Pseudodifferential Operators
Helmut Abels
February 27, 2011
Contents
1 Introduction
2
2 Summary of Some Basic Results
2.1 Functions on Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Lebesgue Integral and Lp -spaces . . . . . . . . . . . . . . . . . . . . .
5
5
6
3 Fourier Transformation
3.1 Definition and Basic Properties . . . . . . . . . . . .
3.2 Rapidly decreasing functions – S(Rn ) . . . . . . . . .
3.3 Ex-course: Fréchet spaces . . . . . . . . . . . . . . .
3.4 Inverse Fourier transformation and Parseval’s formula
3.5 Tempered Distributions and Fourier Transformation .
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10
10
12
14
17
20
4 Basic Calculus of Pseudodifferential Operators on Rn
4.1 Symbol Classes and Basic Properties . . . . . . . . . . . . . . . . . .
4.2 Composition of Pseudodifferential Operators: Motivation . . . . . . .
4.3 Oscillatory Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4 Double Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5 Composition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6 Application: Elliptic Pseudodifferential Operators and Parametrices .
4.7 Boundedness on Cb∞ (Rn ) and Uniqueness of the Symbol . . . . . . . .
4.8 Adjoints of Pseudodifferential Operators and Operators in (x,y)-Form
4.9 L2 -Continuity and Bessel Potential Spaces . . . . . . . . . . . . . . .
4.10 Summary and Final Remarks . . . . . . . . . . . . . . . . . . . . . .
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5 Extensions and Applications
5.1 Resolvents and Parameter-Elliptic Differential Operators . . . . . . .
5.2 Kernel Representation of a Pseudodifferential Operator . . . . . . . .
5.3 Coordinate Transformations and Pseudodifferential Operators on Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.4 Boundedness on Hölder-Zygmund and Besov Spaces . . . . . . . . . .
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1
Introduction
One of the leading ideas in the theory of pseudodifferential operators is to reduce the
study of properties of a linear differential operator
X
P =
cα (x)∂xα ,
|α|≤m
which is a polynomial in the derivatives ∂x = (∂x1 , . . . , ∂xn ) with constants cα depending on x, to its symbol
X
p(x, ξ) =
cα (x)(iξ)α ,
|α|≤m
which is a polynomial in the phase variable ξ ∈ Rn with constants depending on the
space variable x. Then the question arises how properties of the differential operator
p(x, Dx ) – as for example invertibility – are related to properties of the symbol p(x, ξ).
The main tool in the theory is the Fourier transformation
Z
e−ix·ξ f (x)dx,
ξ ∈ Rn ,
(1.1)
F[f ](ξ) := fˆ(ξ) :=
Rn
which is defined for suitable functions f : Rn → C. Here (1.1) can be interpreted as
scalar product of the function f (x) and the function x 7→ eix·ξ :
Z
f (x)eix·ξ dx = (f, ei.·ξ )L2 (Rn )
F[f ](ξ) =
Rn
for fixed ξ ∈ Rn , where (., .) denotes the scalar product on L2 (Rn ). Hence fˆ(ξ)
can be interpreted as the contribution of the complex (multidimensional) oscillation
x 7→ eix·ξ to the function, where the phase variable ξ = (ξ1 , . . . , ξn ) ∈ Rn describes
the frequency of the oscillation.
Knowing the Fourier transform g(ξ) = fˆ(ξ), the function f can be reconstructed
with the aid of the inverse Fourier transformation
Z
1
−1
F [g](x) :=
eix·ξ g(ξ)dξ;
(2π)n Rn
indeed it holds that
F −1 [F[f ]] ≡ F −1 [fˆ] = f
(Inversion formula).
Here F −1 [g] can be interpreted as infinitesimal linear combination of x 7→ eix·ξ with
coefficients g(ξ) (and a correcting factor 1/(2π)n ).
2
Using the inversion formula,
Z
1
Pu =
eix·ξ û(ξ)dξ
(2π)n Rn
|α|≤m
Z
X
1
=
cα (x)
(iξ)α eix·ξ û(ξ)dξ
n
(2π) Rn
|α|≤m
Z
1
=
eix·ξ p(x, ξ)û(ξ)dξ,
(2π)n Rn
X
cα (x)∂xα
where we have used ∂xj eix·ξ = iξj eix·ξ and therefore ∂xα eix·ξ = (iξ)α eix·ξ . This motivates the definition of the symbol p(x, ξ) of P .
In the following let p(x, ξ) = p(ξ) be independent of x ∈ Rn . Then
P u = F −1 [p(ξ)û] = F −1 [p(ξ)F[u]].
Hence application of P to u acts as multiplication of û(ξ) by the symbol p(ξ). This
suggests that inversion of P corresponds to multiplication 1/p(ξ) on the side of the
Fourier transformed functions. Therefore we define
Z
1 ˆ
1
−1
−1 ˆ
f (ξ)dξ
eix·ξ
Qf := F [p(ξ) f (ξ)] =
n
(2π) Rn
p(ξ)
assuming that p(ξ) 6= 0 for all ξ ∈ Rn . Then F[Qf ] = p(ξ)−1 fˆ(ξ) since FF −1 = I
(inversion formula) and therefore
P Qf = F −1 [p(ξ)F[Qf ]] = F −1 [p(ξ)p(ξ)−1 fˆ] = F −1 [fˆ(ξ)] = f,
i.e., Q is indeed the inverse of P .
Of course Q is no longer a differential operator; it belongs to the class of pseudodifferential operators, which are defined as
Z
1
q(x, Dx )f :=
eix·ξ q(x, ξ)fˆ(ξ)dξ
(2π)n Rn
where q(x, ξ) is a suitable function and not necessarily a polynomial in ξ.
In the case that the coefficients of P depend on x the inverse of P is not inverted
that easy. If we define analogously to the constant coefficient case
Z
1
1 ˆ
Qf :=
eix·ξ
f (ξ)dξ,
n
(2π) Rn
p(x, ξ)
then
1
P Qf =
(2π)n
Z
P eix·ξ
Rn
3
1
p(x, ξ)
fˆ(ξ)dξ.
Because of the product rule,
1
1
1
ix·ξ
P e
= (P eix·ξ )
+ r(x, ξ) = p(x, ξ)
+ r(x, ξ) = 1 + r(x, ξ)
p(x, ξ)
p(x, ξ)
p(x, ξ)
where the remainder term r(x, ξ) consists of terms where
respect to x at least once. Hence
1
p(x,ξ)
is differentiated with
P Qf = I + r(x, Dx ),
where r(x, Dx ) 6= 0 if p(x, ξ) is not independent of x. But in some sense r(x, Dx )
is of lower order (order less than 0) and does not play an important rule for many
purposes.
So far all considerations have been made formally. To make them precise and to
prove everything rigorously will be a main part of the lecture.
The structure of the lecture is as follows: First we recall some basic facts from
calculus in Rn and Lebesgue integration in Section 2. Then, in Section 3, we study
some basic properties of the Fourier transformation. In particular, we prove the
inversion formula for functions in L2 (Rn ). The main part of the lecture consists of
Section 4 in which basic properties of pseudodifferential operators are studied. More
precisely, we study compositions, adjoints, and boundedness of pseudodifferential
operators on certain function spaces including L2 (Rn ) and L2 -Sobolev spaces. The
results will be applied to elliptic differential and pseudodifferential equations.
4
2
2.1
Summary of Some Basic Results
Functions on Rn
In this section we briefly summerize some known facts from the basic courses on
calculus in Rn .
In the following N will denote the set of all natural numbers zero not included
and N0 := N ∪ {0}. Moreover, R and C denote the set of real and complex numbers,
respectively. Constants appearing in inequalities will usually be denoted by C sometimes marked with an index as e.g. Cα to denote that C depends on α. In sequences
of inequalities all constants will simply be denoted by C although they may change
from line to line.
We will make use of the multi-indices, which will keep the notation (relatively)
short. A multi-index is a vector α = (α1 , . . . , αn ) ∈ Nn0 . Given α ∈ Nn0 we define the
length of α as |α| = α1 + . . . + αn and its factorial as α! = α1 ! · . . . · αn !.
The multi-indices are used to write polynomials: for x ∈ Rn and α ∈ Nn0 we
define xα = xα1 1 · . . . · xαnn . Then xα is a polynomial of degree |α| and an arbitrary
polynomial p : Rn → C of order m ∈ N0 can be written as
X
p(x) =
cα x α
|α|≤m
P
with coefficients cα ∈ C. Here |α|≤m denotes summation with respect to all multiindices α ∈ Nn0 with length |α| ≤ m. Moreover, if α, β ∈ Nn0 , we write α ≤ β if and
only if αj ≤ βj for all j = 1, . . . , n.
The so-called binomial coefficients are defined by
α
α!
=
β!(α − β)!
β
if β ≤ α and αβ = 0 otherwise. Then it is easy to check that
α
α1
αn
=
· ... ·
.
β
β1
βn
Using this notation and the fundamental relation
n
n
n+1
+
=
k
k+1
k+1
for n, k ∈ N0 , one can prove by induction that
X α
α
(x + y) =
xβ y α−β
β
β≤α
for all x, y ∈ Rn , which generalizes the well-known binomial formula in one-dimension.
5
Recall that the space C k (Rn ) consists of all k-times differentiable functions
f : Rn →
T
C or f : Rn → R with continuous derivatives up to order k and C ∞ (Rn ) = k∈N C k (Rn ).
If α ∈ Nn0 and u ∈ C |α| (Rn ), then we define
∂xα u(x) =
∂ |α|
u(x).
∂xα11 . . . ∂xαnn
If u depends on several variables, say x and y, ∂xα u and ∂yα u denote the derivative
defined above with respect to x and y, respectively.
Using the product rule of differentiation, one can prove the following Leibniz’s
formula:
X α
α
∂x (uv)(x) =
(∂ β u)(x)(∂ α−β v)(x)
(2.1)
β
β≤α
for all α ∈ Nn0 and u, v ∈ C |α| (Rn ).
Finally, we recall Taylor’s formula:
THEOREM 2.1 (Taylor’s formula) Let u ∈ C k (Rn ), k ∈ N. Then for any x
and y ∈ Rn one has
X yα
X yα Z 1
α
u(x + y) =
(1 − t)k−1 ∂xα u(x + ty)dt
∂x u(x) +
k
α!
α! 0
|α|<k
|α|=k
Proof: See basic course or [Ray91, Chapter 1, Theorem 1.1].
Finally, we denote by Cbk (Rn ), k ∈ N0 , the space of all f ∈ C k (Rn ) with bounded
derivatives
up to order k, i.e., ∂ α f is bounded for all |α| ≤ k. Moreover, Cb∞ (Rn ) :=
T∞
k
n
k
n
k=0 Cb (R ). It is well-known from basic courses in functional analysis that Cb (R )
equipped with the norm
kf kCbk = sup sup |∂ α f (x)|
|α|≤k x∈Rn
is a Banach space, i.e., a complete normed vectorspace.
2.2
Lebesgue Integral and Lp -spaces
Recall that a function f : Rn → R is Lebesgue-integrable or just integrable if f is
measurable and
Z
|f (x)|dx < ∞.
(2.2)
Rn
Note that |f | ≥ 0 is measurable if f is measurable and for
R a non-negative measurable
function g : Rn → [0, ∞] there is always a well-defined Rn g(x)dx, which is possibly
infinite. Moreover, if (2.2) holds, f + := max{f, 0} and f − : max{−f, 0} are nonnegative functions with finite integral and
Z
Z
Z
+
f (x)dx :=
f (x)dx −
f − (x)dx
Rn
Rn
Rn
6
is well-defined. More generally, a function f : Rn → C is Lebesgue-integrable if and
only if Re f and Im f are Lebesgue-integrable. Then
Z
Z
Z
f (x)dx :=
Re f (x)dx + i
Im f (x)dx.
Rn
Rn
Rn
The space of all Lebesgue-integrable functions f : Rn → C is denoted by L1 (Rn ). It
is a linear vector space, which can be normed by
Z
|f (x)|dx.
kf k1 :=
Rn
L1 (Rn ) is complete with respect to the norm k.k1 if functions which differ on a set of
measure zero are identified.
In the following almost all functions will be at least continuous, in most cases even
infinitely differentiable. If f and g are continuous and f = g almost everywhere, i.e.,
they differ on a set of measure zero, then f (x) = g(x) for every x ∈ Rn . Therefore we
do not have to pay attention on the possible null-sets on which the functions possibly
differ if we only work with continuous functions.
We need the following theorems from integration theory:
THEOREM 2.2 (Lebesgue’s Theorem on Dominated Convergence)
Let fk ∈ L1 (Rn ) be sequence of functions such that
lim fk (x) = f (x)
a.e.,
|fk (x)| ≤ g(x)
a.e.,
k→∞
where g ∈ L1 (Rn ). Then f ∈ L1 (Rn ) and
Z
Z
fk (x)dx =
lim
k→∞
Rn
f (x)dx.
Rn
As an application of Lebesgue’s Theorem on Dominated Convergence one can prove
the following two theorems on parameter-dependent integrals, which will often be
used in the following.
THEOREM 2.3 Let U ⊂ Rm be an open set and a ∈ U . Moreover, let f : Rn ×U →
C be functions such that:
1. For all x ∈ Rn the function t 7→ f (x, t) is continuous in a.
2. For all t ∈ U the function x 7→ f (x, t) is integrable on Rn .
3. There is an F ∈ L1 (Rn ) such that |f (x, t)| ≤ F (x) for almost all x ∈ Rn and
all t ∈ U .
7
Then the function
Z
g(t) :=
t ∈ U,
f (x, t)dx,
Rn
is continuous in a.
THEOREM 2.4 Let I ⊂ R be an open interval and f : Rn × I → C be such that
1. For every x ∈ Rn the function t 7→ f (x, t) is differentiable on I.
2. For every t ∈ I the function x 7→ f (x, t) is integrable on Rn .
3. There is an F ∈ L1 (Rn ) such that |∂t f (x, t)| ≤ F (x).
Then
Z
f (x, t)dx
g(t) :=
Rn
is a differentiable function on I and for all t ∈ I
Z
0
∂t f (x, t)dx.
g (t) =
Rn
The proofs of these theorems are left to the reader as an exercise. Alternatively, see
[For84, §11, Satz 1/Satz 2].
THEOREM 2.5 (Fubini’s Theorem)
Let fR ∈ L1 (Rk × Rl ), k, l ∈ N. Then f (x, .) ∈ L1 (Rl ) for almost all x ∈ Rk ,
x 7→ Rk f (x, y)dy ∈ L1 (Rk ), and
Z
Z Z
f (x, y)dy dx.
(2.3)
f (x, y)d(x, y) =
Rk ×Rl
Rk
Rl
Conversely, if f (x, .) ∈ L1 (Rl ) for almost all x ∈ Rk and x 7→
then f ∈ L1 (Rk × Rl ) and (2.3) holds.
R
Rk
f (x, y)dy ∈ L1 (Rk ),
The following simple lemma will be essential
Lemma 2.6 Let s > n. Then hxi−s ∈ L1 (Rn ) and (1 + |x|)−s ∈ L1 (Rn ).
Proof: The lemma can easily be proved e.g. by using polar coordinates. An alternative approach can be found in [Ray91, Lemma 1.3].
We will also use the change-of-variable theorem in the following case:
Z
Z
f (Φ(x))| det DΦ(x)|dx =
f (y)dy
Rn
Rn
8
for all f ∈ L1 (Rn ) and C 1 -diffeomorphism Φ : Rn → Rn . In particular, we have
Z
Z
f (Ax + b)| det A|dx =
f (y)dy
Rn
Rn
for all f ∈ L1 (Rn ), b ∈ Rn , and A ∈ Rn×n with det A 6= 0.
We will also use with spaces Lp (Rn ), 1 ≤ p < ∞, which consist of all measurable
functions f : Rn → C such that
Z
kf kp :=
p
p1
< ∞.
|f (x)|
Rn
All these spaces are Banach spaces (again identifying functions which coincide almost
everywhere). We will mainly work with L2 (Rn ) which is also a Hilbert space; its norm
is given by the scalar product:
p
(f, f ),
kf k2 =
Z
f (x)g(x)dx.
(f, g) ≡ (f, g)L2 (Rn ) =
Rn
The L2 -scalar product will play a fundamental role.
An important subspace of Lp (Rn ) is C0∞ (Rn ), which is the set of all smooth
f : Rn → C with compact support
supp f := {x ∈ Rn : f (x) 6= 0}.
Lemma 2.7 Let 1 ≤ p < ∞. Then C0∞ (Rn ) is dense in Lp (Rn ), i.e., for every
f ∈ Lp (Rn ) there is a sequence fk ∈ C0∞ (Rn ) such that limk→∞ kf − fk kp = 0. For
k.kp
short: C0∞ (Rn )
= Lp (Rn ).
Proof: See [Alt85, Satz 2.10] or [For84, §10 Satz 3].
9
3
Fourier Transformation
3.1
Definition and Basic Properties
Let f ∈ L1 (Rn ). Then we define the Fourier transform of f by
Z
ˆ
f (ξ) := F[f ](ξ) :=
e−ix·ξ f (x)dx
(3.1)
Rn
for ξ ∈ Rn . Since |e−ix·ξ f (x)| = |f (x)| for all ξ ∈ Rn , we have e−ix·ξ f (x) ∈ L1 (Rn )
with respect to x and (3.1) is well-defined.
Moreover, we have the following elementary properties:
Lemma 3.1
1. F : L1 (Rn ) → Cb0 (Rn ) is a linear mapping such that
kF[f ]kCb0 (Rn ) ≤ kf kL1 (Rn ) .
2. If f : Rn → C is a continuously differentiable function such that f ∈ L1 (Rn )
and ∂j f ∈ L1 (Rn ), then
F[∂xj f ] = iξj F[f ] = iξj fˆ.
(3.2)
3. If f ∈ L1 (Rn ) such that xj f ∈ L1 (Rn ), then fˆ(ξ) is continuously partial differentiable with respect to ξj and
∂ξj fˆ(ξ) = F[−ixj f (x)].
(3.3)
4. Let f ∈ L1 (Rn ) and (τy f )(x) := f (x + y), y ∈ Rn , denote the translation of f
by y. Then F[τy f ](ξ) = eiy·ξ fˆ(ξ) for all ξ ∈ Rn .
5. If f, g ∈ L1 (Rn ), then
fˆ(ξ)ĝ(ξ) = F [f ∗ g] ,
where
Z
f ∗ g(x) :=
f (x − y)g(y)dy
Rn
denotes the convolution of f and g.
6. Let f ∈ L1 (Rn ) and let (ρε f )(x) := f (εx), ε > 0, denote the dilation of f by
ε. Then
F[ρε f ](ξ) = ε−n fˆ(ξ/ε) = ε−n (ρε−1 fˆ)(ξ).
Proof:
10
1. First of all
kfˆk∞ = sup |fˆ(ξ)| ≤ sup
ξ∈Rn
ξ∈Rn
Z
−ix·ξ
e
f (x) dx = kf k1 .
Rn
Hence fˆ is bounded. Let g(x, ξ) = e−ix·ξ f (x). Then ξ 7→ g(x, ξ) is continuous
for every fixed x ∈ Rn , x 7→ g(x, ξ) is measurable for every ξ ∈ Rn , and
|g(x, ξ)| ≤ |f (x)| ∈ L1 (Rn ). Therefore we can apply Theorem 2.3 which implies
the continuity of f .
2. Use integration by parts.
3. Use Theorem 2.4.
4. Use the change-of-variables theorem.
5. First of all, f (x − y)g(y) ∈ L1 (Rn × Rn ) with respect to (x, y) since
Z
Z Z
|f (z)g(y)|d(y, z) = kf k1 kgk1 .
|f (x − y)g(y)|dydx =
kf ∗ gk1 ≤
Rn
Rn ×Rn
Rn
by Fubini’s theorem and the change-of-variable theorem. Hence
Z Z
e−ix·ξ f (x − y)g(y)dydx
F[f ∗ g](ξ) =
Rn Rn
Z Z
−ix·ξ
e
(τ−y f )(x)dx g(y)dy
=
Rn
Rn
Z
e−iy·ξ fˆ(ξ)g(y)dy = fˆ(ξ)ĝ(ξ),
=
Rn
where we have used Fubini’s theorem and the fourth statement of this lemma.
6. Use the change-of-variables theorem.
The relation (3.2) is the fundamental property of the Fourier transformation from
the point of view of differential equations. Because of the factor i in (3.2), we define
1
Dxj = ∂xj ,
i
Dx = (Dx1 , . . . , Dxn ).
Then (3.2) is equivalent to
F[Dxj f ] = ξj F[f ] = ξj fˆ.
Moreover, if we express a linear differential operator P with constant coefficients as
X
Pf =
cα Dx f
|α|≤m
11
with some constants cα ∈ C, then
F[P f ] =
X
cα ξ α fˆ.
|α|≤m
Hence application of a linear differential operator P to f corresponds to multiplication
of fˆ with the polynomial
X
p(ξ) :=
cα ξ α .
|α|≤m
The function p(ξ) is called the symbol of P . Moreover, we write P = p(Dx ).
Remark 3.2 If f ∈ L1 (Rn ) is continuously differentiable and ∂j f ∈ L1 (Rn ) for all
j = 1, . . . , n, then fˆ(ξ) and ξj fˆ(ξ) are bounded functions. Hence
(1 + |ξ|)|fˆ(ξ)| ≤ C
⇔
|fˆ(ξ)| ≤
C
,
1 + |ξ|
which means that fˆ(ξ) decays as |ξ|−1 as |ξ| → ∞.
More generally, if f ∈ C k (Rn ) such that ∂xα f ∈ L1 (Rn ) for all |α| ≤ k, then one
can show in the same way that
|fˆ(ξ)| ≤
C
.
(1 + |ξ|)k
For short: Differentiability of f implies a polynomial decay of fˆ as |ξ| → ∞.
On the other hand, if (1 + |x|)k f (x) ∈ L1 (Rn ) for some k ∈ N, we can apply
Lemma 3.1.3 successively to conclude that fˆ ∈ C k (Rn ). Hence faster decay of f (x)
as |x| → ∞ yields to a higher differentiability of fˆ.
3.2
Rapidly decreasing functions – S(Rn )
Let f ∈ C0∞ (Rn ). Then ∂xα f ∈ L1 (Rn ) for all α ∈ N0 . As seen in Remark 3.2, fˆ(ξ)
decays faster than any power (1 + |x|)−k , k ∈ N. Moreover, (1 +T|x|)k f ∈ L1 (Rn ) for
k
n
all k ∈ N. Hence fˆ ∈ Cbk (Rn ) for all k ∈ N, i.e., fˆ ∈ Cb∞ (Rn ) = ∞
k=1 Cb (R ). But in
general fˆ(ξ) does not have compact support, cf. Exercise 1. If f ∈ C0∞ (Rn ), then fˆ
belongs to the following function space.
