HARMONIC ANALYSIS Preliminaries jhajlasz/Notatki/Harmonic Analysis4.pdf · A generic open set in Rn...

HARMONIC ANALYSIS

PIOTR HAJ LASZ

1. Preliminaries

Let us first fix notation that will be used throughout the book. We will be workingmostly in the Euclidean spaces and with the Lebesgue measure. The Lebesgue measure of ameasurable set E will be denoted by |E|. The volume of a ball of radius r is |B(0, r)| = ωnr

n.Then it is well known that the (n−1)-dimensional measure of the sphere Sn−1(0, r) equalsnωnr

n−1. The integral average over a set E of positive measure will be denoted by∫E

f(x) dx =1

|E|

∫E

f(x) dx.

The Lebesgue measure on a surface will be denoted by dσ so the integration in the sphericalcoordinates will take the form∫

Rnf(x) dx =

∫ ∞0

∫Sn−1

f(sθ) dσ(θ) ds.

Unless we will explicitly state otherwise, all functions and all function spaces will be com-plex valued.

A generic open set in Rn will usually be denoted by Ω and C∞0 (Ω) will stand for aspace of smooth functions with compact support in Ω. The space of continuous functionsvanishing at infinity will be denote by C0(Rn), i.e. f ∈ C0(Rn) if f is continuous and

lim|x|→∞

f(x) = 0.

C0(Rn) is a Banach space with respect to the supremum norm ‖ · ‖∞ and C∞0 (Rn) is adense subset of C0(Rn).

We will use the multiindex notation

Dαu =∂|α|

∂xα11 . . . ∂xαnn

, xα = xα11 · . . . · xαnn

where α = (α1, . . . , αn), |α| = α1 + . . .+ αn. Then the product rule takes the form

Dα(fg) =∑

β+γ=α

α!

β!γ!DβfDγg, where α! = α1! · . . . · αn!.

Date: February 27, 2017.1

2 PIOTR HAJ LASZ

The characteristic function of a set E will be denoted by χE. By δa we will denote theDirac measure centered at a i.e., ∫

Rnf(x) dδa(x) = f(a).

By an increasing function we will mean a non-decreasing function and increasing func-tions that are actually increasing, will be called strictly increasing. Similar terminologyapplies to decreasing and strictly decreasing functions.

By sgnx we will denote the sign of the number x, i.e. sgnx = 1 for positive x, sgnx = −1for negative x and sgn 0 = 0.

Various constants will usually be denoted by C. By writing C(n,m) we will indicate thethe constant C(n,m) depends on n and m only. We adopt the rule that a constant C maychange its value within one string of estimates. This will allow us to minimize the use ofindices.

HARMONIC ANALYSIS 3

2. The Maximal function

2.1. The maximal theorem.

Definition 2.1. For a locally integrable function f ∈ L1loc(Rn) the Hardy-Littlewood max-

imal function is defined by

Mf(x) = supr>0

∫B(x,r)

|f(y)| dy, x ∈ Rn.

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

|T (f1 + f2)(x)| ≤ |Tf1(x)|+ |Tf2(x)| a.e.

and

|T (kf)(x)| = |k||Tf(x)| for k ∈ C.

The following integrability result, known also as the maximal theorem, plays a fundamentalrole in many areas of mathematical analysis.

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

(a) For f ∈ L1(Rn)

(2.1) |x : Mf(x) > t| ≤ 5n

t

∫Rn|f | for all t > 0.

(b) If f ∈ Lp(Rn), 1 < p ≤ ∞, then Mf ∈ Lp(Rn) and

‖Mf‖p ≤ 2 · 5n/p(

p

p− 1

)1/p

‖f‖p for 1 0, then for |x| > R, B(0, R) ⊂ B(x,R + |x|) so

Mf(x) ≥∫B(x,R+|x|)

|f | ≥ λ

ωn(R + |x|)n,

Since the function on the right hand side is not integrable on Rn, the statement (b) of thetheorem is not true for p = 1.

4 PIOTR HAJ LASZ

Remark 2.4. If g ∈ L1(Rn), then the Chebyschev inequality

(2.2) |x : |g(x)| > t| ≤ 1

t

∫Rn|g| for t > 0

is easy to prove. Thus if an (not necessarily linear) operator T is bounded in L1, ‖Tf‖1 ≤C‖f‖1, it immediately follows from Chebyschev’s inequality that

(2.3) |x : |Tf(x)| > t| ≤ C

t

∫Rn|f(x)| dx.

The maximal function M is not bounded in L1, but it still satisfies the weaker estimate(2.3). For this reason estimate (2.1) is called the weak type estimate.

In the proof of Theorem 2.2 we will need the following two results.

Theorem 2.5 (Cavalieri’s principle). If µ is a σ-finite measure on X and Φ : [0,∞) →[0,∞) is increasing, absolutely continuous and Φ(0) = 0, then∫

X

Φ(|f |) dµ =

∫ ∞0

Φ′(t)µ(|f | > t) dt .

Proof. The result follows immediately from the equality∫X

Φ(|f(x)|) dµ(x) =

∫X

∫ |f(x)|

0

Φ′(t) dt dµ(x)

and the Fubini theorem.

Corollary 2.6. If µ is a σ-finite measure on X and 0 t) dt .

The next result has many applications that go beyond the maximal theorem. Here andin what follows by 5B we denote the ball concentric with B and with five times the radius.

Theorem 2.7 (5r-covering lemma). Let B be a family of balls in a metric space such thatsupdiamB : B ∈ B < ∞. Then there is a subfamily of pairwise disjoint balls B′ ⊂ Bsuch that ⋃

B∈B

B ⊂⋃B∈B′

5B .

More precisely, for every B ∈ B there is B′ ∈ B′ such that B ∩B′ 6= ∅ and B ⊂ 5B′.

If the metric space is separable, then the family B′ is countable (or finite) and we canarrange it as a (possibly finite) sequence B′ = Bi∞i=1 so⋃

B∈B

B ⊂∞⋃i=1

5Bi .

Remark 2.8. Here B can be either a family of open balls or closed balls. In both casesthe proof is the same.

HARMONIC ANALYSIS 5

Proof. Let supdiamB : B ∈ B = R <∞. Divide the balls in the family B according totheir diameters:

Fj =B ∈ B :

R

2j< diamB ≤ R

2j−1

.

Clearly B =⋃∞j=1Fj.

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

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

In the construction of Bj we want to select balls form Fj, but we have to make sure thatballs that we choose do not intersects with any of the previously selected balls. Thus wehave to select balls from

Fj =B ∈ Fj : B ∩B′ = ∅ for all B′ ∈

j−1⋃i=1

Bi.

Then we define Bj to be the maximal family of pairwise disjoint balls in FjFinally we define B′ =

⋃∞j=1 Bj. It follows immediately from the definition of Bj that

every ball B ∈ Fj intersects with a ball in⋃ji=1 Bj. If B ∩B′ 6= ∅, B′ ∈

⋃ji=1 Bi, then

diamB ≤ R

2j−1= 2 · R

2j< 2 diamB′

and hence B ⊂ 5B′.

Proof of Theorem 2.2. (a) Let f ∈ L1(Rn) and let Et = x : Mf(x) > t. For x ∈ Et,there is rx > 0 such that∫

B(x,rx)

|f | > t so |B(x, rx)| < t−1

∫B(x,rx)

|f | .

Observe that supx∈Et rx < ∞, because f ∈ L1(Rn). The family of balls B(x, rx)x∈Etforms a covering of the set Et so applying the 5r-covering lemma there is a sequence ofpairwise disjoint balls B(xi, rxi), i = 1, 2, . . . such that Et ⊂

⋃∞i=1B(xi, 5rxi) and hence

|Et| ≤ 5n∞∑i=1

|B(xi, rxi)| ≤5n

t

∞∑i=1

∫B(xi,rxi )

|f | ≤ 5n

t

∫Rn|f | .

The proof is complete.

(b) Let f ∈ Lp(Rn), 1 t/2, f2 = fχ|f |≤t/2

be a decomposition of |f | into its ‘lower’ and ‘upper’ parts. It is easy to check that f1 ∈L1(Rn) and clearly f2 ∈ L∞(Rn).

The idea now if to apply the weak type estimate from (a) to f1 and the estimate‖Mf‖∞ ≤ ‖f‖∞ to f2. This will give us an estimate for the size of the level set |Mf > t|

6 PIOTR HAJ LASZ

and the result will follow from Corollary 2.6. The same idea is the core of the so called realinterpolation method and we will see it again in the proof of the Marcinkiewcz InterpolationTheorem 8.9.

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

Mf > t ⊂ Mf1 > t/2 .

Thus

|Et| = |Mf > t| ≤ 2 · 5n

t

∫Rn|f1(x)| dx(2.4)

=2 · 5n

t

∫|f |>t/2

|f(x)| dx .

Cavalieri’s principle gives∫Rn|Mf(x)|p dx = p

∫ ∞0

tp−1|Mf > t| dt

≤ p

∫ ∞0

tp−1

(2 · 5n

t

∫|f |>t/2

|f(x)| dx)dt

= 2 · 5np∫Rn|f(x)|

∫ 2|f(x)|

0

tp−2 dt dx

= 2p · 5n p

p− 1

∫Rn|f(x)|p dx

and the results follows.

Remark 2.9. Note that we proved in (2.4) the following inequality

(2.5) |x : Mf(x) > t| ≤ 2 · 5n

t

∫|f |>t/2

|f(x)| dx

which is slightly stronger than (2.1). Later we will see that the measure of the set |Mf >t| has a similar bound from below, see Proposition 2.33.

Remark 2.10. For a positive measure µ on Rn we define the maximal function by

Mµ(x) = supr>0

µ(B(x, r))

|B(x, r)|.

A minor modification of the proof of Theorem 2.2(a) leads to the following result.

Proposition 2.11. If µ is a finite positive Borel measure on Rn, then

|x : Mµ(x) > t| ≤ 5n

tµ(Rn) for all t > 0.

Remark 2.12. It is natural to inquire whether in the Euclidean case, there is a version ofTheorem 2.7 for families of cubes. The answer is in the positive and it follows immediatelyfrom Theorem 2.7 and the fact that cubes with sides parallel to coordinate axes are ballsin Rn with respect to the metric d∞(x, y) = maxi |xi − yi|. In the following result 5Q willdenote a cube concentric with Q and with five times the diameter.

HARMONIC ANALYSIS 7

Corollary 2.13. Let F be a family of open or closed cubes in Rn with sides parallel tocoordinate directions such that supdiamQ : Q ∈ F < ∞. Then there is a subfamilyF ′ ⊂ F of pairwise disjoint cubes such that⋃

Q∈F

Q ⊂⋃Q∈F ′

5Q.

More precisely for every Q ∈ F there is Q′ ∈ F ′ such that Q ∩Q′ 6= ∅ and Q ⊂ 5Q′.

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

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

2.2. Fractional integration theorem. As a first application of the maximal theoremwe will prove a result about integrability of the Riesz potentials.

Definition 2.14. For 0 < α < n and n ≥ 2 we define the Riesz potential by

(Iαf)(x) =1

γ(α)

∫Rn

f(y)

|x− y|n−αdy , where γ(α) =

πn2 2α Γ

(α2

)Γ(n−α

2

) .

At this moment the particular value of the constant γ(α) is not important to us. Wecould even replace this constant by 1.1

Theorem 2.15 (Hardy-Littlewood-Sobolev). Let α > 0, 1 < p < ∞ and αp < n. Thenthere is a constant C = C(n, p, α) such that

‖Iαf‖p∗ ≤ C‖f‖p for all f ∈ Lp(Rn) ,

where p∗ = np/(n− αp).

Remark 2.16. It is not difficult to show directly2 that if Iα : Lp → Lq is bounded, thenq = p∗. The idea is to use the change of variables x 7→ tx, t > 0, and see how both sides ofthe inequality ‖Iαf‖q ≤ C‖f‖p change. This is so called a scaling argument.

We precede the proof with a technical lemma.

Lemma 2.17. If 0 < α < n, and δ > 0, then there is a constant C = C(n, α) such that∫B(x,δ)

|f(y)||x− y|n−α

dy ≤ CδαMf(x) for all x ∈ Rn.

Proof. For x ∈ Rn and δ > 0 consider the annuli

A(k) = B

(x,

δ

2k

)−B

(x,

δ

2k+1

).

1It will play an important role later: I2f is the convolution with the fundamentals solution of −∆, seeTheorem 5.60. Also Riesz potentials will be studied in Section 7 in connection with fractional powers ofthe Laplace operator.

2A nice exercise.

8 PIOTR HAJ LASZ

We have ∫B(x,δ)


dy =∞∑k=0

∫A(k)


dy

≤∞∑k=0

(δ

2k+1

)α−n ∫A(k)

|f(y)| dy

≤ ωn

∞∑k=0

(δ

2k+1

)α−n(δ

2k

)n∫B(x,δ/2k)

|f(y)| dy

≤ ωnδα

(1

2

)α−n( ∞∑k=0

1

2kα

)Mf(x) .


Proof of Theorem 2.15. Fix δ > 0. Holder’s inequality and integration in polar coordinatesyield ∫

Rn\B(x,δ)


dy ≤ ‖f‖p(∫

Rn\B(x,δ)

dy

|x− y|(n−α)p′

)1/p′

= ‖f‖p(nωn

∫ ∞δ

sn−1−(n−α)p′ ds

)1/p′

= C(n, p, α)δα−(n/p)‖f‖p ,because nωn equals the (n− 1)-dimensional measure of the unit sphere Sn−1 and n− (n−α)p′ < 0. This and the lemma give

|Iαf(x)| ≤ C(δαMf(x) + δα−(n/p)‖f‖p

).

Taking3

δ =

(Mf(x)

‖f‖p

)−p/nyields

|Iαf(x)| ≤ C(Mf(x))1−αpn ‖f‖

αpnp

which is equivalent to

|Iαf(x)|p∗ ≤ C(Mf(x))p‖f‖αpnp∗

p .

Integrating both sides over Rn and applying boundedness of the maximal function in Lp

gives the result.

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

For h ∈ Rn we define τhf(x) = f(x+ h).

Lemma 2.18. If f ∈ Lp(Rn), 1 ≤ p <∞, then ‖τhf − f‖p → 0 as h→ 0.

3We apply here a standard trick: we take δ > 0 such that δαMf(x) = δα−(n/p)‖f‖p.

HARMONIC ANALYSIS 9

This result is obvious if f is a compactly supported smooth function, because in thatcase τhf converges uniformly to f . The general case follows from the density of C∞0 in Lp.

Lemma 2.19. Let f ∈ L1(Rn) and let fr(x) =∫B(x,r)

f(y) dy. Then fr → f in L1 as

r → 0. In particular, there is a sequence ri → 0 such that

limi→∞

∫B(x,ri)

f(y) dy = f(x) a.e.

Proof. We have∫Rn|fr(x)− f(x)| dx ≤

∫Rn

∫B(x,r)

|f(y)− f(x)| dy dx

=

∫Rn

∫B(0,r)

|f(x+ y)− f(x)| dy dx

=

∫B(0,r)

‖τyf − f‖1 dy .(2.6)

Since τyf → f in L1 as y → 0, the right hand side of (2.6) converges to 0 as r → 0. Theexistence of a sequence ri follows from the fact that from a sequence convergent in L1 wecan extract a subsequence convergent a.e.

The Lebesgue differentiation theorem gives much more.

Theorem 2.20 (Lebesgue differentiation theorem). If f ∈ L1loc(Rn), then

limr→0

∫B(x,r)

f(y) dy = f(x) a.e.

Proof. Since the theorem is local in nature we can assume that f ∈ L1(Rn). Let fr(x) =∫B(x,r)

f(y) dy and define

Ωf(x) = lim supr→0

fr(x)− lim infr→0

fr(x) .

It suffices to prove that Ωf = 0 a.e. Indeed, this property means that fr → g convergesa.e. to a measurable function g. Since by Lemma 2.19, fri → f a.e. we get g = f a.e.

Observe that Ωf ≤ 2Mf , hence for any ε > 0, Theorem 2.2(a) yields

|x : Ωf(x) > ε| ≤ C

ε

∫Rn|f | .

Let h be a continuous function such that ‖f − h‖1 < ε2. Continuity of h implies Ωh = 0everywhere and hence

Ωf ≤ Ω(f − h) + Ωh = Ω(f − h) ,

so

|Ωf > ε| ≤ |Ω(f − h) > ε| ≤ C

ε

∫Rn|f − h| ≤ Cε .

Since ε > 0 can be arbitrarily small we conclude Ωf = 0 a.e.

10 PIOTR HAJ LASZ

If f ∈ L1loc(Rn), then we can define f at every point by the formula

(2.7) f(x) := lim supr→0

∫B(x,r)

f(y) dy .

According to the Lebesgue differentiation theorem this is a representative of f in the classof functions that coincide with f a.e.

Definition 2.21. Let f ∈ L1loc(Rn). We say that x ∈ Rn is a Lebesgue point of f if

limr→0

∫B(x,r)

|f(y)− f(x)| dy = 0 ,

where f(x) is defined by (2.7).

Theorem 2.22. If f ∈ L1loc(Rn), 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

(2.8) limr→0

∫B(x,r)

|f(y)− c| dy = |f(x)− c|

does not hold. Clearly |Ec| = 0 and hence the set E =⋃c∈QEc has measure zero too. Thus

for x ∈ Rn \ E and all c ∈ Q, (2.8) is satisfied.

If x ∈ Rn \ E and f(x) ∈ R, approximating f(x) by rational numbers one can easilycheck that

limr→0

∫B(x,r)

|f(y)− f(x)| dy = |f(x)− f(x)| = 0 .

Indeed, if ε > 0 is arbitrary and c ∈ Q is such that |f(x) − c| < ε/3, then there is δ > 0such that for 0 < r < δ∫

B(x,r)

|f(y)− f(x)| dy ≤∫B(x,r)

|f(y)− c| dy + |c− f(x)|

<(|f(x)− c|+ ε

3

)+ |c− f(x)| < ε.


Definition 2.23. We say that x ∈ Rn is a p-Lebesgue point of f ∈ Lploc, 1 ≤ p <∞ if

limr→0

∫B(x,r)

|f(y)− f(x)|p dy = 0 .

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

Theorem 2.24. If f ∈ Lploc(Rn), 1 ≤ p <∞, then the set of points that are not p-Lebesguepoints of f has measure zero.

Definition 2.25. Let E ⊂ Rn be a measurable set. We say that x ∈ Rn is a density pointof E if

limr→0

|B(x, r) ∩ E||B(x, r)|

= 1 .

HARMONIC ANALYSIS 11

Applying the Lebesgue theorem to the characteristic function of the set E, i.e., f = χEwe obtain

Theorem 2.26. Almost every point of a measurable set E ⊂ Rn is its density point anda.e. point of Rn \ E is not a density point of E.

The Lebesgue theorem shows that for almost all x, the averages of f over balls centeredas x converge to f(x). It is natural to inquire whether we can replace balls by other setslike cubes or even balls, but not centered at x.

Definition 2.27. We say that a family F of measurable subsets of Rn is regular at x ∈ Rn

if

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

is the smallest closed ball centered at x that contains S.

Note that if |Si| → 0, then the radius of BSi approaches to zero so the sets Si approachto x.

Examples of families regular at x include:

• Family of all balls that contain x.• Family of all cubes centered at x.• Family of all cubes that contain x.

Theorem 2.28. If f ∈ L1loc(Rn), x is a Lebesgue point of f , and if F is a family regular

at x, then

limS∈F|S|→0

∫S

f(y) dy = f(x) .

Proof. We have∣∣∣∣∫S

f(y) dy − f(x)

∣∣∣∣ ≤ ∫S

|f(y)− f(x)| dy ≤ 1

|S|

∫BS

|f(y)− f(x)| dy

≤ |BS||S|

∫BS

|f(y)− f(x)| dy ≤ 1

C

∫BS

|f(y)− f(x)| dy → 0

as |S| → 0.

Note that if F is a family regular at 0 and if we define the maximal function associatedwith F by

MFf(x) = supS∈F

∫S

|f(x− y)| dy ,

then a routine calculation shows that

(2.9) MFf(x) ≤ CMf(x) ,

so MF satisfies the claim of Theorem 2.2 (with different constants). In particular MF isa bounded operator in Lp, 1 < p ≤ ∞.

12 PIOTR HAJ LASZ

Let F be the family of all rectangular boxes in Rn that contain the origin and have sidesparallel to the coordinate axes. With the family we can associate the maximal function

Mf(x) = supS∈F

∫S

|f(x− y)| dy .

Note that the family F is not regular at 0 and hence the boundedness of M in Lp, 1 0 suchthat

(2.10) ‖Mf‖p ≤ C‖f‖p .

Moreover, if f ∈ Lploc, 1 < p <∞, then

(2.11) limdiamS→0S∈F

∫S

f(x− y) dy = f(x) a.e.

Proof. First we will prove how to conclude (2.11) from (2.10). Note that since the familyis not regular at 0, (2.11) is not a consequence of Theorem 2.28 (see also Theorem 2.30).However,

0 ≤ lim supdiamS→0S∈F

∫S

f(x− y) dy − lim infdiamS→0S∈F

∫S

f(x− y) dy ≤ 2Mf(x)

and hence (2.11) follows from (2.10) by almost the same argument as the one used todeduce the Lebesgue Differentiation Theorem from Theorem 2.2. We leave details to thereader.

Thus it remains to prove (2.10). In dimension one

Mf(x) = supa,b>0

∫ x+b

x−a|f(y)| dy

is the so called non-centered Hardy-Littlewood maximal function. Clearly Mf ≤ CMfwhich gives (2.10). For simplicity of notation we will prove the higher dimensional resultfor n = 2 only. The proof in the case of general n > 1 is similar. We have

Mf(x1, x2) = supa1,b1>0a2,b2>0

∫ x2+b2

x2−a2

∫ x1+b1

x1−a1|f(y1, y2)| dy1 dy2

≤ supa2,b2>0

∫ x2+b2

x2−a2

(sup

a1,b1>0

∫ x1+b1

x1−a1|f(y1, y2)| dy1

)dy2 .

On the right hand side we have iteration of one dimensional non-centered maximal func-tions. First we apply the maximal function to variable y1 and evaluate it at x1 and then weapply the maximal function to the variable y2 and evaluate it at x2. It is easy to see nowthat inequality (2.10) follows from the one dimensional case applied twice and the Fubinitheorem.


Surprisingly (2.11) does not hold for p = 1 and hence the maximal function Mf doesnot satisfy the weak type estimate (2.1).4 Namely one can prove the following result thatwe leave without a proof.

Theorem 2.30 (Saks). Let F be the family of all rectangular boxes in Rn that contain theorigin and have sides parallel to the coordinate axes. Then the set of functions f ∈ L1(Rn)such that

lim supdiamS→0S∈F

∫S

f(x− y) dy =∞ for all x ∈ Rn

is a dense Gδ subset of L1(Rn). In particular it is not empty.

2.4. The Calderon-Zygmund decomposition. The following result will play an im-portant role in what follows.

Theorem 2.31 (Calderon-Zygmund decomposition). Suppose f ∈ L1(Rn), f ≥ 0 andα > 0. Then there is an open set Ω and a closed set F such that

(a) Rn = Ω ∪ F , Ω ∩ F = ∅;(b) f ≤ α a.e. in F ;(c) Ω can be decomposed into cubes Ω =

⋃∞k=1Qk with pairwise disjoint interiors such

that

α <

∫Qk

f(x) dx ≤ 2nα , k = 1, 2, 3, . . .

Proof. Decompose Rn into a grid of identical cubes, large enough to have∫Q

f(x) dx ≤ α

for each cube in the grid. Take a cube Q from the grid and divide it into 2n identical cubes.Let Q′ be one of the cubes from this partition. We have two cases:∫

Q′f(x) dx > α or

∫Q′f(x) dx ≤ α .

If the fist case holds we include the open cube Q′ to the family Qk. Note that

α <

∫Q′f = 2n|Q|−1

∫Q′f ≤ 2n

∫Q

f ≤ 2nα

so the condition (c) is satisfied. If the second case holds we divide Q′ into 2n identicalcubes and proceed as above. We continue this process infinitely many times or until it isterminated. We apply it to all the cubes of the original grid. Let Ω =

⋃∞k=1Qk, where the

cubes are defined by the first case of the process. It remains to prove that f ≤ α a.e. inthe set F = Rn \ Ω. The set F consists of faces of the cubes (this set has measure zero)and points x such that there is a sequence of cubes Qi with the property that x ∈ Qi,diam Qi → 0,

∫Qif ≤ α. According to Theorem 2.28 for a.e. such x we have

∫Qif → f(x)

and hence f ≤ α a.e. in F .

Corollary 2.32. Let f , α and Ω be as in Theorem 2.31. Then |Ω| < α−1‖f‖1.

4Such estimate would imply (2.11).

14 PIOTR HAJ LASZ

Proof. We have

|Ω| =∞∑k=1

|Qk| ≤∞∑k=1

α−1

∫Qk

|f | ≤ α−1‖f‖1 .


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

Proposition 2.33. If f ∈ L1(Rn), then

2−nC(n)−1

t

∫|f |>C(n)t

|f(x)| dx ≤ |x : Mf(x) > t| ≤ 2 · 5n

t

∫|f |>t/2

|f(x)| dx.

where we can take C(n) = ωnnn/2.

Proof. For the proof of the right inequality see (2.4). In order to prove the left inequalitywe apply the Calderon-Zygmnud decomposition to the function |f | and α = s > 0. IfΩ =

⋃kQk is an open set as in Theorem 2.31, then

s <

∫Qk

|f(x)| dx ≤ 2ns for all k.

Let `k be the edge-length of the cube Qk. Then for any x ∈ Qk, Qk ⊂ Bx = B(x,√n`k) so

s <

∫Qk

|f(y)| dy ≤ |Bx||Qk|

∫Bx

|f(y)| dy ≤ ωnnn2Mf(x).

Hence

|x : Mf(x) > sω−1n n−

n2 | ≥

∑k

|Qk| ≥2−n

s

∑∫Qk

|f(x)| dx

≥ 2−n

s

∫|f |>s

|f(y)| dy

where the last inequality follows from the fact that |f | ≤ s for almost all x 6∈ Ω which meansthat the set |f | > s is contained in Ω =

⋃kQk. Now it suffices to take s = tωnn

n/2.

2.5. Integrability of the maximal function. In Remark 2.3 we observed that if f ∈L1(Rn), then Mf 6∈ L1(Rn). However, we showed that the function cannot be globallyintegrable. It turns out that, in general, the maximal function need not be even locallyintegrable. We will actually characterize all functions such that the maximal function islocally integrable.

Definition 2.34. We say that a measurable function f belongs to the Zygmund spaceL logL if |f | log(e+ |f |) ∈ L1.

It is easy to see that for a space with finite measure we have⋂p>1

Lp ⊂ L logL ⊂ L1 ,


so the Zygmund space is an intermediate space between all Lp for p > 1 and L1.

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

Proof. Suppose that f ∈ L logL(B).5 Cavalieri’s principle (Theorem 2.5) and weak typeestimates from Proposition 2.33 give∫

B

Mf(x) dx =

∫ ∞0

|x ∈ B : Mf(x) > t| dt

≤ |B|+∫ ∞

1

|x ∈ B : Mf(x) > t| dt

≤ |B|+∫ ∞

1

(2 · 5n

t

∫|f |>t/2

|f(x)| dx)dt

= |B|+ C(n)

∫B

(∫ max2|f(x)|,1

1

dt

t

)|f(x)| dx

= |B|+ C(n)

∫B

|f(x)|max0, log(2|f(x)|) dx

≤ |B|+ 2C(n)

∫B

|f(x)| log(e+ |f(x)|) dx <∞.

To prove the other implication we assume Mf ∈ L1(B) and we want to prove thatf ∈ L logL(B). In the proof of the first implication we used the right inequality fromProposition 2.33, but now we want to use the left inequality. However, the following cal-culations seem to disprove integrability of Mf on B.∫

B

Mf(x) dx =

∫ ∞0

|x ∈ B : Mf(x) > t| dx ≥∫ ∞

0

(C1

t

∫|f(x)|>C2t

|f(x)| dx)dt

= C1

∫B

( ∫ |f(x)|/C2

0

dt

t︸︷︷︸= +∞ if f(x) 6= 0.

)|f(x)| dx = +∞.

These calculations are erroneous. Although the function f is supported in the ball B only,the maximal functionMf is positive on all of Rn and in the left estimate in Proposition 2.33we deal with the measure of the set

x ∈ Rn : Mf(x) > t and not of the set x ∈ B : Mf(x) > t.We will show not how to overcome this difficulty.

Note that Mf ∈ L1loc(Rn). Indeed, Mf is integrable in B and locally bounded outside

the closed ball B, so we need to verify integrability of Mf in a neighborhood of theboundary of B.6 But it is easy to see that if x is near the boundary and outside the ball,

5The proof of integrability ofMf will be similar to the proof of Lp integrability of the maximal functionwhen f ∈ Lp, p > 1.

6Integrability on B and local boundedness outside B does not imply local integrability in Rn. Thefunction f(x) = 0 for x ≤ 0 and f(x) = 1/x for x > 0 is integrable on (−∞, 0] and is locally bounded on(0,∞).

16 PIOTR HAJ LASZ

we can estimate the value ofMf(x) by (constant times) the value of the maximal functionin a point being the reflection of x across the boundary. Thus the integrability ofMf nearthe boundary follows from the integrability ofMf in B. Note also that the set Mf > 1is bounded, because Mf(x) decays to zero as x → ∞. Thus local integrability of Mfand boundedness of the set Mf > 1 implies that the function Mf in integrable inMf > 1. With these remarks we can complete the proof as follows.

∞ >

∫Mf>1

Mf(x) dx

≥∫ ∞

1

|Mf > t| dt

≥∫ ∞

1

2−nC−1

t

∫|f |>Ct

|f(x)| dx dt

= 2−nC−1

∫B

(∫ max|f(x)|/C,1

1

dt

t

)|f(x)| dx

= 2−nC−1

∫B

|f(x)|max0, log(|f(x)|/C) dx .

This easily implies that f ∈ L logL.

Remark 2.36. A more careful analysis leads to the following version of Stein’s theorem.In an open set Ω ⊂ Rn we define the local maximal function by

MΩf(x) = sup∫

Q

|f | : x ∈ Q ⊂ Ω,

where the supremum is taken over all cubes Q in Ω that contain x.

Theorem 2.37 (Stein). Let Q ⊂ Rn be a cube. Then MQf ∈ L1(Q) if and only iff ∈ L logL(Q). Moreover

5−(n+1)

∫Q

MQf ≤∫Q

|f | log

(e+

|f ||f |Q

)≤ 2n+2

∫Q

MQf ,

where

|f |Q =

∫Q

|f | .

