Degree project

The Wave Equation in

Modulation Spaces

Author: Maximilian Reich

Supervisor: Joachim Toft

Examiner: Börje Nilsson

Date: 2013-06-13

Course Code: 4MA11E

Subject: Mathematics

Level: Master

Department of Mathematics

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Abstract. Considering the initial value problem for the homogeneouswave equation we find a general solution for it and investigate propertiesof this solution together with some explicit integral representations. Inorder to obtain better results regarding the regularity of the solution weintroduce modulation spaces. In this context we also consider Sobolevspaces as a special case of modulation spaces.

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Acknowledgment

I want to express my gratitude to all the people who have supported andassisted me throughout my master thesis.I would like to give a special thanks to my supervisor professor JoachimToft, for introducing me to this interesting topic. Moreover, I would alsolike to thank him for his friendly and patient support, his advice andexplanations were always very helpful.Last but not least I would like to thank my family, who has alwayssupported me during my studies. A special thanks to my parents.

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Contents

1. Introduction 52. General Solution of the Wave Equation 53. Integral Representations of the Solution 73.1. Integral Representation in R 73.2. Integral Representation in R2 83.3. Integral Representation in R3 94. Spaces of Distributions 115. Modulation Spaces 125.1. The Short-Time Fourier Transform (STFT) 125.1.1. Introduction and Definition of STFT 125.1.2. Modulation and Translation 135.1.3. Properties of STFT 145.1.4. Extensions of the STFT 185.2. Definition and Properties 195.3. The Fourier Transform 266. Classes of the Factors in the Solution Representation 297. Classes of the Solution of the Wave Equation 317.1. Solution in Modulation Spaces 317.2. Solution in Sobolev Spaces 37References 41

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1. Introduction

The homogeneous wave equation is given by

∂2t u−∆u = 0

in the cylinder [0, T ] × G, where G ⊆ Rd is an arbitrary domain and T > 0a positive number. There is exactly one boundary condition of the

• 1. type (Dirichlet condition)u|∂G = h (providing the solution on ∂G)• 2. type (Neumann condition)∂nu|∂G = h (providing the normal derivative on ∂G)• 3. type (Robin condition)

(h1(x)u+ h2(x)∂nu)|∂G = h (providing a linear combination consistingof the boundary conditions of 1. and 2. type; h1, h2, h are given functionson ∂G)

on (0, T )× ∂G and two initial conditions u(0, x) = f(x), ∂tu(0, x) = g(x) on{t = 0} × G. This holds for interior domains. For exterior domains there isan additional decay condition.A special case is G = Rd. Then there only remains an initial value problem, aso-called Cauchy problem, with the initial conditions u(0, x) = f(x), ∂tu(0, x) =g(x).Within the scope of this paper we only consider the Cauchy problem for thehomogeneous wave equation. The first goal is to find a general solution inorder to determine its properties and reveal some explicit integral represen-tations. Moreover, a very important aspect is the regularity of the solutioncompared to the initial data. Since we only obtain some unsatisfactory resultsfor our purposes, that is a loss of regularity in the solution, we need to intro-duce spaces of distributions. In particular we deeply investigate modulationspaces and consequently sketch a concept of time-frequency analysis. In theend we obtain solutions of our Cauchy problem in modulation spaces andSobolev spaces which do not loose regularity.

2. General Solution of the Wave Equation

We consider the initial value problem for the homogeneous wave equation for(t, x) ∈ Rd+1 which is given by

∂2t u(t, x)−∆u(t, x) = 0, u(0, x) = f(x), ut(0, x) = g(x), (1)

where ∆ denotes the Laplace-operator

∆ =∂2

∂x21+

∂2

∂x22+ . . .+

∂2

∂x2d.

Firstly we are interested in a general solution of (1). For this we stress theFourier transform which is defined by

Ff(ξ) = f(ξ) = (2π)−d2

∫Rdf(x)e−ix·ξdx (x, ξ ∈ Rd)

6

for all Lebesgue integrable functions f ∈ L1(Rd). The inverse Fourier trans-form is defined by

F−1f(x) = f(x) = (2π)−d2

∫Rdf(ξ)eix·ξdξ.

Subsequently we assume that all Fourier transforms exist. We define

F2(u)(t, ξ) = U(t, ξ)

where F2 is the partial Fourier transform of u = u(t, x) in the x-variable. Itfollows

F2(utt)(t, ξ) = Utt(t, ξ),

F2(u)(0, ξ) = f(ξ) and

F2(ut)(0, ξ) = g(ξ).

Note that ut(t, x) = ∂tu(t, x) and utt(t, x) = ∂2t u(t, x), respectively. Further-more by introducing the notation Dj = 1

i∂∂xj

for 1 ≤ j ≤ d, one can easily

prove that

F2(Dju)(t, ξ) = ξj u(t, ξ) (x, ξ ∈ Rd).Thus, we obtain

F2(∆u)(t, ξ) = (−ξ21 − ξ22 − . . .− ξ2d)U(t, ξ) = −|ξ|2U(t, ξ).

Summarizing we transformed the problem (1) by Fourier transform into theproblem

Utt(t, ξ) + |ξ|2U(t, ξ) = 0, U(0, ξ) = f(ξ), Ut(0, ξ) = g(ξ). (2)

This equation is an ordinary differential equation in t with the solution

U(t, ξ) = C1 sin(|ξ|t) + C2 cos(|ξ|t)with arbitrary constants C1 and C2. Due to the initial values we can calculatethem in the following way

U(0, ξ) = C2 = f(ξ),

Ut(0, ξ) = C1|ξ| = g(ξ).

Hence we obtain the solution of (2)

U(t, ξ) = g(ξ)sin(|ξ|t)|ξ|

+ f(ξ) cos(|ξ|t).

Now we apply the inverse Fourier transform on this solution to transform itback from the phase space into the physical space. So we obtain

u(t, x) =F−12 (U)(t, x)

=F−12

(g · sin(|ξ|t)

|ξ|

)(t, x) + F−12 (f · cos(|ξ|t))(t, x)

=F−12

(F2(g)(ξ) · sin(|ξ|t)

|ξ|

)(t, x) + F−12 (F2(f)(ξ) · cos(|ξ|t))(t, x)

(3)

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by linearity of the Fourier transform. This is the general solution of theCauchy problem (1).

3. Integral Representations of the Solution

3.1. Integral Representation in R

Now we can look for explicit integral representations for (3). For that wefirstly consider the one-dimensional case R. By an easy computation it followsthat for the function

h1(t, x) =

√

π2 , −t < x < t√π8 , x = ±t

0, otherwise

the partial Fourier transform in the x-variable h1(t, ξ) is given by

h1(t, ξ) =sin(|ξ|t)|ξ|

.

Moreover, it is easy to check that the partial Fourier transform in x of thefunction

h2(t, x) =

√π

2(δ(x− t) + δ(x+ t)),

where δ = δ(x) is the Dirac function with respect to the x-variable, is givenby

h2(t, ξ) = cos(|ξ|t).

In order to get the desired representation we stress the so-called convolutiontheorem which yields

F((f ∗ g)(x))(ξ) = (2π)d2 f(ξ) · g(ξ) (x, ξ ∈ Rd)

for appropriate functions f and g. Note that we treat the case d = 1. Atthis point we recall the definition of the convolution f ∗ g of two admissiblefunctions f = f(x) and g = g(x). On R it is defined by

(f ∗ g)(x) =

∫Rf(y)g(x− y)dy.

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Now we apply these results on the general representation of the solutionu = u(t, x) of (1). We get

u(t, x) = F−12

(F2(g)(ξ) · sin(|ξ|t)

|ξ|

)(t, x) + F−12 (F2(f)(ξ) · cos(|ξ|t))(t, x)

= F−12 (F2(g)(ξ) · F2(h1)(t, ξ))(t, x)

+F−12 (F2(f)(ξ) · F2(h2)(t, ξ))(t, x)

= F−12 ((2π)−12F2(g ∗ h1)(t, ξ))(t, x)

+F−12 ((2π)−12F2(f ∗ h2)(t, ξ))(t, x)

= (2π)−12 ((g ∗ h1)(t, x) + (f ∗ h2)(t, x))

= (2π)−12

∫Rg(x− y)h1(t, y)dy + (2π)−

12

∫Rf(x− y)h2(t, y)dy

=

∫ t

−t

1

2g(x− y)dy +

1

2(f(x− t) + f(x+ t))

=1

2

∫ x+t

x−tg(y)dy +

1

2(f(x− t) + f(x+ t)).

Thus, we exactly obtained d’Alembert’s representation formula for the waveequation in R.Furthermore it is to mention that in R the solution u of (1) does not looseregularity regarding the initial data. This means given the initial data f ∈C2(R) and g ∈ C1(R), the solution u belongs to the space C2([0,∞)× R).

3.2. Integral Representation in R2

Theorem 3.1. If f ∈ C3(R2) and g ∈ C2(R2), then there exists a uniqueclassical solution u = u(t, x) ∈ C2([0,∞)× R2) of (1) such that

u(t, x) =1

2π

∫Bt(x)

g(y)√t2 − |y − x|2

dy +∂

∂t

(1

2π

∫Bt(x)

f(y)√t2 − |y − x|2

dy

),

where the set Bt(x) = {y ∈ R2 : d(x, y) < t} denotes the ball in R2 withcenter x and radius t.

Proof. The proof will only be roughly sketched. In R2 we basically use theresult of Theorem 3.4 and the so-called method of descent. That is for x =(x1, x2, x3) ∈ R3 we consider the initial data f(x) = f(x1, x2) and g(x) =g(x1, x2) as functions on R3 which are independent of x3. Combining this

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with (7) we obtain

u(t, x) =1

4πt

∫St(x1,x2,0)

g(y)dσt(y) +∂

∂t

(1

4πt

∫St(x1,x2,0)

f(y)dσt(y)

)

=1

2π

∫Bt(x1,x2)

g(y)√t2 − |y − x|2

dy

+∂

∂t

(1

2π

∫Bt(x1,x2)

f(y)√t2 − |y − x|2

dy

).

This followed by conversion of the surface integral into an integral over thedomain {y = (y1, y2) : |y − x|2 ≤ t2} with x ∈ R2. In the end we get therelation

dσt(y) =2tdy√

t2 − |y − x|2.

For details see chapter 4 in [6]. �

3.3. Integral Representation in R3

Now we consider the three-dimensional case R3 and the following initial valueproblem

∂2t u−∆u = 0, u(0, x) = 0, ut(0, x) = g(x), (4)

which is slightly different to (1). Then we obtain the so-called Kirchoff’sformula for the solution u of (4) as stated in chapter 4 in [6].

Theorem 3.2. If the function g = g(x) ∈ Ck(R3) for k ≥ 2, then (4) has aclassical solution u ∈ Ck([0,∞)× R3) which is given by

ug(t, x) =1

4πt

∫St(x)

g(y)dσt(y).

St(x) denotes the sphere in R3 centered in x with radius t and dσt denotesthe corresponding surface element.

Before we can prove this theorem we firstly need Gauss’s theorem.