Definition 3.3 The space S(Rn ) of all rapidly decreasing smooth functions is the
set of all smooth f : Rn → C such that for all α ∈ Nn0 and N ∈ N0 there is a constant
Cα,N such that
|∂xα f (x)| ≤ Cα,N (1 + |x|)−N
(3.4)
uniformly in x ∈ Rn . If f ∈ S(Rn ) and m ∈ N, we define the semi-norm:
|f |m,S :=
sup |xα ∂xβ f (x)|
sup
|α|+|β|≤m x∈Rn
12
2
Obviously, C0∞ (Rn ) ⊂ S(Rn ). The inclusion is strict since f (x) = e−|x| ∈ S(Rn ).
Moreover, S(Rn ) ⊂ Cb∞ (Rn ).
Remark 3.4 At first sight, it may be more natural to use
|f |0m,S :=
sup
sup (1 + |x|)k |∂ α f (x)|
k+|α|≤m x∈Rn
as semi-norms on S(Rn ) since they are more closely related to the inequality (3.4).
But the definition of |.|m,S is more convenient when dealing with Fourier transformation, which will be demonstrated in the proof of the next lemma. Moreover, it does
not matter if we use the semi-norms |.|m,S or |.|0m,S – the semi-norms are equivalent
in the following sense: For every m ∈ N0 there is an k(m) ∈ N0 such that
|f |0m,S ≤ Cm |f |k(m),S
0
and |f |m,S ≤ Cm
|f |0k(m),S
for all f ∈ S(Rn ). (Actually, in this special case we can simply choose k(m) = m.)
2 21
Finally, we note
that
we
can
replace
(1
+
|x|)
in
(3.4)
by
hxi
:=
(1
+
|x|
) since
√
hxi ≤ (1 + |x|) ≤ 2hxi.
For our purposes, the fundamental property of S(Rn ) is the following:
Lemma 3.5 F : S(Rn ) → S(Rn ) is linear mapping. More precisely, |fˆ|m,S ≤ Cm |f |m+n+1,S
for all m ∈ N0 where Cm depends only on n and m.
Proof: First of all, if f ∈ S(Rn ),
Z
kf k1 =
(1 + |x|)−n−1 (1 + |x|)n+1 |f (x)| dx
n
RZ
(1 + |x|)−n−1 dx|f |n+1,S = C|f |n+1,S ,
≤ C
Rn
where C depends only on the dimension n. Thus
|fˆ|0,S = kfˆk∞ ≤ kf k1 ≤ C|f |n+1,S
by Lemma 3.1.1 and the previous estimate. Because of Lemma 3.1.2/3,
ξ α Dξβ fˆ(ξ) = F Dxα xβ f (x) .
Hence
kξ α Dξβ fˆk∞ ≤ C|Dxα xβ f (x) |n+1,S
by (3.5). Using Leibniz’s formula (2.1),
Dxα
X α
x f (x) =
(Dxγ xβ )(Dxα−γ f (x)).
γ
γ≤α
β
13
(3.5)
Since Dxγ xβ is polynomial of degree less than |β|,
|Dxα xβ f (x) |n+1,S ≤ Cα,β |f ||α|+|β|+n+1 .
Collecting all estimates and taking the supremum over α, β ∈ Nn0 with |α| + |β| ≤ m,
we finally conclude
|fˆ|m,S ≤ Cm |f |m+n+1,S
with arbitrary m ∈ N0 where Cm depends only on n and m. Hence F[f ] ∈ S(Rn ) for
all f ∈ S(Rn ).
Example 3.6 Let f (x) = e−|x|
2 /2
2
2
, x ∈ Rn . Then fˆ(ξ) = (2π)n/2 e−|ξ| /2 .
2
Proof: Since f (x) = e−x1 /2 · · · e−xn /2 and e−ix·ξ = e−ix1 ξ1 · · · e−ixn ξn ,
Z
Z
2 /2
2
−ix
ξ
−x
1
1
e
fˆ(ξ) =
e 1 dx1 · · · e−ix1 ξ1 e−xn /2 dxn = ĝ(ξ1 ) · · · ĝ(ξn )
R
R
2
where g(x) = e−x /2 , x ∈ R. Hence it is sufficient to consider the case n = 1.
Because of Lemma 3.1.3, ĝ(ξ) is continuously differentiable and ĝ 0 (ξ) = F[−ixg(x)].
d −x2 /2
Moreover, −xg(x) = dx
e
= g 0 (x). Therefore, using Lemma 3.1.2,
ĝ 0 (ξ) = iF[g 0 (x)] = −ξg(ξ),
ξ ∈ R,
√
R
2
and ĝ(0) = R e−x /2 dx = 2π. Hence ĝ is uniquely determined by the letter initial
√
2
value problem, which has the unique solution ĝ(ξ) = 2πe−ξ /2 .
An alternative proof using the Cauchy integral formula can be found in [Ray91,
Example 1.7].
Exercise 1 Let f ∈ C0∞ (R) such that supp f ⊆ BR (0). Prove that
fˆ(ξ) =
Z
R
eiξx f (x)dx,
ξ ∈ C,
−R
is a holomorphic function in C. Moreover, |fˆ(ξ)| ≤ CeR| Im ξ| .
In particular, fˆ(ξ), ξ ∈ R, is real analytic and supp fˆ is not compact unless
ˆ
f (ξ) ≡ 0.
3.3
Ex-course: Fréchet spaces
Throughout the lecture, we will work with many different spaces of smooth functions
with certain properties. They all have in common that they cannot be normed to
become a Banach space (with their natural topology).
14
As an example we consider the space Cb∞ (Rn ) of all smooth and bounded functions
f : Rn → C with bounded derivatives. Then
Cb∞ (Rn ) =
∞
\
Cbk (Rn ),
k=0
where Cbk (Rn ) is the space of all k-times differentiable functions f : Rn → C with
bounded derivatives up to order k. In contrast to Cb∞ (Rn ) the spaces Cbk (Rn ), k ∈ N0 ,
are Banach spaces equipped with the norm
kf kCbk = sup sup |∂ α f (x)|.
|α|≤k x∈Rn
Since in spaces smooth functions, all (infinitely many) derivatives have to be controlled, in general they cannot be normed by a single norm such that the space is
complete. But these spaces can be “normed” by an infinite sequence of (semi-)norms,
e.g. f ∈ Cb∞ (Rn ) if and only if kf kCbk is finite for all k ∈ N.
Definition 3.7 Let V be a (complex or real) linear space. Then a mapping ρ : V →
[0, ∞) is a semi-norm if
1. ρ(rf ) = |r|ρ(f ) for all f ∈ V and r ∈ R, C, resp.,
2. ρ(f + g) ≤ ρ(f ) + ρ(g) for all f, g ∈ V .
If ρm , m ∈ N, is a sequence of semi-norms on a linear space V and fj ∈ V , j ∈ N,
we say that (fj )j∈N is a Cauchy sequence if limi,j→∞ ρm (fi − fj ) = 0 for all m ∈ N.
Moreover, we say that fj converges to f ∈ V if and only if limj→∞ ρm (fj − f ) = 0
for all m ∈ N.
Definition 3.8 A linear space V is called a Fréchet space if there is a sequence of
semi-norms (ρm )m∈N satisfying the following conditions:
1. (ρm )m∈N is an increasing sequence: ρ1 (f ) ≤ ρ2 (f ) ≤ . . . ≤ ρm (f ) ≤ . . . for all
f ∈V.
2. (ρm )m∈N separates points: for any f 6= 0 there is an m ∈ N such that ρm (f ) 6= 0.
3. V is complete: any Cauchy sequence (fj )j∈N converges to some f ∈ V .
Obviously, every Banach space X with norm k.kX is a Fréchet space with (semi-)
norms ρm = k.kX for all m ∈ N.
Exercise 2 Prove that Cb∞ (Rn ) equipped with the norms ρm = k.kCbm is a Fréchet
space. – It can be used that Cbk (Rn ) is a Banach space.
Lemma 3.9 The space S(Rn ) together with the |.|m,S , m ∈ N, is a Fréchet space.
15
The proof is left as exercise for the interested reader. We just note that it mainly
remains to check the completeness of S(Rn ), which can be done by using the completeness of Cb∞ (Rn ). But the completeness of S(Rn ) will not be used in the following.
Therefore we do not give a proof.
Remark 3.10 (For readers interested in topology)
Every Fréchet space (V, (ρm )m∈N ) is a complete metric space (V, d) where
d(f, g) :=
∞
X
2−m
m=1
Exercise 3
ρm (f − g)
.
1 + ρm (f − g)
(3.6)
1. Prove that d defined in (3.6) is a metric.
2. Prove that limj→∞ d(fj , f ) if and only if ρm (fj − f ) = 0 for all m ∈ N.
Then convergence with respect to the semi-norms coincides with convergence in the
metric sense. Since every metric space carries a natural topology which is defined by
the neighborhood basis
1
Ul (f0 ) = f ∈ V : d(f − f0 ) <
, l ∈ N, f0 ∈ V,
l
every Fréchet space is a topological space. Equipped with this topology, the operations of multiplication by scalars and addition of vectors are continuous, i.e., V is a
linear topological space.
Moreover, every Fréchet space is locally convex, i.e. there is a neighborhood
basis at the origin that consists of convex sets. Finally, the Fréchet spaces can be
characterized as the locally convex linear topological space that can be equipped with
a metric and are complete. See e.g. [Tre67].
Definition 3.11 Let V and W be Fréchet spaces with semi-norms (ρm )m∈N and
(τm )m∈N . Then a linear mapping T : V → W is bounded if for every m ∈ N there is
a k = k(m) ∈ N and a Cm such that
τm (T f ) ≤ Cm ρk(m) (f )
for all f ∈ V.
THEOREM 3.12 Let V and W be Fréchet spaces and T : V → W be a linear mapping. Then T is bounded if and only if T is continuous in the sense that limj→∞ fj = f
implies limj→∞ T fj = T f .
In particular, if W is Banach space, T : V → W is continuous if and only if
kT f kW ≤ Cρm (f )
with some C > 0 and m ∈ N.
16
for all f ∈ V
Proof: Let ρm and τm , m ∈ N, denote the semi-norms of V and W , respectively.
Firstly, let T be bounded and fj be a sequence such that limj→∞ fj = f , i.e.,
limj→∞ ρm (fj − f ) = 0 for all m ∈ N. Since T is bounded for every m ∈ N there
exists a k(m) such that τm (T f ) ≤ Cm ρk(m) (f ), f ∈ V . Hence
τm (T fj − T f ) = τm (T (fj − f )) ≤ Cm ρk(m) (fj − f ) → 0 as j → ∞
for every m ∈ N, which means limj→∞ T fj = T f . Thus T is continuous.
Secondly, let T be continuous. Assume that T is not bounded. Then there is an
m0 ∈ N and a sequence fj ∈ V , j ∈ N, such that for every C > 0 and k ∈ N there is
a j0 with
τm0 (T fj0 ) > Cρk (fj0 ).
In particular, we can choose a subsequence fjk , k ∈ N, such that
τm0 (T fjk ) > kρk (fjk ).
Then f˜k := fjk /τm0 (T fjk ), k ∈ N, is a sequence such that
1
ρm (f˜k ) ≤ ρk (f˜k ) <
k
for all m ≤ k ∈ N.
Thus limk→∞ ρm (f˜k ) = 0 for all m ∈ N, i.e., limk→∞ f˜k = 0. Hence limk→∞ T f˜k = 0
because of the continuity of T . On the other hand τm0 (T f˜k ) = 1 for all k ∈ N which
contradicts limk→∞ f˜k = 0.
Exercise 4 A subset A of a Fréchet space V is called bounded if supf ∈A ρm (f ) < ∞
for every m ∈ N. – Prove that T : V → W is bounded if and only if T (A) is bounded
for every bounded A ⊂ V .
3.4
Inverse Fourier transformation and Parseval’s formula
In the following we will show that the Fourier transformation F : S(Rn ) → S(Rn ) is
invertible and that its inverse is given by
Z
1
−1
F [g](x) :=
eix·ξ g(ξ)dξ,
(2π)n Rn
which is well-defined for all g ∈ L1 (Rn ). – Note that
F −1 [g](x) = (2π)n F[g](−x).
If f (x) = e−|x|
2 /2
is the function discussed in Example 3.6, then
h
i
2
2
F −1 [fˆ(ξ)] = (2π)−n/2 F e−|ξ| /2 (−x) = e−|x| /2 = f (x).
Hence F −1 [fˆ] = f for this special f .
17
(3.7)
Lemma 3.13 (Inversion Formula)
Let f ∈ S(Rn ). Then f (x) = F −1 [fˆ](x) for all x ∈ Rn . In particular, F : S(Rn ) →
S(Rn ) is a linear isomorphism.
Proof: First of all,
F
−1
1
[fˆ](x) =
(2π)n
Z
Z
i(x−y)·ξ
e
Rn
f (y)dy dξ.
Rn
Since ei(x−y)·ξ f (y) 6∈ L1 (R2n ) as function in (y, ξ), we cannot apply Fubini’s theorem
2
2
directly. Therefore we introduce a factor ψε (ξ) := e−ε |ξ| /2 , ε > 0, which assures
absolute integrability. Since limε→0 ψε (ξ) = 1 for all x ∈ Rn , we have by Lebesgue’s
and Fubini’s theorem
Z Z
1
i(x−y)·ξ
−1 ˆ
e
ψε (ξ)f (y)dy dξ
F [f ](x) = lim
ε→0 (2π)n Rn
Rn
Z
1
2
2
ei(x−y)·ξ e−ε |ξ| f (y)d(y, ξ).
= lim
n
ε→0 (2π)
R2n
Using the substitution ξ = η/ε and y = x + εz,
Z
Z Z
1
1
i(x−y)·ξ −ε2 |ξ|2 /2
−iz·η −|η|2 /2
e
e
f (y)d(y, ξ) =
e
e
dη f (x + εz)dz
(2π)n R2n
(2π)n Rn
Rn
Z
1
2
=
e−|z| /2 f (x + εz)dz.
n/2
(2π)
Rn
Since f is continuous, we get
Z
Z
1
1
2
−|z|2 /2
lim
e
f (x + εz)dz = f (x)
e−|z| /2 dz = f (x)
n/2
ε→0 (2π)n/2 Rn
(2π)
Rn
by Lebesgue’s theorem. Hence F −1 [fˆ](x) = f (x), which proves the lemma.
2
2
Remark 3.14 The technique of inserting a rapidly decreasing factor like e−ε |ξ| /2 ,
ε > 0, and passing to the limit afterwards will be fundamental in the following. It is
the basis of the definition of the oscillatory integrals in Section 4.3 below.
Corollary 3.15 (Parseval’s Formula/Plancharel’s Theorem)
For every f, g ∈ S(Rn )
Z
Z
1
f (x)g(x)dx =
fˆ(ξ)ĝ(ξ)dξ.
n
(2π)
n
n
R
R
In particular,
kf k22 =
1
kfˆk22 .
(2π)n
and F extends to an isomorphism F : L2 (Rn ) → L2 (Rn ).
18
(3.8)
Proof: Using Fubini’s theorem, it is easy to check that
Z
Z
1
F[f ](ξ)ĝ(ξ)dξ =
f (x)F −1 [ĝ](x)dx.
(2π)n Rn
n
R
Because of Lemma 3.13, g = F −1 [ĝ], which implies (3.8). In particular, if f = g,
1
kF[f ]k22 = kf k22 .
n
(2π)
Since C0∞ (Rn ) is dense in L2 (Rn ) and obviously C0∞ (Rn ) ⊂ S(Rn ) ⊂ L2 (Rn ), F can
be extended by continuity to a bounded operator F : L2 (Rn ) → L2 (Rn ). By (3.7)
the same is true for F −1 . Moreover, F −1 [F[f ]] = f for all f ∈ L2 (Rn ) since this is
true for all f ∈ S(Rn ).
Remark 3.16 Because of the factor (2π)−n in the definition of the inverse Fourier
transformation and in Parseval’s formula (3.8), we introduce the common notation
dξ
đξ := (2π)
n . Then
Z
−1
F [g](x) =
eix·ξ g(ξ)đξ
(3.9)
Rn
and
Z
Z
f (x)g(x)dx =
fˆ(ξ)ĝ(ξ)đξ.
Rn
Rn
In the following all integrals with respect to a phase variable ξ, η, . . . will be taken
with respect to the scaled Lebesgue measure đξ, đη, . . .. All integrals with respect
to a space variable x, y, z, . . . will be integrals using the usual Lebesgue measure
dx, dy, dz, . . .. Then all constants are part of the measure and we do not have to
worry about them.
Lemma 3.17 (Fourier multipliers on L2 (Rn ))
Let m : Rn → C be a measurable functions. Then
m(Dx )f := F −1 [m(ξ)fˆ(ξ)]
is a well-defined bounded operator m(Dx ) : L2 (Rn ) → L2 (Rn ) if and only if m ∈
L∞ (Rn ). Moreover, km(Dx )kL(L2 (Rn )) = kmk∞ .
Proof: If m ∈ L∞ , obviously m(ξ)fˆ(ξ) ∈ L2 (Rn ) for every f ∈ L2 (Rn ) because of
Corollary 3.15. Moreover,
km(Dx )f kL2 (Rn ) =
1
1
ˆ
ˆ
n km(ξ)f k2 ≤
n kmk∞ kf k2 ≤ kmk∞ kf k2 .
(2π) 2
(2π) 2
Hence m(Dx ) : L2 (Rn ) → L2 (Rn ) is a bounded linear operator and km(Dx )kL(L2 (Rn )) ≤
kmk∞ .
19
For the converse implication it is sufficient to prove that a multiplication operator
M : L2 (Rn ) → L2 (Rn ), (M fˆ)(ξ) = m(ξ)fˆ(ξ) is bounded if and only if m ∈ L∞ (Rn )
and kM kL(L2 (Rn )) = kmk∞ . We skip the details since we will not use the converse
implication. See also [Ste70, Chapter II, Proposition 2].
Examples 3.18
1. If m(ξ) = F[k], where k ∈ L1 (Rn ) then m ∈ Cb0 (Rn ) ⊂
∞
n
L (R ) due to Lemma 3.1. Moreover, m(Dx )f = k ∗ f and
km(Dx )f k2 ≤ kmk∞ kf k2 ≤ kkk1 kf k2 .
(Note that the latter inequality is also a simple consequence of the well-known
estimate kk ∗ f kp ≤ kkk1 kf kp for 1 ≤ p ≤ ∞.)
iξ
2. Let mj (ξ) = |ξ|j , j = 1, . . . , n. Then kmj k∞ = 1 and Rj := mj (Dx ) : L2 (Rn ) →
L2 (Rn ) is a bounded operator. Rj is called Riesz operator. It can be shown
that
Z
x j − yj
f (y)dy = ”kj ∗ f ”,
Rj f = lim cn
n+1
ε→0
Rn |x − y|
where kj (z) = zj /|z|n+1 , cf. [Èsk81, Chapter I, Example 3.4]. Note that the
kernel kj 6∈ L1 (Rn ). These operators are typical examples of singular integral
operators. If n = 1, m1 (ξ) = i sign ξ and R1 is also called Hilbert transformation.
3. The symbol of −∆ is |ξ|2 , i.e., −∆f = F −1 [|ξ|2 fˆ(ξ)]. Hence formally the inverse
of −∆ is given by the operator (−∆)−1 f = F −1 [fˆ(ξ)/|ξ|2 ] and therefore
−1
−1 iξj iξj ˆ
∂j ∂k (−∆) f = F
f (ξ) = Rj Rk f,
|ξ| |ξ|
where Rj , Rk are the Riesz operator defined above. Hence in some sense
∂j ∂k (−∆)−1 f ∈ L2 (Rn ) if f ∈ L2 (Rn ); but the derivatives do not have exist in
the classical sense. (They have to be understood as distributional derivatives,
which will be defined in the next section.)
3.5
Tempered Distributions and Fourier Transformation
Definition 3.19 The space of tempered distributions is S 0 (Rn ) := (S(Rn ))0 – the
space of linear and bounded functionals f : S(Rn ) → C. A sequence fk ∈ S 0 (Rn )
converges to f ∈ S 0 (Rn ) if and only if limk→∞ hfk , ϕi = hf, ϕi for all ϕ ∈ S(Rn ).
Here hf, ϕi := f (ϕ), ϕ ∈ S(Rn ) denotes the duality product.
Remarks 3.20
1. If f : Rn → C is a measurable function such that hxi−N f (x) ∈
L1 (Rn ) for some N ∈ N0 , then f is naturally identified with a tempered distribution Ff ∈ S(Rn )0 defined by
Z
hFf , ϕi :=
f (x)ϕ(x)dx,
ϕ ∈ S(Rn ).
Rn
20
Moreover,
|hFf , ϕi| ≤ khxi−N f k1 |ϕ|0N,S ≤ Ckhxi−N f k1 |ϕ|N,S ,
ϕ ∈ S(Rn ),
cf. Remark 3.4. In the following we will identify the function f with its associated distribution Ff and will denote Ff simply by f .
2. If f : S(Rn ) → C is linear mapping, then by definition f is bounded if an only
if there is an m ∈ N0 and a constant C > 0 such that
|hf, ϕi| ≤ C|ϕ|m,S ,
ϕ ∈ S(Rn ).
Example 3.21 The famous delta distribution δ ∈ S 0 (Rn ) is defined by
hδ, ϕi := ϕ(0),
ϕ ∈ S(Rn ).
One of the most important properties of distributions is that they can be differentiated infinitely many times (in a generalized sense).
Definition 3.22 Let f ∈ S 0 (Rn ). Then the distributional derivative ∂xα f ∈ S 0 (Rn )
of f is the distribution defined by
h∂xα f, ϕi := (−1)|α| hf, ∂xα ϕi
for ϕ ∈ S(Rn ).
Remarks 3.23
1. Recall that the adjoint T 0 : Y 0 → X 0 of a linear operator
T : X → Y is defined by hT 0 y 0 , xi := hy 0 , T xi for y 0 ∈ Y 0 and x ∈ X. Hence the
derivative ∂xα : S 0 (Rn ) → S 0 (Rn ) is the adjoint of (−1)|α| ∂xα : S(Rn ) → S(Rn ).
2. If f : Rn → C is a k-times continuously differentiable function with hxi−N f ∈
L1 (Rn ), then f can be naturally considered as a tempered distribution, cf.
Remark3.20.1, Moreover, the distributional derivatives up to order k coincides
with the usual derivative, i.e.
Z
Z
α
|α|
α
h∂x f, ϕ(x)i = (−1)
f (x)∂x ϕ(x)dx =
∂xα f (x)ϕ(x)dx
Rn
Rn
for all |α| ≤ k due to integration by parts.
3. Since every measurable function f : Rn → C such that hxi−N f (x) ∈ L1 (Rn )
for some N ∈ N0 can be considered as a distribution, it can be differentiated
infinitely many times in the sense of distributions although it may even not be
continuous.
