HARMONIC ANALYSIS
HUIQIANG JIANG
1. Test functions and distributions
Let Ω be a nonempty open set in Rn . The test function space D (Ω) = C0∞ (Ω)
consists of all infinitely differentiable functions with compact support.
For every cpt K ⊂ Ω, we use DK (Ω) ⊂ D (Ω) to define the collection of all
infinitely differentiable functions with compact support in K. We can introduce a
metric in DK (Ω) such that for any f, g ∈ DK (Ω),
dK (f, g) = max
0≤N <∞
where
1 kf − gkN
2N 1 + kf − gkN
kf kN = max kDα f kL∞ , N ≥ 0.
|α|≤N
We use τK to denote the topology of DK (Ω) equipped with such metric.
The topology of D (Ω) can be defined precisely. Let β be the collection of all
convex balanced sets W ⊂ D (Ω) such that DK (Ω) ∩ W ∈ τK for every compact
K ⊂ Ω. Let τ be the collection of all unions of sets of the form φ+W with φ ∈ D (Ω)
and W ∈ β.
Theorem 1. τ is a topology in D (Ω) and β is a local base for τ . The topology τ
makes D (Ω) into a locally convex topological vector space.
Many important properties of D (Ω) are included in the following theorem.
Theorem 2. (a) A convex balanced subset V of D (Ω) is open iff V ∈ β.
(b) The topology τK of any DK (Ω) coincides with the subspace topology that DK
inherits from D (Ω).
(c) If E is bounded subset of D (Ω), then E ⊂ DK (Ω) for some compact K ⊂ Ω
and for each N ≥ 0, there exists MN < ∞ s.t.,
kφkN ≤ MN for any φ ∈ E.
(d) D (Ω) has the Heine-Borel property.
(e) If {φi } is a Cauchy sequence in D (Ω), then {φi } ⊂ DK (Ω) for some compact
K ⊂ Ω and
lim kφi − φj kN = 0 for any N ≥ 0.
i,j→∞
(f ) If φi → 0, then {φi } ⊂ DK (Ω) for some compact K ⊂ Ω and Dα φi → 0
uniformly for every multi-index α.
(g) In D (Ω), every Cauchy sequence converges.
Theorem 3. Every differential operator Dα : D (Ω) → D (Ω) is continuous.
Definition 1. A continuous linear functional on D (Ω) is called a distribution. The
space of all distributions in Ω is denoted by D0 (Ω).
1
2
HUIQIANG JIANG
Theorem 4. If Λ is a linear functional on D (Ω), the following two conditions are
equivalent:
(a) Λ ∈ D0 (Ω).
(b) For every compact K ⊂ Ω, there exists N ≥ 0 and C > 0 s.t.
|Λφ| ≤ C kφkN
holds for every φ ∈ DK (Ω).
Remark 1. If Λ is such that one N will do for all K, then the smallest such N is
called the order of Λ.
Example 1. δx is a distribution of order 0.
Example 2. Every locally integrable complex function is a distribution of order 0.
Example 3. Any locally finite Borel measure is a distribution of order 0.
The space of distribution D (Ω) is a topological vector space with weak-* topology, and it has many other nice structures.
Definition 2. If α is a multi-index and Λ ∈ D0 (Ω), Dα Λ ∈ D0 (Ω) is a linear
functional defined by the formula
|α|
(Dα Λ) (φ) = (−1)
Λ (Dα φ) , φ ∈ D (Ω) .
Proposition 1. For any multi-indices α, β and for any Λ ∈ D0 (Ω),
Dβ Dα Λ = Dα Dβ Λ = Dα+β Λ.
Remark 2. If f has continuous partial derivatives of all orders up to N , then
Dα Λf = ΛDα f for any |α| ≤ N.
Definition 3. Suppose Λ ∈ D0 (Ω) and f ∈ C ∞ (Ω). We can define a distribution
f Λ such that
(f Λ) (φ) = Λ (f φ) , φ ∈ D (Ω) .
Locally, distributions are derivatives of distributions induced by continuous functions:
Theorem 5. Suppose Λ ∈ D0 (Ω) and K is a compact subset of Ω. Then there
is a continuous function f in Ω and there is a multi-index α such that for every
φ ∈ DK (Ω),
Z
|α|
Λφ = (−1)
f (x) Dα φ (x) dx.
Ω
i.e.,
Λ = Dα f on K.
These results on distributions are proved in Analysis 3, please check Chapter 6
of Rudin’s Functional Analysis for details.
HARMONIC ANALYSIS
3
2. Fourier Series
We consider complex valued periodic functions on R with period 2π which can
1
be viewed as functions defined on the the unit circle
¡ S¢ .
The Fourier Coefficients of the function f ∈ L1 S1 ≡ L1 [−π, π] are defined by
Z π
1
an =
f (x) e−inx dx.
2π −π
Formally, the Fourier Series of f is
∞
X
f (x) ∼
an einx .
n=−∞
Since f ∈ L1 [−π, π], we have
1
|an | ≤
2π
Z
π
|f (x)| dx =
−π
1
kf kL1 .
2π
Moreover, we have the Riemann-Lebesgue lemma:
Theorem 6. For every f ∈ L1 [−π, π],
lim an = 0.
n→±∞
¡ ¢
If f ∈ C k S1 , i.e., f ∈ C k (R) is periodic with period 2π, then integration by
parts yields for any n 6= 0,
° k °
° k °
1 °
d f°
1 °
d f°
°
°
° .
|an | ≤
≤ k°
°
°
°
k
k
2πn dx L1
n dxk °L∞
¡ ¢
In particular, if f ∈ C 2 S1 , Weierstrass’s M-test implies that the Fourier Series of
f
∞
X
an einx
n=−∞
¡ ¢
converges uniformly to a continuous function g ∈ C S1 . We will show that g = f
later.
To study the convergence of Fourier series, we define the partial sums
sN (f, x) =
N
X
an einx ,
n=−N
and the Fejer sums
N
1 X
sn (f, x) .
N + 1 n=0
SN (f, x) =
Direct calculations yield
Z
π
sN (f, x) =
f (y) kN (x − y) dy
−π
where
¡
¢
1 sin N + 12 z
kN (z) =
2π
sin z2
and
Z
π
SN (f, x) =
f (y) KN (x − y) dy
−π
4
HUIQIANG JIANG
where
"
¡
¢ #2
sin N + 12 z
1
KN (z) =
.
π (N + 1)
sin z2
If we take f ≡ 1, then for each N ≥ 0, sN (f, x) ≡ SN (f, x) ≡ 1 and hence
Z π
Z π
kN (x) dx =
KN (x) dx = 1.
−π
−π
We observe also that KN (x) ≥ 0 and for each 0 < δ < π, KN converges to 0
uniformly on δ ≤ |x| ≤ π as N → ∞ which implies
Z
KN (x) dx = 0.
lim
N →∞
δ≤|x|≤π
¡ ¢
Theorem 7. For any f ∈ C S1 , the Fejer sum SN (f, x) converge to f uniformly
on S1 .
We leave the proof as an exercise.
Theorem 8. For any f ∈ Lp [−π, π], p ≥ 1, the Fejer sum SN (f, x) converges to
f in Lp [−π, π].
Proof. Young’s inequality implies
1
kSN (f, x)kLp ≤ 2 p kf kLp .
¡ ¢
Since C S1 is dense in Lp [−π, π], the result follows from 7.
¤
¡
¢
Corollary 1. For any f ∈ C 2 S1 , the Fourier series of f converges to f uniformly.
Proof. Since
1
kf 00 kL∞ ,
n2
sN (f, x) converges uniformly, since SN (f, x) converges uniformly to f , we conclude
sN (f, x) converges uniformly to f .
¤
¡
¢
Theorem 9. If f ∈ C α S1 for some α > 0, then for any x,
|an | ≤
lim sN (f, x) = f (x) .
N →∞
Proof.
1
sN (f, x) − f (x) =
2π
Z
π
−π
µ
¶
1
(f (x − y) − f (x))
sin N +
ydy.
sin y2
2
Since
(f (x − y) − f (x))
sin y2
is integrable, the result follows from Riemann-Lebesgue lemma.
¤
¡
¢
Remark 3. Actually, the result can be strengthened to: If f ∈ C α S1 for some
α > 0, then sN (f, x) converges to f uniformly.
¡ ¢
Remark 4. There exists f ∈ C S1 such that sN (f, x) doesn’t converge to f
pointwisely.
HARMONIC ANALYSIS
5
¡ ¢
Theorem 10. If f ∈ BV S1 , then for any x,
lim sN (f, x) =
N →∞
Proof. We first consider
1
λ→∞ π
Z
f (x+) + f (x−)
.
2
π
lim
f (−y)
−π
Let
Z
∞
G (y) =
y
sin λy
dy.
y
sin y
dy.
y
Then we have
Z
1 π
sin λy
f (−y)
dy
π 0
y
Z
1 λπ ³ y ´ sin y
=
f −
dy
π 0
λ
y
Z
1 λπ ³ y ´
=−
f −
dG (y)
π 0
λ
³ y ´¯λπ 1 Z λπ
³ y´
¯
= −G (y) f − ¯ +
G (y) df −
λ 0
π 0
λ
Z
1 π
= −G (λπ) f (−π+) + G (0) f (0−) +
G (λy) df (−y) ,
π 0
hence
1
lim
λ→∞ π
Similarly
1
λ→∞ π
Z
π
f (−y)
sin λy
1
dy = f (0−) .
y
2
f (−y)
sin λy
1
dy = f (0+) .
y
2
0
Z
0
lim
−π
So we have
Z
sin λy
f (0+) + f (0−)
1 π
f (−y)
dy =
.
lim
λ→∞ π 0
y
2
Finally, we have from Riemann-Lebesgue lemma
µ
¶
Z
1 π
sin λy
sin λy
lim
f (−y)
−
dy = 0.
λ→∞ π 0
y
2 sin y2
¤
¡ ¢
¡ ¢
Remark 5. If f ∈ BV S1 ∩ C S1 , then sN (f, x) converges to f uniformly.
Now we consider f ∈ Lp [−π, π], p ≥ 1. When p = 2, we have the well-known
result
Theorem 11. Let f ∈ L2 [−π, π], then sN (f ) → f in L2 [−π, π] as N → ∞.
Moreover,
∞
X
2
2
|an | .
kf kL2 =
n=−∞
6
HUIQIANG JIANG
On the other hand, we can’t expect sN (f ) → f in L1 [−π, π] as N → ∞ if
f ∈ L1 [−π, π]. We define
TN : f 7→ sN (f ) .
If for any f ∈ L1 [−π, π], TN f → f in L1 [−π, π] as N → ∞, then uniform boundedness principle implies the existence of C > 0 such that
kTN f kL1 ≤ C kf kL1
holds for any f ∈ L1 [−π, π] and for any N . By duality, we have
kTN f kL∞ ≤ C kf kL∞
holds for any f ∈ L∞ [−π, π] and for any N which is impossible since
Z π
lim
|kN (x)| dx = ∞.
N →∞
−π
p
Theorem 12. Let f ∈ L [−π, π], 1 < p < ∞, then sN (f ) → f in Lp [−π, π] as
N → ∞.
We will prove this result later.
3. Fourier Transform
¡ d ¢
If f ∈ L R ; C , its Fourier transform F (f ) = fˆ is defined by
Z
1
fˆ (y) =
f (x) e−ix·y dx.
d/2
Rd
(2π)
1
Since
(3.1)
° °
° ˆ°
°f °
1
kf kL1 ,
d/2
(2π)
¡ ¢
¡ ¢
the Fourier transform F : L1 Rd → L∞ Rd .
Now let’s list some basic properties of Fourier transform:
¡
¢
Theorem 13. (i) Translation: Let f ∈ L1 Rd ; C . If g (x) = f (x − a) and h (x) =
eiax f (x) for some a ∈ Rd , then
L∞
≤
ĝ (y) = e−iay fˆ (y) and ĥ (y) = fˆ (y − a) .
¡
¢
(ii) Scaling: Let f ∈ L1 Rd ; C and λ > 0. If h (x) = f (x/λ), then
ĥ (y) = λd fˆ (λt) .
¡
¢
¡
¢
(iii) Convolution: Let f, g ∈ L1 Rd ; C . Then f ∗ g ∈ L1 Rd ; C and
ˆ
d/2
(f ∗ g) = (2π)
fˆ · ĝ.
HARMONIC ANALYSIS
Proof. (iii)
ˆ
(f ∗ g) (y) =
=
=
Z
1
d/2
(2π)
1
(f ∗ g) (x) e−ix·y dx
Rd
Z
d/2
(2π)
1
Rd
Z
d/2
(2π)
= (2π)
7
d/2
µZ
¶
f (x − z) g (z) dz e−ix·y dx
Rd
µZ
¶
f (x − z) e
Rd
−i(x−z)·y
dx e−iz·y g (z) dz
Rd
fˆ (y) ĝ (y) .
¤
¡
¢
A function f ∈ C ∞ Rd ; C is said to be rapidly decreasing if for any n ≥ 0,
³
´n
2
(3.2)
pn (f ) = sup sup 1 + |x|
|(Dα f ) (x)| < ∞.
|α|≤n x∈Rd
i.e., P · Dα f is uniformly bounded for every polynomial P and for every multiindex α. The Schwartz space S = Sd is defined as the function space of rapidly
decreasing functions on Rd . The countable collection of norms pn in (3.2) defines
a locally convex topology for S. The topology of S is compatible with the metric
d (f, g) = max
n≥0
1 pn (f − g)
.
2n 1 + pn (f − g)
It is easy to check f ∈ S implies f ∈ L1 . Moreover, if f ∈ S, then for every multiindex α, Dα f , xα f ∈ S and we have the following property of Fourier transform.
Proposition 2. If f ∈ S and α is a multi-index, then
(Dα f ) ˆ (y) = (iy) fˆ (y) ,
³
´
α
Dα fˆ (y) = ((−ix) f ) ˆ (y) .
α
Proof. Direct calculation.
¤
Theorem 14. (a) S is a Fréchet space.
(b) If P is a polynomial, g ∈ S and α is a multi-index, then each of the three
mappings
f → P f, f → gf, f → Dα f
is a continuous
linear mapping of S into S.
¡ ¢
¡ ¢
(c) D Rd is dense in S and the identity mapping of D Rd into S is continuous.
Proof. (a) Let {fk } ⊂ S be a Cauchy sequence. Then for each n, {fk } is a Cauchy
sequence w.r.t the norm pn . Hence, for any multi-indices α, β, there exists continuous function gαβ such that xα Dβ fk → gαβ uniformly as k → ∞. Moreover,
gαβ (x) = xα Dβ g00 ,
so we have fi → g00 in S. Hence S is complete and it is a Fréchet space.
(b) Since S is a Fréchet space, a mapping Λ : S → S is continuous iff fk → 0
implies Λfk → 0. We also recall that fk → 0 iff for each n, pn (fk ) → 0. The
continuity of the three mappings follows easily.
8
HUIQIANG JIANG
¡ ¢
(c) Let ϕ ∈ D Rd be such that ϕ (x) ≡ 1 for any |x| < 1. For any f ∈ S, we
define
³x´
fk (x) = f (x) ϕ
.
k
Then for any multi-indices α, β,
¯
¯
max ¯xα Dβ (f − fk )¯
x
¯
¯
¯X µ ¶
¯
h
³
´i
¯
¡
¢
x ¯¯
β α
x Dβ−γ f (x) Dγ 1 − ϕ
= max ¯¯
γ
k ¯¯
|x|≥k ¯
γ≤β