Theorem 3.3. Let V ⊆ Rd be compact and let its boundary ∂V = S be piece-wise smooth. If F is a continuously differentiable vector field defined on aneighborhood of V , then it holds∫

V

(∇ · F )dV =

∫S

(F · n)dS,

where n is the outward pointing unit normal field of the boundary ∂V of V .

Now we have all the tools to prove Theorem 3.2.

Proof. This proof is also stated in [6]. Nevertheless it is explained here as wellsince it follows by rather easy computations and shows how we can obtainexplicit integral representations for the solution of the wave equation.Let y = x+ tα where α = (α1, α2, α3) is the unit vector in (y − x)-direction.

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Furthermore we can rewrite the surface element dσt, which describes a ballsurface in R3 with radius t, as dσt = t2dσ1. This yields

ug(t, x) =t

4π

∫S1(0)

g(x+ tα)dσ1

and

∂tug(t, x) =1

4π

∫S1(0)

g(x+ tα)dσ1 +t

4π

∫S1(0)

∇g(x+ tα) · αdσ1.

It is easy to check that the first expression ug = ug(t, x) tends to zero ast → 0 and the second one ∂tug(t, x) tends to g(x) as t → 0. Undoing thesubstitution for y we obtain

∂tug(t, x) =1

tug(t, x) +

1

4πt

∫St(x)

∇g(y) · αdσt(y),

where α is the outward pointing unit normal vector to St(x).Gauss’s theorem, i.e. Theorem 3.3, gives

∂tug(t, x) =1

tug(t, x) +

1

4πt

∫Bt(x)

∆g(y)dy. (5)

The set Bt(x) = {y ∈ R3 : d(x, y) < t} denotes the ball in R3 with center xand radius t. From (5) we easily derive

∂2t ug(t, x) = − 1

t2ug(t, x) +

1

t∂tug(t, x)− 1

4πt2

∫Bt(x)

∆g(y)dy

+1

4πt

∂

∂t

∫Bt(x)

∆g(y)dy. (6)

Combining now the equations (5) and (6) we obtain

∂2t ug(t, x) =1

4πt

∂

∂t

∫Bt(x)

∆g(y)dy

and we decompose the integral expression in the following way

∂

∂t

∫Bt(x)

∆g(y)dy =∂

∂t

∫ t

0

∫Sr(x)

∆g(x+ rα)dσrdr =

∫St(x)

∆g(x+ tα)dσt.

Therefore we get

∂2t ug(t, x) =1

4πt

∫St(x)

∆g(x+ tα)dσt =t

4π

∫S1(0)

∆g(x+ tα)dσ1

which combined with the obvious result

∆ug(t, x) =t

4π

∫S1(0)

∆g(x+ tα)dσ1

yields

∂2t ug −∆ug = 0.

�

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Remark. This theorem also shows that there is no loss of regularity in thesolution, i.e., if the initial function g = g(x) belongs to the space Ck(R3) forsome k ≥ 2, then u ∈ Ck([0,∞)× R3).

But as already mentioned above, Kirchhoff’s formula holds only fora slightly changed Cauchy-problem compared to problem (1). To solve thelatter one we use a more general result which can be found in chapter 4 in[6]. However in this case we have to accept a loss of regularity in the solutionu = u(t, x).

Theorem 3.4. If f ∈ C3(R3) and g ∈ C2(R3), then there exists a uniqueclassical solution u of (1) such that

u(t, x) =1

4πt

∫St(x)

g(y)dσt(y) +∂

∂t

(1

4πt

∫St(x)

f(y)dσt(y)

)(7)

and u = u(t, x) ∈ C2([0,∞)× R3).

Proof. The proof that the given solution u = u(t, x) solves (1) follows anal-ogously to the proof of Theorem 3.2. In addition it is easy to verify thatevery derivate exists in the classical sense. In order to prove uniqueness ofthe solution we assume that there exist two solutions u1 and u2. By defininga function v(t, x) = u1(t, x)− u2(t, x) we can show that v solves the problem

vtt −∆v = 0, v(0, x) = 0, vt(0, x) = 0.

By stressing the so-called energy method it follows that v ≡ 0 and thereforeu1 = u2. For details regarding the energy method see chapter 4 in [6]. �

4. Spaces of Distributions

In the preceding chapter we noticed a loss of regularity in the solution uof the initial value problem (1). In order to avoid this behavior we need tointroduce other spaces, namely spaces of distributions.First of all we have to recall some notations and definitions. In Rd the notationof multi-indices α = (α1, . . . , αd) is used, where |α| =

∑dj=1 αj . Given two

multi-indices α and β, then α ≤ β means αj ≤ βj for 1 ≤ j ≤ d. Furthermorelet f be a function on Rd and x ∈ Rd, then

xα =

d∏j=1

xαjj

and

Dαf(x) =1

i|α|∂α

∂xαf(x) =

1

i|α|

d∏j=1

∂αj

∂xαjj

f(x).

Definition 4.1. A function f ∈ C∞(Rd) is called a Schwartz function if

pα,β(f) := supx∈Rd

|xαDβf(x)| <∞

for all multi-indices α, β.

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The set of all Schwartz functions is referred to as Schwartz space S(Rd),that is

S(Rd) = {f ∈ C∞(Rd) : pα,β(f) <∞}.Moreover, we can define linear and continuous functionals on the Schwartzspace S, the so-called tempered distributions. In fact a linear map F : S(Rd) 7→C satisfying

|F (f)| ≤ C∑

|α|,|β|≤N

pα,β(f), f ∈ S(Rd)

for some N ≥ 0 and a constant C > 0 is called a tempered distribution. Theset of all tempered distributions, that is the dual space of S(Rd), is denotedby S ′(Rd).For now the last space we want to introduce is the so-called Sobolev spaceHps (Rd) of distributions on Rd. This space contains all tempered distributions

u = u(x) whose derivates ∂αu with |α| ≤ s are in Lp(Rd). Another approachto the definition of Sobolev spaces of distributions can be described in termsof the Fourier transform in the following way

Hps (Rd) = {u ∈ S ′(Rd) : ‖u‖Hps (Rd) <∞},

where the norm ‖ · ‖Hps (Rd) is defined by

‖u‖Hps (Rd) =

(∫Rd|F(u)(ξ)(1 + |ξ|2)

s2 |pdξ

) 1p

(p <∞)

or‖u‖Hps (Rd) = ess sup

ξ∈Rd|F(u)(ξ)(1 + |ξ|2)

s2 | (p =∞),

respectively.If p = 2 then we write Hs instead of H2

s . Details regarding this topic can befound in chapter 1 in [6].

5. Modulation Spaces

5.1. The Short-Time Fourier Transform (STFT)

5.1.1. Introduction and Definition of STFT. The short-time Fourier trans-form is a fruitful idea for joint time-frequency representations. Subsequentlythe main goal of introducing time-frequency representations is roughly sketchedto motivate our following deeper researches of modulation spaces. This con-cept is presented in [4] by K. Grochenig. For x ∈ Rd the values f(x) give all

information about the properties of the functions f and f since the Fourier

transform is one-to-one. But one cannot obtain all properties of f by just

looking at f , for example one cannot determine whether f ∈ Lp for p 6= 2.

Summarizing one can say that both f and f are two different representationsof the same object but they show different properties of this object. Thus,in time-frequency analysis the goal is to find representations combining f

and its Fourier transform f . However these time-frequency representations

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have their limits, that is, they cannot be arbitrary exact. For more detailssee chapter 2 in [4].Within the scope of this chapter we will work with a particular time-frequencyrepresentation, the so-called short-time Fourier transform.

Definition 5.1. Let φ 6= 0 be a fixed function, the so-called window function.Then the short-time Fourier transform (STFT) of a function f with respectto φ is defined as

Vφf(x, ξ) = (2π)−d2

∫Rdf(t)φ(t− x)e−it·ξdt (x, ξ ∈ Rd).

It is needed to choose sufficiently smooth window functions to avoidartificial discontinuities of the corresponding STFT Vφf . What this meansfor φ in particular will be shown later on. The window function reveals localproperties of the function f , that is, we just Fourier transform the functionf restricted on an interval, the so-called window, determined by the windowfunction φ. Hence we obtain local information about frequency properties off .Let φ now be supported on a compact set centered in the origin. ThenVφf(x, ·) is the Fourier transform of the function f in a neighborhood ofx. This window can be shifted by choosing different values for x. That iswhy in [4] the STFT is also called the ”sliding window Fourier transform”.Moreover, the STFT is linear in f and conjugate-linear in φ. In the definition5.1 we fixed the window function φ, so that the short-time Fourier transformVφf becomes a linear mapping from functions on Rd to functions on R2d. Butobviously Vφf also depends essentially on φ. Because of this fact we assumedφ to be sufficiently smooth.

5.1.2. Modulation and Translation. In order to simplify subsequent compu-tations we introduce the following operators. For x, ξ ∈ Rd the operators aregiven by

Txf(t) = f(t− x)

and

Mξf(t) = eiξ·tf(t)

and denote a translation by x and a modulation by ξ, respectively. Naturallywe can define the product between both operators Tx and Mξ. The operatorswe obtain are so-called time-frequency shifts MξTx and TxMξ, respectively.Now we can determine some properties of the preceding operators. By thecomputation

MξTxf(t) = eiξ·tTxf(t)

= Tx(eiξ·(t+x)f(t))

= eiξ·xTxMξf(t)

we obtain

MξTxf(t) = eiξ·xTxMξf(t), (8)

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the commutation relation between translation and modulation operator. Weconclude that Tx and Mξ commutes if and only if x · ξ ∈ 2π · Z.Furthermore, the computations

(2π)−d2

∫Rd

(Txf)(t)e−it·ωdt = (2π)−d2

∫Rdf(t)e−i(t+x)·ωdt = e−ix·ω f(ω)

and

(2π)−d2

∫Rd

(Mξf)(t)e−it·ωdt = (2π)−d2

∫Rdf(t)e−it·(ω−ξ)dt = (Tξ f)(ω)

yield

F(Txf)(ω) = e−ix·ω f(ω)

and

F(Mξf)(ω) = (Tξ f)(ω),

respectively. By combining both formulas we obtain

F(TxMξf)(ω) = M−xTξ f(ω) = e−iξ·xTξM−xf(ω). (9)

Moreover it is easy to verify that it holds

‖TxMξf‖Lp = ‖f‖Lp ,

that is, time-frequency shifts are isometries on Lp for 1 ≤ p ≤ ∞.

5.1.3. Properties of STFT. Now we can find different representations for Vφfwhich will help us to prove subsequent theorems. At this point we remarkthat the involution ∗ of a function φ is defined as φ∗(x) = φ(−x).

Lemma 5.2. If f, φ ∈ L2(Rd), then Vφf is uniformly continuous on R2d and

Vφf(x, ξ) = F(f · Txφ)(ξ) (10)

= (2π)−d2 〈f,MξTxφ〉 (11)

= (2π)−d2 〈f , TξM−xφ〉 (12)

= e−ix·ξF(f · Tξ ¯φ)(−x) (13)

= e−ix·ξVφf(ξ,−x) (14)

= (2π)−d2 e−ix·ξ(f ∗Mξφ

∗)(x) (15)

= (2π)−d2 (f ∗M−xφ∗)(ξ) (16)

15

Proof.

