Existence and uniqueness of a Nash equilibrium feedback for a simple nonzero-sum differential game

Pierre Cardaliaguet

Université de Bretagne Occidentale, Département de Mathématiques, 6, avenue Victor-le-Gorgeu, B.P. 809, 29285 Brest cedex, France e-mail : Pierre.Cardaliaguet@univ-brest.fr

Slawomir Plaskacz

Nicholas Copernicus University Faculty of Mathematics and Computer Science Chopina 12/8 87-1000 Toruń, Poland e-mail : plaskacz@mat.uni.torun.pl

February 1, 2002

Abstract : Existence and uniqueness of a Nash equilibrium feedback is established for a simple class nonzero-sum differential games on the line.

Key-words : Nash equilibrium, nonzero-sum differential game.

1 Introduction

The object of this paper is the study of Nash equilibria for some nonzero-sum two players differential games on the line. The dynamics of the differential game is

x′ = u + v, with u ∈ [−1, 1], v = [−1, 1] . (1)

The payoff is the terminal payofff g(x(T)) = (g_{1}(x(T)), g_{2}(x(T))), for some function g : IR → IR^{2}. Player I, playing with u, wants to maximize g_{1}(x(T)), while player II, playing with v, wants to maximize g_{2}(x(T)).

Our goal is to introduce a definition of closed loop Nash equilibrium feedback and of Nash equilibrium payoff for this game, and to prove the existence of a Nash equilibrium feedback as well as some uniqueness result for Nash equilibrium payoff.

For zero-sum differential games, i.e., when g_{2} = −g_{1}, the notion of Nash equilibrium is replaced by the notion of value of the game. The existence and the characterization of the value for such a game is now well-known (see in particular, for the existence, [4, 5], and for the characterization [3]). For the dynamics (1), the value function V (t, x) (for Player I) of the game is simply given by

∀(t, x) ∈ [0, T] × IR, V (t, x) = g_{1}(x) .

Moreover, if g_{1} is sufficiently regular (for instance, if g_{1} = −g_{2} is C^{1} and has a finite number of local extrema), we can define explicitely optimal feedback strategies u^{∗}(t, x) and v^{∗}(t, x) for player I and II given respectively by:

u^{∗}(t, x) = sgn(g′_{1}(x)) and v^{∗}(t, x) = −sgn(g′_{1}(x)) ,

where sgn(s) = 1 if s > 0 , −1 if s < 0, and 0 if s = 0.

In conclusion, for the dynamics (1), the solution of the zero-sum differential game is completely understood and essentially trivial. Surprizingly, this is not at all the case for nonzero-sum differential games, even for dynamics as simple as (1).

For nonzero-sum differential games, played with closed loop strategies or with strategies with memory, one can find in the litterature merely two approaches:

The first one is inspired by Isaacs work for zero-sum differential games. Its main goal is the explicit computation of the Nash equilibrium payoff as a function of the time and the space, i.e., as a function E : [0, T] × IR → IR^{2}, which associates with any initial condition (t_{0}, x_{0}) “some” Nash equilibrium payoff E(t_{0}, x_{)} (see in particular [2], [8] and the references therein). The key idea is that the payoff E should satisfy a system of Hamilton-Jacobi equations given for our game (with dynamics (1)) by:

\{ \begin{matrix}\frac{∂E_{1}}{∂t} + |\frac{∂E_{1}}{∂x} | + \frac{∂E_{1}}{∂x} sgn(\frac{∂E_{2}}{∂x}) = 0 \\ \frac{∂E_{2}}{∂t} + |\frac{∂E_{2}}{∂x} | + \frac{∂E_{2}}{∂x} sgn(\frac{∂E_{1}}{∂x}) = 0 \end{matrix} (2)

with terminal condition E(T, ·) = g. Unfortunately, there is up to now no global theory for such a system. In particular, it does not fit with the assumptions required in the theory viscosity solutions for systems of first order PDEs. A local theory exists - the so-called characteristic method - but it is only be applicable in a neighbourhood of the points (T, x) where g′_{1}(·) and g′_{2}(·) are well defined and do not vanish. Therefore this method says very little in the case we are studying.

However when this method is applicable, it enjoys several desirable properties. First the solution E satisfies a dynamic programming property (also called “time consistency” in [2]): For any time T_{0} ∈ (0, T), the restriction of E to the interval [0, T_{0}] is a Nash equilibrium payoff for the game with horizon T_{0} and terminal payoff E(T_{0}, ·). Second this method provides a Nash equilibrium feedback.

Another theory for the nonzero-sum differential games has been developped by Kleimenov in [6]. Its main result is the characterization of Nash equilibrium payoffs when the game is played with strategies with memory. For the dynamics (1), this result states that a couple (e_{1}, e_{2}) ∈ IR^{2} is a Nash equilibrium payoff for the initial position (t_{0}, x_{0}) if and only if there is some time measurable controlpair (ū(·), v̄(·)) such that the solution of (1) with x(t_{0}) = x_{0} satisfies the following inequalities:

∀t ∈ [t_{0}, T], for j = 1, 2, g_{j}(x(t)) ≤ e_{j} = g_{j}(x(T)) .

Loosely speaking, the strategy for Player I given by Kleimenov’s construction is the following (the strategy for Player II is symmetric): As long as Player II plays v̄(·), Player I plays ū(·). As soon as Player II deviates from v̄(·), i.e., plays some v(t) ≠ v̄(t), Player I punishes Player II by playing −v(t). Thus for Player II the playoff is g_{2}(x(T)) = e_{2} if he does not deviate, while it is g_{2}(x(t_{1})) ≤ e_{2} if he starts deviating at time t_{1}. Hence Player II has a better payoff if he sticks to the control v̄(·). Therefore (e_{1}, e_{2}) is a Nash equilibrium payoff.

Although Kleimenov’s result completely characterizes the Nash equilibrium payoffs, it also shows that there are in general infinitely many Nash equilibria, and that most of them are out of interest. Thus the question of selecting good Nash equilibrium payoffs arises naturally. Kleimenov gives in [6] several selection methods, but none makes the connection with the feedback strategies suggested by the first theory described above, and in particular, none enjoys the dynamic programming property.

The main objective of this paper is to construct, for the dynamics (1) and for a large class G of terminal payoffs g, a pair (u^{∗}, v^{∗}) of feedback strategies, such that, at each point, the associated payoff is a Nash equilibrium (in the sense of Kleimenov) and such that the associated payoff satisfies the dynamic programming property. We call such a pair of strategies a Nash equilibrium feedback.

Unfortunately, the notion of of Nash equilibrium feedback is not enough discriminating: Indeed, even for elementary examples, there are in general a lot a uninteresting Nash equilibrium feedbacks (see Example 2.14 below). So we are led to introduce an additional requirement of being “completely maximal”.

We say that a payoff (e_{1}, e_{2}) ∈ IR^{2} is maximal for a given initial position (t_{0}, x_{0}) if there is some solution x̄(·) of the controlled system (1) (for some time-measurable control (ū(·), v̄(·))) such that: (e_{1}, e_{2}) = g(x̄(T)) and both e_{1} and e_{2} are the maximum of g_{1} and g_{2} among the values that can be reached by the controled system (1) starting from (t_{0}, x_{0}). Of course, such a maximal payoff seldom exists, but when it exists, it is reasonnable to think that both players should prefer it. We say that a Nash equilibrium feedback is maximal if, at any point (t_{0}, x_{0}) where there is a maximal payoff, the payoff of the feedback is equal to this maximal payoff. It is completely maximal, if its restriction to any subinterval [0, T_{0}], with T_{0} < T is also maximal.

Our main result is that such completely maximal Nash equilibrium feedback exists. It turns out that it is even unique and that it is in some sense ”Pareto”. This holds true for a large class of terminal payoffs g, and, in particular, when g is continuous and g_{1} and g_{2} have a finite number of local extrema. To the best of our knowledge, this is the first result of this nature.

However, it is quite difficult to prove, and this is the reason why we had to restrict this study to the dimension one and to the particular dynamics x′ = u + v. We are for the moment unable to say in what extent it can be generalized to more general dynamics and terminal payoffs. In particular, it is possible that the notion of maximal payoff is only interesting in the framework of the dynamics (1) ; it has certainly to be adapted in higher dimensions of space.

Let us finally explain how this paper is organized. In section 2, we introduce the different notions used in the paper, state the main Theorem and give several examples. Section 3 is devoted to the construction of the solution for small intervals of times. In section 4, prove that one can concatenate together two solutions and pass to the limit. Then we give the proof of the main Theorem. We complete the paper by proving that the payoff of the Nash equilibrium feedback we have constructed is at each point an equilibrium payoff in the sense of Kleimenov, and that it is a Pareto one.

2 Nash equilibrium feedbacks

In this section we introduce the main definitions of this paper as well as the main result (Theorem 2.10).

2.1 Strategies and payoffs

The dynamic of the differential game is (1):

x′(t) = u(t) + v(t) u(t) ∈ [−1, 1] and v(t) ∈ [−1, 1] .

The game is of fixed duration and the terminal time is denoted by T. The terminal payoff is a function g = (g_{1}, g_{2}) : IR → IR^{2}: Player I wants to maximize g_{1}(x(T)) while Player II wants to maximize g_{2}(x(T)).

We denote by U the set of functions u : [0, T]×IR → [−1, 1], interpreted as strategies for Payer I, and by V the set of functions v : [0, T] × IR → [−1, 1] interpreted as strategies for Payer II. We call a pair (u, v) ∈ U × V a feedback.

As usual we have to give a sense to the equation

x′(t) = ū(t, x(t)) + v̄(t, x(t))

for discontinuous strategies ū ∈ U and v̄ ∈ V. This has already been done before (see for instance [7]), and we follow more or less the same method.

For any (t_{0}, x_{0}) ∈ [0, T]×IR and any pair of strategies (u, v) ∈ U ×V, we denote by X (t_{0}, x_{0}, u, v) the set of solutions of the differential inclusion

\{ \begin{matrix}ẋ ∈ f̃(t, x, u, v) & on [t_{0}, T] \\ x(t_{0}) = x_{0} \end{matrix}

where f̃(t, x, u, v) is the smallest usc convex and compact set-valued map containing the map (t, x) → u(t, x) + v(t, x). It is well-known that this set of solutions is compact for the uniform convergence, and that it has a closed graph (see [1]).

We also denote by X (t_{0}, x_{0}) the set of all the solutions of of (1).

The lower and upper payoffs of the strategies (u, v) for the initial position (t_{0}, x_{0}) are respectively given by: For j = 1, 2, the lower payoff of Player j, denoted by J_{j}^{♭}(t_{0}, x_{0}, u, v), is

J_{j}^{♭}(t_{0}, x_{0}, u, v) = inf_{x∈X (t_{0},x_{0},u,v)}(g_{j})_{∗}(x(T))

while the upper payoff J_{j}^{♯}(t_{0}, x_{0}, u, v) is

J_{j}^{♯}(t_{0}, x_{0}, u, v) = sup_{x∈X (t_{0},x_{0},u,v)}(g_{j})^{∗}(x(T)) .

Here (g_{j})_{∗} and (g_{j})^{∗} are respectively the lower semi-continuous and the upper semi-continuous envelopes of g_{j} (i.e., respectively the largest lower semi-continuous (lsc) function which is not larger than g_{j} and the the smallest upper semi-continuous (usc) function which is not smaller than g_{j}).

Proposition 2.1 Let us assume that g_{1} and g_{2} are usc. Then, for j = 1 or j = 2, there is some x_{j}(·) ∈ X (t_{0}, x_{0}, u, v) such that

J_{j}^{♯}(t_{0}, x_{0}, u, v) = g_{j}(x_{j}(T)) .

Proof : The proof is a straightforward consequence of the compactness of the set X (t_{0}, x_{0}, u^{∗}, v^{∗}) for the topology of the uniform convergence.

QED

We use below the notations

J^{♯}(t_{0}, x_{0}, u, v) = (J_{1}^{♯}(t_{0}, x_{0}, u, v), J_{2}^{♯}(t_{0}, x_{0}, u, v)) and J^{♭}(t_{0}, x_{0}, u, v) = (J_{1}^{♭}(t_{0}, x_{0}, u, v), J_{2}^{♭}(t_{0}, x_{0}, u, v)) .

2.2 Definition of the Nash equilibrium feedbacks

Definition 2.2 A Nash equilibrium feedback on the time interval [T_{0}, T] for the terminal time T and the terminal payoff g is a feedback (u^{∗}, v^{∗}) ∈ U × V such that for all t_{0} ∈ [T_{0}, T] there exists a set S_{t_{0}} ⊂ IR of zero measure such that for any x_{0} ∈ IR\S_{t_{0}}, we have

∀u ∈ U, inf_{x∈X (t_{0},x_{0},u^{∗},v^{∗})}(g_{1})_{∗}(x(T)) ≥ sup_{x∈X (t_{0},x_{0},u,v^{∗})}(g_{1})^{∗}(x(T))

and

∀v ∈ V, inf_{x∈X (t_{0},x_{0},u^{∗},v^{∗})}(g_{2})_{∗}(x(T)) ≥ sup_{x∈X (t_{0},x_{0},u^{∗},v)}(g_{2})^{∗}(x(T)) .

Remarks :

1. Using the notations of the previous section, the two above inequalities can be rewritten

∀(u, v) ∈ U × V, J_{1}^{♭}(t_{0}, x_{0}, u^{∗}, v^{∗}) ≥ J_{1}^{♯}(t_{0}, x_{0}, u, v^{∗}) and J_{2}^{♭}(t_{0}, x_{0}, u^{∗}, v^{∗}) ≥ J_{2}^{♯}(t_{0}, x_{0}, u^{∗}, v) .

2. It is clear from the definition that, for j = 1, 2, we have

∀x_{0} ∉ S_{t_{0}}, J_{j}^{♭}(t_{0}, x_{0}, u^{∗}, v^{∗}) = J_{j}^{♯}(t_{0}, x_{0}, u^{∗}, v^{∗}) .

In order to simplify the notations, we denote by J_{1}(t_{0}, x_{0}, u^{∗}, v^{∗}) (resp. J_{2}(t_{0}, x_{0}, u^{∗}, v^{∗})) this common value. Then we say that J(t_{0}, x_{0}, u^{∗}, v^{∗}) = (J_{1}(t_{0}, x_{0}, u^{∗}, v^{∗}), J_{2}(t_{0}, x_{0}, u^{∗}, v^{∗})) a Nash equilibrium payoff. It is defined for any t_{0} ∈ [0, T] and for any x_{0} ∈ IR\S_{t_{0}} where S_{t_{0}} has a zero measure in IR.

3. From standard arguments, it is easy to check that the map x → J^{♯}(t, x, u^{∗}, v^{∗}) is continuous at each point (t, x) ∈ IR for which the following equality holds:

J^{♯}(t, x, u^{∗}, v^{∗}) = J^{♭}(t, x, u^{∗}, v^{∗}) .

Before giving the next definition, we introduce a notation : For a map g : IR → IR^{2} and a point x ∈ IR, we set

ess − lim sup_{x′→x} g(x) = {a ∈ IR^{2} | ∀ϵ > 0, ∀r > 0, |g^{−1}(B_{ϵ}(a))∩]x − r, x + r[| > 0} ,

where |A| denotes the outward Lebesgue measure of a subset A of IR and where B_{ϵ}(a) denotes the ball of center a and radius ϵ in IR^{2}.

Definition 2.3 (Maximal payoff) Let (t_{0}, x_{0}) ∈ [0, T] × IR and (e_{1}, e_{2}) ∈ IR^{2}. We say that (e_{1}, e_{2}) is the maximal payoff at the point (t_{0}, x_{0}) for the game with terminal time T and terminal payoff g if there is some x_{0}(·) ∈ X (t_{0}, x_{0}) such that (e_{1}, e_{2}) ∈ ess − lim sup_{x′→x_{0}(T)} g(x′) and such that, for any x(·) ∈ X (t_{0}, x_{0}), we have

for j = 1, 2, e_{j} ≥ (g_{j})^{∗}(x(T)) .

Remarks :

1. Loosely speaking, this only means that both functions g_{1} and g_{2} reach a maximum at the same point x_{0}(T) among the points x(T) that one can reach starting from (t_{0}, x_{0}).

2. For a given point (t_{0}, x_{0}) does not necessarily exist (and in fact seldom exists). However, if it exists, it is clearly unique.

3. If g is continuous, we have of course (e_{1}, e_{2}) = g(x_{0}(T)). However, for later use, we need to have this definition at points where the function g can be discontinuous.

A typical example of what we want to avoid is the following situation. Let g : IR → IR^{2} be defined by

∀x ∈ IR, g(x) = \{ \begin{matrix}(−1, 1) & if x < 0 \\ (1, −1) & if x > 0 \\ (1, 1) & if x = 0 \end{matrix}

Then the value (1, 1) does not belong to ess − lim sup_{x′→0} g(x′), although it could be a good candidate for being a maximal payoff, since at this point both functions g_{1} and g_{2} have a maximum. But examples of differential games can be given for which this would lead to some contradictions (see example 2.16).

For later use, we need the following remark:

Lemma 2.4 Let (t_{0}, x_{0}) ∈ [0, T] × IR and (e_{1}, e_{2}) ∈ IR^{2} be the maximal payoff at the point (t_{0}, x_{0}) for the game with terminal time T and terminal payoff g. Let x_{0}(·) ∈ X (t_{0}, x_{0}) be such that

(e_{1}, e_{2}) ∈ ess − lim sup_{x′→x_{0}(T)} g(x′) .

Then, for any t_{1} ∈ [t_{0}, T), (e_{1}, e_{2}) is the maximal payoff at the point (t_{1}, x_{0}(t_{1})) for the game with terminal time T and terminal payoff g.

Proof : It is a straightforward application of the concatenation property of the solutions of system (1).

QED

Definition 2.5 (Maximal Nash equilibrium feedback) Let (u^{∗}, v^{∗}) ∈ U × V be a Nash equilibrium feedback for the game with terminal time T and terminal payoff g. We say that (u^{∗}, v^{∗}) is maximal on the time interval [T_{0}, T] for the game with terminal time T and terminal payoff g, if (u^{∗}, v^{∗}) is a Nash equilibrium feedback on [T_{0}, T] and if (u^{∗}, v^{∗}) satisfies the additionnal requirement: For any (t_{0}, x_{0}) ∈ [T_{0}, T) × IR, if there is some maximal payoff (e_{1}, e_{2}) for the game with terminal time T and terminal payoff g at the point (t_{0}, x_{0}), then

for j = 1, 2, e_{j} = J_{j}^{♯}(t_{0}, x_{0}, u^{∗}, v^{∗}) .

Remark : The requirement of (u^{∗}, v^{∗}) to be maximal is somehow the weakest requirement we could make. Nash equilibrium feedbacks are not necessarly maximal (see example 2.14 below).

