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Research article

Covering cross-polytopes with smaller homothetic copies

  • Received: 30 November 2023 Revised: 08 January 2024 Accepted: 10 January 2024 Published: 11 January 2024
  • MSC : 52A20, 52C17, 52A15

  • Let Cn be an n-dimensional cross-polytope and Γp(Cn) be the smallest positive number γ such that Cn can be covered by p translates of γCn. We obtain better estimates of Γ2n(Cn) for small n and a universal upper bound of Γ2n(Cn) for all positive integers n.

    Citation: Feifei Chen, Shenghua Gao, Senlin Wu. Covering cross-polytopes with smaller homothetic copies[J]. AIMS Mathematics, 2024, 9(2): 4014-4020. doi: 10.3934/math.2024195

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  • Let Cn be an n-dimensional cross-polytope and Γp(Cn) be the smallest positive number γ such that Cn can be covered by p translates of γCn. We obtain better estimates of Γ2n(Cn) for small n and a universal upper bound of Γ2n(Cn) for all positive integers n.



    Necessary and sufficient conditions for optimality play a crucial role in solving problems in the calculus of variations. The main necessary conditions for such problems are the well-known conditions of Euler, Weierstrass, Legendre and Jacobi. In many cases, depending on the smoothness of the assumptions, Euler's necessary condition corresponds to a second order differential equation which restricts solutions to lie in a family of trajectories with certain uniformity properties. Also, the necessary condition of Jacobi cannot be applied when the extremal has corners and is not nonsingular. This is an unfortunate feature since, in general, the admissible arcs or trajectories which are candidates for solving the problem are neither nonsingular nor smooth.

    On the other hand, one fundamental aspect of the theory of sufficient conditions for optimality consists in slightly strengthening the necessary conditions. Concretely, if an admissible arc satisfies the strengthened conditions of Euler, Legendre and Jacobi, then it is a strict weak minimum. Additionally, if this admissible arc also satisfies the strengthened condition of Weierstrass, then it is a strict strong minimum. Some of the techniques used to obtain sufficiency include the construction of a Mayer field on which the extremals are independent of the path with respect to an invariant integral commonly called the Hilbert integral, the existence of a symmetric solution of the matrix Riccati inequality associated with the problem, a verification function satisfying the Hamilton-Jacobi equation, a quadratic function that satisfies a Hamilton-Jacobi inequality, the nonexistence of conjugate points on the underlying half-closed time interval, or the incorporation of some convexity arguments on the functions delimiting the calculus of variations problem (see for example [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17] and references therein).

    It is important to mention that the smoothness and the nonsingularity assumptions are crucial in the sufficiency theories mentioned above. In other words, the classical sufficiency theory in the calculus of variations, in general, may not give a response when the extremal under consideration is singular or has corners. In fact, as mentioned in [6], there is a gap between the set of necessary and sufficient conditions and [6,Section 3.7] is entirely devoted to the study of some problems for which the nonsingularity assumption fails. There, one finds a method which is only applicable to particular examples and may not hold in general, since "although this algorithm sheds light on the theory, it provides no panacea. Indeed there are no panaceas." Additionally, we refer the reader to [13], where the importance of the nonsingularity assumption in the classical calculus of variations sufficiency theory is fully explained.

    In this paper we derive two new sufficiency results which provide sufficient conditions for strong local minima in certain classes of parametric and nonparametric calculus of variations problems of Bolza with variable or free end-points, inequality and equality nonlinear isoperimetric constrains, and nonlinear mixed pointwise inequality and equality constraints. The main novelty of our new sufficient theorems concerns their applicability to cases in which the extremals under consideration may be singular and nonsmooth, that is, the strengthened condition of Legendre and the continuity of the derivative of the proposed extremal are no longer required. More precisely, given an admissible extremal whose derivative is not continuous nor piecewise continuous but only essentially bounded, the elements comprising the new sufficiency theorems are the classical transversality condition, a crucial inequality which arises from the original algorithm used to prove the main result of the article, the necessary condition of Legendre, but not its strengthened version, the positivity of the second variation over the set of all nonnul admissible variations, and three refined Weierstrass conditions which are related to the functions delimiting the problems.

    Another distinguishing characteristic of the main sufficiency result of the nonparametric calculus of variations problem presented in this paper is the fact that the initial and final end-points of the states are completely free, that is, they are not only variable end-points but they may belong to any set which is not necessarily a smooth manifold described by some functions which usually involve some type of equality or inequality conditions.

    The paper is organized as follows. In Section 2 we pose the parametric calculus of variations problem we shall deal with together with some basic definitions and the statement of the main result of the article. In Section 3 we enunciate the nonparametric calculus of variations problem we shall study together with some basic definitions and a corollary which is also one of the main results of the paper. Section 4 is devoted to state two auxiliary lemmas in which the proof of the theorem is strongly based. Section 5 is dedicated to the proof of the main theorem of the article. In Section 6 we prove the lemmas given in Section 4 and, in the final section, we provide an example which shows how the sufficient theory developed in this paper widens the range of applicability of the classical calculus of variations theory.

    Suppose we are given an interval T:=[t0,t1] in R, and functions l(b):RpR, lγ(b):RpR (γ=1,,K), Ψi(b):RpRn (i=0,1), L(t,x,˙x):T×Rn×RnR, Lγ(t,x,˙x):T×Rn×RnR (γ=1,,K) and φ(t,x,˙x):T×Rn×RnRs. Set

    R:={(t,x,˙x)T×Rn×Rnφα(t,x,˙x)0(αR),φβ(t,x,˙x)=0(βS)}

    where R:={1,,r} and S:={r+1,,s} (r=0,1,,s). If r=0 then R= and we disregard statements involving φα. Similarly, if r=s then S= and we disregard statements involving φβ.

    It will be assumed throughout the paper that L, Lγ (γ=1,,K) and φ=(φ1,,φs) have first and second derivatives with respect to x and ˙x. Moreover, we shall assume that the functions l, lγ (γ=1,,K) and Ψi (i=0,1) are of class C2 on Rp. Also, if we denote by c(t,x,˙x) either L(t,x,˙x), Lγ(t,x,˙x) (γ=1,,K), φ(t,x,˙x) or any of its partial derivatives of order less than or equal to two with respect to x and ˙x, we shall assume that if C is any bounded subset of T×Rn×Rn, then |c(C)| is a bounded subset of R. Additionally, we shall assume that if {(Γq,Λq)} is any sequence in AC(T;Rn)×L1(T;Rn) such that for some UT measurable and some {(Γ0,Λ0)}AC(T;Rn)×L(T;Rn), (Γq(t),Λq(t))(Γ0(t),Λ0(t)) uniformly on U, then for all qN, c(t,Γq(t),Λq(t)) is measurable on U and

    c(t,Γq(t),Λq(t))c(t,Γ0(t),Λ0(t))uniformly on U.

    Note that all conditions above concerning the functions L, Lγ (γ=1,,K) and φ, are satisfied if the functions L, Lγ (γ=1,,K), φ and their first and second derivatives with respect to x and ˙x are continuous on T×Rn×Rn.

    Set

    X:=AC(T;Rn),Us:=L(T;Rs),A:=X×Rp.

    We shall use the notation xb to denote any element xb:=(x,b)A. Let B any subset of Rp which we shall call the set of parameters. The parametric calculus of variations problem we shall deal with, denoted by (P), is that of minimizing the functional

    I(xb):=l(b)+t1t0L(t,x(t),˙x(t))dt

    over all xbA satisfying the constraints

    {c(t,x(t),˙x(t))is integrable on T.bB.x(ti)=Ψi(b) for i=0,1.Ii(xb):=li(b)+t1t0Li(t,x(t),˙x(t))dt0(i=1,,k).Ij(xb):=lj(b)+t1t0Lj(t,x(t),˙x(t))dt=0(j=k+1,,K).(t,x(t),˙x(t))R(a.e. in T).

    Elements b=(b1,,bp) (the notation denotes transpose) in B will be called parameters, elements xb in A will be called arcs or trajectories, and a trajectory xb is admissible if it satisfies the constraints. The notation x0b0 refers to an element (x0,b0)A.

    Let us now introduce some definitions which will be used throughout the paper.

    An arc x0b0 solves (P) if it is admissible and I(x0b0)I(xb) for all admissible arcs xb. For strong local minima, an admissible arc x0b0 is called a strong minimum of (P) if it is a minimum of I relative to the following norm

    xb:=|b|+suptT|x(t)|=|b|+xC,

    that is, if for some ϵ>0, I(x0b0)I(xb) for all admissible arcs satisfying xbx0b0<ϵ.

