In the sequel we denote by ˚B(0, r) (resp. ¯B(0, r)) the open ball
(resp. closed) of center 0 and radius r in the Euclidean space R^{k},
where k is the dimension of M .

Proposition 4.7.1. Let ϕ : ˚B(0, 5) → M be a coordinate chart
and t_{0}> 0 be given. There is a constant K ≥ 0 such that for each
function u ∈ C^{0}(M, R), for each x ∈ ¯B(0, 1), and for each t ≥ t0,
we have:

(1) For each y ∈ ¯B(0, 1) and each extremal curve γ : [0, t] → M with γ(t) = ϕ(x) and

T_{t}^{−}u[ϕ(x)] = u(γ(0)) +
Z _{t}

0

L(γ(s), ˙γ(s))ds, we have

T_{t}^{−}u[ϕ(y)]−T_{t}^{−}u[ϕ(x)] ≤ ∂L

∂v(ϕ(x), ˙γ(t))(Dϕ(x)[y−x])+Kky−xk^{2}.
In particular, if T_{t}^{−}u is differentiable at ϕ(x) then

d_{ϕ(x)}T_{t}^{−}u = ∂L/∂v(ϕ(x), ˙γ(t))
and the curve γ is unique.

(2) For each y ∈ ¯B(0, 1) and each curve γ : [0, t] → M with γ(0) = ϕ(x) and

T_{t}^{+}u[ϕ(x)] = u(γ(t)) −
Z t

0

L(γ(s), ˙γ(s))ds,

we have

T_{t}^{+}u[ϕ(y)]−T_{t}^{+}u[ϕ(x)] ≥ ∂L

∂v(ϕ(x), ˙γ(0))(Dϕ(x)[y−x])−Kky−xk^{2}.
In particular, if T_{t}^{+}u is differentiable at ϕ(x) then

d_{ϕ(x)}T_{t}^{+}u = ∂L/∂v(ϕ(x), ˙γ(0))
and the curve γ is unique.

Proof. We use some auxiliary Riemannian metric on M to have a norm on tangent space and a distance on M .

Since T_{t}^{−}u = T_{t}^{−}_{0}T_{t−t}^{−} _{0}u by the semigroup property, we have
only to consider the case t = t_{0}. By corollary 4.3.2, we can find a
finite constant A, such that any minimizer defined on an interval
of time at least t0 > 0 has speed unifomely bounded in norm
by A.This means that such curves are all A-Lipschitz. We then
pick ǫ > 0 such that for each ball ¯B(y, Aǫ), for y ∈ ϕ( ¯B(0, 1) is
contained in ϕ( ˚B(0, 2). Notice that this ǫ does not depend on u.

Since any curve γ, as in part (1) of the proposition is necessary a minimizer, we therefore have

γ([t_{0}− ǫ, t0]) ⊂ ϕ( ˚B(0, 2)).

We then set ˜γ = ϕ^{−1} ◦ γ and ˜L(x, w) = L(ϕ(x), Dϕ(x)w) for
(x, w) ∈ ˚B(0, 5) × R^{k}. Taking derivatives, we obtain

∂ ˜L(x, w)

∂w = ∂L

∂v(ϕ(x), Dϕ(x)(w))[Dϕ(x)(·)].

The norm of the vector h = y − x is ≤ 2, hence if we define

˜

γ_{h}(s) = ^{s−(t}_{ǫ}^{0}^{−ǫ)}h+˜γ(s), for s ∈ [t_{0}−ǫ, t0], we have ˜γ_{h}(s) ∈ ˚B(0, 4),
and

T_{t}^{−}_{0}u[ϕ(x+h)]−T_{t}^{−}_{0}u[ϕ(x)] ≤
Z t0

t0−ǫ

[ ˜L(˜γ_{h}(s), ˙˜γ_{h}(s))−L(˜γ(s), ˙˜γ(s))] ds.

Since the speed of γ is bounded in norm by A, we see that (˜γ(s), ˙˜γ(s))
is contained in a compact subset of ¯B(0, 2) × R^{k} independent of
u, x and γ. Moreover, we have ˜γ_{h}(s) − ˜γ(s) = ^{s−(t−ǫ)}_{ǫ} h, which
is of norm ≤ 2, and ˙˜γ_{h}(s) − ˙˜γ(s) = ^{1}_{ǫ}h, which is itself of norm

≤ 2/ǫ. We then conclude that there exists a compact convex
sub-set C ⊂ ¯B(0, 4) × R^{k}, independent of u, x and γ, and containing
(˜γ(s), ˙˜γ(s)) and (˜γ_{h}(s), ˙˜γ_{h}(s)), for each s ∈ [t_{0} − ǫ, t_{0}]. Since we
can bound uniformly on the compact subset C the norm of the
second derivative of ˜L, by Taylor’s formula at order 2 applied to
L, we see that there exists a constant ˜˜ K independent of u, x and

