Curs_engl Transo Fourier

8/3/2019 Curs_engl Transo Fourier

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Analiza armonica

Curs scoala doctorala, sem II

2c

Versiune preliminara si incompleta

Eugen Popa

June 4, 2010


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Contents

1 Classical Harmonic Analysis 5

1.1 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.1.1 Generalitati. . . . . . . . . . . . . . . . . . . . . . . . . 51.1.2 Seria Fourier asociata unei functii. . . . . . . . . . . . . 61.1.3 Teorema lui Fejer . . . . . . . . . . . . . . . . . . . . . 10

1.2 Fourier Transform and Distributions . . . . . . . . . . . . . . 121.2.1 Ortogonalitate. . . . . . . . . . . . . . . . . . . . . . . 121.2.2 Cea mai buna aproximare . . . . . . . . . . . . . . . . 131.2.3 Spatii Hilbert . . . . . . . . . . . . . . . . . . . . . . . 14

1.3 Transformata Fourier . . . . . . . . . . . . . . . . . . . . . . . 161.3.1 Definitii si proprietati imediate. . . . . . . . . . . . . . 16

1.3.2 Transformata Fourier pe functii rapid descrescatoare. . 181.3.3 Transformata Fourier pe distributii temperate. . . . . . 20

2 Abstract Harmonic Analysis 212.1 Locally Compact Abelian Groups. . . . . . . . . . . . . . . . . 21

2.1.1 Topological groups . . . . . . . . . . . . . . . . . . . . 212.2 Haar measure . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.3 Convolution of measures and of functions . . . . . . . . . . . . 23

2.3.1 Spaces of functions and of measures . . . . . . . . . . . 242.4 The Dual Group . . . . . . . . . . . . . . . . . . . . . . . . . 252.5 The Fourier transform on GLCA . . . . . . . . . . . . . . . . 26

2.6 The Gelfand Transform . . . . . . . . . . . . . . . . . . . . . . 302.6.1 Banach algebras . . . . . . . . . . . . . . . . . . . . . . 312.6.2 Gelfand transform . . . . . . . . . . . . . . . . . . . . 322.6.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . 34

2.7 Bochners Theorem . . . . . . . . . . . . . . . . . . . . . . . . 352.8 Inversion theorem . . . . . . . . . . . . . . . . . . . . . . . . . 382.9 Plancherels theorem . . . . . . . . . . . . . . . . . . . . . . . 412.10 Duality Theorem of Potryaguine . . . . . . . . . . . . . . . . . 432.11 Bohr Compactification . . . . . . . . . . . . . . . . . . . . . . 45

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4 CONTENTS

2.12 A characterization of positive definite functions . . . . . . . . 46

3 Convolution Semigroups 493.1 Semigroups of kernels . . . . . . . . . . . . . . . . . . . . . . . 493.2 Convolution semigroups . . . . . . . . . . . . . . . . . . . . . 523.3 The symmetric stable semigroup . . . . . . . . . . . . . . . . . 553.4 Excessive measures: the discrete case . . . . . . . . . . . . . . 563.5 Transient convolution semigroups . . . . . . . . . . . . . . . . 583.6 Excessive measures . . . . . . . . . . . . . . . . . . . . . . . . 613.7 The non-commutative case . . . . . . . . . . . . . . . . . . . . 643.8 Heisenberg Group . . . . . . . . . . . . . . . . . . . . . . . . . 66

4 Annexes 694.1 The isoperimetric problem . . . . . . . . . . . . . . . . . . . . 694.2 Construction of the Haar measure . . . . . . . . . . . . . . . . 734.3 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 764.4 The functional equation for a character . . . . . . . . . . . . . 76


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Chapter 1

Classical Harmonic Analysis

1.1 Fourier Series

1.1.1 Generalitati.

Seriile de formaa02

+n=1

(an cos nx + bn sin nx)

se numesc serii Fourier .

Deoarece sumele partiale

Sn(t) := sin t + . . . + sin nt =sin n+1

2t sin nt

2

sin t2

Tn(t) :=1

2+ cos t + . . . + cos nt =

sin

n + 12

t

2sin t2

sunt marginite (chiar uniform) pe fiecare interval [, 2 ], pentru fiecare > 0, deducem o conditie (foarte restrictiva!) de convergenta (tip Abel /Dirichlet):

daca (an

) si (bn

) sunt monoton descrescatoare la 0, atunci seria Fouriereste uniform convergenta pe [, 2 ] si punctual convergenta peste tot.

De asemenea, daca

n=1

(|an| + |bn|) < +

atunci seria Fourier este uniform si absolut convergenta (pe R).Evident, n caz de convergenta, suma unei serii Fourier este o functie

periodica, de perioada 2.

5


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6 CHAPTER 1. CLASSICAL HARMONIC ANALYSIS

Cazul complex.

Fien=0

cnzn o serie de puteri avand raza de convergenta > 1 (cazul razei de

convergenta 1 pune probleme mai delicate (de ex. [3][693], pentru coeficientidescrescatori.. Punand z = eix, x [, ] si cn = an + ibn obtinem

n=0

[an cos nx bn sinnx] + i [bn cos nx + an sin nx]

(seri convergente, conjugate).

De exemplu:

1 +cos x

1!+

cos2x

2!+ . . . + +

cos nx

n!+ . . . = ecosx cos(sin x)

sin x

1!+

sin2x

2!+ . . . + +

sin nx

n!+ . . . = ecosx sin(sin x)

Mai general, putem considera sumekZ

ckeikx, deci o cuplare generala de

doua serii Fourier reale.Invers, nlocuind (formal)

cos nx =einx + einx

2

sin nx =einx einx

2i

se obtine o serie de forma kZ

ckeikx

cu cm = cm.

1.1.2 Seria Fourier asociata unei functii.

Fie f : [, ] R o functie continua, cu proprietatea f() = f() .Functia f se va considera astfel prelungita prin periodicitate la o functiecontinua pe R.

Calculele care urmeaza ar avea de fapt sens pentru orice f L1 ([, ]);Fiind spatiu cu masura finita, din inegalitatea lui Holder deducem Lp ([, ]) L1 ([, ]), pentru orice p [1, +].


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1.1. FOURIER SERIES 7

Vom considera coeficientii Fourier ai functiei f, notati traditional:

an :=1

f(t)cos ntdt, n 0

bn :=1

f(t)sin ntdt, n 1

si vom nota sumele partiale ale seriei Fourier asociata functiei f:

Sn(x) :=a02

+nk=1

(ak cos kx + bk sin kx)

Deocamdata, expresia coeficientiilor Fourier se poate justifica doar dacaseria este uniform convergenta: notand cu f suma sa si integrand termencu termen. Vom vedea ulterior ca rezultatul este general, daca lucram cuintegrala Lebesgue.

Inlocuind, obtinem:

Sn(x) =1

2

f(t)dt +1

nk=1

cos kx

f(t)cos ktdt+

+sin kx

f(t)sin ktdt = 1

12 + cos x cos t + . . . + cos nx cos nt f(t)dt++

1

(sin x sin t + . . . + sin nx sin nt) f(t)dt =

=1

1

2+ cos(x t) + . . . + cos n(x t)

f(t)dt

Formula urmatoare este bine cunoscuta si se poate demonstra n diversemoduri (prin inductie, cu numere complexe, prin transformarea produselorn diferente):

Dn(t) := 12

+ cos t + . . . + cos nt = sin n + 12 t2sin t

2

Functia Dn va fi prelungita prin continuitate n punctele k. Astfel:

Sn(x) =1

Dn(x t)f(t)dt = 1

xx

Dn(s)f(s + x)ds =

=1

x

... +

... +

x

...

=

1

Dn(t)f(x + t)dt


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deoarece

x... =

x

... datorita periodicitatii.

Alegand f 1, se calculeaza imediat Sn(x) 1, de unde

Dn(t)dt = .

Datorita periodicitatii si paritatii, deducem

0

Dn(t)dt =

0

Dn(t)dt adica0

Dn(t)dt =

2.

Sa mai notam x(t) := f(x + t) + f(xt)2f(x). O transformare simplapermite sa scriem

Sn(x) = 10

f(x + t)Dn(t)dt + 10

f(x t)Dn(t)dt

deci:

Sn(x) f(x) = 1

0

x(t)Dn(t)dt

Utilizand aceasta expresie, se obtin rezultatele clasice de convergenta punc-tuala pentru seriile Fourier [3].

Principala sursa de dificultati n acest studiu este semnul oscilant al nu-cleului Dn.

Mai general, n expresia lui se poate scadea 2S0. Convergenta revine laCriteriul lui Dini h

0

|x(t)|t

dt

O conditie suficienta este data deCriteriul lui Lipshitz Daca exista C si 1 such that

|f(x) f(x0)| C|x x0|

pentru toate punctelex dintr-o vecinatate a lui x0, atunci suma seriei este

f(x0).Daca exponentul > 1

2, atunci seria Fourier este uniform si absolut

convergenta, satisfacand conditia

n=1

(|an| + |bn|) < +

(teorema lui Bernstein, cf. [3][pg.539].Pe baza lemei lui Dirichlet:


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Daca f este monoton crescatoare si marginita n [0, h], atunci:

limn+

h0

f(x)sin nx

xdx =

2f(0+)

Din aceasta lema, se obtine:Criteriul lui DirichletJordan Daca f este cu variatie marginita pe o

vecinatate a punctului x0, atunci seria Fourier asociata converge la S0.In particular: Daca f este monotona pe portiuni si are un numar finit de

discontinuitati pe[, ], atunci seria Fourier asociata are suma 12

(f(x0+) + f(x0)).Exista cateva rezultate importante

Lema lui Riemann [3][vol.III, pg. 393] Daca functia f este absolutintegrabila pe [a, b], atunci

limn+

ba

f(x)sin nxdx = 0

limn+

ba

f(x)cos nxdx = 0

** Se va demonstra mai general ulterior **Consecinta imediata este:principiul localizarii Daca f g ntro vecinatate a unui punct x0,

atunci seriile Fourier asociate celor doua functii au aceeasi comportare n x0.De asemenea, rezulta si convergenta la 0 a coeficientilor Fourier asociati

unei functii absolut integrabile.Exemple. 1.

x2

=n=1

sin nx

n

pe (0, 2).Trecand x n 2x si scazand cele doua dezvoltari:

4=

n=1

sin(2n 1)x2n 1

pentru x (0, ). Se continua cu dezvoltarea functiei signum s fenomenulGibbs.

Cazuri particulare: x = 2

:

4= 1 1

3+

1

5 1

7+ . . .


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x = 6

:

4

= 1 + 15

17

111

+ 113

+ 117

. . .