    For all xX, we use the notation (˜x(t)) in order to represent (t,x(t),˙x(t)). Also, (˜x0(t)) represents (t,x0(t),˙x0(t)).

    Given K real numbers λ1,,λK, for any xb admissible define the functional I0 by

    I0(xb):=I(xb)+Kγ=1λγIγ(xb)=l0(b)+t1t0L0(˜x(t))dt,

    where l0:RpR is given by

    l0(b):=l(b)+Kγ=1λγlγ(b),

    and L0:T×Rn×RnR is given by

    L0(t,x,˙x):=L(t,x,˙x)+Kγ=1λγLγ(t,x,˙x).

    Given λ1,,λK, for all (t,x,˙x,ρ,μ)T×Rn×Rn×Rn×Rs, define the Hamiltonian of the problem by

    H(t,x,˙x,ρ,μ):=ρ,˙xL0(t,x,˙x)μ,φ(t,x,˙x),

    where ρRn denotes the adjoint variable and μRs is the associated multiplier of the mixed constraints.

    Given (ρ,μ)X×Us, and λ1,,λK, for all (t,x,˙x)T×Rn×Rn, define the following function associated to the Hamiltonian,

    F0(t,x,˙x):=H(t,x,˙x,ρ(t),μ(t))˙ρ(t),x.

    Given (ρ,μ)X×Us and λ1,,λK, for any xb admissible define the functional J0 by

    J0(xb):=ρ(t1),x(t1)ρ(t0),x(t0)+l0(b)+t1t0F0(˜x(t))dt.

    The notation yβ refers to any element (y,β) in A.

    Given (ρ,μ)X×Us, and λ1,,λK, for any xbA with ˙xL(T;Rn) and any yβA consider the first variations of J0 and Iγ (γ=1,,K) with respect to xb over yβ which are given, respectively, by

    J0(xb;yβ):=ρ(t1),y(t1)ρ(t0),y(t0)+l0(b)β+t1t0{F0x(˜x(t))y(t)+F0˙x(˜x(t))˙y(t)}dt,
    Iγ(xb;yβ):=lγ(b)β+t1t0{Lγx(˜x(t))y(t)+Lγ˙x(˜x(t))˙y(t)}dt.

    For all (t,x,˙x)T×Rn×Rn, denote by

    Ia(t,x,˙x):={αRφα(t,x,˙x)=0},

    the set of active indices of (t,x,˙x) with respect to the mixed inequality constraints.

    For all xbA, denote by

    ia(xb):={i=1,,kIi(xb)=0},

    the set of active indices of xb with respect to the isoperimetric inequality constraints.

    Given xbA, let Y(xb) be the set of all yβA with ˙yL2(T;Rn) satisfying

    {y(ti)=Ψi(b)β(i=0,1),Ii(xb;yβ)0(iia(xb)),Ij(xb;yβ)=0(j=k+1,,K),φαx(˜x(t))y(t)+φα˙x(˜x(t))˙y(t)0(a.e. in T,αIa(˜x(t))),φβx(˜x(t))y(t)+φβ˙x(˜x(t))˙y(t)=0(a.e. in T,βS).

    The set Y(xb) will be called the set of admissible variations along xb.

    Given (ρ,μ)X×Us, and λ1,,λK, for any xbA with ˙xL(T;Rn) and any yβA with ˙yL2(T;Rn), we define the second variation of J0 with respect to xb over yβ, by

    J0(xb;yβ):=l0(b)β,β+t1t02Ω0(x;t,y(t),˙y(t))dt,

    where for all (t,y,˙y)T×Rn×Rn,

    2Ω0(x;t,y,˙y):=y,F0xx(˜x(t))y+2y,F0x˙x(˜x(t))˙y+˙y,F0˙x˙x(˜x(t))˙y.

    Given (ρ,μ)X×Us, λ1,,λK and x0b0A, we say that x0b0 is singular, if for some τT, |H˙x˙x(˜x0(τ),ρ(τ),μ(τ))|=0. It satisfies the Legendre condition if

    F0˙x˙x(˜x0(t))=H˙x˙x(˜x0(t),ρ(t),μ(t))0(a.e. in T)

    and the strengthened Legendre condition, if F0˙x˙x(˜x0(t))>0 (tT).

    Denote by E0 the Weierstrass excess function of F0, given by

    E0(t,x,˙x,u):=F0(t,x,u)F0(t,x,˙x)F0˙x(t,x,˙x)(u˙x).

    Similarly, the Weierstrass excess function of Lγ (γ=1,,K), is given by

    Eγ(t,x,˙x,u):=Lγ(t,x,u)Lγ(t,x,˙x)Lγ˙x(t,x,˙x)(u˙x).

    For all π=(π1,,πn)Rn, set

    V(π):=(1+|π|2)1/21.

    For all xX, define

    D(x):=V(x(t0))+t1t0V(˙x(t))dt.

    As we mentioned above the symbol denotes transpose.

    It is well-known that, under certain normality assumptions (see for example [10]), if x0b0 is a strong minimum of (P), then there exist ρX and μUs with μα(t)0 and μα(t)φα(˜x0(t))=0 (αR,a.e. in T) and multipliers λ1,,λK with λi0 and λiIi(x0b0)=0 (i=1,,k) such that

    ˙ρ(t)=Hx(˜x0(t),ρ(t),μ(t))  and  H˙x(˜x0(t),ρ(t),μ(t))=0(a.e. in T).

    The relations given above are the Euler-Lagrange equations of the constrained problem (P) and if (x0,ρ,μ) satisfies the Euler-Lagrange equations, then (x0,ρ,μ) will be called an extremal.

    The following theorem is the main result of the article. Given an admissible arc x0b0 with ˙x0 neither continuous nor piecewise continuous but only essentially bounded, this theorem gives sufficient conditions assuring that x0b0 is a strong minimum of problem (P). Hypothesis (ⅰ) of Theorem 2.1 is known as the transversality condition, hypothesis (ⅱ) arises from the original proof of the theorem, condition (ⅲ) is the necessary condition of Legendre, but not its strengthened version, hypothesis (ⅳ) is the positivity of the second variation over the set of all nonnull admissible variations and finally, hypothesis (ⅴ) involves three conditions related to the Weierstrass excess functions.

    2.1 Theorem: Let x0b0 be an admissible arc with ˙x0L(T;Rn). Assume that Ia(˜x0()) is piecewise constant on T, and there exist (ρ,μ)X×Us with μα(t)0 and μα(t)φα(˜x0(t))=0 (αR,a.e. in T), two positive numbers h,ϵ, and multipliers λ1,,λK with λi0 and λiIi(x0b0)=0 (i=1,,k) such that (x0,ρ,μ) is an extremal and the following holds:

    (i) l0(b0)+ρ(t1)Ψ1(b0)ρ(t0)Ψ0(b0)=0.

    (ii) ρ(t1)Ψ1(b0;β)ρ(t0)Ψ0(b0;β)0for all βRp.

    (iii) H˙x˙x(˜x0(t),ρ(t),μ(t))0 (a.e. in T).

    (iv) J0(x0b0;yβ)>0 for all nonnull yβY(x0b0).

    (v) For all xb admissible with xx0C<ϵ,

    a. E0(t,x(t),˙x0(t),˙x(t))0 (a.e. in T).

    b. t1t0E0(t,x(t),˙x0(t),˙x(t))dtht1t0V(˙x(t)˙x0(t))dt.

    c. t1t0E0(t,x(t),˙x0(t),˙x(t))dth|t1t0Eγ(t,x(t),˙x0(t),˙x(t))dt| (γ=1,,K).

    Then for some θ1,θ2>0 and all admissible trajectories xb satisfying xbx0b0<θ1,

    I(xb)I(x0b0)+θ2min{|bb0|2,D(xx0)}.

    In particular, x0b0 is a strong minimum of (P).

    Suppose we are given an interval T:=[t0,t1] in R, two sets B0,B1Rn and functions (x1,x2):Rn×RnR, γ(x1,x2):Rn×RnR (γ=1,,K), L(t,x,˙x):T×Rn×RnR, Lγ(t,x,˙x):T×Rn×RnR (γ=1,,K) and ϕ(t,x,˙x):T×Rn×RnRs. Set

    ˉR:={(t,x,˙x)T×Rn×Rnϕα(t,x,˙x)0(αR),ϕβ(t,x,˙x)=0(βS)}

    where R:={1,,r} and S:={r+1,,s} (r=0,1,,s). If r=0 then R= and we disregard statements involving φα. Similarly, if r=s then S= and we disregard statements involving φβ.

