Vietnam J. Math. (2018) 46:127–147 https://doi.org/10.1007/s10013-017-0264-9
Infinite-Horizon Problems Under Periodicity Constraint Jo¨el Blot1
· Abdelkader Bouadi2 · Bruno Nazaret1
Received: 24 January 2017 / Accepted: 7 October 2017 / Published online: 29 November 2017 © Vietnam Academy of Science and Technology (VAST) and Springer Nature Singapore Pte Ltd. 2017
Abstract We study some infinite-horizon optimization problems on spaces of periodic functions, for non periodic Lagrangians with discounting. The main strategy relies on the reduction to finite horizon thanks to the introduction of an averaging operator and a periodic extension operator. We study the properties of these operators, and we then provide existence results and necessary optimality conditions, in which the corresponding averaged Lagrangian appears. Keywords Infinite-horizon variational problem · Weighted Sobolev spaces · Periodic trajectory Mathematics Subject Classification (2010) 49K30 · 49N20
This paper is dedicated to Professor Michel Th´era in celebration of his 70th birthday. Jo¨el Blot
[email protected] Abdelkader Bouadi
[email protected] Bruno Nazaret
[email protected] 1
Laboratoire SAMM EA 4543, Universit´e Paris 1 Panth´eon-Sorbonne, Centre P.M.F., 90 rue de Tolbiac, 75634 Paris cedex 13, France
2
Universit´e D’Alger 1, 2 rue Didouche Morade, 16000 Alger, Alg´erie
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1 Introduction We consider the following problem of Calculus of Variations ⎧ +∞ ⎨ Minimize F (x) := 0 e−rt L(t, x(t), x (t))dt (P 1) when x ∈ W 1,α (R+ , μr ; Rn ), ⎩ x is T -periodic, x(0) = σ, where r, T ∈ (0, +∞), L : R+ × Rn × Rn → R is a function, α ∈ [1, +∞), μr is the positive measure with density er (t) := e−rt with respect to the Lebesgue measure of R+ , W 1,α (R+ , μr , Rn ) is a weighted Sobolev space, and σ ∈ Rn .
Aim We provide assumptions on L to ensure the existence of a solution of (P 1) (Section 5). We establish a necessary optimality condition in the form of an Euler–Lagrange equation (Section 6). When the problem (P 1) comes from economic modeling, this optimization problem defines the rationality of the agent (or of the firm) and the necessary optimality conditions are behavior laws of the rational agent. An admissible function for (P 1) is an x ∈ W 1,α (R+ , μr ; Rn ) which is T -periodic and which satisfies x(0) = σ . On the existence theorems, we want to find a global solution of (P 1), i.e., an admissible function xˆ such that F (x) ˆ ≤ F (x) for every admissible function u. On the necessary optimality conditions, we consider a local solution of (P 1), i.e., an admissible function xˆ such that F (x) ˆ ≤ F (x) for every admissible function which belongs to a neighborhood (with respect to the topology of W 1,α (R+ , μr ; Rn )) of x. ˆ We only consider these two notions of optimality. Motivations In macroeconomic optimal growth theory, the variable x(t) is related to a representative agent (a person and his descendants) at time t. The rationality of the representative agent leads to +∞ Maximize G(x) := 0 e−rt U (t, x(t), x (t))dt (P 2) when x ∈ W 1,α (R+ , μr ; Rn ), x(0) = σ without periodicity condition [16, 23]. The presence of e−rt under the integral gives a stronger importance to the values of x(t) the smaller t is. Consequently (P 2) disadvantages the future generations and fosters the present generation [22]. A better intergenerational justice is to add a periodicity constraint, x(t + T ) = x(t), into (P 2), where T is the lifespan of a generation. Consequently, all the generations will know the same treatment. With the additional periodicity constraint, multiplying the criterion by −1, we obtain (P 1). Notice that similar arguments can be used in problems of optimal management of fisheries or forests (as described in [12]). There exists a literature on the Calculus of Variations and on optimal control theory in continuous time and infinite horizon in presence of a discount rate. Important references on this theory are [11] and [25, Section 4.19]. For existence results, one can quote [21] and the references therein, [16] where the Lagrangian is separated in the variables x and x , and the recent work [18]. The question of the necessary conditions of optimality is treated in [6, 8] and the references therein, on the class of bounded trajectories in [4, 5], while the case of almost-periodic trajectories appears in [3]. Finally, the case of periodic trajectories studied in [13, 14, 17]. In this last paper, the authors deals with criteria of the form is+∞ e−rt g(x(t), u(t))dt, that is g which is not time dependent. In (P 1), L depends on t in 0 a non necessary T -periodic way. Concerning discrete time problems, we refer to [7] and the references therein. We can also quote [1] where questions of periodicity and of optimality appear with different aims.
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Mathematical Tools An important difficulty for the existence result on problem (P 1) without periodicity constraint is that the canonical injection from the Sobolev space W 1,α (R+ , μr , Rn ) into the Lebesgue space Lα (R+ , μr , Rn ) is not compact. This fact is proven in [2] and it is also considered in [19, 21]. Instead of adding strong assumptions on L to compensate this lack of compactness, we avoid this difficulty by introducing two operators. The first, that we call averaging operator, is A defined by
A(f )(t) :=
+∞
e−rkT f (t + kT ),
t ∈ [0, T ],
(1)
k=0
for any f : R+ → Rn , and an auxiliary operator related to A defined by
A1 (L)(·, u, v) := A(L(·, u, v)) for any L : R × Rn × Rn → Rm . T This operator permits to write F (x) = 1−e1−rT 0 A1 (L)(t, x(t), x (t))dt whenever x is T -periodic. Now (P 1) can be reformulated as T Minimize FT (u) := 1−e1−rT 0 A1 (L)(t, u(t), u (t))dt (P 3) when u ∈ W 1,α (0, T ; Rn ), u(0) = σ, u(T ) = σ. We shall also establish an Euler–Lagrange equation in terms of A1 (L) instead of L. Given T ∈ (0, +∞), the second operator, that we call periodic extension to R+ operator, ET is defined by
ET (ϕ)(t) := ϕ(t − kT ) when t ∈ [kT , kT + T ), k ∈ N,
(2)
where ϕ : [0, T ) → Rn . Notice that ET (ϕ) : R → Rn is T -periodic. These operators are not artificial; they are deeply related to the description of orthogonal projections onto the spaces of T -periodic functions, and they also enter in necessary optimality conditions of our variational problems, and they permit to use classical theorems of existence and of necessary optimality conditions of finite-horizon boundary value problems to our infinite-horizon problem under periodicity constraint.
Contents of the Paper In Section 2, we introduce some notations for the used function spaces, and recall some basic results. In Section 3, we study properties of the periodic extension to R+ operator defined in (2). In Section 4, we study the properties of the averaging operator defined in (1). We show that operators A and ET permit to characterize the orthogonal projections in Hilbert Lebesgue spaces onto subspaces of T -periodic functions, and we show that these operators permit to solve a problem issued from Time Series Analysis. Section 5 is devoted to prove existence theorems for (P 1). In Section 6, we establish a necessary optimality condition in the form of an Euler–Lagrange equation.
2 Notation and Preliminaries We set here some notations related to the functional analytic framework and recall some basic Let n ∈ N∗ . For any vectors x = (x i )1≤i≤n and y = (y i )1≤i≤n in Rn , x · y := n facts. i y i will stand for the usual Euclidean inner product and the induced norm will be x i=1 denoted by | · |.
