HAL Id: hal-00519325
https://hal.archives-ouvertes.fr/hal-00519325
Preprint submitted on 20 Sep 2010
HAL
is a multi-disciplinary open access archive for the deposit and dissemination of sci- entific research documents, whether they are pub- lished or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers.
L’archive ouverte pluridisciplinaire
HAL, estdestinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d’enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.
Strictly positive definite functions on the real line
Fabrice Derrien
To cite this version:
Fabrice Derrien. Strictly positive definite functions on the real line. 2010. �hal-00519325�
Strictly positive definite functions on the real line
F. Derrien
Univ Lille Nord de France F-59 000 LILLE, FRANCE UArtois, Laboratoire de Math´ematiques de Lens EA 2462,
F´ed´eration CNRS Nord-Pas-de-Calais FR 2956, F-62 300 LENS,FRANCE
Abstract
We give some necessary or sufficient conditions for a function to be strictly positive definite onR. This problem is intimately linked with the repartition of the zeros of trigonometric polynomials.
Keywords: strictly positive definite functions, Bochner’s theorem, scattered data interpolation 2000 MSC:42A82, 41A05, 41A15 (30C15, 42A75)
1. Introduction and notations 1.1. Introduction
In the theory of scattered data interpolation with linear combinations of translates of a basis function, we have to deal with strictly positive definite functions rather than positive definite ones, or more generally with conditionally strictly positive definite functions [2, 4, 16]. This ensures that the interpolation problem is always solvable. Using Bochner’s characterization of continuous positive definite functions as Fourier transforms of nonnegative finite measures, it is straightfor- ward to see that verifying strict positive definiteness reduces to checking whether the exponen- tials are linearly independent on a certain subset of real numbers. Some years ago, K-F. Chang published a paper on this subject [3] but unfortunately his main result is erroneous. In fact, the author asserts in [3, theorem 3.5] that any nonzero complex polynomialt(z)=PK
k=1ckeiξkzwith ξk∈Ris of sine type that is there are positive constantsa,b,σandτsuch that
aeσ|y|≤ |t(x+iy)| ≤beσ|y|for allx,y∈R,|y| ≥τ.
Taking for instancet(z) = eiz, we see easily that the first inequality can not be satisfied. As a consequence, the author concludes at the end of his paper that there does not exist nonzero trigonometric polynomial vanishing at the pointsxn = n+ 18sgn (n) wheren ∈ Z, but trivially t(x)=sin(8πx) does.
Email address:fabrice.derrien@univ-artois.fr(F. Derrien)
Preprint submitted to Journal of Mathematical Analysis and Applications September 19, 2010
1.2. Notations
Let us fix some usual notations for subspaces of complex valued functions defined on the real line. We denote byCthe set of continuous functions andC0its subspace of functions van- ishing at infinity,Ckthe set of functions withkcontinuous derivatives,Sthe Schwartz space of infinitely differentiable and rapidly decreasing functions,Dthe space of infinitely differentiable functions with compact support,Lp the set ofp-power integrable functions with respect to the Lebesgue measureλ,T the set of trigonometric polynomials that is the functions of the form t(x) =PK
k=1ckeiξkxwhereckare complex numbers andξkreal numbers,AP(R) the set of Bohr almost periodic functions.
We will use also the notationAP(Z) for the set of complex valued Bohr almost periodic functions onZ. The set of real valued Bohr almost periodic functions onR, resp.Z, is denoted byAPR(R), resp.APR(Z).
For a functionφ ∈ C, we note ˇφ, resp. ˜φ, the function defined by ˇφ(ξ) = φ(−ξ), resp.
φ(ξ)˜ =φ(−ξ).
The symbol M+stands for the set of nonnegative finite Borel measure on R. The support of a measureµ ∈ M+ defined as the smallest closed subset ofRwhose complement has µ- measure 0 is denoted by suppµ. The Fourier transform of a measure µ ∈ M+is defined by µ(ξ)ˆ =R
Re−iξxdµ(x).
2. Formulation of the interpolation problem, an analogue for Bochner’s theorem
Given a functionϕ∈C, an arbitrary set of distinct real numbersΞ ={ξ1, . . . , ξK}and complex numbersz1, . . . ,zK, the scattered data interpolation problem consists in finding a function
u(ξ)=
K
X
k=1
ckϕ(ξ−ξk)
such thatu(ξj)=zj,j=1. . .K. Although this problem is equivalent to the nonsingularity of the K×KmatrixAwith entriesAjk=ϕ(ξj−ξk), one wishes to know when the interpolation matrix Ais positive definite for any setΞ. For this purpose, let us recall the following definition.
Definition 2.1. A complex valued continuous functionϕis said positive definite (resp. strictly positive definite) onRif for every finite set of distinct real numbersΞ ={ξ1, . . . , ξK}and every vector (c1, . . . ,cK)∈CK\ {0}, the inequality
K
X
j=1 K
X
k=1
¯
cjckϕ(ξj−ξk)≥0 (resp.>0) (1) holds true.
We denote byP(resp.Ps) the class of such functions. ClearlyPs⊂ P.
The class of positive definite functions is fully characterized by the Bochner’s theorem [1].
Theorem 2.1. A functionϕ∈ Pif and only ifϕ=µˆ whereµ ∈ M+,ϕandµbeing biuniquely determined.
Therefore we can ask for an equivalent characterization of a strictly positive definite function in terms of its Fourier transform. We state before a basic lemma.
2
Lemma 2.1. Letµ∈M+and f ∈C a nonnegative function. We have Z
R
f dµ=0⇔ f =0on suppµ.
Proof. Let us setX =suppµ. SinceR\Xis the largest open subset ofRwithµ-measure 0 and the set{f >0}is open, we have
Z
R
f dµ=0⇔µ({f >0})=0⇔ {f >0} ⊂R\X⇔X⊂ {f =0} ⇔ f =0 onX.
Proposition 2.1. Letϕ= µˆ ∈ P. Thenϕ∈ Ps if and only if there does not exist t ∈ T \ {0}
vanishing on suppµ.
Proof. Lett∈ T \ {0}. Remark first thattcan be written in the formt(x)=PK
k=1ckeiξkxwherexk
are pairwise distinct real numbers andckare complex numbers, not all zero. We have
K
X
j=1 K
X
k=1
¯
cjckϕ(ξj−ξk)=
K
X
j=1 K
X
k=1
¯ cjck
Z
R
e−i(ξj−ξk)xdµ(x)
=Z
R K
X
j=1
¯ cje−iξjx
K
X
k=1
ckeiξkxdµ(x)
=Z
R
K
X
k=1
ckeiξkx
2dµ(x)
=Z
R
|t(x)|2dµ(x).
