Pseudodifferential Operators

Helmut Abels

February 27, 2011

Contents1 Introduction 2

2 Summary of Some Basic Results 52.1 Functions on Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Lebesgue Integral and Lp-spaces . . . . . . . . . . . . . . . . . . . . . 6

3 Fourier Transformation 103.1 Definition and Basic Properties . . . . . . . . . . . . . . . . . . . . . 103.2 Rapidly decreasing functions – S(Rn) . . . . . . . . . . . . . . . . . . 123.3 Ex-course: Fréchet spaces . . . . . . . . . . . . . . . . . . . . . . . . 143.4 Inverse Fourier transformation and Parseval’s formula . . . . . . . . . 173.5 Tempered Distributions and Fourier Transformation . . . . . . . . . . 20

4 Basic Calculus of Pseudodifferential Operators on Rn 234.1 Symbol Classes and Basic Properties . . . . . . . . . . . . . . . . . . 234.2 Composition of Pseudodifferential Operators: Motivation . . . . . . . 274.3 Oscillatory Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 284.4 Double Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324.5 Composition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354.6 Application: Elliptic Pseudodifferential Operators and Parametrices . 364.7 Boundedness on C∞b (Rn) and Uniqueness of the Symbol . . . . . . . . 394.8 Adjoints of Pseudodifferential Operators and Operators in (x,y)-Form 414.9 L2-Continuity and Bessel Potential Spaces . . . . . . . . . . . . . . . 444.10 Summary and Final Remarks . . . . . . . . . . . . . . . . . . . . . . 49

5 Extensions and Applications 515.1 Resolvents and Parameter-Elliptic Differential Operators . . . . . . . 515.2 Kernel Representation of a Pseudodifferential Operator . . . . . . . . 565.3 Coordinate Transformations and Pseudodifferential Operators on Man-

ifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 625.4 Boundedness on Hölder-Zygmund and Besov Spaces . . . . . . . . . . 68

1

1 IntroductionOne of the leading ideas in the theory of pseudodifferential operators is to reduce thestudy of properties of a linear differential operator

P =∑|α|≤m

cα(x)∂αx ,

which is a polynomial in the derivatives ∂x = (∂x1 , . . . , ∂xn) with constants cα de-pending on x, to its symbol

p(x, ξ) =∑|α|≤m

cα(x)(iξ)α,

which is a polynomial in the phase variable ξ ∈ Rn with constants depending on thespace variable x. Then the question arises how properties of the differential operatorp(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

F [f ](ξ) := f(ξ) :=

∫Rne−ix·ξf(x)dx, ξ ∈ Rn, (1.1)

which is defined for suitable functions f : Rn → C. Here (1.1) can be interpreted asscalar product of the function f(x) and the function x 7→ eix·ξ:

F [f ](ξ) =

∫Rnf(x)eix·ξdx = (f, ei.·ξ)L2(Rn)

for fixed ξ ∈ Rn, where (., .) denotes the scalar product on L2(Rn). Hence f(ξ)can be interpreted as the contribution of the complex (multidimensional) oscillationx 7→ eix·ξ to the function, where the phase variable ξ = (ξ1, . . . , ξn) ∈ Rn describesthe frequency of the oscillation.

Knowing the Fourier transform g(ξ) = f(ξ), the function f can be reconstructedwith the aid of the inverse Fourier transformation

F−1[g](x) :=1

(2π)n

∫Rneix·ξg(ξ)dξ;

indeed it holds that

F−1[F [f ]] ≡ F−1[f ] = f (Inversion formula).

Here F−1[g] can be interpreted as infinitesimal linear combination of x 7→ eix·ξ withcoefficients g(ξ) (and a correcting factor 1/(2π)n).

2

Using the inversion formula,

Pu =∑|α|≤m

cα(x)∂αx1

(2π)n

∫Rneix·ξu(ξ)dξ

=∑|α|≤m

cα(x)1

(2π)n

∫Rn

(iξ)αeix·ξu(ξ)dξ

=1

(2π)n

∫Rneix·ξp(x, ξ)u(ξ)dξ,

where we have used ∂xjeix·ξ = iξjeix·ξ and therefore ∂αx eix·ξ = (iξ)αeix·ξ. This moti-

vates 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

Pu = F−1[p(ξ)u] = F−1[p(ξ)F [u]].

Hence application of P to u acts as multiplication of u(ξ) by the symbol p(ξ). Thissuggests that inversion of P corresponds to multiplication 1/p(ξ) on the side of theFourier transformed functions. Therefore we define

Qf := F−1[p(ξ)−1f(ξ)] =1

(2π)n

∫Rneix·ξ

1

p(ξ)f(ξ)dξ

assuming that p(ξ) 6= 0 for all ξ ∈ Rn. Then F [Qf ] = p(ξ)−1f(ξ) since FF−1 = I(inversion formula) and therefore

PQf = F−1[p(ξ)F [Qf ]] = F−1[p(ξ)p(ξ)−1f ] = F−1[f(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 pseudo-

differential operators, which are defined as

q(x,Dx)f :=1

(2π)n

∫Rneix·ξq(x, ξ)f(ξ)dξ

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

Qf :=1

(2π)n

∫Rneix·ξ

1

p(x, ξ)f(ξ)dξ,

thenPQf =

1

(2π)n

∫RnP

(eix·ξ

1

p(x, ξ)

)f(ξ)dξ.

3

Because of the product rule,

P

(eix·ξ

1

p(x, ξ)

)= (Peix·ξ)

1

p(x, ξ)+ r(x, ξ) = p(x, ξ)

1

p(x, ξ)+ r(x, ξ) = 1 + r(x, ξ)

where the remainder term r(x, ξ) consists of terms where 1p(x,ξ)

is differentiated withrespect to x at least once. Hence

PQf = 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 manypurposes.

So far all considerations have been made formally. To make them precise and toprove everything rigorously will be a main part of the lecture.

The structure of the lecture is as follows: First we recall some basic facts fromcalculus in Rn and Lebesgue integration in Section 2. Then, in Section 3, we studysome basic properties of the Fourier transformation. In particular, we prove theinversion formula for functions in L2(Rn). The main part of the lecture consists ofSection 4 in which basic properties of pseudodifferential operators are studied. Moreprecisely, we study compositions, adjoints, and boundedness of pseudodifferentialoperators on certain function spaces including L2(Rn) and L2-Sobolev spaces. Theresults will be applied to elliptic differential and pseudodifferential equations.

4

2 Summary of Some Basic Results

2.1 Functions on Rn

In this section we briefly summerize some known facts from the basic courses oncalculus in Rn.

In the following N will denote the set of all natural numbers zero not includedand 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 some-times marked with an index as e.g. Cα to denote that C depends on α. In sequencesof inequalities all constants will simply be denoted by C although they may changefrom 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) ∈ Nn

0 . 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 ∈ Nn

0 wedefine xα = xα1

1 · . . . · xαnn . Then xα is a polynomial of degree |α| and an arbitrarypolynomial p : Rn → C of order m ∈ N0 can be written as

p(x) =∑|α|≤m

cαxα

with coefficients cα ∈ C. Here∑|α|≤m denotes summation with respect to all multi-

indices α ∈ Nn0 with length |α| ≤ m. Moreover, if α, β ∈ Nn

0 , we write α ≤ β if andonly if αj ≤ βj for all j = 1, . . . , n.

The so-called binomial coefficients are defined by(α

β

)=

α!

β!(α− β)!

if β ≤ α and(αβ

)= 0 otherwise. Then it is easy to check that(

α

β

)=

(α1

β1

)· . . . ·

(αnβn

).

Using this notation and the fundamental relation(n

k

)+

(n

k + 1

)=

(n+ 1

k + 1

)for n, k ∈ N0, one can prove by induction that

(x+ y)α =∑β≤α

(α

β

)xβyα−β

for all x, y ∈ Rn, which generalizes the well-known binomial formula in one-dimension.

5

Recall that the space Ck(Rn) consists of all k-times differentiable functions f : Rn →C or f : Rn → R with continuous derivatives up to order k and C∞(Rn) =

⋂k∈NC

k(Rn).If α ∈ Nn

0 and u ∈ C |α|(Rn), then we define

∂αxu(x) =∂|α|

∂α1x1 . . . ∂αnxn

u(x).

If u depends on several variables, say x and y, ∂αxu and ∂αy u denote the derivativedefined above with respect to x and y, respectively.

Using the product rule of differentiation, one can prove the following Leibniz’sformula:

∂αx (uv)(x) =∑β≤α

(α

β

)(∂βu)(x)(∂α−βv)(x) (2.1)

for all α ∈ Nn0 and u, v ∈ C |α|(Rn).

Finally, we recall Taylor’s formula:

THEOREM 2.1 (Taylor’s formula) Let u ∈ Ck(Rn), k ∈ N. Then for any xand y ∈ Rn one has

u(x+ y) =∑|α|<k

yα

α!∂αxu(x) +

∑|α|=k

kyα

α!

∫ 1

0

(1− t)k−1∂αxu(x+ ty)dt

Proof: See basic course or [Ray91, Chapter 1, Theorem 1.1].

Finally, we denote by Ckb (Rn), k ∈ N0, the space of all f ∈ Ck(Rn) with bounded

derivatives up to order k, i.e., ∂αf is bounded for all |α| ≤ k. Moreover, C∞b (Rn) :=⋂∞k=0C

kb (Rn). It is well-known from basic courses in functional analysis that Ck

b (Rn)equipped with the norm

‖f‖Ckb = sup|α|≤k

supx∈Rn|∂αf(x)|

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 ismeasurable and ∫

Rn|f(x)|dx <∞. (2.2)

Note that |f | ≥ 0 is measurable if f is measurable and for a non-negative measurablefunction g : Rn → [0,∞] there is always a well-defined

∫Rn g(x)dx, which is possibly

infinite. Moreover, if (2.2) holds, f+ := maxf, 0 and f− : max−f, 0 are non-negative functions with finite integral and∫

Rnf(x)dx :=

∫Rnf+(x)dx−

∫Rnf−(x)dx

6

is well-defined. More generally, a function f : Rn → C is Lebesgue-integrable if andonly if Re f and Im f are Lebesgue-integrable. Then∫

Rnf(x)dx :=

∫Rn

Re f(x)dx+ i

∫Rn

Im f(x)dx.

The space of all Lebesgue-integrable functions f : Rn → C is denoted by L1(Rn). Itis a linear vector space, which can be normed by

‖f‖1 :=

∫Rn|f(x)|dx.

L1(Rn) is complete with respect to the norm ‖.‖1 if functions which differ on a set ofmeasure zero are identified.

In the following almost all functions will be at least continuous, in most cases eveninfinitely 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 wedo not have to pay attention on the possible null-sets on which the functions possiblydiffer 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

limk→∞

fk(x) = f(x) a.e.,

|fk(x)| ≤ g(x) a.e.,

where g ∈ L1(Rn). Then f ∈ L1(Rn) and

limk→∞

∫Rnfk(x)dx =

∫Rnf(x)dx.

As an application of Lebesgue’s Theorem on Dominated Convergence one can provethe following two theorems on parameter-dependent integrals, which will often beused 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 andall t ∈ U .

7

Then the function

g(t) :=

∫Rnf(x, t)dx, t ∈ U,

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 |∂tf(x, t)| ≤ F (x).

Theng(t) :=

∫Rnf(x, t)dx

is a differentiable function on I and for all t ∈ I

g′(t) =

∫Rn∂tf(x, t)dx.

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 f ∈ 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∫

Rk×Rlf(x, y)d(x, y) =

∫Rk

(∫Rlf(x, y)dy

)dx. (2.3)

Conversely, if f(x, .) ∈ L1(Rl) for almost all x ∈ Rk and x 7→∫Rk f(x, y)dy ∈ L1(Rk),

then f ∈ L1(Rk × Rl) and (2.3) holds.

The following simple lemma will be essential

Lemma 2.6 Let s > n. Then 〈x〉−s ∈ L1(Rn) and (1 + |x|)−s ∈ L1(Rn).

Proof: The lemma can easily be proved e.g. by using polar coordinates. An alter-native approach can be found in [Ray91, Lemma 1.3].

We will also use the change-of-variable theorem in the following case:∫Rnf(Φ(x))| detDΦ(x)|dx =

∫Rnf(y)dy

8

for all f ∈ L1(Rn) and C1-diffeomorphism Φ: Rn → Rn. In particular, we have∫Rnf(Ax+ b)| detA|dx =

∫Rnf(y)dy

for all f ∈ L1(Rn), b ∈ Rn, and A ∈ Rn×n with detA 6= 0.We will also use with spaces Lp(Rn), 1 ≤ p <∞, which consist of all measurable

functions f : Rn → C such that

‖f‖p :=

(∫Rn|f(x)|p

) 1p

<∞.

All these spaces are Banach spaces (again identifying functions which coincide almosteverywhere). We will mainly work with L2(Rn) which is also a Hilbert space; its normis given by the scalar product:

‖f‖2 =√

(f, f),

(f, g) ≡ (f, g)L2(Rn) =

∫Rnf(x)g(x)dx.

The L2-scalar product will play a fundamental role.An important subspace of Lp(Rn) is C∞0 (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 C∞0 (Rn) is dense in Lp(Rn), i.e., for everyf ∈ Lp(Rn) there is a sequence fk ∈ C∞0 (Rn) such that limk→∞ ‖f − fk‖p = 0. Forshort: C∞0 (Rn)

‖.‖p= 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

f(ξ) := F [f ](ξ) :=

∫Rne−ix·ξf(x)dx (3.1)

for ξ ∈ Rn. Since |e−ix·ξf(x)| = |f(x)| for all ξ ∈ Rn, we have e−ix·ξf(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)→ C0b (Rn) is a linear mapping such that

‖F [f ]‖C0b (Rn) ≤ ‖f‖L1(Rn).

2. If f : Rn → C is a continuously differentiable function such that f ∈ L1(Rn)and ∂jf ∈ L1(Rn), then

F [∂xjf ] = iξjF [f ] = iξj f . (3.2)

3. If f ∈ L1(Rn) such that xjf ∈ L1(Rn), then f(ξ) is continuously partial differ-entiable with respect to ξj and

∂ξj f(ξ) = F [−ixjf(x)]. (3.3)

4. Let f ∈ L1(Rn) and (τyf)(x) := f(x + y), y ∈ Rn, denote the translation of fby y. Then F [τyf ](ξ) = eiy·ξf(ξ) for all ξ ∈ Rn.

5. If f, g ∈ L1(Rn), thenf(ξ)g(ξ) = F [f ∗ g] ,

wheref ∗ g(x) :=

∫Rnf(x− y)g(y)dy

denotes the convolution of f and g.

6. Let f ∈ L1(Rn) and let (ρεf)(x) := f(εx), ε > 0, denote the dilation of f byε. Then

F [ρεf ](ξ) = ε−nf(ξ/ε) = ε−n(ρε−1 f)(ξ).

Proof:

10

1. First of all

‖f‖∞ = supξ∈Rn|f(ξ)| ≤ sup

ξ∈Rn

∫Rn

∣∣e−ix·ξf(x)∣∣ dx = ‖f‖1.

Hence f is bounded. Let g(x, ξ) = e−ix·ξf(x). Then ξ 7→ g(x, ξ) is continuousfor 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 impliesthe 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

‖f ∗ g‖1 ≤∫Rn

∫Rn|f(x− y)g(y)|dydx =

∫Rn×Rn

|f(z)g(y)|d(y, z) = ‖f‖1‖g‖1.

by Fubini’s theorem and the change-of-variable theorem. Hence

F [f ∗ g](ξ) =

∫Rn

∫Rne−ix·ξf(x− y)g(y)dydx

=

∫Rn

(∫Rne−ix·ξ(τ−yf)(x)dx

)g(y)dy

=

∫Rne−iy·ξf(ξ)g(y)dy = f(ξ)g(ξ),

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 fromthe point of view of differential equations. Because of the factor i in (3.2), we define

Dxj =1

i∂xj , Dx = (Dx1 , . . . , Dxn).

Then (3.2) is equivalent to

F [Dxjf ] = ξjF [f ] = ξj f .

Moreover, if we express a linear differential operator P with constant coefficients as

Pf =∑|α|≤m

cαDxf

11

with some constants cα ∈ C, then

F [Pf ] =∑|α|≤m

cαξαf .

Hence application of a linear differential operator P to f corresponds to multiplicationof f with the polynomial

p(ξ) :=∑|α|≤m

cαξα.

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 ∂jf ∈ L1(Rn) for allj = 1, . . . , n, then f(ξ) and ξj f(ξ) are bounded functions. Hence

(1 + |ξ|)|f(ξ)| ≤ C ⇔ |f(ξ)| ≤ C

1 + |ξ|,

which means that f(ξ) decays as |ξ|−1 as |ξ| → ∞.More generally, if f ∈ Ck(Rn) such that ∂αx f ∈ L1(Rn) for all |α| ≤ k, then one

can show in the same way that

|f(ξ)| ≤ C

(1 + |ξ|)k.

For short: Differentiability of f implies a polynomial decay of f as |ξ| → ∞.On the other hand, if (1 + |x|)kf(x) ∈ L1(Rn) for some k ∈ N, we can apply

Lemma 3.1.3 successively to conclude that f ∈ Ck(Rn). Hence faster decay of f(x)as |x| → ∞ yields to a higher differentiability of f .

3.2 Rapidly decreasing functions – S(Rn)

Let f ∈ C∞0 (Rn). Then ∂αx f ∈ L1(Rn) for all α ∈ N0. As seen in Remark 3.2, f(ξ)decays faster than any power (1 + |x|)−k, k ∈ N. Moreover, (1 + |x|)kf ∈ L1(Rn) forall k ∈ N. Hence f ∈ Ck

b (Rn) for all k ∈ N, i.e., f ∈ C∞b (Rn) =⋂∞k=1 C

kb (Rn). But in

general f(ξ) does not have compact support, cf. Exercise 1. If f ∈ C∞0 (Rn), then fbelongs to the following function space.

Definition 3.3 The space S(Rn) of all rapidly decreasing smooth functions is theset of all smooth f : Rn → C such that for all α ∈ Nn

0 and N ∈ N0 there is a constantCα,N such that

|∂αx 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|α|+|β|≤m

supx∈Rn|xα∂βxf(x)|

12

Obviously, C∞0 (Rn) ⊂ S(Rn). The inclusion is strict since f(x) = e−|x|2 ∈ S(Rn).

Moreover, S(Rn) ⊂ C∞b (Rn).

Remark 3.4 At first sight, it may be more natural to use

|f |′m,S := supk+|α|≤m

supx∈Rn

(1 + |x|)k|∂αf(x)|

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 transforma-tion, which will be demonstrated in the proof of the next lemma. Moreover, it doesnot matter if we use the semi-norms |.|m,S or |.|′m,S – the semi-norms are equivalentin the following sense: For every m ∈ N0 there is an k(m) ∈ N0 such that

|f |′m,S ≤ Cm|f |k(m),S and |f |m,S ≤ C ′m|f |′k(m),S

for all f ∈ S(Rn). (Actually, in this special case we can simply choose k(m) = m.)Finally, we note that we can replace (1 + |x|) in (3.4) by 〈x〉 := (1 + |x|2)

12 since

〈x〉 ≤ (1 + |x|) ≤√

2〈x〉.

