Completeness of Random Exponential System in Half-strip

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1 23-24 Prepri for School of Mahemaical Scieces, Beijig Normal Uiversiy Compleeess of Radom Expoeial Sysem i Half-srip Gao ZhiQiag, Deg GuaTie ad Ke SiYu School of Mahemaical Scieces, Laboraory of Mahemaics ad Complex Sysems,Miisry of Educaio, Beijig Normal Uiversiy, Beijig, 875, Chia ( gaozq@bu.edu.c) Absrac Necessary ad sufficie codiios are obaied for he compleeess of some radom expoeial sysem i he Baach space H α, which cosiss of all fucios coiuous o he closed righ half srip, aalyic i is ierior ad vaishig uiformly a ifiiy. Keywords: radom Compleeess, Radom Expoeial Sysem, Half-srip. MSC(2): 3E5, 4A3, 42A6, 42C3. Iroducio The probabilisic approach o classical quesios o expoeial sysems gives a ew isigh ad leads o ew resuls ([-5]) combiig he mehods of probabiliy heory ad fucio heory. Moivaed by heir works, we obai some ecessary ad sufficie codiios for he compleeess of radom expoeial sysem i some Baach space. (O he defiiio of compleeess of a expoeial sysem, he readers ca refer o [6].) Le C = {z = x + iy : x > } ad I α = {z = x+iy : x, y α} (α > ). Deoe by he Baach space H α he class cosisig of all fucios f(z) coiuous o he closed righ half-srip I α, aalyic i is ierior i(i α ) ad lim sup f(x + iy) =, x + y α wih he orm f α = sup{ f(z) : z I α } for f H α. Le Λ = {(λ, ) : λ = λ e iθ } = be a sequece of vecors i C N + saisfyig Reλ Reλ k, k,, k =, 2,, () Θ(Λ) = sup{ θ : =, 2, } < π 2 (2) ad D(Λ) = lim sup (r)/r < +, (3) r + The projec is parially suppored by he Naioal Naural Sciece Foudaio of Chia (NSFC, Gras No. 39, No. 744, No. 2745), by a cooperaio program bewee NSFC ad CNRS of Frace (Gra No. 333), by he Fudameal Research Fuds for he Ceral Uiversiies ad by Hua Provicial Naural Sciece Foudaio of Chia (Gra No. JJ2).

2 GAO ZhiQiag where is regarded as he mulipliciy of he poi λ i he sequece Λ, (r) = ad D(Λ) are called he couig fucio ad he upper desiy of Λ respecively. Le {ξ (ω)} = be a sequece of real radom variables defied o a probabiliy space (Ω, F, P), ad here exis cosas σ, σ 2 ad σ 3 wih σ > σ 2, σ 3 <, such ha { E( ξ Eξ +σ } ) a = sup (Reλ ) σ : =, 2, < + ; (4) 2 { Eξ a 2 = sup (Reλ ) σ 3 } : =, 2, < +, (5) where Eξ deoes he mahemaical expecaio of he radom variable ξ. ad Uder he above assumpio, we defie λ (ω) = λ + iξ (ω), Λ ω = {(λ (ω), )} =, (6) E(Λ ω ) = {z l exp( λ (ω)z)} l=, N +. (7) Obviously, E(Λ ω ) is a subse of H α for each ω Ω, we he ask wheher E(Λ ω ) is complee i H α i he α. Our coclusios are as follows: Theorem Le Λ ω ad E(Λ ω ) be defied by (??) ad (??) respecively, where Λ = {(λ, )} = is a sequece of vecors i C N + saisfyig (??), (??) ad (??), {ξ (ω)} = is a sequece of real radom variables saisfyig (??) ad (??). Defie λ(r) by The (i) If λ(r) = he E(Λ ω ) is complee i H α wih probabiliy. (ii) If he E(Λ ω ) is icomplee i H α wih probabiliy. λ 2. (8) lim sup(λ(r) α log r) = +, (9) r + π lim sup(λ(r) α log r) < +, () r + π Remark Some similar resuls have bee obaied i [3,4,7]. 2. Proof of Theorem I he whole paper, A deoe posiive cosas ad A ω deoe posiive umbers depedig oly o ω which may be differe i differe cases. I order o prove he resul above, we eed he followig lemmas. 2