Before introducing the next definition of this section, we need the following remark:

Proposition 2.6 Let us assume that g_{1} and g_{2} are usc. Let (u^{∗}, v^{∗}) be a Nash equilibrium feedback on the time interval [0, T]. Then, for any T_{0} ∈ (0, T), the restriction of (u^{∗}, v^{∗}) to [0, T_{0}] × IR is a Nash equilibrium feedback for the horizon T_{0} and the terminal payoff J^{♯}(T_{0}, ·, u^{∗}, v^{∗}).

Proof : We only prove that, for any t_{0} ∈ [0, T_{0}), there is some set S_{t_{0}} of zero measure such that, for any u ∈ U, for any x_{0} ∈ IR\S_{t_{0}}, we have

inf_{x^{∗}(·)∈X (t_{0},x_{0},u^{∗},v^{∗})}(J_{1}^{♯})_{∗}(T_{0}, x^{∗}(T_{0}), u^{∗}, v^{∗}) ≥ sup_{x(·)∈X (t_{0},x_{0},u,v^{∗})}(J_{1}^{♯})^{∗}(T_{0}, x(T_{0}), u^{∗}, v^{∗}) . (3)

The inequality for J_{2} can be obtained in a symmetric way.

Let S_{t_{0}} be the set of zero measure associated with (u^{∗}, v^{∗}). Then, for any x_{0} ∈ IR\S_{t_{0}}, for any u ∈ U, let us fix x^{∗}(·) ∈ X (t_{0}, x_{0}, u^{∗}, v^{∗}) and x(·) ∈ X (t_{0}, x_{0}, u, v^{∗}). Let us recall that J_{1}^{♯} is usc. Hence (J_{1}^{♯})^{∗} = J_{1}^{♯}. Moreover, for any t ∈ [t_{0}, T], one has x(t) ∉ S_{t}. In particular x(T_{0}) ∉ S_{T_{0}} and therefore J_{1}^{♯}(T_{0}, ·, u^{∗}, v^{∗}) is continuous at x^{∗}(T_{0}). Thus

(J_{1}^{♯})_{∗}(T_{0}, x^{∗}(T_{0}), u^{∗}, v^{∗}) = J_{1}(T_{0}, x^{∗}(T_{0}), u^{∗}, v^{∗}) = g_{1}(x^{∗}(T)) .

Since g_{1} is usc, there is some solution x_{1}(·) ∈ X (T_{0}, x(T_{0}), u^{∗}, v^{∗}) such that

g_{1}(x_{1}(T)) = J_{1}^{♯}(T_{0}, x(T_{0}), u^{∗}, v^{∗}) .

Let us now define

ũ(t, x) = \{ \begin{matrix}u(t, x) & if t ∈ [0, T_{0}) \\ u^{∗}(t, x) & otherwise \end{matrix} and x̃(t) = \{ \begin{matrix}x(t) & if t ∈ [0, T_{0}) \\ x_{1}(t) & otherwise \end{matrix}

Then we have clearly x̃(·) ∈ X (t_{0}, x_{0}, ũ, v^{∗}). Therefore, since (u^{∗}, v^{∗}) is a Nash equilibrium, and since x_{0} ∈ IR\S_{t_{0}}, we have

g_{1}(x^{∗}(T)) ≥ g_{1}(x̃(T)) .

From the construction of x̃(·), we have on the one hand

g_{1}(x̃(T)) = g_{1}(x_{1}(T)) = J_{1}^{♯}(T_{0}, x(T_{0}), u^{∗}, v^{∗})

and, on the other hand,

g_{1}(x^{∗}(T)) ≤ J_{1}^{♯}(T_{0}, x^{∗}(T_{0}), u^{∗}, v^{∗}) ,

since x^{∗}(·) ∈ X (T_{0}, x^{∗}(T_{0}), u^{∗}, v^{∗}). Therefore (3) is proved.

QED

Definition 2.7 Let (u^{∗}, v^{∗}) ∈ U × V be a Nash equilibrium feedback. We say that (u^{∗}, v^{∗}) is completely maximal on the time interval [T_{0}, T] if, for any T_{1} ∈ (T_{0}, T], (u^{∗}, v^{∗}) is a maximal Nash equilibrium feedback for the game with the horizon T_{1} and terminal payoff J^{♯}(T_{1}, ·, u^{∗}, v^{∗}).

2.3 The main Theorem

In order to state the main results of this paper, we have to define the class G of terminal payoffs for which we can solve the problem.

Definition 2.8 The class of admissible payoffs G is the set of maps g = (g_{1}, g_{2}) : IR → IR^{2} for which there is a partition σ_{0} = −∞ < σ_{1} < … < σ_{k} < σ_{k+1} = +∞ such that

1. for any i = 0, …, k,

- either g_{1} (resp. g_{2}) is continuous and (strictly) increasing or decreasing on (σ_{i}, σ_{i+1}),

- or g_{1} and g_{2} are simultaneously constant on the interval (σ_{i}, σ_{i+1}),

2. g_{1} and g_{2} are usc on IR, and, for i = 1, …, k and j = 1, 2, g_{j} is right or left continuous at σ_{i}.

3. if g_{1} (resp. g_{2}) has a strict local maximum at the point σ_{i}, then g_{1} and g_{2} are continuous at σ_{i}.

Terminology : We say below that Σ = {σ_{0} < σ_{1} < … < σ_{k} < σ_{k+1}} is the partition associated with the map g.

Remark : In particular, if g = (g_{1}, g_{2}) is continuous and g_{1} and g_{2} have a finite number of local extrema, then g belongs to G.

For later use, let us introduce another class of terminal payoffs:

Definition 2.9 The subclass of admissible payoffs G̃ is the set of maps g = (g_{1}, g_{2}) ∈ G which satisfy the additionnal requirement: If g_{1} (resp. g_{2}) has a strict local maximum at some point σ belonging to the partition associated with g, then g_{2} (resp. g_{1}) has a local minimum at σ.

We are now ready to give the main result of this paper:

Theorem 2.10 Assume that the terminal payoff g belongs to the class G. Then the game with terminal time T and terminal payoff g has a completely maximal Nash equilibrium feedback on the time interval [0, T].

Moreover, the payoff of any two completely maximal Nash equilibrium feedbacks coincide almost everywhere.

Remark :

1. The uniqueness part of the result means the following: Let (u^{∗}, v^{∗}) be some completely maximal Nash equilibrium feedback on the time interval [0, T], and let (u_{1}^{∗}, v_{1}^{∗}) be another completely maximal Nash equilibrium feedback on some time interval [T_{1}, T] with T_{1} ∈ [0, T). Then

∀t ∈ [T_{1}, T], J(t, ·, u^{∗}, v^{∗}) = J(t, ·, u_{1}^{∗}, v_{1}^{∗}) a.e. .

2. We prove below that, for any t ∈ [0, T), the function J(t, ·, u^{∗}, v^{∗}) belongs to G̃ in some sense.

We devote the last two sections of this paper to the proof of Theorem 2.10.

2.4 Link with Nash equilibrium payoffs in the sense of Kleimenov

We now make the link between the Nash equilibrium feedbacks defined above and the equilibrium payoffs as defined by Kleimenov in [6].

Let us first recall the definition of Nash equilibrium payoff in the sense of Kleimenov (in fact, we only give here the characterization. For the original definition and the interpretation, see [6]).

Definition 2.11 Let g : IR → IR^{2} be continuous and (t_{0}, x_{0}) ∈ [0, T)×IR be an initial position. We say that (e_{1}, e_{2}) ∈ IR^{2} is a Nash equilibrium payoff in the sense of Kleimenov (in short Kleimenov equilibrium payoff) at (t_{0}, x_{0}), if there is some solution x(·) ∈ X (t_{0}, x_{0}) such that

∀t ∈ [t_{0}, T], for j = 1, 2, g_{j}(x(t)) ≤ e_{j} = g_{j}(x(T)) .

We call the solution x(·) a Kleimenov trajectory.

We say moreover that (e_{1}, e_{2}) is Pareto at the point (t_{0}, x_{0}), if, for any other Kleimenov equilibrium payoff (e′_{1}, e′_{2}) ∈ IR^{2} at (t_{0}, x_{0}), we have:

either e_{1} ≥ e′_{1} or e_{2} ≥ e′_{2} .

In the rest of the section, we assume that g is continuous.

Proposition 2.12 Let (u^{∗}, v^{∗}) be some Nash equilibrium feedback on the time interval [0, T] and (S_{t})_{t∈[0,T]} be its associated set of zero measure. Then, for any t ∈ [0, T) and x ∉ S_{t}, the payoff (e_{1}, e_{2}) = J(t, x, u^{∗}, v^{∗}) is a Kleimenov equilibrium payoff and any solution x^{∗}(·) ∈ X (t, x, u^{∗}, v^{∗}) is a Kleimenov trajectory.

Moreover the payoff of q completely maximal Nash equilibrium feedback is Pareto:

Theorem 2.13 Let (u^{∗}, v^{∗}) be a completely maximal Nash equilibrium feedback and (S_{t}) be its associated set of zero measure. Then, for any t ∈ [0, T) and for any x ∉ S_{t}, the payoff J(t, x, u^{∗}, v^{∗}) is Pareto.

Proposition 2.12 and Theorem 2.13 are proved at the end of the paper.

2.5 Examples of Nash equilibrium feedbacks

We now give several examples of Nash equilibrium feedbacks for various terminal payoffs.

Example 2.14 Let us assume that g_{1} and g_{2} are both strictly increasing and usc. Then the feedback given by

∀(t, x) ∈ [0, T] × IR, u^{∗}(t, x) = v^{∗}(t, x) = 1

is a completely maximal equilibrium feedback. Moreover, the payoff of any other maximal Nash equilibrium feedback coincides almost everywhere with the payoff of (u^{∗}, v^{∗}).

However, there exists infinitely many Nash equilibrium feedbacks for which the associated payoff differs from the the payoff for (u^{∗}, v^{∗}). For instance, for any fixed a ∈ IR at which g is continuous, the feedback (u_{a}, v_{a}) defined by

∀(t, x) ∈ [0, T] × IR, u_{a}(t, x) = v_{a}(t, x) = \{ \begin{matrix}1 & if x > a or x + 2(T − t) < a \\ −1 & otherwise \end{matrix}

is a Nash equilibrium feedback.

Proof : Let S_{T} be the set of points at which g is discontinuous and, for any t ∈ (0, T], S_{t} = S_{T} − 2(T − t). Then, for any t, the set S_{t} has a zero measure since it is enumerable.

Let us now notice that for any (t, x), the payoff g(x + 2(T − t)) is maximal at (t, x) if and only if g is right continuous at this point. Since, for any (t, x), the unique solution x(·) ∈ X (t, x, u^{∗}, v^{∗}) is x(·) = x + 2(· − t), it is clear that the payoff of the strategy (u^{∗}, v^{∗}) equals the maximal payoff as soon as there is one.

This implies that (u^{∗}, v^{∗}) is a Nash equilibrium feedback because, for any t ∈ [0, T), for any x ∉ S_{t}, the payoff J(t, x, u^{∗}, v^{∗}) is maximal.

Let us now prove that (u^{∗}, v^{∗}) is completely maximal. Indeed, for any T_{0} ∈ (0, T], the function J^{♯}(T_{0}, ·, u^{∗}, v^{∗}) = g(· + 2(T − T_{0})) is strictly increasing. Therefore, for any (t, x), with t < T_{0}, there is some maximal payoff (e_{1}, e_{2}) at (t, x) for the game with terminal time T_{0} and terminal payoff J^{♯}(T_{0}, ·, u^{∗}, v^{∗}) if and only if the function J^{♯}(T_{0}, ·, u^{∗}, v^{∗}) is left or right continuous at x + 2(T_{0} − t). Then this payoff is given by: (e_{1}, e_{2}) = J^{♯}(T_{0}, x + 2(T_{0} − t), u^{∗}, v^{∗}). But this is precisely the payoff of the strategy (u^{∗}, v^{∗}) for this game. Thus (u^{∗}, v^{∗}) is completely maximal.

Let us now prove that the payoff of any other maximal Nash equilibrium feedback (ū, v̄) equals the payoff of (u^{∗}, v^{∗}). Indeed, since for any (t, x) such that g(·) is right or left continuous at x + 2(T − t), g(x + 2(T − t)) is a maximal payoff at (t, x). Hence J^{♯}(t, x, ū, v̄) = g(x + 2(T − t)) as such points. Thus J^{♯}(t, x, ū, v̄) = J^{♯}(t, x, u^{∗}, v^{∗}) almost everywhere because, for any t, such points are enumerable.

Let us finally prove that the feedback (u_{a}, v_{a}) is a Nash equilibrium feedback with a payoff different from the payoff of (u^{∗}, v^{∗}). A straightforward computation shows that ∀(t, x), ∀x(·) ∈ X (t, x, u_{a}, v_{a}),

x(T) = \{ \begin{matrix}x + 2(T − t) & if x > a or x ≤ a − 2(T − t) \\ a & if a − 2(T − t) ≤ x < a \end{matrix}

(for x = a, then x(T) ∈ [a, a + 2(T − t)]). Thus the payoff for (u_{a}, v_{a}) differs from the payoff for (u^{∗}, v^{∗}) on the set E = {a − 2(T − t) < x < a}. Let us also point out that, on this set, the payoff of (u_{a}, v_{a}) is well defined (i.e., J^{♯} = J^{♭}) because g is continuous at a. Let us set Σ_{t} = {a} ∪ [S_{t}\(a − 2(T − t), a)].

For proving that (u_{a}, v_{a}) is a Nash equilibrium feedback with associated set of zero measure (Σ_{t})_{t∈[0,T]}, let (u, v) ∈ U × V. We have to prove that, for any t ∈ [0, T) and for any x ∈ IR\Σ_{t}, we have

J_{1}(t, x, u_{a}, v_{a}) ≥ J_{1}^{♯}(t, x, u, v_{a}) and J_{2}(t, x, u_{a}, v_{a}) ≥ J_{2}^{♯}(t, x, u_{a}, v) .

We only do the proof for the first inequality, the proof of the second one being symmetric.

If x ∉ (a − 2(T − t), a) the result is clear, because J(t, x, u_{a}, v_{a}) is the maximal payoff at (t, x). Let us assume that x ∈ (a − 2(T − t), a). Then J_{1}(t, x, u_{a}, v_{a}) = g(a). Let now x(·) ∈ X (t, x, u, v_{a}). We are going to check that x(T) ≤ a. Indeed, let us denote by θ the first time the trajectory x(·) leaves the set E. As long as x(s) ∈ E, v_{a} = −1 in a neighbourhood of (s, x(s)). Hence f̃(s, x(s), u, v_{a}) ≤ 0. Therefore the solution x(·) is non increasing on this interval. Thus x(θ) ≤ x and x(θ) = a − 2(T − θ) because (θ, x(θ)) ∈ ∂E. Therefore x(T) ≤ x(θ) + 2(T − θ) = a. Since g_{1} is strictly increasing, we have

J_{1}^{♯}(t, x, u, v_{a}) ≤ g_{1}(a) = J_{1}(t, x, u_{a}, v_{a}) .

Therefore the desired result is proved for any t and any x ∉ Σ_{t}. Thus (u_{a}, v_{a}) is a Nash equilibium feedback.

QED

Next we give an example where there is no maximal payoff. In this case, there is only one Nash equilibrium payoff.

Example 2.15 Let us assume that g_{1} is strictly increasing while g_{2} is strictly decreasing in IR, and that both functions are usc. Then there is one and only one equilibrium payoff: For any Nash equilibrium feedback (ū, v̄), we have

∀t ∈ [0, T] , J(t, x, ū, v̄) = g(x) a.e. x ∈ IR .

Moreover, there is a least one Nash equilibrium feedback, given by

∀(t, x) ∈ [0, T] × IR, u^{∗}(t, x) = 1 = −v^{∗}(t, x) .

Proof : Let us first notice that there is no maximal payoff at any point, since g_{1} is strictly increasing while g_{2} is strictly decreasing in IR.

We now prove that (u^{∗}, v^{∗}) is a Nash equilibrium feedback. Let S_{T} be the set of points x at which the function g is continuous and S_{t} = S_{T} for any t ∈ [0, T]. Again, for any t, S_{t} is enumerable. For any (t, x) ∈ [0, T] × IR, and for any x(·) ∈ X (t, x, u^{∗}, v^{∗}), we have x(T) = x. Thus

∀t ∈ [0, T] , ∀x ∉ S_{t}, J(t, x, u^{∗}, v^{∗}) = g(x) .

Let u ∈ U and let us prove that

∀t ∈ [0, T], ∀x ∈ IR, J_{1}^{♯}(t, x, u, v^{∗}) ≤ g_{1}(x) .

Indeed, for any x(·) ∈ X (t, x, u, v^{∗}), we have that x(·) is non increasing since v^{∗} = −1. Thus x(T) ≤ x, which proves the desired inequality because g_{1} is increasing. This equality shows that, for any u ∈ U,

∀t ∈ [0, T], ∀x ∉ S_{t}, J_{1}^{♯}(t, x, u, v^{∗}) ≤ J_{1}(t, x, u^{∗}, v^{∗}) .

The symmetric inequality for J_{2} can be proved in the same way. Thus (u^{∗}, v^{∗}) is a Nash equilibrium feedback.

Let us finaly prove that any other Nash equilibrium feedback has the same payoff as (u^{∗}, v^{∗}). Let (ū, v̄) be a Nash equilibrium feedback and let (Σ_{t})_{t∈[0,T]} be its associated set of zero measure. Then, for any (t, x) ∈ [0, T] × IR with x ∉ S_{t}, and any x(·) ∈ X (t, x, ū, v̄) we have, on the one hand:

g_{1}(x(T)) = J_{1}(t, x, ū, v̄) ≥ J^{♯}(t, x, 1, v̄) ≥ g_{1}(x)

because any solution of X (t, x, 1, v̄) is non decreasing and g_{1} is increasing. Inequality g_{1}(x(T)) ≥ g_{1}(x) implies that x(T) ≥ x. On the other hand,

g_{2}(x(T)) = J_{2}(t, x, ū, v̄) ≥ J_{2}^{♯}(t, x, ū, −1) ≥ g_{2}(x)

because any solution of X (t, x, ū, −1) is non increasing and g_{2} is decreasing. Then the inequality g_{2}(x(T)) ≥ g_{2}(x) implies that x(T) ≤ x. Accordingly, x(T) = x, and the desired result is proved: J(t, x, ū, v̄) = g(x).

QED

Next we give an example in which the payoff associated with a completely maximal Nash equilibrium feedback is discontinuous, and which shows that we have to take some care in the definition of the maximal payoffs.