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When X and Y are Banach spaces, C 0 (X, Y ) (resp. C 1 (X, Y )) denotes the space of continuous (respectively continuously Fr´echet differentiable) functions from X to Y . When is R+ , Rn or an interval of R, the Borel σ -algebra on is denoted by B (). For any r > 0, we define the measure μr as
∀B ∈ B (R+ ), μr (B) = e−rt dλ(t), B
where λ stands for the Lebesgue measure. Notice that, for any r > 0, a set is μr -negligible if and only if it is λ-negligible, due to the boundedness and positivity on R+ of the density function. The associated Lebesgue spaces Lα (I, μr ; Rn ), with α ≥ 1 and I any interval in R+ , are the spaces of all (classes of) measurable functions from (I, B (I )) into (Rn , B (Rn )) whose αth-power is μr -integrable and the corresponding Sobolev spaces W 1,α (I, μr ; Rn ) are defined as W 1,α (I, μr ; Rn ) := f ∈ Lα (I, μr ; Rn ); f ∈ Lα (I, μr ; Rn ) , f being understood as the distributional first derivative of f . Endowed respectively with the norms
f
Lα (I,μr ;Rn )
1
:=
|f (t)| dμr (t) α
α
I
and 1 α
f W 1,α (I,μr ;Rn ) := f αLα (I,μr ;Rn ) + f αLα (I,μr ;Rn ) , Lα (I, μr ; Rn ) and W 1,α (I, μr ; Rn ) are Banach spaces, both reflexive if +∞ > α > 1. The usual Lebesgue and Sobolev spaces with respect to the Lebesgue measure will be simply respectively denoted by Lα (I ; Rn ) and W 1,α (I ; Rn ). Notice that if I is bounded (for instance I = [0, T ]), we have ∀r > 0,
Lα (I, μr ; Rn ) = Lα (I ; Rn )
and W 1,α (I, μr ; Rn ) = W 1,α (I ; Rn ).
When I = [a, b] we assimilate Lα ([a, b]; Rn ), Lα ([a, b); Rn ), Lα ((a, b); Rn ) and we denote them by Lα (a, b; Rn ); these assimilations are possible since the singletons are Lebesgue-negligible. We also assimilate W 1,α ([a, b]; Rn ), W 1,α ([a, b); Rn ) and W 1,α ((a, b); Rn ) and we denote them by W 1,α (a, b; Rn ); these assimilations are possible since an absolutely continuous function is uniformly continuous and consequently it is extendable in a unique way from (a, b) into [a, b]. We do the same assimilation for the weighted spaces on [a, b]. Let us now introduce classical spaces of periodic functions. Let, for any T > 0, PT0 (R+ , Rn ) be the space of continuous T -periodic functions from R+ to Rn and PT1 (R+ , Rn ) := PT0 (R+ , Rn ) ∩ C 1 (R+ , Rn ). In addition, we define PT0,0 (R+ , Rn ) (respectively PT1,0 (R+ , Rn )) as the space of functions u ∈ PT0 (R+ , Rn ) n (respectivelyPT1,0 (R+ , Rn )) such that u(0) = 0. We denote by WT1,α ,0 (R+ , μr ; R ) the closure of PT1,0 (R+ , Rn ) in W 1,α (R+ , μr ; Rn ).
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3 The Periodic Extension to R+ Operator In this section, we establish that ET defines an isometric linear operator between Lebesgue spaces and Sobolev spaces. α
Theorem 1 Let α ≥ 1 and let PT0 (R+ , μr ; Rn ) be the closure of PT0 (R+ ; Rn ) in Lα (R+ , μr ; Rn ). Then, α
(i)
PT0 (R+ , μr ; Rn ) = {f ∈ Lα (R+ , μr ; Rn ) : f (t + T ) = f (t) for a.e. t ∈ R+ }.
(ii)
ET is a continuous linear operator from Lα (0, T , μr ; Rn ) into PT0 (R+ , μr ; Rn ) and for all ϕ ∈ Lα (0, T , μr ; Rn ),
α
ET (ϕ) Lα (R+ ,μr ;Rn ) =
1 1 − e−rT
1
α
ϕ Lα (0,T ,μr ;Rn ) .
(3)
Proof We know from [3, Proposition 3] that α PT0 (R+ , μr ; Rn ) ⊂ f ∈ Lα (R+ , μr ; Rn ); f (t + T ) = f (t) for a.e. t ∈ R+ .
Let us now prove the inclusion
α f ∈ Lα (R+ , μr ; Rn ); f (t + T ) = f (t) for a.e. t ∈ R+ ⊂ PT0 (R+ , μr ; Rn ).
Let f ∈ Lα (R+ , μr ; Rn ) satisfying f (t +T ) = f (t) for a.e. t ∈ R+ . Since Cc∞ ((0, T ), Rn ) (here the space of C ∞ functions defined on [0, T ] with compact support in (0, T )) is dense in Lα (0, T ; Rn ) (see [9, Corollary 4.23, p. 109]), we can find, for any positive , some T ϕ ∈ Cc∞ ((0, T ), Rn ) such that 0 |f (t) − ϕ (t)|α dt ≤ ε α . In addition, since the support of ϕε is included in (0, T ), we have ET (ϕ ) ∈ PT0 (R+ , Rn ). Then we get
+∞
e−rt |f (t) − ET (ϕ )(t)|α dt =
0
= =
+∞
(k+1)T
k=0 kT +∞ T
e−rt |f (t) − ET (ϕ )(t)|α dt
e−rt e−rkT |f (t + kT ) − ET (ϕ )(t + kT )|α dt
k=0 0
T +∞ −rkT −rt
e
k=0
e
|f (t) − ϕ (t)|α dt
0
T 1 |f (t) − f (t)|α dt −rT 1−e 0 1 α ≤ ε . 1 − e−rT
=
We have proven that, for all > 0, there exists u ∈ PT0 (R+ , Rn ) such that f − u Lα (R+ ,μr ,Rn ) ≤ (1 − e−rT )−1/α , hence f belongs to the closure of PT0 (R+ , Rn ) and (i) is shown. To prove (ii), notice that when ϕ ∈ Lα (0, T , μr Rn ), ET (ϕ) is clearly well-defined from R+ into Rn . When B ∈ B (Rn ), we have ET (ϕ)−1 (B) = k∈N ([kT , kT + T ) ∩
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(ϕ −1 (B) + kT )) ∈ B (R+ ) as a countable union of elements of B (Rn ), and so ET (ϕ) is Borel measurable from R+ into Rn . Now we can consider the integral
+∞
e−rt |ET (ϕ)(t)|α dλ(t) = e−rt |ET (ϕ)(t)|α dλ(t) R+
= =
k=0 [kT ,kT +T ) +∞
−r(s+kT )
e
k=0 [0,T )
+∞ −kT
e
k=0
1 = 1 − e−rT
[0,T )
T
|ET (ϕ)(s + kT )|α dλ(s)
e−rs |ϕ(s)|α dλ(s)
e−rs |ϕ(s)|α ds < +∞
0
since |ϕ|α ∈ L1 (0, T , μr , Rn ), therefore ET (ϕ) ∈ Lα (R+ , μr Rn ). Now we see, from the α definition of ET (ϕ), that ET (ϕ) ∈ PT0 (R+ , Rn ; μr ). The linearity of ET is clearly satisfied. Using the previous calculations we obtain the following equality
ET (ϕ) Lα (R+ ,μr ;Rn ) =
1
ϕ Lα (0,T ,μr ;Rn ) (1 − e−rT )1/α
which implies the continuity of the linear operator ET , and that it is an isometry. We have proven (ii). We have an analogous result for Sobolev spaces, the space W01,α (0, T , μr ; Rn ) of functions u in W 1,α (0, T , μr ; Rn ) such that u(0) = u(T ) = 0. n Proposition 1 For any α ≥ 1, let WT1,α ,0 (R+ , μr ; μr ; R ) be the closure of the space 1 n 1,α n PT ,0 (R+ , R ) in W (R+ , μr ; R ). Then, n 1,α (R+ , μr ; Rn ) ∩ PT0,0 (R+ , Rn ). WT1,α ,0 (R+ , μr ; R ) = W
Moreover, ET is a continuous linear operator from W01,α (0, T , μr ; Rn ) into 1,α n n WT1,α ,0 (R+ , μr ; R ) and, for any ϕ ∈ W0 (0, T , μr ; R ), we have
ET (ϕ) W 1,α (R+ ,μr ;Rn ) =
1 1 − e−rT
1
α
ϕ W 1,α (0,T ,μr ;Rn ) .
