# Subharmonics with Minimal Periods for Convex Discrete Hamiltonian Systems

Authors: Honghua Bin
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We consider the subharmonics with minimal periods for convex discrete
Hamiltonian systems. By using variational methods and dual functional, we obtain that the
system has apT-periodic solution for each positive integer p, and solution of system has
minimal period pT
as H subquadratic growth both at 0 and infinity.


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## Body

## 1. Introduction

Consider Hamiltonian systems(1)Ju˙(t)+∇H(t,u(t))=0,u(0)=u(pT),
where u(t)∈ℝ2N, t∈ℝ, ∇H stands for the gradient of H with respect to the second variable, and J=(0-ININ0) is the symplectic matrix with IN the identity in ℝN. Moreover, H is T-periodic in the variable t, p∈ℕ∖{0}.Classically, solutions for systems (1) are called subharmonics. The first result concerning the subharmonics problem traced back to Birkhoff and Lewis in 1933 (refer to [1]), in which there exists a sequence of subharmonics with arbitrarily large minimal period, using perturbation techniques. More results can be found in [1–5], where H is convex with subquadratic growth both at 0 and infinity. Using Zp index theory and Clarke duality, Xu and Guo [1, 5] proved that the number of solutions for systems (1) with minimal period pT tends towards infinity as p→∞.For periodic and subharmonic solutions for discrete Hamiltonian systems, Guo and Yu [6, 7] obtained some existence results by means of critical point theory, where they introduced the action functional
(2)F(u)=-12∑n=1pT‍(JΔLu(n-1),u(n))-∑n=1pT‍H(n,Lu(n)).
Using Clarke duality, periodic solution of convex discrete Hamiltonian systems with forcing terms has been investigated in [8]. Clarke duality was introduced in 1978 by Clarke [9], and developed by Clarke, Ekeland, and others, see [10–12]. This approach is different from the direct method of variations; some scholars applied it to consider the periodic solutions, subharmonic solutions with prescribed minimal period of Hamiltonian systems; one can refer to [3, 5, 12–14] and references therein. The dynamical behavior of differential and difference equations was studied by using various methods; see [15–19]. We refer the reader to Agarwal [20] for a broad introduction to difference equations.Motivated by the works of Mawhin and Willem [12] and Xu and Guo [21], we use variational methods and Clarke duality to consider the subharmonics with minimal periods for discrete Hamiltonian systems
(3)JΔu(n)+∇H(n,Lu(n))=0,u(n)=u(n+pT),
where u(n)=(u1(n)u2(n)), Lu(n)=(u1(n+1)u2(n)), ui(n)∈ℝN(i=1,2) with N a given positive integer, and Δu(n)=u(n+1)-u(n) is the forward difference operator. p,T∈ℕ∖{0}. Moreover, hamiltonian function H satisfies the following assumption:(A1)
H:ℤ×ℝ2N→ℝ is continuous differentiable on ℝ2N, H(n,·) convex for each n∈ℤ and H(n+T,u)=H(n,u) for each u∈ℝ2N;(A2)
there exist constantsa1>0, a2>0, 1<θ<2, such that
(4)a1θ|u|θ≤H(n,u)≤a2θ|u|θ,u∈ℝ2N,
which implies H subquadratic growth both at 0 and infinity.Theorem 1.
Assume (A1) holds.H(n,u)→+∞, H(n,u)/|u|2→0, as |u|→∞ uniformly in n∈ℤ. Then there exists a pT-periodic solution up of (3), such that ∥up∥∞≜maxn∈Z[1,pT]{|up(n)|}→∞, and the minimal period Tp of up tends to +∞ as p→∞.Theorem 2.
Under the assumptions (A1) and (A2), if(5)a2a1≤{(14sinπpT)θ/2,whenpTiseven,(12sinπ2pT)θ/2,whenpTisodd
for given integer p>1, then the solution of (3) has minimal period pT.