    It will be assumed throughout this section that L, Lγ (γ=1,,K) and ϕ=(ϕ1,,ϕs) have first and second derivatives with respect to x and ˙x. Moreover, we shall assume that the functions , γ (γ=1,,K) are of class C2 on Rn×Rn.

    Also, if we denote by c(t,x,˙x) either L(t,x,˙x), Lγ(t,x,˙x) (γ=1,,K), ϕ(t,x,˙x) or any of its partial derivatives of order less than or equal to two with respect to x and ˙x, we shall assume that all the assumptions posed in Section 2 in the statement of the problem are satisfied.

    As in Section 2, X will denote the space of absolutely continuous functions mapping T to Rn and Us:=L(T;Rs) the space of essentially bounded functions mapping T to Rs.

    The nonparametric calculus of variations problem we shall deal with, denoted by (ˉP), consists in minimizing the functional

    J(x):=(x(t0),x(t1))+t1t0L(t,x(t),˙x(t))dt

    over all xX satisfying the constraints

    {c(t,x(t),˙x(t))is integrable on T.x(ti)Bi for i=0,1.Ji(x):=i(x(t0),x(t1))+t1t0Li(t,x(t),˙x(t))dt0(i=1,,k).Jj(x):=j(x(t0),x(t1))+t1t0Lj(t,x(t),˙x(t))dt=0(j=k+1,,K).(t,x(t),˙x(t))ˉR(a.e. in T).

    Elements x in X will be called arcs or trajectories, and a trajectory x is admissible if it satisfies the constraints.

    An arc x0 solves (ˉP) if it is admissible and J(x0)J(x) for all admissible arcs x. For strong minima, an admissible arc x0 is called a strong minimum of (ˉP) if it is a minimum of J relative to the norm

    x:=suptT|x(t)|,

    that is, if for some ϵ>0, J(x0)J(x) for all admissible arcs satisfying xx0<ϵ.

    Let Ψ:RnRn×Rn be any function of class C2 such that B0×B1Ψ(Rn). Associate the nonparametric problem (ˉP) with the parametric problem of Section 2, which we denote by (PΨ), that is, (PΨ) will be the parametric problem given in Section 2, with p=n, B=Ψ1(B0×B1), l=Ψ, lγ=γΨ (γ=1,,K), L=L, Lγ=Lγ (γ=1,,K), φ=ϕ and Ψ0,Ψ1 the components of Ψ, that is, Ψ=(Ψ0,Ψ1). Recall that the notation xb means (x,b) where bRn is a parameter.

    3.1 Lemma: The following is satisfied:

    (i) xb is an admissible arc of (PΨ) if and only if x is an admissible arc of (ˉP) and bΨ1(x(t0),x(t1)).

    (ii) If xb is an admissible arc of (PΨ), then

    J(x)=I(xb).

    (iii) If x0b0 is a solution of (PΨ), then x0 is a solution of (ˉP).

    Proof: Conditions (ⅰ) and (ⅱ) follow from the definitions of the problems. Now, let x an admissible arc of (ˉP) and let bΨ1(x(t0),x(t1)). By (ⅰ), x0 is an admissible arc of (ˉP) and xb is an admissible arc of (PΨ). Then by (ⅱ),

    J(x0)=I(x0b0)I(xb)=J(x)

    which shows (ⅲ).

    The following corollary which is a consequence of Theorem 2.1 and Lemma 3.1, provides a set of sufficient conditions of problem (ˉP).

    3.2 Corollary: Let Ψ:RnRn×Rn be any function of class C2 such that B0×B1Ψ(Rn) and let (PΨ) be the parametric problem defined in the previous paragraph of Lemma 3.1. Let x0b0 be an admissible arc of (PΨ) with ˙x0L(T;Rn). Assume that Ia(˜x0()) is piecewise constant on T, and there exist (ρ,μ)X×Us with μα(t)0 and μα(t)φα(˜x0(t))=0 (αR,a.e. in T), two positive numbers h,ϵ, and multipliers λ1,,λK with λi0 and λiIi(x0b0)=0 (i=1,,k) such that (x0,ρ,μ) is an extremal and the following holds:

    (i) l0(b0)+ρ(t1)Ψ1(b0)ρ(t0)Ψ0(b0)=0.

    (ii) ρ(t1)Ψ1(b0;β)ρ(t0)Ψ0(b0;β)0for all βRn.

    (iii) H˙x˙x(˜x0(t),ρ(t),μ(t))0 (a.e. in T).

    (iv) J0(x0b0;yβ)>0 for all nonnull yβY(x0b0).

    (v) For all xb admissible with xx0<ϵ,

    a. E0(t,x(t),˙x0(t),˙x(t))0 (a.e. in T).

    b. t1t0E0(t,x(t),˙x0(t),˙x(t))dtht1t0V(˙x(t)˙x0(t))dt.

    c. t1t0E0(t,x(t),˙x0(t),˙x(t))dth|t1t0Eγ(t,x(t),˙x0(t),˙x(t))dt| (γ=1,,K).

    Then, x0 is a strong minimum of (ˉP).

    In this section we state two auxiliary results which will be used to prove Theorem 2.1. The proof of these results will be given in Section 6. As before X denotes AC(T;Rn).

    In the following two lemmas, we shall assume that we are given x0X and {xq} a sequence in X such that

    limqD(xqx0)=0  and  dq:=[2D(xqx0)]1/2>0(qN).

    For all qN and tT, let

    yq(t):=xq(t)x0(t)dq.

    We say that ˙xq(t)˙x0(t) almost uniformly on T, if for any ϵ>0, there exists UϵT measurable with m(Uϵ)<ϵ such that ˙xq(t)˙x0(t) uniformly on TUϵ.

    4.1 Lemma: For some subsequence of {xq}, again denoted by {xq}, and some y0X with ˙y0L2(T;Rn), ˙xq(t)˙x0(t) almost uniformly on T, yq(t)y0(t) uniformly on T and {˙yq} converges weakly in L1(T;Rn) to ˙y0.

    4.2 Lemma: Let UT measurable, R0L(U;Rn×n) and {Rq} a sequence in L(U;Rn×n). If ˙xq(t)˙x0(t) uniformly on U, Rq(t)R0(t) uniformly on U and R0(t)0 (tU), then

    lim infqURq(t)˙yq(t),˙yq(t)dtUR0(t)˙y0(t),˙y0(t)dt.

    The proof of Theorem 2.1 will be divided in three Lemmas. In Lemmas 5.1, 5.2 and 5.3 below, we shall be assuming that all the hypotheses of Theorem 2.1 are satisfied. Before enunciating the lemmas, we shall introduce some definitions.

    First of all, note that given x=(x1,,xn)Rn and b=(b1,,bp)Rp, if we define xi,bjRn+p by xi:=(x1,,xn,0,,0) and bj:=(0,,0,b1,,bp), then

    xi+bj=(x1,,xn,b1,,bp)=(xb)Rn+p.

    Define ˜F0:T×Rn+p×RnR by

    ˜F0(t,ξ,˙x):=l0(ξn+1,,ξn+p)t1t0+F0(t,ξ1,,ξn,˙x).

    Observe that the Weierstrass excess function ˜E0:T×Rn+p×Rn×RnR of ˜F0 is given by

    ˜E0(t,ξ,˙x,u):=˜F0(t,ξ,u)˜F0(t,ξ,˙x)˜F0˙x(t,ξ,˙x)(u˙x).

    It is clear that for all (t,x,˙x,u)T×Rn×Rn×Rn and all bRp,

    ˜E0(t,xi+bj,˙x,u)=E0(t,x,˙x,u).

    Define

    ˜J0(xb):=ρ(t1),x(t1)ρ(t0),x(t0)+t1t0˜F0(t,x(t)i+bj,˙x(t))dt.

