Asymtotic Behaior

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Chapter I

Asymptotic Behavior of Generalized

Functions

0 Preliminaries

0.1. We denote by Rand Nthe sets of real and natural numbers;

N0=N{0}.The following notation will be used. Ifx= (x1, . . . , xn), y=(y1, . . . , yn) Rn,then xy= x1y1+ + xnyn; x2= x21+ +x2n; x

0

xi

0, i= 1, . . . , n; x

xi

, i=

1, . . . , n; Rn+= {x Rn; x >0}.If x= (x1, . . . , xn) Rn andk= (k1, . . . , kn) Nn0 ,then|k|= k1+ +kn, k! = k1! . . . kn! , xk =xk11 . . . x

knn; D

k = k1/xk11 . . . kn/xknn , f

(k) = Dkf (f(0) = f);

z= (z1, . . . , zn) Cn;D(a, r) denotes the polydisk{z Cn; |ziai|0.

0.2.A conewith vertex at zero in Rn is a non-empty set such that

x andk >0 imply kx .The cone is called solidif int = . Theconjugate cone (dual cone) to the cone is the set{ Rn; x 0for each x}.It is obvious that is also a cone which is convex andclosed. The cone is calledacuteif is a solid cone.

0.3.A function : (a, )R, aR+,is called regularly varying atinfinity[76] if it is positive, measurable, and if there exists a real number

such that for each x >0

limk

(kx)(k)

=x . (0.1)

The numberis called index of regular variation. If= 0,thenis called

slowly varying at infinityand for such a function the letter L will be used.

We then have that any regularly varying function can be written as (x) =

1


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2 Asymptotic Behavior of Generalized Functions

xL(x), x > a,whereLis slowly varying. It is known that the convergence

of (0.1) is uniform on every fixed compact interval [a, b], a < a < b 0,

(i) there exist constants C1, C2 >0 and X > asuch that

C1x L(x)C2x, x X; (0.2)

(ii) limx

xL(x) = +, limx

xL(x) = 0.

We know ([9], p. 16) that if L2(x) , x ,and L1, L2areslowly varying, then L1 L2=L1(L2) is slowly varying, as well. Hence, forx > ,

limh

L(x + h)

L(h) = lim

u

L(log ut)

L(log u)= 1 . (0.3)

The definition of a regular varying function at zero is similar. For the

definition of regularly generalized functions see [156] and [162].

0.4.The class of distributionsf, R,belonging toS+(see0.5.1.)is defined in the following way:

f(t) =

H(t)t1/(), >0,

f(m)+m(t), 0, + m >0,

whereHis the Heaviside function, and the derivativef(m) is taken to be in

the distributional sense (see [192], Chapter I,1). We therefore havef0=,

the Dirac delta distribution; fm=(m), mN; and fp fq=fp+q.Wealso use the notations t+= ( + 1)f+1(t) and t

= (t)+, / N.

Let g S+.We denote g() = f g, R (the is convolutionsymbol).

The sequence mmN C(Rn) is called a -sequence if:a) supp m [m, m], m 0, m ; b) m 0, m N;c) Rn m(t)dt= 1, m N.

If D,then m , m inD,hence{m ; mN}is abounded set inD.

0.5.We will repeat definitions and some basic properties of generalized

functions defined as elements of dual spaces.


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0. Preliminaries 3

0.5.1.The Schwartz spaces of test functions and distributions on Rn are

denoted byD=D(Rn) and D = D(Rn), respectively (Rn will be omittedwherevernis not fixed). The spaceDis a locally convex, barrelled, Monteland complete space. If a filter with a countable basis is weakly convergent

inD,then it is convergent inD with the strong topology, as well (cf.[146]).

Sis thespace of rapidly decreasing functionsand its dualS is thespaceof tempered distributions[146]. For a closed cone Rn, S={fS;supp f}.In the one-dimensional caseS+={f S(R); supp f[0,

)

}.Recall ([189]):

S() ={ C();p< , pN} ,where

p= supx,||p

(1 + x2)p/2|()(x)| .

BySp() is denoted the completion of the setS() with respect to thenorm

p.Note

S() = pN0 Sp() andS

() = pN0 Sp(),where the

intersection and the union have topological meaning. A sequencefnnNinS() converges tof S() if and only if it belongs to someSq() andconverges tof Sq() in the norm ofSq(). The spaceS is isomorphictoS() if is closed convex solid cone ([189], [192]).

E the space of distributions with compact support; it is isomorphic tothe dual space ofE=C(Rn) (cf. [146]).

DLp,1

p

, is the space of smooth functions with all derivatives

belonging to Lp ([146]),DLp DLqifp < q..

Bis a subspace ofB=DL ,defined as follows: .

Bif and only if|()(x)|0 asx for every Nn0 .

DLp ,1< p is the dual space ofDLq ,1 q


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The elements ofKpare called rapidly exponentially decreasing functions.ThenKpis the dual ofKp.

TheconvolutioninD is defined as follows. LetmmNbe a sequenceinD(R2n) such that for every compact set K R2n there exists m0(K)such thatm(x) = 1, x K, mm0(K) and

supxR2n

|()m(x)|< C , Nn0.

The convolution ofT, S D is defined by

T

S,

= lim

mT(x)S(y), m(x, y)(x+ y)

,

D,

if this limit exists for everymm(then, it does not depend onmm).By the BanachSteinhaus theorem we know that T S D.

The spacesD() andS() with the operationare associative andcommutative algebras. The convolution in this case is separately continu-

ous.

We refer also to [63] and [3] for the theory of distributions.

0.5.2 Ultradistribution spaces

We follow the notation and definitions from [79], [81] and [86]. By Mppis denoted a sequence of positive numbers, M0= M1= 1,satisfying some

of the following conditions:

(M.1)M2p Mp1Mp+1, p N;(M.2)Mp/(MqMpq) ABp,0qp, p N;(M.2) Mp+1

ABpMp, p

N0, N0=N

{0

};

(M.3)

q=p+1Mq1/MqApMp/Mp+1, p N;

(M.3)

p=1Mp1/Mp


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S{M}(Rn) S{M} = ind limh0

SMh ,

whereSMh , h >0,is the Banach space of smooth functions on Rwith

finite norm

h() = sup,Nn0

h||+||

M||M||(1 + |x|2)||/2()L.

The Fourier transform is an isomorphism ofS ontoS;D =D(Rn) isdense in S, S is dense in DL1and the inclusion mappings are continuous.

The strong dual ofS, S,is the space of tempered ultradistributions(of Beurling and Roumieu types). There holdsD

S

E

,where

means that the left space is dense in the right one and that the inclusion

mapping is continuous. Thus,E S D.We denoteS+={fS(R); supp f [0, )}.

Let

S[0,)={C[0, ); =|[0,)for some S}with the induced convergence structure fromS; its strong dual is in factS+.

An operator of the form: P(D) =

||=0

aD, a C, Nn0 ,is

called ultradifferential operatorof class (Mp) (of class{Mp}) if there areconstants L >0 and C >0 (for every L >0 there is C >0) such that

|a|C L||/M||,||N0.0.5.3. Fourier hyperfunctions.There are many equivalent definitions

of hyperfunctions, Laplace hyperfunctions and Fourier hyperfunctions (cf.[71], [80], [144], [145], [204]), but we will use definitions and results collected

in [75]. Let Ibe a convex neighborhood of zero inRn and let be a non-

negative constant. A functionF,holomorphic onRn+iI,is said todecrease

exponentiallywith type (), 0,if for every compact subset K Iand every >0,there exists CK, >0 such that

|F(z)| CK,exp(( )|Rez|), zRn + iK . (0.7)The set of all such functions is denoted by

O(Dn + iI),(

O(Dn + iI) for

= 0),where Dn denotes the directional compactification ofRn : Dn =

Rn Sn1 (Sn1 consists of all points at infinity in all direction). SpacePis defined by

P= ind limI0

ind lim0

O(Dn + iI) . (0.8)


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0. Preliminaries 7

The dual space ofP is the spaceQ(Dn) of the Fourier hyperfunctions.Q(Dn) is a space of F S-type. We can give a representation of elementsfromQ(D

n

).LetObe the sheaf of holomorphic functions. DenoteUj= (D

n + iI){Im zj= 0}, j= 1, . . . , n; (Dn + iI)#Dn =U1 Un ,

(Dn + iI)#j Dn =U1 Uj1 Uj+1 Un.

Then

Q(Dn) =O((Dn + iI)#Dn)/n

j=1O((Dn + iI)#j Dn) . (0.9)

Thus, f Q(Dn) is defined as the class [F],whereFO((Dn + iI)#Dn).Fis called a defining functionoffand it is represented by 2n functions

F, F=F,where FO(Dn +iI); Dn + iI=Dn + i(I) is aninfinitesimal wedge of type Rn + i0,are open -th orthants in R

n.

An f L1loc(Rn) is called a function of infra-exponential typeif forevery >0,there exists C >0 such that|f(x)| Cexp(|x|), x Rn.Then by fis denoted the hyperfunction defined by f.

We denote by the set ofn-th variations of{1, 1}.The boundary-value representation off Q(Dn) is:

f= [F] :=

F(x +i0) . (0.10)

F(x + i0) denotes the element of the quotient space given in (0.9); it is

determined by F.

The dual pairing between Pand g= [G] Q(Dn

) is given by

g, =Rn

g(x)(x)dx=

Imw=v

G(w)(w)dw, w=u + iv,

wherev I.Similarly,Q(Dn), >0,is defined usingO instead ofO(cf.

Definition 8.2.5 in [75]).

An infinite-order differential operatorJ(D) of the form

J(D) =|a|0

baDa with lim

|a|

|a|

|ba|a! = 0, ba C ,

is called a local operator. J(D) maps continuouslyO(U) intoO(U), Ubeing an open set in Cn,and alsoQ(Dn) intoQ(Dn).


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TheFourier transformonQ(Dn) is defined by using the functions= 1(z1) . . . n(zn),where k=1, k= 1, . . . , n, = (1, . . . , n)and+(t) =e

t

/(1 + et

), (t) = 1/(1 + et

), tR.Letu(x)=

U(x + i0) =

(U)(x + i0) ,

where U Q(Dn +iI), and Udecreasing exponentiallyalong the real axis outside the closed -th orthant. The Fourier transform

ofuis defined by

F(u)()=

F(U)( i0)

=

Imz=y

eiz(U)(z)dz, y I, =+ i,

where F(U) O(Dn iI) andF(U)(z) =O(e|x|) for a suitable >0 along the real axis outside the closed -orthant.Fis an automor-phism ofQ(Dn).

0.6.Let Ebe a locally convex topological vector space. An absolute

convex closed absorbent subset of Eis called a barrel. If every barrel isa neighborhood of zero in E,then Eis called barreled. Throughout this

bookFgstands for a locally convex barrelled complete Hausdorff space ofsmooth functions (the subscriptgstands for general),Fg E= E(Rn),andFgstands for the strong dual space ofFg; observeE Fg. IfT Fgand Fg ,thenT, is the dual pairing betweenTand.We writeF0if all elements ofFgare compactly supported;F0denotes the dual space of

F0; observe that a notion of support for elements of

F0can be defined in

the usual way. InFg,a weakly bounded set is also a strongly bounded one(MackeyBanachSteinhaus theorem). The spaces of distributions, ultra-

distributions, Fourier hyperfunctions, . . . , are of this kind. Furthermore, we

shall always use the notationF=FgifA F, whereA=D, D,orP;in such caseF A =D, D,orQ(Dn),respectively, and we say thatF is a distribution, ultradistribution, or Fourier hyperfunction space, re-spectively. We setF={T F; supp T }.