## 2. Clarke Duality and Eigenvalue Problem

First we introduce a spaceEpT with dimension 2NpT as follows:
(6)EpT={u={u(n)}∈S∣u(n+pT)=u(n),p∈ℕ∖{0},n∈ℤ},
where
(7)S={u={u(n)}∣u(n)=(u1(n)u2(n))∈ℝ2N,uj(n)∈ℝN,j=1,2,n∈ℤ(u1(n)u2(n))}.
Equipped with inner product 〈·,·〉 and norm ∥·∥ in EpT as
(8)〈u,v〉=∑n=1pT‍(u(n),v(n)),∥u∥=(∑n=1pT‍|u(n)|2)1/2,∀u,v∈EpT,
where (·,·) and |·| denote the usual scalar product and corresponding norm in ℝ2N, respectively. It is easy to see that (EpT,〈·,·〉) is a Hilbert space with 2NpT dimension, which can be identified with ℝ2NpT. Then for any u∈EpT, it can be written as u=(uT(1),uT(2),…,uT(pT))T, where u(j)=(u1(j)u2(j))∈ℝ2N, j∈Z[1,pT], the discrete interval {1,2,…,pT} is denoted by Z[1,pT], and ·T denotes the transpose of a vector or a matrix.Denote the subspaceY-={u∈EpT∣u(1)=u(2)=⋯=u(pT)∈ℝ2N}. Let Y be the direct orthogonal complement of EpT to Y-. Thus EpT can be split as EpT=Y⊕Y-, and for any u∈EpT, it can be expressed in the form u=u~+u-, where u~∈Y,  u-∈Y-.Next we recall Clarke duality and some lemmas.The Legendre transform (see [12]) H*(t,·) of H(t,·) with respect to the second variable is defined by
(9)H*(t,v)=supu∈R2N{(v,u)-H(t,u)},
where (·,·) denotes the inner product in ℝ2N. It follows from (A1) and (A2) that(B1)
H*(n,·) is continuous differentiable on ℝ2N,(B2)
forτ=θ/(θ-1),  v∈ℝ2N,  n∈ℤ, one has
(10)1τ(1a2)τ-1|v|τ≤H*(n,v)≤1τ(1a1)τ-1|v|τ.Associated with variational functional (2), the dual action functional is defined by
(11)χ(v)=∑n=1pT‍12(L(JΔv(n-1)),v(n))+∑n=1pT‍H*(n,Δv(n)),v∈EpT.
Indeed, by (11), we have χ(v+u-)=χ(v) for any u-∈Y- and v∈Y. Therefore, χ can be restricted in subspace Y of EpT. Moreover, in terms of Lemma  2.6 in [8] and the following lemma, the critical points of (11) correspond to the subharmonic solutions of (3).Lemma 3 (see [8, Theorem 1]).
Assume that(H1)
H(n,·)∈C1(ℝ2N,ℝ), H(n,·) is convex in the second variable for n∈ℤ,(H2)
there existsβ:ℤ→ℝ2N such that for all (n,u)∈ℤ×ℝ2N, H(n,u)≥(β(n),u), and β(n+T)=β(n),(H3)
there existα∈(0,2sin(π/pT)) and γ:ℤ→ℝ+, such that for any (n,u)∈ℤ×ℝ2N, H(n,u)≤(α/2)|u|2+γ(n), and γ(n+T)=γ(n),(H4)
for eachu∈ℝ2N, ∑n=1pTH(n,u)→+∞ as |u|→∞.Then system (3) has at least one periodic solution u, such that v=-J[u-(1/pT)∑n=1pTu(n)] minimizes the dual action χ(v)=∑n=1pT(1/2)(LJΔv(n-1),v(n))+∑n=1pTH*(n,Δv(n)).The following lemmas will be useful in our proofs, where Lemma4 can be proved by means of Euler formula eiθ=cosθ+isinθ, and Lemma 5 is a Hölder inequality.Lemma 4.
For anyk∈ℤ, ∑n=1pTsin((2kπ/pT)n)=∑n=1pTcos((2kπ/pT)n)=0.Lemma 5.