    We have that J0(xb)=˜J0(xb) for all xbA, and

    ˜J0(xb)=˜J0(x0b0)+˜J0(x0b0;xbx0b0)+˜K0(x0b0;xb)+˜E0(x0b0;xb) (1)

    where

    ˜E0(x0b0;xb):=t1t0˜E0(t,x(t)i+bj,˙x0(t),˙x(t))dt,˜K0(x0b0;xb):=t1t0{˜M0(t,x(t)i+bj)+˙x(t)˙x0(t),˜N0(t,x(t)i+bj)}dt,˜J0(x0b0;xbx0b0):=ρ(t1),x(t1)x0(t1)ρ(t0),x(t0)x0(t0)+t1t0{˜F0ξ(t,x0(t)i+b0j,˙x0(t))([x(t)x0(t)]i+[bb0]j)+˜F0˙x(t,x0(t)i+b0j,˙x0(t))(˙x(t)˙x0(t))}dt,

    and ˜M0, ˜N0 are given by

    ˜M0(t,xi+bj):=˜F0(t,xi+bj,˙x0(t))˜F0(t,x0(t)i+b0j,˙x0(t))˜F0ξ(t,x0(t)i+b0j,˙x0(t))([xx0(t)]i+[bb0]j),
    ˜N0(t,xi+bj):=˜F0˙x(t,xi+bj,˙x0(t))˜F0˙x(t,x0(t)i+b0j,˙x0(t)).

    We have,

    ˜M0(t,xi+bj)=12[xx0(t)]i+[bb0]j,˜P0(t,xi+bj)([xx0(t)]i+[bb0]j), (2a)
    ˜N0(t,xi+bj)=˜Q0(t,xi+bj)([xx0(t)]i+[bb0]j), (2b)

    where

    ˜P0(t,xi+bj):=210(1λ)˜F0ξξ(t,[x0(t)+λ(xx0(t))]i+[b0+λ(bb0)]j,˙x0(t))dλ,˜Q0(t,xi+bj):=10˜F0˙xξ(t,[x0(t)+λ(xx0(t))]i+[b0+λ(bb0)]j,˙x0(t))dλ.

    5.1 Lemma: For some ν,κ>0 (κϵ) and any admissible arc xb satisfying xbx0b0<κ,

    ˜E0(x0b0;xb)h[D(xx0)V(x(t0)x0(t0))],|˜K0(x0b0;xb)|νxbx0b0[1+D(xx0)].

    Proof: By condition (v)(b) of Theorem 2.1, given xb admissible with xbx0b0<ϵ,

    ˜E0(x0b0;xb)=t1t0˜E0(t,x(t)i+bj,˙x0(t),˙x(t))dt=t1t0E0(t,x(t),˙x0(t),˙x(t))dtht1t0V(˙x(t)˙x0(t))dt=h[D(xx0)V(x(t0)x0(t0))].

    On the other hand, by (2) and using [t,b] in order to denote (t,x(t)i+bj), observe that for some constants c0,c1>0, for all xb admissible with xbx0b0<1 and almost all tT,

    |˜M0[t,b]+˙x(t)˙x0(t),˜N0[t,b]|=|12[x(t)x0(t)]i+[bb0]j,˜P0[t,b]([x(t)x0(t)]i+[bb0]j)+2˜Q0[t,b](˙x(t)˙x0(t))|12|[x(t)x0(t)]i+[bb0]j|[|˜P0[t,b]||[x(t)x0(t)]i+[bb0]j|+2|˜Q0[t,b]||˙x(t)˙x0(t)|]c0|[x(t)x0(t)]i+[bb0]j|[|[x(t)x0(t)]i+[bb0]j|+|˙x(t)˙x0(t)|]c0|[x(t)x0(t)]i+[bb0]j|[|x(t)x0(t)|+|bb0|+|˙x(t)˙x0(t)|]c0|[x(t)x0(t)]i+[bb0]j|[xbx0b0+|˙x(t)˙x0(t)|]c0|[x(t)x0(t)]i+[bb0]j|[1+|˙x(t)˙x0(t)|]c1|[x(t)x0(t)]i+[bb0]j|[1+|˙x(t)˙x0(t)|2]1/2.

    Setting ν:=max{c1,c1(t1t0)}, for all xb admissible with xbx0b0<1,

    |˜K0(x0b0;xb)|c1xbx0b0t1t0[1+V(˙x(t)˙x0(t))]dtνxbx0b0[1+D(xx0)V(x(t0)x0(t0))]νxbx0b0[1+D(xx0)]

    and so the conclusion of the lemma is obtained with κ=min{ϵ,1}.

    5.2 Lemma: If conclusion of Theorem 2.1 is false, then there exists a subsequence {xqbq} of admissible arcs such that

    limqD(xqx0)=0  and  dq:=[2D(xqx0)]1/2>0(qN).

    Proof: If conclusion of Theorem 2.1 is false, then for all θ1,θ2>0, there exists an admissible arc xb such that

    xbx0b0<θ1andI(xb)<I(x0b0)+θ2min{|bb0|2,D(xx0)}. (3)

    Since

    μα(t)0 (αR,a.e. in T)  and  λi0 (i=1,,k),

    if xb is admissible, then I(xb)J0(xb). Also, since

    μα(t)φα(˜x0(t))=0 (αR,a.e. in T)  and  λiIi(x0b0)=0 (i=1,,k),

    then I(x0b0)=J0(x0b0). Therefore (3) implies that, for all θ1,θ2>0, there exists xb admissible with

    xbx0b0<θ1  and  J0(xb)<J0(x0b0)+θ2min{|bb0|2,D(xx0)}.

    Let κ and ν be the positive numbers given in Lemma 5.1. Thus, if conclusion of Theorem 2.1 is false, then for all qN, there exists xqbq admissible such that

    xqbqx0b0<min{κ,1/q},J0(xqbq)J0(x0b0)<min{|bqb0|2q,D(xqx0)q}. (4)

    Clearly, D(xqx0)=0 if and only if xq=x0. Then, by the second relation of (4),

    D(xqx0)=0bqb0.

    Suppose D(xqx0)=0 for infinitely many q's. For i=0,1, we have

    0=xq(ti)x0(ti)=Ψi(bq)Ψi(b0)=10Ψi(b0+λ[bqb0])(bqb0)dλ, (5)
    0=Ψi(bq)Ψi(b0)=Ψi(b0)(bqb0)+10(1λ)Ψi(b0+λ[bqb0];bqb0)dλ. (6)

    Denoting by (bq,b0) the line segment in Rp joining the points bq and b0, by the second relation of (4), by condition (ⅰ) of Theorem 2.1, by (6), and the mean value theorem, there exists Ξq(bq,b0) such that

    0>J0(x0bq)J0(x0b0)=l0(bq)l0(b0)=l0(b0)(bqb0)+12l0(Ξq)(bqb0),bqb0=ρ(t0)Ψ0(b0)(bqb0)ρ(t1)Ψ1(b0)(bqb0)+12l0(Ξq)(bqb0),bqb0=1i=0(1)i+110(1λ)ρ(ti)Ψi(b0+λ[bqb0];bqb0)dλ+12l0(Ξq)(bqb0),bqb0. (7)

    Choose an appropriate subsequence of {(bqb0)/|bqb0|} (without relabeling), such that

    limqbqb0|bqb0|=β0 (8)

    for some β0Rp with |β0|=1. By (5),

    Ψi(b0)β0=0(i=0,1). (9)

    For all (t,ξ,˙x)T×Rn+p×Rn and γ=1,,K, if we set

    ˜Lγ(t,ξ,˙x):=lγ(ξn+1,,ξn+p)t1t0+Lγ(t,ξ1,,ξn,˙x),

    and for all (t,ξ,˙x,u)T×Rn+p×Rn×Rn and γ=1,,K, if we set

    ˜Eγ(t,ξ,˙x,u):=˜Lγ(t,ξ,u)˜Lγ(t,ξ,˙x)˜Lγ˙x(t,ξ,˙x)(u˙x),

    we have that for all xbA and γ=1,,K,

    ˜Iγ(xb)=˜Iγ(x0b0)+˜Iγ(x0b0;xbx0b0)+˜Kγ(x0b0;xb)+˜Eγ(x0b0;xb)

    where

    ˜Eγ(x0b0;xb):=t1t0˜Eγ(t,x(t)i+bj,˙x0(t),˙x(t))dt,˜Kγ(x0b0;xb):=t1t0{˜Mγ(t,x(t)i+bj)+˙x(t)˙x0(t),˜Nγ(t,x(t)i+bj)}dt,˜Iγ(x0b0;xbx0b0):=t1t0{˜Lγξ(t,x0(t)i+b0j,˙x0(t))([x(t)x0(t)]i+[bb0]j)+˜Lγ˙x(t,x0(t)i+b0j,˙x0(t))(˙x(t)˙x0(t))}dt,˜Iγ(xb):=t1t0˜Lγ(t,x(t)i+bj,˙x(t))dt,

    and ˜Mγ, ˜Nγ are given by

    ˜Mγ(t,xi+bj):=˜Lγ(t,xi+bj,˙x0(t))˜Lγ(t,x0(t)i+b0j,˙x0(t))˜Lγξ(t,x0(t)i+b0j,˙x0(t))([xx0(t)]i+[bb0]j),
    ˜Nγ(t,xi+bj):=˜Lγ˙x(t,xi+bj,˙x0(t))˜Lγ˙x(t,x0(t)i+b0j,˙x0(t)).