(4)
n 1,α (R , μ ; Rn ) Proof Let us first take f ∈ WT1,α + r ,0 (R+ , μr ; R ). By definition, f ∈ W and we can find a sequence (fm )m∈N in PT1,0 (R+ , Rn ) such that limm→+∞ fm − f W 1,α (R+ ,μr ;Rn ) = 0. For all K > 0, we have limm→+∞ fm − f W 1,α (0,K,μr ;Rn ) = 0 and consequently we have limm→+∞ fm − f W 1,1 (0,K,μr ;Rn ) = 0. Since, for any K > 0 and any nonnegative measurable function u on R+ , we have
−rK e u(t)dλ(t) ≤ u(t)dλ(t) ≤ u(t)dμr (t), [0,K]
[0,K]
we obtain limm→+∞ fm − f W 1,1 (0,K;Rn ) = 0.
[0,T ]
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Using the completeness of W 1,1 (0, K; Rn ) and the fact that the functions of are absolutely continuous, we obtain that f is continuous on [0, K] and, since the norm of W 1,1 (0, K; Rn ) is equivalent to the norm f AC := |f (0)| +
f L1 (0,K;Rn ) , we have sup0≤t≤T |fm (t) − f (t)| ≤ c · fm − f W 1,1 (0,K;Rn ) . From the last inequality, we obtain that f (0) = 0, and, given t ∈ R+ , by taking K := t + T we also deduce that f (t + T ) = limm→+∞ fm (t + T ) = limm→+∞ fm (t) = f (t), n which proves the T -periodicity of f . We then have proven that WT1,α ,0 (R+ , μr ; R ) ⊂ W 1,α (R+ , μr ; Rn ) ∩ PT0,0 (R+ , Rn ). Let us now prove the inclusion W 1,1 (0, K; Rn )
n W 1,α (R+ , μr ; Rn ) ∩ PT0,0 (R+ , Rn ) ⊂ WT1,α ,0 (R+ , μr ; R ).
Let f ∈ W 1,α (R+ , μr ; Rn ) ∩ PT0,0 (R+ , Rn ), and set fT := f|(0,T ) ∈ W01,α (0, T , Rn ). From [9, Theorem 8.7], for all > 0, there exists a function ϕ ∈ Cc1 ((0, T ), Rn ) such that
ϕ − fT W 1,α (0,T ,Rn ) ≤ . We extend ϕ and ϕ on [0, T ) by setting ϕε (0) = ϕε (0) = 0, 0
we check that ET (ϕε ) = ET (ϕε ) a.e. on R+ , and so we obtain that ET (ϕε ) ∈ PT1,0 (R+ , Rn ). In addition, since f ∈ PT0,0 (R+ , Rn ), we have ET (fT ) = f . Using (ii) of Theorem 1, we finally get 1
α 1
f − ET (ϕε ) W 1,α (R+ ,μr ,Rn ) ≤ , −rT 1−e n and then f ∈ WT1,α ,0 (R+ , μr ; R ). n We have proven that W 1,α (R+ , μr ; Rn ) ∩ PT0,0 (R+ , Rn ) ⊂ WT1,α ,0 (R+ , μr ; R ), and consequently we obtain the equality of these sets. The equality (4) is then an easy consequence of (3).
4 The Averaging Operator In this section, we first establish that A defines a linear continuous operator between Lebesgue spaces. Secondly, we characterize the orthogonal projection in L2 (R+ , μr ; Rn ) on the subspace of T -periodic functions and, thirdly, we solve a problem issued from Time Series Analysis.
4.1 General Properties Theorem 2 Let the operator A be given by (1). Let α ∈ [1, +∞). Then the following assertion holds: A(Lα (R+ , μr ; Rn )) ⊂ Lα (0, T , μr ; Rn ), and A is a linear bounded operator from Lα (R+ , μr ; Rn ) into Lα (0, T , μr ; Rn ). Moreover, for all g ∈ Lα (R+ , μr ; Rn ), we have
1 1 − e−rT α α n
A(g) L (0,T ,μr ;R )) ≤
g Lα (R+ ,μr ;Rn ) . e−rT Proof We treat separately the cases α = 1 and α > 1.
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The Case α = 1 Let g ∈ L1 (R+ , μr ; Rn ). Using the σ -additivity of the positive measure
of density h(t) := e−rt |g(t)| with respect to the Lebesgue measure λ for any integer k, we obtain ⎫
+∞
⎪ ⎪ −rt −rt e |g(t)|dλ(t) = e |g(t)|dλ(t) ⎪ ⎬ [kT ,kT +T ) [kT ,kT +T ) k=0
k∈N ⎪ ⎪ = e−rt |g(t)|dλ(t) < +∞. ⎪ ⎭ R+
Since the singletons are Lebesgue-negligible, we obtain +∞
k=0 [kT ,kT +T ]
e−rt |g(t)|dλ(t) < +∞.
(5)
Since λ is translation invariant, we can easily verify that [kT ,kT +T ] e−rt |g(t)|dλ(t) = [0,T ] e−rkT e−rs |g(s + kT )|dλ(s) and setting φk (s) := e−rs e−rkT |g(s + kT )| for all k ∈ N, we have φk ∈ L1 (0, T ; R+ ). Since the integral of nonnegative is additive and positively homogeneous, we have m measurable functions m k=0 [0,T ] φk (s)dλ(s) = [0,T ] k=0 φk (s)dλ(s), and using the Beppo Levi theorem we +∞ obtain k=0 [0,T ] φk (s)dλ(s) = [0,T ] +∞ k=0 φk (s)dλ(s). Using (5), we obtain
[0,T ]
e−rs
+∞
e−rkT |g(s + kT )|dλ(s) < +∞.