We suppose that the following operations are well defined onF

g:Differentiation:We assume that /xi,Fg Fg, are continuous oper-

ators. LetkNn0 .Then, k

k1x1. . . knxn

T(x), (x)

=

T(x), (1)|k|

k

k1x1. . . knxn

(x)

, Fg.


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0. Preliminaries 9

Change of variables:IfT Fgand k >0,then by definition

T(kx), (x)

= T(x), 1kn xk, Fg.

IfT Fgandh Rn,T(x + h), (x)=T(x), (x h), Fg.

Furthermore, it is always assumed that k (/k), R+ Fg , and h(+ h), Rn Fg, are continuous. Consequently, by the mean valuetheorem, one readily verifies that both maps are indeed C.

Let E,if is a continuous mapping from Fginto Fg , then wesay thatis amultiplierofFg.ThenTis by definition T, =T,.The set of multipliersofFgis denoted by M().

We shall say that Eis a convolutorofFgif the mapping ,Fg Fg,is well defined and continuous, where(t) =(t). We denotebyM()theset of convolutorsofFg.IfM(),then for T Fg,(T )is defined byT , =T, , Fg .

mmN,a sequence in M() Fg, is called a -sequence inFgifm0, mN,and for every Fg, m inFg, m .

We shall say that the convolutionwith compactly supported elements is

well defined in Fgif a notion of support makes sense in Fgand the followingdefinition applies: givenS, T Fg,where supp S= Kis compact in Rn,their convolution is defined by

S T, =S(x) T(y), (x)(x+ y)= Sx Ty, (x)(x+ y) ,where the function Fghas compact support, supp =K, so that theset K

intKand (x) = 1, x

K.For

F it coincides with the usual

definition of convolution.

We assume thatFgcontains regular elementsf L1loc.They are iden-tified with fitself:

f:f, =Rn

f(x)(x)dx, Fg,

iff L1 for every Fgand ifm0 inFimpliesf, m 0.

There is also another situation in which we shall identify locally in-tegrable functions with elements T F. LetA= D, D,orPandsupposeA F. We identify Twith f L1locifT, = f, ,for all A. In such a case we simply write T= f. For example,T(x) = xex

2

sin(ex2

) S(R) is defined in this way, and not as a regularelement ofS(R) in the sense described above.


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1 S-asymptotics in Fg

1.1 Definition

Definition 1.1.Let be a cone with vertex at zero and letcbe a positive

real-valued function defined on .It is said that T Fghas S-asymptoticbehavior related to cwith limit UifT(x + h)/c(h) converges weakly inFgtoUwhen h ,h ,i.e. (w. lim = weak limit):

w.lim

h,h

T(x + h)/c(h) =U in

Fg (1.1)

or

limh,h

T(x + h)/c(h), (x)=U, , Fg. (1.2)

If (1.1) is satisfied, it is also said thatThas S-asymptotics and we write in

short: T(x + h) s c(h)U(x), h .

Remarks.1) If is a convex cone we could use another limit in .

Let h1, h2.We say that h1 h2if and only ifh1 h2+ ; is nowpartially ordered.

For a real-valued function defined on ,we write

limh, h

(h) =A R

if for any >0 there exists h() such that(h) (A , A + ) whenhh(), h .

If is a convex cone, then the S-asymptotics with respect to this limit

might be defined as:

limh, h

T(x + h)/c(h), (t)=u, , Fg. (1.3)

In case n= 1, the limits (1.2) and (1.3) coincide. We will mostly use

Definition 1.1 in this book, for S-asymptotics defined by (1.3) see also [135].

2) If

Fgis a Montel space, then the strong and the weak topologies in

Fgare equivalent on a bounded set. IfBis a filter with a countable basisand if w.lim

hBT(h) = UinFg,then this limit exists in the sense of the

strong topology. Hence, (1.1) is equivalent to

s. limh,h

T(x + h)/c(h) =Uin Fg.


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1. S-asymptotics inFg 11

(Furthermore, from now on, we will omit the symbol s.for the strong

convergence).

For the first ideas of the S-asymptotics see [3] and [146]. The startingpoint of the theory is [127].

1.2 Characterization of comparison functions and limits

Proposition 1.1. Let be a convex cone. Suppose T Fg has the S-asymptoticsT(x + h)

s

c(h)U(x), h

. IfU

= 0, then:

a) There exists a functiond on such that

limh, h

c(h + h0)/c(h) =d(h0) , for every h0 . (1.4)

b) The limitUsatisfies the equation

U( + h) =d(h)U, h .

Proof.a) Since U= 0,there exists a Fgsuch thatU, = 0.Forthis and a fixed h0

limh,h

c(h+ h0)

c(h)

T(x + (h+ h0))

c(h + h0) , (x)

= limh,h

T((x + h0) + h)

c(h) ,(x)

.

(1.5)

Hence, for every h0

limh,h

c(h+ h0)

c(h) =

U(x + h0),(x)U, =d(h0) .

b) Now we can take in (1.5) any function Fginstead of .Then wehave

d(h0)

U,

=

U(x+ h0),

,

Fg

which proves b).

We now restrict the space of generalized functions to a space of distri-

bution, ultradistribution, or Fourier hyperfunction type. So, we have the

following explicit characterization of the comparison function and limit.


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Proposition 1.2. Let be a convex cone withint =(int is the in-terior of). LetT F have S-asymptoticsT(x+h) sc(h)U(x), h,whereU= 0 andc is a positive function defined onR

n

. Then:a) For everyh0Rn there exists

limh(h0+),h

c(h+ h0)/c(h) =d(h0) .

b) There exists Rn such thatd(x) = exp( x), x Rn.c) There existsC R such thatU(x) =Cexp( x).

Proof.a) Let aint.Then there exists r >0 such that B(a, r).Consequently, for every >0, B(a,r) ,as well.

We shall prove that for every h0 Rn and every R >0 the set (h0+) {x Rn; x> R}is not empty. The first step is to prove that(h0+ ) is not empty.

Suppose thatyB(a,r/2) .Then, for every0>2h0/r >0,

a

(h0+ y)

a

y

+

h0

r ,

henceh0+ yB(a,r) .For a fixed R >0,we can choose such thath0+y> R.Then

h0+ yis a common element for (h0+ ), and{x Rn;x> R}.Now we can use the limit (1.5) when h (h0+ ) , and in the same

way as in the proof of Proposition 1.1 a), we obtain

limh(h

0+),h

c(h+ h0)

c(h)

=d(h0) =U(t+ h0),

U, .

From the existence of this limit, it follows:

1)dextends dto the whole Rn;

2)d(0) = 1; d C(Rn) anddsatisfiesd(h + h0) =d(h)d(h0), h , h0 Rn . (1.6)

We can take h0= (0, . . .0, ti, 0, . . . , ) Rn in (1.6); then the limit

limti0

d(h + h0) d(h)ti

=d(h) limti0

d(h0) d(0)ti

exists and gives

hid(h) =

hid(h)

h=0

d(h), for i= 1, . . . , n .


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We introduce the function Vgiven by

d(h) =e(h)

V(h), i= hi d(h)h=0, i= 1, . . . , n .Then

hiV(h) = 0 for everyi= 1, . . . , n .Consequently,V(h) = 1, h Rn.

For the proof of c), we now haveU(x+h) = exp( h)U(x). Differenti-ating with respect to hand then setting h= 0, we obtain that Usatisfies

the differential equations

xjU= j ,1 j n. Proceeding as in the

proof of b), we obtain U(x) =Cexp( x),for some CR. Remarks.

1. We only assumed thatcis a positive function. But if we know that

there existT FgandU= 0 such that T(x + h) s c(h)U(t), h ,thenwe can find a function c C and with the property

limh,h

c(h)/c(h) = 1.

This function ccan be defined as follows: c(h) =T(x +h),(x)/U,,where is chosen so that U, = 0.In this sense we can suppose, wheneverneeded, that c C, and we do not loose generality.

Similarly, we have (see also Lemma 1.4 in 1.12) that

limh,h

c(h)/c(h+ x) = exp( x) in E

if is a convex cone, int

=

,

( is compact int ).

2. If in Proposition 1.2 we replaceF for the general spaceFg, thena) and b) still hold. On the other hand, c) will not be longer true, in

general. We will show this fact in Remark 5 below by constructing explicit

counterexamples.

3. In the one-dimensional case the cone can be only R, R+or R.

In all these three cases int =.Consequently,dfrom Proposition 1.2 hasalways the formd(x) = exp(x),whereR.

Let us write c(x) = L(ex) exp(x), x R.We will show that Lis aslowly varying function. Proposition 1.2 a) gives us the existence of the

limit

limh,h

L(exp(h + h0))/L(exp(h)) = 1, h0R .


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If =R+,then

limxR+,x

L(xp)/L(x) = 1, p

R+

and this defines a slowly varying function (cf. 0.3). Thus if T Fg(R)and T(x+h)

s c(h)U(x),in R+,with U= 0,then it follows that chasthe formc(x) = exp(x)L(exp(x)), xa >0,where Lis a slowly varyingfunction at infinity. Similarly, if = R, then Lis slowly varying at the

origin, while if = R, then Lis slowly varying at both infinity and the

origin.

4. The explicit form of the function cgiven in 3. is not known inthe n-dimensional case, n2. This problem is related to the extensionof the definition of a regularly varying function to the multi-dimensional

case ([135], [162]) and with certain q-admissibleand q-strictly admissible

functions([192]).

5. As mentioned before, c) in Proposition 1.2 does not have to hold for

a general spaceFg. From the proof of Proposition 1.2, we can still obtainthe weaker conclusion U(x+h) = exp(

h)U(x), h

Rn, which in turn

implies the differential equations (/xj)U= j .For distribution, ultra-distribution, and Fourier hyperfunction spaces, these differential equations

imply thatUmust have the form c) of Proposition 1.2. However, the latter

fact is not true in general. We provide two related counterexamples below.

Let

A0(R) ={ C(R); limx

(m)(x) exists and is finite, mN0} ,

it is a Frechet space with seminorms:

k() = supx(,k],mk

|(m)(x)| .

Let us first observe that the definition ofA0(R) does not tell all the trueabout its elements. Notice that if A0(R), then for m= 1, 2, . . . , wehave limx (m)(x) = 0, while limx (x) may not be zero. Indeed,

the proof is easy, it is enough for m= 1, if limx (x) =M, then

(x) =(0) + x0

(t)dt(0) + M x , x ,

but since has limit at, then M= 0. So, we haveA0(R) = { C(R); lim

x(x) exists and lim

x(m)(x) = 0, m N} .


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Its dual spaceA0(R) contains a generalized function concentrated atwhich contradicts c) of Proposition 1.2. Define the Dirac delta concentrated

atby, := lim

x(x), A0(R) .