Vφf(x, ξ) = (2π)−d2

∫Rdf(t)φ(t− x)e−it·ξdt

= (2π)−d2

∫Rdf(t)(Txφ)(t)e−it·ξdt

= F(f · Txφ)(ξ) (gives (10))

= (2π)−d2

∫Rdf(t)(MξTxφ)(t)dt

= (2π)−d2 〈f,MξTxφ〉 (gives (11))

(9)= (2π)−

d2 〈f , TξM−xφ〉 (gives (12))

(8)= (2π)−

d2 〈f , eiξ·xM−xTξφ〉

= (2π)−d2 e−iξ·x〈f ,M−xTξφ〉

(11)= e−iξ·xVφf(ξ,−x) (gives (14))

(10)= e−iξ·xF(f · Tξ ¯

φ)(−x) (gives (13))

= (2π)−d2 e−iξ·x

∫Rdf(t)φ(t− ξ)eix·tdt

= (2π)−d2 e−iξ·x

∫Rdf(t)φ∗(ξ − t)eix·tdt

= (2π)−d2

∫Rdf(t)φ∗(ξ − t)e−ix·(ξ−t)dt

= (2π)−d2 (f ∗M−xφ∗)(ξ) (gives (16))

In order to deduce (12) from (11) we also have used Parseval’s formula.The only missing equation is (15). We easily get this result by the followingcomputation

(2π)−d2 e−iξ·x(f ∗Mξφ

∗)(x) = e−iξ·x(2π)−d2

∫Rdf(t)eiξ·(x−t)φ∗(x− t) dt

= (2π)−d2

∫Rdf(t)φ(t− x)e−iξ·tdt

= Vφf(x, ξ).

Moreover, the continuity of the operators Tx and Mξ for all x, ξ ∈ Rd givesthe uniform continuity of the STFT Vφf . �

By introducing two additional notations we can find another helpfulrepresentation of the STFT. The asymmetric coordinate transform Ta of afunction F on R2d is defined by

(TaF )(x, y) = F (y, y − x)

and F2 denotes the partial Fourier transform of F with respect to the y-variable.

16

Lemma 5.3. If f, φ ∈ L2(Rd), then

Vφf = F2Ta(f ⊗ φ).

Proof.

(F2Ta(f ⊗ φ))(x, ξ) = (F2Ta(f(x)φ(y)))(x, ξ)

= (F2(f(y)φ(y − x)))(x, ξ)

= (2π)−d2

∫Rdf(y)φ(y − x)e−iy·ξdy

= Vφf(x, ξ)

�

Remark. This representation is also well-defined for more function spaces, forinstance tempered distributions.

Furthermore we use the following corollary of the orthogonality relationsfor STFT which are stated in chapter 3 in [4]. It helps us to deduce an inver-sion formula. The ideas of the subsequent argumentation are also presentedin chapter 3 in [4].

Corollary 5.1. If f, φ ∈ L2(Rd), then

‖Vφf‖L2(Rd) = ‖f‖L2(Rd)‖φ‖L2(Rd).

Remark. In particular, if ‖φ‖L2(Rd) = 1, then the STFT is an isometry from

L2(Rd) into L2(R2d) and it holds

‖Vφf‖L2(Rd) = ‖f‖L2(Rd).

Therefore the STFT Vφf of the function f determines f completely. This

means if Vφf(x, ξ) = (2π)−d2 〈f,MξTxφ〉 = 0 for all (x, ξ) ∈ R2d, then f = 0

on Rd. On the other hand this is equivalent to the statement that for everyφ ∈ L2(Rd) the set {MξTxφ : x, ξ ∈ Rd} spans a dense subspace of L2(Rd).At this point the question arises if we can find a particular representation forf . We know that in usual Fourier analysis there is an inversion formula, i.e.,

under sufficient assumptions and by knowing the Fourier transform f of afunction f we can completely recover f . Now we want to find a similar resultfor the short-time Fourier transform. We begin with the very definition ofan integral in Banach spaces. Details are treated in chapter 3 in [7] by W.Rudin. Let B be a Banach space and φ be a mapping of Rd into B. Thenf =

∫Rd φ(x)dx means

〈f, h〉 =

∫Rd〈φ(x), h〉dx for all h ∈ B∗,

where B∗ denotes the dual space of B. Now we define a (conjugate-) linearfunctional

l(h) 7→∫Rd〈φ(x), h〉dx

17

on B∗. Thus l ∈ B∗∗ and if it is bounded, then it defines a unique elementf ∈ B∗∗. This could cause problems if B 6= B∗∗. Therefore we restrict ourconsiderations on reflexive Banach spaces, i.e., it holds B = B∗∗. Now weapply this definition to time-frequency analysis, that is we set

f = (2π)−d2

∫∫R2d

F (x, ξ)MξTxφdx dξ

for F ∈ L2(R2d). Hence we obtain the (conjugate-) linear functional

l(h) = (2π)−d2

∫∫R2d

F (x, ξ)〈h,MξTxφ〉 dx dξ, h ∈ L2(Rd).

Applying Cauchy-Schwartz inequality and Corollary 5.1 yield

|l(h)| ≤ ‖F‖L2(R2d)‖Vφh‖L2(R2d) = ‖F‖L2(R2d)‖φ‖L2(Rd)‖h‖L2(Rd).

So the functional l is bounded and therefore it exists a unique f ∈ L2(Rd)such that

f = (2π)−d2

∫∫R2d

F (x, ξ)MξTxφdx dξ

with

‖f‖L2(Rd) ≤ ‖F‖L2(R2d)‖φ‖L2(Rd) (17)

and l(h) = 〈f, h〉.Now we can state the inversion formula of time-frequency analysis.

Theorem 5.4. Let φ, γ ∈ L2(Rd) such that 〈φ, γ〉 6= 0. Then

f = (2π)−d2

1

〈φ, γ〉

∫∫R2d

Vφf(x, ξ)MξTxγ dξ dx

for every f ∈ L2(Rd).

Proof. Corollary 5.1 shows that Vφf ∈ L2(R2d). Hence

f = (2π)−d2

1

〈φ, γ〉

∫∫R2d

Vφf(x, ξ)MξTxγ dx dξ

is a well-defined function in L2(Rd) which can easily be proved by the Cauchy-Schwartz inequality. Let h ∈ L2(Rd), then we obtain

〈f , h〉 = (2π)−d2

1

〈φ, γ〉

∫∫R2d

Vφf(x, ξ)〈h,MξTxγ〉 dx dξ

=1

〈φ, γ〉〈Vφf, Vγh〉

=1

〈φ, γ〉〈φ, γ〉〈f, h〉

= 〈f, h〉

by using the orthogonality relations for STFT as stated in [4]. Thus, we have

proved f = f and we are done. �

18

After we deduced the inversion formula we are interested in the existenceof an adjoint V ∗φ of the STFT Vφ. On that account we define a linear operatorAφ by

AφF = (2π)−d2

∫∫R2d

F (x, ξ)MξTxφdx dξ, F ∈ L2(R2d)

with an admissible window function φ ∈ L2(Rd) \ {0}. The estimate (17)shows the boundedness of the operator Aφ and that it maps L2(R2d) ontoL2(Rd). If we consider the STFT Vφ mapping L2(Rd) to L2(R2d), then Aφ isits adjoint since

〈AφF, h〉 = (2π)−d2

∫∫R2d

F (x, ξ)〈MξTxφ, h〉 dx dξ

= 〈F, Vφh〉= 〈V ∗φ F, h〉

where h ∈ L2(Rd) and F ∈ L2(R2d). Hence Aφ = V ∗φ .By Theorem 5.4 we obtain

〈f, h〉 = (2π)−d2

1

〈φ, γ〉

∫∫R2d

Vφf(x, ξ)〈MξTxγ, h〉 dξ dx.

This combined with the preceding computations yields

〈f, h〉 =1

〈φ, γ〉〈V ∗γ Vφf, h〉

and thus1

〈φ, γ〉V ∗γ Vφ = I, (18)

where I is the identity operator on L2(Rd).By now we only worked with the function space L2(Rd). We want to extendour considerations of the STFT to other function spaces.

5.1.4. Extensions of the STFT. First of all we easily obtain results for generalLp spaces. Holder’s inequality yields that f ·Txφ ∈ L1(Rd) if φ ∈ Lp(Rd) and

f ∈ Lp′(Rd). The notation p and p′ means that it holds 1

p + 1p′ = 1, that

is, p′ is the conjugate parameter of p. Equality (10) gives now the pointwiseexistence of the STFT Vφf . By using properties of the Fourier transformwe get Vφf ∈ C0(Rd), that is, the STFT belongs to the space consistingof all continuous functions which vanish outside of a compact subset of Rd.The same arguments justified that the STFT Vφf is pointwise defined forφ, f ∈ L2(Rd).So far we assumed that the integrals are defined. At this point we expandour considerations to general Banach spaces. So let B be a Banach space,then we know by chapter 4 in [7] that the dual space B∗ exists and is alsoa Banach space. Additionally we assume that B is invariant under time-frequency shifts. Then the expression 〈·, ·〉 is well defined by duality and dueto (11) the STFT Vφf exists for f ∈ B,φ ∈ B∗ and for f ∈ B∗, φ ∈ B,respectively.

19

Subsequently we focus on the function spaces which were already introducedin Chapter 4.

Proposition 5.1. The mapping (x, ξ) 7→MξTx is strongly continuous on S(Rd)and weak*-continuous on S ′(Rd).

Proof. Cf. chapter 11 in [4]. �

Remark. Let φ ∈ S(Rd) and f ∈ S ′(Rd), then strong continuity of the givenmapping means

lim|x|,|ξ|→0

pα,β(MξTxφ− φ) = lim|x|,|ξ|→0

supy∈Rd

|yαDβ(MξTxφ(y)− φ(y))| = 0

for all multi-indices α, β. Weak*-continuity on S ′(Rd) is shown by

lim|x|,|ξ|→0

〈MξTxf, φ〉 = lim|x|,|ξ|→0

〈f, T−xM−ξφ〉 = 〈f, φ〉,

where the first part of the proof has already been used.

Corollary 5.2. For a fixed window function φ ∈ S(Rd)\{0} and for f ∈ S ′(Rd)the STFT Vφf is both defined and continuous on S ′(R2d).

Theorem 5.5. Let φ ∈ S(Rd) be a fixed window function. If f ∈ S(Rd), thenVφf ∈ S(R2d).

Proof. We use Lemma 5.3 but for f, φ ∈ S(Rd). By chapter 4 in [3] we evenknow that f ⊗ φ ∈ S(R2d). Hence Vφf ∈ S(R2d) because of the invariance ofboth operators F2 and Ta on S(R2d). �

Remark. In chapter 11 in [4] is also shown that if we assume f to be anelement in S ′(Rd) and the STFT Vφf ∈ S(Rd), then f ∈ S(Rd).

