THE VANISHING DISCOUNT PROBLEM FOR MONOTONE SYSTEMS OF HAMILTON … · 2019-06-20 ·...

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THE VANISHING DISCOUNT PROBLEM FOR

MONOTONE SYSTEMS OF HAMILTON-JACOBI EQUATIONS.

PART 2: NONLINEAR COUPLING

HITOSHI ISHII AND LIANG JIN

Abstract. We study the vanishing discount problem for a nonlinear monotone system of

Hamilton-Jacobi equations. This continues the first author’s investigation on the vanishing

discount problem for a monotone system of Hamilton-Jacobi equations. As in Part 1, we

introduce by the convex duality Mather measures and their analogues for the system, which we

call respectively Mather and Green-Poisson measures, and prove a convergence theorem for the

vanishing discount problem. Moreover, we establish an existence result for the ergodic problem.

Contents

1. Introduction 1

2. Preliminaries 5

3. Green-Poisson measures: in a regular case 9

4. Green-Poisson measures: the general case 19

5. A convergence result for the vanishing discount problem 25

6. Ergodic problem 29

Acknowledgement 32

References 33

1. Introduction

We consider the m-system of Hamilton-Jacobi equations

(Pλ) λvλi (x) +Hi(x,Dvλi (x), v

λ(x)) = 0 in Tn, i ∈ I,

where I := {1, . . . , m} with m ∈ N, λ is a nonnegative constant, called the discount factor in

terms of optimal control. Here Tn denotes the n-dimensional flat torus and H = (Hi)i∈I is a

2010 Mathematics Subject Classification. 35B40, 35F50, 49L25 .

Key words and phrases. systems of Hamilton-Jacobi equations, Mather measures, vanishing discount.

1

http://arxiv.org/abs/1906.07979v2

2 H. ISHII AND L. JIN

family of continuous Hamiltonians. The unknown in (Pλ) is an Rm-valued function vλ = (vλi )i∈I

on Tn and the above system can be written in the vector form as follows:

(Pλ) λvλ +H [vλ] = 0 in Tn.

We have used here the abbreviated expression H [vλ] to denote (Hi(x,Dvλi (x), v

λ(x)))i∈I. The

system is weakly coupled in the sense that every i-th equation depends on Dvλ only through

Dvλi but not on Dvλj , with j 6= i.

We are concerned with the vanishing discount problem for (Pλ), that is, the asymptotic

behavior of the solution vλ of (Pλ) as λ → 0+. Notably, the main concern is the convergence

of the whole family (vλ)λ>0 as λ→ 0+.

Recently, there has been a great interest in the vanishing discount problem concerned with

Hamilton-Jacobi equations and, furthermore, fully nonlinear degenerate elliptic PDEs. We

refer to [1, 6, 7, 11, 13, 16, 23, 24, 28–31, 41] for relevant work. The asymptotic analysis in these

papers relies heavily on Mather measures or their generalizations and, thus, it is considered

part of Aubry-Mather and weak KAM theories. For the development of these theories we refer

to [18, 20, 21] and the references therein. We refer to [8, 12, 14, 15, 19, 37–40] for the recent

development in the asymptotic analysis and weak KAM theory for systems of Hamilton-Jacobi

equations.

We are here interested in the case of systems of Hamilton-Jacobi equations. Davini and Zavi-

dovique in [16] have established a general convergence result for the vanishing discount problem

for (Pλ) when the coupling is linear and the coupling coefficients are constant. Adapting the

convex duality argument in [28] to the system, the first author of this paper has treated the

case of linear coupling, with the coupling coefficients depending on the space variable. In this

paper, we extend the scope of the previous work [26] and discuss the case of the system with

nonlinear coupling. Our argument is pretty much parallel to that in [26]. We refer for further

references to [16, 26].

Under our hypotheses described later, the limit function v0 of the solution vλ of (Pλ) satisfies

the system H [v0] = 0 and the principal difficulty in the asymptotic analysis lies in the fact that

the system H [u] = 0, or (P0), usually has multiple solutions of which the structure is not simple

in general. The critical role of Mather measures is indeed to identify the limit function v0 from

the solutions of H [u] = 0.

We assume throughout (see (H2) below) that the functions: Rn × Rm ∋ (p, u) 7→ Hi(x, p, u)

are convex for (x, i) ∈ Tn × I. We have chosen to make the convexity requirement on H in u

simply because of our technical limitations for studying the vanishing discount problem. The

VANISHING DISCOUNT PROBLEM 3

non-convexity issue for the vanishing discount problem has already been addressed in [7,24,45]

in the case of scalar equations. The paper [45] illustrates by examples that, in general, the

convergence of the whole family (vλ)λ>0 does not hold without the convexity of p 7→ Hi(x, p, u),

while [7, 24] indicate some possible generalizations beyond the convexity of u 7→ Hi(x, p, u).

On the other hand, under the convexity assumption, the system (Pλ) may be regarded as

the dynamic programming equation of optimal control of random evolutions, where the state

(x(t), i(t)) at time t is in Tn× I, x(t) is governed by a controlled ordinary differential equation,

and i(t) is a Markov process with controlled transition probability matrix. See [5] for this

application.

In this paper, we adopt the notion of viscosity solution to (Pλ), for which the reader may

consult [2, 4, 9, 10, 33].

Now, we give our main assumptions on the system (Pλ). Throughout we implicitly assume

that the functions Hi are continuous in Tn × R

n × Rm.

We assume that H is coercive, that is, for any i ∈ I and R > 0,

(H1) lim|p|→∞

inf(x,u)∈Tn×Bm

R

Hi(x, p, u) = ∞,

where BmR denotes the m-dimensional open ball with center at the origin and radius R.

This is a standard assumption, under which any upper semicontinuous subsolution of (Pλ)

is Lipschitz continuous on Tn.

We next assume that H is convex in the variables (p, u), that is,

(H2) for any (x, i) ∈ Tn × I, the function (p, u) 7→ Hi(x, p, u) is convex on R

n × Rm.

We assume that the Hamiltonian H is monotone in the variable u, that is, it satisfies

(H3)

{for any (x, p) ∈ T

n × Rn and u = (ui)i∈I, v = (vi)i∈I ∈ R

m, if uk − vk =

maxi∈I(ui − vi) ≥ 0, then Hk(x, p, u) ≥ Hk(x, p, v).