Example 2.16 We assume that

∀x ∈ IR, g_{2}(x) = g_{1}(−x) and g_{1}(x) = \{ \begin{matrix}1 & if x ≤ −2 \\ −x − 1 & if − 2 ≤ x ≤ 1 \\ x − 3 & if 1 ≤ x ≤ 2 \\ −1 & if x ≥ 2 \end{matrix}

Then for any Nash and completely maximal equilibrium feedback (u^{∗}, v^{∗}), we have

• If t ∈ [T − 1, T], then, for almost all x ∈ IR, J_{2}(t, x, u^{∗}, v^{∗}) = J_{1}(t, −x, u^{∗}, v^{∗}) and

J_{1}(t, x, u^{∗}, v^{∗}) = \{ \begin{matrix}1 & if x < −2 + 2(T − t) \\ −x − 1 + 2(T − t) & if − 2 + 2(T − t) ≤ x < −1 + (T − t) \\ −x − 1 & if − 1 + (T − t) ≤ x < 1 − (T − t) \\ x − 3 + 2(T − t) & if 1 − (T − t) ≤ x < 2 − 2(T − t) \\ −1 & if x ≥ 2 − 2(T − t) \end{matrix}

• If t ≤ T − 1, then for almost all x ∈ IR, J_{2}(t, x, u^{∗}, v^{∗}) = J_{1}(t, −x, u^{∗}, v^{∗}) and

J_{1}(t, x, u^{∗}, v^{∗}) = \{ \begin{matrix}1 & if x < 0 \\ −1 & if x ≥ 0 \end{matrix}

We shall not do the proof since existence and uniqueness for such a game are constructed in the next sections. We mainly want to point out that, at time t = T − 1, we have

∀x ∈ IR, J_{1}^{♯}(T − 1, x, u^{∗}, v^{∗}) = \{ \begin{matrix}1 & if x ≤ 0 \\ −1 & if x > 0 \end{matrix}

while

∀x ∈ IR, J_{2}^{♯}(T − 1, x, u^{∗}, v^{∗}) = \{ \begin{matrix}−1 & if x < 0 \\ 1 & if x ≥ 0 \end{matrix}

In particular, J^{♯}(T − 1, 0, u^{∗}, v^{∗}) = (1, 1), and thus the point x = 0 is a point of maximum for both J_{1}^{♯}(T − 1, ·, u^{∗}, v^{∗}) and J_{2}^{♯}(T − 1, ·, u^{∗}, v^{∗}). However, it is not a Nash equilibrium payoff in the sense of Kleimenov, since there no point x ∈ IR such that g(x) = (1, 1). Therefore the payoff (1, 1) cannot be considered as a maximal payoff at (T − 1, 0) without causing trouble. This has lead us to the above Definition of maximal payoffs.

3 Local existence and uniqueness

In this section, we compute explicitely, for any terminal payoff g ∈ G , a completely maximal Nash equilibrium feedback on the time interval [T − τ, T] for some τ > 0. Moreover we prove that the associated payoff belongs to G̃ and that the associated payoff of any completely maximal feedback strategy coincides with it.

3.1 The local existence and uniqueness result

Proposition 3.1 Let us assume that the terminal payoff g belongs to G and that Σ is an associated partition.

A) Then there is some feedback (u^{∗}, v^{∗}) ∈ U × V and some τ > 0 such that

i) (u^{∗}, v^{∗}) is a Nash, completely maximal equilibrium feedback on the time interval [T − τ, T], ii) for any t ∈ [T − τ, T), J^{♯}(t, ·, u^{∗}, v^{∗}) belongs to G̃, iii) for any t ∈ [T − τ, T], there is some partition S_{t} associated with J^{♯}(t, ·, u^{∗}, v^{∗}) such that

∀x ∉ S_{t}, J^{♯}(t, x, u^{∗}, v^{∗}) = J^{♭}(t, x, u^{∗}, v^{∗}) ,

iv) the set-valued map t → S_{t} is 2-Lipschitz continuous on [T − τ, T], with S_{T} = Σ, and there is some fixed M such that the cardinal of S_{t} is not larger than M,

v) (u^{∗}, v^{∗}) is proper with respect to (S_{t})_{t∈[T −τ,T]}, i.e., for any t_{0} ∈ [T − τ, T), for any x_{0} ∉ S_{t_{0}}, for any x^{∗}(·) ∈ X (t_{0}, x_{0}, u^{∗}, v^{∗}), we have

∀t ∈ [t_{0}, T), x^{∗}(t) ∉ S_{t} .

vi) if moreover g belongs to G̃, then one can also choose (u^{∗}, v^{∗}) in such a way that it is proper with respect to the set Σ. Namely, for any t_{0} ∈ [T − τ, T), for any x_{0} ∉ S_{t_{0}}, for any x^{∗}(·) ∈ X (t_{0}, x_{0}, u^{∗}, v^{∗}), we have x^{∗}(T) ∉ Σ.

B) Moreover, the completely maximal Nash equilibrium payoff is unique on [T −τ, T] in the sense that, any other completely maximal Nash equilibrium feedback (u_{1}^{∗}, v_{1}^{∗}) defined on some interval [T − τ_{1}, T] satisfies: for t ∈ [T − min{τ, τ_{1}}, T], for j = 1, 2,

J_{j}(t, x, u^{∗}, v^{∗}) = J_{j}(t, x, u_{1}^{∗}, v_{1}^{∗}) for almost all x ∈ IR .

The proof of Proposition 3.1 is splitted in several steps. We first define explicitely the feedback (u^{∗}, v^{∗}). Then we prove that it is indeed Nash and maximal, and that the payoff of any completely maximal Nash feedback strategy coincides with the payoff of (u^{∗}, v^{∗}). Next we show that condition (ii) is satisfied. We finaly establish that (u^{∗}, v^{∗}) is completely maximal.

3.2 Construction of the maximal Nash equilibrium feedback

Let g ∈ G and Σ = {σ_{0} = −∞, σ_{1}, …, σ_{k}, σ_{k+1} = +∞} be an associated partition. In this section we explain the construction of the strategy (u^{∗}, v^{∗}). For doing so, we first construct some τ > 0 and maps t → σ_{i}^{+}(t) and t → σ_{i}^{−}(t) in such a way that, for any t ∈ [T − τ, T],

σ_{0}(t) = −∞ < σ_{1}^{−}(t) ≤ σ_{1}^{−}(t) < … < σ_{k}^{−}(t) ≤ σ_{k}^{+}(t) < σ_{k+1} = +∞ .

Then, for any t ∈ [T − τ, T], we define (u^{∗}(t, x), v^{∗}(t, x)) in a suitable way according to the interval (σ_{i}^{+}(t), σ_{i+1}^{−}(t)) or (σ_{i}^{−}(t), σ_{i}^{+}(t)) the point x belongs to. The set S_{t} shall be defined below as

S_{t} = {σ_{i}^{−}(t), i = 1, …, k} ∪ {σ_{i}^{+}(t), i = 1, …, k} .

Let us first divide the singularities σ_{i} in 4 classes. For j = 1, 2, and i = 1, …, k, the point σ_{i} is of the type I (resp. II, III or IV) for the map g_{j} if:

• Type I : g_{j} is monotonous on (σ_{i−1}, σ_{i+1}).

• Type II : g_{j} has a strict local maximum at the point σ_{i} and is continuous at σ_{i}.

• Type III : g_{j} has a strict local minimum at the point σ_{i} and is continuous at σ_{i}.

• Type IV : g_{j} is discontinuous at σ_{i} and its lsc enveloppe has a strict local minimum at σ_{i}.

Remark : Let us underline that a point σ_{i} is of one and only one type for g_{j} (j = 1 or j = 2) because g belongs to G.

Notation : If ϕ : IR → IR and x ∈ IR, we denote by ϕ(x^{−}) (resp. ϕ(x^{+})) the limit, if it exists, of ϕ(s) when s → x^{−} (resp. s → x^{+})).

Definition of σ_{i}^{+}(·) and σ_{i}^{−}(·) : For any i = 1, …, k, the singularity σ_{i} propagates via two arcs, denoted σ_{i}^{+}(·) and σ_{i}^{−}(·).

For defining these arcs, we have to consider all the possible cases: For A ∈ {I, II, III, IV } and B ∈ {I, II, III, IV }, the case A-B is the case where σ_{i} is of type A for g_{1} and of type B for g_{2}.

In order to restrict the (lengthy) cases discussion, we use the symmetry of the problem, which implies that the propagation of the singularity in the case (A-B) is the same as in the case (B-A).

• Case I-I :

- if g_{1} and g_{2} are simultaneously non decreasing (resp. non increasing) on the interval (σ_{i−1}, σ_{i+1}) and if either g_{1} or g_{2} is non constant on this interval, then we set σ_{i}^{+}(t) = σ_{i}^{−}(t) = σ_{i} − 2(T − t) (resp. σ_{i}^{+}(t) = σ_{i}^{−}(t) = σ_{i} + 2(T − t)),

- otherwise, we set σ_{i}^{+}(t) = σ_{i}^{−}(t) = σ_{i}.

• Case I-II :

- if g_{1} is strictly increasing, then σ_{i}^{−}(t) = σ_{i} − 2(T − t) and σ_{i}^{+}(t) = σ_{i},

- if g_{1} is strictly decreasing, then σ_{i}^{−}(t) = σ_{i} and σ_{i}^{+}(t) = σ_{i} + 2(T − t),

• Case I-III :

- if g_{1} is strictly increasing, then σ_{i}^{+}(t) = σ_{i}^{−}(t) = σ_{i}(t) where σ_{i}(t) is the unique solution x ∈ (σ_{i} − 2(T − t), σ_{i}) of the equation g_{2}(x) = g_{2}(x + 2(T − t))

- if g_{1} is strictly decreasing, then σ_{i}^{+}(t) = σ_{i}^{−}(t) = σ_{i}(t) where σ_{i}(t) is the unique solution x ∈ (σ_{i} − 2(T − t), σ_{i}) of the equation g_{2}(x) = g_{2}(x + 2(T − t))

• Case I-IV :

- if g_{1} is non decreasing and g_{2}(σ_{i}^{−}) < g_{2}(σ_{i}^{+}), then σ_{i}^{+}(t) = σ_{i}^{−}(t) = σ_{i} − 2(T − t)

- if g_{1} is non decreasing and g_{2}(σ_{i}^{−}) > g_{2}(σ_{i}^{+}), then σ_{i}^{+}(t) = σ_{i}^{−}(t) = σ_{i}

- if g_{1} is non increasing, it is the converse

• Case II-II : σ_{i}^{−}(t) = σ_{i} − 2(T − t) and σ_{i}^{+}(t) = σ_{i} + 2(T − t).

• Case II-III : σ_{i}^{−}(t) = σ_{i}^{+}(t) = σ_{i}.

• Case III-III : For j = 1, 2, let us denote by y_{j}(t) the unique point x ∈ (σ_{i} − 2(T − t), σ_{i} + 2(T − t)) such that g_{j}(x − 2(T − t)) = g_{j}(x + 2(T − t)).

Then σ_{i}^{+}(t) = σ_{i}^{−}(t) = σ_{i}(t) where σ_{i}(·) is the unique absolutely continuous map such that:

i) min{y_{1}(t), y_{2}(t)} ≤ σ_{i}(t) ≤ max{y_{1}(t), y_{2}(t)},

ii) if min{y_{1}(t), y_{2}(t)} < σ_{i}(t) < max{y_{1}(t), y_{2}(t)}, then σ′_{i}(t) = 0

iii) if min{y_{1}(t), y_{2}(t)} = σ_{i}(t) < max{y_{1}(t), y_{2}(t)}, then

σ′_{i}(t) = min\{0, \frac{d}{dt} min\{y_{1}(t), y_{2}(t)\} \}

iv) if min{y_{1}(t), y_{2}(t)} < σ_{i}(t) = max{y_{1}(t), y_{2}(t)}, then

σ′_{i}(t) = max\{0, \frac{d}{dt} max\{y_{1}(t), y_{2}(t)\} \}

• Case III-IV : If g_{2}(σ_{i}^{−}) < g_{2}(σ_{i}^{+}), then σ_{i}^{+}(t) = σ_{i}^{−}(t) = σ_{i}(t) where σ_{i}(t) is the unique solution in (σ_{i} − 2(T − t), σ_{i}) of the equation g_{2}(x) = g_{2}(x + 2(T − t))

- if g_{2}(σ_{i}^{−}) > g_{2}(σ_{i}^{+}), then σ_{i}^{+}(t) = σ_{i}^{−}(t) = σ_{i}(t) where σ_{i}(t) is the unique solution in (σ_{i} − 2(T − t), σ_{i}) of the equation g_{2}(x) = g_{2}(x + 2(T − t))

• Case IV-IV : If g_{1}(σ_{i}^{−}) < g_{1}(σ_{i}^{+}) and g_{2}(σ_{i}^{−}) < g_{2}(σ_{i}^{+}), then σ_{i}^{−}(t) = σ_{i}^{+}(t) = σ_{i} − 2(T − t).

If g_{1}(σ_{i}^{−}) > g_{1}(σ_{i}^{+}) and g_{2}(σ_{i}^{−}) > g_{2}(σ_{i}^{+}), then σ_{i}^{−}(t) = σ_{i}^{+}(t) = σ_{i} + 2(T − t).

Otherwise, σ_{i}^{−}(t) = σ_{i}^{+}(t) = σ_{i}.

Remarks :

1. Let us first point out that the above cases distinction covers all the possibilities because g belongs to G.

2. For the cases (I-III) and (III-III), we prove below that the map σ_{i}(·) is indeed well defined.

3. Let us notice that σ_{i}^{−}(t) = σ_{i}^{+}(t) unless we are in case (I-II) or (II,II). In particular, if g ∈ G̃, then we always have σ_{i}^{−}(t) = σ_{i}^{+}(t).

4. In order to simplify the notations, we set

σ_{0}^{+}(t) = σ_{0}^{−}(t) = −∞ and σ_{k+1}^{+}(t) = σ_{k+1}^{−}(t) = +∞ .

Definition of τ : We choose τ > 0 such that

∀i = 1, …, k, τ ≤ (σ_{i+1} − σ_{i})/24 (4)

and such that, if g_{j} is discontinuous at σ_{i} (for j ∈ {1, 2}, i ∈ {1, …, k}) with, for instance, g_{j}(σ_{i}^{−}) < g_{j}(σ_{i}^{+}), then

sup_{x∈[σ_{i}−2τ,σ_{i})} g_{j}(x) < inf_{x∈(σ_{i},σ_{i}+2τ]} g_{j}(x) . (5)

Proposition 3.2 The maps σ_{i}^{+}(·) and σ_{i}^{−}(·) defined above are well defined and Lispchitz continuous with a Lipschitz constant not larger than 2 on [T − τ, T].

Proof : We do the proof for the cases (I-III) and (III-III), since the result is obvious otherwise. In case (I-III), if, for instance, g_{1} is increasing, then σ_{i}^{+}(t) = σ_{i}^{−}(t) = σ_{i}(t) where σ_{i}(t) is the

unique solution x ∈ (σ_{i} − 2(T − t), σ_{i}) of the equation g_{2}(x) = g_{2}(x + 2(T − t)). Let us first point out that there is indeed a unique solution to this equation in the interval (σ_{i} − 2(T − t), σ_{i}) because the map ϕ_{t} : x → g_{2}(x + 2(T − t)) − g_{2}(x) is increasing on this interval, and satisfies

ϕ_{t}(σ_{i} − 2(T − t)) = g_{2}(σ_{i}) − g_{2}(σ_{i} − 2(T − t)) < 0

and

ϕ_{t}(σ_{i}) = g_{2}(σ_{i} + 2(T − t)) − g_{2}(σ_{i}) > 0

since g_{2} has a minimum on the interval (σ_{i−1}, σ_{i+1}) at σ_{i}.

For proving that the map σ_{i}(·) is 2-Lipschitz continuous, let us show that σ_{i}(·) is increasing and satisfies σ_{i}(s) ≥ σ_{i}(t) − 2(t − s) if t > s. Indeed, the map ϕ_{s}(·) defined above is increasing on (σ_{i} − 2(T − s), σ_{i}) and satisfies

\begin{matrix}ϕ_{s}(σ_{i}(t)) = g_{2}(σ_{i}(t) + 2(T − s)) − g_{2}(σ_{i}(t)) = \\ g_{2}(σ_{i}(t) + 2(T − s)) − g_{2}(σ_{i}(t) + 2(T − t)) > 0 \end{matrix}

because g_{2} is increasing on (σ_{i}, σ_{i+1}), and

\begin{matrix}ϕ_{s}(σ_{i}(t) − 2(t − s)) = g_{2}(σ_{i}(t) + 2(T − t)) − g_{2}(σ_{i}(t) − 2(t − s)) = \\ g_{2}(σ_{i}(t)) − g_{2}(σ_{i}(t) − 2(t − s)) < 0 \end{matrix}

because g_{2} is decreasing on (σ_{i−1}, σ_{i}). Therefore σ_{i}(s) belongs to the interval (σ_{i}(t)−2(t−s), σ_{i}(t)), which is the desired result.

In case (III-III), let us first prove that σ_{i}(·) satisfying the conditions exists. Following the proof above, it is not difficult to check that the maps y_{j}(·) (j = 1, 2) are 2-Lipschitz continuous.

Let us now prove the existence of σ_{i}(·) on intervals of the form (θ_{1}, θ_{2}) on which min{y_{1}(t), y_{2}(t)} < max{y_{1}(t), y_{2}(t)}. In order to fix the ideas, we assume that on this interval, y_{1}(t) < y_{2}(t) and that y_{1}(θ_{2}) = y_{2}(θ_{2}).

For proving the existence, we use a viability Theorem (see for instance [1]). Let F be the set-valued map defined on the locally compact set K, where

K = {(t, x) ∈ (θ_{1}, θ_{2}] × IR | y_{1}(t) ≤ x ≤ y_{2}(t)}

F (t, x) = \{ \begin{matrix}\{−1\} × \{0\} & if y_{1}(t) < x < y_{2}(t) \\ \{−1\} × [0, 2] & if y_{1}(t) = x < y_{2}(t) \\ \{−1\} × [−2, 0] & if y_{1}(t) < x = y_{2}(t) \\ \{−1\} × [−2, 2] & if y_{1}(t) = x = y_{2}(t) \end{matrix}

It is easy to check that F is usc with convex compact values and that K is a locally compact viability domain for F. Accordingly for the initial condition (θ_{2}, y_{1}(θ_{2})) there exists (locally) a solution (t(·), x(·)) to the differential inclusion

\{ \begin{matrix}(t′(s), x′(s)) ∈ F (t(s), x(s)) \\ (t(0), x(0)) = (θ_{2}, y_{1}(θ_{2})) \end{matrix}

which remains in K. Obviously t(s) = θ_{2} − s and it is easily seen that the solution can be extended on the interval [0, θ_{2} − θ_{1}]. Then the map σ_{i}(s) = x(θ_{2} − s) satisfies the required conditions on the interval (θ_{1}, θ_{2}).

Let us now prove that σ_{i}(·) is uniquely defined. Let z_{1}(·) and z_{2}(·) satisfy the required conditions. Since the maps max{z_{1}(·), z_{2}(·)} and min{z_{1}(·), z_{2}(·)} also satisfy these conditions, we can assume without loss of generality that z_{1}(·) ≤ z_{2}(·). Using the definition, one can prove without difficulty that

\frac{d}{ds}(z_{2}(T − s) − z_{1}(T − s))

is always non positive. Since moreover z_{2}(T) = z_{1}(T) = σ_{i}, this proves that z_{1}(·) = z_{2}(·). Therefore σ_{i}(·) is uniquely defined.