    We have

    ˜Mγ(t,xi+bj)=12[xx0(t)]i+[bb0]j,˜Pγ(t,xi+bj)([xx0(t)]i+[bb0]j),
    ˜Nγ(t,xi+bj)=˜Qγ(t,xi+bj)([xx0(t)]i+[bb0]j),

    where

    ˜Pγ(t,xi+bj):=210(1λ)˜Lγξξ(t,[x0(t)+λ(xx0(t))]i+[b0+λ(bb0)]j,˙x0(t))dλ,˜Qγ(t,xi+bj):=10˜Lγ˙xξ(t,[x0(t)+λ(xx0(t))]i+[b0+λ(bb0)]j,˙x0(t))dλ.

    Since x0bq and x0b0 are admissible, for all iia(x0b0), we have

    0Ii(x0bq)=Ii(x0bq)Ii(x0b0)=˜Ii(x0bq)˜Ii(x0b0)=˜Ii(x0b0;x0bqx0b0)+˜Ki(x0b0;x0bq)+˜Ei(x0b0;x0bq)=Ii(x0b0;x0bqx0b0)+˜Ki(x0b0;x0bq)=li(b0)(bqb0)+˜Ki(x0b0;x0bq).

    As one readily verifies, for all γ=1,,K,

    limq˜Kγ(x0b0;x0bq)|bqb0|=0.

    Then, for iia(x0b0),

    0li(b0)β0=Ii(x0b0;0β0). (10)

    On the other hand, once again since x0bq and x0b0 are admissible, for all j=k+1,,K, we have

    0=Ij(x0bq)Ij(x0b0)=˜Ij(x0bq)˜Ij(x0b0)=˜Ij(x0b0;x0bqx0b0)+˜Kj(x0b0;x0bq)+˜Ej(x0b0;x0bq)=Ij(x0b0;x0bqx0b0)+˜Kj(x0b0;x0bq)=lj(b0)(bqb0)+˜Kj(x0b0;x0bq).

    Then, for j=k+1,,K,

    0=lj(b0)β0=Ij(x0b0;0β0). (11)

    Consequently, by (9), (10) and (11), 0β0Y(x0b0).

    By (7), (8) and condition (ⅱ) of Theorem 2.1, it follows that

    012[ρ(t1)Ψ1(b0;β0)ρ(t0)Ψ0(b0;β0)+l0(b0)β0,β0]12l0(b0)β0,β0=12J0(x0b0;0β0)

    which contradicts (ⅳ) of Theorem 2.1. Therefore, we may assume that for all qN,

    dq=[2D(xqx0)]1/2>0.

    Since (x0,ρ,μ) is an extremal, for all qN,

    ˜J0(x0b0;xqbqx0b0)=ρ(t1),xq(t1)x0(t1)ρ(t0),xq(t0)x0(t0)+l0(b0)(bqb0)

    and thus

    limq˜J0(x0b0;xqbqx0b0)=0. (12)

    By (1), the first relation of (4) and Lemma 5.1, for all qN,

    ˜J0(xqbq)˜J0(x0b0)=˜J0(x0b0;xqbqx0b0)+˜K0(x0b0;xqbq)+˜E0(x0b0;xqbq)˜J0(x0b0;xqbqx0b0)νxqbqx0b0+D(xqx0)(hνxqbqx0b0)hV(xq(t0)x0(t0)),

    then, by (4), for all qN,

    D(xqx0)(hνq1q)<νq+hV(xq(t0)x0(t0))˜J0(x0b0;xqbqx0b0).

    By (12),

    limqD(xqx0)=0.

    5.3 Lemma: If conclusion of Theorem 2.1 is false, then condition (iv) of Theorem 2.1 is false.

    Proof: Let {xqbq} be the sequence of admissible arcs given in Lemma 5.2. Then,

    limqD(xqx0)=0  and  dq=[2D(xqx0)]1/2>0(qN).

    Case (1): Suppose first that the sequence {(bqb0)/dq} is bounded in Rp.

    For all qN and tT, define

    yq(t):=xq(t)x0(t)dq  and  ωq(t):=yq(t)i+bqb0dqj.

    By Lemma 4.1, for some y0X with ˙y0L2(T;Rn) and a subsequence of {xq}, again denoted by {xq}, {˙yq} converges weakly in L1(T;Rn) to ˙y0. Once again, by Lemma 4.1,

    limqyq(t)=y0(t)uniformly on  T. (13)

    Since the sequence {(bqb0)/dq} is bounded in Rp, then we may assume that there exists some β0Rp such that

    limqbqb0dq=β0. (14)

    First, we are going to show that for i=0,1,

    y0(ti)=Ψi(b0)β0. (15)

    Note first that for i=0,1 and all qN, we have that

    yq(ti)=10Ψi(b0+λ[bqb0])(bqb0)dqdλ. (16)

    By (13), (14) and (16), we obtain (15). Now, we claim that

    J0(x0b0;y0β0)0  and  y0β0(0,0). (17)

    To prove it, observe that by (2), (13) and (14),

    limq˜M0(t,xq(t)i+bqj)d2q=limq12ωq(t),˜P0(t,xq(t)i+bqj)ωq(t)=12y0(t)i+β0j,˜F0ξξ(t,x0(t)i+b0j,˙x0(t))[y0(t)i+β0j],
    limq˜N0(t,xq(t)i+bqj)dq=limq˜Q0(t,xq(t)i+bqj)ωq(t)=˜F0˙xξ(t,x0(t)i+b0j,˙x0(t))[y0(t)i+β0j]

    both uniformly on T. With this in mind and since {˙yq} converges weakly in L1(T;Rn) to ˙y0, we have

    limq˜K0(x0b0;xqbq)d2q=12t1t0{y0(t)i+β0j,˜F0ξξ(t,x0(t)i+b0j,˙x0(t))[y0(t)i+β0j]+2˙y0(t),˜F0˙xξ(t,x0(t)i+b0j,˙x0(t))[y0(t)i+β0j]}dt. (18)

    Since (x0,ρ,μ) is an extremal and by condition (ⅰ) of Theorem 2.1, it follows that

    limq˜J0(x0b0;xqbqx0b0)d2q=limq1d2q[ρ(t1),xq(t1)x0(t1)ρ(t0),xq(t0)x0(t0)+l0(b0)(bqb0)]=limq1d2q[ρ(t1)(Ψ1(bq)Ψ1(b0)Ψ1(b0)(bqb0))ρ(t0)(Ψ0(bq)Ψ0(b0)Ψ0(b0)(bqb0))]=limq1d2q101i=0(1)i+1(1λ)ρ(ti)Ψi(b0+λ[bqb0];bqb0)dλ=12[ρ(t1)Ψ1(b0;β0)ρ(t0)Ψ0(b0;β0)]. (19)

    Consequently, by (1), (4), (19), and condition (ⅱ) of Theorem 2.1,

    0limq˜K0(x0b0;xqbq)d2q+lim infq˜E0(x0b0;xqbq)d2q. (20)

    Now, let us show that

    lim infq˜E0(x0b0;xqbq)d2q12t1t0˙y0(t),˜F0˙x˙x(t,x0(t)i+b0j,˙x0(t))˙y0(t)dt. (21)

    To this end, let U a measurable subset of T such that ˙xq(t)˙x0(t) uniformly on U. For all qN and tU, we have that

    1d2q˜E0(t,xq(t)i+bqj,˙x0(t),˙xq(t))=12˙yq(t),Rq(t)˙yq(t),

    where

    Rq(t):=210(1λ)˜F0˙x˙x(t,xq(t)i+bqj,˙x0(t)+λ[˙xq(t)˙x0(t)])dλ.

    Clearly,

    limqRq(t)=R0(t):=˜F0˙x˙x(t,x0(t)i+b0j,˙x0(t))uniformly on U.

    By condition (ⅲ) of Theorem 2.1, we have

    ˜F0˙x˙x(t,x0(t)i+b0j,˙x0(t))=R0(t)0(tU).