Previously we showed some extensions of the short-time Fourier trans-form to different spaces. Now we can also reveal properties of the STFTsimilar to Section 5.1.3. Obviously the inversion formula of time-frequencyanalysis holds in the Schwartz space S as well. But it even holds in S ′.

Proposition 5.2. Let φ, γ ∈ S(Rd) such that 〈φ, γ〉 6= 0. Then

f = (2π)−d2

1

〈φ, γ〉

∫∫R2d

Vφf(x, ξ)MξTxγ dξ dx (19)

for every f ∈ S ′(Rd).

Proof. Cf. Corollary 11.2.7 in [4]. �

5.2. Definition and Properties

Our goal is to work in weighted modulation spaces. A detailed concept ofweights can be found in chapter 11 in [4]. Subsequently however we will onlyuse particular weights.

20

Definition 5.6. The so-called integrability parameters are given by 1 ≤ p, q ≤∞. Let φ ∈ S(Rd) \ {0} be a fixed window and assume s, σ ∈ R to be theweight parameters. Then the weighted modulation space Mp,q

s,σ (Rd) is the set

Mp,qs,σ (Rd) := {f ∈ S ′(Rd) : ‖f‖Mp,q

s,σ (Rd) <∞},

where the norm is defined as

‖f‖Mp,qs,σ (Rd) =

(∫Rd

(∫Rd|Vφf(x, ξ)〈x〉σ〈ξ〉s|pdx

) qp

dξ

) 1q

.

Furthermore, the weighted modulation space W p,qs,σ (Rd) consists of all tem-

pered distributions f ∈ S ′(Rd) such that their norm

‖f‖Wp,qs,σ (Rd) =

(∫Rd

(∫Rd|Vφf(x, ξ)〈x〉σ〈ξ〉s|qdξ

) pq

dx

) 1p

is finite.

Remark. Note that

〈x〉σ = (1 + |x|2)σ2 and 〈ξ〉s = (1 + |ξ|2)

s2 .

If s = σ = 0 then we obtain the so-called standard modulation space Mp,q(Rd),that is the modulation space without any weights. Subsequently the spaceMp,qs,σ (Rd) is just referred to as modulation space. Furthermore if p = q we

write Mps,σ(Rd) instead of Mp,p

s,σ (Rd).The same notations apply to the modulation space W p,q

s,σ (Rd). Additionally

all following results hold analogously for W p,qs,σ (Rd).

Note that the weight expression with respect to x in the precedingdefinition corresponds to some growth or decay properties of a function fin the modulation space Mp,q

s,σ . On the other hand the weight expression withrespect to ξ corresponds to regularity properties of f in Mp,q

s,σ . The followingproposition shows these facts in a mathematically more precise way. Here werecall that Dj = 1

i∂∂xj

for 1 ≤ j ≤ d.

Proposition 5.3. Let s, s0, σ, σ0 ∈ R and 1 ≤ p, q ≤ ∞ be the integrabilityparameters. Then it holds:

• the map f 7→ 〈·〉σ0f is a homeomorphism from Mp,qs,σ+σ0

(Rd) to Mp,qs,σ (Rd)

and• the map f 7→ 〈D〉s0f is a homeomorphism from Mp,q

s+s0,σ(Rd) to Mp,qs,σ (Rd).

Proof. Cf. Corollary 3.3 in [9]. �

In order to show properties of modulation spaces we need to introducethe space of Lebesgue measurable functions on R2d.

21

Definition 5.7. Let 1 ≤ p, q ≤ ∞ be the integrability parameters and s, σ bereal numbers. Then the weighted mixed-norm space of all Lebesgue measur-able functions F on R2d is denoted by Lp,qs,σ(R2d) and consists of all F ∈ R2d

such that its norm

‖F‖Lp,qs,σ(R2d) =

(∫Rd

(∫Rd|F (x, ξ)〈x〉σ〈ξ〉s|pdx

) qp

dξ

) 1q

.

is finite.

Remark. Analogously to the previous remark, if p = q then we simply writeLps,σ(R2d) instead of Lp,ps,σ(R2d). If additionally s = σ = 0 the latter space

coincides with the well-known Lp space on R2d.

K. Grochenig investigated some properties of Lp,qs,σ(R2d) in [4]. First werecall the structure of this space.

Lemma 5.8. Let 1 ≤ p, q ≤ ∞ and s, σ ∈ R, then Lp,qs,σ(R2d) is a Banachspace.

Proof. Cf. Lemma 11.1.2 in [4]. �

By Lemma 5.8 and chapter 4 in [7] we know that there exists a dualspace of Lp,qs,σ which is a Banach space itself.

Lemma 5.9. Let s, σ ∈ R and p, q ∈ [1,∞) where p′, q′ such that 1p + 1

p′ =1q + 1

q′ = 1. Then the dual space (Lp,qs,σ)∗ of Lp,qs,σ is given by (Lp,qs,σ)∗ = Lp′,q′

−s,−σ.

Proof. Cf. Lemma 11.1.2 in [4] �

Remark. Duality is obtained in the following sense

〈F,H〉 =

∫∫R2d

F (x)H(x)dx

for F ∈ Lp,qs,σ(R2d) and H ∈ Lp′,q′

−s,−σ(R2d). This integral is pointwise defined

due to Holder’s inequality, i.e., F ·H ∈ L1(R2d).

In Section 5.1.3 we defined an adjoint of the STFT on L2. Now we wantto extend our considerations to more general cases.

Definition 5.10. Let F be a function on R2d and φ ∈ S(Rd) be a fixed windowfunction which is not identically zero. Then the adjoint STFT V ∗φ is definedby

V ∗φ F = (2π)−d2

∫∫R2d

F (x, ξ)MξTxφdx dξ

in a weak sense. That means

〈V ∗φ F, f〉 = (2π)−d2

∫∫R2d

F (x, ξ)〈MξTxφ, f〉 dx dξ

=

∫∫R2d

F (x, ξ)Vφf(x, ξ) dx dξ

= 〈F, Vφf〉,

22

where f is an admissible function on Rd.

Proposition 5.4. Given a window γ ∈ S(Rd) \ {0}, the adjoint STFT V ∗γ is a

mapping from Lp,qs,σ(R2d) into Mp,qs,σ (Rd).

Proof. Cf. Proposition 11.3.2 in [4]. �

In [4] K. Grochenig also shows that the inversion formula (19) is valid inMp,qs,σ . Thus it holds (18) in modulation spaces with a window φ ∈ S(Rd)\{0}.

Another helpful result is the following lemma.

Lemma 5.11. If φ0, φ, γ ∈ S(Rd) \ {0} such that 〈γ, φ〉 6= 0 and f ∈ S ′(Rd),then it holds

|Vφ0f(x, ξ)| ≤ (2π)−d2

1

|〈γ, φ〉|(|Vφf | ∗ |Vφ0γ|)(x, ξ)

for all (x, ξ) ∈ R2d.

Proof. By (18) we get

Vφ0f =

1

|〈γ, φ〉|Vφ0

V ∗γ (Vφf).

In [4] it is shown that V ∗γ (Vφf) ∈ S ′(Rd). Therefore Proposition 5.1 yieldsthe continuity of its STFT.The transformation

Vγ(MηTuφ0)(x, ξ) = (2π)−d2

∫RdMηTuφ0(t)γ(t− x)e−it·ξdt

= (2π)−d2

∫Rdeiη·tφ0(t− u)γ(t− x)e−it·ξdt

=

[z = t− xdz = dt

]= (2π)−

d2

∫Rde−i(z+x)·(η−ξ)φ0(z + x− u)γ(z)dz

= (2π)−d2 e−ix·(η−ξ)

∫Rdγ(z)φ0(z − (u− x))e−iz·(η−ξ)dz

= Vφ0γ(u− x, η − ξ)e−ix·(η−ξ)

is needed for the following computation

Vφ0V ∗γ F (u, η)

(11)= (2π)−

d2 〈V ∗γ F,MηTuφ0〉

= (2π)−d2

∫∫R2d

F (x, ξ)Vγ(MηTuφ0)(x, ξ) dx dξ

= (2π)−d2

∫∫R2d

F (x, ξ)Vφ0γ(u− x, η − ξ)e−ix·(η−ξ)dxdξ.

Taking absolute values in the latter calculation we obtain

|Vφ0V ∗γ F (u, η)| ≤ (2π)−

d2 (|F | ∗ |Vφ0

γ|)(u, η).

23

Combining all results we obtain the inequality

|Vφ0f(x, ξ)| ≤ (2π)−

d2

1

|〈γ, φ〉|(|Vφf | ∗ |Vφ0

γ|)(x, ξ).

�

Theorem 5.12. Let φ ∈ S(Rd) be a fixed non-zero window function. Then thefunction f belongs to the modulation space Mp,q

s,σ (Rd) if and only if Vφf ∈Lp,qs,σ(R2d) for 1 ≤ p, q ≤ ∞ and s, σ ∈ R. Moreover, different window func-tions yield equivalent norms.

Proof. By Lemma 5.11 with a fixed window function φ0 ∈ S(Rd) \ {0} andby setting γ = φ0 it follows

‖Vφf‖Lp,qs,σ(R2d) ≤ (2π)−d2

1

‖φ0‖2L2(Rd)‖|Vφ0

f | ∗ |Vφφ0|‖Lp,qs,σ(R2d)

= (2π)−d2

1

‖φ0‖2L2(Rd)

∥∥∥∥∫∫R2d

|Vφ0f(x− y, ξ − η)||Vφφ0(y, η)| dydη

∥∥∥∥Lp,qs,σ(R2d)

≤ (2π)−d2

1

‖φ0‖2L2(Rd)

∫∫R2d

‖Vφ0f(· − y, · − η)‖Lp,qs,σ(R2d)|Vφφ0(y, η)| dydη

= (2π)−d2

1

‖φ0‖2L2(Rd)‖Vφ0

f‖Lp,qs,σ(R2d)‖Vφφ0‖L1(R2d) (20)

= C0‖Vφ0f‖Lp,qs,σ(R2d).

By interchanging φ and φ0 we additionally obtain

‖Vφ0f‖Lp,qs,σ(R2d) ≤ C1‖Vφf‖Lp,qs,σ(R2d).

This shows the equivalence of norms.Furthermore we conclude

‖Vφ0f‖Lp,qs,σ(R2d) <∞⇐⇒ ‖Vφf‖Lp,qs,σ(R2d) <∞

where ‖Vφ0f‖Lp,qs,σ(R2d) corresponds to ‖f‖(0)Mp,qs,σ (Rd) and ‖Vφf‖Lp,qs,σ(R2d) corre-

sponds to ‖f‖(1)Mp,qs,σ (Rd). So we have shown the statements of the theorem. �

We already mentioned in Lemma 5.8 that Lp,qs,σ is a Banach space. Thisgives rise to investigate whether Mp,q

s,σ is also a Banach space.

Theorem 5.13. Assume s, σ ∈ R and the integrability parameters 1 ≤ p, q ≤∞, then the modulation space Mp,q

s,σ (Rd) is a Banach space.