k=0
+∞
Consequently, we have k=0 φk ∈ L1 (0, T ; R+ ) which implies that this sum is λ-a.e. finite on [0, T ]. Since the absolute convergence of series implies the convergence, we can write, for λ-a.e. s ∈ [0, T ], +∞ +∞ −rs −rkT e g(s + kT ) ≤ e−rs e−rkT |g(s + kT )|, (6) e k=0
k=0
and since er A(g) is Lebesgue measurable, it also belongs to L1 (0, T ; R+ ). Since er is bounded on [0, T ], we obtain that A(g) ∈ L1 (0, T , μr ; R+ ). Moreover from (6), we obtain +∞ +∞
e−rT e−rkT g(s + kT ) dλ(s) ≤ e−rs e−rkT g(s + kT ) dλ(s) [0,T ] [0,T ] k=0
≤
[0,T ]
e−rs
k=0 +∞
e−rkT |g(s + kT )|dλ(s)
k=0
and using (5),
A(g) L1 (0,T ,μr ;Rn ) ≤
1 − e−rT
g L1 (R+ ,μr ;Rn ) . e−rT
n The Case α > 1 Let g ∈ Lα (R+ , μ r ; R ). Since μr (R+ ) < +∞, we have −rkT g(s + kT ) is well defined in Rn Lα (R+ , μr ; Rn ) ⊂ L1 (R+ , μr ; Rn ), and so +∞ k=0 e α 1 n for λ-a.e. s ∈ [0, T ]. Since |g| ∈ L (R+ , μr ; R ), we can assert from the case treated −rkT |g(s + kT )|α is well defined in R for λ-a.e. s ∈ [0, T ]. Now, using above that +∞ k=0 e
Infinite-Horizon Problems Under Periodicity Constraint
the H¨older inequality, with β > 0 such that
1 α
135
+
1 β
= 1, we obtain, for λ-a.e. s ∈ [0, T ],
+∞ +∞ α1 +∞ β1 e−rkT g(s + kT ) ≤ e−rkT |g(s + kT )|α e−rkT k=0
k=0
k=0
which leads to α +∞ 1 +∞ α−1 1 −rkT e g(s + kT ) ≤ e−rkT |g(s + kT )|α . 1 − e−rT k=0
(7)
k=0
Applying the case α = 1 to |g|α , we obtain that the function ψ defined by ψ(s) := +∞ −rkT |g(s + kT )|α , belongs to L1 (0, T ; R ), and consequently from (7) we obtain + k=0 e −rkT g(s + kT ), belongs to Lα (0, T ; Rn ) that the function χ , defined by χ (s) := +∞ e k=0 which implies that A(g) ∈ Lα (0, T ; Rn ). Now we prove the announced inequality. First notice that, using the Beppo Levi theorem, we have +∞
+∞
−rs −rkT −rs −rkT e e |g(s + kT )|dλ(s) = e e |g(s + kT )| dλ(s), k=0 [0,T ]
[0,T ]
k=0
and using (7) and (5) we obtain
1 |A(g)(s)|α dλ(s) e−rT −rT )α [0,T ] (1 − e α +∞
−rT −rkT = e e g(s + kT ) dλ(s) [0,T ] k=0
+∞ α e−rs e−rkT g(s + kT ) dλ(s) ≤ [0,T ] k=0 1
+∞ α−1 1 −rs −rkT α e e |g(s + kT )| dλ(s) ≤ 1 − e−rT [0,T ]
=
=
=
1 1 − e−rT 1 1 − e−rT
1 1 − e−rT e−rT
1 α−1
+∞
k=0 [0,T ]
1 α−1
1 α−1
k=0
e−rs e−rkT |g(s + kT )|α dλ(s)
|g|α L1 (R+ ,μr ;Rn )
g αLα (R+ ,μr ;Rn )
which implies (1−e−rT )α [0,T ] |A(g)(s)|α dλ(s) ≤ obtain the announced inequality.
1
g αLα (R+ ,μr ) , (1−e−rT )α−1
and then we
4.2 Relation with the Orthogonal Projection in L2 Theorem 3 Let f ∈ L2 (R+ , μr ; Rn ). The orthogonal projection of f on the subspace PT0 (R+ , μr ; Rn ) is (ET ◦ A)(f ).
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Proof From Theorem 2 and Theorem 1, we obtain that (ET ◦ A)(f ) belongs to PT0 (R+ , μr ; Rn ). Since PT0 (R+ , μr ; Rn ) is a closed vector subspace of the Hilbertian space L2 (R+ , μr ; Rn ), the orthogonal projection of f on PT0 (R+ , μr ; Rn ) exists and is unique. Let us denote by p this orthogonal projection of f . It is characterized by the following property ∀q ∈ PT0 (R+ , μr ; Rn ),
(f − p | q)L2 (R+ ,μr ;Rn ) = 0,
that is ∀q ∈ PT0 (R+ , μr ; Rn ),
(f | q)L2 (R+ ,μr ;Rn ) = (p | q)L2 (R+ ,μr ;Rn ) .
(8)
Let q ∈ PT0 (R+ , μr ; Rn ). Using the σ -additivity of the measures with density and using the periodicity of q we obtain
e−rt f (t) · q(t)dλ(t) (f | q)L2 (R+ ,μr ;Rn ) = k∈N [kT ,kT +T )
+∞
=
e−rt f (t) · q(t)dλ(t)
k=0 [kT ,kT +T ) +∞ T −rs −rkT
=
e
e
f (s + kT ) · q(s)ds,
k=0 0
and using the inequality in Theorem 2 and the Cauchy–Schwarz–Bunyakovski inequality, we can use the dominated convergence theorem of Lebesgue to obtain +∞
T −rs −rkT (f | q)L2 (R+ ,μr ;Rn ) = e e f (s + kT ) · q(s)ds. 0
k=0
This last equality implies that, for any q ∈ (f | q)L2 (R+ ,μr ;Rn ) =
PT0 (R+ , μr ; Rn ),
1 1 − e−rT
T
e−rs (Af )(s) · q(s)ds.
(9)
0
Replacing f by an arbitrary g ∈ PT0 (R+ , μr ; Rn ) in the previous computation, since g is T -periodic, we have A(g) = g and so we obtain
T 1 (g | q)L2 (R+ ,μr ;Rn ) = e−rs g(s) · q(s)ds, 1 − e−rT 0 that implies, for any g and q ∈ PT0 (R+ , μr ; Rn ), (g | q)L2 (R+ ,μr ;Rn )
1 = 1 − e−rT
T
e−rs g(s) · q(s)ds.
0
Taking g = (ET ◦ A)(f ) in (10) leads, for any q ∈ PT0 (R+ , μr ; Rn ), to
T 1 ((ET ◦ A)(f ) | q)L2 (R+ ,μr ;Rn ) = e−rs A(f )(s) · q(s)ds. 1 − e−rT 0 Finally, we get from (9) that, for all q ∈ PT0 (R+ , μr ; Rn ), (f | q)L2 (R+ ,μr ;Rn ) = (ET ◦ A)(f ) | q)L2 (R+ ,μr ;Rn ) ,
(10)
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137
and using (8) we obtain (p | q)L2 (R+ ,μr ;Rn ) = (ET ◦ A)(f ) | q)L2 (R+ ,μr ;Rn ) for all q ∈ PT0 (R+ , μr ; Rn ), which permits to conclude. Corollary 1 When f ∈ L2 (R+ , μr ; Rn ) the two following assertions are equivalent. (i) (ii)
f is orthogonal to PT0 (R+ , μr ; Rn ) in L2 (R+ , μr ; Rn ). A(f ) = 0.
Proof (i) ⇔ [(ET ◦ A)(f ) = 0] ⇔ [ (ET ◦ A)(f ) L2 (R+ ,μr ;Rn ) = 0]. Using (3) we have [ (ET ◦ A)(f ) L2 (R+ ,μr ;Rn ) = 0] ⇔ [ A(f ) L2 (0,T ,μr ;Rn ) = 0] ⇔ (ii). Remark 1 If f is orthogonal to PT0 (R+ , μr ; Rn ) in L2 (R+ , μr ; Rn ) and if f ≥ 0 on R+ , then using Corollary 1 we have f (s + kT ) = 0 for all k ∈ N and λ-a.e. s ∈ [0, T ) that implies f = 0. Now we denote by L(R+ , Rn ) the space of functions from R+ into Rn of the form a(t) = ta] where a ∈ Rn . L(R+ , Rn ) is a vector subspace of L2 (R+ , μr ; Rn ) which is isomorphic to Rn , and so it has finite dimension. Lemma 1 The direct sum PT0 (R+ , μr ; Rn ) ⊕ L(R+ , Rn ) is a closed vector subspace of L2 (R+ , μr ; Rn ). Proof It is an immediate consequence of [24, Corollaire, p. 229].