µ ¶
¯³
¯
´ ¡
¢
1 X β
¯
¯
2
≤ max
kDγ ϕkL∞ ¯ 1 + |x| xα Dβ−γ f (x)¯ → 0.
γ
|x|≥k 1 + k 2
γ≤β
¡ ¢
Hence, fk → f in S. So D Rd ¡ is dense
in S.
¢
d
The identity mapping of ¡D R
¢ into S is continuous iff for each K compact,
d
the ¡identity
mapping
of
D
R
into S is continuous. ¡Now
K
¢ for K compact, since
¢
d
DK Rd is a Fréchet
space,
the
identity
mapping
of
D
R
into S is continuous iff
K
¡ ¢
fk → 0 in DK Rd implies fk → 0 in S. The latter statement is easy to verify. ¤
Theorem 15. The Fourier transform is a continuous linear mapping of S into S.
Proof. For any f ∈ S and for any multi-index α, β,
³
´
³
´
β
y α Dβ fˆ (y) = y α (−ix) f ˆ (y)
α+β ¡ α ¡ β ¢¢
= (−i)
D x f ˆ (y)
which is uniformly bounded, hence fˆ ∈ S. Let fk → 0 in S, then for any multi-index
α, β,
° ³
´ °
° α ¡ β ¢°
°¡
°
¡
¢¢
1
° α
°
° D x fk ° 1
°y Dβ fˆk (y)° ∞ = ° Dα xβ fk ˆ (y)°L∞ ≤
L
d/2
L
(2π)
°
°
°
°³
° °
°
´d
°
¡ β ¢° °
1
1
°
°
2
α
° 1 + |x|
≤
D x fk °
° → 0.
³
´
° ∞°
d
d/2 °
°
°
2
(2π)
L ° 1 + |x|
°
L1
Hence fˆk → 0 in S. So the Fourier transform is continuous.
¤
¡ d¢
We use C0 R to denote continuous functions converging to zero at infinity.
¡ ¢
¡ ¢
Corollary 2. If f ∈ L1 Rd , then fˆ ∈ C0 Rd .
¡ ¢
¡ ¢
Proof. For each f ∈ L1 Rd , there exists fn ∈ S such that fn → f in L1 Rd . Now
¡
¢
¡ ¢
(3.1) implies that fˆn → fˆ in L∞ Rd . Since fˆn ∈ S, we conclude fˆ ∈ C0 Rd . ¤
Before introducing the inverse transform, we need the following lemma on approximate delta function.
¡ ¢
Proposition 3. Let {kn } be a sequence of functions in L1 Rd satisfying
(1) kn (x) ≥ 0 a.e. x ∈ Rd ;
(2)
Z
Rd
kn (x) dx = 1;
HARMONIC ANALYSIS
(3) For any δ > 0,
9
Z
lim
kn (x) dx = 0.
n→∞
|x|>δ
¡ ¢
¡ ¢
Then for any f ∈ C Rd ∩ L∞ Rd ,
Z
lim
f (y) kn (x − y) dx = f (x) .
And for any f ∈ L
p
n→∞
¡
d
Rd
¢
R , 1 ≤ p < ∞,
Z
¡ ¢
f (y) kn (x − y) dx converges to f in Lp Rd .
fn (y) =
Rd
We also recall that
Z
2
e−(x+it) dx =
√
π
R
holds for any t ∈ R.
Theorem 16. The Fourier transform F : S → S is invertible. And for any f ∈ S,
Z
1
f (x) =
fˆ (y) eix·y dy.
d/2
Rd
(2π)
Proof. We compute
Z
1
d/2
(2π)
= lim
ε→0
= lim
ε→0
= lim
fˆ (y) eix·y dy
Rd
Z
1
d/2
(2π)
Z
1
d
(2π)
1
ε→0
(2πε)
= f (x) .
Here we used the fact that
2
ε
fˆ (y) eix·y e− 2 |y| dy
Rd
Z
ε
Rd
Z
d/2
2
f (z) e−iz·y eix·y e− 2 |y| dydz
Rd
f (z) e−
|x−z|2
2ε
dz
Rd
2
|z|
1
e− 2ε
(2πε)d/2
is a family of approximate delta functions.
¤
Hence, we can define inverse Fourier transform F −1 : S → S s.t. for any g ∈ S,
Z
1
F −1 (g) (x) = ǧ (x) =
g (y) eix·y dy.
d/2
d
R
(2π)
Apparently, F −1 : S → S is also continuous.
Corollary 3. If f, g ∈ S, then f ∗ g ∈ S.
Proof. f, g ∈ S imply fˆ, ĝ ∈ S, hence fˆĝ ∈ S and
d/2
(f ∗ g) ˆ = (2π)
fˆ · ĝ ∈ S.
Hence,
f ∗ g = F −1 ((f ∗ g) ˆ) ∈ S.
¤
10
HUIQIANG JIANG
¡ ¢
Proposition 4. If f ∈ L1 Rd , then as ε → 0, fε defined by
Z
2
ε
1
fε (x) =
fˆ (y) eix·y e− 2 |y| dy
d/2
Rd
(2π)
¡ ¢
converges to f in L1 . If in addition fˆ ∈ L1 Rd , then
Z
1
f (x) =
fˆ (y) eix·y dy a.e. x ∈ Rd .
d/2
Rd
(2π)
Proof. We compute
Z
1
d/2
(2π)
Since
Z
1
2
ε
fˆ (y) eix·y e− 2 |y| dy =
d/2
(2πε)
Rd
f (z) e−
|x−z|2
2ε
dz
Rd
2
|z|
1
e− 2ε
(2πε)d/2
is a family of approximate delta functions,
Z
|x−z|2
1
− 2ε
f
(z)
e
dz → f (x) in L1 as ε → 0.
d/2
Rd
(2πε)
¡ ¢
If in addition fˆ ∈ L1 Rd , Lebesgue dominated convergence implies,
Z
1
f (x) =
fˆ (y) eix·y dy a.e. x ∈ Rd .
d/2
Rd
(2π)
¤
Remark 6. If f ∈ L
¡
1
¢
¡ ¢
¡ ¢
Rd and fˆ ∈ L1 Rd , then actually f, fˆ ∈ C0 Rd .
Now we prove the well known Parseval formula
Theorem 17. If f, g ∈ S, then
Z
Z
fˆĝdy.
f ḡdx =
Rd
Rd
¡ ¢
Using the inner product of L2 Rd , we can write
³
´
(f, g) = fˆ, ĝ .
In particular, when f = g, we have the Plancherel identity
° °
° ˆ°
°f ° 2 = kf kL2 .
L
Proof. Using Fubini theorem, we have
Z
Z Z
f ḡdx =
fˆ (y) eix·y ḡ (x) dxdy
d
d
d
R
ZR R
=
fˆ (y) ĝ (y) dy.
Rd
¤
Corollary 4. If f ∈ L
¡
1
¢
¡ ¢
¡ ¢
Rd ∩ L2 Rd , then fˆ ∈ L2 Rd , and
° °
° ˆ°
°f ° 2 = kf kL2 .
L
HARMONIC ANALYSIS
11
¡ ¢
¡ ¢
Proof. Choose, fn ∈ S, such that fn → f in L1 Rd ∩ L2 Rd . Since
°
°
°ˆ
°
°fn − fˆm ° 2 = kfn − fm kL2 ,
L
n o
¡ ¢
fˆn is a Cauchy sequence in L2 Rd , hence there exists g ∈ L2 , s.t. fˆn → g
¡ ¢
¡ ¢
¡ ¢
in L2 Rd . On the other hand, fn → f in L1 Rd implies, fˆn → fˆ in L∞ Rd .
¡ ¢
Hence fˆ = g ∈ L2 Rd .
¤
¡ ¢
Hence we can extend Fourier¡ transform
as a linear
isometry
from
L2 Rd onto
¢
¡
¢
¡
¢
itself. ¡Actually,
for any f ∈ L2 Rd , if {fn } ⊂ L1 Rd ∩ L2 Rd satisfies fn → f
¢
2
d
in L R , then
fˆ = lim fˆn in L2 -sense.
n→∞
¡ ¢
Theorem 18. Let f ∈ L Rd . Define
Z
1
gn =
f (x) e−ix·y dx,
d/2
|x|<n
(2π)
2
then
fˆ = lim gn in L2 -sense.
n→∞
Moreover, if
hn =
Z
1
(2π)
d/2
fˆ (x) eix·y dx,
|x|<n
then
f = lim hn in L2 -sense.
n→∞
Proof. We leave the proof as an exercise.
¤
¡ ¢
Remark 7. If f ∈ Lp Rd , 1 < p < 2, then we can write f = f1 + f2 where
¡
¢
¡ ¢
f1 ∈ L1 Rd and f2 ∈ L2 Rd , and hence we can define fˆ = fˆ1 + fˆ2 .
To define Fourier transform on more general function, we need to introduce
tempered distributions.
Definition 4. A continuous linear functional on S is called a tempered distribution.
And we use S 0 to denote the collection of all tempered distributions.
¡ ¢
¡ d¢
Since D Rd is dense in S and the identity
mapping
I
of
D
R into S is
¡ ¢
continuous, every Λ ∈ S 0 gives rise to Λ◦I ∈ D0 Rd . Moreover, the map Λ → Λ◦I
is one-to-one. Hence, we can view tempered distributions as distributions which
have continuous extension to S.
Example 4. Every distribution with compact support is tempered.
Example 5. Suppose µ is a positive Borel measure on Rd such that
³
´−k
2
1 + |x|
dµ (x) < ∞.
Then µ is a tempered distribution.
12
HUIQIANG JIANG
Example 6. Suppose 1 ≤ p < ∞, N > 0 and g is a measurable function on Rd
such that
Z ³
´−k
2
p
|g (x)| dx < ∞.
1 + |x|
Rd
Then g is a tempered distribution. In particular, every Lp function is a tempered
distribution and so is every polynomial.
2
Example 7. e|x| is not a tempered distribution.
¡ ¢
Proof. Let φ ∈ D Rd be a radial symmetry function such that ³
φ (r)´ > 0 when
2
1 < r < 2 and φ (r) = 0 otherwise. One can check that e−|x| /2 φ |x|
→ 0 in S
n
while
µ ¶
Z
|x|
|x|2 −|x|2 /2
e e
φ
dx = ∞.
n
Rd
¤
Theorem 19. If α is a multi-index, P is a polynomial, g ∈ S and Λ is a tempered
distribution, then so are Dα Λ, P Λ and gΛ.
Proof. For any f ∈ S,
(Dα Λ) (f ) = (−1)
|α|
Λ (Dα f ) ,
since Dα : S → S is continuous, Dα Λ is continuous. Continuity of P Λ and gΛ
follows similarly.
¤
Now we can define Fourier transform of tempered distributions.
Definition 5. For any u ∈ S 0 , we define for any f ∈ S,
³ ´
û (f ) = u fˆ .
Then û ∈ S 0 .
If f ∈ L1 , then f is a tempered distribution which we denote by uf . And for
any g ∈ S, we have
Z
ûf (g) = uf (ĝ) =
f (x) ĝ (x) dx
Rd
Z Z
Z
=
f (x) e−ix·y g (y) dydx =
fˆ (y) g (y) = ufˆ (g) .
Rd
Rd
Rd
Hence, the definition of Fourier transform of tempered distributions is consistent
with the Fourier transform of integrable functions. Similarly, one
¡ can
¢ show that it
is also consistent with the Fourier transform of functions in L2 Rd .
Theorem 20. The Fourier transform is a continuous, linear bijection from S 0 onto
S 0 . Its inverse is also continuous.
Proof. Let W be a nbhd of 0 in S 0 . Then there exists functions f1 , · · · , fn ∈ S s.t.
{u ∈ S 0 : |u (fk )| < 1, 1 ≤ k ≤ n} ⊂ W.
Define,
¯ ³ ´¯
n
o
¯
¯
V = u ∈ S 0 : ¯u fˆk ¯ < 1, 1 ≤ k ≤ n .
Then V is a nbhd of 0 in S 0 , and F (V ) ⊂ W . Hence F : S 0 → S 0 is continuous.
Since F 4 = I, F is a bijection and F −1 = F 3 is continuous.
¤
HARMONIC ANALYSIS
13
Theorem 21. If u ∈ S 0 , and α is a multi-index, then
α
(Dα u) ˆ = (iy) û,
α
Dα û = ((−ix) u) ˆ.
Proof. By definition.
¤
d
Example 8. 1̂ = (2π) 2 δ and δ̂ =
1
d
(2π) 2
.
Proof. For any f ∈ S,
³ ´ Z
1̂ (f ) = 1 fˆ =
d
fˆ (y) dy = (2π) 2 f (0) .
Rd
An for any f ∈ S,
³ ´
δ̂ (f ) = δ fˆ = fˆ (0) =
Z
1
d
(2π) 2
f (x) dx.
Rd
¤
Example 9. Let α be a multi-index, then
d
(xα ) ˆ = (2π) 2 iα Dα δ.
Since any distribution supported at 0 is a linear combination of distributions of the
form Dα δ, the Fourier transform of f is supported at 0 iff f is a nonzero polynomial.
Definition 6. Let u ∈ S 0 , and ϕ ∈ S. We define for any x ∈ Rd
(u ∗ ϕ) (x) = u (ϕ (x − ·)) .
¡ ¢
Theorem 22. Let u ∈ S 0 , and ϕ ∈ S. Then u ∗ ϕ ∈ C ∞ Rd . Moreover, for any
multi-index α,
Dα (u ∗ ϕ) = Dα u ∗ ϕ = u ∗ Dα ϕ.
Proof. Since the mapping x 7→ ϕ (x − ·) is continuous from Rd into S, u ∗ ϕ is
continuous. Now
µ
¶
(u ∗ ϕ) (x + tei ) − (u ∗ ϕ) (x)
ϕ (x + tei − ·) − ϕ (x − ·)
=u
,
t
t
since
ϕ (x + tei − y) − ϕ (x − y)
→ ∂i ϕ (x − y) in S,
t
we have
∂i (u ∗ ϕ) (x) = lim
t→0
(u ∗ ϕ) (x + tei ) − (u ∗ ϕ) (x)
= u (∂i ϕ (x − ·)) ,
t
¡ ¢
hence u ∗ ϕ has a continuous i-th partial derivative. By induction, u ∗ ϕ ∈ C ∞ Rd .
¤
¡ ¢
Lemma 1. Let u ∈ S 0 , and ϕ, φ ∈ D Rd , we have
(u ∗ ϕ) ∗ φ = u ∗ (ϕ ∗ φ) .
14
HUIQIANG JIANG
Proof. For any h > 0, we define the Riemann sum
X
ϕ (x − kh) φ (kh)
fh (x) = hd
k
where for each h and x, the summation has finitely many nonzero terms. It is
straight forward to verify that fh → ϕ ∗ φ in S. Hence,
u ∗ (ϕ ∗ φ) = lim+ u ∗ fh
h→0
X
= lim+ hd
(u ∗ ϕ) (x − kh) φ (kh)
h→0
k
= (u ∗ ϕ) ∗ φ.
Here the last equality holds because (u ∗ ϕ) (x − y) φ (y) is Riemann integrable in
y.
¤
¡ d¢
Now let ϕ ∈ D R be the standard mollifier and for any ε > 0, we define
³x´
ϕε (x) = ε−n ϕ
.
ε
Lemma 2. For any φ ∈ S, we have
φ ∗ ϕε → φ in S.
Proof. Direct calculation.
¤
Theorem 23. If u ∈ S 0 , then u ∗ ϕε → u in the weak-* topology of S 0 .
Proof. For any φ ∈ S, we have
³
´
u (φ) = u ∗ φ̃ (0)
¡ ¢
where φ̃ (x) = φ (−x). For any φ ∈ D Rd ,
(u ∗ ϕε ) (φ)
³
´
= (u ∗ ϕε ) ∗ φ̃ (0)
³
³
´´
= u ∗ ϕε ∗ φ̃ (0)
µ³
´˜ ¶
= u ϕε ∗ φ̃
.
µ³
Since the expression u
ϕε ∗ φ̃
´˜ ¶
is continuous for φ ∈ S, we conclude u∗ϕε ∈ S 0
and the above formula holds for φ ∈ S. Since ϕε ∗ φ̃ → φ̃ in S, we have
lim (u ∗ ϕε ) (φ) = u (φ) .
ε→0
¤
¢
Remark 8. Similar proof
¡ ¢ will yield that if u ∈ D R , then u ∗ ϕε → u in the
weak-* topology of D0 Rd .
0
¡
d
Remark 9. The formulas for Fourier transforms of regular functions continue to
hold for tempered distributions as long as they are well defined. For example, if
u ∈ S 0 and ϕ ∈ S, then
d
(u ∗ ϕ) ˆ = (2π) 2 ϕ̂û.
HARMONIC ANALYSIS
15
Definition 7. Let Ω ⊂ Rd be open and k be a nonnegative integer and 1 ≤ p ≤ ∞.
Sobolev space W k,p (Ω) is the collection of functions f ∈ Lp and the distributional
derivatives Dα f ∈ Lp for any |α| ≤ k. W k,p (Ω) is a Banach space with norm