Examples 3.24
1. Let f be the Heavyside function, i.e. f (x) = 1 for x ≥ 0 and
f (x) = 0 for x < 0. Then f ∈ S 0 (R) and the distributional derivative f 0 is
Z
Z ∞
0
0
hf , ϕi = − f (x)ϕ (x)dx = −
ϕ0 (x)dx = ϕ(0)
0
R
for all ϕ ∈ S(R). Hence f 0 = δ is the delta distribution.
21
2. By definition the distributional derivative of the delta distribution is
h∂xα δ, ϕi = (−1)|α| hδ, ∂xα ϕi = (−1)|α| (∂xα ϕ)(0).
Definition 3.25 Let f ∈ S 0 (Rn ). The Fourier transform F[f ] of f is defined as the
distribution
hF[f ], ϕi = hf, F[ϕ]i
for ϕ ∈ S(Rn ).
Proposition 3.26 The Fourier transformation is a continuous linear mapping F : S 0 (Rn ) →
S 0 (Rn ). Moreover, F is a bijection on S 0 (Rn ).
Proof: Since F : S(Rn ) → S(Rn ) is a continuous linear operator and
hF[f ], ϕi = hf, F[ϕ]i = (f ◦ F)(ϕ),
F[f ] = f ◦ F : S(Rn ) → C is a continuous linear operator. Moreover, if fk → f in
S 0 (Rn ) as k → ∞, then
hF[fk ], ϕi = hfk , F[ϕ]i → hf, F[ϕ]i = hF[f ], ϕi
as k → ∞ for all ϕ ∈ S(Rn ). Hence F : S 0 (Rn ) → S 0 (Rn ) is continuous.
Finally, if we define F −1 : S 0 (Rn ) → S 0 (Rn ) by
hF −1 [f ], ϕi := hf, F −1 [ϕ]i,
ϕ ∈ S(Rn ),
F −1 has the same properties as F and obviously F −1 [F[f ]] = f for all f ∈ S 0 (Rn ).
Remark 3.27 As differentiation the Fourier transformation is defined by duality,
i.e., F : S 0 (Rn ) → S 0 (Rn ) is defined as the adjoint of F : S(Rn ) → S(Rn ).
Example 3.28 The Fourier transform of the delta distribution δ can be simply
calculated:
Z
hF[δ], ϕi = hδ, F[ϕ]i = F[ϕ](0) =
ϕ(x)dx = h1, ϕi
Rn
for all ϕ ∈ S(Rn ). Hence F[δ] = 1.
22
4
Basic Calculus of Pseudodifferential Operators on
Rn
4.1
Symbol Classes and Basic Properties
m
. There are many
We now introduce the basic pseudodifferential symbol class S1,0
other more general or modified symbol classes, which are used in the literature and
research for different purposes. But the following symbol class is the most simple and
most common. It is a natural symbol class that contains the symbols of differential
operators with smooth coefficients.
m
Definition 4.1 Let m ∈ R, n, N ∈ N. Then S1,0
(RN × Rn ) is the vector-space of all
smooth functions p : RN × Rn → C such that
|Dξα Dxβ p(x, ξ)| ≤ Cα,β (1 + |ξ|)m−|α|
(4.1)
N
n
holds for all α ∈ Nn0 , β ∈ NN
0 , where Cα,β is independent of x ∈ R , ξ ∈ R . The
function p is called pseudodifferential symbol and m is called the order of p. Moreover,
[
\
∞
m
−∞
m
S1,0
(RN × Rn ) :=
S1,0
(RN × Rn ) and S1,0
(RN × Rn ) :=
S1,0
(RN × Rn ).
m∈R
m∈R
m
m
(RN × Rn ).
instead of S1,0
For short we also write S1,0
Remark 4.2 In the following we usually deal with the case N = n. But sometimes
it is useful to have defined the symbol classes for general N, n ∈ N.
m
(Rn × Rn ) is a symbol, then
If p ∈ S1,0
Z
p(x, Dx )f := OP(p)f :=
eix·ξ p(x, ξ)fˆ(ξ)đξ
(4.2)
Rn
defines the associated pseudodifferential operator, where f : Rn → C is a suitable
function. If f ∈ S(Rn ), then fˆ ∈ S(Rn ) and therefore p(x, ξ)fˆ(ξ) ∈ S(Rn ) with
respect to ξ for every fixed x ∈ Rn . Therefore the integral in (4.2) exists and p(x, Dx )f
is well-defined. In the following we will prove that p(x, Dx ) : S(Rn ) → S(Rn ) is a
continuous mapping. But before we prove this fact, we discuss some examples and
make some simple observations.
P
α
Examples 4.3
1. Let p(x, ξ) =
|α|≤m cα (x)ξ be a polynomial in ξ of order
m
m ∈ N0 with smooth coefficients cα (x) ∈ Cb∞ (Rn ). Then p ∈ S1,0
and
X
p(x, Dx )f =
cα (x)Dxα f
|α|≤m
for every f ∈ S(Rn ). Hence every linear differential operator with smooth
and bounded coefficients is a pseudodifferential operator. In particular the
Laplacian ∆ = ∂12 +. . .+∂n2 is a pseudodifferential operator with symbol −|ξ|2 =
−ξ12 − . . . − ξn2 .
23
p
2. The function hξi := 1 + |ξ|2 is a pseudodifferential symbol of order 1, see
Exercise 5. Since 1+|ξ|2 is the symbol of 1−∆, the associated pseudodifferential
operator
Z
p
hDx if =
eix·ξ 1 + |ξ|2 fˆ(ξ)đξ
Rn
can be considered as the square root of 1 − ∆. For short: hDx i =
m
More generally, hξi ∈
m
S1,0
m
√
1 − ∆.
m
2
for every m ∈ R and hDx i = (1 − ∆) .
p
m
for every m ∈ R.
Exercise 5 Let hξi = 1 + |ξ|2 . Prove that hξim ∈ S1,0
2
2 m
2
Hint: Consider the function f (a, x) := (a + |x| ) , where a ∈ R, x ∈ Rn . Use
that f is homogeneous of degree m, i.e., f (ra, rx) = rm f (a, x) for all r > 0, a ∈ R,
x ∈ Rn .
Exercise 6 Let p : Rn × Rn → C be a smooth function which is homogeneous in ξ
of degree m ∈ R for |ξ| ≥ 1, i.e. p(x, rξ) = rm p(x, ξ) for all |ξ| ≥ 1 and r ≥ 1.
m
(Rn ×
Moreover, let ∂ξα ∂xβ p ∈ Cb∞ (Rn ) w.r.t. x for all α, β ∈ Nn0 . Prove that p ∈ S1,0
Rn ).
m
We can define a sequence of semi-norms on S1,0
, which is related to the family of
inequalities (4.1) in a natural way. Let
(m)
|p|k
:= max
sup |Dξα Dxβ p(x, ξ)|(1 + |ξ|)−m+|α|
|α|,|β|≤k x,ξ∈Rn
(4.3)
for k ∈ N. Here supx,ξ∈Rn |Dξα Dxβ p(x, ξ)|(1 + |ξ|)−m+|α| is the smallest constant Cα,β
such that (4.1) holds for all x, ξ ∈ Rn and fixed α, β ∈ N0 .
m
It is not difficult to check that S1,0
with these semi-norms is a Fréchet space.
Remark 4.4 In the
p literature the function (1 + |ξ|) in the estimates (4.1) is often
replaced by hξi = 1 + |ξ|2 . This can be done without changing the symbol classes
since
p
√ p
1 + |ξ|2 ≤ (1 + |ξ|) ≤ 2 1 + |ξ|2 ,
cf. proof of Lemma 4.7 below. Using hξi instead of (1 + |ξ|), the notation becomes
a bit shorter.
m
Proposition 4.5 Let pj ∈ S1,0j (Rn × Rn ), mj ∈ R, j = 1, 2. Then p(x, ξ) :=
(m +m )
(m )
(m )
m1 +m2
p1 (x, ξ)p2 (x, ξ) ∈ S1,0
(Rn × Rn ) and |p|k 1 2 ≤ Ck |p1 |k 1 |p2 |k 2 , where Ck
depends only on k and n.
The lemma is a simple consequence of the Leibniz formula. The main result of this
section is
24
m
THEOREM 4.6 Let p ∈ S1,0
(Rn × Rn ), m ∈ R, be a pseudodifferential symbol.
Then
p(x, Dx ) : S(Rn ) → S(Rn )
(m)
is a continuous mapping. More precisely, |p(x, Dx )f |k,S ≤ Ck |p|k |f |m+2(n+1)+k,S for
all k ∈ N and f ∈ S(Rn ).
Proof: Since f ∈ S(Rn ), fˆ ∈ S(Rn ) due to Lemma 3.5. First of all,
Z
1
hξi−n−1 |hξi−m p(x, ξ)||hξin+m+1 fˆ(ξ)| dξ
sup |p(x, Dx )f (x)| ≤ sup
n
x∈Rn
x∈Rn (2π)
Rn
Z
1
(m)
≤
hξi−n−1 dξ|p|0 khξin+m+1 fˆk∞
n
(2π) Rn
(m)
(m)
≤ C|p| |fˆ|m+n+1,S ≤ C|p| |f |m+2n+2,S
(4.4)
0
0
by Lemma 2.6 and Lemma 3.5, where C depends only on the dimension. Moreover,
m
(Rn × Rn ) and m ∈ R — not just for
we note that (4.4) holds for arbitrary p ∈ S1,0
the one in the assumption of this theorem.
In order to estimate the derivatives, we calculate
Z
∂xj (p(x, Dx )f (x)) = ∂xj eix·ξ p(x, ξ)fˆ(ξ)đξ
Z
Z
ix·ξ
ˆ
=
e p(x, ξ)iξj f (ξ)đξ + eix·ξ ∂xj p(x, ξ)fˆ(ξ)đξ
= p(x, Dx )(∂j f )(x) + (∂xj p)(x, Dx )f,
where we have applied Theorem 2.4. Hence using (4.4) first with f replaced by ∂j f
and then p replaced by ∂xj p, we obtain
(m)
(m)
sup |∂j (p(x, Dx )f (x)| ≤ C |p|0 |∂j f |m+2n+2,S + |∂xj p|0 |f |m+2n+2,S
x∈Rn
(m)
≤ C|p|1 |f |m+2n+3,S .
(4.5)
Similarly,
Z
ixj p(x, Dx )f (x) =
(∂ξj eix·ξ )p(x, ξ)fˆ(ξ)đξ
Z
=
e
ix·ξ
p(x, ξ)∂j fˆ(ξ)đξ +
Z
eix·ξ (∂ξj p)(x, ξ)fˆ(ξ)đξ
= p(x, Dx )(ixj f (x)) + (∂ξj p)(x, Dx )f
and therefore
(m)
(m−1)
sup |xj p(x, Dx )f (x)| ≤ C |p|0 |xj f |m+2n+2,S + |∂ξj p|0
|f |m+2n+1,S
x∈Rn
(m)
≤ C|p|1 |f |m+2n+3,S
25
(4.6)
(m−1)
by (4.4), where we note that ∂ξj p(x, ξ) is of order m − 1 and |∂ξj p|0
definition of the semi-norms.
Using (4.5) and (4.6), one can easily prove by induction that
(m)
≤ |p|1
by
(m)
sup |xα ∂xβ p(x, Dx )f (x)| ≤ Cα,β |p||α|+|β| |f |m+2(n+1)+|α|+|β|,S
x∈Rn
(m)
uniformly in x ∈ Rn for all α, β ∈ Nn0 . Hence |p(x, Dx )f |k,S ≤ Ck |p|k |f |m+2(n+1)+k,S ,
which proves the theorem.
Finally, we prove the following simple but important inequality:
Lemma 4.7 (Peetre’s inequality)
1
Let hξi := (1 + |ξ|2 ) 2 , ξ ∈ Rn . Then for all s ∈ R
hξis ≤ 2|s| hξ − ηi|s| hηis ,
ξ, η ∈ Rn .
Proof: First of all we have
hξi2 = (1 + |ξ|2 ) ≤ (1 + |ξ|)2 ≤ (1 + |ξ|)2 + (1 − |ξ|)2 = 2(1 + |ξ|2 ).
Hence
hξi ≤ (1 + |ξ|) ≤
√
2hξi.
(4.7)
First let s ≥ 0. Then the triangle inequality implies
(1 + |ξ|) ≤ (1 + |ξ − η| + |η|) ≤ (1 + |ξ − η|)(1 + |η|)
and therefore
hξis ≤ (1 + |ξ − η|)s (1 + |η|)s ≤ 2s hξ − ηis hηis
by (4.7). If s < 0, we can use the previous inequality with interchanged role of ξ and
η and s replaced by −s to conclude
hηi−s ≤ 2−s hξ − ηi−s hξi−s
which is equivalent to
hξis ≤ 2|s| hηi|s| hξ − ηi−s .
26
4.2
Composition of Pseudodifferential Operators: Motivation
Because of Theorem 4.6, the composition of two pseudodifferential operators p1 (x, Dx )
and p2 (x, Dx ) is a well-defined bounded operator
p1 (x, Dx )p2 (x, Dx ) : S(Rn ) → S(Rn ).
The natural question arises if this operator is again a pseudodifferential operator,
∞
i.e., if there is a symbol p ∈ S1,0
(Rn × Rn ) such that
p(x, Dx ) = p1 (x, Dx )p2 (x, Dx ).
If this is the case, it is of interest how the symbol p(x, ξ) is related to the symbols
p1 (x, ξ) and p2 (x, ξ).
The behavior of pseudodifferential operators under composition is of particular
interest for calculation inverses or at least approximate inverses of pseudodifferential
operators, which are also called parametrices.
In order to motivate the following sections, we calculate the composition of
p1 (x, Dx )p2 (x, Dx ) formally, ignoring all technical difficulties. First of all,
ZZ
p1 (x, Dx )g =
ei(x−y)·η p1 (x, η)g(y)dyđη
and
ZZ
p2 (x, Dx )f |x=y =
ei(y−z)·ξ p2 (y, ξ)f (z)dzđξ.
Hence we get for g(y) = p2 (x, Dx )f |x=y
Z Z
ZZ
i(x−y)·η
i(y−z)·ξ
p1 (x, Dx )p2 (x, Dx )f =
e
p1 (x, η)
e
p2 (y, ξ)f (z)dzđξ dyđη
ZZZZ
=
ei(x−z)·ξ e−i(x−y)·(ξ−η) p1 (x, η)p2 (y, ξ)f (z)dyđηdzđξ.
Using the substitution x0 = x − y and ξ 0 = ξ − η, we obtain
p1 (x, Dx )p2 (x, Dx )f
Z Z
ZZ
i(x−z)·ξ
−ix0 ·ξ 0
0
0
0
0
=
e
e
p1 (x, ξ + ξ )p2 (x + x , ξ)dx đξ f (z)dzđξ
Hence formally the symbol of p1 (x, Dx )p2 (x, Dx ) is
ZZ
0 0
(p1 #p2 )(x, ξ) :=
e−ix ·ξ p1 (x, ξ + ξ 0 )p2 (x + x0 , ξ)dx0 đξ 0 .
(4.8)
But the main problem is that the latter integral in general does not exists in the
classical sense. We will define it as so called oscillatory integral :
ZZ
0 0
Os–
e−ix ·ξ p1 (x, ξ + ξ 0 )p2 (x + x0 , ξ)dx0 đξ 0 :=
ZZ
0 0
lim
χ(εx0 , εξ 0 )e−ix ·ξ p1 (x, ξ + ξ 0 )p2 (x + x0 , ξ)dx0 đξ 0 ,
ε→0
27
where χ ∈ S(Rn × Rn ) with χ(0, 0) = 1. In the following section we prove that the
oscillatory integral are well-defined for suitable integrands. Moreover, we will observe
some elementary properties, which justify our formal calculations above.
4.3
Oscillatory Integrals
n
n
Definition 4.8 The space of amplitudes Am
τ (R ×R ), m, τ ∈ R, is the set of smooth
functions a : Rn × Rn → C such that
|∂ηα ∂yβ a(y, η)| ≤ Cα,β (1 + |η|)m (1 + |y|)τ
uniformly in y, η ∈ Rn for all α, β ∈ Nn0 . Moreover, let
|a|Am
:= max
τ ,k
sup (1 + |η|)−m (1 + |y|)−τ |∂ηα ∂yβ a(y, η)|,
|α|+|β|≤k y,η∈Rn
k ∈ N, be the associated sequence of monotone increasing semi-norms.
n
n
n
m
It is not difficult to check that Am
τ (R × R ) is a Fréchet space. Moreover, S1,0 (R ×
n
n
n
m
R ) ⊂ A0 (R × R ) with continuous embedding.
n
n
n
n
THEOREM 4.9 Let a ∈ Am
τ (R × R ), m, τ ∈ R, and let χ ∈ S(R × R ) with
χ(0, 0) = 1. Then
ZZ
ZZ
−iy·η
Os–
e
a(y, η)dyđη := lim
χ(εy, εη)e−iy·η a(y, η)dyđη
ε→0
exists and
ZZ
ZZ
0
0 −iy·η
Os–
e
a(y, η)dyđη =
e−iy·η hyi−2l hDη i2l hηi−2l hDy i2l a(y, η) dyđη,
(4.9)
where l, l ∈ N0 are chosen such that 2l > n + m and 2l > n + τ and the integrand is
in L1 (Rn × Rn ). In particular, the definition does not depend on the choice of χ and
ZZ
−iy·η
≤ Cm,τ kakAm ,2(l+l0 )
Os–
e
a(y,
η)dyđη
(4.10)
τ
0
0
where Cm,τ > 0 is independent of a.
Proof: Using Dyα e−iy·η = (−η)α e−iy·η and Dηβ e−iy·η = (−y)β e−iy·η for α, β ∈ Nn0 , we
have
0
0
hηi−2l hDy i2l e−iy·η and hyi−2l hDη i2l e−iy·η .
(4.11)
Since χ(εy, εη) ∈ S(Rn × Rn ) for fixed ε > 0, we can integrate by parts and obtain
ZZ
Iε :=
χ(εy, εη)e−iy·η a(y, η)dyđη
ZZ
=
e−iy·η hηi−2l hDy i2l (χ(εy, εη)a(y, η)) dyđη
ZZ
0
0 =
e−iy·η hyi−2l hDη i2l hηi−2l hDy i2l (χ(εy, εη)a(y, η)) dyđη
28
On the other hand, {χ(εy, εη)}0<ε<1 ≡ {χε (y, η)}0<ε<1 is bounded in A00 = Cb∞ (Rn ×
Rn ) and limε→0 χ(εy, εη) = 1 uniformly on compact sets and limε→0 ∂yα ∂ηβ χε (y, η) =
0 uniformly in (y, η) ∈ Rn × Rn if (α, β) 6= 0. Hence there are constants Cα,β
independent of both 0 < ε < 1 and a ∈ Am
τ such that
α β
∂y ∂η (χε (y, η)a(y, η)) ≤ Cα,β |a|Am ,|α|+|β| hηim hyiτ .
(4.12)
τ
s
Moreover, since hξis ∈ S1,0
,
|∂ηα hηis | ≤ Cs,α hηis−|α| .
(4.13)
Combining (4.12) and (4.13), there are constants Cl,α independent of 0 < ε < 1 such
that
α −2l
∂η hηi hDy i2l (χε (y, η)a(y, η)) ≤ Cl,α |a|Am ,2l+|α| hηim−2l hyiτ .
τ
Consequently there are constants Cl,l0 independent of 0 < ε < 1 and a such that
−2l
0
−2l0
2l0
2l
m−2l
0 hηi
hyi
hD
i
hηi
hD
i
(χ
(y,
η)a(y,
η))
hyiτ −2l (. 4.14)
≤ Cl,l0 |a|Am
η
y
ε
τ ,2(l+l )
0
If we now choose 2l > n + m and 2l0 > n + τ , then hηim−2l hyiτ −2l ∈ L1 (Rn × Rn ), cf.
Lemma 2.6. Hence the Lebesgue dominated convergence theorem, limε→0 χ(εy, εη) =
1, and limε→0 ∂yα ∂ηβ χ(εy, εη) = 0 for α + β > 0 imply
ZZ
0
0 lim Iε =
e−iy·η hyi−2l hDη i2l hηi−2l hDy i2l a(y, η) dyđη.
ε→0
Thus the limit in the definition of the oscillatory integral exists and (4.9) holds.
Moreover, the representation (4.9) shows that the definition does not depend on the
choice of χ. Finally, passing to the limit ε → 0 in (4.14), (4.10) follows from (4.9)
and Lemma 2.6.
α β
Corollary 4.10 Let aj ∈ Am
τ be a bounded sequence such that limj→∞ ∂y ∂η aj (y, η) =
m
n
n
α β
∂y ∂η a(y, η) for every y, η ∈ R , α, β ∈ N0 and a ∈ Aτ . Then
ZZ
ZZ
−iy·η
lim Os–
e
aj (y, η)dyđη = Os–
e−iy·η a(y, η)dyđη.
j→∞
Proof: The assumptions imply that
0
0 0
0 lim hyi−2l hDη i2l hηi−2l hDy i2l aj (y, η) = hyi−2l hDη i2l hηi−2l hDy i2l a(y, η)
j→∞
for every y, η ∈ Rn (pointwise). Moreover, (4.14) implies
−2l
0
−2l0
2l0
2l
hyi hDη i hηi hDy i (χ(εy, εη)aj (y, ε)) ≤ Cl,l0 |aj |2(l+l0 ) hηim−2l hyiτ −2l .
Since the sequence aj is bounded in Am
τ , |aj |2(l+l0 ) ≤ C uniformly in j ∈ N. Hence
the representation (4.9) and the Lebesgue dominated convergence theorem imply the
statement of the theorem.
29
Example 4.11 Let u ∈ Cb∞ (Rn ). Then a(y, η) = eix·η u(y) ∈ A00 and
ZZ
Os–
ei(x−y)·η u(y)dyđη
is well-defined. We can calculate the oscillatory integral explicitely: Let χ(y, η) =
ψ(y)ψ(η), where ψ ∈ S(Rn ) with ψ(0) = 1. Then
Z
ZZ
Z
i(x−y)·η
i(x−y)·η
Os–
e
u(y)dyđη = lim ψ(εy)u(y)
e
ψ(εη)đη dy
ε→0
Z
x−y
−n −1
dy
= lim ψ(εy)u(y)ε F [ψ]
ε→0
ε
Z
= lim ψ(ε(x − εy 0 ))u(x − εy 0 )F −1 [ψ](y 0 )dy 0
ε→0
Z
=
u(x)F −1 [ψ](y 0 )dy 0 = u(x)F[F −1 [ψ]](0)
= u(x)ψ(0) = u(x)
since limε→0 ψ(ε(x − εy 0 ))u(x − εy 0 ) = ψ(0)u(x) = u(x). Thus formally we have
F −1 [F[u]] = u for all u ∈ Cb∞ (Rn ).
n
Lemma 4.12 Let a ∈ Am
τ , m, τ ∈ R, and let α ∈ N0 . Then
ZZ
ZZ
−iy·η α
Os–
e
y a(y, η)dyđη = Os–
e−iy·η Dηα a(y, η)dyđη,
ZZ
ZZ
−iy·η α
Os–
e
η a(y, η)dyđη = Os–
e−iy·η Dyα a(y, η)dyđη.
m+|α|
α
, and
Proof: First of all we note that y α a(y, η) ∈ Am
τ +|α| , η a(y, η) ∈ Aτ
m
α
α
Dη a(y, η), Dy a(y, η) ∈ Aτ . Therefore the oscillatory integrals are well-defined.