Proof. Theorem 5.12 shows

Mp,qs,σ (Rd) = {f ∈ S ′(Rd) : Vφf ∈ Lp,qs,σ(R2d)}

for an appropriate non-zero window function φ. Now we define the set

V := {F ∈ Lp,qs,σ(R2d) : F = Vφf}

which is a subspace of Lp,qs,σ(R2d). Additionally we can show that V is iso-

metrically isomorphic to Mp,qs,σ (Rd). For that reason we define a mapping

24

Φ : Mp,qs,σ (Rd) 7→ V by Φ(f) = Vφf for f ∈ Mp,q

s,σ (Rd). Obviously Φ is ontoand Theorem 5.12 yields that Φ is one-to-one. In fact for Φ(f) = Φ(f ′) wehave

0 = ‖Vφf − Vφf ′‖Lp,qs,σ(R2d)

= ‖f − f ′‖Mp,qs,σ (Rd)

and thus f = f ′. The isometry between Mp,qs,σ and V follows by definition,

i.e., ‖f‖Mp,qs,σ (Rd) = ‖Vφf‖Lp,qs,σ(R2d). Hence the linearity of Lp,qs,σ is inherited by

Mp,qs,σ .

We already defined a norm on Mp,qs,σ . So the only thing we still have to prove

is completeness. For that let {fj}j≥1 be a Cauchy sequence in Mp,qs,σ that is

‖fj − fn‖Mp,qs,σ (Rd) → 0 as j, n→∞.

By Theorem 5.12 this is equivalent to

‖Vφfj − Vφfn‖Lp,qs,σ(R2d) → 0 as j, n→∞.

From Lemma 5.8 it follows that there exists a function F ∈ Lp,qs,σ(R2d) suchthat

limj→∞

‖Vφfj − F‖Lp,qs,σ(R2d) = 0. (21)

Now we have to show that F has the form Vφf . For that we define the functionf by

f = V ∗φ F.

Furthermore let the projection operator Pφ be defined by

Pφ = Vφ ◦ V ∗φwith the STFT Vφ : S ′(Rd) 7→ S ′(R2d) and its adjoint V ∗φ : S ′(R2d) 7→ S ′(Rd).Then we get the following equality

PφF (x, ξ) = (Vφ ◦ V ∗φ )F (x, ξ)

= (2π)−d2

∫Rd

((2π)−

d2

∫∫R2d

F (y, η)φ(t− y)eit·ηdy dη

)φ(t− x)e−it·ξdt

= (2π)−d∫Rd

∫∫R2d

F (y, η)φ(t− y)φ(t− x)e−it·(ξ−η) dy dη dt

=

z = t− yt = z + ydz = dt

= (2π)−

d2

∫∫R2d

((2π)−

d2

∫Rdφ(z)φ(z − (x− y))e−iz·(ξ−η) dz

)F (y, η)e−iy·(ξ−η)dy dη

= (2π)−d2

∫∫R2d

F (y, η)Vφφ(x− y, ξ − η)e−iy·(ξ−η)dy dη

= (F \ Vφφ)(x, ξ).

25

Here \ denotes the so-called twisted convolution which is defined by

(F \G)(x, ξ) = (2π)−d2

∫∫R2d

F (y, η)G(x− y, ξ − η)e−iy·(ξ−η) dy dη

for admissible functions F,G on R2d.It is easy to check that the following estimate holds

|F \ (Vφφ)| ≤ (2π)−d2 (|F | ∗ |Vφφ|)

with the ordinary convolution ∗ on R2d.These computations give

‖PφF‖Lp,qs,σ(R2d) = ‖F \ Vφφ‖Lp,qs,σ(R2d)

≤ (2π)−d2 ‖|F | ∗ |Vφφ|‖Lp,qs,σ(R2d)

(20)

≤ (2π)−d2 ‖Vφφ‖L1(R2d)‖F‖Lp,qs,σ(R2d)

≤ C0‖F‖Lp,qs,σ(R2d) (22)

for some constant C0 > 0. Thus, the projection operator Pφ is a mappingfrom Lp,qs,σ(R2d) to Lp,qs,σ(R2d).Now we have all tools to show the completeness of Mp,q

s,σ . In fact it holds

‖fj − f‖Mp,qs,σ (Rd) = ‖Vφfj − Vφf‖Lp,qs,σ(R2d)

= ‖Vφfj − Vφ(V ∗φ F )‖Lp,qs,σ(R2d)

(18)=

∥∥∥∥∥ 1

‖φ‖2L2(Rd)

Vφ(V ∗φ Vφ)fj − Vφ(V ∗φ F )

∥∥∥∥∥Lp,qs,σ(R2d)

=

∥∥∥∥∥ 1

‖φ‖2L2(Rd)

PφVφfj − PφF

∥∥∥∥∥Lp,qs,σ(R2d)

(22)

≤ C1‖Vφfj − F‖Lp,qs,σ(R2d)

for some constant C1 > 0. Together with (21) this gives the convergence ofthe Cauchy sequence {fj}j≥1 to f . Hence Mp,q

s,σ is complete. �

As already mentioned for Lp,qs,σ spaces we conclude from this result theexistence of a dual space (Mp,q

s,σ )∗ of the modulation space Mp,qs,σ .

Theorem 5.14. Let s, σ ∈ R and p, q ∈ [1,∞) where p′, q′ such that 1p + 1

p′ =1q + 1

q′ = 1. Then (Mp,qs,σ )∗ = Mp′,q′

−s,−σ, where the duality is given by

〈f, g〉 =

∫∫R2d

Vφf(x, ξ)Vφg(x, ξ) dx dξ (23)

for f ∈Mp,qs,σ (Rd) and g ∈Mp′,q′

−s,−σ(Rd).

26

Proof. Let l ∈ (Mp,qs,σ )∗ be a linear functional on Mp,q

s,σ . As already shown inthe proof of Theorem 5.13 the set

V := {F ∈ Lp,qs,σ(R2d) : F = Vφf}

is isometrically isomorphic to Mp,qs,σ (Rd). Thus, the element l induces a linear

functional l on V by l(f) = l(Vφf) which is bounded. Therefore we can apply

the Hahn-Banach theorem which extends l to a linear functional on Lp,qs,σ(R2d).

For details see chapter 3 in [7]. Lemma 5.9 yields (Lp,qs,σ)∗ = Lp′,q′

−s,−σ. By this

duality we know that for a function f ∈ Mp,qs,σ (Rd), i.e. Vφf ∈ Lp,qs,σ(R2d),

there exists a function G ∈ Lp′,q′

−s,−σ(R2d) such that

l(f) = l(Vφf)

=

∫∫R2d

Vφf(x, ξ)G(x, ξ) dx dξ

=

∫Rdf(x)(V ∗φG)︸ ︷︷ ︸

=:g

(x) dx

=

∫Rdf(x)g(x) dx

= 〈f, g〉,

where g ∈ Mp′,q′

−s,−σ(Rd) which follows by Proposition 5.4. Hence every linearfunctional l on Mp,q

s,σ is of the form (23) and we obtain the statement of thetheorem. �

5.3. The Fourier Transform

Proposition 5.5. The set of all Fourier transforms on Mp,qs,σ is equal to the

modulation space W q,pσ,s for 1 ≤ p, q ≤ ∞ and s, σ ∈ R.

Proof. It holds

‖f‖Mp,qs,σ (Rd) =

(∫Rd

(∫Rd|Vφf(x, ξ)〈x〉σ〈ξ〉s|pdx

) qp

dξ

) 1q

(14)=

(∫Rd

(∫Rd|Vφf(ξ,−x)〈x〉σ〈ξ〉s|pdx

) qp

dξ

) 1q

=

(∫Rd

(∫Rd|Vφf(x,−ξ)〈ξ〉σ〈x〉s|pdξ

) qp

dx

) 1q

=

(∫Rd

(∫Rd|Vφf(x, ξ)〈x〉s〈ξ〉σ|pdξ

) qp

dx

) 1q

= ‖f‖W q,pσ,s.

27

Note that if we assume φ ∈ S(Rd)\{0} to be the window, then φ ∈ S(Rd)\{0}is also a window function. We have to take into account that these two normsare taken with respect to different window functions. However by Theorem5.12, that is the equivalence of norms, the given norm remains finite fordifferent window functions. Hence it follows

f ∈Mp,qs,σ (Rd) ⇐⇒ f ∈W q,p

σ,s (Rd).

�

Remark. Analogous to this proof we can show

f ∈Mp,qs,σ (Rd) ⇐⇒ f ∈W q,p

σ,s (Rd).

Proposition 5.6. The spaces Mp,qs,σ and W q,p

s,σ coincide if p = q.

Proof. By using equality (14) and the norm equivalence as mentioned inTheorem 5.12, we obtain

‖f‖Mps,σ(Rd) = ‖Vφf‖Lps,σ(R2d)

≤ C0‖Vφf‖Lps,σ(R2d)

= C0

(∫∫R2d

|Vφf(x, ξ)〈x〉σ〈ξ〉s|p dx dξ) 1p

= C0

(∫∫R2d

|Vφf(−ξ, x)〈x〉σ〈ξ〉s|p dx dξ) 1p

= C0

(∫∫R2d

|Vφf(x, ξ)〈x〉s〈ξ〉σ|p dξ dx) 1p

= C0

(∫∫R2d

|Vφf(x, ξ)〈x〉s〈ξ〉σ|p dx dξ) 1p

= C0‖f‖Mpσ,s(Rd)

for some constant C0 > 0. Conversely we get the same estimate, i.e.,

‖f‖Mpσ,s(Rd) ≤ C1‖f‖Mp

s,σ(Rd)

for some constant C1 > 0. All in all this gives f ∈Mpσ,s(Rd)⇔ f ∈Mp

s,σ(Rd).By the remark of Proposition 5.5 we know that

W pσ,s(Rd) = {f : f ∈Mp

s,σ(Rd)}

and combining these results we obtain W ps,σ(Rd) = Mp

s,σ(Rd). �

Proposition 5.7. Let s, σ ∈ R. For p ≤ q it holds W p,qs,σ ⊆Mp,q

s,σ . Analogously,q ≤ p gives Mp,q

s,σ ⊆W p,qs,σ .

Proof. Cf. Theorem 3.2 in [2] for the unweighted case. However the weightsdo not change the argumentation of the proof. �

28

Once we introduced the Fourier transform and some of its mappingproperties on Mp,q

s,σ , we now mention an important result from chapter 2 in[8] for our subsequent considerations. For our purposes a special case of it issufficient. But firstly it is needed to prove the validity of Holder’s inequalityfor weighted Lebesgue spaces on Rd. Here Lp0,σ(Rd) denotes the space of all

functions f such that f(x) · 〈x〉σ ∈ Lp(Rd), where x ∈ Rd and σ ∈ R.

Lemma 5.15. Let σ1, σ2 be real numbers and p, q, r ∈ [1,∞] such that 1p + 1

q =1r . If f ∈ Lp0,σ1

(Rd) and g ∈ Lq0,σ2(Rd), then f · g ∈ Lr0,σ1+σ2

(Rd) and it holds

‖f · g‖Lr0,σ1+σ2(Rd) ≤ ‖f‖Lp0,σ1 (Rd)‖g‖L

q0,σ2

(Rd). (24)

Proof. By the introduced notation it follows

‖f ·g‖Lr0,σ1+σ2(Rd) = ‖f ·g · 〈x〉σ1 · 〈x〉σ2‖Lr(Rd) = ‖(f · 〈x〉σ1) ·(g · 〈x〉σ2)‖Lr(Rd).