QED

Definition of the completely maximal equilibrium feedback: Let (u^{∗}, v^{∗}) ∈ U × V be the pair of strategies defined by: for i = 0, …, k and t ∈ [T − τ, T],

∀x ∈ [σ_{i}^{+}(t), σ_{i+1}^{−}(t)], u^{∗}(t, x) = \{ \begin{matrix}1 & if g_{1} is increasing on (σ_{i}, σ_{i+1}) \\ −1 & if g_{1} is decreasing on (σ_{i}, σ_{i+1}) \end{matrix}

and

∀x ∈ [σ_{i}^{+}(t), σ_{i+1}^{−}(t)], v^{∗}(t, x) = \{ \begin{matrix}1 & if g_{2} is increasing on (σ_{i}, σ_{i+1}) \\ −1 & if g_{2} is decreasing on (σ_{i}, σ_{i+1}) \end{matrix}

and

∀x ∈ (σ_{i}^{−}(t), σ_{i}^{+}(t)), u^{∗}(t, x) = v^{∗}(t, x) = 0 .

When g_{1} and g_{2} are constant on the interval (σ_{i}, σ_{i+1}), we have to be more carefull. Let x ∈ [σ_{i}^{+}(t), σ_{i+1}^{−}(t)] and let us assume that g_{1} and g_{2} are constant on the interval (σ_{i}, σ_{i+1}). Then we define u^{∗}(t, x) and v^{∗}(t, x) in the following way:

Let us set a_{i}(t) = \frac{2}{3}σ_{i} + \frac{1}{3}σ_{i+1} − 2(T − t) and b_{i}(t) = \frac{1}{3}σ_{i} + \frac{2}{3}σ_{i+1} + 2(T − t). Then

u^{∗}(t, x) = v^{∗}(t, x) = \{ \begin{matrix}1 & if x ≤ a_{i}(t) \\ −1 & if x ≥ b_{i}(t) \\ 0 & otherwise \end{matrix}

Remark : From assumption (4) on τ and since σ_{i}^{−}(·) and σ_{i}^{+}(·) are 2-Lipschitz, we have

σ_{0}(t) = −∞ < σ_{1}^{−}(t) ≤ σ_{1}^{+}(t) < … < σ_{k}^{−}(t) ≤ σ_{k}^{+}(t) < σ_{k+1} = +∞

and a_{i}(t) < b_{i}(t) on [T − τ, T]. Hence (u^{∗}, v^{∗}) is well defined.

Lemma 3.3 Let us set

∀t ∈ [T − τ, T], S_{t} = {σ_{i}^{−}(t), i = 1, …, k} ∪ {σ_{i}^{+}(t), i = 1, …, k} .

Then for any t_{0} ∈ [T − τ, T), for any x_{0} ∉ S_{t_{0}}, there is a unique solution x^{∗}(·) ∈ X (t_{0}, x_{0}, u^{∗}, v^{∗}). This solution satisfies

∀t ∈ [t_{0}, T), x^{∗}(t) ∉ S_{t} .

If moreover, g ∈ G̃, then

x^{∗}(T) ∉ Σ .

Proof : The proof is a straightforward (but quite tiedous !) verification by listing the various cases.

QED

3.3 The pair of strategy (u^{∗}, v^{∗}) is a maximal Nash equilibrium feedback

We prove here the parts (i), (iii), (iv), (v) and (B) of Proposition 3.1. For that purpose, let us set

∀t ∈ [T − τ, T], S_{t} = {σ_{i}^{−}(t), i = 1, …, k} ∪ {σ_{i}^{+}(t), i = 1, …, k} .

For this choice of S_{t}, (iv) is clear with M = 2k and (v) is given by Lemma 3.3.

Let (t_{0}, x_{0}) belong to [T − τ, T] × IR. From the definition of τ (see (4)), the interval I = [x_{0} − 2τ, x_{0} + 2τ] contains at most one σ_{i}. Let us point out that, for any feedback (u, v) ∈ U × V, for any solution x(·) ∈ X (t_{0}, x_{0}, u, v), we have

∀t ∈ [t_{0}, T], x(t) ∈ [x_{0} − 2(t − t_{0}), x_{0} + 2(t − t_{0})] .

Hence only the restriction of the strategy to the set R(t_{0}, x_{0}) defined by

R(t_{0}, x_{0}) = ⋃_{t∈[t_{0},T]}{t} × [x_{0} − 2(t − t_{0}), x_{0} + 2(t − t_{0})]

is important for the point (t_{0}, x_{0}).

Let us first assume that the interval I does not contain any σ_{i}. Then the fact that the restriction of (u^{∗}, v^{∗}) is a maximal and Nash equilibrium to the set R(t_{0}, x_{0}) and that the payoff J^{♯}(t, x, u^{∗}, v^{∗}) is the unique Nash and maximal equilibrium payoff on this set can be proved as in Example 2.14 and 2.15.

Let us now assume that the interval I contains σ_{i} for some i = 1, …, k. For proving that the pair (u^{∗}, v^{∗}) a maximal Nash equilibrium, we discuss according to the type of σ_{i}. In order to avoid lengthy cases discussions, we only give a complete proof for the cases I-III and III-III which are more delicate and leave the other cases to the reader.

Case I-III : We assume for instance that g_{1} is increasing. From the definition, we have σ_{i}^{+}(t) = σ_{i}^{−}(t) = σ_{i}(t) where σ_{i}(t) is the unique solution in (σ_{i} − 2(T − t), σ_{i}) of the equation g_{2}(x) = g_{2}(x + 2(T − t)). The pair of feedback (u^{∗}, v^{∗}) is given by

u^{∗}(t, x) = 1 and v^{∗}(t, x) = \{ \begin{matrix}−1 & if x < σ_{i}(t) \\ +1 & otherwise \end{matrix}

Let us first prove that (u^{∗}, v^{∗}) is a Nash equilibrium feedback. To this end, let (u, v) ∈ U × V be an arbitrary pair of feedbacks. We only prove that

J_{2}(t_{0}, x_{0}, u^{∗}, v^{∗}) ≥ J_{2}^{♯}(t_{0}, x_{0}, u^{∗}, v) , (6)

since the symmetric inequality for J_{1} is clear.

Let us first assume that x_{0} > σ_{i}. Let x^{∗}(·) ∈ X (t_{0}, x_{0}, u^{∗}, v^{∗}) and x(·) ∈ X (t_{0}, x_{0}, u^{∗}, v). Then σ_{i} < x_{0} ≤ x(T) ≤ x_{0} + 2(T − t_{0}) = x^{∗}(T), because u^{∗} = 1. Since g_{2} is increasing on (σ_{i}, σ_{i+1}), this inequality proves (6) in the case x_{0} > σ_{i}.

We now assume that x_{0} ∈ (σ_{i} − 2(T − t_{0}), σ_{i}). Let us denote by x^{∗}(·) the unique solution in X (t_{0}, x_{0}, u^{∗}, v^{∗}). We also fix some x(·) ∈ X (t_{0}, x_{0}, u^{∗}, v). Let us notice that, since u^{∗} = 1, we always have x′(·) ≥ 0. Hence x(T) ≥ x_{0}. Moreover, we always have x(T) ≤ x_{0} + 2(T − t_{0}).

- Let us first assume that x_{0} ∈ (σ_{i}(t), σ_{i}). Then x^{∗}(·) = x_{0} + 2(· − t_{0}) on [t_{0}, T]. If, on the one hand, x(T) ≥ σ_{i}, then we have

g_{2}(x(T)) ≤ g_{2}(x_{0} + 2(T − t_{0})) = g_{2}(x^{∗}(T)) ,

because g_{2} is increasing on (σ_{i}, σ_{i+1}). If, on the other hand, we have x(T) < σ_{i}, then since g_{2} is decreasing on (σ_{i−1}, σ_{i}) and x(T) ≥ x_{0}, we have

g_{2}(x(T)) ≤ g_{2}(x_{0}) < g_{2}(σ_{i}(t_{0})) = g_{2}(σ_{i}(t_{0}) + 2(T − t_{0}))

and since σ_{i}(t_{0}) + 2(T − t_{0}) ≥ 0 and g_{2} is increasing on (σ_{i}, σ_{i+1}), we also have

g_{2}(σ_{i}(t_{0}) + 2(T − t_{0})) < g_{2}(x_{0} + 2(T − t_{0})) = g_{2}(x^{∗}(T)) .

Therefore we have proved that g_{2}(x(T)) ≤ g_{2}(x^{∗}(T)).

- Let us now assume that x_{0} < σ_{i}(t). Then x^{∗}(T) = x_{0}. If on the one hand x(T) ≤ σ_{i}, then, since g_{2} is decreasing on (σ_{i−1}, σ_{i}) and since x(T) ≥ x_{0}, we have

g_{2}(x(T)) ≤ g_{2}(x_{0}) = g_{2}(x^{∗}(T)) .

If on the other hand x(T) > σ_{i}, then, since g_{2} is increasing on (σ_{i}, σ_{i+1}) we have

g_{2}(x(T)) ≤ g_{2}(x_{0} + 2(T − t_{0})) < g_{2}(σ_{i}(t_{0}) + 2(T − t_{0})) .

From the definition of σ_{i}(·) and the fact that g_{2} is decreasing on (σ_{i−1}, σ_{i}), we have

g_{2}(σ_{i}(t_{0}) + 2(T − t_{0})) = g_{2}(σ_{i}(t_{0})) < g_{2}(x_{0}) = g_{2}(x^{∗}(T)) .

Therefore we have proved that g_{2}(x(T)) ≤ g_{2}(x^{∗}(T)).

In conclusion, inequality J_{2}(t_{0}, x_{0}, u^{∗}, v^{∗}) ≥ J_{2}^{♯}(t_{0}, x_{0}, u^{∗}, v) is proved in any case. This shows that (u^{∗}, v^{∗}) is a Nash equilibrium.

Let us now prove that (u^{∗}, v^{∗}) is maximal. For doing so, let (t_{0}, x_{0}) belong to [T −τ, T]×IR, with x_{0} ≠ σ_{i}(t_{0}), and let us assume that (e_{1}, e_{2}) is a maximal payoff at (t_{0}, x_{0}). Let also x(·) ∈ X (t_{0}, x_{0}) be such that

for j = 1, 2, g_{j}(x(T)) = e_{j} .

Since g_{1} is strictly increasing on (σ_{i−1}, σ_{i+1}), we necessarily have x(T) = x_{0}+2(T −t_{0}). In particular, we have: (e_{1}, e_{2}) = g(x_{0} + 2(T − t_{0})). Moreover, from the definition of a maximal payoff, g_{2} has a maximum at x_{0} + 2(T − t_{0}) on the interval [x_{0} − 2(T − t_{0}), x_{0} + 2(T − t_{0})]. Therefore, from the definition of σ_{i}(t_{0}), the following inequality holds: x_{0} > σ_{i}(t_{0}). Since

J(t_{0}, x_{0}, u^{∗}, v^{∗}) = g(x_{0} + 2(T − t_{0})) = (e_{1}, e_{2}) ,

the desired equality is proved.

We have just proved that (u^{∗}, v^{∗}) is maximal. We now prove that, for any completely maximal Nash equilibrium feedback (u_{1}^{∗}, v_{1}^{∗}) defined on [T_{1}, T], we have for any t_{0} ∈ [max{T − τ, T_{1}}, T),

J(t_{0}, x_{0}, u^{∗}, v^{∗}) = J(t_{0}, x_{0}, u_{1}^{∗}, v_{1}^{∗}) for almost every x_{0} ∈ IR .

In the sequel, we denote by (Σ_{t}) the set of zero measure associated with (u_{1}^{∗}, v_{1}^{∗}).

Let us first assume that x_{0} < σ_{i}(t_{0}). We claim that

(7) J(t_{0}, x_{0}, u^{∗}, v^{∗}) = J(t_{0}, x_{0}, u_{1}^{∗}, v_{1}^{∗}) = g(x_{0}) for almost every x_{0} < σ_{i}(t_{0}) .

The equality J(t_{0}, x_{0}, u^{∗}, v^{∗}) = g(x_{0}) being clear, let us prove the other one. Let x_{0} < σ_{i}(t_{0}) be such that J^{♭}(t_{0}, x_{0}, u_{1}^{∗}, v_{1}^{∗}) = J^{♯}(t_{0}, x_{0}, u_{1}^{∗}, v_{1}^{∗}), and let x(·) ∈ X (t_{0}, x_{0}, u_{1}^{∗}, v_{1}^{∗}). It is clearly enough to show that x(T) = x_{0}. We first claim that x(T) ≥ x_{0}. Indeed,

J_{1}(t_{0}, x_{0}, u_{1}^{∗}, v_{1}^{∗}) = g_{1}(x(T)) ≥ J_{1}^{♯}(t_{0}, x_{0}, −v_{1}^{∗}, v_{1}^{∗}) = g_{1}(x_{0}) .

Since g_{1} is strictly increasing, these inequalities imply that x(T) ≥ x_{0}. We finaly prove that x(T) = x_{0}. Indeed, since x_{0} < σ_{i}(t_{0}), we have, on the one hand

∀z ∈ (x_{0}, x_{0} + 2(T − t_{0})], g_{2}(z) < g_{2}(x_{0}) .

In particular, g_{2}(x(T)) < g_{2}(x_{0}) unless x(T) = x_{0}. On the other hand, since (u_{1}^{∗}, v_{1}^{∗}) is a Nash equilibrium feedback, we have

J_{2}(t_{0}, x_{0}, u_{1}^{∗}, v_{1}^{∗}) = g_{2}(x(T)) ≥ J_{2}^{♯}(t_{0}, x_{0}, u_{1}^{∗}, −u_{1}^{∗}) = g_{2}(x_{0}) .

Therefore x(T) = x_{0}, and (7) is established.

Let us now assume that x_{0} > σ_{i}(t_{0}). We are going to prove that ∀x(·) ∈ X (t_{0}, x_{0}, u_{1}^{∗}, v_{1}^{∗}), we have x(T) = x_{0} + 2(T − t_{0}). For this let us set

\begin{aligned}∀t ∈ [T_{1}, T), α(t) = & inf\{x ∈ (σ_{i} − 2(T − t), σ_{i} + 2(T − t)) | ∀y ∈ (x, σ_{i} + 2(T − t)), \\ & ∀y(·) ∈ X (t, y, u_{1}^{∗}, v_{1}^{∗}), y(T) = y + 2(T − t)\} . \end{aligned}

Since g_{1} and g_{2} are strictly increasing on (σ_{i}, σ_{i+1}), the payoff g(y + 2(T − t)) is maximal for the points (t, y) in a neighborhood of (t, σ_{i} + 2(T − t)). Since (u_{1}^{∗}, v_{1}^{∗}) is maximal, for such points y, we must have y(T) = y + 2(T − t) for any y(·) ∈ X (t, y, u_{1}^{∗}, v_{1}^{∗}). Therefore α(·) is well defined. Moreover, we have already proved that α(t) ≥ σ_{i}(t) because if y < σ_{i}(t) and y(·) ∈ X (t, y, u_{1}^{∗}, v_{1}^{∗}), then y(T) = y.

Our aim is to prove that α(t) = σ_{i}(t) on [T_{1}, T], which clearly entails that

J(t, x, u^{∗}, v^{∗}) = J(t, x, u_{1}^{∗}, v_{1}^{∗}) = g(x + 2(T − t))) for almost every x_{0} > σ_{i}(t_{0}) .

We divide the proof of the equality α(·) = σ_{i}(·) in several steps.

Claim 1 : We first claim that α(·) is 2-Lipschitz continuous on [T_{1}, T].

Proof of claim 1: Let us first point out that, for any x > α(t), and for any x(·) ∈ X (t, x, u_{1}^{∗}, v_{1}^{∗}), we have x(·) = x + 2(T − ·). Hence

∀T_{1} ≤ s < t < T, α(t) ≤ α(s) + 2(t − s) .

From the definition of α(·), we have that J(t, x, u_{1}^{∗}, v_{1}^{∗}) = g(x + 2(T − t)) on (α(t), σ_{i} + 2(T − t))\Σ_{t}, so that J_{1}^{♯}(t, x, u_{1}^{∗}, v_{1}^{∗}) and J_{2}^{♯}(t, x, u_{1}^{∗}, v_{1}^{∗}) are strictly increasing on the interval (α(t), σ_{i} + 2(T − t)). Therefore, for any s < t and for any x ∈ (α(t) + 2(t − s), σ_{i} + 2(T − s)), the payoff J(t, x + 2(t − s), u_{1}^{∗}, v_{1}^{∗}) is a maximal payoff at (s, x) for the game with terminal time t and terminal payoff J^{♯}(t, ·, u_{1}^{∗}, v_{1}^{∗}). Since (u_{1}^{∗}, v_{1}^{∗}) is completely maximal, we have, for any x ∈ (α(t)+2(t−s), σ_{i}+2(T −s)) and any x(·) ∈ X (s, x, u_{1}^{∗}, v_{1}^{∗}), the equality x(t) = x + 2(t − s), which entails, since x(t) > α(t), that x(T) = x + 2(T − s). Thus α(s) ≤ α(t) + 2(t − s), and claim 1 is proved.

Claim 2 : Let us now assume that α(t_{0}) > σ_{i}(t_{0}). Then we claim that there is some ϵ > 0 such that α(t) = α(t_{0}) − 2(t_{0} − t) for t ∈ (t_{0} − ϵ, t_{0}).

Proof of claim 2 : Since α(·) is 2-Lipschitz continuous, it is enough to prove that α(t) ≤ α(t_{0}) − 2(t_{0} − t). This is also equivalent with saying that, for any x > α(t_{0}) − 2(t_{0} − t) and any x(·) ∈ X (t, x, u_{1}^{∗}, v_{1}^{∗}), we have x(t_{0}) = x + 2(t_{0} − t) (since x + 2(t_{0} − t) > α(t_{0})).

For proving this, we first establish that

∀x ∈ (σ_{i}(t_{0}), α(t_{0})), for j = 1, 2, J_{j}^{♯}(t_{0}, x, u_{1}^{∗}, v_{1}^{∗}) < J_{j}^{♯}(t_{0}, α(t_{0}), u_{1}^{∗}, v_{1}^{∗}) . (8)

This inequality is obvious for j = 1, since g_{1} is strictly increasing and since J_{1}^{♯}(t_{0}, α(t_{0}), u_{1}^{∗}, v_{1}^{∗}) = g_{1}(α(t_{0}) + 2(T − t_{0})). For j = 2, let x(·) ∈ X (t_{0}, x, u_{1}^{∗}, v_{1}^{∗}). Then, since g_{1} is increasing, it is not difficult to check that x(T) ≥ x. Hence, x(T) > σ_{i}(t_{0}) and therefore,

g_{2}(x(T)) ≤ g_{2}(x + 2(T − t)) < g_{2}(α(t_{0}) + 2(T − t))

because

σ_{i} < x + 2(T − t) < α(t_{0}) + 2(T − t_{0})

and g_{2} is increasing on (σ_{i}, σ_{i+1}). Thus (8) is proved.