We only prove the first identity since the proof of the second is done in the same
way. Moreover, it is sufficient to consider the case |α| = 1. (Then the general case
follows by induction.)
If |α| = 1, then y α = yj for 1 ≤ j ≤ n. Moreover, we choose χ in the definition of
2
the oscillatory integral as χ(y, η) = e−|(y,η)| /2 . Then
ZZ
ZZ
−iy·η
χ(εy, εη)e
yj a(y, η)dyđη = −
χ(εy, εη)(Dηj e−iy·η )a(y, η)dyđη
ZZ
=
e−iy·η Dηj (χ(εy, εη)a(y, η)) dyđη
Using Dyj χ(εy, εη) = iε2 ηj χ(εy, εη), we obtain
Dyj (χ(εy, εη)a(y, η)) = χ(εy, εη)Dyj a(y, η) + iε2 χ(εy, εη)ηj a(y, η).
30
Therefore
ZZ
χ(εy, εη)e−iy·η yj a(y, η)dyđη =
ZZ
ZZ
−iy·η
χ(εy, εη)e
Dηj a(y, η)dyđη + iε
χ(εy, εη)e−iy·η ηj a(y, η)dyđη.
Passing to the limit ε → 0 yields the first equality.
THEOREM 4.13 (Fubini’s theorem for oscillatory integrals)
n+k
Let a ∈ Am
× Rn+k ), m, τ ∈ R, n, k ∈ N. Then
τ (R
ZZ
0 0
n
n
b(y, η) := Os–
e−iy ·η a(y, y 0 , η, η 0 )dy 0 đη 0 ∈ Am
τ (R × R ),
where integration is with respect to Rk × Rk , and
ZZ
0 0
α β
∂y ∂η b(y, η) = Os–
e−iy ·η ∂yα ∂ηβ a(y, y 0 , η, η 0 )dy 0 đη 0
(4.15)
Moreover,
ZZZZ
0
0
e−iy·η−iy ·η a(y, y 0 , η, η 0 )dydy 0 đηđη 0
ZZ
ZZ
−iy·η
−iy 0 ·η 0
0
0
0
0
= Os–
e
Os–
e
a(y, y , η, η )dy đη dyđη.
Os–
Proof: Because of Peetre’s inequality,
h(η, η 0 )im h(y, y 0 )iτ ≤ 2|m|+|τ | hηim hyi|τ | hη 0 i|m| hy 0 iτ .
|m|
Hence ∂yα ∂ηβ a(y, ., η, .) ∈ A|τ | (R2k ) with respect to (y 0 , η 0 ) and
m
τ
|∂yα ∂ηβ a(y, ., η, .)|A|m|
≤ Cj,k,m |a|Am
2(k+n) ),j+|α|+|β| hηi hyi .
2k
τ (R
τ (R ),j
Therefore (4.10) implies
ZZ
−iy 0 ·η 0 α β
0
0
0
0
m
τ
Os–
e
∂y ∂η a(y, y , η, η )dy đη ≤ Cm |a|Am
2(k+n) ),2(l+l0 )+|α|+|β| hηi hyi ,
τ (R
where 2l > |m| + k and 2l0 > k + τ . Moreover, we choose l, l0 so large that 2l > m + n
and 2l0 > n + τ . Because of the representation (4.9) and Theorem 2.4, we can take
∂yα ∂ηβ out of the oscillatory integral, which proves (4.15). Thus
α β
∂y ∂η b(y, η) ≤ Cm |a|Am (R2(n+k) ),2(l+l0 )+|α|+|β| hηim hyiτ .
τ
31
0 n
n
−2l0
Hence b(y, η) ∈ Am
hDη i2l hηi−2l hDy i2l b(y, η) ∈ L1 (Rn ×Rn )
τ (R ×R ). Thus hyi
because of Theorem 4.9. Moreover,
ZZ
0 0
0
0 b(y, η) =
e−iy ·η hy 0 i−2l hDη0 i2l hη 0 i−2l hDy0 i2l a(y, y 0 , η, η 0 ) dy 0 đη 0
and therefore
0
0 hyi−2l hDη i2l hηi−2l hDy i2l b(y, η)
ZZ
0 0
0
0 =
e−iy ·η (hyihy 0 i)−2l (hDη ihDη0 i)2l (hηihη 0 i)−2l (hDy ihDy0 i)2l a dy 0 đη 0 ,
where
0
0 (hyihy 0 i)−2l (hDη ihDη0 i)2l (hηihη 0 i)−2l (hDy ihDy0 i)2l a(y, y 0 , η, η 0 ) ∈ L1 (R2(n+k) ).
Hence we can apply Fubini’s theorem and get
ZZ
ZZ
−iy·η
−iy 0 ·η 0
0
0
0
0
Os–
e
Os–
e
a(y, y , η, η )dy đη dyđη
Z
d(y 0 y, η, η 0 )
0 0
0
0 =
e−iy·η−iy ·η (hyihy 0 i)−2l (hDη ihDη0 i)2l (hηihη 0 i)−2l (hDy ihDy0 i)2l a
(2π)n+k
ZZZZ
0 0
= Os–
e−iy·η−iy ·η (hηihη 0 i)−2l (hDy ihDy0 i)2l a(y, y 0 , η, η 0 )dy 0 đη 0 dyđη
ZZZZ
0 0
= Os–
e−iy·η−iy ·η a(y, y 0 , η, η 0 )dy 0 đη 0 dyđη,
where we have used Lemma 4.12.
4.4
Double Symbols
The composition p1 (x, Dx )p2 (x, Dx ) calculated in Section 4.2 is an example of a
pseudodifferential operator in more general form – a pseudodifferential operator with
a double symbol :
ZZZZ
0
0
00
0
p(x, Dx , x, Dx )u =
ei(x−x )·ξ+i(x −x )·ξ p(x, ξ, x0 , ξ 0 )u(x00 )dx00 đξ 0 dx0 đξ
for u ∈ S(Rn ), where the integral has to be understood as iterated integral. See
[KG74, Chapter 2, §2] for a more detailed discussion. Here p(x, ξ, x0 , ξ 0 ) = p1 (x, ξ)p2 (x0 , ξ 0 ) ∈
m1 ,m2
S1,0
(Rn × Rn × Rn × Rn ) defined as follows:
Definition
4.14 Let m, m0 ∈ R. Then the space of double pseudodifferential symm,m0
bols S1,0 (Rn × Rn × Rn × Rn ) is the space of all smooth functions p : Rn × Rn ×
Rn × Rn → C such that
0
0
0
0
|Dξα Dxβ Dξα0 Dxβ0 p(x, ξ, x0 , ξ 0 )| ≤ Cα,β,α0 ,β 0 (1 + |ξ|)m−|α| (1 + |ξ 0 |)m −|α |
uniformly in x, ξ, x0 , ξ 0 ∈ Rn for arbitrary α, β, α0 , β 0 ∈ Nn0 .
32
max(m,m0 ,m+m0 )
0
m,m
(Rn × Rn × Rn × Rn ) ⊂ A0
(R2n × R2n ).
Note that S1,0
The statements on composition of two pseudodifferential operators will be a corollary to the following more general theorem:
0
m,m
THEOREM 4.15 Let m, m0 ∈ R and let p ∈ S1,0
(Rn × Rn × Rn × Rn ) be a double
symbol. Then
ZZ
m+m0
(Rn × Rn ).
pL (x, ξ) := Os–
e−iy·η p(x, ξ + η, x + y, ξ)dyđη ∈ S1,0
Moreover,
pL (x, ξ) ∼
X 1
∂ξα Dxα0 p(x, ξ, x0 , ξ 0 )|x0 =x,ξ0 =ξ
α!
α∈Nn
0
in the sense that
pL (x, ξ) −
X 1
m+m0 −N −1
∂ξα ∂ξα Dxα0 p(x, ξ, x0 , ξ 0 )|x0 =x,ξ0 =ξ ∈ S1,0
(Rn × Rn )
α!
|α|≤N
for all N ∈ N0 .
Proof: First of all, let ax,ξ (y, η) := p(x, ξ + η, x + y, η). Using Peetre’s inequality,
|∂ηα ∂yβ ax,ξ (y, η)| = |∂ηα ∂yβ p(x, ξ + η, x + y, ξ)|
0
0
≤ Cα,β hξ + ηim−|α| hξim ≤ Cα,β hξ + ηim hξim
0
≤ Cα,β 2|m| hηi|m| hξim+m .
|m|
Hence ax,ξ (y, η) ∈ A0
0
with |ax,ξ |A|m| ,|m|+2n+2 ≤ C(1 + |ξ|)m+m . Therefore
0
ZZ
0
−iy·η
|pL (x, ξ)| = Os–
e
p(x, ξ + η, x + y, ξ)dyđη ≤ C(1 + |ξ|)m+m
(4.16)
because (4.10).
2n
Since p(x, ξ, x0 , η) ∈ Am̃
× R2n ), m̃ = max(m1 , m2 , m1 + m2 ), with respect
0 (R
2n
to (x, x0 ), (ξ, η), we also have p(x, ξ + η, x + y, η) ∈ Am̃
× R2n ) with respect to
0 (R
(x, y), (ξ, η). Hence we can apply (4.15) to conclude that
ZZ
α β
∂ξ ∂x pL (x, ξ) = Os–
e−iy·η ∂ξα ∂xβ [p(x, ξ + η, x + y, ξ)] dyđη.
(4.17)
Combining (4.16) and (4.17) yields
|∂ξα ∂xβ pL (x, ξ)|
ZZ
0
= Os–
e−iy·η ∂ξα ∂xβ [p(x, ξ + η, x + y, ξ)] dyđη ≤ C(1 + |ξ|)m+m −|α|(4.18)
33
In order to prove the asymptotic expansion, we use the Taylor series expansion:
X ηα
pα (x, ξ, x + y, ξ)
p(x, ξ + η, x + y, ξ) =
α!
|α|≤N
X ηα Z 1
+ (N + 1)
(1 − θ)N pα (x, ξ + θη, x + y, ξ)dθ,
α! 0
|α|=N +1
where pα (x, ξ, y, η) = ∂ξα p(x, ξ, y, η). Hence
ZZ
X 1
pL (x, ξ) =
Os–
e−iy·η η α pα (x, ξ, x + y, ξ)dyđη
α!
|α|≤N
ZZ
X 1
+(N + 1)
Os–
e−iy·η η α rα (x, ξ, y, η)dyđη,
α!
|α|=N +1
where
Z
rα (x, ξ, y, η) =
1
(∂ξα p)(x, ξ + θη, x + y, ξ)(1 − θ)N dθ.
0
Because of Lemma 4.12 and Example 4.11,
ZZ
ZZ
−iy·η α
Os–
e
η pα (x, ξ, x + y, ξ)dyđη = Os–
e−iy·η Dyα pα (x, ξ, x + y, ξ)dyđη
= ∂ξα Dyα p(x, ξ, y, η)|y=x,η=ξ .
Therefore it remains to estimate rα (x, ξ, y, η). As in the beginning of the proof,
β γ
∂η ∂y (∂ξα Dyα p)(x, ξ + θη, x + y, ξ) ≤ Cα,β,γ 2|m| (1 + |θη|)|m| (1 + |ξ|)m+m0 −|α|
0
≤ Cα,β,γ 2|m| (1 + |η|)|m| (1 + |ξ|)m+m −|α| ,
where Cα,β,γ does not depend on θ ∈ [0, 1]. Hence {p(x, ξ + θ., x + ., ξ)}0≤θ≤1 is
|m|
|m|
uniformly bounded in A0 as amplitudes in (y, η). Therefore rα (x, ξ, ., .) ∈ A0 and
β γ
∂η ∂y (Dyα rα )(x, ξ, η, y) ≤ Cα,β,γ 2|m| (1 + |η|)|m| (1 + |ξ|)m+m0 −|α| ,
This implies
ZZ
−iy·η
α
Os–
e
η
r
(x,
ξ,
η,
y)dyđη
α
ZZ
0
−iy·η
α
= Os–
e
Dy rα (x, ξ, η, y)dyđη ≤ Cα (1 + |ξ|)m+m −|α|
(4.19)
because of (4.10). Finally, the derivatives ∂ξβ ∂xγ rα,θ (x, ξ) are estimated in the same
way as before.
34
4.5
Composition
With the aid of the oscillatory integrals, we can make the formal calculations in
Section 4.2 rigorous.
m
(Rn × Rn ), then
First of all, if p ∈ S1,0
Z Z
i(x−y)·ξ
p(x, Dx )u =
e
p(x, ξ)u(y)dy đξ
ZZ
0
= Os–
e−ix ·ξ p(x, ξ)u(x + x0 )dx0 đξ
(4.20)
for all u ∈ S(Rn ). The proof is left to the reader as an exercise.
Exercise 7 Prove (4.20).
Using this representation and Theorem 4.13, we easily get
p1 (x, Dx )p2 (x, Dx )u
ZZ
ZZ
−ix0 ·ξ
−ix00 ·ξ 0
0 0
0
00
00
0
= Os–
e
p1 (x, ξ) Os–
e
p2 (x + x , ξ )u(x + x + x )dx đξ dx0 đξ
ZZZZ
0
00 0
= Os–
e−ix ·ξ−ix ·ξ p1 (x, ξ)p2 (x + x0 , ξ 0 )u(x + x0 + x00 )dx00 đξ 0 dx0 đξ
ZZZZ
0
0
= Os–
e−ix ·η−iy·ξ p1 (x, ξ 0 + η)p2 (x + x0 , ξ 0 )u(x + y)dx0 đξ 0 dyđη
ZZ
ZZ
−iy·ξ 0
−ix0 ·η
0
0 0
0
= Os–
e
Os–
e
p1 (x, ξ + η)p2 (x + x , ξ )dx đη u(x + y)dyđξ 0
ZZ
0
= Os–
e−iy·ξ p1 #p2 (x, ξ 0 )u(x + y)dyđξ
where we have used η = ξ − ξ 0 and y = x0 + x00 and p1 #p2 is defined as in (4.8).
m
THEOREM 4.16 Let pj ∈ S1,0j (Rn × Rn ), j = 1, 2, be two pseudodifferential symm1 +m2
bols. Then there is some p1 #p2 ∈ S1,0
(Rn × Rn ) such that
p1 (x, Dx )p2 (x, Dx ) = (p1 #p2 )(x, Dx ).
Moreover, p1 #p2 has the following asymptotic expansion:
X 1
∂ξα p1 (x, ξ)Dxα p2 (x, ξ)
p1 #p2 (x, ξ) ∼
α!
α∈Nn
0
in the sense that
p1 #p2 (x, ξ) −
X 1
m1 +m2 −N
∂ α p1 (x, ξ)Dxα p2 (x, ξ) ∈ S1,0
(Rn × Rn )
α! ξ
|α|<N
for all N ∈ N.
35
m1 ,m2
(Rn × Rn × Rn × Rn )
Proof: Let p(x, ξ, x0 , ξ 0 ) = p1 (x, ξ)p2 (x0 , ξ 0 ). Then p ∈ S1,0
and p1 #p2 (x, ξ) = pL (x, ξ) and the theorem is a consequence of Theorem 4.15.
Hence the composition of two pseudodifferential operators is again a pseudodifferential operator. In particular, the asymptotic expansions implies:
p1 #p2 (x, ξ) = p1 (x, ξ)p2 (x, ξ) + r(x, ξ),
m1 +m2 −1
(Rn × Rn ). Hence the symbol of the composition coincides
where r(x, ξ) ∈ S1,0
with the product of the symbols modulo terms of lower order. This is the essential fact
needed for the construction of a parametrix to an elliptic pseudodifferential operator.
Finally, we note that, if p2 (x, ξ) = p2 (ξ) is independent of x, we simply have
p1 (x, Dx )p2 (Dx ) = OP(p1 (x, ξ)p2 (ξ)) since F[p2 (Dx )u] = p2 (ξ)û(ξ). Moreover, if
p1 (x, Dx ) is a differential operator of order m ∈ N0 with coefficients in Cb∞ (Rn ), then
p1 (x, Dx )p2 (x, Dx ) = (p1 #p2 )(x, Dx )
where
(p1 #p2 )(x, ξ) =
X 1
∂ α p1 (x, ξ)Dxα p2 (x, ξ).
α! ξ
|α|≤m
Hence the asymptotic expansion of p1 #p2 (x, ξ) consists only of finitely many terms.
This is an easy consequence of Leibniz’s formula. (See exercises.) Alternatively, it
can be observed by modifying the proof of Theorem 4.15. (In this case there is no
remainder term rα,θ if N > m.)
4.6
Application: Elliptic Pseudodifferential Operators and Parametrices
m
Definition 4.17 A symbol p ∈ S1,0
(Rn × Rn ), m ∈ R, is called elliptic if there is an
R > 0 such that
|p(x, ξ)| ≥ C|ξ|m
for all |ξ| ≥ R, x ∈ Rn
(4.21)
Examples 4.18
1. Let p(ξ) = |ξ|2 be the symbol of −∆. Then p is an elliptic
symbol of order 2. Moreover, q(ξ) = hξim , m ∈ R, is an elliptic symbol of order
m.
2. Let A(x) ∈ Cb∞ (Rn )n×n be a matrix that is uniformly positive definite, i.e.,
there is a c > 0 such that
ξ T A(x)ξ ≥ c|ξ|2 ,
for all x, ξ ∈ Rn .
Then p(x, ξ) = ξ T A(x)ξ is an elliptic symbol of order 2.
36
m
Lemma 4.19 Let p ∈ S1,0
(Rn × Rn ), m ∈ R, be an elliptic symbol and R > 0 as in
(4.21). Then
−m
(Rn × Rn )
q(x, ξ) := ψ(ξ)p(x, ξ)−1 ∈ S1,0
where ψ ∈ Cb∞ (Rn ) such that ψ(ξ) = 1 for |ξ| ≥ R + 1 and ψ(ξ) = 0 for |ξ| ≤ R.
Proof: Since q(x, ξ) = ψ(ξ) = 0 for |ξ| ≤ R, q(x, ξ) is obviously smooth in (x, ξ)
and it suffices to consider |ξ| ≥ R. Because of the chain rule,
∂ξj p(x, ξ)−1 = −p(x, ξ)−2 ∂ξj p(x, ξ)
and the same identity with ∂ξj replaced by ∂xj . Using (4.21),
|∂ξj p(x, ξ)−1 | ≤ C|ξ|−2m hξim−1 ≤ Chξi−m−1
and
|∂xj p(x, ξ)−1 | ≤ C|ξ|−2m hξim ≤ Chξi−m
since |ξ| ≥ R > 0. In the same way one can easily prove by induction that
|∂ξα ∂xβ p(x, ξ)−1 | ≤ Cα,β hξi−m−|α|
for all |ξ| ≥ R. On the otherhand q : Rn × Rn → C is smooth and therefore
0
|∂ξα ∂xβ q(x, ξ)| ≤ Cα,β
≤ Cα,β hξi−m−|α|
for all |ξ| ≤ R + 1. Since q(x, ξ) = p(x, ξ)−1 , we conclude that for every α, β ∈ Nn0
there is some Cα,β > 0 such that
|∂ξα ∂xβ q(x, ξ)| ≤ Cα,β hξi−m−|α|
for all ξ ∈ Rn .
m
(Rn × Rn ) be an elliptic symbol. Then there is an
Corollary 4.20 Let p ∈ S1,0
−m
n
n
q ∈ S1,0 (R × R ) such that
q(x, Dx )p(x, Dx ) = I + r0 (x, Dx )
p(x, Dx )q(x, Dx ) = I + r(x, Dx ),
−1
with r, r0 ∈ S1,0
(Rn × Rn ).
Proof: Let q be defined as in Lemma 4.19. Then, because of Theorem 4.16,
p(x, Dx )q(x, Dx ) = (pq)(x, Dx ) + r̃(x, Dx ),
−1
where r̃ ∈ S1,0
(Rn × Rn ). Moreover, p(x, ξ)q(x, ξ) = 1 for all |ξ| ≥ R + 1. Hence
−∞
p(x, ξ)q(x, ξ) − 1 ∈ S1,0
(Rn × Rn ) and
p(x, Dx )q(x, Dx ) = I + r(x, Dx )
−1
with r(x, ξ) = p(x, ξ)q(x, ξ) − 1 + r̃(x, ξ) ∈ S1,0
(Rn × Rn ). The identity
q(x, Dx )p(x, Dx ) = I + r0 (x, Dx )
−1
with r0 ∈ S1,0
(Rn × Rn ) is proved the some way.
37
m
THEOREM 4.21 Let p ∈ S1,0
(Rn × Rn ), m ∈ R. Then the following conditions
are equivalent:
1. p is elliptic.
−m
2. There is a q ∈ S1,0
(Rn × Rn ) such that p(x, Dx )q(x, Dx ) = I + r(x, Dx ), where
−1
r ∈ S1,0 (Rn × Rn ).
−m
3. For every N ∈ N there is a qN ∈ S1,0
(Rn × Rn ) such that p(x, Dx )qN (x, Dx ) =
−N
n
I + rN (x, Dx ), where rN ∈ S1,0 (R × Rn ).
Proof: 1. implies 2.: This is a consequence of Corollary 4.20.
−m
2. implies 3.:
By the assumptions there are some q ∈ S1,0
(Rn × Rn ), r ∈
−1
S1,0
(Rn × Rn ) such that
p(x, Dx )q(x, Dx ) = I − r(x, Dx ).
−m
Now let qN ∈ S1,0
(Rn × Rn ) be such that
qN (x, Dx ) = q(x, Dx )
N
−1
X
r(x, Dx )k .
k=0
Then
p(x, Dx )qN (x, Dx ) = (I − r(x, Dx ))
N
−1
X
(−r(x, Dx ))k = I + (−r(x, Dx ))N ,
k=0
−N
where (−r(x, Dx ))N = rN (x, Dx ) with rN ∈ S1,0
(Rn × Rn ) due to Theorem 4.16.
