ∞ ϕ(x) Z Z 1 Z ϕ(x) p.v. dx = ϕ0(tx) dt dx + dx . −∞ x |x|≤1 0 |x|≥1 x Since

Z ϕ(x) Z dx dx ≤ sup |xϕ(x)| = 2kxϕk 2 ∞ |x|≥1 x x∈R |x|≥1 x and Z Z 1 0 0 ϕ (tx) dt dx ≤ 2kϕ k∞ |x|≤1 0 the theorem follows. 2 78 PIOTR HAJLASZ

Exercise. Prove that for ϕ ∈ S(R) Z ∞ Z R ϕ(x) 0 |x| p.v. dx = lim ϕ (x) ln dx . −∞ x R→∞ −R R

Since p.v. 1/x is a tempered distribution, we may try to compute its Fourier transform. Theorem 5.2. 1 1 ∧ p.v. (ξ) = −i sgn (ξ) . π x

Proof. For ϕ ∈ S(R) we have 1 Z ϕ(x) p.v. [ϕ] = lim dx , x ε→0 |x|≥ε x so 1 1 g (x) = χ → p.v. in S0( ), ε x {|x|≥ε} x R and hence 1 ∧ (5.3)g ˆ → p.v. in S0( ) as ε → 0. ε x R

2 24 Since gε ∈ L we can compute its Fourier transform using Theorem 2.31. Z R −2πixξ gˆε(ξ) = lim gε(x)e dx R→∞ −R Z R 1 1 = lim e−2πixξ + e−2πi(−x)ξ dx R→∞ ε x (−x) Z R e−2πixξ − e2πixξ = lim dx R→∞ ε x Z R sin(2πxξ) = lim −2i dx R→∞ ε x Z R sin(2πx|ξ|) = lim −2i sgn (ξ) dx R→∞ ε x Z 2π|ξ|R sin y = lim −2i sgn (ξ) dy R→∞ 2π|ξ|ε y Z ∞ sin y = −2i sgn (ξ) dy , 2π|ξ|ε y

24The limit is understood in the L2 sense, but we will prove that the limit also exists in the pointwise sense, so the pointwise limit must be equal to the L2 one. HARMONIC ANALYSIS 79 where the limit exists as an improper integral, see (5.1). Note that the above computation gives also

(5.4) |gˆε(ξ)| ≤ M independntly of ε and π (5.5) lim gˆε(ξ) = −2i sgn (ξ) = −πi sgn (ξ) . ε→0 2 Thus (5.3) yields the result. 2 Definition. The Hilbert transform of a function f is deﬁned by 1 Z ∞ f(x − y) Hf(x) = p.v. dy π −∞ y 1 Z f(x − y) = lim dy π ε→0 |y|≥ε y 1 Z f(y) = lim dy π ε→0 |x−y|≥ε x − y and the question is for what functions and in what sense the limit exists.

0 Note that if ϕ ∈ S(R) and u = p.v. 1/x ∈ S (R), then 1 1 Hϕ(x) = u[ϕ(x − ·)] = (u ∗ ϕ)(x) , π π so Hϕ ∈ C∞, and Hϕ and all its derivatives are slowly increasing (see Theorem 2.36). Note also that 1 (Hϕ)∧(ξ) = (ˆuϕˆ)(ξ) = −i sgn (ξ)ϕ ˆ(ξ) π and hence Hϕ(x) = − i sgn (·)ϕ ˆ(·)∨(x) . ∞ Since m(ξ) = −i sgn (ξ) ∈ L (R), the Hilbert transform deﬁnes a translation invariant operator on L2. Actually we have. 2 Theorem 5.3. If f ∈ L (R), then (5.6) Hf(x) = − i sgn (·)fˆ(·)∨(x) . 2 2 2 2 Hence H : L (R) → L (R) is an isometry of L onto L , 2 kHfk2 = kfk2 for f ∈ L (R) with the inverse operator satisfying H−1 = −H. Moreover if 1 Z f(y) Hεf = dy , π |x−y|≥ε x − y then Hεf → Hf in L2 as ε → 0. 80 PIOTR HAJLASZ

Proof. We proved (5.6) for f = ϕ ∈ S(R). Since m(ξ) = −i sgn (ξ) is bounded, H uniquely extends to a translation invariant operator H ∈ M2,2 and (5.6) follows from the Plancherel theorem. Another application of the Plancherel theorem gives ˆ ˆ kHfk2 = kHfdk2 = k − i sgn (·)f(·)k2 = kfk2 = kfk2 . Moreover H2f = (−i sgn (ξ))2fˆ(ξ)∨ = −f , so H2 = −I and hence H−1 = −H. Finally if Z ε 1 f(y) 1 H f(x) = dy = gε ∗ f (x) , π |x−y|≥ε x − y π where 1 g (x) = χ ∈ L2 ε x {|x|≥ε} formulas (5.4) and (5.5) give

|gˆε(ξ)| ≤ M independently of ε and gˆε(ξ) → −πi sgn (ξ) as ε → 0 for all ξ ∈ R. Hence 1 Hεf∧(ξ) = fˆ(ξ)ˆg (ξ) → −i sgn (ξ)fˆ(ξ) π ε as ε → 0 and ε ∧ M (H f) (ξ) ≤ |fˆ(ξ)| . π Thus the dominated convergence theorem yields Hεf∧ → −i sgn (ξ)fˆ(ξ) in L2 as ε → 0 and hence Hεf → − i sgn (ξ)fˆ(ξ)∨ = Hf in L2 as ε → 0 by Plancherel’s theorem. 2

5.1. Conjugate harmonic functions. We will see now that the Hilbert transform arises naturally in connection with boundary behavior of holomorphic functions. Let us recall that the Poisson kernel t n Pt(x) = P (x, t) = cn , x ∈ R , t > 0 . (t2 + |x|2)(n+1)/2 n+1 The function P (x, t) is harmonic in R+ . One can check it by a direct computation, but it is worth to note that this also follows the fact that for n ≥ 2 c ∂ 1 P (x, t) = − n n − 1 ∂t (t2 + |x|2)(n−1)/2 HARMONIC ANALYSIS 81 and 1 = k(t, x)k2−(n+1) (t2 + |x|2)(n−1)/2 n+1 25 is the standard radial harmonic function in R \{0}. Similar argument works also for n = 1.

p n If f ∈ L (R ), 1 ≤ p < ∞, then the Poisson integral Z u(x, t) = P (x − y, t)f(y) dy = Pt ∗ f(x) Rn n+1 is harmonic in R+ . Indeed, it is easy to see that we can diﬀerentiate u under the sign of the integral and hence harmonicity of u follows from that of P (x, t). Note that (2.7) and Theorem 2.28 imply that (5.7) u(·, t) → f(·) p n + both in L (R ) and a.e. as t → 0 . Thus the convolution with the Poisson kernel solves the following Dirichlet problem.  n+1  ∆(x,t)u(x, t) = 0 in R+ ,  u(x, 0) = f(x), where the boundary condition u(x, 0) = f(x) is understood in the sense of the limit (5.7). Now let us restrict to the case n = 1, so 1 y P (x) = P (x, y) = , x ∈ , y > 0 , y π y2 + x2 R where for convenience reasons we use variable y instead of t. Thus for a real p valued function f ∈ L (R), 1 ≤ p < ∞, y Z ∞ f(t) u(x, y) = (f ∗ Py)(x) = 2 2 dt π −∞ (x − t) + y is a solution to a corresponding Dirichlet problem in the upper half-plane 2 R+ = {(x, y): y > 0}. 2 Every harmonic function in R+ is a real part of a holomorphic one. Clearly the function i Z ∞ f(t) F (z) = dt π −∞ z − t 2 26 is holomorphic in R+ = {im z > 0} and re F (z) = u(x, y) .

25Up to a constant it is the fundamental solution to the Laplace operator ∆. 26Since re (i/(z − t)) = y/((x − t)2 + y2). 82 PIOTR HAJLASZ

Also 1 Z ∞ (x − t)f(t) im F (z) = 2 2 dt = (f ∗ Qy)(x) , π −∞ (x − t) + y where 1 x Q (x) = y π x2 + y2 is called the conjugate Poisson kernel. The functions

u(x + iy) = (f ∗ Py)(x), v(x + iy) = (f ∗ Qy)(x) 2 are conjugate harmonic functions in R+ since they are real and imaginary parts of the holomorphic function F (z). We know that u(·, y) → f(·) as y → 0+ both in Lp and a.e. and it is natural to ask what is the limit of v(x, y) = + (f ∗ Qy)(x) as y → 0 ? Formally 1 lim Qy(x) = y→0+ πx + so it should not be surprising that the limit of f ∗ Qy as y → 0 equals to the Hilbert transform Hf. p Theorem 5.4. For f ∈ L (R), 1 ≤ p < ∞ we have ε + f ∗ Qε − H f → 0 as ε → 0 p ε both in L (R) and a.e. More precisely f ∗ Qε(x) − H f(x) → 0 whenever x is a Lebesgue point of f.

Proof. Note that Z ∞ ε 1 1 f ∗ Qε(x) − H f(x) = ψε(x − t)f(t) dt = (f ∗ ψε)(x) , π −∞ π where x 1 ψ (x) = − χ . ε x2 + ε2 x {|x|≥ε} Note also that if x 1 ψ(x) = ψ (x) = − χ . 1 x2 + 1 x {|x|≥1} −1 then ψε(x) = ε ψ(x/ε). The function 1 Ψ(x) = x2 + 1 is an integrable radially decreasing majorant of ψ, i.e. |ψ| ≤ |Ψ| and R ψ = R 0. Hence Theorem 2.28 with a = 0 implies that ε f ∗ Qε − H f → 0 a.e. HARMONIC ANALYSIS 83

The convergence to 0 in Lp follows from Corollary 2.12. 2

ε If ϕ ∈ S(R), then H ϕ → Hϕ everywhere by Theorem 5.1. Since every point of ϕ is Lebesgue, Theorem 5.4 gives Corollary 5.5. If ϕ ∈ S(R), then + ϕ ∗ Qε(x) → Hϕ(x) as ε → 0 for every x ∈ R.

Since for f ∈ L2, Hεf → Hf in L2 we immediately get 2 Corollary 5.6. If f ∈ L (R), then 2 + f ∗ Qε → Hf in L as ε → 0 .

Later we will see that Hεf → Hf both in Lp and a.e. for any f ∈ Lp, 1 < p < ∞ and hence the corollary extends to 1 < p < ∞. This will show p that if f ∈ L (R), 1 < p < ∞ is the boundary value of the real part of the 2 holomorphic function F in R+, then the boundary value of the imaginary part is Hf ∈ Lp.

5.2. Lp estimates. The boundedness of the Hilbert transform in Lp, 1 < p < ∞ plays a fundamental role in harmonic analysis and its applications. In this section we will show two diﬀerent proofs of this result and in one of the following sections we will provide one more proof. The Hilbert transform is the simplest example of a singular integral and the result is a special case of Lp estimates for singular integrals that will be discussed in later sections. Theorem 5.7 (Riesz). If 1 0 such that for all ϕ ∈ S(R)

kHϕkp ≤ C(p)kϕkp . Moreover C(p) = C(p0) and C(2) = 1. Thus H ∈ Mp,p for all 1 < p < ∞ and hence m(ξ) = sgn (ξ) ∈ Mp(R).

Proof. We proved the result for p = 2 with C(2) = 1 in Theorem 5.3. Since p,p p0,p0 T ∈ M if and only if T ∈ M with kT kMp,p = kT kMp0,p0 (Theorem 4.6) is suﬃces to prove the result for 2 < p < ∞. We will need the following lemmas which are of independent interest. 1 n p 1 Lemma 5.8. If u ∈ C (Ω), Ω ⊂ R and 1 < p < ∞, then |u| ∈ C (Ω) and ∇|u|p = p|u|p−2u∇u. In particular ∇|u|p(x) = 0 if u(x) = 0.