3 Compleeess of Radom Expoeial Sysem Lemma (Chebyshev Iequaliy [7]) Le ξ be a real radom variable ad f(x) be a icreasig posiive coiuous fucio o (, + ). The, for each a >, P{ ξ > a} E(f( ξ )). f(a) Lemma 2 (Borel-Caelli Lemma [7]) Le {E } = be a sequece of eves from a probabiliy space ad E = k= =k E. If = P(E ) < +, he P(E) = ; if he eves E are idepede ad = P(E ) = +, he P(E) =. Lemma 3 Le Λ = {(λ ; )} = be a sequece of vecors i C N + saisfyig (??), (??) ad (??), he for each posiive umber b, he fucio ( ) m ( z/λ m z G b (z) = Q b (z) exp + m ) z () + z/λ λ λ Reλ >2b is meromorphic ad aalyic i he righ half-plae C b = {z = x + iy : x > b} wih zeros of orders a he pois λ, ad saisfies he followig iequaliy G b (z) exp{2 x φ(r) + A x + A log + r + A}, z = x + iy C b, (2) where Q b (z) = φ(r) = Reλ 2b λ r ( ) m z λ, z + b + ( Reλ λ 2 Reλ ) r 2, (3) ad log + r = {log r, }. Lemma 4 Le Λ = {(λ ; )} = be a sequece of vecors i C N + saisfyig (??), (??) ad (??), {ξ (ω)} = be a sequece of real radom variables o he probabiliy space (Ω, F, P) saisfyig (??) ad (??), he here exiss Ω Ω wih P(Ω ) =, such ha for each ω Ω ad each posiive umber b, he fucio ( ) m ( z/λ (ω) m z G b,ω (z) = Q b,ω (z) exp + z/λ (ω) λ (ω) + m ) z (4) λ (ω) Reλ >2b is meromorphic ad aalyic i he righ half-plae C b wih zeros of orders a he pois λ (ω), ad saisfies he followig iequaliy G b,ω (z) exp{2 x λ(r) + A ω x + A ω log + r + A ω }, z = x + iy C b, (5) where ad λ(r) is defied by (??). Q b,ω (z) = Reλ 2b ( ) m z λ (ω), z + b + 3

4 GAO ZhiQiag Proof of Lemma 3 Sice (??) ad (??) guaraee ha he se { : Reλ 2b} is fiie, Q b (z) is a raioal fucio, so Q b (z) is aalyic ad bouded i C b. Therefore we ca suppose ha Reλ > 2b for all posiive iegers. Le r = z, ad wrie he produc G b (z) = Π (z)π 2 (z), where Π 2 (z) coais he erms wih λ 2r. Le E(z) = ( z)e z. The log Π 2 (z) = λ 2r ( ) ( (log z E log E z ) ) λ λ Sice E(z) = k=2 k k! z k, By (??), ( ) z E ( λ ( ) E z 2 E z ) λ λ k=2 ( 2 z ). λ z 2 λ z + λ exp = m + λ 2 d() r 2 < +, k k! ( ) k z λ where r = mi{reλ : =, 2, }. Thus G b (z) is he quoie of coverge caoical producs. As a resul, he produc (??) defies a meromorphic fucio i he complex plae C, which has zeros of orders a he pois λ. We also observe ha, for z = re iθ, λ = λ e iθ, Re ( ( z λ ) k ( z ) ) k = 2 si k(θ + π λ 2 ) si k(θ + π ( ) k r 2 ) λ ad si k(θ + π 2 ) k si(θ + π 2 ) = k cos θ. The for λ 2 z = 2r, = = ( ) ( z log E log λ E z ) λ ( (log Re z ) ( log + z λ λ 2 k si k(θ + π 2 ) si k(θ + π 2 ) k=2 4 x r cos θ λ 2. ) + z ( r λ + z ) λ λ ) k Therefore, + d() log Π 2 (z) 4 x r 2r 2 A x, z C. 4