From lemma 2.1 the last integral is 0, i.e.ϕ<Ps, if and only iftvanishes on suppµ.
3. Elementary properties of strictly positive definite functions
The next result shows in particular that the classPsis a convex cone, closed under multipli- cation.
Theorem 3.1. Letϕ∈ Psandψ∈ P. Then (i) ϕ+ψ∈ Ps,
(ii) ϕψ∈ Psprovided thatψ,0.
Proof. Lett∈ T and writeϕ=µ,ˆ ψ=νˆso thatϕ+ψ=µ[+νandϕψ=µ ? ν. It is clear that[ µ+νandµ ? νare inM+. We have from proposition 2.1
Z
R
|t|2d(µ+ν)=0 =⇒ Z
R
|t|2dµ=0 =⇒ t=0.
For the second assertion, we remark by Fubini’s theorem that Z
R
|t|2d(µ ? ν) :=Z
R2
|t(x+y)|2dµ(x)dν(y)=Z
R
dν(y)Z
R
|t(x+y)|2dµ(x) 3
and if we supposeψ,0, then suppν,∅. From lemma 2.1 and proposition 2.1, it follows that Z
R
|t|2d(µ ? ν)=0 =⇒ Z
R
|t(x+y)|2dµ(x)=0 fory∈suppν =⇒ t=0.
Theorem 3.2. Letϕbe a function inPs. Then
(i) ϕis hermitian i.e.ϕˇ =ϕ,¯
(ii) ϕ(0)>0and|ϕ(ξ)|< ϕ(0)for allξ,0, (iii) ϕ(a·)∈ Psprovided that a,0,
(iv) the functionsϕ,¯ <ϕand|ϕ|2are inPs, (v) h◦ϕ∈ Psif h(z)=P∞
n=0anznis a nonconstant power series with nonnegative coefficients, converging for z=ϕ(0).
Proof. By theorem 2.1 we haveϕ=µˆ whereµ∈M+so thatϕ(−ξ)=R
Reiξxdµ(x)=ϕ(ξ).
In the definition 1, letn=1 to obtainϕ(0)>0 and then letn=2, ξ1=ξ, ξ2=0,c1=1,c2 =−1 which shows thatϕ(0)2−ϕ(ξ)ϕ(−ξ)=ϕ(0)2− |ϕ(ξ)|2>0 wheneverξ,0.
The third assertion follows immediately from the definition 1.
Takinga=−1 shows that ¯ϕ=ϕˇ ∈ Psand so are the functions<ϕ= 12(ϕ+ϕ) and¯ |ϕ|2 =ϕ¯ϕby the preceding theorem.
Since|anϕn(ξ)| ≤anϕn(0) for allξ∈Rand alln≥0 andP∞
n=0anϕn(0)<∞, the seriesP∞ n=0anϕn converges uniformly onRby the Weierstrass M-test and so its sumh◦ϕis continuous. Sincehis nonconstant, there exists a coefficientan0>0 withn0 ≥1 and hence by theorem 3.1,an0ϕn0∈ Ps and each partial sumPN
n,n0anϕnis either 0 or inPs. It follows that its pointwise limit is inPand thenh◦ϕ=P∞
n,n0anϕn+an0ϕn0is inPs.
4. Reproducing kernel Hilbert space
Given a strictly positive definite functionϕonR, the construction of the associated repro- ducing kernel Hilbert space is standard. We denote byH0the complex linear space spanned by the functionsϕξ, ξ∈Rwhereϕξ(η)=ϕ(η−ξ). Ifu=PK
k=1ckϕξk andv=PJ
j=1djϕηj belong to H0, then we define a sesquilinear form (u,v) onH0by the formula
(u,v)=
J
X
j=1 K
X
k=1
d¯jckϕ(ηj−ξk). (2)
Since (u,v)=PK
k=1ckv(ξk)=PJ
j=1d¯ju(ηj), the definition of (u,v) does not depend on the chosen representations ofuandv. We have (u,u) = PK
j=1
PK
k=1c¯jckϕ(ξj−ξk) ≥0 by assumption and (u,v)=(v,u) sinceϕis hermitian. An immediate consequence of (2) is the reproducing property (u, ϕξ)=u(ξ) for allu∈H0andξ∈R (3) which implies in particular (ϕη, ϕξ)=ϕ(ξ−η). From the Cauchy-Schwartz inequality, we have
|u(ξ)|2=|(u, ϕξ)|2≤(u,u)(ϕξ, ϕξ)=(u,u)ϕ(0) (4) 4
so that (u,u) =0 if and only ifu = 0, hence the form (·,·) is an inner product onH0. In the pre-Hilbert spaceH0, we see from (4) that norm convergence implies uniform convergence on R, thus we can adjoin all limits of Cauchy sequences to obtain a Hilbert spaceHof continuous functions usually called the reproducing kernel Hilbert space associated withϕ. AsH0is a dense subspace ofH, equations (3) and (4) remain valid inH.
Remark 4.1. Conversely, for each Hilbert spaceHof complex valued continuous functions on R, norm invariant under translations and such that point evaluation functionals are continuous and linearly independent, we can associate a unique functionϕ∈ Psverifying (u, ϕ(· −ξ))=u(ξ).
Let us give the proof. Since every point evaluationδξ : u 7→ u(ξ) is continuous, by the Riesz theorem, there exists uniquelyKξ∈Hsuch that
(u,Kξ)=u(ξ) foru∈H.
Denote byeξ : u 7→ u(· −ξ) the translation operator on H. For everyu ∈ H, the function e−ξu =u(·+ξ) belongs toHand hence we have also (e−ξu,K0)=u(ξ). Since||e−ξu||=||u||for u∈H, we see by the formula
4(u,v)=(u+v,u+v)−(u−v,u−v)+i(u+iv,u+iv)−i(u−iv,u−iv) that (e−ξu,e−ξv)=(u,v) foru,v∈H. We have therefore
(u,eξK0)=(e−ξu,K0)=u(ξ)=(u,Kξ)
which shows thatKξ=eξK0. Let us writeϕ=K0so thatKξ=eξϕ=ϕ(· −ξ). We have (u, ϕ(· −ξ))=(u,Kξ)=u(ξ) foru∈H.