For anyuj>0, vj>0, k∈ℤ, one has ∑j=1kujvj≤(∑j=1kujp)1/p(∑j=1kvjq)1/q, where p>1, q>1 and 1/p+1/q=1.Lemma 6 (see [12, proposition 2.2]).
LetH:ℝm→ℝ be of C1 and convex functional, -β≤H(u)≤αq-1|u|q+γ, where u∈ℝm, α>0, q>1, β≥0, γ≥0. Then α-p/qp-1|∇H(u)|p≤(∇H(u),u)+β+γ, where 1/p+1/q=1.In order to know the form ofu∈EpT, we consider eigenvalue problem
(12)LJΔu(n-1)=λu(n),u(n+pT)=u(n),
where u(n)=(u1(n)u2(n)), Lu(n-1)=(u1(n)u2(n-1))∈ℝ2N, n∈ℤ, λ∈ℝ. We can rewrite (12) as the following form:
(13)u1(n+1)=(1-λ2)u1(n)+λu2(n),u2(n+1)=-λu1(n)+u2(n),u1(n+pT)=u1(n),u2(n+pT)=u2(n).
Denoting
(14)M(λ)=((1-λ2)INλIN-λININ),
then problem (12) is equivalent to
(15)u(n+1)=M(λ)u(n),u(n+pT)=u(n).
Letting u(n)=μnc be the solution of (15), for some c∈ℂ2N, we have μc=M(λ)c and μpT=1. Consider the polynomial |M(λ)-μI2N|=0 and condition μpT=1; it follows that
(16)μ=e2kπi/pT,λ=2sinkπpT,k∈Z[-pT+1,pT-1].In the following we denote byμk=e2kπi/pT, λk=2sin(kπ/pT), k∈Z[-pT+1,pT-1], and ρ∈ℝN. By (M(λk)-μkI2N)c=0, it follows that
(17)ck=(ρie(-kπi/pT)ρ).
Thus
(18)uk(n)=μknck=e2kπni/pT(ρie(-kπi/pT)ρ)=(cos(2kπpTn)ρ-sin(2kπpT(n-12))ρ)+i(sin(2kπpTn)ρcos(2kπpT(n-12))ρ).Let(19)ξk(n)=(cos(2kπpTn)ρ-sin(2kπpT(n-12))ρ),ηk=(sin(2kπpTn)ρcos(2kπpT(n-12))ρ).
Obviously, ξk(n) and ηk(n) satisfy (15). Moreover LJΔξk(n-1)=λkξk(n), LJΔηk(n-1)=λkηk(n), ξ2pT+k(n)=ξk(n), η2pT+k(n)=ηk(n), ξpT-k(n)=ξk(n), ηpT-k(n)=-ηk(n).Fork≠pT/2, subspace Yk is defined by(20)Yk={span{ξk(n),ηk+(pT/2)(n)},k∈Z[-pT2+1,pT2-1]∖{0},n∈ℤ,ifpTiseven,span{ξk(n),ηk+((pT+1)/2)(n)},k∈Z[[-pT2],[pT2]]∖{0},n∈ℤ,ifpTisodd,where [·] denotes the greatest-integer function and
(21)YpT/2=span{ξpT/2(n),n∈ℤ},Y-pT/2=span{ξ-pT/2(n),n∈ℤ}.
Therefore,
(22)Y=⊕Yk,k∈Z[-pT2,pT2]∖{0},ifpTiseven,Y=⊕Yk,k∈Z[[-pT2],[pT2]]∖{0},ifpTisodd.