 p1
X
p
kf kW k,p (Ω) = 
kDα f kLp (Ω)  if 1 ≤ p < ∞,
|α|≤k
kf kW k,∞ (Ω) = max kDα f kL∞ (Ω) .
|α|≤k
When p = 2, W k,2 (Ω) = H k (Ω) is a Hilbert space.
¡ ¢
Let f ∈ H k Rd . Then we have for any |α| ≤ k,
¡ ¢
α f = (iy)α fˆ ∈ L2 Rd .
[
D
³
´k/2
¡ ¢
¡ ¢
2
fˆ ∈ L2 Rd . Moreover, the norm
Hence, f ∈ H k Rd iff 1 + |y|
°³
´k/2 °
°
°
2
°
kf kH̃ k = ° 1 + |y|
fˆ°
°
L2
is equivalent to the
¡ d ¢original norm. This expression also allows us to extend
¡ d ¢the
k
k
definition
of
H
R
to
k
∈
R.
One
can
identify
the
dual
space
of
H
R as
¡ d¢
−k
H
R .
Theorem 24 (Sobolev). Let Ω ⊂ Rd be open and k > d2 . Any f ∈ H k (
) can be
represented by a continuous function.
¡ ¢
Proof. For any x ∈ Ω, let ϕ ∈ D Rd¡ has
¢ compact support in Ω and ϕ = 1 in
Br (x) for some r > 0. Then ϕf ∈ H k Rd and¡ϕf ¢= f in Br (x). Since continuity
is a local property, we assume f itself is in H k Rd . Now
°³
°
Z ¯ ¯
´k/2 °
´−k/2 °
°
° °³
°
¯ ˆ¯
2
2
ˆ
°
°
°
° < ∞.
f ° ° 1 + |y|
¯f ¯ dx ≤ ° 1 + |y|
°
Rd
L2
So we have fˆ ∈ L
¡
1
³ ´∨
¢
¡ ¢
Rd and f = fˆ ∈ C0 Rd .
L2
¤
4. The maximal function
4.1. Distribution function. We first recall the concept of distribution function.
Definition 8. Let g be defined on Rn . For each t ≥ 0, we define
λ (t) = |{x ∈ Rn : |g (x)| > t}| .
The function λ is called the distribution function of |g|.
The function λ is monotone decreasing on [0, ∞) and hence measurable. Any
quantity dealing solely with the size of g can be expressed in terms of the distribution function λ (α). If g ∈ Lp (Rn ), then
Z
Z ∞
p
|g (y)| = −
tp dλ (t) .
Rn
And if g ∈ L
∞
0
n
(R ), then
kgk∞ = inf {t : λ (t) = 0} .
16
HUIQIANG JIANG
Proposition 5. If g ∈ Lp (Rn ), p ≥ 1, then for any t > 0,
Z
1
p
λ (t) ≤ p
|g (y)| dy.
t Rn
Proof.
Z
Z
p
|g (y)| dy ≥ tp
Rn
dy = tp λ (t) .
|g(x)|>t
¤
4.2. A covering lemma.
Lemma 3. Let E be a measurable subset of Rn which is covered by the union of a
family of balls {Bα }α∈Λ of bounded radius. Then from this family, we can select a
disjoint sequence, B1 , B2 , · · · , s.t.
X
|Bk | ≥ 5−n |E| .
Proof. We choose B1 s.t.
1
sup r (Bα ) .
2 α∈Λ
When B1 , B2 , · · · , Bk are chosen, we choose Bk+1 s.t.,
r (B1 ) ≥
1
sup {r (Bα ) : α ∈ Λ, Bα ∩ Bi = ∅, 1 ≤ i ≤ k} .
2
This P
procedure terminates if the set in the above expression
is empty.
P
If
|Bk | = ∞, we are done. So we now assume
|Bk | < ∞. Since r (Bk ) ≥
1
∗
2 r (Bj ) for any j ≥ k, either {Bk } is finite or limk→∞ r (Bk ) = 0. where Bk is the
ball with the same center as Bk and with radius 5r (Bk ). Given any α ∈ Λ such
that Bα is not one of {Bk }. If {Bk } is finite, then for some j, Bj ∩ Bα 6= ∅ and
r (Bj ) ≥ 12 r (Bα ) which implies Bα ⊂ Bj∗ . If {Bk } is not finite, then there exists
first k, s.t. r (Bk+1 ) < 12 r (Bα ). Hence, for some 1 ≤ j ≤ k, Bj ∩ Bα 6= ∅ and
S
r (Bj ) ≥ 21 r (Bα ) which implies Bα ⊂ Bj∗ . Hence, in any case, Bα ⊂ Bk∗ . So we
r (Bk+1 ) ≥
k
have
E⊂
[
Bα ⊂
[
α∈Λ
and
|E| ≤
X
|Bk∗ | = 5n
Bk∗
k
X
|Bk | .
¤
4.3. Hardy-Littlewood maximal function. We recall that if f ∈ L1loc (Rn ),
then
Z
1
f (y) dy
f (x) = lim+
r→0 |Br (x)| Br (x)
holds for a.e. x ∈ Rn . To study such limit, we introduce:
Definition 9. Let f ∈ L1loc (Rn ). The maximal function of f is defined by
Z
1
M (f ) (x) = sup
|f (y)| dy, x ∈ Rn .
r>0 |Br (x)| Br (x)
HARMONIC ANALYSIS
17
Theorem 25. Let f be a given function defined on Rn
(a) If f ∈ Lp (Rn ), 1 ≤ p ≤ ∞, then M f is finite a.e..
(b) If f ∈ L1 (Rn ), then for any t > 0,
A
m {x : (M f ) (x) > t} ≤
t
Z
|f | dx,
Rn
where we can take A = 5n .
(c) If f ∈ Lp (Rn ), with 1 < p ≤ ∞, then M f ∈ Lp (Rn ) and
kM f kp ≤ Ap kf kp
where
µ
Ap = 2
5n p
p−1
¶ p1
, 1 < p < ∞ and A∞ = 1.
Proof. (b) Let Et = {x : (M f ) (x) > t}. For any x ∈ Et , there exits a ball Bx
centered at x, s.t.,
Z
|f (y)| dy > t |Bx | .
Bx
The family {Bx } is a cover of set Et , hence the covering lemma implies the existence
of disjoint sub-collection {Bxk } s.t.
X
|Bxk | ≥ 5−n |Et | .
Hence,
|Et | ≤ 5n
X
|Bxk | <
Z
Z
5n X
5n
|f | dx ≤
|f | dx.
t
t Rn
Bx
k
(c) The case p = ∞ is trivial with A∞ = 1. Now we assume 1 < p < ∞. For
any t > 0, we define f1 (x) = f (x) if |f (x)| > 2t , f1 (x) = 0 otherwise. Then
|f (x)| ≤ |f1 (x)| + 2t which yields
t
M f (x) ≤ M f1 (x) + .
2
Hence,
½
Et = {x : (M f ) (x) > t} ⊂
t
x : (M f1 ) (x) >
2
¾
.
So we have
½
¾
Z
t
2A
2A
λ (t) = |Et | ≤ m x : (M f1 ) (x) >
≤
kf1 kL1 =
|f | .
2
t
t |f (x)|> 2t
18
HUIQIANG JIANG
Hence,
Z
p
kM f kp
∞
Z
∞
t dλ (t) = p
tp−1 λ (t) dt
0
0
ÃZ
!
Z ∞
p−1 2A
≤p
t
|f | dx dt
t
0
|f |> 2t
ÃZ
!
Z
2|f |
p−2
t dt dx
= 2pA
|f (x)|
=−
p
Rn
0
Z
2pA
p−1
|f (x)| (2 |f |)
dx
p − 1 Rn
2p pA
p
p
=
kf kp = App kf kp .
p−1
=
¤
Remark 10. Ap → ∞ as p → 1 which suggests that part (c) will not hold for
p = 1. Actually, M f is never integrable unless f = 0 a.e.. The estimate in (c) is
called type (p, p), while the estimate in (b) is called type weak (1, 1).
Corollary 5. If f ∈ L1loc (Rn ), then
f (x) = lim
r→0+
1
|Br (x)|
Z
f (y) dy
Br (x)
holds for a.e. x ∈ Rn .
Proof. We assume f ∈ L1 (Rn ). For any x ∈ Rn , we define
Z
1
fr (x) =
f (y) dy
|Br (x)| Br (x)
and
¯
¯
¯
¯
Ωf (x) = ¯¯lim sup fr (x) − lim sup fr (x)¯¯ .
r→0+
r→0+
n
For any ε > 0, we choose, for any δ > 0, h ∈ C (R ) and denote g = f − h s.t.
kgk1 = kf − hk1 < δ.
Then Ωf (x) = Ωg (x) ≤ 2M g. Now
m {x : Ωf (x) > ε} = m {x : Ωg (x) > ε}
n
ε o 2 · 5n
≤m x : M g (x) >
≤
kgk1 ≤ 2 · 5n δ.
2
ε
Since δ can be arbitrarily small, we have
m {x : Ωf (x) > ε} = 0.
Hence Ωf (x) = 0 a.e. x ∈ Rn . Hence limr→0+ fr (x) exists a.e.. On the other
hand, since fr → f in L1 , there exists a subsequence rk s.t. frk → f a.e. x ∈ Rn .
Hence,
f (x) = lim+ fr (x)
r→0
holds for a.e. x ∈ Rn .
¤
HARMONIC ANALYSIS
19
4.4. Generalized maximal function.
Definition 10. Let F be a family of measurable subsets of Rn . We say F is regular,
if there exists c > 0 s.t. if S ∈ F, then there exists a ball B centered at 0, s.t. S ⊂ B,
m (S) ≥ cm (B).
Here are some examples:
Example 10. The family F = {δU }δ>0 is regular if U is bounded and measurable
with m (U ) > 0.
Example 11. The family F of all cubes with the property that their distance from
the origin is bounded by a constant multiple of their diameter.
Example 12. The family F of all balls containing the origin.
Example 13. Any subfamily of a regular family is regular.
Given a regular family F of measurable subsets of Rn . We define the maximal
function
Z
1
MF (f ) (x) = sup
|f (x − y)| dy.
S∈F m (S) S
Let c > 0 be the number in the Definition of regular family, then we have
1
MF (f ) (x) ≤ M (f ) (x) .
c
Hence strong (p, p), 1 < p ≤ ∞ and weak (1, 1) estimates hold for MF .
Corollary 6. Suppose F is a regular family of measurable subsets of Rn such that
inf m (S) = 0.
S∈F
If f ∈ L1loc (Rn ), then
f (x) =
lim
S∈F
m(S)→0
1
m (S)
Z
f (x − y) dy
S
holds for a.e. x ∈ Rn .
Definition 11. Let f ∈ L1loc (Rn ). A point x ∈ Rn is said to be a Lebesgue point
of f if
Z
1
lim+
|f (y) − f (x)| dy = 0.
r→0 |Br (x)| Br (x)
The Lebesgue set of f is the collection of all Lebesgue points of f .
Theorem 26. Let f ∈ L1loc (Rn ).
(a) The complement of Lebesgue set is of measure zero.
(b) Suppose F is a regular family of measurable subsets of Rn such that
inf m (S) = 0.
S∈F
Then
f (x) =
lim
S∈F
m(S)→0
n
1
m (S)
holds for every Lebesgue point x ∈ R .
Z
f (x − y) dy
S
20
HUIQIANG JIANG
Proof. (a) For any s ∈ Q,
1
lim+
r→0 |Br (x)|
Z
|f (y) − s| dy = |f (x) − s|
Br (x)
holds for a.e. x ∈ Rn . We define the exceptional set Es , then |Es | = 0. Let
E = ∪s∈Q Es , then |E| = 0. For any x 6∈ E, we have
Z
1
lim+
|f (y) − s| dy = |f (x) − s|
r→0 |Br (x)| Br (x)
holds for every s ∈ Q. Since Q is dense in R,
Z
1
lim
|f (y) − f (x)| dy = 0,
r→0+ |Br (x)| Br (x)
i.e., x is a Lebesgue point.
(b) Let x be a Lebesgue point. Then
¯
¯
Z
Z
¯ 1
¯
¯
¯≤ 1
f
(x
−
y)
dy
−
f
(x)
|f (x − y) − f (x)| dy
¯ m (S)
¯ m (S)
S
Z S
1
≤
|f (x − y) − f (x)| dy → 0
c |B| B
as |B| → 0.
¤
4.5. An interpolation theorem for Lp . Let 1 ≤ p, q ≤ ∞ and T be a mapping
from Lp (Rn ) to measurable functions in Rn . Then T is of type (p, q) if for some
A > 0,
kT f kq ≤ A kf kp holds for every f ∈ Lp (Rn ) .
Similarly, if in addition, q < ∞, T is of weak type (p, q) if for some A > 0,
µ
¶q
A kf kp
m {x : |T f (x)| > t} ≤
holds for every f ∈ Lp (Rn ) .
t
When q = ∞, T is of weak type (p, q) if it is of type (p, q).
If T is of type (p, q), then T is of weak type (p, q). Assume q < ∞ and T is of
type (p, q). For every f ∈ Lp (Rn ),
µ
¶q
A kf kp
1
q
m {x : |T f (x)| > t} ≤ q kT f kq ≤
.
t
t
Hence, T is of weak type (p, q).
The following is a special case of Marcinkiewicz interpolation theorem:
Theorem 27 (Marcinkiewicz). Suppose 1 < r ≤ ∞ and T maps L1 (Rn ) + Lr (Rn )
into the space of measurable functions in Rn . Suppose that T is
(i) subadditive: For any f, g ∈ L1 (Rn ) + Lr (Rn ),
|T (f + g) (x)| ≤ |T f (x)| + |T g (x)| ;
(ii) of weak type (1, 1): For any f ∈ L1 (Rn ),
m {x : |T f (x)| > t} ≤
A1
kf k1 ;
t
HARMONIC ANALYSIS
21
(iii) of weak type (r, r): If r < ∞, for any f ∈ Lr (Rn ),
µ
¶r
Ar kf kr
m {x : |T f (x)| > t} ≤
t
and if r = ∞,
kT f k∞ ≤ A∞ kf k∞ .
Then for any 1 < p < r,
kT f kp ≤ Ap kf kp holds for every f ∈ Lp (Rn )
where
·µ
Ap =
r¶
2r A1
(2Ar )
+
p−1
r−p
¸ p1
p
if r < ∞.
Proof. The case r = ∞ follows from the proof of (p, p) estimate for maximal function. Now we assume 1 < r < ∞. Fix f ∈ Lp . For any t > 0, we define
f1 = f χ|f |>t and f2 = f χ|f |≤t .
Then f1 ∈ L1 and f2 ∈ Lr . Since
|T (f1 + f2 ) (x)| ≤ |T f1 (x)| + |T f2 (x)| ,
we have
½
¾ ½
¾
t
t
{x : |T f (x)| > t} ⊂ x : |T f1 (x)| >
∪ x : |T f2 (x)| >
.
2
2
Hence,
λ (t) = |{x : |T f (x)| > t}|
¯½
¾¯ ¯½
¾¯
¯
t ¯¯ ¯¯
t ¯¯
≤ ¯¯ x : |T f1 (x)| >
+
x
:
|T
f
(x)|
>
2
2 ¯ ¯
2 ¯
µ
¶r
2Ar kf2 kr
2A1
kf1 k1 +
≤
t
t
µ
¶r Z
Z
2A1
2Ar
r
=
|f | dx +
|f | dx.
t
t
|f |>t
|f |≤t
Recall that
Z
p
kT f kp
Now we have
Z
∞
t
p−1
0
and
Z
p−1
t
0
Hence,
µ
p
kT f kp
≤
tp−1 λ (t) dt.
0
à Z
!
1
1
p
|f | dx dt =
kf kp
t |f |>t
p−1
Ã
∞
∞
=p
1
tr
!
Z
r
|f | dx dt =
|f |≤t
r¶
2A1
(2Ar )
+
p−1
r−p
1
p
kf kp .
r−p
p
p
p kf kp = App kf kp .
¤
22
HUIQIANG JIANG
4.6. Behavior near general points of measurable sets. Suppose E is a given
measurable subset of Rn . We say x ∈ Rn is a point of density of E if
lim
r→0
|E ∩ Br (x)|
= 1.
|Br (x)|
Proposition 6. Almost every point x ∈ E is a point of density of E, and almost
every point x 6∈ E is not a point of density of E.
Now let E be a closed set. For any x ∈ Rn , we denote δ (x) = dist (x, E).
Proposition 7. Let E be a closed set. Then for almost every x ∈ E, δ (x + y) =
o (|y|). This holds in particular if x is a point of density of E.
Proof. Let x be a point of density of F and suppose ε > 0 is given. If δ (x + y) >
ε |y|, then
¯
¯
µ
¶n
¯E ∩ B(1+ε)|y| (x)¯
ε
¯
¯ ≤1−
¯B(1+ε)|y| (x)¯
1+ε
which is impossible if |y| is sufficiently small.
¤
Now we consider the integral of marcinkiewicz:
Z
δ (x + y)
I (x) =
n+1 dy.
|y|≤1 |y|
We need the following lemma:
Lemma 4. Let E be a closed set such that |E c | < ∞ and
Z
δ (x + y)
I∗ (x) =
n+1 dy.
n
|y|
R
Then I∗ (x) < ∞ for a.e. x ∈ E. Moreover,
Z
I∗ (x) dx ≤ nωn |E c |
E
where ωn is the volume of the unit ball in Rn .
Proof.
Z
Z Z
I∗ (x) dx =
E
δ (x + y)
E
Rn
E
Ec
n+1
|y|
δ (y)
Z Z
=
Z
=
|x − y|
ÃZ
δ (y)
Ec
E
ÃZ
Z
≤
dydx
n+1 dydx
1
= nωn
Z
Ec
n+1 dx
|x − y|
δ (y)
Ec
!
|x|≥δ(y)
1
|x|
n+1 dx
dy
!
dy
dy = nωn |E c | .
¤
HARMONIC ANALYSIS
23
Theorem 28. Let E be a closed set.
(a) When x 6∈ E, I (x) = ∞;
(b) For almost every x ∈ F , I (x) < ∞.
Remark 11. We can also define for λ > 0,
Z
δ λ (x + y)
(λ)
I (x) =
dy.
n+λ
|y|
|y|≤1
4.7. Decomposition in cubes of open sets in Rn .
Theorem 29 (Calderón Zygmund decomposition). Let f be a nonnegative integrable function on Rn and α > 0. Then there exists a decomposition of Rn so that
(i) Rn = F ∪ Ω, F ∩ Ω = ∅;
(ii) f (x) ≤ α a.e. on F .
(iii) Ω is the union of closed cubes, Ω = ∪k Qk , whose interiors are disjoint and for
each k,
Z
1
α<
f (x) dx ≤ 2n α.
|Qk | Qk
Moreover,
1
kf k1 .
α
Proof. We decompose Rn into a mesh of equal cubes such that
Z
Z
1
1
f (x) dx ≤
f (x) dx ≤ α.
|Q| Q
|Q| Rn
|Ω| ≤
Let Q0 be a fixed cube, we divide it into 2n congruent cubes. For each new cube
Q00 ,
Case 1:
Z
1
f (x) dx ≤ α.
|Q00 | Q00
Then we continue the process.
Case 2:
1
|Q00 |
Z
f (x) dx > α.
Q00
We select Q00 as one of Qk .
Let Ω = ∪k Qk , and F = Rn \Ω. Then f (x) ≤ α a.e. on F .
¤
Theorem 30. Let F be a nonempty closed set in Rn . Then its complement Ω = F c
can be written as a union of cubes
Ω=
∞
[
Qk ,
k=1
whose interiors are disjoint and for each k,
diam Qk ≤ dist (Qk , F ) ≤ 4 diam Qk .
Proof. Consider a mesh M0 of cubes with unit length whose vertices are integral
points. Let Mk = 2−k M0 . We also define for some c > 0,
ª
©
Ωk = x : c2−k < dist (x, F ) < c2−k+1 .
24
HUIQIANG JIANG
Let
F0 =
√
Then with c = 2 n, we have
[
{Q ∈ Mk : Q ∩ Ωk 6= ∅} .
k
diam Q ≤ dist (Q, F ) ≤ 4 diam Q
S
for each Q ∈ F0 . We have Ω =
Q. We observe that if Q◦1 ∩ Q◦1 6= ∅, then one
Q∈F0
cube contains the other. For each Q ∈ F0 , let Q0 be the maximal cube in F0 which
contain Q. Such maximal cube is unique and all the interiors of maximal cubes are
disjoint. Let {Qk } be the collection of all maximal cubes in Ω. We have the desired
properties.
¤
With the help of Theorem 30, we can improve certain aspects of the Calderon
Zygmund decomposition.
Theorem 31. Let f be a nonnegative integrable function on Rn and α > 0. Let
F = {x : M f (x) ≤ t}
and
Ω = F c = {x : M f (x) > t} .
We can write Ω as the union of closed cubes, Ω = ∪k Qk , whose interiors are disjoint
and for each k,
diam Qk ≤ dist (Qk , F ) ≤ 4 diam Qk .
Moreover, for each k,
1
|Qk |
Z
f (x) dx ≤ cα
Qk
where c is a number depending only on n.
Proof. Since F is closed, its complement Ω = F c can be written as a union of cubes
Ω=
∞
[
Qk ,
k=1
whose interiors are disjoint and for each k,
diam Qk ≤ dist (Qk , F ) ≤ 4 diam Qk .
For each cube Qk , let pk ∈ F be such that dist (Qk , F ) = dist (Qk , pk ). Let
rk = max dist (x, pk ) ≤ 5 diam Qk ,
x∈Qk
then Qk ⊂ Brk (pk ). So we have
Z
1
|f (x)| dx
|Brk (pk )| Brk (pk )
Z
f (x) dx
t ≥ M f (pk ) ≥
1
|Brk (pk )| Qk
Z
1
≥
f (x) dx.
c |Qk | Qk
≥
HARMONIC ANALYSIS
Hence
1
|Qk |
25
Z
f (x) dx ≤ ct.
Qk
¤
Remark 12. We also have
|Ω| ≤
2n
kf k1 .
t
5. Singular Integral
5.1. Convolution of measures. The dual space of C0 (Rn ) is M (Rn ), the space
of all finite Borel measures. For any µ1 , µ2 ∈ M (Rn ), µ = µ1 ∗ µ2 is defined by
Z Z
f (x + y) dµ1 (x) dµ2 (y) .
µ (f ) =
Rn
Rn
We have µ1 ∗ µ2 = µ2 ∗ µ1 and kµk ≤ kµ1 k kµ2 k.
Let 1 ≤ p < ∞. If f ∈ Lp (Rn ), then f ∗ µ ∈ Lp (Rn ) and
kf ∗ µkp ≤ kf kp kµk .
The map T : f 7→ f ∗ µ is a bounded linear operator from Lp (Rn ) to Lp (Rn )
which commutes with translation. Such mappings can be classified.
Theorem 32. Suppose T : Lp (Rn ) → Lq (Rn ) is linear, bounded and commutes
with translations. Then there exists a unique tempered distribution u ∈ S 0 such that
T (ϕ) = ϕ ∗ u for any ϕ ∈ S.
Before presenting its proof, we consider special cases when p = q where we can
say more on the tempered distribution u.
Theorem 33. Suppose T : L1 (Rn ) → L1 (Rn ) is linear, bounded and commutes
with translations. Then there exists a unique µ ∈ M (Rn ) such that T (f ) = f ∗ µ
for any f ∈ L1 (Rn ). Moreover, kT k = kµk.
Proof. Let u ∈ S 0 be such that T (ϕ) = ϕ ∗ u for any ϕ ∈ S. We have for any ϕ ∈ S,
kϕ ∗ uk1 ≤ kT k kϕk1 .
Suppose ϕε is the standard mollifier, we have
kϕε ∗ uk1 ≤ kT k .
Hence, ϕεk ∗ u * µ for a subsequence εk → 0. On the other hand, we also have
ϕεk ∗ u → u in S 0 . So we have u = µ. kT k = kµk can be verified since
lim kϕε ∗ µk = kµk .
ε→0
¤
Theorem 34. Suppose T : L2 (Rn ) → L2 (Rn ) is linear, bounded and commutes
with translations. Then there exists a multiplier m ∈ L∞ (Rn ), such that
Tcf = m · fˆ
for any f ∈ L2 (Rn ).
26
HUIQIANG JIANG
Proof. Let u ∈ S 0 be such that T (ϕ) = ϕ ∗ u for any ϕ ∈ S. Then for ϕ = e−
x2
2
,
n/2
Tcϕ = (2π)
ϕ̂ · û.
For any φ ∈ D (Rn ), since ϕ̂ = e−
x2
2
we have
φ
ϕ̂
∈ S and
φc
n/2
T ϕ = (2π)
φ · û ∈ L2 .
ϕ̂
So we have û ∈ L2loc (Rn ). Now for any ϕ ∈ S, we have
° °
° °
n/2
(2π)
kϕ̂ · ûk2 = °Tcϕ° = kT ϕk2 ≤ A kϕ̂k2 .
2
Hence û ∈ L∞ (Rn ). Let m = (2π)
n/2
û ∈ L∞ (Rn ). We have for any ϕ ∈ S,
Tcϕ = mϕ̂.
Since S is dense in L2 (Rn ), the continuities of T and Fourier transform imply
Tcf = m · fˆ
for any f ∈ L2 (Rn ).
¤
Lemma 5. Suppose 1 < p < ∞ and T : Lp (Rn ) → Lp (Rn ) is linear, bounded
and commutes with translations. Then T , restricted to Lp (Rn ) ∩ Lq (Rn ), extends
to a map T : Lq (Rn ) → Lq (Rn ) which is linear, bounded and commutes with
translations.
Proof. Let u ∈ S 0 be such that T (ϕ) = ϕ ∗ u for any ϕ ∈ S. We observe that for
any ψ ∈ C0∞ (Rn ), if u ∈ C0∞ (Rn ), we have
Z
(ψ ∗ u) (x) ϕ (x) dx
Rn
Z Z
=
ψ (y) u (x − y) ϕ (x) dxdy
Rn Rn
Z Z
=
ψ (y) u (z) ϕ (y + z) dxdy
n
n
ZR R
=
ψ (y) (u ∗ ϕ̃) (−y) dxdy.
Rn
Using Lemma 1, we can show that for any ϕ, ψ ∈ C0∞ (Rn ),
Z
Z
(ψ ∗ u) (x) ϕ (x) dx =
ψ (y) (ϕ̃ ∗ u) (−y) dxdy,
Rn
i.e.,
Rn
Z
Z
(T ψ) (x) ϕ (x) dx =
Rn
ψ (y) (T ϕ̃) (−y) dxdy.
Rn
For fixed ψ ∈ C0∞ (Rn ),
¯Z
¯
¯
¯
¯
¯
¯ n (T ψ) (x) ϕ (x) dx¯ ≤ kψkq kT ϕ̃kp ≤ kT k kψkq kϕkp
R
holds for any ϕ, ψ ∈ C0∞ (Rn ), hence
kT ψkq ≤ kT k kψkq .
So T can be extended to a continuous linear functional T : Lq (Rn ) → Lq (Rn ). ¤
HARMONIC ANALYSIS
27
Theorem 35. Suppose 1 < p < ∞ and T : Lp (Rn ) → Lp (Rn ) is linear, bounded