2

3= 1 1

5+

1

7 1

11+

1

13 . . .

x = 2n=1

(1)n1 sin nxn

pe (, ).2.

x2 =

2

3 + 4

n=1

(1)ncos nxn2

pe [, ]. Convergenta este uniforma.

1.1.3 Teorema lui Fejer

Sa aplicam procedeul de sumare generalizata prin medii aritmetice. Notam

n(x) :=S0(x) + S1(x) + . . . + Sn(x)

n + 1

Daca functia f este continua, atunci sirul n este uniform convergent peR la f.

Demonstratie. Introducand noul nucleu

Kn(t) :=1

n + 1

nk=0

Dk(t) =1

2(n + 1)

sin(n + 1) t

2

sin t2

2putem scrie:

n(x)

f(x) =

1

0

x(t)Kn(t)dt

Functia f fiind presupusa continua pe [, ], rezulta uniform continua peR, adica

> 0 astfel ncat |x(t)| < , 0 < t < independent de x R.

Astfel, putem scrie:

|n(x)f(x)| 1

0

|x(t)|Kn(t)dt = 1

0

|x(t)|Kn(t)dt+ 1

|x(t)|Kn(t)dt


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0

Kn(t)dt +1

1

2(n + 1)

1

sin2 2

0 |x(t)

|dt

2+

1

1

2(n + 1)

1

sin2 24M

de unde concluzia.

Consecinte ale teoremei lui Fejer.1) Fie f L2 ([, ]) cu proprietatea ca

f(t)sin ntdt = 0,

f(t)cos ntdt = 0, n 0

Atunci f = 0 a.p.t.

Demonstratie. Consideram functia F(x) := C+

x

f(t)dt, unde con-

stanta C este aleasa astfel ncat

F(x)dx = 0, adica

C := 12

dx

x

f(t)dt

Functia F este (absolut) continua pe [, ]. Acum F() = C iar F() =C, deoarece prin ipoteza

20

f(t)dt = 0. Astfel functia satisface ipotezele

teoremei lui Fejer. Coeficient ii Fourier ai functiei F sunt:

F(t)cos ntdt =

cos nt C+

t

f(x)dx dt =

cos ntt

f(x)dx dt =

=

x

cos ntdt

dx =

f(x)sin nt

n|x dt =

1

n

f(x)sin nxdx = 0

si analog:

F(t)sin ntdt =1

n

f(x)cos nxdx = 0

Deducem acum Sn(x) = 0, n(x) = 0, deci F 0. Deoarece F(x) = f(x)a.p.t.[?], urmeaza ca f = 0 a.p.t.

Corolar (Weierstrass). Pentru fiecare f : [a, b] R continua, existaun sir de polinoame (trigonometrice) pn

f uniform pe [a, b].

Teorema lui Fejer justifica aproximarea uniforma a functiilor continue cupolinoame trigonometrice. Functiile sin si cos se aproximeaza uniform cupolinoame chiar din dezvoltarea n serie de puteri. Iar functiile continue suntdense n L2.

Reciproc, n cazul intervalului compact, teorema lui (Stone) Weierstrassarata ca spatiul liniar al polinoamelor este uniform dens n C(I), iar acestaeste dens n L2. Aceasta proprietate se verifica, pe baza teoremei StoneWeierstrass, pentru intervale compacte de familia ortogonala clasica sin nx,cos nx.


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1.2 Fourier Transform and Distributions

Vom considera peste tot un spatiu vectorial (real) H, pe care sa definit unprodus scalar

< , >: H H RSunt asadar satisfacute axiomele:

1. < x + x, y >=< x, y > + < x, y >, x, x, y H2. < x,y >= < x,y >, x, y H, R3. < x, y >=< y, x >,

x, y

H4. < x, x > 0, x H5. < x, x >= 0 x = 0Observatie. In cazul spatiilor vectoriale complexe, cerinta 3. se modifica:

< x, y >= < y, x >, x, y H.Din axiome rezulta imediat

< x,y + y >= < x, y > + < x, y >, x,y,y H, , R

(reformulare corespunzatoare n cazul complex).Reamintim ca are loc inegalitatea lui Cauchy :

| < x, y > |2 < x, x >< y, y >, x, y H

ceea ce arata ca x := < x, x > defineste o norma pe H. In plus, pentrux = 0, y = 0 are loc | < x, y > |xy [1, 1], astfel ca, n prezenta unui produsscalar se poate defini unghiul a doi vectori x = 0, y = 0 ca unicul [0, ]pentru care cos =

< x, y >

xy .

1.2.1 Ortogonalitate.

Fie I o multime cel mult numarabila.Definitie. Dat fiind un spatiu cu produs scalar (real), familia (ei)iI se

numeste ortogonala daca < ei, ej >= 0, i = j.Daca, n plus, < ei, ei >= 1, i, familia se numeste ortonormala .

Observatii. Orice familie ortogonala furnizeaza o familie ortonormala eliminand

eventualele elemente 0 si normalizandei

ei .


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1.2. FOURIER TRANSFORM AND DISTRIBUTIONS 13

Orice familie ortonormala este liniar independenta: este suficient ca nni=1

iei = 0 sa facem produsul scalar cu ej, pentru a obtine j = 0.

Exemple. 1) In Rn cu produsul scalar uzual < x, y >:=nk=1

xkyk, o fam-

ilie ortonormala este: (ek)k=1,n, ek := (0, 0, . . . , 0, 1, 0, . . . , 0)

2) In spatiul liniar C([, ]) al functiilor continue pe [, ], cu produsulscalar < f, g >:=

f(t)g(t)dt, o familie ortogonala este {cos nx, sin mx |n 0, m 1}. Familia ortonormala corespunzatoare este:

12

,1

cos x,1

sin x , . . . ,1

cos nx,1

sin n x . . .

Familia { 12

eikx}kZ este familie (chiar baza) ortonormala n L2C([, ]).3) Exemplul fundamental (si tipic) este L2(), masura pozitiva (-

finita) pe un spatiu masurabil, cu produsul scalar

< f, g >:=

f.gd

1.2.2 Cea mai buna aproximare

Fie (ei)1in o familie ortonormala, care genereaza subspatiul liniar X0. Fiex un element oarecare. Ne propunem sa gasim elementul de cea mai bunaaproximare (pentru x n subspatiul X0, adica y0 X0 cu proprietatea cadistanta x y este minima, cand y X0. Altfel spus:

x y0 = minyX0

x y

Este comod sa introducem piciorul perpendicularei din x pe subspatiulX

0, adica elementul y

0 X

0care are proprietatea ca x

y0

este ortogonalpe orice element din X0. Altfel spus:

< x y0, y >= 0, y X0Notarea cu aceeasi litera a celor doua elemente nu este o greseala: exista unelement unic n subspatiul X0, care este simultan elementul de cea mai bunaaproximare si piciorul perpendicularei din x pe subspatiul X0:

Propozitie. y0 =ni=1

< x, ei > .ei.


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Demonstratie. Scriind ca y := ni=1 i.ei si punand conditia sa fie

ortogonal pe fiecare ei, se gaseste imediat y0 := ni=1 < x,ei > .ei X0.Acum se verifica usor ca y0 este si elementul de cea mai buna aproximare:x y2 = x y0 + y0 y2 = x y2 + y0 y2 x y02 (caci x y0este ortogonal pe y0 y X0). Rezulta ca distanta de la x la subspatiulX0 este x y02 =< x y0, x y0 >= x02 2 < x , y0 > +y02 =x2 2ni=1 < x, y0 >2 +ni=1 < x, y0 >2= x2 ni=1 < x, y0 >2.Aceasta distanta se poate exprima cu determinanul Gram:

Aplicatie. Patratul distantei de la x la subspatiul liniar generat de vectoriiliniar independenti (xi) este

G(x, x1, . . . , xn)G(x1, . . . , xn)

Este suficient sa observam ca matricea de trecere de la x, x1, . . . , xn lax, e1, . . . , en are acelasi determinant ca si A, iar

G(x, e1, . . . , en) =

< x, x > < x, e1 > . . . < x, en >< e1, x > 1 . . . 0

. . . . . . . . . . . .< en, x > 0 . . . 1

=

= x2 ni=1

< x, ei >2

Calculul precedent demonstreaza si inegalitatea lui Bessel:Propozitie. Fie (ei) o familie ortonormala, iar x H oarecare. Atunci

ni=1

| < x, ei > |2 x2

cu egalitate iff x este n nf asur atoarea liniar a a familiei (ei).

1.2.3 Spatii Hilbert

Fie (ek)k=1,n o familie ortonormala. Sa observat ca, dintre toate elementele

de forma y =nk=1

kek, elementul y0 =nk=1

< x, ek > ek este n acelasi timp

piciorul perpendicularei (din x pe subspatiul liniar generat de (ek)k=1,n), catsi elementul de cea mai buna aproximare.


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1.2. FOURIER TRANSFORM AND DISTRIBUTIONS 15

In cazul spatiilor vectoriale infinit dimensionale, este normal sa con-

sideram familii ortonormale numarabile (en)nN si sa asociem seria Fouriera elementului x H.

Daca spatiul vectorial cu produs scalar H este complet (fata de normaindusa de produsul scalar), atunci seria Fourier este totdeauna convergenta.

Lema. Serian=1

xn de elemente ortogonale este convergenta daca si

numai dacan=1

xn2 < +

Demonstratie. Deoarece

N

n=1 xn2 =N

n=1 xn2, o implicatie este ime-diata.

Invers, dacan=1

xn2 < +, atunci xn+1 + . . . + xm2 = xn+12 +. . . + +xm2 arata ca este ndeplinita conditia de sir Cauchy (pentru sirulsumelor partiale).

Corolar. Seria Fourier asociata elementului x H este convergenta.Demonstratie.

n=1

< x, en > en fiind o serie de elemente ortogonale,

convergenta este echivalenta cu n=1

| < x, en > |2 < +, ceea ce rezulta din

inegalitatea lui BesselNn=1

| < x, en > |2 x2.

Urmatorul rezultat indica o proprietate echivalenta cu faptul ca seriaFourier are ca suma chiar x.

Teorema (Parseval). Fie (en)nN o familie ortonormala. Sunt echiva-lente:

1) x =

n=1 < x, en > en;2) x2 =

n=1

| < x, en > |2

Demonstratie. 1) 2) rezulta simplu din linearitatea si continuitateaprodusului scalar.

2) 1) Consecinta a inegalitatii lui Bessel:

0 x ni=1

< x, ei > ei2 = x2 ni=1

| < x, ei > |2 0


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Urmatorul rezultat indica de asemenea o conditie necesara si suficienta

ca seria Fourier a oricarui element x H sa aiba ca suma x. Spre deosebirede teorema lui Parseval, este vorba de o condit ie globala.

