Z b

a

[L(γ(t), ˙γ(t)) − C_{γ}]ds + C_{γ}(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·k_{x} 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 C^{1} 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|T_{x}M : T_{x}M → 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 C^{1} Lagrangian
convex in the fibers, and superlinear above compact subsets of M .
If a sequence of curves γ_{n} ∈ C^{ac}([a, b], M ) converges uniformly to
the curve γ : [a, b] → M and

lim inf

n→∞

L(γ_{n}) < +∞,

then the curve γ is also absolutely continuous and lim inf

n→∞ L(γ_{n}) ≥ L(γ).

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

Consider the set K = γ([a, b]) ∪S

nγn([a, b]), it is compact
because γ_{n}converges uniformly to γ. By superlinearity of L above
each compact subset of M , we can find a constant C_{0} such that

∀x ∈ K, ∀v ∈ TxM, L(x, v) ≥ C_{0}.

If [a^{′}, b^{′}] is a subinterval of [a, b], taking C_{0} as a lower bound of
L(γ_{n}(s), ˙γ_{n}(s)) on [a, b] \ [a^{′}, b^{′}], we see that

∀n, L(γn|[a^{′}, b^{′}]) ≤ L(γ_{n}) − C_{0}[(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 < a_{1} < · · · < a_{p} = b and a sequence of domains of coordinate
charts U_{1}, . . . , U_{p} such that γ([a_{i−1}, a_{i}]) ⊂ U_{i}, for i = 1, . . . , p.

Since γ_{n} converges uniformly to γ, forgetting the first curves γ_{n}if
necessary, we can assume that γ_{n}([a_{i−1}, a_{i}]) ⊂ U_{i}, for i = 1, . . . , p.

91
By condition (∗), we know that lim inf_{n→∞}L(γ_{n}|[ai−1, a_{i}]) < ∞, it
is then enough to show that this condition implies that γ|[a_{i−1}, a_{i}]
is absolutely continuous and that

L(γ|[ai−1, ai]) ≤ lim inf

n→∞ L(γn|[ai−1, ai]),

because we have lim inf(α_{n} + β_{n}) ≥ lim inf α_{n} + lim inf β_{n}, for
sequences of real numbers α_{n} and β_{n}. As the U_{i} 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 R^{k}.

In the sequel of the proof, we will thus suppose that M = U is
an open subset of R^{k}and thus T U = U × R^{k}and that γ([a, b]) and
all the γ_{n}([a, b]) are contained in the same compact subset K_{0} ⊂ U .
Let us set ℓ = lim inf_{γ}_{n}_{→∞}L(γ_{n}). Extracting a subsequence such
that L(γn) → ℓ < +∞ and forgetting some of the first curves γn,
we can suppose that

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

We denote by k·k a norm on R^{k}.

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 ∈ R^{k} with kvk ≤ C, we have

L(y, w) ≥ L(x, v) +∂L

∂v(x, v)(w − v) − ǫ.

Proof. Let us choose η_{0} > 0 such that the set

V¯_{η}_{0}(K) = {y ∈ R^{k}| ∃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 C_{1} > −∞ such that

∀y ∈ ¯V_{η}_{0}(K), ∀w ∈ R^{k}, L(y, w) ≥ (A + 1)kwk + C_{1}.

We then set

C_{2}= sup{L(x, v) −∂L

∂x(x, v)(v) | x ∈ K, kvk ≤ C}.

By compactness the constant C_{2}is finite. We remark that if kwk ≥
C_{2}− C1, then for y ∈ ¯V_{η}_{0}(K), x ∈ K and kvk ≤ C, we have

L(y, w) ≥ (A + 1)kwk + C_{1}

≥ Akwk + (C_{2}− C_{1}) + C_{1}

= Akwk + C_{2}

≥∂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 ≤ η_{0}, 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 ≤ η_{0} and such that, if
x ∈ K, ky − xk ≤ η, kvk ≤ C et kwk ≤ C_{2}− C_{1}, we have

L(y, w) > L(x, v) + ∂L

∂x(x, v)(w − v) − ǫ.

We return to the sequence of curves γ_{n} : [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 K_{0} ⊂ U with U an open subset of R^{k}. We
now show that the derivatives ˙γ_{n} 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 ∈ R^{k}, L(x, v) ≥ C(0).

We recall that

∀n ∈ N, L(γ_{n}) ≤ 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 ∈ R^{k}, 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(γ_{n}(s), ˙γ_{n}(s)) ds
and also

C(0)(b − a − m(E)) ≤ Z

[a,b]\E

L(γ_{n}(s), ˙γ_{n}(s)) ds.

Adding the inequalities and using L(γ_{n}) ≤ 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 ˙γ_{n}.
We can then conclude by proposition 3.1.4 that ˙γ_{n}converges to ˙γ
in the σ(L^{1}, L^{∞}) topology. We must show that lim_{n→∞}L(γ_{n}) ≥
L(γ).

Let C be a fixed constant, we set

E_{C} = {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 = K_{0}⊃ γ([a, b])∪S

n∈Nγ_{n}[a, b]

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

L(γ_{n}(t), ˙γ_{n}(t)) ≥ L(γ(t), ˙γ(t)) + ∂L

∂v(γ(t), ˙γ(t))( ˙γ_{n}(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(γ_{n}) ≥

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 E_{C} 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(γ_{n}) ≥
Z b

a

L(γ(t), ˙γ(t)) dt = L(γ).

Corollary 3.2.3. Suppose L : T M → R is a C^{1} Lagrangian
convex in the fibers, and superlinear above every compact
sub-set of M . Then the action L : C^{ac}([a, b], M ) → R ∪ {+∞} is
lower semi-continuous for the topology of uniform convergence on .
C^{ac}([a, b], M ). In particular, on any compact subset of C^{ac}([a, b], M ),
the action L achieves its infimum.

Proof. It is enough to see that if γ_{n} → γ uniformly, with all the
γ_{n} and γ in C^{ac}([a, b], M ), then, we have

lim inf

γn→∞ L(γ_{n}) ≥ L(γ).

If lim inf_{n→∞}L(γ_{n}) = +∞, there is nothing to show. The case
where we have lim inf_{n→∞}L(γ_{n}) < +∞ results from theorem
3.2.1.