Notice that the constant multiples ofare the only elements ofA0(R)satisfying the differential equation U = 0. This generalized function is

translation invariant, i.e., ( +h) = , for all hR; in particular,it has the S-asymptotics

(x + h) s

(x), h

=R .

Therefore, we have found an example of a non-constant limit for S-

asymptotics related to the constant functionc(h) = 1.

We can go beyond the previous example and give a counterexample

for the failure of conclusion c) of Proposition 1.2 with a general cand S-

asymptotics inFg. Letc(h) = exp(h)L(exp h), where Lis slowly varyingat infinity and R. We assume that cis C and, for all m N,c(m)(h)

mc(h), h

(otherwise replace cby cgiven in Remark 1).

Next, we define

Ac(R) ={ C(R); limx

(m)(x)

c(x) exists and is finite, mN0} .

It is a Frechet space with seminorms:

k,c() = supx(,k],mk

|(m)(x)|c(x) .

Note thatAc(R) =c(x) A0(R), consequently,(x) Cc(x), x ,

where C= limx (x)/c(x). An inductive argument shows thatfor all m, (m)(x)(1)mmCc(x), x . Set now gc,=(1/c(x)) Ac(R), a generalized function concentrated at infinityand given by

gc,,

= (x),

(x)

c(x) = limx(x)

c(x),

Ac(R) .

It is easy to show that the constant multiples ofgc,are the only element

ofAc(R) satisfying the functional equationU(x + h) = exp(h)U(x) (andhence the differential equation U =U); in particular,

gc,(x+ h) s ehgc,(x), h =R .


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In addition, there is an infinite number of elements ofAc(R) having S-asymptotics in the cone = R+related to c(h) with limit of the form

Cgc,(x), C R. In fact, consider (m)

, the derivatives of the Diracdelta concentrated at the origin. We have that

(m)(x + h) sc(h)mgc,(x), h =R+ ,

since

limh

(m)(x+ h), (x)

c(h) = lim

h(1)m

(m)(h)c(h)

=m gc,(x), (x) .

1.3 Equivalent definitions of the S-asymptotics inF

Theorem 1.1. LetT F and letint =. The following assertions areequivalent:

a) w.limh,h

T(x + h)

c(h)

=U(x) =Mexp(x) in

F, M

= 0 . (1.7)

b) For a-sequencemm (cf. 0.6) there exists a sequenceMmm inR, such thatMmM= 0, m , and

w.limh,h

(T m)(x + h)c(h)

=Mmexp(x), inF, uniformly inm N .(1.8)

c) For a-sequencemm, (cf. 0.6),lim

h,h

(T m)(h)c(h)

=pm, mN , (1.9)

wherepm= 0 for somem, and for every F,

suph,h0

(T )(h)

c(h)


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Proof. a) b).Letmmbe a -sequence. For any F,{m ;m N} is a compact set in F.We have by the properties of the convolution(cf. 0.6) and the BanachSteinhaus theorem

limh,h

(T m)(x+ h)

c(h) , (x)

= lim

h,h

(T m)(x), (x h)c(h)

= limh,h

T, (m )( h)c(h)

= limh,h

T(x + h)

c(h) , (m )(x)

=

M ex,(m

)(x)=

M

mex, (x)

, (1.11)

uniformly in m. Now (1.11) implies (1.8) and b).

b) a).Let Fand

am,h=

(T m)(x + h)

c(h) , (x)

=(T (m ))(h)

c(h) , mN, h .

We have am,ham, h ,h ,uniformly for mN,where

am=Mmexp( x), (x), m N,ama=Mexp( x), (x), m .

Also am,hah, m ,where

ah=

T(x + h)

c(h) , (x)

, h .

This implies ah

a, h,

h

,what is in fact a).

a)c).From (1.7) it follows that (T )(h)/c(h) converges for every F,when h, h .Hence, (T )(h)/c(h) is bounded, h,h 0; (1.9) follows directly from (1.7).

c) a).First, we shall prove that the setG={m(+ x), m N, xRn}is dense inF.Suppose that T F and that

T, m( + x)= 0, m N, x Rn .It follows that (Tm)(x) = 0, m N, x Rn.Then, for any F,T m, = 0, mN,and consequently

T, = limm

T, m = limm

T m, = 0 .


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This implies that T= 0 and hence, the setGis dense inDby the Hahn-Banach theorem. Thus, by (1.10) and the BanachSteinhaus theorem, c)

implies a).a) d).Note that (1.1) implies the strong convergence ofT( +h)/c(h)

to UinD andD respectively. Since the convolution inD andD ishypocontinuous, it follows the following equality in the sense of convergence

inE(Rn)

limh,h

T( + h)

c(h)

=

lim

h,h

T( + h)c(h)

=U

in both cases ( D,respectively D

). The paper which can be consulted for related results is [128].

1.4 Basic properties of the S-asymptotics

Theorem 1.2. LetT Fg.a) IfT(x+h)

s

c(h)U(x), h

, then for everyk

Nn0 , T

(k)(x+h) s

c(h)U(k)(x), h .b) Assume additionally thatFg is a Montel space. Let g M() (set

of multipliers ofFg (cf. 0.6)); let c, c1 be positive functions. If for every Fg, (g(x+h)/c1(h))(x) converges to G(x)(x) inFg when h,h and ifT(x +h) sc(h)U(x), h, theng(x +h)T(x +h) sc1(h)c(h)G(x)U(x), h .

c) IfT

F0 andsuppT is compact, then T(x+h)

s

c(h)

0, h

,

for every positive functionc.

d) Suppose thatFg is a Montel spaces in which the convolution withcompactly supported elements is well defined and hypocontinuous (cf. 0.6).

LetS Fg ,suppS being compact. IfT(x+h) s c(h)U(x), h, then(S T)(x + h) sc(h)(S U)(x), h.

e) LetT F = D andT(x+h) s c(h) U(x), h , inD. Assumethat (M.2) holds. LetP(D)be an ultradifferential operator of class.Then

(P(D)T) (x + h) sc(h) (P(D)U)(x), h , inD .

f ) Let T F = Q(Dn). Let P(D) be a local operator and let T(x+h)

s c(h)U(x), h, h inQ(Dn), then(P(D)T)(x+h) sc(h) (P(D)U)(x), h ,h inQ(Dn), as well.


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Proof.a) The assertion is a consequence of the definition of the deriva-

tive of a generalized function. Namely,

limh,h

T(k)(x + h)

c(h) , (x)

= lim

h,h

T(x + h)

c(h) , (1)k(k)(x)

= (1)kU(x), (k)(x)=U(k), , Fg.

b) Since Fgis a Montel space, we have that limh,hT(x + h)

c(h) =U(x)inFgwith respect to strong topology. As (g(x+h)/c1(h)G(x))(x), h, h 0,is a bounded set inFg,it follows:

limh,h

g(x + h)T(x + h)/(c1(h)c(h)), (x)

= limh,h

T(x + h)/c(h),

g(x + h)

c1(h) G(x)

(x)

+ lim

h,hT(x + h)/c(h), G(x)(x)

=

U(x), lim

h,h

g(x + h)

c1(h) G(x)

(x)

+ U(x), G(x)(x)

= 0 + U(x)G(x), (x)= UG,, Fg.

c) For each F0there exists r >0 such that suppB(0, r) ={x Rn; x< r}.The support of T(x+h) is (suppT h).Thus, byour assumption, there exists rsuch that for allh,h> rthe set(suppTh)B(0, r) is empty and consequentlyT(x+h), (x)= 0, h, h r .

d) By definition of the convolution

(S T)(x + h), (x)= (S T)(x), (x h)= St Ty, (t)(t+ y h)= St Ty( + h), (t)(t+ y) .

(1.12)

Hence, (S T)(x + h) = (S T( + h))(x).


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Since in a Montel space the weak and the strong convergence are equiv-

alent, we have by (1.12)

limh,h

(ST)(x+h)/c(h) = limh,h

S T(+h)

c(h)

(x) = (SU)(x) .

For the proof of e) and f), we note that an ultradifferential operator

P(D) maps continuouslyD intoD and a local operator maps continu-ouslyQ(Dn) intoQ(Dn).

Remark.From assertion a) of Theorem 1.2, a natural question arises

for spacesF: The limit Ucan be a constant generalized function, henceU = 0.Is there a positive functioncsuch thatT has S-asymptotics related

to thisc, but with a limit different from zero?

In general the answer is negative as shown by the following example: Let

Tbe defined byx2+ sin(expx2), x R.Then,T(x + h) sh2 1, h R+.But T(x) = 2x(1 + exp(x2) cos(expx2)).The same situation is obtained

with the distribution f(x) =x2 + x sin x.

We can now formulate an open problem: Suppose that T(x+

h) s c(h)U(x), h. If U(x) is a constant generalized function, then

the problem is to find some additional conditions onT which guarantee the

existence of a functioncsuch thatT has the S-asymptotics in related to

c.

More generally, let S F and T= (/xk)S. If T(x+ h) sc(h)U(x), h . The question is what we can say about the S-asymptoticsofS.

Recall the well known result: Let hbe a real-valued function which

has the first derivative h(x) = 0, x x0 and h(x) , x .If a function F has its first derivative on(x0, ) such that there existslim

xF(x)/h(x) =A, then there existslim

xF(x)/h(x) =A, as well.

We know that an opposite assertion does not hold, and this is at the

basis of the open problem quoted above.

We give a theorem to illustrate the relation between the S-asymptoticsof a distribution and the S-asymptotics of its primitive. We refer to [114]

for the proof.

Theorem 1.3. 1) Letf, g D(R) and for somem N, g(m) =f .


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a) If f(x+h) s hL(h) 1, h R+, where >1, then g(x+h) s

h+mL(h)

1, h

R+.

b) Iff(x + h) s exp(h)L(exp h)exp(x), hR+, R, andx0

exp(h)L(exph)dh , whenx , then

g(x+h) s h

0

hm10

. . .

h10

exp(t)L(exp t)dtdh1 . . . d hm1

exp(x), h R+ .2) Let0

D(R) such that 0(t)dt= 1. If

limh

g(i)(x + h)

exp(h)L(exph), 0(x)

=iexp(x), 0(x), i= 0, 1, . . . , m 1

and

f(x + h) sexp(h)L(exph)m exp(x), h R+,

then

g(x + h) s

exp(h)L(exp h)exp(x), h

R+.

3) Suppose that T D, ={x Rn; x= (0, . . . , xk, 0, . . . , 0)} andT= (/xk)S. IfT(x+h)

sc(h)U(x), h andc(h)is locally integrableinhk such that

c1(hk) =

hkh0k

c(v)dvk as hk , h0k 0,

thenS(x + h) s c1(h)U(x), h .4) Suppose thatS D and that for anm {1, 2, . . . , n},

(DtmS)(x + h) sc(h) U(x), h .

LetV D, DtmV=U and0 D(R),R

0()d= 1. Let

limh,h

S(x + h)/c(h), 0(xm)m(x)= V, 0m ,

wherex= (x1, . . . , xm1, xm+1, . . . , xn) and

m(x) =

R

(x1, . . . , xm, . . . , xn)dxm, D .

ThenS(x + h) sc(h)V(x), h.


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The following theorem asserts that the S-asymptotics is a local property

if the elements ofFgare compactly supported.