153

supposing that ǫ < 1. As ˜γ is an extremal curve for the Lagrangian L, it satisfies the Euler-Lagrange equation˜

∂ ˜L

Since we can also apply the inequality above with −v instead of v, we conclude that If we divide by δ > 0 and we let δ go to 0, we obtain

∀v ∈ R^{n}d_{ϕ(x)}T_{t}^{−}uDϕ(x)[v] = ∂L

∂v(ϕ(x), ˙γ(t))(Dϕ(x)[v]),

by the linearity in v of the involved maps. Since ϕ is a
diffeomor-phism this shows that d_{ϕ(x)}T_{t}^{−}u∂L/∂v(ϕ(x), ˙γ(t)). By the
bijec-tivity of the Legendre Transform the tangent vector ˙γ(t) is unique.

Since γ is necessarily a minimizing extremal, it is also unique, since both its position x and its speed ˙γ(t) at t are uniquely determined.

To prove (2), we can make a similar argument for T_{t}^{+}, or,
more simply, apply what we have just done to the symmetrical
Lagrangian ˇL of L.

Exercise 4.7.2. 1) Let u_{−} : M → R be a weak KAM solution.

Show that u_{−} has a derivative at x if and only if there is one and
only one curve γ_{−}^{x} :] − ∞, 0] → M such that γ_{−}^{x}(0) = x and

∀t ≥ 0, u−(γ_{−}^{x}(0)) − u_{−}(γ^{x}_{−}(−t)) =
Z 0

−t

L(γ_{−}^{x}(s), ˙γ_{−}^{x}(s)) ds + c[0]t.

In that case we have d_{x}u_{−}= ^{∂L}_{∂v}(x, ˙γ_{−}^{x}(0)).

2) Suppose that x ∈ M , and that γ_{−}^{x} :] − ∞, 0] → M satisfies
γ_{−}^{x}(0) = x and

∀t ≥ 0, u−(γ_{−}^{x}(0)) − u_{−}(γ^{x}_{−}(−t)) =
Z 0

−t

L(γ_{−}^{x}(s), ˙γ_{−}^{x}(s)) ds + c[0]t.

Show that necessarily H(x,∂L

∂v(x, ˙γ_{−}^{x}(0)) = c[0].

We will need a criterion to show that a map is differentiable with a Lipschitz derivative. This criterion has appeared in dif-ferent forms either implicitly or explicitly in the literature, see [CC95, Proposition 1.2 page 8], [Her89, Proof of 8.14, pages 63–

65], [Kni86], [Lio82, Proof of Theorem 15.1, pages 258-259], and also [Kis92] for far reaching generalizations. The simple proof given below evolved from discussions with Bruno Sevennec.

Proposition 4.7.3 (Criterion for a Lipschitz Derivative). Let ˚B be the open unit ball in the normed space E. Fix a map u : ˚B → R.

If K ≥ 0 is a constant, denote by AK,uthe set of points x ∈ ˚B, for
which there exists ϕ_{x} : E → R a continuous linear form such that

∀y ∈ ˚B, |u(y) − u(x) − ϕ_{x}(y − x)| ≤ Kky − xk^{2}.

155
Then the map u has a derivative at each point x ∈ A_{K,u},
and d_{x}u = ϕ_{x}. Moreover, the restriction of the map x 7→ d_{x}u
to {x ∈ A_{K,u} | kxk < ^{1}_{3}} is Lipschitzian with Lipschitz constant

≤ 6K.

More precisely

∀x, x^{′}∈ A_{K,u}, kx−x^{′}k < min(1−kxk, 1−kx^{′}k) ⇒ kd_{x}u−d_{x}^{′}uk ≤ 6Kkx−x^{′}k.

Proof. The fact that dxu = ϕx, for x ∈ AK,u, is clear. Let us fix
x, x^{′} ∈ AK,u, with kx − x^{′}k < min(1 − kxk, 1 − kx^{′}k). If x = x^{′}
there is nothing to show, we can then suppose that kx − x^{′}k > 0.

