We consider Trigonometric series with real exponents $\lambda_k$: $$\sum_{k=1}^{+\infty} x_ke^{i\lambda_kt}.$$ Under an assumption on the gap $\gamma_M$ between $\lambda_k$, we show the inequality \begin{equation*}\label{conf000} \frac {2\pi}{\gamma_M(2-c_M)}\sum_{n=1}^M\vert x_n\vert^2 \leq \int_{-\pi/\gamma_M}^{\pi/\gamma_M}\vert \sum_{k=1}^{M} x_ke^{i \lambda_kt}\vert^2dt\leq  \frac {2\pi}{c_M\gamma_M} \sum_{n=1}^M\vert x_n\vert^2 \end{equation*} and  we show for a class of  problems that the limit as $M\to + \infty$ leads to the Parseval's equality. The role of constants $c_M$  in the above formula is one of the key points of the pape

Avantaggiati, A.

Loreti, P.

English

ESE - Salento University Publishing

Note di Matematica ISSN 1123-2536, e-ISSN 1590-0932Note Mat. 36 (2016) no. 1, 67–77. doi:10.1285/i15900932v36n1p67Ingham type inequalities towards ParsevalequalityAntonio AvantaggiatiVia Bartolomeo Maranta 7300156 Roma, ItalyPaola LoretiDipartimento di Scienze di Base e Applicate per l’IngegneriaSapienza Universita´ di RomaVia A. Scarpa 16, 00161 Roma, Italypaola.loreti@sbai.uniroma1.itReceived: 20.2.2016; accepted: 30.3.2016.Abstract. We consider Trigonometric series with real exponents λk:+∞∑k=1xkeiλkt.Under an assumption on the gap γM between λk, we show the inequality2piγM (2− cM )M∑n=1|xn|2 ≤∫ pi/γM−pi/γM|M∑k=1xkeiλkt|2dt ≤ 2picMγMM∑n=1|xn|2and we show for a class of problems that the limit as M → +∞ leads to the Parseval’s equality.The role of constants cM in the above formula is one of the key points of the paper.Keywords: Trigonometric polynomials, inequalities, Parseval equalityMSC 2000 classification: 42A05,41A581 IntroductionIn this paper we establish direct and inverse inequalities involving Trigono-metric series∑+∞k=1 xkeiλkt under assumptions that ensure Parseval’s equality asa limiting case.The inequalities are here obtained for finite M - sums∑Mk=1 xkeiλkt, then weinvestigate the behavior of the constants c1,M , c2,M and of the gap γM appearingon them, as M → +∞.c1,MM∑k=−M|xk|2 ≤∫ pi/γM−pi/γM|+∞∑k=−Mxkeiλkt|2dt ≤ c2,MM∑k=−M|xk|2. (1.1)http://siba-ese.unisalento.it/ c© 2016 Universita` del Salento68 A. Avantaggiati, P. LoretiThe method used here takes into account the arguments in [2]. In that paperthe author is interested to obtain estimates for the single coefficient xk in thecritical time T = pi/γ. However many of his arguments have to be adapted tothe proofs of this paper. The role of the function k in (2.13) is also inspired by[2], as well as the choice of the polynomials P and Q: here we introduce a newcouple of polynomials R and S helpful to define the constant cM , that appearsin the main result of the paper (see formula (2.2) in Theorem 2), and whoselimit as M → +∞ gives the expected result. During the development of thetheoretical part of the paper, we had in mind an application that we describein Section 3. In the application we prove Parseval’s equality in the limit.For any fixed M ∈ N and any set (xk)Mk=−M ∈ C2M+1, we consider gM : R→ Cdefined bygM (t) =M∑k=−Mxkeikt. (1.2)The orthogonality of the set {eikt}Mk=−M in L2(−pi, pi) leads to the identity∫ pi−pi|gM (t)|2dt = 2piM∑k=−M|xk|2. (1.3)For any g ∈ L2(−pi, pi), the limit as M → +∞ gives Parseval’s equality,∫ pi−pi|g(t)|2dt = 2pi∞∑k=−∞|xk|2, (1.4)valid for any sequence (xk)k∈Z such that∑∞k=−∞ |xk|2 < +∞ withg(t) =∞∑k=−∞xkeikt. (1.5)Let (λk)k∈Z be a sequence of real numbers. We considerfM (t) =M∑k=−Mxkeiλkt, f(t) =∞∑k=−∞xkeiλkt, (1.6)and we investigate the problem to find inverse and direct inequalities, that isc1,MM∑k=−M|xk|2 ≤∫ pi/γM−pi/γM|fM (t)|2dt ≤ c2,MM∑k=−M|xk|2, (1.7)Ingham type inequalities towards Parseval equality 69where γM , c1,M , c2,M are real positive constants. The novelty of this note is toestablish inverse and direct inequalities as (1.7) with accurate estimates of theconstants c1,M , c2,M . The estimates narrow, as M → +∞, to the equality∫ pi−pi|f(t)|2dt = 2pi∞∑k=−∞|xk|2. (1.8)However we are unable to give general conditions under which the theoremholds. We give an application of general interest to support the validity of theestimates we obtain.As a reference in this study we recall Ingham’s theorem with explicit con-stants on non-harmonic Fourier series (see [1], see also the Young’s textbook [6],and [3].)Here we recall Ingham’s theorem. The result is largely used in control theoryand it is the basis of the Fourier series method in observability problems. Theexplicit constant on the right hand side of formula (1.9) may be deduced from[3] pg 63 by a change of variables. For the explicit constant on the left handside of formula (1.9) we refer to [1] pg. 369.Theorem 1. Assume the following gap condition: there exists γ > 0 suchthatλk+1 − λk ≥ γ.Then for any fixed T > 0 the following inequality holds for all square summablesequences (xk)k∈Z ∈ C :4Tpi(1− pi2T 2γ2) ∞∑k=−∞|xk|2 ≤∫ T−T|∞∑k=−∞xkeiλkt|2 dt ≤(4T +4piγ) ∞∑k=−∞|xk|2(1.9)Remark 1. To get positive constants in (1.9) we need1− pi2T 2γ2> 0 ⇐⇒ T > piγIngham’s theorem does not generalize Parseval’s equality since ifT → pi we obtain4(1− 1γ2) ∞∑k=−∞|xk|2 ≤∫ pi−pi|∞∑k=−∞xkeiλkt|2 dt ≤ 4pi(1 +1γ) ∞∑k=−∞|xk|2As γ = 10 ≤∫ pi−pi|∞∑k=−∞xkeiλkt|2 dt ≤ 8pi∞∑k=−∞|xk|2.70 A. Avantaggiati, P. Loreti2 The resultTo simplify, we assume that k ∈ N, the arguments used here can be adaptedto k ∈ Z. LetfM (t) =M∑k=1xkeiλkt. (2.1)The main result of the paper is the