3. implies 1.: Since p(x, Dx )q(x, Dx ) = OP (p(x, ξ)q(x, ξ)) + r(x, Dx ) and
p(x, Dx )q(x, Dx ) = I + r0 (x, Dx )
−1
with r, r0 ∈ S1,0
(Rn × Rn ), we obtain
−1
p(x, ξ)q(x, ξ) − 1 = −r(x, ξ) + r0 (x, ξ) ∈ S1,0
(Rn × Rn ),
where we have used that the pseudodifferential operator determines its symbol uniquely,
which is proved in Corolloray 4.24 below. In particular
|p(x, ξ)q(x, ξ) − 1| ≤ Chξi−1
for all x, ξ ∈ Rn . Hence there is some R > 0 such that |p(x, ξ)q(x, ξ) − 1| ≤ 21 for
all |ξ| ≥ R and x ∈ Rn . Thus (p(x, ξ)q(x, ξ))−1 ≤ 2 for all |ξ| ≥ R and x ∈ Rn .
Therefore we finally conclude that
|p−1 (x, ξ)| = |(p(x, ξ)q(x, ξ))−1 ||q(x, ξ)| ≤ 2Chξi−m
38
−m
for all x ∈ Rn , |ξ| ≥ R since q ∈ S1,0
(Rn × Rn ).
For completeness we note that for an elliptic symbol p there is also an q∞ ∈
× Rn ) such that
−m
S1,0
(Rn
p(x, Dx )q∞ (x, Dx ) = I + r∞ (x, Dx )
(4.22)
−∞
with r ∈ S1,0
(Rn × Rn ). In order to prove this statement, the following lemma is
needed:
m
Lemma 4.22 Let pj ∈ S1,0j (Rn × Rn ) with m1 ≥ . . . ≥ mj → −∞ as j → ∞. Then
m1
there is some p ∈ S1,0
(Rn × Rn ) such that
p(x, ξ) ∼
∞
X
pj (x, ξ) :⇔ p(x, ξ) −
j=1
N
−1
X
mN
(Rn × Rn ).
pj (x, ξ) ∈ S1,0
j=1
Proof: See [KG74, Lemma 3.2] or [Ray91, Lemma 2.2].
In order to prove (4.22), we can define
q∞ (x, Dx ) = q(x, Dx )q 0 (x, Dx ),
where
0
q (x, ξ) ∼
∞
X
r#k (x, ξ),
k=0
r is the same as in the proof of Theorem 4.21, and
−k
r#k (x, ξ) = r# . . . #r(x, ξ) ∈ S1,0
(Rn × Rn ).
|
{z
}
k-times
4.7
Boundedness on Cb∞ (Rn ) and Uniqueness of the Symbol
As seen above,
ZZ
p(x, Dx )u = Os–
0
e−ix ·ξ p(x, ξ)u(x + x0 )dx0 đξ
(4.23)
for all u ∈ S(Rn ). But the latter oscillatory integral is defined for all u ∈ Cb∞ (Rn ).
Therefore we can extend the definition of p(x, Dx ) to Cb∞ (Rn ).
m
THEOREM 4.23 Let p ∈ S1,0
(Rn × Rn ). Then p(x, Dx ) defined by (4.23) for
∞
n
u ∈ Cb (R ) is a bounded operator
p(x, Dx ) : Cb∞ (Rn ) → Cb∞ (Rn ).
39
Proof: Consider ax (x0 , ξ) = p(x, ξ)u(x + x0 ). Since for every α, β ∈ Nn0
(m)
|∂xα0 ∂ξβ p(x, ξ)u(x + x0 )| ≤ Cα,β |p||β| kukC |α| hξim ,
b
(m)
we have ax (x0 , ξ) ∈ Am
≤ C|p|k kukCbk uniformly in x ∈ Rn . Thus
0 with |ax |Am
0 ,k
(m)
0
0
0 ≤ C |p|
sup |p(x, Dx )u| ≤ C|ax |Am
2(l+l0 ) kukC 2(l+l )
0 ,2(l+l )
x∈Rn
(4.24)
b
by Theorem 4.9, where 2l > m + n, 2l0 > n. In order to estimate ∂xα p(x, Dx )u we
2n
observe that a(x, x0 , ξ, ξ 0 ) := p(x, ξ)u(x + x0 ) ∈ Am
× R2n ) and therefore
0 (R
ZZ
0
∂xj p(x, Dx )u = Os–
e−ix ·ξ (∂xj p)(x, ξ)u(x + x0 )dx0 đξ
ZZ
0
+ Os–
e−ix ·ξ p(x, ξ)(∂xj u)(x + x0 )dx0 đξ
= (∂xj p)(x, Dx )u + p(x, Dx )∂xj u
by (4.15) in Theorem 4.13. Applying this formula successively and using (4.24), it
(m)
follows immediately that kp(x, Dx )ukCbk ≤ C|p|2(l+l0 )+k kukC 2(l+l0 )+k , which proves the
b
theorem.
Using this extension, we can prove that the symbol of a pseudodifferential operator
is uniquely defined by its values on Cb∞ (Rn ):
Exercise 8 Show that for all x, ξ ∈ Rn
e−ix·ξ p(x, Dx )ei.·ξ = p(x, ξ).
(4.25)
∞
Corollary 4.24 Let p, q ∈ S1,0
(Rn × Rn ). Then p(x, Dx )u = q(x, Dx )u for all u ∈
n
S(R ) implies p(x, ξ) = q(x, ξ).
Proof: Because of the latter exercise, we only need to prove
for all u ∈ Cb∞ (Rn ).
p(x, Dx )u = q(x, Dx )u
(4.26)
To this end let u ∈ Cb∞ (Rn ) and uε (x) := ψ(εx)u(x) for all x ∈ Rn , ε > 0 and some
ψ ∈ S(Rn ) with ψ(0) = 1. Then uε ∈ S(Rn ) and
∂xα uε (x) →ε→0 ∂xα u(x)
for every x ∈ Rn , α ∈ Nn0 .
Therefore Corollary 4.10 implies
ZZ
0 0
p(x, Dx )uε (x) = Os–
e−ix ·ξ p(x, ξ)uε (x + x0 ) dx0 đξ 0 →ε→0 p(x, Dx )u(x)
|
{z
}
=:aε,x (x0 ,ξ)
40
for every x ∈ Rn since
∂xα ∂ξβ aε,x (x0 , ξ) →ε→0 ∂xα ∂ξβ ax (x0 , ξ)
for all x0 , ξ ∈ Rn , α, β ∈ Nn0
n
n
with ax (x0 , ξ) = p(x, ξ)u(x + x0 ) and since (aε,x )0<ε≤1 is bounded in Am
0 (R × R ) for
n
∞
n
every x ∈ R , which follows from the boundedness of (uε )0<ε≤1 in Cb (R ). In the
same way, we obtain the same statement for q(x, Dx ). Therefore
p(x, Dx )u(x) = lim p(x, Dx )uε (x) = lim q(x, Dx )uε (x) = q(x, Dx )u(x)
ε→0
ε→0
for every x ∈ Rn since p(x, Dx )uε = q(x, Dx )uε due to uε ∈ S(Rn ). Since u ∈ Cb∞ (Rn )
was arbitrary, (4.26) follows and the corollary is proved.
More generally, if P : Cb∞ (Rn ) → Cb∞ (Rn ) is an arbitrary bounded linear map,
then we can define a symbol of P by
p(x, ξ) := e−ix·ξ P ei.·ξ
This gives an alternative approach for determining the symbol of p1 (x, Dx )p2 (x, Dx ):
Exercise 9 Calculate
e−ix·ξ p1 (x, Dx )p2 (x, Dx )ei.·ξ
without using Theorem 4.16.
4.8
Adjoints of Pseudodifferential Operators and Operators
in (x,y)-Form
Definition 4.25 Let A, A∗ : S(Rn ) → S(Rn ). Then A∗ is called formal adjoint of A
if
(Au, v)L2 (Rn ) = (u, A∗ v)L2 (Rn )
for all u, v ∈ S(Rn ).
The adjoint of an operator plays an important role in many purposes. (For example:
solvability of equations).
In the following, we will prove that every pseudodifferential operator p(x, Dx )
possesses a formal adjoint p∗ (x, Dx ), which is again a pseudodifferential operator (of
the same order). This will allows us to extend p(x, Dx ), up to now only defined on
S(Rn ), to the a linear operator defined on S 0 (Rn ).
Now we calculate the formal adjoint of p(x, Dx ):
ZZ
(p(x, Dx )u, v)L2 (Rn ) =
eix·ξ p(x, ξ)û(ξ)đξv(x)dx
ZZ
=
eix·ξ p(x, ξ)v(x)dxû(ξ)đξ
Z
Z
=
û(ξ) e−ix·ξ p(x, ξ)v(x)dxđξ,
41
where have used Fubini’s theorem. – Note that v, û ∈ S(Rn ) which implies that
eix·ξ p(x, ξ)û(ξ)v(x) ∈ L1 (Rn × Rn ) w.r.t. (x, ξ).
m
Lemma 4.26 Let p ∈ S1,0
(Rn × Rn ), m ∈ R. Then
Z
w(ξ) := e−ix·ξ p(x, ξ)v(x)dx ∈ S(Rn ).
Proof: Exercise. - It is also a consequence of Lemma 4.28 below.
Hence we can use (F[u], v) = (u, F −1 [v]) and get
ZZ
∗
(p(x, Dx ) v)(x) =
ei(x−y)·ξ p(y, ξ)v(y)dyđξ.
(4.27)
This operator is a pseudodifferential operator in the so called y-form (also called
R-form), which is a special case of operators of the form
ZZ
p(x, Dx , x)u :=
ei(x−y)·ξ p(x, y, ξ)u(y)dyđξ
(4.28)
m
for u ∈ S(Rn ), where p ∈ S1,0
(R2n × Rn ), i.e.,
|∂ξα ∂xβ ∂yγ p(x, y, ξ)| ≤ Cα,β,γ (1 + |ξ|)m−|α|
for all α, β, γ ∈ Nn0 . As before we obtain
ZZ
0
p(x, Dx , x)u = Os–
e−ix ·ξ p(x, x + x0 , ξ)u(x + x0 )dx0 đξ
for all u ∈ S(Rn ).
m
THEOREM 4.27 Let p ∈ S1,0
(R2n × Rn ), m ∈ R. Then p(x, Dx , x)u = pL (x, Dx )u
∞
n
for all u ∈ Cb (R ), where
ZZ
m
pL (x, ξ) = Os–
e−iy·η p(x, x + y, ξ + η)dyđη ∈ S1,0
(Rn × Rn ).
Moreover,
pL (x, ξ) ∼
X 1
∂ξα Dyα p(x, y, ξ)|y=x
α!
α∈Nn
0
in the sense that for all N ∈ N0
pL (x, ξ) −
X 1
m−N −1
∂ α Dα p(x, y, ξ)|y=x ∈ S1,0
(Rn × Rn ).
α! ξ y
|α|≤N
42
Proof: Using
0
u(x + x ) = Os–
ZZ
−iy·η
e
ZZ
0
u(x + x + y)dyđη = Os–
0
ei(x+x −y)·η u(y)dyđη
due to Example 4.11, we get
ZZZZ
0
0
p(x, Dx , x)u = Os–
e−ix ·ξ ei(x+x −y)·η p(x, x + x0 , ξ)u(y)dyđηdx0 đξ
ZZ
ZZ
−ix0 ·(ξ−η)
0
0
−i(x−y)·η
Os–
e
p(x, x + x , ξ)dx đξ u(y)dyđη
= Os–
e
ZZ
= Os–
e−i(x−y)·η pL (x, ξ)u(y)dyđη.
Hence application of Theorem 4.15 finishes the proof.
As a direct consequence we obtain:
m
Lemma 4.28 Let p ∈ S1,0
(R2n × Rn ), m ∈ R. Then p(x, Dx , x) : S(Rn ) → S(Rn ) is
a bounded operator. Moreover, if the definition (4.28) is replaced by
ZZ
0
p(x, Dx , x)u := Os–
e−ix ·ξ p(x, x + x0 , ξ)u(x + x0 )dx0 đξ
for u ∈ Cb∞ (Rn ), then p(x, Dx , x) : Cb∞ (Rn ) → Cb∞ (Rn ) is a bounded operator.
Proof: Since p(x, Dx , x)u = pL (x, Dx )u for all u ∈ Cb∞ (Rn ), the lemma is a consequence of the corresponding statements for pL (x, Dx ).
m
(Rn ×Rn ), then the formal adjoint of p(x, Dx ) is p∗ (x, Dx )
Corollary 4.29 If p ∈ S1,0
where
ZZ
∗
m
p (x, ξ) = Os–
e−iy·ξ p(x + y, ξ + η)dyđη ∈ S1,0
(Rn × Rn ).
Moreover,
p∗ (x, ξ) ∼
X 1
∂ξα Dxα p(x, ξ)
α!
α∈Nn
0
in the sense that for every N ∈ N0
X 1
m−N −1
p∗ (x, ξ) −
∂ α Dα p(x, ξ) ∈ S1,0
(Rn × Rn ).
α! ξ x
|α|≤N
Proof: The corollary is a direct consequence of (4.27) and the latter theorem.
m
Definition 4.30 Let p ∈ S1,0
(Rn × Rn ). Then we define p(x, Dx ) : S 0 (Rn ) → S 0 (Rn )
by
hp(x, Dx )u, vi := hu, p∗ (x, Dx )vi
u ∈ S 0 (Rn ), v ∈ S(Rn ).
43
Since S(Rn ) ⊆ S 0 (Rn ) (identifying functions with functional in the way described in
Section 3.5), it is important to notice that, if u ∈ S(Rn ),
Z
∗
∗
hp(x, Dx )u, vi = hu, p (x, Dx )vi = (u, p (x, Dx )v)L2 (Rn ) =
p(x, Dx )u(x)v(x) dx
Rn
for all v ∈ S(Rn ). I.e., the definition of p(x, Dx )u in sense of S 0 (Rn ) coincides with
the starting definition of p(x, Dx )u for u ∈ S(Rn ).
L2 -Continuity and Bessel Potential Spaces
4.9
0
THEOREM 4.31 Let p ∈ S1,0
(Rn ×Rn ). Then p(x, Dx ) (defined on S(Rn )) extends
to a bounded operator p(x, Dx ) : L2 (Rn ) → L2 (Rn ).
The proof of Theorem 4.31 is divided into several parts.
−n−1
Lemma 4.32 Theorem 4.31 holds for p ∈ S1,0
(Rn × Rn ).
−n−1
Proof: If p ∈ S1,0
(Rn × Rn ), then p(x, ξ) ∈ L1 (Rn ) with respect to ξ and therefore
Z
ZZ
ix·ξ
ˆ
p(x, Dx )f =
e p(x, ξ)f (ξ)đξ =
ei(x−y)·ξ p(x, ξ)đξf (y)dy
Z
=
k(x, x − y)f (y)dy
for all f ∈ S(Rn ) by Fubini’s theorem, where k(x, z) = Fξ7−1
→z [p(x, ξ)]. This kernel
satisfies
Z
(−n−1)
α
iz·ξ α
|z k(x, z)| = e ∂ξ p(x, ξ)đξ ≤ Cα |p||α|
(−n−1)
since |∂ξα p(x, ξ)| ≤ |p||α|
(−n−1)
C|p|2n
hξi−n−1−|α| ∈ L1 (Rn ) w.r.t. ξ. Hence |(1 + |z|2 )n k(x, z)| ≤
and therefore
g(z) := sup |k(x, z)| ≤ (1 + |z|2 )−n ∈ L1 (Rn ).
x∈Rn
Thus
Z
kp(x, Dx )f k2 ≤ C 0
g(x − y)|f (y)|dy ≤ C kgk1 kf k2
n
2
R
because of kg ∗ f k2 ≤ kgk1 kf k2 , cf. Example 3.18.1.
−m
Lemma 4.33 Theorem 4.31 holds for p ∈ S1,0
(Rn × Rn ) with m > 0.
44
Proof: In order to prove kp(x, Dx )f k2 ≤ Ckf k2 for f ∈ S(Rn ), it is sufficient to
show
kp∗ (x, Dx )p(x, Dx )f k2 ≤ Ckf k2
since
kp(x, Dx )f k22 = (p∗ (x, Dx )p(x, Dx )f, f ) ≤ kp∗ (x, Dx )p(x, Dx )f k2 kf k2 .
−m
−2m
But, if p ∈ S1,0
(Rn × Rn ), then p∗ (x, Dx )p(x, Dx ) = p0 (x, Dx ) with p0 ∈ S1,0
(Rn ×
Rn ). Hence, using the previous lemma, we obtain by mathematical inductions that
kp(x, Dx )f k2 ≤ Ckf k2
−mk
with mk = (n + 1)/2k , k ∈ N. This proves the lemma since for every
for all p ∈ S1,0
m > 0 there is a k ∈ N such that −mk > −m.
In order to finish the proof of Theorem 4.31, we need
0
(Rn × Rn ) with p(x, ξ) ∈ R, x, ξ ∈ Rn , and F ∈ C ∞ (R),
Lemma 4.34 If p ∈ S1,0
n
0
then F (p(x, ξ)) ∈ S1,0 (R × Rn ).
Proof: First of all |p(x, ξ)| ≤ R for some R > 0 and F is bounded on compact sets.
Hence |F (p(x, ξ))| ≤ sup|z|≤R |F (z)|. Moreover,
|∂ξj F (p(x, ξ))| ≤
sup |F 0 (z)||∂ξj p(x, ξ)| ≤ Chξi−1 ,
|z|≤R
|∂xj F (p(x, ξ))| ≤
sup |F 0 (z)||∂xj p(x, ξ)| ≤ C.
|z|≤R
Finally, the estimate of |∂ξα ∂xβ F (p(x, ξ))| can be proved using an induction and the
chain rule as before.
0
(Rn × Rn ), then |p(x, ξ)| ≤ M for M :=
Proof of Theorem 4.31: If p(x, ξ) ∈ S1,0
(0)
0
(Rn × Rn ) and p0 (x, ξ) ≥ 0.
|p|0 ≥ 0. Therefore p0 (x, ξ) := M 2 − p(x, ξ)p(x, ξ) ∈ S1,0
1
Now let F ∈ C ∞ (R) be defined by F (x) = (1 + x) 2 for x ≥ 0. Then q(x, ξ) :=
0
F (p0 (x, ξ)) ∈ S1,0
(Rn × Rn ) and
q ∗ (x, Dx )q(x, Dx )f = OP(F (p0 (x, ξ))2 )f + r(x, Dx )f
= (1 + M 2 )f − OP(p(x, ξ)p(x, ξ))f + r(x, Dx )f
= (1 + M 2 )f − p∗ (x, Dx )p(x, Dx )f + r0 (x, Dx )f
−1
for f ∈ S(Rn ), where r, r0 ∈ S1,0
(Rn × Rn ) because of Theorem 4.16 and Corollary 4.29. Hence
kp(x, Dx )f k2L2 (Rn )
≤ (p∗ (x, Dx )p(x, Dx )f, f )L2 (Rn ) + (q ∗ (x, Dx )q(x, Dx )f, f )L2 (Rn )
≤ (1 + M 2 )kf k2L2 (Rn ) + (r0 (x, Dx )f, f ).
45
−1
Since r0 ∈ S1,0
(Rn × Rn ), kr0 (x, Dx )f k2 ≤ Ckf k2 due to Lemma 4.33. Hence
kp(x, Dx )f k22 ≤ (1 + M 2 + C)kf k22 for all f ∈ S(Rn ).
Remark 4.35 Checking the previous proofs, it can be observed that the estimates
(0)
of kp(x, Dx )kL(L2 (Rn )) depend only on |p|k for some suitably large k ∈ N (and not p
directly). Since the mapping p 7→ p(x, Dx ) is linear, we have
(0)
kp(x, Dx )kL(L2 (Rn )) ≤ C|p|k ;
(0)
(4.29)
(0)
this can seen as follows: Let q(x, ξ) = p(x, ξ)/|p|k . Then |q|k = 1 and
kq(x, Dx )kL(L2 (Rn )) ≤ C
(0)
0
(Rn × Rn ). Hence multiplication by |p|k yields (4.29).
independent of p ∈ S1,0
Definition 4.36 Let s ∈ R. Then the Bessel potential space H2s (Rn ) is defined as
H2s (Rn ) = {u ∈ S 0 (Rn ) : hDx is u ∈ L2 (Rn )},
kuks,2 := khDx is uk2 .
Here a tempered distribution u ∈ S 0 (Rn ) is said to be in L2 (Rn ) if there is a
U ∈ L2 (Rn ) such that
Z
U (x)v(x)dx
for all v ∈ S(Rn )
hu, vi =
Rn
(where h., .i denotes the duality product and (., .)L2 denotes the L2 -scalar product).
For simplicity we will identify U and u.
Note: By Riesz’ representation theorem and since S(Rn ) is dense in L2 (Rn ), u ∈
0
S (Rn ) is in L2 (Rn ) if and only if there is a constant C > 0 such that |hu, vi| ≤ Ckvk2
for all v ∈ S(Rn ).
Remark 4.37 By definition hDx is is an isomorphism from H2s (Rn ) to L2 (Rn ). Hence
H2s (Rn ) normed by k.k2,s is Banach space. Moreover,
(u, v)H2s := (hDx is u, hDx is v)L2 ,
u, v ∈ H2s (Rn )
is scalar product on H2s (Rn ) and kuk22,s = (u, u)H2s . Thus H2s (Rn ) is a Hilbert space.
Since S(Rn ) is dense in L2 (Rn ) and hDx is : S(Rn ) → S(Rn ) for all s ∈ R, S(Rn )
is dense in H2s (Rn ) for all s ∈ R.
Finally, H2s1 (Rn ) ⊆ H2s2 (Rn ) for s1 ≥ s2 since
kuks2 ,2 = khDx is2 uk2 = khDx is2 −s1 hDx is1 uk2 ≤ CkhDx is1 uk2 = Ckuks1 ,2 ,
0
(Rn × Rn ) and therefore hDx is2 −s1 : L2 (Rn ) →
where we have used that hξis2 −s1 ∈ S1,0
L2 (Rn ) is a bounded operator. – The parameter s ∈ R of H2s (Rn ) determines the
regularity of the functions u ∈ H2s (Rn ) (“how many derivatives of u are in L2 (Rn )”),
cf. Lemma 4.42, below.
46
m
Lemma 4.38 Let p ∈ S1,0
(Rn × Rn ). Then p(x, Dx ) : H2s+m (Rn ) → H2s (Rn ). More(m)
over, there is some k ∈ N0 such that kp(x, Dx )kL(H2s+m (Rn ),H2s (Rn )) ≤ Cs,m |p|k .
Proof: First of all, we note that, if u ∈ L2 (Rn ) ⊂ S 0 (Rn ), p(x, Dx )u is defined in
the sense of S 0 (Rn ), cf. Definition 4.30, and m = 0, then
|hp(x, Dx )u, vi| = (u, p∗ (x, Dx )v)L2 (Rn ) ≤ kukL2 (Rn ) kp∗ (x, Dx )kL2 (Rn ) kvkL2 (Rn )
for all v ∈ S(Rn ) due to Theorem 4.31 and Corollary 4.29. Hence p(x, Dx )u ∈
L2 (Rn ) ∼
= L2 (Rn )0 .