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, |f |m,S ≤ Cm|f |m+n+1,Sfor all m ∈ N0 where Cm depends only on n and m.

Proof: First of all, if f ∈ S(Rn),

‖f‖1 =

∫Rn

(1 + |x|)−n−1((1 + |x|)n+1|f(x)|

)dx

≤ C

∫Rn

(1 + |x|)−n−1dx|f |n+1,S = C|f |n+1,S ,

where C depends only on the dimension n. Thus

|f |0,S = ‖f‖∞ ≤ ‖f‖1 ≤ C|f |n+1,S (3.5)

by Lemma 3.1.1 and the previous estimate. Because of Lemma 3.1.2/3,

ξαDβξ f(ξ) = F

[Dαx

(xβf(x)

)].

Hence‖ξαDβ

ξ f‖∞ ≤ C|Dαx

(xβf(x)

)|n+1,S

by (3.5). Using Leibniz’s formula (2.1),

Dαx

(xβf(x)

)=∑γ≤α

(α

γ

)(Dγ

xxβ)(Dα−γ

x f(x)).

13

Since Dγxx

β is polynomial of degree less than |β|,

|Dαx

(xβf(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|f |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) forall f ∈ S(Rn).

Example 3.6 Let f(x) = e−|x|2/2, x ∈ Rn. Then f(ξ) = (2π)n/2e−|ξ|

2/2.

Proof: Since f(x) = e−x21/2 · · · e−x2n/2 and e−ix·ξ = e−ix1ξ1 · · · e−ixnξn ,

f(ξ) =

∫Re−ix1ξ1e−x

21/2dx1 · · ·

∫Re−ix1ξ1e−x

2n/2dxn = g(ξ1) · · · g(ξn)

where g(x) = e−x2/2, x ∈ R. Hence it is sufficient to consider the case n = 1.

Because of Lemma 3.1.3, g(ξ) is continuously differentiable and g′(ξ) = F [−ixg(x)].Moreover, −xg(x) = d

dxe−x

2/2 = g′(x). Therefore, using Lemma 3.1.2,

g′(ξ) = iF [g′(x)] = −ξg(ξ), ξ ∈ R,

and g(0) =∫R e−x2/2dx =

√2π. Hence g is uniquely determined by the letter initial

value problem, which has the unique solution g(ξ) =√

2πe−ξ2/2.

An alternative proof using the Cauchy integral formula can be found in [Ray91,Example 1.7].

Exercise 1 Let f ∈ C∞0 (R) such that supp f ⊆ BR(0). Prove that

f(ξ) =

∫ R

−Reiξxf(x)dx, ξ ∈ C,

is a holomorphic function in C. Moreover, |f(ξ)| ≤ CeR| Im ξ|.In particular, f(ξ), ξ ∈ R, is real analytic and supp f is not compact unless

f(ξ) ≡ 0.

3.3 Ex-course: Fréchet spaces

Throughout the lecture, we will work with many different spaces of smooth functionswith certain properties. They all have in common that they cannot be normed tobecome a Banach space (with their natural topology).

14

As an example we consider the space C∞b (Rn) of all smooth and bounded functionsf : Rn → C with bounded derivatives. Then

C∞b (Rn) =∞⋂k=0

Ckb (Rn),

where Ckb (Rn) is the space of all k-times differentiable functions f : Rn → C with

bounded derivatives up to order k. In contrast to C∞b (Rn) the spaces Ckb (Rn), k ∈ N0,

are Banach spaces equipped with the norm

‖f‖Ckb = sup|α|≤k

supx∈Rn|∂αf(x)|.

Since in spaces smooth functions, all (infinitely many) derivatives have to be con-trolled, in general they cannot be normed by a single norm such that the space iscomplete. But these spaces can be “normed” by an infinite sequence of (semi-)norms,e.g. f ∈ C∞b (Rn) if and only if ‖f‖Ckb 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) = 0for all m ∈ N.

Definition 3.8 A linear space V is called a Fréchet space if there is a sequence ofsemi-norms (ρm)m∈N satisfying the following conditions:

1. (ρm)m∈N is an increasing sequence: ρ1(f) ≤ ρ2(f) ≤ . . . ≤ ρm(f) ≤ . . . for allf ∈ V .

2. (ρm)m∈N separates points : for any f 6= 0 there is anm ∈ 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 ‖.‖X is a Fréchet space with (semi-)norms ρm = ‖.‖X for all m ∈ N.

Exercise 2 Prove that C∞b (Rn) equipped with the norms ρm = ‖.‖Cmb is a Fréchetspace. – It can be used that Ck

b (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 mainlyremains to check the completeness of S(Rn), which can be done by using the com-pleteness of C∞b (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) :=∞∑m=1

2−mρm(f − g)

1 + ρm(f − g). (3.6)

Exercise 3 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 themetric sense. Since every metric space carries a natural topology which is defined bythe neighborhood basis

Ul(f0) =

f ∈ V : d(f − f0) <

1

l

, l ∈ N, f0 ∈ V,

every Fréchet space is a topological space. Equipped with this topology, the opera-tions of multiplication by scalars and addition of vectors are continuous, i.e., V is alinear topological space.

Moreover, every Fréchet space is locally convex, i.e. there is a neighborhoodbasis at the origin that consists of convex sets. Finally, the Fréchet spaces can becharacterized as the locally convex linear topological space that can be equipped witha 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 isa k = k(m) ∈ N and a Cm such that

τm(Tf) ≤ Cmρk(m)(f) for all f ∈ V.

THEOREM 3.12 Let V and W be Fréchet spaces and T : V → W be a linear map-ping. Then T is bounded if and only if T is continuous in the sense that limj→∞ fj = fimplies limj→∞ Tfj = Tf .

In particular, if W is Banach space, T : V → W is continuous if and only if

‖Tf‖W ≤ Cρm(f) for all f ∈ V

with some C > 0 and m ∈ N.

16

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 thereexists a k(m) such that τm(Tf) ≤ Cmρk(m)(f), f ∈ V . Hence

τm(Tfj − Tf) = τm(T (fj − f)) ≤ Cmρk(m)(fj − f)→ 0 as j →∞

for every m ∈ N, which means limj→∞ Tfj = Tf . 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 isa j0 with

τm0(Tfj0) > Cρk(fj0).

In particular, we can choose a subsequence fjk , k ∈ N, such that

τm0(Tfjk) > kρk(fjk).

Then fk := fjk/τm0(Tfjk), k ∈ N, is a sequence such that

ρm(fk) ≤ ρk(fk) <1

kfor all m ≤ k ∈ N.

Thus limk→∞ ρm(fk) = 0 for all m ∈ N, i.e., limk→∞ fk = 0. Hence limk→∞ T fk = 0because of the continuity of T . On the other hand τm0(T fk) = 1 for all k ∈ N whichcontradicts limk→∞ fk = 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 boundedfor 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) isinvertible and that its inverse is given by

F−1[g](x) :=1

(2π)n

∫Rneix·ξg(ξ)dξ,

which is well-defined for all g ∈ L1(Rn). – Note that

F−1[g](x) = (2π)nF [g](−x). (3.7)

If f(x) = e−|x|2/2 is the function discussed in Example 3.6, then

F−1[f(ξ)] = (2π)−n/2F[e−|ξ|

2/2]

(−x) = e−|x|2/2 = f(x).

Hence F−1[f ] = f for this special f .

17

Lemma 3.13 (Inversion Formula)Let f ∈ S(Rn). Then f(x) = F−1[f ](x) for all x ∈ Rn. In particular, F : S(Rn) →S(Rn) is a linear isomorphism.

Proof: First of all,

F−1[f ](x) =1

(2π)n

∫Rn

(∫Rnei(x−y)·ξf(y)dy

)dξ.

Since ei(x−y)·ξf(y) 6∈ L1(R2n) as function in (y, ξ), we cannot apply Fubini’s theoremdirectly. Therefore we introduce a factor ψε(ξ) := e−ε

2|ξ|2/2, ε > 0, which assuresabsolute integrability. Since limε→0 ψε(ξ) = 1 for all x ∈ Rn, we have by Lebesgue’sand Fubini’s theorem

F−1[f ](x) = limε→0

1

(2π)n

∫Rn

(∫Rnei(x−y)·ξψε(ξ)f(y)dy

)dξ

= limε→0

1

(2π)n

∫R2n

ei(x−y)·ξe−ε2|ξ|2f(y)d(y, ξ).

Using the substitution ξ = η/ε and y = x+ εz,

1

(2π)n

∫R2n

ei(x−y)·ξe−ε2|ξ|2/2f(y)d(y, ξ) =

1

(2π)n

∫Rn

(∫Rne−iz·ηe−|η|

2/2dη

)f(x+ εz)dz

=1

(2π)n/2

∫Rne−|z|

2/2f(x+ εz)dz.

Since f is continuous, we get

limε→0

1

(2π)n/2

∫Rne−|z|

2/2f(x+ εz)dz = f(x)1

(2π)n/2

∫Rne−|z|

2/2dz = f(x)

by Lebesgue’s theorem. Hence F−1[f ](x) = f(x), which proves the lemma.

Remark 3.14 The technique of inserting a rapidly decreasing factor like e−ε2|ξ|2/2,ε > 0, and passing to the limit afterwards will be fundamental in the following. It isthe 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)∫

Rnf(x)g(x)dx =

1

(2π)n

∫Rnf(ξ)g(ξ)dξ. (3.8)

In particular,

‖f‖22 =

1

(2π)n‖f‖2

2.

and F extends to an isomorphism F : L2(Rn)→ L2(Rn).

18

Proof: Using Fubini’s theorem, it is easy to check that

1

(2π)n

∫RnF [f ](ξ)g(ξ)dξ =

∫Rnf(x)F−1[g](x)dx.

Because of Lemma 3.13, g = F−1[g], which implies (3.8). In particular, if f = g,

1

(2π)n‖F [f ]‖2

2 = ‖f‖22.

Since C∞0 (Rn) is dense in L2(Rn) and obviously C∞0 (Rn) ⊂ S(Rn) ⊂ L2(Rn), F canbe 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 istrue for all f ∈ S(Rn).

Remark 3.16 Because of the factor (2π)−n in the definition of the inverse Fouriertransformation and in Parseval’s formula (3.8), we introduce the common notationđξ := dξ

(2π)n. Then

F−1[g](x) =

∫Rneix·ξg(ξ)đξ (3.9)

and ∫Rnf(x)g(x)dx =

∫Rnf(ξ)g(ξ)đξ.

In the following all integrals with respect to a phase variable ξ, η, . . . will be takenwith respect to the scaled Lebesgue measure đξ, đη, . . .. All integrals with respectto a space variable x, y, z, . . . will be integrals using the usual Lebesgue measuredx, dy, dz, . . .. Then all constants are part of the measure and we do not have toworry 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)f(ξ)]

is a well-defined bounded operator m(Dx) : L2(Rn) → L2(Rn) if and only if m ∈L∞(Rn). Moreover, ‖m(Dx)‖L(L2(Rn)) = ‖m‖∞.

Proof: If m ∈ L∞, obviously m(ξ)f(ξ) ∈ L2(Rn) for every f ∈ L2(Rn) because ofCorollary 3.15. Moreover,

‖m(Dx)f‖L2(Rn) =1

(2π)n2

‖m(ξ)f‖2 ≤1

(2π)n2

‖m‖∞‖f‖2 ≤ ‖m‖∞‖f‖2.

Hencem(Dx) : L2(Rn)→ L2(Rn) is a bounded linear operator and ‖m(Dx)‖L(L2(Rn)) ≤‖m‖∞.

19

For the converse implication it is sufficient to prove that a multiplication operatorM : L2(Rn) → L2(Rn), (Mf)(ξ) = m(ξ)f(ξ) is bounded if and only if m ∈ L∞(Rn)and ‖M‖L(L2(Rn)) = ‖m‖∞. We skip the details since we will not use the converseimplication. See also [Ste70, Chapter II, Proposition 2].

Examples 3.18 1. If m(ξ) = F [k], where k ∈ L1(Rn) then m ∈ C0b (Rn) ⊂

L∞(Rn) due to Lemma 3.1. Moreover, m(Dx)f = k ∗ f and

‖m(Dx)f‖2 ≤ ‖m‖∞‖f‖2 ≤ ‖k‖1‖f‖2.

(Note that the latter inequality is also a simple consequence of the well-knownestimate ‖k ∗ f‖p ≤ ‖k‖1‖f‖p for 1 ≤ p ≤ ∞.)

2. Let mj(ξ) =iξj|ξ| , j = 1, . . . , n. Then ‖mj‖∞ = 1 and Rj := mj(Dx) : L2(Rn)→

L2(Rn) is a bounded operator. Rj is called Riesz operator. It can be shownthat

Rjf = limε→0

cn

∫Rn

xj − yj|x− y|n+1

f(y)dy = ”kj ∗ f”,

where kj(z) = zj/|z|n+1, cf. [Èsk81, Chapter I, Example 3.4]. Note that thekernel kj 6∈ L1(Rn). These operators are typical examples of singular integraloperators. If n = 1,m1(ξ) = i sign ξ andR1 is also called Hilbert transformation.

3. The symbol of−∆ is |ξ|2, i.e., −∆f = F−1[|ξ|2f(ξ)]. Hence formally the inverseof −∆ is given by the operator (−∆)−1f = F−1[f(ξ)/|ξ|2] and therefore

∂j∂k(−∆)−1f = F−1

[iξj|ξ|

iξj|ξ|f(ξ)

]= RjRkf,

where Rj, Rk are the Riesz operator defined above. Hence in some sense∂j∂k(−∆)−1f ∈ L2(Rn) if f ∈ L2(Rn); but the derivatives do not have exist inthe 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 ′(Rn) := (S(Rn))′ – thespace of linear and bounded functionals f : S(Rn) → C. A sequence fk ∈ S ′(Rn)converges to f ∈ S ′(Rn) if and only if limk→∞〈fk, ϕ〉 = 〈f, ϕ〉 for all ϕ ∈ S(Rn).

Here 〈f, ϕ〉 := f(ϕ), ϕ ∈ S(Rn) denotes the duality product.

Remarks 3.20 1. If f : Rn → C is a measurable function such that 〈x〉−Nf(x) ∈L1(Rn) for some N ∈ N0, then f is naturally identified with a tempered distri-bution Ff ∈ S(Rn)′ defined by

〈Ff , ϕ〉 :=

∫Rnf(x)ϕ(x)dx, ϕ ∈ S(Rn).

20

Moreover,

|〈Ff , ϕ〉| ≤ ‖〈x〉−Nf‖1|ϕ|′N,S ≤ C‖〈x〉−Nf‖1|ϕ|N,S , ϕ ∈ S(Rn),

cf. Remark 3.4. In the following we will identify the function f with its asso-ciated 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 onlyif there is an m ∈ N0 and a constant C > 0 such that

|〈f, ϕ〉| ≤ C|ϕ|m,S , ϕ ∈ S(Rn).

Example 3.21 The famous delta distribution δ ∈ S ′(Rn) is defined by

〈δ, ϕ〉 := ϕ(0), ϕ ∈ S(Rn).

One of the most important properties of distributions is that they can be differenti-ated infinitely many times (in a generalized sense).

Definition 3.22 Let f ∈ S ′(Rn). Then the distributional derivative ∂αx f ∈ S ′(Rn)of f is the distribution defined by

〈∂αx f, ϕ〉 := (−1)|α|〈f, ∂αxϕ〉 for ϕ ∈ S(Rn).

Remarks 3.23 1. Recall that the adjoint T ′ : Y ′ → X ′ of a linear operatorT : X → Y is defined by 〈T ′y′, x〉 := 〈y′, Tx〉 for y′ ∈ Y ′ and x ∈ X. Hence thederivative ∂αx : S ′(Rn)→ S ′(Rn) is the adjoint of (−1)|α|∂αx : S(Rn)→ S(Rn).

2. If f : Rn → C is a k-times continuously differentiable function with 〈x〉−Nf ∈L1(Rn), then f can be naturally considered as a tempered distribution, cf.Remark3.20.1, Moreover, the distributional derivatives up to order k coincideswith the usual derivative, i.e.

〈∂αx f, ϕ(x)〉 = (−1)|α|∫Rnf(x)∂αxϕ(x)dx =

∫Rn∂αx f(x)ϕ(x)dx

for all |α| ≤ k due to integration by parts.

3. Since every measurable function f : Rn → C such that 〈x〉−Nf(x) ∈ L1(Rn)for some N ∈ N0 can be considered as a distribution, it can be differentiatedinfinitely many times in the sense of distributions although it may even not becontinuous.

Examples 3.24 1. Let f be the Heavyside function, i.e. f(x) = 1 for x ≥ 0 andf(x) = 0 for x < 0. Then f ∈ S ′(R) and the distributional derivative f ′ is

〈f ′, ϕ〉 = −∫Rf(x)ϕ′(x)dx = −

∫ ∞0

ϕ′(x)dx = ϕ(0)

for all ϕ ∈ S(R). Hence f ′ = δ is the delta distribution.

21

2. By definition the distributional derivative of the delta distribution is

〈∂αx δ, ϕ〉 = (−1)|α|〈δ, ∂αxϕ〉 = (−1)|α|(∂αxϕ)(0).

Definition 3.25 Let f ∈ S ′(Rn). The Fourier transform F [f ] of f is defined as thedistribution

〈F [f ], ϕ〉 = 〈f,F [ϕ]〉 for ϕ ∈ S(Rn).

Proposition 3.26 The Fourier transformation is a continuous linear mapping F : S ′(Rn)→S ′(Rn). Moreover, F is a bijection on S ′(Rn).

Proof: Since F : S(Rn)→ S(Rn) is a continuous linear operator and

〈F [f ], ϕ〉 = 〈f,F [ϕ]〉 = (f F)(ϕ),

F [f ] = f F : S(Rn) → C is a continuous linear operator. Moreover, if fk → f inS ′(Rn) as k →∞, then

〈F [fk], ϕ〉 = 〈fk,F [ϕ]〉 → 〈f,F [ϕ]〉 = 〈F [f ], ϕ〉

as k →∞ for all ϕ ∈ S(Rn). Hence F : S ′(Rn)→ S ′(Rn) is continuous.Finally, if we define F−1 : S ′(Rn)→ S ′(Rn) by

〈F−1[f ], ϕ〉 := 〈f,F−1[ϕ]〉, ϕ ∈ S(Rn),

F−1 has the same properties as F and obviously F−1[F [f ]] = f for all f ∈ S ′(Rn).

Remark 3.27 As differentiation the Fourier transformation is defined by duality,i.e., F : S ′(Rn)→ S ′(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 simplycalculated:

〈F [δ], ϕ〉 = 〈δ,F [ϕ]〉 = F [ϕ](0) =

∫Rnϕ(x)dx = 〈1, ϕ〉

for all ϕ ∈ S(Rn). Hence F [δ] = 1.