followingTheorem 2. We assume that the sequence (λk)k∈N is increasing and itsatisfies a gap condition∃γM > 1 such that λk+1 − λk > γM , ∀k ∈ {1, 2, . . . ,M − 1}Then there exists a positive constant cM < 1 such that2piγM (2− cM )M∑n=1|xn|2 ≤∫ pi/γM−pi/γM|fM (t)|2dt ≤ 2picMγMM∑n=1|xn|2 (2.2)Proposition 1. Assume thatlimM→+∞cM = 1 and limM→+∞γM = 1then Parseval’s equality holds∫ pi−pi|f(t)|2dt = 2pi∞∑n=1|xn|2.The proof of the proposition follows from formula (2.2).Proof of the theorem 2. Let j, q ∈ {1, 2, . . . ,M}, and setµj,q = λj − λqwith µq,j = −µj,q.We consider the integers mj,q uniquely defined by the relationsmj,q =bµj,qγM c if q < j ≤M−mq,j if j < q,0 j = q, q ∈ {1, . . . ,M}.(2.3)The sequence of integer numbers m1,q,m2,q, . . . ,mM,q is strictly increasing, forany fixed q ∈ {1, . . . ,M}. It is useful to observe that, by the definition (2.3),the numbers mj,q and µj,q have the same sign for j 6= q, and for j > q we haveγMµj,q≤ γMmj,qµj,q≤ 1, (2.4)Ingham type inequalities towards Parseval equality 71We denote by IM the set {(j, q) ∈ {1, 2, . . . ,M}2 : j 6= q} and we consider thepolynomialsR(u) =∏(j,q)∈IM( uµj,q− 1)S(u) = u∏(j,q)∈IM( uγMmj,q− 1). (2.5)We setcM = lim|u|→∞uR(u)S(u)=∏(j,q)∈IMγMmj,qµj,q(2.6)We see thatR(u)S(u)= cMP (u)Q(u), (2.7)withP (u) =∏(j,q)∈IM(u− µj,q)Q(u) = u∏(j,q)∈IM(u− γMmj,q). (2.8)Assuming simple roots, we look for the decompositioncMP (u)Q(u)=B0u+∑(j,q)∈IMBl.qu− γMml,q . (2.9)we find the coefficients B0 = 1 andBl,r = cMP (γMml,r)Q′(γMml,r)= cMγMml,r − µl,rγMml,r∏(j,q)∈IM ,(j,q) 6=(l,r)γMml,r − µj,qγM (ml,r −mj,q)(2.10)By the previous computation we get1 +∑(j,q)∈IMBl,q = cM (2.11)To end the proof we observe that for every (l, r) ∈ IM each Bl,r is negative.Indeed sinceγMml,r < µj,q ⇐⇒ γMml,r < γMmj,qall the factorsγMml,r−µj,qγMml,r−γMmj,q are positive, on the contraryγMml,r−µl,rγMml,ris negative,then for every (l, r) ∈ IM each Bl,r is negative.Hence by (2.11)1− cM = −∑(j,q)∈IMBl,q =∑(j,q)∈I|Bl,q| (2.12)72 A. Avantaggiati, P. LoretiWe are going to define the functions k, whose Fourier transform verifies theproperties:kˆ(λn − λq) = kˆ(µn,q) = 0, ∀n 6= q, kˆ(0) = kˆ(µq,q) > 0.Inspired by the Ingham paper [2], we can set as k function the followingk(ξ) ={1 +∑(j,q)∈IM (−1)mj,qBj,qeimj,qξ if |ξ| ≤ pi/γ.0 if |ξ| > pi/γ. (2.13)The Fourier transform of the function k iskˆ(u) =12pi∫ pi/γM−pi/γM(1 +∑(l,q)∈IM(−1)ml,qBl,qeiml,qξ)e−iξudξ =1pi[sin piuγMu+∑(l,q)∈IM(−1)ml,qBl,qsin( piγM (u− γMml,q))u− γMml,q]=1pi[sin piuγMu+∑(l,q)∈IMBl,qu− γMml,q sin(piγMu)]=1pi[1u+∑(l,q)∈IMBl,qu− γMml,q]sinpiγMu ==1picMP (u)Q(u)sinpiγMu =1piR(u)S(u)sinpiγMuThenlimu→0kˆ(u) =1γM.Moreover, since the functionsin piγuQ(u) is regular in all the zeros of the polinomialQ(u), and P (λj − λq) = 0 for (j, q) ∈ IM , it follows{kˆ(λj − λq) = 0 ∀(j, q) ∈ IMkˆ(0) = 1γM .