Completeness of Random Exponential Systems in the weighted Banach Space

Yan Feng Department of Mathematics

Handan College Handan, China

e-mail:yan_tian310@126.com

Liu Xiaoling Department of Mathematics

Handan College Handan, China

e-mail: hdsz_lxl@163.com

Abstract—In this paper,the authors investigate the completeness of random exponential systems }{ )( tk net ωλ in

αC , where αC is the weighted Banach space of complex

continuous functions of f on R with ))(exp()( ttf α− vanishing at infinity.

Keywords- Random Exponential Systems; Weighted Banach Space; Completeness (key words)

I. INTRODUCTION

Let )(tα be a nonnegative convex function on R ,henceforth called a weight, satisfying

.)(lim 1 +∞=−

+∞→tt

tα (1)

Given a weight )(tα ,we take a weighted Banach space

0|)(|lim:)({ )(

||=∈= −

∞→

t

tetfRCfC α

α and }|||| ∞<αf ,

where )(RC is the set of complex functions continuous on R and

}|:)(sup{||||| )( Rtetff t ∈= −αα .

Suppose that },2,1,||:;{ ===Λ nem ninnnn

θλλλ is a complex sequence in the open right half-plane

:{C iyxz +==+ }0>x satisfying

,0},2,1:)inf{Re()( 1 >=−=Λ + nnn λλδ∞<=Λ }{sup)( n

nmK (2)

and

.2

},2,1|:sup{|)( πθ <==ΛΘ nn (3)

Let )}({ ωξn be a sequence of independent real random

variables defined on a probability space ),,( PFΩ , satis-fying

+∞<=Ε−Ε= + },2,1:||sup{ 11 nc nn

σξξ , (4)

+∞<=Ε= },2,1|:|/||sup{2 nc nn λξ , (5)

where 0>σ and ξΕ denotes the mathematical expectation of ξ . Under the above assumption, we define

),()( ωξλωλ nnn i+= },2,1:)({ ==Λ nn ωλω ,

},2,1,1,1,0:{)( )( =−==Λℑ nmket ntk n ωλ

ω .(6)

The condition (1) guarantees that αω C)( ⊂Λℑ ,we

then ask whether )( ωΛℑ is complete in αC . We will use a similar method in [1] and [2], the main

conclusions are as follows: Theorem Let )(tα be a nonnegative convex function

satisfying (1), and )( ωΛℑ be defined by (6), where

},2,1,||:;{ ===Λ nem ninnnn

θλλλ is a complex

sequence satisfying (2) and (3), )}({ ωξn is a sequence of independent real random variables satisfying (4) and (5). Define )(rλ as follows:

⎪⎩

⎪⎨

⎧ ≥Ε+= ∑

≤

.,0

Re,Re)( Re

1

otherwise

rifim

r r nn

n

nλλ

ξλλ (7)

Then (i) If Ra∈∀ ,

,1

))((0 2 +∞=

+−

∫∞

dttatλα

(8)

then )( ωΛℑ is complete in αC with probability 1. (ii) If there is some Rb∈ , such that

,1

))((0 2 +∞<

+−

∫∞

dttbtλα

(9)

then )( ωΛℑ is incomplete in αC with probability 1.

II. LEMMA In order to prove Theorem, we need the following

lemmas. And we note that in the whole paper A denotes the constant and ωA denotes the positive number only de-

2010 First ACIS International Symposium on Cryptography, and Network Security, Data Mining and Knowledge Discovery,

E-Commerce and Its Applications, and Embedded Systems

978-0-7695-4332-1/10 $26.00 © 2010 IEEE

DOI 10.1109/CDEE.2010.15

34

pending on ω , whose values may be different in different cases.

Lemma 1 ]3[ Let ξ be a real-valued random variable and )(xf be a non-decreasing positive continuous function. Then 0>∀a ,

.)(|))(|(}|{|P

affa ξξ Ε≤>

Lemma 2 ]3[ Let ,,,, 21 nEEE be a sequence of

events from a probability space and nknk EE ∞=

∞== ∪∩ 1 .