    With this in mind, and since by (v)(a) of Theorem 2.1 for all qN,

    E0(t,xq(t),˙x0(t),˙xq(t))0(a.e. in T),

    by Lemma 4.2,

    lim infq˜E0(x0b0;xqbq)d2q=lim infq1d2qt1t0˜E0(t,xq(t)i+bqj,˙x0(t),˙xq(t))dt=lim infq1d2qt1t0E0(t,xq(t),˙x0(t),˙xq(t))dtlim infq1d2qUE0(t,xq(t),˙x0(t),˙xq(t))dt=lim infq1d2qU˜E0(t,xq(t)i+bqj,˙x0(t),˙xq(t))dt=12lim infqU˙yq(t),Rq(t)˙yq(t)dt12U˙y0(t),R0(t)˙y0(t)dt.

    As U can be chosen to differ from T by a set of an arbitrarily small measure and the function

    t˙y0(t),R0(t)˙y0(t)

    belongs to L1(T;R), this inequality holds when U=T, and this establishes (21). With this in mind, by (18) and (20), we have

    0t1t0{˙y0(t),˜F0˙x˙x(t,x0(t)i+b0j,˙x0(t))˙y0(t)+2˙y0(t),˜F0˙xξ(t,x0(t)i+b0j,˙x0(t))[y0(t)i+β0j]+y0(t)i+β0j,˜F0ξξ(t,x0(t)i+b0j,˙x0(t))[y0(t)i+β0j]}dt=l0(b0)β0,β0+t1t0{˙y0(t),F0˙x˙x(˜x0(t))˙y0(t)+2˙y0(t),F0˙xx(˜x0(t))y0(t)+y0(t),F0xx(˜x0(t))y0(t)}dt=l0(b0)β0,β0+t1t02Ω0(x0;t,y0(t),˙y0(t))dt=J0(x0b0;y0β0).

    Now, let us show that y0β0(0,0). By (20), the first conclusion of Lemma 5.1, the fact that V(π)|π|2/2 for all πRn,

    0limq˜K0(x0b0;xqbq)d2q+h2h2lim supq|xq(t0)x0(t0)|2d2q=limq˜K0(x0b0;xqbq)d2q+h2h2lim supq|Ψ0(bq)Ψ0(b0)|2d2q=limq˜K0(x0b0;xqbq)d2q+h2h2lim supq|10Ψ0(b0+λ[bqb0])(bqb0dq)dλ|2=limq˜K0(x0b0;xqbq)d2q+h2h2|Ψ0(b0)β0|2=limq˜K0(x0b0;xqbq)d2q+h2h2|y0(t0)|2.

    With this in mind and (18), the fact that y0β0(0,0) contradicts the positivity of h and this establishes (17).

    Let us now show that

    Ii(x0b0;y0β0)0(iia(x0b0)). (22)

    To this end, note that, for all γ=0,1,,K,

    limq˜Mγ(t,xq(t)i+bqj)dq=limq12[xq(t)x0(t)]i+[bqb0]j,˜Pγ(t,xq(t)i+bqj)ωq(t)=0,
    limq˜Nγ(t,xq(t)i+bqj)=limq˜Qγ(t,xq(t)i+bqj)([xq(t)x0(t)]i+[bqb0]j)=0,

    all uniformly on T and {˙yq} converges weakly in L1(T;Rn) to ˙y0, then for all γ=0,1,,K,

    limq˜Kγ(x0b0;xqbq)dq=0. (23)

    As in (19), we have

    limq˜J0(x0b0;xqbqx0b0)dq=limq1dq1i=0(1)i+110(1λ)ρ(ti)Ψi(b0+λ[bqb0];bqb0)dλ=0. (24)

    By (4), (23) and (24),

    0lim supq˜J0(xqbq)˜J0(x0b0)dq=lim supq˜E0(x0b0;xqbq)dq.

    Since ˜E0(x0b0;xqbq)0 (qN), then

    limq˜E0(x0b0;xqbq)dq=0.

    Therefore, by condition (v)(c) of Theorem 2.1, for all γ=1,,K,

    limq˜Eγ(x0b0;xqbq)dq=0. (25)

    As for all qN and iia(x0b0),

    0Ii(xqbq)=Ii(xqbq)Ii(x0b0)=˜Ii(xqbq)˜Ii(x0b0)=˜Ii(x0b0;xqbqx0b0)+˜Ki(x0b0;xqbq)+˜Ei(x0b0;xqbq)=Ii(x0b0;xqbqx0b0)+˜Ki(x0b0;xqbq)+˜Ei(x0b0;xqbq),

    then, by (23) and (25), for all iia(x0b0),

    0limqIi(x0b0;xqbqx0b0)dq.

    Therefore, since yq(t)y0(t) uniformly on T, {˙yq} converges weakly in L1(T;Rn) to ˙y0 and (bqb0)/dqβ0, then for all iia(x0b0),

    0limqIi(x0b0;xqbqx0b0)dq=Ii(x0b0;y0β0)

    which establishes (22).

    Now, let us show that

    Ij(x0b0;y0β0)=0(j=k+1,,K). (26)

    Indeed, as for all qN and j=k+1,,K,

    0=Ij(xqbq)Ij(x0b0)=˜Ij(xqbq)˜Ij(x0b0)=˜Ij(x0b0;xqbqx0b0)+˜Kj(x0b0;xqbq)+˜Ej(x0b0;xqbq)=Ij(x0b0;xqbqx0b0)+˜Kj(x0b0;xqbq)+˜Ej(x0b0;xqbq),

    by (23) and (25), for all j=k+1,,K,

    0=limqIj(x0b0;xqbqx0b0)dq=Ij(x0b0;y0β0)

    which is precisely (26).

    We claim that, for all αIa(˜x0(t)),

    φαx(˜x0(t))y0(t)+φα˙x(˜x0(t))˙y0(t)0(a.e. in T). (27)

    Indeed, for all αR, qN, almost all tT and λ[0,1], define

    Gαq(t;λ):=φα(t,x0(t)+λ[xq(t)x0(t)],˙x0(t)+λ[˙xq(t)˙x0(t)]),
    Wαq(t):=[φα(˜xq(t))]1/2,
    Zα0(t):=φαx(˜x0(t))y0(t)φα˙x(˜x0(t))˙y0(t),

    where as usual (˜xq(t)):=(t,xq(t),˙xq(t)) (a.e. in T). Given t[t0,t1) a point of continuity of Ia(˜x0()) and αIa(˜x0(t)), since Ia(˜x0()) is piecewise constant on T, there exists an interval [t,ˉt]T with t<ˉt such that φα(˜x0(τ))=0 for all τ[t,ˉt]. Using the notation

    [τ]:=(τ,x0(τ)+λ[xq(τ)x0(τ)],˙x0(τ)+λ[˙xq(τ)˙x0(τ)]),

    we have

    0limq[t,ˉt]U(Wαq(τ))2dqdτ=limq1dq[t,ˉt]U{φα(˜xq(τ))+φα(˜x0(τ))}dτ=limq1dq[t,ˉt]U{Gαq(τ;1)Gαq(τ;0)}dτ=limq1dq[t,ˉt]U10λGαq(τ;λ)dλdτ=limq1dq[t,ˉt]U10{φαx[τ](xq(τ)x0(τ))+φα˙x[τ](˙xq(τ)˙x0(τ))}dλdτ=limq[t,ˉt]U10{φαx[τ]yq(τ)+φα˙x[τ]˙yq(τ)}dλdτ=[t,ˉt]U{φαx(˜x0(τ))y0(τ)φα˙x(˜x0(τ))˙y0(τ)}dτ=[t,ˉt]UZα0(τ)dτ.

    Since U can be chosen to differ from T by a set of an arbitrarily small measure, then

    0ˉttZα0(τ)dτ.

    If Zα0(τ)<0 on a measurable set Θ such that Θ[t,ˉt] and m(Θ)>0, then

    0>ΘUZα0(τ)dτ=limqΘU(Wαq(τ))2dqdτ0

    which is a contradiction. Therefore, Zα0(τ)0 a.e. in [t,ˉt] with t[t0,t1) an arbitrary point of continuity of Ia(˜x0()) and, therefore, Zα0(t)0 for almost all tT which shows (27).

    Now, let us show that for all βS,

    φβx(˜x0(t))y0(t)+φβ˙x(˜x0(t))˙y0(t)=0 (a.e. in T). (28)

    Indeed, for all βS, qN, almost all tT and λ[0,1], define

    Hβq(t;λ):=φβ(t,x0(t)+λ[xq(t)x0(t)],˙x0(t)+λ[˙xq(t)˙x0(t)]).