If h is such that khk = kx − x^{′}k, then the two points x + h et x^{′}+ h
are in ˚B. This allows us to write

|u(x + h) − u(x) − ϕ_{x}(h)| ≤ Kkhk^{2}

|u(x) − u(x^{′}) − ϕ_{x}^{′}(x − x^{′}) ≤ Kkx − x^{′}k^{2}

|u(x + h) − u(x^{′}) − ϕ_{x}^{′}(x − x^{′}+ h)| ≤ Kkx − x^{′}+ hk^{2}.
As khk = kx − x^{′}k, we obtain from the last inequality

|u(x^{′}) − u(x + h) + ϕ_{x}^{′}(x − x^{′}+ h)| ≤ 4Kkx − x^{′}k^{2}.
Adding this last inequality with the first two above, we find that,
for each h such that khk = kx − x^{′}k, we have

|ϕ_{x}^{′}(h) − ϕ_{x}(h)| ≤ 6Kkx − x^{′}k^{2},
hence

kϕ_{x}^{′} − ϕ_{x}k = sup

khk=kx−x^{′}k

|ϕ_{x}^{′}(h) − ϕ_{x}(h)|

kx − x^{′}k ≤ 6Kkx − x^{′}k.

Exercise 4.7.4. If A_{k,u}is convex, for example if A_{K,U} = ˚B, show
that the derivative is Lipschitzian on A_{K,u}with Lipschitz constant

≤ 6K.

Theorem 4.7.5. If ǫ > 0 is given, then there are constants A ≥ 0
and η > 0, such that any map u : M → R, with u ≺ L + c, is
differentiable at every point of the set A_{ǫ,u} formed by the x ∈

M for which there exists a (continuous) piecewise C^{1} curve γ :
[−ǫ, ǫ] → M with γ(0) = x and

u(γ(ǫ)) − u(γ(−ǫ)) = Z ǫ

−ǫ

L(γ(s), ˙γ(s)) ds + 2cǫ.

Moreover we have

(1) Such a curve γ is a minimizing extremal and
d_{x}u = ∂L

∂x(x, ˙γ(0));

(2) the set A_{ǫ,u} is closed;

(3) the derivative map A_{ǫ,u}→ T^{∗}M, x 7→ (x, d_{x}u) is Lipschitzian
with Lipschitz constant ≤ A on each subset A_{ǫ,u}with
diam-eter ≤ η.

Proof. The fact that γ is a minimizing extremal curve results from u ≺ L + c. This last condition does also imply that

u(γ(ǫ)) − u(x) = Z ǫ

0

L(γ(s), ˙γ(s)) ds + cǫ, u(x) − u(γ(−ǫ)) =

Z 0

−ǫ

L(γ(s), ˙γ(s)) ds + cǫ. (*)
Since ǫ > 0 is fixed, by the Corollary of A Priori Compactness, we
can find a compact subset K_{ǫ} ⊂ T M such that for each minimizing
extremal curve γ : [−ǫ, ǫ] → M , we have (γ(s), ˙γ(s)) ∈ K_{ǫ}. It is
not then difficult to deduce that Aǫ,uis closed. It also results from
(∗) that

T_{ǫ}^{+}u(x) ≥ u(x) + cǫ

T_{ǫ}^{−}u(x) ≤ u(x) − cǫ. (**)
As u ≺ L + c, we have

T_{ǫ}^{+}u ≤ u + cǫ,
T_{ǫ}^{−}u ≥ u − cǫ.

157 We then obtain equality in (∗∗). Subtracting this equality from the inequality above, we find

∀y ∈ M, T_{ǫ}^{+}u(y) − T_{ǫ}^{+}u(x) ≤ u(y) − u(x) ≤ T_{ǫ}^{−}u(y) − T_{ǫ}^{−}u(x).

(∗∗∗)
Let us then cover the compact manifold M by a finite number of
open subsets of the form ϕ_{1}( ˚B(0, 1/3)), · · · , ϕ_{ℓ}( ˚B(0, 1/3)), where
ϕ_{i} : ˚B(0, 5) → M, i = 1, . . . , ℓ, is a C^{∞} coordinate chart. By
Proposition 4.7.1, there exists a constant K, which depends only
on ǫ and the fixed ϕ_{p}, p = 1, . . . , ℓ such that, if x ∈ ϕ_{i}( ˚B(0, 1/3)),
setting x = ϕ_{i}(˜x), for each y ∈ ˚B(0, 1), we have

T_{ǫ}^{−}u(ϕ_{i}(y)) − T_{ǫ}^{−}u(ϕ_{i}(˜x)) ≤ ∂L

∂v(x, ˙γ(0)) ◦ Dϕ_{i}(˜x)(y − ˜x) + Kky − ˜xk^{2}
T_{ǫ}^{+}u(ϕ_{i}(y) − T_{ǫ}^{+}u(ϕ_{i}(˜x)) ≥ ∂L