Propozitie. Pentru o famile ortonormala (en)n sunt echivalente:

(i) x H: x =m=1

< x, en > en.

(ii) x H, < x, en >= 0, n = x = 0.Demonstratie. (i) = (ii) este evident.(ii) = (ii). Notam y :=

n=1

< x, en > en. Rezulta imediat ca < x

y, en >= 0, n. Conform ipotezei x y = 0, c.c.t.d.Observatie. Conditia (ii) este implicata (de fapt, echivalenta cu) de:

subspatiul liniar generat de (en)n este dens n H.Definitie. Daca proprietatile echivalente din proprozitie au loc, atunci

spunem ca familia (en) este o baza ortonormala pentru spatiul Hilbert H.Ca o consecinta a teoremei lui Fejer, familia

{ 12

,1

cos x,1

sin x , . . . ,1

cos nx,1

sin n x . . .}

este o baza ortonormala n L2(

, ].

1.3 Transformata Fourier

1.3.1 Definitii si proprietati imediate.

Pentru f L1(Rd) definim transformata Fourier prin

f(y) :=

Rd

eix.yf(x)dx

Definitia are sens si pentru f Lp

(Rd

), p > 1, deoarece eix.y

este dinfiecare Lq(Rd), cu q (1, +].Sa observa ca nu exista incluziuni ntre spatiile Lp(Rd).Definitia se extinde si la masuri pe (Rd, B) prin

(y) :=

Rd

eix.y(dx)

se numeste transformata FourierStieltjes a masurii .Se noteaza traditional cu A(Rd) multimea functiilor f, cand f L1(Rd).


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1.3. TRANSFORMATA FOURIER 17

Direct din teorema de convergenta dominata, deducem ca f este continua.

Lema lui Riemann-Lebesgue

limy

f(y) = 0

Demonstratie. Pentru fiecare interval compact [a, b] si orice functie fcu derivata continua pe [a, b], o inegrare prin parti arata ca:

f(y) =

ba

f(x)eixydx = f(x)1

iyeixy

ba

1

iy

ba

f(x)eixydx 0

Afirmatia este deci adevarata pentru toate functiile cu derivata continuasi cu suport compact.

Functiile de clasa C1 cu suport compact fiind dense n L1(R), se obtinerezultatul general.

(sau: Din teorema de convergenta dominata, rezultatul se extinde la toatefunctiile pozitive si integrabile pe [a, b]. Fiecare functie f integrabila pe [a, b]se scrie f = f+ f.)

Theorem. A(Rd) este o subalgebra, densa n C0(Rd).Demonstratie. Pentru demonstratie se folosesste teorema lui Stone

Weierstrass (forma complexa) pe compactul [, +]. Astfel apare adjunctiaunitatii, caci 1 = 0.

Daca f, g

L1(R, atunci

(f g)(x) :=R

f(x y)g(y)dy

este bine definita si f g L1(R.(i) Daca f, g sunt masurabile, atunci (x, y) f(xy)g(y) este masurabila

(pe RR, ca fiind compunere de aplicatii masurabile: (x, y) (x y, y) f(x y)g(y).

(ii) Daca F(x, y) 0 este masurabila, atunci x

F(x, y)(dy) este

masurabila.

Pentru demonstratie, se considera f(x, y) = f(x)g(y), pentru care rezul-tatul este evident. Aceasta include AB. Prin clase monotone, rezultatulse extinde la toate functiile caracteristice de multimi masurabile din R R.Deci masurabilitatea se pastraza pentru toate functiile masurabile pozitive.Rezultatul se extinde la toate functiile integrabile luand partea pozitiva sinegativa.

(iii) Pentru existenta convolutiei, mai ramane sa asiguram ca, macar a.p.t.are loc

(x) :=

R

|f(x y)g(y)|dy < +


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Conform celor de mai sus, este masurabila. Pe baza teoremei lui Tonelli:R

(x)dx =R

R

|f(x y)g(y)|dydx = f1g1

Deci L1(R), n particular este finita a.p.t. Astfel, f g este bine definitaa. p. t.

Mai mult: |(fg)(x)| (x), ceea ce arata ca fg L1(R) si fg1 f1g1.

(iv) Proprietatile de comutativitate si asociativitate rezultand prin calculdirect, deducem ca L1(R) este algebra Banach comutativa. Observam ca nuadmite unitate.

Faptul ca A(Rd) este o subalgebra rezulta din formula f g = f .g, careva fi demonstrata ulterior.

Separarea decurge din urmatoarele observatii:R

ex2

eixydx =

R

e(x+iy2 )

2

ey2

4 dx =

ey2

4

Fie a R si fa(x) := f(x + a). Atunci

fa(y) = eiayf(y)

In sfarsit, conjugata transformatei este transformata functiei x f(x).Problema esentiala a transformatei Fourier este cea a inversarii.

1.3.2 Transformata Fourier pe functii rapid descrescatoare.

Se noteaza cu S(Rd) multimea functiilor f, indefinit diferentiabile pe Rd,care verifica

p,(f) := supxRd

xDf(x) < +pentru orice multiindici si .

S se considera nzestrat cu topologia (local convexa, metrizabila, estecomplet / fata de orice metrica compatibila, invarianta la translatii/ spatiuFrechet) data de familia de seminorme (p,),.

Observam ca Scontine, pe langa functiile indefinit diferentiabile cu su-port compact si x ea|x|2, pentru a > 0.

Teorema. Transformata Fourier este un izomorfism al lui S.Demonstratie. 1. Transformata Fourier aplica continu Sn S.Mai mult, au loc urmatoarele fapte:* transformata Fourier a functiei Djf este yj f(y);


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1.3. TRANSFORMATA FOURIER 19

* transformata Fourier a functiei xjf(x) este

Dj f.

2. Are loc formula de inversare a transformatei Fourier:

f(x) = (2)n

eif(y)dy

Interpretare: f(y) este densitatea frecventei y, n descompunerea armonicaa lui f.

Pentru demonstratie, ar trebui sa calculam integrala iterata

eidy f(t)e

idt

pentru f S. Dar integrala dubla nu este absolut convergenta, iar ordineade integrare nu poate fi inversata.

Se introduce nca un factor g(y) cu g S; (de fapt g(y) := e|y|2).Astfel, (y, t) g(y)eif(t)ei devine absolut integrabila si putem

transforma:eig(y)dy

f(t)eidt =

f(y)eig(y)dy =

(schimband ordinea de integrare)f(t)dt

eig(y)dy =

g(t x)f(t)dt =

g(t)f(x + t)dt

Schimband g cu g(y), care are transformata Fourier ng (y/) se obtine:f(y)g(y)eidy =

g(y)f(x + y)dy

Trecerea la limita 0 sub integrala este permisa, deoarece f, g suntintegrabile, iar f, g sunt continue si marginite. Se obtine astfel

g(0)

f(y)eidy = f(x)

g(y)dy

In sfarsit, deoarece pentru g(x) := e|x|2/2 avem g(y) = (2)n/2e|y|

2/2 iarg(y)dy = (2)n/2, se obtine formula propusa:

f(x) = (2)n

f(y)eidy


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1.3.3 Transformata Fourier pe distributii temperate.

Ca de obicei, vom nota cu S sptiul dual. u S se numeste distributietemperata . Adica u : S C este o functionala liniara si continua. Aceastanseamna ca exista o constanta C > 0 si multiindicii , such that

u(f) C supxRd

xDf(x) , f SS se considera cu topologia slab.

Denumirea de distributie se justifica astfel. Este evident ca D = C0 S. Mai mult, incluziunea este continua. Astfel, restritia oricarei distributiitemperate este o distributie. In plus, daca doua distributii temperate coincidca distributii, atunci coincid si ca distributii temperate. Aceasta afirmatierezulta din faptul ca imaginea D este densa n S. n adevar, fie f S. Fixam D cu (x) = 1, x 1. Notand f(x) := f(x)(x), avem: f D,iar

f(x) f(x) = f(x) ((x) 1) = 0daca x < 1

. Un calcul direct arata ca f f n S.

Acum are sens sa definim transformata Fourier a unei distributii temper-ate u ca u(f) := u(f), f S.

Ce au n comun seriile Fourier cu transformata Fourier?


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Chapter 2

Abstract Harmonic Analysis

2.1 Locally Compact Abelian Groups.

2.1.1 Topological groups

Let (G, +) be a commutative group with neutral element denoted 0 and be a compatible, locally compact topology on G. This means that the maps(x, y) x + y and x (x) are continuous. Equivalently: (x, y) x yis continuous. Particularly, for each x0 G the map x x + x0 is a

homeomorphism ofG. This remark shows that the topology is completelydefined by a fundamental system of neighbourhoods of 0, i.e. a family V=(V) such that:

0 V

V, V V such that V V V

V W V such that W + W V.W may be chosen symmetric and compact.

Let H G be a closed subgroup. : G G/H will denote the canonicalprojection. We define a topology on G/H: D G/H is called an open setwhen 1(D) is open in (G, ).

Theorem G/H is a locally compact commutative group.Proof Clearly we have defined a topology on G/H. This topology is

separated. Indeed, let x + H = y + H. Since y + H is closed, there existsa neighborhood V for 0, such that (x + V) (y + H) = . Let W be asymmetric, compact neighborhood for 0, such that W + W V. Then(x + H+ W) (y + H+ W) = .

21


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22 CHAPTER 2. ABSTRACT HARMONIC ANALYSIS

Let us remark that is not only continuous, but also open, since 1 ((D1)) =

D1 + H. We conclude that the topology on G/H is also locally compact.Let us prove the compatibility. Let D G/H be an open set and consider

D := {(x + H, y + H)|(x y) + H D}. We have ( )1(D) ={(x, y)|x y 1(D)}. This set is open, since:

D is open in G/H G/H iff ( )1(D) is open in G G is surjective, hence f(f1(D)) = D (where f = ); is an open mapping.Examples 1. Finite (commutative) groups;

2. (Rn

, +) with its (discrete) subgroup (Zn

, +) and the torus Rn

/Zn

.3. (R, ) with the subgroup (0, +), which is open and closed.4. (C, ) and the closed subgroup T of complex numbers of modulus 1.We have the isomorphisms C/T (0, +); C/(0, +) T; C/R is

called the (one dimensional) projective space.5. GLn(R) the (noncommutative) group of invertible matrices, with

several remarkable subgroups.6. The Heisenberg group Hn := C

nR with (z, t)(z, t) = (z + z, t + t +12

Im (zz )).