(i) If ∞<∑∞

=1)(P

n nE , then 0)(P =E ;

(ii) If the events nE are independent and

∞=∑∞

=1)(P

n nE , then 1)(P =E .

Lemma 3 ]3[ A sufficient condition for the convergence

of the series ∑∞

=1n nξ of independent random variables with

probability 1 is that both series ∑∞

=1)(D

n nξ and

∑∞

=1)(E

n nξ converge, where )(E nξ and )(D nξ denote

the mathematical expectation and variance of nξ respectively.

Lemma 4 ]4[ If ,1,||:;{ ===Λ nem ninnnn

θλλλ },2 is a sequence of complex numbers satisfying (2) and

(3), then the function

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎟

⎠

⎞

⎜⎜

⎝

⎛

+−

= ∏∞

= n

n

n

n

m

nz

z zmzmzn

n

n

λλλ

λ exp11

)(G1

is analytic in the closed right half-plane }0:{ ≥+==+ xiyxzC , and satisfies the following

inequalities:

+∈+≤ CzAxrxz },)(exp{|)(G| 1λ ,

)(},)(exp{|)(G| 1 ΛΩ∈−≥ zAxrxz λ , where

},2,1,4

)(|:|{)(|,| =Λ≥−∈=ΛΩ= + nzCzzr nδλ ,

⎪⎩

⎪⎨

⎧ ≥= ∑

≤

.,0

Re,||

cosRe)( Re

11

otherwise

rifmr r n

nn

nλλ

λθ

λ

Lemma 5 ]5[ Let )(tβ be a nonnegative convex

function on R satisfying ,/)(lim +∞=∞→

ttt

β and assume

that )(},:)(sup{)(* RtRxxxtt ∈∈−= ββ is the

Young transform ]8[ of the function )(xβ . Suppose that )(rλ is an increasing function on ],0[ ∞ satisfying

).1()1log(log)()( >>+−≤− rRrRArR λλ (10)

Then there exists an analytic function 0)( ≠zf in +C satisfying

)(|)}(|)(exp{|)(| +∈+=−+≤ CiyxzzxxAxAzf λβ (11) if and only if there exists Ra∈ such that

.1

))((0 2

*

+∞<+

−∫

∞+dt

tatλβ

(12)

Lemma 6 Assuming that the hypothesis of Theorem holds. Then there is Ω⊂Ω′ such that 1)P( =Ω′ and for every Ω′∈ω , the function

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎟

⎠

⎞

⎜⎜

⎝

⎛

+−

= ∏∞

= )()(exp

11

)(G1 )(

)(

ωλωλωλ

ωλω

n

n

n

n

m

nz

z zmzmzn

n

n (13)

is analytic in the closed right half-plane +C and satisfies the following inequalities:

+∈+≤ CzxArxz },)(exp{|)(G| ωω λ , (14)

,)(},)(exp{|)(G| ∑ Λ∈−≥ ωωω λ zxArxz (15) where

},2,1,4

)(|)(:|{)(|,| =Λ≥−∈=Λ= +∑ nzCzzr nδωλω

and )(rλ is defined as (7). Proof. For a fixed positive number ,0>τ consider

the truncated random variables

⎩⎨⎧ ≤Ε−Ε−

=.,0

||||,*

otherwiseif nnnnn

n

λτξξξξξ （16）

Let ,**nnnn ii ξξλλ +Ε+= by the Lemma 1 and (4),

{ |}|||P)}(P{ *nnnnnn λτξξξξξ >Ε−=Ε−≠

.|||)|(

)|-(|1

n

)1(1

1n

1nn

σ

σ

σ

σ

λτ

λτξξ

+

+−

+

+

≤ΕΕ≤ c

The separation condition (2) yields ,||

11

1 ∞<∑∞

=+

n nσλ

hence

.)}-(P{1

nn*n ∞<Ε≠∑

∞

=nξξξ （17）

By the Lemma 2 we will have .0)}})(()({P{ *

0 =Ε−≠∞=

∞= nnnknk ξωξωξω：∪∩

Set )},)(()({ *

01 nnnknk ξωξωξω Ε−≠−Ω=Ω ∞=

∞= ：∪∩

then 1)(P 1 =Ω ，and 1Ω∈∀ω ，

35

,)}({#)}()(:{# ** ∞<Ε−≠=≠ nnnnn nn ξξξωλωλ ： (18) where E# denotes the number of elements in the set E .