Moreover, for any u={u(n)}∈EpT, we may express u(n) as
(23)u(n)=∑k=-pT+1pT-1‍[(cos(2kπpTn)ak-sin(2kπpT(n-12))ak)+(sin(2kπpTn)bkcos(2kπpT(n-12))bk)],
where ak,bk∈ℝN.Since(Δu(n),Δu(n))=-(Δ2u(n-1),u(n)), we consider eigenvalue problem
(24)-Δ2u(n-1)=λu(n),u(n+pT)=u(n),u(n)∈ℝN,
where Δ2u(n-1)=Δu(n)-Δu(n-1)=u(n+1)-2u(n)+u(n-1). The second order difference equation (24) has complexity solution u(n)=einθc for c∈ℂN, where θ=2kπ/pT. Moreover, λ=2-e-iθ-eiθ=2(1-cosθ)=4sin2(θ/2); that is, λ=4sin2(kπ/pT),  k∈Z[0,pT-1].By the previous, it follows Lemma7.Lemma 7.
For anyu∈EpT, one has -λmax∥u∥2≤∑n=1pT(LJΔu(n-1),u(n))≤λmax∥u∥2, and 0≤∑n=1pT|Δu(n)|2≤λmax2∥u∥2, where
(25)λmax=maxk∈[0,pT-1]{2sinkπpT}={2,ifpTiseven,2cosπ2pT,ifpTisodd.
Moreover, if u∈Y, then 4sin2(π/pT)∥u∥2≤∑n=1pT|Δu(n)|2≤λmax2∥u∥2.


## 3. Proofs of Main Results

Lemma 8.
Consider(26)∑n=1pT‍(LJΔu(n-1),u(n))≥-(2sinπpT)-1∑n=1pT‍|Δu(n)|2,∀u∈EpT.Proof.
Lettingu~(n)=u(n)-(1/pT)∑n=1pTu(n), then u~∈Y. By Lemmas 5 and 7, we have
(27)∑n=1pT‍(LJΔu(n-1),u(n))=∑n=1pT‍(LJΔu(n-1),u~(n))≥-(∑n=1pT‍|LJΔu(n-1)|2)1/2·(∑n=1pT‍|u~(n)|2)1/2≥-(∑n=1pT‍|Δu(n)|2)1/2·(2sinπpT)-1(∑n=1pT‍|Δu~(n)|2)1/2=-(2sinπpT)-1∑n=1pT‍|Δu(n)|2.Lemma 9.
If there existα∈(0,sin(π/pT)), β≥0 and δ>0, such that
(28)δ|u|-β≤H(n,u)≤α2|u|2+γ
for all n∈[1,pT] and u∈ℝ2N, then each solution of (3) satisfies the inequalities
(29)∑n=1pT‍|Δu(n)|2≤2α(β+γ)pTsin(π/pT)sin(π/pT)-α,∑n=1pT‍|Lu(n)|≤(β+γ)pTsin(π/pT)δ(sin(π/pT)-α).Proof.
Letu be the solution of (3). By Lemma 6, we have
(30)12α|∇H(n,Lu(n))|2≤(∇H(n,Lu(n)),Lu(n))+β+γ=-(JΔu(n),Lu(n))+β+γ.
Obviously, |JΔu(n)|2=(-∇H(n,Lu(n)),JΔu(n))=|∇H(n,Lu(n))|2 by (3), and it follows that (1/2α)∑n=1pT|JΔu(n)|2+∑n=1pT(JΔu(n),Lu(n))≤(β+γ)pT; that is,
(31)12α∑n=1pT‍|Δu(n)|2+∑n=1pT‍(LJΔu(n-1),u(n))≤(β+γ)pT.
By means of Lemma 8, we have
(32)[12α-(2sinπpT)-1]∑n=1pT‍|Δu(n)|2≤(β+γ)pT,
which gives first conclusion.
Now,H(n,0)≤γ in view of (28); therefore by convex and Lemma 8, we have
(33)δ∑n=1pT‍|Lu(n)|-βpT≤∑n=1pT‍H(n,Lu(n))≤∑n=1pT‍[H(n,0)+(∇H(n,Lu(n)),Lu(n))]≤γpT-∑n=1pT‍(JΔu(n),Lu(n))=γpT-∑n=1pT‍(JLΔu(n-1),u(n))≤γpT+(2sinπpT)-1∑n=1pT‍|Δu(n)|2≤γpT+α(β+γ)pTsin(π/pT)-α,
which gives the second conclusion. The proof is completed.Proof of Theorem1.