Next the functionϕis positive definite becauseϕ(ξ−η) = K0(ξ−η) = Kη(ξ) = (Kη,Kξ) which gives
K
X
j=1 K
X
k=1
¯
cjckϕ(ξj−ξk)=(
K
X
k=1
ckKξk,
K
X
j=1
cjKξj)≥0.
Suppose thatϕis not strictly positive definite. Then there exist distinct pointsξ1, . . . , ξK and coefficientsc1, . . . ,cKnot all zero such that
K
X
j=1 K
X
k=1
¯
cjckϕ(ξj−ξk)=0⇔
K
X
k=1
ckKξk =0 inH
⇔
K
X
k=1
¯
ck(u,Kξk)=0 for everyu∈H
⇔
K
X
k=1
¯
cku(ξk)=0 for everyu∈H
⇔
K
X
k=1
¯
ckδξk =0 inH0.
which shows that point evaluation functionalsδξare linearly dependent inH0the dual space of H.
5
LetΞ = {ξ1, . . . , ξK}be a set of distinct real numbers andz=(z1, . . . ,zK)∈ CK. We define the linear mapl : H → CK byl(v) =(v(ξj))j=1,...,K. Since the matrix Awith entriesϕ(ξj−ξk) is positive definite, the functionu = PK
k=1ckϕξk is uniquely determined inl−1({z}) in solving the linear systemAc = zwherec = (c1, . . . ,cK). The next theorem shows that this function minimizes the norm inHunder interpolation constraints.
Theorem 4.1. The function u is the unique solution ofmin{||v||, l(v)=z}.
Proof. From the reproducing property, the linear maplis continuous since each coordinate func- tion is continuous. We have furthermorel−1({z})=u+l−1({0}), hence the setl−1({z}) is a closed affine hyperplane in H. By the projection theorem, there exists an unique solution minimiz- ing the norm onl−1({z}). This solution is simply given byusince for allv ∈ l−1({z}), we have (v−u,u)=PK
k=1c¯k(v−u)(ξk)=0 and hence||v||2=||v−u+u||2=||v−u||2+||u||2 ≥ ||u||2. 5. Some inequalities
We prove some inequalities which are consequences of definition 2.1. We recall that for ϕ∈ PsandΞ ={ξ1, . . . , ξK}a set of distinct real numbers, there exists a functionw∈ H0such thatw(ξ1) = 1 andw(ξk) = 0 for anyk , 1. Equivalently, ifΞ = {ξ1, . . . , ξK}is a set of (not necessarily distinct) real numbers andξ<Ξ, there exists a functionw∈ H0vanishing onΞand such thatw(ξ)=1.
Theorem 5.1. Letϕbe a function inPs. Let us consider{ξ1 < . . . < ξK}and{η1 < . . . < ηJ} two sets of pairwise distinct real numbers,(c1, . . . ,cK)and(d1, . . . ,dJ)two complex vectors with nonzero entries. Then
(i) |ϕ(ξ)−ϕ(η)|2<2ϕ(0)[ϕ(0)− <ϕ(ξ−η)]unlessξ=η,
(ii) |ϕ(0)ϕ(ξ+η)−ϕ(ξ)ϕ(η)|2 < [ϕ(0)2− |ϕ(ξ)|2][ϕ(0)2− |ϕ(η)|2]unlessξ = 0 orη = 0 or ξ+η=0,
(iii) |PK
k=1ckϕ(−ξk)|2< ϕ(0)PK
j,k=1c¯jckϕ(ξj−ξk)unless K=1andξ1=0, (iv) |PJ
j=1
PK
k=1d¯jckϕ(ηj−ξk)|2<PK
j,k=1c¯jckϕ(ξj−ξk)PJ
j,k=1d¯jdkϕ(ηj−ηk)
unless J=K, ξk= ηkand ck=αdkfor a numberα∈C\ {0}.
Proof. Letu=PK
k=1ckϕξkandv=PJ
j=1djϕηjbe two functions in the reproducing kernel Hilbert spaceHassociated withϕ. Then we have by the Cauchy-Schwartz inequality
|(u,v)|2≤ ||u||2||v||2⇐⇒ |
J
X
j=1 K
X
k=1
d¯jckϕ(ηj−ξk)|2≤
K
X
j,k=1
¯
cjckϕ(ξj−ξk)
J
X
j,k=1
d¯jdkϕ(ηj−ηk) and equality holds if and onlyuandvare linearly dependent that is,u=αvfor a complex number α,0 sinceuandvare different from 0. But this is equivalent to say that
K
X
k=1
ckw(ξk)=
J
X
j=1
(αdj)w(ηj) forw∈H0.
Suppose there existsξk0 in{ξ1, . . . , ξK} but not in{η1, . . . , ηJ}and pick up a function w ∈ H0 vanishing on{ξ1, . . . , ξK, η1, . . . , ηJ} \ {ξk0}and such thatw(ξk0) = 1. This leads us to the con- tradictionck0 =0. Similarly, if there existsηj0in{η1, . . . , ηJ}but not in{ξ1, . . . , ξK}, we would
6
obtainαdj0=0 and hencedj0 =0. We thus have{ξ1< . . . < ξK}={η1< . . . < ηJ}which shows that
K
X
k=1
ckw(ξk)=
K
X
k=1
αdkw(ξk) forw∈H0.
Choosingw ∈ H0 vanishing on{ξ1, . . . , ξK} \ {ξk}and such thatw(ξk) = 1, we conclude that ck=αdkfor anyk=1, . . . ,K.
Setting J =1, η1 = 0,d1 = 1 in (iv), we get (iii). The inequality (i) follows from (iii) by settingK=2, ξ1=−ξ, ξ2=−η,c1=1,c2=−1.
To prove (ii) we first remark that if a matrix of the form
1 a b
¯
a 1 c
b¯ c¯ 1
is positive definite then its determinant is positive. Computing this determinant we obtain 1+ abc¯ +ab¯¯ c> |a|2+|b|2+|c|2 or, equivalently|c−ab|¯ 2 <(1− |a|2)(1− |b|2). Assume first that ϕ(0)=1. Applying this last inequality to the matrix
(ϕ(ξj−ξk))3j,k=1 =
1 ϕ(−ξ) ϕ(η)
ϕ(ξ) 1 ϕ(ξ+η)
ϕ(−η) ϕ(−ξ−η) 1
whereξ1 =0, ξ2 =ξandξ3=−ηare pairwise distinct, we obtain (ii). The caseϕ(0),1 can be reduced to the caseϕ(0)=1 by consideringϕ/ϕ(0).
