It is proved that if the increasing sequence {kn} n=0..∞
n=0 of nonnegative integers has density greater than 1/2 and D is an arbitrary simply
connected subregion of C\R then the system of Hermite associated functions
Gkn(z) n=0..∞ is complete in the space H(D) of complex functions holomorphic in D

Rusev, Peter

Bulgarian Digital Mathematics Library at IMI-BAS

Serdica Math. J. 26 (2000), 221-228COMPLETE SYSTEMS OF HERMITE ASSOCIATEDFUNCTIONSPeter RusevCommunicated by S. L. TroyanskiAbstract. It is proved that if the increasing sequence {kn}∞n=0 of non-negative integers has density greater than 1/2 and D is an arbitrary simplyconnected subregion ofC\R then the system of Hermite associated functions{Gkn(z)}∞n=0is complete in the space H(D) of complex functions holomor-phic in D.1. Completeness in spaces of holomorphic functions.1.1. A subset X of a topological vector space V is called complete in Vif its linear span is everywhere dense in the space V .Suppose that X is at most denumerable, i.e X = {xn}ωn=0, 0 ≤ ω ≤ ∞.Then X is complete in V iff for every v ∈ V and every neighbourhood U of theorigin of V there exists a linear combination x =∑Nn=0 anxn(0 ≤ N ≤ ω if ω <∞and 0 ≤ N <∞ if ω =∞; a1, a2, a3, . . . , aN are scalars) such that v − x ∈ U .Remark. If {λn}∞n=0 are nonzero scalars then the systems {xn}∞n=0 and{λnxn}∞n=0 are simultaneously complete or incomplete.2000 Mathematics Subject Classification: 30B60, 33C45.Key words: Hermite polynomials, Hermite associated functions, completeness.222 Peter Rusev1.2. Let B be a nonempty open subset of the complex plane C and H(B)be the C-vector space of complex functions holomorphic in B. As usual, weconsider H(B) with the topology of uniform convergence on compact subsets ofB.A sequence {fn(z)}∞n=0 ⊂ H(B) is complete in H(B) iff vor every f ∈H(B), every compact set K ⊂ B and every ε > 0 there exists a “polynomial”p(z) =∑Nn=0 anfn(z)(0 ≤ N < ∞; an ∈ C, n = 0, 1, 2, . . . , N) such that |f(z) −p(z)| < ε whenever z ∈ K.Let γ ⊂ C be a Jordan curve and Hγ be the C-vector space of complexfunctions holomorphic on the closed set Cγ = C\intγ and vanishing at the infinitepoint. That is, Hγ consists of functions F everyone of which is holomorphic inan open set containing Cγ and, in addition, F (∞) = 0.The following statement is a criterion for completeness in spaces of thekind H(B) [4, p. 211, Theorem 17]:(CC) Let D ⊂ C be a simply connected region. A sequence {fn(z)}∞n=0 ⊂H(D) is complete in the space H(D) iff for every rectifiable Jordan curve γ ⊂ Dthe only function F ∈ Hγ, which is orthogonal on γ to each of the functions{fn(z)}∞n=0, is identically zero, i.e. the equalities∫γ fn(z)F (z) dz = 0, n =0, 1, 2, . . . imply F ≡ 0.2. Series in Hermite polynomials.2.1. It is known that the region of convergence of a series of the kind(2.1)∞∑n=0anHn(z), an ∈ C, n = 0, 1, 2, . . . ,where {Hn(z)}∞n=0 are the Hermite polynomials, is either the whole complex planeor a horizontal strip symmetrically situated with respect to the real axis. Moreprecisely, if(2.2) τ0 := − lim supn→∞(2n + 1)−1/2 log |(2n/e)n/2an| > 0,then the series (2.1) is absolutely uniformly convergent on every compact subsetof the region S(τ0) := {z : |ℑz| < τ0} and diverges at every point of the openset C \ S(τ0) [8, 9.2, (5)]. This statement is a corollary of a particular case ofG. Szego¨’s asymptotic formula for the Hermite polynomials [8, Theorem 8.22.7],namely,(2.3) Hn(z) =√2 