(2.14)Moreover we have the estimatek(t) ≥ 1−∑(j,q)∈IM|Bj,q| = cM (2.15)Ingham type inequalities towards Parseval equality 73k(t) ≤ 1 +∑(j,q)∈IM|Bl,q| = 2− cM (2.16)Therefore∫ pi/γM−pi/γMk(t)|fM (t)|2dt ≥(1−∑(j,q)∈IM|Bj,q|)∫ pi/γM−pi/γM|fM (t)|2dt =cM∫ pi/γM−pi/γM|fM (t)|2dt (2.17)and∫ pi/γM−pi/γMk(t)|fM (t)|2dt ≤(1 +∑(j,q)∈IM|Bj,q|)∫ pi/γM−pi/γM|fM (t)|2dt =(2− cM )∫ pi/γM−pi/γM|fM (t)|2dt (2.18)By the properties of kˆ, see (2.14)∫ pi/γM−pi/γMk(t)|fM (t)|2dt = 2pikˆ(0)M∑n=1|xn|2we get the estimate2piγM (2− cM )M∑n=1|xn|2 ≤∫ pi/γM−pi/γM|fM (t)|2dt ≤ 2picMγMM∑n=1|xn|2. (2.19)3 Application3.1 Perturbed wave equationAt first, consider the one dimensional interval, (0, T ), T ≥ 2pi, and the waveequation with Dirichlet boundary conditions:ftt(x, t)− fxx(x, t) = 0 in (0, pi)× (0, T ),f(x, 0) = f0(x) in (0, pi),ft(x, 0) = f1(x) in (0, pi),f(0, t) = f(pi, t) = 0 in (0, T )(3.1)74 A. Avantaggiati, P. LoretiThe solution of (3.1) is given in terms of Fourier series byf(x, t) =∑k∈Z∗xkeikt sin kx.The system (3.1) is the dual observability problem of the exact controllabilityboundary control problem for the wave equation, the observation functions arefx in x = 0, or/and in x = pi. We havefx(0, t) =∑k∈Z∗kxkeiλkt fx(pi, t) =∑k∈Z∗(−1)kkxkeiλkt.We write g  h if there exist two positive constants α, β such that αg ≤ h ≤ βg.With this notation we have‖fx(0, t)‖2L2(0,T ) ∑k∈Z∗|kxk|2 ‖fx(pi, t)‖2L2(0,T ) ∑k∈Z∗|kxk|2.Due to the orthogonality of trigonometric polynomials, T = 2pi gives‖fx(0, t)‖2L2(0,T ) = 2pi∑k∈Z∗|kxk|2.However Ingham’s theorem does not include the critical time T = 2pi as γ = 1.We consider the wave equation with a perturbation of zero-order.For any c such that |c| < 1, we considerftt(x, t)− fxx(x, t)− c2f(x, t) = 0 in (0, pi)× (0, T ),f(x, 0) = f0(x) in (0, pi),ft(x, 0) = f1(x) in (0, pi),f(0, t) = f(pi, t) = 0 in (0, T ).(3.2)The solution of (3.2) is given in terms of Fourier series byf(x, t) =∑n∈Z∗xneiλnt sinnxwithλn = sgn(n)√n2 − c2 = n√1− c2n2.For large n we haveλn ≈ n− c22n.Ingham type inequalities towards Parseval equality 753.2 An application of the main resultFor any real positive number a we consider the set of real numbersλn(a) = λn n = 1 . . .M , withλn = n− 1anλn+1 − λn = 1 + 1an(n+ 1)(3.3)Thenλn+1 − λn > γM , ∀n ∈ {1, 2, . . . ,M − 1}withγM = 1 +1aM(M + 1)(3.4)IM ={(j, q) ∈ {1, 2, . . . ,M}2 such that j > q}andI =+∞⋃M=1IMThen, with the same meaning of µj,q and mj,q of previous section we set aspreviouslyR(u) =∏(j,q)∈IM( uµj,q− 1). (3.5)S(u) = u∏(j,q)∈IM( uγMmj,q− 1)(3.6)and can be determined ascM = limu→∞uR(u)S(u)= limu→∞∏(j,q)∈IM(uµj,q− 1)(uγMmj,q− 