Proof. Clearly |u|p is C1 on the open set where u 6= 0 and the formula for ∇|u|p is easy to verify on that set. Thus it remains to prove that |u|p is diﬀerentiable on the set where u = 0 with ∇|u|p = 0 on that set. We leave details as an exercise. 2 84 PIOTR HAJLASZ

Now if p > 2 and u ∈ C2(Ω), then |u|p ∈ C2(Ω). Indeed, ∂ ∂u ∂u |u|p = p|u|p−2u = p sgn (u)|u|p−1 . ∂xi ∂xi ∂xi p−1 1 The lemma gives |u| ∂u/∂xi ∈ C and ∂ ∂u |u|p−1 = 0 if u(x) = 0 , ∂xj ∂xi so it is easy to see that 2 ∂ p ∂ p−1 ∂u ∂ p−1 ∂u ∂xi |u| = p sgn (u)|u| = p sgn (u) |u| ∂xj ∂xj ∂xi ∂xj ∂xi and hence the second derivatives of |u|p are continuous. In particular if u ∈ C2(Ω) and p > 2 we can compute ∆|u|p. Easy calculations give 2 n 2 2 Lemma 5.9. If u ∈ C (Ω), Ω ⊂ R and p > 2, then |u| ∈ C (Ω) and ∆|u|p = p|u|p−2u∆u + (p − 1)|∇u|2 .

Using a variant of the above argument combined with the Cauchy- Riemann equations one can easily prove

Lemma 5.10. If F (z) = u + iv is holomorphic in Ω ⊂ C and p > 2, then |F |p ∈ C2(Ω) and p 2 p−2 2 2 ∆|F | = p |F | (ux + uy) .

∞ Now we can return to the proof of the theorem. By density of C0 (R) in ∞ S(R) it suﬃces to assume that ϕ ∈ C0 (R). This assumption will simplify some estimates. Let i Z ∞ ϕ(t) F (z) = dt . π −∞ z − t Then as we have seen in Section 5.1

re F (z) = u(x, y) = (ϕ ∗ Py)(x) im F (z) = v(x, y) = (ϕ ∗ Qy)(x) , where 1 y 1 x P (x) = ,Q (x) = . y π y2 + x2 y π x2 + y2 Moreover + v(x, y) = (ϕ ∗ Qy)(x) → Hϕ(x) as y → 0 for every x ∈ R, see Corollary 5.5. Since v is harmonic Lemma 5.9 gives p p−2 2 2 p−2 2 2 ∆|v| = p(p − 1)|v| (vx + vy) = p(p − 1)|v| (ux + yy) , where in the last equality follows from the Cauchy-Riemann equations. This and Lemma 5.10 imply p (5.8) ∆ |F |p − |v|p = p2(|F |p−2 − |v|p−2)(u2 + u2) ≥ 0 . p − 1 x y HARMONIC ANALYSIS 85

1 2 Let us recall that if f ∈ C (Ω), where Ω ⊂ R is a bounded domain with piecewise smooth boundary and ~n is the outer normal vector to the boundary, then Z ∂f Z ZZ (5.9) ds = ∇f · ~nds = ∆f dx dy . ∂Ω ∂~n ∂Ω Ω This formula is known as Green’s identity. We want to apply it to the function p f = |F |p − |v|p p − 1 and Ω as on the picture.

The integral of ∇f · ~n along the boundary is nonnegative by (5.9) and (5.8). Elementary but tedious estimates27 show that the part of the integral corresponding to the semi-circle converges to 0 as R → ∞. Thus for every y > 0 we obtain28 ∂ ∂ Z ∞ p (5.10) I(y) = |F (x, y)|p − |v(x, y)|p dx ≤ 0 ∂y ∂y −∞ p − 1 Hence the function Z ∞ p I(y) = |F (x, y)|p − |v(x, y)|p dx −∞ p − 1 is decreasing. The same estimates imply also that I(y) → 0 as y → ∞, so I(y) ≥ 0 for all y > 0 and thus Z ∞ p Z ∞ |F (x, y)|p dx ≥ |v(x, y)|p dx . −∞ p − 1 −∞

27Since ϕ has compact support, the integral formulas that deﬁne u and v show that the growth of u, v, ∇u, ∇v can be estimated by C/(x2 +y2)1/2 and hence |f| ≤ C/(x2 +y2)p/2, |∇f| ≤ C/(x2 + y2)p/2. 28Note that ∂/∂y = −∂/∂~n along this line. 86 PIOTR HAJLASZ

Observe that Z ∞ 2/p p 2 2 2 2 |F (x, y)| dx = ku (·, y)+v (·, y)kp/2 ≤ ku (·, y)kp/2+kv (·, y)kp/2 . −∞ Hence p 2/p kv2(·, y)k ≤ ku2(·, y)k + kv2(·, y)k , p − 1 p/2 p/2 p/2 1 (5.11) kv2(·, y)k ≤ ku2(·, y)k . p/2 2/p p/2 p p−1 − 1 p + Note that u(·, y) = ϕ∗Py → ϕ in L as y → 0 by (2.7) and v(x, y) → Hϕ(x) + for every x ∈ R. Thus letting y → 0 in (5.11) and applying Fatou’s lemma we obtain Z ∞ 1 Z ∞ |Hϕ(x)|p dx ≤ |ϕ(x)|p dx . 2/p p/2 −∞ p −∞ p−1 − 1 The proof is complete. 2 Now we will present another proof of Theorem 5.7 based on the following interesting identity.

Lemma 5.11 (Cotlar). If ϕ ∈ S(R) is a real valued function, then H(ϕ)2 = ϕ2 + 2H(ϕH(ϕ)) .

Proof. Before we give a rigorous proof we will show a heuristic argument that leads to this identity. As we have already seen, the function ϕ + iH(ϕ) has a holomorphic extension to the upper half-plane. Namely i Z ∞ ϕ(t) F (z) = dt . π −∞ z − t Hence also the function (ϕ + iH(ϕ))2 = ϕ2 − H(ϕ)2 + i2ϕH(ϕ) has a holomorphic extension F (z)2. Thus we may expect that29 2ϕH(ϕ) = H(ϕ2 − H(ϕ)2) Applying H to both sides and using the fact that H2 = −I we get 2H(ϕH(ϕ)) = −ϕ2 + H(ϕ)2

29This, however, would require a proof. The function ϕ2 − H(ϕ)2 does not belong to S(R) in general and we proved that the boundary value of the imaginary part is the Hilbert transform of the boundary value of the real part only in a specific situation when the real part is in S(R) and the extension is defined by the integral F (z). We will not clarify this issue now as we will present in a moment a different, rigorous, proof based on the Fourier transform. HARMONIC ANALYSIS 87 which implies the claim. Now we present a different and rigorous proof. Let m(ξ) = −i sgn (ξ) be the symbol of the Hilbert transform. We have30 ∧ ϕc2(ξ) + 2 H(ϕH(ϕ)) (ξ) = (ϕ ˆ ∗ ϕˆ)(ξ) + 2m(ξ)(ϕ ˆ ∗ Hϕd)(ξ) Z ∞ Z ∞ = ϕˆ(ζ)ϕ ˆ(ξ − ζ) dζ + 2m(ξ) ϕˆ(ζ)ϕ ˆ(ξ − ζ)m(ξ − ζ) dζ = ♥ . −∞ −∞ Since Z ∞ Z ∞ ϕˆ(ζ)ϕ ˆ(ξ − ζ)m(ξ − ζ) dζ = ϕˆ(ξ − ζ)ϕ ˆ(ζ)m(ζ) dζ −∞ −∞ we have Z ∞ 2m(ξ) ϕˆ(ζ)ϕ ˆ(ξ − ζ)m(ξ − ζ) dζ −∞ Z ∞ = m(ξ) ϕˆ(ζ)ϕ ˆ(ξ − ζ)m(ζ) + m(ξ − ζ) dζ −∞ and hence Z ∞ ♥ = ϕˆ(ζ)ϕ ˆ(ξ − ζ)1 + m(ξ)(m(ζ) + m(ξ − ζ)) = ♦ −∞ Since m(ξ) = −i sgn (ξ) one easily verifies that 1 + m(ξ)(m(ζ) + m(ξ − ζ)) = m(ζ)m(ξ − ζ) everywhere except ξ = ζ = 0 and hence Z ∞ ♦ = ϕˆ(ζ)ϕ ˆ(ξ − ζ)m(ζ)m(ξ − ζ) dζ −∞ Z ∞ = Hϕd(ξ − ζ)Hϕd(ζ) dζ −∞ = Hϕd ∗ Hϕd (ξ) = H(ϕ)2∧(ξ) . Taking the inverse Fourier transform yields the result. 2 Proof of Theorem 5.7. First observe that it suffices to prove the inequality

(5.12) kHϕkpk ≤ C(pk)kϕkpk k for p = pk = 2 , k = 1, 2, 3,... Indeed, this and the duality argument (Theorem 4.6) will imply that

kHϕk 0 ≤ C(p )kϕk 0 pk k pk

30In the arguments below we use Plancherel’s theorem many times, because all functions for which we compute Hilbert’s transform, Fourier’s transform and convolution are 2 in L and not always in S(R). 88 PIOTR HAJLASZ and hence the Riesz-Thorin theorem will yield

kHϕkp ≤ C(p)kϕkp 0 for all pk < p < pk. Since pk can be arbitrarily large, boundedness of the Hilbert transform in Lp for all 1 < p < ∞ will follow. We already proved (5.12) for k = 1 with C(p) = 1 in Theorem 5.3. Suppose now that we established the inequality for p = pk and we will show how to deduce the inequality for 2p = pk+1. For 0 6= ϕ ∈ S(R) Cotlar’s identity yields 2 1/2 kHϕk2p = k(Hϕ) kp 2 1/2 ≤ kϕ kp + k2H(ϕH(ϕ))kp 2 1/2 ≤ kϕk2p + 2C(p)kϕH(ϕ)kp 2 1/2 ≤ kϕk2p + 2C(p)kϕk2pkHϕk2p and hence kHϕk 2 kHϕk 2p − 2C(p) 2p − 1 ≤ 1 . kϕk2p kϕk2p This is a quadratic inequality which immediately yields

kHϕk2p p ≤ C(p) + C(p)2 + 1 kϕk2p and hence (5.12) for pk+1 = 2p follows with p 2 C(pk+1) = C(pk) + C(pk) + 1 . The proof is complete. 2 Remark. The proof gives also good estimates for the norm of the Hilbert transform in Lp.

5.3. Lp multipliers. Theorem 5.7 implies that m(ξ) = −i sgn (ξ) is an Lp multiplier m ∈ Mp(R), 1 < p < ∞. Hence also

χ[0,∞)(ξ) = (1 + i sgn (ξ))/2 ∈ Mp(R) . Since translations, reﬂections and product of multipliers is a multiplier we conclude that χ[a,b] ∈ Mp(R) . One dimensional multipliers generate n-dimensional ones. Indeed, if m ∈ Mp(R) and

(5.13)m ˜ (ξ1, . . . , ξn) = m(ξi) , the operator Tm˜ ϕ(x) = Tmϕ(x1, . . . , xi−1, ·, xi+1, . . . , xn) (xi) HARMONIC ANALYSIS 89 is bounded in Lp. This easily follows from the Fubini theorem Z p |Tm˜ ϕ(x)| dx Rn Z Z p = | Tmϕ(x1,..., ·, . . . , xn) (xi)| dxi dx1 . . . dxi−1 dxi+1 . . . dxn Rn−1 R Z Z p ≤ C |ϕ| dxi dx1 . . . dxi−1 dxi+1 . . . dxn . Rn−1 R n Thusm ˜ ∈ Mp(R ).

Corollary 5.12. If m1, . . . , mn ∈ Mp(R), then n m(ξ1, . . . , ξn) = m1(ξ1) . . . mn(ξn) ∈ Mp(R ) .