This is a natural assumption implying that (Pλ) should possess the comparison principle be-

tween a subsolution and a supersolution.

When the coupling is linear, that is, when H has the form

(1) Hi(x, p, u) = Gi(x, p) +∑

j∈Ibij(x)uj for i ∈ I,

condition (H3) is valid if and only if for each x ∈ Tn, the matrix B(x) is monotone matrix in

the following sense

(2) bij(x) ≤ 0 if i 6= j and∑

j∈Ibij(x) ≥ 0 for all i ∈ I,


which is equivalent to that for each x ∈ Tn,

(3) if u = (ui) ∈ Rm, k ∈ I, and uk = maxi∈I ui ≥ 0, then (B(x)u)k ≥ 0,

where (B(x)u)k denotes the k-th component of the m-vector B(x)u. We refer, for instance, to

[26, Lemma 3] for the equivalence of the last two conditions (2) and (3), while it is obvious in

the linear coupling case (1) that (H3) and (3) are equivalent each other.

When we deal with problem (P0), we use the assumption that

(H4) problem (P0) has a solution in C(Tn)m.

In the scalar case, that is, the case when m = 1 and the case when H(x, p, u) is independent of

u, a natural problem which replaces (P0) is the so-called ergodic problem that seeks a pair of

a constant c ∈ R and a function u ∈ C(Tn) such that u is a solution of

(4) H(x,Du) = c in Tn.

This problem is well-posed under (H1), which means that there exists such a pair (c, u) ∈R×Lip(Tn) and the constant c is unique. See for this [34]. For such (c, u), if we set Hc = H−c,then u is a solution of (P0), with H replaced by Hc. If vλ is a solution of (Pλ), with current

scalar Hamiltonian H(x, p), then the function vλ + λ−1c is a solution of (Pλ), with H replaced

by Hc. This way, if m = 1 and H(x, p, u) is independent of u, then the vanishing discount

problem can be transferred to the case where the limit problem (P0) admits a solution. Even in

the scalar case and much more in the case where m > 1, if H(x, p, u) depends genuinely on u,

then the reduction argument above does not work and the solutions of H [u] = c, where c 6= 0,

do not help investigate the vanishing discount problem for (Pλ).

The rest of this paper is organized as follows. In Section 2, we present some basic properties

concerning (H3) and a standard comparison and existence result of solutions of (Pλ) for λ > 0.

In Section 3, under additional hypotheses on the continuity of the Lagrangian L and the

compactness of the domain of L, we study Green-Poisson measures for our system, which

are crucial in our asymptotic analysis. Section 4 establishes the compactness of the support

of the Green-Poisson measures and gives a representation theorem for the solution of (Pλ),

with λ > 0, using the Green-Poisson measures. We establish the main result for the vanishing

discount problem in Section 5. In Section 6, we establish an existence theorem for the ergodic

problem.


2. Preliminaries

We use the symbol u ≤ v (resp., u ≥ v) for m-vectors u, v ∈ Rm to indicate ui ≤ vi (resp.,

ui ≥ vi) for all i ∈ I. Let ei denote the unit vector in Rm with unity as its i-th entry and let 1

denote the m-vector (1, . . . , 1) ∈ Rm.

Concerning the monotonicity of H , we give a basic lemma.

Lemma 1. Assume that H satisfies (H3). Let α ≥ 0.

(i) For all i ∈ I and (x, p, u) ∈ Tn × R

n × Rm,

Hi(x, p, u+ α1) ≥ Hi(x, p, u).

(ii) For all i, j ∈ I and (x, p, u) ∈ Tn × R

n × Rm, if i 6= j, then

Hj(x, p, u+ αei) ≤ Hj(x, p, u)

Proof. (i) Fix k ∈ I and (x, p, u) ∈ Tn × R

n × Rm. Set v = u+ α1 and note that

(v − u)k = α = maxi∈I

(v − u)i > 0.

By the monotonicity, we have

Hk(x, p, v) ≥ Hk(x, p, u).

That is,

Hk(x, p, u+ α1) ≥ Hk(x, p, u). �

(ii) Fix i, j ∈ I so that i 6= j. Let (x, p, u) ∈ Tn×R

n×Rm. Note that uk − (u+αei)k = 0 if

k 6= i and = −α < 0 if k = i, which can be stated that uj−(u+αei)j = maxk∈I[uk−(u+αei)k] =

0. By (H3), we have

Hj(x, p, u) ≥ Hj(x, p, u+ αei).

The following theorem is well-known: see [17, 27] for instance for a general background and

[26, the proof of Theorem 1] for some details how to adapt general results to (Pλ).

Theorem 2. Assume (H1) and (H3). Let λ > 0. Then there exists a unique solution vλ ∈Lip(Tn)m of (Pλ). Also, if v = (vi), w = (wi) are, respectively, upper and lower semicontinuous

on Tn and a subsolution and a supersolution of (Pλ), then v ≤ w on T

n.

We remark that if H satisfies (H1) and (H3), then so does the Hamiltonian Hε(x, p, u) :=

H(x, p, u) + εu, with ε > 0 and that if our problem (Pλ) has Hε in place of H , then the limit

problem (P0) reads εu +H [u] = 0 in Tn and has a unique solution due to Theorem 2. Thus,

the asymptotic analysis for the vanishing discount problem in such cases is fairly easy.


With reference to [27], we outline the proof of the theorem above.

Outline of proof. We choose a constant C > 0 so that

max(x,i)∈Tn×I

|Hi(x, 0, 0)| ≤ C.

Note by Lemma 1 that

Hi(x, 0,−λ−1C1) ≤ Hi(x, 0, 0) ≤ Hi(x, 0, λ−1C1) for x ∈ T

n.

By using this, it is easily checked that the functions f(x) = λ−1C1 and g(x) = −λ−1C1 are a

supersolution and subsolution of (Pλ) and satisfy f ≥ g on Tn.

Our assumption (H3) implies the quasi-monotonicity ofH in [27] as was shown in [27, Lemma

4.8]. By [27, Theorem3.3], the function z = (zi)i∈I on Tn given by

zi(x) = sup{ζi(x) : g ≤ ζ ≤ f in Tn, ζ is a subsolution of (Pλ)} for (x, i) ∈ T

n × I,

is a solution of (Pλ), in the sense that z∗ = (z∗i )i∈I and z∗ = (zi∗)i∈I, where each z∗i and zi∗ are

respectively the upper and lower semicontinuous envelope of zi, are respectively a subsolution

and a supersolution of (Pλ).