and commutes with translations. Then there exists a multiplier m ∈ L∞ (Rn ), such
that
Tcϕ = m · ϕ̂
for any ϕ ∈ S.
Proof. T : Lp (Rn ) → Lp (Rn ) and T can be extended to T : Lq (Rn ) → Lq (Rn )
with
1 1
+ = 1.
p q
Hence Marcinkiewicz interpolation theorem implies T : L2 (Rn ) → L2 (Rn ). The
conclusion now follows from Theorem 34.
¤
Now we prepare to prove Theorem 32.
Lemma 6. If f ∈ W n+1,p (Rn ), 1 ≤ p ≤ ∞, then f equals a.e. a continuous
function g satisfying
X
kgk∞ ≤ C
kDα f kp .
|α|≤n+1
Proof. We first assume f ∈ D (Rn ). Then we have
¯
¯
³
´− n+1
X
2
¯ˆ ¯
2
¯f (x)¯ ≤ c 1 + |x|
¯
¯
¯
α¯
|x| ¯fˆ (x)¯
|α|≤n+1
³
= c 1 + |x|
2
´− n+1
2
2
´− n+1
2
¯
¯
¯[
¯
¯Dα f (x)¯
X
|α|≤n+1
³
≤ c 1 + |x|
X
kDα f k1 .
|α|≤n+1
Hence
X
kf k∞ ≤ c2
Z
kDα f k1
|α|≤n+1
³
1 + |x|
2
´− n+1
2
X
dx ≤ C
Rn
kDα f k1 .
|α|≤n+1
Hence the case p = 1 follows from the density of D (Rn ) in W n+1,1 (Rn ).
For 1 < p ≤ ∞, f ∈ W n+1,p (Rn ) implies ϕf ∈ W n+1,1 (Rn ) where ϕ ∈ D (Rn )
is a cutoff function s.t. ϕ = 1 on B1 (0) and ϕ = 0 on Rn \B2 (0). Then ϕf is
continuous, and
X
X
|f (0)| = |ϕf (x)| ≤ C
kDα (ϕf )k1 ≤ C2
kDα f kL1 (B2 (x))
≤ C3
X
|α|≤n+1
|α|≤n+1
kDα f kp .
|α|≤n+1
Since the choice of origin is arbitrary, we have
X
kgk∞ ≤ C3
kDα f kp .
|α|≤n+1
¤
Now we prove Theorem 32:
28
HUIQIANG JIANG
Proof of Theorem 32. Let ϕ ∈ S. Since
τte1 ϕ − ϕ
→ ∂x1 ϕ in Lp as t → 0+ ,
t
we have
µ
¶
τte1 ϕ − ϕ
(T ϕ) (x + te1 ) − (T ϕ) (x)
=T
(x) → T (∂x1 ϕ)
t
t
in Lq as t → 0+ . Hence ∂x1 (T ϕ) = T (∂x1 ϕ) ∈ Lq . Similar calculation yields
Dα (T ϕ) = T (Dα ϕ) ∈ Lq for any α. Hence T ϕ is continuous and
X
X
|(T ϕ) (0)| ≤ C
kDα (T ϕ)kq = C
kT (Dα ϕ)kq
|α|≤n+1
≤ C kT k
|α|≤n+1
X
α
kD ϕkp .
|α|≤n+1
Hence the map ϕ → (T ϕ) (0) is a continuous map on S, and there exists a unique
ũ ∈ S 0 s.t.
(T ϕ) (0) = ũ (ϕ) .
0
Now we define u ∈ S s.t u (ϕ) = ũ (ϕ̃), then we have
(T ϕ) (x) = T (τx ϕ) (0) = ũ (ϕ (x + ·)) = u (ϕ (x − ·)) = (ϕ ∗ u) (x) .
¤
5.2. Singular integrals: the heart of the matter. In this section, we will study
a typical singular integral.
Theorem 36. Let K ∈ L2 (Rn ). We suppose:
(a)
¯
¯
¯
¯
¯K̂ (x)¯ ≤ B.
(b) K ∈ C 1 (Rn \ {0}) and
|∇K (x)| ≤ B |x|
1
n+1
.
p
For f ∈ L ∩ L , 1 < p < ∞, define
Z
(T f ) (x) =
K (x − y) f (y) dy.
Rn
Then
kT f kp ≤ Ap kf kp
where Ap = c (n, p) B.
Proof. We assume B = 1.
Step 1. T is of type (2, 2). Since T f = K ∗ f ,
n
Tcf = (2π) 2 K̂ · fˆ,
we have
° °
n
° °
kT f k2 = °Tcf ° ≤ (2π) 2
2
° °
n
° ˆ°
°f ° = (2π) 2 kf k2 .
2
Step 2. T is of weak type (1, 1). Fix α > 0, we define
F = {x : M f (x) ≤ t}
and
Ω = F c = {x : M f (x) > t} .
HARMONIC ANALYSIS
29
From Theorem 31, we can write Ω as the union of closed cubes, Ω = ∪k Qk , whose
interiors are disjoint and for each k,
diam Qk ≤ dist (Qk , F ) ≤ 4 diam Qk .
Moreover, for each k,
1
|Qk |
Z
f (x) dx ≤ cα.
Qk
Moreover,
c
kf k1 .
α
Here we use c to denote different constants depending only on n.
Now we define
½
for x ∈ F,
R f (x)
g (x) =
1
◦
|Qk | Qk f (x) dx for x ∈ Qk .
|Ω| ≤
Then g is defined a.e. Let b = f − g, then we have b (x) = 0 for x ∈ F and
Z
b (x) dx = 0
Qk
for each cube Qk .
Since
Z
Z
Z
g 2 (x) dx =
Rn
f 2 (x) dx +
g 2 (x) dx
F
Ω
Z
2
≤α
|f (x)| dx + (cα) |Ω|
F
≤ cα kf k1 .
We have
¯n
c
c
4
α o¯¯
¯
2
2
¯ ≤ 2 kT gk2 ≤ 2 kgk2 ≤ kf k1 .
¯ x : |T g (x)| >
2
α
α
α
Now we estimate T b. Write
bk (x) = b (x) χQk .
We have for any x ∈ F ,
¯Z
¯ ¯Z
¯
¯
¯ ¯
¯
¯
¯
¯
|T bk (x)| = ¯
K (x − y) b (y) dy ¯ = ¯
[K (x − y) − K (x)] b (y) dy ¯¯
Qk
Qk
Z
|b (y)|
≤ c diam Qk
n+1 dy.
Qk |x − y|
Since
Z
Z
|b (y)| dy ≤
Qk
|f (y)| dy + cα |Qk | ≤ cα |Qk | ,
Qk
and
diam Qk ≤ dist (Qk , F ) ≤ 4 diam Qk .
we have for any x ∈ F ,
Z
|T bk (x)| ≤ cα
Qk
δ (y)
n+1 dy
|x − y|
30
HUIQIANG JIANG
where δ (y) = dist (y, F ). Hence,
Z
δ (y)
|T b (x)| ≤ cα
Ω
|x − y|
n+1 dy.
Applying Lemma 4, we have
Z
|T b (x)| dx ≤ cα |Ω| ≤ c kf k1 .
F
Hence,
¯n
α o¯¯
¯
¯ x ∈ Rn : |T b (x)| >
¯
2 o¯
¯n
α
¯
¯
≤ ¯ x ∈ F : |T b (x)| >
¯ + |Ω|
2
c
≤ kf k1 .
α
Hence
|{x : |T f (x)| > α}|
¯n
α o¯¯ ¯¯n
α o¯¯
¯
≤ ¯ x : |T g (x)| >
¯ + ¯ x : |T b (x)| >
¯
2
2
c
≤ kf k1 .
α
Step 3. Interpolation theorem implies the case 1 < p < 2.
Step 4. The case 2 < p < ∞ comes from the standard duality argument since
Z
(K ∗ f ) (x) ϕ (x) dx
Rn
Z
=
(K ∗ ϕ̃) (−x) ϕ (x) dx.
Rn
¤
2
n
Remark 13. The assumption K ∈ L (R ) can be dropped.
Remark 14. We could use standard Calderón Zygmund decomposition instead of
Theorem 31.
5.3. Singular integrals: Some extension and variants.
Theorem 37. Let K ∈ L2 (Rn ). We suppose:
(a)
¯
¯
¯
¯
¯K̂ (x)¯ ≤ B.
(b) For any |y| > 0,
Z
|K (x − y) − K (x)| dx ≤ B.
|x|≥2|y|
For f ∈ L1 ∩ Lp , 1 < p < ∞, define
Z
(T f ) (x) =
K (x − y) f (y) dy.
Rn
Then
kT f kp ≤ Ap kf kp
where Ap = c (n, p) B.
HARMONIC ANALYSIS
31
Proof. We only need to establish weak (1, 1) estimate. Let F, Ω and
√ Qk be defined
in the Calderon Zygmund decomposition to |f |. We define Q∗k = 2 nQk ,
[
Ω∗ =
Q∗k and F ∗ = Rn \Ω∗ .
k
Let yk be the center of Qk . We have for any x 6∈ Q∗k and y ∈ Qk ,
|x − yk | ≥ 2 |y − yk | .
Since
Z
T bk (x) =
Z
K (x − y) b (y) dy =
[K (x − y) − K (x − yk )] b (y) dy,
Qk
Qk
we have for any x ∈ F ∗
Z
XZ
|T b (x)| ≤
F∗
=
k
x6∈Q∗
k
k
Qk
XZ
Z
≤
Ω
Hence,
Z
|[K (x − y) − K (x − yk )] b (y)| dydx
y∈Qk
|b (y)| dy
|b (y)| dy ≤ c kf k1 .
¯n
α o¯¯
¯
¯ x ∈ Rn : |T b (x)| >
¯
2 o¯
¯n
α
¯
¯
≤ ¯ x ∈ F ∗ : |T b (x)| >
¯ + |Ω∗ |
2
c
c
≤ kf k1 + c |Ω| ≤ kf k1 .
α
α
¤
Remark 15. Theorem 36 is a special case of the above theorem.
Theorem 38. Suppose the kernel K satisfies
(a) For any |x| > 0,
−n
|K (x)| ≤ B |x| .
(b) For any |y| > 0,
Z
|K (x − y) − K (x)| dx ≤ B.
|x|≥2|y|
(c) For any 0 < R1 < R2 < ∞,
Z
K (x) dx = 0
R1 <|x|<R2
For f ∈ Lp , 1 < p < ∞, define for ε > 0,
Z
(Tε f ) (x) =
f (x − y) K (y) dy.
|y|≥ε
Then
kTε f kp ≤ Ap kf kp
where Ap = c (n, p) B. Also,
lim Tε f = T f
ε→0
32
HUIQIANG JIANG
exists in Lp for each f ∈ Lp and
kT f kp ≤ Ap kf kp .
Lemma 7. Let
½
K (x) for |x| ≥ ε,
0
for |x| < ε.
¯
¯
¯
¯c
sup ¯K
ε (y)¯ ≤ cB
Kε (x) =
Then
y
where c = c (n).
Proof. We first assume ε = 1. We observe that K1 satisfies
(a) For any |x| > 0,
−n
|K1 (x)| ≤ B |x| .
(b) For any |y| > 0,
Z
|K1 (x − y) − K1 (x)| dx ≤ cB.
|x|≥2|y|
(c) For any 0 < R1 < R2 < ∞,
Z
K1 (x) dx = 0.
R1 <|x|<R2
Z
To see this,
|K1 (x − y) − K1 (x)| dx
|x|≥2|y|
Z
Z
≤
|K1 (x − y) − K1 (x)| dx +
|x|≥2|y|
Z
≤B+
1≤|x−y|≤ 23
Z
Z
|x|≥2|y|
|x|≤1
|K1 (x − y)| dx +
|K1 (x − y)| dx +
|x|≥2|y|
|x−y|≤1
|K1 (x)| dx
|K1 (x)| dx ≤ cB.
1≤|x|≤2
Now since K1 ∈ L2 (Rn ), for any |y| > 0,
Z
1
K̂1 (y) = lim
K1 (x) e−ix·y dx
R→∞ (2π)n/2 |x|≤R
Z
Z
1
1
−ix·y
=
K
(x)
e
dx
+
lim
K1 (x) e−ix·y dx.
1
n/2
n/2 2π
2π
R→∞
|x|≤ |y|
≤|x|≤R
(2π)
(2π)
|y|
We estimate
¯Z
¯
¯
¯
¯
¯
−ix·y
K1 (x) e
dx¯
¯
¯
¯ |x|≤ 2π
|y|
¯Z
¯
¯
£ −ix·y
¤ ¯¯
¯
=¯
K1 (x) e
− 1 dx¯
¯ |x|≤ 2π
¯
|y|
Z
|x| |K1 (x)| dx
≤ |y|
2π
|x|≤ |y|
Z
1
≤ B |y|
2π
|x|≤ |y|
= 2nωn πB.
|x|
n−1 dx
HARMONIC ANALYSIS
πy
.
|y|2
On the other hand, let z =
33
We have
e−iz·y = e−iπ = −1
and
Z
2π
≤|x|≤R
|y|
K1 (x − z) e−ix·y dx
Z
=−
2π
≤|x+z|≤R
|y|
Hence,
K1 (x) e−ix·y dx.
Z
lim
R→∞
2π
≤|x|≤R
|y|
K1 (x) e−ix·y dx
Z
1
lim
(K1 (x) − K1 (x − z)) e−ix·y
2π
2 R→∞ |y|
≤|x|≤R
Z
Z
1
1
−ix·y
K
(x)
e
K1 (x) e−ix·y dx
−
dx
+
2π
2π
1
2 2π|x|≤ |y|
2 2π|x|≥ |y|
=
|y|
≤|x+z|
|y|
≥|x+z|
= I + II + III.
We have
|II| ≤
1
2
1
|III| ≤
2
and
Z
Z
π
2π
≤|x|≤ |y|
|y|
2π
3π
≤|x|≤ |y|
|y|
Z
|K1 (x)| dx ≤ cB,
|K1 (x)| dx ≤ cB
Z
|I| ≤
2π
≤|x|≤∞
|y|
|K1 (x) − K1 (x − z)| dx =
|K1 (x) − K1 (x − z)| dx ≤ cB
|x|≥2|z|
Hence, we conclude
¯
¯
¯c
¯
sup ¯K
1 (y)¯ ≤ cB.
y
The general ε case can be obtained by a scaling.
Now we are ready to prove Theorem 38:
Proof of Theorem 38. Since
¯
¯
¯c
¯
sup ¯K
ε (y)¯ ≤ cB,
y
we have
kTε f kp ≤ Ap kf kp
where Ap = c (n, p) B. Now assume f ∈ C0∞ (Rn ), then we have
Z
(Tε f ) (x) =
f (x − y) K (y) dy
|y|≥ε
Z
Z
=
f (x − y) K (y) dy +
(f (x − y) − f (x)) K (y) dy
|y|≥1
= I + II.
ε≤|y|≤1
¤
34
HUIQIANG JIANG
Since f ∈ L1 and K ∈ Lp (Rn \B1 (0)), I ∈ Lp . Now II is compactly supported
and since
|f (x − y) − f (x)| ≤ A |y| ,
II converges uniformly to
Z
(f (x − y) − f (x)) K (y) dy.
|y|≤1
Hence,
Z
Z
(f (x − y) − f (x)) K (y) dy
f (x − y) K (y) dy +
Tε f →
|y|≤1
|y|≥1
in Lp . Now standard density argument implies Tε f → T f for each f ∈ Lp .
¤
Let
1
, x ∈ R1 .
πx
It is easy to check that K satisfies the assumptions in Theorem 38. Hence for any
f ∈ Lp , the Hilbert transform
Z
f (x − y)
1
T f = lim
dy
+
y
ε→0 π |y|≥ε
K (x) =
exists in the Lp norm.
5.3.1. Singular integral operators which commute with dilations. Let
Tf = K ∗ f
p
for f ∈ L . T commutes with dilation iff K is homogeneous of degree −n. i.e.,
K (x) =
Ω (x)
n ,
|x|
where Ω is homogeneous of degree 0, i.e.,
Ω (εx) = Ω (x) for any ε > 0.
When K is homogeneous of degree −n, the cancellation property in Theorem 38
is equivalent to
Z
Ω (x) dσ = 0.
Sn−1
For the assumption b, we have
Lemma 8. Assume that Ω satisfies the following ”Dini-type” condition
Z 1
ω (δ)
dδ < ∞
(5.1)
δ
0
where the modulus of continuity
ω (δ) =
sup |Ω (x) − Ω (x0 )| .
|x−x0 |≤δ,
|x|=|x0 |=1
Then there exists B > 0, such that for any y > 0,
Z
|K (x − y) − K (x)| dx ≤ B.
|x|≥2|y|
HARMONIC ANALYSIS
Proof.
35
·
¸
Ω (x − y) − Ω (x)
1
1
K (x − y) − K (x) =
+ Ω (x)
n
n −
n .
|x − y|
|x − y|
|x|
Now
¯
¯
¯
1
1 ¯¯
¯
n −
n dx ≤ c.
¯
|x| ¯
|x|≥2|y| |x − y|
Z
Since
¯
¯
¯y¯
¯ x−y
x ¯¯
¯ ¯
¯
≤
c
−
¯ ¯
¯ |x − y| |x| ¯
x
holds for any |x| ≥ 2 |y|, we have
¯ µ
µ
¶
µ ¶¯
¶
¯
x−y
|y|
x ¯¯
|Ω (x − y) − Ω (x)| = ¯¯Ω
−Ω
≤
ω
c
|x − y|
|x| ¯
|x|
and
¯
¯
¯ Ω (x − y) − Ω (x) ¯
¯
¯ dx
n
¯
¯
|x − y|
|x|≥2|y|
µ
¶
Z
|y|
1
≤c
ω c
n dx
|x|
|x|
|x|≥2|y|
Z 2c
ω (δ)
=c
dδ < ∞.
δ
0
Z
Hence,
Z
Z
c
2
|K (x − y) − K (x)| dx ≤ c
|x|≥2|y|
ω (δ)
dδ + c kΩk∞
δ
0
µZ
1
≤c
0
¶
ω (δ)
dδ + kΩk∞ .
δ
¤
Remark 16. The Dini-type condition is satisfied if Ω is Hölder continuous.
Theorem 39. Let Ω be homogeneous of degree 0. Suppose that
Z
Ω (x) dσ = 0
Sn−1
and Dini condition (5.1) holds. For 1 < p < ∞ and f ∈ Lp (Rn ), define
Z
Ω (y)
Tε f =
n f (x − y) dy.
|y|≥ε |y|
(a) There exists constant Ap independent of ε, s.t.
kTε f kp ≤ Ap kf kp .
(b)
lim Tε f = T f
ε→0
exists in Lp norm, and
kT f kp ≤ Ap kf kp .
36
HUIQIANG JIANG
(c) If f ∈ L2 (Rn ), then the Fourier transforms of f and T f are related by
Tcf (y) = m (y) fˆ (y)
where m is a homogeneous function of degree 0.
Proof. (a) (b) follows from Theorem 38.
(c) Let
(
Ω(x)
|x|n
Kε,η (x) =
0
Then we have
K̂ε,η (y) =
if ε ≤ |x| ≤ η,
otherwise.
Z
1
n/2
(2π)
ε≤|x|≤η
Ω (x) −ix·y
dx.
n e
|x|
Writing r = |x|, R = |y|, x = |x| x0 and y = |y| y 0 for x, y 6= 0, we have for any
y 6= 0,
Z
1
Ω (x0 ) −iRrx0 ·y0
K̂ε,η (y) =
e
dx
n/2
rn
ε≤|x|≤η
(2π)
Z
Z η
1
Ω (x0 ) −iRrx0 ·y0
=
e
drdσ (x0 )
n/2
r
n−1
S
ε
(2π)
Z
Z η
´
1
Ω (x0 ) ³ −iRrx0 ·y0
=
e
−
cos
(Rr)
drdσ (x0 )
n/2
r
Sn−1 ε
(2π)
Z
1
=
Ω (x0 ) Iε,η (x0 , y) dσ (x0 )
n/2
Sn−1
(2π)
where
Z
Iε,η (x0 , y) =
ε
η
´
1 ³ −iRrx0 ·y0
e
− cos (Rr) dr.
r
Now,
Z
η
1
sin (Rrx0 · y 0 ) dr
ε r
Z ∞
sin t
π
→−
dt sign (x0 · y 0 ) = − sign (x0 · y 0 )
t
2
0
Im Iε,η = −
as ε → 0 and η → ∞. Moreover, |Im Iε,η | is uniformly bounded for 0 < ε < η < ∞.
On the other hand,
Z η
1
Re Iε,η =
[cos (Rrx0 · y 0 ) − cos (Rr)] dr
ε r
1
→ ln 0 0
|x · y |
HARMONIC ANALYSIS
37
as ε → 0, η → ∞ and |Re Iε,η | ≤ 2 ln |x01·y0 | . To see this, we consider for λ, µ > 0,
Z η
cos (λr) − cos (µr)
dr
r
ε
Z λη
Z µη
cos r
cos r
=
dr −
dr
r
r
λε
µε
Z λη
Z µε
cos r
cos r
=
dr +
dr
r
r
µη
λε
µ ¶
Z λ
Z µε
Z µε
1
r
1
cos r − 1
=
cos
dr +
dr +
dr
r
η
r
r
µ
λε
λε
µ ¶
Z λ
Z µε
1
µ
cos r − 1
r
=
cos
dr + ln +
dr.
r
η
λ
r
µ
λε
Since
¯Z
¯
¯
¯
µε
λε
we have
Z
lim
ε→0,
η→∞
Moreover
¯Z
¯
¯
¯
ε
η
¯
cos r − 1 ¯¯
dr¯ ≤ |µ − λ| ε,
r
cos (λr) − cos (µr)
µ
dr = ln .
r
λ
¯
cos (λr) − cos (µr) ¯¯
dr¯
r
ε
¯
¯Z
¯ λη cos r ¯ ¯¯Z µε cos r ¯¯
¯ µ¯
¯
¯
¯
¯
dr¯ + ¯¯
dr¯¯ ≤ 2 ¯ln ¯ .
≤¯
¯
¯ µη
r
r
λ
λε
η
Hence, we have
lim Iε,η (x0 , y) = −
ε→0,
η→∞
πi
1
sign (x0 · y 0 ) + ln 0 0
2
|x · y |
and for any 0 < ε < η < ∞,
µ
|Iε,η (x0 , y)| ≤ c ln
1
+1
0
|x · y 0 |
¶
for some universal constant c. Since for any |y 0 | = 1,
Z
1
ln 0 0 dσ (x0 ) ≤ c,
|x
·y |
n−1
S
we have for some c = c (n) ,
sup
0<ε<η<∞
°
°
°
°
°K̂ε,η (y)°
∞
≤ c,
and Lebesgue dominated convergence implies
Z
1
lim K̂ε,η (y) =
Ω (x0 ) I (x0 , y 0 ) dσ (x0 )
n/2
ε→0,
n−1
S
(2π)
η→∞
where
I (x0 , y 0 ) = −
πi
1
sign (x0 · y 0 ) + ln 0 0 .
2
|x · y |
38
HUIQIANG JIANG
Now for any f ∈ L2 , we have
n
ˆ
ˆ
2
T[
ε,η f (y) = (2π) K̂ε,η f (y) → m (y) f (y)
in L2 as ε → 0, η → ∞, where
Z
Ω (x0 ) I (x0 , y 0 ) dσ (x0 ) .
m (y) =
Sn−1
Hence, Tε,η f converges to some function g ∈ L2 . Now fix ε, it is easy to see
Tε,η f → Tε f in L2 . So we conclude g = T f . Hence
Tcf (y) = m (y) fˆ (y) .
¤
From the proof, we see
Z
Ω (x0 ) I (x0 , y 0 ) dσ (x0 )
m (y) =
Sn−1
with
I (x0 , y 0 ) = −
πi
1
sign (x0 · y 0 ) + ln 0 0 .
2
|x · y |
of Hilbert transform, we have
1
In particular, for the kernel πx
Z
i
m (y) =
− x0 sign (x0 · y 0 ) dσ (x0 ) = −i sign (y) .
2
n−1
S
Theorem 40. Suppose Ω satisfies the conditions of the previous theorem. Let
f ∈ Lp (Rn ), 1 ≤ p < ∞.
(a) limε→0 Tε f (x) exists for a.e. x;
(b) Let
T ∗ f (x) = sup |Tε f (x)| .
ε>0
Then the map f 7→ T ∗ f is of weak type (1, 1).
(c) If 1 < p < ∞, then kT ∗ f kp ≤ Ap kf kp .
Proof. Check the proof of (b) and (c) in Stein’s book if you are interested. We now
prove (a) assuming that (b) and (c) hold.
We define for any f ∈ Lp (Rn ),
¯
¯
¯
¯
Λf (x) = ¯¯lim sup Tε f (x) − lim inf Tε f (x)¯¯ .
ε→0
ε→0
Then Λf (x) ≤ 2T ∗ f (x). We also observe that if f ∈ D (Rn ), then Λf (x) ≡ 0. If
1 < p < ∞, writing f = f1 + f2 , where f1 ∈ D (Rn ), we have
kΛf kp = kΛf2 kp ≤ 2 kT ∗ f2 kp ≤ 2Ap kf2 kp .
Since we can make kf2 kp arbitrarily small, we have kΛf kp = 0. Hence, Λf (x) = 0
a.e. in Rn . If p = 1, we have for any t > 0,
¯½
¾¯
¯ ∗
t ¯¯
¯
|{Λf > t}| = |{Λf2 > t}| ≤ ¯ T f2 >
2 ¯
2c
kf2 k1 .
t
Since we can make kf2 k1 arbitrarily small, we have |{Λf > t}| = 0 for every t > 0,
i.e., Λf (x) = 0 a.e. in Rn .
¤
≤
HARMONIC ANALYSIS
39
5.4. Hilbert transform on a circle and Fourier series. For any f ∈ Lp [−π, π],
we can define the Hilbert transform
Z
cot (x/2)
Sf (y) = lim
f (x − y) dx.
ε→0 ε≤|x|≤π
2π
Comparing this with the modified Hilbert transform
Z
1
f (x − y) dx.
S̃f (y) = lim
ε→0 ε≤|x|≤π πx
We see that Sf (y) exists for a.e. y ∈ [−π, π] and S is of type (p, p) for any
1 < p < ∞.
Lemma 9. For any n ∈ Z,
Proof. Since
1
2π
Rπ
−π
¡
¢
S einx = −i sign (n) einx .
sin[(N + 12 )y ]
dy
sin y2
= 1 for N = 0, 1, 2, · · · , we have
Z
¡ ¡ inx ¢¢
S e
y = lim
ε→0
= −ie
ε≤|x|≤π
Z π
iny
−π
π
Z
= −ieiny
−π
cos (x/2) in(y−x)
e
dx
2π sin (x/2)
cos (x/2)
sin (nx) dx
2π sin (x/2)
¡¡
¢ ¢
¡¡
¢ ¢
sin n + 12 x + sin n − 12 x
dx
4π sin (x/2)
= −i sign (n) einy .
¤
Hence, if
f (x) =
then
iSf (y) =
X
P0 f =
then
Tf =
1
2π
Z
an einx ,
an einx −
n≥1
We define
X
X
an einx .
n≤−1
π
f (x) dx = a0 (f ) ,
−π
X
1
(iS + I + P0 ) f =
an einx .
2
n≥0
We also define for any integer k,
Mk f = eikx f
It is easy to verify that
sN = M−N T MN − MN +1 T M−(N +1) .
Now we apply the singular integral theory to prove the Lp convergence of Fourier
series:
Theorem 41. Let f ∈ Lp [−π, π], 1 < p < ∞, then sN (f ) → f in Lp [−π, π] as
N → ∞.
40
HUIQIANG JIANG
Proof. Since S is of type (p, p) and P0 is of type (p, p), we conclude T is of type
(p, p). Obviously MN is of type (p, p),
sN is of type (p, p). Since sN (f ) → f
¡ hence
¢
in Lp [−π, π] as N → ∞ for f ∈ C 2 S 1 , standard density argument implies that
sN (f ) → f in Lp [−π, π] as N → ∞ for every f ∈ Lp [−π, π].
¤
6. Riesz transforms, Poisson Integrals and Spherical Harmonics
6.1. The Riesz transforms. Recall that for f ∈ Lp (R), 1 ≤ p < ∞, the Hilbert
transform of f is defined by
Z
1
f (x − y)
Hf (y) = lim+
dy, y ∈ R.
y
ε→0 π |y|≥ε
Since
d = −i sign (y) fˆ,
Hf
Hilbert transform is unitary on L2 (R) and H 2 = −I.
For any δ 6= 0, we define the dilation operator τδ (f ) (x) = f (δx). For any δ > 0,
Hτδ = τδ H and for any δ < 0, Hτδ = −τδ H.
Theorem 42. Suppose T is a bounded operator on L2 (R) which satisfies
(a) T commutes with translations;
(b) for any δ > 0, T τδ = τδ T and for any δ < 0, T τδ = −τδ T .
Then T is a constant multiple of the Hilbert transform.
Proof. There exists m ∈ L∞ (R), s.t.
Tcf (y) = m (y) fˆ (y) for a.e. y ∈ R.
Since
−1
Fτδ = |δ|
τδ−1 F.
We have
¢
−1
−1 ¡
T[
τδ f (y) = m (y) τd
τδ−1 fˆ (y) = m (y) |δ| fˆ δ −1 y ,
δ f (y) = m (y) |δ|
and
−1
T[
τδ f (y) = sign (δ) τ[
τδ−1 Tcf (y)
δ T f (y) = sign (δ) |δ|
¡
¢
¡
¢ ¡
¢
−1
−1
= sign (δ) |δ| Tcf δ −1 y = sign (δ) |δ| m δ −1 y fˆ δ −1 y .
Hence
¡
¢
m (y) = sign (δ) m δ −1 y
which implies that for some constant c, m (y) = −ci sign (y) a.e..
¤
Now we generalize the Hilbert transform to multi-dimensions. For any f ∈
Lp (Rn ), 1 ≤ p < ∞, the Riesz transform of f is defined by
Z
yj
Rj f (x) = lim cn
n+1 f (x − y) dy, j = 1, 2, · · · , n
ε→0
|y|≥ε |y|
where
¡
¢
Γ n+1
2
cn = (n+1)/2 .
π
HARMONIC ANALYSIS
Ω (x)
41
x
j
For the Riesz kernel Kj (x) = |x|
and Ωj (x) = cn |x|j , we have
n
Z
m (y) =
Ωj (x0 ) I (x0 , y 0 ) dσ (x0 )
Sn−1
·
¸
Z
πi
1
= cn
x0 − sign (x0 · y 0 ) + ln 0 0 dσ (x0 )
2
|x · y |
Sn−1
Z
πi
= − cn
x0 sign (x0 · y 0 ) dσ (x0 ) .
2
Sn−1
y
Since m commutes with rotations, we have m (y) = c |y|
. Choose y = en , we have
for n ≥ 2,
Z
πi
c = − cn
x0n sign (x0n ) dσ (x0 )
2
Sn−1
Z
πi
= − cn
|x0n | dσ (x0 )
2
Sn−1
Z 1
¡
¢ n−2
dt
= −πicn
t 1 − t2 2 (n − 1) ωn−1 √
1 − t2
0
¡ n+1
¢
Γ 2
= −πi (n+1)/2
ωn−1
π
¡ n+1 ¢
n−1
Γ 2
π 2
¡ n+1 ¢
= −πi (n+1)/2
π
Γ 2
= −i.
Here we used
ωn =
π n/2
¡
¢.