5 Compleeess of Radom Expoeial Sysem For x, sice log( x) x ( x ), ( 2xReλ log Π (z) = λ log( 4xReλ ) z + λ ) 2 By (??) ad (??), he we have φ(r) φ(r) λ <2r 2x λ <2r 2x R λ <2r 2xφ(3r). r cos Θ(Λ) d() ( Reλ λ 2 Reλ ) z + λ 2 ( Reλ λ 2 Reλ ) (3r) 2 A(log R log r) + A (R > r > r ), (6) log Π (z) 2xφ(r) + Ax. For b x, sice log( + x) x (x ) ad (??) holds, ( log Π (z) 2 log + 4 x Reλ ) z + λ 2 λ <2r 2 x 8b λ <2r 2r r d() z + λ 8b 2 A log + r + A. λ <2r Reλ Therefore, (??) holds from he iequaliies above. This complees he proof of Lemma 3. Proof of Lemma 4 By (??), for each posiive umber σ, we have = m + (Reλ ) +σ = r For a fixed posiive umber τ, cosider he rucaed radom variables d() < +. (7) +σ { ξ ξ Eξ, if ξ Eξ τreλ ; =, oherwise. (8) Le λ (ω) = λ + ieξ + iξ (ω). By he Chebyshev iequaliy, where a is defied by (??). Hece by (??), P{ξ (ξ Eξ )} = P{ ξ Eξ > τreλ } E ξ Eξ +σ τ +σ (Reλ ) +σ a τ σ (Reλ ) +σ σ2, P{ξ (ξ Eξ )} < +. = 5

6 GAO ZhiQiag By he Borel-Caelli Lemma, we have { } P {ω : ξ(ω) (ξ (ω) Eξ )} =. Se k= =k Ω = Ω\ he P(Ω ) = ad for each ω Ω, k= =k {ω : ξ(ω) (ξ (ω) Eξ )}, { : λ (ω) λ (ω)} = { : (ξ (ω) Eξ ) ξ (ω)} < +, (9) where E deoes he umber of elemes i he se E. By (??) ad (??), he se E = { : Reλ } is fiie, he we have ad by (??), A = max{ Eξ / Reλ : E } < +, (2) { } Eξ sup Reλ : E { } Eξ sup (Reλ ) : E σ3 a 2, (2) where a 2 is defied by (??). Hece by (??), (??) ad (??), arg λ (ω) = arca Imλ + Eξ + ξ Reλ ( Imλ arca Reλ + Eξ Reλ + ξ ) Reλ τ arca(a Θ(Λ) + A + a 2 + cos Θ(Λ) ) A 2 < π 2, where A 2 is a posiive cosa idepede of. The akig accou of (??), for each ω Ω, here exiss a posiive umber Θ(Λ ω ) such ha arg λ (ω) max{{ arg λ (ω) : λ (ω) λ (ω)} {A 2 }} = Θ(Λ ω ) < π 2. Noe ha Reλ (ω) = Reλ (ω Ω), he by Lemma 3, for each ω Ω ad each posiive umber b, he fucio G b,ω (z) defied by (??) is meromorphic ad aalyic i he righ halfplae C b wih zeros of orders a he pois λ (ω), ad saisfies he followig iequaliy where G b,ω (z) exp{2 x φ ω (r) + A ω x + A ω log + r + A ω }, z = x + iy C b, φ ω (r) = λ (ω) r ( Reλ λ (ω) 2 Reλ ) r 2. (22) To complee he proof, we oly eed o prove ha φ ω (r) λ(r) A ω holds for almos every ω Ω. Le b = a Θ(Λ) + τ, b 2 = / + b 2, he b 2 λ (ω) Reλ λ (ω). (23) 6