2.2 Haar measureWe consider G with the Borel algebra B. Also, all measures are Radon(positive, finite on compacts and regular). Let us recall that a Radon measurecan be characterized as a linear and positive functional on K(G).

We accept that, on each locally compact (commutative) group there existsa (non zero) measure which is invariant for translations meaning that

m(x + A) = m(A), A B, x Gequivalently:

G

f(x + y)dm(x) = G

f(x)dm(x)

for any measurable f.Proposition. This measure is unique up to a multiplicative constant

and is called the Haar measure on the group G.Proof. Indeed, let m be another invariant measure and let us fix f0

K(G) with G

f0dm = 1. For any f K(G) we have:G

f dm =

f0(y)dm(y)

G

f(x)dm(x) =


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2.3. CONVOLUTION OF MEASURES AND OF FUNCTIONS 23

= G f0(y)dm(y) G f(x + y)dm(x) ==

G

G

f0(y x)f(y)dm(y)dm(x) =

=

G

f(y)dm(y)

G

f0(y x)dm(x) = CG

f dm

Properties. a. Let us remark that the invariance to the opposite results:indeed, defining m(A) := m(A), for each A B, we obtain clearly aninvariant measure and taking for A a symmetric, compact neighborhood of0 we deduce that the constant is 1.

b. Let D be a non-void, open set. Then m(D) > 0. Indeed, let supposeby contradiction that m(D) = 0. For any compact set K we have m(K) = 0,since it can be covered by a finite number of negligible sets of the form x+D.From the regularity of the Radon measure, it follows that m(A) = 0, for anyBorel set.

Comment. The result remains true for non-commutative, (locally com-pact) groups; but there appear right and leftinvariant measures, connectedby the modular function.There exist important examples of non-commutativegroups (called unimodular, for example the Heisenberg group) such that theHaar measure is both right and leftinvariant.

Usually, m is normalized: for compact G, we may take m(G) = 1; ifG isdiscrete, then m({x}) = 1(and conversely, if m({0}) > 0, then m({x}) > 0,x G; thus any compact neighborhood of 0 contains only a finite numberof points). For finite groups, these requirements are contradictory, so one ofthe possibilities will be adopted.

Examples of Haar measure

on the discrete groups, the Haar measure isxG

x. It is -finite iff the

group is at most countable.

on Rd

the Lebesgue measure. Also on the torus

Exercise: describe the Haar measure on (0, +) and on GLn(Rd).

2.3 Convolution of measures and of functions

Let f be a function on G and y G. We denote fy(x) := f(x y). Due tothe invariance to translation, we deduce that f and fy has the same norm,in each Lp(G), p [1, ].


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Lemma The map y

fy is (uniformly) continuous from G to L

1(G).

The result hold in fact for any Lp and C0; but does not holds for L,unless G is discrete.

The uniform continuity is to be understood in the following sense : > 0V neighborhood of 0, such that x y V fx fy < .

ProofFor the proof, let us remark that fyfy = (f fyy)y. Thus, wemay suppose y = 0 and prove the continuity. Moreover, as the continuousfunctions with compact support are dense in L1, we may suppose that fis continuous and has support K. Now, > 0 being given, there exists a(compact) neighborhood V for 0, such that f fx < /m(K), x V.Hence f fx1 < , x V.

Convolution of functions Let us recall that, for f, g L1

(G), thefollowing definition is correct

(f g)(x) :=G

f(x y)g(y)dy

and turns L1(G) into a (commutative) Banach algebra. In particular f g1 f1g1.

The definition can be extended to measures from M(G) as:

G f(x)( )(dx) := G G f(x + y)(dx)(dy)This definition turns M(G) into a (commutative) Banach algebra. Moreover,we have (f.m) (g.m) = (f g).m .

Remark. The convolution of unbounded functions or measures may ormay not exist.

Let m be the Lebesgue measure on R. Then m m does not exists.Indeed, for any measurable part A, with m(A) > 0 we have

(m m)(A) =R

m(A y)m(dy) = +

Let := [0,+).m (the Lebesgue measure on [0, +)). In this case, exists: this amounts to the convolution of functions [0,+) [0,+), whichequals x.[0,+)(x).

2.3.1 Spaces of functions and of measures

We will encounter several spaces of functions and measures.If K is compact, the Banach space C(K) of continuous functions on K

whose dual is the space of Radon measures on K. If X is locally compact,


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2.4. THE DUAL GROUP 25

then we may consider the Banach space

C0(X) of continuous functions, with

limit 0 at . Also, we have the locally convex space K(X) of continuousfunctions, with compact support. Again, its dual is the space of Radonmeasures.

With any space with measure (X, B, ), one associate the Banach spacesLp(X), for p [1, ]. The dual of Lp is (canonically identified with ) Lq,where p (1, ) and 1/p + 1/q = 1. The dual of L1 is L. Particularly, L2is a Hilbert space.

2.4 The Dual Group

Definition. Let G be a LCA group. We call a character on G any function : G C with the properties:

|(x)| = 1, x G

(x + y) = (x)(y), x, y GNotation. The set of continuous characters on G will be denoted as

and is endowed with the structure of a commutative group, given by ( 1 +2)(x) := 1(x)2(x), x G

Definition. We define a topology on as follows. For each compactK G and r > 0 let us denote:

N(K, r) := { | |(x) 1| < r,x K}

Proposition. (N(K, r))K,r is a fundamental system of (open) neighbor-hoods for 0 (compact open topology).

Proof. Clearly 0 N(K, r) and

N(K1, r1) N(K2, r2) N(K1 K2, min(r1, r2))

Let

N(K, r) and choose 0 < < r

supK |

(x)

1|. We have then

+ N(K, ) N(K, r).This topology is compatible with the group structure, since

[1 + N(K, r)] [2 + N(K, r)] 1 2 + N(K, 2r)

Comment. It does not seem to be easy to prove directly that we havedefined a locally compact topology. Instead, we consider another topology on, which is locally compact and prove the coincidence of the two topologies.

This definition is based on the Fourier transform.


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2.5 The Fourier transform on GLCA

Definition. For each f L1(G), we denote by f the function on , definedas:

f() :=

G

f(x)(x)dx

The definition is correct, since by Holders inequality we have:

|f()| f1Proposition. f is a continuous function on .Proof. Indeed, let 0

be fixed and let

+ N(K, r), where the

compact K G and r > 0 are to be chosen. It means: |( 0)(x) 1| < r |(x)0(x) 1| < r |(x) 0(x)| < r, x K. Now:

f() f(0) =G

f(x)(x)dx G

f(x)0(x)dx =

=

K

f(x)[(x) 0(x)]dx +G\K

f(x)[(x) 0(x)]dx

f() f(0)

r.f1 + 2

G\K|f(x)|dx

Since f L1(G) and dx is regular, we can choose the compact K such thatG\K |f(x)|dx becomes arbitrarily small.

Notation. Let us denote by A() the set of all functions f (when f L1(G)). We show that this set separates . Let 1 = 2 . From thecontinuity, there exists a nonvoid, open set U G such that 1(x) = 2(x),x U. We can find a compact set K U, with m(K) > 0. Now f(x) :=K(x)(1(x)2(x)) is in L1(G) and f(1) f(2) =

K

|1(x)2(x)|2dx >0.

Theorem The map (x, ) (x) is continuous on G .Proof Let us compute:

fx() = G

fx(y) (y) dy =

G

f(y) (x) (y) dy = (x) f()

Since the open sets [f = 0] cover , it suffices to prove the continuity of themap (x, ) fx() on G . Let (x0, 0) G and > 0 be fixed. Now,we have the inequality:

|

fx()

fx0(0)| fx fx01 + |

fx0()

fx0(0)|


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28/79


Comment. Of course, some facts follow now more easily. For example,

the continuity of the map f (from the very definition of the weak topology);also the fact that A() separates the points (using HahnBanach theorem).

Properties of the Fourier transform.1. (f )(x) = (x)f(), f L1(G), , x G. In particular, the

Fourier transform can be interpreted as a convolution: f() = (f )(0),f L1(G), .

2. If0 and g(x) = 0(x)f(x), then g() = f( 0).If g = fx, then g() = (x) f().Thus, A() is translation invariant.3. Let f(x) := f(x). Then g is the complex conjugate of f.Combining with the fact, already proved, that

f g = f g, we deduce

that A() is a dense subalgebra in C0().FourierStieltjes transform Let M(G). We define a function on

, denoted and called the FourierStieltjes transform of the measure by:

() :=

G

(x) (dx)

The set of all such functions will be denoted by B(). We have f = f.m.The following results hold true:

1. Each is a bounded and uniformly continuous function on .

2. = . Hence, the map () is a complex homomorphism,for each .

3. B() is invariant under: translation; multiplication by (x); and com-plex conjugation.

Topology on G Let K be a compact subset, r > 0 and defineN(K, r) := {x G | |(x) 1| < r, K}. We prove that N(K, r) is anopen set in G (with the original topology). Indeed, let x0 N(K, r). Foreach 0 K, there exists V neighborhood for x0 and W neighborhood for 0,such that |(x) 0(x0)| < r, x V, W. Since K is compact, we coverit with a finite family of such V. Let us denote by W the intersection of the

corresponding W. Then W is a neighborhood for 0 and W N(K, r).We shall prove that (N(K, r))K,r form even a base for the topology on G,

but we need some more preparation.Theorem Let G be discrete. Then is compact.Proof Indeed, L1(G) has a unit {0}. Now {0}() = (0) = 1, ,

hence {0} 1. But {0} C0() and the conclusion follows.Theorem Let G be compact; then is discrete.Proof Indeed, now f 1 belongs to L1(G), while f = {0}. Indeed:

f(0) =G

dx = 1; while for each = 0, there exists x0 G such that


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2.5. THE FOURIER TRANSFORM ON GLCA 29

(x0)

= 1. Then

G

(x)dx = (x0)

(x x0)dx = (x0)(x)dxhence f() =

G

(x)dx = 0. Since f is continuous, we deduce that {0} isan open set.

Examples 1. Let G := R. Each character can be identified with a realnumber t through the correspondence (x) = eixt. It is obvious that eachsuch is indeed a character and to different t correspond different characters.

Conversely, let : R T be a continuous map, with the property (x +y) = (x)(y). Then there exists t R such that (x) = eixt. The proof isleft as an exercise.Moreover, the compactopen topology is the usual topology on R, sinceN([n, n], r) corresponds with the set of those t for which |1 eitx| < r,x [n, n], which is equivalent to |t| < 2

narcsin r

2.

Thus, the Fourier transform particularizes to:

f(x) =

R

f(t)eitxdt

2. Let now G := R/Z T. Similar computations show that is of theform above, with the additional property that (x + 1) = (x). This comesto t

2

Z, thus identifying canonically with Z.