Let us now choose ϵ > 0 in such a way that α_{i}(t_{0}) − 4ϵ > σ_{i}(t_{0}). let x > α(t_{0}) − 2(t_{0} − t), with x ∉ Σ_{t}, and x(·) ∈ X (t, x, u_{1}^{∗}, v_{1}^{∗}). The inequalities (8) and the fact that the functions J_{j}^{♯}(t_{0}, ·, u_{1}^{∗}, v_{1}^{∗}) are strictly increasing on (α(t_{0}), σ_{i} + 2(T − t_{0})) prove that the payoff J(t_{0}, x + 2(t_{0} − t), u_{1}^{∗}, v_{1}^{∗}) is a maximal payoff at (t, x) for the game with terminal time t_{0} and terminal payoff J^{♯}(t_{0}, ·, u_{1}^{∗}, v_{1}^{∗}). Since (u_{1}^{∗}, v_{1}^{∗}) is maximal, we obtain that x(t_{0}) = x + 2(t_{0} − t). Thus x(T) = x + 2(T − t) because x(t_{0}) > α(t_{0}). Since this is true for any x > α(t_{0}) − 2(t_{0} − t), claim 2 is proved.

Claim 3 : We finally claim that α(t) = σ_{i}(t) for t ∈ [T_{1}, T].

Proof of claim 3 : We already know that α(·) ≥ σ_{i}(·). Let us assume that, contrary to our claim, there is some t for which α(t) > σ_{i}(t). Let (t_{0}, t_{1}) be the largest intervall containing t on which we have the inequality α(·) > σ_{i}(·). Then α(t_{1}) = σ_{i}(t_{1}), and, from the previous step, α(s) = α(t) − 2(t − s) for any t_{0} < s < t < t_{1}. Letting t → t_{1} implies that

∀t_{0} < s < t_{1}, σ_{i}(s) < α(s) = α(t_{1}) − 2(t_{1} − s) = σ_{i}(t_{1}) − 2(t_{1} − s) .

This later inequality is impossible since σ_{i}(·) is 2-Lipschitz continuous. This completes the proof of claim 3.

In conclusion, we have proved that, in any case, the payoff of the strategy (u_{1}^{∗}, v_{1}^{∗}) coincides almost everywhere with the payoff of the strategy (u^{∗}, v^{∗}).

Case III-III : We first prove that (u^{∗}, v^{∗}) is a Nash equilibrium feedback. We only do the proof for x_{0} ∈ (min{y_{1}(t), y_{2}(t)}, max{y_{1}(t), y_{2}(t)}) since the other cases can be treated as above. To fix the ideas, we consider the case y_{1}(t_{0}) < y_{2}(t_{0}) and x_{0} < σ_{i}(t_{0}). In particular, these assumptions imply that σ_{i}(t_{0}) > y_{1}(t_{0}).

Let us fix (u, v) ∈ U × V and let us start by proving that

J_{2}(t_{0}, x_{0}, u^{∗}, v^{∗}) ≥ J_{2}(t_{0}, x_{0}, u^{∗}, v) . (9)

Let us denote by x^{∗}(·) the unique solution in X (t_{0}, x_{0}, u^{∗}, v^{∗}) and let us fix some x(·) ∈ X (t_{0}, x_{0}, u^{∗}, v). Then x^{∗}(T) = x_{0} − 2(T − t_{0}). If on the one hand, x(T) ≥ σ_{i}, then

g_{2}(x(T)) ≤ g_{2}(x_{0} + 2(T − t_{0})) < g_{2}(x_{0} − 2(T − t_{0})) = g_{2}(x^{∗}(T))

because g_{2} is increasing on (σ_{i}, σ_{i+1}) and x_{0} < y_{2}(t_{0}). If, on another hand, x(T) < σ_{i}, then

g_{2}(x(T)) ≤ g_{2}(x_{0} − 2(T − t_{0})) = g_{2}(x^{∗}(T))

because g_{2} is decreasing on (σ_{i−1}, σ_{i}). Therefore we have proved that g_{2}(x(T)) ≤ g_{2}(x^{∗}(T)), which entails (9).

Let us now prove that J_{1}(t_{0}, x_{0}, u^{∗}, v^{∗}) ≥ J_{1}(t_{0}, x_{0}, u, v^{∗}) . (10)

Let x(·) belong to X (t_{0}, x_{0}, u, v^{∗}). Since g_{1} is decreasing on (σ_{i−1}, σ_{i}) and since x(T) ≥ x_{0} − 2(T − t_{0}) = x^{∗}(T), if x(T) ≤ σ_{i} then we have g_{1}(x(T)) ≤ g_{1}(x^{∗}(T)). We assume from now on that x(T) > σ_{i}. In particular, there is some t ∈ (t_{0}, T) such that x(t) = σ_{i}(t). Let us set

θ = min{t ≥ t_{0} | x(t) ≥ σ_{i}(t)} .

Since v^{∗} = −1 in a neighbourhood of x(t) for t ∈ [t_{0}, θ), x(·) is non increasing on this interval.

We claim that there is some θ_{0} ∈ [t_{0}, θ] such that x(θ_{0}) = y_{1}(θ_{0}). Indeed, let us assume on the contrary that x(t) > y_{1}(t) on [t_{0}, θ]. This implies that σ_{i}(t) > y_{1}(t) on [t_{0}, θ]. Therefore, the map s → σ_{i}(θ − s) is non increasing on the interval [0, θ − t_{0}] while s → x(θ − s) is non decreasing. Therefore

σ_{i}(θ) − x(θ) ≥ σ_{i}(t_{0}) − x(t_{0}) > 0

which is in contradiction with the definition of θ. Therefore we have proved the existence of some θ_{0} ∈ [t_{0}, θ] such that x(θ_{0}) = y_{1}(θ_{0}).

Since g_{1} is increasing on (σ_{i}, σ_{i+1}) and since x(T) ≤ x(θ_{0}) + 2(T − θ_{0}) = y_{1}(θ_{0}) + 2(T − θ_{0}), we have that

g_{1}(x(T)) ≤ g_{1}(y_{1}(θ_{0}) + 2(T − θ_{0})) = g_{1}(y_{1}(θ_{0}) − 2(T − θ_{0}))

from the very definition of y_{1}(·). Moreover, we have

y_{1}(θ_{0}) = x(θ_{0}) ≥ x_{0} − 2(θ_{0} − t_{0}) ,

which entails that

g_{1}(y_{1}(θ_{0}) − 2(T − θ_{0})) ≤ g_{1}(x_{0} − 2(T − t_{0})) = g_{1}(x^{∗}(T))

since g_{1} is decreasing on (σ_{i−1}, σ_{i}). Thus we have proved that g_{1}(x(T)) ≤ g_{1}(x^{∗}(T)), which entails (10).

Therefore (u^{∗}, v^{∗}) is a Nash equilibrium feedback. We now prove that (u^{∗}, v^{∗}) is maximal. Let us fix t_{0} ∈ [T − τ, T) be such that there is some maximal payoff (e_{1}, e_{2}) at (t_{0}, x_{0}). We have to prove that J^{♯}(t_{0}, x_{0}, u^{∗}, v^{∗}) = (e_{1}, e_{2}) . (11)

For this, let us consider some x(·) ∈ X (t_{0}, x_{0}) such that (e_{1}, e_{2}) = g(x(T)). Let us first assume that x_{0} > y_{1}(t_{0}). Then the map g_{1} has a strict maximum at x_{0} + 2(T − t_{0}) on the interval [x_{0} − 2(T − t_{0}), x_{0} + 2(T − t_{0})]. Therefore we have x(T) = x_{0} + 2(T − t_{0}). Since (e_{1}, e_{2}) is a maximal payoff at the point (t_{0}, x_{0}), g_{2} has also a maximum at x_{0} + 2(T − t_{0}) on the interval [x_{0} − 2(T − t_{0}), x_{0} + 2(T − t_{0})]. Hence x_{0} ≥ y_{2}(t_{0}). So we have proved that, if x_{0} > y_{1}(t_{0}), then x_{0} ≥ max{y_{1}(t_{0}), y_{2}(t_{0})}. For such a point, we have already computed J^{♯}(t_{0}, x_{0}, u^{∗}, v^{∗}) = g(x_{0} + 2(T − t_{0})), and equality (11) is proved in this case.

The cases x_{0} < y_{1}(t_{0}), x_{0} > y_{2}(t_{0}) and x_{0} < y_{2}(t_{0}) can be treated in the same way. If x_{0} = y_{1}(t_{0}) = y_{2}(t_{0}), then (e_{1}, e_{2}) = g(x_{0} − 2(T − t_{0})) = g(x_{0} + 2(T − t_{0})) and the solution x(·) = x_{0}−2(T −·) belongs to X (t_{0}, x_{0}, u^{∗}, v^{∗}), so that J^{♯}(t_{0}, x_{0}, u^{∗}, v^{∗}) = g(x_{0}−2(T −t_{0})) = (e_{1}, e_{2}). So we have proved that (u^{∗}, v^{∗}) is maximal.

It remains to prove that any completely maximal equilibrium feedback has the same payoff as (u^{∗}, v^{∗}). For this, let (u_{1}^{∗}, v_{1}^{∗}) be some completely maximal Nash equilibrium feedback and (Σ_{t})_{t∈[0,T]} its associated set of zero measure.

Let us first underline that, if x_{0} < min{y_{1}(t_{0}), y_{2}(t_{0})}, then any x(·) ∈ X (t_{0}, x_{0}, u_{1}^{∗}, v_{1}^{∗}) satisfies x(T) = x_{0} − 2(T − t_{0}). Indeed the payoff g(x_{0} − 2(T − t_{0})) is the maximal payoff at such a point (t_{0}, x_{0}), and it is attained only at x_{0} − 2(T − t_{0}). The conclusion follows from the maximality of (u_{1}^{∗}, v_{1}^{∗}). In the same way, if x_{0} > max{y_{1}(t_{0}), y_{2}(t_{0})}, then any x(·) ∈ X (t_{0}, x_{0}, u_{1}^{∗}, v_{1}^{∗}) satisfies x(T) = x_{0} + 2(T − t_{0}).

Let us now define, for t ∈ [T − τ, T],

\begin{matrix}α(t) = inf\{x ∈ [σ_{i} − 2(T − t), σ_{i} + 2(T − t)] | \\ ∀x(·) ∈ X (t, x, u_{1}^{∗}, v_{1}^{∗}), x(T) > x_{0} − 2(T − t_{0})\} . \end{matrix}

and

\begin{matrix}β(t) = sup\{x ∈ [σ_{i} − 2(T − t), σ_{i} + 2(T − t)] | \\ ∀x(·) ∈ X (t, x, u_{1}^{∗}, v_{1}^{∗}), x(T) < x_{0} + 2(T − t_{0})\} . \end{matrix}

The largest part of the proof amounts to establish that α(t) = β(t) = σ_{i}(t) for t ∈ [T − τ, T]. We split the proof into several claims.

Claim 1 We first prove that

∀t ∈ [T − τ, T], min{y_{1}(t), y_{2}(t)} ≤ α(t) ≤ β(t) ≤ max{y_{1}(t), y_{2}(t)} .

Proof of claim 1 : We have already proved that

α(t) ≥ min{y_{1}(t), y_{2}(t)}  and  β(t) ≤ max{y_{1}(t), y_{2}(t)} .

Let us now show that α(t) ≤ β(t) holds for all t. Indeed, if, on the contrary, we had β(t) < α(t) for some t, then, for any x ∈ (β(t), α(t)), there would exist x_{1}(·) ∈ X (t, x, u_{1}^{∗}, v_{1}^{∗}) with x_{1}(t) = x + 2(T − t) and x_{2}(·) ∈ X (t, x, u_{1}^{∗}, v_{1}^{∗}) with x_{2}(t) = x − 2(T − t). Since g(x_{1}(T)) ≠ g(x_{2}(T)) (unless x = y_{1}(t) = y_{2}(t))), there is a contradiction with the fact that

J^{♯}(t, x, u_{1}^{∗}, v_{1}^{∗}) = J^{♭}(t, x, u_{1}^{∗}, v_{1}^{∗}) for almost all x ∈ (β(t), α(t)) .

So we have proved the inequality α(t) ≤ β(t) for all t and the proof of claim 1 is complete.

Claim 2 : Let us now prove that α(·) is usc while β(·) is lsc, and that

∀s < t, α(s) ≥ α(t) − 2(t − s) , while β(s) ≤ β(t) + 2(t − s) . (12)

In particular, α(·) and β(·) are left continuous.

Proof of claim 2 : We only do the proof for α(·). Let (t_{n}) be a sequence converging to some t. For proving that lim sup_{n} α(t_{n}) ≤ α(t), let us consider x ≤ lim sup_{n} α(t_{n}). Then we can find some sequence (x_{n}) such that x_{n} → x (up to some subsequence) and x_{n} < α(t_{n}). From the definition of α(t_{n}), the solution x_{n}(·) = x_{n} − 2(· − t_{n}) belongs to X (t_{n}, x_{n}, u_{1}^{∗}, v_{1}^{∗}). Therefore, the solution x(·) = x − 2(· − t) belongs to X (t, x, u_{1}^{∗}, v_{1}^{∗}). Since this holds true for any x ≤ lim sup_{n} α(t_{n}), we have proved that lim sup_{n} α(t_{n}) ≤ α(t). Thus α(·) is ucs.

Let us now prove (12). For this, we consider the game with terminal time t and terminal payoff J^{♯}(t, ·, u_{1}^{∗}, v_{1}^{∗}). Since (u_{1}^{∗}, v_{1}^{∗}) is completely maximal, (u_{1}^{∗}, v_{1}^{∗}) is a maximal Nash equilibrium strategy for this game. Since both g_{1} and g_{2} are decreasing on (σ_{i−1}, σ_{i}), the functions J_{j}(t, ·, u_{1}^{∗}, v_{1}^{∗}) are decreasing on (σ_{i} − 2(T − t), α(t)] because they coincide with g_{j}(· − 2(T − t)) on this interval. Therefore, for any s < t and any x < α(t) − 2(t − s) the payoff J_{j}(t, x − 2(t − s), u_{1}^{∗}, v_{1}^{∗}) is maximal for the game with terminal time t and terminal payoff J^{♯}(t, ·, u_{1}^{∗}, v_{1}^{∗}) and is only attained at the point x − 2(t − s). This implies that any solution x(·) ∈ X (s, x, u_{1}^{∗}, v_{1}^{∗}) satisfies x(t) = x − 2(t − s). From this one derives easily the desired inequality (12). Hence the proof of claim 2 is complete.

Let us keep in mind that the previous proof shows that the maps J_{j}^{♯}(t, ·, u_{1}^{∗}, v_{1}^{∗}) are decreasing on (σ_{i} − 2(T − t), α(t)) for any t.

Claim 3 : Let us now assume that, for some t_{0}, we have α(t_{0}) < max{y_{1}(t_{0}), y_{2}(t_{0})}. We claim that there is some positive ϵ such that α(t) ≥ α(t_{0}) for t ∈ [t_{0} − ϵ, t_{0}].

In particular this implies that α(·) is non increasing on the intervalls where α(·) < max{y_{1}(·), y_{2}(·)}. Proof of claim 3 : To fix the ideas, we assume that α(t_{0}) < y_{2}(t_{0}). Since α(t_{0}) − 2(T − t_{0}) is

the unique point of maximum of g_{2} on the interval [α(t_{0}) − 2(T − t_{0}), α(t_{0}) + 2(T − t_{0})], and since

J_{2}^{♯}(t_{0}, α(t_{0}), u_{1}^{∗}, v_{1}^{∗}) = g_{2}(α(t_{0}) − 2(T − t_{0})) ,

and since one can easily check that there is some γ > 0 such that

∀x ∈ (α(t_{0}), α(t_{0}) + γ), J_{2}^{♯}(t_{0}, x, u_{1}^{∗}, v_{1}^{∗}) < J_{2}^{♯}(t_{0}, α(t_{0}), u_{1}^{∗}, v_{1}^{∗}) .

We set ϵ = γ/2. Let us now prove the intermediate result: ∀t ∈ [t_{0} − ϵ, t_{0}],

∀x < α(t_{0}) with x ∉ Σ_{t}, ∀x(·) ∈ X (t, x, u_{1}^{∗}, v_{1}^{∗}), x(t_{0}) ≤ α(t_{0}) . (13)

Indeed, if on the contrary we had x(t_{0}) > α(t_{0}), then

J_{2}(t, x, u_{1}^{∗}, v_{1}^{∗}) = J_{2}(t_{0}, x(t_{0}), u_{1}^{∗}, v_{1}^{∗}) < J_{2}^{♯}(t_{0}, α(t_{0}), u_{1}^{∗}, v_{1}^{∗}) < J_{2}^{♯}(t_{0}, x, u_{1}^{∗}, v_{1}^{∗}) ,

because x(t_{0}) < α(t_{0}) + γ. This leads to a contradiction with the fact that (u_{1}^{∗}, v_{1}^{∗}) is Nash since the strategy

v(s, y) = \{ \begin{matrix}−u_{1}^{∗}(s, y) & if s < t_{0} \\ v_{1}^{∗}(s, y) & otherwise \end{matrix}

gives

J_{2}^{♯}(t, x, u_{1}^{∗}, v) = J_{2}^{♯}(t_{0}, x, u_{1}^{∗}, v_{1}^{∗}) > J_{2}^{♯}(t, x, u_{1}^{∗}, v_{1}^{∗}) .

Thus we have proved (13).

Since the minimum of two solutions is still a solution and since the functions J_{j}^{♯}(t_{0}, ·, u_{1}^{∗}, v_{1}^{∗}) are decreasing on (σ_{i} − 2(T − t_{0}), α(t_{0})), one easily derive from (13) that, for any t ∈ [t_{0} − ϵ, t_{0}], the functions J_{j}^{♯}(t, ·, u_{1}^{∗}, v_{1}^{∗}) are non increasing on the interval [σ_{i} − 2(T − t), α(t_{0})].

For completing the proof of the claim, we argue by contradiction by assuming that there is some t_{1} ∈ (t_{0} − ϵ, t_{0}) such that α(t_{1}) < α(t_{0}). Then for any t < t_{1} such that α(t_{1}) + 2(t_{1} − t) ≤ α(t_{0}), and for any x < α(t_{1}), the payoff J^{♯}(t_{1}, x − 2(t_{1} − t), u_{1}^{∗}, v_{1}^{∗}) is the unique maximal payoff at the point (t, x) for the game with terminal time t_{1} and terminal payoff J^{♯}(t_{1}, ·, u_{1}^{∗}, v_{1}^{∗}), because the function J_{j}^{♯}(t_{1}, ·, u_{1}^{∗}, v_{1}^{∗}) are non increasing on the interval (σ_{i} − 2(T − t_{1}), α(t_{0})) and strictly decreasing on (σ_{i} − 2(T − t_{1}), α(t_{1})). Therefore, for any x(·) ∈ X (t, x, u_{1}^{∗}, v_{1}^{∗}), we have x(t_{1}) = x − 2(t_{1} − t). Hence, since x(t_{1}) < α(t_{1}), we have x(T) = x − 2(T − t). Thus α(t) ≥ α(t_{1}) for any t such that α(t_{1}) + 2(t_{1} − t) ≤ α(t_{0}). Since α(·) is usc and left continuous, one derives easily a contradiction. So claim 3 is proved.