4.3 Relation with Time Series Analysis In Time Series Analysis, one question is to split the evolution variable X(t) as a sum X(t) = T (t)+S(t)+I (t), where T (t) is called the trend and it is an affine function of the time, S(t) is called seasonal variation and it is a periodic function of the time (with a fixed period), and I (t) is called the irregular component of X(t), i.e., the part of X(t) that we do not try to explain, see [20]. There exist methods to find the affine part and the T -periodic part of the evolution variable: the periodogram of frequencies, the moving average. In these methods, the affine part and the periodic part are found one after the other. Here we show that we can find simultaneously the trend and the seasonal variation. For any fixed function x ∈ L2 (R+ , μr ; Rn ), we consider the following minimization problem +∞ Minimize E(p, a) := 0 e−rt |x(t) − p(t) − ta|2 dt (P 4) when p ∈ PT0 (R+ , μr ; Rn ), a ∈ Rn . Theorem 4 For any x ∈ L2 (R+ , μr ; Rn ), the problem (P4) admits a unique solution (p, ˆ a) ˆ ∈ PT0 (R+ , μr ; Rn ) × Rn . In addition, it is given by
r T −rs ˜ aˆ = e (A(x)(s) − A(x)(s))ds, (11) T 0 ˆ (12) pˆ = (ET ◦ A)(x − a), where
+∞ (1 − e−rT )2 −rkT A˜ (x)(s) := ke x(s + kT ). e−rT k=0
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Proof First, in order to work in a function space, we formulate a problem which is equivalent to (P 4) +∞ Minimize E(p, a) := 0 e−rt |x(t) − p(t) − a(t)|2 dt (13) when (p, a) ∈ PT0 (R+ , μr ; Rn ) × L(R+ , Rn ). Since, by Lemma 1, PT0 (R+ , μr ; Rn ) ⊕ L(R+ , Rn ) is a closed vector subspace of the Hilbert space L2 (R+ , μr ; Rn ), the existence and the uniqueness of a solution (p, ˆ a) ˆ of problem (13) are simply due to the theorem of the orthogonal projection on a closed vector subspace in a Hilbert space. Now we want to obtain more information on this unique solution. We denote by π1 : L2 (R+ , μr ; Rn ) → L2 (R+ , μr ; Rn ) the orthogonal projector on PT0 (R+ , μr ; Rn );
recall that π1 = E ◦ A. We denote by PT0 (R+ , μr ; Rn )⊥ the orthogonal complement of
PT0 (R+ , μr ; Rn ) in L2 (R+ , μr ; Rn ), and by π2 the orthogonal projector on PT0 (R+ , μr ; Rn )⊥ . Notice that we have E(p, ˆ a) ˆ = inf E(p, a) : p ∈ PT0 (R+ , μr ; Rn ), a ∈ Rn ⎞ ⎛ = infn ⎝ a∈R
and we have E(p, ˆ a) ˆ = inf
p∈PT0 (R+ ,μr ;Rn )
x − a − p 2L2 (R
+ ,μr ;R
inf p∈PT0 (R+ ,μr ;Rn )
n)
E(p, a)⎠
inf
p∈PT0 (R+ ,μr ;Rn )
E(p, a). ˆ Moreover, for all a ∈ Rn , E(p, a) =
, and then
x − a − p 2L2 (R
+ ,μr ;R
= x − a − π1 (x − a) 2L2 (R
+ ,μr ;R
n)
n)
= π2 (x − a) 2L2 (R
+ ,μr ;R
=
n) 2
π2 (x) − π2 (a) L2 (R ,μ ;Rn ) . + r
In particular, for a = a, ˆ we have inf p∈PT0 (R+ ,μr ;Rn )
x − aˆ − p 2L2 (R
+ ,μr ;R
n)
= π2 (x) − π2 (a)
ˆ 2L2 (R
+ ,μr ;R
n)
,
hence E(p, ˆ a) ˆ = infn π2 (x) − π2 (a) 2L2 (R
+ ,μr ;R
a∈R
n)
(14)
.
Consequently, we obtain inf π2 (x) − π2 (a) 2L2 (R
a∈Rn
+ ,μr ;R
n)
= π2 (x) − π2 (a)
ˆ 2L2 (R
+ ,μr ;R
n)
.
From this equality, we deduce that π2 (a) ˆ = π3 (π2 (x)),
(15)
where π3 denotes the orthogonal projector on the subspace π2 (L(R+ , Rn )). Notice that the restriction of π2 to L(R+ , Rn ) is injective since when a ∈ Rn is such that π2 (a) = 0 then a is T -periodic and linear which implies a = 0. Consequently, the relation (15) determinates the unique value a. ˆ
Infinite-Horizon Problems Under Periodicity Constraint
139
It remains to prove the explicit formula for a. ˆ Let us write, for any a ∈ Rn , for any s ∈ [0, T ) π1 (x − a)(s) = =
= =
⎫ ⎪ ⎪ A(x)(s) − (1 − e−rT ) e−rkT (s + kt)a ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ k=0 ⎪ ⎪ +∞ ⎪ ⎪ ⎪ −rT −rkT ⎪ A(x)(s) − (1 − e ) e sa ⎪ ⎪ ⎪ ⎪ k=0 ⎬ +∞ ⎪ −(1 − e−rT ) e−rkT (kT )a ⎪ ⎪ ⎪ ⎪ k=0 ⎪ ⎪ −rT ⎪ T e ⎪ −rT ⎪ A(x)(s) − sa − (1 − e ) a, ⎪ ⎪ −rT 2 ⎪ (1 − e ) ⎪ ⎪ −rT ⎪ aT e ⎪ ⎪ ⎭ A(x)(s) − as − . (1 − e−rT ) +∞
(16)
Let us introduce the function F : Rn → R defined by F (a) := E(π1 (x − a)), a). Using (14), we see that
F (a) ˆ = infn F (a). a∈R
We use here again that π1 (x − a) = (ET ◦ A)(x) − (ET ◦ A)(a) and (16) to write
+∞
F (a) = =
e−rt |x(t) − (ET ◦ A)(x)(s) + (ET ◦ A)(a)(t) − ta|2 dt
0 +∞ T
e−rs e−rkT |x(s + kT ) − A(x)(s) +
k=0 0
T e−rT a − kT a|2 ds. (1 − e−rT )
The function F is a nonnegative quadratic function on Rn . Its minimizer aˆ then can be T −rs −rkT T e−rT characterized as a critical point, that is +∞ e ( (1−e−rT ) − kT )(x(s + kT ) − k=0 0 e
A(x)(s) +
T e−rT aˆ (1−e−rT )
− kT a)ds ˆ = 0, which is equivalent to
T e−rT − kT (A(x)(s) − x(s + kT ))ds (1 − e−rT ) 0 k=0
2
+∞ T T e−rT −rs −rkT − kT ds a. e e ˆ = (1 − e−rT ) 0
T
e−rs
+∞
e−rkT
(17)
k=0
To simplify the formula (17), we use the formulas +∞
−rT )
T 2 e−rT (1+e (1−e−rT )3
−rkT (kT )2 ) = k=0 (e −kz . Doing a direct k=0 e
+∞
0
T
e−rs
+∞ k=0
+∞
k=0 (e
−rkT kT )
=
T e−rT (1−e−rT )2
and
obtained by differentiating the power series S(z) :=
calculation, for the coefficient of aˆ in (17), we obtain e−rkT
T e−rT − kT (1 − e−rT )
2 ds =
T 2 e−rT r(1 − e−rt )2
(18)
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and for the left-hand side of (17), we obtain
T
e 0
=
−rs
+∞
e
−rkT
k=0 −rT Te
(1 − e−rT )2
T
T e−rT − kT (1 − e−rT )
(A(x)(s) − x(s + kT ))ds
e−rs A˜ (x)(s) − A(x)(s) ds.