Theorem 1.4. LetT1, T2 F0.Let the open set Rn have the followingproperty: for everyr >0 there exists ar >0 such that the ballB(0, r) =

{x Rn; x< r} is in{h;h, h r}. If T1= T2 on andT1(x + h)

sc(h)U(x), h , thenT2(x + h) s c(h)U(x), h , as well.

Proof.Let F0with supp B(0, r).We shall prove that

limh,hT1(x + h) T2(x + h)c(h) , (x) = 0 . (1.13)The complement of the set supp(T1(x+h) T2(x+h)) contains the set{ h, h}.By our supposition the number ris fixed in such a waythat the sets{h;h, h r}contain B(0, r) and consequentlysupp.Since T1has the S-asymptotics related tocand with the limit U,

(1.13) implies

limh,h

T2(x + h), (x)

c(h) = lim

h,h T1(x + h), (x)

c(h) =

U,

.

Remark.The open set ,by its property, has to contain a set 0{xRn; x> R}where 0 is an open acute cone such that 0 andRis apositive number.

1.5 S-asymptotic behavior of some special classes of

generalized functions

1.5.1 Examples with regular distributions

1.exp(a (x + h)) s exp(a h) exp(a x), h Rn.

2.exp((x +h)2 + (x+ h))

sexph exp

x +1

2

, h R+.

3. Let w Sn1

={x Rn

; x= 1}, p= (p1, . . . , pn) Rn

+and ={qw; q R+}.Denote, J={k {1, . . . , n};wk= 0}and =iJ

pi.

Then

(x + h)p s q

iJ

wpii 1, h ,


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and

(1 + x + h)p s

q iJ wpii 1, h ,

((x+ h)p = (x1+ h1)p1 . . . (xn+ hn)

pn) .

4. For a slowly varying functionL(t), t >0 we haveL(t + h)

sL(h) 1, hR+ .Namely,

limh

L(t + h)/L(h), (t)= limh

rr

(t)L(t+ h)/L(h)dt

= limq

erer

(log y)L(log(yq))/L(log q)dy

y=

R

(t)dt, D(R) .

We used above that L(log y) is also a slowly varying function (cf. 0.3)and thatL(log(uh))/L(logh) converges uniformly to 1 as h ifustaysin compact intervals [1, 2],0< 1 < 2 0 (cf. Theorem 1.2 c)).

For >0,

f(x + h) s(1/())h1 1, h R+ .


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In the case 0, hh0.Then

limh

f(x)/h1, (x h) = limh

a

1

()

x

h

1(x h)dx

= limh

R

1

()

u + h

h

1 (u)du=

R

1

()(u)du .

Hence,f(x + h) sh1 1

(), h R+,


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Property (*) is not equivalent to the existence of the S-asymptotics.

The next example illustrates that this condition is not necessary.

Assume that S C(R)L1(R) but not being bounded on R.Thisfunction has S-asymptotics equal to zero related to c= 1 (see 5.). For T

we take 1 + S(x).Then, T(x + h) s 1 1, hR+and

limh

[(1 + S) m](h) = limh

1 + S(x + h),m(x)=1, m(x)=pm .

This limit is not uniform in m Nbecause limm

[(1+S) m](h) does notnecessarily exist.

9.Every distribution in DLp , 1 p < has S-asymptotic behavior relatedtoc= 1,and =Rn,with limit U= 0.Let us show this.

By Theorem XXV, Chapter VI in [146] it follows that (T ) Lp(Rn)for every D.Every derivative of (T ),

hk(T)(h) = (T

xk)(h), h Rn,is also in Lp(Rn).Hence (T) DLp.We know

that every element of

DLp,1

p 0, c(h) 0, h, such that

|T(x + h)/c(h), (x)|M, h> D ,where andM are positive constants depending on.

The answer to this question is also negative (cf. [153]).

10.LetT Kp.Then there exists k0 N0such that Thas S-asymptoticbehavior with limit U= 0 related toc(h) exp(k0hp),wherec(h) tends toinfinity ash .


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First, we prove that there exists a positive integerk,such that the set

{T(+ h) exp(khp), hRn}is bounded inD.We start by giving a bound for the seminorms k(( h)), Kp:

k[( h)] = supxRn,|a|k

exp(kxp)|Da(x h)|

= supxRn,|a|k

exp(kx + hp)|Da(x)|

exp(2pk

h

p) sup

xRn,|a|2pk

exp(2pk

x

p)

|Da(x)

|exp(2pkhp)2pk().

By assumption, Tis a continuous linear functional onKp.Note, thesequence of normskkis increasing. Thus, there exist >0 and k0 N0such that

|T, |1 for Kp, k0() .This inequality holds for all k k0.Hence

|T, |1k(), kk0 for every Kp.

We know thatD Kpand that the inclusion is continuous. Let D,Then

|exp(2pkhp)T(x + h), (x)|=|T(x), exp(2pkhp)(x h)|

1 exp(2pkhp)k[(x h)] 12pk(), k > k0.

We can choose k02pk.The set{exp(k0hp)T(x+h);h Rn}isbounded inD (and weakly bounded inD,as well).

Now, for every

D:

limh

exp(k0hp)T(x + h)/c(h), (x)

= limh

1

c(h)exp(k0hp)T(x + h), (x)= 0.


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1.5.3 S-asymptotics of ultradistributions and Fourier hyperfunc-

tions Comparisons with the S-asymptotics of distributions

11.If a distribution has S-asymptotics inD (see Definition 1.1), it has thesame S-asymptotics inD,as well. But the opposite is not true. This isillustrated by the following example:

T(x) = 1 +

n=1

(n)(x n)/Mn, x R,

has S-asymptotics inD(Mp)(R),but it does not have S-asymptotics inD(R).We will show this. Let D(Mp)(R).Then,

n=1

(n)(x+hn)/Mn, (x)=

n=1

(1)n(n)(nh)/Mn0, h .

This is a consequence of the property

supxR

|(n)(x)|/knMn0, n for every k >0 .

Suppose that there exists a function csuch that for every D(R)

n=1

(1)n(n)(n h)/Mnc(h) ,

converges, as h .Taking h= nthis implies that (n)(0)/Mnc(n)converges to zero, as n ,for every D(R).However, such a func-tion cdoes not exist (Borels theorem). Namely, for every given sequence

Mnc(n), n

N,there exists

C0(R) such that

(n)(0)/Mnc(n)

,

as n .This example is the motivation for the following assertion.

12. Let T D(R) be such that limh

T(x+h)/c(h) exists inD(R).Assume that for some s Nand D(R) with the property

R

(t)dt= 1,

R

tj (t)dt= 0, j= 1, . . . , s ,

the following limit

limh

T(t + h)

c(h) ,

ts

ps+1

t

p

, p (0, 1],

exists uniformly in p.Then, limh

T(x + h)/c(h) exists inD(R) as well.


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Proof.Let D(R), h h0 >0.The function Fh(t) =T(x+h+t), (x), t R,is smooth and by the Taylor formula, we have:

Fh(t) =Fh(0) + tddt

Fh(0) + + ts1(s 1)!

ds1dts1

Fh(0)

+ ts

(s 1)!

10

(1p)s1F(s)h(pt)dp, s 1,0< p 1 .

This implies

T(x + h), (x)=T(x + h + t), (x) T(x + h), (x)

(1)s1ts1(s 1)! T(x + h),

(s1)(x)

(1)sts

(s 1)!

10

(1 p)s1T(x + h +pt), (s)(x)dp,

h h0, tR .Multiplying both sides of the last equality by(t),integrating with respect

tot, and using Fubinnis theorem, we obtain

T(x + h), (x)=T(x + h + t), (t), (x) (1)s

(s 1)!1

0

(1p)s1T(x + h +pt), ts(t), (s)(x)dp .

Set

G(x,h,p) =T(x + h+pt), ts

(t), x R, h h0, p [0, 1] .We have G(x,h, 0) = 0, x R, hh0and

limh

G(x,h,p)/c(h) =Ceaxeapt, ts(t), x R, p (0, 1] ,for some a R(cf. Proposition 1.2 a)), where the limit is uniform in

p(0, 1] and xsupp. The limit function is continuous in x Randp[0, 1],because ofeapt, ts(t) 0,as p0.Because of that, weobtain

limh

1

c(h) T(x + h +pt), ts

(t), (s)

(x)

=

limh

1

c(h)T(x + h+pt), ts(t), (s)(x)

=Ceax+apt, ts(t), (s)(x)=Ceax, (s)(x)eapt, ts(t) .


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This implies

limh1c(h)T(x + h), (x) = limh T(x + h+ t)c(h) , (t), (x) (1)

s

(s 1)!

10

(1p)s1eapt, ts(t)dpCeax, (s)(x) ,

which proves the assertion.

Such an assertion can be proved in the multi-dimensional case by ad-

justing the previous argument. Butan open problemis to find necessaryand sufficient conditions for a distribution T, which has S-asymptotics in

D, to have the S-asymptotics in D as well.13.We shall construct an ultradistribution out of the space of Schwartz

distributions and having S-asymptotics.

Assume (M.2) holds. LetP(D) be an ultradifferential operator of class

of infinite order (a= 0 for infinitely ) (see 0.5.2). Then, P(D)is anelement of

D which is not a distribution and which has S-asymptotics in

D equal to zero related to any c.If T D and T(x+h) s11, h inD,the ultradistribution

T+ P(D)is not a distribution, but (T+ P(D))(x+ h) s1 1, h in

D :lim

h,h((T(x + h), (x) + P(D)(x+ h), (x))

= 1, + limh,h(x + h), (1)mam(m)(x)= 1, , D .14. Let P(D) be a local operator

||0 b

D, b = 0, ||0 (see0.5.3.). The Fourier hyperfunction f= 1 + P(D)has S-asymptotics

related to c= 1 in any cone and with limit U= 1, but fis not a

distribution. For the S-asymptotics offit is enough to prove that

limh,h

P(D)(x+ h), (x)= 0, P .

SinceP(D) mapsPintoP,P(D)(x+ h), (x)=(x + h), P(D)(x)=(h) ,


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where=P(D).By the property of the elements ofP(see0.5.3), wehave lim

h,h(h) = 0,for every cone .

A hyperfunction gsupported by the origin is uniquely expressible as

g=P(D),whereP(D) is a local operator. In such a way, the above

proof implies that every Fourier hyperfunction with support at{0}hasS-asymptotics with limit equal zero.

Since P(D)=||0

bDis a distribution if and only ifb= 0 for a

finite number of,we note that 1+P(D)is not a distribution, but it has

the S-asymptotics related to c= 1.We can also find coefficients bof a local operator P(D) such that

f= 1+P(D)is not defined by an ultradistribution belonging to the Gevrey

classD(s) orD{s} , s >1 (see [81]). For the sake of simplicity, we shallconsider the one-dimensional case. Choose P(D) such that the coefficients

of P(D) are: bn= (n! )(1+cn), n N,where cn= (log logn)1.With

these coefficients, P(D) is a local operator. Namely,

limn

nbnn! = limn(n! ) 1

n log logn = 0 .