∂v(x, ˙γ(0)) ◦ Dϕ_{i}(˜x)(y − ˜x) − Kky − ˜xk^{2},
Using the inequalities (∗∗∗) we get

|u(ϕ_{i}(y)) − u(ϕ_{i}(˜x)) −∂L

∂v(x, ˙γ(0)) ◦ Dϕ_{i}(˜x)(y − ˜x)| ≤ Kky − ˜xk^{2}.
By the Criterion for a Lipschitz Derivative 4.7.3, we find that
d_{x}u exists and is equal to ^{∂L}_{∂v}(x, ˙γ(0)). Moreover the restriction of
x 7→ dxu on Aǫ,U∩ ϕi( ˚B(0, 1/3)) is Lipschitzian with a constant of
Lipschitz which depends only on ǫ. It is then enough to choose for
η > 0 a Lebesgue number for the open cover (ϕ_{i}( ˚B(0, 1/3))_{i=1,··· ,ℓ}
of the compact manifold M .

Definition 4.7.6 (The sets S−and S+). We denote by S−(resp.

S+) the set of weak KAM solutions of the type u_{−} (resp. u_{+}), i.e.

the continuous functions u : M → R such that T_{t}^{−}u + c[0]t = u
(resp. T_{t}^{+}u − c[0]t = u).

Exercise 4.7.7. If c ∈ R, show that u ∈ S_{−} (resp. u ∈ S_{+}) if and
only if u + c ∈ S_{−} (resp. u + c ∈ S_{+}). If x_{0} ∈ M is fixed, show
that the set {u ∈ S_{−}| u(x_{0}) = 0} (resp. {u ∈ S_{+} | u(x_{0}) = 0}) is
compact for the topology of uniform convergence.

Theorem 4.7.8. Let u : M → R be a continuous function. The following properties are equivalent

(1) the function u is C^{1} and belongs to S_{−},
(2) the function u is C^{1} and belongs to S_{+}.

(3) the function u belongs to the intersection S_{−}∩ S+. (4)] the
function u is C^{1}and there exists c ∈ R such that H(x, dxu) =
c, for each x ∈ M .

In all the cases above, the derivative of u is (locally) Lipschitzian.

Proof. Conditions (1) or (2) imply (4). It is enough, then to show
that (4) implies (1) and (2) and that (3) implies that u is C^{1}, and
that its derivative is (locally) Lipschitzian. Thus let us suppose
condition (3) satisfied. Indeed, in this case, if x ∈ M , we can find
extremal curves γ_{−}^{x} :] − ∞, 0] → M and γ_{+}^{x} : [0, ∞[→ M , with
γ_{−}^{x}(0) = γ_{+}^{x}(0) = x and

∀t ≥ 0,u(γ_{+}^{x}(t)) − u(x) =
Z t

0

L(γ_{+}^{x}(s), ˙γ_{+}^{x}(s)) ds + c[0]t,
u(x) − u(γ_{−}^{x}(−t)) =

Z 0

−t

L(γ_{−}^{x}(s), ˙γ_{−}^{x}(s)) ds + c[0]t.

The curve γ : [−1, 1] → M , defined by γ|[−1, 0] = γ^{x}_{−}and γ|[0, 1] =
γ_{+}^{x}, shows that x ∈ A_{1,u}. By the previous theorem, the function
u is of class C^{1} and its derivative is (locally) Lipschitzian.

Let us suppose that u satisfies condition (4). By the Fenchel inequality, we have

∀(x, v) ∈ T M, dxu(v) ≤ H(x, d_{x}u) + L(x, v)

= c + L(x, v).

Consequently, if γ : [a, b] → M is a C^{1} curve, we obtain

∀s ∈ [a, b], d_{γ(s)}u( ˙γ(s)) ≤ c + L(γ(s), ˙γ(s)), (*)
and by integration on the interval [a, b]

u(γ(b)) − u(γ(a)) ≤ c(b − a) + Z b

a

L(γ(s), ˙γ(s)) ds, (∗∗)
which gives us u ≺ L + c. If γ : [a, b] → M is an integral curve
of grad_{L}u, we have in fact equality in (∗) and thus in (∗∗). Since

159
u ≺ L + c, it follows that γ is a minimizing extremal curve. If two
solutions go through the same point x at time t_{0}, they are, in fact,
equal on their common interval of definition, since they are two
ex-tremal curves which have the same tangent vector (x, grad_{L}u(x))
at time t_{0}. As we can find local solutions by the Cauchy-Peano
Theorem, we see that grad_{L}u is uniquely integrable. Since M
is compact, for any point x ∈ M , we can find an integral curve
γ^{x} :] − ∞, +∞[→ M of grad_{L}u with γ^{x}(0) = x. This curve gives
at the same time a curve of the type γ_{−}^{x} and one of the type γ_{+}^{x}
for u. This establishes that u ∈ S−∩ S+.