Since G is compact, the topology on is the discrete one.As for the Fourier transform, let us interpret the measurable space G as

[, ) . Then the Haar measure is the (restriction of the) Lebesgue measureand, for f L1 we have

f(k) =

f(x)eikxdx , k Z

Thus, one obtains the Fourier coefficients corresponding to the (periodic)function f.

3. Finally, let G := Z. Each is uniquely defined by the value(1) = e

i

, with R. Thus is a (group) isomorphism between and R/Z. Moreover, the map R R/Z being continuous and andR/Z having compact topologies, we obtain the coincidence of the topologiesalso.

Finally, let f : Z C; its Fourier transform acts as:

f(ei) =

n=f(n)ein

which is exactly the Fourier series associated to the coefficients (f(n))nZ.


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2.6 The Gelfand Transform

Let G be a LCA group and let A = L1(G) be the (commutative, Banach)algebra with respect to the convolution.

Theorem There exists a bijective correspondence between the non-zerohomomorphisms h of the algebra L1(G) and the characters , throughthe relation h(f) = f(), f L1(G).

Proof Indeed, let . We associate the homomorphism on L1(G):f f(); it is obviously non-zero, linear and continuous and

f g

() = f()g()

Conversely, let h be a non-zero homomorphism. As h (L1(G)) L(G),there exists L(G) such that

h(f) =

G

f(x)(x)dx, f L1(G)

We have to prove that (a choice for) is a character on G. First, let usprove that there exists a representative of , which is continuous. In orderto do that, we write the fact that h is a homomorphism, that is: for anyf, g L1(G) we have h(f g) = h(f)h(g).

Explicitly:G

(x)G

f(x y)g(y)dy

dx =G

h(f)g(y)(x)dy

Using Fubinis theorem (since |(x)f(x y)g(y)| = |f(x y)g(y)|):G

g(y)

G

f(x y)(x)dx

dy =

G

h(f)g(y)(x)dyG

g(y)h(fy)dy =

G

h(f)g(y)(x)dy

Thus h(fy) = h(f)(x), a. e. in y. We can fix f L1(G) such that h(f) = 0,hence the choice

(y) =h(fy)

h(f)

gives a continuous function on G and the equality h(fy) = h(f)(x) holdseverywhere. Let us prove that is a character. Using the definition, wehave:

h(f)(x + y) = h(fx+y) = h((fx)y) = h(fx)(y) = h(f)(x)(y)


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2.6. THE GELFAND TRANSFORM 31

hence (x + y) = (x)(y),

x, y

G.

Next, let us prove that |(x)| = 1, x G. Since, by the identification = h, we compute h. Now, there exists a sequence fn L1(G),such that fn1 = 1 and |h(fn)| h. Thus |h(fn fn)| = |h(fn)|2 h2.Since |h(fn fn)| hfn fn1 = h, we get h 1, since clearly h = 0.Hence |(x)| 1, x G. Now, it remains to remark that (0) = 1 (fromthe definition), hence 1 = (0) = (x)(x) shows that |(x)| = 1, x G.

The last step is to prove that the correspondences are inverse one another.Let and let h be the corresponding homomorphism: h(f) = f(). Wewrite:

G f(x)(x)dx = h(f) = f() = G f(x)(x)dxf L1(G). It follows = , a.e. on G; the continuity proves the equalityeverywhere. Conversely, let h be a homomorphism; let be the characterassociated with h (i. e. h(f) =

G

f(x)(x)dx) and let h1 be the homomor-phism associated with . We have then:

h1(f) = f() =

G

f(x)(x)dx = h(f)

f L1(G). Hence h h1.

2.6.1 Banach algebras

A will be a commutative Banach algebra (over C). IfA has a unit, then wecan speak of invertible elements. Let us remark that, for x < 1 we have(e x)1 =

n=0

xn. It follows that the set of invertible elements is open. If

A has a unit, then x A is invertible iff x does not belongs to any maximalideal. Clearly, if x is invertible, then it cannot belong to any (maximal)ideal. Conversely, if x is not invertible, then the ideal generated by x isproper, hence contained in a maximal ideal.

Let us remark that any maximal ideal is closed: its closure is an ideal, andis proper, since the unit has a neighborhood, formed of invertible elements.

For each x A we define the resolvent function () := (xe)1, on theresolvent set of all C such that (x e) is invertible. The complement ofthe resolvent set is called the spectrum of the element x. The resolvent set isopen, non-void and the resolvent function is holomorphic (in the weak sense).Due to the explicit expression of the inverse (for > x), the spectrum isbounded (by x, hence compact) and the resolvent function has limit 0 at.


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The spectrum of any element is non-void by Liouville theorem: otherwise

the resolvent function would be entire and with limit 0 at . Using thepower series for f

1z

and the uniqueness of the Laurent development, we

obtain a contradiction.Now we can prove Mazurs theorem:Theorem (Mazur). If each non-zero element from A is invertible, then

A is isomorphic with CProof. For each x A there exists C such that x e is not

invertible; the only possibility is x e = 0. Thus, the correspondencex is well defined and bijective: A is isomorphic with C through x = e.

We denote by the set of non-zero homomorphisms h : A CFor each h , Ker h is a maximal ideal such that A/Ker h hasa unit (Ker h is a regular ideal). Of course, if A has a unit, then each

(maximal) ideal is regular. Conversely, by Gelfand-Mazur theorem, for anysuch maximal ideal I, we have an isomorphism A/I C, thus there is abijective correspondence between and the set of regular maximal ideals ofA (the spectrum of A).

Thus, Ghelfands theorem identifies the spectrum of the algebra L1(G)with the dual group and the Gelfand transform with the Fourier transform.

Each h has norm 1. Hence, is identified canonically with a partof the unit sphere S from the dual A. We consider on the (trace of)the weak topology: the coarsest topology, such that all maps h

h(x) are

continuous.Let us remark that {0} is closed in S with the weak topology. In-

deed, any weak adherent point for is obviously a linear and multiplicativefunctional. The continuity follows from the Banach-Steinhaus theorem. Itfollows that is locally compact.

2.6.2 Gelfand transform

The Gelfand transform of an element x A is the map x : C, definedas x(h) := h(x).

Moreover:A has a unit = is compactsince e(h) = h(e) = 1, h and (1 C0() is compact).The Gelfand transform x x is a homomorphism ofA onto a subalgebra

A C0() and x x.IfA has a unit, then x A is invertible iff h(x) = 0, for all h (as the

last part is the same as: x does not belong to any maximal ideal).If A does not have a unit, then we replace it by A

C.e. This is again

a commutative Banach algebra, with unit. Moreover, the homomorphisms


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2.6. THE GELFAND TRANSFORM 33

are the same. Indeed, any homomorphism h : A

C extends uniquely to

AC.e as: h(x + .e) := h(x) + . Conversely, any homomorphism onAC.e has as restriction to A a homomorphism.

Let x A be fixed. Then the equation xy = x + y has a solution y A iffh(x) = 1, for all h . Indeed, working in the algebra with unit, the problemcomes to the invertibility of x e. Moreover, the result y = x(e x)1belongs to the initial algebra.

For each x A we define its spectrum as the set of all complex numbersof the form h(x), when h {0} (ifA has no unit). Thus the spectrum ofeach element is a compact, nonvoid part of the complex plane. Moreover,for the case A = L1(G), using the Gelfand theorem, it follows that f isthe radius of the smallest disk, centered in 0, which contains the spectrum.

Let us apply the above in the following setting: x A is fixed and C.The equation x = y + xy has a solution, denoted by R(x; ) A iff isnot in the spectrum ofx. Moreover, the map R(x; ) is holomorphic inthe complement of the spectrum of x. Indeed subtracting the two equationsx = y + xy and x = 0y0 + xy0 we obtain (working in the algebra withunit):

y y0 0 = y0(x e)

1

(hence, the derivative may belong to the algebra with unit). Moreover, for

|| > x we haveR(x; ) =

n=1

xn

n

Theorem (Beurlings lemma)

limn

xn1/n = x

Proof. The fact that the limit exists is an elementary exercise. Let usdenote

l := lim infn

xn

1/n

L := lim supn

xn1/n

Since |h(x)|n = |h(xn)| xn for any homomorphism, we have x l.The inequality L x is the important one. Using the residue theorem,we obtain that, for any R > x we have:

xn = 12i

||=R

n1R(x; )d


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for any n 1.

Since the function under the integral sign is holomorphic outside thespectrum, we can use in fact any R > x, without changing the result.Thus

xnRn

0, which shows that L R, hence in fact L x.

2.6.3 Examples

1. Let K be a compact space and let us consider the Banach algebra C(K).In this case, we have a commutative Banach algebra with unit.

Let us prove that the spectrum is (homeomorphic with) K and theGelfand transform is simply the evaluation map f f(x). For each x Kwe define x : C(K) C as x(f) := f(x). Clearly, x is a nonzero homomor-phism, hence x . Moreover, the map x x is injective. This map is alsocontinuous, since carries the weak topology and x f(x) is continuous.It remains to prove that this map is surjective.

Let h and consider the maximal closed ideal I = Ker h. Let usdenote ZI := {x K| f(x) = 0, f I}. The set ZI is not empty; if not,for each x K, there exists fx C(K) such that fx(x) = 0. Thus, we can

cover K with a finite number of sets of the form [f = 0]. Now k |fk|2 Iis an invertible element in a proper ideal, a contradiction. In fact, ZI is asingleton: if it contains at least two distinct points, this would contradict themaximality ofI, since Ker x is already a maximal ideal.

2. Let X be a locally compact space and let us consider the Banachalgebra (without unit) C0(X).

Let K := X{} be its onepoint compactification. Each h Kinduces by restriction an element from the spectrum of C0(X). This restric-tion is also nonzero, except for

. Conversely, as each f C(K) has a

unique expression as f0 + f(), with f0 C0(X), we easily obtain that eachh0 from the spectrum of C0(X) has a unique extension to an element from. Thus, the spectrum of C0(X) is canonically identified with the locallycompact space X.

3. Let A = L on a measurable space, with the usual operations. Wehave again a commutative Banach algebra, with unit.

4. Let now consider A = L1([0, +)), again with convolution. We iden-tify the with the (closed) righthalf plane {z| Re z 0} and the Gelfandtransform with the Laplace transform.


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Theorem A measure

M() is positive iff the function (x) := (x)(d) is positive definite and continuous.

We have just proved the trivial implication of the theorem.Conversely, let be a continuous, positive definite function. We have to

construct a measure on G, prove that its FourierStieltjes transform is and that the measure is positive. The most difficult part is the constructionof a convenient linear form on A().