Γ 1 + n2
Hence,
yj ˆ
d
R
f (y) .
j f (y) = −i
|y|
To understand the interaction of dilations and rotations with the Riesz transform,
we consider multipliers which commute with dilations and rotations.
Lemma 10. Suppose m : Rn → Rn is homogeneous of degree 0, and for any
rotation ρ,
m (ρx) = ρm (x) .
Then
x
m (x) = c
|x|
for some constant c.
Proof. Fix any unit vector e, we have
m (ρe) = ρm (e) .
Hence, ρe = e implies ρm (e) = m (e), so there exists constant c (e) s.t. m (e) =
c (e) e. Now let e1 , e2 be two unit vectors, and ρ be a rotation such that ρe1 = e2 ,
then
m (ρe1 ) = ρm (e1 ) = c (e1 ) e2 ,
m (ρe1 ) = m (e2 ) = c (e2 ) e2 ,
42
HUIQIANG JIANG
hence c (e1 ) = c (e2 ). Hence m (e) = ce. Since m is homogeneous of degree 0,
m (x) = c
x
|x|
for some constant c.
¤
To translate the above lemma in terms of Riesz transforms, we consider the
interaction of dilations and rotations with Fourier transform. For any δ > 0, we
have
Fτδ = δ −n τδ−1 F.
And for any rotation about the origin ρ, we define ρf (x) = f (ρx), then
Z
1
(Fρf ) (y) =
e−ix·y f (ρx) dx
n/2
Rn
(2π)
Z
−1
1
e−iρ x·y f (x) dx
=
n/2
n
R
(2π)
Z
1
=
e−ix·ρy f (x) dx
n/2
Rn
(2π)
= (ρFf ) (y) .
Proposition 8. Let T = (T1 , T2 , · · · , Tn ) be an n-tuple of bounded transformations
on L2 (Rn ). Suppose
(a) Each Tj commutes with the translation in Rn ;
(b) Each Tj commutes with the dilations in Rn ;
(c) For every rotation ρ of Rn ,
ρTj ρ−1 f =
n
X
ρjk Tk f.
k=1
Then there exists constant c such that Tj = cRj , j = 1, 2, · · · , n.
We now consider a function u ∈ C 2 (Rn ) with compact support. We have
u[
xi xj = −yi yj û = −
yi yj
2
|y| û = −R\
i Rj ∆u.
|y| |y|
Hence,
∂2u
= −Ri Rj ∆u.
∂xi ∂xj
i.e., once we know ∆u, we know all second order partial derivatives of u.
Theorem 43. Suppose u ∈ C 2 (Rn ) is compact supported. Then for any 1 < p <
∞, we have
° 2 °
° ∂ u °
°
°
° ∂xi ∂xj ° ≤ Ap kukp
p
where Ap is a constant depending only on n and p.
HARMONIC ANALYSIS
43
6.2. Poisson Integral. We write Rn+1
= {(x, t) : x ∈ Rn , t > 0}. Given a func+
n
tion f defined on R , we want to find harmonic function u on Rn+1
such that u = f
+
n
n
˜
on R . We use ∆ to denote the Laplacian on R , and ∆ the Laplacian on Rn+1 .
Then
˜ = ∆u + utt = 0.
∆u
We write û = û (y, t) the Fourier transform of u in x-variable, then
2
− |y| û + ûtt = 0.
We also impose boundary condition û (y, 0) = fˆ and û (y, ∞) = 0. Then formally
û = e−|y|t fˆ.
Hence,
u (x, t) =
Z
1
(2π)
eix·y e−|y|t fˆ (y) dy.
n/2
Rn
If f ∈ L2 (Rn ), u defined above is a smooth harmonic function in Rn+1
and
+
u (·, t) → f in L2 (Rn ) norm as t → 0+ . Hence, indeed u is a solution to the
Dirichlet problem in Rn+1
+ .
Now we write
µZ
¶
Z
1
ix·y −|y|t
−iyz
u (x, t) =
e
e
e
f
(z)
dz
dy
n
(2π) Rn
Rn
Z
=
P (x − z, t) f (z) dz
Rn
where
P (x, t) =
1
n
(2π)
Z
eix·y e−|y|t dy
Rn
is called the Poisson kernel in Rn+1
+ .
Lemma 11. For any γ > 0,
e
−γ
1
=√
π
Z
0
∞
e−u − γ 2
√ e 4u du.
u
Proof.
e
−γ
1
=
π
Z
∞
−∞
eiγs
1
ds =
2
1+s
π
Z
µZ
∞
e
−∞
iγs
∞
¶
−(1+s2 )u
e
du ds.
0
¤
Proposition 9.
P (x, t) = ³
cn t
2
|x| + t2
with cn =
´ n+1
2
Γ
¡ n+1 ¢
2
π
n+1
2
.
44
HUIQIANG JIANG
Proof.
Z
1
eix·y e−|y|t dy
n
(2π) Rn
µ
¶
Z
Z ∞ −u
|y|2 t2
1
e
1
ix·y
− 4u
√
√
e
e
du
dy
=
n
π 0
u
(2π) Rn
¶
Z ∞ −u µZ
|y|2 t2
1
1
e
ix·y − 4u
√
√
=
e
e
dy
du.
n
π 0
u
(2π)
Rn
P (x, t) =
Now
Z
eix·y e−
Rn
Z
=
e
dy
¯
¯2
√
¯
¯ √
|x|2 u
−¯ 2ty
−i tu x¯ − t2
u
Rn
Hence
|y|2 t2
4u
= e−
|x|2 u
t2
= e−
|x|2 u
t2
= e−
|x|2 u
t2
Z
dy
¯
¯
¯ √ ¯2
−¯ 2ty
¯
u
e
dy
Rn
µ √ ¶n µZ ∞
¶n
2 u
−s2
e ds
t
−∞
µ √ ¶n
2 πu
.
t
√ n Z ∞ −u
|x|2 u
1 (2 π)
e
√
√ t−n un/2 e− t2 du
n
π
u
(2π)
0
Z ∞
n−1
t
1
= n+1 ³
e−s s 2 ds
n+1
´
2
π 2
0
2
|x| + t2
¡
¢
Γ n+1
t
2
=
.
n+1
³
´ n+1
2
π 2
2
2
|x| + t
P (x, t) =
¤
Proposition 10. The Poisson kernel P (x, t) satisfies:
(i) P > 0 for any x ∈ Rn , t > 0;
(ii) for any t > 0,
Z
P (x, t) dx = 1.
Rn
(iii) For any δ > 0,
Z
lim+
t→0
P (x, t) dx = 0.
|x|≥δ
Proof. Direct verification.
¤
Theorem 44. Suppose f ∈ Lp (Rn ), 1 ≤ p ≤ ∞, and let u (x, t) = P (·, t) ∗ f.
(i) u is a smooth harmonic function in Rn+1
+ ;
(ii) for each x ∈ Rn ,
sup |u (x, t)| ≤ M f (x) ;
t>0
HARMONIC ANALYSIS
45
(iii) If p < ∞, u (·, t) → f in Lp (Rn ) as t → 0+ ;
(iv) For a.e. x ∈ Rn ,
lim+ u (x, t) = f (x) ;
t→0
(v) If p = ∞ and f is continuous in Rn , then u (x, t) → f (x) uniformly on any
compact subset of Rn as t → 0+ .
Proof. (i) Trivial.
(ii) Assume x = 0 and M f (0) < ∞. Define
Z
g (r) =
|f (x)| dx ≤ ωn rn M f (0) ,
Br (0)
then
¯Z
¯ Z
¯
¯
¯
P (−x, t) |f (x)| dx
P (−x, t) f (x) dx¯¯ ≤
|u (0, t)| = ¯
Rn
Rn
µZ
¶
Z ∞
=
P (r, t)
|f (x)| dσ (x) dr
0
∂Br
Z ∞
Z ∞
∞
0
=
P (r, t) g (r) dr = g (r) P (r, t)|0 −
Pr (r, t) g (r) dr
0
Z ∞
Z ∞ 0
=−
Pr (r, t) g (r) dr ≤ −M f (0)
Pr (r, t) ωn rn dr
0
0
Z ∞
∞
= − M f (0) ωn rn P (r, t)|0 + M f (0)
P (r, t) nωn rn−1 dr
0
= M f (0) .
Here we used the fact that P is radial and monotone decreasing in |x|.
(iii) It follows from the fact that P (·.t) is an approximate delta function as
t → 0+ .
(iv) First assume 1 ≤ p < ∞. Define for any x,
¯
¯
¯
¯
Ωf (x) = ¯¯lim sup u (x, t) − lim inf u (x, t)¯¯ .
t→0+
t→0+
Then
|Ωf (x)| ≤ 2M f (x)
and hence for any δ > 0,
¯½
¾¯
¯
δ ¯¯
¯
|{x : Ωf (x) > δ}| ≤ ¯ x : Ωf (x) >
≤ Ap kf kp .
2 ¯
It is easy to verify that Ωf (x) = 0 if f is continuous and compact supported. For
any f ∈ Lp , writing f = f1 + f2 such that f1 is continuous and compact supported,
and kf2 kp small. Then we have
|{x : Ωf (x) > δ}| = |{x : Ωf2 (x) > δ}| ≤ Ap kf2 kp
which can be made arbitrarily small. Hence
|{x : Ωf (x) > δ}| = 0.
46
HUIQIANG JIANG
i.e., limt→0+ u (x, t) exists a.e. x ∈ Rn . Because of (iii)
lim u (x, t) = f (x)
t→0+
a.e..
Now for f ∈ L∞ (Rn ), if suffices to show that for any R > 0,
lim u (x, t) = f (x)
t→0+
a.e. x ∈ BR (0) ⊂ Rn . We write f = f1 + f2 where f1 = f χB2R , then
lim f2 ∗ P = 0
t→0
uniformly on BR (0). and the result follows from the L1 case for f1 .
(v) Similar.
¤
6.3. Spherical harmonics. Let Pk , k ≥ 0 be the set of all homogeneous polynomials in Rn of order k. Hence p ∈ Pk iff
X
p (x) =
cα xα .
|α|=k
Pk is a vector space of dimension
µ
¶
n+k−1
(n + k − 1)!
dk =
=
.
n−1
(n − 1)!k!
We introduce an inner product on Pk by
hp, qi = p (D) q̄.
Definition 12. A solid spherical harmonic of degree k is a homogeneous harmonic
polynomial in Rn of order k. A (surface) spherical harmonic of degree k is the
restriction on Sn−1 of solid spherical harmonic of degree k in Rn . We use Hk to
denote all spherical harmonics of degree k and we will not distinguish solid spherical
harmonic and surface spherical harmonic.
Lemma 12. For k ≥ 2, the linear map
ϕ : Pk → Pk−2
defined by ϕ (p) = ∆p is onto.
Proof. If not, there exists 0 6= q ∈ Pk−2 , q⊥∆p for any p ∈ Pk , hence
D
E
2
0 = hq, ∆pi = |x| q, p
2
which is a contradiction if we take p = |x| q.
¤
Remark 17. As a direct consequence of the lemmas, for k ≥ 2, the dimension of
Hk is
µ
¶ µ
¶
n+k−1
n+k−3
dk − dk−2 =
−
.
n−1
n−1
Moreover, the dimension of H0 is 1 and the dimension of H1 is n. When n = 3,
the dimension of Hk is 2k + 1 for every k ≥ 0.
HARMONIC ANALYSIS
47
Remark 18. Let ∆S denote the Laplace-Beltrami operator on Sn−1 , i.e., the spherical Laplacian. Then we have for any f defined on Sn−1 ,
∆S f = ∆g
³
where g = f
x
|x|
´
. In polar coordinates, we have for any f defined on Rn ,
∆f = frr +
1
n−1
fr + 2 ∆S f.
r
r
Hence, if p ∈ Hk , we have
∆S p = −k (k + n − 2) p.
i.e., p is an eigenfunction of the spherical Laplacian.
Theorem 45. If p ∈ Pk , then
2
2l
p (x) = p0 (x) + |x| p1 (x) + · · · + |x| pl (x)
where l =
£k¤
2
and pj (x) ∈ Hk−2j .
Proof. Since any polynomial of order ≤ 1 is harmonic, we can assume k ≥ 2. We
2
claim Pk is the direct sum of |x| Pk−2 and Hk . For any q ∈ Pk s.t.
D
E
2
0 = |x| p, q
holds for any p ∈ Pk−2 , we have
0 = hp, ∆qi ,
hence ∆q = 0 and q ∈ Hk .
The theorem then follows from induction.
¤
Corollary 7. The restriction of any polynomial on Sn−1 is a sum of spherical
harmonics.
Corollary 8.¡ The ¢collection¡ of finite
¢ linear combinations of spherical harmonics
is dense in C Sn−1 and L2 Sn−1 .
¡
¢
¡
¢
Proof. Density in C Sn−1¡ follows
from Weierstrass theorem. Density in L2 Sn−1
¢
follows from density in C Sn−1 .
¤
¢
¡
The inner product in L2 Sn−1 is defined by
Z
(f, g) =
f (x) ḡ (x) dσ (x) .
Sn−1
∞
Theorem
¡
¢ 46. The finite dimensional spaces {Hk }k=0 are mutually orthogonal in
L2 Sn−1 . Hence
∞
¡
¢ X
Hk .
L2 Sn−1 =
k=0
48
HUIQIANG JIANG
Proof. If k 6= j and p ∈ Hk , q ∈ Hj , then
Z
(k − j) (p, q) = (k − j)
p (x) q̄ (x) dσ (x)
Sn−1
Z
∂p
∂ q̄
=
q̄ −
pdσ (x)
∂ν
n−1 ∂ν
ZS
=
∆pq̄ − ∆q̄pdσ (x) = 0.
|x|≤1
Hence (p, q) = 0.
¤
When n = 2, the above decomposition gives rise to the Fourier series on S1 .
¡
¢
Proposition 11. Suppose f ∈ L2 Sn−1 and
f=
∞
X
pk
k=0
where pk ∈ Hk . Then f is smooth iff for any N > 0,
Z
¡
¢
2
|pk | = O k −N
S n−1
as k → ∞.
Proof. If f is smooth,for any positive integer r,
Z
Z
Z
r
2
r
r
|pk | ,
pk ∆s f = ∆s pk f = [−k (k + n − 2)]
hence for k large,
Z
2
|pk | ≤
1
r
|k (k + n − 2)|
So we have
µZ
2
¶ 21 µZ
|pk |
Z
2
|pk | ≤
|∆rs f |
2
¶ 12
≤
c
³R
|pk |
k 2r
2
´ 12
.
c
.
k 4r
On the other hand, if for any N > 0,
Z
¡
¢
2
|pk | = O k −N
S n−1
as k → ∞, we can show that
Z
2
|∆rs f | ≤ c.
¢
¡
Hence, f ∈ H 2r Sn−1 . Sobolev emdedding implies f is smooth.
¤
Theorem 47 (Identity of Hecke). Suppose p is a solid spherical harmonic of degree
k in Rn , then
2
−|x|2 /2 = (−i)k p (y) e−|y| /2 .
pe\
HARMONIC ANALYSIS
Proof. We have
−|x|2 /2 (y) =
pe\
=
=
Z
1
(2π)
1
(2π)
1
(2π)
49
n/2
p (x) e−|x|
Rn
2
e−|y|
n/2
2
/2
e
dx
p (x) e−|x+iy|
Z
2
/2 −ix·y
Z
e−|y|
n/2
= q (y) e−|y|
2
2
/2
dx
Rn
p (x − iy) e−|x|
/2
2
/2
dx
Rn
/2
where q is a polynomial. Replacing y by iy, we also have
Z
2
1
p (x + y) e−|x| /2 dx = p (y)
q (iy) =
n/2
n
R
(2π)
since p is harmonic. Hence
k
q (y) = p (−iy) = (−i) p (y) .
¤
6.4. Higher Riesz Transforms. We consider for k ≥ 1 the singular integral with
p(x)
the kernel |x|
n+k where p ∈ Hk . When k = 1, we recover Riesz transform. Such
singular integrals are hence called higher Riesz transforms.
The main theorem we want to establish is:
Theorem 48. Let p ∈ Hk , k ≥ 1. Then the multiplier corresponding to the
transform
Z
p (x)
T f (y) = lim
f (y − x) dx
ε→0 |x|≥ε |x|n+k
is
p (x)
γk
k
|x|
where
¡ ¢
n
π 2 Γ k2
¢.
¡
γk = (−i)
Γ k+n
2
k
Remark 19. When k = 1, we have
¡ ¢
n
π 2 Γ 12
π (n+1)/2
γ1 = −i ¡ 1+n ¢ = −i ¡ n+1 ¢
Γ 2
Γ 2
which recovers the multiplier for Riesz transform.
We first prove several lemmas.
Lemma 13. Let p ∈ Hk , k ≥ 0. Then
2
|y|
n
−δ|x|2 /2 = δ −k− 2 (−i)k p (y) e− 2δ .
pe\
Proof. Simple scaling.
¤
50
HUIQIANG JIANG
Lemma 14. Let p ∈ Hk , k ≥ 0. For any α ∈ (0, n) and for any ϕ ∈ S, we have
Z
Z
p (x)
p (x)
ϕ̂
(x)
dx
=
γ
ϕ (x) dx
k,α
n+k−α
k+α
Rn |x|
Rn |x|
where
k
α− n
2
γk,α = (−i) 2
¡
¢
Γ k+α
2
¢.
¡
Γ k+n−α
2
Proof. For any δ > 0, we have
Z
Z
2
n
k
p (x) e−δ|x| /2 ϕ̂ (x) dx = δ −k− 2 (−i)
Rn
p (x) e−
|x|2
2δ
ϕ (x) dx.
Rn
Multiplying both sides with δ β−1 with
k+n−α
,
2
β=
and integrate δ from 0 to ∞, we have
¶
Z µZ ∞
Z
2
k
δ β−1 e−δ|x| /2 dδ p (x) ϕ̂ (x) dx = (−i)
Rn
Rn
0
Now
Z ∞
Ã
δ
µZ
β−1 −δ|x|2 /2
e
dδ =
0
and
|x|
2
Z
0
2
Ã
=2
where we used u =
Hence,
Z
p (x)
Rn
|x|2
2δ
n+k−α
|x|
∞
δ−
k+α
2 −1
|x|
2
2
k+α
2
e−
|x|2
2δ
e−
|x|2
2δ
¶
dδ p (x) ϕ (x) dx.
k+n−α
2
Γ
k+n−α
2
¶
1
|x|
k+n−α
dδ
∞
k+α
2 −1
u
µ
k+α
2 −1
µ
uβ−1 e−u dδ = 2
!− k+α
Z
2
Γ
δ−
0
0
∞
=
!−β Z
∞
e−u du
0
k+α
2
¶
1
|x|
k+α
2
and dδ = − |x|
2u2 du.
ϕ̂ (x) dx =
=
2
k+α
2
Γ
¡ k+α ¢
Z
p (x)
(−i) k+n−α ¡ 2
ϕ (x) dx
¢
k+α
k+n−α
2 2 Γ
Rn |x|
2
¡
¢ Z
Γ k+α
p (x)
k α− n
2
2
¡ k+n−α ¢
(−i) 2
ϕ (x) dx.
k+α
Γ
Rn |x|
2
k
¤
Lemma 15. Let p ∈ Hk , k ≥ 1. For any α ∈ (0, n) and for any ϕ ∈ S, we have
Z
Z
p (x)
p (x)
lim+
ϕ̂
(x)
dx
=
lim
ϕ̂ (x) dx.
n+k−α
n+k
+
α→0
ε→0
Rn |x|
|x|≥ε |x|
,
HARMONIC ANALYSIS
Proof.
Z
p (x)
Rn
Z
|x|
n+k−α
ϕ̂ (x) dx
Z
p (x)
=
Z
=
Hence,
ϕ̂ (x) dx +
n+k−α
[ϕ̂ (x) − ϕ̂ (0)] dx +
|x|≥1
|x|
|x|≤1
Z
Z
n+k−α
|x|
p (x)
Rn
=
|x|
Z
|x|≤1
= lim+
ε→0
ϕ̂ (x) dx
n+k−α
Z
p (x)
|x|
n+k−α
ϕ̂ (x) dx.
ϕ̂ (x) dx
Z
n+k
p (x)
[ϕ̂ (x) − ϕ̂ (0)] dx +
|x|
|x|≥1
p (x)
n+k
|x|
|x|≥ε
|x|
|x|≥1
p (x)
lim
α→0+
p (x)
n+k−α
|x|
p (x)
|x|≤1
51
n+k
ϕ̂ (x) dx
ϕ̂ (x) dx.
¤
Now we can prove the main theorem.
Proof of Theorem 48. Let f ∈ S. Then for any fixed x, there exists ϕ such that
f (x − y) = ϕ̂ (y)
and we have
Z
1
ϕ (y) =
f (x − z) eiy·z dz
n/2
(2π)
1
=
Rn
Z
n/2
Rn
(2π)
= fˆ (y) eix·y .
Hence
Z
lim+
ε→0
p (x)
|x|
|x|≥ε
Z
Rn
= lim+ γk,α
α→0
Z
= γk,0
|x|
Z
n+k−α
Rn
|x|
k
ϕ̂ (x) dx
p (x)
p (x)
Rn
ϕ̂ (x) dx
p (x)
= lim
α→0+
n+k
f (u) eiy·(x−u) du
|x|
k+α
ϕ (x) dx
Z
ϕ (x) dx = γk,0
where
k
γk,0 = (−i) 2
−n
2
So we have
Z
T f (y) = lim+
ε→0
|x|≥ε
p (x)
|x|
Rn
p (x)
k
|x|
eix·y fˆ (x) dx
¡ ¢
Γ k2
¡
¢.
Γ k+n
2
f (y − x) dx = γk,0
n+k
Z
Rn
p (x)
k
|x|
eix·y fˆ (x) dx.
52
HUIQIANG JIANG
Hence,
p (x)
Tcf (x) = (2π) 2 γk,0 k fˆ (x) = m (x) fˆ (x)
|x|
n
where
n
m (x) = (2π) 2 γk,0
p (x)
= γk
k
|x|
Since S is dense in L2 , the theorem is proved.
p (x)
|x|
k
.
¤
Theorem 49. Let c be a constant and Ω be a homogeneous function on Rn of
degree 0, which is smooth on Sn−1 and
Z
Ωdσ = 0.
Sn−1
Define
Z
T f (x) = cf (x) + lim
ε→0
|y|≥ε
Ω (y)
n f (x − y) dy.
|y|
Then there exists a homogeneous function m of degree 0 which is smooth on Sn−1
such that
Tcf = mfˆ.
Moreover, the reverse is also true.
Proof. Write
Ω (x) =
∞
X
pk (x) ,
k=1
then on Sn−1
m (x) = c +
∞
X
γk pk (x)
k=1
n−1
which is smooth whenever Ω is smooth on S
.
¤
7. Differentiability Properties in Terms of Function Spaces
7.1. Riesz Potential. Let f ∈ S. Recall
2
(−∆f ) ˆ (x) = |x| fˆ.
We can define for any β > −n,
³
´
β
β
(−∆) 2 f ˆ (x) = |x| fˆ.
β
β
Since |x| fˆ ∈ L1 (Rn ) for any β > −n, (−∆) 2 f is well defined as inverse Fourier
transform of integrable function.
For any α ∈ (0, n), we define the Riesz potential of f ∈ S by
−α
2
(Iα f ) (x) = (−∆)
f.
Hence,
−α ˆ
Id
f
α f (x) = |x|
and
(Iα f ) (x) =
³
1
(2π)
n/2
|x|
Apparently, Iα f is well defined as long as |x|
−α
−α
´∨
∗ f.
fˆ ∈ L1 .
HARMONIC ANALYSIS
53
Proposition 12. For any f ∈ S.
(a)
Iα (Iβ f ) = Iα+β f
if α > 0, β > 0 and α + β < n.
(b) If n ≥ 3, then
∆ (Iα f ) = Iα (∆f ) = −Iα−2 f
holds for any α ∈ (2, n). And when α = 2, we have
∆ (I2 f ) = I2 (∆f ) = −f.
Proof. (a) Since α + β < n
−α−β ˆ
−α
\
I\
f = |x| Id
α+β f = |x|
β f = Iα (Iβ f ).
(b) For any α ∈ (2, n),
2
2−α b
∆\
(Iα f ) = − |x| Id
f = −I\
α f = − |x|
α−2 f ,
Iα\
(∆f ) = |x|
−α
c = − |x|2−α fb = −I\
∆f
α−2 f .
When α = 2,
∆ (I2 f ) = I2 (∆f ) = −f
follows similarly.
¤
³
−α
Now we calculate |x|
´∨
:
Lemma 16. Let α ∈ (0, n). Then for any ϕ ∈ S,
Z
Z
−n+α
−α
|x|
ϕ̂ (x) dx = cα
|x| ϕ (x) dx
Rn
Rn
where
γ0,α = 2
Hence,
³
1
n/2
(2π)
where
−α
|x|
¡ ¢
Γ α2
¢.
¡
Γ n−α
2
α− n
2
´∨
=
1
−n+α
|x|
γα
¡ ¢
2α π n/2 Γ α2
¡
¢ .
γα =
Γ n−α
2
Proof. Take k = 0 in Lemma 14, we have
Z
Z
−n+α
|x|
ϕ̂ (x) dx = γ0,α
Rn
where
γ0,α = 2
Hence,
γ0,α
Z
|x|
Rn
α− n
2
ϕ (x) dx
¡ ¢
Γ α2
¡
¢.
Γ n−α
2
Z
−α
−α
|x|
Rn
Z
ϕ (x) dx =
|x|
−n+α
ϕ̂ (x) dx =
Rn
i.e.,
−n+α
−α
\
|x|
= γ0,α |x|
Rn
−n+α
\
|x|
(y) ϕ (y) dy,
54
HUIQIANG JIANG
and
³
1
|x|
n/2
(2π)
−α
´∨
=
1
n/2
γ0,α (2π)
|x|
−n+α
.
¤
Corollary 9. For any α ∈ (0, n) and f ∈ S,
Z
1
1
(7.1)
(Iα f ) (x) =
f (y) dy
γα Rn |x − y|n−α
where
¡ ¢
2α π n/2 Γ α2
¡
¢ .
γα =
Γ n−α
2
We now ask for what pairs p and q, we have
kIα f kq ≤ A kf kp .
A necessary condition follows from a scaling argument. Let (p, q) be such a pair.
Since for any δ > 0,
Iα (τδ f ) = δ −α τδ Iα f
and
kτδ f kp = δ −n/p kf kp ,
we have
kIα (τδ f )kq = δ −α kτδ Iα f kq
n
n
=δ −α− q kIα f kq ≤ δ −α− q A kf kp
n
n
=δ −α− q + p A kτδ f kp .
Hence, we must have
1
1 α
= − .
q
p n
p
n
Theorem 50. Let 0 < α < n, 1 ≤ p < q < ∞, 1q = p1 − α
n and f ∈ L (R ).
n
(a) Iα f defined in (7.1) converges absolutely for a.e. x ∈ R .
(b) If p > 1, then
kIα f kq ≤ Ap,q kf kp
where Ap,q is a constant independent of f .
(c) If p = 1, then for any t > 0,
µ
|{x : Iα f (x) > t}| ≤
A kf k1
t
¶q
where A is a constant independent of f .
Proof. (a) Fix µ > 0. Let K =
1
K1 ∈ L and K2 ∈ L
0
p
1
,
γα |x|n−α
where
Moreover,
Z
kK1 k1 =
|x|≤µ
K1 = Kχ{|x|≤µ} and K2 = K − K1 . Then
1
1
+ = 1.
0
p
p
1
γα |x|
n−α dx
=
nωn α
µ ≡ c1 µα ,
αγα
HARMONIC ANALYSIS
and
ÃZ
kK2 kp0 =
µ
1
¶p0
55
! p10
dx
n−α
γα |x|
µ
¶1
n
1 ωn q p0 − nq
=
µ
≡ c2 µ− q .
0
γα
p
p
p
Hence for any f ∈ L , K1 ∗ f ∈ L , K2 ∗ f ∈ L∞ and
|x|>µ
Iα f = K ∗ f = K1 ∗ f + K2 ∗ f.
converges absolutely a.e. in Rn .
(b) (c) Assume kf kp = 1. For each µ > 0,
kK1 ∗ f kp ≤ kK1 k1 kf kp = c1 µα
and
n
kK2 ∗ f k∞ ≤ kK2 kp0 kf kp = c2 µ− q .
For any t > 0, we have
|{x : |Iα f (x)| > t}|
¯½
¾¯ ¯½
¾¯
¯
t ¯¯ ¯¯
t ¯¯
¯
≤ ¯ x : |(K1 ∗ f ) (x)| >
+ x : |(K2 ∗ f ) (x)| >
2 ¯ ¯
2 ¯
p
p
kK1 ∗ f kp
c 2p
≤ 1 p µαp
≤
p
t
(t/2)
if we choose
Hence,
n
t
= c2 µ− q .
2
1
.
tq
i.e., Iα is of weak type (p, q). In particular, (c) holds. Now the following Marcinkiewicz
interpolation theorem implies (b).
¤
|{x : |Iα f (x)| > t}| ≤ c
Theorem 51 (Marcinkiewicz). Assume 1 ≤ pi ≤ qi ≤ ∞, i = 1, 2 and p1 < p2 ,
q1 6= q2 . Let T be a subadditive transformation defined on Lp1 (Rn ) + Lp2 (Rn ).
Suppose that T is of weak type (pi , qi ), i = 1, 2. Then for any θ ∈ (0, 1) and
1
(1 − θ)
θ 1
(1 − θ)
θ
=
+ , =
+ ,
p
p1
p2 q
q1
q2
T is of type (p, q).
7.2. The Sobolev spaces. For any nonnegative integer k, the Sobolev space
W k,p (Rn ) = Lpk (Rn ) is defined as the space of functions f with Dα f ∈ Lp (Rn )
whenever |α| ≤ k. W k,p (Rn ) is a Banach space with the norm