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

(2.12)

(∫Y

(∑k

|F (k, y)|)pdν(y)

)1/p

≤∑k

(∫Y

|F (k, y)|p dν(y)

)1/p

.

Since the sum over k can be regarded as an integral with respect to the counting measure,the following result is a natural generalization of the Minkowski inequality.


Theorem 2.38 (Minkowski’s integral inequality). If µ and ν are σ-finite measures on Xand Y respectively and if F : X × Y → R is measurable, then for 1 ≤ p <∞ we have(∫

Y

(∫X

|F (x, y)| dµ(x)

)pdν(y)

)1/p

≤∫X

(∫Y

|F (x, y)|p dν(y)

)1/p

dµ(x) .

Remark 2.39. It would be natural to try approximate F by simple functions in x andapply the discrete version of the inequality (2.12). The proof given below uses however, avery different and much more elegant argument.

Proof. If p = 1 then we actually have equality by Fubini’s theorem. Thus assume that 1 <p <∞. We will use the fact that the Lp(Y, ν) norm of the function y 7→

∫X|F (x, y)| dµ(x)

equals to the norm of the corresponding functional on Lq(Y, ν), p−1 + q−1 = 1.(∫Y

(∫X

|F (x, y)| dµ(x)

)pdν(y)

)1/p

= suph∈Lq(ν)‖h‖q=1

∫Y

h(y)

(∫X

|F (x, y)| dµ(x)

)dν(y)

= suph∈Lq(ν)‖h‖q=1

∫X

(∫Y

h(y)|F (x, y)| dν(y)

)dµ(x)

≤ suph∈Lq(ν)‖h‖q=1

∫X

(∫Y

|h(y)|q dν(y)︸︷︷︸1

)1/q(∫

Y

|F (x, y)|p dν(y)

)1/p

dµ(x)

=

∫X

(∫Y

|F (x, y)|p dν(y)

)1/p

dµ(x) .


2.7. Convergence of averages. In this section we will show that a large class of averagesof f converge to f . Our approach will be based on the maximal function estimates. Thiswill lead to far reaching generalizations of the Lebesgue Differentiation Theorem 2.20. Notethat in the proofs of Theorem 2.20 and its generalizations discussed above (Theorems 2.28and 2.29), maximal function also plays an important role. The results of this section willbe used later in Section 3.3

Since many averages are constructed with the help of convolution, let us recall that theconvolution of measurable functions f and g is defined as

(f ∗ g)(x) =

∫Rnf(x− y)g(y) dy =

∫Rnf(y)g(x− y) dy.

The equality of the integrals follows from a linear change of variables. This equality actuallymeans that f ∗ g = g ∗ f . One can also easily check that (f ∗ g) ∗ h = f ∗ (g ∗ h).

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

(2.13) ‖f ∗ g‖p ≤ ‖f‖p‖g‖1.

18 PIOTR HAJ LASZ

If f ∈ C0(Rn), then f ∗ g ∈ C0(Rn) and

(2.14) ‖f ∗ g‖∞ ≤ ‖f‖∞‖g‖1.

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

‖f ∗ g‖p =(∫

Rn

∣∣∣ ∫Rnf(x− y)g(y) dy

∣∣∣p dx)1/p

≤(∫

Rn

(∫Rn|f(x− y)| |g(y)| dy︸︷︷︸

dµ

)pdx︸︷︷︸dν

)1/p

≤∫Rn

(∫Rn|f(x− y)|p dx

)1/p

|g(y)| dy

= ‖f‖p‖g‖1 .

The proof in the case of f ∈ C0(Rn) is left to the reader as an exercise.

Remark 2.41. In the above proof one could use the classical Minkowski inequality insteadof the integral one via the following estimate∫

Rn|f(x− y)| |g(y)| dy =

∫Rn|f(x− y)| |g(y)|1/p|g(y)|1/q dy

≤(∫

Rn|f(x− y)|p|g(y)| dy

)1/p(∫Rn|g(y)| dy

)1/q

.

We leave details to the reader.

Theorem 2.40 is a special case of a more general Young’s inequality. We leave the proofas an exercise.

Theorem 2.42 (Young’s inequality). If 1 ≤ p, q, r ≤ ∞ and q−1 = p−1 + r−1 − 1, then

‖f ∗ g‖q ≤ ‖f‖p‖g‖r .

For ϕ ∈ L1(Rn) and t > 0 we define

ϕt(x) = t−nϕ(xt

).

Note that∫Rn ϕt =

∫Rn ϕ.

Example 2.43. If ϕ = ω−1n χB(0,1), then ϕt = (ωnt

n)−1χB(0,t) so

f ∗ ϕt(x) =1

ωntn

∫B(0,t)

f(x− y) dy =

∫B(x,t)

f(y) dy.

Thus7 in that case f ∗ ϕt → f a.e. and in L1, provided f ∈ L1. Note also that

(2.15) supt>0

(|f | ∗ ϕt)(x) =Mf(x).

7See Lemma 2.19 and Theorem 2.20.


Actually this identity8 plays a crucial role in the proof of the pointwise convergence f ∗ϕt →f a.e. In what follows we will study convergence of f ∗ ϕt for more general functions ϕ.Not surprisingly we will be looking for conditions that will guarantee an estimate similarto (2.15).9

Theorem 2.44. Let ϕ ∈ L1(Rn) and let ϕt(x) = t−nϕ(x/t) for t > 0. If f ∈ Lp(Rn),1 ≤ p <∞ or f ∈ C0(Rn) and p =∞, then

(2.16) limt→0+‖f ∗ ϕt − af‖p = 0, where a =

∫Rnϕ(x) dx.

Remark 2.45. The two most interesting cases are when a = 1 and when a = 0.

Remark 2.46. If f ∈ C0(Rn), then f ∗ ϕt ∈ C0(Rn)10 and (2.16) with p =∞ means thatf ∗ ϕt converges to af uniformly.

Remark 2.47. If a = 1, then (2.16) means that f ∗ ϕt → f in Lp so the operatorsf 7→ f ∗ ϕt converge pointwise, i.e. for every f , to the identity operator f 7→ f . For thatreason the family of functions ϕt : t > 0 is often called an approximation of identity(provided a = 1).

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

|f ∗ ϕt(x)− af(x)| =∣∣∣ ∫

Rn(f(x− y)− f(x))ϕt(y) dy

∣∣∣ .Given δ > 0, the Minkowski Integral Inequality yields

‖f ∗ ϕt − af‖p ≤∫Rn

(∫Rn|f(x− y)− f(x)|p dx

)1/p

|ϕt(y)| dy

=

∫Rn‖τ−yf − f‖p|ϕt(y)| dy

=

∫Rn‖τ−tyf − f‖p|ϕ(y)| dy,

where the last equality follows from a simple linear change of variables.

Note that ‖τ−tyf −f‖p|ϕ(y)| ≤ 2‖f‖p|ϕ(y) ∈ L1 and for every y ∈ Rn, ‖τ−tyf −f‖p → 0as t→ 0 (Lemma 2.18) so

‖f ∗ ϕt − af‖p ≤∫Rn‖τ−tyf − f‖p|ϕ(y)| dy → 0 as t→ 0

by the Dominated Convergence Theorem.

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

8Along with weak type estimates for the maximal function, see Theorem 2.20.9See Theorem 2.49.10See Theorem 2.40.

20 PIOTR HAJ LASZ

Definition 2.48. Let ϕ ∈ L1(Rn). We say that Ψ is a radially decreasing majorant of ϕ if

(a) Ψ(x) = η(|x|) for some11 η : [0,∞)→ [0,∞].12

(b) η is decreasing.13.(c) |ϕ(x)| ≤ Ψ(x) a.e.

Every function ϕ ∈ L1 has the least radially decreasing majorant14

Ψ0(x) = ess sup|y|≥|x|

|ϕ(y)|.

Thus the existence of an integrable radially decreasing majorant Ψ of ϕ is equivalent toΨ0 ∈ L1.

Theorem 2.49. Suppose that ϕ ∈ L1(Rn) has an integrable radially decreasing majorant

Ψ(x) = η(|x|) ∈ L1(Rn) .

Then for f ∈ L1loc(Rn) and all x ∈ Rn

(2.17)

∣∣∣∣supt>0

(f ∗ ϕt)(x)

∣∣∣∣ ≤ ‖Ψ‖1Mf(x) .

If in addition f ∈ Lp(Rn), 1 ≤ p <∞, and∫Rn ϕ(x) dx = a, then

(2.18) f ∗ ϕt → af a.e. as t→ 0.

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

(2.19) |f | ∗Ψ(x) ≤ ‖Ψ‖1Mf(x).

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

|f ∗ ϕt(x)| ≤ |f | ∗Ψt(x) ≤ ‖Ψt‖1Mf(x) = ‖Ψ‖1Mf(x).

Thus it remains to prove (2.19).

For a positive integer k let

sk(x) =k·2k∑i=1

1

2kχΨ≥i/2k(x) =

k·2k∑i=1

1

2kχBi(x) .

Since Ψ is radially symmetric the sets

Bi =x : Ψ(x) ≥ i

2k

are balls15 centered at zero.

11Functions of this form, i.e. functions constant on spheres Sn−1(0, r) are called radially symmetric.12It may happen that η(t) = +∞ for all t.13i.e. non-increasing14However, it may happen that Ψ0 = +∞ everywhere.15Open or closed – find an example such that for some t the balls are open while for other values of t

they are closed.


The sequence sk is increasing and convergent to Ψ almost everywhere. Indeed, forx 6= 0, Ψ(x) <∞16 so Ψ(x) < k0, for some k0 ∈ N. Then for k ≥ k0

`

2k≤ Ψ(x) <

`+ 1

2k, for some ` = 0, 1, 2, . . . k · 2k − 1.

Since the inequality Ψ ≥ i/2k is satisfied for i = 1, 2, . . . , `, the definition of the functionsk gives that sk(x) = `/2k so

Ψ(x)− 1

2k< sk(x) ≤ Ψ(x).

That means sk(x) → Ψ(x) for all all x 6= 0. Observe also that the sequence sk isincreasing, sk ≤ sk+1. Indeed, if sk(x) = `/2k, then Ψ(x) ≥ `/2k = (2`)/(2k+1) so thedefinition of sk+1 gives

sk+1(x) ≥ 1

2k+1· 2` =

`

2k= sk(x).

Thus if we can prove

(2.20) |f | ∗ sk(x) ≤ ‖Ψ‖1Mf(x),

inequality (2.19) will follow from the Monotone Convergence Theorem.

We have

|f | ∗ sk(x) =k·2k∑i=1

|Bi|2k|f | ∗ χBi

|Bi|(x).

Since17

|f | ∗ χBi|Bi|

(x) =

∫Bi

|f(x− y)| dy ≤Mf(x) andk·2k∑i=1

|Bi|2k

= ‖sk‖1 ≤ ‖Ψ‖1,

inequality (2.20) follows. The proof of (2.17) is complete.

It remains now to prove convergence (2.18).

If p = 1, then

(2.21) Ωf(x) := lim supt→0+

f ∗ ϕt − lim inft→0+

f ∗ ϕt ≤ 2‖Ψ‖1Mf(x)

and almost the same argument as in the proof of Theorem 2.20 yields that18 Ωf = 0 a.e.so f ∗ϕt converges a.e. to a measurable function. Since f ∗ϕt → af in L1 (Theorem 2.44),(2.18) follows.

The case 1 t| = |x : |Mf(x)|p > tp| ≤ 1

tp

∫Rn|Mf(x)|p dx

≤ C(n, p)

tp

∫Rn|f(x)|p dx.

16Because Ψ ∈ L1.17Because the balls Bi are centered at the origin.18We will actually repeat this argument in the case of 1 0, inequality (2.21) yields

|x : Ωf(x) > ε| ≤ C

εp

∫Rn|f(x)|p dx,

where the constant C depends on n, p and ‖Ψ‖1. If h ∈ C0(Rn), then Ωh=0.19 Takingh ∈ C0(Rn) such that ‖f − h‖pp ≤ εp+1 we conclude that

|x : Ωf(x) > ε| = |x : Ω(f − h)(x) > ε| < Cε

so Ωf = 0 a.e. and exactly as in the case of p = 1, (2.18) follows.

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


3. The Fourier transform

3.1. Fourier transform.

Definition 3.1. For f ∈ L1(Rn) we define the Fourier transform as

F(f)(x) = f(ξ) =

∫Rnf(x)e−2πix·ξ dx where x · ξ =

n∑j=1

xjξj .

Theorem 3.2. The Fourier transform has the following properties

(a) ∧ : L1(Rn)→ C0(Rn) is a bounded linear operator with ‖f‖∞ ≤ ‖f‖1.

(b) (f ∗ g) = f g for f, g ∈ L1(Rn).

(c) If f, g ∈ L1(Rn), then f g, f g ∈ L1(Rn) and∫Rnf(x)g(x) dx =

∫Rnf(x)g(x) dx.

(d) If τhf(x) = f(x+ h), then

(τhf ) (ξ) = f(ξ)e2πih·ξ and (f(x)e2πih·x) (ξ) = f(ξ − h).

(e) If ft(x) = t−nf(x/t), then

(ft) (ξ) = f(tξ) and (f(tx)) (ξ) = (f)t(ξ) .

(f) If ρ ∈ O(n) is an orthogonal transformation, then

(f(ρ ·)) (ξ) = f(ρξ).

Proof. (a) Clearly, ∧ : L1(Rn)→ L∞(Rn) is a bounded linear mapping with ‖f‖∞ ≤ ‖f‖1.Indeed,

|f(ξ)| ≤∫Rn

∣∣f(x)e−2πix·ξ∣∣ dx = ‖f‖1.

The Dominated Convergence Theorem implies that the function f is continuous. It remainsto prove that f(ξ)→ 0 as |ξ| → ∞. Let ξ 6= 0. Since eπi = −1 we have

f(ξ) =

∫Rnf(x)e−2πix·ξ dx = −

∫Rnf(x)e−2πix·ξeπi dx

= −∫Rnf(x) exp

(− 2πi

(x− ξ

2|ξ|2)· ξ)dx

24 PIOTR HAJ LASZ

= −∫Rnf(x+

ξ

2|ξ|2)e−2πix·ξ dx .

Hence

f(ξ) =1

2

∫Rn

(f(x)− f

(x+

ξ

2|ξ|2))

e−2πix·ξ dx

and thus Lemma 2.18 yields

|f(ξ)| ≤ 1

2

∥∥∥f − τ ξ

2|ξ|2f∥∥∥

1→ 0 as |ξ| → ∞.

(b)

(f ∗ g) (ξ) =

∫Rn

(∫Rnf(x− y)g(y) dy

)e−2πix·ξ dx

=

∫Rn

(∫Rnf(x− y)e−2πi(x−y)·ξ dx

)g(y)e−2πiy·ξ dy

= f(ξ)g(ξ) .

(c) f g, f g ∈ L1, because the functions f , g are bounded and the equality of the integralseasily follows from the Fubini theorem.

(d)

(τhf ) (ξ) =

∫Rnf(x+ h)e−2πix·ξ dx =

∫Rnf(x)e−2πi(x−h)·ξ dx

= e2πih·ξ∫Rnf(x)e−2πix·ξ dx = e2πih·ξf(ξ) .

The second equality follows from a similar argument.

(e)

(ft) (ξ) =

∫Rnt−nf

(xt

)e−2πix·ξ dx =

∫Rnf(y)e−2πi(ty)·ξ dy

=

∫Rnf(y)e−2πiy·(tξ) dy = f(tξ) .

The second formula follows from the first one if we replace t by t−1.

(f)20

(f(ρ ·)) (ξ) =

∫Rnf(ρx)e−2πix·ξ dx =

∫Rnf(y)e−2πi(ρ−1y)·ξ dy

=

∫Rnf(y)e−2πiy·(ρξ) dy = f(ρξ) .

Equality (ρ−1y) · ξ = y · (ρξ) follows from the fact that the mapping x 7→ ρx is anisometry.

20This property is in some sense obvious: the Fourier transform does not depend on the choice of thecoordinate system since it depends on the scalar product x · ξ. Thus the Fourier transform should stay thesame if we rotate the coordinate system by ρ.


Theorem 3.3. Suppose f ∈ L1(Rn) and xkf(x) ∈ L1(Rn), where xk is the k-th coordinate

function. Then f is differentiable with respect to ξk and

(−2πixkf(x)) =∂f

∂ξk(ξ) .

Proof. Let ek be the unit vector along the k-th coordinate. Then the second part of The-orem 3.2(d) gives

f(ξ + hek)− f(ξ)

h=

(e−2πi(hek)·x − 1

hf(x)

)∧(ξ)→ (−2πixkf(x)) (ξ) .

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

Definition 3.4. We say that f ∈ L1loc(Rn) is Lp-differentiable with respect to xk if there is

g ∈ Lp(Rn) such that∫Rn

∣∣∣∣f(x+ hek)− f(x)

h− g(x)

∣∣∣∣p dx→ 0 as h→ 0.

The function g is called the Lp-partial derivative of f with respect to xk and is denoted byg = ∂f/∂xk.

Theorem 3.5. If f ∈ L1(Rn) is L1-differentiable with respect to xk, then(∂f

∂xk

)∧= 2πiξkf(ξ) .

Proof. The first part of Theorem 3.2(d) and the L1-differentiability of f give(∂f

∂xk

)∧− f(ξ)

e2πi(hek)·ξ − 1

h=

(∂f

∂xk− f(x+ hek)− f(x)

h

)∧→ 0

as h→ 0, so (∂f

∂xk

)∧(ξ) = lim

h→0f(ξ)

e2πi(hek)·ξ − 1

h= 2πiξkf(ξ) .


Remark 3.6. With each polynomial

P (x) =∑|α|≤m

aαxα

of variables x1, . . . , xn we associate a differential operator

P (D) =∑|α|≤m

aαDα =

∑|α|≤m

aα∂|α|

∂xα11 . . . ∂xαnn

.

Then under suitable assumptions Theorems 3.3 and 3.5 have the following higher ordergeneralizations

(3.1) P (D)f(ξ) = (P (−2πix)f(x)) (ξ), (P (D)f ) (ξ) = P (2πiξ)f(ξ).

The next result is our first (and actually one of the most important) example of theFourier transform.

26 PIOTR HAJ LASZ

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

2.

Proof. First observe that it suffices to prove the result in dimension 1. Indeed, assuming itfor n = 1, the case of general n follows from Fubini’s theorem

f(ξ) =

∫Rne−π|x|

2

e−2πix·ξ dx =n∏k=1

∫ ∞−∞

e−πx2ke−2πixkξk dxk =

n∏k=1

e−πξ2k = e−π|ξ|

2

.

In the case n = 1 we will show two different proofs. The first one will use the contourintegration while the second one will use differential equations.

Thus in the remaining part of the proof we will assume that f(x) = e−πx2

is a functionof one variable.

Proof by contour integration. The function e−πz2

is holomorphic and hence its integralalong the following curve equals zero.

Letting R→∞ we obtain∫ ∞−∞

e−πx2

dx =

∫ ∞−∞

e−π(x+iξ)2 dx .

The left hand side equals 1, so

1 =

∫ ∞−∞

e−πx2

e−2πixξeπξ2

dx .

Hence

e−πξ2

=

∫ ∞−∞

e−πx2

e−2πixξ dx = f(ξ) .


Proof by differential equations. We have

f ′(x) = −2πxf(x), −i(−2πixf(x)) = f ′(x)

so Theorems 3.3 and 3.5 yield

−i(− 2πixf(x)

)∧(ξ)︸︷︷︸

(f)′(ξ)

= f ′(ξ)︸︷︷︸2πiξf(ξ)

, (f)′(ξ) = −2πξf(ξ).


Solving this differential equation for the unknown function ξ 7→ f(ξ) yields

f(ξ) = f(0)e−πξ2

= e−πξ2

, because f(0) =

∫ ∞−∞

e−πx2

dx = 1.

3.2. Measures of finite total variation. Let us first recall basic facts regarding mea-sures of finite total variation.

If µ is a complex (Borel) measure on Rn, then there is a unique positive measure |µ|called the total variation of measure µ and a Borel function h : Rn → C, |h(x)| = 1 for allx ∈ Rn such that

µ(E) =

∫E

h(x) d|µ|(x) for all Borel sets E ⊂ Rn.

The space B(Rn) of complex measures of finite total variation ‖µ‖ := |µ|(Rn) is a Banachspace with the norm ‖ · ‖.

Note that f ∈ L1(Rn) defines a measure of finite total variation by

µ(E) =

∫E

f(x) dx.

In that case

|µ|(E) =

∫E

|f(x)| dx so ‖µ‖ =

∫Rn|f(x)| dx = ‖f‖1 <∞.

This proves that L1(Rn) is a closed subspace of B(Rn).

In a special case when µ is a real valued measure, called signed measure, we have h :Rn → ±1 so if we define h+ = hχh=1 and h− = hχh=−1, then

µ±(E) =

∫E

h±(x) d|µ|(x)

are positive measures such that

(3.2) µ = µ+ − µ−

and

|µ| = µ+ + µ−.

The representation (3.2) is called the Hahn decomposition of µ. Note that the measuresµ+ and µ− are supported on disjoint sets.

Theorem 3.8 (Riesz representation theorem). The dual space to C0(Rn) is isometri-cally isomorphic to the space of measures of finite total variation. More precisely, ifΦ ∈ (C0(Rn))∗, then there is a unique measure µ of finite total variation such that

Φ(f) =

∫Rnf dµ for f ∈ C0(Rn).

Moreover ‖Φ‖ = ‖µ‖ = |µ|(Rn).

28 PIOTR HAJ LASZ

This result allows us to define convolution of measures.

If f, g ∈ L1(Rn), then f ∗ g ∈ L1(Rn) ⊂ B(Rn) and hence it acts as a functional onC0(Rn) by the formula

Φ(h) =

∫Rnh(x)(f ∗ g)(x) dx =

∫Rnh(x)

(∫Rnf(x− y)g(y) dy

)dx

=

∫Rn

(∫Rnh(x)f(x− y) dx

)g(y) dy

=

∫Rn

∫Rnh(x+ y)f(x)g(y) dx dy .

This suggests how to define convolution of measures.

If µ1, µ2 ∈ B(Rn), then

Φ(h) =

∫Rn

∫Rnh(x+ y) dµ1(x) dµ2(y)

defines a functional on C0(Rn) and hence there is a unique measure µ ∈ B(Rn) such that∫Rn

∫Rnh(x+ y) dµ1(x) dµ2(y) =

∫Rnh(x) dµ(x) for all h ∈ C0(Rn).

Definition 3.9. We denote the measure µ by

µ = µ1 ∗ µ2

and call it convolution of measures.

Clearly

µ1 ∗ µ2 = µ2 ∗ µ1 and ‖µ1 ∗ µ2‖ ≤ ‖µ1‖ ‖µ2‖.If dµ1 = f dx, dµ2 = g dx, then

d(µ1 ∗ µ2) = (f ∗ g) dx,

so the convolution of measures extends the notion of convolution of functions. If dµ1 = f dxand µ ∈ B(Rn), then∫

Rn

∫Rnh(x+ y) dµ1(x) dµ(y) =

∫Rn

∫Rnh(x+ y)f(x) dx dµ(y)

=

∫Rn

∫Rnh(x)f(x− y) dx dµ(y)

=

∫Rnh(x)

(∫Rnf(x− y) dµ(y)

)dx.

Thus µ1 ∗ µ can be identified with a function

x 7→∫Rnf(x− y) dµ(y) ∈ L1(Rn),

so we can write

f ∗ µ =

∫Rnf(x− y) dµ(y), ‖f ∗ µ‖1 ≤ ‖f‖1 ‖µ‖.


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

‖f ∗ µ‖p ≤ ‖f‖p ‖µ‖.

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

Definition 3.11. The Fourier transform of a measure of finite total variation µ ∈ B(Rn)is defined by

µ(x) =

∫Rne−2πix·ξ dµ.

Proposition 3.12. If µ ∈ B(Rn), then µ is a bounded and continuous function such that‖µ‖∞ ≤ ‖µ‖.

Proof. Boundedness of µ is obvious and continuity follows from the Dominated Conver-gence Theorem.

Proposition 3.13. If µ1, µ2 ∈ B(Rn), then µ1 ∗ µ2 = µ1 · µ2.

We leave the proof as an exercise.

Remark 3.14. Convolution of measures and the Fourier transform of measures play afundamental role in probability.

3.3. Summability methods. The answer to an important question of how to reconstructa function f from its Fourier transform f is provided by the inversion formula21

(3.3) f(x) =

∫Rnf(ξ) e2πix·ξ dξ .

It seems that this formula requires both f and f to be integrable. Integrability of f isneeded for the existence of f and integrability of f is needed for the integral in (3.3) to

make sense. We will actually prove in Corollary 3.26 that if f, f ∈ L1, then (3.3) is true a.e.There is, however, a problem here. Equality (3.3) means that f(x) is the Fourier transform

of f ∈ L1 evaluated at −x. Since the Fourier transform of an integrable function f ∈ L1

belongs to C0(Rn), it follows that f ∈ C0(Rn). That means, if f ∈ L1 \ C0, then f cannotbe integrable so the right hand side of (3.3) does not make much sense. Is there any chanceto give meaning to the integral at (3.3), beyond the class of integrable Fourier transformsso that the equality at (3.3) would be true for all f ∈ L1? The answer is in the positivebut we need to interpret the integral at (3.3) using suitable summability methods.22

21Compare it with the formula for Fourier series:

f(n) =

∫ 1

0

f(x)e−2πixn dx, f(x) =∑n∈Z

f(n)e2πixn.

22An example of a summability method for sequences is given by the so called Cesaro method: witheach sequence an we associate the limit of (a1 + . . .+an)/n. This extends the notion of limit far beyondthe class of convergent sequences.

30 PIOTR HAJ LASZ

If Φ ∈ C0(Rn) is such that Φ(0) = 1, then for t > 0 we define the Φ-mean of a functionf as

MΦ,tf = Mtf =

∫Rnf(x)Φ(tx) dx.

If f ∈ L1(Rn), then MΦ,tf →∫Rn f(x) dx as t→ 0+, but the limit of Mtf might also exist

for some non-integrable functions f .

Definition 3.15. We say that∫Rn f is Φ-summable to ` ∈ R if limt→0+ MΦ,tf = `.

In the case of Φ(x) = e−|x| and Φ(x) = e−|x|2

we obtain the so called the Abel and theGauss summability methods. More precisely we have

Definition 3.16. For each t > 0 the Abel mean and the Gauss mean of a function f are

At(f) =

∫Rnf(x)e−t|x| dx and Gt(f) =

∫Rnf(x) e−t|x|

2

dx .

We say that∫Rn f(x) dx is Abel summable23 (Gauss summable) to ` ∈ R if limt→0At(f) = `

(limt→0Gt(f) = `).

Remark 3.17. If Φ(x) = e−|x|2, then Φ(tx) = e−t

2|x|2 so MΦ,tf = Gt2(f). This however,does not cause any problems when you compare notions of the Φ-summability and theGauss summability because limt→0+ MΦ,tf = limt→oGt2(f).

Remark 3.18. Clearly, in the case of integrable functions, the Abel and the Gauss methodsgive the usual integral. However, the means At(f) and Gt(f) are finite whenever f isbounded so for some bounded, but non-integrable f , the integral

∫Rn f(x) dx might still

be Abel or Gauss summable to a finite value. For example if f ∈ L1(Rn) \ C0(Rn), the

the function ξ 7→ f(ξ)e2πix·ξ is in C0(Rn) but it is not integrable. However, as we will see

(Theorems 3.23 and 3.30) for almost all x ∈ Rn, the integral∫Rn f(ξ)e2πix·ξ dξ is both Abel

and Gauss summable to f(x). This gives some meaning of the inversion formula (3.3) forgeneric f ∈ L1(Rn).

We want to find a reasonably large class of Φ-means (including the Abel and the Gaussmeans) such that ∫

Rnf(ξ)e2πix·ξ dξ

is Φ-summable to f(x) i.e.

Mt(x) =

∫Rnf(ξ)e2πix·ξΦ(tξ) dξ → f(x) as t→ 0+

and we can consider both, convergence Mt → f almost everywhere and in L1 (with respectto variable x). The main result reads as follows

Theorem 3.19. Let Φ ∈ L1(Rn) be such that ϕ = Φ ∈ L1(Rn) and∫Rn ϕ(x) dx = 1. For

f ∈ L1(Rn) let

Mt(x) =

∫Rnf(ξ)e2πix·ξ Φ(tξ) dξ

23Exercise. Prove that if lima→∞∫ a

0f(x) dx = `, then At =

∫∞0f(x)e−tx dx converges to ` as t→ 0.


be the Φ-mean of the integral (3.3). Then

(3.4) Mt → f in L1 as t→ 0.

If in addition ϕ = Φ has an integrable radially decreasing majorant, then

(3.5) Mt → f a.e. as t→ 0.

Proof. The lemma below is a link between the Φ-mean and the approximation by convo-lution discussed in Section 2.7.

Lemma 3.20. If f,Φ ∈ L1(Rn) and ϕ = Φ, then∫Rnf(ξ)e2πix·ξΦ(tξ) dξ =

∫Rnf(x− y)ϕt(−y) dy = f ∗ ϕt(x), where ϕ(y) = ϕ(−y).

Proof. For h ∈ Rn define

g(h)(x) = e2πix·hΦ(tx) ∈ L1(Rn) .

Recall that24

(w(x)e2πih·x)∧(ξ) = w(ξ − h) and (w(tx))∧(ξ) = (w)t(ξ).

Hence(g(h))∧(ξ) = (Φ)t(ξ − h) = ϕt(ξ − h)

so replacing h by x we get(g(x))∧(ξ) = ϕt(ξ − x).

We have∫Rnf(ξ)e2πix·ξΦ(tξ) dξ =

∫Rnf(ξ)g(x)(ξ) dξ =

∫Rnf(ξ)(g(x))∧(ξ) dξ

=

∫Rnf(ξ)ϕt(ξ − x) dξ =

∫Rnf(x− y)ϕt(−y) dx.

We used here Theorem 3.2(c) and in the last equality the change of variables−y = ξ−x.

Now (3.4) follows directly from Theorem 2.44 and (3.5) follows from Theorem 2.49.

Remark 3.21. Theorem 3.19 is not easy to apply since given Φ ∈ L1, we have to computeΦ and show its integrability. In the case of the Gauss mean Φ(x) = e−|x|

2it can be easily

done with the help of Theorem 3.7, but the case of the Abel mean Φ(x) = e−|x| is muchharder due to difficulty of computing the Fourier transform of Φ(x) = e−|x|.