    For all βS, qN and almost all tT, we have

    0=Hβq(t;1)Hβq(t;0)=10λHβq(t;λ)dλ=10{φβx[t](xq(t)x0(t))+φβ˙x[t](˙xq(t)˙x0(t))}dλ.

    Therefore, for all βS, qN and almost all tT,

    0=10{φβx[t]yq(t)+φβ˙x[t]˙yq(t)}dλ. (29)

    By (29), for all tT and βS,

    0=[t0,t]U{φβx(˜x0(τ))y0(τ)+φβ˙x(˜x0(τ))˙y0(τ)}dτ.

    Since, as before, U can be chosen to differ from T by a set of an arbitrarily small measure, then for all tT and βS,

    0=tt0{φβx(˜x0(τ))y0(τ)+φβ˙x(˜x0(τ))˙y0(τ)}dτ

    and so (28) is verified. Consequently, from (15), (22), (26), (27) and (28), y0β0Y(x0b0). This fact together with (17) contradicts condition (ⅳ) of Theorem 2.1.

    Case (2): Now, suppose that the sequence {(bqb0)/dq} is not bounded. Then,

    limq|bqb0dq|=+. (30)

    Choose an appropriate subsequence of {(bqb0)/|bqb0|} (without relabeling), and ˉβ0Rp with |ˉβ0|=1, such that

    limqbqb0|bqb0|=ˉβ0. (31)

    For all qN and tT, define

    ˉωq(t):=xq(t)x0(t)|bqb0|i+bqb0|bqb0|j.

    By Lemma 4.1 and (30),

    limqxq(t)x0(t)|bqb0|=limqyq(t)dq|bqb0|=y0(t)0=0uniformly on  T. (32)

    For i=0,1 and all qN, we have

    xq(ti)x0(ti)|bqb0|=10Ψi(b0+λ[bqb0])(bqb0|bqb0|)dλ. (33)

    By (31), (32) and (33), for i=0,1,

    Ψi(b0)ˉβ0=0. (34)

    Now, by (2), (31) and (32),

    limq˜M0(t,xq(t)i+bqj)|bqb0|2=limq12ˉωq(t),˜P0(t,xq(t)i+bqj)ˉωq(t)=120ˉβ0,˜F0ξξ(t,x0(t)i+b0j,˙x0(t))0ˉβ0=ˉβ0,l0(b0)ˉβ02(t1t0),
    limq˜N0(t,xq(t)i+bqj)|bqb0|=limq˜Q0(t,xq(t)i+bqj)ˉωq(t)=˜F0˙xξ(t,x0(t)i+b0j,˙x0(t))0ˉβ0=0

    both uniformly on T. Together with Lemma 4.1, this implies that

    limq˜K0(x0b0;xqbq)|bqb0|2=12ˉβ0,l0(b0)ˉβ0+limqt1t0dq|bqb0|˙yq(t),˜N0(t,xq(t)i+bqj)|bqb0|dt=12ˉβ0,l0(b0)ˉβ0. (35)

    As in (19), we have

    limq˜J0(x0b0;xqbqx0b0)|bqb0|2=12[ρ(t1)Ψ1(b0;ˉβ0)ρ(t0)Ψ0(b0;ˉβ0)]. (36)

    Even more, by (1), (4), (36), and condition (ⅱ) of Theorem 2.1,

    0limq˜K0(x0b0;xqbq)|bqb0|2+lim infq˜E0(x0b0;xqbq)|bqb0|2. (37)

    Consequently, since ˜E0(x0b0;xqbq)0 (qN), by (35) and (37),

    012ˉβ0,l0(b0)ˉβ0=12J0(x0b0;0ˉβ0). (38)

    Let us now show that for all iia(x0b0),

    Ii(x0b0;0ˉβ0)0. (39)

    To prove it, note that since

    limqxq(t)x0(t)|bqb0|=0 uniformly on T,

    and {˙yq} converges weakly in L1(T;Rn) to ˙y0, for all γ=0,1,,K,

    limq˜Kγ(x0b0;xqbq)|bqb0|=0. (40)

    As in (24),

    limq˜J0(x0b0;xqbqx0b0)|bqb0|=0. (41)

    Then, by (4), (40) and (41),

    0lim supq˜J0(xqbq)˜J0(x0b0)|bqb0|=lim supq˜E0(x0b0;xqbq)|bqb0|0.

    Thus, by condition (v)(c) of Theorem 2.1, for all γ=1,,K,

    limq˜Eγ(x0b0;xqbq)|bqb0|=0. (42)

    As for all qN and iia(x0b0),

    0Ii(xqbq)=Ii(xqbq)Ii(x0b0)=˜Ii(xqbq)˜Ii(x0b0)=˜Ii(x0b0;xqbqx0b0)+˜Ki(x0b0;xqbq)+˜Ei(x0b0;xqbq)=Ii(x0b0;xqbqx0b0)+˜Ki(x0b0;xqbq)+˜Ei(x0b0;xqbq),

    then, by (40) and (42), for all iia(x0b0),

    0limqIi(x0b0;xqbqx0b0)|bqb0|.

    Hence, since

    {˙yq} converges weakly to  ˙y0 in L1(T;Rn),dq|bqb0|0,
    limqxq(t)x0(t)|bqb0|=0  uniformly on  T,  and  bqb0|bqb0|ˉβ0,

    for all iia(x0b0),

    0limqIi(x0b0;xqbqx0b0)|bqb0|=li(b0)ˉβ0=Ii(x0b0;0ˉβ0)

    which establishes (39). Finally, let us show that for all j=k+1,,K,

    Ij(x0b0;0ˉβ0)=0. (43)

    Indeed, as for all qN and j=k+1,,K,

    0=Ij(xqbq)Ij(x0b0)=˜Ij(xqbq)˜Ij(x0b0)=˜Ij(x0b0;xqbqx0b0)+˜Kj(x0b0;xqbq)+˜Ej(x0b0;xqbq)=Ij(x0b0;xqbqx0b0)+˜Kj(x0b0;xqbq)+˜Ej(x0b0;xqbq),

    then, by (40) and (42), for all j=k+1,,K,

    0=limqIj(x0b0;xqbqx0b0)|bqb0|=Ij(x0b0;0ˉβ0)

    which is precisely (43). Consequently, (34), (38), (39) and (43) contradicts condition (ⅳ) of Theorem 2.1 and this completes the proof of Theorem 2.1.

    Proof of Lemma 4.1: For all qN and almost all tT, define

    cq:=[1+12V(xq(t0)x0(t0))]1/2  and  Wq(t):=[1+12V(˙xq(t)˙x0(t))]1/2.

    For all qN, note that

    |yq(t0)|2c2q+t1t0|˙yq(t)|2W2q(t)dt=|xq(t0)x0(t0)|2d2q[1+12V(xq(t0)x0(t0))]+1d2qt1t0|˙xq(t)˙x0(t)|21+12V(˙xq(t)˙x0(t))dt=|xq(t0)x0(t0)|2D(xqx0)[2+V(xq(t0)x0(t0))]+1D(xqx0)t1t0|˙xq(t)˙x0(t)|22+V(˙xq(t)˙x0(t))dt=1D(xqx0)(V(xq(t0)x0(t0))+t1t0V(˙xq(t)˙x0(t))dt)=D(xqx0)D(xqx0)=1.

    Clearly limqcq=1. Then, there exist some subsequence of {xq}, again denoted by {xq}, some ˉy0Rn and some σ0L2(T;Rn) such that

    limqyq(t0)cq=limqyq(t0)=ˉy0,
    {˙yq/Wq}converges weakly in  L2(T;Rn) to σ0.

    As W2q(t)Wq(t)1 for all qN and for almost all tT, we have

    0t1t0[Wq(t)1]dtt1t0[W2q(t)1]dt=12t1t0V(˙xq(t)˙x0(t))dtD(xqx0).

    Thus, it follows that

    limqt1t0[Wq(t)1]dt=limqt1t0[W2q(t)1]dt=0.

    Note also that

    t1t0[Wq(t)1]2dt=t1t0[W2q(t)1]dt2t1t0[Wq(t)1]dt.

    Then for any hL(T;Rn),

    limqhWqh2=0

    and so

    limqt1t0h(t),˙yq(t)dt=limqt1t0h(t)Wq(t),˙yq(t)Wq(t)dt=t1t0h(t),σ0(t)dt.