 p1
X
p
kf kW k,p (Rn ) = 
kDα f kp  .
|α|≤k
An equivalent norm can be given by
X
|α|≤k
kDα f kp .
56
HUIQIANG JIANG
When p = 2, H k (Rn ) = W k,p (Rn ) is a Hilbert space with inner product
X Z
(f, g) =
Dα f Dα gdx.
|α|≤k
Rn
Proposition 13. Let 1 ≤ p < ∞. Then f ∈ W k,p (Rn ) iff there exists a sequence
{fm } ⊂ D (Rn ) s.t.
lim kf − fm kp = 0
m→∞
and for each |α| ≤ k, {Dα fm } converges in Lp norm. In particular, D (Rn ) is
dense in W k,p (Rn ).
Proof. Let {ϕε } be the standard mollifier. If f ∈ W k,p (Rn ), then gε = ϕε ∗ f ∈
C ∞ (Rn ) and
lim kf − gε kp = 0.
ε→∞
Since Dα gε = ϕε ∗ Dα f , for each |α| ≤ k
lim kDα f − Dα gε kp = 0.
ε→∞
Let η ∈ D (Rn ) be a smooth cutoff function s.t η (x) = 1 on B1 (0). For each
g ∈ C ∞ (Rn ) ∩ W k,p (Rn ), define hδ (x) = g (x) η (δx), then
lim khδ − gkp = 0.
ε→∞
The existence of {fm } follows from a Diagonal argument. The reverse is trivial.
¤
Lemma 17. Suppose f ∈ D (Rn ), then
n Z
1 X
∂f
yj
f (x) =
(x − y) n dy.
nωn j=1 Rn ∂xj
|y|
Proof. For any ξ ∈ Sn−1 , we have
Z
f (x) =
∞
ξ · ∇f (x − ξt) dt.
0
Integrating over ξ ∈ Sn−1 , we have
Z
Z ∞
1
f (x) =
ξ · ∇f (x − ξt) dtdσ (ξ)
nωn Sn−1 0
n Z
yj
1 X
∂f
(x − y) n dy.
=
nωn j=1 Rn ∂xj
|y|
¤
Lemma 18. Suppose f ∈ W 1,1 (Rn ), then

