# Lagrangian Convex in the Fibers

Nel documento Weak KAM Theorem in Lagrangian Dynamics Seventh Preliminary Version (pagine 103-109)

Z b

a

[L(γ(t), ˙γ(t)) − Cγ]ds + Cγ(b − a).

This quantity is clearly in R ∪ {+∞}. It is not difficult to see that the action L(γ) is well defined (i.e. independent of the choice of lower bound Cγ).

### 3.2 Lagrangian Convex in the Fibers

In this section, we consider a manifold M provided with a Rie-mannian metric of reference. For x ∈ M , we denote by k·kx the norm induced on TxM by the Riemannian metric. We will denote by d the distance induced on M by the Riemannian metric. The canonical projection of T M on M is as usual π : T M → M .

We will consider a C1 Lagrangian L on M which is convex in the fibers, and superlinear above every compact subset of M , see

definitions 1.3.12 and 1.3.14. Note that, for Lagrangians convex in the fibers, the superlinearity above compact subset is equivalent to the restriction L|TxM : TxM → R issuperlinear, for each x ∈ M , see 1.3.15. It is clear that a Lagrangian superlinear above every compact subset of M is also bounded below above every compact subset of M , therefore we can define the action for absolutely continuous curves.

Theorem 3.2.1. Suppose that L : T M → R is a C1 Lagrangian convex in the fibers, and superlinear above compact subsets of M . If a sequence of curves γn ∈ Cac([a, b], M ) converges uniformly to the curve γ : [a, b] → M and

lim inf

n→∞

L(γn) < +∞,

then the curve γ is also absolutely continuous and lim inf

n→∞ L(γn) ≥ L(γ).

Proof. Let us start by showing how we can reduce the proof to the case where M is an open subset of Rk, where k = dim M .

Consider the set K = γ([a, b]) ∪S

nγn([a, b]), it is compact because γnconverges uniformly to γ. By superlinearity of L above each compact subset of M , we can find a constant C0 such that

∀x ∈ K, ∀v ∈ TxM, L(x, v) ≥ C0.

If [a, b] is a subinterval of [a, b], taking C0 as a lower bound of L(γn(s), ˙γn(s)) on [a, b] \ [a, b], we see that

∀n, L(γn|[a, b]) ≤ L(γn) − C0[(b − a) − (b− a)].

It follows

∀[a, b] ⊂ [a, b], lim inf

n→∞ L(γn|[a, b]) < +∞. (*) By continuity of γ : [a, b] → M , we can find a finite sequence a0= a < a1 < · · · < ap = b and a sequence of domains of coordinate charts U1, . . . , Up such that γ([ai−1, ai]) ⊂ Ui, for i = 1, . . . , p.

Since γn converges uniformly to γ, forgetting the first curves γnif necessary, we can assume that γn([ai−1, ai]) ⊂ Ui, for i = 1, . . . , p.

91 By condition (∗), we know that lim infn→∞L(γn|[ai−1, ai]) < ∞, it is then enough to show that this condition implies that γ|[ai−1, ai] is absolutely continuous and that

L(γ|[ai−1, ai]) ≤ lim inf

n→∞ L(γn|[ai−1, ai]),

because we have lim inf(αn + βn) ≥ lim inf αn + lim inf βn, for sequences of real numbers αn and βn. As the Ui are domains of definition of coordinates charts, we do indeed conclude that it is enough to show the theorem in the case where M is an open subset of Rk.

In the sequel of the proof, we will thus suppose that M = U is an open subset of Rkand thus T U = U × Rkand that γ([a, b]) and all the γn([a, b]) are contained in the same compact subset K0 ⊂ U . Let us set ℓ = lim infγn→∞L(γn). Extracting a subsequence such that L(γn) → ℓ < +∞ and forgetting some of the first curves γn, we can suppose that

L(γn) → ℓ and ∀n ∈ N, L(γn) ≤ ℓ + 1 < +∞.

We denote by k·k a norm on Rk.

Lemma 3.2.2. If C ≥ 0 is a constant, K ⊂ U is compact, and ǫ > 0, we can find η > 0 such that for each x, y ∈ U with x ∈ K and ky − xk < η, and for each v, w ∈ Rk with kvk ≤ C, we have

L(y, w) ≥ L(x, v) +∂L

∂v(x, v)(w − v) − ǫ.

Proof. Let us choose η0 > 0 such that the set

η0(K) = {y ∈ Rk| ∃x ∈ K, ky − xk ≤ η0} is a compact subset of U .

We denote by A the finite constant A = sup{k∂L

∂x(x, v)k | x ∈ K, kvk ≤ C}.

Since L is uniformly superlinear above every compact subset of M = U , we can find a constant C1 > −∞ such that

∀y ∈ ¯Vη0(K), ∀w ∈ Rk, L(y, w) ≥ (A + 1)kwk + C1.

We then set

C2= sup{L(x, v) −∂L

∂x(x, v)(v) | x ∈ K, kvk ≤ C}.

By compactness the constant C2is finite. We remark that if kwk ≥ C2− C1, then for y ∈ ¯Vη0(K), x ∈ K and kvk ≤ C, we have

L(y, w) ≥ (A + 1)kwk + C1

≥ Akwk + (C2− C1) + C1

= Akwk + C2

≥∂L

∂v(x, v)(w) + L(x, v) − ∂L

∂v(x, v)(v)

= L(x, v) +∂L

∂v(x, v)(w − v).