Now we consider a general s, m ∈ R. Since hDx is+m : H2s+m (Rn ) → L2 (Rn ) and
hDx i−s : L2 (Rn ) → H2s (Rn ) are linear isomorphisms, p(x, Dx ) : H2s+m (Rn ) → H2s (Rn )
is a linear bounded operator if and only if
q(x, Dx ) := hDx is p(x, Dx )hDx i−s−m : L2 (Rn ) → L2 (Rn )
0
is a bounded operator. Because of Theorem 4.16, q ∈ S1,0
(Rn × Rn ). Hence the first
statement is a consequence of Theorem 4.31. The second statement is a consequence
of Remark 4.35 and the fact that the mapping
m1
m2
m1 +m2
S1,0
× S1,0
3 (p1 , p2 ) 7→ p1 #p2 ∈ S1,0
is bounded/continuous applied to p1 (x, ξ) = hξis , p2 (x, ξ) = p(x, ξ)hξi−s−m .
m
(RnS× Rn ) be an elliptic symbol
Corollary 4.39 (Elliptic regularity) Let p ∈ S1,0
and f ∈ H2s (Rn ), s, m ∈ R. Then, if u ∈ H2−∞ (Rn ) := s∈R H2s (Rn ) is a solution of
the pseudodifferential equation
p(x, Dx )u = f,
then u ∈ H2s+m (Rn ).
Proof: Let u ∈ H2−∞ (Rn ). Then u ∈ H2s+m−N (Rn ) for some N ∈ N. By the same
−m
construction as in Theorem 4.21, there is an qN (x, ξ) ∈ S1,0
(Rn × Rn ) such that
!
N
−1
X
k
qN (x, Dx ) =
(−r(x, Dx )) q(x, Dx ),
k=0
where q is the same as in Corollary 4.20. (I.e., p(x, Dx )q(x, Dx ) = I + r(x, Dx ) and
−1
q(x, Dx )p(x, Dx ) = I +r0 (x, Dx ) where r, r0 ∈ S1,0
(Rn ×Rn ). Then as in Theorem 4.21
qN (x, Dx )p(x, Dx ) = I + rN (x, Dx ),
−N
where rN ∈ S1,0
(Rn × Rn ). Therefore
qN (x, Dx )f = qN (x, Dx )p(x, Dx )u = u + rN (x, Dx )u.
Since qN (x, Dx )f ∈ H2s+m (Rn ) and rN (x, Dx )u ∈ H2s+m (Rn ) by Lemma 4.38, we have
u ∈ H s+m (Rn ).
47
Remark 4.40 The latter corollary says that the solution u is “as smooth as the
right-hand side allows”. This property is also called elliptic regularity.
Definition 4.41 Let m ∈ N0 . Then the Sobolev space W2m (Rn ) of order m is the
space of all u ∈ L2 (Rn ) such that the distributional derivative ∂ α u ∈ L2 (Rn ) for all
α ∈ Nn0 with |α| ≤ m. Moreover, W2m (Rn ) is normed by
 21

X
kukW2m := 
k∂xα uk22  .
|α|≤m
Lemma 4.42 Let m ∈ N0 . Then W2m (Rn ) = H2m (Rn ) with equivalent norms.
Proof: If f ∈ H2m (Rn ), then
Dxα f = OP(ξ α hξi−m )hDx im f,
0
(Rn × Rn ) if |α| ≤ m. Hence
where qα (ξ) := ξ α hξi−m ∈ S1,0
 21

X

kDxα f k22  ≤ Cm khDx im f k2 = Cm kf km,2 .
|α|≤m
Conversely, if f ∈ W2m (Rn ), ∂xα f ∈ L2 (Rn ) for all |α| ≤ m. Moreover,
n
X ξj
hξi2
1
hξi =
=
+
ξj
.
hξi
hξi j=1 hξi
Thus
m
hξi =
N
X
pk (ξ)qk (ξ),
k=1
0
where qk (ξ) is a polynomial of order m and pk ∈ S1,0
(Rn × Rn ), N = N (m) ∈ N.
Hence
N
N
X
X
m
0
kf km,2 = khDx i f k2 = pk (Dx )qk (Dx )f ≤ Cm
kqk (Dx )f k2 ≤ Cm
kf kW2m ,
k=1
2
k=1
where we have used that pk (Dx ) : L2 (Rn ) → L2 (Rn ) are bounded operators and that
kqk (Dx )f k2 ≤ Ckf kW2m since qk (Dx ) is a differential operator of order m.
Lemma 4.43 Let s > n2 . Then there is some Cs > 0 such that
kuk∞ ≤ Cs khDx is uk2 ,
kuk2 ≤ Cs khxis uk∞
for all u ∈ S(Rn ). In particular, H2s (Rn ) ,→ Cb0 (Rn ).
48
Proof: See exercises.
Corollary 4.44 Let
|u|00k,S :=
sup
|α|+|β|≤k
kxα Dxβ uk2
|.|00k,S ,
n
for u ∈ S(R ), k ∈ N0 . Then
k ∈ N, is a decreasing sequence of semi-norms
on S(Rn ) which is equivalent to the semi-norms |.|k,S defined above. More precisely,
|u|00k,S ≤ Ck |u|k+2n,S
and |u|k,S ≤ Ck |u|00k+2n,S
for all u ∈ S(Rn ) and k ∈ N0 .
Proof: Since hDx i2n = (1 − ∆)n is a differential operator of order 2n,
|u|k,S =
sup
|α|+|β|≤k
kxα Dxβ uk∞ ≤ C
sup
|α|+|β|≤k
khDx i2n xα Dxβ uk2 ≤ Ck |u|00k+2n,S
by Lemma 4.43. Similarly, since hxi2n = (1 + |x|2 )n is a polynomial of order 2n,
|u|00k,S =
4.10
sup
|α|+|β|≤k
kxα Dxβ uk2 ≤ C
sup
|α|+|β|≤k
khxi2n xα Dxβ uk∞ ≤ Ck |u|k+2n,S .
Summary and Final Remarks
m
In this chapter we have studied pseudodifferential symbols p ∈ S1,0
(Rn ×Rn ), m ∈ Rn ,
and the relation to the associated operators p(x, Dx ), first defined on S(Rn ) and later
extended to larger spaces.
m
(Rn ×
Finally, we note the following structural properties: The symbol space S1,0
n
R ) is a topological vector space, which topology is defined by a sequence of (increasing and separating) semi-norms (a Fréchet space). The space of continuous linear
mappings T : S(Rn ) → S(Rn ) denoted by L(S(Rn )) is also a topological vector space
if equipped with the topology of pointwise convergence. In these terms
m
OP : S1,0
(Rn × Rn ) → L(S(Rn )) : p 7→ p(x, Dx )
is a homomorphism of topological vector spaces (i.e., linear and continuous). Moreover, on the level of symbol we have the pointwise multiplication of symbols:
m1
m2
m1 +m2
· : S1,0
(Rn × Rn ) × S1,0
(Rn × Rn ) → S1,0
(Rn × Rn ) : (p1 , p2 ) 7→ p1 (x, ξ) · p2 (x, ξ),
which is a continuous bilinear mapping. On the level of operators we have the
composition. Therefore the question arises how these operations on the symbol and
49
on the operator level are related. We have proved that there is a continuous bilinear
mapping
m1 +m2
m2
m1
(Rn × Rn ) : (p1 , p2 ) 7→ p1 #p2
(Rn × Rn ) → S1,0
(Rn × Rn ) × S1,0
# : S1,0
such that
OP(p1 #p2 ) = OP(p1 ) ◦ OP(p2 ).
Although in general (p1 #p2 )(x, ξ) 6= p1 (x, ξ) · p2 (x, ξ), we have observed that
(p1 #p2 )(x, ξ) = p1 (x, ξ) · p2 (x, ξ) + r(x, ξ),
m1 +m2 −1
(Rn × Rn ).
r ∈ S1,0
Hence the relation (p1 #p2 )(x, ξ) ≡ p1 (x, ξ) · p2 (x, ξ) holds modulo symbols of lower
order. This relation shows that inversion of an operator modulo lower order operators
coincides with inversion of the symbol (for large |ξ|). This was used to construct a
parametrix for an elliptic pseudodifferential operator.
After studying compositions of pseudodifferential operators, it was shown that
every pseudodifferential operator p(x, Dx ) possesses a formal adjoint p∗ (x, Dx ) and
∗
m
m
: S1,0
(Rn × Rn ) → S1,0
(Rn × Rn ) : p 7→ p∗
is a continuous involution, i.e. (p∗ )∗ = p, which coincides with complex conjugation
modulo lower order terms:
p∗ (x, ξ) = p(x, ξ) + r(x, ξ),
m−1
r ∈ S1,0
(Rn × Rn ).
Hence the pseudodifferential operators are a rich class of operators with many algebraic properties. These properties make pseudodifferential operators to an important
tool in index theory, cf. [RS82] or [WRL95].
Finally, we studied the pseudodifferential operator p(x, Dx ) on Cb∞ (Rn ), S 0 (Rn ),
and as operator between Bessel potential spaces. We note that
S(Rn ) ⊂ Cb∞ (Rn ) ⊂ S 0 (Rn ),
S(Rn ) ⊂ H2s (Rn ) ⊂ S 0 (Rn ).
In order to study p(x, Dx ) (initially defined on S(Rn )) on these spaces, the operator
has to be extended in a suitable way. But, since all spaces are contained in S 0 (Rn ),
it remains to check that the extension is consistent with the restriction of p(x, Dx )
defined on S 0 (Rn ) by duality to the corresponding space. This was done for S(Rn )
and L2 (Rn ). But it remains to prove this for the definition of p(x, Dx ) on Cb∞ (Rn )
due to (4.23), which can be done using the approximation used in the proof of
Corollary 4.24.
50
5
Extensions and Applications
5.1
Resolvents and Parameter-Elliptic Differential Operators
The study parabolic initial value problems
∂t u(x, t) + (Au)(x, t) = 0
u(x, 0) = u0
for (x, t) ∈ Rn × (0, ∞),
for x ∈ Rn ,
where A is a suitable elliptic operator as e.g. −∆, can be reduced to the resolvent
equation
(λ + A)u(x) = f
for x ∈ Rn .
This is the content of the theory of analytic semi-groups, cf. e.g. [Paz83],[RR93].
We only state one of the main results
THEOREM 5.1 Let X be a Banach space and let A : D(A) ⊆ X → X be a linear
and closed operator such that D(A) is dense in X. If λ + A is invertible for every
λ ∈ λ0 + Σδ ,
Σδ := {z ∈ C \ {0} : | arg z| < δ},
where δ ∈ (π/2, π) and λ0 ∈ R, and satisfies
k(λ + A)−1 kL(X) ≤
C
|λ − λ0 |
for λ ∈ λ0 + Σδ ,
(5.1)
then
Z
1
eλt (λ + A)−1 dλ
for t ∈ Σδ−π/2
e
:=
2πi Γ
is a bounded linear operator, which depends analytically on t ∈ Σδ−π/2 and which
satisfies
−At
1. e−tA e−sA = e−(t+s)A for all t, s > 0,
2.
d −At
e
dt
= −Ae−At for t ∈ Σδ−π/2 ,
3. limt→0 e−At u0 = u0 for all u0 ∈ X,
4. ke−At kL(X) + |t|kAe−At kL(X) ≤ Ceλ0 t for t ∈ Σδ−π/2 .
Here Γ is the negatively orientated boundary of (λ0 + Σδ ) \ B1 (0).
This theorem is a consequence of the characterization of the so called analytic and
strongly continuous semi-group, cf. [Paz83, Theorem 3.1, Chapter 1] or [RR93, Theorem 11.17].
P
In the following let a(x, ξ) = |α|≤m cα (x)ξ α , cα ∈ Cb∞ (Rn ), m ∈ N. Moreover,
P
let a0 (x, ξ) = |α|=m cα (x)ξ α denote its principle symbol
51
Definition 5.2 Let δ ∈ (0, π) and let a(x, ξ) be a differential symbol as above. Then
a is said to be (uniformly) parameter elliptic on Σδ if
{a0 (x, ξ) : x ∈ Rn , |ξ| = 1} ⊂ Σπ−δ
for all x ∈ Rn , |ξ| = 1
We note that, since 0 6∈ Σπ−δ , a uniformly elliptic differential symbol satisfies
|a0 (x, ξ)| ≥ c > 0
for all x ∈ Rn , |ξ| = 1
for some c > 0.
Example 5.3 Let a(x, ξ) = |ξ|2 . Then m = 2 and |a0 (x, ξ)| = 1 for all x ∈ Rn and
|ξ| = 1. Hence a is parameter elliptic on Σδ for every δ ∈ (0, π).
Moreover, if A(x) ∈ Cb∞ (Rn )n×n is a real symmetric matrix, which is uniformly
positive definite, i.e., ξ T A(x)ξ ≥ c > 0, then a(x, ξ) = ξ T A(x)ξ is parameter elliptic
on Σδ for every δ ∈ (0, π).
Finally, if n = 1, a(x, ξ) = iξ is uniformly elliptic on Σδ only if δ ∈ (0, π/2).
Proposition 5.4 Let a be parameter elliptic on Σδ , δ ∈ (0, π), and let a0 (x, ξ) be
its principal symbol. Then
1
|∂ξα ∂xβ (λ + a0 (x, ξ))−1 | ≤ C(1 + |λ| m + |ξ|)−m−|α|
(5.2)
for all λ ∈ Σδ , x, ξ ∈ Rn , such that |λ| + |ξ|m ≥ 1.
Proof: The proof relies essentially on the following statement:
Claim:
|∂ξα ∂xβ (λ + a0 (x, ξ))−1 | ≤ C
(5.3)
for all λ ∈ Σδ and ξ ∈ Rn with max(|ξ|m , |λ|) = 1.
Proof of the claim: First let max(|ξ|m , |λ|) = |ξ|m = 1. Since cα ∈ Cb∞ (Rn ), the
set
A := {−a0 (x, ξ) : x ∈ Rn , |ξ| = 1}
is bounded. Moreover, A ⊂ −Σπ−δ = C \ Σδ . Hence dist(A, ∂Σδ ) > 0 since A is
compact, ∂Σδ is closed, and A ∩ ∂Σδ = ∅. Thus
|(λ + a0 (x, ξ))−1 | ≤
1
dist(A, ∂Σδ )
for all |ξ| = 1 and λ ∈ Σδ . Thus
|∂z (λ + a0 (x, ξ))−1 | = |(λ + a0 (x, ξ))−2 ∂z a0 (x, ξ)| ≤ Cα,β
for |ξ| = 1, x ∈ Rn , λ ∈ Σδ ,
for z = xj and z = ξj , j = 1, . . . , n. Similarly it is easy to prove by mathematical
induction that
|∂ξα ∂xβ (λ + a0 (x, ξ))−1 | ≤ Cα,β
for |ξ| = 1, x ∈ Rn , λ ∈ Σδ ,
52
since all derivatives of a0 (x, ξ) are uniformly bounded on |ξ| = 1, x ∈ Rn .
Now we consider the case max(|ξ|m , |λ|) = |λ| = 1. Because of the homogeneity of
the principal symbol, i.e., a0 (x, rξ) = rm a0 (x, ξ) for r > 0,
[
{a0 (x, ξ) : x ∈ Rn , |ξ| ≤ 1} =
rm · {a0 (x, ξ) : x ∈ Rn , |ξ| = 1} ⊂ Σπ−δ ∪ {0},
0≤r≤1
S
where 0≤r≤1 rm · {a0 (x, ξ) : x ∈ Rn , |ξ| = 1} is a compact set since it is the image of
[0, 1] × {a0 (x, ξ) : x ∈ Rn , |ξ| = 1} under the multiplication (r, z) 7→ rm · z. Now let
A := {−a0 (x, ξ) : x ∈ Rn , |ξ| ≤ 1}.
Then A ⊂ C \ Σδ and dist(A, ∂Σδ ∩ {|λ| = 1}) > 0. Thus
|(λ + a0 (x, ξ))−1 | ≤
1
dist(A, ∂Σδ ∩ {|λ| = 1})
and therefore
|∂ξα ∂xβ (λ + a0 (x, ξ))−1 | ≤ Cα,β
for all |ξ| ≤ 1 and λ ∈ Σδ with |λ| = 1. This completes the proof of the claim.
Finally, let aλ (x, ξ) := (λ + a0 (x, ξ))−1 . We note that aλ (x, ξ) is homogeneous of
1
degree −m with respect to (λ− m , ξ) in the sense that
arm λ (x, rξ) = r−m aλ (x, ξ)
for all r > 0, x, ξ ∈ Rn , λ ∈ Σδ .
(5.4)
This implies that ∂ξα ∂xβ aλ (x, ξ) is homogeneous of degree −m − |α| with respect to
1
(λ− m , ξ), i.e.,
(∂ξα ∂xβ arm λ )(x, rξ) = r−m−|α| ∂ξα ∂xβ aλ (x, ξ)
for all r > 0, x, ξ ∈ Rn , λ ∈ Σδ ,
by differentiating (5.4). Replacing (λ, ξ) by (r−m λ, r−1 ξ), we have
ξ
α β
−m−|α| α β
∂ξ ∂x aλ (x, ξ) = r
∂ξ ∂x ar−m λ x,
.
r
1
Now choosing r = max(|ξ|, |λ| m ), (5.3) implies
1
1
|∂ξα ∂xβ (λ + a0 (x, ξ))−1 | ≤ C max(|ξ|, |λ| m )−m−|α| ≤ C 0 (|λ| m + |ξ|)−m−|α|
for all ξ ∈ Rn and λ ∈ Σδ , which proves the proposition.
Lemma 5.5 Let a(x, ξ) be a parameter-elliptic symbol on Σδ , δ ∈ (0, π), as defined
above. Then A = a(x, Dx ) : D(A) → H2s (Rn ), s ∈ R, with D(A) = H2s+m (Rn ) ⊂
H2s (Rn ) is closed and densely defined linear operator. Moreover, there is some R > 0
such that λ + A is invertible for all λ ∈ Σδ with |λ| ≥ R > 0 and
k(λ + A)−1 kL(H2s (Rn )) ≤ C|λ|−1 .
53
Proof: Let λ ∈ Σδ with |λ| ≥ 1. Moreover, let aλ (x, ξ) = (λ + a0 (x, ξ))−1 and let
m0 ∈ [0, m]. Then for all α, β ∈ Nn0
−m−|α|
m−m0
1
0
α β
m
|∂ξ ∂x aλ (x, ξ)| ≤ Cα,β 1 + |λ| + |ξ|
(5.5)
≤ Cα,β |λ|− m hξi−m −|α|
(−m0 )
0
−m
for all λ ∈ Σδ , |λ| ≥ 1, and ξ ∈ Rn . Hence aλ ∈ S1,0
(Rn × Rn ) with |aλ |k
0
− m−m
m
Ck |λ|
≤
uniformly in λ ∈ Σδ with |λ| ≥ 1 for every k ∈ N. Therefore
kaλ (x, Dx )kL(H s (Rn ),H s+m0 (Rn )) ≤ C|λ|−
m−m0
m
(5.6)
2
2
uniformly in λ ∈ Σδ with |λ| ≥ 1 because of Lemma 4.38. Moreover, because of
the asymptotic expansion stated in Theorem 4.16 and the fact that all terms depend
continuously on p1 and p2
(λ + a0 (x, Dx ))aλ (x, Dx ) = λaλ (x, Dx ) + a0 (x, Dx )aλ (x, Dx ) = I + rλ (x, Dx ),
where for every k ∈ N0 there are n(k) ∈ N0 and Ck > 0 such that
(0)
(m)
(−m+1)
|rλ |k ≤ Ck |a0 |n(k) ||aλ |n(k)
1
≤ Ck |λ|− m .
(5.7)
– More precisely, since a0 (x, Dx ) is a differential operator,
rλ (x, ξ) =
X
|α|≤m,α6=0
1 α
∂ξ a0 (x, ξ)Dxα aλ (x, ξ).
α!
We note that one can prove (5.7) directly using this identity. – Thus
(0)
1
krλ (x, Dx )kL(H2s (Rn )) ≤ C|rλ |k ≤ C 0 |λ|− m
for all |λ| ≥ 1, where k is as in Lemma 4.38 for m = 0. Furthermore,
(λ + a(x, Dx ))aλ (x, Dx ) = I + rλ (x, Dx ) + Rλ0 ,
where Rλ0 :=
α
|α|<m cα (x)Dx aλ (x, Dx )
P
satisfies
1
kRλ0 kL(H2s (Rn )) ≤ Ckaλ (x, Dx )kL(H2s (Rn ),H2s+m−1 (Rn )) ≤ C|λ|− m
due to (5.6) for m0 = m − 1 and Lemma 4.38 again.
Hence there is some R > 0 such that krλ (x, Dx )+Rλ0 kL(H2s (Rn )) ≤
Therefore I + rλ (x, Dx ) + Rλ0 is invertible and
k(I + rλ (x, Dx ) + Rλ0 )−1 kL(H2s (Rn )) ≤ 2
for all |λ| ≥ R. Thus
(λ + a(x, Dx ))aλ (x, Dx )(I + rλ (x, Dx ) + Rλ0 )−1 = I
54
1
2
for all |λ| ≥ R.
for all λ ∈ Σδ , |λ| ≥ R. Therefore λ + a(x, Dx ) is surjective and has a continuous
right inverse. By the same arguments as before one shows that also
aλ (x, Dx )(λ + a(x, Dx )) = I + Rλ00 ,
1
where kRλ00 kL(H2s+m (Rn )) ≤ C|λ| m for all λ ∈ Σδ , |λ| ≥ 1. This show that there
λ + a(x, Dx ) is injective if |λ| ≥ R, λ ∈ Σδ and R ≥ 1 is sufficiently large. Therefore
λ + a(x, Dx ) is invertible for all |λ| ≥ R, λ ∈ Σδ and
k(λ + a(x, Dx ))−1 kL(H2s (Rn ))
≤ kaλ (x, Dx )kL(H2s (Rn )) k(I + rλ (x, Dx ) + Rλ0 )−1 kL(H2s (Rn )) ≤ C|λ|−1 .
This proves the lemma.
Remark 5.6 The latter lemma implies that the spectrum of −A, where A is a
parameter elliptic differential operator, is contained in a key-hole region (C \ Σδ ) ∪
BR (0) for some R > 0 and that the resolvent (λ + A)−1 decays as |λ|−1 as |λ| → ∞
in Σδ .
Corollary 5.7 Let A = a(x, Dx ) and δ ∈ (π/2, π) be as in Lemma 5.5. Then
there is some λ0 > 0 such that A satisfies the conditions of Theorem 5.1 with X =
H s (Rn ), D(A) = H s+m (Rn ) for every s ∈ Rn . In particular, A generates a strongly
continuous, analytic semi-group e−At , t ≥ 0 on H2s (Rn ), which satisfies
ke−At kL(H2s (Rn )) + tkAe−At kL(H2s (Rn )) ≤ Cs eλ0 t
for all t ≥ 0.