6. Sufficient conditions
6.1. Fourier transform of a nonnegative finite continuous measure Proposition 6.1. Ifϕ=µˆ∈ Pand suppµis not a discrete set thenϕ∈ Ps.
Proof. Ift∈ T vanishes on suppµwhich contains an accumulation point, it must vanish identi-
cally as an analytic function onR.
We recall that a measureµ∈M+is called atomic ifµ=Panδxn(an≥0 andPan <+∞) and continuous ifµ(X)=0 for every countable setX ⊂R. Every measureµ∈M+can be uniquely decomposed to a sumµ=µc+µatwhereµcis continuous andµatis atomic.
Corollary 6.1. Ifµ∈M+is not atomic thenµˆ ∈ Ps.
Proof. It is enough to show thatX =suppµis not a discrete set. Letµc ,0 be the continuous part and suppose on the contrary thatXis discrete. SinceXis countable we haveµ(X)=0 and thereforeµc(R)=µc(X)+µc(R\X)=0 i.e.µc=0, a contradiction.
Theorem 6.1. Letϕ=µˆ∈ P. Thenϕ∈AP(R)if and only ifµis atomic.
7
Proof. Ifµ =Panδxn, then ˆµ(ξ)= Paneiξxn is almost periodic. Conversely, suppose thatϕ∈ AP(R) is almost periodic and let use the decompositionµ=µc+µat. It results that ˆµc=ϕ−µˆat is also almost periodic. From the inversion formula, we have
T→lim+∞
1 2T
Z T
−T
e−ixξµˆc(ξ)dξ=µc({x})=0 forx∈R
which shows that ˆµc=0 and henceµc=0.
Corollary 6.2. Ifϕ=µˆ ∈ Pis not almost periodic, in particular iflim sup|ξ|→+∞|ϕ(ξ)|< ϕ(0), thenϕ∈ Ps.
6.2. Derivatives
Theorem 6.2. Letϕ=µˆ andψ =νˆ withµ, ν∈ M+. Suppose there exists a point x0such that suppµ\ {x0} ⊂suppν. Thenϕ∈ Ps =⇒ ψ∈ Ps.
Proof. Fort∈ T, we have alsot(·) sin(· −x0)∈ T and therefore
t=0 on suppν =⇒ t=0 on suppµ\ {x0} =⇒ t(·) sin(· −x0)=0 on suppµ
=⇒ t(·) sin(· −x0)=0 =⇒ t=0.
Corollary 6.3. Letϕ=µˆ∈ P.
(i) If suppµ⊂[0,+∞)and xµ∈M+, thenϕ∈ Ps ⇐⇒ iϕ0∈ Ps. (ii) If x2µ∈M+, thenϕ∈ Ps ⇐⇒ −ϕ00∈ Ps
Proof. By the hypothesis on the moment of the measureµ, it follows thatϕ∈C1withiϕ0=xµb in the first case andϕ∈C2with−ϕ00=dx2µin the second case. It suffices now to apply theorem 6.2 firstly withν = xµand secondly with ν= x2µsince in both cases, we have suppµ\ {0} ⊂
suppν⊂suppµ.
Remark 6.1. In fact, it is well known thatx2µ∈M+is equivalent toϕ∈C2[11, p. 21–22].
For a functionφ∈Cn,n≥1, its Maclaurin series of degreen−1 with integral remainder is given by
φ(ξ)=
n−1
X
k=0
ξk
k!ϕ(k)(0)+ ξn (n−1)!
Z 1
0
ϕ(n)(ξx)(1−x)n−1dx.
We can therefore define the continuous functionTnϕby the formula
Tnϕ(ξ)=
φ(ξ)−Pn−1 k=0
ξk k!ϕ(k)(0)
ξn ifξ,0,
ϕ(n)(0)
n! ifξ=0.
Corollary 6.4. Letϕ=µˆbe a nonconstant function inPand n a positive integer.
(i) If suppµ⊂[0,+∞)and x2n−1µ∈M+, then(−1)n−1iT2n−1ϕ∈ Ps. 8
(ii) If x2nµ∈M+, then(−1)nT2nϕ∈ Ps.
Proof. Let us prove the first item, the same reasoning is valid for the second one. We have ϕ∈C2n−1andϕ(2n−1)=(−i)2n−1x\2n−1µ, hence the function (−1)n−1iϕ(2n−1)is inP. Furthermore this function is noncontant otherwiseϕshould be a polynomial and hence a constant function since it is bounded. With the help of corollary 6.6, the proof is done.
The same proof withϕ0in place ofϕgives the next result.
Corollary 6.5. Letϕ=µˆbe a nonconstant function inPand n a positive integer.
(i) If x2nµ∈M+, then(−1)nT2n−1ϕ0∈ Ps.
(ii) If suppµ⊂[0,+∞)and x2n+1µ∈M+, then(−1)n−1iT2nϕ0∈ Ps. 6.3. Integration against a nonnegative measure
Theorem 6.3. Letψ∈ Psandµ∈M+such thatµ(R\ {0})>0. Then the functionϕgiven by ϕ(ξ)=Z
R
ψ(ξx)dµ(x) is inPs.
Proof. We remark that the integral is well defined since the function ψ is bounded. For an arbitrary vectorc=(c1, . . . ,cK)∈CKand arbitrary real numbersξ1, . . . , ξK, we have
K
X
j=1 K
X
k=1
¯
cjckϕ(ξj−ξk)=Z
R K
X
j=1 K
X
k=1
¯
cjckψ((ξj−ξk)x)dµ(x).
This integral is nonnegative sinceψis positive definite. If it vanishes, the integrand equals 0 on suppµby lemma 2.1 and hence there exists ana,0 such thatPK
j=1
PK
k=1c¯jckψ((ξj−ξk)a)=0 which impliesc=0 since the functionψ(a·) is inPsby theorem 3.2.
As an application we give two important classes of strictly positive definite functions. Let µ∈M+be a measure not concentrated at 0.
(i) The functionψ(ξ)=e−ξ2is inPssince e−ξ2 = 1
2√ π
Z
R
e−x2/4e−iξxdx, hence the function
ϕ(ξ)=Z +∞ 0
e−ξ2x2dµ(x)
is strictly positive definite. By Bernstein–Widder theorem [17, Th. 12a, p. 160], a function admits such an integral representation if and only ifϕ(√
·) is not constant and completely monotone on [0,+∞) meaning thatϕ∈C[0,∞)∩C∞(0,∞) and (−1)kϕ(k)(√
ξ)≥0 for all k∈N0and allξ >0.