22

4 Basic Calculus of Pseudodifferential Operators onRn

4.1 Symbol Classes and Basic Properties

We now introduce the basic pseudodifferential symbol class Sm1,0. There are manyother more general or modified symbol classes, which are used in the literature andresearch for different purposes. But the following symbol class is the most simple andmost common. It is a natural symbol class that contains the symbols of differentialoperators with smooth coefficients.

Definition 4.1 Let m ∈ R, n,N ∈ N. Then Sm1,0(RN ×Rn) is the vector-space of allsmooth functions p : RN × Rn → C such that

|DαξD

βxp(x, ξ)| ≤ Cα,β(1 + |ξ|)m−|α| (4.1)

holds for all α ∈ Nn0 , β ∈ NN

0 , where Cα,β is independent of x ∈ RN , ξ ∈ Rn. Thefunction p is called pseudodifferential symbol andm is called the order of p. Moreover,

S∞1,0(RN × Rn) :=⋃m∈R

Sm1,0(RN × Rn) and S−∞1,0 (RN × Rn) :=⋂m∈R

Sm1,0(RN × Rn).

For short we also write Sm1,0 instead of Sm1,0(RN × Rn).

Remark 4.2 In the following we usually deal with the case N = n. But sometimesit is useful to have defined the symbol classes for general N, n ∈ N.

If p ∈ Sm1,0(Rn × Rn) is a symbol, then

p(x,Dx)f := OP(p)f :=

∫Rneix·ξp(x, ξ)f(ξ)đξ (4.2)

defines the associated pseudodifferential operator, where f : Rn → C is a suitablefunction. If f ∈ S(Rn), then f ∈ S(Rn) and therefore p(x, ξ)f(ξ) ∈ S(Rn) withrespect to ξ for every fixed x ∈ Rn. Therefore the integral in (4.2) exists and p(x,Dx)fis well-defined. In the following we will prove that p(x,Dx) : S(Rn) → S(Rn) is acontinuous mapping. But before we prove this fact, we discuss some examples andmake some simple observations.

Examples 4.3 1. Let p(x, ξ) =∑|α|≤m cα(x)ξα be a polynomial in ξ of order

m ∈ N0 with smooth coefficients cα(x) ∈ C∞b (Rn). Then p ∈ Sm1,0 and

p(x,Dx)f =∑|α|≤m

cα(x)Dαxf

for every f ∈ S(Rn). Hence every linear differential operator with smoothand bounded coefficients is a pseudodifferential operator. In particular theLaplacian ∆ = ∂2

1 +. . .+∂2n is a pseudodifferential operator with symbol −|ξ|2 =

−ξ21 − . . .− ξ2

n.

23

2. The function 〈ξ〉 :=√

1 + |ξ|2 is a pseudodifferential symbol of order 1, seeExercise 5. Since 1+|ξ|2 is the symbol of 1−∆, the associated pseudodifferentialoperator

〈Dx〉f =

∫Rneix·ξ

√1 + |ξ|2f(ξ)đξ

can be considered as the square root of 1−∆. For short: 〈Dx〉 =√

1−∆.

More generally, 〈ξ〉m ∈ Sm1,0 for every m ∈ R and 〈Dx〉m = (1−∆)m2 .

Exercise 5 Let 〈ξ〉 =√

1 + |ξ|2. Prove that 〈ξ〉m ∈ Sm1,0 for every m ∈ R.Hint: Consider the function f(a, x) := (a2 + |x|2)

m2 , where a ∈ R, x ∈ Rn. Use

that f is homogeneous of degree m, i.e., f(ra, rx) = rmf(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 ξof degree m ∈ R for |ξ| ≥ 1, i.e. p(x, rξ) = rmp(x, ξ) for all |ξ| ≥ 1 and r ≥ 1.Moreover, let ∂αξ ∂βxp ∈ C∞b (Rn) w.r.t. x for all α, β ∈ Nn

0 . Prove that p ∈ Sm1,0(Rn ×Rn).

We can define a sequence of semi-norms on Sm1,0, which is related to the family ofinequalities (4.1) in a natural way. Let

|p|(m)k := max

|α|,|β|≤ksupx,ξ∈Rn

|DαξD

βxp(x, ξ)|(1 + |ξ|)−m+|α| (4.3)

for k ∈ N. Here supx,ξ∈Rn |DαξD

βxp(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.It is not difficult to check that Sm1,0 with these semi-norms is a Fréchet space.

Remark 4.4 In the literature the function (1 + |ξ|) in the estimates (4.1) is oftenreplaced by 〈ξ〉 =

√1 + |ξ|2. This can be done without changing the symbol classes

since √1 + |ξ|2 ≤ (1 + |ξ|) ≤

√2√

1 + |ξ|2,

cf. proof of Lemma 4.7 below. Using 〈ξ〉 instead of (1 + |ξ|), the notation becomesa bit shorter.

Proposition 4.5 Let pj ∈ Smj1,0 (Rn × Rn), mj ∈ R, j = 1, 2. Then p(x, ξ) :=

p1(x, ξ)p2(x, ξ) ∈ Sm1+m21,0 (Rn × Rn) and |p|(m1+m2)

k ≤ Ck|p1|(m1)k |p2|(m2)

k , where Ckdepends only on k and n.

The lemma is a simple consequence of the Leibniz formula. The main result of thissection is

24

THEOREM 4.6 Let p ∈ Sm1,0(Rn × Rn), m ∈ R, be a pseudodifferential symbol.Then

p(x,Dx) : S(Rn)→ S(Rn)

is a continuous mapping. More precisely, |p(x,Dx)f |k,S ≤ Ck|p|(m)k |f |m+2(n+1)+k,S for

all k ∈ N and f ∈ S(Rn).

Proof: Since f ∈ S(Rn), f ∈ S(Rn) due to Lemma 3.5. First of all,

supx∈Rn|p(x,Dx)f(x)| ≤ sup

x∈Rn

1

(2π)n

∫Rn〈ξ〉−n−1|〈ξ〉−mp(x, ξ)||〈ξ〉n+m+1f(ξ)| dξ

≤ 1

(2π)n

∫Rn〈ξ〉−n−1dξ|p|(m)

0 ‖〈ξ〉n+m+1f‖∞

≤ C|p|(m)0 |f |m+n+1,S ≤ C|p|(m)

0 |f |m+2n+2,S (4.4)

by Lemma 2.6 and Lemma 3.5, where C depends only on the dimension. Moreover,we note that (4.4) holds for arbitrary p ∈ Sm1,0(Rn × Rn) and m ∈ R — not just forthe one in the assumption of this theorem.

In order to estimate the derivatives, we calculate

∂xj(p(x,Dx)f(x)) = ∂xj

∫eix·ξp(x, ξ)f(ξ)đξ

=

∫eix·ξp(x, ξ)iξj f(ξ)đξ +

∫eix·ξ∂xjp(x, ξ)f(ξ)đξ

= p(x,Dx)(∂jf)(x) + (∂xjp)(x,Dx)f,

where we have applied Theorem 2.4. Hence using (4.4) first with f replaced by ∂jfand then p replaced by ∂xjp, we obtain

supx∈Rn|∂j(p(x,Dx)f(x)| ≤ C

(|p|(m)

0 |∂jf |m+2n+2,S + |∂xjp|(m)0 |f |m+2n+2,S

)≤ C|p|(m)

1 |f |m+2n+3,S . (4.5)

Similarly,

ixjp(x,Dx)f(x) =

∫(∂ξje

ix·ξ)p(x, ξ)f(ξ)đξ

=

∫eix·ξp(x, ξ)∂j f(ξ)đξ +

∫eix·ξ(∂ξjp)(x, ξ)f(ξ)đξ

= p(x,Dx)(ixjf(x)) + (∂ξjp)(x,Dx)f

and therefore

supx∈Rn|xjp(x,Dx)f(x)| ≤ C

(|p|(m)

0 |xjf |m+2n+2,S + |∂ξjp|(m−1)0 |f |m+2n+1,S

)≤ C|p|(m)

1 |f |m+2n+3,S (4.6)

25

by (4.4), where we note that ∂ξjp(x, ξ) is of order m − 1 and |∂ξjp|(m−1)0 ≤ |p|(m)

1 bydefinition of the semi-norms.

Using (4.5) and (4.6), one can easily prove by induction that

supx∈Rn|xα∂βxp(x,Dx)f(x)| ≤ Cα,β|p|(m)

|α|+|β||f |m+2(n+1)+|α|+|β|,S

uniformly in x ∈ Rn for all α, β ∈ Nn0 . Hence |p(x,Dx)f |k,S ≤ Ck|p|(m)

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)Let 〈ξ〉 := (1 + |ξ|2)

12 , ξ ∈ Rn. Then for all s ∈ R

〈ξ〉s ≤ 2|s|〈ξ − η〉|s|〈η〉s, ξ, η ∈ Rn.

Proof: First of all we have

〈ξ〉2 = (1 + |ξ|2) ≤ (1 + |ξ|)2 ≤ (1 + |ξ|)2 + (1− |ξ|)2 = 2(1 + |ξ|2).

Hence〈ξ〉 ≤ (1 + |ξ|) ≤

√2〈ξ〉. (4.7)

First let s ≥ 0. Then the triangle inequality implies

(1 + |ξ|) ≤ (1 + |ξ − η|+ |η|) ≤ (1 + |ξ − η|)(1 + |η|)

and therefore〈ξ〉s ≤ (1 + |ξ − η|)s(1 + |η|)s ≤ 2s〈ξ − η〉s〈η〉s

by (4.7). If s < 0, we can use the previous inequality with interchanged role of ξ andη and s replaced by −s to conclude

〈η〉−s ≤ 2−s〈ξ − η〉−s〈ξ〉−s

which is equivalent to〈ξ〉s ≤ 2|s|〈η〉|s|〈ξ − η〉−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 ∈ S∞1,0(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 symbolsp1(x, ξ) and p2(x, ξ).

The behavior of pseudodifferential operators under composition is of particularinterest for calculation inverses or at least approximate inverses of pseudodifferentialoperators, which are also called parametrices.

In order to motivate the following sections, we calculate the composition ofp1(x,Dx)p2(x,Dx) formally, ignoring all technical difficulties. First of all,

p1(x,Dx)g =

∫∫ei(x−y)·ηp1(x, η)g(y)dyđη

andp2(x,Dx)f |x=y =

∫∫ei(y−z)·ξp2(y, ξ)f(z)dzđξ.

Hence we get for g(y) = p2(x,Dx)f |x=y

p1(x,Dx)p2(x,Dx)f =

∫∫ei(x−y)·ηp1(x, η)

(∫∫ei(y−z)·ξp2(y, ξ)f(z)dzđξ

)dyđη

=

∫∫∫∫ei(x−z)·ξe−i(x−y)·(ξ−η)p1(x, η)p2(y, ξ)f(z)dyđηdzđξ.

Using the substitution x′ = x− y and ξ′ = ξ − η, we obtain

p1(x,Dx)p2(x,Dx)f

=

∫∫ei(x−z)·ξ

(∫∫e−ix

′·ξ′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

(p1#p2)(x, ξ) :=

∫∫e−ix

′·ξ′p1(x, ξ + ξ′)p2(x+ x′, ξ)dx′đξ′. (4.8)

But the main problem is that the latter integral in general does not exists in theclassical sense. We will define it as so called oscillatory integral :

Os–

∫∫e−ix

′·ξ′p1(x, ξ + ξ′)p2(x+ x′, ξ)dx′đξ′ :=

limε→0

∫∫χ(εx′, εξ′)e−ix

′·ξ′p1(x, ξ + ξ′)p2(x+ x′, ξ)dx′đξ′,

27

where χ ∈ S(Rn × Rn) with χ(0, 0) = 1. In the following section we prove that theoscillatory integral are well-defined for suitable integrands. Moreover, we will observesome elementary properties, which justify our formal calculations above.

4.3 Oscillatory Integrals

Definition 4.8 The space of amplitudes Amτ (Rn×Rn),m, τ ∈ R, is the set of smoothfunctions a : Rn × Rn → C such that

|∂αη ∂βy a(y, η)| ≤ Cα,β(1 + |η|)m(1 + |y|)τ

uniformly in y, η ∈ Rn for all α, β ∈ Nn0 . Moreover, let

|a|Amτ ,k := max|α|+|β|≤k

supy,η∈Rn

(1 + |η|)−m(1 + |y|)−τ |∂αη ∂βy a(y, η)|,

k ∈ N, be the associated sequence of monotone increasing semi-norms.

It is not difficult to check that Amτ (Rn×Rn) is a Fréchet space. Moreover, Sm1,0(Rn×Rn) ⊂ Am0 (Rn × Rn) with continuous embedding.

THEOREM 4.9 Let a ∈ Amτ (Rn × Rn), m, τ ∈ R, and let χ ∈ S(Rn × Rn) withχ(0, 0) = 1. Then

Os–

∫∫e−iy·ηa(y, η)dyđη := lim

ε→0

∫∫χ(εy, εη)e−iy·ηa(y, η)dyđη

exists and

Os–

∫∫e−iy·ηa(y, η)dyđη =

∫∫e−iy·η〈y〉−2l′〈Dη〉2l

′ [〈η〉−2l〈Dy〉2la(y, η)]dyđη,

(4.9)where l, l′ ∈ N0 are chosen such that 2l > n+m and 2l′ > n+ τ and the integrand isin L1(Rn×Rn). In particular, the definition does not depend on the choice of χ and∣∣∣∣Os–

∫∫e−iy·ηa(y, η)dyđη

∣∣∣∣ ≤ Cm,τ‖a‖Amτ ,2(l+l′) (4.10)

where Cm,τ > 0 is independent of a.

Proof: Using Dαy e−iy·η = (−η)αe−iy·η and Dβ

η e−iy·η = (−y)βe−iy·η for α, β ∈ Nn

0 , wehave

〈η〉−2l〈Dy〉2le−iy·η and 〈y〉−2l′〈Dη〉2l′e−iy·η . (4.11)

Since χ(εy, εη) ∈ S(Rn × Rn) for fixed ε > 0, we can integrate by parts and obtain

Iε :=

∫∫χ(εy, εη)e−iy·ηa(y, η)dyđη

=

∫∫e−iy·η〈η〉−2l〈Dy〉2l (χ(εy, εη)a(y, η)) dyđη

=

∫∫e−iy·η〈y〉−2l′〈Dη〉2l

′ [〈η〉−2l〈Dy〉2l (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 = C∞b (Rn×

Rn) and limε→0 χ(εy, εη) = 1 uniformly on compact sets and limε→0 ∂αy ∂

βηχε(y, η) =

0 uniformly in (y, η) ∈ Rn × Rn if (α, β) 6= 0. Hence there are constants Cα,βindependent of both 0 < ε < 1 and a ∈ Amτ such that∣∣∂αy ∂βη (χε(y, η)a(y, η))

∣∣ ≤ Cα,β|a|Amτ ,|α|+|β|〈η〉m〈y〉τ . (4.12)

Moreover, since 〈ξ〉s ∈ Ss1,0,

|∂αη 〈η〉s| ≤ Cs,α〈η〉s−|α|. (4.13)

Combining (4.12) and (4.13), there are constants Cl,α independent of 0 < ε < 1 suchthat ∣∣∂αη [〈η〉−2l〈Dy〉2l (χε(y, η)a(y, η))

]∣∣ ≤ Cl,α|a|Amτ ,2l+|α|〈η〉m−2l〈y〉τ .

Consequently there are constants Cl,l′ independent of 0 < ε < 1 and a such that∣∣∣〈y〉−2l′〈Dη〉2l′ [〈η〉−2l〈Dy〉2l (χε(y, η)a(y, η))

]∣∣∣ ≤ Cl,l′ |a|Amτ ,2(l+l′)〈η〉m−2l〈y〉τ−2l′ .(4.14)

If we now choose 2l > n+m and 2l′ > n+ τ , then 〈η〉m−2l〈y〉τ−2l′ ∈ L1(Rn×Rn), cf.Lemma 2.6. Hence the Lebesgue dominated convergence theorem, limε→0 χ(εy, εη) =1, and limε→0 ∂

αy ∂

βηχ(εy, εη) = 0 for α + β > 0 imply

limε→0

Iε =

∫∫e−iy·η〈y〉−2l′〈Dη〉2l

′ [〈η〉−2l〈Dy〉2la(y, η)]dyđη.

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 thechoice of χ. Finally, passing to the limit ε → 0 in (4.14), (4.10) follows from (4.9)and Lemma 2.6.

Corollary 4.10 Let aj ∈ Amτ be a bounded sequence such that limj→∞ ∂αy ∂

βη aj(y, η) =

∂αy ∂βη a(y, η) for every y, η ∈ Rn, α, β ∈ Nn

0 and a ∈ Amτ . Then

limj→∞

Os–

∫∫e−iy·ηaj(y, η)dyđη = Os–

∫∫e−iy·ηa(y, η)dyđη.

Proof: The assumptions imply that

limj→∞〈y〉−2l′〈Dη〉2l

′ [〈η〉−2l〈Dy〉2laj(y, η)]

= 〈y〉−2l′〈Dη〉2l′ [〈η〉−2l〈Dy〉2la(y, η)

]for every y, η ∈ Rn (pointwise). Moreover, (4.14) implies∣∣∣〈y〉−2l′〈Dη〉2l

′ [〈η〉−2l〈Dy〉2l (χ(εy, εη)aj(y, ε))]∣∣∣ ≤ Cl,l′ |aj|2(l+l′)〈η〉m−2l〈y〉τ−2l′ .

Since the sequence aj is bounded in Amτ , |aj|2(l+l′) ≤ C uniformly in j ∈ N. Hencethe representation (4.9) and the Lebesgue dominated convergence theorem imply thestatement of the theorem.

29

Example 4.11 Let u ∈ C∞b (Rn). Then a(y, η) = eix·ηu(y) ∈ A00 and

Os–

∫∫ei(x−y)·ηu(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

Os–

∫∫ei(x−y)·ηu(y)dyđη = lim

ε→0

∫ψ(εy)u(y)

(∫ei(x−y)·ηψ(εη)đη

)dy

= limε→0

∫ψ(εy)u(y)ε−nF−1[ψ]

(x− yε

)dy

= limε→0

∫ψ(ε(x− εy′))u(x− εy′)F−1[ψ](y′)dy′

=

∫u(x)F−1[ψ](y′)dy′ = u(x)F [F−1[ψ]](0)

= u(x)ψ(0) = u(x)

since limε→0 ψ(ε(x − εy′))u(x − εy′) = ψ(0)u(x) = u(x). Thus formally we haveF−1[F [u]] = u for all u ∈ C∞b (Rn).

Lemma 4.12 Let a ∈ Amτ , m, τ ∈ R, and let α ∈ Nn0 . Then

Os–

∫∫e−iy·ηyαa(y, η)dyđη = Os–

∫∫e−iy·ηDα

η a(y, η)dyđη,

Os–

∫∫e−iy·ηηαa(y, η)dyđη = Os–

∫∫e−iy·ηDα

y a(y, η)dyđη.