Proof We begin by defining a linear functional, associated with , onL1(G).

T(f) :=

G

f(x)(x)dx

The definition is correct, since is bounded. We prove the inequality

(1) |T(f)| fThe proof of (1) is divided in several steps.

(i) Since is positive defined, we haveG

G

f(x)f(y)(x y)dxdy 0

f L1(G). Indeed, it suffices to suppose f K(G). Thus if K de-notes the support of f, then (x, y) f(x)f(y)(x y) is uniformly con-tinuous on K K. The integral can be approximated by positive sums:ni,j=1

f(xi)f(xj)(xi xj)m(Ei)m(Ej) where xi Ei and (Ei)i is a partitionof K with measurable sets.

(ii) In the usual way, we associate the bilinear form

[f, g] :=

G

G

f(x)g(y)(x y)dxdy = T(f g)

where g(x) := g(x). Due to the positivity, we obtain a Cauchytype in-equality:

|[f, g]|2 [f, f][g, g]A convenient choice for g gives:

|T(f)|2 (0)[f, f] = (0)T(f f)f L1(G).

Let V be a symmetric neighbourhood of 0 and denote g := 1m(V)

V. Then:

[f, g] T(f) =G

f(x)1

m(V)

V

[(x y) (x)] dydx


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2.7. BOCHNERS THEOREM 37

and [g, g]

(0) =

1

m(V)2 V V [(x y) (0)] dxdySince is uniformly continuous, > 0 there exists a neighbourhood Vof the origin, such that for any x y V we have:

|[f, g] T(f)| < ; |[g, g] (0)| <

Let us denote h := f f and hn := hn1 h. Of course hn L1(G).We have proved that |T(f)|2 (0)T(h). Hence we have also T(h) (T(h

2)(0))1/2

. By induction:

|T(f)

|2 (0)1+

12+...+

12n T(h2n)

12n

Combining with |T(f)| (0)f1 we obtain

|T(f)|2 (0)1+ 12+...+ 12n h2n12n

1 (0)2hby Beurlings lemma. Since h = f2, we get |T(f)| (0)f.

This shows that T(f) depends only on f. In fact, the map f f isinjective, but this will be proved on the Bochners theorem basis. Hence, T isa linear and continuous functional on A(). From (1) and the density ofA()it follows that T extends (uniquely) to a linear and continuous functional

on C0(), that is a measure M(). Thus, we have T(f) =

f()(d).

We will denote by the measure image of by the inversion. It means that

f()(d) =

f()(d)

We recall that ()(x) = (x) = (x).(iii) This is the sought measure. Comparing the two formulas, we obtain:

f()(d) = G f(x)(x)dxThe first member transforms to:

G

f(x)(x)dx(d) =

G

f(x)

(x)(d)

dx

Since the equality holds for any f L1(G), it follows that (x) =

(x)(d),for almost all x G. Both functions being continuous, we obtain the equalityeverywhere.


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It remains to prove that the measure is positive:

(0) =

(d) = () (0)

Thus = () which shows that is positive, recalling the definitionof the norm as:

:= supnk=1

|(Ak)|

where the supremum is taken over all measurable, finite partitions (Ak)k ofG.

2.8 Inversion theorem

Before proving the duality theorem of Pontryaghin, we need some additionalfacts. The most important is that the (original) topology on G can be de-scribed as that of . This can be achieved by establishing an inversion resultfor the Fourier transform. Let us denote by B(G) the set of all functions fon G, of the form

f(x) =

(x)d()

for some M(). By Bochners theorem, B(G) is exactly the set of allfinite linear combinations of continuous positive definite functions on G.Theorem. If f L1(G) B(G), then f L1() and the following

inversion formula for the Fourier transform holds:

f(x) =

f()(x)d, x G

for a conveniently (fixed) Haar measure on .Proof. Let us denote f M() the measure associated with f

B(G). Hence we have f(x) =

(x)df(). We have to prove that f is

absolutely continuous with respect to d and for a convenient choice of theHaar measure, the density is f.

The first remark is that, iff and g are functions from L1(G)B(G), thenf dg = gdf. Indeed, for any h L1(G) we have:

(h f)(0) =G

h(x)f(x)dx =G

h(x)

(x)df()dx =

=

G

h(x)(x)dx

df() =

h()df()


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2.8. INVERSION THEOREM 39

Since g belongs to the same space, the equality ((h

g)

f)(0) = ((h

f)

g)(0)

becomes

hgdf =

hf dg

Now h is an arbitrary element from A(), which is dense in C0() andthe proposed equality follows.

The next step is to construct a non-zero positive, invariant measure (aHaar measure) on . This measure will be constructed as a (positive, linear)functional T on K() using the formula:

T() :=

g

dg

We must prove that this definition is correct. The previous remark showsthe independence of g. We construct, for each K() a function g L1(G) B(G), such that g > 0 on the support of .

Let K denote the support of K(G). Since K(G) is dense in L1(G), wecan find, for each 0 K a function u K(G) such that u(0) = 0. Henceu u(0) > 0 and is 0 at any other point. Since K is compact, we can findsuch functions u1, . . . , un such that g := u1 u1 + . . . + un u1 has g > 0 onK. We have to prove that g B(G). By Bochners theorem, it is sufficientto show that g is positive defined. If := f

f then

cncm(xn xm) =

cncmG

f(xn xm y)f(y)dy =

=

cncm

G

f(xn y)f(xm y)dy =G

cnf(xn y)2 dyThus T is well defined and is clearly linear. It is also positive, since

g 0, also by Bochners theorem. T is non-zero, since we can find and fsuch that

df = 0. Then:

T(f) = fg dg = df = 0Let us prove now that T is translation invariant. Let 0 ; we can constructg as above (associated with with support K), such that g > 0 on bothcompacts K and 0 + K. Let us consider the function f(x) := 0(x)g(x).Then f() = g(+ 0) and f(A) = g(A 0).

Finally, if 0() := ( 0) then:

T 0 =

( 0)g()

dg() =

()

f()df() = T


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Thus, T is translation invariant, hence a Haar measure on . We shall write:

T =

()d

for any K(G).Now, for any f L1(G) B(G) we have

df = T(f) =

()f()d

This equality can be written as: f d = df. Since f is a measure with

bounded total variation, it follows that

f L1

(). Recalling the definitionof f we obtain the inversion formula:

f(x) =

(x)df() =

f(x)(x)d

Theorem. For each neighborhood V of 0 in G, there exists a compactK in and r > 0 such that N(K, r) V.

Proof. Let us choose a compact neighborhood W for 0 such that W W V. Let us denote f := 1

m(W)1/2W and g := f f. Then g is positive

definite and 0 outside W

W. It is also continuous:

|(f g)(x) (f g)(z)| G

|f(x y) f(z y)| |g(y)|dy

fx fz1gThe inversion formula applies to g, hence

g(x) =

g()(x)d

Moreover, g has also the properties: g =

|f

|2 0 and 1 = g(0) = g()d.By the regularity, for each < 1 there exists a compact set K such

thatK

g()d > . x N(K, r) means |1 (x)| < r, K and this isequivalent with Re (x) > 1 r2. Hence

K

g()(x)d

K

g()Re (x)d > (1 r2)

Now

g(x)

K

g()(x)d

\K

g()(x)d

>


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2.9. PLANCHERELS THEOREM 41

> (1

r2)

(1

)

Taking = 2/3 (near to 1) and r = 1/3 (near to 0), we obtain g(x) > 7/27(near to 1). Since g(x) = 0 if x W W V, it results x V.

Corollary. The topology ofG has as a fundamental system of neighbor-hoods for 0 the sets (N(K, r)), for compact K and r > 0.

In particular separates G. Indeed, let x1 = x2 G; there exists aneighborhood V for 0 such that x1 x2 V. Hence, there exists suchthat (x1 x2) = 1, so (x1) = (x2).

Examples. Let us find out the corresponding Haar measure on the dualin some particular cases.

1. IfG is compact and m(G) = 1, then taking f

1 (obviously positive

defined and from L1), we have f = {0}. By the inversion formula:

1 = f(0) =

f()d = m({0})

2. Let G be a discrete group, with m({0}) = 1. Now we consider f = {0}(which is positive defined and form L1), then, again by the inversion formula:

m() =

f()d = f(0) = 1

3. Finally, for G = R we have already found the constant 12

, using the

function ex2/2 (other functions, such as e|x| work as well). Since R is itsown dual, we can either choose on both spaces the measure 1

2dx; or on one

space the Lebesgue measure, while on the dual 12

dx.

2.9 Plancherels theorem

Theorem. The Fourier transform, restricted to L1(G)L2(G) is an isometry,with respect to the L2norm, onto a dense linear subspace of L2().

Hence it extends uniquely to an isometry of L2(G) onto L2().

Proof. Let f L1

(G) L2

(G) and denote g := f f. We know thatg L1(G), g = |f|2. Moreover, g is positive defined and continuous, hencethe inversion theorem works for g:

g(0) =

g()d =

|f()|2d = f22

On the other hand, from the definition we have:

g(0) =

G

f(x)f(x)dx =G

|f(x)|2dx = f22


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We have to prove that the subspace

H:=

{f

|f

L1(G)

L2(G)

}is dense in L2(). We show that any element from L2(), which is orthog-onal on H, is 0. Hence, let L2() be such that

f()()d = 0,

f L1(G) L2(G). Since L1(G) L2(G) is invariant to translations (byx G), it follows that H contains also the function f()(x), hence also

f()(x)()d = 0, f L1(G) L2(G), x G.

Let us consider the measure := fd. We know that

(x)d() = 0.

Since both f and are from L2(), it follows that f L1(), thus M(). We prove now that = 0. Indeed, for any h L

1

(G) we have:

h()d() =

G

f(x)(x)dxd() =

=

G

f(x)dx

(x)d() = 0

As A() is dense in C0(), we have = 0, hence f = 0, a.e.Finally, since the sets [f = 0] cover , it follows that = 0 a.e.Warning. 1. The bijective isometry between L2(G) and L2() will also

be denoted by f, however no explicit formula, as f() =

G

f(x)(x)dx canbe expected.

2. In a space with finite measure, we have L2 L1 due to Holder in-equality.

3. We have l1 l2. Taking functions equal to bn on (n an, n + an) and0 elsewhere, we show that L1(R) L2(R) and L2(R) L1(R).

However, at least in the case of Radon measures, L1 L2 is dense bothin L1 and in L2, as in contains K.

Parseval formula For any f, g L2(G) we have the identity4f g = |f + g|2 |f g|2 + i|f + ig|2 i|f ig|2

Using the isometric character of Fourier-Plancherel transform, we obtain the

Parseval formula: G

f(x)g(x)dx =

f()g()d

Theorem A() consists precisely of the convolutions F1F2, with F1, F2 L2().