We apply the ordinary Holder’s inequality to the right hand side and obtain

‖(f · 〈x〉σ1) · (g · 〈x〉σ2)‖Lr(Rd) ≤ ‖f · 〈x〉σ1‖Lp(Rd)‖g · 〈x〉σ2‖Lq(Rd)

since f ∈ Lp0,σ1(Rd), i.e. f · 〈x〉σ1 ∈ Lp(Rd), and g ∈ Lq0,σ2

(Rd). Obviously

this is exactly (24). �

Theorem 5.16. Let pj , qj ∈ [1,∞], where j = 0, 1, 2, such that

1

p1+

1

p2= 1 +

1

p0and

1

q1+

1

q2=

1

q0.

Furthermore, let s, σ0, σ1, σ2 ∈ R be numbers such that σ1 + σ2 = σ0. Thenit holds

W q1,p1s,σ1

·W q2,p2s,σ2

⊆W q0,p0s,σ0

. (25)

Now assume that

1

p1+

1

p2=

1

p0and

1

q1+

1

q2= 1 +

1

q0.

Then

Mp1,q1s,σ1

·Mp2,q2s,σ2

⊆Mp0,q0s,σ0

(26)

and

W p1,q1s,σ1

·W p2,q2s,σ2

⊆W p0,q0s,σ0

. (27)

Proof. In order to prove (25) we basically use the proof of Theorem 2.4 in [8]but instead of taking Holder’s inequality we apply its modification (24) withrespect to the x-variable. The inclusions (26) and (27) follow by analogousarguments. �

29

6. Classes of the Factors in the Solution Representation

We recall our Cauchy problem (1) which is given by

∂2t u−∆u = 0, u(0, x) = f(x), ut(0, x) = g(x)

on Rd with respect to the x-variable and on the interval [0, T ] with respect tothe time t for some T > 0. The general solution (3) of this problem containson one hand the factor

V1(t, ξ) =sin(|ξ|t)|ξ|

and on the other hand the factor

V2(t, ξ) = cos(|ξ|t).In order to know in which function space the solution (3) lies we need to inves-tigate these factors, that is, determine which function spaces contain them.Naturally there exists Taylor expansions for the trigonometrical functionscosine and sine, i.e.,

cos(x) =

∞∑k=0

(−1)k

(2k)!x2k and

sin(x) =

∞∑k=0

(−1)k

(2k + 1)!x2k+1.

Therefore we obtain Taylor expansions for V1 = V1(t, ξ) and V2 = V2(t, ξ) asfollows

V1(t, ξ) =sin(|ξ|t)|ξ|

=

∞∑k=0

(−1)k|ξ|2k

(2k + 1)!t2k+1

and

V2(t, ξ) = cos(|ξ|t) =

∞∑k=0

(−1)k|ξ|2k

(2k)!t2k.

Hence, V1(t, ·), V2(t, ·) ∈ C∞(Rd). Note that V1(t, ·) ∈ C∞(Rd) is interpretedas V1 ∈ C∞ with respect to the ξ-variable. This is because firstly we wantto gather information about the functions with respect to ξ. Until it is notstated different we continue using this notation.Moreover, both functions are bounded since it holds

|V1(t, ξ)| ≤ t

and

|V2(t, ξ)| ≤ 1.

Thus, we have V1(t, ·), V2(t, ·) ∈ L∞(Rd). In fact there is a more generalstatement for V1, namely V1(t, ·) ∈ Lp(Rd) if p > 1.At this point we mention two embedding results to get more informationabout the spaces containing the functions V1 = V1(t, ξ) and V2 = V2(t, ξ)

30

with respect to the ξ-variable. In the following proposition the space L∞Nconsists of all functions which have bounded derivatives up to order N , i.e.,if f ∈ L∞N (Rd) then ∂αf ∈ L∞(Rd) for |α| ≤ N .

Proposition 6.1. Assume N to be an integer. Then it holds for every ε > 0

H∞N+ε ⊆ L∞N ⊆ H∞N−ε.

Proof. Let Bp,qs be the Besov space with Lebesgue exponents p, q and Sobolevparameter s. For details see [1]. By standard argumentations we obtain thefollowing inclusions

Bp,q1s+ε ⊆ Bp,q2s

andBp,1s ⊆ Hp

s ⊆ Bp,∞s .

The embedding L∞N ⊆ B∞,∞N is implicitly given in chapter 6 in [1], the other

embedding is explicitly given in [1]. Hence it follows

L∞N ⊆ B∞,∞N ⊆ B∞,1N−ε ⊆ H

∞N−ε.

Analogously we get HN+ε ⊆ L∞N . �

Proposition 6.2. Let s ∈ R and 1 ≤ p, q ≤ ∞. Define θ1 = θ1(p, q) andθ2 = θ2(p, q) as follows

θ1(p, q) = max(0, q−1 −min(p−1, p′−1)),

θ2(p, q) = min(0, q−1 −max(p−1, p′−1)),

where p′ is the conjugate exponent of p. Then

Hps+µdθ1(p,q)

(Rd) ⊆Mp,qs,0 (Rd) ⊆ Hp

s+µdθ2(p,q)(Rd)

for µ > 1.

Proof. Cf. (0.4) in [9]. �

It is easy to check that for every integer N

ξ 7→ sin(|ξ|t)|ξ|

· 〈ξ〉 ∈ L∞N (Rd)

and

ξ 7→ cos(|ξ|t) ∈ L∞N (Rd).

Since this holds for every N it follows V1(t, ·)〈·〉 ∈ H∞N (Rd) and V2(t, ·) ∈H∞N (Rd) due to Proposition 6.1. By applying Proposition 6.2 with p = ∞and q = 1 we get H∞s+µd(Rd) ⊆M

∞,1s,0 (Rd) for every µ > 1. Again because of

the fact V1(t, ·)〈·〉 ∈ H∞N (Rd) and V2(t, ·) ∈ H∞N (Rd) for all N , we obtain

ξ 7→ sin(|ξ|t)|ξ|

· 〈ξ〉 ∈M∞,1N,0 (Rd)

and

ξ 7→ cos(|ξ|t) ∈M∞,1N,0 (Rd).

31

Finally Proposition 5.3 gives

ξ 7→ sin(|ξ|t)|ξ|

∈M∞,1N,1 (Rd).

It is also possible to extend this result with the help of the followingproposition.

Proposition 6.3. Let p1, p2, q1, q2 ∈ [1,∞] such that p1 ≤ p2 and q1 ≤ q2.Additionally assume s1, s2, σ1, σ2 ∈ R are real numbers, where s2 ≤ s1 andσ2 ≤ σ1. Then it holds Mp1,q1

s1,σ1⊆Mp2,q2

s2,σ2.

Proof. Cf. Theorem 12.2.2 in [4]. �

Remark. An immediate consequence of the latter proposition is that thestandard modulation space increases with its integrability parameters p andq.

We can now conclude that the functions V1 = V1(t, ξ) and V2 = V2(t, ξ)belong to all modulation spaces M∞,qs,σ (Rd) for all q ≥ 1, s ≤ N and σ ≤ 0

with respect to ξ and even V1(t, ·) ∈ M∞,qs,σ (Rd) for all q ≥ 1, s ≤ N andσ ≤ 1.

7. Classes of the Solution of the Wave Equation

The general solution u = u(t, x) of the Cauchy problem (1) is given by

u(t, x) = F−12

(g(ξ) · sin(|ξ|t)

|ξ|

)(t, x) + F−12

(f(ξ) · cos(|ξ|t)

)(t, x).

In order to determine the space which contains the solution u = u(t, x)we need to introduce the following notation. The space C([0, T ],Mp,q

s,σ (Rd))denotes the function space with the following properties:

• for all t ∈ [0, T ] it holds u(t, ·) ∈Mp,qs,σ (Rd)),

• limt1→t2 ||u(t1, ·)− u(t2, ·)||Mp,qs,σ (Rd)) = 0 (that is continuity in t) and

• the norm is defined by maxt∈[0,T ] ||u(t, ·)||Mp,qs,σ (Rd)).

Moreover, u ∈ Cn([0, T ],Mp,qs,σ (Rd))) if ∂ltu ∈ C([0, T ],Mp,q

s,σ (Rd))) for all0 ≤ l ≤ n. Naturally this notation can be applied to every other space withrespect to t and to x, respectively.

7.1. Solution in Modulation Spaces

Now assume the initial data f ∈Mp,qs+1,N (Rd) and g ∈Mp,q

s,N (Rd), where N isan arbitrary integer, s is real number and 1 ≤ p, q ≤ ∞. By the definitions we

obtain f(t, ·) ∈ W q,pN,s+1(Rd) and g(t, ·) ∈ W q,p

N,s(Rd), respectively. Recallingthat

V1(t, ·) =sin(| · |t)| · |

∈M∞,1N,1 (Rd)

32

and

V2(t, ·) = cos(| · |t) ∈M∞,1N,0 (Rd)

from Chapter 6, by Proposition 5.7 it follows that V1(t, ·) ∈ W∞,1N,1 (Rd) and

V2(t, ·) ∈ W∞,1N,0 (Rd) for every N . Now we can apply Theorem 5.16 on ourresults and we obtain

ξ 7→ g(ξ) · sin(|ξ|t)|ξ|

∈ W q,pN,s+1(Rd),

ξ 7→ f(ξ) · cos(|ξ|t) ∈ W q,pN,s+1(Rd).

This in turn yields

F−12

(g(ξ) · sin(|ξ|t)

|ξ|

)(t, x) ∈ Mp,q

s+1,N (Rd),

F−12 (f(ξ) · cos(|ξ|t))(t, x) ∈ Mp,qs+1,N (Rd).