Claim 4 : The next step consists in proving that α(t) ≥ σ_{i}(t) and that β(t) ≤ σ_{i}(t) for any t ∈ [T − τ, T].

Proof of claim 4 : We only do the proof for α(·). Let us assume on the contrary that there is some maximal interval (t_{0}, t_{1}) on which inequality α(t) < σ_{i}(t) hold. Since α(·) is left continuous, we have α(t_{1}) = σ_{i}(t_{1}). Since, on the one hand, α(t) < σ_{i}(t) ≤ max{y_{1}(t), y_{2}(t)}, we deduce from the previous step that α(·) is non increasing on (t_{0}, t_{1}). On the other hand, since min{y_{1}(t), y_{2}(t)} ≤ α(t) < σ_{i}(t) on (t_{0}, t_{1}), the map σ_{i}(·) is non decreasing on (t_{0}, t_{1}) from its construction. Therefore the map σ_{i}(·) − α(·) is non decreasing on (t_{0}, t_{1}). Since it is positive, vanishes at t_{1} and is left-continuous at this point, we have found a contradiction. So claim 4 is proved.

Using claim 4 and claim 1, we get the desired result: α(·) = β(·) = σ_{i}(·) on [T −τ, T]. From this one can prove without difficulty that the payoff for (u^{∗}, v^{∗}) is the same as for (u_{1}^{∗}, v_{1}^{∗}). Therefore the proof of the uniqueness is complete.

3.4 The payoff belongs to the class G̃ on [T − τ, T)

The aim of this subsection is to prove part (ii) of Proposition 3.1. Namely, we want to show that, for any t ∈ [T − τ, T), the map J^{♯}(t, ·, u^{∗}, v^{∗}) belongs to G̃.

A straightforward computation shows that, for any t ∈ [T − τ, T] and any x ∉ S_{t},

for j = 1, 2, J_{j}(t, x, u^{∗}, v^{∗}) = g_{j}(y(x))

where, if x ∈ (σ_{i}^{+}(t), σ_{i+1}^{−}(t)), for some i = 0, …, k,

y(x) = \{ \begin{matrix}x + 2(T − t) & if g_{1} and g_{2} are increasing on (σ_{i}, σ_{i+1}) \\ x − 2(T − t) & if g_{1} and g_{2} are decreasing on (σ_{i}, σ_{i+1}) \\ min\{max\{x, σ_{i}\}, σ_{i+1}\} & otherwise \end{matrix}

and, if x ∈ (σ_{i}^{−}(t), σ_{i}^{+}(t)) for some i = 1, …, k,

y(x) = σ_{i} .

Let us point out that y(x) ≠ x^{∗}(T) (where x^{∗}(·) ∈ X (t, x, u^{∗}, v^{∗})) only in the case g_{1} and g_{2} constant on the interval (σ_{i}, σ_{i+1}).

Therefore, for any i = 0, …, k, for t ∈ [T − τ, T],

- either J_{1}^{♯}(t, ·, u^{∗}, v^{∗}) (resp. J_{2}^{♯}(t, ·, u^{∗}, v^{∗})) is continuous and strictly increasing or decreasing on (σ_{i}(t)^{+}, σ_{i+1}^{−}(t)),

- or J_{1}^{♯}(t, ·, u^{∗}, v^{∗}) and J_{2}^{♯}(t, ·, u^{∗}, v^{∗}) are simultaneously constant on the interval (σ_{i}(t)^{+}, σ_{i+1}^{−}(t)) or on the interval (σ_{i}(t)^{−}, σ_{i}^{+}(t)).

Moreover J_{1}^{♯}(t, ·, u^{∗}, v^{∗}) and J_{2}^{♯}(t, ·, u^{∗}, v^{∗}) are usc on IR and left- or right-continuous at each point. Finally, if for instance J_{1}^{♯}(t, ·, u^{∗}, v^{∗}) has a strict local maximum at the point σ_{i}^{−}(t), then we are necessarily in case (II-III), which implies that J_{1}^{♯}(t, ·, u^{∗}, v^{∗}) and J_{2}^{♯}(t, ·, u^{∗}, v^{∗}) are continuous at σ_{i}^{−}(t) and that J_{2}^{♯}(t, ·, u^{∗}, v^{∗}) has a strict local minimum. Therefore J^{♯}(t, ·, u^{∗}, v^{∗}) belongs to G̃.

QED

3.5 The feedback (u^{∗}, v^{∗}) is completely maximal

It only remains to prove that the feedback (u^{∗}, v^{∗}) is also completely maximal. To this end, let us denote by ϕ(·, ·, g) the map

∀t ∈ [T − τ, T], ∀x ∈ IR, ϕ(t, T, x, g) = J^{♯}(t, x, u^{∗}, v^{∗}) ,

where (u^{∗}, v^{∗}) is the feedback strategy defined above for the game for terminal time T and terminal payoff g. We have already proved that ϕ(t, ·, g) belongs to G̃. A simple (but again tiedous) verification shows that the map ϕ(t, ·, g) satisfies the semi-group property:

If T − τ ≤ s ≤ t ≤ T, then ϕ(s, T, x, g) = ϕ(s, t, x, ϕ(t, T, ·, g)) .

In particular, this implies that, for any t ∈ (T − τ, T), the payoff ϕ(s, x, g) is a Nash and maximal equilibrium payoff for the terminal time t and the terminal payoff ϕ(t, ·, g) on the time interval [T − τ, t].

This proves that the feedback (u^{∗}, v^{∗}) is completely maximal, and completes the proof of Proposition 3.1.

4 Global existence and uniqueness

In the previous section, we have obtained short time existence and uniqueness. Our aim is on the one hand, to prove that it is possible to concatenate such solutions (Proposition 4.6), and, on the other hand, to show that a solution defined on some interval of the form (T_{0}, T] can naturally be extended to the interval [T_{0}, T] (Proposition 4.7).

Once these results obtained, it is not difficult to prove Theorem 2.10.

4.1 Some preliminary results

We give in this part a list of results which are needed later on.

Lemma 4.1 Let (u, v) ∈ U × V, (t_{0}, x_{0}) ∈ [0, T) × IR. For any x(·) ∈ X (t_{0}, x_{0}, u, v), h > 0, there exists some feedback u_{h} ∈ U, some trajectories x_{h}^{+}(·) ∈ X (t_{0}, x_{0} + h, u_{h}, v) and x_{h}^{−}(·) ∈ X (t_{0}, x_{0} − h, u_{h}, v) such that

• either x_{h}^{+}(T) = x(T) or x_{h}^{−}(T) = x(T)

• or |x_{h}^{+}(T) − x(T)| + |x_{h}^{−}(T) − x(T)| ≤ 2h.

Proof of Lemma 4.1 : Let u_{h}^{1} ∈ U be the strategy defined by

u_{h}^{1}(t, x) = \{ \begin{matrix}−1 & if x > x(t) \\ 1 & otherwise \end{matrix}

Let us fix y_{h}^{+}(·) ∈ X (t_{0}, x_{0} + h, u_{h}^{1}, v) and y_{h}^{−}(·) ∈ X (t_{0}, x_{0} − h, u_{h}^{1}, v). We set

τ_{h} = inf{t > t_{0} | y_{h}^{+}(t) = x(t) or y_{h}^{−}(t) = x(t)} ,

with the convention τ_{h} = T if the set in the right-hand side is empty.

Let us notice that, from the definition of u_{h}^{1}, y_{h}^{+}(·) is non increasing on [t_{0}, τ_{h}] while y_{h}^{−}(·) is non decreasing on this interval.

We consider now two cases:

First case : τ_{h} = T. Then

∀t ∈ [t_{0}, T], x_{0} − h ≤ y_{h}^{−}(t) ≤ x(t) ≤ y_{h}^{+}(t) ≤ x_{0} + h .

Hence, for t = T, the second inequality of the Lemma holds true if we set u_{h} = u_{h}^{1}, x_{h}^{+}(·) = y_{h}^{+}(·) and x_{h}^{−}(·) = y_{h}^{−}(·).

Second case : If, on the contrary, τ_{h} < T, let us assume for instance that x_{h}^{+}(τ_{h}) = x(τ_{h}). Then we define a new strategy u_{h} ∈ U and a new solution x_{h}^{+}(·) by setting

u_{h}(t, x) = \{ \begin{matrix}u_{h}^{1}(t, x) & if t ≤ τ_{h} \\ u^{∗}(t, x) & otherwise \end{matrix}

and

x_{h}^{+}(t) = \{ \begin{matrix}y_{h}^{+}(t) & if t ≤ τ_{h} \\ x(t) & otherwise \end{matrix}

It is clear that x_{h}^{+}(·) belongs to X (t_{0}, x_{0}, u_{h}, v) and that x_{h}^{+}(T) = x(T). Therefore the proof is complete.

QED

Corollary 4.2 Let g = (g_{1}, g_{2}) : IR → IR^{2} be such that g_{1} (resp. g_{2}) is, at each point, either rightor left-continuous. Let (ū, v̄) ∈ U × V some Nash equilibrium feedback on the interval [T − τ, T] for the terminal payoff g and let (Σ_{t}) be its associated set of zero measure. Then for any (u, v) ∈ U ×V, for any t_{0} ∈ [T − τ, T] and any x_{0} ∈ IR, we have

J_{1}^{∗}(t_{0}, x_{0}, ū, v̄) ≥ J_{1}^{♯}(t_{0}, x_{0}, u, v̄)

and

J_{2}^{∗}(t_{0}, x_{0}, ū, v̄) ≥ J_{2}^{♯}(t_{0}, x_{0}, ū, v) .

where J_{1}^{∗}(t, ·, ū, v̄) and J_{2}^{∗}(t, ·, ū, v̄) are the upper semi-continuous envelopes of the functions J_{1}(t, ·, ū, v̄) and J_{2}(t, ·, ū, v̄) (defined as functions on IR\Σ_{t}).

Remark : If g ∈ G, then g satisfies the regularity conditions of the Corollary.

Proof of Corollary 4.2 : We only prove the first inequality, the proof of the other being similar. The inequalitiy we have to prove is obvious for x_{0} ∉ S_{t_{0}}, from the definition of a Nash equilibrium since at such a point, J^{♯} = J is continuous. Let now x_{0} ∈ S_{t_{0}} and x(·) ∈ X (t_{0}, x_{0}, u, v̄) be such that

g_{1}(x(T)) = J_{1}^{♯}(t_{0}, x_{0}, u, v̄) .

Since S_{t_{0}} is of zero measure, we can find h_{n} → 0^{+} such that x_{0} + h_{n} ∉ S_{t_{0}} and x_{0} − h_{n} ∉ S_{t_{0}}. We apply Lemma 4.1 to h_{n}, to the set of strategies (u, v̄) and to the trajectory x(·). Let u_{h_{n}} ∈ U, x_{h_{n}}^{+}(·) and x_{h_{n}}^{−}(·) satisfying the conclusions of Lemma 4.1. Since g_{1} is either left continuous or right continuous at the point x(T), we have clearly, up to a subsequence,

either lim_{n} g_{1}(x_{h_{n}}^{−}(T)) = g_{1}(x(T)) or lim_{n} g_{1}(x_{h_{n}}^{+}(T)) = g_{1}(x(T)) .

Let us assume for instance that lim_{n} g_{1}(x_{h_{n}}^{−}(T)) = g_{1}(x(T)). Then, since x_{0} − h_{n} ∉ S_{t_{0}}, we have

J_{1}(t_{0}, x_{0} − h_{n}, ū, v̄) ≥ J_{1}^{♯}(t_{0}, x_{0} − h_{n}, u_{h_{n}}, v̄) ≥ g_{1}(x_{h_{n}}^{−}(T)) .

Taking the lim sup in this inequalities gives the desired result:

J_{1}^{∗}(t_{0}, x_{0}, ū, v̄) ≥ g_{1}(x(T)) = J_{1}^{♯}(t_{0}, x_{0}, u, v̄) .

QED

Corollary 4.3 Let g and (ū, v̄) be as in Corollary 4.2, Then we have, for any t_{0} ∈ [T − τ, T] and any x_{0} ∈ IR,

for j = 1, 2, J_{j}^{∗}(t_{0}, x_{0}, ū, v̄) = J_{j}^{♯}(t_{0}, x_{0}, ū, v̄) .

Proof of Corollary 4.3 : Inequality J_{j}^{∗} ≤ J_{j}^{♯} is clear since the map x_{0} → J_{j}^{♯}(t_{0}, x_{0}, ū, v̄) is usc and coincides with J_{j}^{∗} ouside of the set S_{t_{0}} which has a zero measure. The converse inequality is given by Corollary 4.2 applied to (u, v) = (ū, v̄).

QED

Lemma 4.4 Let g = (g_{1}, g_{2}) : IR → IR^{2} be such that g_{1} (resp. g_{2}) is usc and is, at each point, either right- or left-continuous. If (ū, v̄) is a prefect Nash equilibrium feedback on the time interval [T − τ, T] for some τ > 0, then, for j = 1, 2, for any x ∈ IR, the map t → J_{j}^{♯}(t, x, ū, v̄) is non increasing.

Proof of the Lemma 4.4 : We only do the proof for j = 1. Let us fix t and t_{0} such that T − τ ≤ t < t_{0} ≤ T. From Proposition 2.6, (ū, v̄) is a Nash equilibrium feedback for the time horizon t_{0} and the terminal payoff J^{♯}(t_{0}, ·, ū, v̄). Hence for any x ∈ IR, Corollary 4.2 states that the upper payoff for the feedback (ū, v̄) is not smaller than the upper payoff for the feedback (−v̄, v̄): Namely, there is some x^{∗}(·) ∈ X (t, x, ū, v̄) such that, for any x(·) ∈ X (t, x, −v̄, v̄), we have:

J_{1}^{♯}(t_{0}, x^{∗}(t_{0}), ū, v̄) ≥ J_{1}^{♯}(t_{0}, x(t_{0}), −v̄, v̄) .

Since the set X (t, x, −v̄, v̄) only contains the constant solution x(·) = x, the previous inequality becomes:

J_{1}^{♯}(t_{0}, x^{∗}(t_{0}), ū, v̄) ≥ J_{1}^{♯}(t_{0}, x, ū, v̄) .

Since

J_{1}^{♯}(t_{0}, x^{∗}(t_{0}), ū, v̄) ≤ J_{1}^{♯}(t, x, ū, v̄) ,

the two previous inequalities give the desired result:

J_{1}^{♯}(t, x, ū, v̄) ≥ J_{1}^{♯}(t_{0}, x, ū, v̄) .

QED

4.2 Concatenation

Definition 4.5 Let g ∈ G̃ and Σ = {σ_{0} = −∞, σ_{1}, …, σ_{k}, σ_{k+1} = +∞} be an associated partition. We say that a Nash equilibrium feedback (u^{∗}, v^{∗}), defined on some time interval [T − τ, T] for some τ > 0 for the terminal payoff g, is proper with respect to terminal set Σ if

∀t ∈ [T − τ, T], ∀x ∉ S_{t}, ∀x^{∗}(·) ∈ X (t, x, u^{∗}, v^{∗}), x^{∗}(T) ∉ Σ .

We also say that the Nash equilibrium feedback (u^{∗}, v^{∗}) is proper if there is an associated set of zero measure S_{t} (t ∈ [T − τ, T]) such that, for any t_{0} ∈ [T − τ, T] and any x_{0} ∉ S_{t_{0}}, we have:

∀x^{∗}(·) ∈ X (t_{0}, x_{0}, u^{∗}, v^{∗}), ∀t ∈ [t_{0}, T], x^{∗}(t) ∉ S_{t} .

Remark : From Proposition 3.1, if g ∈ G̃ and if Σ is an associated partition, then there exists a proper Nash equilibrium feedback, which is completely maximal.

Proposition 4.6 Let g ∈ G. Let us assume

• that (u_{1}^{∗}, v_{1}^{∗}) is a Nash equilibrium feedback on the time interval [T − τ_{1}, T] (for some τ_{1} > 0),

• that J^{♯}(T − τ_{1}, ·, u_{1}^{∗}, v_{1}^{∗}) belongs to G̃ and that there is an associated partition Σ such that, for

x ∉ Σ,

J^{♯}(T − τ_{1}, x, u_{1}^{∗}, v_{1}^{∗}) = J^{♭}(T − τ_{1}, x, u_{1}^{∗}, v_{1}^{∗}) ,

• and that (u_{2}^{∗}, v_{2}^{∗}) is a Nash equilibrium feedback on the interval [T − τ_{2}, T − τ_{1}] (for some τ_{2} > τ_{1}) for the terminal time T − τ_{1} and the terminal payoff J^{♯}(T − τ_{1}, ·, u_{1}^{∗}, v_{1}^{∗}), proper with respect to the terminal set Σ.

A) Then the feedback (u^{∗}, v^{∗}) defined by

(u^{∗}, v^{∗})(t, x) = \{ \begin{matrix}(u_{1}^{∗}, v_{1}^{∗})(t, x) & if t ∈ (T − τ_{1}, T] \\ (u_{2}^{∗}, v_{2}^{∗})(t, x) & if t ∈ [T − τ_{2}, T − τ_{1}] \end{matrix} (14)

is a Nash equilibrium feedback on the time interval [T − τ_{2}, T] for the terminal time T and the terminal payoff g.

B) If moreover (u_{1}^{∗}, v_{1}^{∗}) and (u_{2}^{∗}, v_{2}^{∗}) are proper, then so is (u^{∗}, v^{∗}).

C) If (u_{1}^{∗}, v_{1}^{∗}) and (u_{2}^{∗}, v_{2}^{∗}) are completely maximal, then so is (u^{∗}, v^{∗}).

Proof of Proposition 4.6 : A) For any t ∈ [T − τ_{2}, T − τ_{1}), let S_{t} be the set of zero measure associated with (u_{2}^{∗}, v_{2}^{∗}). We only prove that, if u ∈ U, if t_{0} ∈ [T − τ_{2}, T − τ_{1}) and x_{0} ∉ S_{t_{0}}, then

J_{1}^{♭}(t_{0}, x_{0}, u^{∗}, v^{∗}) ≥ J_{1}^{♯}(t_{0}, x_{0}, u, v^{∗}) .

Indeed the corresponding inequality for t_{0} ∈ [T − τ_{1}, T] is clear and the corresponding inequality for J_{2} can be proved in a similar way.

Let x^{∗}(·) ∈ X (t_{0}, x_{0}, u^{∗}, v^{∗}) and x(·) ∈ X (t_{0}, x_{0}, u, v^{∗}). Since (u^{∗}, v^{∗}) = (u_{2}^{∗}, v_{2}^{∗}) on [T −τ_{2}, T −τ_{1}) and since (u_{2}^{∗}, v_{2}^{∗}) is a Nash equilibrium feedback for the terminal time T − τ_{1} and the terminal payoff J^{♯}(T − τ_{1}, ·, u_{1}^{∗}, v_{1}^{∗}) on this interval, we have

J_{1}^{♯}(T − τ_{1}, x^{∗}(T − τ_{1}), u_{1}^{∗}, v_{1}^{∗}) ≥ J_{1}^{♯}(T − τ_{1}, x(T − τ_{1}), u_{1}^{∗}, v_{1}^{∗}) .