7 Compleeess of Radom Expoeial Sysem By (??) ad (??), + φ ω (r) λ(r) ( Reλ λ λ (ω) 2 Reλ ) r 2 Reλ λ 2 (ω) r Reλ r + A ω Reλ λ Reλ r (ω) 2 Reλ λ 2 + Reλ λ Reλ r λ (ω) 2 (ω) r r 2 + A ω Reλ λ (ω) 2 Reλ λ 2 Reλ r Reλ r + b 2r< λ (ω) 2 + = Roma + Roma 2 + A ω. r 2 + A ω Nex we esimae Roma ad Roma 2. By (??), (??), (??) ad (??), Roma = Reλ r (2 a Θ(Λ) + τ) + ( ξ ( ξ + 2 Imλ ) + 2 Eξ ( Imλ + ξ ) + Eξ 2 ) λ 2 λ (ω) 2 Reλ r ξ + (2 a Θ(Λ) + 2τ) (Reλ ) 2 Eξ 2 (Reλ ) 3 (2 a Θ(Λ) + τ) + (2 a Θ(Λ) + 2τ)a 2 (2 a Θ(Λ) + τ) By (??) ad Hölder iequaliy, By (??), (??), (??) ad Fubii Theorem, (Reλ ) 2 σ 3 + a2 2 ξ (Reλ ) 2 + A. ξ (Reλ ) 2 Reλ r (Reλ ) 3 σ 3 Eξ (Reλ ) 2 E ξ E ξ Eξ (E ξ Eξ +σ ) +σ. (24) E(Roma ) (2 a Θ(Λ) + τ) (2 a Θ(Λ) + τ)a < +. Reλ r The here exiss Ω 2 Ω wih P(Ω 2 ) = such ha +σ = E ξ (Reλ ) 2 + A Roma A ω, ω Ω 2. (Reλ ) 2 σ 2/(+σ ) + A 7

8 GAO ZhiQiag By (??) ad (??), for each ω Ω, Roma 2 b 2 2 (2 + b 2 ) b 2r< Reλ r r 2 + r Therefore, we ca ake Ω = Ω Ω 2, he P(Ω ) =, ad which complees he proof of Lemma 4. = (2 + b2 ) (r) r r 2 A. φ ω (r) λ(r) A ω, ω Ω, (25) Proof of Theorem (i): Suppose ha here exiss E Ω wih P(E) > such ha for each ω E, E(Λ ω ) is icomplee i H α. By Hah-Baach Theorem, i is equivale o say ha for each ω E, here exiss a bouded liear fucioal T ω o H α, such ha T ω =, ad T ω (z l e λ(ω)z ) =, l =,,, ; =, 2,. So by he Riesz represeaio Theorem, here exiss a complex measure µ ω saisfyig µ ω (I α ) = T ω = ad T ω (f) = f()dµ ω (), f H α, I α where I α is he boudary of I α ad ω E. Le Γ(z) be he Euler-Gamma fucio, he log Γ( 2 + 2α π z) = 2α π x log+ r α y + C(z), where C(z) A 3 x+a 3 (x ), ad A 3 is a posiive cosa. Because P(E Ω ) = P(E) >, we ca ake oe ϖ E Ω, where Ω is meioed i Lemma 4. Hece he fucio f ϖ (z) = Γ( 2 + 2α π z)e A 3z A 3 I α e zs dµ ϖ (s) is aalyic i he ope righ half-plae C, coiuous o he closed righ half-plae C, ad saisfies f (l) ϖ (λ (ϖ)) =, l =,,, ; =, 2, ; f ϖ (z) exp{ 2α π x log+ r}, z C. By he Carlema formula [8], whe r >, φ ϖ (r) π 2 log f ϖ (re iθ ) cos θdθ πr + 2π π 2 r ( y 2 r 2 ) log f ϖ(iy)f ϖ ( iy) dy + A ϖ, where φ ϖ (r) is defied by (??), he φ ϖ (r) α π log r is bouded above o [, + ). So by (??), (??) holds, which complees he proof of (i). (ii): Suppose (??) holds. For each ω Ω, where Ω is meioed i Lemma 4, le 8 g ω (z) = z2 G b,ω (z) exp( M ω z M ω ) Γ( 2 + 2α π (z + b))(z + b +, )M ω