It then remains to find η ≤ η0, so that we satisfy the sought inequality when kwk ≤ C2− C1. But the set

{(x, v, w) | x ∈ K, kvk ≤ C, kwk ≤ C2− C1}

is compact and L(x, w) ≥ L(x, v) + ∂L∂v(x, v)(w − v) by convexity of L in the fibers of the tangent bundle T U . It follows that for ǫ > 0 fixed, we can find η > 0 with η ≤ η0 and such that, if x ∈ K, ky − xk ≤ η, kvk ≤ C et kwk ≤ C2− C1, we have

L(y, w) > L(x, v) + ∂L

∂x(x, v)(w − v) − ǫ.

We return to the sequence of curves γn : [a, b] → U which converges uniformly to γ : [a, b] → U . We already have reduced the proof to the case where γ([a, b]) ∪S

n∈Nγn([a, b]) is included in a compact subset K0 ⊂ U with U an open subset of Rk. We now show that the derivatives ˙γn are uniformly integrable. Since L is superlinear above each compact subset of U , we can find a constant C(0) > −∞ such that

∀x ∈ K0, ∀v ∈ Rk, L(x, v) ≥ C(0).

We recall that

∀n ∈ N, L(γn) ≤ ℓ + 1 < +∞.

93 We then fix ǫ > 0 and we take A > 0 such that

ℓ + 1 − C(0)(b − a)

A < ǫ/2.

Again by the superlinearity of L above compact subsets of the base M , there exists a constant C(A) > −∞ such that

∀x ∈ K0, ∀v ∈ Rk, L(x, v) ≥ Akvk + C(A).

Let E ⊂ [a, b] be a measurable subset, we have C(A)m(E) + A

Z

E

k ˙γn(s)kds ≤ Z

E

L(γn(s), ˙γn(s)) ds and also

C(0)(b − a − m(E)) ≤ Z

[a,b]\E

L(γn(s), ˙γn(s)) ds.

Adding the inequalities and using L(γn) ≤ ℓ + 1, we find [C(A) − C(0)]m(E) + C(0)(b − a) + A

Z

E

k ˙γn(s)kds ≤ ℓ + 1, this in turn yields

Z

E

k ˙γn(s)kds ≤ ℓ + 1 − C(0)(b − a)

A +[C(0) − C(A)]

A m(E)

≤ ǫ/2 +[C(0) − C(A)]

A m(E).

If we choose δ > 0 such that C(0)−C(A)A δ < ǫ/2, we see that m(E) < δ ⇒ ∀n ∈ N,

Z

E

k ˙γn(s)k ds < ǫ.

This finishes to proves the uniform integrability of the sequence ˙γn. We can then conclude by proposition 3.1.4 that ˙γnconverges to ˙γ in the σ(L1, L) topology. We must show that limn→∞L(γn) ≥ L(γ).

Let C be a fixed constant, we set

EC = {t ∈ [a, b] | k ˙γ(t)k ≤ C}.

We fix ǫ > 0 and we apply lemma 3.2.2 with this ǫ, the constant C fixed above and the compact set K = K0⊃ γ([a, b])∪S

n∈Nγn[a, b]

to find η > 0 as in the conclusion of the lemma 3.2.2. Since γn→ γ uniformly, there exists an integer n0 such that for each n ≥ n0, we have kγn(t) − γ(t)k < η, for each t ∈ [a, b]. Lemma 3.2.2 then shows that for each n ≥ n0 and almost all t ∈ EC, we have

L(γn(t), ˙γn(t)) ≥ L(γ(t), ˙γ(t)) + ∂L

∂v(γ(t), ˙γ(t))( ˙γn(t) − ˙γ(t)) − ǫ, hence using this, together with the inequality L(γn(t), ˙γn(t)) ≥ C(0) which holds almost everywhere, we obtain

L(γn) ≥

Going to the limit in the inequality (*), we find ℓ = lim Since the derivative ˙γ(t) exists and is finite for almost all t ∈ [a, b], we find EC ր E, when C ր +∞, with [a, b] \ E of zero Lebesgue measure. Since L(γ(t), ˙γ(t)) is bounded below by C(0), we have by the monotone convergence theorem

Z

95 If we let C ր +∞ in (∗∗), we find

ℓ = lim

n→∞L(γn) ≥ Z b

a

L(γ(t), ˙γ(t)) dt = L(γ).

Corollary 3.2.3. Suppose L : T M → R is a C1 Lagrangian convex in the fibers, and superlinear above every compact sub-set of M . Then the action L : Cac([a, b], M ) → R ∪ {+∞} is lower semi-continuous for the topology of uniform convergence on . Cac([a, b], M ). In particular, on any compact subset of Cac([a, b], M ), the action L achieves its infimum.

Proof. It is enough to see that if γn → γ uniformly, with all the γn and γ in Cac([a, b], M ), then, we have

lim inf

γn→∞ L(γn) ≥ L(γ).

If lim infn→∞L(γn) = +∞, there is nothing to show. The case where we have lim infn→∞L(γn) < +∞ results from theorem 3.2.1.

Nel documento Weak KAM Theorem in Lagrangian Dynamics Seventh Preliminary Version (pagine 103-109)