(19)
0
From (18) and (19), we obtain (11).
5 Existence Results for Problem (P 1) In this section, we establish two existence theorems for (P 1), when α = 1 and when α > 1. We start by establishing some properties on the operator A1 , defined for any function L : R+ × Rn × Rn by ∀(s, x, y) ∈ [0, T ] × Rn × Rn , A1 (L)(s, x, y) = A (L(·, x, y)) (s). Lemma 2 Let L : R+ × Rn × Rn → R+ . (α) If, for a.e. t ∈ R+ , L(t, ·, ·) is lower semi-continuous on Rn × Rn , then, for a.e. s ∈ [0, T ], A1 (L)(s, ·, ·) is lower semi-continuous on Rn × Rn . (β) If, for a.e. t ∈ R+ , L(t, ·, ·) is upper semi-continuous on Rn × Rn , then, for a.e. s ∈ [0, T ], A1 (L)(s, ·, ·) is upper semi-continuous on Rn × Rn . (γ ) If, for a.e. t ∈ R+ , L(t, ·, ·) is continuous on Rn × Rn , then, for a.e. s ∈ [0, T ], A1 (L)(s, ·, ·) is continuous on Rn × Rn . (δ) If, for a.e. t ∈ R+ and for all x ∈ Rn , the function L(t, x, ·) is convex, then, for a.e. s ∈ [0, T ], and for all x ∈ Rn , the function A1 (L)(s, x, ·) is convex. () Let ρ : Rn → R+ be such that, for all (t, x, y) ∈ R+ × Rn × Rn , L(t, x, y) ≥ ρ(y), then we have, for all (s, x, y) ∈ [0, T ] × Rn × Rn , A1 (L)(s, x, y) ≥ ρ(y). (ζ ) If L is measurable from (R+ ×Rn ×Rn , B(R+ )⊗B(Rn )⊗B(Rn )) into (R+ , B(R+ ), then A1 (L) is measurable from ([0, T ] × Rn × Rn , B([0, T ]) ⊗ B(Rn ) ⊗ B(Rn )) into ([0, +∞], B([0, +∞])). (η) If L is a Caratheodory function (i.e., L(t, ·, ·) is continuous on Rn × Rn for a.e. t ∈ R+ , and L(·, x, y) is Borel measurable on R+ for all (x, y) ∈ Rn × Rn ) then A1 (L) is a Caratheodory function. Proof (α) We first establish the following assertion ∀A ∈ B(R+ ), ∀p ∈ R+ s.t. A − p ⊂ R+ ,
μr (A − p) ≤ e−rp μr (A).
Indeed, we have
μr (A − p) =
1dμr (t) =
=
A−p
R+
R+
1A−p (t)dμr (t)
1A (t + p)dμr (t) = 0
+∞
e−rt 1A (t + p)dt
(20)
Infinite-Horizon Problems Under Periodicity Constraint
=
+∞ p
≤ e
−rp
141
e−rs e−rp 1A (s)ds
+∞
e−rs 1A (s)ds = e−rp μr (A)
0
and (20) follows. As noticed in Section 2, we have ∀B ∈ B([0, T ]),
μr (B) = 0
=⇒
λ(B) = 0.
Consequently, when N ⊂ [0, T ] is μr -negligible then it is also λ-negligible. Let us now set S := {t ∈ R+ : L(t, ·, ·) is l.s.c.}, and S1 := {s ∈ [0, T ] : ∀k ∈ N, s + kT ∈ S} =
([0, T ] \ (S − kT )). k∈N
From the invariance under translation of the Lebesgue measure we obtain that [0, T ] \ (S − kT ) is negligible, and so [0, T ] \ S1 is negligible as a countable union of negligible sets. Consequently, for a.e. s ∈ [0, T ], L(s + kt, ·, ·) is l.s.c. for all k ∈ N. Let χ0 : 2N → [0, +∞] be the counting measure on N and let χ be the positive measure with density ξ with respect to χ0 , where ξ = [k → ξk ], from N into R+ , is defined by ξk := (1 − e−rT )e−rkT . We then have A1 (L)(s, x, y) = N L(s + kT , x, y)dχ (k). Let us arbitrarily fix (s, x, y) ∈ [0, T ] × Rn × Rn . Let (xq , yq ) be a sequence in Rn × Rn which converges to (x, y). For all q ∈ N, we set ϕq (k) := L(s + kT , xq , yq ). We have
A1 (L)(s, xq , yq ) =
N
ϕq (k)dχ (k).
(21)
The Fatou theorem provides
lim inf ϕq (k)dχ (k) ≤ lim inf
N q→+∞
q→+∞ N
ϕq (k)dχ (k).
(22)
Since L(t, ·, ·) is l.s.c., we have lim inf ϕq (k) = lim inf L(s + kT , xq , yq ) ≥ L(s + kT , x, y),
q→+∞
q→+∞
therefore A1 (L)(s, x, y) ≤ N lim infq→+∞ ϕq (k)dχ (k). Using (21) and (22), we obtain A1 (L)(s, x, y) ≤ lim infq→+∞ A1 (L)(s, xq , yq ), and the conclusion. (β) Since A1 (−L) = −A1 (L), it is a mere consequence of (α). (γ ) immediately follows from (α) and (β). (δ) We set C1 := {s ∈ [0, T ] : ∀k ∈ N, ∀x ∈ Rn , L(s + kT , x, ·) is convex}. Arguing as in the proof of (α), we obtain that for λ-a.e. s ∈ [0, T ], for all k ∈ N and for all x ∈ Rn ,
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L(s +kT , x, ·) is convex. Let (s, x) ∈ C1 ×Rn . Let y, y1 ∈ Rn and θ ∈ (0, 1). Then we have
A1 (L)(s, x, (1 − θ)y + θy1 ) = (1 − e−rT )
+∞
e−rkT L(t + kT , x, (1 − θ)y + θy1 )
k=0
≤ (1 − e−rT )
+∞
e−rkT ((1 − θ)L(t + kT , x, y) + θL(t + kT , x, y1 ))
k=0
= (1 − θ)(1 − e−rT )
+∞
e−rkT L(t + kT , x, y)
k=0
+θ(1 − e−rT )
+∞
e−rkT L(t + kT , x, y1 )
k=0
= (1 − θ)A1 (L)(s, x, y) + θ A1 (L)(s, x, y1 ) and the convexity is proven. () For any (s, x, y) ∈ [0, T ] × Rn × Rn , we have
A1 (L)(s, x, y) = (1 − e−rT )
+∞
e−rkT L(t + kT , x, y)
k=0
≥ (1 − e−rT )
+∞
e−rkT ρ(y) = ρ(y).