By the definition of z, it is easy to infer that for all i ∈ I, zi = z∗i and the function zi is

upper semicontinuous on Tn. By (H1) (the coercivity of H), we deduce that the function z is

Lipschitz continuous on Tn.

To see that the comparison between v and w, we apply [27, Theorem 4.7] to v and z as

well as z and w, to conclude that v ≤ z and z ≤ w on Tn, which implies that v ≤ w on T

n.

Here, the comparison theorem ([27, Theorem 4.7]) requires the regularity of H (see [27, (A.2)]),

which can be reduced just to the continuity of H since z is Lipschitz continuous on Tn. This

reduction of regularity of H is a standard observation and we leave it to the interested reader

to adapt the proof of [27, Theorem 4.7] to this case. �

Setting

Li(x, ξ, η) := sup(p,u)∈Rn×Rm

[ξ · p+ η · u−Hi(x, p, u)] for (x, i, ξ, η) ∈ Tn × I× R

n × Rm,

by the convex duality we have

Hi(x, p, u) = sup(ξ,η)∈Rn×Rm

[ξ · p+ η · u− Li(x, ξ, η)] for (x, i, p, u) ∈ Tn × I× R

n × Rm.

We call Li (resp., (Li)i∈I) the Lagrangian of Hi (resp., the Lagrangian of (Hi)i∈I). Similarly,

we call Hi (resp., (Hi)i∈I) the Hamiltonian of Li (resp., the Hamiltonian of (Li)i∈I).

It should be remarked that the functions Li are lower semicontinuous on Tn × R

n × Rm.


We examine the Lagrangian in the linear coupling case (1). By the definition of Li and a

simple manipulation, we deduce that

(5)

Li(x, p, ξ, η) = supp∈Rn

(ξ · p−Gi(x, p)) + supu∈Rm

(η · u− (B(x)u)i)

= supp∈Rn

(ξ · p−Gi(x, p)) + 0{bi(x)}(η),

where for x ∈ Tn, bi(x) := (bij)j∈I ∈ R

m and, for any sets X ⊂ Y , 0X denotes the indicator

function of X on Y given by

0X(y) =

0 if y ∈ X,

+∞ if y ∈ Y \X.

Lemma 3. Assume (H1)–(H2).

(i) We have

(6) Li(x, ξ, η) ≥ −Hi(x, 0, 0) for (x, i, ξ, η) ∈ Tn × I× R

n × Rm.

(ii) For any A > 0 there exists a constant CA such that

(7) Li(x, ξ, η) ≥ A|ξ| − CA for (x, i, ξ, η) ∈ Tn × I× R

n × Rm.

Proof. Fix i ∈ I. We have

Li(x, ξ, η) = sup(p,u)∈Rn×Rm

(ξ · p+ η · u−Hi(x, p, u)) ≥ −Hi(x, 0, 0),

and

Li(x, ξ, η) = sup(p,u)∈Rn×Rm

(ξ · p+ η · u−Hi(x, p, u)) ≥ A|ξ| −Hi(x,Aξ/|ξ|, 0) if ξ 6= 0.

Hence, setting

CA = sup(x,i,ξ)∈Tn×I×Bn

A

Hi(x, ξ, 0),

we obtain

Li(x, ξ, η) ≥ A|ξ| − CA. �

Lemma 3, (ii) asserts that the functions Li(x, ξ, η) have a superlinear growth as |ξ| → ∞.

We give a characterization of the monotonicity (H3) of (Hi)i∈I through (Li)i∈I. For k ∈ I,

we write

Yk := {(ηi)i∈I ∈ Rm : ηi ≤ 0 if i 6= k,

∑

i∈Iηi ≥ 0},

and domLk = {(x, ξ, η) : Lk(x, ξ, η) <∞}.


Proposition 4. Assume (H1)–(H2). Then (Hi)i∈I satisfies (H3) if and only if

domLi ⊂ Tn × R

n × Yi for all i ∈ I.

One can check directly that, in the linear coupling case (1), if (H1)–(H3) hold, then the

inclusion above is valid. Indeed, in this case, the coupling matrix B(x) = (bij(x)) satisfies

(2), which implies that (bij(x))j∈I ∈ Yi for all (x, i) ∈ Tn × I, and, by (5), we conclude that

domLi ⊂ Tn × R

n × Yi for all i ∈ I.

Proof. We assume first that (Hi)i∈I satisfies (H3). Fix any (x, k, ξ, η) ∈ Tn × I× R

n × Rm and

suppose that η = (ηi)i∈I 6∈ Yk. We have either ηj > 0 for some j 6= k or∑

i∈I ηi < 0.

Consider the case when ηj > 0 for some j 6= k. Let t > 0, and, by (ii) of Lemma 1, we have

Hk(x, p, 0) ≥ Hk(x, p, tej) for p ∈ Rn,

and hence,

ξ · p + η · tej −Hk(x, p, tej) ≥ ξ · p+ tηj −Hk(x, p, 0) for p ∈ Rn,

which implies that Lk(x, ξ, η) = ∞.

Consider next the case when∑

i∈I ηi < 0. For t > 0, we observe by (i) of Lemma 1 that

Hk(x, p, 0) ≥ Hk(x, p, t1) for all p ∈ Rn. Consequently,

ξ · p+ η · t1−Hk(x, p, t1) ≥ ξ · p− t∑

i∈Iηi −Hk(x, p, 0) for p ∈ R

n,

which shows that Lk(x, ξ, η) = ∞. We thus conclude that domLk ⊂ Tn × R

n × Yk.

Next, we assume that domLi ⊂ Tn × R

n × Yi for all i ∈ I. It is obvious that for any

(x, i, p, u) ∈ Tn × I× R

n × Rm,

Hi(x, p, u) = sup(ξ,η)∈Rn×Yi

[ξ · p+ η · u− Li(x, ξ, η)].

Fix any (x, p) ∈ Tn × R

n and u, v ∈ Rm. Assume that for some k ∈ I,

(u− v)k = maxi∈I

(u− v)i ≥ 0,

which can be stated as

(u− v)k − (u− v)i ≥ 0 for i ∈ I and (u− v)k ≥ 0.