Proof. Indeed, m is a product of multipliers of the form (5.13). 2

n n Since χ[0,∞) ∈ Mp(R ), the characteristic function of the half-space R+ n belongs to Mp(R ). Rotations, translations and product of such characteristic functions is also an Lp multiplier and hence we have. Theorem 5.13. The characteristic function of any convex polyhedron in n p R is an L multiplier for 1 < p < ∞.

5.4. Pointwise convergence. We will prove that Hεf → Hf a.e. for f ∈ p L (R), 1 ≤ p < ∞. In this section we will treat the case 1 0 p deﬁned for all f ∈ L (R), 1 ≤ p < ∞. p Lemma 5.14. For f ∈ L (R), 1 < p < ∞ and all x ∈ R we have

(5.14) (f ∗ Qε)(x) = (H(f) ∗ Pε)(x) .

Proof. It suffices to prove the equality for f = ϕ ∈ S(R). Indeed, if p p p f ∈ L and ϕk ∈ S(R), ϕk → f in L , then H(ϕk) → H(f) in L and hence ϕk ∗Qε(x) → f ∗Qε(x), H(ϕk)∗Pε(x) → H(f)∗Pε(x) because of the fact that p0 −1 Pε,Qε ∈ L and the Hölderinequality. Moreover since Qε(x) = ε Q1(x/ε) −1 and Pε(x) = ε P1(x/ε) it suffices to prove the equality for ε = 1.

Taking the Fourier transform of (5.14) for with f = ϕ ∈ S(R) and ε = 1 we see that it is equivalent to31 −2π|ξ| ϕˆ(ξ)Qˆ1(ξ) = −i sgn (ξ)ϕ ˆ(ξ)e

31See Corollary 2.24. 90 PIOTR HAJLASZ

Observe that this identity follows from ∨ 1 x (5.15) −i sgn (·)e−2π|·| (x) = . π x2 + 1 The proof of (5.15) goes as follows ∨ Z ∞ −i sgn (·)e−2π|·| (x) = −i sgn (ξ)e−2π|ξ|e2πixξ dξ −∞ Z ∞ = −i e−2πξe2πixξ − e−2πξe2πx(−ξ) dξ 0 Z ∞ = 2 e−2πξ sin(2πxξ) dξ 0 1 x = , π x2 + 1 where the last equality follows from the twice integration by parts. 2 p Lemma 5.15 (Cotlar). If f ∈ L (R), 1 < p < ∞, then for all x ∈ R we have |H∗f(x)| ≤ Mf(x) + M(Hf)(x) .

Proof. Following notation from the proof of Theorem 5.4 we have 1 f ∗ Q (x) − Hεf(x) = (f ∗ ψ )(x) ε π ε and the function 1 Ψ(x) = x2 + 1 is an integrable radially decreasing majorant of ψ. Hence Theorem 3.12 yields 1 |f ∗ Q (x) − Hεf(x)| ≤ kΨk Mf(x) = Mf(x) ε π 1 Another application of the same theorem combined with Lemma 5.14 gives

sup |f ∗ Qε(x)| = sup |H(f) ∗ Pε(x)| ≤ kP1k1M(Hf)(x) = M(Hf)(x) . ε>0 ε>0 Thus ∗ |H f(x)| ≤ sup |f ∗ Qε(x)| + Mf(x) ≤ M(Hf)(x) + Mf(x) . ε>0 The proof is complete. 2 Corollary 5.16. The operator H∗ is of strong type (p, p) for all 1 < p < ∞, p i.e. for f ∈ L (R) ∗ kH fkp ≤ Cpkfkp .

Proof. It follows immediately from Lemma 5.15, boundedness of the Hilbert transform in Lp (Theorem 5.7) and boundedness of the maximal function in Lp (Theorem 3.10). 2 HARMONIC ANALYSIS 91

p ε Corollary 5.17. For f ∈ L (R), 1 < p < ∞, H f → Hf as ε → 0 both in Lp and a.e.

Remark. Compare the proof with that of Theorem 1.2.

ε Proof. If ϕ ∈ S(R), then H ϕ → Hϕ everywhere. Since |Hεϕ − Hϕ| ≤ H∗ϕ + |Hϕ| ∈ Lp we conclude that (5.16) Hεϕ → Hϕ in Lp. Now we will prove that for f ∈ Lp, Hεf converges a.e. to some measurable function g. To this end it suﬃces to show that (5.17) Ωf(x) = 0 a.e. where Ωf(x) = lim sup Hεf(x) − lim inf Hεf(x). ε→0 ε→0 Note that 0 ≤ Ωf(x) ≤ 2H∗f(x) and hence Z Z 2 ∗ p c p |{x :Ωf(x) > t}| ≤ p |H f| ≤ p |f| . t R t R To prove (5.17) it suﬃces to show that for any t > 0 (5.18) |{x :Ωf(x) > t}| = 0 .

Given any γ > 0 let ϕ ∈ S(R) be such that kf − ϕkp < γt. It is easy to see that Ωf ≤ Ω(f − ϕ) + Ωϕ = Ω(f − ϕ) , where the last equality follows from the fact that Hεϕ → Hϕ everywhere. Thus Z c p p |{x :Ωf(x) > t}| ≤ |{x : Ω(f − ϕ)(x) > t}| ≤ p |f − ϕ| ≤ cγ . t R It is true for any γ > 0, so (5.18) and hence (5.17) follows. Since |Hεf| ≤ H∗f and Hεf → g a.e. we conclude that |g| ≤ H∗f a.e. Now |Hεf − g| ≤ 2H∗f ∈ Lp and the dominated convergence theorem yields Hεf → g in Lp. It remains to prove that g = Hf a.e.

Given γ > 0 let ϕ ∈ S(R) be such that kf − ϕkp < γ and let ε > 0 be such that ε kHϕ − H ϕkp < γ (see (5.16)) ε kH f − gkp < γ . 92 PIOTR HAJLASZ

We have ε ε ε kHf − gkp ≤ kH(f − ϕ)kp + kHϕ − H ϕkp + kH (f − ϕ)kp + kH f − gkp ∗ 0 ≤ Ckf − ϕkp + γ + kH (f − ϕ)kp + γ ≤ C γ and hence g = Hf a.e. 2 The proofs of Theorem 1.2 and Corollary 5.17 are based on the same method. Not surprisingly, the same argument appears in other similar situations, so it is wise to present it in a form of an abstract and general result which will allow us to apply the result directly and avoid repeating the same argument over and over again.

Theorem 5.18. Let (X, µ) be a measure space and let {Tt}t>0 be a family of linear operators from Lp(µ), 1 ≤ p < ∞ into the space of measurable functions on X. Suppose that the limit

(5.19) lim Ttf(x) t→0 exists a.e. for all functions f in a dense subset A ⊂ Lp(µ). Deﬁne the maximal operator associated with the family {Tt} by ∗ T f(x) = sup |Ttf(x)| . t>0 If T ∗ is of weak type (p, q), 1 ≤ q < ∞, then the limit (5.19) exists a.e. for all f ∈ Lp(µ). Denote the limit by

T f(x) = lim Ttf(x) a.e. t→0 for all f ∈ Lp(µ). If in addition T ∗ is of strong type (p, q), then T is of strong type (p, q) and q Ttf → T f in L (µ) as t → 0 for all f ∈ Lp(µ).

Proof. We can assume that the functions are real valued as otherwise we can consider the real and imaginary parts separately. Suppose T is of weak type (p, q). We will prove now the ﬁrst part of the theorem which says that p Ttf converges a.e. as t → 0 for all f ∈ L (µ). To this end it suﬃces to show that Ωf = 0 a.e., where

Ωf(x) = lim sup Ttf(x) − lim inf Ttf(x) . t→0 t→0 Note that 0 ≤ Ωf(x) ≤ 2T ∗f(x) and hence Ckfk q |{x ∈ X :Ωf(x) > t}| ≤ p . t

For ε > 0 let ϕ ∈ A be such that kf − ϕkp < εt. It is easy to see that Ωf ≤ Ω(f − ϕ) + Ωϕ = Ω(f − ϕ) , HARMONIC ANALYSIS 93 because Ωϕ = 0 a.e. since Ttϕ converges a.e. as t → 0. Thus Cεtq |{Ωf > t}| ≤ |{Ω(f − ϕ) > t}| ≤ = Cqεq . t Since it is true for any t > 0 and ε > 0 we conclude that Ωf = 0 a.e. which completes the proof of the ﬁrst part of the theorem.

∗ ∗ Suppose now that T is of strong type (p, q). Since |Ttf| ≤ T f we conclude that |T f| ≤ T ∗f and hence ∗ q |Ttf − T f| ≤ 2T f ∈ L .

This inequality, the fact that Ttf → T f a.e. and the dominated convergence q theorem imply that Ttf → T f in L . 2

6. The Riesz transforms

The Hilbert transform is, up to a constant, convolution with the principal value of 1/x = x/|x|2. Thus a natural generalization to the n-dimensional case would be convolution with the principal value of x/|x|n+1. However, x/|x|n+1 is a vector valued function, so we should consider its components n+1 xj/|x| separately. For ϕ ∈ Sn we deﬁne Z xj Wj[ϕ] = cn p.v. n+1 ϕ(x) dx Rn |x| Z xj = cn lim n+1 ϕ(x) dx , ε→0 |x|≥ε |x| n+1 (n+1)/2 where cn = Γ 2 /π . Note that the constant is the same as the one in the deﬁnition of the Poisson kernel.

0 Exercise. Prove that the above limit exists for any ϕ ∈ Sn and that Wj ∈ Sn. Definition. For 1 ≤ j ≤ n the Riesz transform of a function f is deﬁned by

(Rjf)(x) = (Wj ∗ f)(x) Z yj = cn p.v. n+1 f(x − y) dy Rn |y| Z xj − yj = cn lim n+1 f(y) dy . ε→0 |x−y|≥ε |x − y|

The Riesz transform is well deﬁned for ϕ ∈ Sn and the question is how to p n extend it to L (R ), 1 ≤ p < ∞. The following results generalizes Theorem 5.2. Theorem 6.1. ξj Wcj(ξ) = −i . |ξ| 94 PIOTR HAJLASZ

Proof. 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 spherical coordinates we have Z ε−1 Z n−1 −2πix·(sθ) sθj I = s e n+1 dθ ds ε Sn−1 s Z ε−1 Z ds = cos(2πsx · θ) − i sin(2πsx · θ) θj dσ(θ) ε Sn−1 s −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 equality Z ∞ sin t π dt = . 0 t 2 Thus it remains to prove that Z π xj (6.1) cn sgn (x · θ)θj dσ(θ) = . 2 Sn−1 |x| We will need the following result. n n Lemma 6.2. If m : R → R 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. (6.2) m(ρ(x)) = ρ(m(x)) n for all x ∈ R and ρ ∈ O(n), then there is a constant c such that x (6.3) m(x) = c for all x 6= 0. |x| HARMONIC ANALYSIS 95

Before we will prove the lemma we show how to use it to establish (6.1). n n it is obvious that the function m : R → R deﬁned as Z m(x) = sgn (x · θ)θ dσ(θ) Sn−1 is homogeneous of degree 0. It also commutes with orthogonal transformations since 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)(6.4) Sn−1 Z (6.5) = ρ sgn (x · θ)θ dσ(θ) Sn−1 (6.6) = ρ(m(x)) .

Note that (6.5) follows from the fact that ρ induces a volume preserving change of variables on Sn−1, while (6.6) is a direct consequence of linearity of ρ. thus the lemma yields

Z x sgn (x · θ)θ dσ(θ) = c Sn−1 |x| and hence looking at the jth component we have Z xj (6.7) sgn (x · θ)θj dσ(θ) = c . Sn−1 |x| now it remains to prove that

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

Taking x = ej in (6.7) 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 n−1 into two half spheres S± . Thus Z c = 2 θj dσ(θ) . n−1 S+ 96 PIOTR HAJLASZ

n 1 n−1 Recall that if M ⊂ R is a graph of a C function f :Ω → R,Ω ⊂ R , then for a measurable function g on M we have Z Z p g dσ = g(x, f(x)) 1 + |∇f(x)|2 dx . M Ω n−1 In our situation we parametrize S+ as a graph of the function p f(x) = 1 − |x|2 , x ∈ Bn−1 . Form the picture

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. Thus we are left with the proof of the lemma. HARMONIC ANALYSIS 97

Proof of Lemma 6.2. Let e1, e2, . . . , en be the standard orthogonal basis n of R . If [ρjk] is the matrix representation of ρ ∈ O(n), then the condition (6.2) reads as n X (6.8) 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 matric 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 (6.8) 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 n−1 orthogonal transformation of R , so it must be a zero vector, i.e.