Also, any ultradistribution in the Gevrey class s >1,supported by{0},is of the form

J(D)=

n=0

anDn,necessarily with|an|C kn/(n! )s

for some constantskand C(Beurlings type) or for anyk >0 with a con-

stant C(Roumieus type). But the coefficients bn= (n! )(1+cn) do not

satisfy these conditions, therefore P(D)cannot represent an ultradistri-

bution. Namely, since cn0 when n ,for any s >1,there exists n0such that 1 + cn< s, n n0.Thus,

(n! )(1+cn) > Ckn/(n! )s, nn0, k >0 .Consequently, P(D)does not represent an ultradistribution of Gevrey

type.

On the other hand, if we suppose that g= P(D)is an ultradistribu-tion with support{0}in the Gevrey class s >1,then, we would have anultradifferential operatorJ1(D) such that

g=J1(D)=

n=0

enDn,|en|C kn/(n! )s .


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But in this case J1(D) is also a local operator and J1(D)= P(D).Thiscontradicts the fact that a hyperfunction with support at{0}is given by aunique local operator.

1.6 S-asymptotics and the asymptotics of a function

We suppose in this subsection that elements ofFare compactly supportedand, as usual, that the topology inFis stronger than the topology inE.Recall, we use the notationF0in this case.

Every locally integrable function fdefines an element ofF0(regulargeneralized functions).

We shall compare the asymptotic behavior of a locally integrable func-

tionfand the S-asymptotic behavior of the generalized function generated

by it.

A functionfhas asymptotics at infinity if there exists a positive function

c such thatlimx

f(x)/c(x) =A = 0, (in shortf(x) Ac(x), x ).1.The following example points out that a continuous andL1-integrable

function can have S-asymptotics as a distribution without having an ordi-

nary asymptotics. Suppose thatg L1(R) C(R) has the property thatg(n) =n, nNand that it is equal to zero outside suitable small intervalsIn n, n N.Denote byf(t) =et

t0

g(x)dx, t R.It is easy to see that

f(t + h) seh et

0

g(x)dx, hR+ .

By Theorem 1.2 a)f(t) has S-asymptotics related to eh and with the same

limit. But, in view of the properties ofg, f(t) =f(t) +etg(t) has not the

same asymptotics (in the ordinary sense). Moreover, gcan be chosen so

thatf has no asymptotics at all.

2.The following example shows that a function fcan have asymptoticbehavior without having S-asymptotics with limitUdifferent from zero. An

example is xexp(x2), x R.Suppose that exp(x2) has S-asymptoticsrelated to a c(h) >0, h R+with a limit Udifferent from zero. ByProposition 1.2 c), Uhas the form U(x) = Cexp(ax), C >0.Then, for


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every F0such that >0 we havelim

h

1

c(h) exp[(x + h + h0)2](x)dx=eah0Ceax, (x) .Therefore,

eah0U, = exp(h20) limh

1

c(h)

e(x+h)

2

e2h0(x+h)(x)dx

exp(h20)U, , for every h0>0 .But this inequality is absurd. Consequently, exp(x2) cannot have such an

S-asymptotic behavior.

One can prove a more general assertion.

Proposition 1.3. LetfL1loc(R) F0(R)have one of the four propertiesfor >1, >0, x x0, h >0, M >0 andN >0 :

a) f(x + h) Mexp(h)f(x)0,a)f(x + h) Mexp(h)f(x) 0,b)0f(x + h) Nexp(h)f(x),b)0 f(x + h) Nexp(h)f(x).

Thenfcannot have S-asymptotics with limitU= 0, but the functionf canhave asymptotics.

For the proof see ([135], p. 89).

It is easy to show that for some classes of real functions fon Rthe

asymptotic behavior at infinity implies the S-asymptotics.

Proposition 1.4. a) Let c be a positive function and let T L1loc(Rn).Suppose that there exist locally integrable functionsU(x)andV(x), x Rn,such that for every compact setKRn

|T(x + h)/c(h)| V(x), x K,h> rK,lim

h,hT(x +h)/c(h) =U(x), x K .

Then, T(x + h) s

c(h)U(x), h

in

F0.

b) LetT L1loc(R) have the ordinary asymptotic behaviorT(x) exp(x)L(exp x), x , R ,

whereL is a slowly varying function. Then,

T(x + h) sexp(h)L(exph) exp(x), h R+, inF0(R) .


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Proof.a) For every F0lim

h,hT(x + h)c(h) , (x) = limh,h Rn

T(x + h)

c(h) (x)dx . (1.14)

Since supp K Rn and Thas all the listed properties, Lebesguestheorem implies the result.

b) It is enough to use in (1.14) that L(yt)/L(t)1, t uniformlyin y,when ystays in compact interval contained inR+.

A more general result is the following one ([135], pp. 8990).

Proposition 1.5. Letbe a cone and letRn be an open set such thatfor everyr >0 there exists ar such thatB(0, r) { h; h,hr}. Suppose that G L1loc() and it has the following properties: Thereexist locally integrable functionsU andV inRn such that for everyr >0

we have

|G(x + h)/c(h)|V(x), x B(0, r), h ,h r;lim

h,h

G(x + h)/c(h) =U(x), x

B(0, r).

IfG0 F0 coincides withG on, thenG0(x + h)

sc(h)U(x), h .

Proof.By Theorem 1.4, it is enough to proof that

limh,h

G(x + h)

c(h) (x)dx=

Rn

U(x)(x)dx;

but as in the proof of Proposition 1.4 a), the exchange of the limit and theintegral sign is justified by our assumptions and Lebesgues theorem.

The following proposition gives a sufficient condition under which the

S-asymptotics of f L1loc(R), inD(R), implies the ordinary asymptoticbehavior off.

Proposition 1.6. Let f L1loc(R), c(h) = hL(h), where > 1and L be a slowly varying function. If for some m

N, xmf(x), x >

0, is monotonous and f(x+ h) s c(h)1, h R+,, inD(R), thenlim

hf(h)/c(h) = 1. If we suppose thatLis monotonous, then we can omit

the hypothesis >1.

For the proof see [115].


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1.7 Characterization of the support of T F0

We suppose in this subsection that the topology inF0is defined in sucha way that a sequence{n}inF0converges if and only if there exists acompact set KRn and(k)mconverges to (k) uniformly on Kfor everyk Nn0asm .

We already proved in Theorem 1.4 a relation between the support of a

distribution and its S-asymptotics. Now, we shall complete this result.

We need a property of the S-asymptotics given in the next lemma.

Lemma 1.1. Let be a cone and let be a convex cone (it is partially

ordered). A necessary and sufficient condition that for every c(h) >0,

h a) w.lim

h,hT(x + h)/c(h) = 0 inF0,

b) w.limh,h

T(x + h)/c(h) = 0 inF0,

is that for every F0 the following holds:In case a): There exists()>0 such that

T(x + h), (x)= 0, h (), h . (1.15)

In case b): There existsh such that

T(x +h), (x)

= 0, h

h, h

. (1.16)

Proof.We have to prove only that the condition is necessary in both

cases. It is obvious that the condition is sufficient. Let us suppose the

opposite, that is, the condition is not necessary. Then, we could find a

sequenceammin such that in case a)am in ,and in case b)am in,(m ) and that

T(x + am), (x)

=m

= 0, m

N .

Let find csuch that c(h) = mfor h= am, m N.Clearly, for sucha c(h) the function h T(x+h)/c(h), (x)cannot converge to zero ash , h,or h , h.This contradicts our supposition thatthe S-asymptotics equals zero in both cases.


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Theorem 1.5. Letbe a cone and letT F0.A necessary and sufficientcondition that for everyr >0 there existsr such that the sets

suppT B(h, r), h , h r are emptyis thatT(x + h)

sc(h) 0, h for every positive functionc on.

Proof.Theorem 1.2 c) and Theorem 1.4 assert that the condition in

Theorem 1.5 is necessary. We have to prove only that this condition is also

sufficient.

Let us suppose that

w.limh,h

T(x + h)/c(h) = 0 inF0 .

Let F0.By Lemma 1.1 a), we know that there exists 0() =inf(),where () is such that (1.15) holds. We shall prove that the

set{0(); F0,supp K}is bounded for every compact set KRn.Let us suppose the opposite. Then we could find a sequencekkinF0, suppkK, kN,and a sequencehkkin , hk such that

T(t + hk), p(t)=Ak,p= ak= 0, p=k0, p < k .

We give the construction of sequences {k}and {hk}.Let k F0, suppkK, kN,be such that0(k)kis a strictly increasing sequence whichtends to infinity. Then, there existhkkin andk >0, kN,such that0(k1) + k hk0(k) k.Now, we shall construct the sequencep(t)pinF0, supppK, pN,for which we have

T(t + hk), p(t)=0, p=kak, p=k .

()

Put

p(t) =p(t) p11(t) pp1p1(t), p >1, t Rn .The sequencepi iwill be determined in such a way thatp(t) satisfies (*)the sought property, p

N.

It is easy to see thatT(t+hk), k(t)= akandT(t+hk), p(t)=0, k > p.For a fixed pandk < pwe can find pi , i= 1, . . . , p 1,such thatfork= 1, . . . , p 1,

0 = T(t + hk), p(t)=Ak,p p1Ak,1 pp1Ak,p1 .


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Hence

p1Ak,1+

+ pp1Ak,p1=Ak,p, k= 1, . . . , p

1, p >1 .

SinceAk,k= 0 for every k,the system always has a solution.We introduce a sequence of numbersbkk,

bk= sup{2k|(i)k(t)|; i k}, k N .Then, the function

(t) =

p=1p(t)/bp, t Rn, is inF0 and supp K .

Thus, we have

T(t + hk), (t)=

p=1

T(t + hk), p(t)/bp =ak/bk .

Now, if we choosec(h) such that c(hk) =ak/bk, k N,then

T(t + h), (t)

c(h) does not converge to zero whenh , h .

This proves that for every compact setKthere exists a0(K) such that

T(t + h), (t)= 0, h 0(K), h , F0,supp K.

It follows thatT(t+h) = 0 overB(0, r), h(r), h andT(t) = 0over B(h, r), h (r), h .

Remark.The condition of Theorem 1.5 implies that the support of

Thas the following property: The distance from supp Tto a point h, d(suppT, h),tends to infinity whenh , h.

The next proposition shows that if in Definition 1.1 we take the limit

(1.3) instead of the limit (1.2), then a more precise result is obtained.

Theorem 1.6. Let T F0 and be an acute, open and convex cone(partially ordered, see Remark 1 after Definition 1.1). A necessary and

sufficient condition for

suppT

CRn(a+) for somea (1.17)

is that

limh,h

T(x + h)/c(h) = 0 inF0 for everyc(h) (1.18)

(CRnA=Rn \ A).


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Proof.If (1.17) holds, then for any ball B(0, 1),there exists a hrsuch that B(h, r)

(a+) for h

hr, h

.This implies (1.18), and by

Lemma 1.1, (1.17) follows.

Let us suppose now that (1.16) and consequently (1.18) hold, but for

anya,suppTCRn(a+).We fix such an a=a0 >0.There existsan a1(a0+) suppT.Since suppT CRn(2a1+),there exists ana2(2a1+ suppT).In such a way, we construct a sequenceakkinsuch thatak (kak1+) suppTandakka0, kN.

Since aksuppT, k N,it follows that there exists a sequencekkinF0such that suppkB(0, 1), k N,and

T(x + ak), k(x) = 0, k N.