9
(ii) We consider now the functionψ(ξ)=(1− |ξ|)+wherex+denotes the greater ofxand 0. As before the function
ϕ(ξ)=Z +∞ 0
(1− |ξx|)+dµ(x) is strictly positive definite since we have
(1− |ξ|)+= 1 2π
Z
R
sinc2(x/2)e−iξxdx.
In this case, we describe the class of even continuous functions which are nonconstant, nonnegative, bounded and convex on (0,+∞) (see [11, p. 87]).
Corollary 6.6. Letψbe a nonconstant function inPandµ ∈ M+such thatµˆ ∈ Ps. Then the function
ϕ(ξ)=Z
R
ψ(ξx)dµ(x) is inPs.
Proof. By assumption onψthere existsν∈M+such thatψ=νˆandν(R\ {0})>0. The Fubini’s theorem gives next
ϕ(ξ)=Z
R
ν(ξx)ˆ dµ(x)=Z
R
Z
R
e−iξxydν(y)dµ(x)=Z
R
Z
R
e−iξyxdµ(x)dν(y)=Z
R
µ(ξy)ˆ dν(y),
hence the proof is done by theorem 6.3.
Corollary 6.7. Letψbe a nonconstant function inPsandα >0a real number. The function ϕ(ξ)= α
ξα Z ξ
0
ψ(x)xα−1dx is inPs.
Proof. The corollary follows by puttingdµ(x)=αxα−1χ(0,1)(x)dxin corollary 6.6.
6.4. Limit at infinity
Theorem 6.4. Letϕ=µˆ ∈ P. Iflim|ξ|→+∞ϕ(ξ)=a inC, then a≥0andϕ−a∈ P. Furthermore, we have a< ϕ(0) ⇐⇒ ϕ∈ Ps ⇐⇒ ϕ−a∈ Ps.
Proof. From the inversion formula [11, p. 35], we have for every realx µ({x})= lim
T→+∞
1 2T
Z T
−T
ϕ(ξ)eixξdξ.
From the hypothesis lim|ξ|→+∞ϕ(ξ)=a, we obtain µ({x})=
a ifx=0, 0 ifx,0.
10
Hence we havea≥0 andµ=aδ0+µcwhereµcis a continuous measure inM+. This shows that ϕ−a=µˆcis positive definite. In particular, we have|ϕ−a| ≤ϕ(0)−a, hence
ϕ(0)=a ⇐⇒ ϕ=a ⇐⇒ µc=0 or equivalently
a< ϕ(0) ⇐⇒ µc,0.
This is clearly equivalent toϕ∈ Psorϕ−a∈ Ps.
6.5. Integrable positive definite functions
Theorem 6.5. Ifϕ∈ P ∩Lp\ {0}with0<p<+∞thenϕ∈ Ps.
Proof. It is sufficient to show thatϕ ∈ C0. Letα > 0 and consider the closed set Fα = {ξ ∈ R; |ϕ(ξ)| ≥2α}. Sinceϕis uniformly continuous, there exists a neighbourhoodVαof 0 such that
|ϕ(ξ)| ≥αfor everyξ∈Fα+Vα. Ifϕ∈Lp, the Chebyshev’s inequality λ({ξ∈R; |ϕ(ξ)| ≥α})≤α−p||ϕ||pp
ensures that the setFα+Vαhas a finite Lebesgue measure. HenceFαis bounded otherwise there would exist a sequence (ξn) ∈ Fα such that{ξn+Vα} ∩ {ξn+1+Vα} ,and we would have the contradiction
λ(Fα+Vα)≥X
n
λ(ξn+Vα)=X
n
λ(Vα)= +∞.
Finally we have proved that for anyα > 0, there is a compact set such that |ϕ| < 2αon its
complement which means thatϕ∈C0.
Theorem 6.6. Letϕ∈C∩L1. Thenϕ∈ Ps ⇐⇒ ϕˆ ≥0andϕ,0.
Proof. Suppose thatϕ=µˆ ∈ Ps. We have evidentlyϕ,0 and the Riemann–Lebesgue lemma tells us that ˆϕ∈C0. For every test functionuin the Schwartz spaceS, we have
Z
R
ϕ(−x)u(x)ˆ dx=Z
R
ϕ(−ξ) ˆu(ξ)dξ
=Z
R
Z
R
eixξdµ(x) ˆu(ξ)dξ
=Z
R
Z
R
eixξu(ξ)ˆ dξdµ(x)
=2πZ
R
u(x)dµ(x).
Sinceµis nonnegative, the last integral is nonnegative for everyu ≥ 0 inSwhich permits to conclude that ˆϕ≥0. Asµis nonnegative, we conclude that ˆϕ≥0. Inversely, ifϕis integrable and continuous with a nonnegative Fourier transform, then ˆϕ ∈ L1 and we have the inversion formula [14, p. 15]
ϕ(x)= 1 2π
Z
R
ϕ(ξ)eˆ ixξdξ.
This shows thatϕ ∈ P. From the hypothesisϕ ∈ L1, we conclude thatϕ ∈ Ps provided that
ϕ,0.
11
Remark 6.2. In the necessary part of the proof, we have also ˆϕ∈L1and the equality Z
R
ϕ(−ˆ x)u(x)dx=2πZ
R
u(x)dµ(x)
remains valid for every functionu∈C0by a density argument. It follows by the Riesz uniqueness theorem thatdµ(x)=(2π)−1ϕ(−x)dx, henceˆ µis absolutely continuous.
Before stating a result about convolution of positive definite functions, we give some lemma.
Lemma 6.1. Ifϕ∈ Pandρ∈ Dthenϕ∗ρ∗ρ˜∈ P ∩C∞.
Proof. The regularity of the convolution product is standart. Moreover it is known that a con- tinuous functionϕis positive definite if and only ifhϕ,ˇ u∗ui ≥˜ 0 for everyu ∈ D. For such a functionu, we have
h(ϕ∗ρ∗ρ)ˇ,˜ u∗ui˜ =hϕ, ρˇ ∗ρ˜∗u∗ui˜ =hϕ, ρˇ ∗u∗(ρ∗u)˜i ≥0.
Lemma 6.2. Letϕ∈ P ∩Lp,1≤p<+∞. There exists a sequence(ϕn)inP ∩ Swhich tends to ϕin Lp.
Proof. Suppose before thatϕ∈C∞. Then the functionϕn(x)=ϕ(x)e−x2/nis inPas the product of two positive definite functions and belongs toSsinceϕis bounded and infinitely differentiable.