In Theorems 3.23 and 3.30 we will investigate the case of the the Gauss and the Abeland methods.

Theorem 3.22. Let f(x) = e−4π2t|x|2, t > 0. Then

(a) W (x, t) := f(x) = (4πt)−n/2e−|x|2/(4t).

(b) The function W has the following scaling property with respect to t: if ϕ(x) =W (x, 1), then W (x, t) = ϕt1/2(x).

24see Theorem 3.2.

32 PIOTR HAJ LASZ

(c) ∫RnW (x, t) dx = 1 for all t > 0.

Proof. We can write

f(x) = e−4π2t|x|2 = u((4πt)1/2x), where u(x) = e−π|x|2

.

Since (ψ(tx))∧(ξ) = (ψ)t(ξ) and u(ξ) = u(ξ) = e−π|ξ|2

we obtain25

W (ξ, t) = f(ξ) = (u)(4πt)1/2(ξ) = u(4πt)1/2(ξ) = (4πt)−n/2e−|ξ|2/(4t)

which is (a). In particular

ϕ(ξ) = W (ξ, 1) = u(4π)1/2(ξ).

Since (ψt1)t2(x) = ψt1t2(x) we get

W (ξ, t) = u(4πt)1/2(ξ) =(u(4π)1/2

)t1/2

(ξ) = ϕt1/2(ξ)

which is (b). Finally equality∫Rn ψt1 =

∫Rn ψt2 , t1, t2 > 0 along with W (ξ, t) = ϕt1/2(ξ)

yields (c): ∫RnW (ξ, t) dξ =

∫RnW(ξ,

1

4π

)dξ =

∫Rne−π|ξ|

2

dξ = 1.

As an application we obtain

Theorem 3.23 (The Gauss-Weierstrass summability method). If f ∈ L1(Rn), then

(3.6)

∫Rnf(ξ)e2πix·ξe−4π2t|ξ|2 dξ =

∫Rnf(y)W (x− y, t) dy → f as t→ 0+

both in L1(Rn) and almost everywhere.

Proof. First we will prove equality. Let Φ(x) = e−4π2|x|2 . Using notation from Theorem 3.22

Φ(x) = W (x, 1) = ϕ(x). Since ϕ(x) = ϕ(−x), ϕt1/2(x) = ϕt1/2(x) = W (x, t). ThusLemma 3.20 yields∫

Rnf(ξ)e2πix·ξe−4π2t|ξ|2 dξ =

∫Rnf(ξ)e2πix·ξΦ(t1.2ξ) dξ

= f ∗ ϕt1/2(x) =

∫RnW (x− y, t)f(y) dy.

Theorem 3.22 shows that the function Φ satisfies the assumptions of Theorem 3.19 so bothconvergences in L1 and almost everywhere follow.

Remark 3.24. The function W (x, t) = ϕt1/2(x) satisfies the assumptions of Theorem 2.49with a = 1 so for f ∈ Lp, 1 ≤ p <∞

(3.7)

∫Rnf(y)W (x− y, t) dy → f a.e. and in Lp as t→ 0+.

25The function W is defined as a Fourier transform so it will be more convenient for us to prove (a)-(c)with the variable x replaced by ξ.


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

w(x, t) =

∫RnW (x− y, t)f(y) dy =

1

(4πt)n/2

∫Rne−|x−y|2

4t f(y) dy

satisfies the heat equation ∂w∂t

= ∆xw in the interior of Rn+1+ . As we showed, the function

w(x, t) converges to f(x) as t→ 0, so w is a solution to∂w∂t

= ∆xw on Rn+1+ ,

w(x, 0) = f(x), x ∈ Rn.

Corollary 3.26. If both f and f are integrable26, then

(3.8) f(x) =

∫Rnf(ξ)e2πix·ξ dξ a.e.

Proof. Since f ∈ L1, the left hand side of (3.6) converges to∫Rn f(ξ)e2πix·ξ dξ so the result

follows from Theorem 3.23.

The above inversion formula motivates the following definitions.

Definition 3.27. The inverse Fourier transform is defined by

f(x) = F−1(f)(x) =

∫Rnf(ξ)e2πix·ξ dξ.

Corollary 3.28. If f1, f2 ∈ L1(Rn) and f1 = f2 on Rn, then f1 = f2 a.e.

Proof. Let f = f1 − f2. Then f ∈ L1 and f = 0 ∈ L1 so (3.8) yields that f = 0 a.e.

The following result provides another example of a function that satisfies the assumptionsof Theorem 3.19.

Theorem 3.29. Let f(x) = e−2π|x|t, t > 0. Then

(a)

P (x, t) := f(x) = cnt

(t2 + |x|2)(n+1)/2,

where

cn =Γ(n+1

2

)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) ∫RnP (x, t) dx = 1 for all t > 0.

By the same arguments as before we obtain.

26The assumption about integrability of f is very strong. As already observed, the equality (3.8) impliesthat f equals a.e. to a function in C0(Rn).

34 PIOTR HAJ LASZ

Theorem 3.30 (The Abel summability method). If f ∈ L1(Rn), then∫Rnf(ξ)e2πix·ξe−2π|ξ|t dξ =

∫Rnf(y)P (x− y, t) dy → f as t→ 0+

both in L1(Rn) and almost everywhere.

Remark 3.31. The function P (x, t) = ϕt(x) satisfies the assumptions of Theorem 2.49with a = 1 so for f ∈ Lp, 1 ≤ p <∞∫

Rnf(y)P (x− y, t) dy → f a.e. and in Lp as t→ 0+.

Remark 3.32. P (x, t) is called the Poisson kernel. Under suitable assumptions about f ,differentiation under the sign of the integral shows that the function

u(x, t) =

∫RnP (x− y, t)f(y) dy

satisfies the Laplace equation ∆(x,t)u = 0 in the interior of Rn+1+ . As we showed u(x, t)

converges to f as t→ 0+ so u(x, t) is a solution to the Dirichlet problem in the half-space∆(x,t)u = 0 on Rn+1

+ ,u(x, 0) = f(x), x ∈ Rn

The proof of Theorem 3.29 is substantially more difficult than that of Theorem 3.22.Since the formula for the Fourier transform involves the Γ function we need to recall itsbasic properties.

Definition 3.33. For 0 < x <∞ we define

Γ(x) =

∫ ∞0

tx−1e−t dt .

Theorem 3.34.

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

√π.

Proof. (a) follows from the integration by parts. Since Γ(1) = 1, (b) follows from (a) byinduction. The substitution t = s2 gives

Γ(x) =

∫ ∞0

tx−1e−t dt =

∫ ∞0

s2(x−1)e−s2

2s ds = 2

∫ ∞0

s2x−1e−s2

ds

and hence

Γ

(1

2

)= 2

∫ ∞0

e−s2

ds =√π .

The formula Γ(x) = Γ(x + 1)/x allows us to define Γ(x) for all negative non-integervalues of x. For example

Γ(− 3

2

)=

Γ(−1/2)

−3/2=

Γ(1/2)

(−3/2)(−1/2)=

4√π

3.


Definition 3.35. For x ∈ (−∞, 0) \ −1,−2,−3, . . ., the function Γ(x) is defined by

Γ(x) =Γ(x+ k)

x(x+ 1) · . . . · (x+ k − 1),

where k is a positive integer such that x+ k ∈ (0, 1).

Lemma 3.36. ∫ π/2

0

sinn θ dθ =

√πΓ(n+1

2

)nΓ(n2

) for n = 1, 2, 3, . . .

Proof. Denote the left hand side by an and the right hand side by bn. Easy one timeintegration by parts27 shows that

an+2 = (n+ 1)(an − an+2), an+2 =n+ 1

n+ 2an .

Also elementary properties of the Γ function show that

bn+2 =n+ 1

n+ 2bn

and now it is enough to observe that a1 = 1 = b1 = 1, a2 = π/4 = b2.

Lemma 3.37.

(a) The volume of the unit ball in Rn equals

(3.9) ωn =2πn/2

nΓ(n/2)=

πn/2

Γ(n2

+ 1) .

(b) The (n− 1)-dimensional measure of the unit sphere in Rn equals nωn

Proof.

It follows from the picture and the Fubini theorem that the volume of the upper half ofthe ball equals

1

2ωn =

∫ 1

0

ωn−1r(h)n−1 dh .

The substitutionh = 1− cos θ, dh = sin θ dθ, r(h) = sin θ

27Use sinn+2 Θ = (− cos θ)′ sinn+1 Θ.

36 PIOTR HAJ LASZ

and Lemma 3.36 give

1

2ωn = ωn−1

∫ π/2

0

sinn θ dθ = ωn−1

π1/2Γ(n+1

2

)nΓ(n2

) ,

so

ωn =2π1/2Γ

(n+1

2

)nΓ(n2

) ωn−1 .

If

an =2πn/2

nΓ(n2

)then a direct computation shows that a1 = 2 = ω1 and that an satisfies the same recur-rence relationship as ωn, so ωn = an for all n. The second equality in (3.9) follows fromTheorem 3.34(a).

(b) follows from the fact that the (n−1)-dimensional measure of a sphere of radius r equalsto the derivative with respect to r of the volume of an n-dimensional ball of radius r. 2

We will also need the following result.

Lemma 3.38. For β > 0 we have

e−β =1√π

∫ ∞0

e−u√ue−β

2/(4u) du .

Proof. Applying the theory of residues to the function eiβz/(1 + z2) one can easily provethat

e−β =2

π

∫ ∞0

cos βx

1 + x2dx .

This and an obvious identity

1

1 + x2=

∫ ∞0

e−(1+x2)u du

yields

e−β =2

π

∫ ∞0

cos βx

1 + x2dx

=2

π

∫ ∞0

cos βx(∫ ∞

0

e−ue−ux2

du)dx

=2

π

∫ ∞0

e−u(∫ ∞

0

e−ux2

cos βx dx)du

=2

π

∫ ∞0

e−u(1

2

∫ ∞−∞

e−ux2

eiβx dx)du

(x=−2πy)=

2

π

∫ ∞0

e−u(π

∫ ∞−∞

e−4π2uy2e−2πiβy dy︸︷︷︸W (β,u)

)du

=2

π

∫ ∞0

e−u(1

2

√π

ue−β

2/(4u))du


=1√π

∫ ∞0

e−u√ue−β

2/(4u) du .


Proof of Theorem 3.29. (a) By a change of variables formula it suffices to prove the formulafor t = 1. We have∫

Rne−2π|x|e−2πix·ξ dx =

∫Rn

( 1√π

∫ ∞0

e−u√ue−4π2|x|2/(4u) du

)e−2πix·ξ dx

=1√π

∫ ∞0

e−u√u

(∫Rne−4π2|x|2/(4u)e−2πix·ξ dx︸︷︷︸

W (ξ,(4u)−1)

)du

=1√π

∫ ∞0

e−u√u

(√u

π

)ne−u|ξ|

2

du

=1

π(n+1)/2

∫ ∞0

e−u(1+|ξ|2)u(n−1)/2 du

=1

π(n+1)/2

1

(1 + |ξ|2)(n+1)/2

∫ ∞0

e−ss(n−1)/2 ds︸︷︷︸Γ[(n+1)/2]

.

(b) is obvious.

(c) Because of the scaling property (b) it suffices to consider t = 1. We have∫RnP (x, 1) dx = cn

∫Rn

dx

(1 + |x|2)(n+1)/2

polar= cn

∫ ∞0

(∫Sn−1(0,1)

dσ

(1 + r2)(n+1)/2

)rn−1 dr

= cn nωn

∫ ∞0

rn−1

(1 + r2)(n+1)/2dr

r=tan θ= cn nωn

∫ π/2

0

sinn−1 θ dθ

=Γ(n+1

2

)π(n+1)/2

n2πn/2

nΓ(n2

) π1/2Γ(n2

)(n− 1)Γ

(n−1

2

)= 1 .


Corollary 3.26 applied to f(x) = e−4π2t|x|2 and to f(x) = e−2πt|x| yields

Corollary 3.39.∫RnW (ξ, t)e2πix·ξ dξ = e−4π2t|x|2 and

∫RnP (ξ, t)e2πix·ξ dξ = e−2π|x|t

for all x ∈ Rn.

38 PIOTR HAJ LASZ

The Weierstrass and Poisson kernels have the following semigroup property.

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

(Wt1 ∗Wt2)(x) = Wt1+t2(x) and (Pt1 ∗ Pt2)(x) = Pt1+t2(x)

for all x ∈ Rn.

Proof. It follows from Corollary 3.39 with x replaced by −x that

Wt(x) = e−4π2t|x|2 and Pt(x) = e−2π|x|t .

Hence Theorem 3.2(b) yields

(Wt1 ∗Wt2 ) (x) = Wt1(x)Wt2(x) = Wt1+t2(x)

(Pt1 ∗ Pt2 ) (x) = Pt1(x)Pt2(x) = Pt1+t2(x)

and the result follows from Corollary 3.28. 2

3.4. The Schwarz class and the Plancherel theorem.

Definition 3.41. We say that f belongs to the Schwarz class S (Rn) = Sn if f ∈ C∞(Rn)and for all multiindices α, β

supx∈Rn|xαDβf(x)| = pα,β(f) <∞.

That means all derivatives of f rapidly decrease to zero as |x| → ∞, faster than the inverseof any polynomial.

Clearly C∞0 (Rn) ⊂ Sn, but also e−|x|2 ∈ Sn so functions in the Schwarz class need not

be compactly supported.

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

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

limk→∞

pα,β(fk − f) = 0 for all multiindices α, β.

This convergence is metrizable. Indeed, dα,β(f, g) = pα,β(f − g) is a metric and if wearrange all these metrics in a sequence d′1, d

′2, . . ., then

d(f, g) =∞∑k=1

2−kd′k(f, g)

1 + d′k(f, g)

defines a metric in Sn such that fn → f in Sn if and only if fn → f in the metric d.

Proposition 3.43. The space Sn has the following properties.

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


(d) The mapping

Sn 3 ϕ 7→ xαDβϕ(x) ∈ Sn

is continuous.(e) If ϕ ∈ Sn, then

ϕ(x+ hek)− ϕ(x)

h→ ∂ϕ

∂xk(x) as h→ 0.

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.


Theorem 3.44. The Fourier transform is a continuous, one-to-one mapping of Sn ontoSn such that

(a) (∂f

∂xj

)∧(ξ) = 2πiξj f(ξ),

(b)

(−2πixjf ) (ξ) =∂f

∂ξj(ξ) ,

(c)

(f ∗ g) = f g,

(d) ∫Rnf(x)g(x) dx =

∫Rnf(x)g(x) dx ,

(e)

f(x) =

∫Rnf(ξ)e2πix·ξ dξ .

Proof. We already proved (a), (b), (c) and (d). Now we will prove that the Fouriertransform is a continuous mapping from Sn into Sn. Formulas (a) and (b) imply that

ξαDβ f(ξ) = C(Dα(xβf)) (ξ).

Since ‖g‖∞ ≤ ‖g‖1 we get

pα,β(f) = ‖ξαDβ f(ξ)‖∞ ≤ C‖Dα(xβf)‖1 .

An application of the Leibnitz rule28 implies that Dα(xβf) equals a finite sum of expressionsof the form xβiDαif . Since

‖xβiDαif‖1 =

∫Rn|(1 + |x|2)nxβiDαif(x)|(1 + |x|2)−n dx

28Product rule.

40 PIOTR HAJ LASZ

≤ C(n) supx∈Rn|(1 + |x|2)nxβiDαif(x)| <∞

it follows that f ∈ Sn. One can also easily deduce that the mapping ˆ : Sn → Sn iscontinuous.

If f ∈ Sn than f ∈ Sn and hence both f and f are integrable, so (e) follows from theinversion formula. This formula also shows that the Fourier transform applied four timesis an identity on Sn and hence the Fourier transform is a bijection on Sn. 2

Theorem 3.45 (Plancherel). The Fourier transform is an L2 isometry on a dense subsetSn of L2

‖f‖2 = ‖f‖2, f ∈ Sn,

and hence it uniquely extends to an isometry of L2

‖f‖2 = ‖f‖2, f ∈ L2(Rn) .

Moreover for f ∈ L2(Rn)

f(ξ) = limR→∞

∫|x|<R

f(x)e−2πix·ξ dx

in the L2 sense, i.e. ∥∥∥f − ∫|x|<R

f(x)e−2πix·ξ dx∥∥∥

2→ 0 as R→∞

and similarly

f(x) = limR→∞

∫|ξ|<R

f(ξ)e2πix·ξ dξ

in the L2 sense.

Proof. Given f ∈ Sn let g =¯f , so g = f . Indeed,

g(ξ) =

∫Rn

¯f(x)e−2πix·ξ dx =

∫Rnf(x)e2πix·ξ dx = f(x) .

Hence Theorem 3.44(d) gives

‖f‖2 =

∫Rnff =

∫Rnfg =

∫Rnf g =

∫Rnf

¯f = ‖f‖2 .

Thus the Fourier transform is an L2 isometry on Sn. Since Sn is a dense subset of L2 ituniquely extends to an isometry of L2. Now for f ∈ L2(Rn) we have

L1 3 fχB(0,R)L2

−→ f as R→∞and hence

(fχB(0,R)) (ξ) =

∫|x|≤R

f(x)e−2πix·ξ dxL2

−→ f(ξ)

as R→∞. Similarly ∫|ξ|<R

f(ξ)e2πix·ξ dξL2

−→ f(x) as R→∞.


Proposition 3.46. If f, g ∈ L2(Rn), then∫Rnf(x)g(x) dx =

∫Rnf(x)g(x) dx .

Proof. Approximate f and g in L2 by functions in Sn, apply Theorem 3.44(d) and pass tothe limit.

Consider the class L1(Rn) + L2(Rn) consisting of functions of the form f = f1 + f2,f1 ∈ L1, f2 ∈ L2. Then we define

f = f1 + f2.

In order to show that the Fourier transform is well defined in the class L1 +L2 we need toshow that it does not depend on the particular choice of the representation f = f1 + f2.Indeed, if we also have f = g1 + g2, g1 ∈ L1, g2 ∈ L2, then f1− g1 = g2− f2 ∈ L1 ∩L2 andhence

f1 − g1 = (f1 − g1) = (g2 − f2) = g2 − f2,

f1 + f2 = g1 + g2 .

It is an easy exercise to show that

Lp(Rn) ⊂ L1(Rn) + L2(Rn), for 1 ≤ p ≤ 2,

and hence the Fourier transform is well defined on Lp(Rn), 1 ≤ p ≤ 2 and

ˆ: Lp(Rn)→ L2(Rn) + C0(Rn), for 1 ≤ p ≤ 2.

Later we will prove the Hausdorff-Young Inequality which implies that

ˆ: Lp(Rn)→ Lp′(Rn), for 1 ≤ p ≤ 2,

where p′ is the Holder conjugate to p.

42 PIOTR HAJ LASZ

4. Miscellaneous results about the Fourier transform

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

4.1. The Poisson summation formula.

Theorem 4.1 (Poisson summation formula). If f ∈ C1(R) and

(4.1) |f(x)|+ |f ′(x)| ≤ C

1 + x2for x ∈ R,

then∞∑

n=−∞

f(n) =∞∑

n=−∞

f(n).

Remark 4.2. Here f(n) stand for the Fourier transform of f evaluated at n – do not getconfused with the Fourier series.

Note that it follows from the growth condition (4.1) that the series∑

n f(n) convergesand that any function f ∈ Sn satisfies (4.1).

Proof. It follows from (4.1) that

(4.2) g(x) =∞∑

k=−∞

f(x+ k)

is a periodic function of class C1 with period 1. Therefore g is can be represented as aFourier series29

g(x) =∞∑

n=−∞

g(n)e2πinx

where

g(n) =

∫ 1

0

g(x)e−2πinx dx =∞∑

k=−∞

∫ 1

0

f(x+ k)e−2πinx dx =∞∑

k=−∞

∫ k+1

k

f(x)e−2πinx dx

=

∫ ∞−∞

f(x)e−2πinx dx = f(n)

29Now g(n) are Fourier series coefficients – do not get confused with the Fourier transform!


so∞∑

k=−∞

f(k) = g(0) =∞∑

n=−∞

g(n) =∞∑

n=−∞

f(n).

As an application we will prove the Jacobi identity.

Definition 4.3. The Jacobi ‘theta’ function is defined by

ϑ(t) =∞∑

n=−∞

e−πn2t, t > 0.

Clearly ϑ ∈ C∞(R).

Corollary 4.4 (Jacobi identity). For t > 0 we have ϑ(t) = t−12ϑ(1/t).

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

f(x) = t−1/2u(xt−1/2) where u(x) = e−πx2

.

Hence30

f(ξ) = t−1/2(u)t−1/2(ξ) = t−1/2ut−1/2(ξ) = u(ξt1/2) = e−πξ2t.

Thus the Poisson summation formula yields

t−1/2

∞∑n=−∞

e−πn2/t =

∞∑n=−∞

e−πn2t i.e., t−1/2ϑ(1/t) = ϑ(t).

Corollary 4.5. For t > 0 we have

∞∑n=−∞

1

t2 + n2=π

t

1 + e−2πt

1− e−2πt.

Remark 4.6. It is easy to show by taking the limit as t→ 0+ in the above identity that

∞∑n=1

1

n2=π2

6.

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

f(x) =t

π(t2 + x2)

30We use here two facts: (ϕ(sx))∧(ξ) = (ϕ)s(ξ) = s−nϕ(ξ/s) with n = 1 and u(ξ) = u(ξ).31We are in dimension n = 1 and c1 = 1.

44 PIOTR HAJ LASZ

so the Poisson summation formula gives32

t

π

∞∑n=−∞

1

t2 + n2=

∞∑n=−∞

e−2π|n|t =1 + e−2πt

1− e−2πt

from which the theorem follows.

4.2. The Heisenberg inequality. Celebrated Heisenberg’s uncertainty principle assertsthat position and momentum of a particle cannot be measured at the same time and itcan be written as an inequality

σxσp ≥h

4πwhere σx and σp are standard deviations of position and momentum and h is the Plankconstant. Translating it into our language the above inequality can be formulated as

Theorem 4.7 (Heisenberg’s inequality). For any f ∈ L2(R) and a, b ∈ R,√∫ ∞−∞

(x− a)2|f(x)|2 dx

√∫ ∞−∞

(ξ − b)2|f(ξ)|2 dξ ≥ ‖f‖22

4π.

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

Proof. We will prove the result under the assumption that f ∈ S1. The case of generalf ∈ L2 is left to the reader as an exercise. Let f(x) = e2πibxg(x− a). Since

f(ξ) = g(ξ − b)e−2πa(ξ−b)

we easily obtain that the Heisenberg inequality is equivalent to

(4.3)

(∫ ∞−∞|x|2|g(x)|2 dx

)1/2 (∫ ∞−∞|ξ|2|g(ξ)|2 dξ

)1/2

≥ ‖g‖22

4π.

Note that (|g|2)′ = (gg)′ = 2 re g′ g so integration by parts gives∫ β

α

|g(x)|2 dx =

∫ β

α

x′|g(x)|2 dx = x|g(x)|2∣∣∣βα− 2 re

∫ β

α

xg′(x)g(x) dx.

Letting α→ −∞, β → +∞ and using the fact that g ∈ S1 yields∫ ∞−∞|g(x)|2 dx = −2 re

∫ ∞−∞

xg′(x)g(x) dx

≤ 2

∫ ∞−∞|x| |g(x)| |g′(x)| dx(4.4)

≤ 2

(∫ ∞−∞|x|2|g(x)|2 dx

)1/2(∫ ∞−∞|g′(x)|2 dx

)1/2

(4.5)

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


= 2

(∫ ∞−∞|x|2|g(x)|2 dx

)1/2(∫ ∞−∞|g′(ξ)|2 dξ

)1/2

= 2

(∫ ∞−∞|x|2|g(x)|2 dx

)1/2(∫ ∞−∞|2πiξg(ξ)|2 dξ

)1/2

from which (4.3) follows. Finally, let us investigate the case of equality. For the equalityin the Schwarz inequality (4.5) we need |g′(x)| to be proportional to |xg(x)|, say |g′(x)| =k|xg(x)|, for some k > 0 so g′(x) = keiϕ(x)xg(x) for some real valued function ϕ(x). Butthen equality in (4.4) implies that

−re e−ϕ(x) = 1 so eiϕ(x) = −1

and we have g′(x) = −kxg(x). Solving this differential equation yields g(x) = Ce−kx2

and

hence f(x) = Ce2πibxe−k(x−a)2 .

4.3. Eigenfunctions of the Fourier transform. The Fourier transform applied fourtimes is identity on L2(Rn) so if f is an eigenfunction of the Fourier transform,

f = λf, then f = |λ|4f so λ ∈ −1, 1,−i, i.

We know already one eigenfunction with the eigenvalue 1, namely f(x) = e−πx2

and nowwe will find now all eigenfunctions in dimension n = 1.

It is natural to search among functions of the form p(x)e−πx2, where p(x) is a polyno-

mial.33

Definition 4.8. Hermite functions are defined by the formula

hn(x) =(−1)n

n!eπx

2( ddx

)ne−2πx2 , for all integers n ≥ 0.

Note that h0(x) = e−πx2.

Lemma 4.9. hn(x) = pn(x)e−πx2

for a polynomial pn(x) of order n of the form

pn(x) =(4π)n

n!xn + lower order terms.

In particular hn ∈ S1.

This lemma easily follows from the definition of hn.

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

Proof. Applying the product rule to

hn(x) =[eπx

2]·[(−1)n

n!

( ddx

)ne−2πx2

]33Note that evaluation of the Fourier transform of p(x)e−πx

2

is straightforward: since F(e−πx2

)(ξ) =

e−πξ2

it suffices to apply (3.1).

46 PIOTR HAJ LASZ

gives

h′m(x) = 2πxhn(x) +[eπx]·[ (−1)n

n!︸︷︷︸−(n+1)

(−1)n+1

(n+1)!

( ddx

)n+1

e−2πx2]

= 2πxhn(x)− (n+ 1)hn+1(x).

Lemma 4.11. If f ∈ C∞(R) and n is a positive integer, then

(4πx)( ddx

)nf(x)−

( ddx

)n(4πxf(x)) = −4πn

( ddx

)n−1

f(x) for all x ∈ R.

The result follows from a simple induction.

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

Proof. For n = 0 we check the equality directly. For n ≥ 1 Lemma 4.10 gives

h′n(x) + 2πxhn(x) = 4πxhn(x)− (n+ 1)hn+1(x)

= 4πx(−1)n

n!eπx

2( ddx

)ne−2πx2 −(n+ 1)

(−1)n+1

(n+ 1)!︸︷︷︸+(−1)n/n!

eπx2( ddx

)n+1

e−2πx2︸︷︷︸( ddx

)n(−4πxe−2πx2 )

=(−1)n

n!eπx

2[4πx

( ddx

)ne−2πx2 −

( ddx

)n(4πxe−2πx2)

]︸︷︷︸

−4πn( ddx

)n−1e−2πx2by Lemma 4.11

= 4π(−1)n−1

(n− 1)!eπx

2( ddx

)n−1

e−2πx2 = 4πhn−1(x).

Lemma 4.13. hn = (−i)nhn and hn = inhn.

Remark 4.14. The lemma yields

h4n = h4n, h4n+2 = −h4n+2, h4n+3 = ih4n+3, h4n+1 = −ih4n+1

Thus the lemma provides examples of eigenfunctions corresponding to all possible eigen-values 1,−1, i,−i.

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

h′n − 2πxhn = −(n+ 1)hn+1, h′n − 2πxhn = −(n+ 1)hn+1.

Since (−2πixf)∧ = (f)′ and f ′ = 2πiξf(ξ) we have

(4.6) 2πiξhn(ξ)− i(hn)′(ξ) = −(n+ 1)hn+1(ξ)

We will prove the equality hn = (−i)nhn by induction.

For n = 0, h0 = e−πx2

so h0 = h0 = (−i)0h0.


Suppose now that the equality holds for n and we will prove it for n + 1. The equality(4.6) yields.

2πiξ(−i)nhn(ξ)− i((−i)nhn(ξ))′ = −(n+ 1)hn+1(ξ),

−(−i)n+12πξhn(ξ) + (−i)n+1h′n(ξ) = −(n+ 1)hn+1(ξ),

(−i)n+1[h′n(ξ)− 2πξhn(ξ)

]︸︷︷︸−(n+1)hn+1(ξ) by Lemma 4.10

= −(n+ 1)hn+1(ξ).

This proves that hn+1 = (−i)n+1hn+1(ξ) which completes the proof of the claim that

hn = (−i)nhn(ξ) fro all n. Finally

hn = (hn)∨ = ((−i)nhn)∨ = (−i)nhn, hn = inhn.

Let Kf = f ′′ − 4π2x2f .

Lemma 4.15. Khn = −4π(n+ 12)hn

Proof. Lemmas 4.10 and 4.12 yield

h′n − 2πxhn = −(n+ 1)hn+1,

h′′n − 2πhn − 2πxh′n = −(n+ 1)h′n+1,

h′′n = 2πhn + 2πxh′n − (n+ 1)h′n+1.

Hence

Khn = h′′n − 4π2x2hn = 2πhn + 2πxh′n − (n+ 1)h′n+1 − 4π2x2hn

= 2πhn − (n+ 1)h′n+1 + 2πx (h′n − 2πxhn)︸︷︷︸−(n+1)hn+1

= 2πhn − (n+ 1)(h′n+1 + 2πxhn+1)

= −4π(n+

1

2

)hn.

Lemma 4.16. For real valued functions f, g ∈ S1 we have34

〈Kf, g〉 = 〈f,Kg〉

Proof. The integration by parts yields

〈Kf, g〉 =

∫R(f ′′ − 4π2x2f)g =

∫Rf ′′g −

∫R

4π2x2fg

=

∫Rfg′′ −

∫R

4π2x2fg =

∫f(g′′ − 4π2x2g) = 〈f,Kg〉.

Lemma 4.17. The Hermite functions hn are orthogonal,

〈hn, hm〉 = 0, for n 6= m.

34Here 〈·, ·〉 denotes the L2 scalar product.

48 PIOTR HAJ LASZ

Proof. We have

−4π(n+

1

2

)〈hn, hm〉 = 〈Khn, hm〉 = 〈hn,Khm〉 = −4π

(m+

1

2

)〈hn, hm〉

and the result easily follows.

Lemma 4.18.

‖hn‖22 =

(4π)n√2n!

.

Proof. Since by Lemma 4.10 we have (n+1)hn+1 = 2πxhn−h′n, integration by parts yields

(n+ 1)‖hn+1‖22 = 2π

∫Rxhnhn+1 −

∫Rh′nhn+1 = 2π

∫Rxhnhn+1 +

∫Rhnh

′n+1

=

∫Rhn (h′n+1 + 2πxhn+1)︸︷︷︸

4πhn

= 4π‖hn‖22.