Letc1=maxn∈ℤ|H(n,0)|. By assumption in Theorem 1, there exists R>0, such that H(n,u)≥1+c1, for n∈ℤ and |u|≥R. Moreover, there exist α∈(0,2sin(π/pT)), γ>0 such that
(34)H(n,u)≤α2|u|2+γ,∀(n,u)∈ℤ×ℝ2N.
Thus, by convex of H, for all (n,u)∈ℤ×ℝ2N with |u|≥R, we have
(35)1+c1≤H(n,R|u|u)≤H(n,0)+R|u|(H(n,u)-H(n,0))≤R|u|H(n,u)+c1.
Therefore there exist β>0 and δ>0, such that
(36)H(n,u)≥δ|u|-β,∀(n,u)∈ℤ×ℝ2N.
Combining the previous argument, by Lemma 3, the system (3) has a pT-periodic solution up such that vp=-J[up-(1/pT)∑n=1pTup(n)]∈Y minimizes the dual action
(37)χp(vp)=∑n=1pT‍12(LJΔvp(n-1),vp(n))+∑n=1pT‍H*(n,Δvp(n))onEpT.
It follows that Δup(n)=JΔvp(n) and Jvp(n)=up(n)-(1/pT)∑n=1pTup(n).
We next prove that∥up∥∞→∞ as p→∞.
Suppose not, there existc2>0 and a subsequence {pk} such that
(38)pk→∞,∥upk∥∞≤c2ask⟶∞.
In terms of (3), it follows that ∥Δupk∥∞≤c3 for some c3>0, and ∥vpk∥∞≤2c2, ∥Δvpk∥∞≤c3. Consequently, by H*(n,v)≥-H(n,0)≥-c1, we have
(39)cpk=χpk(vpk)=∑n=1pkT‍12(LJΔvpk(n-1),vpk(n))+∑n=1pkT‍H*(n,Δvpk(n))≥-12∑n=1pkT‍|LJΔvpk(n-1)||vpk(n)|-c1pkT≥-(2c2c3+c1)pkT,
where n∈Z[1,pkT] and
(40)|LJΔvpk(n-1)|=(|Δv2,pk(n)|2+|Δv1,pk(n-1)|2)1/2≤2∥Δvpk∥∞≤2c3.
By (36), if |v|≤δ, we have (v,u)-H(n,u)≤(v,u)-δ|u|+β≤β, and H*(n,v)≤β. Letting ρ∈ℝN and |ρ|=1, in terms of (12), hp associated with λ-1=-2sin(π/pT) is given by
(41)hp(n)=δ4sin(π/pT)·((cos2πpTn-sin2πpTn)ρ(sin2πpT(n-12)+cos2πpT(n-12))ρ)
which belongs to EpT, and(42)|Δhp(n)|2=(δ4sin(π/pT))2·|2sinπpT((-sin2πpT(n+12)-cos2πpT(n+12))ρ(cos2πpTn-sin2πpTn)ρ)|2=14[2+sin2πpT(2n+1)-sin2πpT(2n)]·|ρ|2δ2≤δ2.Moreover, by Lemma 4 we have
(43)∑n=1pT‍|hp(n)|2=∑n=1pT‍(δ4sin(π/pT))2·(2+sin2πpT(2n-1)-sin2πpT(2n))|ρ|2=(δ4sin(π/pT))22|ρ|2pT=δ2pT8sin2(π/pT).
Thus cp=χp(hp)≤∑n=1pT(1/2)(LJΔhp(n-1),hp(n))+βpT=∑n=1pT(1/2)(-2sin(π/pT))|hp(n)|2+βpT=-δ2pT/8sin(π/pT)+βpT. Combining (39), we have 8(2c2c3+c1+β1)sin(π/pkT)≥δ2, which is impossible as k large. So the claim limp→∞∥up∥∞=∞ is valid.