By the dominated convergence theorem, the conclusion holds.
In the general case, let us prove that the functionϕis the limit inLpof a sequence belonging to P ∩Lp∩C∞. Letua function inDsuch thatu≥0 andR
Ru=1. Thenρ=u∗u˜shares the same properties asuand the sequence (ρn) defined byρn(x)=nρ(nx) is then an approximate identity.
By the preceding lemma, we haveϕn=ϕ∗ρn∈ P ∩C∞and it is well known thatϕn→ϕinLp. Theorem 6.7. Letϕ ∈ P ∩ Lp and ψ ∈ P ∩ Lq with1 < p,q < +∞and 1p +1q = 1. Then ϕ ? ψ∈ Psifϕ ? ψ,0.
Proof. It is known thatϕ ? ψ∈C0 and we have the Young’s inequality||ϕ ? ψ||∞ ≤ ||ϕ||p||ψ||q. From theorem 6.4 it is enough to prove thatϕ ? ψ∈ P.
Suppose first thatϕ, ψ∈ S. Thenϕ ? ψ∈ Sand it is clearly inPsinceϕ ? ψ[ =ϕˆψˆ ≥0.
Now ifϕ∈Lpandψ∈Lq, 1<p,q<+∞, lemma 6.2 tells us that there exists some sequences (ϕn) and (ψn) inP ∩ Ssuch thatϕn→ϕinLpandψn→ψinLq. But we know from above that ϕn? ψn∈ P ∩ S. The Young’s inequality says that the bilinear application (ϕ, ψ)7→ϕ ? ψfrom Lp×Lq → L∞is continuous and soϕn? ψn → ϕ ? ψuniformly. From definition 2.1, we see
easily that the uniform limit belongs toP.
Recall that a functionϕ∈ Lp∩Lq, 1 ≤ p ≤q ≤+∞, belongs toLr for p ≤r ≤q. This elementary fact permits to give a more general result.
Corollary 6.8. Letϕ ∈ P ∩Lp and ψ ∈ P ∩Lq with1 ≤ p,q < +∞and 1p +1q ≥ 1. Then ϕ ? ψ∈ Psifϕ ? ψ,0.
Proof. We know that a positive definite function is bounded thenϕ∈Lp0andψ∈Lq0withp0≥p andq0≥q. It is easy to verify that we can take in particular p10 = 12
1+1p−1q
andq10 = 12
1+1q−1p so that the two exponents are conjugate and we can therefore apply the preceding theorem.
12
7. Unicity sets
From corollary 6.1 we can restrict the problem of strict positive definiteness on atomic mea- suresµat∈ M+and we obtain by proposition 2.1 that ˆµat ∈ Psif and only if every trigonometric polynomial vanishing on the countable setX=suppµatvanishes identically onR.
In the following we call such a setXa unicity set and we denote byUthe class of all unicity sets.
Proposition 7.1. Let X∈ Uand Y be a countable set. We have then (i) αX+β∈ Uforα, β∈Rsuch thatα,0,
(ii) X⊂Y implies Y ∈ U(hence X∪Y∈ U), (iii) X+Y ∈ U.
Proof.
(i) Lett ∈ T and consider the transformationψ(x) =αx+βfromRto itself. It is clear that t◦ψ∈ T. So we havet◦ψ(X)={0} =⇒ t◦ψ=0 and hencet=0 sinceψis onto.
(ii) Fort∈ T we havet(X)⊂t(Y) and hencet(Y)={0} =⇒ t(X)={0} =⇒ t=0.
(iii) We can supposeYnonempty otherwise the result is trivial. Fory∈Y we haveX+y∈ U by (i) and sinceX+yis a subset of the countable setX+YthenX+Y∈ Uby (ii).
Proposition 7.2. X∪Y ∈ Uimplies X∈ Uor Y ∈ U.
Proof. IfX∪Yis countable so areXandY. Suppose thatXandYare not inU. Then there exists t1,t2∈ P\{0}such thatt1(X)=t2(Y)=0. Hencet=t1t2∈ P\{0}andt(X∪Y)=t(X)∪t(Y)={0}
which lead toX∪Y <U.
Since the ring of trigonometric polynomials is an integral domain, we can easily construct unicity sets from knowing one.
Proposition 7.3. Let X = {xn,n ∈ N} ∈ U and α, β ∈ R such that α , 0. Then the set Y={yn,n∈N}where yn∈ {xn, αxn+β}is also of unicity.
Proof. Let t ∈ T such thatt(Y) = {0} i.e. t(xn) = 0 or t(αxn +β) = 0. The trigonometric polynomialt·t◦ψwhereψ(x)=αx+βvanishes onX∈ Uand hence onR. We have thent=0
ort◦ψ=0 which means thatt=0.
We give now trivial sets which are not of unicity forT. Proposition 7.4.
(i) If Xk=αkZ+βkwhereαk, βk∈RthenSn
k=1Xk<U, (ii) Any finite set F is not inU.
Proof.
(i) Since sin(πZ)=0 thenZ<Uand henceα1Z+β1 <Uby proposition 7.1. The conclusion follows inductively by proposition 7.2.
13
(ii) Any finite setFis a subset ofS
α∈FαZwhich is not inU, thereforeF<U.
Corollary 7.1. X\F∈ Uwhenever X∈ Uand F is finite.
Proof. WriteX=X\F∪Fand use propositions 7.2 and 7.4.
IfXis a set of reals numbers, we denote by{X}the subset{{x},x∈X}of [0,1) where{x}is the fractional part ofx.
Corollary 7.2. If X∈ Uthen for everyα,0the set{αX}has an accumulation point in[0,1].
Proof. It is enough to considerα=1 sinceX∈ U =⇒ αX∈ Ufor everyα,0. If it is not the case, ({X}) is a finite setF and we have thenX ⊂S
β∈FZ+βwhich is not inUby proposition
7.4.
Remark 7.1. The set X = N∪πNis such that{αX}is dense in [0,1] for every α , 0 but is manifestly not inU.
8. Thick set, syndetic set, Hartman sequences Definition 8.1. LetEbe a set inN.
(i) E is called thick if it contains arbitrarily long intervals i.e. for everyk ∈ N, there exists n∈Esuch that [n,n+k]⊂E,
(ii) Eis called syndetic if it has bounded gaps i.e. there existsk∈Nsuch thatE∩[n,n+k],∅ for everyn∈N
Let us now define the notion of Hartman sequences.