Proof Let f, g L2(G) and write Parseval formula with g replaced byx 0(x)g(x):

f g(0) =

G

f(x)g(x)0(x)dx =

f()g(0 )d = (f g)(0)


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2.10. DUALITY THEOREM OF POTRYAGUINE 43

On the one hand, every h

L1(G) can be written as h = |h|h

|h| witheach factor from L2(G). The above formula shows that h = f g, withf , g L2(). Conversely, starting with f , g L2(), the formula shows thatf g A().

Theorem For each nonempty open A , there exists f L1(G), suchthat f = 0, but f() = 0, A.

Proof We choose K A a compact set, with m(K) > 0; and V acompact neighborhood of 0, such that K+ V A. Since the characteristicfunctions of K and V belong to L2(), we can find g, h L2(G) such thatg = K and h = V. By the preceding theorem, there exists f L1(G) suchthat f = g h. Thus f A(), f = 0 outside K+ V, while

f()d = m(K)m(V) > 0

so that f is not identically 0.

2.10 Duality Theorem of Potryaguine

Let G be a LCA group; its dual being also a LCA group, has a dual,denoted temporarily . There exists a canonical map : G

, defined as

follows: for each x G, (x) is the map from to T, [(x)] () := (x).Theorem (Pontryagin). is an isomorphism of groups and a homeo-

morphism.Thus, G is canonically identified (through ), with the dual of .Proof. The proof will consists in several steps.1. (x) since

(x)(1 + 2) = (1 + 2)(x) = 1(x)2(x) = (x)(1)(x)(1)

while (x)() = (x) is continuous, since carries the weak topology.Moreover, is a homomorphism of groups, since

(x + y)() = (x + y) = (x)(y) = (x)()(y)() =

= [(x) + (y)] ()

2. We proved, as a consequence of the inversion theorem, that(i) separates G, hence is one-to-one;(ii) the topology of G can be described as that of . This can be written

as: (N(K, r)) = N(K, r) (G), for any compact K and r > 0. Thisshows that is a homeomorphism between G and its image.


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We have to prove that:

3. (G) is dense in ;4. (G) is closed in .3. Let us suppose, on the contrary, that (G) = . Using the consequence

of Plancherel theorem, there exists then F L1(), such that F 0, but is0 on (G). That means:

0 = F((x)) =

F()(x)d

x G. As proved during Plancherel theorem, this implies that the measureF()d = 0. Hence F = 0 a.e. and F

0, contradiction.

4. We prove that (G) is open. Since \ (G) is a union of translates of(G), it results that (G) is also closed.

Let 0 (G); let V be an open neighborhood of 0 in (G). Since (G)is homeomorphic with G, it is locally compact; we may suppose that theclosure of V in (G) is compact. Let U denote an open set in such thatV = U (G). Now we have

U = U (G) U (G) = V

Thus U

(G) and 0 is an interior point for (G).

Any result obtained for the dual pair (G, ) is now valid also for the pair(, G).

Theorem IfG is a compact group, then G is the dual of a discrete group.If G is a discrete group, then it is the dual of a compact group.Theorem L1(G) has a unit iffG is discrete.Theorem[uniqueness] Let M(G) be such that = 0. Then = 0.Finally, let us prove the following consequence of the dual of the inversion

theorem.Theorem Let M(G) be such that L1(). Then there exists

f L1(G) such that = f.dx and

(1) f(x) =

()(x)d

Proof By the inversion theorem applied in the dual pair (, G) for we know that the function defined by the formula (1) is in L1(G) and thefollowing relation holds:

() =

G

f(x)(x)dx


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A continuous function f : R

R is said to be almost periodic if, for every

> 0, there exists a number L > 0 such that for every interval I of lengthL there exists a number I I such that f(x + I) f(x)| < wheneverx R (Bohr).

A function f is almost periodic if every sequence (f(tn+T) of translationsof f has a subsequence that converges uniformly for T R (Bochner).

A function on a LCA group is called almost periodic iff the set of itstranslates is pre-compact (compact after completion).

Thus, the almost periodic functions on G are precisely those which havecontinuous extensions to G. So, G may also be described as the maximal idealspace of the Banach algebra of all almost periodic functions on G (Loomis).

2.12 A characterization of positive definitefunctions

Let G be a LCA group and its dual. Let us recall that B() denotedthe set of all Fourier-Stieltjes transforms () :=

G

(x)d(x) of measures M(G). The norm on B()will be = if = .

Theorem. Let : C; the following statements are equivalent:(a) B() and A.

(b) is continuous and ni=1

ci(i) Af for every f : G C of theform f(x) =

ni=1

cii(x) (trigonometric polynomial).

Proof. (a) (b) is clear, sinceni=1

ci(i) =ni=1

ci

G

i(x)d(x) =

G

f(x)d(x)

(b)

(a) We extend each trigonometric polynomial f to G by the formula

f(x) :=ni=1

cix(i)

Moreover, since G is dense in G, the sup norm of this extension remains thesame.

Hence, if we define a linear functional T on the space of all trigonometric

polynomials f by T f :=ni=1

ci(i) we have |T f| Af. Thus, T can


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2.12. A CHARACTERIZATION OF POSITIVE DEFINITE FUNCTIONS47

be extended to a bounded (by A) linear functional on

C(G). T is in fact a

measure M(G), such that A and ni=1

ci(i) =G

f(x)d(x).We apply this formula for the particular case f(x) = x(), for arbitrary . We obtain

() () =G

f(x)d(x)

The proof is ended if we prove that the measure is concentrated on G(i.e. (G \ (G)) = 0).

By the RadonNikodym theorem, there exists g L1(||), of absolutevalue 1, such that gd = d

|

|. By StoneWeierstrass theorem, the trigono-

metric polynomials are (uniformly) dense in C(G), which is dense in L1(||).Hence, there is a sequence (fn) of trigonometric polynomials on G, such that

G

|fn g|d|| 0

Let us introduce

n() :=

G

x()fn(x)d(x)

Recalling that each fn is a finite linear combination of characters, it resultsthat each n is a finite linear combination of translates of functions as in (),which are continuous on . Moreover (n) is uniformly convergent to

() =

G

x()d||(x)

Thus is continuous and also positive definite. By Bochners theorem, isthe Fourier-Stieltjes transform of a measure on the dual of , that is, on G.By the uniqueness result for the Fourier-Stieltjes transform, it results that|| is concentrated on G, and so is .

Corollary. Ifn B(), n A and C() is such that () = limn n(),for all , then B() and A.


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49/79

Chapter 3

Convolution Semigroups

3.1 Semigroups of kernels

Let X be a Banach space. A family of linear, continuous operators Pt : X X is called a semigroup if Pt Ps = Pt+s, t , s > 0. Also some measur-ability condition must be imposed. Usually, one considers C0semigroups:limt0

Ptx = x, x X. In potential theory, one is interested in semigroups ofcontractions Pt 1, t > 0. Moreover, most of the classical Banach spaceshave a natural order relation, so we can consider positive operators.

For example, the system x = A x has the solution Pt := etA.Let (X, X) be a measurable space. We denote by Fthe (ordered convex

cone of) numerical, positive, measurable functions. A map V : F F iscalled a kernel if V0 = 0 and V fn V f, for any fn f from F. Thus,V is in fact a family (Vx)xX of (positive) measures on (X, X), such thatx Vx(A) is measurable, for each A X. The most interesting kernels areof the form

V f(x) =

X

v(x, y)f(y)d(y)

where v : X X [0, +] is a measurable function and is a positivemeasure (absolute continuous).

A semigroup of kernels is a family (Pt)t>0 of kernels such that

(t, x) Ptf(x) is measurable, for any f F; Pt Ps = Pt+s, t, s > 0.With a semigroup of kernels one can associate a stochastic process, essen-

tially a family of measures Px on the space of trajectories : [0, +) Ewhere E is the state space (usually a locally compact space).

49


50/79

50 CHAPTER 3. CONVOLUTION SEMIGROUPS

Let

P= (Pt)t>0 be a measurable, submarkovian semigroup of kernels on

a measurable space (E, E).1) s, t 0 : Ps Pt = Ps+t2) The map (t, x) Ptf(x) is measurable on (0, +) E, for all f F.3) Pt1 1, t 0.s F is called Psupramedian if Pts s, t > 0. The set of P

supramedian functions is denoted by SP.Clearly Pis submarkovian iff 1 SP.The set SP is a convex cone.The function t Pts(x) is decreasing, ifs SP, since Pt+ts = Pt(Pts)

Pts. This remark allows us to define

s(x) = supt>0

Pts(x) = limt0

Pts(x)

s is positive, measurable and s(x) s(x). Moreover s SP.A Eis called Pnegligible if Pt(A) 0, t > 0.s Fis called Pexcessive if: s SP; s = s;

s is finite

Pa. e.

The set ofPexcessive functions is denoted by EP.The next result presents the basic properties of the set of excessive func-

tions (with respect to a submarkovian semigroup of kernels),Theorem

If s, t EP and , 0, then s + t EP. If s,t,u EP and s + u t + u, then s t. Let (sn) be any sequence from EP. Then n

sn exists and:

s +n

sn =n

(s + sn)

Let (sn) be an increasing and dominated sequence from EP. Thenn

sn

exists and:s +

n

sn =n

(s + sn)


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3.1. SEMIGROUPS OF KERNELS 51

Riesz splitting property (G. Mokobodzki) Let s, s1, s2

EP with s

s1 + s2. There exists then s1, s2 EP such that s1 s1, s2 s2 ands = s1 + s

2.

If the semigroup is absolutely continuous, then the lattice properties holdfor arbitrary families (standard Hcones of N. Boboc, Gh. Bucur and A.Cornea).

Let U Rd be an open set. h : U R is called a harmonic function inU if h C2(U) and

h :=n

i=12h

x2i= 0

For each sphere S(a, r) := {x Rd|x a < r} centered in a Rdand of radius r > 0 and for each x S(a, r), we define the Poisson kernelPx : S(a, r) R as:

Px(y) = rd2 r

2 x a2x yd

The Poisson kernel solves the Dirichlet problem on the sphere.For any continuous function f on S(a, r) we denote:

Hf(x) = S(a,r)

Px(y)f(y)a,r(dy) = Sd1

Px(a + r.z)f(a + r.z)(dz)

(for each x S(a, r)).Then:Theorem

Hf is defined and harmonic in S(a, r);

y S(a, r) limx y

x S(a, r)

Hf(x) = f(y);

Let g be defined and continuous on S(a, r), harmonic on S(a, r). If wedenote f := g

S(a,r) , then Hf = g in S(a, r).

s : U (, +] is called a superharmonic function in U if:

it is l. s. c. in U;

s is finite on a dense set;


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Sd1

s(x + r.z)(dz) s(x), x U, r (0, Rx)

Lemma. If s C2(U) and s 0 in U, then s is superharmonic in U.Theorem A function s Fis Pexcessive iff it is superharmonic.The proof is based on the formula

1

2Ptf =

Ptf

t

valid for any bounded f

F.