It remains only to prove that the minimal periodTp of up tends to +∞ as p→∞.
If not, there existsT>0 and a sequence {pk} such that the minimal period Tpk of upk satisfies 1≤Tpk≤T. By assumption in Theorem 1, there exists α∈(0,sin(π/T)) and γ>0 such that
(44)H(n,u)≤α2|u|2+γ,∀(n,u)∈ℤ×ℝ2N.
By (36) and Lemma 9 with pT replaced by Tpk, we get
(45)∑n=1Tpk‍|Δupk(n)|2≤2α(β+γ)Tpksin(π/Tpk)sin(π/Tpk)-α≤2α(β+γ)Tsin(π/T)sin(π/T)-α,(46)∑n=1Tpk‍|Lupk(n)|≤(β+γ)Tpksin(π/Tpk)δ(sin(π/Tpk)-α)≤(β+γ)Tpksin(π/T)δ(sin(π/T)-α).
Writeupk=u~pk+u-pk, where u-pk=(1/Tpk)∑n=1Tpkupk(n)=(1/Tpk)∑n=1TpkLupk(n)∈Y-. Inequality (46) implies that
(47)∥u-pk∥∞≜maxn∈Z[1,Tpk]{|u-pk|}≤1Tpk∑n=1Tpk‍|Lupk(n)|≤(β+γ)sin(π/T)δ(sin(π/T)-α).
By Lemma 7 and (45), it follows that
(48)∥u~pk∥2=∑n=1Tpk‍|u~pk(n)|2≤(2sinπTpk)-1∑n=1Tpk‍|Δupk(n)|2≤(2sin(π/T))-12α(β+γ)Tsin(π/T)sin(π/T)-α≤α(β+γ)Tsin(π/T)-α,
which implies that {∥u~pk∥∞} is bounded, therefore {∥upk∥∞} is bounded; a contradiction with the second claim limp→∞∥up∥∞=∞. This completes the proof.Proof of Theorem2.
Under the assumptions (A1) and (A2), all conditions in Theorem1 are satisfied. Then, for each integer p>1, there exists a pT-periodic solution u of (3) such that v=-J[u-(1/pT)∑n=1pTu(n)]∈Y minimizes the dual action
(49)χ(v)=∑n=1pT‍12(LJΔv(n-1),v(n))+∑n=1pT‍H*(n,Δv(n))onEpT.
If the critical pointv of dual action functional χ has minimal period pT/l∈ℕ∖{0}, where l∈ℕ∖{0}, then by Lemma 7 with pT replaced by pT/l, we have the following estimate:
(50)4sin2lπpT∑n=1pT‍|v(n)|2≤∑n=1pT‍|Δv(n)|2.
By Lemma 5 and the previous inequality, we have
(51)∑n=1pT‍(LJΔv(n-1),v(n))≥-(∑n=1pT‍|LJΔv(n-1)|2)1/2·(∑n=1pT‍|v(n)|2)1/2≥-(∑n=1pT‍|Δv(n)|2)1/2·(2sinlπpT)-1(∑n=1pT‍|Δv(n)|2)1/2=-(2sinlπpT)-1∑n=1pT‍|Δv(n)|2≥-(2sinlπpT)-1(pT)(1-2/τ)(∑n=1pT‍|Δv(n)|τ)2/τ,
where τ=θ/(θ-1)>2 for 1<θ<2. It follows from assumption (B2) that
(52)H*(n,Δv(n))≥1τ(1a2)τ-1|Δv(n)|τ,
thus
(53)χ(v)≥-(2sinlπpT)-1(pT)(τ-2)/τ(∑n=1pT‍|Δv(n)|τ)2/τ+1τ(1a2)τ-1∑n=1pT‍|Δv(n)|τ(54)≥(1/τ-1/2)pT(a22)(τ-1)/(τ-2)(sin(lπ/pT))τ/(τ-2).