By linearity of the modulation space Mp,qs+1,N (Rd) we finally conclude that

the solution u = u(t, x) belongs to the modulation space Mp,qs+1,N (Rd) for

every fixed time t ∈ [0, T ]. In fact, this is the same space as the initial dataf(x) = u(0, x).By now we just considered u = u(t, x) as a function only dependent onthe physical space, i.e., as a function with respect to the x-variable. At thispoint we want to investigate the regularity of u with respect to the timet. We already showed that u(t, ·) ∈ Mp,q

s+1,N (Rd). Therefore it immediately

follows u ∈ L∞([0, T ],Mp,qs+1,N (Rd)). But it can also be proved that u ∈

C([0, T ],Mp,qs+1,N (Rd)). For that we take

limt1→t2

‖u(t1, ·)− u(t2, ·)‖Mp,qs+1,N (Rd) =

limt1→t2

(∫Rd

(∫Rd|Vφ(u(t1, ·)− u(t2, ·))(y, ω)〈y〉N 〈ω〉s+1|pdy

) qp

dω

) 1q

and it is easy to check that

limt1→t2

|Vφ(u(t1, x)− u(t2, x))(y, ω)| = 0

due to the continuity of cosine and sine, respectively. Note that the sameresult holds in any weighted modulation space since the weight functions areindependent of the time t.The next step is to differentiate u = u(t, x) with respect to t. Naturally this is

possible since both functions sin(|ξ|t)|ξ| and cos(|ξ|t) belong to the space C∞(R)

with respect to the t-variable. Even more precisely, their Taylor expansions

33

exist with respect to t. It follows

ut(t, x) = ∂t(F−12

(g(ξ) · sin(|ξ|t)

|ξ|

)(t, x)) + ∂t(F−12 (f(ξ) · cos(|ξ|t))(t, x))

= F−12 (∂t

(g(ξ) · sin(|ξ|t)

|ξ|)

)(t, x) + F−12 (∂t(f(ξ) · cos(|ξ|t)))(t, x)

= F−12 (g(ξ) · cos(|ξ|t))(t, x)−F−12 (f(ξ) · sin(|ξ|t)|ξ|)(t, x)

and

utt(t, x) = ∂t(F−12 (g(ξ) · cos(|ξ|t))(t, x))− ∂t(F−12 (f(ξ) · sin(|ξ|t)|ξ|)(t, x))

= F−12 (∂t(g(ξ) · cos(|ξ|t)))(t, x)−F−12 (∂t(f(ξ) · sin(|ξ|t)|ξ|))(t, x)

= −F−12 (g(ξ) · sin(|ξ|t)|ξ|)(t, x)−F−12 (f(ξ) · cos(|ξ|t)|ξ|2)(t, x).

Now we need to analyze in which spaces the functions ut and utt are con-tained. Obviously it arises a new problem since it does not hold sin(|ξ|t)|ξ| ∈L∞(Rd) anymore. Even more precisely, the factor sin(|ξ|t)|ξ| does not belongto any Lp-space with p ≥ 1. However applying appropriate weight functionssolves the problem. In fact, it is easy to verify that sin(|ξ|t)|ξ|〈ξ〉−1 ∈ L∞N (Rd)with respect to ξ. Analogously it follows cos(|ξ|t)|ξ|2〈ξ〉−2 ∈ L∞N (Rd). Thesame argumentation as in Chapter 6 gives

ξ 7→ sin(|ξ|t)|ξ| ∈M∞,1N,−1(Rd) ⊂W∞,1N,−1(Rd)

and

ξ 7→ cos(|ξ|t)|ξ|2 ∈M∞,1N,−2(Rd) ⊂W∞,1N,−2(Rd),where we also used Proposition 5.7. By Theorem 5.16 we get

g(ξ) · cos(|ξ|t) ∈ W q,pN,s(R

d),

f(ξ) · sin(|ξ|t)|ξ| ∈ W q,pN,s(R

d),

g(ξ) · sin(|ξ|t)|ξ| ∈ W q,pN,s−1(Rd) and

f(ξ) · cos(|ξ|t)|ξ|2 ∈ W q,pN,s−1(Rd).

By definition these results yield

F−12 (g(ξ) · cos(|ξ|t))(t, ·) ∈ Mp,qs,N (Rd),

F−12 (f(ξ) · sin(|ξ|t)|ξ|)(t, ·) ∈ Mp,qs,N (Rd),

F−12 (g(ξ) · sin(|ξ|t)|ξ|)(t, ·) ∈ Mp,qs−1,N (Rd) and

F−12 (f(ξ) · cos(|ξ|t)|ξ|2)(t, ·) ∈ Mp,qs−1,N (Rd).

So the preceding computations show

ut(t, ·) ∈ Mp,qs,N (Rd) and

utt(t, ·) ∈ Mp,qs−1,N (Rd).

As already mentioned above it immediately follows

ut ∈ L∞([0, T ],Mp,qs,N (Rd)) and utt ∈ L∞([0, T ],Mp,q

s−1,N (Rd)),

34

respectively. Furthermore the weights do not affect the continuity propertiesof ut(t, x) and utt(t, x) with respect to t. Summarizing we showed

u ∈ C1([0, T ],Mp,qs,N (Rd)) and

u ∈ C2([0, T ],Mp,qs−1,N (Rd)).

Naturally the question arises whether our obtained solution is a classicalsolution.

Proposition 7.1. Assume s, σ are real numbers such that s ≥ 0. Then the mod-ulation space M∞,1s,σ is contained in the space C of all continuous functions,

i.e. M∞,1s,σ (Rd) ⊂ C(Rd).

Proof. Let f ∈ M∞,1s,σ (Rd) and φ ∈ S(Rd) \ {0} be a fixed window. This

means Vφf(x, ξ) is an element of L∞0,σ(Rd) with respect to the x-variable

and an element of L1s,0(Rd) with respect to the ξ-variable. The equality (10)

ensures that we can rewrite Vφf as follows

Vφf(x, ξ) = F(f · Txφ)(ξ)

and as already mentioned this expression is contained in L1s,0(Rd) with respect

to the ξ-variable. Furthermore the inversion formula yields

f(t)φ(t− x) = (2π)−d2

∫RdF(f(t)φ(t− x))(ξ)eit·ξdξ. (28)

The estimate∫RdF(f(t)φ(t− x))(ξ)eit·ξdξ ≤

∫RdF(f(t)φ(t− x))(ξ)(1 + |ξ|2)

s2 eit·ξdξ

and the information Vφf(x, ·) ∈ L1s,0(Rd) gives the existence of (28). More-

over, this integral can be interpreted as a Fourier transform. Due to themapping properties of the Fourier transform we know that f · Txφ is contin-uous. Since it holds for every window φ the function f is continuous, that isf ∈ C(Rd). �

Remark. Proposition 6.3 yields Mp,1n,l ⊂ C for all p ∈ [1,∞] and all integers

l ≥ σ and n ≥ s ≥ 0.

Moreover, we mention a special case of Theorem 3.9 in [10].

Proposition 7.2. Let s, s0, σ, σ0 be real numbers and 1 ≤ p, q ≤ ∞ be the inte-grability parameters. Assume that f ∈ S ′(Rd), then the following conditionsare equivalent:

1. f ∈Mp,qs+s0,σ+σ0

(Rd),

2. xβ∂αf ∈ Mp,qs,σ (Rd) for all multi-indices α, β, where |α| ≤ s0 and |β| ≤

σ0,3. ∂α(xβf) ∈ Mp,q

s,σ (Rd) for all multi-indices α, β, where |α| ≤ s0 and|β| ≤ σ0.

35

With the help of these tools we can investigate the solution u = u(t, x)whether it is classical. The function u is called classical solution if all neces-sary derivatives exist in the classical sense. Since we already showed that allderivatives of u with respect to t exist in a classical sense, we only need toshow this with respect to the x-variable. Due to Proposition 7.2 and assumingu(t, ·) ∈Mp,q

s+1,N (Rd) where s ≥ 1, then it follows ∂αx u(t, ·) ∈Mp,qs−1,N (Rd) for

all multi-indices α such that |α| ≤ 2. If u(t, ·) ∈ Mp,1s+1,N (Rd), then Propo-

sition 7.1 gives ∂αx u(t, ) ∈ C(Rd) for all multi-indices α such that |α| ≤ 2.Hence, u(t, ·) ∈ C2(Rd) and ∆u exist in the classical sense. Once we haveshown that there exists a classical solution it is easy to check uniqueness ofthe solution due to the energy method. For details see the proof of Theorem3.4.Furthermore, we obtain a so-called a priori estimate for the solution u =u(t, x). For that we firstly consider the function

F2(u)(t, ξ) = v(t, ξ) = v1(t, ξ) + v2(t, ξ),

where v1 and v2 are defined by

v1(t, ξ) = g(ξ) · sin(|ξ|t)|ξ|

,

v2(t, ξ) = f(ξ) · cos(|ξ|t).

Note that Minkowski’s inequality holds for weighted Lebesgue spaces aswell. Here the space Lps,0(Rd) denotes the space of all functions f such that

f(ξ)〈ξ〉s ∈ Lp(Rd).

Lemma 7.1. Assume that s, σ are real numbers and p ∈ [1,∞] is an integra-bility parameter. If f, g ∈ Lps,0(Rd), then the sum f + g is also an element of

Lps,0(Rd). In fact it holds

‖f + g‖Lps,0(Rd) ≤ ‖f‖Lps,0(Rd) + ‖g‖Lps,0(Rd). (29)

Proof. This proof is very similar to the proof of Lemma 5.15. We have

‖f + g‖Lps,0(Rd) = ‖〈ξ〉s(f + g)‖Lp(Rd) = ‖〈ξ〉sf + 〈ξ〉sg‖Lp(Rd).

Thus, by the ordinary Minkowski inequality it follows

‖〈ξ〉sf + 〈ξ〉sg‖Lp(Rd) ≤ ‖〈ξ〉sf‖Lp(Rd) + ‖〈ξ〉sg‖Lp(Rd)= ‖f‖Lps,0(Rd) + ‖g‖Lps,0(Rd).

�

36

The following computations yield an estimate for v = v1 + v2 in thespace W q,p

σ,s for some real numbers s, σ, i.e.,

‖v(t, ·)‖W q,pσ,s (Rd) =

(∫Rd

(∫Rd|Vφv(t, x, ω)〈x〉s〈ω〉σ|pdω

) qp

dx

) 1q

=

(∫Rd

((∫Rd|(Vφv1(t, x, ω) + Vφv2(t, x, ω))〈ω〉σ|pdω

) 1p

〈x〉s)q

dx

) 1q

=

(∫Rd

(‖Vφv1(t, x, ω) + Vφv2(t, x, ω)‖Lpσ,0(Rdω)〈x〉

s)qdx

) 1q

≤(∫

Rd

((‖Vφv1(t, x, ω)‖Lpσ,0(Rdω) + ‖Vφv2(t, x, ω)‖Lpσ,0(Rdω)

)〈x〉s

)qdx

) 1q

=∥∥∥‖Vφv1(t, x, ω)‖Lpσ,0(Rdω) + ‖Vφv2(t, x, ω)‖Lpσ,0(Rdω)

∥∥∥Lq0,s(Rdx)

≤∥∥∥‖Vφv1(t, x, ω)‖Lpσ,0(Rdω)

∥∥∥Lq0,s(Rdx)

+∥∥∥‖Vφv2(t, x, ω)‖Lpσ,0(Rdω)

∥∥∥Lq0,s(Rdx)

= ‖v1(t, ·)‖W q,pσ,s (Rd) + ‖v2(t, ·)‖W q,p

σ,s (Rd).

Here we used the Minkowski inequality (29). Moreover, by previous compu-tations and Theorem 5.16 we obtain

‖v1(t, ·)‖W q,pσ,s (Rd) =

∥∥∥∥g · sin(| · |t)| · |

∥∥∥∥W q,pσ,s (Rd)

≤ C‖g‖W q,pσ,s (Rd)

∥∥∥∥ sin(| · |t)| · |

∥∥∥∥W∞,1σ,0 (Rd)︸ ︷︷ ︸

=C1(t)

≤ C2(t)‖g‖W q,pσ,s (Rd)

for some constants C,C1, C2 > 0, where C1 = C1(t) and C2 = C2(t). Thesame estimate holds for v2 = v2(t, x), that is, we get

‖v2(t, ·)‖W q,pσ,s (Rd) ≤ C

′2(t)‖f‖W q,p

σ,s (Rd).