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

Now formula (6.8) for any ρ ∈ O(n) and x = e1, takes the form

mj(ρ(e1)) = ρj1m1(e1) = cρj1 . By homogeneity it suﬃces to prove (6.3) 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 and hance that of Theorem 6.1. 2 98 PIOTR HAJLASZ

Corollary 6.3. For ϕ ∈ Sn and 1 ≤ j ≤ n iξ ∨ (R ϕ)(x) = − j ϕˆ(ξ) (x) . j |ξ|

2,2 Since the function m(ξ) = −iξj/|ξ| is bounded, Rj ∈ M and 2 n kRjfk2 ≤ kfk2 for f ∈ L (R ). Moreover since the Riesz transform is a convolution with a tempered dis- ∞ tribution, for every ϕ ∈ Sn, Rjϕ ∈ C is slowly increasing and all its derivatives are slowly increasing. Corollary 6.4. The Riesz transforms satisfy n X 2 2 n Rj = −I on L (R ). j=1

Proof. Applying the previous corollary and the Plancherel theorem, for f ∈ L2 we have n n 2 X ∧ X iξj R2f (ξ) = − fˆ(ξ) = −fˆ(ξ) j |ξ| j=1 j=1 which yields the claim. 2 An amazing property of the Riesz transforms is that they allow to compute mixed partial derivatives ∂j∂ku is we only know ∆u. More precisely we have

Proposition 6.5. If ϕ ∈ Sn, then for 1 ≤ j, k ≤ n we have ∂ϕ = −RjRk∆ϕ(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) and the result follows by taking the inverse Fourier transform in L2. 2

0 It is quite convincing the the above argument applied to u ∈ Sn such that 2 ∆u = f ∈ L , gives ∂j∂ku = −RjRkf. However this is not true. For example 0 if u = xy, then as a slowly increasing function u ∈ S2. Clearly ∆u = 0 = f, but ∂x∂yu = 1 6= 0 = −RxRyf. In general we have HARMONIC ANALYSIS 99

0 Theorem 6.6. If u ∈ Sn satisﬁes 2 n (6.9) ∆u = f ∈ L (R ) , then for any 1 ≤ j, k ≤ n ∂2u = −RjRkf + Pjk , ∂xj∂xk where Pjk is a polynomial.

Proof. Taking the Fourier transform of (6.9) we have −4π2|ξ|2uˆ = fˆ . ∞ n Hence if λ ∈ C (R ) and all its derivatives are slowly increasing, then (6.10) −λ(ξ)(−4π2|ξ|2)û = −λ(ξ)fˆ . According to Corollary 2.51 it suffices to prove that the tempered distribution ∂2u ∧ + RjRkf ∂xj∂xk has support contained in {0}. Let ϕ ∈ Sn be such that 0 6∈ supp ϕ, say ϕ(x) = 0 for |x| ≤ r. We need to show that ∧ ∧ (∂j∂ku) [ϕ] = (−RjRkf) [ϕ] . ∞ n Let η ∈ C (R ) 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 [ηϕ] = ♥ 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. Thus ♥ = − λ(ξ)(−4π2|ξ|2)û[ϕ] = (−λ(ξ)fˆ(ξ))[ϕ] −iξ −iξ = −η(ξ) j k fˆ(ξ) [ϕ] |ξ| |ξ| −iξ −iξ = − j k fˆ(ξ) [ϕ] |ξ| |ξ| ∧ = (−RjRkf) [ϕ] . The proof is complete. 2 100 PIOTR HAJLASZ

Roughly speaking, one dimensional directional sections of the kernel of the Riesz transform deﬁnes one dimensional Hilbert transforms and it is possible p to use this fact to prove boundedness of Rj in L by a so called method of rotations. Since the same method works for a larger class of operators we p postpone the proof of the boundedness of Rj in L to the next section where a more general result will be proved.

6.1. Homogeneous distributions. In this section we will present a different and shorter proof of Theorem 6.1.

n Recall that a function f is homogeneous of degree a if for all 0 6= x ∈ R and t > 0 f(tx) = taf(x).

For such a function and ϕ ∈ Sn we have Z Z a f(x)ϕt(x) dx = t f(x)ϕ(x) dx , Rn Rn −n where ϕt(x) = t ϕ(x). This suggests the following deﬁnition. 0 Definition. We say that a distribution u ∈ Sn is homogeneous of degree a if for any ϕ ∈ Sn a u[ϕt] = t u[ϕ] . Proposition 6.7. If u ∈ S0 is homogeneous of degree a, then uˆ is homogeneous of degree −n − a.

Proof. Recall that for ϕ ∈ Sn we have n ϕbt(ξ) =ϕ ˆ(tξ) and (ϕ ˆ)t−1 = t ϕˆ(tξ) , so −n ϕbt(ξ) = t (ϕ ˆ)t−1 (ξ) . Thus −n −n−a −n−a uˆ[ϕt] = u[ϕbt] = t u[(ϕ ˆ)t−1 ] = t u[ϕ ˆ] = t uˆ[ϕ] . The proof is complete. 2 As an application of this result we will prove the following Proposition 6.8. For n/2 < a < n n a− 2 n−a −a∧ π Γ 2 a−n |x| (ξ) = a |ξ| . Γ 2

Proof. Observe that |x|−a ∈ L1 + L2. Indeed, −a 1 −a 2 |x| χ{|x|≤1} ∈ L and |x| χ{|x|>1} ∈ L , −a 2 −a so the Fourier transform of |x| is function in C0 +L . Since |x| is radially symmetric, the fact that the Fourier transform commutes with rotations HARMONIC ANALYSIS 101 implies that its Fourier transform is radially symmetric too. Lastly, |x|−a is homogeneous of degree −a, so Proposition 6.7 implies that the Fourier transform is homogeneous of degree −n + a. Thus −a∧ a−n |x| (ξ) = ca,n|ξ| and it remains to compute the coeﬃcient ca,n. Employing the fact that e−π|x|2 is a ﬁxed point of the Fourier transform we have Z Z −π|x|2 −a −π|x|2 a−n (6.11) e |x| dx = ca,n e |x| dx . Rn Rn The integrals in this identity are easy to compute. Indeed, for γ > −n we have ∞ Z 2 Z 2 e−π|x| |x|γ dx = sn−1|Sn−1|e−πs sγ ds Rn 0 Z ∞ −πs2 n+γ−1 = nωn e s ds 0 Z ∞ t=πs2 nωn n+γ −t 2 −1 = n+γ e t dt 2π 2 0 n+γ nωn n + γ Γ 2 = n+γ Γ = γ . 2 2 n 2π 2 π Γ 2 Applying this formula to both sides of (6.11) we have n−a a Γ 2 Γ 2 a = ca,n − n a−n n π 2 Γ 2 2 π Γ 2 and hence n−a n Γ a− 2 2 ca,n = π a . Γ 2 The proof is complete. 2

The distribution Wj arises naturally as an attempt to diﬀerentiate the function 1/|x|n−1. Namely we have

1−n 0 Proposition 6.9. If n ≥ 2, then |x| ∈ Sn and its distributional partial derivatives satisfy ∂ 1−n xj |x| [ϕ] = (1 − n) p.v. n+1 . ∂xj |x|

Before we prove the proposition let us recall the integration by parts n n formula for functions deﬁned in a domain in R . If Ω ⊂ R is a bounded domain with piecewise C1 boundary and f, g ∈ C1(Ω), then Z Z (6.12) (∇f(x)g(x) + f(x)∇g(x)) dx = fg~νdσ , Ω ∂Ω 102 PIOTR HAJLASZ where ~ν = (ν1, . . . , νn) is the unit outer normal vector to ∂Ω. Comparing jth components on both sides of (6.12) we have Z ∂f Z ∂g Z (6.13) (x) g(x) dx = − f(x) (x) dx + fg νj dσ . Ω ∂xj Ω ∂xj ∂Ω

Proof of Proposition 6.9. 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 with R → ∞, because the part of the second integral corresponding to the integration over {|x| = R} clearly con- 1−n verges 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 → 0 as ε → 0, |x|=ε |x|=ε because 1−n n ϕ(x) − ϕ(0) |x| νj ≤ Cε , and the result follows. 2 Proof of Theorem 6.1. Applying Proposition 6.9 and then Proposition 6.8 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 2 n−1− 2 −1 = π n−1 |ξ| 1 − n Γ 2 n+1 π 2 −iξj = n+1 , Γ 2 |ξ| Where we used facts that n − 1 n − 1 n + 1 Γ(1/2) = π1/2 and Γ = Γ . 2 2 2 HARMONIC ANALYSIS 103

The proof is complete. 2

6.2. The Bochner-Hecke formula. Now we will generalize Theorem 6.1. n+1 The distribution Wj is, up to a constant, the principal value of xj/|x| . The function xj is harmonic and thus Theorem 6.1 follows also from a more general formula for the Fourier transform of P (x) p.v. k , |x|n+k where Pk(x) is a homogeneous harmonic polynomial of degree k ≥ 1.

Definition. We say that Pk(x) is a homogeneous harmonic polynomial of degree k if X α Pk(x) = aαx and ∆Pk = 0 . |α|=k Let us start with a general observation. For Ω ∈ L1(Sn−1) with Z (6.14) Ω(θ) dσ(θ) = 0 Sn−1 we define Ω(x/|x|) K (x) = , x 6= 0 . Ω |x|n 1 As in the case of the Riesz transform KΩ 6∈ L , so in order to define KΩ as a tempered distribution we need to consider the principal value of the integral. For ϕ ∈ Sn we define Z WΩ[ϕ] = p.v. KΩ(x)ϕ(x) dx Rn Z = lim KΩ(x)ϕ(x) dx ε→0 |x|≥ε Z = lim KΩ(x)ϕ(x) dx ε→0 ε≤|x|≤ε−1

As in the case of Theorem 5.1 one can prove that for ϕ ∈ Sn the limit exists 0 and deﬁnes WΩ ∈ Sn. Note that the condition (6.14) plays an essential role in the proof.

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 −k where the function Ω(x) = Pk(x)|x| satisﬁes Z Z (6.15) Ω(θ) dσ(θ) = Pk(θ) dσ(θ) = 0 . Sn−1 Sn−1 104 PIOTR HAJLASZ

Indeed, 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 formula yields Z Z Z ∂Pk k Pk(θ) dσ(θ) = (θ) dσ(θ) = ∆Pk dx = 0 . Sn−1 Sn−1 ∂ν Bn Thus 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 6.10. 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 Γ k γ = (−i)kπn/2 2 . k k+n Γ 2

Note that this result immediately implies Theorem 6.1. Let us start with an alternative proof of the following fact (see Theo- rem 2.14).

2 ∧ 2 Proposition 6.11. e−πx (ξ) = e−πξ .