Since

|Re

Imarctan||)(arg|*

*

n

nnnn λ

ξξλωλ +Ε+=

|),Re

||Re

||ReImarctan(|

*

n

n

n

n

n

n

λξ

λξ

λλ +Ε+≤

and )(|arg| ΛΘ≤nλ yields

,)(cos||

1|Re

1|),(tan|ReIm|

ΛΘ≤ΛΘ≤

nnn

n

λλλλ

we obtain that for all 1Ω∈ω

,2

))(cos

)(arctan(tan|)(arg| 12* πτωλ <≤

ΛΘ++ΛΘ≤ Ac

n

where 1A is a positive constant independent of n . Then by (18), 1Ω∈∀ω , there exists an positive

number )( ωΛΘ , such that

}}{)}()(:)(argmax{{||)(arg| 1* Annnn ∪≠≤ ωλωλωλωλ

.2

)( πω <ΛΘ=

Besides, .0)()}{Re(inf))}()({Re(inf 11 >Λ=−=− ++ δλλωλωλ nnnnnn

Then by Lemma 4 , we have the following estimates:

1Ω∈∀ω , the function )(G zω defined by (13) is

analytic in the closed right half-plane +C and satisfies:

+∈+≤ CzxArxz },)(exp{|)(G| ωωω λ ,

,)(},)(exp{|)(G| ∑ Λ∈−≥ ωωωω λ zxArxz Where

⎪⎩

⎪⎨

⎧ ≥= ∑

≤

.,0

)(Re,)(

Re)( Re

1

otherwise

rifmr r n

n

nλω

ωλωλλ

To complete the proof of Lemma 6 , we only need to prove that ωω λλ Arr ≤− |)()(| holds for almost every

1Ω∈ω . For every 1Ω∈ω ，since (18) holds, then

|)()(| rr λλω −

∑≤

−+Ε+

≤r nnn

n

nii

mλ ξξλRe

*Re| ∑≤

+Ε+r nn

n

n

Aim

λωξλRe

|Re

∑≤

−+Ε+

=r nnn

nn

nii

mλ ξξλ

λRe

2* ||Re[| ωξλ

λ Ai

m

nn

nn +Ε+

|]||

Re2

∑≤ ++Ε++Ε+

+Ε+=r nnnnnnn

nnnnnn

n

mλ λξξλλξλ

ξλξξλRe

2*222

*2*

])Im()][(Re)Im()[(Re])Im(2)[(Re[|

+ ωA

ωλ λ

λξξξ Amr n

nnnnn

n

++Ε+≤ ∑≤Re

3

*2*

)(Re|))Im||(|||2|(|[

ωλξλξλξλτ Acm

n n

nnnnnnn +++ΛΘ

≤ ∑∞

=13

**2

*

3 |||)|||2||||2||||(

)(cos1

ωλξτ AcK

n n

n +ΛΘ

++Λ≤ ∑∞

=12

*

32

||||

)(cos)22)(( .

By (4) and HÖlder inequality , |)(||| *

nnn ξξξ Ε−Ε≤Ε

,)]|(|[ 11

11

11 ∞<≤Ε−Ε≤ +++ σσσξξ cnn

so we have

,||

1||

||1

211

2

*1

1

∞<≤Ε ∑∑∞

=

∞

=

+

n nn n

n cλλ

ξσ

And

≤∑∞

=

)||||(

12

*

n n

nDλξ ∑

∞

=

Ε1

4

2*

||||

n n

n

λξ

.||

1||

||1

311

3

*1

1

∞<≤Ε≤ ∑∑∞

=

∞

=

+

n nn n

n cλ

τλ

ξτσ

Then according to Lemma 3 ,there exists Ω⊂Ω2 such that

1)(P 2 =Ω , and ∞<∑∞

=12

*

||||

n n

n

λξ holds for every 2Ω∈ω .