We put now:

ck,i=T(x + ai), k(x), k , i N;

bk= supp{2k|(j)(x)|;jk, x Rn}, kN.

We have to prove that there exists a sequenceckksuch that ck1, kNand

k1

ck,i/(bkck) = 0, i N . (1.19)

First, we notice that ck,k= 0.If

k=1

ck,1/bk= 0,then we take c1 >1,

and if this series is different from 0,then we put c1= 1.Leti= 2.If

c1,2/(b1c1) +k2

ck,2/bk= 0 (= 0),we takec2 >1 (c2= 1)

such that

c1,1/(c1b1) + c2,1/(c2b2) +k3

ck,1/bk= 0 .

Leti= 3,

c1,3/(b1c1) + c2,3/(b2c2) +

k=3 ck,3/bk= 0 or = 0 .Then we take c3 >1 and c3= 1 respectively such that

c1,1/(b1c1) + c2,1/(b2c2) + c3,1/(b3c3) +

k=4

ck,1/bk= 0,


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c1,2/(b1c1) + c2,2/(b2c2) + c3,2/(b3c3) +

k=4ck,2/bk= 0.

Continuing in this way, we construct a sequence ckkfor which (1.19)holds.

Let us put

k(x) =k(x)/(bkck), k Nand =ki

k.

From the properties of sequences

bk

kand

ck

k ,we can easily show

that

Nk=1

kinF0when N .

Relation (1.18) implies

T(x + ai), (x)=

k=1 ck,1/(bkck)= 0,i N.We obtain that (1.18) does not hold for .This completes the proof.

In Theorem 1.6 the support ofTcan be justCRn(a +).The question

is: Is it possible to obtain a similar proposition for the S-asymptotics given

by Definition 1.1?This question is analyzed in the next examples. We take

F0=Din which the S-asymptotics by the weak and strong convergenceare equivalent.

Examples.i) Let T(x, y) =m1 m(xm, y).The given series convergesinD(R2).Since for a D(R2), supp B(0, r),we have

limn

n

m=1

m(x m, y), (x, y)=

1mr

m(m, 0).

It follows that Tis a distribution onR2 (see Theorem XIII, Chapter 3

in [146]).

Let us remark that the support ofTlies on the half line {(p, 0)R2; p >0}.We can take for the cone R2+ {(, )R2; >0, >0}.It is a convex, open and acute cone in R2.We shall show that the limit

limh,h

T(u + h), (u)


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does not exist. To do this, it is enough to take the limit over the half line

{(0, 0) +}, 0 >0,which belongs to .If we choose such that >0 and (0, 0) = 1,then for h= (p, 0)

T(u + h), (u)=m1

m(m p, 0) p.

Note, ifh ,then p ,as well. Consequently, the answer to theposed question is negative.

ii) The following example shows that if

limh,h

T(x + h)/c(h) = 0 inD

for every positive c(h),and every ={pw, p >0}, w,then this doesnot imply that

limh,h

T(x + h)/c(h) = 0 inD.Let us remark that both limits on a , whenh orh ,are equal.

LetTbe given byT(x, y) =

m1

m(x m, y 1m) on R2.The support

ofTlies on the curve

{(x, 1/x);x >0

}.Let =R2+, w= (cos, sin), 00}.Then, for a D(R2)T(x+h1, y+h2), (x, y)=

m1

m

mh1,1

mh2

0, h , h.

In order to show that

limh,h

T(x + h1, y+ h2), (x, y)does not exist, takehto belong to ={(x, a); x >0}for a fixeda >0,as

we did in example i).Remarks.As a consequence of Theorem 1.5 and Theorem 1.6, we have

some results concerning the convolution product (see [135], p. 102).

Let

G1 {f C; supp fCRn{h ; h f} ,G2 {f C; supp fCRn{h ;h hf}.

Corollary 1.1. For a fixedT D

, the convolutionT mapsD into G1if and only if the support ofThas the property given in Theorem 1.5.

Corollary 1.2. For a fixed T D and for a convex, open and partiallyordered conethe convolutionTmapsDintoG2 if and only ifsuppTCRn(a+) for somea.


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1.8 Characterization of some generalized function spaces

Theorem 1.7. A necessary and sufficient condition for a distributionT tobelong to:

a)E is thatT(x + h) sc(h) 0, h Rn for every positivec.b)Oc is that T has S-asymptotic behavior related to every c(h) =

h, R+ and with limitU= 0.c)B is that T has S-asymptotic behavior related to every positive

c, c(h)

,

h

, and with limitU= 0.

Proof.a) It is a direct consequence of Theorem 1.5.

b) It is enough to apply Theorem IX Chapter VII in [146] which states:

A necessary and sufficient condition that a distribution T belongs to Ocis that for every D the function(T )(h) is continuous and of fastdescent at infinity.(see0.5.1).

c) By Theorem XXV, Chapter VI in [146] a distribution T

B if and

only ifT L(Rn) for every D.Suppose that T B.Then forevery Dand c(h) , h

limh,h

T(x + h)

c(h) , (x)

= lim

h,h

(T)(h)c(h)

= 0 .

Suppose that (T)(h)/c(h)0, h ,for every Dand for

every c, c(h) ash .We will show that (T)(h) L

(R

n

)for every D.Then, by the same theorem, it follows that T B.Let us assume the contrary, i.e., that (To)(h) is not bounded for

a o D.Then, we could find two sequenceshmmin Rn andcmmin Rsuch that|cm| as m ,hm mand (To)(hm) = cm.Now, for co(h) such that co(hm) =

|cm|, m N,the limit < T(x+h)/co(h), o(x) >would not exist, ash .This is in a contradictionwith our assumption that Thas the S-asymptotics related to every c(h)

which tends to infinity ash .

Proposition 1.7. a) If for every rapidly decreasing functionc, T has the

S-asymptotic behavior related toc1 and with limitUc(Uc= 0is included),

thenT S.


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b) If for every rapidly exponentially decreasing function c (for every

k >0, c(h)exp(kh) 0, h ) a distributionThas the S-asymptoticbehavior related to c

1

with limitUc(Uc= 0 is included), thenT K1.

Proof.a) Letcbe given. There exists0such that for everyh,h 0,and for every D

|< T(x + h) c(h), (x)>| |< Uc, >|+M+ .Therefore, the set{T(x+h) c(h);h 0}is weakly bounded and thusbounded inD.By Theorem VI 4o,Chapter VII in [146] if{c(h)T(x+

h); h> 0}is bounded inD

,for every cof fast descent, thenT S

.b) The proof is similar to that of a), if we use the following theorem

proved which will be showed below (cf. Theorem 1.15):

LetT D. If for every rapidly exponentially decreasing functionr onRn the set{r(h)T(x + h);h Rn} is bounded inD, thenT K1.

1.9 Structural theorems for S-asymptotics inF

In the analysis of the S-asymptotics and its applications, it is useful to

know analytical expression for generalized functions having S-asymptotics,

especially if it is given via continuous functions and their derivatives. The

next theorems are of this kind. These types of theorems are usually referred

as structural theorems.

Our first result concerns the one-dimensional case, a refinement will be

obtained in Theorem 1.9 below.

Theorem 1.8. Let R and letL be a slowly varying function. Supposethatf D(R). Iff(x + h) sc(h) ex, hR+, then there is anm0Nsuch that for everym m0 the following holds:

(1) Let = 0 and c(h) = ehL(eh), h >0. Then there are gm,iC(1, ), i= 0, 1, . . . , m such that

f(x) =m

i=0 g(i)m,i(x), x (1, ),and

gm,i(x)Cixm exp(x)L(exp x), x , i= 0, . . . , m ,whereCi are suitable constants.


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(2) Let= 0 andc(h) =hL(h), h >0. Then,

a) If >

1, then there is Fm

C(1,

) such that f= F

(m)m and

Fm(x) xm+L(x), x ;b) If 1, then there are fm,i C(1, ) and Am,i = 0, i=

0, 1, . . . , m , such that

fm,i(x)Am,ixm+iL(x), i= 0, 1, . . . , mand

f(x) =m

i=0 f(mi)m,i (x), x (1, ) .Proof.Case= 0.By the remark 1. after Proposition 1.2, there exists

a function cC(R) such that c(h)/c(x +h)exp(x), h in Rand c(i)(x + h)/c(h) i exp(x), i N0.

Since f(x+h)/c(h)exp(x) with strong convergence and the set{c(h)(x)/c(x + h);h A}is bounded inD(R),we can then use TheoremXI,Chapter III in [146] to obtain

limh

f(x + h)c(x + h)

, (x)= limh

f(x + h)c(h)

, c(h)

c(c + h)(x)

=1, (x), D(R) .Let C, (x) = 0 for x 1.We have

(x + h)f(x + h)

c(x + h) 1 in D(R) as h .

Thus,{(x+h)f(x+h)/c(x+h);h >0} is a bounded subset ofD(R).Thisimplies that this set is bounded inS(R) as well (cf. Theorem XXV, Chap-ter VI in [146]). SinceD(R) is dense inS(R),by the BanachSteinhaustheorem, we obtain

(x + h)f(x + h)

c(x + h) 1 in S(R) as h , i.e.,

limh

(f /c)(x + h)d(h)

, (x) =1, , for every S(R),where d(h) = 1, h > A.We shall now borrow two theorems about quasi-

asymptotics (see Definition 2.2) which will be shown later. Since the S-

asymptotics inS+with >1 implies the quasi-asymptotics of f /c


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(see Theorem 2.47), the structural theorem for the quasi-asymptotics (see

Theorem 2.2) implies that there is m0 Nsuch that for every m > m0there is FmC(R) such that

(f /c)(x) =F(m)m (x), x R,andFm(x)xm as x .Thus, we obtain

f(x) = c(x)F(m)m (x), x (1, ).The Leibnitz formula implies

f(x) =

mi=0mi (1)i(c(i)(x)Fm(x))(mi), x (1, ).

Since

c(i)(x+ h)

c(h) iex, h , x R,

we obtain

c(i)(h)

ic(h), h

.

This implies the result when = 0 and c(h) =ehL(eh), h >0.In case= 0 andc(h) =hL(h), h >0, > 1,the S-asymptotics of

frelated to c(h) =hL(h), h >0, >1,implies the quasi-asymptoticsoffrelated to the same c(h) (see Theorem 2.47). The assertion follows

now from Theorem 2.2.

If 1,then we take k >0 such that k+ >1.With as in thepreceding proof and by Theorem 1.2 b), we have

(1 + (x+ h)2)k/2(x+ h)f(x + h) s hk+L(h) 1, h R+.

By the same arguments as in the preceding proof, we have that there is

m0 Nsuch that for every m > m0there is an Fm C(R), suppFm(0, )

Fm(x)x+k+mL(x), x

(1 + x2)k/2(x)f(x) =F(m)m (x), x R.Thus, for x (1, ),

f(x) =

mi=0

(1)i

m

i

1

(1 + x2)k/2

(i)Fm(x)

(mi).


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The result now follows from the fact

1(1 + x2)k/2(i)

Cixki

, x ,whereCi= 0 are suitable constants,i= 0, . . . , m .

We can show a more general result which describes the precise structure

of a distribution having S-asymptotics.