Proof: First of all we note that yαa(y, η) ∈ Amτ+|α|, ηαa(y, η) ∈ Am+|α|

τ , andDαη a(y, η), Dα

y a(y, η) ∈ Amτ . 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 casefollows by induction.)

If |α| = 1, then yα = yj for 1 ≤ j ≤ n. Moreover, we choose χ in the definition ofthe oscillatory integral as χ(y, η) = e−|(y,η)|2/2. Then∫∫

χ(εy, εη)e−iy·ηyja(y, η)dyđη = −∫∫

χ(εy, εη)(Dηje−iy·η)a(y, η)dyđη

=

∫∫e−iy·ηDηj (χ(εy, εη)a(y, η)) dyđη

Using Dyjχ(εy, εη) = iε2ηjχ(εy, εη), we obtain

Dyj (χ(εy, εη)a(y, η)) = χ(εy, εη)Dyja(y, η) + iε2χ(εy, εη)ηja(y, η).

30

Therefore∫∫χ(εy, εη)e−iy·ηyja(y, η)dyđη =∫∫

χ(εy, εη)e−iy·ηDηja(y, η)dyđη + iε

∫∫χ(εy, εη)e−iy·ηηja(y, η)dyđη.

Passing to the limit ε→ 0 yields the first equality.

THEOREM 4.13 (Fubini’s theorem for oscillatory integrals)Let a ∈ Amτ (Rn+k × Rn+k), m, τ ∈ R, n, k ∈ N. Then

b(y, η) := Os–

∫∫e−iy

′·η′a(y, y′, η, η′)dy′đη′ ∈ Amτ (Rn × Rn),

where integration is with respect to Rk × Rk, and

∂αy ∂βη b(y, η) = Os–

∫∫e−iy

′·η′∂αy ∂βη a(y, y′, η, η′)dy′đη′ (4.15)

Moreover,

Os–

∫∫∫∫e−iy·η−iy

′·η′a(y, y′, η, η′)dydy′đηđη′

= Os–

∫∫e−iy·η

(Os–

∫∫e−iy

′·η′a(y, y′, η, η′)dy′đη′)dyđη.

Proof: Because of Peetre’s inequality,

〈(η, η′)〉m〈(y, y′)〉τ ≤ 2|m|+|τ |〈η〉m〈y〉|τ |〈η′〉|m|〈y′〉τ .

Hence ∂αy ∂βη a(y, ., η, .) ∈ A|m||τ | (R2k) with respect to (y′, η′) and

|∂αy ∂βη a(y, ., η, .)|A|m|τ (R2k),j≤ Cj,k,m|a|Amτ (R2(k+n)),j+|α|+|β|〈η〉m〈y〉τ .

Therefore (4.10) implies∣∣∣∣Os–

∫∫e−iy

′·η′∂αy ∂βη a(y, y′, η, η′)dy′đη′

∣∣∣∣ ≤ Cm|a|Amτ (R2(k+n)),2(l+l′)+|α|+|β|〈η〉m〈y〉τ ,

where 2l > |m|+k and 2l′ > k+ τ . Moreover, we choose l, l′ so large that 2l > m+nand 2l′ > n + τ . Because of the representation (4.9) and Theorem 2.4, we can take∂αy ∂

βη out of the oscillatory integral, which proves (4.15). Thus∣∣∂αy ∂βη b(y, η)

∣∣ ≤ Cm|a|Amτ (R2(n+k)),2(l+l′)+|α|+|β|〈η〉m〈y〉τ .

31

Hence b(y, η) ∈ Amτ (Rn×Rn). Thus 〈y〉−2l′〈Dη〉2l′ [〈η〉−2l〈Dy〉2lb(y, η)

]∈ L1(Rn×Rn)

because of Theorem 4.9. Moreover,

b(y, η) =

∫∫e−iy

′·η′〈y′〉−2l′〈Dη′〉2l′ [〈η′〉−2l〈Dy′〉2la(y, y′, η, η′)

]dy′đη′

and therefore

〈y〉−2l′〈Dη〉2l′ [〈η〉−2l〈Dy〉2lb(y, η)

]=

∫∫e−iy

′·η′(〈y〉〈y′〉)−2l′(〈Dη〉〈Dη′〉)2l′[(〈η〉〈η′〉)−2l(〈Dy〉〈Dy′〉)2la

]dy′đη′,

where

(〈y〉〈y′〉)−2l′(〈Dη〉〈Dη′〉)2l′[(〈η〉〈η′〉)−2l(〈Dy〉〈Dy′〉)2la(y, y′, η, η′)

]∈ L1(R2(n+k)).

Hence we can apply Fubini’s theorem and get

Os–

∫∫e−iy·η

(Os–

∫∫e−iy

′·η′a(y, y′, η, η′)dy′đη′)dyđη

=

∫e−iy·η−iy

′·η′(〈y〉〈y′〉)−2l′(〈Dη〉〈Dη′〉)2l′[(〈η〉〈η′〉)−2l(〈Dy〉〈Dy′〉)2la

] d(y′y, η, η′)

(2π)n+k

= Os–

∫∫∫∫e−iy·η−iy

′·η′(〈η〉〈η′〉)−2l(〈Dy〉〈Dy′〉)2la(y, y′, η, η′)dy′đη′dyđη

= Os–

∫∫∫∫e−iy·η−iy

′·η′a(y, y′, η, η′)dy′đη′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 apseudodifferential operator in more general form – a pseudodifferential operator witha double symbol :

p(x,Dx, x,Dx)u =

∫∫∫∫ei(x−x

′)·ξ+i(x′−x′′)·ξ′p(x, ξ, x′, ξ′)u(x′′)dx′′đξ′dx′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, x′, ξ′) = p1(x, ξ)p2(x′, ξ′) ∈Sm1,m2

1,0 (Rn × Rn × Rn × Rn) defined as follows:

Definition 4.14 Let m,m′ ∈ R. Then the space of double pseudodifferential sym-bols Sm,m

′

1,0 (Rn × Rn × Rn × Rn) is the space of all smooth functions p : Rn × Rn ×Rn × Rn → C such that

|DαξD

βxD

α′

ξ′Dβ′

x′p(x, ξ, x′, ξ′)| ≤ Cα,β,α′,β′(1 + |ξ|)m−|α|(1 + |ξ′|)m′−|α′|

uniformly in x, ξ, x′, ξ′ ∈ Rn for arbitrary α, β, α′, β′ ∈ Nn0 .

32

Note that Sm,m′

1,0 (Rn × Rn × Rn × Rn) ⊂ Amax(m,m′,m+m′)0 (R2n × R2n).

The statements on composition of two pseudodifferential operators will be a corol-lary to the following more general theorem:

THEOREM 4.15 Let m,m′ ∈ R and let p ∈ Sm,m′

1,0 (Rn×Rn×Rn×Rn) be a doublesymbol. Then

pL(x, ξ) := Os–

∫∫e−iy·ηp(x, ξ + η, x+ y, ξ)dyđη ∈ Sm+m′

1,0 (Rn × Rn).

Moreover,

pL(x, ξ) ∼∑α∈Nn0

1

α!∂αξD

αx′p(x, ξ, x

′, ξ′)|x′=x,ξ′=ξ

in the sense that

pL(x, ξ)−∑|α|≤N

1

α!∂αξ ∂

αξD

αx′p(x, ξ, x

′, ξ′)|x′=x,ξ′=ξ ∈ Sm+m′−N−11,0 (Rn × Rn)

for all N ∈ N0.

Proof:First of all, let ax,ξ(y, η) := p(x, ξ + η, x+ y, η). Using Peetre’s inequality,

|∂αη ∂βy ax,ξ(y, η)| = |∂αη ∂βy p(x, ξ + η, x+ y, ξ)|≤ Cα,β〈ξ + η〉m−|α|〈ξ〉m′ ≤ Cα,β〈ξ + η〉m〈ξ〉m′

≤ Cα,β2|m|〈η〉|m|〈ξ〉m+m′ .

Hence ax,ξ(y, η) ∈ A|m|0 with |ax,ξ|A|m|0 ,|m|+2n+2≤ C(1 + |ξ|)m+m′ . Therefore

|pL(x, ξ)| =∣∣∣∣Os–

∫∫e−iy·ηp(x, ξ + η, x+ y, ξ)dyđη

∣∣∣∣ ≤ C(1 + |ξ|)m+m′ (4.16)

because (4.10).Since p(x, ξ, x′, η) ∈ Am0 (R2n × R2n), m = max(m1,m2,m1 + m2), with respect

to (x, x′), (ξ, η), we also have p(x, ξ + η, x + y, η) ∈ Am0 (R2n × R2n) with respect to(x, y), (ξ, η). Hence we can apply (4.15) to conclude that

∂αξ ∂βxpL(x, ξ) = Os–

∫∫e−iy·η∂αξ ∂

βx [p(x, ξ + η, x+ y, ξ)] dyđη. (4.17)

Combining (4.16) and (4.17) yields

|∂αξ ∂βxpL(x, ξ)|

=

∣∣∣∣Os–

∫∫e−iy·η∂αξ ∂

βx [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:

p(x, ξ + η, x+ y, ξ) =∑|α|≤N

ηα

α!pα(x, ξ, x+ y, ξ)

+ (N + 1)∑

|α|=N+1

ηα

α!

∫ 1

0

(1− θ)Npα(x, ξ + θη, x+ y, ξ)dθ,

where pα(x, ξ, y, η) = ∂αξ p(x, ξ, y, η). Hence

pL(x, ξ) =∑|α|≤N

1

α!Os–

∫∫e−iy·ηηαpα(x, ξ, x+ y, ξ)dyđη

+(N + 1)∑

|α|=N+1

1

α!Os–

∫∫e−iy·ηηαrα(x, ξ, y, η)dyđη,

where

rα(x, ξ, y, η) =

∫ 1

0

(∂αξ p)(x, ξ + θη, x+ y, ξ)(1− θ)Ndθ.

Because of Lemma 4.12 and Example 4.11,

Os–

∫∫e−iy·ηηαpα(x, ξ, x+ y, ξ)dyđη = Os–

∫∫e−iy·ηDα

y pα(x, ξ, x+ y, ξ)dyđη

= ∂αξDαy 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,∣∣∂βη ∂γy ((∂αξDαy p)(x, ξ + θη, x+ y, ξ)

)∣∣ ≤ Cα,β,γ2|m|(1 + |θη|)|m|(1 + |ξ|)m+m′−|α|

≤ Cα,β,γ2|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 isuniformly bounded in A|m|0 as amplitudes in (y, η). Therefore rα(x, ξ, ., .) ∈ A|m|0 and∣∣∂βη ∂γy ((Dα

y rα)(x, ξ, η, y))∣∣ ≤ Cα,β,γ2

|m|(1 + |η|)|m|(1 + |ξ|)m+m′−|α|,

This implies∣∣∣∣Os–

∫∫e−iy·ηηαrα(x, ξ, η, y)dyđη

∣∣∣∣=

∣∣∣∣Os–

∫∫e−iy·ηDα

y rα(x, ξ, η, y)dyđη∣∣∣∣ ≤ Cα(1 + |ξ|)m+m′−|α| (4.19)

because of (4.10). Finally, the derivatives ∂βξ ∂γxrα,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 inSection 4.2 rigorous.

First of all, if p ∈ Sm1,0(Rn × Rn), then

p(x,Dx)u =

∫ (∫ei(x−y)·ξp(x, ξ)u(y)dy

)đξ

= Os–

∫∫e−ix

′·ξp(x, ξ)u(x+ x′)dx′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

= Os–

∫∫e−ix

′·ξp1(x, ξ)

(Os–

∫∫e−ix

′′·ξ′p2(x+ x′, ξ′)u(x+ x′ + x′′)dx′′đξ′)dx′đξ

= Os–

∫∫∫∫e−ix

′·ξ−ix′′·ξ′p1(x, ξ)p2(x+ x′, ξ′)u(x+ x′ + x′′)dx′′đξ′dx′đξ

= Os–

∫∫∫∫e−ix

′·η−iy·ξ′p1(x, ξ′ + η)p2(x+ x′, ξ′)u(x+ y)dx′đξ′dyđη

= Os–

∫∫e−iy·ξ

′(

Os–

∫∫e−ix

′·ηp1(x, ξ′ + η)p2(x+ x′, ξ′)dx′đη)u(x+ y)dyđξ′

= Os–

∫∫e−iy·ξ

′p1#p2(x, ξ′)u(x+ y)dyđξ

where we have used η = ξ − ξ′ and y = x′ + x′′ and p1#p2 is defined as in (4.8).

THEOREM 4.16 Let pj ∈ Smj1,0 (Rn × Rn), j = 1, 2, be two pseudodifferential sym-

bols. Then there is some p1#p2 ∈ Sm1+m21,0 (Rn × Rn) such that

p1(x,Dx)p2(x,Dx) = (p1#p2)(x,Dx).

Moreover, p1#p2 has the following asymptotic expansion:

p1#p2(x, ξ) ∼∑α∈Nn0

1

α!∂αξ p1(x, ξ)Dα

xp2(x, ξ)

in the sense that

p1#p2(x, ξ)−∑|α|<N

1

α!∂αξ p1(x, ξ)Dα

xp2(x, ξ) ∈ Sm1+m2−N1,0 (Rn × Rn)

for all N ∈ N.

35

Proof: Let p(x, ξ, x′, ξ′) = p1(x, ξ)p2(x′, ξ′). Then p ∈ Sm1,m2

1,0 (Rn × Rn × Rn × Rn)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 pseudodiffer-ential operator. In particular, the asymptotic expansions implies:

p1#p2(x, ξ) = p1(x, ξ)p2(x, ξ) + r(x, ξ),

where r(x, ξ) ∈ Sm1+m2−11,0 (Rn ×Rn). Hence the symbol of the composition coincides

with the product of the symbols modulo terms of lower order. This is the essential factneeded 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 havep1(x,Dx)p2(Dx) = OP(p1(x, ξ)p2(ξ)) since F [p2(Dx)u] = p2(ξ)u(ξ). Moreover, ifp1(x,Dx) is a differential operator of order m ∈ N0 with coefficients in C∞b (Rn), then

p1(x,Dx)p2(x,Dx) = (p1#p2)(x,Dx)

where(p1#p2)(x, ξ) =

∑|α|≤m

1

α!∂αξ p1(x, ξ)Dα

xp2(x, ξ).

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, itcan be observed by modifying the proof of Theorem 4.15. (In this case there is noremainder term rα,θ if N > m.)

4.6 Application: Elliptic Pseudodifferential Operators and Para-metrices

Definition 4.17 A symbol p ∈ Sm1,0(Rn×Rn), m ∈ R, is called elliptic if there is anR > 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 ellipticsymbol of order 2. Moreover, q(ξ) = 〈ξ〉m, m ∈ R, is an elliptic symbol of orderm.

2. Let A(x) ∈ C∞b (Rn)n×n be a matrix that is uniformly positive definite, i.e.,there is a c > 0 such that

ξTA(x)ξ ≥ c|ξ|2, for all x, ξ ∈ Rn.

Then p(x, ξ) = ξTA(x)ξ is an elliptic symbol of order 2.

36

Lemma 4.19 Let p ∈ Sm1,0(Rn × Rn), m ∈ R, be an elliptic symbol and R > 0 as in(4.21). Then

q(x, ξ) := ψ(ξ)p(x, ξ)−1 ∈ S−m1,0 (Rn × Rn)

where ψ ∈ C∞b (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,

∂ξjp(x, ξ)−1 = −p(x, ξ)−2∂ξjp(x, ξ)

and the same identity with ∂ξj replaced by ∂xj . Using (4.21),

|∂ξjp(x, ξ)−1| ≤ C|ξ|−2m〈ξ〉m−1 ≤ C〈ξ〉−m−1

and|∂xjp(x, ξ)−1| ≤ C|ξ|−2m〈ξ〉m ≤ C〈ξ〉−m

since |ξ| ≥ R > 0. In the same way one can easily prove by induction that

|∂αξ ∂βxp(x, ξ)−1| ≤ Cα,β〈ξ〉−m−|α|

for all |ξ| ≥ R. On the otherhand q : Rn × Rn → C is smooth and therefore

|∂αξ ∂βxq(x, ξ)| ≤ C ′α,β ≤ Cα,β〈ξ〉−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

|∂αξ ∂βxq(x, ξ)| ≤ Cα,β〈ξ〉−m−|α|

for all ξ ∈ Rn.

Corollary 4.20 Let p ∈ Sm1,0(Rn × Rn) be an elliptic symbol. Then there is anq ∈ S−m1,0 (Rn × Rn) such that

p(x,Dx)q(x,Dx) = I + r(x,Dx), q(x,Dx)p(x,Dx) = I + r′(x,Dx)

with r, r′ ∈ S−11,0(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) + r(x,Dx),

where r ∈ S−11,0(Rn × Rn). Moreover, p(x, ξ)q(x, ξ) = 1 for all |ξ| ≥ R + 1. Hence

p(x, ξ)q(x, ξ)− 1 ∈ S−∞1,0 (Rn × Rn) and

p(x,Dx)q(x,Dx) = I + r(x,Dx)

with r(x, ξ) = p(x, ξ)q(x, ξ)− 1 + r(x, ξ) ∈ S−11,0(Rn × Rn). The identity

q(x,Dx)p(x,Dx) = I + r′(x,Dx)

with r′ ∈ S−11,0(Rn × Rn) is proved the some way.

37

THEOREM 4.21 Let p ∈ Sm1,0(Rn × Rn), m ∈ R. Then the following conditionsare equivalent:

1. p is elliptic.

2. There is a q ∈ S−m1,0 (Rn×Rn) such that p(x,Dx)q(x,Dx) = I + r(x,Dx), wherer ∈ S−1

1,0(Rn × Rn).

3. For every N ∈ N there is a qN ∈ S−m1,0 (Rn×Rn) such that p(x,Dx)qN(x,Dx) =

I + rN(x,Dx), where rN ∈ S−N1,0 (Rn × Rn).

Proof: 1. implies 2.: This is a consequence of Corollary 4.20.

2. implies 3.: By the assumptions there are some q ∈ S−m1,0 (Rn × Rn), r ∈S−1

1,0(Rn × Rn) such that

p(x,Dx)q(x,Dx) = I − r(x,Dx).

Now let qN ∈ S−m1,0 (Rn × Rn) be such that

qN(x,Dx) = q(x,Dx)N−1∑k=0

r(x,Dx)k.

Then

p(x,Dx)qN(x,Dx) = (I − r(x,Dx))N−1∑k=0

(−r(x,Dx))k = I + (−r(x,Dx))

N ,

where (−r(x,Dx))N = rN(x,Dx) with rN ∈ S−N1,0 (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 + r′(x,Dx)

with r, r′ ∈ S−11,0(Rn × Rn), we obtain

p(x, ξ)q(x, ξ)− 1 = −r(x, ξ) + r′(x, ξ) ∈ S−11,0(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| ≤ C〈ξ〉−1

for all x, ξ ∈ Rn. Hence there is some R > 0 such that |p(x, ξ)q(x, ξ) − 1| ≤ 12for

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)| ≤ 2C〈ξ〉−m

38

for all x ∈ Rn, |ξ| ≥ R since q ∈ S−m1,0 (Rn × Rn).