Since, from the assumption, (u_{2}^{∗}, v_{2}^{∗}) is proper with respect to the set Σ, the point x^{∗}(T − τ_{1}) does not belong to Σ. Therefore, from the definition of Σ, we have

J_{1}^{♯}(T − τ_{1}, x^{∗}(T − τ_{1}), u_{1}^{∗}, v_{1}^{∗}) = J_{1}(T − τ_{1}, x^{∗}(T − τ_{1}), u_{1}^{∗}, v_{1}^{∗}) = g_{1}(x^{∗}(T)) .

Since (u_{1}^{∗}, v_{1}^{∗}) is a Nash equilibrium feedback on [T − τ_{1}, T] and since, from the construction of (u^{∗}, v^{∗}), the solution x(·) belongs to X (T − τ_{1}, x(T − τ_{1}), u, v_{1}^{∗}), we have, from Corollary 4.2,

J_{1}^{♯}(T − τ_{1}, x(T − τ_{1}), u_{1}^{∗}, v_{1}^{∗}) ≥ J_{1}^{♯}(T − τ_{1}, x(T − τ_{1}), u, v_{1}^{∗}) ≥ g_{1}(x(T)) .

Therefore we have proved that

g_{1}(x^{∗}(T)) ≥ g_{1}(x(T)) ,

which is the desired result.

B) If (u_{1}^{∗}, v_{1}^{∗}) and (u_{2}^{∗}, v_{2}^{∗}) are proper, then it is clear that so is (u^{∗}, v^{∗}).

C) Let us now assume that (u_{1}^{∗}, v_{1}^{∗}) and (u_{2}^{∗}, v_{2}^{∗}) are completely maximal. We have to prove that (u^{∗}, v^{∗}) is also completely maximal, i.e., that, for any T_{0} ∈ (T − τ_{2}, T], if the pair (e_{1}, e_{2}) is a maximal Nash equilibrium payoff for some point (t_{0}, x_{0}) ∈ [T − τ_{2}, T_{0}) for the game with terminal time T_{0} and terminal payoff J^{♯}(T_{0}, ·, u^{∗}, v^{∗}), then we have

J^{♯}(t_{0}, x_{0}, u^{∗}, v^{∗}) = (e_{1}, e_{2}) . (15)

Without loss of generality, we can assume that T_{0} ∈ (T − τ_{1}, T] and t_{0} ∈ [T − τ_{2}, T − τ_{1}) since otherwise the result is a straightforward consequence of the fact that (u_{1}^{∗}, v_{1}^{∗}) and (u_{2}^{∗}, v_{2}^{∗}) are completely maximal. By using Proposition 2.6 and again the fact that (u_{1}^{∗}, v_{1}^{∗}) is completely maximal, we can also assume, in order to simplify the notation, that T_{0} = T.

In order to apply the assumption that (u_{2}^{∗}, v_{2}^{∗}) is completely maximal, we are going to prove that (e_{1}, e_{2}) is a maximal Nash equilibrium at the point (t_{0}, x_{0}) for the differential game with terminal time T − τ_{1} and terminal payoff J^{♯}(T − τ_{1}, ·, u^{∗}, v^{∗}).

From the definition of a maximal equilibrium, there is some solution x̄(·) ∈ X (t_{0}, x_{0}) such that (e_{1}, e_{2}) ∈ ess − lim sup_{x′→x̄(T)} g(x′) and, for any x(·) ∈ X (t_{0}, x_{0}), we have

for j = 1, 2, g_{j}(x(T)) ≤ e_{j} .

Lemma 2.4 states that (e_{1}, e_{2}) is the maximal equilibrium at any point (t, x̄(t)). In particular, for t = T − τ_{1}, we have, since (u_{1}^{∗}, v_{1}^{∗}) is maximal:

(e_{1}, e_{2}) = J^{♯}(T − τ_{1}, x̄(T − τ_{1}), u_{1}^{∗}, v_{1}^{∗}) . (16)

For simplicity, we set ȳ = x̄(T − τ_{1}). Let us now prove that

(17) (e_{1}, e_{2}) belongs to ess − lim sup_{x′→ȳ} J^{♯}(T − τ_{1}, x′, u_{1}^{∗}, v_{1}^{∗}) .

Proof of (17) : We consider two cases.

Case 1 : Let us first assume that, for any t ∈ [T − τ_{1}, T], ȳ = x̄(t) = x̄(T). From Lemma 4.4, we have

∀x ∈ IR, J_{j}^{♯}(T − τ_{1}, x, u_{1}^{∗}, v_{1}^{∗}) ≥ g_{j}(x)

for j = 1, 2. Since (e_{1}, e_{2}) ∈ ess − lim sup_{x′→x̄(T)} g(x′) and since J_{j}^{♯}(T − τ_{1}, ·, u_{1}^{∗}, v_{1}^{∗}) is usc for j = 1, 2, one concludes easily from (16) that (17) holds true in this case.

Case 2 : Let us now assume that there is some t ∈ [T − τ_{1}, T] with x̄(t) ≠ x̄(T). We can assume for instance that there is some ϵ > 0 with:

[ȳ − ϵ, ȳ] ⊂ x̄([T − τ_{1}), T]) . (18)

Since, from Lemma 2.4, (e_{1}, e_{2}) is a maximal payoff at the point (t, x̄(t)) for any t ∈ [T − τ_{1}, T] and since (u_{1}^{∗}, v_{1}^{∗}) is maximal, we have

∀t ∈ [T − τ_{1}, T], J^{♯}(t, x̄(t), u_{1}^{∗}, v_{1}^{∗}) = (e_{1}, e_{2}) .

Combining Lemma 4.4, the previous equality and inclusion(18) gives:

∀x ∈ [ȳ − ϵ, ȳ], J_{j}^{♯}(T − τ_{1}, x, u_{1}^{∗}, v_{1}^{∗}) ≥ e_{j} ,

for j = 1, 2. This inequality together with the fact that J_{j}^{♯}(t, ·, u_{1}^{∗}, v_{1}^{∗}) is usc, leads to (17). Therefore (17) holds in any case.

Let x(·) ∈ X (t_{0}, x_{0}). We now prove that

for j = 1, 2, J_{j}^{♯}(T − τ_{1}, x(T − τ_{1}), u_{1}^{∗}, v_{1}^{∗}) ≤ e_{j} .

We do the proof for j = 1 for instance. From Corollary 4.3, there is x_{1}(·) belonging to X (T − τ_{1}, x(T − τ_{1}), u_{1}^{∗}, v_{1}^{∗}) such that

g_{1}(x_{1}(T)) = J_{1}^{♯}(T − τ_{1}, x(T − τ_{1}), u_{1}^{∗}, v_{1}^{∗}) .

The trajectory x_{2}(·) defined by

x_{2}(t) = \{ \begin{matrix}x(t) & if t ∈ [t_{0}, T − τ_{1}] \\ x_{1}(t) & if t ∈ [T − τ_{1}, T] \end{matrix}

belongs to X (t_{0}, x_{0}). Since (e_{1}, e_{2}) is the maximal payoff at (t_{0}, x_{0}), we have

e_{1} ≥ g_{1}(x_{2}(T)) = g_{1}(x_{1}(T)) = J_{1}^{♯}(T − τ_{1}, x(T − τ_{1}), u_{1}^{∗}, v_{1}^{∗}) .

Thus we have proved that (e_{1}, e_{2}) is the maximal payoff at the point (t_{0}, x_{0}) for the game with terminal time T − τ_{1} and terminal payoff J_{1}^{♯}(T − τ_{1}, ·, u_{1}^{∗}, v_{1}^{∗}).

Then equality (15) holds true because (u_{2}^{∗}, v_{2}^{∗}) is a maximal equilibrium feedback for this game.

QED

4.3 Extension

Proposition 4.7 Let us assume that the terminal payoff g belongs to G and that there are some (u^{∗}, v^{∗}) ∈ U × V and some time T_{0} ∈ [0, T) such that

i) ∀t ∈ (T_{0}, T), (u^{∗}, v^{∗}) is a completely maximal Nash equilibrium feedback on [t, T], ii) for any t ∈ (T_{0}, T), J^{♯}(t, ·, u^{∗}, v^{∗}) belongs to G̃, iii) for any t ∈ (T_{0}, T), there is some partition S_{t} associated with J^{♯}(t, ·, u^{∗}, v^{∗}) such that

∀x ∉ S_{t}, J^{♯}(t, x, u^{∗}, v^{∗}) = J^{♭}(t, x, u^{∗}, v^{∗}) ,

iv) the set-valued map t → S_{t} is 2-Lipschitz continuous and there is some M such that the cardinal of S_{t} is not larger than M,

v) (u^{∗}, v^{∗}) is proper with respect to (S_{t})_{t∈(T_{0},T]}, i.e., for any t_{0} ∈ (T_{0}, T), for any x_{0} ∉ S_{t_{0}}, for any x^{∗}(·) ∈ X (t_{0}, x_{0}, u^{∗}, v^{∗}), we have

∀t ∈ [t_{0}, T], x^{∗}(t) ∉ S_{t} .

Then

a) (u^{∗}, v^{∗}) is still a Nash and completely maximal equilibrium on [T_{0}, T],

b) J^{♯}(T, ·, u^{∗}, v^{∗}) belongs to G̃,

c) the limit S_{T_{0}} of S_{t} when t → T_{0} exists and the cardinal of S_{T_{0}} is not larger than M,

d) ∀x ∉ S_{T_{0}}, J^{♯}(T_{0}, x, u^{∗}, v^{∗}) = J^{♭}(T_{0}, x, u^{∗}, v^{∗}) ,

e) (u^{∗}, v^{∗}) is proper with respect to (S_{t})_{t∈[T_{0},T]} up to time T_{0}.

Remark : We have proved in Proposition 3.1 that, if g belongs to G, then there is some T_{0} < T and some feedback (u^{∗}, v^{∗}) for which (i),…,(v) hold.

Proof of Proposition 4.7 : We divide the proof in two steps. In the first step, we prove that (a), (c), (d) and (e) hold. We complete the proof by establishing the more delicate point (b).

First step : Point (c) of the conclusion is immediate. Let x_{0} ∉ S_{T_{0}} be fixed. From the definition of S_{T_{0}} we can find some positive ϵ such that

∀x ∈ [x_{0} − ϵ, x_{0} + ϵ], ∀t ∈ [T_{0}, T_{0} + ϵ], x ∉ S_{t} .

Let us set t_{0} = T_{0} + ϵ/4. Since J^{♯}(t_{0}, ·, u^{∗}, v^{∗}) belongs to G̃ and since the set S_{t_{0}} has an empty intersection with the interval [x_{0} − ϵ, x_{0} + ϵ], the function J^{♯}(t_{0}, ·, u^{∗}, v^{∗}) can only have two types of behaviour on [x_{0} − ϵ, x_{0} + ϵ]:

- either J_{1}^{♯}(t_{0}, ·, u^{∗}, v^{∗}) (resp. J_{2}^{♯}(t_{0}, ·, u^{∗}, v^{∗})) is continuous and strictly increasing or decreasing on [x_{0} − ϵ, x_{0} + ϵ],

- or J_{1}^{♯}(t_{0}, ·, u^{∗}, v^{∗}) and J_{2}^{♯}(t_{0}, ·, u^{∗}, v^{∗}) are constant on [x_{0} − ϵ, x_{0} + ϵ].

From Proposition 2.6, we know that (u^{∗}, v^{∗}) is a completely maximal Nash equilibrium feedback for the terminal time t_{0} and the terminal payoff J^{♯}(t_{0}, ·, u^{∗}, v^{∗}). Then it is not very difficult to check that we necessarilly have, for any t ∈ (T_{0}, t_{0}) and any x ∈ I_{ϵ} = [x_{0} − ϵ/2, x_{0} + ϵ/2],

J^{♯}(t, x, u^{∗}, v^{∗}) = J^{♯}(t_{0}, x + 2γ(t_{0} − t), u^{∗}, v^{∗})

where

γ = \{ \begin{matrix}1 & if J_{1}^{♯}(t_{0}, ·, u^{∗}, v^{∗}) and J_{2}^{♯}(t_{0}, ·, u^{∗}, v^{∗}) are increasing on I_{ϵ} \\ −1 & if J_{1}^{♯}(t_{0}, ·, u^{∗}, v^{∗}) and J_{2}^{♯}(t_{0}, ·, u^{∗}, v^{∗}) are decreasing on I_{ϵ} \\ 0 & otherwise \end{matrix}

In particular, the map (t, x) → J^{♯}(t_{0}, x, u^{∗}, v^{∗}) is uniformly continuous on (t_{0}, T_{0}] × I_{ϵ}.

In order to prove that (u^{∗}, v^{∗}) is a Nash equilibrium feedback, let us now fix u ∈ U, x^{∗}(·) ∈ X (T_{0}, x_{0}, u^{∗}, v^{∗}) and x(·) ∈ X (T_{0}, x_{0}, u, v^{∗}).

Let us first notice that ∀t ∈ [T_{0}, T], x^{∗}(t) ∉ S_{t} . (19)

Indeed, for any t ∈ (T_{0}, t_{0}], the point x^{∗}(t) does not belong to S_{t}. Since, moreover, x^{∗}(·) belongs to X (t, x^{∗}(t), u^{∗}, v^{∗}), the fact that the feedback (u^{∗}, v^{∗}) is proper implies (19). Hence (e) is proved.

Let us now fix η > 0. Since the map (t, x) → J^{♯}(t, x, u^{∗}, v^{∗}) is uniformly continuous on (t_{0}, T_{0}]× [x_{0} − ϵ/2, x_{0} + ϵ/2], we can choose t sufficiently close to T_{0} such that

g_{1}(x^{∗}(T)) = J_{1}(t, x^{∗}(t), u^{∗}, v^{∗}) ≥ J_{1}(t, x(t), u^{∗}, v^{∗}) − η .

Moreover, since (u^{∗}, v^{∗}) is a Nash equilibrium feedback, we also have

J_{1}(t, x(t), u^{∗}, v^{∗}) ≥ J_{1}^{♯}(t, x(t), u, v^{∗}) ≥ g_{1}(x(T)) .

Hence inequality g_{1}(x^{∗}(T)) ≥ g_{1}(x(T)) − η holds true for any η > 0. This implies that:

∀x_{0} ∉ S_{T_{0}}, J_{1}^{♭}(T_{0}, x_{0}, u^{∗}, v^{∗}) ≥ J_{1}^{♯}(T_{0}, x_{0}, u, v^{∗}) .

Thus (d) is proved and (u^{∗}, v^{∗}) is a Nash equilibrium feedback on [T_{0}, T].

Let us finally prove that (u^{∗}, v^{∗}) is completely maximal on [T_{0}, T]. For simplicity we only prove that (u^{∗}, v^{∗}) is maximal on [T_{0}, T], the proof for the complete maximality being the same.

Let (T_{0}, x_{0}) be a fixed point for which there is some maximal payoff (e_{1}, e_{2}) ∈ IR^{2}. There is some solution x̄(·) ∈ X (T_{0}, x_{0}) such that (e_{1}, e_{2}) belongs to ess − lim sup_{x→x̄(T)} g(x) and for any x(·) ∈ X (T_{0}, x_{0}),

for j = 1, 2, g_{j}(x(T)) ≤ e_{j} .

Since (u^{∗}, v^{∗}) is completely maximal on [T_{0} + ϵ, T] for any ϵ > 0, and since (e_{1}, e_{2}) is a maximal payoff at the point (t, x̄(t)), we have

∀t ∈ (T_{0}, T), J^{♯}(t, x̄(t), u^{∗}, v^{∗}) = (e_{1}, e_{2}) .

Hence we have

for j = 1, 2, J_{j}^{♯}(T_{0}, x_{0}, u^{∗}, v^{∗}) ≥ lim sup_{t→T_{0}^{+}} J_{j}^{♯}(t, x̄(t), u^{∗}, v^{∗}) = e_{j} .

Since obviously

for j = 1, 2, J_{j}^{♯}(T_{0}, x_{0}, u^{∗}, v^{∗}) ≤ e_{j} ,

we have proved the desired equality

J^{♯}(T_{0}, x_{0}, u^{∗}, v^{∗}) = (e_{1}, e_{2}) .

Second step : Let us finally prove that (b) holds true, i.e., that J^{♯}(T_{0}, ·, u^{∗}, v^{∗}) belongs to G̃. Let us set as usual

S_{T_{0}} = {σ_{0} = −∞, σ_{1}, …, σ_{k}, σ_{k+1} = +∞}

From the first step, it is immediate that, on the interval (σ_{i}, σ_{i+1}) for i = 1, …, k, we have - either J_{1}^{♯}(T_{0}, ·, u^{∗}, v^{∗}) (resp. J_{2}^{♯}(T_{0}, ·, u^{∗}, v^{∗})) is strictly increasing or strictly decreasing,

- or J_{1}^{♯}(T_{0}, ·, u^{∗}, v^{∗}) and J_{2}^{♯}(T_{0}, ·, u^{∗}, v^{∗}) are constant.

From Corollary 4.3, it is also clear that J^{♯}(T_{0}, ·, u^{∗}, v^{∗}) is usc and left or right continuous at each point σ_{i}.

It now remains to prove that, if σ_{i} is a strict local maximum of (say) J_{1}^{♯}(T_{0}, ·, u^{∗}, v^{∗}), then J_{1}^{♯}(T_{0}, ·, u^{∗}, v^{∗}) is continuous at σ_{i} and J_{2}^{♯}(T_{0}, ·, u^{∗}, v^{∗}) has a strict local minimum and is continuous at this point.

From Corollary 4.3, there is some sequence (x_{n}) which converges to σ_{i}, with x_{n} ∉ S_{T_{0}} and such that (J_{1}(T_{0}, x_{n}, u^{∗}, v^{∗})) converges to J_{1}^{♯}(T_{0}, σ_{i}, u^{∗}, v^{∗}). Let x_{n}(·) ∈ X (T_{0}, x_{n}, u^{∗}, v^{∗}).

From standard arguments, the sequence (x_{n}(·)) converges, up to some subsequence, to some solution x(·) ∈ X (T_{0}, σ_{i}, u^{∗}, v^{∗}) uniformly on [T_{0}, T]. Since x_{n} ∉ S_{T_{0}}, we have

∀t ∈ [T_{0}, T], J_{1}^{♯}(t, x_{n}(t), u^{∗}, v^{∗}) = J_{1}^{♯}(T_{0}, x_{n}, u^{∗}, v^{∗}) .

Taking the lim sup in this equality gives

∀t ∈ [T_{0}, T], J_{1}^{♯}(t, x(t), u^{∗}, v^{∗}) ≥ J_{1}^{♯}(T_{0}, σ_{i}, u^{∗}, v^{∗}) .