exp(z2/2)(2n/e)n/2{cos ((2n + 1)1/2z − npi/2) + hn(z)}.Complete systems of Hermite associated functions 223In the above representation, {hn(z)}∞n=1 are entire functions and, more-over, the sequence {n1/2 exp(−|ℑz|√2n+ 1)hn(z)}∞n=1 is uniformly bounded onevery compact subset of the complex plane.2.2. Let the complex function f have a representation by a series inHermite polynomials in the region S(τ0)(0 < τ0 ≤ ∞), i.e. f(z) =∑∞n=0 anHn(z)for every z ∈ S(τ0). Then it is holomorphic in S(τ0) and, moreover,(2.4) an = (√pin!2n)−1∫∞−∞exp(−t2)Hn(t)f(t) dt, n = 0, 1, 2, . . .This statement is a corollary of (2.2) as well as of the following inequalityfor the Hermite polynomials(2.5) |Hn(x)| ≤ (√pin!2n)1/2 exp(x2/2), x ∈ R, n = 0, 1, 2, . . . .Remark. The above inequality is due to O. Sza´sz [7].2.3. Let E(τ0)(0 < τ0 ≤ ∞) be the C-vector space of complex functions fholomorphic in the region S(τ0) and having the property that for every τ ∈ [0, τ0)there exists a constant B = B(f, τ) ≥ 0 such that if z = x + iy ∈ S(τ) := {z :|ℑz| ≤ τ}, then |f(z)| = |f(x+ iy)| ≤ B exp{x2/2− |x|(τ2 − y2)1/2}.A remarkable result due to E. Hille [3] is that a function f ∈ H(S(τ0))can be represented in S(τ0) by a series in Hermite polynomials iff f ∈ E(τ0).3. Hermite associated functions.3.1. A well-known fact is that the system of Hermite polynomials{Hn(z)}∞n=0 is a solution of the difference equation yn+1 − 2zyn + 2nyn−1 =0(n ≥ 1).The system of complex functions {Gn(z)}∞n=0, defined for z ∈ C \ R bythe equalities(3.1) Gn(z) = −∫∞−∞exp(−t2)Hn(t)t− z dt, n = 0, 1, 2, . . . ,is another solution of the same equation. Indeed, for every n ≥ 1,Gn+1(z)− 2zGn(z) + 2nGn−1(z)= −∫∞−∞exp(−t2)t− z {Hn+1(t)− 2zHn(t) + 2nHn−1(t)} dt= −∫∞−∞exp(−t2)t− z {Hn+1(t)− 2tHn(t) + 2nHn−1(t)} dt224 Peter Rusev−2∫∞−∞exp(−t2)Hn(t) dt = 0.We call the functions (3.1) Hermite associated functions. Since they aredefined by means of Cauchy-type integrals it is clear that all they are holomorphicin the open set C \ R.3.2. Let us define the system of complex functions(3.2) G˜n(z, ζ) := (√pin!2n)−1Hn(ζ)Gn(z), n = 0, 1, 2, . . . ,where ζ is an arbitrary complex number. Then Mehler’s generating function[1, II, 10.13,(22)] implies that if |w| < 1 then∞∑n=0G˜n(z, ζ)wn = −∞∑n=0{(√pin!2n)−1Hn(ζ)∫∞−∞exp(−t2)Hn(t)t− z dt}wn= −pi−1/2∫∞−∞exp(−t2)t− z{ ∞∑n=0(n!2n)−1Hn(ζ)Hn(t)wn}dt(3.3) = −pi−1/2 exp{− ζ2w21−w2}√1− w2∫∞−∞(t− z)−1 exp{− t21− t2 +2wtζ1− w2}dt.Let us note that if |w| < 1 then ℜ{(1−w2)−1} > 0 and, therefore, the lastintegral in (3.3) is (absolutely) convergent. Moreover, the change of summationand integration we have just performed is allowed. Indeed, as a corollary of theasymptotic formula (2.3), Sza´sz’s inequality (2.5) as well as Stirling’s formula, wecan conclude that there exists a positive constant M independent of n, t, ζ andw and such that the inequality(n!2n)−1|Hn(ζ)Hn(t)wn| ≤M exp(t2/2 + |ℑζ|√2n+ 1)|w|nholds for each t ∈ R and each n = 0, 1, 2, . . ..Further, we define(3.4) ϕ(t, ζ, w) = exp{−(1 − w2)−1t2 + 2ζwt(1− w2)−1}provided that t ∈ R, ζ ∈ C and w ∈ C \ {−1, 1}.Let W be a region in C determined by the inequality ℜ{(1−w2)−1} > 0.It is easy to see that W is a subregion of C \ {(−∞,−1]⋃[1,∞)}. Indeed, theinequality ℜ{(1−w2)−1} > 0 is equivalent to ℜw2 < 1, i.e. W is nothing but theexterior of the hyperbola with Cartesian equation u2 − v2 = 1(w = u+ iv).Complete systems of Hermite associated functions 225Let