Proof: Just choose some λ0 > 0 such that λ0 + Σδ ⊆ Σδ \ BR (0). Then apply
Theorem 5.1.
55
5.2
Kernel Representation of a Pseudodifferential Operator
A classical theorem due to Schwartz states that for every bounded linear operator
T : S(Rn ) → S 0 (Rn )
there is a unique k ∈ S 0 (Rn × Rn ) such that
for all u, v ∈ S(Rn ),
hT u, viS(Rn ) = hK, u ⊗ viS(Rn ×Rn )
(5.8)
where (u ⊗ v)(x, y) = u(x)v(y) and hf, ϕiS(Rm ) := f (ϕ) denotes the duality product
of f ∈ S 0 (Rm ), and ϕ ∈ S(Rm ), cf. e.g. [Tre67, Corollary
R to Theorem 51.6]. Here
K ∈ S 0 (Rn × Rn ) is called kernel of T . Formally “T u = Rn K(x, y)u(y)dy”.
Obviously, this result applies to pseudodifferential operators since p(x, Dx ) : S(Rn ) →
m
S(Rn ) ⊂ S 0 (Rn ) for p ∈ S1,0
(Rn × Rn ) and m ∈ R arbitrary.
m
For the case that p ∈ S1,0 (Rn × Rn ) with m ≤ −n − 1, we have seen in the proof
of Lemma 4.32 that
Z
p(x, Dx )f =
k(x, x − y)f (y)dy
for all f ∈ S(Rn ),
Rn
n
where k(x, z) = Fξ7−1
→x [p(x, .)] is a continuous function in z ∈ R . This shows that in
this case the kernel of T = p(x, Dx ) due to (5.8) is the continuous function
K(x, y) = k(x, x − y).
In general the kernel of a pseudodifferential operator does not have to be continuous.
For example if p(x, ξ) = 1 and therefore p(x, Dx ) = I, the kernel is the distribution
K ∈ S 0 (Rn × Rn ) defined by
Z
hK, u ⊗ viS(Rn ×Rn ) =
u(x)v(x)dx
Rn
Formally “K(x, y) = δ0 (x − y)” and
p(x, Dx )u(x) = δ0 ∗ u(x) := hδ0 , u(x − ·)iS(Rn ) = u(x)
for all u ∈ S(Rn ).
The main result of this section gives a precise statement of the form of the Schwartz
kernel of a pseudodifferential operator away from the diagonal {(x, y) ∈ Rn × Rn :
x = y}.
m
THEOREM 5.8 Let p ∈ S1,0
(Rn × Rn ), m ∈ R. Then there is a smooth function
n
n
k : R × (R \ {0}) such that
Z
p(x, Dx )u(x) =
k(x, x − y)u(y) dy
for all x 6∈ supp u
(5.9)
Rn
56
for all u ∈ S(Rn ). Moreover, for every α, β ∈ Nn0 , N ∈ N0 , k satisfies

−n−m−|α|

hzi−N
if n + m + |α| > 0,
Cα,β,N |z|
β α
−N
|∂x ∂z k(x, z)| ≤ Cα,β,N (1 + | log |z||)hzi
if n + m + |α| = 0,


−N
Cα,β,N hzi
if n + m + |α| < 0
(5.10)
uniformly in x, z ∈ Rn , z 6= 0. In particular, we that
Z
hp(x, Dx )u, viS(Rn ) =
k(x, x − y)u(y)v(x)d(x, y),
Rn ×Rn
for all u, v ∈ S(Rn ) with supp u∩supp v = ∅ and the Schwartz kernel K ∈ S 0 (Rn ×Rn )
of p(x, Dx ) is a smooth function on {(x, y) ∈ Rn × Rn : x 6= y}.
In order to prove this theorem we need some preliminaries.
Dyadic Partition of Unity
In this section we introduce an important tool in modern harmonic analysis and the
theory of function spaces - the dyadic decomposition of the phase space.
Let ϕ0 ∈ C0∞ (Rn ) such that ϕ0 (ξ) = 1 for |ξ| ≤ 1 and ϕ0 (ξ) = 0 for |ξ| ≥ 2.
Moreover, let ϕj (ξ) = ϕ0 (2−j ξ) − ϕ0 (2−j+1 ξ) for j ∈ N. Then
supp ϕj ⊆ {ξ ∈ Rn : 2j−1 ≤ |ξ| ≤ 2j+1 },
and
∞
X
ϕj (ξ) =
k
X
j=0
j ∈ N,
supp ϕ0 ⊆ B2 (0)
ϕj (ξ) = ϕ0 (2−k ξ) = 1
j=0
for ξ ∈ Rn and k ∈ N such that |ξ| ≤ 2k . Hence ϕj , j ∈ N, is a partition of unity on
Rn subordinated to the dyadic rings {ξ ∈ Rn : 2j−1 ≤ |ξ| ≤ 2j+1 }, j ∈ N, and B2 (0).
Remark 5.9 The asymptotic behavior of a function f (ξ) as |ξ| → ∞ can be described with the aid of this partition of unity an alternative way as follows:
|f (ξ)| ≤ Chξim
⇔
sup |ϕj (ξ)f (ξ)| ≤ C 0 2jm
ξ∈Rn
for all j ∈ N0
where m ∈ R and C 0 > 0 does not depend on j. - See exercises.
−∞
Obviously, ϕj ∈ S1,0
(Rn × Rn ) since ϕj ∈ C0∞ (Rn ). Moreover,
N
X
ϕj (ξ) = ϕ0 (2−N ξ) →N →∞ 1
pointwise and
j=0
N
X
∂ξα ϕj (ξ) →N →∞ 0
uniformly if α 6= 0
j=0
57
since for every ξ ∈ Rn there are at most two non-zero terms in the sums and
∂ξα ϕj (ξ) = 2−|α|(j−1) ∂ξα ϕ1 (2−j+1 ξ) for all α ∈ N0 , j ∈ N.
Hence
∞
X
ϕj (ξ)fˆ = fˆ and
j=0
∞
X
ϕj (Dx )f = f
(5.11)
in S(Rn )
j=0
since | j=0 ϕj (ξ)fˆ − fˆ|00k,S →N →∞ 0 for all k ∈ N by dominated convergence, where
|.|00k,S , k ∈ N, is the equivalent sequence of semi-norms on S(Rn ) which was defined
in Corollary 4.44 by replacing k.k∞ by k.k2 .
With the aid of the dyadic decomposition we decompose a pseudodifferential
operator as
PN
p(x, Dx )f =
∞
X
p(x, Dx )ϕj (Dx )f =
j=0
∞
X
pj (x, Dx )f
for all f ∈ S(Rn ),
(5.12)
j=0
−∞
where pj (x, ξ) = p(x, ξ)ϕj (ξ) ∈ S1,0
(Rn × Rn ) and the series converges in S(Rn )
since p(x, Dx ) : S(Rn ) → S(Rn ) is continuous. Moreover, since pj (x, ξ) is compactly
supported in ξ,
Z
kj (x, x − y)f (y)dy,
pj (x, Dx )f =
(5.13)
Rn
where kj (x, z) = Fξ7−1
→z [pj (x, ξ)] as in the proof of Lemma 4.32.
m
(Rn × Rn ), m ∈ R, and let kj (x, z) be defined as above.
Lemma 5.10 Let p ∈ S1,0
Then
|∂xβ ∂zα kj (x, z)| ≤ Cα,β,M |z|−M 2j(n+m−M +|α|)
(5.14)
for all α, β ∈ Nn0 , M ∈ N0 , where Cα,β,M does not depend on j ∈ N0 .
Proof: First of all,
z γ ∂xβ Dzα kj (x, z)
Z
=
Rn
eix·ξ Dξγ [ξ α ∂xβ pj (x, ξ)]đξ
for all α, β, γ ∈ Nn0 . We now make the most direct estimates on the above integral.
Firstly, the integrand is supported in the ball {|ξ| ≤ 2j+1 }, which has volume bounded
by a multiple of 2nj . Secondly, since the support is even contained in the set {2j−1 ≤
|ξ| ≤ 2j+1 } (when j 6= 0) and c2j ≤ hξi ≤ C2j on {2j−1 ≤ |ξ| ≤ 2j+1 },
γ α β
D [ξ ∂x pj (x, ξ)] ≤ Cα,β,γ 2j(m+|α|−|γ|)
ξ
m+|α|
due to the symbol estimates of ξ α ∂xβ pj (x, ξ) ∈ S1,0
|z γ Dxβ Dzα kj (x, z)| ≤ Cα,β,γ 2j(n+m+|α|−M )
58
(Rn × Rn ). Hence
whenever |γ| = M.
Taking the supremum over all γ with |γ| = M , gives (5.14), and proves the lemma.
Proof of Theorem 5.8: First of all, because of (5.12) and (5.13)
∞ Z
X
p(x, Dx )u(x) =
kj (x, x − y)u(y) dy
for all x ∈ Rn .
j=0
(5.15)
Rn
We will show (5.9) and (5.10) by showing that
∞
X
∂zα ∂xβ kj (x, z)
j=0
converges absolutely and uniformly w.r.t (x, ξ) ∈ Rn × (Rn \ Bε (0)) for every ε > 0
to a function k(x, z) satisfying (5.10).
First let 0 < |z| ≤ 1. Then we split the sum into
X
X
∂zα ∂xβ kj (x, z).
and
∂zα ∂xβ kj (x, z)
2j >|z|−1
2j ≤|z|−1
In order to estimate the first sum, we use Lemma 5.10 with M = 0 and obtain
bld(|z|−1 )c
X
X
|∂zα ∂xβ kj (x, z)| ≤ Cα,β
2j ≤|z|−1
2j(n+m+|α|)
j=0

−(m+n+|α|)

if m + n + |α| > 0,
Cα,β |z|
−1
≤
Cα,β (1 + | log |z| |) if m + n + |α| = 0,


Cα,β
if m + n + |α| < 0
where ld = log2 . For the second term we use Lemma 5.10 with M > n + m + |α| and
estimate
X
|∂zα ∂xβ kj (x, z)|
∞
X
−M
≤ Cα,β |z|
2j(n+m+|α|−M )
j=bld(|z|−1 )c+1
2j >|z|−1
≤ Cα,β |z|−(n+m+|α|) .
Finally, if |z| ≥ 1, we choose M > n + m + |α| + N in Lemma 5.10 to conclude
∞
X
|∂zα ∂xβ kj (x, z)| ≤ |z|−M
j=0
∞
X
2j(n+m+|α|−M )
j=0
−M
≤ C|z|
≤ C|z|−m−n−|α|−N .
P∞
Hence
j=0 kj (x, z) converges absolutely and uniformly with respect to (x, ξ) ∈
n
n
R × (R \ Bε (0)) for every ε > 0 to a function k(x, z) satisfying (5.10). Using the
59
uniform convergence and (5.15), we conclude that (5.9) holds for all x ∈ Rn with
dist(x, supp u) < ε for arbitrary ε > 0. Hence (5.9) follows for all x 6∈ supp u.
We have the following consequences of Theorem 5.8:
m
(Rn × Rn ), m < 0, then supx∈Rn |k(x, z)| ∈ L1 (Rn ) and
1. If p ∈ S1,0
Z
p(x, Dx )u(x) =
k(x, x − y)u(y) dy
for a.e. x ∈ Rn
Rn
for all u ∈ Lq (Rn ), 1 ≤ q ≤ ∞. Therefore
kp(x, Dx )f kq ≤ sup kk(x, ·)kL1 (Rn ) kf kq ,
f ∈ Lq (Rn ),
x∈Rn
for all 1 ≤ q ≤ ∞; cf. exercises.
0
(Rn ×Rn ). Because of the boundedness p(x, Dx ) : L2 (Rn ) → L2 (Rn )
2. Let p ∈ S1,0
and the kernel estimates
sup |k(x, z)| ≤ C|z|−n ,
x∈Rn
sup |∇x,z k(x, z)| ≤ C|z|−n−1 ,
x∈Rn
a general result from the theory of singular integral operators, cf. [Ste93, Chapter 5, Section 5.1] or [Tay91, §0.11], implies that
p(x, Dx ) : Lq (Rn ) → Lq (Rn )
(5.16)
is a bounded operator for all 1 < q ≤ 2. We note that this is a (non-translationinvariant) generalization of [Abea, Theorem 5.5], which is proved in the same
way.1
0
The same statement is true for the formal adjoint p∗ (x, Dx ) since p∗ ∈ S1,0
if
0
p ∈ S1,0
. For 2 < q < ∞, this implies that
Z
kp(x, Dx )ukLq (Rn ) =
sup
p(x,
D
)u(x)v(x)
dx
x
v∈Lq0 (Rn ),kvk
Rn
=1
0
Lq (Rn )
=
sup
v∈S(Rn ),kvk
=1
0
Lq (Rn )
≤
sup
kvk
Z
R
u(x)p∗ (x, Dx )v(x) dx
n
kukLq (Rn ) kp∗ (x, Dx )vkLq0 (Rn ) ≤ CkukLq (Rn )
=1
0
Lq (Rn )
for all u ∈ S(Rn ), where we have used the Riesz representation theorem for
Lq (Rn )0 . Therefore we have that the mapping (5.16) is bounded for every
1 < q < ∞.
1
A complete proof can also be found in [Abeb, Chapter IV].
60
3. If ϕ, ψ ∈ Cb∞ (Rn ) with dist(supp ϕ, supp ψ) > 0, then ϕ(x)p(x, Dx )ψ(x) is a
smoothing operator in the sense that
ϕ(x)p(x, Dx )ψ(x) : S 0 (Rn ) → C ∞ (Rn ).
This statement can be proved as follows: If f ∈ S 0 (Rn ), then there is an N ∈ N
such that f 0 := hxi−2N hDx i−2N f ∈ L2 (Rn ), cf. exercises. Hence
Z
ϕ(x)p(x, Dx )ψf = ϕ(x)
k(x, x − y)ψ(y)hDy i2N (hyi2N f 0 (y))dy
Rn
since supp ϕ ∩ supp ψ = ∅ implies ϕ(x)k(x, x − y)ψ(y) = 0 in a neighborhood
of the diagonal x = y. Moreover,
Z
ϕ(x)p(x, Dx )(ψf ) =
hDy i2N (ϕ(x)k(x, x − y)ψ(y)) hyi2N f 0 (y)dy,
Rn
where Theorem 5.8 implies that hDy i2N ϕ(x)k(x, x − y)ψ(y) is a smooth kernel
which is rapidly decreasing in y. This implies the statement.
4. If f ∈ S 0 (Rn ), then the singular support of f denoted by sing supp f is the
complement of the set U of all x ∈ Rn such that f |V coincides with a smooth
function for an open neighborhood V of x, i.e.,
Z
f˜(x)ϕ(x)dx
for all ϕ ∈ C0∞ (V ),
hf, ϕi =
Rn
where f˜ ∈ C ∞ (V ). It is easy to show that U is open and therefore sing supp f
is closed.
m
(Rn × Rn ), it holds that
If p ∈ S1,0
sing supp p(x, Dx )f ⊆ sing supp f
for all f ∈ S 0 (Rn ).
An operator satisfying this property is called pseudo-local. The statement can
be proved using the definitions and the previous statement.
61
5.3
Coordinate Transformations and Pseudodifferential Operators on Manifolds
In this section we will show that for every suitable smooth diffeomorphism κ : Rn →
Rn the operator Q : S(Rn ) → S(Rn ) defined by
Qu(x) = (p(x, Dx )w)(κ(x)),
w(x) = u(κ−1 (x)), u ∈ S(Rn )
(5.17)
is again a pseudodifferential operator. This is important to obtain a definition of
pseudodifferential operators on a manifold in a way that is essentially independent
of the choice of local charts. We note that
Qu = κ∗ p(x, Dx )κ∗,−1 u,
where (κ∗ v)(x) = v(κ(x)) and (κ∗,−1 u)(x) = u(κ−1 (x)).
More precisely, we assume that κ : Rn → Rn is a smooth function such that
∂xj κ ∈ Cb∞ (Rn ) for all j = 1, . . . , n and
0 < c ≤ | det Dκ(x)| ≤ C < ∞
(5.18)
for all x ∈ Rn and some constants c, C > 0. In particular, this implies that κ−1 : Rn →
Rn is a again smooth and ∂xj κ−1 ∈ Cb∞ (Rn ) for all j = 1, . . . , n. In the following
∇x κ the total derivative of κ : Rn → Rn .
The main result of this section is:
m
THEOREM 5.11 Let p ∈ S1,0
(Rn × Rn ), m ∈ R, let κ : Rn → Rn be as above and
m
(Rn × Rn )
Q : S(Rn ) → S(Rn ) be defined by (5.17). Then there is some q ∈ S1,0
n
such that Qu = q(x, Dx )u for all u ∈ S(R ). Moreover, q(x, Dx ) has the asymptotic
expansion
q(x, ξ) ∼
X 1
∂ξα Dyα q̃(x, y, ξ)|y=x ,
α!
α∈Nn
where
0
q̃(x, y, ξ) = p(κ(x), A(x, y)−T ξ)| det A(x, y)|−1 | det ∇y κ(y)| and
Z 1
A(x, y) =
∇y κ(x + t(y − x)) dt
for all x, y close enough
0
in the sense that for any N ∈ N0
q(x, ξ) −
X 1
m−N −1
(Rn × Rn ).
∂ α Dα q̃(x, y, ξ)|y=x ∈ S1,0
α! ξ y
|α|≤N
In particular, we have
q(x, ξ) = p(κ(x), (∇y κ(x))−T ξ) + r(x, ξ)
62
m−1
with r ∈ S1,0
.
Proof: We only prove the theorem under the additional assumption that
1
sup |∇κ(x) − I| ≤ .
2
x∈Rn
For the proof in the general case we refer to [KG74, Theorem 6.3]. Here |.| is any
matrix norm, which is induced by some vector norm. This ensures that
|A(x, y) − I| ≤
1
2
for all x, y ∈ Rn
and therefore A(x, y)−T = (A(x, y)−1 )T exists for all x, y ∈ Rn . Moreover, let χ ∈
S(Rn ) with χ(0) = 1 and χε (ξ) = χ(εξ) for ε > 0, ξ ∈ Rn .
Under these assumptions we have
Z
d(y, ξ)
ei(κ(x)−y)·ξ χε (ξ)p(κ(x), ξ)u(κ−1 (y))
Qu = lim
ε→0 Rn ×Rn
(2π)n
Z
d(x0 , ξ)
0
= lim
ei(κ(x)−κ(x ))·ξ χε (ξ)p(κ(x), ξ)u(x0 )| det ∇x0 κ(x0 )|
ε→0 Rn ×Rn
(2π)n
for all u ∈ S(Rn ). On the other hand, since
Z 1
0
0
0
∇x0 κ(x + t(x − x )) dt (x − x0 ) = A(x0 , x)(x − x0 ),
κ(x) − κ(x ) =
0
we obtain by the change of variable ξ = A(x, x0 )−T η
Z
1
0
lim
ei(x−x )·η χε (A(x0 , x)−T η)q̃(x, x0 , η)u(x0 ) d(x0 , η)
Qu =
n
(2π) ε→0 Rn ×Rn
ZZ
= lim Os–
e−iy·η χε (A(x + y, x)−T η)q̃(x, x + y, η)u(x + y) dx0 đη
ε→0
due to (κ(x) − κ(x0 )) · ξ = (x − x0 ) · A(x, x)−1 ξ, where
q̃(x, x0 , η) = p(κ(x), A(x0 , x)−T η)| det A(x, x0 )|−1 | det ∇x0 κ(x0 )|.
Since {χε (A(x, x0 )−T η) : ε ∈ (0, 1)} is bounded in A00 (Rn × Rn ) with respect to (x0 , η)
and
χε (A(x0 , x)−T η) →ε→0 1
for all x, x0 , η ∈ Rn
∂xα0 ∂ηβ χε (A(x0 , x)−T η) →ε→0 0
for all x, x0 , η ∈ Rn , |α| + |β| > 0,
we can apply Corollary 4.10 to conclude that
ZZ
Qu(x) = Os–
e−iy·η q̃(x, x + y, η)u(x + y) d(x0 , η).
Finally, an application of Theorem 4.27 finishes the proof.
Next we define pseudodifferential operators on smooth compact manifolds. To
this end recall:
63
Definition 5.12 M is a smooth compact manifold if M is a topological compact
space and the following conditions hold:
S
1. There are finitely many open sets Ω1 , . . . , ΩN ⊂ M such that M = N
j=1 Ωj .
2. For every j ∈ {1, . . . , N } there is an open set Uj ⊂ Rn and a continuous
bijective functions κj : Ωj → Uj with continuous inverse, called charts.
3. For every j, k ∈ {1, . . . , N } such that Ωj,k := Ωj ∩ Ωk 6= ∅ the mapping
κj,k : κk (Ωj,k ) → κj (Ωj,k )
can be extended to some is C ∞ -diffeomorphism with κ̃j,k : Vx → Vy for some
open Vx ⊃ κk (Ωj,k ), Vy ⊃ κj (Ωj,k ).
A function f : M → C is called smooth if for every j = 1, . . . , N as above uj :=
∞
u ◦ κ−1
j ∈ C (κj (Ωj )). Moreover, if Ω ⊂ M is open
C0∞ (Ω) = {f ∈ C ∞ (M ) : supp f ⊂ Ω}.
Before we define pseudodifferential operators on a compact manifold we the definition of certain remainder operators:
Definition 5.13 Let M be a smooth compact manifold as above and let P : C ∞ (M ) →
C ∞ (M ) be a linear operator. Then P has a C ∞ -kernel representation if for every
j, k ∈ {1, . . . , N } there is some Kj,k ∈ C ∞ (Uk × Uj ) such that for every u ∈ C0∞ (Ωj )
Z
−1
Kj,k (x, x0 )uj (x0 ) dx0
for all x ∈ Uk ,
(P u)(κk (x)) =
Uj
where uj (x) = u(κ−1
j (x)) in Uj .
For the following let Φj , j = 1, . . . , N be a smooth partition of unity subordinate to
Ωj , j = 1, . . . , N , i.e.,
0 ≤ Φj ∈
C0∞ (Ωj ),
N
X
Φj (x) = 1 for all x ∈ M
(5.19)
j=1
Moreover, let Ψj ∈ C0∞ (Ωj ), j = 1, . . . , N , be such that
Ψj (x) = 1
for all x ∈ supp Φj , j = 1, . . . , N.