Proof. The function e−πz2 is holomorphic and hence its integral along the following curve equals zero. HARMONIC ANALYSIS 105

Letting R → ∞ we obtain ∞ ∞ Z 2 Z 2 e−πx dx = e−π(x+iξ) dx . −∞ −∞ The left hand side equals 1, so ∞ Z 2 2 1 = e−πx e−2πixξeπξ dx . −∞ Hence ∞ 2 Z 2 2 ∧ e−πξ = e−πx e−2πixξ dx = e−πx (ξ) . −∞ The proof is complete. 2 By the same argument involving the same contour integration, for any polynomial P we have ∞ ∞ Z 2 Z 2 (6.16) P (x)e−π(x+iξ) dx = P (x − iξ)e−πx dx . −∞ −∞ If P is a polynomial in n variables, then (6.16) and the Fubini theorem yield Z Z −π P (x +iξ )2 2 (6.17) P (x)e j j j dx = P (x − iξ)e−π|x| dx . Rn Rn

Theorem 6.12 (Bochner-Hecke). If Pk(x) is a homogeneous harmonic polynomial of degree k, then ∧ −π|·|2 k −π|ξ|2 Pk(·)e (ξ) = (−i) Pk(ξ)e .

Proof. Applying the diﬀerential operator Pk(Dξ) to both sides of the identity Z 2 2 e−π|x| e−2πix·ξ dx = e−π|ξ| Rn we see that Z −π|x|2 −2πix·ξ −π|ξ|2 (6.18) Pk(x)e e = Q(ξ)e Rn for some polynomial Q and it remains to prove that Q(ξ) = Pk(−iξ). Mul- tiplying both sides of (6.18) by eπ|ξ|2 and applying (6.17) we have

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 106 PIOTR HAJLASZ 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

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. 2

Using homogeneity of Pk and the second part of Theorem 2.7(e) one can easily deduce from the Bochner-Hecke formula the folloiwing result.

Corollary 6.13. If Pk is a homogeneous harmonic polynomial of degree k, then for any t > 0 we have ∧ −πt|·|2 −k− n k −π|ξ|2/t Pk(·)e (ξ) = t 2 (−i) Pk(ξ)e .

We leave details as an easy exercise. We will deduce Theorem 6.10 from the following result.

Theorem 6.14. If Pk is a homogeneous harmonic polynomial of order k and 0 < α < n, then P (x) ∧ P (ξ) k (ξ) = γ k , |x|k+n−α k,α |ξ|k+α where k+α k n −α Γ 2 γ = (−i) π 2 . k,α k+n−α Γ 2

Remark. Note that the function

Pk(x) |x|k+n−α HARMONIC ANALYSIS 107 is a tempered L1 function, so it deﬁnes a tempered distribution without necessity of taking the principal value of the integral.

Proof. For any t > 0 and ϕ ∈ Sn Corollary 6.13 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 2 e−πt|x| tβ−1 dt = (π|x|2)−βΓ(β) 0 the integral on the left hand side will be equal to Z ∧ Γ(β) Pk(x) Γ(β) Pk(·) (6.19) β k+n−α ϕˆ(x) dx = β k+n−α [ϕ] . π Rn |x| π | · | 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. 2

Proof of Theorem 6.10. 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| and let α → 0. The right hand side converges to k Z n Γ Pk(x) k 2 2 (−i) π k+n k ϕ(x) dx . n |x| Γ 2 R 108 PIOTR HAJLASZ

To compute the limit on the left hand side we observe that the integral of −(k+n−α) Pk(x)|x| on the unit ball equals zero, see (6.15) 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. 2

6.3. How to diﬀerentiate functions. We plan to generalize Proposi- tion 6.9 to a class of more general functions. Let’s start with the following elementary result. 1 n Theorem 6.15. Suppose that K ∈ C (R \{0}) is such that both K and |∇K| have polynomial growth 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 distributions 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 (6.13) 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 HARMONIC ANALYSIS 109 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 K(x)ϕ(x) ν dσ ≤ εn−1 = Cεα → 0 as ε → 0. j n−1−α |x|=ε ε The proof is complete. 2 An interesting problem is the case α = 0, i.e. when K and ∇K satisfy the estimates C C (6.21) |K(x)| ≤ , |∇K(x)| ≤ for x 6= 0. |x|n−1 |x|n One such situation was described in Proposition 6.9. Here we make an additional assumption about K. We assume that K ∈ 1 n C (R \{0}) is homogenelus of degree 1 − n, i.e. K(x/|x|) K(x) = for x 6= 0. |x|n−1 Since K is bounded in {|x| ≥ 1} and integrable in {|x| ≤ 1} we have K ∈ 0 Sn and there is no need to interpret K through the principal value of the integral. The first estimate at (6.21) 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 (6.21) 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. To do this we have to check if the condition (6.14) is satisfied. 1 n Theorem 6.16. Suppose that K ∈ C (R \{0}) is homogeneous of degree 1 − n. Then ∇K(x) is homogeneous of degree −n. Moreover Z (6.22) ∇K(θ) dσ(θ) = 0 . Sn−1 110 PIOTR HAJLASZ

Hence the condition (6.14) is satisﬁed and thus

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

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

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|

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 (6.12) 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 (6.22). Therefore p.v. ∇K(x) ∈ Sn is a well deﬁned tempered distribution. We are left with the proof that the distributional gradient ∇K HARMONIC ANALYSIS 111 satisﬁes (6.23). 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| The last equality follows from a simple change of variables and homogeneity of K. We have Z x K(x)ϕ(x) dσ(x)(6.24) |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 (6.24) converges to 0 with ε → 0. 2

6.4. Integral representations of functions. A straightforward application of the above theorem gives a well known formula for the fundamental solution to the Laplace equation. Theorem 6.17. For n ≥ 2 we have

∆Φ = δ0 , where  1 log |x| if n = 2,  2π Φ(x) = 1 1  − n−2 if n ≥ 3. n(n−2)ωn |x| 112 PIOTR HAJLASZ

Proof. We will prove the theorem for n ≥ 3, but a similar argument works for n = 2. According to Theorem 6.15 1 x (6.25) ∇Φ = n nωn |x| in the sense of distributions.32 Note that the function Φ(x) is harmonic in n R \{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 6.16 gives a formula for the distributional Laplacean 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. 2

33 ∞ 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 to the Poisson equation ∆u = ϕ . This explains the importance of the fundamental solution in partial diﬀer- ential 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| 32Formula (6.25) is also true for n = 2. 33 0 As a convolution of Φ ∈ Sn with ϕ ∈ Sn. HARMONIC ANALYSIS 113

n for every x ∈ R . In the last equality we employed (6.25). Thus we proved Theorem 6.18. 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 6.19. 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

n Proof. Fix x ∈ R and deﬁne X (x − y)β ψ(y) = Dβϕ(y) . β! |β|≤m−1 Then ψ(x) = ϕ(x) and34 ∂ψ X (x − y)β (y) = Dβ+δj ϕ(y) ∂yj β! |β|=m−1 35 where δj = (0,..., 1,..., 0). Hence Theorem 6.18 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. 2 For 0 < α < n and n ≥ 2 we deﬁne the Riesz potentials by 1 Z f(y) (Iαf)(x) = n−α dy , γ(α) Rn |x − y| 34 We compute ∂ψ/∂yj using the Leibniz rule and observe that the lower order terms cancel out. 351 on jth coordinate, 0 otherwise. 114 PIOTR HAJLASZ where n 2 α α π 2 Γ 2 γ(α) = n−α . Γ 2 In particular, when n ≥ 3, I2f is the convolution with the fundamental solution to the Laplace operator taken with the minus sign, so for ϕ ∈ Sn

−∆(I2ϕ)(x) = ϕ(x) .

If α = 1, then, up to a constant, I1f is a convolution with the distribution 1−n 0 |x| ∈ Sn. n−1 Z Γ 2 f(y) (I1f)(x) = n+1 n−1 dy . 2 π 2 Rn |x − y| ∞ Hence for ϕ ∈ Sn, I1ϕ ∈ C and Proposition 6.9 gives n−1 Z ∂ Γ 2 xj − yj (I1ϕ)(x) = (1 − n) n+1 p.v. n+1 ϕ(y) dy = −Rjϕ(x). ∂xj 2 π 2 Rn |x − y| Thus the Riesz operators appear naturally as derivatives of the integral operator I1ϕ. We proved

Proposition 6.20. If n ≥ 2 and ϕ ∈ Sn, then for 1 ≤ j ≤ n n−1 Z Γ 2 ∂ ϕ(y) n+1 n−1 dy = −Rjϕ(x) . 2 π 2 ∂xj Rn |x − y|

7. Singular integrals I

For Ω ∈ L1(Sn−1) with Z (7.1) Ω(θ) dσ(θ) = 0 Sn−1 we deﬁne Ω(x/|x|) K (x) = , x 6= 0 . Ω |x|n

Then for ϕ ∈ Sn we deﬁne the tempered distribution Z WΩ[ϕ] = p.v. KΩ(x)ϕ(x) dx Rn Z = lim KΩ(x)ϕ(x) dx ε→0 |x|≥ε

As in the case of Theorem 5.1 one can prove that for ϕ ∈ Sn the limit exists 0 and deﬁnes WΩ ∈ Sn. Note that the condition (7.1) plays an essential role in the proof. The following result generalizes Theorem 6.1. HARMONIC ANALYSIS 115

Theorem 7.1. Let n ≥ 2 and Ω ∈ L1(Sn−1) be such that Z Ω(θ) dσ(θ) = 0 . Sn−1 Then the Fourier transform of the distribution WΩ is a ﬁnite a.e. function given by the formula Z 1 iπ WdΩ(ξ) = Ω(θ) log − sgn (ξ · θ) dσ(θ)(7.2) 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|.

Before we prove the theorem we start with some auxiliary results. Lemma 7.2. 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 n for all x ∈ R \{0}.

This result follows from arguments similar to those used to establish (6.7); we leave details to the reader as an exercise.

Lemma 7.3. 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 (7.3) ds ≤ 2khk∞ log . ε s λ Moreover Z N h(λs) − h(µs) µ (7.4) 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 (7.3). Since Z µε h(s) µ ds → h(0) log as ε → 0 λε s λ 116 PIOTR HAJLASZ and Z µN h(s) ds → 0 as N → ∞ λN s (7.4) follows. 2 Corollary 7.4. 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 the integral is bounded by a constant independent of ε and N.

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 and the result follows from Lemma 7.3. 2 Proof of Theorem 7.1. First observe that the last equality in (7.2) follows from 1 1 1 log = log + log , |ξ · θ| |ξ| |ξ0 · θ| and the fact that Z 1 Ω(θ) log dσ(θ) = 0 . Sn−1 |ξ| Let F (ξ) be the function defined by the integral in the second line of (7.2). We will show first that F is finite a.e. and that it actually defines a tempered distribution. Note that Z iπ Ω(θ) sgn (ξ0 · θ) dσ(θ) Sn−1 2 is a bounded function of ξ, 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| We need to show that the integral is finite for a.e. ξ and that Z G[ϕ] = G(ξ)ϕ(ξ) dξ Rn is a tempered distribution, ϕ ∈ Sn. To this end it suffices to show that the function G(ξ)ϕ(ξ) is integrable along with suitable estimates for its integral. HARMONIC ANALYSIS 117

We have Z Z 1 |G[ϕ]| ≤ |ϕ(ξ)| |Ω(θ)| log 0 dσ(θ) dξ Rn Sn−1 |ξ · θ| Z Z ∞ Z n−1 0 1 0 = |Ω(θ)| s |ϕ(sξ )| log 0 dσ(ξ ) ds dσ(θ) Sn−1 0 Sn−1 |ξ · θ| = ♥ . Lemma 7.2 gives Z 1 0 log 0 dσ(ξ )(7.5) Sn−1 |ξ · θ| Z 1 1 2 n−3 = (n − 1)ωn−1 log (1 − s ) 2 ds = cn < ∞ −1 |s| and hence Z Z ∞ n−1 ♥ ≤ cn |Ω(θ)| s sup |ϕ(ξ)| ds dσ(θ) Sn−1 0 |ξ|=s ! n+1 ≤ CkΩkL1(Sn−1) kϕk∞ + sup |ξ| |ϕ(ξ)| . ξ∈Rn This proves that G is ﬁnite a.e. and deﬁnes a tempered distribution, so does F . Now we are ready to prove formula (7.2). Let ξ0 = ξ/|ξ|. We have

WdΩ[ϕ] = WΩ[ϕ ˆ] Z Ω(x/|x|) = lim n ϕˆ(x) dx ε→0 |x|≥ε |x| 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| 2 118 PIOTR HAJLASZ by Corollary 7.4 and hence 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, i.e. Ω(θ) = −Ω(−θ), then the integral of Ω against log(1/|ξ · θ|) vanishes and hence Z iπ 0 WdΩ(ξ) = − Ω(θ) sgn (ξ · θ) dσ(θ) . 2 Sn−1

In particular the Fourier transform WdΩ is bounded. More generally any function Ω on Sn−1 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 7.5. 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.