Replacing n by n− 1 yields

‖hn‖22 =

4π

n‖hn−1‖2

2

and a simple induction argument yields

‖hn‖22 =

(4π)n

n!‖h0‖2

2.

Now the result follows from a simple observation that

‖h0‖22 =

∫R(e−πx

2

)2 dx =

∫Re−2πx2 dx =

1√2

∫Re−πy

2

dy =1√2.

The last two lemmas show that the scaled Hermite functions

en =((4π)n√

2n!

)−1/2

hn

form an orthonormal family in L2(R).

Theorem 4.19. The fmaily en∞n=0 is an orthonormal basis of L2(Rn).

Proof. It suffices to prove that if f ∈ L2(R) and∫R fhn = 0 for all n ≥ 0, then f = 0 a.e.

Thus suppose that∫R fhn = 0 for all n ≥ 0. Since

hn(x) =((4π)2

n!xn + . . .

)e−πx

2

,

it easily follows that the functions xne−πx2, n ≥ 0, are linear combinations of the functions

h0, h1, . . . , hn. Hence ∫Rf(x)xne−πx

2

dx = 0 for all n ≥ 0.


Since by the Schwarz inequality f(x)e−πx2 ∈ L1 we can compute the Fourier transform

directly(f(x)e−πx

2)∧=

∫Rf(x)e−πx

2

e−2πixξ dx =

∫Rf(x)e−πx

2∞∑n=0

(−2πixξ)n

n!dx

=∞∑n=0

(−2πiξ)n

n!

∫Rf(x)xne−πx

2

dx = 0.

Vanishing of the Fourier transform of f(x)e−πx2

implies that f(x)e−πx2

= 0 a.e. and hencef(x) = 0 a.e.

Since en∞n=0 is an orthonormal basis of L2(R) any f ∈ L2 can be written as

f =∞∑n=0

〈f, en〉 en

This and the fact that en = (−i)nen yield

Theorem 4.20 (Wiener’s definition of the Fourier transform). For f ∈ L2(R) we have

f =∞∑n=0

〈f, en〉(−i)nen.

Let Hk, k = 0, 1, 2, 3 be subspaces of L2(R) defined by

Hk = span e4n+k∞n=0.

The subspaces Hk are orthogonal and

L2(R) = H0 ⊕H1 ⊕H2 ⊕H3.

Clearly, the subspaces Hk are eigenspaces of the Fourier transform:

f = (−i)kf for f ∈ Hk and k = 0, 1, 2, 3.

4.4. The Bochner-Hecke formula. As was pointed out in Section 4.3 computing theFourier transform of P (x)e−π|x|

2is straightforward due to formula (3.1) and Theorem 3.7,

but the computations are often algebraically complicated and it is not always obvious howto write the answer in a compact and nice form. We have see examples of such computationsin Section 4.3 when we found eigenfunctions of the Fourier transform and now we will seeanother example of this type. The results of this section will be used later in Section 6.3

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

Pk(x) =∑|α|=k

aαxα and ∆Pk = 0 .

Theorem 4.22 (Bochner-Hecke). If Pk(x) is a homogeneous harmonic polynomial of de-gree k, then

F(Pk(x)e−π|x|

2)

(ξ) = (−i)kPk(ξ)e−π|ξ|2

.

50 PIOTR HAJ LASZ

Proof. Since the polynomial Pk is homogeneous of degree k,

Pk(−2πix) =∑|α|=k

aα(2πix)α = (−2πi)kPk(x).

Hence (3.1) applied to f(x) = e−π|x|2

yields35∫RnPk(x)e−π|x|

2

e−2πix·ξ dx = F(Pk(x)e−π|x|

2)

(ξ)(4.7)

= (−2πi)−kPk(D)e−π|ξ|2

= Q(ξ)e−π|ξ|2

for some polynomial Q and it remains to prove that Q(ξ) = Pk(−iξ).

Lemma 4.23. If P (x) is a polynomial36 in Rn, then

(4.8)

∫RnP (x)e−π

∑j(xj+iξj)

2

dx =

∫RnP (x− iξ)e−π|x|2 dx .

Proof. By the same argument involving the same contour integration as in the proof ofTheorem 3.7 for any polynomial P of one variable we have

(4.9)

∫ ∞−∞

P (x)e−π(x+iξ)2 dx =

∫ ∞−∞

P (x− iξ)e−πx2 dx .

If P is a polynomial in n variables, then (4.8) follows from (4.9) and the Fubini theorem.

Multiplying both sides of (4.7) by eπ|ξ|2

and applying (4.8) we obtain

Q(ξ) =

∫RnPk(x)e−π

∑j(xj+iξj)

2

dx =

∫RnPk(x− iξ)e−π|x|

2

dx .

We can write

Pk(x− η) =∑α

ηαPα(x)

and then ∫RnPk(x− η)e−π|x|

2

dx =∑α

ηα∫RnPα(x)e−π|x|

2

dx ,

so clearly the integral is a polynomial is η. With this notation we obtain

Q(ξ) =∑α

(iξ)α∫RnPα(x)e−π|x|

2

dx

and hence

Q(ξ/i) =∑α

ξα∫RnPα(x)e−π|x|

2

dx =

∫RnPk(x− ξ)e−π|x|

2

dx .

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

e−π|x|2

e−2πix·ξ dx = e−π|ξ|2

and differentiate under the sign of the integral.36Any polynomial, not necessarily harmonic or homogeneous.


Since Pk is a harmonic function, it has the mean value property37∫Sn−1

Pk(sθ − ξ) dσ(θ) = |Sn−1|Pk(−ξ) .

Thus integration in polar coordinates gives

Q(ξ/i) =

∫ ∞0

sn−1

(∫Sn−1

Pk(sθ − ξ) dσ(θ)

)e−πs

2

ds

= Pk(−ξ)∫ ∞

0

sn−1|Sn−1|e−πs2 ds

= Pk(−ξ)∫Rne−π|x|

2

dx = Pk(−ξ)

and hence Q(ξ) = Pk(−iξ). The proof is complete.

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

Corollary 4.24. If Pk is a homogeneous harmonic polynomial of degree k, then for anyt > 0 we have

F(Pk(x)e−πt|x|

2)

(ξ) = t−k−n2 (−i)kPk(ξ)e−π|ξ|

2/t .

We leave details as an easy exercise.

4.5. Symmetry of the Fourier transform. If f is radially symmetric i.e., f = f ρ for

all ρ ∈ O(n), then f ρ = f ρ = f for all ρ ∈ O(n) so f is radially symmetric too. Nowwe will prove a more delicate result about the symmetry of the Fourier transform.

Proposition 4.25. Let f ∈ L1(Rn) be a real valued function of the form

f(x) =xj|x|g(|x|).

Then there is a continuous function h : [0,∞) → R, h(0) = 0, h(t) → 0 as t → ∞ suchthat

f(ξ) = iξj|ξ|h(|ξ|), ξ ∈ Rn.

While the function f is not radially symmetric a more symmetric object is the map

F (x) =x

|x|g(|x|) ∈ L1(Rn,Rn), x ∈ Rn

so f is the jth component of F . We will prove that

F (ξ) = iξ

|ξ|h(|ξ|), ξ ∈ Rn

from which Proposition 4.25 will readily follow.

In the proof we will use the following result which is interesting on its own. It ill be usedlater one more time in a proof of Theorem 6.2.

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

52 PIOTR HAJ LASZ

Lemma 4.26. If m : Rn → Rn is a measurable function that is homogeneous of degree 0,i.e. m(tx) = m(x) for t > 0, and commutes with the orthogonal transformations, i.e.

(4.10) m(ρ(x)) = ρ(m(x))

for all x ∈ Rn and ρ ∈ O(n), then there is a constant c such that

(4.11) m(x) = cx

|x|for all x 6= 0.

Proof. Let e1, e2, . . . , en be the standard orthogonal basis of Rn. If [ρjk] is the matrixrepresentation of ρ ∈ O(n), then the condition (4.10) reads as

(4.12) mj(ρ(x)) =n∑k=1

ρjkmk(x), j = 1, 2, . . . , n,

where m(x) = (m1(x), . . . ,mn(x)).

Let m1(e1) = c. Consider all ρ ∈ O(n) such that ρ(e1) = e1. This condition means thatthe first column of the matrix [ρjk] equals e1, i.e. ρ11 = 1, ρj1 = 0, for j > 1. Since columnsare orthogonal, for k > 1 we have

0 =n∑j=1

ρj1ρjk = ρ1k .

Thus

ρ =

1 0 . . . 00 ρ22 . . . ρ2n...

.... . .

...0 ρn2 . . . ρnn

,

where [ρjk]nj,k=2 is the matrix of an arbitrary orthogonal transformation in the (n − 1)-

dimensional subspace orthogonal to e1.

For x = e1 = ρ(e1) = ρ(x) and j ≥ 2 identity (4.12) yields

mj(e1) =n∑k=1

ρjkmk(e1) =n∑k=2

ρjkmk(e1) ,

and hence m2(e1)...

mn(e1)

=

ρ22 . . . ρ2n...

. . ....

ρn2 . . . ρnn

m2(e1)...

mn(e1)

.

That means the vector [m2(e1), . . . ,mn(e1)]T is fixed under an arbitrary orthogonal trans-formation of Rn−1, so it must be a zero vector, i.e.

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

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

mj(ρ(e1)) = ρj1m1(e1) = cρj1 .


By homogeneity it suffices to prove (4.11) for |x| = 1. Let ρ ∈ O(n) be such that ρ(e1) = x.Then ρj1 = xj, j = 1, 2, . . . , n and hence

mj(x) = cρj1 = cxj = cxj|x|

.

This completes the proof of the lemma.

Proof of Proposition 4.25. Since the function F is odd

iF (ξ) =

∫RniF (x)(cos(2πx · ξ)− i sin(2πx · ξ)) dx =

∫Rn

sin(2πx · ξ)F (x) dx

so the mapping ξ 7→ iF (ξ) takes values in Rn. Fix k > 0 and define mk : Sn−1(0, k)→ Rn

by mk(ξ) = iF (ξ) for |ξ| = k. Extend mk to mk : Rn\0 → Rn as a function homogeneousof degree 0, i.e.

mk(ξ) = i F(kξ|ξ|

)for ξ 6= 0.

We claim that

(4.13) mk(ρ(ξ)) = ρ(mk(ξ)) for ρ ∈ O(n) and ξ 6= 0.

Since mk is homogeneous of degree 0, it suffices to check (4.13) for |ξ| = k. We have38

mk(ρ(ξ)) =

∫Rn

sin(2πx · ρ(ξ))x

|x|g(|x|) dx =

∫Rn

sin(2πρ−1(x) · ξ) x|x|g(|x|) dx

=

∫Rn

sin(2πx · ξ) ρ(x)

|ρ(x)|g(|ρ(x)|) dx =

∫Rn

sin(2πx · ξ)ρ(x)

|x|g(|x|) dx

= ρ(∫

Rnsin(2πx · ξ) x

|x|g(|x|) dx

)= ρ(mk(ξ)).

According to Lemma 4.26 there is a constant h(k) ∈ R such that mk(ξ) = −h(k)ξ/|ξ|. Inparticular for |ξ| = k we have

iF (ξ) = mk(ξ) = − ξ

|ξ|h(|ξ|).

Clearly h is continuous, h(0) = 0 and h(t)→ 0 as t→∞, because iF ∈ C0(Rn,Rn).

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

54 PIOTR HAJ LASZ

5. Tempered distributions

5.1. Basic constructions.

Definition 5.1. The space S ′n of all continuous linear functionals on Sn is called the

space of tempered distributions. The evaluation of u ∈ S ′n on ϕ ∈ Sn will usually be

denoted by u[ϕ]. The space S ′n is equipped with the weak-∗ convergence (usually called

weak convergence or just convergence). Namely uk → u in S ′n if uk[ψ] → u[ψ] for every

ψ ∈ Sn.

Here are examples:

1. If f ∈ Lp(Rn), 1 ≤ p ≤ ∞, then

Lf [ϕ] =

∫Rnf(x)ϕ(x) dx defines Lf ∈ S ′

n.

2. If µ is a measure of finite total variation, then

Lµ[ϕ] =

∫Rnϕdµ defines Lµ ∈ S ′

n.

3. We say that a function f is a tempered Lp function if f(x)(1+ |x|2)−k ∈ Lp(Rn) for somenonnegative integer k. If p =∞ we call f a slowly increasing function. Then

Lf [ϕ] =

∫Rnf(x)ϕ(x) dµ defines Lf ∈ S ′

n for all 1 ≤ p ≤ ∞.

Note that slowly increasing functions are exactly measurable functions bounded by poly-nomials.

4. A tempered measure is a Borel measure µ such that∫Rn

(1 + |x|2)−k dµ <∞

for some integer k ≥ 0. As before Lµ ∈ S ′n.

5. L[ϕ] = Dαϕ(x0) is a tempered distribution L ∈ S ′n.

The distributions generated by a function or by a measure will often be denoted byLf [ϕ] = f [ϕ], Lµ[ϕ] = µ[ϕ].


Suppose that u ∈ S ′n. If there is a tempered Lp 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 . Theidentification is possible, because the function f is uniquely defined (up to a.e. equivalence).This follows from a well known result.

Lemma 5.2. If Ω ⊂ Rn is open and f ∈ L1loc(Ω) satisfies

∫Ωfϕ = 0 for all ϕ ∈ C∞0 (Ω),

then f = 0 a.e.

Note that not every function f ∈ C∞(Rn) defines a tempered distribution, because itmay happen that form some ϕ ∈ Sn the function fϕ is not integrable and hence theintegral f [ϕ] =

∫Rn f(x)ϕ(x) dx does not make sense.

Theorem 5.3. A linear functional L on Sn is a tempered distribution if and only if thereis a constant C > 0 and a positive integer m such that

|L(ϕ)| ≤ C∑

|α|,|β|≤m

pα,β(ϕ) for all ϕ ∈ Sn.

Proof. If a linear functional L satisfies the given estimate, then clearly it is continuouson Sn, so it remains to prove the converse implication. Let L ∈ S ′

n. We claim that thereis a positive integer m such that |L(ϕ)| ≤ 1 for all

ϕ ∈ϕ ∈ Sn :

∑|α|,|β|≤m

pα,β(ϕ) ≤ 1

m

:= Nm .

To the contrary suppose that there is a sequence ϕk ∈ Sn such that |L(ϕk)| > 1 and

(5.1)∑|α|,|β|≤k

pα,β(ϕk) ≤1

k, k = 1, 2, 3, . . .

Note that (5.1) implies that ϕk → 0 in Sn, so the inequality |L(ϕk)| > 1 contradictscontinuity of L. This proves the claim. Denote

‖ϕ‖ =∑

|α|,|β|≤m

pα,β(ϕ) .

Observe that ‖ · ‖ is a norm. For an arbitrary 0 6= ϕ ∈ Sn, ϕ = ϕ/(m‖ϕ‖) satisfies‖ϕ‖ ≤ 1/m, so ϕ ∈ Nm and hence

|L(ϕ)| = m‖ϕ‖|L(ϕ)| ≤ m‖ϕ‖which proves the theorem. 2

For any function g on Rn we define g(x) = g(−x). Then it easily follows from the Fubinitheorem that for u, ϕ, ψ ∈ Sn∫

Rn(u ∗ ϕ)(x)ψ(x) dx =

∫Rnu(x)(ϕ ∗ ψ)(x) dx .

Regarding the functions u and u ∗ ϕ as distributions we can rewrite this equality as

(u ∗ ϕ)[ψ] = u[ϕ ∗ ψ]

If u ∈ S ′n and ϕ ∈ Sn, then ψ 7→ u[ϕ ∗ ψ] is a tempered distribution so it motivates the

following definition.

56 PIOTR HAJ LASZ

Definition 5.4. If u ∈ S ′n and ϕ ∈ Sn, then the convolution of u and ϕ is a tempered

distribution defined by the formula

(u ∗ ϕ)[ψ] := u[ϕ ∗ ψ] .

The following two results are left as an easy exercise.

Proposition 5.5. If u ∈ S ′n, ϕ, ψ ∈ Sn, then

(u ∗ ϕ) ∗ ψ = u ∗ (ϕ ∗ ψ) .

Proposition 5.6. If µ ∈ B(Rn) and ϕ ∈ Sn, then µ ∗ ϕ = ϕ ∗ µ where µ ∗ ϕ is theconvolution of a distribution µ ∈ S ′

n with a function ϕ ∈ Sn and ϕ ∗ µ is the convolutionof a function with a measure.

Theorem 5.7. If u ∈ S ′n and ϕ ∈ Sn, then the tempered distribution u∗ϕ can be identified

with a slowly increasing function f defined by

f(x) = u[τ−xϕ] = u[ϕ(x− ·)] for all x ∈ Rn.

Moreover f ∈ C∞(Rn) and all its derivatives are slowly increasing.

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

It follows from Theorem 3.43(c) that f is continuous. Observe that for a fixed x ∈ Rn

f(x+ hek)− f(x)

h= u

[ϕ(x+ hek − ·)− ϕ(x− ·)

h

]→ u[(∂kϕ)(x− ·)] as h→ 0

because it is easy to check using Theorem 3.43(e) that for every x ∈ Rn

ϕ(x+ hek − ·)− ϕ(x− ·)h

→ (∂kϕ)(x− ·) as h→ 0

in the topology of Sn. Thus the partial derivatives

∂kf(x) = u[(∂kϕ)(x− ·)]exist and are continuous by Theorem 3.43(c). Since ∂kϕ ∈ Sn, iterating the above processfor any multiindex α we obtain

(5.2) Dαf(x) = u[(Dαϕ)(x− ·)] .This implies that f ∈ C∞(Rn). Since u is a tempered distribution, Theorem 5.3 gives theestimate

(5.3) |f(x)| = |u[τ−xϕ]| ≤ C∑

|α|,|β|≤m

pα,β(τ−xϕ) ≤ C ′(1 + |x|)m .

Indeed,

pα,β(τ−xϕ) = supz∈Rn|zα(Dβϕ)(z − x)| = sup

z∈Rn|(z + x)α(Dβϕ)(z)|

≤ C(1 + |x||α|

)supz∈Rn

(1 + |z||α|

)|Dβϕ(z)| ≤ C ′(1 + |x|)m .


Hence f is slowly increasing, and so it defines a tempered distribution. Since the derivativesof f satisfy (5.2), which is an expression of the same type as the one in the definition off , we conclude that all derivatives Dαf are slowly increasing.

It remains to prove that f = u ∗ ϕ in the sense of tempered distributions i.e.,

(5.4) (u ∗ ϕ)[ψ] = f [ψ] for all ψ ∈ Sn.

Using f(y) = u[ϕ(y − ·)], (5.4) reads as

u[ ∫

Rnϕ(y − ·)ψ(y) dy

]=

∫Rnu[ϕ(y − ·)]ψ(y) dy.

The idea of proving this equality is to approximate the integral on on the left hand side byRiemann sums, use linearity of u to go under the sign of the sum and to pass to the limit.

However, the details of this approximation argument are not obvious. They are elemen-tary and tedious, but not obvious and so in most of the textbooks they are summarizedas ‘easy to see’. For the sake of completeness we decided to include details.

First of all observe that is suffices to prove (5.4) under the assumption that ϕ, ψ ∈ C∞0 .Indeed, suppose we know (5.4) for compactly supported functions, but now ϕ, ψ ∈ Sn.Let39 C∞0 3 ϕk → ϕ in Sn and C∞0 3 ψ` → ψ in Sn. Observe that fk(x) = u[τ−xϕk]→ f(x)pointwise. It follows from the proof of (5.3) that all the functions fk have a commonestimate independent of k

|fk(x)| ≤ C(1 + |x|)m

and hence fk → f in S ′n by the Dominated Convergence Theorem. It is also clear from

the definition of the convolution that u ∗ ϕk → u ∗ ϕ in S ′n, so

(u ∗ ϕ)[ψ`] = limk→∞

(u ∗ ϕk)[ψ`] = limk→∞

fk[ψ`] = f [ψ`] .

The second equality follows from assuming (5.4) for all ϕk, ψ` ∈ C∞0 . Now letting ` → ∞yields (5.4).

If the support of ψ is contained in a cube Q with integer edge-length we divide Q intocubes Qkii of edge-length 2−k and centes yki. Then the Riemann sums converge in thetopology of Sn (as a function of x) to the integral∑

i

ϕ(yki − x)ψ(yki)|Qki| →∫Rnϕ(y − x)ψ(y) dy

and hence

u[ ∫


]= lim

k→∞

∑i

u[ϕ(yki − ·)]ψ(yki)|Qki| =∫Rnu[ϕ(y − ·)]ψ(y) dy.

Boring (but elementary) details of the convergence of the Riemann sums in the topologyof Sn are moved to Appendix 5.7.

Note that formula (5.2) implies

Theorem 5.8. If u ∈ S ′n and ϕ ∈ Sn, then for any multiindex α we have

Dα(u ∗ ϕ)(x) = (u ∗ (Dαϕ))(x) .

39See Theorem 3.43(b).

58 PIOTR HAJ LASZ

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

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

Proof. In order to be able to use triangle inequality to prove the convergence we need toknow uniform estimates for the sequence uk. Convergence uk → u in S ′

n means pointwiseconvergence on a complete metric space Sn so it is natural to use the Banach-SteinhausTheorem to get uniform estimates for uk. We state it as a lemma.

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

supi∈I|fi(x)| <∞ for x ∈ X,

then there is an open set U ⊂ X and a constant M > 0 such that

supi∈I|fi(x)| ≤M for all x ∈ U .

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

supk|uk[ϕ]| ≤M for all ϕ ∈ B(ϕ0, r0).

Fix ε > 0. There is k1 such that |uk[ϕ0]| < ε for k ≥ k1. Since ϕk → 0 in Sn, then alsoε−1ϕk → 0 and hence

d(ϕ0 + ε−1ϕk, ϕ0) = d(ε−1ϕk, 0)→ 0 ,

so there is k2 such that for k ≥ k2, ϕ0 + ε−1ϕk ∈ B(ϕ0, r0). Hence for k ≥ maxk1, k2 wehave

|uk[ε−1ϕk]| ≤ |uk[ϕ0 + ε−1ϕk]|+ |uk[ϕ0]| ≤M + ε , so |uk[ϕk]| ≤ ε(M + ε).

Theorem 5.11. If uk → u in S ′n, then there is a constant C > 0 and a positive integer

m such that

supk|uk[ϕ]| ≤ C

∑|α|,|β|≤m


Proof. Let vk = uk − u. Clearly vk → 0 in S ′n. It suffices to prove that there is C > 0 and

a positive integer m such that

supk|vk[ϕ]| ≤ C

∑|α|,|β|≤m


Indeed, this estimate, the inequality

supk|uk[ϕ]| ≤ |u[ϕ]|+ sup

k|vk[ϕ]|

and Theorem 5.3 readily yield the result.


As in the proof of Theorem 5.3 it suffices to prove that there is a positive integer m suchthat supk |vk[ϕ]| ≤ 1 for

ϕ ∈ Nm :=ϕ ∈ Sn :

∑|α|,|β|≤m

pα,β(ϕ) ≤ 1

m

.

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

(5.5) |vki [ϕi]| > 1 for some ki.

Clearly ϕi → 0 in Sn. If the sequence ki is bounded, (5.5) contradicts continuity offunctionals vk. If ki is unbounded, then there is a subsequence kij → ∞ and again (5.5)contradicts Theorem 5.9.

Theorem 5.12. C∞0 (Rn) is dense in S ′n, i.e. if u ∈ S ′

n, then there is a sequence wk ∈C∞0 (Rn) such that

wk[ψ]→ u(ψ) for all ψ ∈ Sn.

Proof. Let u ∈ S ′n. If ϕ ∈ C∞0 (Rn),

∫Rn ϕ = 1, then for every ψ ∈ Sn, ϕε ∗ ψ → ψ in Sn

40

so(u ∗ ϕε)[ψ] = u[ϕε ∗ ψ]→ u[ψ] .

Since uε = u ∗ ϕε ∈ C∞ we obtain a sequence of smooth functions that converge to u inS ′n. If we take a cut-off function η ∈ C∞0 (Rn), 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 5.941 we obtain that wk(x) = η(x/k)(u∗ϕk−1) ∈ C∞0 convergesto u in S ′

n. Indeed,wk[ψ] = (u ∗ ϕk−1)[η(·/k)ψ(·)]→ u[ψ].

If ϕ, ψ ∈ Sn, then the integration by parts gives∫RnDαϕ(x)ψ(x) dx = (−1)|α|

∫Rnϕ(x)Dαψ(x) dx, Dαϕ[ψ] = ϕ[(−1)αψ].

For h ∈ Rn we have∫Rn

(τhϕ)(x)ψ(x) dx =

∫Rnϕ(x)(τ−hψ)(x) dx, τhϕ[ψ] = ϕ[τ−hψ].

Moreover ∫Rnϕ(x)ψ(x) dx =

∫Rnϕ(x)ψ(x) dx, ϕ[ψ] = ϕ[ψ],∫

Rnϕ(x)ψ(x) dx =

∫Rnϕ(x)ψ(x) dx, ϕ[ψ] = ϕ[ψ],∫

Rnϕ(x)ψ(x) dx =

∫Rnϕ(x)ψ(x) dx, ϕ[ψ] = ϕ[ψ].

40Why?41Instead of using Theorem 5.9 which is somewhat not constructive as it is based on the Banach-

Steinhaus theorem, we could prove convergence by a brute force estimates.

60 PIOTR HAJ LASZ

If η ∈ C∞ is slowly increasing and all derivatives of η are slowly increasing, i.e. everyderivative Dαη is bounded by a polynomial, then for ψ ∈ Sn, ηψ ∈ Sn and the mappingψ 7→ ηψ is continuous in Sn. In particular xαψ(x) ∈ Sn. Moreover

(ηϕ)[ψ] =

∫Rn

(η(x)ϕ(x)

)ψ(x) dx =

∫Rnϕ(x)

(η(x)ψ(x)

)dx = ϕ[ηψ].

This motivates the following definition.

Definition 5.13. For u ∈ S ′n we define

• The distributional partial derivative Dαu is a tempered distribution defined by theformula

Dαu[ψ] = u[(−1)|α|Dαψ].

• The translation τhu ∈ S ′n is defined by

(τhu)[ψ] = u[τ−hψ].

• The reflection u ∈ S ′n is defined by

u[ψ] = u[ψ].

• The Fourier transform F(u) = u ∈ S ′n and the inverse Fourier transform u ∈ S ′

n

are

u[ψ] = u[ψ] and u[ψ] = u[ψ].

• If η ∈ C∞ is slowly increasing and all derivatives of η are slowly increasing, thenwe define

(ηu)[ψ] = u[ηψ]. In particular (xαu)[ψ] = u[xαψ].

The formulas preceding the definition show that on the subclass Sn ⊂ S ′n the partial

derivative, the translation, the reflection, the Fourier transform and the multiplication bya function defined in the distributional sense coincide with those defined in the classicalway.

If f ∈ L1(Rn)+L2(Rn), in particular, if f ∈ Lp(Rn), 1 ≤ p ≤ 2, then the classical Fouriertransform coincides with the distributional one. That easily follows from Theorem 3.2(c)and Proposition 3.46.

The basic properties of the Fourier transform, distributional derivative and convolutionin S ′

n are collected in the next result whose easy proof is left to the reader.

Theorem 5.14. The Fourier transform in a homeomorphism of S ′n onto itself.42 Moreover

for u ∈ S ′n and ϕ ∈ Sn we have

(a) (u)∨ = u,(b) (u ∗ ϕ) = ϕu,(c) Dα(u ∗ ϕ) = Dαu ∗ ϕ = u ∗Dαϕ,(d) (Dαu) = (2πix)αu,(e) ((−2πix)αu) = Dαu.

42With respect to the weak convergence in S ′n.


Just for the illustration we will prove the property (d). For any ψ ∈ Sn we have

(Dαu)∧[ψ] = u[(−1)|α|Dαψ] = u[(−1)|α|((−2πix)αψ)∧]

= u[((2πix)αψ)∧] =((2πix)αu

)[ψ].

Note that in the case (c), u ∗ ϕ ∈ S ′n and Dα(u ∗ ϕ) is understood in the distributional

sense. On the other hand u ∗ ϕ ∈ C∞ and Theorem 5.8 shows that the distributionalderivative of u ∗ ϕ coincides with the classical one.

Example 5.15. The Dirac measure δa defines a tempered distribution so

δa[ψ] = δa[ψ] = ψ(a) =

∫Rnψ(x)e−2πix·a dx

and hence δa = e−2πix·a.

Example 5.16. Let µ be a measure on R defined by µ(E) = #(E ∩Z), where #A standsfor the cardinality of a set A. In other words µ is the counting measure on Z triviallyextended to R. Clearly, µ is a tempered measure so it defines a tempered distribution.Moreover µ = µ in S ′

1. Indeed, this is a simple consequence of the Poisson summationformula Theorem 4.1 that

µ[ϕ] = µ[ϕ] =∞∑

n=−∞

ϕ(n) =∞∑

n=−∞

ϕ(n) = µ[ϕ].

5.2. Tempered distributions as derivatives of functions. Using the notion of distri-butional derivative we can provide a new class of examples of a distributions in S ′

n. If fα,|α| ≤ m are slowly increasing functions and aα ∈ C for |α| ≤ m, then

(5.6) u =∑|α|≤m

aαDαfα ∈ S ′

n .

where the derivative Dαfα is understood in the distributional sense.

The aim of this section is to prove a surprising result that every distribution in S ′n can

be represented in the form (5.6), see Theorem 5.21.

Proposition 5.17. Let N be a positive integer, then the operators43

(I −∆)N : Sn → Sn and (I −∆)N : S ′n → S ′

n

are homeomorphisms. The inverse operator in both cases is given by the operator

(I −∆)−N := F−1((1 + 4π2|ξ|2)−NF

).

Remark 5.18. That means for every v ∈ Sn (v ∈ S ′n) the equation (I −∆)Nu = v has

a unique solution u ∈ Sn (u ∈ S ′n) given by

u = (I −∆)Nv = F−1((1 + 4π2|ξ|2)−NF(v)

).

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

62 PIOTR HAJ LASZ

Proof. Suppose that u, v ∈ Sn satisfy (I −∆)Nu = v. Taking the Fourier transform yields

(1 + 4π2|ξ|2)NF(u) = F(v)

and henceF(u) = (1 + 4π2|ξ|2)−NF(v).

Since F(u) ∈ Sn, the right hand side is also in Sn so we can take the inverse Fouriertransform and we obtain

(5.7) u = F−1((1 + 4π2|ξ|2)−NF(v)

).