(5.20)
Finally, we define ϕj , ψj ∈ C0∞ (Uj ) by
ϕj = Φj ◦ κ−1
j ,
ψj = Ψj ◦ κ−1
j
64
(5.21)
Definition 5.14 Let M be a smooth compact manifold as above and let P : C ∞ (M ) →
m
C ∞ (M ) be a linear operator. Then P is a pseudodifferential operator of class S1,0
(M )
N
with symbols (pj )j=1 with respect to the charts κj , j = 1, . . . , N , if the following holds
true: Let Φj , Ψj ∈ C ∞ (M ), j = 1, . . . , N satisfy (5.19) and (5.20) and let
N
N
X
X
(P u)(x) =
(Φj P Ψj u)(x) +
(Φj P (1 − Ψj )u)(x)
j=1
for all x ∈ M.
(5.22)
j=1
Then, using ϕj , ψj defined in (5.21), Φj P Ψj u can be written in the form
(Φj P Ψj u)(κ−1
j (x)) = ϕj pj (x, Dx )(ψj uj )(x)
for all x ∈ Uj
∞
where uj (x) = u(κ−1
j (x)), x ∈ Uj , for all u ∈ C (M ), j = 1, . . . , N and (Φj P (1 −
∞
Ψj )u) has a C -kernel representation.
Remark 5.15 If P is a differential operator, supp P (1 − Ψj )u ⊆ supp(1 − Ψj ).
Therefore (Φj P (1 − Ψj )u)(x) = 0 for all x ∈ M . Hence
N
X
(P u)(x) =
(Φj P Ψj u)(x)
for all x ∈ M
j=1
in this case.
The following theorem shows that the definition of pseudodifferential operators on
M are independent of the choice of charts κ1 , . . . , κN .
THEOREM 5.16 Let M be a smooth compact manifolds as above and let Ω0j , Uj0 , κ0j , κ0j,k ,
j, k = 1, . . . , N 0 be another sequence satisfying conditions 1.-3 in Definition 5.12 such
that
0
0
0
0
0
κ̃j,j 0 := κj ◦ κ−1
j 0 : κj 0 (Ωj ∩ Ωj 0 ) → κj (Ωj ∩ Ωj 0 ), j ∈ {1, . . . , N }, j ∈ {1, . . . , N },
extends to a C ∞ -diffeomorphim on some open neighborhoods of κ0j 0 (Ωj ∩ Ω0j 0 ) and
m
κj (Ωj ∩ Ω0j 0 ). If P : C ∞ (M ) → C ∞ (M ) is a pseudodifferential operator in S1,0
(M )
m
with respect to {Ωj , Uj , κj , κj,k : j, k = 1, . . . , N }, then P is in S1,0 (M ) with respect
to {Ω0j , Uj0 , κ0j , κ0j,k : j, k = 1, . . . , N 0 }.
Proof: For simplicity we assume that κ̃j,j 0 can be extended to a C ∞ -diffeormphism
∞
n
κ̃j,j 0 : Rn → Rn such that ∇κ̃j,j 0 , ∇κ̃−1
j,j 0 ∈ Cb (R ). For the proof in the general case
we refer to [KG74, Theorem 7.3].
First of all, we prove that for every Ψ, Φ ∈ C ∞ (M ) we have:
supp Φ ∩ supp Ψ = ∅
⇒
ΦP Ψ has a C ∞ -kernel representation.
65
(5.23)
We write
ΦP Ψ =
N
X
Φ(Φj P Ψj )Ψ +
j=1
N
X
Φ(Φj P (1 − Ψj ))Ψ.
j=1
Then Φ(Φj P (1 − Ψj ))Ψ has a C ∞ -kernel representation since Φj P (1 − Ψj ) has one.
Now we have
(Φ(Φj P Ψj )Ψu)(κ−1
j (x)) = (ϕϕj )(x)pj (x, Dx )(ψj ψuj )(x)
for all x ∈ Uj ,
for all u ∈ C ∞ (M ), j = 1, . . . , N , where uj = u ◦ κ−1
and ϕ = Φ ◦ κ−1
j
j , ψ =
−1
Ψ ◦ κj . Since supp(ψj ψ) ∩ supp(ϕj ϕ) = ∅, Theorem 5.8 implies that there are
smooth kj : Rn × (Rn \ {0}) → C such that
Z
−1
(Φ(Φj P (1 − Ψj ))Ψu)(κj (x)) =
(ϕϕj )(x)kj (x, x − y)(1 − ψj (y))ψ(y)uj (y) dy.
Rn
Hence (ϕϕj )(x)kj (x, x − y)ψj (y)ψ(y) ∈ C ∞ (Rn × Rn ) and Φ(Φj P (1 − Ψj ))Ψ has a
C ∞ -kernel representation. This proves (5.23).
Let Φ0j 0 , Ψ0j 0 ∈ C ∞ (M ), j 0 = 1, . . . , N 0 , satisfy (5.19)-(5.20) with Ωj replaced by
Ω0j 0 . Now let us write P as
0
P =
N
X
j 0 =1
0
Φ0j 0 P Ψ0j 0 +
N
X
Φ0j 0 P (1 − Ψ0j 0 ).
j 0 =1
Because of (5.23), the second term has a C ∞ -kernel representation. For the first term
we use that
Φ0j 0 P Ψ0j 0 =
N
X
Φ0j 0 (Φj P Ψj )Ψ0j 0 +
j=1
N
X
Φ0j 0 (Φj P (1 − Ψj ))Ψ0j 0 ,
j=1
where again the second term has a C ∞ -representation. Moreover, if Ωj ∩ Ω0j 0 6= ∅ by
assumption
0
0
(Φ0j 0 Φj P (Ψj Ψ0j 0 u))(κ−1
j (x)) = ϕj 0 (x)ϕj (x)(pj (x, Dx )(ψj 0 ψj uj ))(x) for all x ∈ Uj .
Hence
(Φ0j 0 Φj P (Ψj Ψ0j 0 u))((κ0j 0 )−1 (x))
−1 ∗
−1 ∗
0
0
0
◦
κ
)
(
ψ̃
u
))
(x) for all x ∈ Uj .
)
ϕ
p
(x,
D
)(ψ
(κ
= ϕ̃0j 0 (x) (κ0j 0 ◦ κ−1
0
0
0
j
j
x
j
j
j
j
j
j
where u0j 0 (x) = u((κ0j 0 )−1 (x)), (κ∗ v)(x) = v(κ(x)) for κ : Rn → Rn , and ϕ̃0j 0 = Φ0j 0 ◦
−1
−1
0
0
0
can be extended to some C ∞ κ−1
j 0 , ψ̃j 0 = Ψj 0 ◦ κj 0 . Now, since κ̃j,j 0 := κj 0 ◦ κj
∞
n
diffeomorphism on Rn with ∇x κ̃j,j 0 , ∇x κ̃−1
j,j 0 ∈ Cb (R ), Theorem 5.11 implies that
m
there is some qj,j 0 ∈ S1,0
(Rn × Rn ) such that
−1
−1 ∗
0 ∗
qj,j 0 (x, Dx )u = (κ0j 0 ◦ κ−1
)
ϕ
p
(x,
D
)
ψ
((κ
◦
κ
)
u
.
0
j
j
x
j
j
j
j
66
If we now define
qj (x, ξ) =
X
qj,j 0 (x, ξ),
j 0 :Ω0j 0 ∩Ωj 6=∅
we have that
(Φ0j 0 P Ψ0j 0 u)((κ0j )−1 (x)) = ϕ̃0j qj (x, Dx )(ψj0 u0j 0 )(x)
which proves the theorem.
67
for all x ∈ Uj0 ,
5.4
Boundedness on Hölder-Zygmund and Besov Spaces
Definition 5.17 Let s > 0. Then the (Hölder-)Zygmund space C∗s (Rn ) is defined as
C∗s (Rn ) = {u ∈ Cb0 (Rn ) : kϕj (Dx )uk∞ ≤ C2−js , j ∈ N0 },
kukC∗s = sup 2js kϕj (Dx )uk∞ .
j∈N0
Remark 5.18 If s 6∈ N, then C∗s (Rn ) coincides with the usual Hölder space C s (Rn )
normed by
|Dα u(x) − Dα u(y)|
.
|α|≤[s] x,y∈Rn ,x6=y
|x − y|s−[s]
kukC s = max kDα uk∞ + max
|α|≤[s]
sup
This is a consequence of Theorem 5.27 below.
m
(Rn ×Rn ), m ∈ R, and let s > 0 such that s+m > 0.
THEOREM 5.19 Let p ∈ S1,0
s+m
n
s
Then p(x, Dx ) : C∗ (R ) → C∗ (Rn ) is a bounded linear operator.
m
(Rn × Rn ), m ∈ R, and let pj (x, Dx ) = p(x, Dx )ϕj (Dx ) be
Lemma 5.20 Let p ∈ S1,0
defined as above. Then
kpj (x, Dx )kL(Lq (Rn )) ≤ C2jm ,
j ∈ N0 ,
(5.24)
for 1 ≤ q ≤ ∞ where C does not dependent on j.
Proof: As seen in Section 5.2,
Z
kj (x, x − y)f (y)dy,
pj (x, Dx )f =
Rn
where the kj satisfy (5.14). According to these estimates
Z
Z
Z
j(n+m)
−n−1 j(m−1)
|kj (x, z)|dz ≤ C
2
dz +
|z|
2
dz ,
|z|≤2−j
Rn
|z|>2−j
the first coming from (5.14) for M = 0, and the second from the case M = n + 1.
Hence we get by a simple calculation
Z
|kj (x, z)|dz ≤ C2jm ,
Rn
which proves (5.24) since
Z
Z
sup
n kj (x, x − y)f (y)dy ≤ x∈R
n
R
Rn
q
68
|kj (x, z)|dzkf kq .
Remark 5.21 If pj (Dx , x) = ϕj (Dx )p(Dx , x) is the corresponding operator in yform, then
Z Z
Z
iξ·(x−y)
pj (Dx , x)f =
e
pj (y, ξ)f (y)đξdy =
kj (y, x − y)f (y)dy,
Rn
Rn
Rn
where kj (y, z) = Fξ7−1
→z [pj (y, ξ)] is the same function as for the operator in x-form.
Moreover,
kpj (Dx , x)kL(Lq (Rn )) ≤ C2jm ,
j ∈ N0
(5.25)
holds for 1 ≤ q ≤ ∞. This inequality is proved in precisely the same way as (5.24).
m
Lemma 5.22 Let p ∈ S1,0
(Rn × Rn ), m ∈ R, and let pi (x, Dx ) be defined as above.
Then for every l ∈ N
kϕj (Dx )pi (x, Dx )kL(Lq (Rn )) ≤ Cl 2min(i,j)m 2−|i−j|l ,
i, j ∈ N0 ,
(5.26)
for 1 ≤ q ≤ ∞ where Cl does not dependent on i, j.
Proof: Similarly as in the proof of Lemma 4.42,
n
1=
X ξj
1
+
ξj
1 + |ξ|2 j=1 1 + |ξ|2
!l
=
X
p(α) (ξ)ξ α ,
(5.27)
|α|≤l
−l
where p(α) ∈ S1,0
(Rn × Rn ). Hence
ϕj (Dx )pi (x, Dx ) = ϕj (Dx )
X
p(α) (Dx )Dxα pi (x, Dx ) =
|α|≤l
X
p(α) (Dx )ϕj (Dx )Dxα pi (x, Dx )
|α|≤l
−l
since ϕj (Dx ) and p(α) (Dx ) commute. Since p(α) (ξ) ∈ S1,0
(Rn ×Rn ) and Dxα pi (x, Dx ) =
m+|α|
q(x, Dx )ϕi (Dx ) = qi (x, Dx ) with q ∈ S1,0 (Rn × Rn ), we can now invoke (5.24) for
(α)
pj (Dx ) = p(α) (Dx )ϕj (Dx ) and qi (x, Dx ) to get
kϕj (Dx )pi (x, Dx )kL(Lq (Rn ))
X
≤
kp(α) (Dx )ϕj (Dx )kL(Lq (Rn )) kqi (x, Dx )kL(Lq (Rn )) ≤ C2−jl · 2i(l+m) ,
|α|≤l
which proves (5.26) for the case i ≤ j.
In order to prove the case i > j, we use some kind of symmetry in i and j. First of
all, we have p(x, Dx ) = (p(x, Dx )∗ )∗ = (p∗ (x, Dx ))∗ = q(Dx , x) for q(y, ξ) = p∗ (y, ξ).
Hence
X
ϕj (Dx )pi (x, Dx ) = ϕj (Dx )q(Dx , x)ϕi (Dx ) =
qj (Dx , x)Dxα p(α) (Dx )ϕi (Dx ).
|α|≤l
69
(α)
Now using (5.25) for qj (Dx , x)Dxα and p(α) (Dx )ϕi (Dx ) = pi (x, Dx ), we conclude in
the same way as before that
kϕj (Dx )pi (x, Dx )kL(Lq (Rn ))
X
≤
kqj (x, Dx )Dxα kL(Lq (Rn )) kp(α) (Dx )ϕi (Dx )kL(Lq (Rn )) ≤ C2−il · 2j(l+m) ,
|α|≤l
which proves (5.26) for the case i > j.
Proof of Theorem 5.19: First of all,
f=
∞
X
fj ,
j=0
where fj = ϕj (Dx )f , f ∈ S 0 (Rn ). Since supp ϕj ∩ supp ϕk = ∅ of |j − k| > 1,
p(x, Dx )f =
∞
X
pk (x, Dx )f =
k=0
∞
X
pk (x, Dx )(fk−1 + fk + fk+1 ) =
k=0
∞
X
pk (x, Dx )f˜k ,
k=0
where f˜k = fk−1 + fk + fk+1 and we have set f−1 = 0. If f ∈ C∗s+m (Rn ), then
2(s+m)k kf˜k k∞ ≤ 2(s+m)k (kfk−1 k∞ + kfk k∞ + kfk+1 k∞ ) ≤ Ckf kC∗s+m
Now using (5.26),
2sj kϕj (Dx )pk (x, Dx )f˜k k∞ ≤ C2sj+min(j,k)m−|j−k|l−k(s+m) 2k(s+m) kf˜k k∞ ,
where
2sj+min(j,k)m−|j−k|l−k(s+m)
(
2j(s−l)+k(l−s) = 2−|j−k|(l−s)
=
2−k(m+s+l)+j(m+s+l) = 2−|j−k|(m+s+l)
if j ≥ k,
if j < k.
Choosing l ∈ N0 so large that l − s ≥ 1 and m + s + l ≥ 1,
kp(x, Dx )f kC∗s = sup 2sj kϕj (Dx )p(x, Dx )f k∞
j∈N0
≤ C sup
j∈N0
∞
X
2−|j−k| 2k(s+m) kf˜k k∞ ≤ Ckf kC∗s+m .
k=0
THEOREM 5.23 Let 0 < s < 1. Then C∗s (Rn ) = C s (Rn ) with equivalent norms,
where C s (Rn ) is normed by
2kf kC s = kf k∞ + sup
x,y∈Rn
70
|f (x) − f (y)|
|x − y|s
Proof: First let f ∈ C s (Rn ). Then
sup |f (x − y) − f (x)| ≤ kf kC s |y|s
x∈Rn
for all y ∈ Rn . Moreover, ϕj (Dx ) for j ∈ N is the convolution with ψ2−j (x) :=
2jn ψ(2j x), ψ(x) = Fξ7−1
→x [ϕ0 (ξ) − ϕ0 (2ξ)], where
Z
Z
ψ2−j (y)dy =
ψ(y)dy = F[ψ](0) = 0.
Rn
Rn
Hence
Z
Z
f (x − y)ψ2−j (y)dy =
ϕj (Dx )f =
(f (x − y) − f (x))ψ2−j (y)dy
Rn
Rn
and therefore
Z
−js
s
kϕj (Dx )f k∞ ≤ kf kC s
|y| |ψ2−j (y)|dy = 2
Z
|z|s |ψ(z)|dz = C2−js kf kC s
kf kC s
Rn
Rn
for all j ∈ N. The latter inequality implies kf kC∗s ≤ kf kC s since also kϕ0 (Dx )f k∞ ≤
Ckf k∞ .
P
Conversely let f ∈ C∗s (Rn ). Then f = ∞
j=0 fj , fj = ϕj (Dx )f , where the sum
0
n
−sj
converges in Cb (R ) since kfj k∞ ≤ 2 kf kC∗s . Now, if |y| ≤ 1,
X
X
f (x − y) − f (y) =
(fj (x − y) − fj (y)) +
(fj (x − y) − fj (y)) .
2j ≤|y|−1
2j >|y|−1
For the first sum we use the mean value theorem to conclude that
|fj (x − y) − fj (y)| ≤ |y|k∇fj k∞ .
Moreover, since
∂xk fj = ∂xk ϕj−1 (Dx )fj + ∂xk ϕj−1 (Dx )fj + ∂xk ϕj+1 (Dx )fj
1
we can use (5.24) for p(x, ξ) = ξk ∈ S1,0
(Rn × Rn ), k = 1, . . . , n, to obtain
X
X
|fj (x − y) − fj (y)| ≤ C
|y|2j k∇fj k∞
2j ≤|y|−1
2j ≤|y|−1
X
≤ C|y|
2j(1−s) kf kC∗s ≤ C|y|s kf kC∗s .
2j ≤|y|−1
The second sum is simply estimated by
X
X
X
|fj (x − y) − fj (y)| ≤ C
kfj k∞ ≤ Ckf kC∗s
2−js = C|y|s kf kC∗s
2j >|y|−1
2j >|y|−1
2j >|y|−1
Altogether kf kC s ≤ Ckf kC∗s .
71
We now come to a generalization of the Zygmund-Hölder spaces. Let X be an
arbitrary Banach space. Then the vector-valued variant of the classical spaces `q (N0 ),
1 ≤ q ≤ ∞, are defined as
`q (N0 , X) = {(ak )k∈N0 : ak ∈ X, k(ak )k∈N0 k`q (N0 ,X) < ∞},
! 1q
∞
X
k(ak )k∈N0 k`q (N0 ,X) =
for 1 ≤ q < ∞,
kak kqX
where
k=0
k(ak )k∈N0 k`∞ (N0 ,X) = sup kak kX .
k∈N0
s
(Rn ) is
Definition 5.24 Let 1 ≤ q, r ≤ ∞ and let s ∈ R. Then the Besov space Bqr
defined as
s
Bqr
(Rn ) = {u ∈ S 0 (Rn ) : (2sj ϕj (Dx )u)j∈N0 ∈ `r (N0 , Lq (Rn ))},
s
kukBqr
= k(2sj ϕj (Dx )u)j∈N0 k`r (N0 ,Lq (Rn ))

r1
 P∞
sj
r
k2
ϕ
(D
)uk
if 1 ≤ r < ∞,
j
x
Lq
j=0
=
sup
sj
if r = ∞.
j∈N0 k2 ϕj (Dx )ukLq
s
Here s is called order of Bq,r
(Rn ), q is called integral exponent, and r is called sum
exponent.
Remark 5.25
s
1. If s > 0, then B∞∞
(Rn ) = C∗s (Rn ).
s
(Rn ) = H2s (Rn ), cf. [BL76].
2. For the case q = r = 2 it can be shown that B22
s
3. The Besov space Bqq
(Rn−1 ), s > 0, appear naturally as the trace spaces of
Bessel potential: If s > 1q , 1 < q < ∞, then there is a bounded trace operator
s− 1
γ : Hqs (Rn ) → Bqq q (Rn−1 )
such that (γu)(x0 ) = u(x0 , 0) for all u ∈ S(Rn ). For a proof of this fact and
for people which interested in Bessel potential and Besov spaces we refer to
[BL76].
m
THEOREM 5.26 Let p ∈ S1,0
(Rn × Rn ), m ∈ R, and let s ∈ R, 1 ≤ q, r ≤ ∞.
s+m
s
Then p(x, Dx ) : Bqr
(Rn ) → Bqr
(Rn ) is a bounded linear operator.
The proof is a simple modification of the proof of 5.19. For the convenience of the
reader we include the complete modified proof here:
Proof: First of all,
f=
∞
X
j=0
72
fj ,
where fj = ϕj (Dx )f , f ∈ S 0 (Rn ). Since supp ϕj ∩ supp ϕk = ∅ of |j − k| > 1,
p(x, Dx )f =
∞
X
pk (x, Dx )f =
k=0
∞
X
pk (x, Dx )(fk−1 + fk + fk+1 ) =
k=0
∞
X
pk (x, Dx )f˜k ,
k=0
s+m
(Rn ), then
where f˜k = fk−1 + fk + fk+1 and we have set f−1 = 0. If f ∈ Bqr
k(2(s+m)k f˜k )k∈N0 k`r (N0 ,Lq (Rn )) = k(2(s+m)k (fk−1 + fk + fk+1 ))k∈N0 k`r (N0 ,Lq (Rn ))
s+m
= Ckf kBqr
≤ C 2(s+m)k ϕk (Dx )f k∈N0 r
q
n
` (N0 ,L (R ))
Now using (5.26),
2sj kϕj (Dx )pk (x, Dx )f˜k kLq ≤ C2sj+min(j,k)m−|j−k|l−k(s+m) 2k(s+m) kf˜k kLq ,
where
2sj+min(j,k)m−|j−k|l−k(s+m)
(
2j(s−l)+k(l−s) = 2−|j−k|(l−s)
=
2−k(m+s+l)+j(m+s+l) = 2−|j−k|(m+s+l)
if j ≥ k,
if j < k.
Choosing l ∈ N so large that l − s ≥ 1 and m + s + l ≥ 1,
sj
2 kϕj (Dx )p(x, Dx )f kLq ≤ C
∞
X
2−|j−k| 2k(s+m) kf˜k kLq
k=0
Now let aj = 2−|j| , bj = 2(s+m)j kf˜j kLq for j ∈ N0 and aj = bj = 0 for j ∈ Z \ N0 .
Then
s
kp(x, Dx )f kBqr
≤ C k(aj )j∈Z ∗ (bj )j∈Z k`r ≤ Ck(aj )j∈Z k`1 k(bj )j∈Z k`r
= Ck(2(s+m)j f˜j )j∈N k`r (N ,Lq (Rn )) ≤ Ckf k s+m ,
0
0
Bqr
where the sequence (cj )j∈Z = (aj )j∈Z ∗(bj )j∈Z is the convolution of (aj )j∈Z and (bj )j∈Z
defined by
∞
X
aj−k bk .
cj =
k=−∞
Here we have used the discrete version of Young’s inequality k(aj )j∈Z ∗ (bj )j∈Z k`r ≤
k(aj )j∈Z k`1 k(bj )j∈Z k`r , which can be proved in the same way as for the usual convolution using Hölder’s inequality.
For completeness we state the following theorem, which shows that the Besov spaces
can be considered as generalization of Hölder spaces:
THEOREM 5.27 Let 1 ≤ q < ∞ and let 0 < s < 1. Then
1q
Z Z
|f (x) − f (y)|q
0
q
kf kBqq
kf kq +
dxdy
s =
|x − y|n+sq
Rn Rn
s
is an equivalent norm on Bqq
(Rn ).
See [BL76] for the proof.
73
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