Let η = (ηi)i∈I ∈ Yk. Multiplying the first inequality above by ηi, with i 6= k, we get

0 ≥∑

i 6=kηi[(u− v)k − (u− v)i] =

∑

i∈Iηi[(u− v)k − (u− v)i]

= (u− v)k∑

i∈Iηi −

∑

i∈Iηi(u− v)i.

Since (u− v)k ≥ 0 and∑

i∈I ηi ≥ 0, we infer from the above that η · u ≥ η · v. Thus, we have

Hk(x, p, u) = sup(ξ,η)∈Rn×Yk

[ξ · p+ η · u− Lk(x, ξ, η)]

≥ sup(ξ,η)∈Rn×Yk

[ξ · p+ η · v − Lk(x, ξ, η)] = Hk(x, p, v).

This shows that (Hi)i∈I is monotone, which completes the proof. �

3. Green-Poisson measures: in a regular case

In what follows, given a topological space X , B(X) denotes the σ-algebra of Borel sets in X ,

and M(X) and M+(X) denote, respectively, the spaces of Borel measures on X having bounded

variation and of nonnegative finite Borel measures on X . Also, Cb(X) denotes the space of

bounded continuous functions on X .

For any ν ∈ M(Tn × Rn × R

m) and integrable function φ on Tn × R

n × Rm with respect to

ν, we write

〈ν, φ〉 =∫

Tn×Rn×Rm

φ(x, ξ, η) ν(dxdξdη).

Similarly, for any ν = (νi)i∈I ∈ M(Tn × Rn × R

m)m and Borel function φ = (φi)i∈I on Tn ×

Rn × R

m, we write

〈ν, φ〉 =∑

i∈I〈νi, φi〉

if φi is integrable with respect to νi for every i ∈ I.

For λ > 0, we define the function Sλ : Rm → R by Sλ(η) = λ+∑

i∈I ηi for η = (ηi)i∈I ∈ Rm,

and, we write Pλ for the set of all µ = (µi)i∈I ∈ M+(T

n × Rn × R

m)m such that

(8) 〈µi, |ξ|+ |η|〉 <∞ for all i ∈ I and⟨µ, Sλ 1

⟩= 1,

where |ξ| + |η| denotes the function: Rn × R

m ∋ (ξ, η) 7→ |ξ| + |η| ∈ R. Note that Sλ1 is the

function: Rm ∋ η 7→ Sλ(η)1 ∈ Rm, which can be regarded as a function of (x, ξ, η) ∈ T

n×Rn+m.

We write P0 for the set of all µ = (µi)i∈I ∈ M+(T

n × Rn × R

m)m such that

(9) 〈µi, |ξ|+ |η|〉 <∞ for all i ∈ I and 〈µ, 1〉 ≤ 1.


Similarly to the standard definition [11,16] (see also [35,36]) of Mather measures, we introduce

the closed measures as follows. We call any µ ∈ P0 a closed measure (associated with λ = 0)

provided it satisfies

(10) 〈µ, ξ ·Dψ + η · ψ 1〉 = 0 for all ψ ∈ C1(Tn)m if λ = 0,

and denote by C(0) the set of all such closed measures associated with λ = 0. In (10) above, and

henceforth, we use the notation that the functions: Tn × R

n ∋ (x, ξ) 7→ (ξ · Dψi(x))i∈I ∈ Rm

and Tn × R

m ∋ (x, η) 7→ η · ψ(x) ∈ R are denoted by ξ · Dψ and η · ψ, respectively. For

(z, k, λ) ∈ Tn × I × (0, ∞), we call any µ ∈ P

λ a closed measure (associated with (z, k, λ))

provided it satisfies

(11) 〈µ, ξ ·Dψ + η · ψ 1+ λψ〉 = ψk(z) for all ψ = (ψi)i∈I ∈ C1(Tn)m if λ > 0.

The set of all closed measures associated with (z, k, λ) is denoted by C(z, k, λ).

We now introduce the following working hypothesis.

(H5)

There exist nonempty, compact, convex sets K1 ⊂ Rn and K2 ⊂ R

m such that K1 is

a neighborhood of the origin and such that for i ∈ I,

Li ∈ C(Tn ×K1 × (K2 ∩ Yi)),

Hi(x, p, u) = sup(ξ,η)∈K1×(K2∩Yi)

[ξ · p+ η · u− Li(x, ξ, η)] for (x, p, u) ∈ Tn × R

n × Rm.

Assuming (H5) in addition, we have

Li(x, ξ, η) = Li(x, ξ, η) + 0K1×(K2∩Yi)(ξ, η) for (x, ξ, η) ∈ Tn × R

n × Rm.

We remark that under (H5), the functions Hi(x, p, u) grows at most linearly as |(p, u)| → ∞.

Theorem 5. Assume (H1)– (H3) and (H5). Let (z, k, λ) ∈ Tn × I × (0, ∞) and let vλ =

(vλi )i∈I ∈ C(Tn)m be the solution of (Pλ). Then there exists µ ∈ C(z, k, λ) such that

(12) vλk (z) = 〈µ, L〉 = minν∈C(z,k,λ)

〈ν, L〉 .

Remark that, thanks to Proposition 4, if (Hi)i∈I satisfies (H1), (H2) and (H5), then it has

the property (H3) as well.

The theorem above and Theorem 12, stated later, are generalizations of the previous results

in [6, 11, 16, 23, 24, 26, 28, 29, 41]

Our proof of Theorem 5 is close to the ones in [28, 29] in the technicality and is similar to

the ones in [22, 30] in the use of duality. It depends crucially on a minimax theorem, and, in


the application of the minimax theorem, it is essential to make the set compact on which the

measures µ ∈ C(z, k, λ), having the property 〈µ, L〉 < ∞, are supported. The condition (H5)

realizes all such measures µ to be supported on the compact set∏

i∈I Tn×K1×(K2∩Yi), which

we mean that supp µi ⊂ Tn ×K1 × (K2 ∩ Yi) for all i ∈ I.

In the next section, we remove the restriction (H5) on L adopted in Theorem 5 by appealing

the fact that the solution vλ of (Pλ) is Lipschitz continuous.