Using now Lemma 4.4, we also have:

∀t ∈ [T_{0}, T], J_{1}^{♯}(t, x(t), u^{∗}, v^{∗}) ≤ J_{1}^{♯}(T_{0}, x(t), u^{∗}, v^{∗}) .

Since J_{1}^{♯}(T_{0}, ·, u^{∗}, v^{∗}) has a strict local maximum at σ_{i}, putting together the previous inequalities yields that ∀t ∈ [T_{0}, T], x(t) = σ_{i} and J_{1}^{♯}(t, σ_{i}, u^{∗}, v^{∗}) = J_{1}^{♯}(T_{0}, σ_{i}, u^{∗}, v^{∗}) . (20)

Using again Lemma 4.4 gives that, for any x ≠ σ_{i} sufficiently close to σ_{i},

\begin{matrix}J_{1}^{♯}(T_{0}, σ_{i}, u^{∗}, v^{∗}) = J_{1}^{♯}(t, σ_{i}, u^{∗}, v^{∗}) \\ > J_{1}^{♯}(T_{0}, x, u^{∗}, v^{∗}) ≥ J_{1}^{♯}(t, x, u^{∗}, v^{∗}) . \end{matrix} (21)

Therefore, for any t ∈ [T_{0}, T], the map J_{1}^{♯}(t, ·, u^{∗}, v^{∗}) has a strict local maximum at σ_{i}.

From the assumption, for t ∈ (T_{0}, T), J^{♯}(t, ·, u^{∗}, v^{∗}) belongs to G̃. Therefore J_{1}^{♯}(t, ·, u^{∗}, v^{∗}) is continuous at σ_{i}. Using again (20) and (21) proves that J_{1}^{♯}(T_{0}, ·, u^{∗}, v^{∗}) is lsc at σ_{i}. Since it is also usc, we have proved that J_{1}^{♯}(T_{0}, ·, u^{∗}, v^{∗}) is continuous at σ_{i}.

Let us finally establish that J_{2}^{♯}(T_{0}, ·, u^{∗}, v^{∗}) is continuous at σ_{i} and has a strict local minimum at this point. For that purpose, let us fix some t_{0} ∈ (T_{0}, T). Since J^{♯}(t_{0}, ·, u^{∗}, v^{∗}) belongs to G̃ and since J_{1}^{♯}(t_{0}, ·, u^{∗}, v^{∗}) has a strict local maximum at σ_{i}, J_{2}^{♯}(t_{0}, ·, u^{∗}, v^{∗}) has a strict local minimum at σ_{i} and it is continuous at this point.

Let x_{n} → σ_{i}, with, for instance, x_{n} < σ_{i}. Then, for n sufficiently large, the point x_{n} does not belong to S_{T_{0}}. Let x_{n}(·) be some solution in X (T_{0}, x_{n}, u^{∗}, v^{∗}). Since (u^{∗}, v^{∗}) is proper, and since the point σ_{i} belongs to S_{t} for any t (because σ_{i} is a strict local maximum of J_{1}^{♯}(t, ·, u^{∗}, v^{∗})) we always have x_{n}(t) < σ_{i}. Since x_{n} ∉ S_{T_{0}}, we have, for j = 1, 2

∀t ∈ [T_{0}, T], J_{j}^{♯}(T_{0}, x_{n}, u^{∗}, v^{∗}) = J_{j}^{♯}(t, x_{n}(t), u^{∗}, v^{∗}) .

Moreover, from Lemma 4.4, we have

∀t ∈ [T_{0}, T], J_{j}^{♯}(t, x_{n}, u^{∗}, v^{∗}) ≤ J_{j}^{♯}(T_{0}, x_{n}, u^{∗}, v^{∗}) .

Therefore, for j = 1, 2, J_{j}^{♯}(t_{0}, x_{n}(t_{0}), u^{∗}, v^{∗}) ≥ J_{j}^{♯}(t_{0}, x_{n}, u^{∗}, v^{∗}) . (22)

Using arguments already developped above and the fact that J_{1}^{♯}(t_{0}, ·, u^{∗}, v^{∗}) is continuous at σ_{i}, we can prove that the sequence x_{n}(·) converges uniformly to the constant solution x(t) = σ_{i}. Therefore the point x_{n}(t_{0}) belongs to some interval (σ_{i} − η, σ_{i}) for n large enough. We choose η > 0 such that J_{1}^{♯}(t_{0}, ·, u^{∗}, v^{∗}) is stricly decreasing and J_{2}^{♯}(t_{0}, ·, u^{∗}, v^{∗}) is strictly increasing on (σ_{i} − η, σ_{i}). Then inequality (22) imply that x_{n}(t_{0}) = x_{n}. Hence we have:

\begin{matrix}J_{2}^{♯}(T_{0}, x_{n}, u^{∗}, v^{∗}) = J_{2}^{♯}(t_{0}, x_{n}, u^{∗}, v^{∗}) \\ > J_{2}^{♯}(t_{0}, σ_{i}, u^{∗}, v^{∗}) = J_{2}^{♯}(T_{0}, σ_{i}, u^{∗}, v^{∗}) \end{matrix}

and

\begin{matrix}lim_{n} J_{2}^{♯}(T_{0}, x_{n}, u^{∗}, v^{∗}) = lim_{n} J_{2}^{♯}(t_{0}, x_{n}, u^{∗}, v^{∗}) \\ = J_{2}^{♯}(t_{0}, σ_{i}, u^{∗}, v^{∗}) = J_{2}^{♯}(T_{0}, σ_{i}, u^{∗}, v^{∗}) \end{matrix}

because J_{2}^{♯}(t_{0}, ·, u^{∗}, v^{∗}) has a strict local minimum at σ_{i} and is continuous at this point. Therefore we have proved that J_{2}^{♯}(T_{0}, ·, u^{∗}, v^{∗}) has a strict local minimum at σ_{i} and is continuous at this point.

This completes the proof of Proposition 4.7.

QED

4.4 Proof of the main Theorem

We are now ready to prove Theorem 2.10. Using Proposition 3.1 and Zorn Lemma shows that there is some maximal interval (T̄, T], some feedback (u^{∗}, v^{∗}) and some associated set of zero measure (S_{t})_{t∈(T̄,T]} such that conditions (i),…,(v) of Proposition 4.7 are satisfied. Then the conclusion of Proposition 4.7 holds, which implies that the map J^{♯}(T̄, ·, u^{∗}, v^{∗}) belongs to G̃.

Let us now assume that T̄ > 0. Then using Proposition 3.1 again shows that there is some τ > 0 and some feedback (u_{1}^{∗}, v_{1}^{∗}), and some sets of zero measure (S_{t})_{t∈[T̄−τ,T̄]} satisfying the conditions (i),…,(v) of Proposition 3.1 for the terminal time T̄ and the terminal payoff J^{♯}(T̄, ·, u^{∗}, v^{∗}) on the time interval [T̄ − τ, T̄]. Then Proposition 4.6 implies that the new feedback

(u_{2}^{∗}, v_{2}^{∗})(t, x) = \{ \begin{matrix}(u^{∗}, v^{∗})(t, x) & if t ∈ (T̄, T] \\ (u_{1}^{∗}, v_{1}^{∗})(t, x) & otherwise \end{matrix}

satisfies conditions (i),…,(v) of Proposition 4.7 on the times interval [T̄ − τ, T]. This is in contradiction with the maximality of the interval [T̄, T].

Hence we have proved that there is at least one solution (u^{∗}, v^{∗}) defined on [0, T].

Uniqueness of this solution is a straightforward application of the uniqueness result of Proposition 3.1 and of the fact that the map J^{♯}(t, ·, u^{∗}, v^{∗}) belongs to G̃ for t ∈ [0, T).

QED

4.5 Proof of Proposition 2.12 and of Theorem 2.13

We complete this paper by the proof of the results on the link between Nash equilibrium feedbacks and equilibrium payoffs in the sense of Kleimenov.

Proof of Proposition 2.12 : Let t, x, and x^{∗}(·) be as in the Proposition. We have on the one hand that

∀s ∈ [t, T], J(s, x^{∗}(s), u^{∗}, v^{∗}) = g(x^{∗}(T)) = J(t, x, u^{∗}, v^{∗})

and, on the other hand,

∀s ∈ [t, T], for j = 1, 2, J_{j}(s, x^{∗}(s), u^{∗}, v^{∗}) ≥ g_{j}(x^{∗}(s)) ,

thanks to Lemma 4.4. Whence the result.

QED

Proof of Theorem 2.13 : Let t_{0} ∈ [0, T) and x_{0} ∉ S_{t_{0}} be fixed. Let us assume that (e_{1}, e_{2}) is a Kleimenov equilibrium at (t_{0}, x_{0}) such that

for j = 1, 2, e_{j} ≥ J_{j}(t_{0}, x_{0}, u^{∗}, v^{∗}) .

We have to prove that J(t_{,}x_{0}, u^{∗}, v^{∗}) = (e_{1}, e_{2}).

From the definition of Kleimenov equilibrium payoffs, there is some x(·) ∈ X (t_{0}, x_{0}, u^{∗}, v^{∗}) such that

∀t ∈ [t_{0}, T], for j = 1, 2, g_{j}(x(t)) ≤ e_{j} = g_{j}(x(T)) .

In order to fix the ideas, we assume that x(T) ≥ x_{0}. It is easy to see that one can choose x(·) such that x′(·) = 0 on [t_{0}, θ_{1}) and x′(·) = 2 on [θ_{1}, T], where θ_{1} = T − (x(T) − x_{0})/2.

Let now x^{∗}(·) ∈ X (t_{0}, x_{0}, u^{∗}, v^{∗}). We know from Proposition 2.12 that x^{∗}(·) is a Kleimenov trajectory and moreover that

∀t ∈ [t_{0}, T], J(t, x^{∗}(t), u^{∗}, v^{∗}) = J(t_{0}, x_{0}, u^{∗}, v^{∗}) = g(x^{∗}(T)) .

Hence, if x^{∗}(T) ≥ x(T), there exists some t ∈ [t_{0}, T] such that x^{∗}(t) = x(T), and thus

J(t_{0}, x_{0}, u^{∗}, v^{∗}) = J(t, x^{∗}(t), u^{∗}, v^{∗}) ≥ g(x^{∗}(t)) = g(x(T)) = (e_{1}, e_{2})

because x^{∗}(·) is a Kleimenov trajectory (here and throughout the proof we endow IR^{2} with the usual partial ordering). Therefore, if x^{∗}(T) ≥ x(T), the result is proved.

Let us assume from now on that x^{∗}(T) < x(T). Let us set

θ_{2} = max{t ∈ [θ_{1}, T] | x^{∗}(t) = x(t)} ,

with θ_{2} = θ_{1} if there is no t ∈ [θ_{1}, T] with x^{∗}(t) = x(t). Let us point out that ∀t ∈ (θ_{2}, T], x^{∗}(t) < x(t). We claim that

∀t ∈ (θ_{2}, T], ∀y ∈ [x^{∗}(t), x(t)], J^{♯}(t, y, u^{∗}, v^{∗}) ≤ (e_{1}, e_{2}) . (23)

Proof of (23) : Let us first prove that

∀t ∈ (θ_{2}, T], ∀y ∈ (x^{∗}(t), x(t)) with y ∉ S_{t}, J(t, y, u^{∗}, v^{∗}) ≤ (e_{1}, e_{2}) . (24)

Let t and y as above, and let y(·) ∈ X (t, y, u^{∗}, v^{∗}). Then y_{1}(·) = max{y(·), x^{∗}(·)} also belongs to X (t, y, u^{∗}, v^{∗} and y_{1}(T) ∈ [x^{∗}(T), x(T)]. Using now the fact that x(·) and x^{∗}(·) are Kleimenov trajectories with x(t_{0}) = x^{∗}(t_{0}) = x_{0}, and that g(x^{∗}(T)) ≤ (e_{1}, e_{2}), it is easy to see that, for any z ∈ [x^{∗}(T), x(T)], g(z) ≤ (e_{1}, e_{2}). Therefore

J(t, y, u^{∗}, v^{∗}) = g(y_{1}(T)) ≤ (e_{1}, e_{2}) ,

so that (24) is proved.

Thanks to Corollary 4.3, inequality (24) implies inequality (23) for y ∈ (x^{∗}(t), x(t)). Let us notice that the result is obvious for x^{∗}(t) since J(t, x^{∗}(t), u^{∗}, v^{∗}) = J(t_{0}, x_{0}, u^{∗}, v^{∗}) ≤ (e_{1}, e_{2}) from the assumption. Let us finally prove the result for x(t). Corollary 4.3 states that there two sequences (x_{k}^{j})_{k} (for j = 1, 2) converging to x(t), with x_{k}^{j} ∉ S_{t}, such that

lim_{k} J_{j}(t, x_{k}^{j}, u^{∗}, v^{∗}) = J_{j}^{♯}(t, x(t), u^{∗}, v^{∗}) .

If x_{k}^{j} ≤ x(t) for infinitely many k, then, since for such a k we already know that J_{j}(t, x_{k}^{j}, u^{∗}, v^{∗}) ≤ e_{j}, we have J_{j}^{♯}(t, x(t), u^{∗}, v^{∗}) ≤ e_{j}. If on the contrary, x_{k}^{j} > x(t) for any k sufficiently large, let us consider some solutions x_{k}^{j}(·) ∈ X (t, x_{k}^{j}, u^{∗}, v^{∗}). If x_{k}^{j}(T) ≤ x(T), then we can complete the proof as above. If x_{k}^{j}(T) ≥ x(T), then lim_{k} x_{k}^{j}(T) = x(T) because x(T) = x(t) + 2(T − t) and the x_{k}^{j}(·) are 2-Lipschitz continuous. Then the result follows from the continuity of g. Therefore, (23) holds in any case.

Let us now set

θ_{3} = inf{t ∈ [θ_{2}, T] | ∀s ∈ [t, T], J^{♯}(s, x(s), u^{∗}, v^{∗}) = (e_{1}, e_{2})} ,

with θ_{3} = T if there is no such a t. Our aim is to prove that θ_{3} = θ_{2}.

For doing so, we argue by contradiction by assuming that θ_{3} > θ_{2}. Let us prove that in this case, for any ϵ > 0 sufficiently small, (e_{1}, e_{2}) is a maximal payoff at the point (θ_{3} − ϵ, x(θ_{3} − ϵ)) for the game with horizon θ_{3} and terminal payoff J^{♯}(θ_{3}, ·, u^{∗}, v^{∗}) (let us notice that x(θ_{3} − ϵ) = x(θ_{3}) − 2ϵ).

Indeed, we have, thanks to (23) and the fact that the J_{j}^{♯} are usc, that:

∀z ∈ [x(θ_{3}) − 4ϵ, x(θ_{3})), J^{♯}(θ_{3}, z, u^{∗}, v^{∗}) ≤ (e_{1}, e_{2}) and J^{♯}(θ_{3}, x(θ_{3}), u^{∗}, v^{∗}) = (e_{1}, e_{2}) .

If θ_{3} < T, Lemma 4.4 shows that

∀s ∈ [θ_{3}, T], J^{♯}(θ_{3}, x(s), u^{∗}, v^{∗}) ≥ J^{♯}(s, x(s), u^{∗}, v^{∗}) = (e_{1}, e_{2}) ,

with x(s) = x(θ_{3}) + 2(s − θ_{3}). Using the fact that the maps J_{j}^{♯}(θ_{3}, ·, u^{∗}, v^{∗}) (for j = 1, 2) are usc then gives that

(e_{1}, e_{2}) ∈ ess − lim sup_{x′→x(θ_{3})} J^{♯}(θ_{3}, x′, u^{∗}, v^{∗}) .

If θ_{3} = T, the above formula is also clear because g is continuous.

Hence we have proved that (e_{1}, e_{2}) is a maximal payoff at the point (θ_{3} − ϵ, x(θ_{3} − ϵ)) for the game with horizon θ_{3} and terminal payoff J^{♯}(θ_{3}, ·, u^{∗}, v^{∗}). Since (u^{∗}, v^{∗}) is completely maximal, this implies that

J^{♯}(θ_{3}, x(θ_{3} − ϵ), u^{∗}, v^{∗}) = (e_{1}, e_{2})

for any ϵ > 0 sufficiently small, which is in contradiction with the definition of θ_{3}. Therefore we have proved that θ_{3} = θ_{2}.

Hence

J^{♯}(θ_{2}, x(θ_{2}), u^{∗}, v^{∗}) ≥ (e_{1}, e_{2}) ,

because the maps J_{j}^{♯} are usc. So, if θ_{2} > θ_{1}, we have x^{∗}(θ_{2}) = x(θ_{2}), and thus

J(t_{0}, x_{0}, u^{∗}, v^{∗}) = J(θ_{2}, x^{∗}(θ_{2}), u^{∗}, v^{∗}) ≥ (e_{1}, e_{2}) ,

and equality J(t_{0}, x_{0}, u^{∗}, v^{∗}) = (e_{1}, e_{2}) is proved. If, on the contrary, θ_{2} = θ_{1}, then x(θ_{1}) = x_{0}, and Lemma 4.4 implies that

J(t_{0}, x_{0}, u^{∗}, v^{∗}) ≥ J^{♯}(θ_{1}, x_{0}, u^{∗}, v^{∗}) ≥ (e_{1}, e_{2}) ,

which completes the proof of Theorem 2.13.

QED

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[3] EVANS L.C. & SOUGANIDIS P.E. (1984) Differential games and representation formulas for solutions of Hamilton-Jacobi Equations, Indiana Univ. Math. J., 33, 773-797.

[4] FLEMING W. H. (1967) The convergence problem for differential games, J. Math. Anal. and Appl., Vol. 3, pp. 102-116

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Contents

1 Introduction 1

2 Nash equilibrium feedbacks 3 2.1 Strategies and payoffs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 2.2 Definition of the Nash equilibrium feedbacks . . . . . . . . . . . . . . . . . . . . . . . 4 2.3 The main Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.4 Link with Nash equilibrium payoffs in the sense of Kleimenov . . . . . . . . . . . . . 8 2.5 Examples of Nash equilibrium feedbacks . . . . . . . . . . . . . . . . . . . . . . . . . 8

3 Local existence and uniqueness 11 3.1 The local existence and uniqueness result . . . . . . . . . . . . . . . . . . . . . . . . 12 3.2 Construction of the maximal Nash equilibrium feedback . . . . . . . . . . . . . . . . 12 3.3 The pair of strategy (u^{∗}, v^{∗}) is a maximal Nash equilibrium feedback . . . . . . . . . 16 3.4 The payoff belongs to the class G̃ on [T − τ, T) . . . . . . . . . . . . . . . . . . . . . 23 3.5 The feedback (u^{∗}, v^{∗}) is completely maximal . . . . . . . . . . . . . . . . . . . . . . . 24

4 Global existence and uniqueness 24 4.1 Some preliminary results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 4.2 Concatenation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 4.3 Extension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 4.4 Proof of the main Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 4.5 Proof of Proposition 2.12 and of Theorem 2.13 . . . . . . . . . . . . . . . . . . . . . 33