Theorem 1.9. If T D has S-asymptotics related to the open coneandthe continuous and positive function c(h), h, then for the ball B(0, r)there exist continuous functions F

i,|i|

m, such that Fi(x+h)/c(h) con-

verges uniformly forxB(0, r) whenh, h , and the restrictionof the distributionT onB(0, r)+can be given in the formT=

|i|m

DiFi.

We need the following lemma for the proof of the theorem:

Lemma 1.2. Let T D, T(x+h) s c(h)U(x), h. Then, for anopen ball B(0, r) and a relatively compact open neighborhood of zero in

Rn

, there exists an m N0 such that for every , Dm the function

(T)(x)is continuous forx B(0, r)+.Moreover, the set of functions{(Th )(x);h}converges uniformly forx B(0, r)to(U)(x),ash , h ;Th=T(x +h)/c(h).

Proof.Suppose thatThas S-asymptotics related toc(h).Then, the set

{T(x+h)/c(h) Th;h}is weakly bounded inD and consequently,bounded inD.A necessary and sufficient condition that a setB D isbounded inD

is: for every Dthe set of functions{T ;T B

}is bounded on every compact set K Rn (see7, Chapter VI in [146]).Moreover{T ;T B}defines a bounded set of regular distributions.

LetCl() =K(Cl() is the closure of ); Kis a compact set. For a

fixed C0 , supp K,the linear mappings (Th ) , h,are continuous mappings ofDKintoEbecause of the separate continuityof the convolution. Since the set{Th ; h}is a bounded set inD,forevery ball B(0, r) the set of mappings

{(Th

)

;h

}is the set

of equicontinuous mappings ofDKintoLB,whereB=B(0, r).Now thereexists anmN0such that the linear mappings (, )Th whichmapDK DKinto LBcan be extended toDm Dmin such a way that(, )Th , h,are equicontinuous mappings ofDm DmintoLB(see for example the proof of Theorem XXII, Chapter VI in [146]).


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We proved that for every , Dmand every h the functionsx (Th )(x) are continuous in x B(0, r).From the relation(Th )(x) = (T )(x+h)/c(h) and from the properties ofcitfollows thaty(T )(y) is a continuous function for y B(0, r) + and, Dm .

It remains to prove thatTh converges toUas h , h,in LBfor , Dm .We know thatDis a dense subset ofDm, m0.We can construct a subset AofDKto be dense inDm. The set of functionsThconverges inLBfor, A,whenh , h .Taking care ofthe equicontinuity of the mappings

Dm

DmintoL

B ,defined byTh

,

we can use the BanachSteinhaus theorem to prove thatThconvergesin LBwhen h , h .

Proof of the Theorem.We shall use (VI, 6; 23) from [146].

2k (E E T) 2k (E T) + ( T) =T , (1.20)

whereEis a solution of the iterated Laplace equation kE=; , D.We have only to choose the natural number klarge enough so that Ebelongs

toDm .Now, it is possible to take F1= E E T, F2=E TandF3= T.All of these functions are of the formFi=Ti , i, , Dm , i= 1, 2, 3.

The following holds:

Fi(x + h)/c(h) = (Fi(x)/c(h)) h= (T i i, )(x) h/c(h)= ((T(x) h)/c(h) (i i) = (Th i i)(x), x B(0, r), h .

Hence, by Lemma 1.2 it follows that Fi(x+h)/c(h) converges uniformly

forx B(0, r) whenh, h .

Consequences of Theorem 1.9

a) If the functions Fi, |i| m,have the property given in Theorem 1.9and if =Rn,then the regular distributions defined by the functionsFi/c

have the S-asymptotic behavior related to c1(h) 1.

b) If T S,then all functions Fi, |i| m,are continuous for xB(0, r) + ,and of slow growth (Fi= (1 +r2)qfi, |i| m, q R,wherer= xand fi,|i|m,are continuous and bounded functions).

c) The converse of Theorem 1.9 is also true. Therefore, it completely

characterizes those distributions having S-asymptotics.


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Proof.a) Suppose that = Rn.Then by the properties ofFi, |i| m,the functions Fi/care continuous andFi(x)/c(x) converge to the numbers

Ciasx .Now, for|i| mwe havelim

hFi(x + h)/c(x + h), (x)

= limh

Rn

(Fi(x + h)/c(x + h))(x)dx= Ci, , D.

b) IfT S,then there exists a q Rsuch that the set of distributions

{T(x+h)/(1 +

h2)q;h

Rn

}= Wis bounded in

D (Theorem VI,

Chapter VII in [146]). We can now repeat the first part of the proof of

Lemma 1.2 but with c(h) = (1 +h2)q, h = Rn.In this may weobtain that there exists a pN0such that for , Dpand xRn thefunction x(T)(x) is continuous and (T)(x)/(1+x2)q, x Ris bounded. It remains only to choose the numberkin (1.20) large enough

so thatE Dmax(m,p) .c) It follows directly from Theorem 1.2 a) and Proposition 1.5.

We can also characterize the structure of ultradistributions having S-asymptotics.

Theorem 1.10.([132]). Let T D . Suppose: (M.1), (M.2) and (M.3)are satisfied byMp; = 1+1,where1Rn is open set and1 Rn isconvex cone;is a subcone of1.ThenThas the S-asymptotics related to c

andif and only if for a given open and relatively compact setA(A ),there exist an ultradifferential operator P(D) of class

and continuous

functionsf1 andf2 onA + such that

limh,h

fi(x + h)/c(h), i= 1, 2,

exist uniformly inx A, i= 1, 2, andT=P(D)f1+ f2 onA + .

Proof.One can easily prove that the condition is sufficient.

The condition is necessary. Suppose that T(x+ h) s

c(h)

U(x), h

inD.Denote by Th= hT /c(h) and by (CB)Athe space of continuousand bounded functions onA.By Theorem 6.10 in [79],Fh: Th, h,are continuous mappings: DBr+ (CB)A.Consequently, Fhare thecontinuous mappings:DK(CB)A,where K=B r .The set {Th; h,h}, >0,is bounded inDbecause of the S-asymptotics ofT.


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Thus, for a fixed DK,the set{Fh(); h,h }is bounded in(CB)A.Since DKis a barrelled space, by the Banach theorem it follows thatthe family of functions{Fh}={Fh; h, h }is equicontinuous.Therefore, there exists Hp, p Nof the form (0.1) and >0 such thatany function in{Fh}maps the neighborhood of zero

V=

DK; sup

xK,||N0

|()(x)|/H||M||<

(1.21)

into the unit ball B(0, 1) in (CB)A.Denote byDHpMpK the completion of

DKunder the normqHpMp(see (0.5)).

Using the extension of a function through its continuity (see [12]), we

shall show that the family{Fh}can be extended onDHpMpK , keeping theuniform continuity; let us denote it by{Fh}.

Let DHpMpK and letjjbe a sequence inDKwhich converges tobeing a Cauchy sequence in the norm qHpMp .We shall prove thatTh jconverges, as j ,in (CB)Auniformly in h,h.It is enoughto prove thatTh jjis a Cauchy sequence in (CB)A.

Every neighborhood of zero Win (CB)Acontains the ball B(0, ) forsome >0.The neighborhood of zero V ,given by (1.21), satisfies ThV W,when h ,h .

Letj0 N0be such that i jVifi, jj0.Then,Th i Th j=Th (i j) W, h ,h .

This proves the existence of the family{Fh}and that limj

Th j=gh (CB)A,h .We shall prove thatgh=Th .The sequencejjconverges to also inEBr+ , as well. So, Th jconverges to Th as

j ,inD,where = Br+(see [79], p. 73). Thus, (Th )|Amust be gh.

It remains to prove that for DHpMpK , Th converges in (CB)Aash,h .Since{Fh}is an equicontinuous family of functions, forevery DHpMpK the set{Th ; h,h }is bounded in (CB)A.By the BanachSteinhaus theorem Th converges in (CB)A.

Now we will give the analytic form of T.Let DQsuch that Qisa compact subset of the interior ofKand qHpMp()


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byf1andw Tby f2.The properties of these two functions follow by theprevious part of the proof.

We pass now to the case of Fourier hyperfunctions.

Theorem 1.11. Letf= [F] Q(Dn), F O((Dn+iI#Dn).Iffhas theS-asymptotics related to c and, then there exist an elliptic local operator

J(D) and functionsqs C(Rn), s , is the set ofn-vectors with entry{1, 1}), of infra exponential type such that:

1. qs(z) O(Dn + iIs),whereDn + iIs is an infinitesimal wedge of type

Dn

+ is0, s.2. f=J(D)

s

qs(x).

3. For everys there exists0>0 such thatqs(x + is)/c(x), s converge asx , x , for every fixed,0< < 0.

Proof.Let

f= F(x + i), F= sgnF

and

F(f) = [R]=

F(F)( i0) .

Then there exists a monotone increasing continuous positive valued function

(r), r 0,which satisfies (0) = 1, (r) , r and such that

|R()|Ckexp(||/(||),1k |Im j |1 ,

wherej= 1, . . . , nand k N(cf. [74], p. 652).By Lemma 1.2 in [73], we can choose an entire function Jof infra

exponential type on Cn which satisfies the estimate:

|J()|Cexp(||/(||)), |Im |1 .Then J2()O(Dn + i{||


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Denote byg=F1(1/J2).By Theorem 8.2.6 in [75], g Q1(Dn).ByProposition 8.4.3. in [75]

f g= R

f( t)g(t)dt Q(Dn)

and

F(f g) =F(f)F(g) .By the Corollaries of Proposition 2 in [165]

f=J0(D)(g f), J0=J2

. (1.22)

We can always assume that there exists R+such thatI= (, )n.Then we denote by I=I , .

By Proposition 8.3.2 in [75] the following assertions are true:

F O(x+iI) and it decreases exponentially outside any conecontaining as a proper subcone.

F(F)

O(x

iI) and it decreases exponentially outside any cone

containing as a proper subcone.

F(F)J2 has the same cited properties asF(F).F(F)J2sO(x iIs) and it decreases exponentially outside any

cone containing and sproper subcone.

F1(F(F)J2s) O(x+ i(I Is)) and it decreases exponentiallyoutside any cone containing sas a proper subcone.

Consider now the Fourier hyperfunctionf ggiven in (1.22):f g=F1(F(f)F(g))

= 1(2)n

Rn

eizF(F)()/J2()d ,

where Iandz Rn + iI.Let be fixed. Then for and zRn + iI,we have

S,(z) = 1(2)n

Rn

eizF(F)()/J2()d;

|S,(z)| 1(2)n

Rn

exyF(F)()/J2()d .


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One can see that S,(z), z Rn +Iare continuable to the realaxis. The corresponding functionsxS,(x) are continuous and of infraexponential type on R

n

.By Lemma 8.4.7 in [75], for x Rn

,S,(x)=S(x + i0),,

(f g)(x) =

S,(x), x Rn .(1.23)

FunctionsS,can be written in the form

S,

(z) =

1

(2)n Rn

eiz

F(

F

)(

)/J2(

)

s(

)d, z

Rn+I

.