k=0
(ζ ) [(s, x, y) → (s + kT , x, y) → L(s + kT , x, y)] is measurable as a composition of Borel measurable functions. Since a linear combination of Borel measurable functions is Borel measurable, and since a limit of Borel measurable functions is Borel measurable, A1 (L) is Borel measurable. (η) From (γ ), it is sufficient to prove that, for all (x, y) ∈ Rn × Rn , A1 (L)(·, x, y) is measurable when L(·, x, y) is measurable. Notice that [s → L(s + kT , x, y)] is measurable as a composition of Borel measurable functions, hence [s → e−rkT L(s + kT , x, y)] is Borel measurable as a product of Borel measurable functions. Finally, for any integer , the map [s → k=0 e−rkT L(s + kT , x, y)] is Borel measurable as a finite sum of Borel measurable functions, and A1 (L)(·, x, y) is Borel measurable as a limit of Borel measurable functions. We can now state our first main result on the existence of solutions for the problem (P 1) with α = 1 that we formulate in the following form +∞ Minimize F (x) = 0 e−rt L(t, x(t), x (t))dt (23) n when x ∈ x0 + WT1,1 ,0 (R+ , μr ; R ), 1
where σ ∈ Rn is fixed and x0 ∈ W 1,1 (R+ , μr ; Rn ) ∩ PT0 (R+ , Rn ) satisfies x0 (0) = σ . Theorem 5 Let L : R+ × Rn × Rn → R+ be a function which satisfies the following conditions (a) L is a Caratheodory function. (b) For a.e. t ∈ R+ , for all x ∈ Rn , L(t, x, ·) is convex on Rn .
Infinite-Horizon Problems Under Periodicity Constraint
143
There exists ρ : Rn → R+ such that lim|y|→+∞ ρ(y) |y| = +∞ and, for all (t, x, y) ∈ R+ × Rn × Rn , L(t, x, y) ≥ ρ(y). n (d) There exists x˜ ∈ x0 + WT1,1 ,0 (R+ , μr ; R ) such that
+∞ e−rt L(t, x(t), ˜ x˜ (t))dt < +∞. (c)
0
Then Problem (23) possesses a solution. Proof We consider the following problem T Minimize FT (u) = 0 A1 (L)(s, u(s), u (s))ds
(24)
when u ∈ W 1,1 (0, T ; Rn ), u(0) = u(T ) = σ.
From (a), using (η), (δ) and () in Lemma 2, we get that A1 (L) has the following properties: – – –
A1 (L) is a Caratheodory function. For a.e. t ∈ [0, T ], and for all x ∈ Rn , the function A1 (L)(t, x, ·) is convex. For all (s, x, y) ∈ [0, T ] × Rn × Rn , A1 (L)(s, x, y) ≥ ρ(y), where ρ is superlinear.
By (d), the criterion of Problem (24) is not constant equal to +∞ and according to [10, Remark 1, p.115], it admits a solution u. ˆ Finally, since
T
1 e−rt L(t, x(t), x (t))dt = e−rs A1 (L)(s, x(s), x (s))ds, (25) 1 − e−rT 0 R+ ˆ is a solution of Problem (23). we obtain that xˆ := ET (u) This existence result can be extended to the Sobolev spaces W 1,α (R+ , μr ; Rn ) with α ∈ (1, +∞). We formulate (P 1) with α ∈ (1, +∞) in the following form ⎫ +∞ Minimize F (x) = 0 e−rt L(t, x(t), x (t))dt ⎬ (26) n ⎭ when x ∈ x0 + WT1,α ,0 (R+ , μr ; R ), α
where σ ∈ Rn is fixed and x0 ∈ W 1,α (R+ , μr ; Rn ) ∩ PT0 (R+ , Rn ) satisfies x0 (0) = η. Theorem 6 Let L : R+ × Rn × Rn → R+ be a function which satisfies the following conditions (a) L is a Caratheodory function. (b) For all (t, x) ∈ R+ × Rn , L(t, x, ·) is convex. (c) There exist a ∈ L1 (R+ , μr ; R+ ) and b ∈ (0, +∞) such that L(t, x, y) ≥ a(t) + b|y|α for a.e. t ∈ R+ and for all (x, y) ∈ Rn × Rn . n (d) There exists x˜ ∈ x0 + WT1,α ˜ < +∞. ,0 (R+ , μr ; R ) such that I (x) Then Problem (25) possesses a solution. Proof We consider the following problem T Minimize FT (u) = 0 A1 (L)(s, u(s), u (s))ds when u ∈ W 1,α (0, T , μr ; Rn ), u(0) = u(T ) = η.
(27)
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As in the proof of Theorem 5, the assumptions (a), (b), and (d) imply the same properties for A1 on [0, T ] × Rn × Rn . From assumption (c), we obtain that A1 (L)(s, x, y) ≥ A(a)(s) + b · |y|α for a.e. s ∈ [0, T ] and for all (x, y) ∈ Rn × Rn , and using Theorem 2 we know that A(a) ∈ L1 (0, T ; R). Consequently, all the assumptions of [15, Theorem 4.1, p. 82] are fulfilled for Problem (27) which allows us to assert that there exists uˆ a solution of Problem (27). To conclude, ˆ is a solution of Problem (25). using (25) we verify that xˆ := ET (u) Remark 2 In Theorem 6, the assumption (d) is ensured as soon as L is supposed to have a polynomial growth in the third variable. In this case, L can even take nonpositive values, since upper integrals turn into regular integrals. Notice that such arguments can not extend to the W 1,1 setting. According to [10], in both Theorems 5 and 6, the assumption (a) can be replaced by requiring that L is globally measurable and that for a.e. t ∈ R+ , the map L(t, ·, ·) is lower semi-continuous.
6 Necessary Conditions of Optimality We do not treat the question of the Euler–Lagrange equation in the W 1,1 setting. Indeed, for boundary value problems, the authors of [10] say, in the point (b) in p. 139, that even the Euler–Lagrange equation may fail for solutions issued from a Tonelli’s partial regularity theorem which provides conditions which ensure that an optimal solution possesses a (possibly infinite) classical derivative in the whole interval of the problem. We will here only consider the case W 1,α with α ∈ (1, +∞). The Euler–Lagrange equation appears via a regularity result, under strictly stronger assumptions on the Lagrangian. As in the previous section, we first prove some preliminary results about the continuity and differentiability properties of the averaged Lagrangian. For any finite dimensional normed real vector space E and for any map : R+ × Rn × n R → E, we consider the following properties: (P1)
∀(x, y) ∈ Rn × Rn , ∀ > 0, ∃δ > 0, ∀t, t1 ∈ R+ , ∀(x1 , y1 ) ∈ Rn × Rn s.t. (|t − t1 | ≤ δ, |x − x1 | ≤ δ, |y − y1 | ≤ δ) =⇒ |(t, x, y) − (t1 , x1 , y1 )| ≤ .
(P2)
(P3)
∈ C 1 (R+ × Rn × Rn , E) and ∀(x, y) ∈ Rn × Rn , ∀ > 0, ∃η > 0, ∀t, t1 ∈ R+ , ∀(x1 , y1 ) ∈ Rn × Rn , (|t − t1 | ≤ η, |x − x1 | ≤ η, |y − y1 | ≤ η) =⇒ |(t1 , x1 , y1 ) − (t, x, y) − D(t, x, y)(t1 − t, x1 − x, y1 − y)| ≤ (|t − t1 | + |x − x1 | + |y − y1 |). The partial differential D3 φ(t, x, y) exists for any (t, x, y) ∈ R+ × Rn × Rn and satisfies the following condition: ∀(x, y) ∈ Rn × Rn , ∀ > 0, ∃β > 0, ∀z ∈ Rn , |z| ≤ β
=⇒
∀t ∈ R+ , |(t, x, y + z) − (t, x, y) − D3 (t, x, y)z| ≤ |z|.