3.2 Convolution semigroups

Let G be a LCA group. A family (t)t>0 of positive, bounded measures onG, with the properties:

t s = t+s, t, s > 0

t(G) 1, t > 0

limt0 t = 0 vaguely(it means that

G

f(x)dt(x) f(0), f K(G)).

is called a (vaguely continuous) convolution semigroup on G. Each posi-tive and bounded measure defines a kernel through convolution: V f(x) := f(x) =

Gf(x y)d(y). In the special case = v.dx we have V f(x) =

Gv(x y)f(y)dy.This kernel is translation invariant V fx = (V f)x and continuous (in the

sense that V f is continuous, for each f K(G)). Conversely, if V is acontinuous, translation invariant kernel, then the formula

(f) = V(f)(0)

defines a measure, which induces the original kernel (where f(x) := f(x)).Moreover, in the same way, there is a bijection between measures

bounded by 1 and translation invariant operators T on Lp(G) (p [1, +)),which satisfy 0 f 1 0 T f 1. The case p = 2 is especially inter-esting, as any such operator is unitarily equivalent with the multiplicationoperator on L2() (multiplication with ).


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3.2. CONVOLUTION SEMIGROUPS 53

A function :

C is called negative definite if for each n 1 and

c1, . . . , cn C, 1, . . . , n , the following relation holds:ni,j=1

cicj((i) + (j) (i j)) 0

The next connection with positive definite functions holds:Theorem. (Schoenberg) A function : C is negative definite iff

((0) 0 and the function et() is positive definite, t > 0).Proof. The proof is based on the following remark: let A := (aij) be a

n n matrix of complex numbers. Thenni,j=1

aijcicj 0 (c1, . . . , cn) Cn

iff (aij = aji and the eigenvalues of A are positive).Theorem There is a one-to-one correspondence between convolution

semigroups on G and continuous, negative definite functions on .Explicitly, if (t)t>0 is a convolution semigroup on G, then there exists a

uniquely determined continuous, negative definite function on such that

(1) t() = et(), t > 0,

Conversely, given a continuous, negative definite function on , then (1)determines a convolution semigroup (t)t>0 on G.

Proof Let (t)t>0 be a convolution semigroup on G and consider, forfixed , the function (t) := t(). Let us prove that this function iscontinuous on (0, +). We remark first that, for any continuous, positivef 1, with f(0) = 1 we have:

1 = f(0) = limt0

t(f) lim inft0

t(G) lim supt0

t(G) 1

Let us prove next that, for any bounded and continuous f we have limt0

t(f) = f(0).

Let us fix h K(G), with h(0) = 1. We have limt0

t(1 h) = 0 and alsolimt0

t(f h) = f(0). Moreover |t(f(1 h))| f t(1 h). The conclu-sion follows from the inequality

|t(f) f(0)| |t(f(1 h))| + |t(f h) f(0)|Now we obtain the continuity at any t0 > 0 from the inequality

|t() t0()| |tt0|() 1


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Moreover, we have (s + t) = (s)(t) and limt0

(t) = 1. Hence, there

exists a uniquely determined complex number () such that

(t) = et(), t > 0

Let us prove the continuity of . We consider the measure , defined onf K(G) as

(f) :=

0

ett(f)dt

This measure is positive and of total mass 1. Moreover

() = 0

ett()dt = 0

et(1+())dt = 11 + ()

Since is continuous it follows that t() is continuous and positivedefinite. Obviously (0) 0 hence from Schoenbergs theorem it followsthat is a negative definite function.

Conversely, let be a continuous, negative definite function. Then, foreach t > 0, the function et() is positive definite by Schoenbergstheorem, and there exists consequently a positive, bounded measure t onG such that t() = e

t(), . We have to prove that (t)t>0 is aconvolution semigroup.

We have

t s() = t()s() = et()et() = e(t+s)() = t+s()

hence t s = t+s. Also t(G) = t(0) = et(0) 1 since (0) 0. Westill have to prove that lim

t0t(f) = f(0), for any f K(G).Let > 0; by

the density, we find K() such that f < . Now, we have theinequality:

|t(f) f(0)| |t(f )| + |t( (0)| + |(0) f(0)|

(1 + t(1)) +

|t() 1| |()| d

The conclusion follows, since, as is continuous, hence bounded on compactsets, we have

limt0

t() = limt0

et() = 1

uniformly on compact subsets of .


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3.3. THE SYMMETRIC STABLE SEMIGROUP 55

3.3 The symmetric stable semigroup

The symmetric stable semigroup of order (0, 2] is associated with thecontinuous and negative definite function y y. Hence, it is character-ized by

t (y) = ety

Since the function t is integrable, we have

t = gt (x)dx

where

g

t (x) =

1

(2)d Rd eietydyThis function belongs to C+0 (Rd).

The only fact that needs a proof is that y y is negative definite.This follows from the representation

y2 =0

1 ety2

d(t)

for some positive, (unbounded) measure on [0, +). Indeed, we start withthe integral (for x > 0):

0

(1 ext)t1dt

which is convergent exactly when (0, 1). It is transformed successively:

x0

(1 et)t1dt = x0

t0

eudu

t1dt =

= x0

eu

t

|u du = x1

0

euudu = x1

(1 )

Hence, the measure is d(t) = (1 ) t1dt. It remains to prove that

1 ety2 is negative defined. This is obtained as follows. First, we havethe following characterization (we can take it as the definition; the proof isessentially computational):

Theorem A function : C is negative definite iff: (0) 0; = ;


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For all n N

, 1, . . . , n and c1, . . . , cn C

,

n

i=1 ci = 0 impliesni,j=1

cicj(i j) 0.

We obtain:Corollary Let be a positive defined function on . The function

(0) () is negative definite.Proof This function clearly satisfy the first two requirements from the

theorem, while

ni,j=1

cicj ((0) (i j)) = ni,j=1

cicj(i j) 0

ifni=1

ci = 0.

In the case = 1 we find, using the results about subordinated semi-groups and Bernstein functions:

g1t (x) = t(4) d+12

0

sd+32 e

x2+t2

4s ds =

= (d + 1

2)t

x2 + t2 d+12

For d = 1 one obtains the density of the Cauchy semigroup:1

t

x2 + t2.

3.4 Excessive measures: the discrete case

Let be a positive measure on G, such that (G) 1. D+() will denotethe set of all positive measure, such that

is defined.

A positive measure on G is called superharmonic (resp. harmonic)if D+() and (resp. = ).

Theorem. (i) The Haar measure dx is in D+() and dx = (G)dx.(ii) The infimum of an arbitrary set of superharmonic measures is

superharmonic.(iii) The supremum of an upward filtering and bounded set ofsuperharmonic

(resp. harmonic) measures is superharmonic (resp. harmonic).(iv) Let be a superharmonic (resp. harmonic) measure and let

D+(). Then is superharmonic (resp. harmonic).


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3.4. EXCESSIVE MEASURES: THE DISCRETE CASE 57

Theorem. There exists a superharmonic measure which is not

harmonic iff the seriesn=0

n is (vaguely) convergent.

In this case, the measure :=n=0

n is superharmonic but not

harmonic. is called the elementary kernel determined by .Proof. Let be a superharmonic measure which is not harmonic.

Then := is a positive, non-zero measure and

N

n=0 n = N+1 Hence, the increasing sequence

Nn=0

n

N

is bounded, therefore (vaguely)

convergent.Let us suppose that exists. Then = 0 which shows the

conclusion.Theorem. Let be the elementary kernel determined by . Every

superharmonic measure can be written = + , where D+()and is harmonic.

In such a decomposition, and are uniquely determined and given bythe formulas = ; = lim

nn vaguely.

The measures and are called respectively the potential part andthe harmonic part of .

Proof. Let us start with such a decomposition = + .We obtain = ( ) + hence = ( ) = 0 = . Moreovern = (n ) + . Since

n =

k=nk 0

we get = limn

n vaguely. We obtained the uniqueness of and .Conversely, if is superharmonic, then (n )n is decreasing and thevague limit = lim

nn exists. Moreover, is harmonic. Now, the

measure := is positive andNn=0

n

= N+1

Letting N we obtain that = .


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3.5 Transient convolution semigroups

Let (t)t>0 be a convolution semigroup. When it exists, the measure

(f) :=

0

t(f)dt < +, f K(G)

is called the potential kernel of the semigroup and the semigroup ois calledtransient. When this measure is finite, the semigroup is called integrable.

The next sufficient condition for integrability was obtained by Berg.Theorem. Let (t)t>0 be a convolution semigroup on G, with associated

continuous negative definite function on . If the function1

is locally

integrable, then (t)t>0 is integrable and its potential kernel is the Fourier

transform of the measure1

d.

Let d 1 be fixed and denote Sd1 := {x Rd| x = 1 }. Let Gd bethe set of all linear and isometric maps T : Rd Rd. Each such map is infact a unitary operator, hence Gd is a group. Moreover, each T Gd is abijection on Sd1.

Theorem. There exists a unique measure on Sd1, denoted , such that: T1 = , T Gd and (Sd1) = 1.

Proof. Gd

Rd

2is a compact group which acts continuously on Sd1,

in the following sense (T, x) T x is a continuous map Gd Sd1 Sd1.Let be the Haar measure on Gd, such that (Gd) = 1.

Let be a measure with the asserted properties. Necessarily :

(A) =

T1(A)

=

Gn

T1(A)

d(T)

for any A B and T Gd. Indeed, T (T1(A)) is constant by hypothe-ses and has total mass 1.

Since:

T1(A) = T1(A)(x)d(x) = (A T) d(x)we obtain:

=

(A T) (x)d(T)d(x)

But Gd acts also tranzitively (meaning that x, y Sd1 there exists T0 Gdsuch that T0(x) = y). This allows us to write:

(A T) d(T) =

(A(T T0(x))) d(T) =


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2) We define the measure on (0, +

) through (B) := m i(B S

d1).In fact, since((0, t)) = m

i((0, t) Sd1) = td((0, 1)) = c.td

it follows that is the Stieltjes measure corresponding with the functiong(t) = c.td, that is:

0

f(t)(dt) = dc

0

f(t).td1dt

3) Let us prove the formula; since the sets of the form (0, t) A gen-erate the p

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