Proof. A calculation similar to (7.5) implies that log(1/|ξ0 ·θ|) is integrable over Sn−1 with any positive exponent. In particular it belongs to Lq0 (Sn−1) and hence Z 1 iπ 0 Ωe(θ) log 0 − sgn (ξ · θ) dσ(θ) ≤ CkΩekLq(Sn−1) , Sn−1 |ξ · θ| 2 with a constant C independent of ξ. Now the formula (7.2) yields Z iπ 0 WdΩ(ξ) = − Ωo(θ) sgn (ξ · θ) dσ(θ) 2 Sn−1 Z 1 iπ 0 + Ωe(θ) log 0 − sgn (ξ · θ) dσ(θ) Sn−1 |ξ · θ| 2 from which we have

kWdΩk∞ ≤ C(kΩokL1(Sn−1) + kΩekLq(Sn−1)) . The proof is complete. 2

Convolutions with WΩ, i.e. operators of the form Z Ω(y/|y|) TΩf(x) = (WΩ ∗ f)(x) = lim n f(x − y) dy ε→0 |y|≥ε |y| are called singular integrals. Examples include the Hilbert transform and the Riesz transforms. HARMONIC ANALYSIS 119

If Ω satisfies assumptions of Corollary 7.5, then the singular integral TΩ 2 uniquely extends from Sn to a bounded operator on L . However we are interested to see whether there is a more direct formula to define TΩf(x) when f ∈ L2 without using the density argument. Moreover we want to know if the operator is bounded in Lp for p 6= 2. More precisely we want to know if Corollary 5.17 proved for the Hilbert transform generalizes to the more general class of singular integrals TΩ. Definition. For a function Ω ∈ L1(Sn−1) with vanishing integral and f ∈ Lp(Sn−1), 1 ≤ p < ∞ we define the truncated singular integral Z (ε,N) Ω(y/|y|) TΩ f(x) = n f(x − y) dy, 0 < ε < N . ε≤|y|≤N |y| Note that (ε,N) kTΩ fkp ≤ kΩkL1(Sn−1) log(N/ε)kfkp (ε,N) and hence TΩ f is well defined. The maximal singular integral is defined by ∗ (ε,N) TΩf(x) = sup |TΩ f(x)| . 0<ε0 and hence in the case in which Ω is bounded we could define the maximal singular operator as the left hand side of (7.6).

7.1. The method of rotations. The following result proves boundedness of singular integrals in Lp, 1 < p < ∞. The proof is based on the so called method of rotations. 1 n−1 ∗ Theorem 7.6. If Ω ∈ L (S ) is an odd function, then TΩ is of strong type (p, p) for all 1 < p < ∞. In particular TΩ uniquely extends from Sn to p n a bounded operator in L (R ), 1 < p < ∞.

36This covers the case of the Hilbert transform, the Riesz transform and more generally n+k the case of transforms with the kernel Pk(x)/|x| . 120 PIOTR HAJLASZ

Proof. The idea is to show that the singular integral TΩf is an aver- n age over all possible directions in R of one dimensional directional Hilbert transforms and then the result will follow from the corresponding results about boundedness of one dimensional Hilbert transform.

If e1 is the direction of the first coordinate, then the operator Z ∞ 1 f(x − te1) He1 f(x) = p.v. dt π −∞ t p n n−1 is bounded in L (R ), 1 < p < ∞. Indeed, for a.e. (x2, . . . , xn) ∈ R the p function z 7→ f(z, x2, . . . , xn) belongs to L (R) and hence the one dimensional truncated Hilbert transform applied to the first coordinate Z ε 1 f(z − t, x2, . . . , xn) H f(z, x2, . . . , xn) = dt π |t|≥ε t p converges a.e. with respect to z and in L (R) to the Hilbert transform applied to the first coordinate, which is He1 f(z, x2, . . . , xn), see Corollary 5.17. Hence the Fubini theorem yields Z Z p kHe1 fkp = ... |He1 f(z, x2, . . . , zn)| dz dx2, . . . dxn R R Z Z p p ≤ C(p) ... |f(z, x2, . . . , xn)| dz dx2 . . . dxn R R p p = C(p) kfkp , p where C(p) stands for the norm of the Hilbert transform in L (R). Now for a direction θ ∈ Sn−1 we define the directional Hilbert transform as 1 Z ∞ f(x − tθ) Hθf(x) = p.v. dt . π −∞ t The directional Hilbert transform commutes with the orthogonal transformations, i.e. for ρ ∈ O(n) −1 Hρ(e1)f(x) = He1 (f ◦ ρ)(ρ x) .

This identity follows immediately from the deﬁnition of Hθf. Hence Hθ is bounded in Lp with the norm independent of θ

kHθfkp ≤ C(p)kfkp . p n Similarly for f ∈ L (R ), 1 ≤ p < ∞ we deﬁne Z (ε,N) 1 f(x − tθ) Hθ f(x) = dt , π ε≤|t|≤N t

∗ (ε,N) (7.7) Hθ f(x) = sup |Hθ f(x)| . 0<ε

The two operators also commute with the orthogonal transformations and ∗ p hence boundedness of Hθ in L , 1 < p < ∞ will follow from the boundedness ∗ if He1 . Observe that for the one dimensional Hilbert transform |H(ε,N)g(x)| ≤ |HN g(x)| + |Hεg(x)| and hence sup |H(ε,N)g(x)| ≤ 2H∗g(x) 0<ε0 ∗ p Thus boundedness of He1 in L , 1 < p < ∞ follows from Corollary 5.16 and ∗ p the Fubini theorem. Then also Hθ is bounded in L , 1 < p < ∞ with a constant independent of θ ∗ kHθ fkp ≤ 2A(p)kfkp , p ∗ p p where A(p) is the L norm of the operator H : L (R) → L (R).

Now we will show how to represent the singular integral TΩf as an average p n of directional Hilbert transforms. For f ∈ L (R ), 1 ≤ p < ∞ we have Z Ω(y/|y|) Z Z N f(x − tθ) n f(x − y) dy = + Ω(θ) dt dσ(θ) ε≤|y|≤N |y| Sn−1 ε t Z Z N f(x + tθ) = − Ω(θ) dt dσ(θ) . Sn−1 ε t The ﬁrst equality is just the representation of the integral in polar coordinates, while the second one follows from the change of variables θ 7→ −θ and the fact that Ω(−θ) = −Ω(θ). Hence Z Ω(y/|y|) (7.8) n f(x − y) dy ε≤|y|≤N |y| 1 Z Z N f(x − tθ) − f(x + tθ) = Ω(θ) dt dσ(θ) 2 Sn−1 ε t Z π (ε,N) = Ω(θ)Hθ f(x) dσ(θ) . 2 Sn−1 Thus Z ∗ π ∗ TΩf(x) ≤ |Ω(θ)|Hθ f(x) dσ(θ) 2 Sn−1 and for 1 < p < ∞ the Minkowski integral inequality yields ∗ kTΩfkp ≤ πA(p)kΩkL1(Sn−1)kfkp .

37 ∗ ∗ Observe that the definition of H g is not consistent with the definition of Hθ f. In the first definition we take supremum over ε > 0, while in the second one supremum over 0 < ε < N. However, we need the maximal function H∗g, because we want to apply Corollary 5.16. 122 PIOTR HAJLASZ

The proof is complete. 2

(ε,N) Note that for ϕ ∈ Sn, Hθ ϕ(x) is bounded by a constant independent of ε, N and θ, so passing to the limit in (7.8) yields π Z TΩϕ(x) = Ω(θ)Hθϕ(x) dσ(θ) . 2 Sn−1 1 n−1 p n Corollary 7.7. If Ω ∈ L (S ) is an odd function, and f ∈ L (R ), (ε,N) p 1 < p < ∞, then TΩ f → TΩf a.e. and in L as ε → 0 and N → ∞. If in ε p addition Ω is a bounded function, then TΩf → TΩf a.e. and in L as ε → 0.

The proof is almost the same as that for Corollary 5.17; we leave details to the reader. One can actually prove a stronger result. 1 n−1 q n−1 Theorem 7.8. If Ω ∈ L (S ) has vanishing integral and Ωe ∈ L (S ) for some q ≥ 1, then for any 1 0 such that

kTΩϕkp ≤ Ckϕkp, ϕ ∈ Sn . p Hence TΩ uniquely extends to a bounded operator on L . Moreover for f ∈ p n (ε,N) p L (R ), TΩ f → TΩf a.e. and in L as ε → 0, N → ∞.

The estimate for the odd part of Ω was done above, so we can assume q that Ω is even, i.e. Ω = Ωe ∈ L . In this case the method of rotations cannot be applied directly. However since n X 2 Rj = −I j=1 we have, at least formally, n n X 2 X TΩf = − Rj (TΩf) = − Rj(RjTΩf) . j=1 j=1

The operator RjTΩ is odd as a composition of an even and an odd operator, so we can apply the method of rotations to estimate it. However, the details are not as easy as they seem and we will prove it.

8. Singular integrals II

So far we investigated boundedness of singular integrals for 1 < p < ∞, without investigating the case p = 1, but it turns out that a more powerful method is based on the weak (1, 1) estimates and the Marcinkiewicz interpolation theorem. First we will consider the simplest case of the Hilbert transform. HARMONIC ANALYSIS 123

8.1. Hilbert transform again. We proved that the Hilbert transform is bounded in Lp, 1 < p < ∞ by two diﬀerent methods. We will add one more method now. Namely we will show that the Hilbert transform is of weak type (1, 1). This fact, the Marcinkiewicz interpolation theorem and a duality argument will easily imply boundedness of the Hilbert transform in Lp for all 1 t}| ≤ kfk . R t 1

In the proof we will use boundedness of the Hilbert transform in L2, Theorem 5.3, and we will not refer to any result proved after Theorem 5.3. In particular the Hilbert transform in Kolmogorov’s theorem is deﬁned on L2 as an extension from S(R). As we already mentioned Kolmogorov’s theorem implies Corollary 8.2 (Riesz). The Hilbert transform is bounded in Lp, 1 < p < ∞,

kHfkp ≤ Ckfkp for f ∈ S(R).

Proof. Since the Hilbert transform is bounded in L2 (Theorem 5.3) and of weak type (1, 1), the Marcinkiewicz interpolation theorem implies that for all 1 < p < 2, kHfkp ≤ Cpkfkp, f ∈ S(R) . Now the duality argument, Theorem 4.6, yields that for all 2 0 we have Z Z Hεf · g = f · Hεg . R R Letting ε → 0 and using the fact that Hεu → Hu in L2 for u ∈ L2 (Theo- rem 5.3) we have Z Z Hf · g = f · g . R R Now if 2 < p < ∞, then Z

kHfkp = sup Hf · g : kgkp0 ≤ 1 R Z

= sup f · Hg : kgkp0 ≤ 1 R ≤ kfkp sup{kHgkp0 : kgkp0 ≤ 1}

≤ Cp0 kfkp . 124 PIOTR HAJLASZ

The proof is complete. 2

Proof of Theorem 8.1. We can assume that f ≥ 0. Indeed, every complex- valued function is a linear combination of nonnegative functions: positive and negative parts of real and imaginary part of the function.