Lets .Then functionsS,,s(z) =

1

(2)n

Rn

eizF(F)()/J2()s()d ,

zRn + iI, ,, s ,are also continuable to the real axis. The corre-sponding functions x S,,s(x) are continuous and of infra exponentialtype onR

n

.Moreover, on Rn

,S,,s(x)=S,,s(x + i0) ,

S,(x) =s

S,,s(x) .(1.24)

Let us analyze functions

Is() =J2()ess(), (Rn + i{||0.These functions are elements ofPbecause of

|Is,()|=|J2()|exp

n

i=1sii

ni=1|si(i)|

|J2()|ni=1|si(i) exp(sii)Cexp(

n

i=1|i|), ||


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Denote by

qs(x) = S,,s(x)

=

F1(F(F)J2)(x + i(s)0), x Rn . (1.25)

We prove that functions qs, s,have the properties cited in Theorem1.11.

Property 1 follows from (1.24) and (1.25).

By (1.22) and (1.23) property 2 is satisfied.

It remains to prove property 3.

If f Q(Dn) and P,then f O(Dn +iI),where I isan interval containing zero. We shall use this fact and the properties of

functionsIs,,already analyzed.

For a fixed s there exists 0 >0 such that sbelongs to all in-finitesimal wedges of the formDn + i(

s)0 which appear in (1.25). For

(0, 0],we have

qs(x + is) =

1

(2)n

Rn

ei(x+is)F(F )()J2()si()d

=

1

(2)n

Rn

eixF(F )()F(s,)()d

=

((F ) s,)(x)

=

F

s,

(x) = (f s,)(x).

Now, for every fixed (0, 0],ands,

limx,x

qs(x + is)/c(x) = limx,x

(f

s,)(x)/c(x)

= limx,x

f(t+ x)/c(x), s,(t).

We cite some papers related to this problem: ([131], [132], [155],[151], [123]).


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1.10 S-asymptotic expansions inFg

A sequencennof positive real-valued functionsn(t), t(t0, ), t0 0(defined on (0, t0), t0 >0) is said to be asymptoticif and only ifn+1(t) =o(n(t)), t (t0).The formal series

n1

un(t) is an asymptotic

expansion of the function urelated to the asymptotic sequence{p(t)}if

u(t) k

n=1

un(t) =o(k(t)), t (t0) (1.26)

for everykN.We write in this case:

u(t)

n=1

un(t)|{n(t)}, t (t0). (1.27)

Ifun(t) =cnn(t),for every nN, where cnare complex numbers, thenexpansion (1.27) is unique. Indeed, the numbers cncan be unambiguously

computed from (1.26). In this case, we omit from the notation{

n(t)}in

(1.27). A series which is an asymptotic expansion of a function fcan be

also convergent. However, the series is divergent in general; nevertheless,

several terms of it can give valuable information, and very often its good

approximation properties come actually from the fact that the series is

divergent. Sometimes, if we take more terms from the asymptotic series,

we obtain a worse approximation; consequently, the determination of the

optimal number of terms for a good approximation depends on a careful

analysis of the problem under consideration.

An asymptotic expansion does not determine only one function. The

following example illustrates this fact:

exp

1

x

k=0

xk

k! and exp

1

x

+ exp(x2)

k=0

xk

k! , x .

In many problems of applied mathematics one is led to the use of asymp-

totic series. (See [44], [10], [15], [14], [192], [69]). A clear exposition of thetheory and the use of asymptotic series of functions and distributions can

be found in ([50][56]).

We shall discuss in this section the S-asymptotic expansion of general-

ized functions belonging toFg.


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1.10.1 General definitions and assertions

In this section will be a convex cone with the vertex at zero belonging toRd and () the set of all real-valued and positive functions c(h), h.We shall consider the asymptotic expansion whenh , h .

Definition 1.2.A distribution T Fghas a S-asymptotic expansion re-lated to the asymptotic sequencecn(h)n (),if for every F

T(t + h), (t)

n=1Un(t, h), (t)|{cn(h)}, h , h , (1.28)

where Un(t, h) Fg(with respect to t) for nNand h.We write inshort:

T(t + h) s

n=1

Un(t, h)|{cn(h)}, h , h . (1.29)

Remarks.1) In the special caseUn(t, h) =un(t)cn(h), un Fg, nN,we simply write

T(t + h) s

n=1

un(t)cn(h), h , h . (1.30)

In this case the given S-asymptotic expansion is unique.

2) Brychkovs general definition is inS(R) and slightly different fromours ([15], [16] and [20]); his idea reformulated inFg(R) gives the followingdefinition. Suppose thatf

Fg(R) and that the function exp(ixh),where

his a real parameter, is a multiplier inFg(R).

Definition 1.3.Suppose that f Fg(R).It is said that f(x)eixh has anasymptotic expansion related to the asymptotic sequencen(h)nif forevery Fg(R)

f(x)eixh, (x)

n=1Cn(x, h), (x)|{n(h)}, h ,

where Cn(x, h) Fg(R) (with respect to x), nN, hh0.We write inshort:

f(x)eixh

n=1

Cn(x, h)|{n(h)}, h .


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To obtain an equivalent definition of this asymptotic expansion,

Brychkov has supposed thatFg(R) =S(R) andFg(R) =S(R).Then, byputtingg=f=F(f), cn(, t) =F(Cn(, t)) and =F(),Definition 1.3reduces to:

Definition 1.4.A distribution g S(R) has an asymptotic expansionrelated to the sequencen(h)nif for every S(R)

g(h t), (t)

n=1

cn(t, h), (t)|{n(h)}, h ,

wherecn(, h) S(R), n Nandhh0.

Definition 1.4 is, of course, a particular case of Definition 1.2 if we use

T(t) =g(t) and the cone =R.In [15] and [20] authors studied asymptotic expansions of tempered dis-

tributions given by Definition 1.3. In [16] Brychkov extended Definition 1.4

to then-dimensional case, but only on a ray{y; >0}for a fixedy Rn.We study in this section the asymptotic expansion not only inS(R)

and not only on a ray, but on a cone in Rn.The next remarks state some

motivations for such investigations.

Remarks.1) A distribution inS(R) can have an S-asymptotic expan-sion inD(R) without having the same S-asymptotic expansion inS(R).Such an example is the regular distribution fdefined by

f(t) =H(t) exp(1/(1 + t2))exp(

t), t

R ,

whereHis the Heaviside function.

It is easy to prove that forh R+

f(t+h) s

n=1

1

(n 1)! (1+(t+h)2)1n exp(th)|{ehh2(1n)}, h ,

while

tUn(t, h) = (1 + (t+ h)2

)1n

exp(t h), n N, h >0do not belong toS(R).

2) The regular distribution gdefined by

g(t) = exp(1/(1 + t2))exp(t), tR


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belongs toD(R) but not toS(R).It has the S-asymptotic expansion inD(R) :

g(t+ h) s

n=1

1

(n 1)! (1 + (t+ h)2)1n exp(t+ h)|{ehh2(1n)}, h ,

where =R+.

3) We can distinguish two cases of S-asymptotic expansions. If in

Definition 1.2,

Un(t, h) =un(t)cn(h), nN ,

then the S-asymptotic expansion iscalled of second type. (See Remark

after Definition 1.2). IfUn(t, h) =un(t+h), n N,then theS-asymptoticexpansion is of first type.

The following example illustrates the difference between these two types

of S-asymptotic expansions.

Let f(x) = x2 + x , x >0 and f(x) = 0, x0.A S-asymptoticexpansion of the first type for this distribution is

f(x + h) s

n=1

1/2

n 1(x + h)2n|{h2n}, h ,

but the sequenceun(x) =

1/2

n1x2n cannot give a S-asymptotic expansion

of the second type for f.S-asymptotic expansions have similar properties as those of S-

asymptotics.

Theorem 1.12. LetT Fg and

T(t + h) s

n=1 Un(t, h)|{cn(h)},h , h .Then:

a) T(k)(t+ h) s

n=1

U(k)n(t, h)|{cn(h)},h , h .


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b) Let the open set have the property: for everyr >0 there exists a

0 such that the closed ball B(0, r) ={xRn, x r} is in{ h, h, h 0}. IfT, T1 F

0 andT1=T over, then

T1(t + h) s

n=1

Un(t, h)|{cn(h)},h , h ,

as well.

c) Assume additionally thatFg is a Montel space and that inFg theconvolution is well defined (see 0.6) and hypocontinuous. Let S Fg,supp S being compact. Then

(S T)(t+ h) s n=1

(S Un)(t, h)|{cn(h)},h , h .

Proof.We prove only a). The proofs of b) and c) are the same as in the

proof of Theorem 1.2. We have

limh,h

T(k)(t+ h)

mn=1

U(k)n(t, h), (t)

cm(h)

= limh,h

T(t + h)

mn=1

Un(t, h), (1)|k|(k)(t)

cm(h) = 0 .

A relation between the asymptotic expansion of a locally integrable func-

tion fand its S-asymptotic expansion when seen as a regular generalized

function is provided in the following proposition (see [152]).

Proposition 1.8. Let f(t), Un(t, h) and Vn(t), t Rn, n N, h,be locally integrable functions such that for every compact setKRn the

following ordinary asymptotic expansion holds,

f(t + h)

n=1

Un(t, h)|{cn(h)},h , h, t K

and for everykN1

ck(h) f(t + h) k

n=1 Un(t, h) Vk(t), t K, h , h r(k, K) .Then forf F0, we have

f(t + h) s

n=1

Un(t, h)|{cn(h)},h , h .


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Proposition 1.9. Suppose that has the nonempty interior. LetT Fand

T(t + h) s

n=1

un(t)cn(h),h , h .

Ifum= 0, m N, thenum has the form

um(t) =

mk=1

Pmk(t) exp(ak t), tRn, mN ,

where ak = (ak1 , . . . , akn) Rn and Pmk are polynomials with degrees less

thank at every ti, i= 1, . . . , n .

Proof.Definition 1.2 and the given asymptotics implies

limh,h

T(t + h)/c1(h) =u1(t) = 0 in F .

Now Proposition 1.2 implies the explicit form ofu1.

The following limit givesu2:

limh,h

T(t + h), (t) u1(t), (t)c1(h)c2(h)

=

u2, ,

F.

Note,

limh,h

(Dti a1i )T(t + h), (t)c2(h)

=(Dti a1i )u2(t), (t), F.

Two cases are possible.

a) If (Dti a1i )u2= 0, i= 1, . . . , n ,then u2(t) =C2exp(a1 t).

b) If (Dti a1

i )u2= 0 for some i,then by Proposition 1.2, (Dtia1i )u2(t) =c exp(a2 t) and u2has the formC2exp(a

1 t) + P22 (t1, . . . , tn)exp(a2 t) ,where P22is a polynomial of the degree less than 2 with respect to each

ti, i= 1, . . . , n .

In the same way, we prove the assertion for every um.

We will give an example of a function which has S-asymptotic expansion

of the first type but does not have the asymptotic expansion as a function:

Let (t) = 1, t(n 2n, n+ 2n), n N,and (t) = 0 outside ofthese intervals. Let

(x) =ex

x0

(t)dt, x R, R .


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Sincex0

(t)dt2 as x ,we have that (x)2ex, x but(x) does not have an ordinary asymptotic behavior (see Example 5. in

1.5.1).Letjjbe a strictly decreasing sequence of positive numbers. Let

be a function, C, 1 for x >1, 0 for x


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We write in short

Asymtotic Behaior

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