1
° n
°
n °
n °
Y
X
° ∂f °
° ∂f °
1
°
°
°
°
 ≤
kf k n ≤ 
° ∂xj °
° ∂xj ° .
n−1
n
1
1
j=1
j=1
HARMONIC ANALYSIS
57
Proof. We assume f ∈ D (Rn ). If n = 1, we have
¯Z x
¯
¯
¯
|f (x)| = ¯¯
f 0 (y) dy ¯¯ ≤ kf 0 k1 ,
−∞
hence
kf k∞ ≤ kf 0 k1 .
Suppose the inequality holds for n = k, then for n = k + 1, we write x = (x0 , xn ),
and
Z
n
n
n−1
kf k n =
|f (x)| n−1 dx
n−1
Rn
!
1
¯
µZ ¯
¶ n−1
Z ÃZ
¯ ∂f ¯
¯
¯ dxn
≤
|f (x)| dx0 dxn
¯
¯
R
Rn−1
R ∂xn
!
1
¯
¶ n−1
µZ
¶ n−2
Z õZ
Z ¯
n−1
¯ ∂f ¯
n−1
0
0
¯ dxn dx
¯
|f (x)| n−2 dx
≤
dxn
¯
¯
R
Rn−1 R ∂xn
Rn−1
1

 n−1
°
° 1 Z
°
n−1
Y°
° ∂f ° n−1
° ∂f °
°
°
°


≤°
dxn
° ∂xn °
° ∂xj ° 1 n−1
R
1
L (R
)
j=1

°
° 1
YZ
° ∂f ° n−1 n−1
°

=°
° ∂xn °
1
j=1
R
°
°
° ∂f °
°
°
° ∂xj °
1
 n−1
dxn 
L1 (Rn−1 )

 1
° n−1
n °
Y
° ∂f °
°
° 
=
.
° ∂xj °
j=1
1
Since D (Rn ) is dense in W 1,1 (Rn ), standard density argument implies that the
inequality holds for f ∈ W 1,1 (Rn ).
¤
The following theorem is essential for Sobolev spaces.
Theorem 52. Suppose k is a positive integer and 1q = p1 − nk .
(a) If 1 ≤ p < nk , then W k,p (Rn ) ⊂ Lq (Rn ) and the natural inclusion is continuous.
(b) If p = nk ≥ 1, then W k,p (Rn ) ⊂ Lrloc (Rn ) for any 1 ≤ r < ∞.
(c) If p > nk and p ≥ 1, then W k,p (Rn ) ⊂ C (Rn ).
Proof. It suffices to prove the Theorem when k = 1 and the general case follows
from an induction.
(a) Let f ∈ D (Rn ), then
µ
¶
n
\
X
∂f
xj d
∂f
Rj
= −i
= |x| fˆ.
∂x
|x|
∂x
j
j
j=1
Hence,

f = I1 
n
X
j=1
µ
Rj

¶
∂f 
∂xj
58
HUIQIANG JIANG
if n ≥ 2. If 1 < p, q < ∞, then n ≥ 2, so part (i) when p > 1 follows from the
properties of Riesz transform and Riesz potential. When p = 1, part (i) becomes

1
° n
n °
Y
° ∂f °
°
° 
kf k n ≤ 
° ∂xj °
n−1
1
j=1
which is the lemma above.
(b) It follows from (a).
(c) Assume f ∈ D (Rn ) with support in BR (0), then for any x ∈ BR (0)
¯
¯
¯
¯
n Z
¯ 1 X
¯
y
∂f
j
¯
(x − y) n dy ¯¯
|f (x)| = ¯
|y|
¯ nωn j=1 Rn ∂xj
¯
¯
¯
¯
¯
n Z
¯ 1 X
¯
y
∂f
j
¯
=¯
(x − y) n dy ¯¯
|y|
¯ nωn j=1 |x−y|≤R ∂xj
¯
°
°
°
°
n
° °
°
1 X°
° ∂f ° ° yjn °
≤
°
°
°
nωn j=1 ∂xj p |y| °Lp0 (B2R (0))
°
n °
X
° ∂f °
°
°
≤c
° ∂xj °
p
j=1
since
°
° 0
° yj °p
° n°
° |y| ° p0
Z
1
≤
|y|
B2R (0)
L (B2R (0))
Z
2R
= nωn
(n−1)p0
dy
0
r−(n−1)p +n−1 dr
0
p−n
nωn (p − 1)
=
(2R) p−1 ,
p−n
here we used
(n − 1) p
p−n
+n=
> 0.
p−1
p−1
Density argument yields the conclusion for f ∈ W 1,p (Rn ).
− (n − 1) p0 + n = −
7.3. Bessel Potential and Bessel Spaces. Bessel potential Jα is defined by
−α
2
Jα = (I − ∆)
Hence, for any f ∈ S,
.
³
´− α2
2
Jd
f
=
1
+
|x|
fˆ
α
and
Jα f = Gα ∗ f
where
Gα (x) =
·³
1
n
1 + |x|
2
´− α2 ¸∨
(2π) 2
Z ³
´− α2
1
2
=
1 + |y|
eix·y dy.
n
(2π) Rn
¤
HARMONIC ANALYSIS
59
³
´− α2
2
If α > n, then 1 + |x|
∈ L1 , and hence
Gα (x) =
Z
1
n
(2π)
³
´− α2
2
1 + |y|
eix·y dy.
Rn
Applying the identity
³
1 + |x|
2
´− α2
=
Γ
Z
1
¡α¢
∞
2
α
e−t(1+|x| ) t 2 −1 dt,
0
2
we have
1
Gα (x) =
n ¡α¢
(2π) Γ 2
=
1
n ¡α¢
(2π) Γ 2
n
2
µZ
Z
∞
e
Rn
Z ∞
−t(1+|y|2 )
0
¶
dt eix·y dy
µZ
α
e−t t 2 −1
e−t|y|
2
+ix·y
¶
dy dt
Rn
0
Z
t
α
2 −1
∞
α−n
π
t 2 −1 dt
e
n ¡ ¢
(2π) Γ α2 0
Z ∞
|x|2 α−n
1
=
e−t− 4t t 2 −1 dt.
¡α¢
n
(4π) 2 Γ 2 0
=
|x|2
−t− 4t
Now we claim that the above expression holds whenever α > 0.
Proposition 14. For any α > 0, define
Gα (x) =
Z
1
n
2
(4π) Γ
¡α¢
∞
e−t−
|x|2
4t
t
α−n
2 −1
dt.
0
2
Then Gα ∈ L1 (Rn ) and
cα (y) =
G
1
(2π)
n
2
³
´− α2
2
1 + |y|
.
Proof. Since
Z
e−
|x|2
4t
Z
Rn
n
n
2
e−|u| (4t) 2 du = (4πt) 2 ,
dx =
Rn
we have
Z
Rn
Gα (x) dx =
1
n
2
(4π) Γ
1
Z
¡α¢
2
∞
µZ
∞
µZ
−t−
e
¡ ¢
n
(4π) 2 Γ α2 0
Rn
Z ∞
α
1
= ¡α¢
e−t t 2 −1 dt = 1.
Γ 2 0
=
|x|2
4t
Rn
0
Z
e−t−
|x|2
4t
¶
α−n
dx t 2 −1 dt
¶
α−n
dx t 2 −1 dt
60
HUIQIANG JIANG
Now,
Z
n
cα (y) =
(2π) 2 G
=
=
=
=
Gα (x) eix·y dx
¶
Z ∞ µZ
|x|2
α−n
1
e− 4t eix·y dx e−t t 2 −1 dt
¡α¢
n
(4π) 2 Γ 2 0
Rn
¶
Z ∞ µZ
α−n
1
1
|x−2ity|2 −t|y|2
− 4t
dx e−t t 2 −1 dt
e
¡α¢
n
2
(4π) Γ 2 0
Rn
Z ∞
n
2
α−n
1
(4πt) 2 e−t(1+|y| ) t 2 −1 dt
¡α¢
n
(4π) 2 Γ 2 0
Z ∞
³
´− α2
2
α
1
2
¡α¢
e−t(1+|y| ) t 2 −1 dt = 1 + |y|
.
Γ 2 0
Rn
¤
Proposition 15. (a) If 0 < α < n,
³
´
|x|
−n+α
+ o |x|
as |x| → 0.
γ (α)
−n+α
Gα (x) =
(b) For any α > 0,
³ 1 ´
Gα (x) = O e− 2 |x| as |x| → ∞.
Proof. (a). Let
2
2
|x|
|x|
, dt = − 2 du.
4t
4u
u=
Hence,
Z
1
n
2
(4π) Γ
=
¡α¢
2
1
n
2
∞
e−
|x|2
4t
t
α−n
2 −1
dt
0
Z
¡α¢
∞
Ã
−u
e
2
|x|
4u
! α−n
2 −1
(4π) Γ 2 0
¯ x ¯α−n Z ∞
¯ ¯
n−α
2
=
e−u u 2 −1 du
¡α¢
n
2
(4π) Γ 2 0
−n+α
=
where
|x|
γ (α)
¡ ¢
n
2α π 2 Γ α2
¡
¢ .
γ (α) =
Γ n−α
2
2
|x|
du
4u2
HARMONIC ANALYSIS
61
So we have
1
n−α
− |x|
Gα (x)
γ (α)
Z ∞
n−α
¡
¢ |x|2 α−n
|x|
=
1 − e−t e− 4t t 2 −1 dt
¡α¢
n
(4π) 2 Γ 2 0
Ã
! α−n
¶
Z ∞µ
2 −1
n−α
2
2
|x|2
|x|
|x|
|x|
− 4u
−u
=
e
1−e
du
¡α¢
n
4u
4u2
(4π) 2 Γ 2 0
¶
Z ∞µ
|x|2
n−α
1
1 − e− 4u e−u u 2 −1 du → 0 as x → 0.
= α n ¡α¢
2 π2Γ 2 0
(b) When |x| ≥ 1
Gα (x) =
≤
≤
1
n
2
¡α¢
n
2
¡α¢
(4π) Γ
1
(4π) Γ
1
n
2
(4π) Γ
≤ ce−
Z
|x|
2
2
2
¡α¢
2
∞
e−t−
|x|2
4t
t
α−n
2 −1
dt
0
Z
∞
t
1
t
t
1
|x|
2
e− 2 − 8t − 2 −
|x|2
8t
t
α−n
2 −1
dt
0
Z
∞
e− 2 − 8t −
t
α−n
2 −1
dt
0
.
¤
For any α > 0, since Gα ∈ L1 (Rn ), for any f ∈ Lp (Rn ), 1 ≤ p ≤ ∞, the Bessel
potential of f ,
Jα f = Gα ∗ f ∈ Lp (Rn )
is well defined. We also define J0 f = f for any f ∈ Lp (Rn ).
Proposition 16. For any α, β ≥ 0,
Jα Jβ = Jα+β .
Lemma 19. Let α ≥ 0.
(a) Jα is a bijection from S onto S.
(b) Let 1 ≤ p ≤ ∞ and g1 , g2 ∈ Lp (Rn ). Then Jα (g1 ) = Jα (g2 ) implies g1 = g2 .
Proof. (a) Since Fourier transform is a bijection from S onto S, it follows from the
definition that for any f ∈ S,
³
´− α2
2
Jd
fˆ.
α f = 1 + |x|
(b) Let g = g1 − g2 . Jα (g) = 0 implies that for any ϕ ∈ S,
Z
Z
0=
Jα (g) ϕ =
gJα (ϕ) .
Rn
Rn
Since Jα (ϕ) can be any function in S, g = 0.
Definition 13. The potential space Lpα (Rn ) is defined as
Lpα = Jα (Lp (Rn )) .
¤
62
HUIQIANG JIANG
The Lpα -norm of f is defined by
kf kp,α = kgkp if f = Jα (g) .
Remark 20. k·kp,α is indeed well defined since Jα (g1 ) = Jα (g2 ) implies g1 = g2 .
Obviously Lpα is a Banach space.
Lemma 20. Let α > 0.
(i) There exists a finite measure µα on Rn s.t. its Fourier transform is given by
α
|x|
µ
cα (x) = ³
1 + |x|
2
´ α2 .
(ii) There exists a finite measure να on Rn s.t.
³
´α
α
2 2
= νc
1 + |x|
α (x) (1 + |x| ) .
Proof. (a)
³
|x|
Ã
α
1 + |x|
2
´ α2 =
1−
=1+
! α2
1
1 + |x|
∞
X
2
k
µα¶ ³
2
(−1)
k
k=1
"
= (2π)
n
2
δ0 +
∞
X
k=1
1 + |x|
2
´−k
#ˆ
µα¶
2
(−1)
G2k (x)
k
k
=µ
cα (x) .
Here
"
µα = (2π)
n
2
δ0 +
∞
X
k=1
#
µα¶
2 G
(−1)
2k (x)
k
k
is a finite measure.
(b) We need Weiner’s theorem: If ϕ ∈ L1 (Rn ) satisfies ϕ̂ + 1 is nowhere zero,
then there exists ψ ∈ L1 (Rn ) s.t.
−1
(ϕ̂ + 1)
Let
"
ϕ = (2π)
n
2
∞
X
k=1
= ψ̂ + 1.
#
µα¶
2
(−1)
G2k (x) + Gα (x) ,
k
k
then
α
ϕ̂ + 1 = ³
|x| + 1
´α
2 2
1 + |x|
never vanishes, hence, there exists ψ ∈ L1 (Rn ) s.t.
³
´α
2 2
1 + |x|
ψ̂ + 1 =
,
α
|x| + 1
HARMONIC ANALYSIS
i.e.,
³
1 + |x|
2
´ α2
α
= ψ̂ + 1 + |x|
³
63
´
ψ̂ + 1 .
n
Thus we can take να = ψ + (2π) 2 δ0 .
¤
Lemma 21. Suppose 1 < p < ∞, and α ≥ 1. Then f ∈ Lpα (Rn ) iff f ∈ Lpα−1 (Rn )
∂f
and for each j, ∂x
∈ Lpα−1 (Rn ). Moreover, the two norms kf kp,α and
j
°
n °
X
° ∂f °
°
°
kf kp,α−1 +
° ∂xj °
p,α−1
j=1
are equivalent.
Proof. 1. If f ∈ Lpα (Rn ), then there exists g ∈ Lp (Rn ), f = Jα (g). Since
f = Jα (g) = Jα−1 (J1 g)
and J1 g ∈ Lp (Rn ), we have f ∈ Lpα−1 (Rn ) with
kf kp,α−1 = kJ1 gkp ≤ kgkp = kf kp,α .
Now for each j,
³
´− α2
d
∂f
2
ĝ
= −ixj fˆ = −ixj 1 + |x|
∂xj