Thus if a solution u ∈ Sn exists, it is unique and given by (5.7). To see that the formulaindeed, defines a solution u ∈ Sn observe that the function (1 + 4π2|ξ|2)−N and all itsderivatives are slowly increasing so if f ∈ Sn, then (1 + 4π2|ξ|2)−Nf ∈ Sn. Hence for anyv ∈ Sn, the right hand side of (5.7) defines a function in Sn and backward computationsshows that (I −∆)Nu = v.

Similarly, if u, v ∈ S ′n, then (I − ∆)Nu = v if and only if the Fourier transforms are

equal i.e.,(1 + 4π2|ξ|2)NF(u) = F(v)

which is equivalent to

F(u) = (1 + 4π2|ξ|2)−NF(v), u = F−1((1 + 4π2|ξ|2)−NF(v)

).

Note that we were allowed to multiply both sides by (1 + 4π2|ξ|2)−N because this functionand all its derivatives are slowly increasing.

The operator (I−∆)N maps functions of class Ck to functions of class Ck−2N so it lowersregularity of functions. That means the inverse operator (I − ∆)−N increases regularity.Hence we should expect that it also increases regularity of distributions. Since every tem-pered distribution has finite order in a sense that it can be estimated by a finite sum as inTheorem 5.3, one could expect that for u ∈ S ′

n and sufficiently large N , the distribution(I −∆)−Nu is actually a regular function. As we will see this intuition is correct.

Remark 5.19. The operator (I−∆)−N is called a Bessel potential and one can prove thatit can be represented as an integral operator. In Section ?? we will study Bessel potentialsin detail. We will find an explicit integral formula for the Bessel potentials and we willshow how use them to characterize Sobolev spaces.

Definition 5.20. By Ck,1loc we will denote the class of k times continuously differentiable

function whose derivatives of order k are locally Lipschitz continuous. In particular C0,1loc

denotes the class of locally Lipschitz continuous functions.

Theorem 5.21. Suppose u ∈ S ′n satisfies the estimate44

|u(ϕ)| ≤ C∑

|α|,|β|≤m

pα,β(ϕ) .

If k is a nonnegative integer and N > (n+m+ k)/2, then the tempered distribution

v = (I −∆)−Nu

has the following properties:

44See Theorem 5.3.


(a) If k = 0, then v is a slowly increasing function.

(b) If k ≥ 1, then v is a slowly increasing function of the class Ck−1,1loc . Also all deriva-

tives of v or order |α| ≤ k are slowly increasing.

Proof. We will need

Lemma 5.22. If P (x) is a polynomial in Rn of degree p and N > (n + p)/2, then allderivatives of

f(x) =P (x)

(1 + |x|2)N

belong to L1(Rn).

Proof. Since 2N − p > n, f ∈ L1(Rn). We have

∂f

∂xi=

Q(x)

(1 + |x|2)2N, degQ = 2N + p− 1.

The function on the right hand side is of the same form as f . Since

2N >n+ (2N + p− 1)

2

we conclude that ∂f/∂xi ∈ L1. The integrability of higher order derivatives follows byinduction. 2

We will prove that the distributional derivatives of v of orders |γ| ≤ k are slowly in-creasing functions. We have

(−1)|γ|Dγv[ψ] = u[F((1 + 4π2|ξ|2)−NF−1(Dγψ)

)].

Hence

|Dγv[ψ]| ≤ C∑

|α|,|β|≤m

supx∈Rn

∣∣∣xαDβ(F((1 + 4π2|ξ|2)−NF−1(Dγψ)

))∣∣∣= C

∑|α|,|β|≤m

Cα,β,γ supx∈Rn

∣∣∣∣F (Dα

(ξβ+γ

(1 + 4π2|ξ|2)NF−1(ψ)

))∣∣∣∣≤ C ′

∑|α|,|β|≤m

∥∥∥∥Dα

(xβ+γ

(1 + 4π2|x|2)NF−1(ψ)

)∥∥∥∥1

.

Note that

Dα

(xβ+γ

(1 + 4π2|x|2)NF−1(ψ)

)=

∑αi+βi=α

α!

α1!βi!Dαi

(xβ+γ

(1 + 4π2|x|2)N

)Dβi(F−1(ψ)) .

Since deg xβ+γ ≤ m+ k and N > (n+m+ k)/2, Lemma 5.22 gives

Dαi

(xβ+γ

(1 + 4π2|x|2)N

)∈ L1(Rn) ,

64 PIOTR HAJ LASZ

so

|Dγv[ψ]| ≤ C∑i

‖F−1(xβiψ)‖∞

≤ C∑i

‖xβiψ‖1

≤ C ′‖(1 + |x|2)m/2ψ(x)‖1 .

This proves that for |γ| ≤ k the functional

ψ 7→ Dγv[ψ]

is bounded on L1((1 + |x|2)m/2 dx). Thus there are functions gγ ∈ L∞ such that

Dγv[ψ] =

∫Rnψ(x)gγ(x)(1 + |x|2)m/2 dx ,

i.e.Dγv = gγ(x)(1 + |x|2)m/2

in the distributional sense. In particular, if γ = 0, v(x) = g0(x)(1 + |x|2)m/2 is slowlyincreasing which proves (a). The part (b) follows from the following result whose proof ispostponed to Section ??, see Theorem ??

Proposition 5.23. If a function u is slowly increasing and its distributional derivativesof order less than or equal k, k ≥ 1, are slowly increasing functions, then u ∈ Ck−1,1

loc .

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

Lemma 5.24. Let Ω ⊂ Rn be open and u ∈ S ′n. If u[ϕ] = 0 for all ϕ ∈ C∞0 (Ω), then

u[ϕ] = 0 for all ϕ ∈ Sn such that suppϕ ⊂ Ω.

Proof. Let ϕ ∈ Sn, suppϕ ⊂ Ω and let η be a cut-off function, i.e. η ∈ C∞0 (Rn), 0 ≤ η ≤ 1,η(x) = 1 for |x| ≤ 1 and η(x) = 0 for |x| ≥ 2. One can easily prove that η(x/R)ϕ(x)→ ϕ(x)in Sn as R→∞. Since η(x/R)ϕ(x) ∈ C∞0 (Ω), the lemma follows.

Definition 5.25. Let u ∈ S ′n. The support of u (suppu) is the intersection of all closed

sets E ⊂ Rn such thatϕ ∈ C∞0 (Rn \ E) ⇒ u[ϕ] = 0.

Proposition 5.26. If u ∈ S ′n and ϕ ∈ C∞0 (Rn \ suppu), then u[ϕ] = 0.

Proof. Indeed, complement of suppu is the union of all open sets Ui such that if ϕ ∈C∞0 (Ui), then u[ϕ] = 0, and we need to prove that if ϕ ∈ C∞0 (

⋃i Ui), then still u[ϕ] = 0.

Since the support of ϕ is compact we can select a finite subfamily of open sets Ui coveringthe support and the result follows from a simple argument involving partition of unity.

Thus the support of u is the smallest closed set such that the distribution vanishes onC∞0 functions supported outside that set.

The lemma shows that we can replace ϕ ∈ C∞0 (Rn\E) in the above definition by ϕ ∈ Sn

with suppϕ ⊂ Rn \ E.


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

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

(5.8) f(x) =∑α

aα(x− x0)α, aα ∈ C,

i.e. if in a neighborhood of any point f equals to its Taylor series.

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

f(z) =∑α

aα(z − z0)α .

Rn has a natural embedding into Cn, just like R into C.

Rn 3 x = (x1, . . . , xn) 7→ (x1 + i · 0, . . . , xn + i · 0) = x+ i · 0 ∈ Cn .

It is easy to see that an R-analytic function f in Ω ⊂ Rn extends to a C-analytic functionf in an open set Ω ⊂ Cn, Ω ⊂ Ω. Namely, if f satisfies (5.8), we set

f(z) =∑α

aα(z − z0)α, z0 = x0 + i · 0.

On the other hand, if f is C-analytic in Ω ⊂ Cn, then the restriction f of f to Ω = Ω∩Rn

is R-analytic.

For example for any ξ ∈ Rn, f(x) = ex·ξ is R-analytic and f(z) = ez·ξ is its C-analyticextension.

Definition 5.28. We say that a continuous function f : Ω → C defined in an open setΩ ⊂ Cn is holomorphic if

∂f

∂zi= 0 for i = 1, 2, . . . , n.

It is easy to see that C-analytic function are holomorphic, but the converse implicationis also true.

Lemma 5.29 (Cauchy). If f is holomorphic in45

Dn(w, r) = D1(w1, r1)× . . .×D1(wn, rn) ⊂ Cn

and continuous in the closure Dn(x, r), then

(5.9) f(z) =1

(2πi)n

∫∂D1(w1,r1)

. . .

∫∂D1(wn,rn)

f(ξ) dξ1 . . . dξn(ξ1 − z1) . . . (ξn − zn)

for all z ∈ Dn(w, r).

45Product of one dimensional discs.

66 PIOTR HAJ LASZ

Proof. The function f is holomorphic in each variable separately, so (5.9) follows fromone dimensional Cauchy formulas and the Fubini theorem. 2

Just like in the case of holomorphic functions of one variable one can prove that theintegral on the right hand side of (5.9) can be expanded as a power series and hencedefines a C-analytic function. We proved

Theorem 5.30. A function f is holomorphic in Ω ⊂ Cn if and only if it is C-analytic.

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

Theorem 5.31. If u ∈ S ′n has compact support, then u is a slowly increasing C∞ function

and all derivatives of u are slowly increasing. Moreover u is R-analytic on Rn and has aholomorphic extension to Cn.

Proof. Let η ∈ C∞0 (Rn) be such that η(x) = 1 in a neighborhood of suppu. Then u = ηuin S ′

n and hence for ψ ∈ Sn we have

u[ψ] = (ηu) [ψ] = u

[∫Rnη(·)ψ(x)e−2πix·(·) dx

]=

∫Rnu[η(·)e−2πix·(·)]ψ(x) dx .

We could pass with u under the sign of the integral, because of an argument with approx-imation of the integral by Riemann sums.46

Note that the function

F (x1, . . . , xn) = u[η(·)e−2πix·(·)]

is C∞ smooth and

DαF (x1, . . . , xn) = u[(−2πi(·))αη(·)e−2πix·(·)] .

Indeed, we could differentiate under the sign of u because the corresponding differencequotients converge in the topology of Sn. It also easily follows from Theorem 5.3 that Fand all its derivatives are slowly increasing. Thus we may identify u with F , so u ∈ C∞.

Moreover F has a holomorphic extension to Cn by the formula

F (z1, . . . , zn) = u[η(·)e−2πiz·(·)]

so in particular F (x1, . . . , xn) is R-analytic. 2

Remark 5.32. Note that if u ∈ S ′n has compact support, then we can reasonably define

u[e−2πix·(·)] by the formula

u[e−2πix·(·)] := u

[η(·)e−2πix·(·)] ,

where η ∈ C∞0 (Rn) is such that η = 1 in a neighborhood of suppu. Note that this con-struction does not depend on the choice of η. In the same way we can define u[f ] for anyf ∈ C∞(Rn)

46Compare with the proof of Theorem 5.7.


Theorem 5.33. If u ∈ S ′n and suppu = x0, then there is an integer m ≥ 0 and complex

numbers aα, |α| ≤ m such that

u =∑|α|≤m

aαDαδx0 .

Proof. Without loss of generality we may assume that x0 = 0. According to Theorem 5.3the distribution u satisfies the estimate

|u[ϕ]| ≤ C∑

|α|,|β|≤m

pα,β(ϕ) .

First we will prove that for ϕ ∈ Sn we have

(5.10) Dαϕ[0] = 0 for |α| ≤ m =⇒ u[ϕ] = 0 .

Indeed, it follows from Taylor’s formula that

ϕ(x) = O(|x|m+1) as x→ 0

and hence also

(5.11) Dβϕ(x) = O(|x|m+1−|β|) as x→ 0 for all |β| ≤ m.

Let η ∈ C∞0 (Rn) be a cut-off function, i.e. 0 ≤ η ≤ 1, η(x) = 1 for |x| ≤ 1, η(x) = 0 for|x| ≥ 2 and define ηε(x) = η(x/ε). The estimate (5.11) easily implies that

pα,β(ηεϕ)→ 0 as ε→ 0

for all |α|, |β| ≤ m.47 Note that ϕ− ηεϕ = 0 in a neighborhood of 0, so u[ϕ− ηεϕ] = 0 andhence

|u[ϕ]| ≤ |u[ϕ− ηεϕ]|+ |u[ηεϕ]| ≤ 0 +∑

|α|,|β|≤m

pα,β(ηεϕ)→ 0 as ε→ 0.

This completes the proof of (5.10).

Let now ψ ∈ Sn be arbitrary and let

h(x) = ψ(x)−∑|α|≤m

Dαψ(0)

α!xα .

Clearly

(5.12) Dαh(0) = 0 for |α| ≤ m.

We have

ψ(x) = η(x)

∑|α|≤m

Dαψ(0)

α!xα

+ η(x)h(x) + (1− η(x))ψ(x) .

Since (1−η)ψ vanishes in a neighborhood of 0 we have u[(1−η)ψ] = 0. The equality (5.12)implies that ϕ = ηh ∈ C∞0 (Rn) satisfies the assumptions of (5.10), so u[ηh] = 0. Hence

u[ψ] = u[η(x)

( ∑|α|≤m

Dαψ(0)

α!xα)]

47Check it!

68 PIOTR HAJ LASZ

=∑|α|≤m

(−1)|α|u

[η(x)xα

α!

]︸︷︷︸

aα

(−1)|α|Dαψ(0)︸︷︷︸Dαδ0[ψ]

.

The proof is complete. 2

Corollary 5.34. Let u ∈ S ′n. If supp u = ξ0, then u = P (x)e2πix·ξ0 for some polynomial

P (x) in Rn. In particular if supp u = 0, then u is a polynomial.


Corollary 5.35. If u ∈ S ′n satisfies ∆u = 0, then u is a polynomial.

Proof. We have−4π2|ξ|2u = (∆u) = 0 in S ′

n.

This implies that supp u = 0, so u is a polynomial by Corollary 5.34. 2

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

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

is homogeneous of degree a ∈ R if

f(tx) = taf(x) for almost all (x, t) ∈ Rn × (0,∞).

If f ∈ C(Rn \ 0) is homogeneous of degree a > −n, then it defines a tempereddistribution. Indeed,

|f(x)| =∣∣∣∣ |x|af ( x

|x|

)∣∣∣∣ ≤ C|x|a

and |x|a is integrable in a neighborhood of the origin (because a > −n).

If f is a measurable homogeneous function of degree a we can assume (by changing fon a set of measure zero) that f is continuous on almost all open rays going out of 0. Thenon the remaining rays (which form a set of measure zero) we can make f homogeneous.This is to say that we can find a representative of f (in class of functions that are equala.e.) such that

f(tx) = taf(x) for all x 6= 0 and t > 0.

If a measurable function that is not a.e. equal to zero is homogeneous of degree a ≤ −n,then the integration in the spherical coordinate system shows that it is not integrable inany neighborhood of 0 and hence it cannot define a tempered distribution.

Lemma 5.37. Suppose that f ∈ L1loc(Rn) defines a tempered distribution. Then the func-

tion f is homogeneous of degree a ∈ R if and only if 48

(5.13) f [ϕt] = taf [ϕ] for all ϕ ∈ Sn and all t > 0.

Remark 5.38. If f is not equal to zero a.e., then necessarily a > −n.

48As always ϕt(x) = t−nϕ(x/t).


Proof. For ϕ ∈ Sn we have

f [ϕt] =

∫Rnf(x)ϕt(x) dx =

∫Rnf(tx)ϕ(x) dx and taf [ϕ] =

∫Rntaf(x)ϕ(x) dx.

Hence condition (5.13) is equivalent to the equality

(5.14)

∫Rnf(tx)ϕ(x) dx =

∫Rntaf(x)ϕ(x) dx for all ϕ ∈ Sn and all t > 0

which, in turn, is equivalent to the fact that for all t > 0, equality f(tx) = taf(x) holdsa.e.

The above result motivates the following definition.

Definition 5.39. We say that a distribution u ∈ S ′n is homogeneous of degree a ∈ R if 49

u[ϕt] = tau[ϕ] for all ϕ ∈ Sn and all t > 0.

Proposition 5.40. u ∈ S ′n is homogeneous of degree a ∈ R if and only if u ∈ S ′

n ishomogeneous of degree −n− a.

Proof. Let ϕ ∈ Sn. Then

ϕt(ξ) = ϕ(tξ) = t−n(t−1)−nϕ(ξ/t−1) = t−n(ϕ)t−1(ξ)

and hence

u[ϕt] = u[ϕt] = t−nu[(ϕ)t−1 ] = t−n−au[ϕ] = t−n−au[ϕ].

If u ∈ S ′n is homogeneous of degree a, the above equality shows that u is homogeneous of

degree −n− a. Now if u is homogeneous of degree −n− a, then its Fourier transform i.e.,u is homogeneous of degree −n− (−n− a) = a and hence also u is homogeneous of degreea.

Example 5.41. u(x) = |x|a, a ≥ 0, is homogeneous of degree a and it defines a tempereddistribution. However, u is homogeneous of degree −n − a ≤ −n and hence it cannot berepresented as a function.

The notion of a homogeneous distribution will play a significant role in the proof of thenext result.

Theorem 5.42. For 0 < a < n,

F(|x|−a

)(ξ) =

πa−n2 Γ(n−a

2

)Γ(a2

) |ξ|a−n .

Remark 5.43. |x|−a is homogeneous of degree −a > −n and |ξ|a−n is homogeneous ofdegree a − n > −n so both |x|−a and |ξ|a−n are tempered distributions represented asfunctions.

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

70 PIOTR HAJ LASZ

Proof. First we will prove the theorem under the assumption that n/2 < a < n. In thatcase

|x|−a = |x|−aχ|x|≤1 + |x|−aχ|x|>1 ∈ L1 + L2 .

Thus the Fourier transform of |x|−a is a function in C0 + L2.

The Fourier transform commutes with orthogonal transformations

f ρ = f ρ, for ρ ∈ O(n) and f ∈ L1 + L2.

If f is radially symmetric i.e., f = f ρ for all ρ ∈ O(n), then f ρ = f ρ = f for all

ρ ∈ O(n) so f is radially symmetric too. Thus the Fourier transform of |x|−a ∈ L1 + L2 isa radially symmetric function in C0 + L2, homogeneous of degree a− n as a distribution.Hence Lemma 5.37 yields

F(|x|−a

)(ξ) = ca,n|ξ|a−n

and it remains to compute the coefficient ca,n. Employing the fact that ϕ(x) = e−π|x|2

is afixed point of the Fourier transform we have

ca,n|x|a−n[ϕ] = F(|x|−a)[ϕ] = |x|−a[ϕ] = |x|−a[ϕ],

i.e.,

(5.15) ca,n

∫Rne−π|x|

2 |x|a−n dx =

∫Rne−π|x|

2|x|−a dx.

The integrals in this identity are easy to compute.

Lemma 5.44. For γ > −n we have∫Rne−π|x|

2|x|γ dx =Γ(n+γ

2

)πγ2 Γ(n2

) .Proof. For γ > −n we have50∫

Rne−π|x|

2|x|γ dx = nωn

∫ ∞0

e−πs2

sγ+n−1 ds

t=πs2=

nωn

2πn+γ2

∫ ∞0

e−ttn+γ2−1 dt

=nωn

2πn+γ2

Γ

(n+ γ

2

)=

Γ(n+γ

2

)πγ2 Γ(n2

) .

Applying the lemma to both sides of (5.15) yields

ca,nΓ(a2

)πa−n2 Γ

(n2

) =Γ(n−a

2

)π−

a2 Γ(n2

) so ca,n = πa−n2

Γ(n−a

2

)Γ(a2

) .

This completes the proof when n/2 < a < n.

50Recall that the volume of the unit sphere in Rn equals nωn so the first equality is just integration in

the spherical coordinates. In the last equality we use ωn = 2πn/2

nΓ(n/2) , see (3.9).


Suppose now that 0 < a < n/2. Since n/2 < n− a < n we have

F(|x|−(n−a)

)(ξ) =

πn2−aΓ

(a2

)Γ(n−a

2

) |ξ|−aand applying the Fourier transform to this equality yields

| − x|a−n =πn2−aΓ

(a2

)Γ(n−a

2

) F (|ξ|−a) (x).

Replacing x by ξ and ξ by x yields the result when 0 < a < n/2.

We are left with the case of a = n/2. Take a sequence n/2 < ak < n, ak → a = n/2.Since the result is true for ak we have∫

Rn|x|−akϕ(x) dx =

πak−n2 Γ(n−ak

2

)Γ(ak2

) ∫Rn|x|ak−nϕ(x) dx.

Passing to the limit ak → a = n/2 yields the result for a = n/2.

Remark 5.45. Clearly |ξ|ak−n → |ξ|a−n and |x|−ak → |x|−a in S ′n. The last convergence

and the continuity of the Fourier transform yields that F(|x|−ak)→ F(|x|−a) in S ′n. Hence

the result for a = n/2 follows by passing to the limit in the equality

F(|x|−ak

)(ξ) =

πak−n2 Γ(n−ak

2

)Γ(ak2

) |ξ|ak−n in S ′n.

Remark 5.46. In the case a = n/2 we obtain a particularly simple formula

F(|x|−n2 ) = |x|−

n2 , i.e.,

∫Rn|x|−

n2 ϕ(x) dx =

∫Rn|x|−

n2ϕ(x) dx

for all ϕ ∈ Sn. This formula can be regarded as a version of the Poisson summationformula.51

5.5. The principal value.

Proposition 5.47. If u ∈ S ′n is homogeneous of degree a, then Dαu is homogeneous of

degree a− |α|.

We leave the proof as an (easy) exercise.

In particular if f ∈ C∞(Rn \ 0) is homogeneous of degree 1 − n, then it definesa distribution, but its first order distributional partial derivatives are homogeneous ofdegree −n and hence they cannot be represented as functions. It is however, still possibleto represent these distributional derivatives as improper integrals, the so called principalvalue of the integral.

Definition 5.48. If for any ε > 0, a function f is integrable in Rn \B(0, ε), then we definethe principal value of the integral as

p.v.

∫Rnf(x) dx = lim

ε→0

∫|x|≥ε

f(x) dx

provided the limit exists.

51cf. Theorem 4.1.

72 PIOTR HAJ LASZ

If a function f is slowly increasing, but has singularity at 0, then we can try to define adistribution p.v. f by

(p.v. f)[ϕ] = p.v.

∫Rnf(x)ϕ(x) dx = lim

ε→0

∫|x|≥ε

f(x)ϕ(x) dx for ϕ ∈ Sn.

In many interesting cases p.v. f defines a tempered distribution, but not always.

Theorem 5.49. Let

KΩ(x) =Ω(x/|x|)|x|n

, x 6= 0,

where Ω ∈ L1(Sn−1) satisfies ∫Sn−1

Ω(θ) dσ(θ) = 0.

Then

p.v. KΩ ∈ S ′n

and

|(p.v. KΩ)[φ]| ≤ C(n)‖Ω‖L1(Sn−1)

(‖∇ϕ‖∞ + ‖|x|ϕ(x)‖∞

)for all ϕ ∈ Sn.

Remark 5.50. Distributions p.v. KΩ ∈ S ′n described in the above result will be called

distributions of the p.v. type or simply p.v. distributions.

Remark 5.51. Note that the function KΩ is homogeneous of degree −n so the functionKΩ does not define a tempered distribution.

Remark 5.52. It is not difficult to verify that if∫Sn−1 Ω(θ) dσ(θ) 6= 0, then p.v. KΩ

does not define a tempered distribution because p.v. KΩ[ϕ] will diverge for some functionsϕ ∈ Sn.

Proof. For ϕ ∈ Sn we have∣∣(p.v. KΩ)[ϕ]∣∣ =

∣∣∣∣limε→0

∫ε≤|x|≤1

Ω(x/|x|)|x|n

(ϕ(x)− ϕ(0)) +

∫|x|≥1

Ω(x/|x|)|x|n

ϕ(x) dx

∣∣∣∣≤ ‖∇ϕ‖∞

∫|x|≤1

|Ω(x/|x|)||x|n−1

dx+ ‖ϕ(y)|y|‖∞∫|x|≥1

|Ω(x/|x|)||x|n+1

dx

and the result follows upon integration in spherical coordinates. We could subtract ϕ(0)

in the first integral because∫ε≤|x|≤1

Ω(x/|x|)|x|n dx = 0. In the last step used the following

estimates

|ϕ(x)− ϕ(0)| =∣∣∣∣∫ 1

0

d

dtϕ(tx) dt

∣∣∣∣ =

∣∣∣∣∫ 1

0

∇ϕ(tx) · x dt∣∣∣∣ ≤ |x| ‖∇ϕ‖∞

and

|ϕ(x)| = |ϕ(x)| |x||x|

≤ ‖ϕ(y)|y|‖∞|x|

.


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

Proposition 5.53. For n ≥ 2, the distributional partial derivatives of |x|1−n ∈ S ′n satisfy

∂

∂xj|x|1−n = (1− n) p.v.

(xj|x|n+1

)∈ S ′

n,

i.e.,

(5.16)

(∂

∂xj|x|1−n

)[ϕ] = (1− n) lim

ε→0

∫|x|≥ε

xj|x|n+1

ϕ(x) dx. for ϕ ∈ Sn.

Before we prove the proposition let us recall the integration by parts formula for functionsdefined in a domain in Rn. If Ω ⊂ Rn is a bounded domain with piecewise C1 boundaryand f, g ∈ C1(Ω), then

(5.17)

∫Ω

(∇f(x)g(x) + f(x)∇g(x)) dx =

∫∂Ω

fg~ν dσ ,

where ~ν = (ν1, . . . , νn) is the unit outer normal vector to ∂Ω. Comparing jth componentson both sides of (5.17) we have

(5.18)

∫Ω

∂f

∂xj(x) g(x) dx = −

∫Ω

f(x)∂g

∂xj(x) dx+

∫∂Ω

fg νj dσ .

Proof of Proposition 5.53. Let A(ε, R) = x : ε ≤ |x| ≤ R. We have(∂

∂xj|x|1−n

)[ϕ] = −

∫Rn

∂ϕ

∂xj|x|1−n dx

= limε→0

(limR→∞

−∫ε≤|x|≤R

∂ϕ

∂xj|x|1−n dx

)= lim

ε→0

(limR→∞

(∫ε≤|x|≤R

ϕ(x)∂

∂xj|x|1−n︸︷︷︸

(1−n)xj/|x|n+1

dx−∫∂A(ε,R)

ϕ(x)|x|1−n νj dσ))

= limε→0

((1− n)

∫|x|≥ε

ϕ(x)xj|x|n+1

−∫|x|=ε

ϕ(x)|x|1−n νj dσ)

Indeed, we could pass to the limit as R→∞, because the part of the integral over ∂A(ε, R)corresponding to |x| = R converges to zero. Since the integral of the function |x|1−n νjover the sphere |x| = ε equals zero we have∣∣∣∣∫

|x|=εϕ(x)|x|1−n νj dσ

∣∣∣∣ =

∣∣∣∣∫|x|=ε

(ϕ(x)− ϕ(0))|x|1−n νj dσ∣∣∣∣ ≤ Cε→ 0 as ε→ 0,

because ∣∣(ϕ(x)− ϕ(0))|x|1−n νj

∣∣ ≤ Cε · ε1−n = ε2−n

and the surface area of the sphere |x| = ε equals Cεn−1.

74 PIOTR HAJ LASZ

Remark 5.54. We proved that p.v. (xj/|x|n+1) defines a tempered distribution somewhatindirectly by showing that for ϕ ∈ Sn equality (5.16) is satisfied. Since ∂|x|1−n/∂xj ∈ S ′

n,equality (5.16) implies that p.v. (xj/|x|n+1) ∈ S ′

n.

Remark 5.55. When n = 1 the counterpart of Theorem 5.53 is that

d

dx(log |x|) = p.v.

x

|x|2= p.v.

1

xin S ′

n.


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

Theorem 5.56. Suppose that K ∈ C1(Rn \ 0) is such that both K and |∇K| havepolynomial growth52 for |x| ≥ 1 and there are constants C, α > 0 such that

|K(x)| ≤ C

|x|n−1−α for 0 < |x| < 1,

|∇K(x)| ≤ C

|x|n−αfor 0 < |x| < 1.

Then K ∈ S ′n and the distributional partial derivatives ∂K/∂xj, 1 ≤ j ≤ n, coincide with

pointwise derivatives, i.e. for ϕ ∈ Sn

∂K

∂xj[ϕ] := −

∫RnK(x)

∂ϕ

∂xj(x) dx =

∫Rn

∂K

∂xj(x)ϕ(x) dx .

Proof. Let A(ε) = x : ε ≤ |x| ≤ ε−1. From (5.18) we have

∂K

∂xj[ϕ] = lim

ε→0−∫ε≤|x|≤ε−1

K(x)∂ϕ

∂xj(x) dx

= limε→0

(∫ε≤|x|≤ε−1

∂K

∂xj(x)ϕ(x) dx−

∫∂A(ε)

K(x)ϕ(x) νj dσ(x)

).

Since

limε→0

∫ε≤|x|≤ε−1

∂K

∂xj(x)ϕ(x) dx =

∫Rn

∂K

∂xj(x)ϕ(x) dx

it remains to show that

limε→0

∫∂A(ε)

K(x)ϕ(x) νj dσ = 0 .

The integral on the outer sphere |x| = ε−1 converges to 0 since K has polynomial growthand ϕ rapidly converges to 0 as |x| → ∞ and on the inner sphere |x| = ε we have∣∣∣∣∫

|x|=εK(x)ϕ(x) νj dσ

∣∣∣∣ ≤ C

εn−1−α εn−1 = Cεα → 0 as ε→ 0.


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


An interesting problem is the case α = 0, i.e. when K and ∇K satisfy the estimates

(5.19) |K(x)| ≤ C

|x|n−1, |∇K(x)| ≤ C

|x|nfor x 6= 0.

One such situation was described in Proposition 5.53.

Here we make an additional assumption about K. We assume that K ∈ C1(Rn \ 0) ishomogenelus of degree 1− n, i.e.

K(x) =K(x/|x|)|x|n−1

for x 6= 0.

Since K is bounded in |x| ≥ 1 and integrable in |x| ≤ 1 we have K ∈ S ′n and there

is no need to interpret K through the principal value of the integral. The first estimate at(5.19) is satisfied. To see that the second estimate if satisfied too we observe that ∇K ishomogeneous of degree −n. Indeed, for 1 ≤ j ≤ n and t > 0

∂K

∂xj(tx)t =

∂

∂xj(K(tx)) = t1−n

∂

∂xjK(x)

and hence

(∇K)(tx) = t−n∇K(x) .