We call a Green-Poisson measure any measure µ ∈ C(z, k, λ) that is a minimizer of the most

right hand of (12).

For the proof of Theorem 5, we assume henceforth (H1)–(H3) and (H5), and introduce the

following notation. Set Zi = K1 × (K2 ∩ Yi) for i ∈ I and note that for any i ∈ I, Zi is

a compact convex subset of Rn × Rm. Let λ > 0 and F(λ) denote the set of all (φ, u) ∈

∏i∈I C(T

n×Zi)×C(Tn)m such that for any (x, i) ∈ Tn× I, φi(x, ξ, η) is convex in the variable

(ξ, η) on Zi and such that u is a subsolution of λu+Hφ[u] = 0 in Tn, where Hφ = (Hφ,i)i∈I is

given by

(13) Hφ,i(x, p, u) = max(ξ,η)∈Zi

(ξ · p+ η · u− φi(x, ξ, η)).

Here, we note by the compactness of Zi that Hφ,i is continuous on Tn × R

n × Rm. Also, if

we identify φi with the function φi on Tn × R

n × Rm given by

(14) φi(x, ξ, η) =

φi(x, ξ, η) if (ξ, η) ∈ Zi,

+∞ otherwise,

then (13) reads

Hφ,i(x, p, u) = max(ξ,η)∈Rn×Rm

(ξ · p+ η · u− φi(x, ξ, η)).

Notice that the functions φi(x, ξ, η) defined by (14) are lower semicontinuous in the variable

(x, ξ, η) and convex in the variable (ξ, η). By the convex duality, we see that φi is the Lagrangian

of Hφ,i. Arguing similarly to the proof of (ii) of Lemma 3, with (φi, Hφ,i) in place of (Hi, Li),

we see that Hφ satisfies (H1). By Proposition 4, we easily see that Hφ satisfies (H3). Moreover,

it is clear that Hφ satisfies (H5).

In what follows, for any (φ, u) ∈ F(λ), we identify the function φi on Tn × Zi with φi

according to the situation. In particular, if vλ ∈ C(Tn) is the solution of (Pλ), then we have

(L, vλ) ∈ F(λ). Note in addition that if φ = 0 ∈ ∏i∈I C(T

n × Zi), then φi(x, ξ, η) is convex in

(ξ, η) for all i ∈ I and Hφ(·, 0, 0) = 0. Hence, (0, 0) ∈ F(λ).


For (z, k, λ) ∈ Tn × I× (0, ∞), we set

G(z, k, λ) = {φ− uk(z)Sλ(η)1 : (φ, u) ∈ F(λ)},

Pλ(K1, K2) = {µ ∈ P

λ : supp µi ⊂ Tn × Zi for all i ∈ I},

G ′(z, k, λ) = {µ ∈ Pλ(K1, K2) : 〈µ, f〉 ≥ 0 for all f ∈ G(z, k, λ)}.

We recall that, by definition, the support of measure µ ∈ M(X) for a topological space X is

defined as the closed set

supp µ = X \⋃

{U ⊂ X : |µ|(U) = 0, U is open},

where |µ| denotes the total variation of the measure µ. Accordingly, if X has a countable basis

of its topology, we have

µ(X \ suppµ) = 0,

and ∫

X

φ(x)µ(dx) =

∫

supp µ

φ(x)µ(dx) for all φ ∈ Cb(X), µ ∈ M(X).

In what follows, when Q ⊂ X is a closed subset of a topological space X with a countable

basis, we may identify µ ∈ M(X) satisfying supp µ ⊂ Q with its restriction µ|Q to Q, defined

by

µ|Q(A) = µ(A) for all A ∈ B(Q).

Then we have

(15)

∫

X

φ(x)µ(dx) =

∫

Q

φ(x)µ|Q(dx) for all φ ∈ Cb(X).

Recalling that (L, vλ) ∈ F(λ), where vλ is a solution of (Pλ), we easily infer that

vλk (z) ≤ 〈µ, L〉 for all µ ∈ G ′(z, k, λ).

Since the functions Li are bounded from below, for any µ ∈ Pλ, the inequality 〈µ, L〉 < ∞

always makes sense and, by (H5), we have 〈µ, L〉 < ∞ if and only if supp µi ⊂ Tn × Zi for all

i ∈ I. Hence, we have

(16) Pλ(K1, K2) = {µ ∈ P

λ : 〈µ, L〉 <∞}.

Lemma 6. The set F(λ) is a convex cone in∏

i∈I C(Tn × Zi) × C(Tn)m with vertex at the

origin.


Proof. Recall [3, Remark 2.5] that for any u ∈ Lip(Tn)m, u is a subsolution of

λu+H [u] = 0 in Tn

if and only if for all i ∈ I,

λui(x) +Hi(x,Dui(x), u(x)) ≤ 0 a.e. in Tn,

and by the coercivity (H1) that for any (φ, u) ∈ F(λ), we have u ∈ Lip(Tn)m.

Fix (φ, u), (ψ, v) ∈ F(λ) and t, s ∈ [0,∞). Fix i ∈ I and observe that

λui(x) +Hφ,i(x,Dui(x), u(x)) ≤ 0 a.e. in Tn,

λvi(x) +Hψ,i(x,Dvi(x), v(x)) ≤ 0 a.e. in Tn,

which imply that there is a set N ⊂ Tn of Lebesgue measure zero such that

λui(x) + ξ ·Dui(x) + η · u(x) ≤ φi(x, ξ, η) for all (x, ξ, η) ∈ (Tn \N) × Zi,

λvi(x) + ξ ·Dvi(x) + η · u(x) ≤ ψi(x, ξ, η) for all (x, ξ, η) ∈ (Tn \N) × Zi.

Multiplying the first and second by t and s, respectively, adding the resulting inequalities and

setting w = tu+ sv, we obtain

λwi(x) + ξ ·Dwi(x) + η · w(x) ≤ (tφi + sψi)(x, ξ, η) for all (x, ξ, η) ∈ (Tn \N) × Zi,

which implies that t(φ, u) + s(ψ, v) ∈ F(λ). �

Lemma 7. Let (z, k, λ) ∈ Tn × I × (0, ∞) and µ = (µi)i∈I ∈ P

λ(K1, K2). Then, we have

µ ∈ C(z, k, λ) if and only if µ ∈ G ′(z, k, λ).