³
´− α−1
|x|
−ixj 
2

2
= 1 + |x|
³
´ 12 |x| ĝ 
2
1 + |x|
³
´
³
´− α−1
2
2
d
µ
c1 · R
= 1 + |x|
jg
³
´− α−1
n
2
−\
2
(2π) 2 µ1 ∗ Rj g.
= 1 + |x|
Hence,
³
´
∂f
−n
= Jα−1 (2π) 2 µ1 ∗ Rj g ∈ Lpα−1 (Rn ) .
∂xj
Moreover,
°
°
° ∂f °
°
°
° ∂xj °
p,α−1
°
°
°
°
−n
= °(2π) 2 µ1 ∗ Rj g ° ≤ c kRj gkp ≤ C kgkp .
2. If f ∈ Lpα−1 (Rn ) and for each j,
gj in Lp (Rn ), such that
p
∂f
∂xj
∈ Lpα−1 (Rn ). Then there exists g and
f = Jα−1 (g) and for each j,
∂f
= Jα−1 (gj ) .
∂xj
³
´− α−1
2
2
Since fˆ = 1 + |x|
ĝ, we have
³
´− α−1
³
´− α−1
d
d
∂f
∂g
2
2
2
2
= −ixj fˆ = 1 + |x|
.
(−ixj ) ĝ = 1 + |x|
∂xj
∂xj
64
HUIQIANG JIANG
However,
³
´− α−1
³
´− α−1
d
d
∂f
∂g
2
2
2
2
= 1 + |x|
.
ĝj = 1 + |x|
∂xj
∂xj
Hence, we have
∂g
= gj .
∂xj
Now, the following formal calculation can be justified,
³
´− α−1
2
2
fˆ = 1 + |x|
ĝ
³
´− α2 ³
´1
2
2 2
= 1 + |x|
1 + |x|
ĝ
α
³
´− 2
2
= 1 + |x|
νc
α (x) (1 + |x|) ĝ
µ
³
´− α2 µ
\ µ ∂g ¶¶¶
2
−n
= 1 + |x|
(2π) 2 να ∗ g + Rj
.
∂xj
Hence, f ∈ Lpα (Rn ) and
kf kp,α
°
µ
¶¶°
µ
°
°
∂g
−n
2
°
=°
ν
∗
g
+
R
(2π)
α
j
°
°
∂xj
p
°
¶°
µ
°
°
∂g
°
≤ c°
°g + Rj ∂xj °
p
Ã
° !
X°
° ∂g °
°
°
≤ c kgkp +
° ∂xj ° .
p
¤
Theorem 53. Suppose k is a nonnegative integer and p ∈ (1, ∞). Then
Lpk (Rn ) = W k,p (Rn ) .
Proof. By induction.
¤
Suppose that f ∈ Lp (Rn ). We define the Lp modulus of continuity:
ωp (t) = kf (x + t) − f (x)kp .
Then
lim ωp (t) = 0.
|t|→0
Let’s first recall the Minkowski integral inequality.
Theorem 54. Let (X, dµ) and (Y, dν) be two σ-finite measure spaces and 1 ≤ p <
∞. Then for any measurable function F (x, y) on X × Y ,
µZ µZ
¶p
¶ p1
¶ p1
Z µZ
p
|F (x, y)| dµx dνy
≤
|F (x, y)| dνy
dµx .
Y
X
X
Y
If (X, dµ) is the counting measure for {1, 2} and (Y, dν) is the Lebesgue measure
on Rn . We deduce
kf1 + f2 kp ≤ kf1 kp + kf2 kp .
HARMONIC ANALYSIS
65
Proposition 17. Suppose 1 < p < ∞. Then f ∈ Lp1 (Rn ) iff f ∈ Lp (Rn ) and
ωp (t) = O (|t|) as |t| → 0.
Proof. Suppose f ∈ D (Rn ). Let t = |t| t0 .
Z |t|
f (x + t) − f (x) =
(∇f, t0 ) (x + st0 ) ds.
0
Hence, Minkowski’s inequality implies
kf (x + t) − f (x)kp ≤ |t|
°
n °
X
° ∂f °
°
°
° ∂xj ° .
p
j=1
n
Since D (R ) is dense in
Lp1
n
n
(R ), for any f ∈ D (R ), we have
°
n °
X
° ∂f °
°
°
kf (x + t) − f (x)kp ≤ |t|
° ∂xj °
p
j=1
and hence ωp (t) = O (|t|) as |t| → 0.
On the other hand, suppose f ∈ Lp (Rn ) and ωp (t) = O (|t|) as |t| → 0. Let ej
be the unit vector in j-th direction. Since
°
°
° f (x + sej ) − f (x) °
°
° = O (1) ,
°
°
s
p
there exists a subsequence, sk → 0, s.t.,
f (x + sk ej ) − f (x)
* fj weakly in Lp (Rn ) as k → ∞.
sk
Since for any ϕ ∈ D (Rn ),
Z
Z
f (x + sk ej ) − f (x)
ϕ (x) − ϕ (x − sk ej )
ϕ (x) dx = −
f (x)
dx,
sk
sk
Rn
Rn
we have, letting k → ∞,
Z
Z
∂ϕ
fj ϕ (x) dx = −
f (x)
dx.
∂xj
Rn
Rn
Hence,
∂f
= fj ∈ Lp (Rn )
∂xj
and f ∈ Lp1 (Rn ).
¤
8. Extensions and Restrictions
8.1. Decomposition of open sets into cubes. Let’s recall the following theorem
we proved.
Theorem 55. Let F be a nonempty closed set in Rn s.t. F 6= Rn . Then its
complement Ω = F c can be written as a union of cubes
∞
[
Qk ,
Ω=
k=1
whose interiors are disjoint and for each k,
diam Qk ≤ dist (Qk , F ) ≤ 4 diam Qk .
66
HUIQIANG JIANG
∞
Let F be the collection {Qk }k=1 . We say that two cubes Q1 , Q2 in F touch if
Q1 ∩ Q2 6= ∅.
Proposition 18. Suppose two cubes Q1 , Q2 ∈ F touch. Then
1
diam Q1 ≤ diam (Q1 ) ≤ 4 diam Q2 .
4
Proof. Since Q1 ∩ Q2 6= ∅,
diam Q2 ≤ dist (Q2 , F )
≤ dist (Q1 , F ) + diam Q1 ≤ 5 diam Q1 .
Since diam Q2 = 2k diam Q1 for some integer k, we deduce
diam Q2 ≤ 4 diam Q1 .
¤
n
Proposition 19. Suppose Q ∈ F . Then there are at most N = 12 cubes in F
which touch Q.
¡
¢
Let ε ∈ 0, 14 be fixed. For any Qk ∈ F with center xk , we define
¡
¢
Q∗k = (1 + ε) Qk − xk + xk .
Proposition 20. Each point of Ω is contained in at most N of the cubes Q∗k .
Proof. Let Q and Qk be two cubes in F. We consider the union of Qk with all
cubes in F which touches Qk , then this union contains Q∗k as its interior because
of the proposition 18. Hence, Q∗k ∩ Q 6= ∅ iff Qk ∩ Q 6= ∅.
¤
Let Q0 be the unit cube centered at the origin. Fix ϕ ∈ D (Rn ), s.t. 0 ≤ ϕ ≤ 1,
ϕ = 1 on Q0 and ϕ = 0 on Rn \Q∗0 . Let
µ
¶
x − xk
ϕk (x) = ϕ
lk
where xk is the center of Qk and lk is the side length of Qk .
We now define
ϕk (x)
ϕ∗k (x) = P∞
, x ∈ Ω.
k=1 ϕk (x)
Then we have
∞
X
ϕ∗k (x) ≡ 1 for x ∈ Ω = F c .
k=1
8.2. Extension theorems of Whitney type. We first consider the regularized
distance. Let F be a nonempty closed set in Rn s.t. F is not the whole space. Let
δ (x) be the distance of x from F . Then δ is a Lipschitz function.
Theorem 56. There exists a function ∆x = ∆ (x, F ) defined in F c s.t.
(a) c1 δ (x) ≤ ∆x ≤ c2 δ (x), x ∈ F c ;
(b) ∆x is smooth in F c and
¯
¯ α
¯
¯ ∂
1−|α|
¯
¯
¯ ∂xα ∆ (x)¯ ≤ Bα (δ (x))
where Bα , c1 , c2 are constants independent of F .
HARMONIC ANALYSIS
67
Proof. Let
∆x =
∞
X
(diam Qk ) ϕk (x) .
k=1
If x ∈ Qk , then
δ (x) = dist (x, F ) ≤ dist (Qk , F ) + diam Qk ≤ 5 diam Qk ,
and hence
∆x =
∞
X
(diam Qk ) ϕk (x) ≥ (diam Qk ) ϕk (x) = diam Qk ≥
k=1
Also if x ∈
Q∗k ,
1
δ (x) .
5
then
1
3
diam Qk ≥ diam Qk .
4
4
c
∗
Since any x ∈ F lies in at most N of the Qk , we have
X
4N
∆x ≤
diam Qk ≤
δ (x) .
3
∗
δ (x) ≥ dist (Qk , F ) −
x∈Qk
For any α, |α| ≥ 1, there exists Aα such that,
|Dα ϕk (x)| ≤ Aα (diam Qk )
−|α|
∆x is smooth in F c and for any x ∈ F c ,
X
|Dα ∆ (x)| ≤
diam Qk |Dα ϕk (x)|
x∈Q∗
k
≤
X
x∈Q∗
k
≤ N Aα
1−|α|
Aα (diam Qk )
µ
¶1−|α|
1
δ (x)
6
1−|α|
= Bα (δ (x))
where we used the fact that if x ∈ Q∗k , then
δ (x) = dist (x, F ) ≤ dist (Qk , F ) + 2 diam Qk ≤ 6 diam Qk .
¤
∞
Given closed F and the cube decomposition {Qk }k=1 of F c . For each Qk , fix a
point pk ∈ F s.t.
dist (Qk , F ) = dist (Qk , pk ) .
Let f be a continuous function on F . We define

f (x)
if x ∈ F,

∞
P
E0 (f ) (x) =
∗
f (pk ) ϕk (x) if x ∈ F c .

k=1
Proposition 21. E0 (f ) is continuous on Rn and smooth in F c .
68
HUIQIANG JIANG
Proof. E0 (f ) is smooth in F c because for each x ∈ F c , there exists a nbhd of x
such that the summation is finite. Now we need to show the continuity of E0 (f )
for each y ∈ F . If x ∈ Q∗k , then
|y − pk | ≤ |y − x| + |x − pk | ≤ c |y − x| .
For each ε > 0, there exists δ, s.t.
|f (z) − f (y)| < ε
whenever |z − y| < δ. Now for any x ∈ F c , s.t. |y − x| < δc ,
|E0 (f ) (x) − f (y)|
≤
∞
X
|f (pk ) − f (y)| ϕ∗k (x)
k=1
=
X
|f (pk ) − f (y)| ϕ∗k (x) < ε
x∈Q∗
k
X
ϕ∗k (x) ≤ ε.
x∈Q∗
k
Hence E0 (f ) is continuous at y.
¤
For any 0 < γ ≤ 1, we define
γ
Lip (γ, Rn ) = {f : |f (x)| ≤ M, |f (x) − f (y)| ≤ M |x − y| , x, y ∈ Rn } .
Then Lip (γ, Rn ) is a Banach space with norm
kf kC γ = kf k∞ + [f ]γ = kf k∞ + inf
x,y∈R
x6=y
n
|f (x) − f (y)|
.
γ
|x − y|
We remark here that if γ > 1, then Lip (γ, Rn ) defined as above contains only
constant functions.
If F ⊂ Rn is a nonempty closed set, we can define the Banach space Lip (γ, F )
in a similar way.
Theorem 57. The linear extension operator E0 maps Lip (γ, F ) continuously into
Lip (γ, Rn ) if 0 < γ ≤ 1. The norm of this mapping has a bound independent of the
closed set F .
Proof. First, we observe that for any x ∈ F c ,
−|α|
|Dα ϕ∗k (x)| ≤ A0α (diam Qk )
.
c
Now we claim for any x ∈ F ,
¯
¯
¯ ∂
¯
γ−1
¯
¯
.
¯ ∂xi E0 (f ) (x)¯ ≤ c [f ]γ (δ (x))
In fact, for any x ∈ F c , choose y ∈ F s.t. δ (x) = |x − y|, we have
¯
¯
¯ ¯¯ ∞
¯
∗
¯ ∂
¯ ¯X
(x)
∂ϕ
¯
k
¯
¯=¯
E
(f
)
(x)
(f
(p
)
−
f
(y))
¯
0
k
¯ ∂xi
¯ ¯
∂xi ¯
k=1
¯ ∗
¯
∞
X
¯ ∂ϕk (x) ¯
¯
¯
≤
|f (pk ) − f (y)| ¯
∂xi ¯
k=1
≤ c [f ]γ (δ (x))
where c only depends on n.
γ−1
HARMONIC ANALYSIS
69
Now for any x ∈ F c , y ∈ F ,
¯
¯∞
¯X
¯
¯
¯
∗
(f (pk ) − f (y)) ϕk (x)¯
|E0 (f ) (x) − E0 (f ) (y)| = ¯
¯
¯
k=1
X
≤c
|f (pk ) − f (y)|
x∈Q∗
k
≤ c [f ]γ
X
γ
γ
|pk − y| ≤ c [f ]γ |x − y| .
x∈Q∗
k
For any x, y ∈ F c , let L be the line segement from x to y.
Case I: |x − y| ≤ dist (L, F ). Then
|E0 (f ) (x) − E0 (f ) (y)| ≤ |x − y| sup |∇E0 (f ) (x0 )|
x0 ∈L
≤ c [f ]γ |x − y| (δ (x))
γ−1
γ
≤ c [f ]γ |x − y| .
Case II: |x − y| > dist (L, F ). Then there exists x0 ∈ L and y 0 ∈ F s.t.
|x0 − y 0 | < |x − y| ,
hence
|x − y 0 | < 2 |x − y| and |y − y 0 | < 2 |x − y| .
So we have
|E0 (f ) (x) − E0 (f ) (y)|
≤ |E0 (f ) (x) − E0 (f ) (y 0 )| − |E0 (f ) (y) − E0 (f ) (y 0 )|
γ
γ
γ
≤ c [f ]γ |x − y 0 | + c [f ]γ |y − y 0 | ≤ c [f ]γ |x − y| .
¤
More generally, let ω (δ) be a modulus of continuity satisfying
(i) ω(δ)
is increasing as δ → 0;
δ
(ii) ω (2δ) ≤ cω (δ) for any δ > 0.
We define Lip (ω, F ) to define the collection of continuous functions f such that
kf kC ω = kf k∞ + [f ]ω = kf k∞ + inf
x,y∈Rn
x6=y
|f (x) − f (y)|
< ∞.
ω (|x − y|)
Then Lip (ω, F ) is a Banach space.
Corollary 10. The linear extension operator E0 maps Lip (ω, F ) continuously into
Lip (ω, Rn ).
Let γ > 1 and k be an integer s.t. k < γ ≤ k + 1. We say that f ∈ Lip (γ, F ) if
f is defined on F and there exists f (α) , |α| ≤ k, such that, for any 0 ≤ |α| ≤ k, if
f (α) (x) =
X
|α+β|≤k
then
f (α+β) (y)
β
(x − y) + Rα (x, y) ,
β!
¯
¯
¯ (α)
¯
γ−|α|
¯f (x)¯ ≤ M and |Rα (x, y)| ≤ M |x − y|
70
HUIQIANG JIANG
©
ª
holds for any x, y ∈ F . Since for f ∈ Lip (γ, F ), the collection f (α) may not be
© (α) ª
unique, we could view f
as an element in Lip (γ, F ) and take the smallest
|α|≤k
M as its
© norm.
ª
Let f (α) |α|≤k ∈ Lip (γ, F ). We define

f (0) (x)
if x ∈ F,
³n
o´ 
∞
(α)
P
Ek f
=
∗
P (x, pj ) ϕj (x) if x ∈ F c

j=1
where
P (x, y) =
X f α (y)
α
(x − y) , x ∈ F c , y ∈ F.
α!
|α|≤k
Theorem 58. The linear extension operator Ek maps Lip (γ, F ) continuously into
Lip (γ, Rn ) if k < γ ≤ k + 1. The norm of this mapping has a bound independent
of the closed set F .
8.3. Extension theorem for Sobolev functions in Lipschitz domain. Let
k ≥ 0 be an integer and 1 ≤ p ≤ ∞. Similar to W k,p (Rn ), we can define W k,p (Ω)
for any open subset Ω ⊂ Rn . The goal of this section is to find an extension operator
from W k,p (Ω) to W k,p (Rn ).
We first consider special Lipschitz domains whose boundary are represented by
uniform Lipschitz graphs.
Let ϕ : Rn → R be a function satisfying
|ϕ (x) − ϕ (x0 )| ≤ M |x − x0 | for any x, x0 ∈ Rn .
Let
©
ª
Ω = (x, y) ∈ Rn+1 : x ∈ Rn , y ∈ R, y > ϕ (x) .
The smallest such M is called the bound of the specail Lipschitz domain Ω.
Lemma 22. There exists a continuous function ψ defined on [1, ∞) which is rapidly
decreasing at ∞ and satisfies
Z ∞
ψ (t) dt = 1,
1
Z ∞
tk ψ (t) dt = 0 for k ∈ N.
1
Proof. Let
ψ (t) =
µ
¶
1
e
Im e−ω(t−1) 4
πt
iπ
where ω = e− 4 .
¤
Lemma 23. Let F = D and ∆ (x, y) be the regularized distance from F . Then for
any (x, y) ∈ F c ,
ϕ (x) − y ≤ c∆ (x, y)
where c is a constant depending only on M .
Proof. For any (x, y) ∈ F c ,
µ
¶− 12
1
δ (x, y) ≥ 1 + 2
(ϕ (x) − y) .
M
HARMONIC ANALYSIS
Since
∆ (x, y) ≥
the lemma holds with
71
1
δ (x, y) ,
5
µ
¶1
1 2
c=5 1+ 2
M
¤
Let
δ ∗ (x, y) = 2c∆ (x, y) .
¡ ¢
We define for any f ∈ W k,p (Ω) ∩ C ∞ Ω such that f and all its partial derivatives
are bounded in Ω,
½
f (x, y)
if y ≥ ϕ (x) ,
R∞
E (f ) (x, y) =
∗
ψ
(t)
f
(x,
y
+
tδ
(x,
y))
dt
if y < ϕ (x) .
1
¡
¢
Proposition 22. E (f ) ∈ C ∞ Rn+1 and
kE (f )kW k,p (Rn+1 ) ≤ Ak,n (M ) kf kW k,p (Ω) .
Proof. First,
¤
¡
¢
Theorem 59. The extension operator E maps W k,p (Ω) continuously into W k,p Rn+1 .
Proof. Let η be a nonnegative smooth function with compact support in the interior
of the cone
Γ− = {(x, y) : M |x| < |y| , y < 0} .
s.t.
Z
η = 1.
Rn+1
For any ε > 0, define
ηε (u) =
1
εn+1
For any f ∈ W k,p (Ω), we define
fε (u) = f ∗ ηε =
¡ ¢
Then fε ∈ C ∞ Ω and
η
³u´
ε
, u ∈ Rn+1 .
Z
Rn+1
f (u − v) ηε (v) dv.
kfε kW k,p (Ω) ≤ kf kW k,p (Ω) .
Now if 1 ≤ p < ∞, then
lim kfε − f kW k,p (Ω) = 0.
ε→0+
Let
Eε (f ) = E (fε ) .
¡
¢
Then Eε (f ) is a Cauchy sequence in W k,p Rn+1 and hence
E (f ) = lim+ Eε (f )
ε→0
defines an extension operator satisfying
kE (f )kW k,p (Rn+1 ) ≤ Ak,n (M ) kf kW k,p (Ω) .
72
HUIQIANG JIANG
And if p = ∞, k ≥ 1,
lim kfε − f kW k−1,∞ (Ω) = 0.
¡
¢
So Eε (f ) is a Cauchy sequence in W k−1,∞ Rn+1 and hence
ε→0+
E (f ) = lim Eε (f )
ε→0+
defines an extension operator satisfying
kE (f )kW k,∞ (Rn+1 ) ≤ Ak,n (M ) kf kW k,∞ (Ω) .
¤