Thus

∇K(x) =(∇K)(x/|x|)|x|n

, x 6= 0

from which the second estimate at (5.19) follows.

Observe that ∇K(x) is not integrable at any neighborhood of 0, but we may try to con-sider the principal value of ∇K(x), i.e. the principal value of each of the partial derivatives∂K/∂xj.

Theorem 5.57. Suppose that K ∈ C1(Rn \ 0) is homogeneous of degree 1 − n. Then∇K(x) is homogeneous of degree −n. Moreover

(5.20)

∫Sn−1

∇K(θ) dσ(θ) = 0

and

p.v.∇K(x) ∈ S ′n

is a well defined tempered distribution, i.e. for each 1 ≤ j ≤ n

p.v.∂K

∂xj(x) ∈ S ′

n .

Finally the distributional gradient ∇K satisfies

(5.21) ∇K︸︷︷︸dist.

= cδ0 + p.v. ∇K(x)︸︷︷︸pointwise

,

where

c =

∫Sn−1

K(x)x

|x|dσ(x) .

76 PIOTR HAJ LASZ

In other words for ϕ ∈ Sn and 1 ≤ j ≤ n we have

∂K

∂xj[ϕ] := −

∫RnK(x)

∂ϕ

∂xj(x) dx = cjϕ(0) + lim

ε→0

∫|x|≥ε

∂K

∂xj(x)ϕ(x) dx ,

where

cj =

∫Sn−1

K(x)xj|x|

dσ(x) .

Remark 5.58. Note that Proposition 5.53 follows from this result: since K(x) = |x|1−n,the constant c =

∫Sn−1 |x|−nx dσ(x) = 0.

Remark 5.59. The fact that p.v.∇K ∈ S ′n is a straightforward consequence of (5.20) and

Theorem 5.49. However, our argument will be direct without referring to Theorem 5.49.

Proof. We already checked that ∇K(x) is homogeneous of degree −n. For r > 1 letA(1, r) = x : 1 ≤ |x| ≤ r. From the integration by parts formula (5.17) we have∫

1≤|x|≤r∇K(x) dx =

∫∂A(1,r)

K(x)~ν(x) dσ(x)

= −∫|x|=1

K(x)x

|x|dσ(x) +

∫|x|=r

K(x)x

|x|dσ(x) = 0 .

Indeed, the last two integrals are equal by a simple change of variables and homogeneity ofK. Thus the integral on the left hand side equals 0 independently of r. Hence its derivativewith respect to r is also equal zero.

0 =d

dr

∣∣∣r=1+

∫1≤|x|≤r

∇K(x) dx =

∫|x|=1

∇K(θ) dσ(θ) .

This proves (5.20). As we will see (5.20) implies that p.v.∇K(x) ∈ S ′n is a well defined

tempered distribution as well as that the distributional gradient ∇K satisfies (5.21). Letϕ ∈ Sn. We have

∇K[ϕ] := −∫RnK(x)∇ϕ(x) dx

= limε→0

limR→∞

−∫ε≤|x|≤R

K(x)∇ϕ(x) dx

= limε→0

limR→∞

(∫ε≤|x|≤R

∇K(x)ϕ(x) dx−∫∂A(ε,R)

K(x)ϕ(x)~ν(x) dσ(x)

)= lim

ε→0

(∫|x|≥ε∇K(x)ϕ(x) dx+

∫|x|=ε

K(x)ϕ(x)x

|x|dσ(x)

).

It remains to prove that

limε→0

∫|x|=ε

K(x)ϕ(x)x

|x|dσ(x) = ϕ(0)

∫|x|=1

K(x)x

|x|dσ(x) .

Let

c =

∫|x|=1

K(x)x

|x|dσ(x) =

∫|x|=ε

K(x)x

|x|dσ(x) .


The last equality follows from a simple change of variables and homogeneity of K. We have∫|x|=ε

K(x)ϕ(x)x

|x|dσ(x)(5.22)

= cϕ(0) +

∫|x|=ε

K(x)(ϕ(x)− ϕ(0)

) x

|x|dσ(x)

→ cϕ(0)

as ε→ 0. Indeed, for |x| = ε∣∣∣∣K(x)(ϕ(x)− ϕ(0)

) x

|x|

∣∣∣∣ ≤ Cε1−nε = Cε2−n .

Since the surface area of the sphere |x| = ε is nωnεn−1, the integral on the right hand

side of (5.22) converges to 0 as ε→ 0.

5.6. Fundamental solution of the Laplace equation. A straightforward applicationof Theorem 5.57 gives a well known formula for the fundamental solution to the Laplaceequation.

Theorem 5.60. For n ≥ 2 we have53

∆Φ = δ0 ,

where

Φ(x) =

1

2πlog |x| if n = 2,

− 1n(n−2)ωn

1|x|n−2 if n ≥ 3.

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

(5.23) ∇Φ(x) =1

nωn

x

|x|nin S ′

n.

Note that the function Φ(x) is harmonic in Rn \ 0 and hence

0 = ∆Φ(x) = div∇Φ(x) =1

nωn

n∑j=1

∂

∂xj

xj|x|n

for x 6= 0.

Now Theorem 5.57 gives a formula for the distributional Laplacian

∆Φ = div∇Φ = cδ0 +1

nωnp.v.

n∑j=1

∂

∂xj

xj|x|n︸︷︷︸

0

= cδ0 ,

where

c =n∑j=1

∫|x|=1

1

nωn

xj|x|n

xj|x|

dσ(x) =1

nωn

∫|x|=1

dσ(x) = 1 .


53Note that when n ≥ 3, then Φ ∗ f = −I2f , where I2 is the Riesz potential, see Definition 2.14.54Formula (5.23) is also true for n = 2.

78 PIOTR HAJ LASZ

For ϕ ∈ Sn let u(x) = (Φ ∗ ϕ)(x). Then55 u ∈ C∞(Rn) 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 anexplicit solution of the Poisson equation

∆u = ϕ .

This explains the importance of the fundamental solution in partial differential equations.

Observe that the above calculation gives also

ϕ(x) = ∆(Φ ∗ ϕ)(x)

=n∑j=1

∂2

∂x2j

(Φ ∗ ϕ)(x)

=n∑j=1

(∂Φ

∂xj∗ ∂ϕ∂xj

)(x)

=

∫Rn∇Φ(x− y) · ∇ϕ(y) dy

=1

nωn

∫Rn

(x− y) · ∇ϕ(y)

|x− y|ndy

for every x ∈ Rn. In the last equality we employed (5.23). Thus we proved

Theorem 5.61. For ϕ ∈ Sn, n ≥ 2 we have

ϕ(x) =1

nωn

∫Rn

(x− y) · ∇ϕ(y)

|x− y|ndy for all x ∈ Rn.

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

Theorem 5.62. For ϕ ∈ Sn, n ≥ 2 and m ≥ 1 we have

ϕ(x) =m

nωn

∫Rn

∑|α|=m

Dαϕ(y)

α!

(x− y)α

|x− y|ndy for all x ∈ Rn.

Proof. Fix x ∈ Rn and define

ψ(y) =∑

|β|≤m−1

Dβϕ(y)(x− y)β

β!.

Then ψ(x) = ϕ(x) and56

∂ψ

∂yj(y) =

∑|β|=m−1

Dβ+δjϕ(y)(x− y)β

β!

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


where57 δj = (0, . . . , 1, . . . , 0). Hence Theorem 5.60 applied to ψ yields

ϕ(x) = ψ(x)

=1

nωn

∫Rn

n∑j=1

∂ψ

∂yj(y)

xj − yj|x− y|n

dy

=1

nωn

n∑j=1

∑|β|=m−1

∫RnDβ+δjϕ(y)

(x− y)β

β!

xj − yj|x− y|n

dy

=m

nωn

∑|α|=m

∫Rn

Dαϕ(y)

α!

(x− y)α

|x− y|ndy ,

because for α with |α| = m ∑j,β:β+δj=α

1

β!=m

α!.


5.7. Appendix. We will present here details of the last step in the proof of Theorem 5.7.Namely we will show that if u ∈ S ′

n and ϕ, ψ ∈ C∞0 (Rn), then

u[ ∫


]=

∫Rnu[ϕ(y − ·)ψ(y)] dy.

To approximate the integral∫Rn η(y) dy, η ∈ C∞0 by Riemann sums, we fix a cube

Q = [−N,N ]n so large that supp η ⊂ Q and divide the cube into cubes Qkimki=1 ofsidelength 2−k. Denote the centers of these cubes by yki. Clearly

mk∑i=1

η(yki)|Qki| →∫Rnη(y) dy as k →∞,

and

(5.24)

mk∑i=1

∫Qki

|η(y)− η(yki)| dy → 0 as k →∞.

Suppose ϕ, ψ ∈ C∞0 (Rn), so ϕ ∗ ψ ∈ C∞0 (Rn). Hence for every x ∈ Rn

wk(x) =

mk∑i=1

ϕ(x− yki)ψ(yki)|Qki| →∫Rnϕ(x− y)ψ(y) dy as k →∞.

The functions wk(x) belong to Sn.58 We will prove that

wk(x)→∫Rnϕ(x− y)ψ(y) dy = (ϕ ∗ ψ)(x) as k →∞

in the topology of Sn. First let us show that

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

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

80 PIOTR HAJ LASZ

We have

supx∈Rn|wk(x)− (ϕ ∗ ψ)(x)| ≤

mk∑i=1

supx∈Rn

∫Qki

|ϕ(x− yki)ψ(yki)− ϕ(x− y)ψ(y)| dy

≤ ‖ϕ‖∞mk∑i=1

∫Qki

|ψ(yki)− ψ(y)| dy

+

mk∑i=1

supx∈Rn

∫Qki

|ϕ(x− yki)− ϕ(x− y)| |ψ(y)| dy → 0.

The convergence of the first sum follows from (5.24) and the convergence of the secondone follows from uniform continuity of ϕ. Since

Dβwk(x) =

mk∑i=1

Dβϕ(x− yki)ψ(yki)|Qki| ,

exactly the same argument as above shows that

supx∈Rn|Dβwk(x)− (Dβϕ ∗ ψ)(x)︸︷︷︸

Dβ(ϕ∗ψ)(x)

| → 0 as k →∞.

Thereforepα,β(wk − ϕ ∗ ψ)→ 0 as k →∞,

because |x|α is bounded as the functions have compact support.

This proves that wk → ϕ ∗ ψ in Sn.

Since the function f(y) = u[ϕ(· − y)] is smooth, u[ϕ(· − y)]ψ(y) ∈ C∞0 and hence

u[ ∫


]= u[ϕ ∗ ψ] = lim

k→∞u[wk] = lim

k→∞

mk∑i=1

u[ϕ(· − yki)]ψ(yki)|Qki|

=

∫Rnu[ϕ(· − y)]ψ(y) dy =

∫Rnu[ϕ(y − ·)]ψ(y) dy.


6. Convolution with principal value distributions

In Theorem 5.49 we proved that if Ω ∈ L1(Sn−1) satisfies∫Sn−1

Ω(θ) dσ(θ) = 0,

then

WΩ = p.v. KΩ, where KΩ(x) =Ω(x/|x|)|x|n

defines a tempered distribution. In this chapter we will study convolution operators of theform

TΩϕ(x) = WΩ ∗ ϕ(x) = limε→0

∫|y|≥ε

Ω(y/|y|)|y|n

ϕ(x− y) dy.

Since TΩϕ = (ϕWΩ)∨ for ϕ ∈ Sn, the main objective will be to compute the Fourier

transform WΩ and conclude properties of TΩ from those of WΩ. We will focus on thesetting of tempered distributions and L2 spaces only.

Operators TΩ are called singular integrals and the main objective is to study their bound-edness in Lp spaces for 1 < p <∞. We will return to this problem in Chapters ?? and ??.

First we will consider the Riesz transforms which are amongst the most important sin-gular integrals, but have a very simple structure. Finally we will consider WΩ in its generalform.

6.1. Riesz transforms: L2 theory. In Proposition 5.53 we proved that for n ≥ 2

∂

∂xj|x|1−n = (1− n) p.v.

(xj|x|n+1

)∈ S ′

n.

In this section we will investigate the Riesz transforms which are convolutions with thedistributions

Wj = cn p.v.

(xj|x|n+1

), cn =

Γ(n+1

2

)πn+12

, n ≥ 2.

The choice of the coefficient cn will be justified in Theorem 6.2; with this choice the Fouriertransform of Wj has a particularly simple form.

Note that when n = 1, Wj becomes59

W =1

πp.v.

x

|x|2=

1

πp.v.

1

x

59cf. Remark 5.55.

82 PIOTR HAJ LASZ

and convolution with W is known as the Hilbert transform. Thus the Riesz transforms arehigher dimensional generalizations of the Hilbert transform. The Hilbert transform willbe carefully studied in Section 10. The reader may choose to read Section 10.1 now. Thissection contains elementary properties of the Hilbert transform and it fits well into thecontent of Chapter 6.

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

(Rjϕ)(x) = (Wj ∗ ϕ)(x)

= cn p.v.

∫Rn

yj|y|n+1

ϕ(x− y) dy

= cn limε→0

∫|x−y|≥ε

xj − yj|x− y|n+1

ϕ(y) dy .

Theorem 6.2.

Wj(ξ) = −i ξj|ξ|

.

Proof. Applying Proposition 5.53 and then Theorem 5.42 we have(p.v.

xj|x|n+1

)∧=

1

1− n

(∂

∂xj|x|1−n

)∧=

2πiξj1− n

(|x|1−n

)∧=

2πiξj1− n

πn−1−n2

Γ(n−(n−1)

2

)Γ(n−1

2

) |ξ|−1

=πn+12

Γ(n+1

2

) −iξj|ξ|

.

We used here equalities

Γ(1/2) = π1/2 andn− 1

2Γ

(n− 1

2

)= Γ

(n+ 1

2

).


Corollary 6.3. For ϕ ∈ Sn and 1 ≤ j ≤ n

(6.1) (Rjϕ)(x) =

(−iξj|ξ|

ϕ(ξ)

)∨(x) .

Rjϕ ∈ C∞ is slowly increasing and all its derivatives are slowly increasing. Moreover

‖Rjϕ‖2 ≤ ‖ϕ‖2 for ϕ ∈ Sn.

Remark 6.4. One has to be careful here. Although ϕ ∈ Sn, in general −i ξj|ξ| ϕ 6∈ Sn so

the inverse Fourier transform in (6.1) has to be understood in the distributional or in theL2 sense.


Proof. Equality (6.1) follows from

(Rjϕ)∧ = (Wj ∗ ϕ)∧ = ϕWj = −i ξj|ξ|

ϕ.

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

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

Definition 6.5. The Riesz transforms are bounded operators on L2(Rn), n ≥ 2,

‖Rjf‖2 ≤ ‖f‖2 for f ∈ L2(Rn) and j = 1, 2, . . . , n,

defined by the formula

(Rjf)(x) =

(−iξj|ξ|

f(ξ)

)∨(x) for f ∈ L2(Rn).

Remark 6.6. The Riesz transform on L2 was defined as an extension of a bounded operatorRj defined on a subspace Sn ⊂ L2 by the formula Rjϕ = cn(p.v. xj/|x|n+1) ∗ ϕ. Aninteresting question is whether the L2 extension can also be defined by the same p.v.integral formula and if yes, whether the limit in the definition of the p.v. integral existsin the L2 sense or in the a.e. sense. These highly non-trivial questions will be answered inSection ??.

Corollary 6.7. The Riesz transforms satisfyn∑j=1

R2j = −I on L2(Rn).

Proof. For f ∈ L2 we have( n∑j=1

R2jf)∧

(ξ) =n∑j=1

(− iξj|ξ|

)2

f(ξ) = −f(ξ)

and the result follows upon taking the inverse Fourier transform.

The next general and elementary result applies to Riesz transforms.

Theorem 6.8. If m ∈ L∞(Rn), then the operator

Tmf = (mf)∨, f ∈ L2

is bounded in L2,‖Tmf‖2 ≤ ‖m‖∞‖f‖2.

Moreover, there is a tempered distribution u ∈ S ′n such that

Tmϕ = u ∗ ϕ for all ϕ ∈ Sn.

60cf. Theorem 6.8 and estimates (6.3).

84 PIOTR HAJ LASZ

Hence Tmϕ is smooth, slowly increasing, all its derivatives are slowly increasing and

(6.2) Dα(Tmϕ) = Tm(Dαϕ) for all multiindices α.

If m1,m2 ∈ L∞, then Tm1 Tm2 = Tm2 Tm1 = Tm1m2.

Proof. Boundedness of Tm on L2 is an obvious consequence of the Plancherel theorem

(6.3) ‖Tmf‖2 = ‖Tmf‖2 = ‖mf‖2 ≤ ‖m‖∞‖f‖2 = ‖m‖∞‖f‖2.

To prove existence of u ∈ S ′n, observe that m ∈ L∞ is a tempered distribution and

u = m ∈ S ′n satisfies u = m. Thus (u ∗ ϕ)∧ = ϕm = (Tmϕ)∧ so Tmϕ = u ∗ ϕ.

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

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

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

Proposition 6.9. If ϕ ∈ Sn, then for 1 ≤ j, k ≤ n we have

∂2ϕ

∂xj∂xk= −RjRk∆ϕ(x) = −∆(RjRkϕ)(x), .

Proof. For ϕ ∈ Sn we have

(∂j∂kϕ)∧(ξ) = (2πiξj)(2πiξk)ϕ(ξ)

= −(−iξj|ξ|

)(−iξk|ξ|

)(−4π2|ξ|2)ϕ(ξ)

= −(−iξj|ξ|

)(−iξk|ξ|

)∆ϕ(ξ)

= (−RjRk∆ϕ)∧(ξ).

Thus taking the inverse Fourier transform in L2 yields

∂2ϕ

∂xj∂xk= −RjRk∆ϕ(x).

The composition of the Riesz transforms RjRk is understood as composition of boundedoperators in L2 and according to Theorem 6.8 there is u ∈ S ′

n such that

RjRkϕ =

((−iξj|ξ|

)(−iξk|ξ|

)ϕ

)∨= u ∗ ϕ.

so RjRkϕ ∈ C∞ is slowly increasing and all its derivatives are slowly increasing. Also

∂2ϕ

∂xj∂xk= −RjRk∆ϕ(x) = −u ∗∆ϕ = −∆(u ∗ ϕ) = −∆(RjRkϕ).



Remark 6.10. One can prove that if j 6= k, then RjRk can be written as a convolutionwith a p.v. distribution. However, R2

j cannot be written as a convolution with a p.v.distribution. Fortunately, it is still a convolution with a tempered distribution, but not ofa p.v. type.61

As a direct consequence of Proposition 6.9 and of boundedness of the Riesz transformsin L2 we obtain

Corollary 6.11. ∥∥∥∥ ∂ϕ

∂xj∂xk

∥∥∥∥2

≤ ‖∆ϕ‖2 for ϕ ∈ Sn.

Remark 6.12. The same estimate can be proved directly using integration by parts. Laterwe will see that a similar estimate is also true for the Lp norm, 1 < p < ∞. But this is amuch harder result and a simple integration by parts is not enough.

Remark 6.13. It is quite convincing that the proof of Proposition 6.9 applied to u ∈ S ′n

such that ∆u = f ∈ L2, gives ∂j∂ku = −RjRkf . However this is not true. For exampleif u = xy, then, as a slowly increasing function, u ∈ S ′

2. Clearly ∆u = 0 ∈ L2, but∂x∂yu = 1 6= 0 = −RxRy0. The problem is of a very delicate nature and it is important tounderstand it well. Remember that u is a tempered distribution, not a function so in theequality

(2πiξj)(2πiξk)u(ξ) = −(−iξj|ξ|

)(−iξk|ξ|

)[(−4π2|ξ|2)u(ξ)]

the left hand side is well defined since we are allowed to multiply distributions by(2πiξj)(2πiξk). However, the right hand side is not well defined: (−4π2|ξ|2)u(ξ) is a dis-tribution and we are not allowed to multiply this distribution by a non-smooth function.

It is actually still confusing. After all, (−4π2|ξ|2)u = ∆u = f ∈ L2 so why are we notallowed to multiply it by a bounded function? We are, but if we do, the above equalityis no longer true, because we switch from the distributional context to the L2 context.In fact, the distribution u may have a non-trivial part that is supported at 0 such thatthe multiplication by (2πiξj)(2πiξk) does not ‘kill’ this part. However, multiplication by−4π2|ξ|2 does ‘kill’ it so it may happen that

(2πiξj)(2πiξk)u(ξ) = −(−iξj|ξ|

)(−iξk|ξ|

)f + v

where v is a distribution supported at 0. This will be carefully explained in the next result.

Theorem 6.14. If u ∈ S ′n satisfies

(6.4) ∆u = f ∈ L2(Rn) ,

then for any 1 ≤ j, k ≤ n

∂2u

∂xj∂xk= −RjRkf + Pjk ,

where Pjk is a harmonic polynomial.

61c.f. Problems ?? and ?? and Example 6.41.

86 PIOTR HAJ LASZ

Proof. We will only prove that Pjk are polynomials and we leave the proof that they areharmonic to the reader. According to Corollary 5.34 it suffices to prove that the tempereddistribution (

∂2u

∂xj∂xk+RjRkf

)∧has support contained in 0. To this end we need to show that if ϕ ∈ Sn, supp ϕ ⊂Rn \ 0, say ϕ(x) = 0 for |x| ≤ r, then(

∂2u

∂xj∂xk+RjRkf

)∧[ϕ] = 0, i.e., (∂j∂ku)∧[ϕ] = (−RjRkf)∧[ϕ] .

Let η ∈ C∞(Rn) be such that η(x) = 0 for |x| ≤ r/2 and η(x) = 1 for |x| ≥ r. The functionη and its derivatives are slowly increasing and ηϕ = ϕ. Hence

(∂j∂ku)∧[ϕ] =((2πiξi)(2πiξj)u

)[ηϕ]

=(η(ξ)(2πiξi)(2πiξj)u

)[ϕ] = ♥.

Observe that

η(ξ)(2πiξi)(2πiξj) = − η(ξ)

(−iξj|ξ|

)(−iξk|ξ|

)︸︷︷︸

λ(ξ)

(−4π2|ξ|2)

and λ ∈ C∞ and its derivatives are slowly increasing since η vanishes in a neighborhoodof ξ = 0. Thus62

♥ =(− λ(ξ)(−4π2|ξ|2)u

)[ϕ]

= (−λ(ξ)f(ξ))[ϕ]

=

(−η(ξ)

(−iξj|ξ|

)(−iξk|ξ|

)f(ξ)

)[ϕ](6.5)

=

(−(−iξj|ξ|

)(−iξk|ξ|

)f(ξ)

)[ηϕ]

=

(−(−iξj|ξ|

)(−iξk|ξ|

)f(ξ)

)[ϕ]

= (−RjRkf)∧[ϕ] .


6.2. The Beurling-Ahlfors transform. We proved that

∂2ϕ

∂xj∂xk= −RjRk∆ϕ = −∆(RjRkϕ) for ϕ ∈ Sn

62Taking the Fourier transform of (6.4) yields −4π2|ξ|2u = f in S ′n. We should understand this equality

in S ′n and not in L2, because despite the fact that −4π2|ξ|2u = f ∈ L2, it may happen that the distributionu is not a function. If λ ∈ C∞(Rn) and all its derivatives are slowly increasing, then −λ(ξ)(−4π2|ξ|2)u =

−λ(ξ)f in S ′n. Now −λf ∈ L2 and f ∈ L2 are functions. Functions that are equal a.e. define the samedistribution. Hence equality (6.5) is true because the functions that define distributions on the left handside and on the right hand side of (6.5) are equal for all ξ 6= 0 so they are equal a.e.


so the Riesz transforms, in some sense, can change the direction of partial derivatives. Nowwe will find an operator S such that

S

(∂ϕ

∂z

)=∂ϕ

∂zfor ϕ ∈ S (R2) = S (C).

We will be using complex notation for points in R2 = Cz = x+ iy and ξ = ξ1 + iξ2,

and also for the Lebesgue measure

dz = dz + idy, dz = dx− idy sodz ∧ dz

2i= dx ∧ dy = dx dy.

Definition 6.15. The operator

S = (iR1 +R2)2 defined on L2(C).

is called the Beurling-Ahlfors transform.

The next result collects basic properties of the Beurling-Ahlfors transform.

Theorem 6.16. The Beurling-Ahlfors transform has the following properties

(1) S : L2(C)→ L2(C) is a surjective isometry,

‖Sf‖2 = ‖f‖2 for f ∈ L2(C).

(2) The inverse operator is

S−1f = Sf .

(3) For f ∈ L2(C) we have

Sf(ξ) =

(ξ

ξf(ξ)

)∨, S−1f(ξ) =

(ξ

ξf(ξ)

)∨.

(4)

S

(∂ϕ

∂z

)=

∂

∂z(Sϕ) =

∂ϕ

∂zfor ϕ ∈ S (R2) = S (C).

Proof. Since

ξ

ξ=

(ξ

|ξ|

)2

=

(ξ1 − iξ2

|ξ|

)2

=

(i

(−iξ1

|ξ|

)+

(−iξ2

|ξ|

))2

we conclude that

Sf = (iR1 +R2)2f =

(ξ

ξf

)∨for f ∈ L2(C).

Clearlyξ

ξ· ξξ

= 1 so S−1f =

(ξ

ξf

)∨which proves (3). The fact that S is an isometry easily follows from the Plancherel theorem

‖Sf‖2 = ‖Sf‖2 =

∥∥∥∥ ξξ f∥∥∥∥

2

= ‖f‖2 = ‖f‖2.

88 PIOTR HAJ LASZ

Since S has the inverse operator, it maps L2(C) onto L2(C). This establishes (1). A simplecomputation based on the fact that

F(f(ξ)) = F(f)(−ξ) and F−1(f(ξ)) = F−1(f)(−ξ)yields

Sf =

(ξ

ξf

)∨= S−1f

which is (2). Equality

S

(∂ϕ

∂z

)=

∂

∂z(Sϕ) for ϕ ∈ S (R2) = S (C)

follows from Theorem 6.8, see (6.2). Recall that

∂

∂z=

1

2

(∂

∂x+ i

∂

∂y

)and

∂

∂z=

1

2

(∂

∂x− i ∂

∂y

)so (

∂ϕ

∂z

)∧=

1

2(2πiξ1 + i(2πiξ2))ϕ = πiξϕ

and (∂ϕ

∂z

)∧=

1

2(2πiξ1 − i(2πiξ2))ϕ = πiξϕ.

Therefore (S

(∂ϕ

∂z

))∧=ξ

ξπiξϕ = πiξϕ =

(∂ϕ

∂z

)∧.

This completes the proof of (4).

Corollary 6.17. For ϕ ∈ S (C) we have∥∥∥∥∂ϕ∂z∥∥∥∥

2

=

∥∥∥∥∂ϕ∂z∥∥∥∥

2

.

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

Similarly as in the case of the Riesz transforms, the Beurling-Ahlfors transform canbe represented as a convolution with a principal value distribution. In order to find aformula we need to investigate a connection between the Beurling-Ahlfors transform andthe fundamental solution to the Laplace equation. Recall that63

Φ(z) =1

2πlog |z| = 1

4πlog |z|2

satisfies

∆Φ =

(∂2

∂x2+

∂2

∂y2

)Φ = 4

∂2

∂z∂zΦ = δ0.

Hence

(6.6)1

π

(∂

∂z(4πΦ) ∗ ∂ϕ

∂z

)=

1

π

∂

∂z

(∂

∂z(4πΦ) ∗ ϕ

)=

1

π

(∂2

∂z∂z(4πΦ) ∗ ϕ

)= ϕ.

63Theorem 5.60.


An argument similar to that used in the proof of Theorem 5.56 allows us to differentiateΦ once in the pointwise sense without necessity of taking the principal value distributionso64

(6.7)∂

∂z(4πΦ(z)) =

∂

∂zlog |z|2 =

1

|z|2∂

∂z(zz) =

z

|z|2=

1

zfor z 6= 0

and hence∂

∂z

(1

πz

)=

∂

∂z

(∂

∂z(4Φ)

)= ∆Φ = δ0.

We proved

Proposition 6.18. The function G(z) = 1πz

is a fundamental solution of the operator∂/∂z, i.e.

∂

∂z

(1

πz

)= δ0 in S (C).

Also (6.7) allows us to rewrite (6.6) as65

(6.8)1

2πi

∫C

ϕz(ξ)

z − ξdξ ∧ dξ =

1

2πi

∂

∂z

∫C

ϕ(ξ)

z − ξdξ ∧ ξ = ϕ(z).

This is a representation formula similar to that in Theorem 5.61. It allows us to reconstructthe function ϕ from its derivative ϕz. In paricular it follows immediately that the onlyholomorphic function in S (C) is the zero function.

Definition 6.19. The Cauchy transform of a function f is defined by

Cf(z) = (G ∗ f)(z) =1

2πi

∫C

f(ξ)

z − ξdξ ∧ ξ.

Thus (6.8) reads as

Corollary 6.20.∂

∂z(Cϕ) = C

(∂ϕ

∂z

)= ϕ for ϕ ∈ S (C).

The next result allows us to identify the Beurling-Ahlfors transform with the convolutionwith a principal value distribution. A different proof will be presented in Section 6.3, seeExample 6.27.

Theorem 6.21. For ϕ ∈ S (C) we have(6.9)

S(ϕ) =∂

∂z(Cϕ) = C

(∂ϕ

∂z

)= − 1

2πilimε→0

∫|ξ−z|≥ε

ϕ(ξ)

(z − ξ)2dξ ∧ dξ = − 1

π

(p.v.

1

z2

)∗ ϕ.

Proof. First we will prove that for ϕ ∈ S (C) we have

(6.10)∂

∂z(Cϕ) = C

(∂ϕ

∂z

)= − 1

2πilimε→0

∫|ξ−z|≥ε

ϕ(ξ)

(z − ξ)2dξ ∧ dξ = − 1

π

(p.v.

1

z2

)∗ ϕ.

64If f is holomorphic and g is smooth in the real sense, then ∂∂z (f g)(z) = ∂f

∂z (g(z))∂g∂z (z).65Recall that dξ1 dξ2 = 1

2i dξ ∧ dξ.

90 PIOTR HAJ LASZ

We could actually conclude this equality from Theorem 5.57 but we prefer to provide adirect proof. Note that

(6.11) C(∂ϕ

∂z

)=

∂

∂z(Cϕ), for ϕ ∈ S (C)

is a consequence of a general rule how we differentiate convolution of a tempered distribu-tion with ϕ ∈ S (C).