For completeness we note that for an elliptic symbol p there is also an q∞ ∈S−m1,0 (Rn × Rn) such that

p(x,Dx)q∞(x,Dx) = I + r∞(x,Dx) (4.22)

with r ∈ S−∞1,0 (Rn × Rn). In order to prove this statement, the following lemma isneeded:

Lemma 4.22 Let pj ∈ Smj1,0 (Rn × Rn) with m1 ≥ . . . ≥ mj → −∞ as j →∞. Then

there is some p ∈ Sm11,0 (Rn × Rn) such that

p(x, ξ) ∼∞∑j=1

pj(x, ξ) :⇔ p(x, ξ)−N−1∑j=1

pj(x, ξ) ∈ SmN1,0 (Rn × Rn).

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′(x,Dx),

where

q′(x, ξ) ∼∞∑k=0

r#k(x, ξ),

r is the same as in the proof of Theorem 4.21, and

r#k(x, ξ) = r# . . .#r(x, ξ)︸ ︷︷ ︸k-times

∈ S−k1,0 (Rn × Rn).

4.7 Boundedness on C∞b (Rn) and Uniqueness of the Symbol

As seen above,

p(x,Dx)u = Os–

∫∫e−ix

′·ξp(x, ξ)u(x+ x′)dx′đξ (4.23)

for all u ∈ S(Rn). But the latter oscillatory integral is defined for all u ∈ C∞b (Rn).Therefore we can extend the definition of p(x,Dx) to C∞b (Rn).

THEOREM 4.23 Let p ∈ Sm1,0(Rn × Rn). Then p(x,Dx) defined by (4.23) foru ∈ C∞b (Rn) is a bounded operator

p(x,Dx) : C∞b (Rn)→ C∞b (Rn).

39

Proof: Consider ax(x′, ξ) = p(x, ξ)u(x+ x′). Since for every α, β ∈ Nn0

|∂αx′∂βξ p(x, ξ)u(x+ x′)| ≤ Cα,β|p|(m)

|β| ‖u‖C|α|b 〈ξ〉m,

we have ax(x′, ξ) ∈ Am0 with |ax|Am0 ,k ≤ C|p|(m)k ‖u‖Ckb uniformly in x ∈ Rn. Thus

supx∈Rn|p(x,Dx)u| ≤ C|ax|Am0 ,2(l+l′) ≤ C ′|p|(m)

2(l+l′)‖u‖C2(l+l′)b

(4.24)

by Theorem 4.9, where 2l > m + n, 2l′ > n. In order to estimate ∂αx p(x,Dx)u weobserve that a(x, x′, ξ, ξ′) := p(x, ξ)u(x+ x′) ∈ Am0 (R2n × R2n) and therefore

∂xjp(x,Dx)u = Os–

∫∫e−ix

′·ξ(∂xjp)(x, ξ)u(x+ x′)dx′đξ

+ Os–

∫∫e−ix

′·ξp(x, ξ)(∂xju)(x+ x′)dx′đξ

= (∂xjp)(x,Dx)u+ p(x,Dx)∂xju

by (4.15) in Theorem 4.13. Applying this formula successively and using (4.24), itfollows immediately that ‖p(x,Dx)u‖Ckb ≤ C|p|(m)

2(l+l′)+k‖u‖C2(l+l′)+kb

, which proves thetheorem.

Using this extension, we can prove that the symbol of a pseudodifferential operatoris uniquely defined by its values on C∞b (Rn):

Exercise 8 Show that for all x, ξ ∈ Rn

e−ix·ξp(x,Dx)ei.·ξ = p(x, ξ). (4.25)

Corollary 4.24 Let p, q ∈ S∞1,0(Rn × Rn). Then p(x,Dx)u = q(x,Dx)u for all u ∈S(Rn) implies p(x, ξ) = q(x, ξ).

Proof: Because of the latter exercise, we only need to prove

p(x,Dx)u = q(x,Dx)u for all u ∈ C∞b (Rn). (4.26)

To this end let u ∈ C∞b (Rn) and uε(x) := ψ(εx)u(x) for all x ∈ Rn, ε > 0 and someψ ∈ S(Rn) with ψ(0) = 1. Then uε ∈ S(Rn) and

∂αxuε(x)→ε→0 ∂αxu(x) for every x ∈ Rn, α ∈ Nn

0 .

Therefore Corollary 4.10 implies

p(x,Dx)uε(x) = Os–

∫∫e−ix

′·ξ′ p(x, ξ)uε(x+ x′)︸ ︷︷ ︸=:aε,x(x′,ξ)

dx′đξ′ →ε→0 p(x,Dx)u(x)

40

for every x ∈ Rn since

∂αx∂βξ aε,x(x

′, ξ)→ε→0 ∂αx∂

βξ ax(x

′, ξ) for all x′, ξ ∈ Rn, α, β ∈ Nn0

with ax(x′, ξ) = p(x, ξ)u(x+x′) and since (aε,x)0<ε≤1 is bounded in Am0 (Rn×Rn) forevery x ∈ Rn, which follows from the boundedness of (uε)0<ε≤1 in C∞b (Rn). In thesame way, we obtain the same statement for q(x,Dx). Therefore

p(x,Dx)u(x) = limε→0

p(x,Dx)uε(x) = limε→0

q(x,Dx)uε(x) = q(x,Dx)u(x)

for every x ∈ Rn since p(x,Dx)uε = q(x,Dx)uε due to uε ∈ S(Rn). Since u ∈ C∞b (Rn)was arbitrary, (4.26) follows and the corollary is proved.

More generally, if P : C∞b (Rn) → C∞b (Rn) is an arbitrary bounded linear map,then we can define a symbol of P by

p(x, ξ) := e−ix·ξPei.·ξ

This gives an alternative approach for determining the symbol of p1(x,Dx)p2(x,Dx):

Exercise 9 Calculatee−ix·ξp1(x,Dx)p2(x,Dx)e

i.·ξ

without using Theorem 4.16.

4.8 Adjoints of Pseudodifferential Operators and Operatorsin (x,y)-Form

Definition 4.25 Let A,A∗ : S(Rn)→ S(Rn). Then A∗ is called formal adjoint of Aif

(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 (ofthe same order). This will allows us to extend p(x,Dx), up to now only defined onS(Rn), to the a linear operator defined on S ′(Rn).

Now we calculate the formal adjoint of p(x,Dx):

(p(x,Dx)u, v)L2(Rn) =

∫∫eix·ξp(x, ξ)u(ξ)đξv(x)dx

=

∫∫eix·ξp(x, ξ)v(x)dxu(ξ)đξ

=

∫u(ξ)

∫e−ix·ξp(x, ξ)v(x)dxđξ,

41

where have used Fubini’s theorem. – Note that v, u ∈ S(Rn) which implies thateix·ξp(x, ξ)u(ξ)v(x) ∈ L1(Rn × Rn) w.r.t. (x, ξ).

Lemma 4.26 Let p ∈ Sm1,0(Rn × Rn), m ∈ R. Then

w(ξ) :=

∫e−ix·ξp(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

(p(x,Dx)∗v)(x) =

∫∫ei(x−y)·ξp(y, ξ)v(y)dyđξ. (4.27)

This operator is a pseudodifferential operator in the so called y-form (also calledR-form), which is a special case of operators of the form

p(x,Dx, x)u :=

∫∫ei(x−y)·ξp(x, y, ξ)u(y)dyđξ (4.28)

for u ∈ S(Rn), where p ∈ Sm1,0(R2n × Rn), i.e.,

|∂αξ ∂βx∂γy p(x, y, ξ)| ≤ Cα,β,γ(1 + |ξ|)m−|α|

for all α, β, γ ∈ Nn0 . As before we obtain

p(x,Dx, x)u = Os–

∫∫e−ix

′·ξp(x, x+ x′, ξ)u(x+ x′)dx′đξ

for all u ∈ S(Rn).

THEOREM 4.27 Let p ∈ Sm1,0(R2n×Rn), m ∈ R. Then p(x,Dx, x)u = pL(x,Dx)ufor all u ∈ C∞b (Rn), where

pL(x, ξ) = Os–

∫∫e−iy·ηp(x, x+ y, ξ + η)dyđη ∈ Sm1,0(Rn × Rn).

Moreover,

pL(x, ξ) ∼∑α∈Nn0

1

α!∂αξD

αy p(x, y, ξ)|y=x

in the sense that for all N ∈ N0

pL(x, ξ)−∑|α|≤N

1

α!∂αξD

αy p(x, y, ξ)|y=x ∈ Sm−N−1

1,0 (Rn × Rn).

42

Proof: Using

u(x+ x′) = Os–

∫∫e−iy·ηu(x+ x′ + y)dyđη = Os–

∫∫ei(x+x′−y)·ηu(y)dyđη

due to Example 4.11, we get

p(x,Dx, x)u = Os–

∫∫∫∫e−ix

′·ξei(x+x′−y)·ηp(x, x+ x′, ξ)u(y)dyđηdx′đξ

= Os–

∫∫e−i(x−y)·η

(Os–

∫∫e−ix

′·(ξ−η)p(x, x+ x′, ξ)dx′đξ)u(y)dyđη

= Os–

∫∫e−i(x−y)·ηpL(x, ξ)u(y)dyđη.

Hence application of Theorem 4.15 finishes the proof.As a direct consequence we obtain:

Lemma 4.28 Let p ∈ Sm1,0(R2n ×Rn), m ∈ R. Then p(x,Dx, x) : S(Rn)→ S(Rn) isa bounded operator. Moreover, if the definition (4.28) is replaced by

p(x,Dx, x)u := Os–

∫∫e−ix

′·ξp(x, x+ x′, ξ)u(x+ x′)dx′đξ

for u ∈ C∞b (Rn), then p(x,Dx, x) : C∞b (Rn)→ C∞b (Rn) is a bounded operator.

Proof: Since p(x,Dx, x)u = pL(x,Dx)u for all u ∈ C∞b (Rn), the lemma is a conse-quence of the corresponding statements for pL(x,Dx).

Corollary 4.29 If p ∈ Sm1,0(Rn×Rn), then the formal adjoint of p(x,Dx) is p∗(x,Dx)where

p∗(x, ξ) = Os–

∫∫e−iy·ξp(x+ y, ξ + η)dyđη ∈ Sm1,0(Rn × Rn).

Moreover,

p∗(x, ξ) ∼∑α∈Nn0

1

α!∂αξD

αxp(x, ξ)

in the sense that for every N ∈ N0

p∗(x, ξ)−∑|α|≤N

1

α!∂αξD

αxp(x, ξ) ∈ Sm−N−1

1,0 (Rn × Rn).

Proof: The corollary is a direct consequence of (4.27) and the latter theorem.

Definition 4.30 Let p ∈ Sm1,0(Rn×Rn). Then we define p(x,Dx) : S ′(Rn)→ S ′(Rn)by

〈p(x,Dx)u, v〉 := 〈u, p∗(x,Dx)v〉 u ∈ S ′(Rn), v ∈ S(Rn).

43

Since S(Rn) ⊆ S ′(Rn) (identifying functions with functional in the way described inSection 3.5), it is important to notice that, if u ∈ S(Rn),

〈p(x,Dx)u, v〉 = 〈u, p∗(x,Dx)v〉 = (u, p∗(x,Dx)v)L2(Rn) =

∫Rnp(x,Dx)u(x)v(x) dx

for all v ∈ S(Rn). I.e., the definition of p(x,Dx)u in sense of S ′(Rn) coincides withthe starting definition of p(x,Dx)u for u ∈ S(Rn).

4.9 L2-Continuity and Bessel Potential Spaces

THEOREM 4.31 Let p ∈ S01,0(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.

Lemma 4.32 Theorem 4.31 holds for p ∈ S−n−11,0 (Rn × Rn).

Proof: If p ∈ S−n−11,0 (Rn×Rn), then p(x, ξ) ∈ L1(Rn) with respect to ξ and therefore

p(x,Dx)f =

∫eix·ξp(x, ξ)f(ξ)đξ =

∫∫ei(x−y)·ξp(x, ξ)đξf(y)dy

=

∫k(x, x− y)f(y)dy

for all f ∈ S(Rn) by Fubini’s theorem, where k(x, z) = F−1ξ 7→z[p(x, ξ)]. This kernel

satisfies

|zαk(x, z)| =∣∣∣∣∫ eiz·ξ∂αξ p(x, ξ)đξ

∣∣∣∣ ≤ Cα|p|(−n−1)|α|

since |∂αξ p(x, ξ)| ≤ |p|(−n−1)|α| 〈ξ〉−n−1−|α| ∈ L1(Rn) w.r.t. ξ. Hence |(1+ |z|2)nk(x, z)| ≤

C|p|(−n−1)2n and therefore

g(z) := supx∈Rn|k(x, z)| ≤ (1 + |z|2)−n ∈ L1(Rn).

Thus

‖p(x,Dx)f‖2 ≤ C

∥∥∥∥∫Rng(x− y)|f(y)|dy

∥∥∥∥2

≤ C ′‖g‖1‖f‖2

because of ‖g ∗ f‖2 ≤ ‖g‖1‖f‖2, cf. Example 3.18.1.

Lemma 4.33 Theorem 4.31 holds for p ∈ S−m1,0 (Rn × Rn) with m > 0.

44

Proof: In order to prove ‖p(x,Dx)f‖2 ≤ C‖f‖2 for f ∈ S(Rn), it is sufficient toshow

‖p∗(x,Dx)p(x,Dx)f‖2 ≤ C‖f‖2

since

‖p(x,Dx)f‖22 = (p∗(x,Dx)p(x,Dx)f, f) ≤ ‖p∗(x,Dx)p(x,Dx)f‖2‖f‖2.

But, if p ∈ S−m1,0 (Rn × Rn), then p∗(x,Dx)p(x,Dx) = p′(x,Dx) with p′ ∈ S−2m1,0 (Rn ×

Rn). Hence, using the previous lemma, we obtain by mathematical inductions that

‖p(x,Dx)f‖2 ≤ C‖f‖2

for all p ∈ S−mk1,0 with mk = (n+ 1)/2k, k ∈ N. This proves the lemma since for everym > 0 there is a k ∈ N such that −mk > −m.In order to finish the proof of Theorem 4.31, we need

Lemma 4.34 If p ∈ S01,0(Rn × Rn) with p(x, ξ) ∈ R, x, ξ ∈ Rn, and F ∈ C∞(R),

then F (p(x, ξ)) ∈ S01,0(Rn × 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,

|∂ξjF (p(x, ξ))| ≤ sup|z|≤R

|F ′(z)||∂ξjp(x, ξ)| ≤ C〈ξ〉−1,

|∂xjF (p(x, ξ))| ≤ sup|z|≤R

|F ′(z)||∂xjp(x, ξ)| ≤ C.

Finally, the estimate of |∂αξ ∂βxF (p(x, ξ))| can be proved using an induction and thechain rule as before.

Proof of Theorem 4.31: If p(x, ξ) ∈ S01,0(Rn × Rn), then |p(x, ξ)| ≤ M for M :=

|p|(0)0 ≥ 0. Therefore p′(x, ξ) := M2 − p(x, ξ)p(x, ξ) ∈ S0

1,0(Rn ×Rn) and p′(x, ξ) ≥ 0.Now let F ∈ C∞(R) be defined by F (x) = (1 + x)

12 for x ≥ 0. Then q(x, ξ) :=

F (p′(x, ξ)) ∈ S01,0(Rn × Rn) and

q∗(x,Dx)q(x,Dx)f = OP(F (p′(x, ξ))2)f + r(x,Dx)f

= (1 +M2)f −OP(p(x, ξ)p(x, ξ))f + r(x,Dx)f

= (1 +M2)f − p∗(x,Dx)p(x,Dx)f + r′(x,Dx)f

for f ∈ S(Rn), where r, r′ ∈ S−11,0(Rn × Rn) because of Theorem 4.16 and Corol-

lary 4.29. Hence

‖p(x,Dx)f‖2L2(Rn)

≤ (p∗(x,Dx)p(x,Dx)f, f)L2(Rn) + (q∗(x,Dx)q(x,Dx)f, f)L2(Rn)

≤ (1 +M2)‖f‖2L2(Rn) + (r′(x,Dx)f, f).

45

Since r′ ∈ S−11,0(Rn × Rn), ‖r′(x,Dx)f‖2 ≤ C‖f‖2 due to Lemma 4.33. Hence

‖p(x,Dx)f‖22 ≤ (1 +M2 + C)‖f‖2

2 for all f ∈ S(Rn).

Remark 4.35 Checking the previous proofs, it can be observed that the estimatesof ‖p(x,Dx)‖L(L2(Rn)) depend only on |p|(0)

k for some suitably large k ∈ N (and not pdirectly). Since the mapping p 7→ p(x,Dx) is linear, we have

‖p(x,Dx)‖L(L2(Rn)) ≤ C|p|(0)k ; (4.29)

this can seen as follows: Let q(x, ξ) = p(x, ξ)/|p|(0)k . Then |q|(0)

k = 1 and

‖q(x,Dx)‖L(L2(Rn)) ≤ C

independent of p ∈ S01,0(Rn × Rn). Hence multiplication by |p|(0)

k yields (4.29).

Definition 4.36 Let s ∈ R. Then the Bessel potential space Hs2(Rn) is defined as

Hs2(Rn) = u ∈ S ′(Rn) : 〈Dx〉su ∈ L2(Rn),‖u‖s,2 := ‖〈Dx〉su‖2.

Here a tempered distribution u ∈ S ′(Rn) is said to be in L2(Rn) if there is aU ∈ L2(Rn) such that

〈u, v〉 =

∫RnU(x)v(x)dx for all v ∈ S(Rn)

(where 〈., .〉 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 ∈S ′(Rn) is in L2(Rn) if and only if there is a constant C > 0 such that |〈u, v〉| ≤ C‖v‖2

for all v ∈ S(Rn).

Remark 4.37 By definition 〈Dx〉s is an isomorphism fromHs2(Rn) to L2(Rn). Hence

Hs2(Rn) normed by ‖.‖2,s is Banach space. Moreover,

(u, v)Hs2

:= (〈Dx〉su, 〈Dx〉sv)L2 , u, v ∈ Hs2(Rn)

is scalar product on Hs2(Rn) and ‖u‖2

2,s = (u, u)Hs2. Thus Hs

2(Rn) is a Hilbert space.Since S(Rn) is dense in L2(Rn) and 〈Dx〉s : S(Rn)→ S(Rn) for all s ∈ R, S(Rn)

is dense in Hs2(Rn) for all s ∈ R.

Finally, Hs12 (Rn) ⊆ Hs2

2 (Rn) for s1 ≥ s2 since

‖u‖s2,2 = ‖〈Dx〉s2u‖2 = ‖〈Dx〉s2−s1〈Dx〉s1u‖2 ≤ C‖〈Dx〉s1u‖2 = C‖u‖s1,2,

where we have used that 〈ξ〉s2−s1 ∈ S01,0(Rn×Rn) and therefore 〈Dx〉s2−s1 : L2(Rn)→

L2(Rn) is a bounded operator. – The parameter s ∈ R of Hs2(Rn) determines the

regularity of the functions u ∈ Hs2(Rn) (“how many derivatives of u are in L2(Rn)”),

cf. Lemma 4.42, below.