One can obtain the previous inequality by minimizing in (53) with respect to (∑n=1pT|Δv(n)|τ)1/τ, and the minimum is attained at (pT)1/τ(a2)(τ-1)/(τ-2)/(sin(lπ/pT))1/(τ-2).
On the other hand, let(55)v(n)=1pT(cos2kπpTn·ak-sin2kπpT(n-12)·ak),
where ak∈ℝN, k∈Z[[-pT/2],[pT/2]]∖{0}. Then v∈Yk, and
(56)Δv(n)=-2sinkπpT1pT(sin2kπpT(n+12)·akcos2kπpTn·ak).
Taking ak=(d,0,…,0)T∈ℝN, where d∈ℝ, by Lemma 4, it follows that
(57)∑n=1pT‍(LJΔv(n-1),v(n))=∑n=1pT‍[-Δv2(n)v1(v)+Δv1(n-1)v2(n)]=∑n=1pT‍1pT·2sinkπpT·(cos22kπpTn·|d|2+sin22kπpT(n-12)·|d|2)=λk·|d|2,
where λk=2sin(kπ/pT) and
(58)∑n=1pT‍|Δv(n)|τ=∑n=1pT‍|λk|τ(pT)-τ/2·(sin22kπpT(n+12)+cos22kπpTn)τ/2|d|τ≤λmaxτ·(pT)1-(τ/2)·2τ/2|d|τ.
Therefore, taking k=-[pT/2], by eigenvalue problem (24) and (B2), it follows that
(59)χ(v)=12∑n=1pT‍(LJΔv(n-1),v(n))+∑n=1pT‍H*(n,Δv(n))≤-12λmax·|d|2+1τ(1a1)τ-1∑n=1pT‍|Δv(n)|τ≤-12λmax·|d|2+1τ(1a1)τ-1λmaxτ·(pT)1-(τ/2)·2τ/2|d|τ.
Letf(ρ) equal the right-hand side of (59) where ρ=|d|. It is easy to see that the absolute minimum of f is attained at ρmin=(a1)(τ-1)/(τ-2)(pT)1/2/[λmax(τ-1)/(τ-2)·2τ/2(τ-2)] and given by fmin=(1/τ-1/2)pT(a12)(τ-1)/(τ-2)/(2λmax)τ/(τ-2). Hence, by (19), let
(60)ξ(n)=ξ-[pT/2](n)=(cos2kπpTn·ρ-sin2kπpT(n-12)·ρ),
where ρ∈ℝN,k=-[pT/2].
IfpT is even, then ξ(n)=(1,1)T·(-1)nρ. Set
(61)Yρmin={v∈Y-[pT/2]:v(n)=ξ(n),ρ=(d,0,…,0)T∈ℝN,d∈ℝY-[pT/2]}.
For v∈Yρmin, we have
(62)χ(v)≤fmin.
Combining (54), (59), and (62), we have
(63)(1/τ-1/2)pT(a22)(τ-1)/(τ-2)(sin(lπ/pT))τ/(τ-2)≤(1/τ-1/2)pT(a12)(τ-1)/(τ-2)(2λmax)τ/(τ-2).
By τ>2, and θ=τ/(τ-1), it follows that
(64)sin(lπ/pT)(2λmax)≤(a2/a1)2/θ.
For integer p>1, T≥1, l∈ℕ∖{0}, pT/l∈ℕ∖{0}, we have 0<lπ/pT≤π, 0<π/pT≤π/2.
IfpT is even, then λmax=2. By assumption a2/a1≤((1/4)sin(π/pT))θ/2 we have sin(lπ/pT)≤sin(π/pT), which implies that l=1 or l=pT-1. If pT>2, then pT/l=pT/(pT-1)∉ℕ. So we have l=1.
IfpT is odd, then λmax=2cos(π/2pT). By assumption a2/a1≤((1/2)sin(π/2pT))θ/2, we have sin(lπ/pT)≤sin(π/pT), so l=1. This completes the proof.



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*Source: 508247-2013-04-03.xml*