It is easy to check that in the complex notation Green’s theorem can be rewritten asfollows

Lemma 6.22 (Green’s theorem). If Ω ⊂ C is a bounded domain with the piecewise C1

boundary and f, g ∈ C1(Ω), then∫Ω

(∂f

∂z+∂g

∂z

)dz ∧ dz =

∫∂Ω

g dz − f dz.

Here ∫∂Ω

f dz =

∫∂Ω

f (dx− idy) =

∫∂Ω

f dz.

In particular ∫Ω

∂

∂z(fg) dz ∧ dz = −

∫∂Ω

fg dz

i.e. we have the following version of the integration by parts formula

(6.12)

∫Ω

∂f

∂zg dz ∧ dz = −

∫Ω

f∂g

∂zdz ∧ dz −

∫∂Ω

fg dz.

Applying (6.12) to our situation yields66

2πi C(∂ϕ

∂z

)=

∫C

∂ϕ(ξ)/∂ξ

z − ξdξ ∧ dξ

= limε→0

∫|ξ−z|≥ε

∂ϕ(ξ)/∂ξ

z − ξdξ ∧ dξ

= limε→0−∫|ξ−z|≥ε

ϕ(ξ)∂

∂ξ

(1

z − ξ

)dξ ∧ dξ + lim

ε→0

∫|ξ−z|=ε

ϕ(ξ)

z − ξdξ = ♥.

The sign ‘+’ in the last limit is opposite to the sign ‘-’ in the boundary integral in (6.12).This is because the positive orientation of the boundary |ξ− z| = ε of the exterior domain|ξ − z| ≥ ε is clockwise.

Note that ∫|ξ−z|=ε

dξ

z − ξ= −

∫|ξ|=ε

dξ

ξ= −

∫|ξ|=ε

dξ

ξ= −ε−2

∫|ξ|=ε

ξ dξ = 0

and hence

limε→0

∫|ξ−z|=ε

ϕ(ξ)

z − ξdξ = lim

ε→0

∫|ξ−z|=ε

ϕ(ξ)− ϕ(z)

z − ξdξ = 0.

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


Since∂

∂ξ

(1

z − ξ

)=

1

(z − ξ)2

we obtain

♥ = − limε→0

∫|ξ−z|≥ε

ϕ(ξ)

(z − ξ)2dξ ∧ dξ.

This completes the proof of (6.10).

Note that (6.10) was a quite straightforward consequence of Green’s formula. It remainsto prove that S(ϕ) = ∂

∂z(Cϕ) and this is not obvious.67 Let us start with an interesting

remark.

Remark 6.23. The space

Sz = ϕz : ϕ ∈ S (C)is dense in L2(C). For if not, we would find 0 6= f ∈ L2 which is orthogonal to Sz, i.e.,∫

Cϕzf dz ∧ dz = 0 for all ϕ ∈ S (C).

That means fz = 0 in S ′(C) and hence ∆f = 4fzz = 0 in S ′(C). According to Corol-lary 5.35, f is a polynomial. Since f ∈ L2, we conclude that f = 0 which is a contradiction.Observe also that

S(ϕz) = ϕz and∂

∂z(Cϕz) = ϕz.

This shows that equality (6.9) holds on the subspace Sz ⊂ S (C) which is dense in L2.Since the operator S is an isometry on L2, we conclude that ϕ 7→ ∂

∂zC(ϕ) is an isometry on

a dense subset Sz of L2. We could conclude the theorem if we could prove that ϕ 7→ ∂∂zC(ϕ)

is bounded in the L2 norm on S (C). After all, boundedness of this operator follows fromTheorem 6.21, but we cannot use it before we prove Theorem 6.21. Vicious circle! In ourproof given below we will follow the idea described in this remark by constructing a suitableL2 approximation of ϕ ∈ S (C) by functions of the class Sz(C).

Let’s return to the proof. Let ϕ ∈ S (C) and let ψ = Cϕ. As we know68 ψz = ϕ. However,in general ψ 6∈ S (C).69 Since the Cauchy transform is bounded by the Riesz potential I1

and ϕ ∈ Lp for all p, the Fractional Integration Theorem 2.15 implies that ψ ∈ Lp for all2 < p <∞, but in general ψ 6∈ L2.70

67The function ξ/ξ ∈ L∞ defines a tempered distribution. If u = (ξ/ξ)∨ ∈ S ′(C), then S(ϕ) = u ∗ ϕ(c.f. Theorem 6.8) and it remains to prove that u is the p.v. distribution given by (6.10). However, findingthe Fourier transform of ξ/ξ is not easy, see Example 6.27. On the other hand since S = (iR1 +R2)2 is acomposition of convolutions with p.v. distributions, S should be easy to represent as a convolution witha p.v. distribution. However, finding a formula for the composition is not easy either. For example R2

j

cannot be represented as a composition with a p.v. distribution (see Problem ?? and Example 6.41.) Wewill discuss this topic in detail in Section 6.6.

68Corollary 6.20.69cf. Problem ??70cf. Problem ??

92 PIOTR HAJ LASZ

Let η ∈ C∞(C), η(z) = 1 for |z| ≤ 1 and η(z) = 0 for |z| ≥ 2 be a standard cut-off function. Let ψR(z) = ψ(z)η(z/R). Since the function has compact support we haveS(ψRz ) = ψRz . We claim that

(6.13) ψRz → ϕ in L2 as R→∞.

The chain rule gives

ψRz = ψzη( zR

)+ ψ(z)ηz

( zR

) 1

R= ϕ(z)η

( zR

)+ ψ(z)ηz

( zR

) 1

R.

Clearly

ϕ(z)η( zR

)→ ϕ in L2.

The function ηz(z/R) is bounded an supported in B(0, 2R). Hence for any 2 < p <∞ wehave71 ∥∥∥∥ψ(z)ηz

( zR

) 1

R

∥∥∥∥2

≤ C(∫

B(0,2R)

|ψ|2)1/2

≤ C(∫

B(0,2R)

|ψ|p)1/p

≤ CR−2/p‖ψ‖p → 0 as R→∞.

This proves (6.13) and thus72

(6.14) S(ψRz )→ S(ϕ) in L2.

On the other hand by the arguing in the same way as in the proof of (6.13) we have

(6.15) S(ψRz ) = ψRz = ψz(z)η( zR

)+ ψ(z)ηz

( zR

) 1

R→ ψz =

∂

∂z(Cϕ) in L2 as R→∞.

Comparing (6.14) and (6.15) yields S(ϕ) = ∂∂z

(Cϕ). The proof is complete.

6.3. Higher Riesz transforms.

Definition 6.24. Higher Riesz transforms of degree k are defined as a convolution with

(6.16) p.v.Pk(x)

|x|n+k,

where Pk(x) is a homogeneous harmonic polynomial of degree k ≥ 1.

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

If Pk is a homogeneous harmonic polynomial of degree k ≥ 1 we can write

Pk(x)

|x|n+k=Pk(x)|x|−k

|x|n=Pk(x/|x|)|x|n

=Ω(x/|x|)|x|n

and in order to prove that (6.16) defines a tempered distribution we need to prove that

(6.17)

∫Sn−1

Ω(θ) dσ(θ) =

∫Sn−1

Pk(θ) dσ(θ) = 0 .

71In the first inequality we use the fact that the area of the disc B(0, 2R) equals CR2. We also use hereHolder’s inequality and the fact that ψ ∈ Lp.

72S is bounded in L2.


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

∂Pk∂~ν

=d

dt

∣∣∣t=1Pk(tx) = ktk−1Pk(x)

∣∣∣t=1

= kPk(x)

and hence Green’s formula73 yields

k

∫Sn−1

Pk(θ) dσ(θ) =

∫Sn−1

∂Pk∂~ν

(θ) dσ(θ) =

∫Bn

∆Pk dx = 0 .

Thus according to Theorem 5.49

WΩ[ϕ] = p.v.

∫Rn

Pk(x)

|x|n+kϕ(x) dx

is a well defined tempered distribution.

Our aim is to prove the following result.

Theorem 6.25. If Pk is a homogeneous harmonic polynomial of degree k ≥ 1, then(p.v.

Pk(x)

|x|n+k

)∧(ξ) = γk

Pk(ξ)

|ξ|k,

where

γk = (−i)kπn/2Γ(k2

)Γ(k+n

2

) .Example 6.26. This result immediately implies Theorem 6.2.

Example 6.27. We will show now that the formula for the Beurling-Ahlfors transform

(6.18) S(ϕ) = − 1

π

(p.v.

1

z2

)∗ ϕ for ϕ ∈ S (C)

originally proved in Theorem 6.21 is a straightforward consequence of Theorem 6.25.

Let n = 2 and let74 P2(ξ) = ξ2 = ξ21−2iξ1ξ2− ξ2

2 . Clearly P2 is a homogeneous harmonicpolynomial of degree 2. Since γ2 = −π we obtain(

p.v.1

z2

)∧(ξ) =

(p.v.

P2(z)

|z|4

)∧(ξ) = −π P2(ξ)

|ξ|2= −π ξ

ξ

so (− 1

π

(p.v.

1

z2

)∗ ϕ)∧

=ξ

ξϕ = S(ϕ)

which proves (6.18).

We will deduce Theorem 6.25 from the following result.

Theorem 6.28. If Pk is a homogeneous harmonic polynomial of degree k and 0 < α < n,then (

Pk(x)

|x|k+n−α

)∧(ξ) = γk,α

Pk(ξ)

|ξ|k+α,

73∫

Ω(f∆g − g∆f) dx =

∫∂Ω

(f ∂g∂~ν − g∂f∂~ν ) dσ where ~ν is the outward normal.

74We use complex notation in R2 = C.

94 PIOTR HAJ LASZ

where

γk,α = (−i)kπn2−α Γ

(k+α

2

)Γ(k+n−α

2

) .Remark 6.29. Note that the function

Pk(x)

|x|k+n−α

is a tempered L1 function, so it defines a tempered distribution without necessity of takingthe principal value of the integral.

Proof. For any t > 0 and ϕ ∈ Sn, Corollary 4.24 gives75∫RnPk(x)e−πt|x|

2

ϕ(x) dx = (−i)k∫RnPk(x)e−π|x|

2/tt−k−n2ϕ(x) dx .

Now we multiply both sides by

tβ−1, where β =k + n− α

2> 0

and integrate with respect to 0 < t <∞. Since∫ ∞0

e−πt|x|2

tβ−1 dt = (π|x|2)−βΓ(β)

the integral on the left hand side will be equal to

(6.19)Γ(β)

πβ

∫Rn

Pk(x)

|x|2βϕ(x) dx =

Γ(k+n−α

2

)πk+n−α

2

(Pk(·)| · |k+n−α

)∧[ϕ] .

Similarly ∫ ∞0

e−π|x|2/tt−k−

n2 tβ−1 dt =

∫ ∞0

e−π|x|2/tt−

k+α2−1 dt

s=π|x|2/t= (π|x|2)−

k+α2

∫ ∞0

e−ssk+α2−1 ds

= (π|x|2)−k+α2 Γ

(k + α

2

).

Thus the integral on the right hand side equals

(6.20) (−i)kΓ(k+α

2

)πk+α2

∫Rn

Pk(x)

|x|k+αϕ(x) dx .

Since integrals at (6.19) and (6.20) are equal one to another, the theorem follows.

Proof of Theorem 6.25. For ϕ ∈ Sn we take the identity∫Rn

Pk(x)

|x|k+n−α ϕ(x) dx = γk,α

∫Rn

Pk(x)

|x|k+αϕ(x) dx

75∫Rn f(x)ϕ(x) dx =

∫Rn f(x)ϕ(x) dx.


and let α→ 0. The right hand side converges to

(−i)kπn2

Γ(k2

)Γ(k+n

2

) ∫Rn

Pk(x)

|x|kϕ(x) dx .

To compute the limit on the left hand side we observe that the integral of Pk(x)|x|−(k+n−α)

on the unit ball equals zero, see (6.17), and hence∫Rn

Pk(x)

|x|k+n−α ϕ(x) dx

=

∫|x|≤1

Pk(x)

|x|k+n−α (ϕ(x)− ϕ(0)) dx+

∫|x|≥1

Pk(x)

|x|k+n−α ϕ(x) dx

α→0+−→∫|x|≤1

Pk(x)

|x|k+n(ϕ(x)− ϕ(0)) dx+

∫|x|≥1

Pk(x)

|x|k+nϕ(x) dx

= limε→0

∫ε≤|x|≤1

Pk(x)

|x|k+n(ϕ(x)− ϕ(0)) dx+

∫|x|≥1

Pk(x)

|x|k+nϕ(x) dx

= limε→0

∫|x|≥ε

Pk(x)

|x|k+nϕ(x) dx

=

(p.v.

Pk(x)

|x|k+n

)∧[ϕ] .

Comparing the above limits yields the result.

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

For ϕ ∈ Sn we have

Wj[ϕ] = Wj[ϕ] = limε→0

cn

∫|ξ|≥ε

ϕ(ξ)ξj|ξ|n+1

dξ

= limε→0

cn

∫ε≤|ξ|≤ε−1

(∫Rnϕ(x)e−2πix·ξ dx

)ξj|ξ|n+1

dξ

= limε→0

cn

∫Rnϕ(x)

(∫ε≤|ξ|≤ε−1

e−2πix·ξ ξj|ξ|n+1

dξ︸︷︷︸I

)dx = ♥ .

Expressing the integral I is the spherical coordinates yields

I =

∫ ε−1

ε

sn−1

(∫Sn−1

e−2πix·(sθ) sθjsn+1

dσ(θ)

)ds

=

∫ ε−1

ε

∫Sn−1

(cos(2πsx · θ)− i sin(2πsx · θ)

)θj dσ(θ)

ds

s

76The second proof was given in Remark 6.26.

96 PIOTR HAJ LASZ

= −i∫ ε−1

ε

∫Sn−1

sin(2πsx · θ)θj dσ(θ)ds

s

= −i∫Sn−1

(∫ ε−1

ε

sin(2πs|x · θ|)s

ds

)sgn (x · θ)θj dσ(θ)

= −i∫Sn−1

(∫ 2π|x·θ|ε−1

2π|x·θ|ε

sin t

tdt

)sgn (x · θ)θj dσ(θ) .

Thus

♥ =

∫Rnϕ(x)

(−icn

π

2

∫Sn−1

sgn (x · θ)θj dσ(θ)

)dx .

Indeed, we could pass to the limit under the sign of the integral using the DominatedConvergence Theorem and we employed the equality77∫ ∞

0

sin t

tdt =

π

2.

Thus it remains to prove that

(6.21) cnπ

2

∫Sn−1

sgn (x · θ)θj dσ(θ) =xj|x|

.

It is obvious that the function m : Rn → Rn defined as

m(x) =

∫Sn−1

sgn (x · θ)θ dσ(θ)

is homogeneous of degree 0. We will use Lemma 4.26 so we need to check that m commuteswith orthogonal transformations. It does because

m(ρ(x)) =

∫Sn−1

sgn (ρ(x) · θ)θ dσ(θ)

=

∫Sn−1

sgn (x · ρ−1(θ))θ dσ(θ)

=

∫Sn−1

sgn (x · θ)ρ(θ) dσ(θ)(6.22)

= ρ

(∫Sn−1

sgn (x · θ)θ dσ(θ)

)(6.23)

= ρ(m(x)) .

Note that (6.22) follows from the fact that ρ induces a volume preserving change of variableson Sn−1, while (6.23) is a direct consequence of linearity of ρ. Thus the Lemma 4.26 yields∫

Sn−1

sgn (x · θ)θ dσ(θ) = cx

|x|and hence looking at the jth component we have

(6.24)

∫Sn−1

sgn (x · θ)θj dσ(θ) = cxj|x|

.

77Recall that this is understood as an improper integral which is consistent with our setting since wepass to the limit as ε→ 0.


Thus it remains to prove that

c =(cnπ

2

)−1

=2π(n−1)/2

Γ(n+1

2

) = 2ωn−1 .

Taking x = ej in (6.24) we have ∫Sn−1

|θj| dσ(θ) = c .

The unit ball Bn−1 in coordinates perpendicular to xj split the sphere Sn−1 into two halfspheres Sn−1

± . Thus

c = 2

∫Sn−1+

θj dσ(θ) .

Recall that if M ⊂ Rn is a graph of a C1 function f : Ω → R, Ω ⊂ Rn−1, then for ameasurable function g on M we have∫

M

g dσ =

∫Ω

g(x, f(x))√

1 + |∇f(x)|2 dx .

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

f(x) =√

1− |x|2 , x ∈ Bn−1 .

Form the picture

98 PIOTR HAJ LASZ

we conclude∫Sn−1+

θj dσ(θ) =

∫Sn−1+

(1− h) dσ(θ) =

∫Bn−1

(1− h)√

1 + tan2 α dx

=

∫Bn−1

dx = ωn−1 ,

because √1 + tan2 α =

1

cosα=

1

1− hand the result follows. 2

The proof of the next result is based on an argument similar to that used to prove (6.24);we leave details to the reader as an exercise.

Lemma 6.30. Let K be a function of one variable, then for n ≥ 2 we have∫Sn−1

K(x · θ) dσ(θ) = (n− 1)ωn−1

∫ 1

−1

K(s|x|)(1− s2)n−32 ds

for all x ∈ Rn \ 0.

6.5. The general case. In this section we will consider operators TΩϕ = WΩ ∗ ϕ in themost general form where Ω ∈ L1(Sn−1),

∫Sn−1 Ω(θ) dσ(θ) = 0, and

WΩ = p.v. KΩ, where KΩ(x) =Ω(x/|x|)|x|n

.

The following lemma will play an important role.

Lemma 6.31. If F is a function homogeneous of degree zero and F∣∣Sn−1 ∈ L1(Sn−1), then

F is a tempered L1 function∫Rn|F (x)|(1 + |x|n+1)−1 dx <∞.

Hence F ∈ S ′n. Moreover

|F [ϕ]| ≤ C(n)‖F‖L1(Sn−1) supx∈Rn

(1 + |x|n+1)|ϕ(x)|.

Proof. We have

|F [ϕ]| ≤∫Rn|F (x)ϕ(x)| dx =

∫ ∞0

sn−1

∫Sn−1

|F (θ)ϕ(sθ)| dσ(θ) ds

=

∫Sn−1

|F (θ)|∫ ∞

0

sn−1|ϕ(sθ)| ds dσ(θ) = ♥

∫ ∞0

sn−1|ϕ(sθ)| ds ≤∫ ∞

0

sn−1

1 + sn+1sup|x|=s

(1 + |x|n+1)|ϕ(x)| ds

≤ C(n) supx∈Rn

(1 + |x|n+1)|ϕ(x)|.

Hence♥ ≤ C(n)‖F‖L1(Sn−1) sup

ξ∈Rn(1 + |ξ|n+1)|ϕ(ξ)|.


In particular taking78 ϕ(x) = (1 + |x|n+1)−1 yields∫Rn|F (x)|(1 + |x|n+1)−1 ≤ C‖F‖L1(Sn−1) <∞

so F is a tempered L1 function.

The main result of this section generalizes Theorem 6.2.

Theorem 6.32. Let n ≥ 2 and Ω ∈ L1(Sn−1) be such that∫Sn−1

Ω(θ) dσ(θ) = 0 .

Then the Fourier transform of the distribution WΩ is a tempered L1 function, homogeneousof degree zero such that

(6.25) ‖WΩ‖L1(Sn−1) ≤ C‖Ω‖L1(Sn−1) and

∫Sn−1

WΩ(ξ) dσ(ξ) = 0.

Therefore WΩ is a tempered distribution satisfying

(6.26) |WΩ[ϕ]| ≤ C‖Ω‖L1(Sn−1)

(supξ∈Rn

(1 + |ξ|n+1)|ϕ(ξ)|).

Moreover

WΩ(ξ) =

∫Sn−1

Ω(θ)

(log

1

|ξ · θ|− iπ

2sgn (ξ · θ)

)dσ(θ)(6.27)

=

∫Sn−1

Ω(θ)

(log

1

|ξ′ · θ|− iπ

2sgn (ξ′ · θ)

)dσ(θ) ,

where ξ′ = ξ/|ξ|.

Finally, if Ω is odd, then

(6.28) WΩ(ξ) = −iπ2

∫Sn−1

Ω(θ) sgn (ξ′ · θ) dσ(θ) is odd

and if Ω is even, then

(6.29) WΩ(ξ) =

∫Sn−1

Ω(θ)

(log

1

|ξ′ · θ|

)dσ(θ) is even.

We will precede the proof with some technical lemmas.

Lemma 6.33. For θ ∈ Sn−1 define a function on Sn−1 by fθ(ξ′) = log 1

|ξ′·θ| . Then fθ ∈Lp(Sn−1) for all 1 ≤ p <∞ and

supθ∈Sn−1

‖fθ‖Lp(Sn−1) = C(n, p) <∞.

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

100 PIOTR HAJ LASZ

Proof. According to Lemma 6.3079

(6.30)

∫Sn−1

logp(

1

|ξ′ · θ|

)dσ(ξ′) = (n− 1)ωn−1

∫ 1

−1

logp(

1

|s|

)(1− s2)

n−32 dx <∞

The last integral has three singularities80

logp1

|s|at s = 0, (1 + s)

n−32 at s = −1, (1− s)

n−32 at s = 1

and all these singularities are integrable.

Lemma 6.34. If h : [0,∞)→ R is continuous, bounded and the improper integral∫ ∞1

h(s)

sds

converges, then for µ > λ > 0 and N > ε > 0 we have

(6.31)

∣∣∣∣∫ N

ε

h(λs)− h(µs)

sds

∣∣∣∣ ≤ 2‖h‖∞ log(µλ

).

Moreover

(6.32) limN→∞ε→0

∫ N

ε

h(λs)− h(µs)

sds = h(0) log

(µλ

).

Proof. We have∫ N

ε

h(λs)− h(µs)

sds =

∫ λN

λε

h(s)

sds−

∫ µN

µε

h(s)

sds

=

∫ µε

λε

h(s)

sds−

∫ µN

λN

h(s)

sds .

Estimating the absolute value of the last two integrals gives (6.31). Since∫ µε

λε

h(s)

sds→ h(0) log

(µλ

)as ε→ 0

and ∫ µN

λN

h(s)

sds→ 0 as N →∞

(6.32) follows. 2

Corollary 6.35. For a 6= 0 we have

limN→∞ε→0

∫ N

ε

e−isa − cos s

sds = log

1

|a|− iπ

2sgn a

and ∣∣∣∣∫ N

ε

e−isa − cos s

sds

∣∣∣∣ ≤ 2

∣∣∣∣log1

|a|

∣∣∣∣+ C,

where C is some positive constant independent of a, ε and N .

79Since |ξ′ · θ| ≤ 1, log 1|ξ′·θ| ≥ 0 and we do not have to take the absolute value of the log term.

80The last two are singularities only when n = 2.


Proof. We have∫ N

ε

e−isa − cos s

sds =

∫ N

ε

cos(sa)− cos s

sds− i

∫ N

ε

sin(sa)

sds

=

∫ N

ε

cos(s|a|)− cos s

s− i sgn (a)

∫ N

ε

sin(s|a|)s

ds

=

∫ N

ε

cos(s|a|)− cos s

s− i sgn (a)

∫ N |a|

ε|a|

sin s

sds

and the result follows from Lemma 6.34.

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

(1) We will prove that the two integrals at (6.27) are equal.(2) Define

F (ξ) =

∫Sn−1

Ω(θ)

(log

1

|ξ′ · θ|− iπ

2sgn (ξ′ · θ)

)dσ(θ) ,

Clearly F is homogeneous of degree zero. We will prove that

‖F‖L1(Sn−1) ≤ C‖Ω‖L1(Sn−1),

∫Sn−1

F (ξ) dσ(ξ) = 0.

Hence according to Lemma 6.31, F is a tempered L1 function so F ∈ S ′n and

|F [ϕ]| ≤ C‖Ω‖L1(Sn−1)

(supξ∈Rn

(1 + |ξ|n+1)|ϕ(ξ)|).

(3) Since ξ 7→ log 1|ξ′·θ| is even and ξ 7→ sgn (ξ′ · θ) is odd, it follows that if Ω is odd or

even, then

F (ξ) = −iπ2

∫Sn−1

Ω(θ) sgn (ξ′ · θ) dσ(θ) or F (ξ) =

∫Sn−1

Ω(θ)

(log

1

|ξ′ · θ|

)dσ(θ)

are odd and even respectively.

(4) Finally we will prove that WΩ(ξ) = F (ξ).

First observe that the equality of the two integrals in (6.27) follows from

log1

|ξ · θ|= log

1

|ξ|+ log

1

|ξ′ · θ|and

∫Sn−1

Ω(θ) log1

|ξ|dσ(θ) = 0 .

This proves (1).

Let

F (ξ) =

∫Sn−1

Ω(θ)

(log

1

|ξ′ · θ|− iπ

2sgn (ξ′ · θ)

)dσ(θ)

be the function defined by the integral in the second line of (6.27).

It is clear that F (ξ) is homogeneous of degree zero. We will show that it is integrableon the unit sphere and

‖F‖L1(Sn−1) ≤ C‖Ω‖L1(Sn−1).

102 PIOTR HAJ LASZ

Note that

ξ 7→∫Sn−1

Ω(θ)iπ

2sgn (ξ′ · θ) dσ(θ)

is a bounded by π2‖Ω‖L1(Sn−1), so this component of F (ξ) does not cause any troubles and

hence we only need to estimate

G(ξ) =

∫Sn−1

Ω(θ) log1

|ξ′ · θ|dσ(θ) .

Lemma 6.33 with p = 1 yields

‖G‖L1(Sn−1) ≤∫Sn−1

∣∣∣∣∫Sn−1

Ω(θ) log1

|ξ′ · θ|dσ(θ)

∣∣∣∣ dσ(ξ′)

≤∫Sn−1

|Ω(θ)|(∫

Sn−1

log1

|ξ′ · θ|dσ(ξ′)

)dσ(θ) = C(n)‖Ω‖L1(Sn−1).

It easily follow now from the Fubini theorem that∫Sn−1 F (ξ) dσ(ξ) = 0. That completes

the proof of (2).

Since F ∈ L1(Sn−1), the step (3) is obvious.

To prove (4) we need to show that WΩ = F in the sense of distributions i.e.,

WΩ[ϕ] =

∫Rnϕ(ξ)

∫Sn−1

Ω(θ)

(log

1

|ξ′ · θ|− iπ

2sgn (ξ′ · θ)

)dσ(θ) dξ.

We have

WΩ[ϕ] = WΩ[ϕ]

= limε→0N→∞

∫ε≤|x|≤N

Ω(x/|x|)|x|n

ϕ(x) dx

= limε→0N→∞

∫Rnϕ(ξ)

∫ε≤|x|≤N

Ω(x/|x|)|x|n

e−2πix·ξ dx dξ

= limε→0N→∞

∫Rnϕ(ξ)

∫Sn−1

Ω(θ)

∫ N

ε

e−2πisθ·ξ ds

sdσ(θ) dξ

= limε→0N→∞

∫Rnϕ(ξ)

∫Sn−1

Ω(θ)

∫ N

ε

(e−2πisθ·ξ − cos(2πs|ξ|)

) dssdσ(θ) dξ

= ♦ .The last equality follows from the fact that the integral of Ω over the sphere vanishes. Wehave ∫ N

ε


) dss

=

∫ 2π|ξ|N

2π|ξ|ε

e−isθ·ξ′ − cos s

sds

−→ log1

|θ · ξ′|− iπ

2sgn (θ · ξ′)

by Corollary 6.35. Also Corollary 6.35 and integrability of G easily imply that∣∣∣∣ϕ(ξ)Ω(θ)

∫ N

ε


) dss

∣∣∣∣


≤ C|ϕ(ξ)| |Ω(θ)|(

1 + log1

|ξ′ · θ|

)∈ L1(Rn × Sn−1)

so the Dominated Convergence Theorem gives

♦ =

∫Rnϕ(ξ)

∫Sn−1

Ω(θ)

(log

1

|θ · ξ′|− iπ

2sgn (θ · ξ′)

)dσ(θ) dξ .

The proof is complete. 2

If Ω is an odd function, then

WΩ(ξ) = −iπ2

∫Sn−1

Ω(θ) sgn (ξ′ · θ) dσ(θ) .

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

can be decomposed into its even and odd parts

Ωe(θ) =1

2(Ω(θ) + Ω(−θ)), Ωo(θ) =

1

2(Ω(θ)− Ω(−θ)) .

Corollary 6.36. Let Ω ∈ L1(Sn−1) be such that∫Sn−1

Ω(θ) dσ(θ) = 0 .

If Ωo ∈ L1(Sn−1) and Ωe ∈ Lq(Sn−1) for some q > 1, then the Fourier transform of WΩ

is a bounded function. In particular the operator TΩϕ = WΩ ∗ ϕ for ϕ ∈ Sn extends to abounded operator in L2.

Proof. Lemma 6.33 implies that log(1/|ξ′ · θ|) belongs to Lq′(Sn−1) and hence∣∣∣∣∫

Sn−1

Ωe(θ)

(log

1

|ξ′ · θ|

)dσ(θ)

∣∣∣∣ ≤ C‖Ωe‖Lq(Sn−1) ,

with a constant C independent of ξ. Now the formulas (6.28) and (6.29) yield

WΩ(ξ) = −iπ2

∫Sn−1

Ωo(θ) sgn (ξ′ · θ) dσ(θ) +

∫Sn−1

Ωe(θ)

(log

1

|ξ′ · θ|

)dσ(θ)

from which we have

‖WΩ‖∞ ≤ C(‖Ωo‖L1(Sn−1) + ‖Ωe‖Lq(Sn−1)) .

Since TΩϕ = (ϕWΩ)∨ and multiplication by a bounded function constitutes a boundedoperator in L2 it follows that

‖TΩϕ‖2 ≤ ‖WΩ‖∞‖ϕ‖2 for ϕ ∈ Sn.

It is natural to expect that if Ω in Theorem 6.32 is smooth, then WΩ is smooth awayfrom the origin. As we will see this it true. The next result is quite surprising since it gives

a complete characterization of such Fourier transforms WΩ.

104 PIOTR HAJ LASZ

Theorem 6.37. If Ω ∈ C∞(Sn−1) satisfies

(6.33)

∫Sn−1

Ω(θ) dσ(θ) = 0,

then

(6.34) m = WΩ ∈ C∞(Rn \ 0)is homogeneous of degree zero and

(6.35)

∫Sn−1

m(ξ) dσ(ξ) = 0.

Conversely if m ∈ C∞(Rn \0) is homogeneous of degree zero and it satisfies (6.35), then

there is Ω ∈ C∞(Sn−1) satisfying (6.33) such that m = WΩ.