Proof. Assume first that µ ∈ C(z, k, λ). Fix any (φ, u) ∈ F(λ) and recall that, in the viscosity

sense,

λu+Hφ[u] ≤ 0 in Tn.

Thanks to the coercivity property (H1) of Hφ, u is Lipschitz continuous on Tn. In view of the

continuity of Hφ and the convex property (H2) of Hφ, mollifying u, we may choose, for each

ε > 0, a function uε ∈ C1(Tn)m such that λuε+Hφ[uε] ≤ ε 1 in T

n and ‖u− uε‖∞ < ε. Hence,

we have

λuε + ξ ·Duε(x) + η · uε(x) 1 ≤ φ(x, ξ, η) + ε 1.

Using (11) and integrating the inequality above with respect to µ, we obtain

uεk(z) = 〈µ, ξ ·Duε + η · uε 1 + λuε〉 ≤ 〈µ, φ+ ε 1〉


and, after sending ε→ 0,

0 ≤ 〈µ, φ〉 − uk(z) =⟨µ, φ− uk(z)S

λ1⟩,

which implies, together with the assumption that supp µi ⊂ Tn × Zi for all i ∈ I, that µ ∈

G ′(z, k, λ).

Next, we assume that µ ∈ G ′(z, k, λ). Fix any ψ ∈ C1(Tn)m, set φ = ξ ·Dψ + η · ψ + λψ,

which is a function on Tn ×R

n+m, and observe that λψ+Hφ[ψ] ≤ 0 in Tn, i.e., (φ, ψ) ∈ F(λ).

Hence, by the definition of G ′(z, k, λ), we have

0 ≤⟨µ, φ− ψk(z)S

λ1⟩= 〈µ, ξ ·Dψ + η · ψ1+ λψ〉 − ψk(z).

The inequality above holds also for −ψ in place of ψ, which reads

0 ≥ 〈µ, ξ ·Dψ + η · ψ1+ λψ〉 − ψk(z).

Thus, we have

〈µ, ξ ·Dψ + η · ψ1+ λψ〉 = ψk(z),

and conclude that µ ∈ C(z, k, λ). �

Lemma 8. Let i ∈ I, λ > 0, and (x, ξ, η) ∈ Tn × Zi, and let δ(x,ξ,η) denote the Dirac measure

at (x, ξ, η). Then,(Sλ

)−1δ(x,ξ,η)ei is a member of Pλ(K1, K2).

Proof. Note first that Sλ(η) ≥ λ > 0 for all η ∈ K2 ∩ Yi. It follows immediately that(Sλ

)−1δ(x,ξ,η) ∈ M

+(Tn × Zi), suppSλδ(x,ξ,η) = {(x, ξ, η)} ⊂ T

n × Zi,

⟨(Sλ

)−1δ(x,ξ,η)ei, (|ξ|+ |η|)1

⟩=

(Sλ(η)

)−1(|ξ|+ |η|) <∞,

and⟨(Sλ

)−1δ(x,ξ,η)ei, S

λ1⟩= 1. Thus, we we see that

(Sλ

)−1δ(x,ξ,η)ei ∈ P

λ(K1, K2). �

For the reader’s convenience, we state a minimax theorem ([44, Corollary 2]).

Proposition 9. Let K and Y be convex subsets of vector spaces. Assume in addition that K

is a compact space. Let f : K × Y → R be a function satisfying:

(i) For each y ∈ Y , the function: x 7→ f(x, y) is lower semicontinuous and convex on K.

(ii) For each x ∈ K, the function: y 7→ f(x, y) is concave on Y .

Then

supy∈Y

minx∈K

f(x, y) = minx∈K

supy∈Y

f(x, y).


We remark that in [44, Corollary 2], it is assumed that K is a convex compact subset of a

topological vector space X , but in its proofs, the compatibility of the linear structure and the

topological structure (i.e., the continuity of addition and scalar multiplication) of X is not used

and the proposition above is valid.

In the application below of Proposition 9, we take K to be a bounded subset of the Banach

space M(Σ) with the total variation norm, where Σ is a compact subset of Tn × Rn+m.

Let Σ be a compact subset of Tn × Rn+m. By the Riesz representation theorem, for each

F ∈ C(Σ)∗, there exists a unique (regular) Borel measure µ on Σ such that for all φ ∈ C(Σ),

F (φ) =

∫

s∈Σφ(s)µ(ds) = 〈µ, φ〉 .

The mapping ιΣ of F ∈ C(Σ)∗ to µ ∈ M(Σ), given above, is an isomorphism between two Ba-

nach spaces. Through the mapping ιΣ : C(Σ)∗ → M(Σ), the weak star convergence corresponds

to the weak convergence of measures.

Thanks to the Banach-Alaoglu theorem, we know that any closed ball B (in the strong

topology) of C(Σ)∗, equipped with the weak star topology, is a compact metrizable space.

Moreover, if B is such a ball and N is a closed subset (in the weak star topology) of B, then

N is a compact subset of B. These say that if D is a closed ball (in the total variation norm)

of M(Σ), then D is a compact metrizable space with the topology of the weak convergence

of measures and so is any K ⊂ D that is sequentially closed in the topology of the weak

convergence of measures.

The following lemma is a simple consequence of the discussion above.

Lemma 10. Let a > 0 and Σ = (Σi)i∈I be a collection of compact subsets Σi of Tn × R

n+m.

Let P(Σ, a) denote the collection of µ = (µi) ∈ M+(Tn×R

n+m)m such that supp µi ⊂ Σi for all

i ∈ I and such that 〈µ, 1〉 ≤ a. Then, P(Σ, a) is a compact metrizable space with the topology

of weak convergence of measures.

It is to be noticed that in the lemma above, P(Σ, a) is sequentially compact.

Proof. In view of (15), it is clear that any sequence of measures µq = (µqi ) ∈ P(Σ, a) on

Tn × R

n+m, with q ∈ N, converges to µ = (µi) weakly in the sense of measures (i.e., in the

topology of weak convergence of measures) if and only if, for each i ∈ I, the sequence of measures

µqi |Σion Σi converges to µi|Σi

and µi|Tn×Rn+m\Σi= 0. Note also that µi|Tn×Rn+m\Σi

= 0 if and

only if supp µi ⊂ Σi.