We ﬁx t > 0 and apply the Calder´on-Zygmund decomposition (Theo- rem 1.13 and Corollary 1.14) to f and α = t. We obtain non-overlapping intervals {Ij} such that S f(x) ≤ t for a.e. x 6∈ Ω = j Ij,

1 |Ω| ≤ kfk , t 1 Z t ≤ f ≤ 2t , j = 1, 2, 3,... Ij This allows us to represent f as a sum of two functions f = g + b (good and bad) that are deﬁned as follows

 f(x) if x 6∈ Ω,   g(x) = Z  f if x ∈ I  j Ij and ∞ X b(x) = bj(x) , j=1 where Z !

bj(x) = f(x) − f χIj (x) . Ij Note that g ∈ L∞ ∩ L2. Indeed, 0 ≤ g ≤ 2t and Z Z 2 (8.1) g(x) dx ≤ 2t g(x) dx = 2tkfk1 , R R 2 Hence also b ∈ L (R). Actually it is easy to see that Z Z 2 2 |bj(x)| dx ≤ 4 |f(x)| dx , R Ij P 2 so the series j bj converges to b in L . Since Hf = Hg + Hb we have |{|Hf| > t}| ≤ |{|Hg| > t/2}| + |{|Hb| > t/2}| . HARMONIC ANALYSIS 125

The estimate for the ﬁrst term on the right hand side is easy, which explains the name “good” for g. We have 22 Z |{|Hg| > t/2}| ≤ |Hg(x)|2 dx t R 22 Z = |g(x)|2 dx t R 8 ≤ kfk . t 1 We used here the fact that H is an isometry on L2 and inequality (8.1). The estimate for the second term is more involved (so the name “bad” for b). Let 2Ij be the interval concentric with Ij of twice the length. Deﬁne ∗ [ Ω = 2Ij . j Clearly 2 |Ω∗| ≤ 2|Ω| ≤ kfk . t 1 We have ∗ ∗ |{x ∈ R : |Hb(x)| > t/2}| ≤ |Ω | + |{x 6∈ Ω : |Hb(x)| > t/2}| 2 2 Z ≤ kfk1 + |Hb(x)| dx . t t R\Ω∗ Thus to complete the proof it remains to show that Z |Hb(x)| dx ≤ Ckfk1 . R\Ω∗ Observe that ∞ X (8.2) |Hb(x)| ≤ |Hbj(x)| a.e. j=1 Indeed, for every k we have   k k ∞ X X X H  bj ≤ |Hbj| ≤ |Hbj| a.e.

j=1 j=1 j=1 Pk 2 and it remains to use the fact that j=1 bj converges to b in L as k → ∞ and that H is bounded in L2. Inequality (8.2) gives Z X Z X Z |Hb(x)| ≤ |Hbj(x)| dx ≤ |Hbj(x)| dx . ∗ ∗ R\Ω j R\Ω j R\2Ij

Note that for x 6∈ 2Ij Z 1 bj(y) Hbj(x) = dy , π Ij x − y 126 PIOTR HAJLASZ because bj vanishes outside Ij, x is away from Ij and hence there is no singularity in the denominator. Let cj be the center of the interval Ij. Since the integral of bj equals 0 we have

Z Z Z b (y) |Hb (x)| dx = j dy dx j x − y R\2Ij R\2Ij Ij

Z Z 1 1 = b (y) − dy dx j x − y x − c R\2Ij Ij j ! Z Z |y − c | ≤ |b (y)| j dx dy j |x − y| |x − c | Ij R\2Ij j ! Z Z |I | ≤ |b (y)| j dx dy . j |x − c |2 Ij R\2Ij j

The last inequality follows from a simple geometric observation that |y−cj| < |Ij|/2 and |x − y| > |x − cj|/2. Since Z |I | j dx = 2 |x − c |2 R\2Ij j we have Z X Z |Hb(x)| dx ≤ |Hbj(x)| dx ∗ R\Ω j R\2Ij X Z ≤ 2 |bj(y)| dy j Ij

≤ 4kfk1 . The proof is complete. 2 In order to prove pointwise and Lp convergence of Hεf to Hf for f ∈ Lp, 1 < p < ∞ we needed to prove boundedness of the maximal Hilbert transform in Lp. We complement this result by showing that the maximal Hilbert transform is of weak type (1, 1). Theorem 8.3. H∗ is of weak type (1, 1). More precisely C |{x : : H∗f(x) > t}| ≤ kfk for all f ∈ L1( ). R t 1 R

Proof. The proof follows similar steps to those employed in the proof of Theorem 8.1. We can assume that f ≥ 0. Fix t > 0 and apply the Calder´on- Zygmund decomposition to f and α = t. As in the proof of Theorem 8.1 we decompose f = g + b, so |{H∗f > t}| ≤ |{H∗g > t/2}| + |{H∗b > t/2}| . HARMONIC ANALYSIS 127

Note that g ∈ L1 ∩ L∞ and hence g ∈ L2. Since 0 ≤ g ≤ 2t we have Z Z 2 g(x) dx ≤ 2t g(x) dx = 2tkfk1 . R R Since H∗ is bounded in L2 (Theorem 5.16) we get C |{H∗g > t/2}| ≤ kfk t 1 by the same argument as in the proof of Theorem 8.1 and in order to estimate |{H∗b > t/2}| it suﬃces to show that C |{x 6∈ Ω∗ : H∗b(x) > t/2}| ≤ kfk . t 1 ∗ Fix x 6∈ Ω , ε > 0 and a function bj. Recall that the function vanishes outside Ij. Denote by cj the center of Ij. Clearly x∈ / 2Ij and one of the following conditions is satisﬁed

(a) (x − ε, x + ε) ∩ Ij = Ij; (b) (x − ε, x + ε) ∩ Ij = ∅; (c) x − ε ∈ Ij or x + ε ∈ Ij.

ε In the ﬁrst case H bj(x) = 0 and in the second one Z Z ε bj(y) 1 1 H bj(x) = dy = − bj(y) dy . Ij x − y Ij x − y x − cj In either the ﬁrst or second case we have Z Z ε |Ij| 2|Ij| (8.3) |H bj(x)| ≤ 2 |bj(y)| dy ≤ 2 |f(y)| dy . |x − cj| Ij |x − cj| Ij

In the third case, since x 6∈ 2Ij, we have Ij ⊂ (x−3ε, x+3ε) and |x−y| > ε/3 for all y ∈ Ij, so Z Z x+3ε ε |bj(y)| 3 (8.4) |H bj(x)| ≤ dy ≤ |bj(y)| dy . Ij |x − y| ε x−3ε P 1 Since b = j bj ∈ L and the function 1 y 7→ χ x − y {y: |x−y|≥ε} is bounded we have Z X bj(y) X Hεb(x) = dy = Hεb (x) x − y j j |x−y|≥ε j for every x, so ε X ε |H b(x)| ≤ |H bj(x)| j 128 PIOTR HAJLASZ everywhere. Adding up the estimates (8.3) and (8.4) for all j we obtain

Z Z x+3ε X 2|Ij| 3 |Hεb(x)| ≤ |f(y)| dy + |b(y)| dy . |x − c |2 ε j j Ij x−3ε | {z } | {z } ≤18Mb(x) h(x)

Since this estimate is valid for every x and every ε > 0 taking the supremum over ε we have H∗b(x) ≤ h(x) + 18Mb(x) . Thus

|{x 6∈ Ω∗ : H∗b(x) > t/2}| ≤ |{x 6∈ Ω∗ : h(x) > t/4}| + |{x ∈ R : Mb(x) > t/72}| 4 Z C Z ≤ |h(x)| dx + |b(x)| dx t R\Ω∗ t R C ≤ kfk , t 1 because Z |b(x)| dx ≤ 2kfk1 R and Z Z Z X 2|Ij| |h(x)| dx ≤ |f(y)| dy 2 dx ≤ 4kfk1 . ∗ |x − c | R\Ω j Ij R\2Ij j | {z } 4 The proof is complete. 2

Applying Theorem 5.18 we immediately obtain

1 Theorem 8.4. For all f ∈ L (R) the limit Hf(x) = lim Hεf(x) ε→0 exists a.e. and C |{x ∈ : Hf(x) > t}| ≤ kfk . R t 1

The last inequality is a slight improvement of Theorem 8.1, because it is true for all f ∈ L1 and not only for f ∈ L1 ∩ L2. Note that in this theorem the Hilbert transform of f ∈ L1 is deﬁned as the pointwise limit and not through the density argument. HARMONIC ANALYSIS 129

8.2. Calder´on-Zygmund theory of singular integrals. The method of the proof of boundedness of the Hilbert transform in Lp presented above easily generalizes to the case of multi-dimensional singular integrals.

0 Definition. Let K ∈ Sn be a tempered distribution such that

∞ n (a) Kˆ ∈ L (R ); n (b) K coincides with a locally integrable function in R \{0}; (c) K(x), x 6= 0 satisﬁes the H¨ormandercondition Z |K(x − y) − K(x)| dx ≤ B |x|>2|y| n for some constant B > 0 and all y ∈ R . Then the convolution operator

T ϕ(x) = (K ∗ ϕ)(x), ϕ ∈ Sn is called a singular integral. At this moment it is not clear how the class of singular integrals defined here is related to that defined in Section 7 as convolution with WΩ. The two classes are not the same, but strongly related. We will investigate this relationship later, but now we will prove the main result about boundedness in Lp of the singular integrals defined here. Observe that the condition (a) implies

kT ϕk2 ≤ kKˆ k∞ kϕk2, ϕ ∈ Sn 2 n so the operator T uniquely extends to a bounded operator in L (R ). Theorem 8.5 (Calder´on-Zygmund). If T is a singular integral as deﬁned above, then 2 p (8.5) kT fkp ≤ Cpkfkp for f ∈ L ∩ L , 1 t}| ≤ kfk for f ∈ L1 ∩ L2. R t 1

Remark. In this theorem T f for f ∈ Lp ∩ L2 is understood as the exten- 2 sion of T from Sn to L . Proof. The proof is similar to that of Theorem 8.1 and 8.2. We will prove first (8.6). We can assume that f ≥ 0. Fix t > 0 and apply the Calderón- Zygmund decomposition to f and α = t. We have S f(x) ≤ t for a.e. x 6∈ Ω = j Qj 1 |Ω| ≤ kfk , t 1 130 PIOTR HAJLASZ Z t ≤ f ≤ 2nt, j = 1, 2, 3,... Qj Next we decompose f = g + b, where  f(x) if x 6∈ Ω,   g(x) = Z  f if x ∈ Q  j Qj and ! X Z b(x) = bj(x) , bj(x) = f(x) − f χQj (x) . j Qj Observe that g ∈ L1 ∩ L∞, 0 ≤ g(x) ≤ 2nt and hence Z Z 2 n n g(x) dx ≤ 2 t g(x) dx = 2 tkfk1 . Rn Rn Moreover Z Z 2 2 |bj| ≤ 4 |f| , n R Qj P 2 2 so the series j bj converges to b in L . Since T is bounded in L we easily conclude that X |T b(x)| ≤ |T bj(x)| a.e. j We have |{|T f| > t}| ≤ |{|T g| > t/2}| + |{|T b| > t/2}| . The estimate for the first term on the right hand side is easy 2 Z Z 0 2 2 C 2 C |{|T g| > t/2}| ≤ |T g| ≤ 2 |g| ≤ kfk1 . t Rn t Rn t ∗ √ √ Let Qj = 2 nQj be a cube concentric with Qj hose sides are 2 n times longer and let ∗ [ ∗ Ω = Qj . j Clearly C0 |Ω∗| ≤ C|Ω| ≤ kfk . t 1 We have n ∗ ∗ |{x ∈ R : |T b(x)| > t/2}| ≤ |Ω | + |{x 6∈ Ω : |T b(x)| > t/2}| C 2 Z ≤ kfk1 + |T b(x)| dx t t Rn\Ω∗ C 2 X Z ≤ kfk1 + |T bj(x)| dx t t n ∗ j R \Qj HARMONIC ANALYSIS 131 and it remains to show that Z Z |T bj(x)| dx ≤ C |f(x)| dx . n ∗ R \Qj Qj ∗ ∗ Denote the common center of the cubes Qj and Qj by cj. If x 6∈ Qj , then Z T bj(x) = K(x − y)bj(y) dx . Qj R Since bj(y) dy = 0 we have Qj Z T bj(x) = K(x − y) − K(x − cj) bj(y) dy Qj and hence Z Z Z ! |T bj(x)| dx ≤ |bj(y)| |K(x − y) − K(x − cj)| dx dy n ∗ n ∗ R \Qj Qj R \Qj Z ≤ B |bj(y)| dy Qj Z ≤ 2B |f(y)| dy , Qj because an easy geometric investigation shows that n ∗ n R \ Qj ⊂ {x ∈ R : |x − cj| ≥ 2|y − cj|} and hence the above estimate follows from Hörmander’scondition. This completes the proof of (8.6). Since the operator is bounded in L2 and of weak type (1, 1), the Marcinkiewicz interpolation theorem implies (8.5) for 1 < p < 2 and then the case 2 < p < ∞ follows from the case 1 < p < 2 by a duality argument, Theorem 4.6. 2 The conditions (a) and (c) in the definition of the singular integral seem difficult to verify, so we will investigate now sufficient and easy to verify conditions that imply (a) or (c). We will also compare the class of singular integrals considered in this section to that considered in Section 7. Let us start with the condition (c). 1 n Proposition 8.6. If K ∈ C (R \{0}) is such that C (8.7) |∇K(x)| ≤ x 6= 0 |x|n+1 then the function K satisfies the Hörmander condition.