1) = ∏(j,q)∈IMγMmj,qµj,q. (3.7)The aim of this section is to showProposition 2. Let cM be as in (3.7). ThenlimM→∞cM = 1.To give the proof of the above proposition we recall some basic facts on thenotion of limit for generalized sequences76 A. Avantaggiati, P. Loreti3.2.1 Limits for generalized sequencesFor any (j, q) and (r, s) belonging to I we write(j, q)  (r, s) ⇐⇒ j ≥ r and q ≥ s.The set I endowed with  is a partial ordered set, the set IM is a direct setand the generalized sequence(j, q) ∈ I → ajqajq + 1∈ R,is increasing and bounded in IM , with maximum value aM(M−1)aM(M−1)+1 .Lemma 1. With the previous notationslimM→∞∏(j,q)∈IM(1 +1aM(M + 1))ajqajq + 1= 1Proof. We consider∏(j,q)∈IM(1+1aM(M + 1))ajqajq + 1=(1+1aM(M + 1))M(M−1)2 ∏(j,q)∈IMajqajq + 1SincelimM→∞(1 +1aM(M + 1))M(M−1)2= e12a .The geometric mean theorem implieslimM→∞( ∏(j,q)∈IMajqajq + 1) 2M(M−1)= limM→∞aM(M − 1)aM(M − 1) + 1hencelimM→∞∏(j,q)∈IMajqajq + 1= limM→∞(( ∏(j,q)∈IMajqajq + 1) 2M(M−1))M(M−1)2=limM→∞(aM(M − 1)aM(M − 1) + 1)M(M−1)2=1e12aPlug the limits in the formulalimM→∞∏(j,q)∈IM(1 +1aM(M + 1))ajqajq + 1= 1QEDIngham type inequalities towards Parseval equality 77Proof of the Proposition.We recall thatcM =∏(j,q)∈IMγMmj,qµj,q.Then we computeγMmj,qµj,q=γM [λn − λm]λn − λm ≥ γMj − q(j − q)(1 + 1ajq) =(1 +1aM(M + 1))ajqajq + 1.It follows1 ≥ cM =∏(j,q)∈IMγMmj,qµj,q≥∏(j,q)∈IM(1 +1aM(M + 1))ajqajq + 1,the result will follow by the lemma 1limM→∞∏(j,q)∈IM(1 +1aM(M + 1))ajqajq + 1= 1. (3.8)References[1] A. E. Ingham: Some trigonometrical inequalities with applications in the theory of series,Math. Z. 41 (1936), 367–379.[2] A. E. Ingham: A further note on trigonometrical inequalities, Mathematical Proceed-ings of the Cambridge Philosophical Society, Volume46, Issue 04, (1950), 535–537.[3] V. Komornik and P. Loreti: Fourier Series in Control Theory, Springer-Verlag, NewYork, 2005.[4] J.-L. Lions: Exact controllability, stabilizability, and perturbations for distributed sys-tems, Siam Rev. 30 (1988), 1–68.[5] P. Loreti: On some gap theorems, European women in mathematics—Marseille 2003,39–45, CWI Tract, 135, Centrum Wisk. Inform., Amsterdam, 2005.[6] Robert M. Young: An Introduction to Non-Harmonic Fourier Series, Revised Edition,Cambridge University Press, 1959.[7] A. Zygmund,: Trigonometric Series I-II, Cambridge University Press, 1959.

Ingham type inequalities towards Parseval equality

We consider Trigonometric series with real exponents
We show for a class of problems that the limit 
leads to the Parseval's equality.
The role of constants in the above formula is one of the key points of the paper

Ingham type inequalities towards Parseval equality

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