Hence it follows

‖v(t, ·)‖W q,pσ,s (Rd) ≤ C1(t)‖g‖W q,p

σ,s (Rd) + C2(t)‖f‖W q,pσ,s (Rd)

for some constants C1, C2 > 0 depending on the time t. Hence we directlyobtain the a priori estimate

‖u(t, ·)‖Mp,qs,σ (Rd) ≤ C1(t)‖g‖Mp,q

s,σ (Rd) + C2(t)‖f‖Mp,qs,σ (Rd) (30)

by definition. Thus, the solution u = u(t, x) is continuously dependent on theinitial data f = f(x) = u(0, x) and g = g(x) = ut(0, x).Summing up we proved the following theorem.

Theorem 7.2. Assume that N is an arbitrary integer and s ∈ R such thats ≥ 1. If f ∈ Mp,1

s+1,N (Rd) and g ∈ Mp,1s,N (Rd), where p ∈ [1,∞], then there

37

exists a unique classical solution u = u(t, x) of the Cauchy problem (1) suchthat

u ∈ C([0, T ],Mp,1s+1,N (Rd))

⋂C1([0, T ],Mp,1

s,N (Rd))⋂C2([0, T ],Mp,1

s−1,N (Rd)).

Furthermore the a priori estimate

‖u(t, ·)‖Mp,1s+1,N (Rd) ≤ C1(t)‖g‖Mp,1

s+1,N (Rd) + C2(t)‖f‖Mp,1s+1,N (Rd)

holds for some constants C1 = C1(t), C2 = C2(t) > 0.

Remark. Obviously the solution u does not loose regularity, that is, for thegiven initial data f which is contained in the modulation space Mp,1

s+1,N the

solution u also belongs to Mp,1s+1,N for every t ∈ [0, T ].

Moreover, this result is independent of the dimension of the physical space,that is the dimension of the x-variable. The condition q = 1 for the modula-tion space Mp,q

s+1,N ensures the existence of a classical solution together withthe condition s ≥ 1. If at least one of these conditions are violated, then westill obtain a solution, but not in the classical sense anymore.

We can easily show that there exists a result for standard modulationspaces, that is for unweighted modulation spaces Mp,q.

Corollary 7.1. Let f, g ∈ Mp,q(Rd), where p, q ∈ [1,∞]. Then the Cauchyproblem (1) has a solution u ∈ C([0, T ],Mp,q(Rd)).Furthermore it holds the a priori estimate (30) but without weights, i.e. s =σ = 0.

Proof. This result immediately follows from the preceding investigations. �

Remark. Obviously we can neither say something about uniqueness nor aboutregularity of the derivatives of the solution with respect to t.

7.2. Solution in Sobolev Spaces

In Chapter 4 we already mentioned Sobolev spaces of distributions. Nowwe want to connect both topics modulations spaces and Sobolev spaces.Therefore we set the integrability parameters p = q = 2. Hence it followsf ∈M2

s,0(Rd) if and only if

‖f‖M2s,0(Rd) =

(∫∫R2d

|Vφf(x, ξ)|2(1 + |ξ|2)s dx dξ

) 12

<∞.

Proposition 7.3. Let φ ∈ S(Rd) \ {0} be a window function. Then M2s,0 =

FL2s,0.

Proof. Cf. Proposition 11.3.1 in [4]. �

Remark. Using only the weight 〈ξ〉s = (1 + |ξ|2)s2 gives

‖f‖FL2s,0(Rd) =

(∫Rd|f(ξ)|2(1 + |ξ|2)sdξ

) 12

= ‖f‖Hs(Rd)

since f ∈ FL2s,0 ⇐⇒ f ∈ L2

s,0. Therefore Proposition 7.3 yields Hs(Rd) =

M2s,0(Rd).

38

In order to obtain classical solutions the following result is needed.

Proposition 7.4. Let s, σ be real numbers and p ∈ [1,∞]. If s > d2 , then the

modulation space Mp,2s,σ is contained in the space C of continuous functions,

i.e. Mp,2s,σ (Rd) ⊆ C(Rd). Even more precisely it holds

Mp,2s,σ ⊆M

p,10,σ ⊆ C.

Proof. Let f ∈Mp,2s,σ (Rd), then by Holder’s inequality we get

‖f‖Mp,10,σ(Rd)

=

∫Rd

(∫Rd|Vφf(x, ξ)〈x〉σ|pdx

) 1p

dξ

=

∫Rd

(∫Rd|Vφf(x, ξ)〈x〉σ〈ξ〉s|pdx

) 1p 1

〈ξ〉sdξ

≤∥∥∥∥ 1

〈ξ〉s

∥∥∥∥L2(Rd)

(∫Rd

(∫Rd|Vφf(x, ξ)〈x〉σ〈ξ〉s|pdx

) 2p

dξ

) 12

= C‖f‖Mp,2s,σ(Rd)

for some constant C > 0. Note that∥∥∥ 1〈ξ〉s

∥∥∥L2(Rd)

is indeed finite since s > d2 .

Hence f ∈Mp,10,σ(Rd) and Mp,1

0,σ(Rd) ⊆ C(Rd) by Proposition 7.1. �

Remark. By applying Proposition 7.2 it easily follows for s, σ ∈ R and 1 ≤p ≤ ∞ that Mp,2

s,σ (Rd) ⊂ C2(Rd) if s > d2 + 2.

Now we have again all the tools to state results for the Cauchy problem(1) assuming that the initial data belong to some Sobolev spaces.

Theorem 7.3. If f ∈ Hs+1(Rd) and g ∈ Hs(Rd), where s is a real numbersuch that s > d

2 + 1, then there exists a unique classical solution

u ∈ C([0, T ], Hs+1(Rd))⋂C1([0, T ], Hs(Rd))

⋂C2([0, T ], Hs−1(Rd))

of the initial value problem (1).

Proof. Note that by Proposition 7.3 it holds Hs+1(Rd) = M2s+1,0(Rd). We

recall from Chapter 6 that

cos(|ξ|t) ∈ M∞,1N,0 (Rd) ⊂W∞,1N,0 (Rd) and

sin(|ξ|t)|ξ|

∈ M∞,1N,1 (Rd) ⊂W∞,1N,1 (Rd).

Furthermore we have f ∈ W 20,s+1(Rd) and g ∈ W 2

0,s(Rd). Thus by settingN = 0 and applying Theorem 5.16 we obtain

f · cos(|ξ|t) ∈ W 20,s+1(Rd) and

g · sin(|ξ|t)|ξ|

∈ W 20,s+1(Rd).

39

Hence it immediately follows by definition that the solution u(t, ·) is containedin the space M2

s+1,0(Rd) = Hs+1(Rd). It is easy to check that differentiating

the factors cos(|ξ|t) and sin(|ξ|t)|ξ| with respect to t yields ut(t, ·) ∈ Hs(Rd) and

utt(t, ·) ∈ Hs−1(Rd), respectively. The proof of continuity of u, ut and uttwith respect to t is also easily shown and is left to the reader. We basicallyhave shown this already in Section 7.1. Hence we get

u ∈ C([0, T ], Hs+1(Rd))⋂C1([0, T ], Hs(Rd))

⋂C2([0, T ], Hs−1(Rd)).

This is a classical solution due to Theorem 7.4. In fact we have shown u(t, ·) ∈Hs+1(Rd) = M2

s+1,0(Rd). Moreover, M2s+1,0(Rd) ⊂ C2(Rd) if s + 1 > d

2 + 2and this holds by the assumption. Uniqueness now follows by the energymethod. For details see the proof of Theorem 3.4. �

Remark. This result is independent of the dimension d and contains no lossof regularity since for f ∈ Hs+1(Rd) the regularity u ∈ C([0, T ], Hs+1(Rd))can be shown.Furthermore we determined that we get classical solutions for s > d

2 + 1. In

chapter 4 in [6] there is a remark about other cases, namely for 0 ≤ s ≤ d2 +1

we obtain Sobolev solutions. Otherwise, that is in more general cases such ass < −1, we obtain solutions in the sense of distributions. For 0 ≥ s ≥ −1 weobtain Sobolev solutions which do not possess an energy.

Finally we mention a more general result considering the Cauchy prob-lem (1) with initial data contained in Sobolev spaces.

Theorem 7.4. Assume that s0, s1 are real numbers and let the initial databe f ∈ Hs0(Rd), g ∈ Hs1(Rd). Then there exists a unique solution u ∈C∞([0, T ],S ′(Rd)). Even more precisely, we obtain

u ∈∞⋂n=0

Cn([0, T ], Hs3−n(Rd))

with s3 = min{s0, s1 + 1}.

Proof. This result can be proved in a similar way to the proof of Theorem7.3. Again setting N = 0 and applying Theorem 5.16 gives

f · cos(|ξ|t) ∈ W 20,s0(Rd) and

g · sin(|ξ|t)|ξ|

∈ W 20,s1+1(Rd).

Hence u(t, ·) is contained in the space Hs3(Rd), where s3 = min{s0, s1 + 1}.Furthermore the derivatives of all orders of u with respect to t exist and arecontinuous with respect to t. However by differentiating the solution u withrespect to t it looses regularity, that is

u ∈∞⋂n=0

Cn([0, T ], Hs3−n(Rd)).

40

Due to the definition of Sobolev spaces of distributions it follows that everysuch space is contained in the space S ′ of tempered distributions. This givesu ∈ C∞([0, T ],S ′(Rd)).For uniqueness we refer to Theorem 10.1 in [5]. �

41

References

[1] J. Bergh, J. Lofstrom, Interpolation Spaces: An Introduction, Springer (1976)

[2] H.G. Feichtinger, Banach spaces of distributions of Wiener’s type and interpo-lation, in: P. Butzer, B.Sz. Nagy, E. Gorlich (Eds.), Functional Analysis andApproximation, Proc. Conf. Oberwolfach (1980), Int. Ser. Num. Math., Vol. 69,Birkhauser (1981)

[3] F.G. Friedlander, M. Joshi, Introduction to the Theory of Distributions, 2ndedition, Cambridge University Press (1998)

[4] K. Grochenig, Foundations of Time-Frequency Analysis, Birkhauser (2001)

[5] M. Reissig, Vorlesung: Distributionentheorie, lecture notes

[6] M. Reissig, Vorlesung: Partielle Differentialgleichungen, lecture notes (2002)

[7] W. Rudin, Functional Analysis, McGraw-Hill, Inc. (1973)

[8] J. Toft, Continuity properties for modulation spaces, with applications to pseudo-differential calculus - I, Journal of Functional Analysis 207 (2004) 399-429

[9] J. Toft, Convolutions and Embeddings for Weighted Modulation Spaces,Birkhauser (2004)

[10] J. Toft, Pseudo-differential operators with smooth symbols on modulationspaces, Cubo - A Mathematical Journal (2009) 99-101

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