Proof. Points on the interval connecting x to (x−y) are of the form x−ty, t ∈ [0, 1], so for |x| > 2|y| |x| |x − ty| > 2 132 PIOTR HAJLASZ and hence the mean value theorem gives |y| |K(x − y) − K(x)| ≤ C . |x|n+1 Thus the Hörmandercondition follows upon integration in polar coordinates. 2 The condition (8.7) is very easy to check and it covers majority of singular integrals that appear in applications. For example if Ω ∈ C1(Sn−1), Z Ω(θ) dσ(θ) = 0 Sn−1 and Ω(x/|x|) K(x) = p.v. |x|n then Kˆ ∈ L∞ by Corollary 7.5 and C |∇K(x)| ≤ , x 6= 0 |x|n+1 since ∇K is homogeneous of degree −(n + 1). Thus Z Ω(y/|y|) TΩϕ(x) = (K ∗ ϕ)(x) = lim n ϕ(x − y) dy ε→0 |y|≥ε |y| is the singular integral in the sense described above. Actually a weaker assumption about Ω implies the Hörmandercondition. Theorem 8.7. Let Ω ∈ C(Sn−1) be such that Z 1 ω (t) (8.8) ∞ dt < ∞ , 0 t where n−1 ω∞(t) = sup{|Ω(θ1) − Ω(θ2)| : |θ1 − θ2| ≤ t, θ1, θ2 ∈ S } . Then the function Ω(x/|x|) K(x) = , x 6= 0 |x|n satisfies the Hörmandercondition. In particular the Hörmandercondition is satisfied if Ω is Höldercontinuous with exponent 0 < α ≤ 1.

Remark. (8.8) is called a Dini-type condition. Proof. With the notation ξ0 = ξ/|ξ| we have 0 0 Ω((x − y) ) Ω(x ) |K(x − y) − K(x)| = − |x − y|n |x|n 0 0 |Ω((x − y) ) − Ω(x )| 0 1 1 (8.9) ≤ + |Ω(x )| − |x − y|n |x − y|n |x|n HARMONIC ANALYSIS 133

The function Ω is bounded and the function 1/|x|n satisfies the Hörmander condition by Proposition 8.6 so the integral of the second term in (8.9) over the region {|x| > 2|y|} is bounded by a constant independent of y. To estimate the first term in (8.9) observe that

x − y x |y| − ≤ 2 . |x − y| |x| |x| The inequality easily follows from the picture

√ |y| x − y x 2 x − y x ≥ h = sin α − ≥ − , |x| |x − y| |x| 2 |x − y| |x| Hence Z 0 0 Z |Ω((x − y) ) − Ω(x )| ω∞(2|y|/|x|) n dx ≤ n dx |x|>2|y| |x − y| |x|>2|y| (|x|/2) Z ∞ n n−1 n−1 ω∞(2|y|/t) = 2 |S | t n dt 2|y| t Z 1 n ω∞(t) = 2 nωn dt < ∞ . 0 t The proof is complete. 2

Thus if Ω ∈ C(S1) satisfies Z Ω(θ) dσ(θ) = 0 Sn−1 134 PIOTR HAJLASZ and the Dini-type condition (8.8), then Z Ω(y/|y|) TΩϕ(x) = p.v. n ϕ(x − y) dy Rn |y| is the singular integral as defined above. In particular it is of strong type (p, p), 1 < p < ∞ and of weak type (1, 1). Observe that in Section 7 we proved strong type (p, p), 1 < p < ∞ of the singular integral associated with odd Ω that is just integrable, Theorem 7.6, and now we require some additional Dini-type regularity. It is natural to inquire if in the setting of Theorem 7.6 we can also prove the weak type (1, 1). Surprisingly, the answer is not known. While we have seen situations where the Hörmandercondition was satisfied, we still need to see when the condition (a) holds true. Now we will investigate sufficient conditions for the convolution with the principal value 1 n of K ∈ Lloc(R \{0}) to be a singular integral. This will include investiga- 0 ∞ tion of conditions for p.v.K ∈ Sn and p\.v.K ∈ L . We will consider the following properties Z (8.10) |K(x)| dx ≤ C1 r<|x|<2r for some C1 > 0 and all r > 0.

(8.11) K(x) dx ≤ C2 r<|x| 0 and all 0 < r < R. Z (8.12) lim K(x) dx exists and is ﬁnite. ε→0 ε<|x|<1 Z (8.13) |K(x − y) − K(x)| dx ≤ C3 |x|>2|y| n for some C3 > 0 and all y ∈ R . The last condition is nothing but the H¨ormandercondition. Observe that (8.10) is equivalent to Z (8.14) |x||K(x)| dx ≤ C4r |x| 0 and all r > 0. Indeed, ∞ Z X Z |x||K(x)| dx = |x||K(x)| dx −(k+1) −k |x|

Proof. For ϕ ∈ Sn we have Z ∞ Z X −k |K(x)ϕ(x)| dx ≤ k|x|ϕk∞ 2 |K(x)| dx k k+1 |x|≥1 k=0 2 ≤|x|<2 ≤ 2C1k|x|ϕk∞ .

Since |ϕ(x) − ϕ(0)| ≤ |x|k∇ϕk∞ Z Z lim K(x)(ϕ(x) − ϕ(0)) dx = K(x)(ϕ(x) − ϕ(0)) dx , ε→0 ε<|x|<1 |x|<1 because of the estimate (8.14) with C4 = 2C1. Denoting Z L = lim K(x) dx ε→0 ε<|x|<1 we have Z Z p.v.K[ϕ] = Lϕ(0) + K(x)(ϕ(x) − ϕ(0)) dx + K(x)ϕ(x) dx |x|<1 |x|≥1 136 PIOTR HAJLASZ and hence

|p.v.K[ϕ]| ≤ |L|kϕk∞ + 2C1(k∇ϕk∞ + k|x|ϕk∞) . The proof is complete. 2

1 n For K ∈ Lloc(R \{0} and 0 < ε < R we deﬁne the truncated kernels by

Kε,R = Kχ{ε<|x|

Kε = Kχ{|x|>ε} .

1 n Theorem 8.9. If K ∈ Lloc(R \{0}) satisﬁes (8.10), (8.11) and (8.13), ∞ n then K[ε,R ∈ L (R ) and

(8.15) kK[ε,Rk∞ ≤ C 0 for some constant C > 0 independent of ε and R. Moreover Kε ∈ Sn, ∞ n Kcε ∈ L (R ) and kKcεk∞ ≤ C with the same constant as in (8.15).

Proof. Observe that for |y| ≤ ε/2 the truncated kernel Kε,R satisﬁes a version of the H¨ormandercondition in a form described below

Kε,R(x − y) − Kε,R(x) = (K(x − y) − K(x))χ{ε<|x|

+ |K(x)| χ{ε<|x|

(8.16) ≤ C3 + 4C1 . Indeed, a simple geometric consideration shows that the symmetric diﬀer- ence of the two annuli {ε < |x| < R} and {ε < |x + y| < R} is contained in {ε − |y| < |x| < ε + |y|} ∪ {R − |y| < |x| < R + |y|} . and the last inequality follows from (8.10), (8.13) and the fact that |y| ≤ ε/2. HARMONIC ANALYSIS 137

To estimate the Fourier transform of Kε,R observe ﬁrst that

|K[ε,R(0)| = K(x) dx ≤ C2 , ε<|x|

K[ε,R(ξ) = Kdε,r(ξ) + K[r,R(ξ) . First we will estimate the second term. Z −2πix·ξ K[r,R(ξ) = Kr,R(x)e dx Rn Z −2πi(x−y)·ξ = Kr,R(x − y)e dx Rn Z 2πiy·ξ −2πix·ξ = e Kr,R(x − y)e dx . Rn 1 −2 Taking y = 2 ξ|ξ| we have exp(2πiy · ξ) = −1 and hence Z 1 −2πix·ξ |K[r,R(ξ)| = Kr,R(x) − Kr,R(x − y) e dx 2 Rn Z 1 ≤ Kr,R(x − y) − Kr,R(x) dx 2 Rn 1 ≤ C + 2C , 2 3 1 by (8.16) since |y| = r/2. For the other term we have

Z Z −2πix·ξ |Kdε,r(ξ)| ≤ K(x) e − 1 dx + K(x) dx ε<|x|

≤ 2π|ξ| · 2C1r + C2

= 4πC1 + C2 . −1 −1 In the case |ξ| ≤ ε or |ξ| ≥ R we directly estimate K[ε,R(ξ) without −1 splitting it into two parts. If |ξ| ≤ ε the estimate goes as that for K[r,R −1 and if |ξ| ≥ R as that for Kdε,r. We leave easy details to the reader. 0 Now it is time to take care of Kε. Observe first that Kε ∈ Sn by an estimate similar to that at the beginning of the proof of Proposition 8.8. We have Kcε[ϕ] = Kε[ϕ ˆ] = lim Kε,R[ϕ ˆ] = lim K[ε,R[ϕ] R→∞ R→∞ 138 PIOTR HAJLASZ and hence |Kcε[ϕ]| ≤ Ckϕk1 0 1 n for all ϕ ∈ Sn. Thus ϕ 7→ Kcε[ϕ] extends to a bounded functional on L (R ), ∞ so Kcε ∈ L and kKcεk∞ ≤ C. The proof is complete. 2 1 n Theorem 8.10 (Calderón-Zygmund). Suppose that K ∈ Lloc(R \{0}) satisfies (8.10), (8.11), (8.12) and (8.13). Then p.v.K is a tempered distribution and the convolution with p.v.K,

T ϕ = (p.v.K) ∗ ϕ, ϕ ∈ Sn is a singular integral as deﬁned in this section. In particular p 2 kT fkp ≤ Cpkfkp, for f ∈ L ∩ L , 1 t}| ≤ kfk , for f ∈ L1 ∩ L2. R t 1

0 Proof. We already proved in Proposition 8.8 that p.v.K ∈ Sn and accord- ∞ ing to Theorem 8.5 it remains to show that p\.v.K ∈ L . For ϕ ∈ Sn we have p\.v.K[ϕ] = lim Kε[ϕ ˆ] = lim Kcε[ϕ] ≤ Ckϕk1 , ε→0 ε→0 ∞ so p\.v.K ∈ L with kp\.v.Kk∞ ≤ C. 2

Department of Mathematics, University of Pittsburgh, 301 Thackeray Hall, Pittsburgh, PA 15260, USA, [email protected]