9 Compleeess of Radom Expoeial Sysem where G b,ω (z) is defied by (??), ad M ω is a large eough posiive umber depedig oly o ω. The g ω (z) is aalyic i he righ half-plae C b, ad saisfies Le g (l) ω (λ (ω)) =, l =,,, ; =, 2, ; (26) g ω (z) h ω,k (z) = + z 2 exp{α y αx}, z C b. (27) + g ω (i k ) exp{zi k }i k d, (28) he h ω,k (z) is aalyic i ope half-plae C k = {z = x + iy : Re(i k z) < ( ) k α k }, coiuous ad bouded i he closed half-plae C k = {Re(i k z) ( ) k α k }, ad h ω, (z) = h ω, (z), z C C ; h ω, (z) = h ω, (z), z C C. The h ω, (z) ca be aalyically coiued o a aalyic fucio h ω (z) i he regio C C C. Le δ = b/2, he for each s R, Γ δ = {ζ : ζ is = δ} C b. By Cauchy formula ad (??), g ω (j) (is) = j! 2π g ω (is + δe iθ ) 2π (δe iθ ) j dθ je α(s+2δ) δ j ( + (s + δ) 2, j =, 2. (29) ) By (??) ad (??), he Thus, Similarly, we have so h(z) L ( I α ). Le h ω, ( + iα) = i α + i = (i α) 2 g ω(is)e s(i α) ds + h ω, ( + iα) πe2αδ δ 2 i α h ω, ( + iα) d < +. h ω, ( iα) d < +, g ω(is)e s(i α) ds, gω(z) = h ω (s)e zs ds, z C. (3) 2πi I α We will prove ha g ω(z) = g ω (z) (z C ). Sice boh g ω(z) ad g ω (z) are aalyic i C, we oly eed o prove ha g ω(x) = g ω (x) (x > ). I fac, by Cauchy formula ad (??), g ω(x) = 2π + h ω (i)e ix d. By Riema-Lebesgue Theorem [9], g ω (x + ) lim R + ϵ > x si R d ( ϵ (, x)). 9

10 GAO ZhiQiag By (??), g ω(x) = lim R 2π = lim R π = lim R = lim R π π x <ϵ R g ω (u) e (u x)i d du R si R(u x) g ω (u) du u x si R g ω (x + ) d si R g ω (x + ) d, he gω(x) g ω (x) = lim R si R (g ω (x + ) g ω (x)) π <ϵ d. Le ϵ, we obai ha gω(x) = g ω (x) (x > ). Thus g ω (z) = h ω (s)e zs ds, 2πi I α z C. (3) So by (??) ad (??), he ozero liear fucioal T ω defied by T ω (f) = h ω (s)f(s)ds, f H α 2πi I α saisfies ad T ω ( l e λ(ω)z ) = ( ) l g (l) ω (λ (ω)) = T ω = h ω (s) ds < +. 2π I α By Hah-Baach Theorem, E(Λ ω ) is icomplee i H α for each ω Ω. Tha is o say, E(Λ ω ) is icomplee i H α wih probabiliy. This complees he proof of Theorem. Refereces [] Chisyakov G. ad Lyubarskii Yu., Radom perurbaio of expoeial Riesz bases i L 2 ( π, π), A. Is. Fourier (Greoble), 47(), 997, [2] Chisyakov G., Lyubarskii Yu. ad Pasur L., O compleeess of radom expoeials i he Bargma-Fock space, J. Mah. Phys., 42(8), 2, [3] Seip K. ad Ulaovskii A. M., Radom expoeial frames, J. Lodo Mah. Soc. (2), 53(3), 996, [4] Seip K. ad Ulaovskii A. M., The Beurlig-Malliavi desiy of a radom sequece, Proc. Amer. Mah. Soc., 25(6), 997,

11 Compleeess of Radom Expoeial Sysem [5] Gao Zhiqiag ad Deg Guaie, O compleeess ad miimaliy of radom expoeial sysem i a weighed Baach space of fucios coiuous o he real lie, Chi. A. Mah., 27B(3), 26, [6] Youg R. M., A iroducio o oharmoic Fourier series, Academic Press, New York, 98. [7] Shiryayev A. N., Probabiliy, Traslaed from Russia by R. P. Boas Jr., Spriger, New York, 984. [8] Boas R. P., Eire Fucios, Academic Press, New York, 954. [9] kazelso Y., A iroducio o harmoic aalysis, Third ediio, Chia Machie Press, Beijig, 25.