Thus, we need only to prove that the set P := {µ|Σ : µ ∈ P(Σ, a)}, where µ|Σ := (µi|Σi)i∈I,

is a compact mterizable space with the topology of weak convergence of measures.


As noted prior to the lemma, it is enough to prove that P is a subset of a closed ball of∏

i∈I M(Σi) in the norm topology and it is closed in the weak convergence of measures.

Recall that the Banach space∏

i∈I M(Σi) has the total variation norm ‖ν‖ =∑

i∈I |νi|(Σi)for ν = (νi). If µ = (µi) ∈ M+(T

n × Rn+m)m and supp µi ⊂ Σi for all i ∈ I, then

〈µ, 1〉 =∑

i∈Iµi(Σi) = ‖µ|Σ‖.

This shows that the closed ball B := {ν = (νi) ∈∏

i∈I M(Σi) : ‖ν‖ ≤ a} contains P.

It remains to show that P is closed in the weak convergence of measures. For this, as noted at

the beginning, we need only to prove that P(Σ, a) is closed in the weak convergence of measures.

Now, let µj = (µji ) ∈ P(Σ, a) for all j ∈ N and assume that the sequence of µj converges

weakly in the sense of measures to µ = (µi) ∈ M(Tn × Rn+m)m. We already know that

suppµi ⊂ Σi for all i ∈ I and that the sequence of µj|Σ converges to µ|Σ ∈ B weakly in the

sense of measures. It follows that

a ≥ ‖µ|Σ‖ =∑

i∈I|µi|(Σi),

and, moreover, that for any i ∈ I and nonnegative function ψ ∈ Cb(Tn × R

n+m),

〈µi, ψ〉 = limj→∞

⟨µji , ψ

⟩≥ 0.

From these, we see that µ ∈ M+(T×Rn+m)m and 〈µ, 1〉 ≤ a, and conclude that µ ∈ P(Σ, a). �

Lemma 11. Let λ > 0 and Σ = (Σi)i∈I be a collection of compact subsets Σi of Tn × R

n ×(Rm ∩ Yi). Let P

λ(Σ) denote the collection of all µ = (µi) ∈ Pλ such that supp µi ⊂ Σi for

every i ∈ I. Then, Pλ(Σ) is a compact metrizable space with the topology of weak convergence

of measures.

Proof. Let P(Σ, a) denote the set defined in Lemma 10 for a > 0. For µ = (µi) ∈ Pλ(Σ), since

suppµi are compact for all i ∈ I, it is clear that 〈µ, (|ξ|+ |η|)1〉 < ∞. Since Sλ(η) ≥ λ for all

η ∈ ⋃i∈I Yi, if µ = (µi) ∈ P

λ(Σ), then

1 =⟨µ, Sλ1

⟩=

∑

i∈I

∫

Σi

Sλ(η)µi(dxdξdη) ≥ λ∑

i∈Iµi(Σi) = λ 〈µ, 1〉 ,

which implies that Pλ(Σ) ⊂ P(Σ, 1/λ).

It remains to prove that Pλ(Σ) is a closed subset of P(Σ, 1/λ). Let µj = (µji ) ∈ P

λ(Σ)

for j ∈ N. Assume that the sequence (µj) converges weakly in the sense of measures to

µ = (µi) ∈ P(Σ, 1/λ).


For the proof of the lemma, we need only to show that⟨µ, Sλ1

⟩= 1. We easily check that

⟨µ, Sλ1

⟩=

∑

i∈I

∫

Σi

Sλ(η)µi(dxdξdη) = limj→∞

∑

i∈I

∫

Σi

Sλ(η)µji (dxdξdη) = limj→∞

⟨µj, Sλ1

⟩= 1,

which finishes the proof. �

It is a consequence of the lemma above that Pλ(K1, K2) is a compact metrizable space with

the topology of weak convergence of measures.

Proof of Theorem 5. In view of (16) and Lemma 7, it is enough to prove that

(17) vλk (z) = minµ∈G ′(z,k,λ)

〈µ, L〉 .

We intend to show that

(18) sup(φ,u)∈F(λ)

infν∈Pλ(K1,K2)

⟨ν, L− φ+ (uk(z)− vλk (z))S

λ1⟩= 0.

We postpone the proof of (18) and, assuming temporarily that (18) is valid, we prove that

(17) holds.

To this end, we see easily that Pλ(K1, K2) is a convex subset of a vector space M(Tn×Rn+m)

and that, by Lemma 6, F(λ) is a convex subset of∏

i∈I C(Tn×Zi)×C(Tn)m. Observe as well

that the functional:

Pλ(K1, K2) ∋ ν 7→


λ1⟩∈ R

is convex and continuous, in the topology of weak convergence of measures for any (φ, u) ∈ F(λ),

and the functional:

F(λ) ∋ (φ, u) 7→⟨ν, L− φ+ (uk(z)− vλk (z))S

λ1⟩∈ R

is concave, as well as continuous, for any ν ∈ Pλ(K1, K2).

By Lemma 11, the set Pλ(K1, K2) is a compact space with the topology of weak convergence

of measures. Hence, we may apply the minimax theorem (Proposition 9 or [43, 44]), to deduce

from (18) that

(19)

0 = sup(φ,u)∈F(λ)

minν∈Pλ(K1,K2)


λ1⟩

= minν∈Pλ(K1,K2)

sup(φ,u)∈F(λ)


λ1⟩.


Observe by using the cone property of F(λ) that

sup(φ,u)∈F(λ)

⟨ν, uk(z)S

λ1− φ⟩=

0 if ν ∈ G ′(z, k, λ),

∞ if ν ∈ Pλ(K1, K2) \ G ′(z, k, λ).

This and (19) yield

0 = minν∈Pλ(K1,K2)

sup(φ,u)∈F(λ)


λ1⟩

= minν∈G ′(z,k,λ)

⟨ν, L− vλk (z)S

λ1⟩= min

ν∈G ′(z,k,λ)〈ν, L〉 − vλk (z),

which proves (17).

It remains to show (18). Note that

sup(φ,u)∈F(λ)

infν∈Pλ(K1,K2)


λ1⟩

≥ infν∈Pλ(K1,K2)


λ1⟩ ∣∣∣

(φ,u)=(L,vλ)= 0.