Definition 8.2. A sequence (xn) inRis Hartman-uniformly distributed (H-u.d.) inRif
N→∞lim 1 N
N
X
n=1
e2πitxn=0 for everyt,0.
A sequence (xn) inZis Hartman-uniformly distributed (H-u.d.) inZif
N→∞lim 1 N
N
X
n=1
e2πitxn=0 for everyt∈R\Z.
Similarly, we can define the notion of Hartman-well distributed (H-w.d.) sequence by requiring that the limit in the summationPN+M
n=1+Mis uniform inM∈N.
Remark 8.1. By using the Weyl criterion [8, Theorem 2.1, p. 7], we observe that a sequence (xn) is H-u.d. inRif and only if the sequence (txn) is uniformly distributed modulo 1 for every realt,0.
14
LetGdenote the groupRorZ. If f is an almost periodic (a.p.) function onG, it admits a mean value denoted byM(f). ForG=R, we have
M(f)= lim
T→∞
1 T
Z a+T a
f(x)dx uniformly ina∈R, whereas forG=Z, the following identity holds
M(f)= lim
N→∞
1 N
n+N
X
k=n+1
f(k) uniformly inn∈Z.
The following result gives rise to an interesting relation between H-u.d. sequences and almost periodic functions onG[8, p. 298].
Theorem 8.1. A sequence(xn)is H-u.d. in G if and only if for every a.p. function f on G, we have
lim
N→∞
1 N
N
X
n=1
f(xn)=M(f).
Similarly, a sequence(xn)is H-w.d. in G if and only if for every a.p. function f on G, we have lim
N→∞
1 N
N+M
X
n=1+M
f(xn)=M(f)uniformly in M∈N.
Corollary 8.1. Let f be an a.p. function on G and E a thick set inNwhich contains intervals IN with card IN → +∞. Then f vanishes identically on G whenever one of the two following conditions is satisfied
(i) (xn)is H-u.d. in G and lim
N→∞
1 N
N
X
n=1
|f(xn)|=0, (ii) (xn)is H-w.d. in G and lim
N→∞
1 card IN
X
n∈IN
|f(xn)|=0.
Proof. It is enough to use the fact thatM(|f|)=0 implies f =0.
Example 8.1. The following sequences are Hartman-u.d. inR[8] :
(i) (nαlogβn)n≥2withα∈R+\N0andβ∈R; (ii) (nklogβn)n≥2withk∈Nandβ,0; (iii) (logβn) withβ >1; (iv) (nklog logn)n≥2withk≥1.
The following theorem provides many examples of H-u.d. sequences inZ[12, Theorem 1].
Theorem 8.2. If(xn)is H-u.d. inRthen the sequence([xn])of integral parts is H-u.d. inZ. Example 8.2. Letp(x)=a0+a1x+· · ·+akxkbe a polynomial overRof degree at least 2 and suppose thata1,· · ·akdo not lie in a singly generated additive subgroup of the reals. Then the sequence (p(n)) is H-w.d. inR[8, Ch. 4, Example 5.4] and the sequence ([p(n)]) is H-w.d. inZ [15]
Let us notice that the sequence (p(n)) is proved to be H-u.d. in [8] but it is even H-w.d. as a consequence of a result of Lawton [9].
15
9. Some unicity sets
We will use a primarily result in the sequel.
Lemma 9.1. Letαbe a real number, f ∈AP(R)andψ∈APR(R). Then the function g defined by g(y)=f(αy+ψ(y))belongs to AP(R).
Proof. The functioneiψis a.p. sinceeiz is uniformly continuous on every compact subset ofC andψ is bounded [5, Th. 1.7]. Hence the functionei(αy+ψ(y)) is a.p. as a product of two a.p.
functions. Ift(y)=PK
k=1ckeixky∈ T thent(αy+ψ(y))=PK
k=1ckeixk(αy+ψ(y))is a.p. as a finite sum of a.p. functions. But f being a.p. is the uniform limit of trigonometric polynomials (tn). Hence gis also a.p. as the uniform limit of the sequences of a.p. functions (tn(αy+ψ(y)).
We give first some results for H-u.d. sequences inR.
Theorem 9.1. Letα∈R\ {0},ψ∈APR(R),(yn)a H-u.d. sequence inRand(εn)a real sequence such thatlimN→∞ 1
N
PN
n=1|εn|=0. Then X ={xn=αyn+ψ(yn)+εn,n∈N}is inU.
Proof. We assume α > 0 without loss of generality. Let t ∈ T such that t(X) = 0 and set zn=αyn+ψ(yn). We have
0=t(xn)=t(zn+εn)=t(zn)+t0(θn)εnwithθn∈R. Sincet0is bounded, we get
N→∞lim 1 N
N
X
n=1
|t(zn)|=0,
i.e. lim
N→∞
1 N
N
X
n=1
|g(yn)|=0 whereg(y)=t(αy+ψ(y)).
Since the functiong(y) = t(αy+ψ(y)) is a.p. we deduce from corollary 8.1 thatg vanishes identically onR. As the function h(y) = αy+ψ(y) is continuous onR, h(−∞) = −∞ and
h(+∞)= +∞, we haveh(R)=Rand hencet=0
Remark 9.1. We can takenin a thick setEprovided that (yn) is H-w.d. and limn→∞,n∈Eεn=0.
We now give results for H-u.d. sequences inG=ZorR.
Theorem 9.2. Letα∈R,ψ∈APR(G),(yn)a Hartman sequence in G and(εn)a real sequence such thatεn,0andlimN→∞ 1
N
PN
n=1|εn|=0. Then X={xn=αyn+ψ(yn)+εn,n∈N}is inU.
Proof. Lett∈ T such thatt(X)=0 and setzn =αyn+ψ(yn). We have 0=t(xn)=t(zn+εn)=t(zn)+t0(θn)εnwithθn∈R. Sincet0is bounded, we get
lim
N→∞
1 N
N
X
n=1
|t(zn)|=0, 16
i.e. lim
N→∞
1 N
N
X
n=1
|g(yn)|=0 whereg(y)=t(αy+ψ(y)),y∈G.
Since the functiongis a.p. onG, we deduce from corollary 8.1 thatt(αy+ψ(y)) = 0 for any y∈G. In particular, we havet(zn)=0 for anyn∈N. Let us now suppose thatt(l)(αy+ψ(y))=0 fory∈Gand 0≤l≤L. The Taylor formula gives
0=t(xn)= t(L+1)(zn)
(L+1)!εLn+1+t(L+2)(θn) (L+2)!εLn+2 withθn ∈Rand dividing byεLn+1,0, we get
0= t(L+1)(zn)
(L+1)! +t(L+2)(θn) (L+2)!εn. Since the functiont(L+2)is bounded, it follows that
N→∞lim 1 N
N
X
n=1
|t(L+1)(zn)|=0,
i.e. lim
N→∞
1 N
N
X
n=1
|g(yn)|=0 whereg(y)=t(L+1)(αy+ψ(y)),y∈G.