46

Lemma 4.38 Let p ∈ Sm1,0(Rn × Rn). Then p(x,Dx) : Hs+m2 (Rn)→ Hs

2(Rn). More-over, there is some k ∈ N0 such that ‖p(x,Dx)‖L(Hs+m

2 (Rn),Hs2(Rn)) ≤ Cs,m|p|(m)

k .

Proof: First of all, we note that, if u ∈ L2(Rn) ⊂ S ′(Rn), p(x,Dx)u is defined inthe sense of S ′(Rn), cf. Definition 4.30, and m = 0, then

|〈p(x,Dx)u, v〉| =∣∣(u, p∗(x,Dx)v)L2(Rn)

∣∣ ≤ ‖u‖L2(Rn)‖p∗(x,Dx)‖L2(Rn)‖v‖L2(Rn)

for all v ∈ S(Rn) due to Theorem 4.31 and Corollary 4.29. Hence p(x,Dx)u ∈L2(Rn) ∼= L2(Rn)′.

Now we consider a general s,m ∈ R. Since 〈Dx〉s+m : Hs+m2 (Rn) → L2(Rn) and

〈Dx〉−s : L2(Rn)→ Hs2(Rn) are linear isomorphisms, p(x,Dx) : Hs+m

2 (Rn)→ Hs2(Rn)

is a linear bounded operator if and only if

q(x,Dx) := 〈Dx〉sp(x,Dx)〈Dx〉−s−m : L2(Rn)→ L2(Rn)

is a bounded operator. Because of Theorem 4.16, q ∈ S01,0(Rn ×Rn). Hence the first

statement is a consequence of Theorem 4.31. The second statement is a consequenceof Remark 4.35 and the fact that the mapping

Sm11,0 × Sm2

1,0 3 (p1, p2) 7→ p1#p2 ∈ Sm1+m21,0

is bounded/continuous applied to p1(x, ξ) = 〈ξ〉s, p2(x, ξ) = p(x, ξ)〈ξ〉−s−m.

Corollary 4.39 (Elliptic regularity) Let p ∈ Sm1,0(Rn × Rn) be an elliptic symboland f ∈ Hs

2(Rn), s,m ∈ R. Then, if u ∈ H−∞2 (Rn) :=⋃s∈RH

s2(Rn) is a solution of

the pseudodifferential equation

p(x,Dx)u = f,

then u ∈ Hs+m2 (Rn).

Proof: Let u ∈ H−∞2 (Rn). Then u ∈ Hs+m−N2 (Rn) for some N ∈ N. By the same

construction as in Theorem 4.21, there is an qN(x, ξ) ∈ S−m1,0 (Rn × Rn) such that

qN(x,Dx) =

(N−1∑k=0

(−r(x,Dx))k

)q(x,Dx),

where q is the same as in Corollary 4.20. (I.e., p(x,Dx)q(x,Dx) = I + r(x,Dx) andq(x,Dx)p(x,Dx) = I+r′(x,Dx) where r, r′ ∈ S−1

1,0(Rn×Rn). Then as in Theorem 4.21

qN(x,Dx)p(x,Dx) = I + rN(x,Dx),

where rN ∈ S−N1,0 (Rn × Rn). Therefore

qN(x,Dx)f = qN(x,Dx)p(x,Dx)u = u+ rN(x,Dx)u.

Since qN(x,Dx)f ∈ Hs+m2 (Rn) and rN(x,Dx)u ∈ Hs+m

2 (Rn) by Lemma 4.38, we haveu ∈ Hs+m(Rn).

47

Remark 4.40 The latter corollary says that the solution u is “as smooth as theright-hand side allows”. This property is also called elliptic regularity.

Definition 4.41 Let m ∈ N0. Then the Sobolev space Wm2 (Rn) of order m is the

space of all u ∈ L2(Rn) such that the distributional derivative ∂αu ∈ L2(Rn) for allα ∈ Nn

0 with |α| ≤ m. Moreover, Wm2 (Rn) is normed by

‖u‖Wm2

:=

∑|α|≤m

‖∂αxu‖22

12

.

Lemma 4.42 Let m ∈ N0. Then Wm2 (Rn) = Hm

2 (Rn) with equivalent norms.

Proof: If f ∈ Hm2 (Rn), then

Dαxf = OP(ξα〈ξ〉−m)〈Dx〉mf,

where qα(ξ) := ξα〈ξ〉−m ∈ S01,0(Rn × Rn) if |α| ≤ m. Hence∑

|α|≤m

‖Dαxf‖2

2

12

≤ Cm‖〈Dx〉mf‖2 = Cm‖f‖m,2.

Conversely, if f ∈ Wm2 (Rn), ∂αx f ∈ L2(Rn) for all |α| ≤ m. Moreover,

〈ξ〉 =〈ξ〉2

〈ξ〉=

1

〈ξ〉+

n∑j=1

ξjξj〈ξ〉

.

Thus

〈ξ〉m =N∑k=1

pk(ξ)qk(ξ),

where qk(ξ) is a polynomial of order m and pk ∈ S01,0(Rn × Rn), N = N(m) ∈ N.

Hence

‖f‖m,2 = ‖〈Dx〉mf‖2 =

∥∥∥∥∥N∑k=1

pk(Dx)qk(Dx)f

∥∥∥∥∥2

≤ Cm

N∑k=1

‖qk(Dx)f‖2 ≤ C ′m‖f‖Wm2,

where we have used that pk(Dx) : L2(Rn)→ L2(Rn) are bounded operators and that‖qk(Dx)f‖2 ≤ C‖f‖Wm

2since qk(Dx) is a differential operator of order m.

Lemma 4.43 Let s > n2. Then there is some Cs > 0 such that

‖u‖∞ ≤ Cs‖〈Dx〉su‖2, ‖u‖2 ≤ Cs‖〈x〉su‖∞

for all u ∈ S(Rn). In particular, Hs2(Rn) → C0

b (Rn).

48

Proof: See exercises.

Corollary 4.44 Let|u|′′k,S := sup

|α|+|β|≤k‖xαDβ

xu‖2

for u ∈ S(Rn), k ∈ N0. Then |.|′′k,S , k ∈ N, is a decreasing sequence of semi-normson S(Rn) which is equivalent to the semi-norms |.|k,S defined above. More precisely,

|u|′′k,S ≤ Ck|u|k+2n,S and |u|k,S ≤ Ck|u|′′k+2n,S

for all u ∈ S(Rn) and k ∈ N0.

Proof: Since 〈Dx〉2n = (1−∆)n is a differential operator of order 2n,

|u|k,S = sup|α|+|β|≤k

‖xαDβxu‖∞ ≤ C sup

|α|+|β|≤k‖〈Dx〉2nxαDβ

xu‖2 ≤ Ck|u|′′k+2n,S

by Lemma 4.43. Similarly, since 〈x〉2n = (1 + |x|2)n is a polynomial of order 2n,

|u|′′k,S = sup|α|+|β|≤k

‖xαDβxu‖2 ≤ C sup

|α|+|β|≤k‖〈x〉2nxαDβ

xu‖∞ ≤ Ck|u|k+2n,S .

4.10 Summary and Final Remarks

In this chapter we have studied pseudodifferential symbols p ∈ Sm1,0(Rn×Rn),m ∈ Rn,and the relation to the associated operators p(x,Dx), first defined on S(Rn) and laterextended to larger spaces.

Finally, we note the following structural properties: The symbol space Sm1,0(Rn ×Rn) is a topological vector space, which topology is defined by a sequence of (increas-ing and separating) semi-norms (a Fréchet space). The space of continuous linearmappings T : S(Rn)→ S(Rn) denoted by L(S(Rn)) is also a topological vector spaceif equipped with the topology of pointwise convergence. In these terms

OP: Sm1,0(Rn × Rn)→ L(S(Rn)) : p 7→ p(x,Dx)

is a homomorphism of topological vector spaces (i.e., linear and continuous). More-over, on the level of symbol we have the pointwise multiplication of symbols:

· : Sm11,0 (Rn ×Rn)× Sm2

1,0 (Rn ×Rn)→ Sm1+m21,0 (Rn ×Rn) : (p1, p2) 7→ p1(x, ξ) · p2(x, ξ),

which is a continuous bilinear mapping. On the level of operators we have thecomposition. 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 bilinearmapping

#: Sm11,0 (Rn × Rn)× Sm2

1,0 (Rn × Rn)→ Sm1+m21,0 (Rn × Rn) : (p1, p2) 7→ p1#p2

such thatOP(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), r ∈ Sm1+m2−11,0 (Rn × Rn).

Hence the relation (p1#p2)(x, ξ) ≡ p1(x, ξ) · p2(x, ξ) holds modulo symbols of lowerorder. This relation shows that inversion of an operator modulo lower order operatorscoincides with inversion of the symbol (for large |ξ|). This was used to construct aparametrix for an elliptic pseudodifferential operator.

After studying compositions of pseudodifferential operators, it was shown thatevery pseudodifferential operator p(x,Dx) possesses a formal adjoint p∗(x,Dx) and

∗ : Sm1,0(Rn × Rn)→ Sm1,0(Rn × Rn) : p 7→ p∗

is a continuous involution, i.e. (p∗)∗ = p, which coincides with complex conjugationmodulo lower order terms:

p∗(x, ξ) = p(x, ξ) + r(x, ξ), r ∈ Sm−11,0 (Rn × Rn).

Hence the pseudodifferential operators are a rich class of operators with many alge-braic properties. These properties make pseudodifferential operators to an importanttool in index theory, cf. [RS82] or [WRL95].

Finally, we studied the pseudodifferential operator p(x,Dx) on C∞b (Rn), S ′(Rn),and as operator between Bessel potential spaces. We note that

S(Rn) ⊂ C∞b (Rn) ⊂ S ′(Rn), S(Rn) ⊂ Hs2(Rn) ⊂ S ′(Rn).

In order to study p(x,Dx) (initially defined on S(Rn)) on these spaces, the operatorhas to be extended in a suitable way. But, since all spaces are contained in S ′(Rn),it remains to check that the extension is consistent with the restriction of p(x,Dx)defined on S ′(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 C∞b (Rn)due to (4.23), which can be done using the approximation used in the proof ofCorollary 4.24.

50

5 Extensions and Applications

5.1 Resolvents and Parameter-Elliptic Differential Operators

The study parabolic initial value problems

∂tu(x, t) + (Au)(x, t) = 0 for (x, t) ∈ Rn × (0,∞),

u(x, 0) = u0 for x ∈ Rn,

where A is a suitable elliptic operator as e.g. −∆, can be reduced to the resolventequation

(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 linearand closed operator such that D(A) is dense in X. If λ + A is invertible for everyλ ∈ λ0 + Σδ,

Σδ := z ∈ C \ 0 : | arg z| < δ,

where δ ∈ (π/2, π) and λ0 ∈ R, and satisfies

‖(λ+ A)−1‖L(X) ≤C

|λ− λ0|for λ ∈ λ0 + Σδ, (5.1)

thene−At :=

1

2πi

∫Γ

eλt(λ+ A)−1dλ for t ∈ Σδ−π/2

is a bounded linear operator, which depends analytically on t ∈ Σδ−π/2 and whichsatisfies

1. e−tAe−sA = e−(t+s)A for all t, s > 0,

2. ddte−At = −Ae−At for t ∈ Σδ−π/2,

3. limt→0 e−Atu0 = u0 for all u0 ∈ X,

4. ‖e−At‖L(X) + |t|‖Ae−At‖L(X) ≤ Ceλ0t 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 andstrongly continuous semi-group, cf. [Paz83, Theorem 3.1, Chapter 1] or [RR93, The-orem 11.17].

In the following let a(x, ξ) =∑|α|≤m cα(x)ξα, cα ∈ C∞b (Rn), m ∈ N. Moreover,

let a0(x, ξ) =∑|α|=m cα(x)ξα denote its principle symbol

51

Definition 5.2 Let δ ∈ (0, π) and let a(x, ξ) be a differential symbol as above. Thena 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) ∈ C∞b (Rn)n×n is a real symmetric matrix, which is uniformlypositive definite, i.e., ξTA(x)ξ ≥ c > 0, then a(x, ξ) = ξTA(x)ξ is parameter ellipticon 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) beits principal symbol. Then

|∂αξ ∂βx (λ+ a0(x, ξ))−1| ≤ C(1 + |λ|1m + |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:|∂αξ ∂βx (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 ∈ C∞b (Rn), theset

A := −a0(x, ξ) : x ∈ Rn, |ξ| = 1is bounded. Moreover, A ⊂ −Σπ−δ = C \ Σδ. Hence dist(A, ∂Σδ) > 0 since A iscompact, ∂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∂za0(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 mathematicalinduction that

|∂αξ ∂βx (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 ofthe principal symbol, i.e., a0(x, rξ) = rma0(x, ξ) for r > 0,

a0(x, ξ) : x ∈ Rn, |ξ| ≤ 1 =⋃

0≤r≤1

rm · a0(x, ξ) : x ∈ Rn, |ξ| = 1 ⊂ Σπ−δ ∪ 0,

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|∂αξ ∂βx (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

degree −m with respect to (λ−1m , ξ) in the sense that

armλ(x, rξ) = r−maλ(x, ξ) for all r > 0, x, ξ ∈ Rn, λ ∈ Σδ. (5.4)

This implies that ∂αξ ∂βxaλ(x, ξ) is homogeneous of degree −m − |α| with respect to(λ−

1m , ξ), i.e.,

(∂αξ ∂βxarmλ)(x, rξ) = r−m−|α|∂αξ ∂

βxaλ(x, ξ) for all r > 0, x, ξ ∈ Rn, λ ∈ Σδ,

by differentiating (5.4). Replacing (λ, ξ) by (r−mλ, r−1ξ), we have

∂αξ ∂βxaλ(x, ξ) = r−m−|α|∂αξ ∂

βxar−mλ

(x,ξ

r

).

Now choosing r = max(|ξ|, |λ| 1m ), (5.3) implies

|∂αξ ∂βx (λ+ a0(x, ξ))−1| ≤ C max(|ξ|, |λ|1m )−m−|α| ≤ C ′(|λ|

1m + |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 definedabove. Then A = a(x,Dx) : D(A) → Hs

2(Rn), s ∈ R, with D(A) = Hs+m2 (Rn) ⊂

Hs2(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

‖(λ+ A)−1‖L(Hs2(Rn)) ≤ C|λ|−1.

53

Proof: Let λ ∈ Σδ with |λ| ≥ 1. Moreover, let aλ(x, ξ) = (λ + a0(x, ξ))−1 and letm′ ∈ [0,m]. Then for all α, β ∈ Nn

0

|∂αξ ∂βxaλ(x, ξ)| ≤ Cα,β

(1 + |λ|

1m + |ξ|

)−m−|α|≤ Cα,β|λ|−

m−m′m 〈ξ〉−m′−|α| (5.5)

for all λ ∈ Σδ, |λ| ≥ 1, and ξ ∈ Rn. Hence aλ ∈ S−m′

1,0 (Rn × Rn) with |aλ|(−m′)

k ≤Ck|λ|−

m−m′m uniformly in λ ∈ Σδ with |λ| ≥ 1 for every k ∈ N. Therefore

‖aλ(x,Dx)‖L(Hs2(Rn),Hs+m′

2 (Rn))≤ C|λ|−

m−m′m (5.6)

uniformly in λ ∈ Σδ with |λ| ≥ 1 because of Lemma 4.38. Moreover, because ofthe asymptotic expansion stated in Theorem 4.16 and the fact that all terms dependcontinuously 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

|rλ|(0)k ≤ Ck|a0|(m)

n(k)||aλ|(−m+1)n(k) ≤ Ck|λ|−

1m . (5.7)

– More precisely, since a0(x,Dx) is a differential operator,

rλ(x, ξ) =∑

|α|≤m,α 6=0

1

α!∂αξ a0(x, ξ)Dα

xaλ(x, ξ).

We note that one can prove (5.7) directly using this identity. – Thus

‖rλ(x,Dx)‖L(Hs2(Rn)) ≤ C|rλ|(0)

k ≤ C ′|λ|−1m

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) +R′λ,

where R′λ :=∑|α|<m cα(x)Dα

xaλ(x,Dx) satisfies

‖R′λ‖L(Hs2(Rn)) ≤ C‖aλ(x,Dx)‖L(Hs

2(Rn),Hs+m−12 (Rn)) ≤ C|λ|−

1m

due to (5.6) for m′ = m− 1 and Lemma 4.38 again.Hence there is some R > 0 such that ‖rλ(x,Dx)+R′λ‖L(Hs

2(Rn)) ≤ 12for all |λ| ≥ R.

Therefore I + rλ(x,Dx) +R′λ is invertible and

‖(I + rλ(x,Dx) +R′λ)−1‖L(Hs

2(Rn)) ≤ 2

for all |λ| ≥ R. Thus

(λ+ a(x,Dx))aλ(x,Dx)(I + rλ(x,Dx) +R′λ)−1 = I

54

for all λ ∈ Σδ, |λ| ≥ R. Therefore λ + a(x,Dx) is surjective and has a continuousright inverse. By the same arguments as before one shows that also

aλ(x,Dx)(λ+ a(x,Dx)) = I +R′′λ,

where ‖R′′λ‖L(Hs+m2 (Rn)) ≤ C|λ| 1m 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. Thereforeλ+ a(x,Dx) is invertible for all |λ| ≥ R, λ ∈ Σδ and

‖(λ+ a(x,Dx))−1‖L(Hs

2(Rn))

≤ ‖aλ(x,Dx)‖L(Hs2(Rn))‖(I + rλ(x,Dx) +R′λ)

−1‖L(Hs2(Rn)) ≤ C|λ|−1.

This proves the lemma.

Remark 5.6 The latter lemma implies that the spectrum of −A, where A is aparameter 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. Thenthere is some λ0 > 0 such that A satisfies the conditions of Theorem 5.1 with X =Hs(Rn), D(A) = Hs+m(Rn) for every s ∈ Rn. In particular, A generates a stronglycontinuous, analytic semi-group e−At, t ≥ 0 on Hs

2(Rn), which satisfies

‖e−At‖L(Hs2(Rn)) + t‖Ae−At‖L(Hs

2(Rn)) ≤ Cseλ0t for all t ≥ 0.

Proof: Just choose some λ0 > 0 such that λ0 + Σδ ⊆ Σδ \ BR(0). Then applyTheorem 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 ′(Rn)

there is a unique k ∈ S ′(Rn × Rn) such that

〈Tu, v〉S(Rn) = 〈K, u⊗ v〉S(Rn×Rn) for all u, v ∈ S(Rn), (5.8)

where (u⊗ v)(x, y) = u(x)v(y) and 〈f, ϕ〉S(Rm) := f(ϕ) denotes the duality productof f ∈ S ′(Rm), and ϕ ∈ S(Rm), cf. e.g. [Tre67, Corollary to Theorem 51.6]. HereK ∈ S ′(Rn × Rn) is called kernel of T . Formally “Tu =

∫Rn K(x, y)u(y)dy”.