Lemma 6.38. If Ω ∈ C∞(Sn−1) and

(6.36)

∫ 1

−1

|K(s)|(1− s2)n−32 ds <∞

then m ∈ C∞(Sn−1) where

m(ξ) =

∫Sn−1

Ω(θ)K(ξ · θ) dσ(θ).

Remark 6.39. We do not assume here that the integral of Ω vanishes.

Remark 6.40. According to Lemma 6.30 inequality (6.36) implies that for every ξ ∈ Sn−1∫Sn−1

|K(ξ · θ)| dσ(θ) = (n− 1)ωn−1

∫ 1

−1

|K(s)|(1− s2)n−32 ds <∞.

Proof. Observe that it suffices to prove that m is smooth in a neighborhood of e1. Indeed,if ξ0 ∈ Sn−1 and ρ(e1) = ξ0 for some ρ ∈ O(n), then

m(ξ) := (m ρ)(ξ) =

∫Sn−1

Ω(θ)K(ρ(ξ) · θ) dσ(θ)

=

∫Sn−1

Ω(θ)K(ξ · ρ−1(θ)) dσ(θ) =

∫Sn−1

Ω(ρ(θ))K(ξ · θ) dσ(θ)

and Ω ρ ∈ C∞(Sn−1). Thus smoothness of m is a neighborhood of e1 will imply smooth-ness of m in a neighborhood of ξ0, because m = m ρ−1 and ρ−1 : Sn−1 → Sn−1 is adiffeomorphism that maps a neighborhood of ξ0 onto a neighborhood of e1, where m issmooth.

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

If ξ is in the right hemisphere generated by e1, we can represent it in a local coordinatesystem

ξ = q(ξ2, . . . , ξn) =

√√√√1−n∑j=2

ξ2j , ξ2, . . . , ξn

, (ξ2, . . . , ξn) ∈ Bn−1(0, 1).

It suffices to prove that (m q)(ξ2, . . . , ξn) is smooth in Bn−1(0, 1).


Given ξ = q(ξ2, . . . , ξn) let

ρ(ξ2, . . . , ξn) ∈ SO(n), ρ(ξ2, . . . , ξn)(ξ) = e1

be a rotation that rotates ξ to e1 in the plane span e1, ξ and fixes the orthogonal com-plement of that plane. Such a rotation is unique and it is easy to see that components ofthe matrix ρ(ξ2, . . . , ξn) smoothly depend in (ξ2, . . . , ξn). By an argument similar to theone used at the beginning of the proof

(m q)(ξ2, . . . , ξn) =

∫Sn−1

Ω(θ)K(ξ · θ) dσ(θ) =

∫Sn−1

(Ω ρ−1(ξ2, . . . , ξn))(θ)K(θ1) dσ(θ).

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

Proof of Theorem 6.37. Let Ω ∈ C∞(Sn−1) satisfy (6.33). For ξ 6= 0,

m(ξ) = WΩ(ξ) =

∫Sn−1

Ω(θ)K(ξ · θ) dσ(θ)

where

K(s) = log1

|s|− iπ

2sgn (s)

and we already proved in (6.30) that∫ 1

−1

|K(s)|(1− s2)n−32 ds <∞.

Thus m ∈ C∞(Sn−1) by Lemma 6.38 and since m is homogeneous of degree zero, m ∈C∞(Rn \0). The fact that the integral of m over the sphere vanishes was already provedin Theorem 6.32, see (6.25).

Suppose now that m ∈ C∞(Rn \ 0) is homogeneous of degree zero and it satisfies(6.35). Since m is a tempered distribution, m is a tempered distribution too and(

∂nm

∂xnj

)∧(ξ) = (2πiξj)

nm(ξ).

The function ∂nm/∂xnj is homogeneous of degree −n. Since

∂nm

∂xnj=

∂

∂xj

(∂n−1m

∂xn−1j

)it is the derivative of a function that is homogeneous of degree 1− n, Theorem 5.57 yield∫

Sn−1

∂nm(θ)

∂xnjdσ(θ) = 0 and

∂nm

∂xnj= cδ0 + p.v.

∂nm

∂xnj.

81Perhaps it will be easier to see it if we rewrite the integral in a way that it does not involve integrationof functions on manifolds (sphere). Let F ∈ C∞(Rn) be a smooth extension of Ω from Sn−1 to Rn. Then

(m q)(ξ2, . . . , ξn) =

∫Rn

F (ρ−1(ξ2, . . . , ξn)(x))dµ(x)

where µ is a measure concentrated on the sphere Sn−1. Namely µ is the extension of K(θ1)dσ(θ) fromSn−1 to Rn by zero, i.e., µ(E) =

∫Sn−1∩E K(θ1) dσ(θ).

106 PIOTR HAJ LASZ

Taking the Fourier transform yields

(2πi)nξnj m(ξ) = c+

(p.v.

∂nm

∂xnj

)∧(ξ).

Note that the function on the right hand side is of class C∞(Rn \ 0), homogeneous ofdegree zero (by the first part of the proof).

The above equality yields

(6.37) (2πi)nn∑j=1

(ξ2nj )m(ξ) =

n∑j=1

ξnj

(c+

(p.v.

∂nm

∂xnj

)∧(ξ)

).

We claim that the distribution m coincides in Rn \ 0 with the function that is smoothin Rn \ 0, homogeneous of degree −n

(6.38) m(ξ) =

∑nj=1 ξ

nj

(c+

(p.v. ∂

nm∂xnj

)∧(ξ)

)(2πi)n

∑nj=1(ξ2n

j )in Rn \ 0.

That means, if ϕ ∈ Sn with suppϕ ⊂ Rn \ 0, then

(6.39) m[ϕ] =

∫Rn

∑nj=1 ξ

nj

(c+

(p.v. ∂

nm∂xnj

)∧(ξ)

)(2πi)n

∑nj=1(ξ2n

j )ϕ(x) dx.

To see this observe that since ϕ vanishes in a neighborhood of 0, the function∑n

j=1 ξ2nj is

positive on the support of ϕ so

ψ =ϕ

(2πi)n∑n

j=1(ξ2nj )∈ Sn

is in the Schwarz class.82 Now it is easy to see that if we evaluate both sides of thedistributional equality (6.37) on ψ we obtain equality (6.39).

We proved that in Rn \0, m coincides with a smooth function, homogeneous of degree−n which is given by the formula on the right hand side of (6.38). We still do not knowwhat happens at 0. There is a possibility that m has a part supported at 0.

Let Ω(ξ) = m(ξ) for ξ ∈ Sn−1. Thus

m(ξ) =Ω(ξ/|ξ|)|ξ|n

for ξ 6= 0.

We claim that

(6.40)

∫Sn−1

Ω(θ) dσ(θ) = 0,

Indeed, let ϕ ∈ Sn be a radial function supported in the annulus 1 ≤ |x| ≤ 2 and positivein the interior of the annulus. Then the integration in the spherial coordinates gives

(6.41) m[ϕ] =

∫Rn

Ω(x/|x|)|x|n

ϕ(x) dx = C

∫Sn−1

Ω(θ) dσ(θ), where C > 0.

82Why?


On the other hand the Fourier transform commutes with orthogonal transformations andhence ϕ(ξ) is also a radial function ϕ(ξ) = f(|ξ|) so

(6.42) m[ϕ] = m[ϕ] =

∫Rnm(ξ)ϕ(ξ) dξ =

∫Sn−1

m(θ)dσ(θ)

∫ ∞0

sn−1f(s) ds = 0.

Now (6.40) follows from the equality between (6.41) and (6.42).

We proved that m(x) coincides with Ω(x/|x|)/|x|n in Rn \ 0. Hence the distribution

(6.43) m− p.v.Ω(x/|x|)|x|n

is supported at the origin. According to Corollary 5.34 its Fourier transform is a polynomial.However, the Fourier transform of (6.43) is a bounded function so the polynomial mustbe constant. The constant must be zero because m and (p.v. Ω(x/|x|)/|x|n)∧ have the zerointegrals over Sn−1. Hence

m = p.v.Ω(x/|x|)|x|n

so m(ξ) =

(p.v.

Ω(x/|x|)|x|n

)∧(ξ)

where Ω(θ) = Ω(−θ).

6.6. Algebra of singular integrals. In this section we will investigate compositions ofsingular integrals

TΩf(x) = WΩ ∗ f(x) = limε→0

∫|y|≥ε

Ω(y/|y|)|y|n

f(x− y) dy where

∫Sn−1

Ω(θ) dσ(θ) = 0.

We will assume that Ω ∈ C∞(Sn−1).

In general we cannot expect that the composition of singular integrals is a singularintegral. This can be seen in the following example

Example 6.41. There is a function Ω ∈ C∞(Sn−1) with the vanishing integral such thatthat the square of a Riesz transform satisfies83

(6.44) R2j = − 1

nI + TΩ

Indeed, ((R2j +

1

nI

)[ϕ]

)∧=

(1

n−

ξ2j

|ξ|2

)ϕ(ξ) = m(ξ)ϕ(ξ)

and ∫Sn−1

m(ξ) dσ(ξ) =

∫Sn−1

(1

n− ξ2

j

)dσ(ξ) =

1

n

∫Sn−1

(1−

n∑k=1

ξ2k

)dσ(ξ) = 0

so the existence of Ω satisfying (6.44) follows from Theorem 6.37.

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

83I stands for the identity.

108 PIOTR HAJ LASZ

Theorem 6.42. If m ∈ C∞(Rn \ 0) is homogeneous of degree zero, then there is a ∈ Cand Ω ∈ C∞(Sn−1) with vanishing integral such that

Tϕ := (mϕ)∨ = aϕ+ TΩϕ for ϕ ∈ Sn.

Proof. Let a ∈ C be such that∫Sn−1(m(θ)− a) dσ(θ) = 0. Then by Theorem 6.37 there is

Ω such that Tf − af = TΩ.

Corollary 6.43. The class A of operators of the form

aI + TΩ

where a ∈ C and Ω ∈ C∞(Sn−1) has vanishing integral coincides with the class of operatorsof the form

Tϕ = (mϕ)∨,

where m ∈ C∞(Rn\0) is homogeneous of degree zero. Hence A is a commutative algebra.

An operator in the class A is invertible if and only if m = a+ WΩ does not vanish at anypoint of Sn−1.


7. Riesz potentials and fractional powers of the Laplacian

Recall that for 0 < α < n the Riesz potantial Iα is defined by84

(Iαf)(x) = (Uα ∗ f)(x) =1

γ(α)

∫Rn

f(y)

|x− y|n−αdy ,

where

γ(α) =πn2 2α Γ

(α2

)Γ(n−α

2

) so Uα = (2π)−απα−

n2 Γ(n−α

2

)Γ(α2

) |x|α−n .

Clearly, Iαϕ is smooth, slowly increasing, and all its derivatives are slowly increasing whenϕ ∈ Sn.85

A relationship between Riesz potentials and Riesz transforms is explained in the nextresult

Proposition 7.1. If n ≥ 2 and ϕ ∈ Sn, then for 1 ≤ j ≤ n

∂

∂xj(I1ϕ)(x) = −Rjϕ(x) .

Proof. This result easily follows from Proposition 5.53

∂

∂xj(I1ϕ) =

(∂U2

∂xj

)∗ ϕ =

Γ(n−1

2

)2π

n+12

(∂

∂xj|x|1−n

)∗ ϕ

= (1− n)Γ(n−1

2

)2π

n+12

(p.v.

xj|x|n+1

)∗ ϕ = −Rjϕ.

According to Theorem 5.42, Uα(ξ) = (2π)−α|ξ|−α = (4π2|ξ|2)−α/2. Since (Iαϕ)∧ = ϕUαwe get

Theorem 7.2. For 0 < α < n and ϕ ∈ Sn,86

(7.1) (Iαϕ)∧(ξ) = (4π2|ξ|2)−α/2ϕ(ξ),

84c.f. Definition 2.14.85However, in general Iαϕ 6∈ Sn, see Remark 7.5.86The Fourier transform in (7.1) is understood as a Fourier transform of a tempered distribution Uα ∗ϕ.

On the other hand (4π2|ξ|2)−α/2ϕ ∈ L1 so the inverse Fourier transform in (7.2) is just a regular inverseFourier transform of an integrable function.

110 PIOTR HAJ LASZ

i.e.,

(7.2) Iαϕ =((4π2|ξ|2)−α/2ϕ(ξ)

)∨.

Recall that when n ≥ 3 the fundamental solution of the Laplace operator is87

Φ(x) = − 1

n(n− 2)ωn

1

|x|n−2= −U2(x).

That means

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

Hence in some sense the Riesz potential I2 is the inverse of −∆ so it is reasonable to usenotation

(−∆)−1ϕ = I2ϕ =((4π2|ξ|2)−1ϕ(ξ)

)∨.

With this notation (7.3) reads as

(−∆)(−∆)−1ϕ = (−∆)−1(−∆)ϕ = ϕ.

Also for ϕ ∈ Sn and a positive integer k we have

−∆ϕ =(4π2|ξ|2ϕ(ξ)

)∨so (−∆)kϕ =

((4π2|ξ|2)kϕ(ξ)

)∨.

This suggests how to define fractional powers of the Laplace operator, including powerswith negative exponents

Definition 7.3. For α > −n and ϕ ∈ Sn we define

(−∆)α/2ϕ =((4π2|ξ|2)α/2ϕ(ξ)

)∨.

Thus

Iαϕ =((4π2|ξ|2)−α/2ϕ(ξ)

)∨= (−∆)−α/2ϕ for 0 < α < n.

Proposition 7.4. For any α > −n there is u ∈ S ′n such that (−∆)α/2ϕ = u ∗ ϕ for

ϕ ∈ Sn. In particular (−∆)α/2ϕ is smooth, slowly increasing, and all its derivatives areslowly increasing.

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

v(ξ) = v(ξ) = (4π2|ξ|2)α/2 ∈ S ′n and u = v so u = v = v.

Then

(u ∗ ϕ)∧ = ϕu = ϕv = (4π2|ξ|2)α/2ϕ(ξ) =((−∆)α/2ϕ

)∧(ξ).

87Theorem 5.60.88Since α > −n, the singularity of v at ξ = 0 is integrable and hence the function v defines a tempered

distribution.


Remark 7.5. Although (−∆)α/2ϕ is always smooth, it is not always in the Schwarz classSn. Namely for every α 6∈ 0, 2, 4, 6, . . ., α > −n, there is ϕ ∈ Sn such that (−∆)α/2ϕ 6∈Sn. Suppose to the contrary that for some α 6∈ 0, 2, 4, 6, . . ., α > −n, and all ϕ ∈ Sn wehave (−∆)α/2ϕ ∈ Sn. Then the Fourier transform of this function also belongs to Sn i.e.,

(7.4) (4π2|ξ|2)α/2ϕ(ξ) ∈ Sn.

However |ξ|α is not C∞ smooth at ξ = 0 so if ϕ(0) 6= 0, the function (7.4) is not C∞

smooth at ξ = 0 and hence it cannot belong to Sn.

On the other hand if α = 2k ∈ 0, 2, 4, 6, . . ., then

(−∆)α/2 = (−∆)k : Sn → Sn

since the Laplace operator maps Sn to Sn and so does the k-fold composition of (−∆).

Proposition 7.6. If α > −n, then for ϕ ∈ Sn we have

(−∆)(−∆)α/2ϕ = (−∆)α/2(−∆)ϕ = (−∆)α+22 ϕ.

Remark 7.7. We need to understand this result as follows. The function (−∆)α/2ϕ issmooth so we can apply the classical Laplace operator −∆ to it. Also since (−∆)α/2ϕ isslowly increasing and all its derivatives are slowly increasing, the classical Laplace operatorcoincides with the distributional one (since there is no problem with the integration byparts). This is how we understand the left hand side. In the middle term (−∆)ϕ ∈ Sn sowe can apply (−∆)α/2 to it. And finally the the term on the right hand side is well definedbecause ϕ ∈ Sn.

Proof. According to Proposition 7.4, (−∆)α/2ϕ = u ∗ ϕ for some u ∈ S ′n. Hence

(−∆)((−∆)α/2ϕ) = −∆(u ∗ ϕ) = u ∗ (− ∆ϕ︸︷︷︸in Sn

) = (−∆)α/2(−∆ϕ)

=(

(4π2|ξ|2)α/2(−∆ϕ))∨

=(

(4π2|ξ|2)α/2(4π2|ξ|2)ϕ)∨

= (−∆)α+22 ϕ.

Remark 7.8. Formally we can extend the above result to composition of operators(−∆)α/2(−∆)β/2 as follows(

(−∆)α/2(−∆)β/2ϕ))∧

= (4π2|ξ|2)α/2(4π2|ξ|2)β/2ϕ(ξ)

= (4π2|ξ|2)α+β2 ϕ(ξ) =

((−∆)

α+β2 ϕ)∧

so formally

(7.5) (−∆)α/2(−∆)β/2 = (−∆)α+β2 on Sn

provided α, β > −n and α + β > −n.

The problem is that in general if ϕ ∈ Sn, then (−∆)β/2ϕ 6∈ Sn so in the composition

(−∆)α/2((−∆)β/2ϕ

)we apply the operator (−∆)α/2 to a function which is not in Sn and Definition 7.3 requiresthe function to which we apply (−∆)α/2 to be in the space Sn. Well, the formula that

112 PIOTR HAJ LASZ

defines (−∆)α/2 actually applies to a much larger class of function than just Sn and withthis extension in mind we can justify the above composition formula. There is however, aprice we have to pay for it. If ϕ ∈ Sn, then89 (−∆)α/2ϕ = uα ∗ ϕ and (−∆)β/2ϕ = uβ ∗ ϕfor some uα, uβ ∈ S ′

n and we would like to write

(7.6) (−∆)α/2(−∆)β/2ϕ = uα ∗ (uβ ∗ ϕ)

and this does not make any sense if uβ ∗ ϕ = (−∆)β/2ϕ 6∈ Sn. An example where thiscreates a problem will be seen in the next result. A formal proof, similar to the argumentsused above will be very short and elegant, but the actual rigorous proof will be quite longand boring.

Theorem 7.9. If α, β > 0, α + β < n, then

(7.7) Iα(Iβϕ) = Iα+βϕ for ϕ ∈ Sn.

Remark 7.10. Formally we can write this equality as

(7.8) (−∆)−α/2(−∆)−β/2ϕ = (−∆)−α+β2 ϕ

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

Uα ∗ (Uβ ∗ ϕ) = Uα+β ∗ ϕ

If we would know that Iβϕ = Uβ ∗ϕ ∈ Sn we could apply the Fourier transform and easilyprove the identity, but in general, without this information, we have no reason to claim

that (Uα ∗ (Uβ ∗ ϕ))∧ = Uβ ∗ ϕ Uα.

Proof.

Lemma 7.11. If α, β > 0, α + β < n, then there is a constant90 C0 = C0(α, β, n) suchthat ∫

Rn

dy

|x− y|n−α|y|n−β=

C0

|x|n−(α+β).

Proof. First observe that the integral on the left hand side is finite whenever x 6= 0. To seethis let R = |x| and divide the integral over Rn into four integrals over mutually disjointregions∫

B(0,R2

)

... dy,

∫B(x,R

2)

... dy,

∫B(0,2R)\(B(0,R

2)∪B(x,R

2))

... dy,

∫Rn\B(0,2R)

... dy.

It is very easy to see that the first three integrals are finite. Indeed, in the first integralwe have a singularity |y|n−β at y = 0 only and this singularity is integrable. In the secondintegral we have a singularity |x− y|n−α at y = x only which is integrable again and in thethird integral we do not have singularities at all. It remains to prove finiteness of the lastintegral.

89Proposition 7.4.90In Corollary 7.12 we will find the exact value of C0.


If |y| ≥ 2R, then |x− y| ≥ |y| − |x| ≥ |y| − |y|/2 = |y|/2 so∫Rn\B(0,2R)

dy

|x− y|n−α|y|n−β≤ 2n−α

∫Rn\B(0,2R)

dy

|y|2n−(α+β)<∞,

because 2n− (α + β) > n.

Note that the integral from the lemma is invariant under rotations and hence∫Rn

dy

|x− y|n−α|y|n−β = f(|x|).

For x 6= 0 let x0 = x/|x| and t = |x| so x = tx0. A simple change of variables gives∫Rn

dy

|x− y|n−α|y|n−β=

∫Rn

dy

|tx0 − y|n−α|y|n−β

= tα+β−n∫Rn

dy

|x0 − y|n−α|y|n−β= |x|α+β−nf(1).

We have

Iα(Iβϕ) =1

γ(α)γ(β)

∫Rn

1

|x− y|n−α

(∫Rn

ϕ(z)

|y − z|n−βdz

)dy

=1

γ(α)γ(β)

∫Rn

(∫Rn

dy

|x− y|n−α|y − z|n−β

)ϕ(z) dz

=1

γ(α)γ(β)

∫Rn

(∫Rn

dy

|(x− z)− y|n−α|y|n−β

)ϕ(z) dz

=C0

γ(α)γ(β)

∫Rn

ϕ(z)

|x− z|n−(α+β)dz

=C0γ(α + β)

γ(α)γ(β)Iα+βϕ(x).

We could use the Fubini theorem here because the function that we integrated over Rn×Rn

was integrable. To see the integrability, repeat the above computations with ϕ replaced by|ϕ|.

It remains to show that

C0 =γ(α)γ(β)

γ(α + β).

To prove this it suffices to verify that Iα(Iβϕ) = Iα+βϕ for just one non-zero function ϕ.To this end let ϕ ∈ Sn be such that ϕ = 0 in a neighborhood of 0. Then

Iα(Iβϕ) = Iα

(((4π2|ξ|2)−β/2ϕ︸︷︷︸

∈Sn

)∨)=((4π2|ξ|2)−α/2(4π2|ξ|2)−β/2ϕ

)∨= Iα+βϕ.


As a corollary we obtain

114 PIOTR HAJ LASZ

Corollary 7.12. If α, β > 0, α + β < n, then∫Rn

dy

|x− y|n−α|y|n−β=γ(α)γ(β)

γ(α + β)

1

|x|n−(α+β).

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

Theorem 7.13. For 0 < α < 2 and ϕ ∈ Sn we have

(−∆)α/2ϕ = limε→0

1

γ(−α)

∫|x−y|≥ε

ϕ(y)− ϕ(x)

|y − x|n+αdy

=1

2γ(−α)

∫Rn

ϕ(x+ y) + ϕ(x− y)− 2ϕ(x)

|y|n+αdy,

where91

γ(−α) =πn2 2−αΓ

(−α

2

)Γ(n+α

2

) .

Proof. First we will prove equality of the integrals

(7.9)1

2

∫Rn

ϕ(x+ y) + ϕ(x− y)− 2ϕ(x)

|y|n+αdy = lim

ε→0

∫|x−y|≥ε

ϕ(y)− ϕ(x)

|y − x|n+αdy.

It easily follows from Taylor’s formula that

|ϕ(x+ y) + ϕ(x− y)− 2ϕ(x)||y|n+α

≤ C‖D2ϕ‖∞|y|n+α−2

,

where

‖D2ϕ‖∞ = supx∈Rn

(n∑

j,k=1

∣∣∣∣ ∂2ϕ

∂xj∂xk(x)

∣∣∣∣2)1/2

.

Since n + α − 2 < n, both sides of this inequality are integrable over the ball |y| ≤ 1.They are also integrable in |y| > 1 because of rapid decay of ϕ. Thus the first integralat (7.9) is well defined and finite. For the second integral we have∫

|x−y|≥ε

ϕ(y)− ϕ(x)

|y − x|n+αdy =

∫|y|≥ε

ϕ(x+ y)− ϕ(x)

|y|n+αdy =

∫|y|≥ε

ϕ(x− y)− ϕ(x)

|y|n+αdy

so ∫|x−y|≥ε

ϕ(y)− ϕ(x)

|y − x|n+αdy =

1

2

∫|y|≥ε

ϕ(x+ y) + ϕ(x− y)− 2ϕ(x)

|y|n+αdy

and hence passing to the limit as ε→ 0+ yields (7.9).

Observe that

(7.10) (x, y) 7→ ϕ(x+ y) + ϕ(x− y)− 2ϕ(x)

|y|n+α∈ L1(Rn × Rn).

91We use here Γ evaluated at a negative number. Actually, Γ(−α2 ) = − 2αΓ(1− α

2 ), where 1− α2 > 0, see

Definition 3.35. Note also that γ(−α) is given by the same formula as the constant in the definition of theRiesz potential.


Indeed, for (x, y) ∈ Rn ×B(0, 1) we have∣∣∣∣ϕ(x+ y) + ϕ(x− y)− 2ϕ(x)

|y|n+α

∣∣∣∣ ≤ C‖D2ϕ‖L∞(B(x,1))

|y|n+α−2≤ C

(1 + |x|)−2n

|y|n+α−2∈ L1(Rn ×B(0, 1))

and ∫|y|≥1

∫Rn

|ϕ(x+ y) + ϕ(x− y)− 2ϕ(x)||y|n+α

dx dy ≤ 3‖ϕ‖1

∫|y|≥1

dy

|y|n+α<∞.

This allows us to compute the Fourier transform of the integrable function

(7.11) f(x) =

∫Rn

ϕ(x+ y) + ϕ(x− y)− 2ϕ(x)

|y|n+αdy

by changing the order of integration. We have

f(ξ) =

∫Rne−2πix·ξ

∫Rn

ϕ(x+ y) + ϕ(x− y)− 2ϕ(x)

|y|n+αdy dx

=

∫Rn

1

|y|n+α

(∫Rne−2πix·ξ(ϕ(x+ y) + ϕ(x− y)− 2ϕ(x)) dx

)dy

= ϕ(ξ)

∫Rn

e2πiy·ξ + e−2πiy·ξ − 2

|y|n+αdy

= −2ϕ(ξ)

∫Rn

1− cos(2πy · ξ)|y|n+α

dy.

Let ρ ∈ O(n) be such that ρ(ξ) = |ξ|e1. Then the change of variables y = ρ(y) gives∫Rn

1− cos(2πy · ξ)|y|n+α

dy =

∫Rn

1− cos(2πρ−1(y) · ξ)|ρ−1(y)|n+α

dy =

∫Rn

1− cos(2πy · ρ(ξ))

|y|n+αdy

=

∫Rn

1− cos(2π|ξ|y1)

|y|n+αdy

= (2π|ξ|)α∫Rn

1− cos y1

|y|n+αdy

where in the last equality we used another change of variables y = 2π|ξ|y. Actually it isa consequence of the above calculation that the function (1 − cos y1)/|y|n+α is integrableon Rn, but one can see it more directly by using the estimate |1 − cos y1| ≤ C|y|2 in aneighborhood of the origin. Let

C(α) =

∫Rn

1− cos y1

|y|n+αdy.

We proved that

f(ξ) = −2ϕ(ξ)(2π|ξ|)αC(α) = −2C(α)((−∆)α/2ϕ

)∧(ξ).

Hence

(−∆)α/2ϕ(x) = − 1

2C(α)f(x) = − 1

2C(α)

∫Rn

ϕ(x+ y) + ϕ(x− y)− 2ϕ(x)

|y|n+αdy

= − 1

C(α)limε→0

∫|y|≥ε

ϕ(x+ y)− ϕ(x)

|y|n+αdy.

116 PIOTR HAJ LASZ

It remains to prove that

(7.12) C(α) = −πn2 2−αΓ

(−α

2

)Γ(n+α

2

) .

Let ϕ(x) = e−π|x|2. Since ϕ = ϕ we have

(7.13) (−∆)α/2ϕ(0) = − 1

C(α)limε→0

∫|y|≥ε

ϕ(0 + y)− ϕ(0)

|y|n+αdy = − 1

C(α)

∫Rn

e−π|y|2 − 1

|y|n+αdy.

Both sides of this equality are easy to compute. Recall that according to Lemma 5.44 forγ > −n we have ∫

Rne−π|x|

2 |x|γ dx = nωn

∫ ∞0

e−πs2

sn+γ−1 ds =Γ(n+γ

2

)πγ2 Γ(n2

) .Since for g ∈ L1, g(0) =

∫Rn g(x) dx, we obtain

(−∆)α/2ϕ(0) =((4π2|ξ|2)α/2ϕ(ξ)

)∨(0) =

∫Rn

(2π|ξ|)αe−π|ξ|2 dξ(7.14)

= (2π)αΓ(n+α

2

)πα2 Γ(n2

) =2απ

α2 Γ(n+α

2

)Γ(n2

) .

On the other hand∫Rn

e−π|y|2 − 1

|y|n+αdy = nωn

∫ ∞0

sn−1 e−πs2 − 1

sn+αds = nωn

∫ ∞0

(e−πs2 − 1)

(s−α

−α

)′ds

=nωnα

∫ ∞0

(e−πs2 − 1)′s−α ds = −2π

nωnα

∫ ∞0

e−πs2

s1−α ds

γ=2−α−n= −2π

αnωn

∫ ∞0

e−πs2

sn+γ−1 ds = −2π

α

Γ(n+γ

2

)πγ2 Γ(n2

)= π

α+n2

Γ(−α

2

)Γ(n2

) .This identity, (7.13) and (7.14) yield

2απα2 Γ(n+α

2

)Γ(n2

) = − 1

C(α)πα+n2

Γ(−α

2

)Γ(n2

)which easily implies (7.12).

Corollary 7.14. For 0 < α < 2∫Rn

1− cos y1

|y|n+αdy = −

πn2 2−αΓ

(−α

2

)Γ(n+α

2

) .

Corollary 7.15. If 0 < α < 2 and ϕ ∈ Sn, then (−∆)α/2ϕ ∈ L1(Rn).

Proof.

(−∆)α/2ϕ =1

2γ(−α)

∫Rn

ϕ(x+ y) + ϕ(x− y)− 2ϕ(x)

|y|n+αdy =

1

2γ(−α)f(x) ∈ L1

since we proved in (7.10) that the function f defined in (7.11) is integrable.


It follows from Proposition 7.6 that is 0 < α < 2 and k ≥ 0 is an integer, then

(−∆)α2

+kϕ = (−∆)α2 (−∆)kϕ = (−1)k(−∆)

α2 (∆kϕ)

and hence Theorem 7.13 can be generalized to the case of higher powers as follows

Corollary 7.16. If 0 < α < 2, k ≥ 0 is an integer and ϕ ∈ Sn, then

(−∆)α2

+kϕ = limε→0

(−1)k

γ(−α)

∫|x−y|≥ε

∆kϕ(y)−∆kϕ(x)

|y − x|n+αdy

=(−1)k

2γ(−α)

∫Rn

∆kϕ(x+ y) + ∆kϕ(x− y)− 2∆kϕ(x)

|y|n+αdy,

where

γ(−α) =πn2 2−αΓ

(−α

2

)Γ(n+α

2

) .

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