The functiongbeing a.p. onG, we obtain from corollary 8.1 thatt(L+1)(αy+ψ(y))=0 for any y∈G. We conclude thatt(l)(αy+ψ(y))=0 for anyy∈Gandl∈N0, hencet=0 astis analytic.
Remark 9.2. We have the same remark as before.
It is easy to show that the sequence (n2) is not H-u.d. inZ. In the following, we will give a result for polynomials.
The following result is classic in diophantine approximation [6, p. 31].
Lemma 9.2. Let p1, . . . ,pKbe real polynomials. For everyε >0, the set of positive integers n such that
max
1≤k≤K|eipk(n)−eipk(0)|< ε is syndetic.
Lemma 9.3. Letε >0, f ∈AP(R)and p a real polynomial. For every n0∈N, the set of positive integers n such that
|f(p(n))−f(p(n0))|< ε is syndetic.
Proof. Letn0∈Nandε >0.
(i) By lemma 9.2 the set
E={n∈N: max
1≤k≤K|eipk(n+n0)−eipk(n0)|< ε}
is syndetic and so is the set
E+n0 ={n∈N: max
1≤k≤K
|eipk(n)−eipk(n0)|< ε}.
17
(ii) Lett(x)=PK
k=1ckeiλkx∈ T. SettingC=PK
k=1|ck|, we have
|t(p(n))−t(p(n0))|=|
K
X
k=1
ckeiλkp(n)−
K
X
k=1
ckeiλkp(n0)| ≤Cmax
1≤k≤K|eiλkp(n)−eiλkp(n0)|.
Hence the set{n∈N:|t(p(n))−t(p(n0))|< ε}contains the syndetic set {n∈N:C max
1≤k≤K|eiλkp(n)−eiλkp(n0)|< ε}.
(iii) Lett∈ T such that|f(x)−t(x)|<ε3 for everyx∈R. We have
|f(p(n))−f(p(n0))| ≤ |f(p(n))−t(p(n))|+|t(p(n))−t(p(n0))|+|t(p(n0))−f(p(n0))|
< ε
3+|t(p(n))−t(p(n0))|+ε 3.
Hence the set{n∈N:|f(p(n))−f(p(n0))|< ε}contains the set{n∈N:|t(p(n))−t(p(n0))|<
ε
3}which is syndetic.
Lemma 9.4. Let f ∈AP(R)and p a real polynomial such thatlimn→∞,n∈E f(p(n))=0where E is a thick set ofN. Then f ◦p vanishes onN.
Proof. Letε >0 andn0∈N. There existsN∈Nsuch that
|f(p(n))|< ε
2 for everyn∈E0={n∈E:n>N}
and a syndetic setE00such that
|f(p(n))−f(p(n0))|< ε
2 for everyn∈E00.
Since E0 is thick, the set E0∩ E00 is nonempty i.e. there exists an integer n which verifies
|f(p(n0))| ≤ |f(p(n))|+|f(p(n))− f(p(n0))| < ε. Hence f(p(n0)) = 0 sinceε is arbitrary.
Theorem 9.3. Let α ∈ R, ψ ∈ APR(R), p a real polynomial and (εn) a real sequence such that εn , 0,n ∈ E and limn→∞,n∈Eεn = 0 where E is a thick set of N. Then X = {xn = αp(n)+ψ(p(n))+εn,n∈E}is inU.
Proof. Lett∈ T such thatt(X)=0 and setzn =αp(n)+ψ(p(n)),n∈E. We have forn∈E, 0=t(xn)=t(zn+εn)=t(zn)+t0(θn)εnwithθn∈R.
Sincet0is bounded, we get
n→∞,n∈Elim t(zn)=0, i.e. lim
n→∞,n∈Eg(p(n))=0 whereg(y)=t(αy+ψ(y)),y∈R. 18
Since the functiongis a.p. onR, we have from lemma 9.4 thatg◦p = 0 onN i.e.t(αp(n)+ ψ(p(n))=0 for anyn∈Nand in particulart(zn)=0 for anyn∈E.
Let us now suppose thatt(l)(zn)=0 forn∈Nand 0≤l≤L. The Taylor formula gives 0=t(xn)= t(L+1)(zn)
(L+1)!εLn+1+t(L+2)(θn) (L+2)!εLn+2 withθn ∈Rand dividing byεLn+1,0, we get
0= t(L+1)(zn)
(L+1)! +t(L+2)(θn) (L+2)!εn. Since the functiont(L+2)is bounded, it follows that
n→∞,n∈Elim |t(L+1)(zn)|=0, i.e. lim
n→∞,n∈Eg(p(n))=0 whereg(y)=t(L+1)(αy+ψ(y)),y∈R.
The functiongbeing a.p. onR, we have from lemma 9.4 thatg◦p=0 onN i.e.t(L+1)(αp(n)+ ψ(p(n)) = 0 for anyn ∈ Nand in particulart(L+1)(zn) = 0 for any n ∈ E. We conclude that t(l)(zn)=0 for anyn∈Eandl∈N0, hencet=0 astis analytic.
Krein and Levin [10] gave a nice result about the repartition of the real part of the zeros of a trigonometric polynomial.
Theorem 9.4. Consider a nonzero trigonometric polynomial with at least one zero inC. Then the real part of its complex zeros forms a nondecreasing sequence(an)∞−∞given by
an=λn+ψ(n) whereλ >0andψ∈APR(Z).
From this theorem we obtain the following corollary.
Corollary 9.1. If t∈ P \ {0}then its number of zeros N(x)occurring in the interval[x−1,x+1]
is bounded by some constant not depending on x.
Proof. Let (an)∞−∞denote the sequence of the real part of its complex zeros as given in theorem 9.4. Sinceψis a almost periodic, there exists some constantM such that|ψ| < M−1 and we have therefore
N(x)≤card{n∈Z :x−1≤an≤x+1}
≤card{n∈Z :x−M< π
∆n<x+M}
≤2∆M π
Corollary 9.2. Let X be a countable subset of reals such thatsupk∈Zcard(X∩[k,k+1])= +∞.
Then X∈ U.
19