Obviously, this result applies to pseudodifferential operators since p(x,Dx) : S(Rn)→S(Rn) ⊂ S ′(Rn) for p ∈ Sm1,0(Rn × Rn) and m ∈ R arbitrary.

For the case that p ∈ Sm1,0(Rn ×Rn) with m ≤ −n− 1, we have seen in the proofof Lemma 4.32 that

p(x,Dx)f =

∫Rnk(x, x− y)f(y)dy for all f ∈ S(Rn),

where k(x, z) = F−1ξ 7→x[p(x, .)] is a continuous function in z ∈ Rn. 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 distributionK ∈ S ′(Rn × Rn) defined by

〈K, u⊗ v〉S(Rn×Rn) =

∫Rnu(x)v(x)dx

Formally “K(x, y) = δ0(x− y)” and

p(x,Dx)u(x) = δ0 ∗ u(x) := 〈δ0, u(x− ·)〉S(Rn) = u(x) for all u ∈ S(Rn).

The main result of this section gives a precise statement of the form of the Schwartzkernel of a pseudodifferential operator away from the diagonal (x, y) ∈ Rn × Rn :x = y.

THEOREM 5.8 Let p ∈ Sm1,0(Rn × Rn), m ∈ R. Then there is a smooth functionk : Rn × (Rn \ 0) such that

p(x,Dx)u(x) =

∫Rnk(x, x− y)u(y) dy for all x 6∈ suppu (5.9)

56

for all u ∈ S(Rn). Moreover, for every α, β ∈ Nn0 , N ∈ N0, k satisfies

|∂βx∂αz k(x, z)| ≤

Cα,β,N |z|−n−m−|α|〈z〉−N if n+m+ |α| > 0,

Cα,β,N(1 + | log |z||)〈z〉−N if n+m+ |α| = 0,

Cα,β,N〈z〉−N if n+m+ |α| < 0

(5.10)

uniformly in x, z ∈ Rn, z 6= 0. In particular, we that

〈p(x,Dx)u, v〉S(Rn) =

∫Rn×Rn

k(x, x− y)u(y)v(x)d(x, y),

for all u, v ∈ S(Rn) with suppu∩supp v = ∅ and the Schwartz kernel K ∈ S ′(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 thetheory of function spaces - the dyadic decomposition of the phase space.

Let ϕ0 ∈ C∞0 (Rn) such that ϕ0(ξ) = 1 for |ξ| ≤ 1 and ϕ0(ξ) = 0 for |ξ| ≥ 2.Moreover, let ϕj(ξ) = ϕ0(2−jξ)− ϕ0(2−j+1ξ) for j ∈ N. Then

suppϕj ⊆ ξ ∈ Rn : 2j−1 ≤ |ξ| ≤ 2j+1, j ∈ N, suppϕ0 ⊆ B2(0)

and∞∑j=0

ϕj(ξ) =k∑j=0

ϕj(ξ) = ϕ0(2−kξ) = 1

for ξ ∈ Rn and k ∈ N such that |ξ| ≤ 2k. Hence ϕj, j ∈ N, is a partition of unity onRn 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 de-scribed with the aid of this partition of unity an alternative way as follows:

|f(ξ)| ≤ C〈ξ〉m ⇔ supξ∈Rn|ϕj(ξ)f(ξ)| ≤ C ′2jm for all j ∈ N0

where m ∈ R and C ′ > 0 does not depend on j. - See exercises.

Obviously, ϕj ∈ S−∞1,0 (Rn × Rn) since ϕj ∈ C∞0 (Rn). Moreover,

N∑j=0

ϕj(ξ) = ϕ0(2−Nξ)→N→∞ 1 pointwise and

N∑j=0

∂αξ ϕj(ξ)→N→∞ 0 uniformly if α 6= 0

57

since for every ξ ∈ Rn there are at most two non-zero terms in the sums and

∂αξ ϕj(ξ) = 2−|α|(j−1)∂αξ ϕ1(2−j+1ξ) for all α ∈ N0, j ∈ N. (5.11)

Hence∞∑j=0

ϕj(ξ)f = f and∞∑j=0

ϕj(Dx)f = f in S(Rn)

since |∑N

j=0 ϕj(ξ)f − f |′′k,S →N→∞ 0 for all k ∈ N by dominated convergence, where|.|′′k,S , k ∈ N, is the equivalent sequence of semi-norms on S(Rn) which was definedin Corollary 4.44 by replacing ‖.‖∞ by ‖.‖2.

With the aid of the dyadic decomposition we decompose a pseudodifferentialoperator as

p(x,Dx)f =∞∑j=0

p(x,Dx)ϕj(Dx)f =∞∑j=0

pj(x,Dx)f for all f ∈ S(Rn), (5.12)

where pj(x, ξ) = p(x, ξ)ϕj(ξ) ∈ S−∞1,0 (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 compactlysupported in ξ,

pj(x,Dx)f =

∫Rnkj(x, x− y)f(y)dy, (5.13)

where kj(x, z) = F−1ξ 7→z[pj(x, ξ)] as in the proof of Lemma 4.32.

Lemma 5.10 Let p ∈ Sm1,0(Rn × Rn), m ∈ R, and let kj(x, z) be defined as above.Then

|∂βx∂αz kj(x, z)| ≤ Cα,β,M |z|−M2j(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,

zγ∂βxDαz kj(x, z) =

∫Rneix·ξDγ

ξ [ξα∂βxpj(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 boundedby 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 ≤ 〈ξ〉 ≤ C2j on 2j−1 ≤ |ξ| ≤ 2j+1,∣∣Dγ

ξ [ξα∂βxpj(x, ξ)]∣∣ ≤ Cα,β,γ2

j(m+|α|−|γ|)

due to the symbol estimates of ξα∂βxpj(x, ξ) ∈ Sm+|α|1,0 (Rn × Rn). Hence

|zγDβxD

αz kj(x, z)| ≤ Cα,β,γ2

j(n+m+|α|−M) whenever |γ| = M.

58

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)

p(x,Dx)u(x) =∞∑j=0

∫Rnkj(x, x− y)u(y) dy for all x ∈ Rn. (5.15)

We will show (5.9) and (5.10) by showing that∞∑j=0

∂αz ∂βxkj(x, z)

converges absolutely and uniformly w.r.t (x, ξ) ∈ Rn × (Rn \Bε(0)) for every ε > 0to a function k(x, z) satisfying (5.10).

First let 0 < |z| ≤ 1. Then we split the sum into∑2j≤|z|−1

∂αz ∂βxkj(x, z) and

∑2j>|z|−1

∂αz ∂βxkj(x, z).

In order to estimate the first sum, we use Lemma 5.10 with M = 0 and obtain

∑2j≤|z|−1

|∂αz ∂βxkj(x, z)| ≤ Cα,β

bld(|z|−1)c∑j=0

2j(n+m+|α|)

≤

Cα,β|z|−(m+n+|α|) if m+ n+ |α| > 0,

Cα,β(1 + | log |z|−1|) 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| andestimate ∑

2j>|z|−1

|∂αz ∂βxkj(x, z)| ≤ Cα,β|z|−M∞∑

j=bld(|z|−1)c+1

2j(n+m+|α|−M)

≤ Cα,β|z|−(n+m+|α|).

Finally, if |z| ≥ 1, we choose M > n+m+ |α|+N in Lemma 5.10 to conclude∞∑j=0

|∂αz ∂βxkj(x, z)| ≤ |z|−M∞∑j=0

2j(n+m+|α|−M)

≤ C|z|−M ≤ C|z|−m−n−|α|−N .

Hence∑∞

j=0 kj(x, z) converges absolutely and uniformly with respect to (x, ξ) ∈Rn × (Rn \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 withdist(x, suppu) < ε for arbitrary ε > 0. Hence (5.9) follows for all x 6∈ suppu.

We have the following consequences of Theorem 5.8:

1. If p ∈ Sm1,0(Rn × Rn), m < 0, then supx∈Rn |k(x, z)| ∈ L1(Rn) and

p(x,Dx)u(x) =

∫Rnk(x, x− y)u(y) dy for a.e. x ∈ Rn

for all u ∈ Lq(Rn), 1 ≤ q ≤ ∞. Therefore

‖p(x,Dx)f‖q ≤ supx∈Rn‖k(x, ·)‖L1(Rn)‖f‖q, f ∈ Lq(Rn),

for all 1 ≤ q ≤ ∞; cf. exercises.

2. Let p ∈ S01,0(Rn×Rn). Because of the boundedness p(x,Dx) : L2(Rn)→ L2(Rn)

and the kernel estimates

supx∈Rn|k(x, z)| ≤ C|z|−n,

supx∈Rn|∇x,zk(x, z)| ≤ C|z|−n−1,

a general result from the theory of singular integral operators, cf. [Ste93, Chap-ter 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-translation-invariant) generalization of [Abea, Theorem 5.5], which is proved in the sameway.1

The same statement is true for the formal adjoint p∗(x,Dx) since p∗ ∈ S01,0 if

p ∈ S01,0. For 2 < q <∞, this implies that

‖p(x,Dx)u‖Lq(Rn) = supv∈Lq′ (Rn),‖v‖

Lq′(Rn)

=1

∣∣∣∣∫Rnp(x,Dx)u(x)v(x) dx

∣∣∣∣= sup

v∈S(Rn),‖v‖Lq′(Rn)

=1

∣∣∣∣∫Rnu(x)p∗(x,Dx)v(x) dx

∣∣∣∣≤ sup

‖v‖Lq′(Rn)

=1

‖u‖Lq(Rn)‖p∗(x,Dx)v‖Lq′ (Rn) ≤ C‖u‖Lq(Rn)

for all u ∈ S(Rn), where we have used the Riesz representation theorem forLq(Rn)′. Therefore we have that the mapping (5.16) is bounded for every1 < q <∞.

1A complete proof can also be found in [Abeb, Chapter IV].

60

3. If ϕ, ψ ∈ C∞b (Rn) with dist(suppϕ, suppψ) > 0, then ϕ(x)p(x,Dx)ψ(x) is asmoothing operator in the sense that

ϕ(x)p(x,Dx)ψ(x) : S ′(Rn)→ C∞(Rn).

This statement can be proved as follows: If f ∈ S ′(Rn), then there is an N ∈ Nsuch that f ′ := 〈x〉−2N〈Dx〉−2Nf ∈ L2(Rn), cf. exercises. Hence

ϕ(x)p(x,Dx)ψf = ϕ(x)

∫Rnk(x, x− y)ψ(y)〈Dy〉2N(〈y〉2Nf ′(y))dy

since suppϕ ∩ suppψ = ∅ implies ϕ(x)k(x, x − y)ψ(y) = 0 in a neighborhoodof the diagonal x = y. Moreover,

ϕ(x)p(x,Dx)(ψf) =

∫Rn

[〈Dy〉2N(ϕ(x)k(x, x− y)ψ(y))

]〈y〉2Nf ′(y)dy,

where Theorem 5.8 implies that 〈Dy〉2Nϕ(x)k(x, x− y)ψ(y) is a smooth kernelwhich is rapidly decreasing in y. This implies the statement.

4. If f ∈ S ′(Rn), then the singular support of f denoted by sing supp f is thecomplement of the set U of all x ∈ Rn such that f |V coincides with a smoothfunction for an open neighborhood V of x, i.e.,

〈f, ϕ〉 =

∫Rnf(x)ϕ(x)dx for all ϕ ∈ C∞0 (V ),

where f ∈ C∞(V ). It is easy to show that U is open and therefore sing supp fis closed.

If p ∈ Sm1,0(Rn × Rn), it holds that

sing supp p(x,Dx)f ⊆ sing supp f for all f ∈ S ′(Rn).

An operator satisfying this property is called pseudo-local. The statement canbe proved using the definitions and the previous statement.

61

5.3 Coordinate Transformations and Pseudodifferential Op-erators 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 ofpseudodifferential operators on a manifold in a way that is essentially independentof the choice of local charts. We note that

Qu = κ∗p(x,Dx)κ∗,−1u,

where (κ∗v)(x) = v(κ(x)) and (κ∗,−1u)(x) = u(κ−1(x)).More precisely, we assume that κ : Rn → Rn is a smooth function such that

∂xjκ ∈ C∞b (Rn) for all j = 1, . . . , n and

0 < c ≤ | detDκ(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 ∂xjκ

−1 ∈ C∞b (Rn) for all j = 1, . . . , n. In the following∇xκ the total derivative of κ : Rn → Rn.

The main result of this section is:

THEOREM 5.11 Let p ∈ Sm1,0(Rn × Rn), m ∈ R, let κ : Rn → Rn be as above andQ : S(Rn) → S(Rn) be defined by (5.17). Then there is some q ∈ Sm1,0(Rn × Rn)such that Qu = q(x,Dx)u for all u ∈ S(Rn). Moreover, q(x,Dx) has the asymptoticexpansion

q(x, ξ) ∼∑α∈Nn0

1

α!∂αξD

αy q(x, y, ξ)|y=x, where

q(x, y, ξ) = p(κ(x), A(x, y)−T ξ)| detA(x, y)|−1| det∇yκ(y)| and

A(x, y) =

∫ 1

0

∇yκ(x+ t(y − x)) dt for all x, y close enough

in the sense that for any N ∈ N0

q(x, ξ)−∑|α|≤N

1

α!∂αξD

αy q(x, y, ξ)|y=x ∈ Sm−N−1

1,0 (Rn × Rn).

In particular, we have

q(x, ξ) = p(κ(x), (∇yκ(x))−T ξ) + r(x, ξ) with r ∈ Sm−11,0 .

62

Proof: We only prove the theorem under the additional assumption that

supx∈Rn|∇κ(x)− I| ≤ 1

2.

For the proof in the general case we refer to [KG74, Theorem 6.3]. Here |.| is anymatrix norm, which is induced by some vector norm. This ensures that

|A(x, y)− I| ≤ 1

2for 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

Qu = limε→0

∫Rn×Rn

ei(κ(x)−y)·ξχε(ξ)p(κ(x), ξ)u(κ−1(y))d(y, ξ)

(2π)n

= limε→0

∫Rn×Rn

ei(κ(x)−κ(x′))·ξχε(ξ)p(κ(x), ξ)u(x′)| det∇x′κ(x′)| d(x′, ξ)

(2π)n

for all u ∈ S(Rn). On the other hand, since

κ(x)− κ(x′) =

(∫ 1

0

∇x′κ(x′ + t(x− x′)) dt)

(x− x′) = A(x′, x)(x− x′),

we obtain by the change of variable ξ = A(x, x′)−Tη

Qu =1

(2π)nlimε→0

∫Rn×Rn

ei(x−x′)·ηχε(A(x′, x)−Tη)q(x, x′, η)u(x′) d(x′, η)

= limε→0

Os–

∫∫e−iy·ηχε(A(x+ y, x)−Tη)q(x, x+ y, η)u(x+ y) dx′đη

due to (κ(x)− κ(x′)) · ξ = (x− x′) · A(x, x)−1ξ, where

q(x, x′, η) = p(κ(x), A(x′, x)−Tη)| detA(x, x′)|−1| det∇x′κ(x′)|.

Since χε(A(x, x′)−Tη) : ε ∈ (0, 1) is bounded in A00(Rn×Rn) with respect to (x′, η)

and

χε(A(x′, x)−Tη)→ε→0 1 for all x, x′, η ∈ Rn

∂αx′∂βηχε(A(x′, x)−Tη)→ε→0 0 for all x, x′, η ∈ Rn, |α|+ |β| > 0,

we can apply Corollary 4.10 to conclude that

Qu(x) = Os–

∫∫e−iy·η q(x, x+ y, η)u(x+ y) d(x′, η).

Finally, an application of Theorem 4.27 finishes the proof.

Next we define pseudodifferential operators on smooth compact manifolds. Tothis end recall:

63

Definition 5.12 M is a smooth compact manifold if M is a topological compactspace and the following conditions hold:

1. There are finitely many open sets Ω1, . . . ,ΩN ⊂M such that M =⋃Nj=1 Ωj.

2. For every j ∈ 1, . . . , N there is an open set Uj ⊂ Rn and a continuousbijective 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 κj,k : Vx → Vy for someopen 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

C∞0 (Ω) = f ∈ C∞(M) : supp f ⊂ Ω.

Before we define pseudodifferential operators on a compact manifold we the defi-nition of certain remainder operators:

Definition 5.13 LetM 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 everyj, k ∈ 1, . . . , N there is some Kj,k ∈ C∞(Uk × Uj) such that for every u ∈ C∞0 (Ωj)

(Pu)(κ−1k (x)) =

∫Uj

Kj,k(x, x′)uj(x

′) dx′ for all x ∈ Uk,

where uj(x) = u(κ−1j (x)) in Uj.

For the following let Φj, j = 1, . . . , N be a smooth partition of unity subordinate toΩj, j = 1, . . . , N , i.e.,

0 ≤ Φj ∈ C∞0 (Ωj),N∑j=1

Φj(x) = 1 for all x ∈M (5.19)

Moreover, let Ψj ∈ C∞0 (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 ∈ C∞0 (Uj) by

ϕj = Φj κ−1j , ψj = Ψj κ−1

j (5.21)

64

Definition 5.14 LetM be a smooth compact manifold as above and let P : C∞(M)→C∞(M) be a linear operator. Then P is a pseudodifferential operator of class Sm1,0(M)with symbols (pj)

Nj=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

(Pu)(x) =N∑j=1

(ΦjPΨju)(x) +N∑j=1

(ΦjP (1−Ψj)u)(x) for all x ∈M. (5.22)

Then, using ϕj, ψj defined in (5.21), ΦjPΨju can be written in the form

(ΦjPΨju)(κ−1j (x)) = ϕjpj(x,Dx)(ψjuj)(x) for all x ∈ Uj

where uj(x) = u(κ−1j (x)), x ∈ Uj, for all u ∈ C∞(M), j = 1, . . . , N and (ΦjP (1 −

Ψj)u) has a C∞-kernel representation.

Remark 5.15 If P is a differential operator, suppP (1 − Ψj)u ⊆ supp(1 − Ψj).Therefore (ΦjP (1−Ψj)u)(x) = 0 for all x ∈M . Hence

(Pu)(x) =N∑j=1

(ΦjPΨju)(x) for all x ∈M

in this case.

The following theorem shows that the definition of pseudodifferential operators onM are independent of the choice of charts κ1, . . . , κN .

THEOREM 5.16 LetM be a smooth compact manifolds as above and let Ω′j, U′j, κ′j, κ′j,k,

j, k = 1, . . . , N ′ be another sequence satisfying conditions 1.-3 in Definition 5.12 suchthat

κj,j′ := κj κ−1j′ : κ′j′(Ωj ∩ Ω′j′)→ κj(Ωj ∩ Ω′j′), j ∈ 1, . . . , N, j′ ∈ 1, . . . , N ′,

extends to a C∞-diffeomorphim on some open neighborhoods of κ′j′(Ωj ∩ Ω′j′) andκj(Ωj ∩ Ω′j′). If P : C∞(M) → C∞(M) is a pseudodifferential operator in Sm1,0(M)with respect to Ωj, Uj, κj, κj,k : j, k = 1, . . . , N, then P is in Sm1,0(M) with respectto Ω′j, U ′j, κ′j, κ′j,k : j, k = 1, . . . , N ′.