    Therefore, {˙yq} converges weakly in L1(T;Rn) to σ0. Hence, {˙yq} is equi-integrable on T and therefore the sequence {yq} is equi-continuous on T. Thus, if y0(t):=ˉy0+tt0σ0(τ)dτ, then, y0X with ˙y0L2(T;Rn) and

    limqyq(t)=limqyq(t0)+limqtt0˙yq(τ)dτ=y0(t)uniformly on  T,
    {˙yq}converges weakly in  L1(T;Rn) to ˙y0=σ0.

    Now, let us show that ˙xq(t)˙x0(t) almost uniformly on T. For almost all tT, define

    W(t):=[1+12V(˙x(t))]1/2.

    Observe that

    t1t0|˙x(t)|22W2(t)dt=t1t0|˙x(t)|22+V(˙x(t))dt=t1t0V(˙x(t))dtD(x),
    t1t02W2(t)dt=2t12t0+t1t0V(˙x(t))dt2t12t0+D(x).

    From these relations, we have

    ˙x21t1t0|˙x(t)|22W2(t)dtt1t02W2(t)dtD(x)[2t12t0+D(x)].

    Consequently, ˙xq˙x010 and so some subsequence of {˙xq} converges almost uniformly to ˙x0 on T.

    Proof of Lemma 4.2: Recall the definition of Wq given in the proof of Lemma 4.1. As Wq(t)1 uniformly on U, then for all hL2(U;Rn),

    limqU˙yq(t),h(t)dt=limqU˙yq(t)Wq(t),Wq(t)h(t)dt=U˙y0(t),h(t)dt,

    that is, {˙yq} converges weakly in L2(U;Rn) to ˙y0. As R0(t)0 (tU), the function

    ˙yUR0(t)˙y(t),˙y(t)dt

    is convex in L2(U;Rn) and since this function is strongly continuous on L2(U;Rn), then this function is weakly lower semicontinuous in L2(U;Rn). Thus,

    lim infqUR0(t)˙yq(t),˙yq(t)dtUR0(t)˙y0(t),˙y0(t)dt.

    Since Rq(t)R0(t) uniformly on U, it follows that

    lim infqURq(t)˙yq(t),˙yq(t)dtUR0(t)˙y0(t),˙y0(t)dt.

    In this section, we show with an example how our sufficiency theory is able to detect optimality even when the proposed extremal to be a strong minimum is singular and its derivative is only essentially bounded. It is worthwhile observing that the initial and final end-points of the states of admissible trajectories are not restricted to belong to any manifold described by any smooth function, in contrast, these boundary points must only lie in the set of real numbers, that is, these boundary points are completely free.

    In Example 7.1 since no isoperimetric constraints occur l0, L0, F0, E0 and J0 correspond simply to l, L, F, E and J respectively.

    7.1 Example: Let x0:[0,1]R be any absolutely continuous function with ˙x0L([0,1];R) and x0(0)=x0(1)=0. Consider the nonparametric problem (ˉP) of minimizing

    J(x):=x2(0)x(1)+10{exp(t(˙x(t)˙x0(t)))+x(t)}dt

    over all arcs xX satisfying the constraints

    {c(t,x(t),˙x(t))is integrable on [0,1].x(0)R,x(1)R.(t,x(t),˙x(t))ˉR(a.e. in [0,1])

    where

    ˉR:={(t,x,˙x)[0,1]×R×Rϕ1(t,x,˙x)0},
    ϕ1(t,x,˙x):=(˙x˙x0(t))2exp(t(˙x˙x0(t)))+t(˙x˙x0(t))+1,

    X:=AC([0,1];R) and c(t,x,˙x) denotes either

    L(t,x,˙x):=exp(t(˙x˙x0(t)))+x,

    ϕ1(t,x,˙x), or any of its partial derivatives of order less than or equal to two with respect to x and ˙x.

    For this problem we shall consider the data of the nonparametric problem given in Section 3 which are given by T=[0,1], n=1, r=1, s=1, k=K=0, B0=R, B1=R, (x1,x2)=x21x2, L(t,x,˙x)=exp(t(˙x˙x0(t)))+x, ϕ1(t,x,˙x)=(˙x˙x0(t))2exp(t(˙x˙x0(t)))+t(˙x˙x0(t))+1.

    Clearly, all the assumptions posed in the statement of the problem are easily verified.

    Also, it is evident that the trajectory x0 is admissible of (ˉP). Let Ψ:RR×R be defined by Ψ(b):=(b,b). Clearly, Ψ is C2 in R and B0×B1Ψ(R). The associated parametric problem of Section 2 denoted by (PΨ) has the following data, p=1, B=Ψ1(B0×B1)=R, l=Ψ, L=L, φ=φ1=ϕ1 and Ψ0, Ψ1 the components of Ψ, that is, Ψ0(b)=b, Ψ1(b)=b (bR). Recall that the notation xb means (x,b) where bR is a parameter.

    Observe that if we set b0:=0, then x0b0 is admissible of (PΨ). Also, clearly Ia(˜x0()) is constant on T. Let (ρ(t),μ1(t)):=(t,0) (tT) and note that (ρ,μ)X×U1, μ1(t)0, μ1(t)φ1(˜x0(t))=0 (a.e. in T).

    Now, observe that the Hamiltonian H is given by

    H(t,x,˙x,ρ(t),μ(t))=t˙xexp(t(˙x˙x0(t)))x.

    Also, note that

    Hx(˜x0(t),ρ(t),μ(t))=1,H˙x(˜x0(t),ρ(t),μ(t))=0(a.e. in T),

    and so (x0,ρ,μ) satisfies the Euler-Lagrange equations which are given by

    ˙ρ(t)=Hx(˜x0(t),ρ(t),μ(t)),H˙x(˜x0(t),ρ(t),μ(t))=0(a.e. in T).

    Thus, (x0,ρ,μ) is an extremal. As Ψ0(b)=b, Ψ1(b)=b, l(b)=b2b (bR), then

    l(b0)+ρ(1)Ψ1(b0)ρ(0)Ψ0(b0)=0

    and so condition (ⅰ) of Corollary 3.2 is satisfied. Also, as one readily verifies,

    ρ(1)Ψ1(b0;β)ρ(0)Ψ0(b0;β)=0 for all  βR

    and so condition (ⅱ) of Corollary 3.2 is fulfilled.

    Additionally, as

    H˙x˙x(˜x0(t),ρ(t),μ(t))=t2(a.e. in T),

    it follows that (ⅲ) of Corollary 3.2 is satisfied and (x0,ρ,μ) is singular since

    H˙x˙x(˜x0(0),ρ(0),μ(0))=0.

    As φ1x(˜x0(t))=φ1˙x(˜x0(t))=0 (a.e. in T), then Y(x0b0) is given by all yβX×R with ˙yL2(T;R) satisfying y(0)=β, y(1)=β. We have,

    J(x0b0;yβ)=2β2+10t2˙y2(t)dt>0

    for all yβY(x0b0), yβ(0,0), and hence (ⅳ) of Corollary 3.2 is satisfied. Now, observe that for all (t,x,˙x)T×R×R,

    F(t,x,˙x)=H(t,x,˙x,ρ(t),μ(t))˙ρ(t)x=t˙x+exp(t(˙x˙x0(t))).

    Hence for almost all tT, if xb is admissible,

    E(t,x(t),˙x0(t),˙x(t))=F(t,x(t),˙x(t))F(t,x(t),˙x0(t))F˙x(t,x(t),˙x0(t))(˙x(t)˙x0(t))=t˙x(t)+exp(t(˙x(t)˙x0(t)))+t˙x0(t)1=exp(t(˙x(t)˙x0(t)))t(˙x(t)˙x0(t))10

    which implies that (v)(a) of Corollary 3.2 is verified with any ϵ>0. Finally, for all xb admissible, we have

    10E(t,x(t),˙x0(t),˙x(t))dt=10{exp(t(˙x(t)˙x0(t)))t(˙x(t)˙x0(t))1}dt10(˙x(t)˙x0(t))2dt210V(˙x(t)˙x0(t))dt.

    Therefore, (v)(b) of Corollary 3.2 is satisfied with any ϵ>0 and h=2. Since k=K=0, it is evident that (v)(c) of Corollary 3.2 is also verified with any ϵ>0 and h=2. By Corollary 3.2, x0 is a strong minimum of (ˉP).

    The author is grateful to Dirección General de Asuntos del Personal Académico, Universidad Nacional Autónoma de México, for the support given by the project PAPIIT-IN102220.

    The author declares no conflict of interest.



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