us define(3.5) Φ(z, ζ, w) =∫∞−∞ϕ(t, ζ, w)t− z dtand(3.6) G˜(z, ζ, w) = pi−1/2(1−w2)−1/2 exp{−ζ2w2(1− w2)−1}Φ(z, ζ, w)where z ∈ C \ R, ζ ∈ C and w ∈W .It is clear that G˜(z, ζ, w), as a function of w, is holomorphic in the regionW . Moreover, as a corollary of (3.2), (3.3), (3.4), (3.5) and (3.6) we obtain thatits Taylor expansion centered at the origin is(3.7) G˜(z, ζ, w) =∞∑n=0G˜n(z, ζ)wn.Since the unit disk is contained in the regionW , the radius of convergenceof the above power series is at least equal to one.4. The main result. The main result in this paper is the followingstatement:Let k = {kn}∞n=0 be an increasing sequence of positive integers with densitygreater than 1/2, i.e. there exists(4.1) limn→∞nkn= δ(k) > 1/2.Then the system {Gkn(z)}∞n=0 is complete in the space H(D) provided thatD is any simply connected subregion of C \R.P r o o f. Suppose the statement we wish to prove is not true. Then, inview of the completeness criterion (CC), there exists a simply connected regionD˜ ⊂ C \R such that the system {Gkn(z)}∞n=0 is not complete in the space H(D˜).Let ζ ∈ C \ R be fixed. Since Hn(ζ) 6= 0 for every n = 0, 1, 2, . . ., the system{G˜kn(z, ζ)}∞n=0 is also not complete in the space H(D˜). That means (because of(CC)) that there exists a rectifiable Jordan curve γ˜ ⊂ D˜ and a function F˜ ∈ Hγ˜which is not identically zero but∫γ˜ G˜kn(z, ζ)F˜ (z) dz = 0, n = 0, 1, 2, . . . . Wecan assume that the curve γ˜ is negatively oriented.We define the function A(F˜ ; ζ, w) in the region W by the equality(4.2) A(F˜ ; ζ, w) =∫γ˜G˜(z, ζ, w)F˜ (z) dz226 Peter Rusevand, as a corollary of (3.7), we obtain that if |w| < 1 then(4.3) A(F˜ ; ζ, w) =∞∑n=0An(F˜ ; ζ)wn,where An(F˜ ; ζ) =∫γ˜ G˜n(z, ζ)F˜ (z) dz, n = 0, 1, 2, . . . . Then (3.1) and (3.2) im-ply that (n = 0, 1, 2, . . .)(4.4) An(F˜ ; ζ) = −(√pin!2n)−1Hn(ζ)∫γ˜{∫∞−∞exp(−t2)Hn(t)t− z dt}F˜ (z) dz.Since γ˜ is a compact subset of C\R, µ := minz∈γ˜ |ℑz| is positive. Moreoverthe inequality |t− z| ≥ µ holds for every t ∈ R and every z ∈ γ˜. Then from (2.5)it follows that |(t − z)−1 exp(−t2)Hn(t)| ≤ µ−1(√pin!2n)1/2 exp(−t2/2), n =0, 1, 2, . . . ; t ∈ R, z ∈ γ˜.Therefore, the improper integral in (4.4) is absolutely uniformly conver-gent with respect to z on the curve γ˜. After changing the order of integrationin (4.4) we obtain that (n = 0, 1, 2, . . .)An(F˜ ; ζ) = (√pin!2n)−1Hn(ζ)∫∞−∞exp(−t2)Hn(t) dt∫γ˜F˜ (z)(z − t) dz.As a connected set D is contained either in the upper or in the lowerhalf-plane. Since F (∞) = 0 in both cases the Cauchy integral formula gives thatfor every t ∈ R we have ∫γ˜F˜ (z)z − t dz = 2piiF˜ (t).In this way we obtain that An(F˜ ; ζ) = 2piiHn(ζ)an(F˜ )(n = 0, 1, 2, . . .),where(4.5) an(F˜ ) = (√pin!2n)−1∫∞−∞exp(−t2)Hn(t)F˜ (t) dt, n = 0, 1, 2, . . . .Let T be the set of all positive τ such that the function F˜ is holomorphicin the strip S(τ). Since F˜ ∈ H(S(µ)), T is not empty. It is clear that τ0(F˜ ) :=supT < ∞. Otherwise F˜ would be holomorphic in the extended complex planeC and, since F˜ (∞) = 0, it would be identically zero.For every 0 < τ < τ0(F˜ ) the function F˜ is continuous on the closed stripeS(τ) and since F˜ (∞) = 0, it is bounded on S˜(τ). Therefore, F˜ is in the spaceE(τ0(F˜ )), i.e. it has an expansion in Hermite polynomials in the strip S(τ0(F˜ )).In view of (2.4) the coefficients of this expansion are given by the equalities (4.5).Now we claim that the radius of convergence R(F˜ ; ζ) of the power seriesComplete systems of Hermite associated functions 227in (4.3) is exactly equal to one. At first we note that, since 0 < τ0(F˜ ) <∞, (4.5)and (2.2) yield(4.6) −∞ < lim supn→∞(2n + 1)−1/2 log |(2n/e)n/2an(F˜ )| < 0Further, the asymptotic formula (2.3) implies that(4.7) log |Hn(ζ)an(F˜ )| = log |(2n/e)n/2an(F˜ )|+ ηn(ζ),where ηn(ζ) = O(n1/2) when n → ∞. Then, as a corollary of (4.5), (4.6) and(4.7) we have thatlim supn→∞n−1 log |An(F˜ )|= lim supn→∞n−1(2n + 1)1/2(2n+ 1)−1/2 log |(2n/e)n/2an(F˜ )| = 0and, therefore,{R(F˜ ; ζ)}−1= lim supn→∞exp{n−1 log |An(F˜ ; ζ)} = exp{lim supn→∞n−1 log |An(F˜ ; ζ)|} = 1.It is clear that the function (4.2) is holomorphic in the region W . SinceW contains the closed unit disk except the points 1 and −1, at least one of thesepoints is a singular point for the function (4.2).Since the power series in (4.3) cannot be reduced to a polynomial, thecomplementary sequence k˜ of the sequence k with respect to the set of nonnegativeintegers is not finite. Moreover, as it is easy to prove, k˜ also has density andδ(k˜) = 1− δ(k).The sequence k∗ of the indices of the nonzero coefficients of the powerseries in (4.3) is also ifinite. If ∆(k∗) is its maximal density, then ∆(k∗) ≤ δ(k˜)[4, Note I, 2] and, in view of (4.1), we have that ∆(k∗) < 1/2.By a Theorem of G. Po´lya [5, p. 625, Satz IV, a] every closed arc of theunit circle with angular measure 2pi∆(k∗) < pi contains at least one singular pointof the function A(F˜ ; ζ, w). But that is a contradiction since each point exp iθ isa regular point for the power series in( 4.3) provided that 0 < |θ| < pi.It is easy to see that the statement just proved is not true when δ(k) = 1/2.Indeed, let a be a nonzero real number, D ⊂ C \ {(−∞,∞) ∪ {−ia, ia}} be asimply connected region and γ ⊂ D be an arbitrary rectifiable negatively orientedJordan curve. Then, as a corollary of the equalities (3.1) as well as of the Cauchy228 Peter Rusevintegral formula, we obtain that(2pii)−1∫γG2n+1(z)(z2 + a2)−1 dz=∫∞−∞exp(−t2)H2n+1(t)(t2 + a2)−1 dt = 0, n = 0, 1, 2, . . .and, therefore, the system {G2n+1(z)}∞n=0 is not complete in the space H(D).REFERENCES[1] H. Bateman, A. Erde´lyi. Higher Transcendental Functions, I, II, III. NewYork, Mc Graw-Hill BC, 1953.[2] V. Bernstein. Lec¸ons sur les progres re´cent de la the´orie des se´ries deDirichlet. Paris, Gauthier-Villars, 1933.[3] PE. Hille. Contribution to the theory of Hermitian series, II: The repre-sentation problem. Trans. Amer. Math. Soc. 47 (1940), 80–94.[4] B. Ja. Levin. Zeros of Entire Functions. Transl. Math. Monogr. vol. 5,Providence RI, 1964.[5] G. Po´lya. Untersuchungen u¨ber Lu¨cken und Singularita¨ten von Potenzrei-hen. Math. Z. 29 (1929), 549–640.[6] S. Saks, A. Zygmund. Analytic Functions. Monografie Matematiczne vol.28, Warszawa, 1952.[7] O. Sza´sz. On the relative extrema of the Hermitian orthogonal functions.J. Indiana Math. Soc. 25 (1951), 129–134.[8] G. Szego¨. Orthogonal polynomials. Colloquium publications vol. 23, Pro-vidence RI, 1939.Bulgarian Academy of SciencesInstitute of Mathematics and InformaticsAcad. G. Bontchev Str., Block 81113 Sofia, Bulgariae-mail: pkrusev@math.bas.bgReceived April 7, 2000Revised August 7, 2000

Complete Systems of Hermite Associated Functions

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Complete Systems of Hermite Associated Functions

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