Proof: For simplicity we assume that κj,j′ can be extended to a C∞-diffeormphismκj,j′ : Rn → Rn such that ∇κj,j′ ,∇κ−1

j,j′ ∈ C∞b (Rn). For the proof in the general casewe refer to [KG74, Theorem 7.3].

First of all, we prove that for every Ψ,Φ ∈ C∞(M) we have:

supp Φ ∩ supp Ψ = ∅ ⇒ ΦPΨ has a C∞-kernel representation. (5.23)

65

We write

ΦPΨ =N∑j=1

Φ(ΦjPΨj)Ψ +N∑j=1

Φ(ΦjP (1−Ψj))Ψ.

Then Φ(ΦjP (1−Ψj))Ψ has a C∞-kernel representation since ΦjP (1−Ψj) has one.Now we have

(Φ(ΦjPΨj)Ψu)(κ−1j (x)) = (ϕϕj)(x)pj(x,Dx)(ψjψuj)(x) for all x ∈ Uj,

for all u ∈ C∞(M), j = 1, . . . , N , where uj = u κ−1j and ϕ = Φ κ−1

j , ψ =

Ψ κ−1j . Since supp(ψjψ) ∩ supp(ϕjϕ) = ∅, Theorem 5.8 implies that there are

smooth kj : Rn × (Rn \ 0)→ C such that

(Φ(ΦjP (1−Ψj))Ψu)(κ−1j (x)) =

∫Rn

(ϕϕj)(x)kj(x, x− y)(1− ψj(y))ψ(y)uj(y) dy.

Hence (ϕϕj)(x)kj(x, x − y)ψj(y)ψ(y) ∈ C∞(Rn × Rn) and Φ(ΦjP (1 − Ψj))Ψ has aC∞-kernel representation. This proves (5.23).

Let Φ′j′ ,Ψ′j′ ∈ C∞(M), j′ = 1, . . . , N ′, satisfy (5.19)-(5.20) with Ωj replaced by

Ω′j′ . Now let us write P as

P =N ′∑j′=1

Φ′j′PΨ′j′ +N ′∑j′=1

Φ′j′P (1−Ψ′j′).

Because of (5.23), the second term has a C∞-kernel representation. For the first termwe use that

Φ′j′PΨ′j′ =N∑j=1

Φ′j′(ΦjPΨj)Ψ′j′ +

N∑j=1

Φ′j′(ΦjP (1−Ψj))Ψ′j′ ,

where again the second term has a C∞-representation. Moreover, if Ωj ∩ Ω′j′ 6= ∅ byassumption

(Φ′j′ΦjP (ΨjΨ′j′u))(κ−1

j (x)) = ϕ′j′(x)ϕj(x)(pj(x,Dx)(ψ′j′ψjuj))(x) for all x ∈ Uj.

Hence

(Φ′j′ΦjP (ΨjΨ′j′u))((κ′j′)

−1(x))

= ϕ′j′(x)((

(κ′j′ κ−1j )−1

)∗ (ϕjpj(x,Dx)(ψ

′j′(κ

′j′ κ−1

j )∗(ψju′j′))))

(x) for all x ∈ Uj.

where u′j′(x) = u((κ′j′)−1(x)), (κ∗v)(x) = v(κ(x)) for κ : Rn → Rn, and ϕ′j′ = Φ′j′

κ−1j′ , ψ

′j′ = Ψ′j′ κ−1

j′ . Now, since κj,j′ := κ′j′ κ−1j can be extended to some C∞-

diffeomorphism on Rn with ∇xκj,j′ ,∇xκ−1j,j′ ∈ C∞b (Rn), Theorem 5.11 implies that

there is some qj,j′ ∈ Sm1,0(Rn × Rn) such that

qj,j′(x,Dx)u =((κ′j′ κ−1

j )−1)∗ (

ϕjpj(x,Dx)(ψj((κ

−1j κ′j′)∗u

)).

66

If we now defineqj(x, ξ) =

∑j′:Ω′

j′∩Ωj 6=∅

qj,j′(x, ξ),

we have that

(Φ′j′PΨ′j′u)((κ′j)−1(x)) = ϕ′jqj(x,Dx)(ψ

′ju′j′)(x) for all x ∈ U ′j,

which proves the theorem.

67

5.4 Boundedness on Hölder-Zygmund and Besov Spaces

Definition 5.17 Let s > 0. Then the (Hölder-)Zygmund space Cs∗(Rn) is defined as

Cs∗(Rn) = u ∈ C0

b (Rn) : ‖ϕj(Dx)u‖∞ ≤ C2−js, j ∈ N0,‖u‖Cs∗ = sup

j∈N0

2js‖ϕj(Dx)u‖∞.

Remark 5.18 If s 6∈ N, then Cs∗(Rn) coincides with the usual Hölder space Cs(Rn)

normed by

‖u‖Cs = max|α|≤[s]

‖Dαu‖∞ + max|α|≤[s]

supx,y∈Rn,x 6=y

|Dαu(x)−Dαu(y)||x− y|s−[s]

.

This is a consequence of Theorem 5.27 below.

THEOREM 5.19 Let p ∈ Sm1,0(Rn×Rn), m ∈ R, and let s > 0 such that s+m > 0.Then p(x,Dx) : Cs+m

∗ (Rn)→ Cs∗(Rn) is a bounded linear operator.

Lemma 5.20 Let p ∈ Sm1,0(Rn×Rn), m ∈ R, and let pj(x,Dx) = p(x,Dx)ϕj(Dx) bedefined as above. Then

‖pj(x,Dx)‖L(Lq(Rn)) ≤ C2jm, j ∈ N0, (5.24)

for 1 ≤ q ≤ ∞ where C does not dependent on j.

Proof: As seen in Section 5.2,

pj(x,Dx)f =

∫Rnkj(x, x− y)f(y)dy,

where the kj satisfy (5.14). According to these estimates∫Rn|kj(x, z)|dz ≤ C

(∫|z|≤2−j

2j(n+m)dz +

∫|z|>2−j

|z|−n−12j(m−1)dz

),

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∫

Rn|kj(x, z)|dz ≤ C2jm,

which proves (5.24) since∥∥∥∥∫Rnkj(x, x− y)f(y)dy

∥∥∥∥q

≤ supx∈Rn

∫Rn|kj(x, z)|dz‖f‖q.

68

Remark 5.21 If pj(Dx, x) = ϕj(Dx)p(Dx, x) is the corresponding operator in y-form, then

pj(Dx, x)f =

∫Rn

∫Rneiξ·(x−y)pj(y, ξ)f(y)đξdy =

∫Rnkj(y, x− y)f(y)dy,

where kj(y, z) = F−1ξ 7→z[pj(y, ξ)] is the same function as for the operator in x-form.

Moreover,‖pj(Dx, x)‖L(Lq(Rn)) ≤ C2jm, j ∈ N0 (5.25)

holds for 1 ≤ q ≤ ∞. This inequality is proved in precisely the same way as (5.24).

Lemma 5.22 Let p ∈ Sm1,0(Rn × Rn), m ∈ R, and let pi(x,Dx) be defined as above.Then for every l ∈ N

‖ϕj(Dx)pi(x,Dx)‖L(Lq(Rn)) ≤ Cl2min(i,j)m2−|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,

1 =

(1

1 + |ξ|2+

n∑j=1

ξj1 + |ξ|2

ξj

)l

=∑|α|≤l

p(α)(ξ)ξα, (5.27)

where p(α) ∈ S−l1,0(Rn × Rn). Hence

ϕj(Dx)pi(x,Dx) = ϕj(Dx)∑|α|≤l

p(α)(Dx)Dαxpi(x,Dx) =

∑|α|≤l

p(α)(Dx)ϕj(Dx)Dαxpi(x,Dx)

since ϕj(Dx) and p(α)(Dx) commute. Since p(α)(ξ) ∈ S−l1,0(Rn×Rn) andDαxpi(x,Dx) =

q(x,Dx)ϕi(Dx) = qi(x,Dx) with q ∈ Sm+|α|1,0 (Rn × Rn), we can now invoke (5.24) for

p(α)j (Dx) = p(α)(Dx)ϕj(Dx) and qi(x,Dx) to get

‖ϕj(Dx)pi(x,Dx)‖L(Lq(Rn))

≤∑|α|≤l

‖p(α)(Dx)ϕj(Dx)‖L(Lq(Rn))‖qi(x,Dx)‖L(Lq(Rn)) ≤ C2−jl · 2i(l+m),

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

ϕj(Dx)pi(x,Dx) = ϕj(Dx)q(Dx, x)ϕi(Dx) =∑|α|≤l

qj(Dx, x)Dαxp

(α)(Dx)ϕi(Dx).

69

Now using (5.25) for qj(Dx, x)Dαx and p(α)(Dx)ϕi(Dx) = p

(α)i (x,Dx), we conclude in

the same way as before that

‖ϕj(Dx)pi(x,Dx)‖L(Lq(Rn))

≤∑|α|≤l

‖qj(x,Dx)Dαx‖L(Lq(Rn))‖p(α)(Dx)ϕi(Dx)‖L(Lq(Rn)) ≤ C2−il · 2j(l+m),

which proves (5.26) for the case i > j.

Proof of Theorem 5.19: First of all,

f =∞∑j=0

fj,

where fj = ϕj(Dx)f , f ∈ S ′(Rn). Since suppϕj ∩ suppϕk = ∅ of |j − k| > 1,

p(x,Dx)f =∞∑k=0

pk(x,Dx)f =∞∑k=0

pk(x,Dx)(fk−1 + fk + fk+1) =∞∑k=0

pk(x,Dx)fk,

where fk = fk−1 + fk + fk+1 and we have set f−1 = 0. If f ∈ Cs+m∗ (Rn), then

2(s+m)k‖fk‖∞ ≤ 2(s+m)k (‖fk−1‖∞ + ‖fk‖∞ + ‖fk+1‖∞) ≤ C‖f‖Cs+m∗Now using (5.26),

2sj‖ϕj(Dx)pk(x,Dx)fk‖∞ ≤ C2sj+min(j,k)m−|j−k|l−k(s+m)2k(s+m)‖fk‖∞,

where

2sj+min(j,k)m−|j−k|l−k(s+m) =

2j(s−l)+k(l−s) = 2−|j−k|(l−s) if j ≥ k,

2−k(m+s+l)+j(m+s+l) = 2−|j−k|(m+s+l) if j < k.

Choosing l ∈ N0 so large that l − s ≥ 1 and m+ s+ l ≥ 1,

‖p(x,Dx)f‖Cs∗ = supj∈N0

2sj‖ϕj(Dx)p(x,Dx)f‖∞

≤ C supj∈N0

∞∑k=0

2−|j−k|2k(s+m)‖fk‖∞ ≤ C‖f‖Cs+m∗ .

THEOREM 5.23 Let 0 < s < 1. Then Cs∗(Rn) = Cs(Rn) with equivalent norms,

where Cs(Rn) is normed by

2‖f‖Cs = ‖f‖∞ + supx,y∈Rn

|f(x)− f(y)||x− y|s

70

Proof: First let f ∈ Cs(Rn). Then

supx∈Rn|f(x− y)− f(x)| ≤ ‖f‖Cs|y|s

for all y ∈ Rn. Moreover, ϕj(Dx) for j ∈ N is the convolution with ψ2−j(x) :=2jnψ(2jx), ψ(x) = F−1

ξ 7→x[ϕ0(ξ)− ϕ0(2ξ)], where∫Rnψ2−j(y)dy =

∫Rnψ(y)dy = F [ψ](0) = 0.

Hence

ϕj(Dx)f =

∫Rnf(x− y)ψ2−j(y)dy =

∫Rn

(f(x− y)− f(x))ψ2−j(y)dy

and therefore

‖ϕj(Dx)f‖∞ ≤ ‖f‖Cs∫Rn|y|s|ψ2−j(y)|dy = 2−js‖f‖Cs

∫Rn|z|s|ψ(z)|dz = C2−js‖f‖Cs

for all j ∈ N. The latter inequality implies ‖f‖Cs∗ ≤ ‖f‖Cs since also ‖ϕ0(Dx)f‖∞ ≤C‖f‖∞.

Conversely let f ∈ Cs∗(Rn). Then f =

∑∞j=0 fj, fj = ϕj(Dx)f , where the sum

converges in C0b (Rn) since ‖fj‖∞ ≤ 2−sj‖f‖Cs∗ . Now, if |y| ≤ 1,

f(x− y)− f(y) =∑

2j≤|y|−1

(fj(x− y)− fj(y)) +∑

2j>|y|−1

(fj(x− y)− fj(y)) .

For the first sum we use the mean value theorem to conclude that

|fj(x− y)− fj(y)| ≤ |y|‖∇fj‖∞.

Moreover, since

∂xkfj = ∂xkϕj−1(Dx)fj + ∂xkϕj−1(Dx)fj + ∂xkϕj+1(Dx)fj

we can use (5.24) for p(x, ξ) = ξk ∈ S11,0(Rn × Rn), k = 1, . . . , n, to obtain∑

2j≤|y|−1

|fj(x− y)− fj(y)| ≤ C∑

2j≤|y|−1

|y|2j‖∇fj‖∞

≤ C|y|∑

2j≤|y|−1

2j(1−s)‖f‖Cs∗ ≤ C|y|s‖f‖Cs∗ .

The second sum is simply estimated by∑2j>|y|−1

|fj(x− y)− fj(y)| ≤ C∑

2j>|y|−1

‖fj‖∞ ≤ C‖f‖Cs∗∑

2j>|y|−1

2−js = C|y|s‖f‖Cs∗

Altogether ‖f‖Cs ≤ C‖f‖Cs∗ .

71

We now come to a generalization of the Zygmund-Hölder spaces. Let X be anarbitrary 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, ‖(ak)k∈N0‖`q(N0,X) <∞, where

‖(ak)k∈N0‖`q(N0,X) =

(∞∑k=0

‖ak‖qX

) 1q

for 1 ≤ q <∞,

‖(ak)k∈N0‖`∞(N0,X) = supk∈N0

‖ak‖X .

Definition 5.24 Let 1 ≤ q, r ≤ ∞ and let s ∈ R. Then the Besov space Bsqr(Rn) is

defined as

Bsqr(Rn) = u ∈ S ′(Rn) : (2sjϕj(Dx)u)j∈N0 ∈ `r(N0, L

q(Rn)),‖u‖Bsqr = ‖(2sjϕj(Dx)u)j∈N0‖`r(N0,Lq(Rn))

=

(∑∞

j=0 ‖2sjϕj(Dx)u‖rLq) 1r if 1 ≤ r <∞,

supj∈N0‖2sjϕj(Dx)u‖Lq if r =∞.

Here s is called order of Bsq,r(Rn), q is called integral exponent, and r is called sum

exponent.

Remark 5.25 1. If s > 0, then Bs∞∞(Rn) = Cs

∗(Rn).

2. For the case q = r = 2 it can be shown that Bs22(Rn) = Hs

2(Rn), cf. [BL76].

3. The Besov space Bsqq(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

γ : Hsq (Rn)→ B

s− 1q

qq (Rn−1)

such that (γu)(x′) = u(x′, 0) for all u ∈ S(Rn). For a proof of this fact andfor people which interested in Bessel potential and Besov spaces we refer to[BL76].

THEOREM 5.26 Let p ∈ Sm1,0(Rn × Rn), m ∈ R, and let s ∈ R, 1 ≤ q, r ≤ ∞.Then p(x,Dx) : Bs+m

qr (Rn)→ Bsqr(Rn) is a bounded linear operator.

The proof is a simple modification of the proof of 5.19. For the convenience of thereader we include the complete modified proof here:

Proof: First of all,

f =∞∑j=0

fj,

72

where fj = ϕj(Dx)f , f ∈ S ′(Rn). Since suppϕj ∩ suppϕk = ∅ of |j − k| > 1,

p(x,Dx)f =∞∑k=0

pk(x,Dx)f =∞∑k=0

pk(x,Dx)(fk−1 + fk + fk+1) =∞∑k=0

pk(x,Dx)fk,

where fk = fk−1 + fk + fk+1 and we have set f−1 = 0. If f ∈ Bs+mqr (Rn), then

‖(2(s+m)kfk)k∈N0‖`r(N0,Lq(Rn)) = ‖(2(s+m)k(fk−1 + fk + fk+1))k∈N0‖`r(N0,Lq(Rn))

≤ C∥∥∥(2(s+m)kϕk(Dx)f

)k∈N0

∥∥∥`r(N0,Lq(Rn))

= C‖f‖Bs+mqr

Now using (5.26),

2sj‖ϕj(Dx)pk(x,Dx)fk‖Lq ≤ C2sj+min(j,k)m−|j−k|l−k(s+m)2k(s+m)‖fk‖Lq ,

where

2sj+min(j,k)m−|j−k|l−k(s+m) =

2j(s−l)+k(l−s) = 2−|j−k|(l−s) if j ≥ k,

2−k(m+s+l)+j(m+s+l) = 2−|j−k|(m+s+l) if j < k.

Choosing l ∈ N so large that l − s ≥ 1 and m+ s+ l ≥ 1,

2sj‖ϕj(Dx)p(x,Dx)f‖Lq ≤ C∞∑k=0

2−|j−k|2k(s+m)‖fk‖Lq

Now let aj = 2−|j|, bj = 2(s+m)j‖fj‖Lq for j ∈ N0 and aj = bj = 0 for j ∈ Z \ N0.Then

‖p(x,Dx)f‖Bsqr ≤ C ‖(aj)j∈Z ∗ (bj)j∈Z‖`r ≤ C‖(aj)j∈Z‖`1‖(bj)j∈Z‖`r

= C‖(2(s+m)j fj)j∈N0‖`r(N0,Lq(Rn)) ≤ C‖f‖Bs+mqr,

where the sequence (cj)j∈Z = (aj)j∈Z∗(bj)j∈Z is the convolution of (aj)j∈Z and (bj)j∈Zdefined by

cj =∞∑

k=−∞

aj−kbk.

Here we have used the discrete version of Young’s inequality ‖(aj)j∈Z ∗ (bj)j∈Z‖`r ≤‖(aj)j∈Z‖`1‖(bj)j∈Z‖`r , which can be proved in the same way as for the usual convo-lution using Hölder’s inequality.For completeness we state the following theorem, which shows that the Besov spacescan be considered as generalization of Hölder spaces:

THEOREM 5.27 Let 1 ≤ q <∞ and let 0 < s < 1. Then

‖f‖′Bsqq =

(‖f‖qq +

∫Rn

∫Rn

|f(x)− f(y)|q

|x− y|n+sqdxdy

) 1q

is an equivalent norm on Bsqq(Rn).

See [BL76] for the proof.

73

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