Let 21 Ω∩Ω=Ω′ , then 1)(P =Ω′ and |)()(| rr λλω −

ωA≤ , for Ω′∈ω , which completes the proof of Lemma 6.

III. PROOF OF THEOREM

(i)Suppose its contrary holds,i.e. there exists Ω⊂E such that 0)E(P > and E∈∀ω , )( ωΛℑ is incomplete

in αC . By Hahn-Banach Theorem, there exists a nonzero

bounded linear functional ωT on αC vanishing on )( ωΛℑ . So by the Riesz representation Theorem, there exists a complex measure ωμ satisfying

1||||)(||)( ==∫∞

∞− ωωα μ Ttde t ,

and

.)()()()( ∫∞

∞−∈= αωω μ ChtdthhT

where E∈ω . Because 0)P()E(P >=Ω′∩ E ,we can

take one Ω′∩∈ Eω , where Ω′ is mentioned in Lemma 6. Then the function

36

∫∞

∞−= )(

)(1)( tdezG

zf tzω

ωω μ

is analytic in the open right half-plane +C , satisfying

},|)(|)(exp{|)(| * xAzxxzf ωω λα +−≤ Where

}:)(sup{)(* Rttxtx ∈−= αα is the Young transform of the function )(xα .

Because the condition (2) yields (10), and from Lemma 5 with )1()( * += xx αβ , we see that there exists Rb∈ such that (9) holds which contradicts to (8). Hence the assertion of (i)follows from the contradiction.

(ii) If (9) holds for some real number b , then according to [1], there exists a function )(zg analytic in +C and satisfying

.1),1())()(1()(Re *1 ≥−−−−≥≥ xxbrxzgxA αλ (19)

Therefore for every Ω′∈ω , taking Lemma 6 and the above inequality into account and properly choosing the number

ωN (only depending on ω ), we can establish that the function

})(exp{)1()()( ωω

ωω ω

NzNzgzzGzg N −−−

+=

satisfies the following inequality:

}.)1(exp{||1

1|)(| *2 xx

zzg −−

+≤ αω (20)

Set

∫∞

∞−

+−+= ,)1(21)( )1( dyeiygth tiy

ωω π

then )(thω is continuous on R . Moreover, by (20) and Cauchy contour theorem,

)0(,)(21)( )( >+= ∫

∞

∞−

+− xdyeiyxgth tiyxωω π

. (21)

Since xteth )(ω can be viewed as Fourier transform of

)( iyxg +ω , so the inverse transform shows

).0(,)(21)( >= ∫

∞

∞−xdtethzg zt

ωω π (22)

Therefor

,0))((2)( )()( ==∫∞

∞−ωλπ ω

ωλω n

ktk gdtetth n

).,2,1,1,1,0( =−= nmk n We obtain from (20), (21) and the formula of the Young

transform αα =**)( that

|},|)(exp{|)(| ttzh −−≤ αω so the linear functional

∫∞

∞−∈= )()()()( αωω ChdtththhT

Satisfies ,0))((2)( )()( == ωλπ ω

ωλω n

ktk getT n

),,2,1,1,1,0( =−= nmk n and

∫∞

∞−>= .0|)(||||| )( dtethT tα

ωω

By Hahn-Banach Theorem, Ω′∈∀ω , the space )( ωΛℑ is incomplete in αC . This completes the proof of

Theorem.

[1] G.T.Deng,“Incompleteness and closure of a linear span of exponential system in a weighted Banach space,” Journal of

Approximation Theory,125,:1-9(2003) [2] G.T.Deng, “On Weighted Polynomial Approximation With Gaps,” Nagoya Math.J,178:55-61(2005) [3] A.N.Shiryayev. Probability,Translated from the Russian by R.P.Boas,New York,Springer(1984) [4] Z.Q.Gao,“Some topics on the completeness of some function systems,”PhD thesis of Beijing Normal University(2006) [5] Malliavin P, “ Sur quelques procédés d'extrapo-lation,” Acta Math, 83: 179--255 (1955) [6] Rockafellar R,“Convex analysis,”Princeton Univ. Press, Princeton, 1970

37