Lemma 3 Let L : R+ × Rn × Rn → R. The following assertions hold. (i) If L satisfies (P1) then A1 (L) ∈ C 0 ([0, T ] × Rn × Rn , R). (ii) If L satisfies (P1) and (P3) and if D3 L satisfies (P1) then D3 A1 (L) ∈ C 0 ([0, T ] × Rn × Rn , L(Rn , R)) and D3 A1 (L) = A1 (D3 L). (iii) We assume that L ∈ C 1 (R+ × Rn × Rn , R), that L satisfies (P1) and (P2) and that DL satisfies (P1).
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Then A1 (L) ∈ C 1 ([0, T ] × Rn × Rn , R) and we have D A1 (L) = A1 (DL). (iv) We assume that L ∈ C 2 (R+ × Rn × Rn , R), that L and DL satisfy (P1) and (P2), and that D 2 L satisfies (P1). Then A1 (L) ∈ C 2 ([0, T ] × Rn × Rn , R) and we have D A1 (L) = A1 (DL) and D 2 A1 (L) = A1 (D 2 L). Proof We first prove (i). Let (x, y) ∈ Rn × Rn and > 0. Let s, s1 ∈ [0, T ], (x1 , y1 ) ∈ Rn × Rn be such that |s − s1 | ≤ δ, |x − x1 | ≤ δ and |y − y1 | ≤ δ. Then we have, for all k ∈ N, |(s + kT ) − (s1 + kT )| = |s − s1 | ≤ δ, and we get from (P1), for any k ∈ N, |L(s + kT , x, y) − L(s1 + kT , x1 , y1 )| ≤ leading to |A1 (L)(s, x, y) − A1 (L)(s1 , x1 , y1 )| +∞ e−rkT |L(s + kT , x, y) − L(s1 + kT , x1 , y1 )| ≤ (1 − e−rT ) k=0
≤ (1 − e−rT )
+∞
e−rkT = .
k=0
Let us now establish (ii). Let (s, x, y) ∈ [0, T ] × Rn × Rn and > 0. Let z ∈ Rn be such that |z| ≤ η. From (P3), we have, for all k ∈ N, |L(s + kT , x, y + z) − L(s + kT , x, y) − D3 L(s + kT , x, y)z| ≤ |z|. Then, setting ρ := (1 − e−rT ), we obtain |A1 (L)(s, x, y + z) − A1 (L)(s, x, y) − A1 (D3 L)(s, x, y) · z| +∞ −rkT e (L(s + kT , x, y + z) − L(s + kT , x, y) − D3 L(s + kT , x, y) · z) =ρ k=0
≤ρ ≤ρ
+∞ k=0 +∞
e−rkT |(L(s + kT , x, y + z) − L(s + kT , x, y) − D3 L(s + kT , x, y) · z)| e−rkT |z| = |z|
k=0
and the conclusion follows. Next, we prove (iii). Let (s, x, y) ∈ [0, T ] × Rn × Rn and > 0, and consider δs, δx and δy such that |δs| ≤ η, |δx| ≤ η and |δy| ≤ η, where η = η(L, , x, y) is provided by (P2). Then we have |A1 (L)(s + δs, x + δx, y + δy) − A1 (L)(s, x, y) − A1 (DL)(s, x, y)(δs, δx, δy)| +∞ e−rkT |L(s + δs + kT , x + δx, y + δy) − L(s + kT , x, y) ≤ k=0
−DL(s + kT , x, y)(δs, δx, δy)| 1 ≤ s(|δs| + |δx| + |δy|). 1 − e−rT These inequalities prove that A1 (L) is Fr´echet differentiable at (s, x, y) and that D A1 (L)(s, x, y) = A1 (DL)(s, x, y). Since DL satisfies (P1), using (i), we infer that A1 (DL) is continuous, and consequently D(A1 (L)) is continuous. As a conclusion, (iv) is just a consequence of (i) applied to L and to DL.
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Theorem 7 Let L : R+ × Rn × Rn → R be a function, and let α ∈ (1, +∞). We assume that the following assumptions are fulfilled. (a) L is of class C 2 on R+ × Rn × Rn , L and DL satisfy the conditions (P1) and (P2) and D 2 L satisfies (P1). (b) There exist constants c0 , c1 ∈ (0, +∞) such that, ∀(t, x, y) ∈ R+ × Rn × Rn , (c)
c0 |y|α ≤ L(t, x, y) ≤ c1 (1 + |y|α ).
There exists a function M : (0, +∞) → (0, +∞) such that ∀(t, x, y) ∈ R+ × Rn × Rn , |x|2 + |y|2 ≤ R 2
(d)
=⇒
|D2 L(t, x, y)| + |D3 L(t, x, y)| ≤ M(R)(1 + |y|2 ).
∀(t, x, y) ∈ R+ × Rn × Rn , ∀ξ ∈ Rn \ {0}, D33 L(t, x, y)(ξ, ξ ) > 0.
Suppose that xˆ is a local solution for the strong topology of the Sobolev space of Problem (26), then xˆ is C 2 on R+ except at most at the points kT where k ∈ N, and satisfies the Euler–Lagrange equation ˆ xˆ (t)) = D2 A1 (L)(t, x(t),
d D3 A1 (L)(t, x(t), ˆ xˆ (t)) dt
for all t ∈ R+ \ T N. Proof From assumption (a) using Lemma 3(iv), we know that ! " A1 (L) ∈ C 2 [0, T ] × Rn × Rn , R .
(28)
From assumption (b), we obtain ∃c0 , c1 ∈ (0, +∞), ∀(s, x, y) ∈ [0, T ] × Rn × Rn , c0 |y|α ≤ A1 (L)(s, x, y) ≤ c1 (1 + |y|α ).
(29)
Since D2 A1 (L) = A1 (D2 L) and D3 A1 (L) = A1 (D3 L) after Lemma 3, from assumption (c), we obtain ∃M : (0, +∞) → (0, +∞), ∀(s, x, y) ∈ [0, T ] × Rn × Rn , |x|2 + |y|2 ≤ R 2 =⇒ |D2 A1 (L)(s, x, y)| + |D3 A1 (L)(s, x, y)| ≤ M(R)(1 + |y|2 ). (30) Since D33 A1 (L) = A1 (D33 L) after Lemma 3, from assumption (d), we obtain ∀(s, x, y) ∈ [0, T ] × Rn × Rn , ∀ξ ∈ Rn \ {0},
D33 A1 (L)(s, x, y)(ξ, ξ ) > 0.
(31)
With (28), (29), (30), (31), all the assumptions of the regularity theorem given in [10, p. 134], are fulfilled for Problem (26) and so when xˆ is a solution of Problem (26), its restriction uˆ to [0, T ] is a solution of Problem (25). Then, using the regularity theorem ([10, Theorem 4.1]) on Problem (26), we obtain that uˆ ∈ C 2 ([0, T ], Rn ) and uˆ satisfies the associated Euler–Lagrange equation at each point of [0, T ]. The conclusion is simply the translation on xˆ of the properties of u. ˆ
Concluding Remark The two operators A and ET have permitted to translate the periodic optimization problem on R+ into a finite-horizon problem and consequently to use classical results (on the existence of optimal solution and on the Euler–Lagrange equation) to obtain results on the periodic optimization problem on R+ . It is possible that this method can be used to establish other results on the periodic optimization problems.
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Acknowledgements We thank Professor Rabah Tahraoui (University of Rouen) for helpful discussions which permit us to improve the paper, and the anonymous referees who have helped us to correct several mistakes and omissions.
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