Hence, we only need to show that

(20) sup(φ,u)∈F(λ)

infν∈Pλ(K1,K2)


λ1⟩≤ 0.

For this, we argue by contradiction and thus suppose that (20) does not hold. Accordingly,

we have

sup(φ,u)∈F(λ)

infν∈Pλ(K1,K2)


λ1⟩> ε

for some ε > 0. We may select (φ, u) ∈ F(λ) so that

infν∈Pλ(K1,K2)


λ1⟩> ε.

That is, for any ν ∈ Pλ(K1, K2), we have


λ1⟩> ε =

⟨ν, εSλ1

⟩.

According to Lemma 8, the measure (Sλ)−1δ(x,ξ,η)ei is in Pλ(K1, K2) for every (x, ξ, η) ∈

Tn × Zi and i ∈ I. Plugging all such ν = (Sλ)−1δ(x,ξ,η)ei ∈ P

λ(K1, K2) into the above, we find

that

(Li − φi)(x, ξ, η) + (uk(z)− vλk (z)− ε)Sλ(η) > 0 for all (x, ξ, η) ∈ Tn × Zi, i ∈ I.


Hence, setting w := u− (uk(z)− vλk (z)− ε)1, we have

λwi(x) + ξ · p+ η · w(x)− Li(x, ξ, η)

= λui(x) + ξ · p+ η · u(x)− (uk(z)− vλk (z)− ε)Sλ(η)− Li(x, ξ, η)

< λui(x) + ξ · p+ η · u(x)− φi(x, ξ, η)

for all (x, p, ξ, η) ∈ Tn × R

n × Zi and i ∈ I. This ensures that w is a subsolution of

λw +H [w] = 0 in Tn.

By Theorem 2, we get u(x)− (uk(z)− vλk (z)− ε)1 ≤ vλ(x) for all x ∈ Tn. The k-th component

of the last inequality, evaluated at x = z, yields an obvious contradiction, which proves that

(20) holds. �

4. Green-Poisson measures: the general case

We now remove the hypothesis (H5) in Theorem 5 and establish the following theorem.

Theorem 12. Assume (H1)–(H3). Let (z, k, λ) ∈ Tn × I × (0, ∞) and vλ ∈ C(Tn)m be the

solution of (Pλ). Then there exists µ ∈ C(z, k, λ) such that

(21) vλk (z) = 〈µ, L〉 = minν∈C(z,k,λ)

〈ν, L〉 .

The theorem above guarantees the existence of a Green-Poisson measure associated with any

(z, k, λ) ∈ Tn × I× (0, ∞).

In what follows we fix a (z, k, λ) ∈ Tn × I × (0, ∞). According to Theorem 2, the unique

solution of (Pλ) is Lipschitz continuous on Tn. With this in mind, we fix a constant C > 0 and

consider the condition that

(22) |vλ(x)| + |Dvλ(x)| ≤ C a.e. x ∈ Tn.

We choose a function h ∈ C1(Rn × Rm) so that

(23)

h is nonnegative and convex on Rn × R

m,

h(p, u) = 0 if and only if |p|+ |u| ≤ C,

lim|p|+|u|→∞

(|p|+ |u|)−1h(p, u) = ∞.

Also, we choose a compact convex set Q ⊂ Rn+m such that for all (x, i, p, u) ∈ T

n × I× Rn+m,

(24) ∂(p,u)Hi(x, p, u) ⊂ Q if |p|+ |u| ≤ C,

where ∂(p,u)Hi denotes the subdifferential of the convex function: (p, u) 7→ Hi(x, p, u).


Theorem 13. Assume (H1)–(H3). Let vλ be the solution of (Pλ) and assume that (22) is

satisfied for some constant C > 0. Let Q be a compact convex subset of Rn+m such that (24)

holds. Assume that there exists µ ∈ C(z, k, λ) such that

vλk (z) = 〈µ, L〉 .

Then

supp µi ⊂ Tn × [Q ∩ (Rn × Yi)] for i ∈ I.

We recall some basic properties related to the subdifferentials of H and L.

Lemma 14. Assume (H2). Let (x, i) ∈ Tn × I.

(i) We have

∂(p,u)Hi(x, p, u) 6= ∅ for (p, u) ∈ Rn × R

m.

(ii) Let (p, u), (ξ, η) ∈ Rn+m. The following three statements are equivalent each other.

(a) (ξ, η) ∈ ∂(p,u)Hi(x, p, u).

(b) (p, u) ∈ ∂(ξ,η)Li(x, ξ, η).

(c) Hi(x, p, u) + Li(x, ξ, η) = ξ · p+ η · u.

Proof. (i) Since (p, u) 7→ Hi(x, p, u) is continuous and convex in Rn+m, it is locally Lipschitz

continuous (see [25, Theorem B.3]) and hence almost everywhere differentiable (see [25, The-

orem F.1]) in Rn+m. Fix any (p, u) ∈ R

n+m and choose a sequence of points (pk, uk) ∈ Rn+m

converging to (p, u) such that (p, u) 7→ Hi(x, p, u) is differentiable at (pk, uk) for all k ∈ N. Set

(ξk, ηk) = Dp,uHi(x, pk, uk) for k ∈ N. The local Lipschitz continuity of Hi(x, ·, ·) allows us to

assume that (ξk, ηk)k∈N is bounded and, moreover, convergent to some (ξ0, η0) ∈ Rn+m after

passing to a subsequence. Since

Hi(x, pk + q, uk + r) ≥ Hi(x, p

k, uk) + ξk · q + ηk · r for (q, r) ∈ Rn+m,

sending k → ∞ yields

Hi(x, p+ q, u+ r) ≥ Hi(x, p, u) + ξ0 · q + η0 · r for (q, r) ∈ Rn+m,

which shows that (ξ0, η0) ∈ ∂(p,u)Hi(x, p, u) and ∂(p,u)Hi(x, p, u) 6= ∅.We here skip to prove (ii) and leave it to the reader to consult [42, Theorem 23.5] or [25,

Theorem B.2]. �

Lemma 15. Assume (H1)–(H3). Let λ ∈ [0, ∞), and let u ∈ Lip(Tn) be a subsolution of

(Pλ). If λ > 0, then uk(z) ≤ 〈µ, L〉 for all (z, k) ∈ Tn × I and µ ∈ C(z, k, λ), and, if λ = 0,

then 0 ≤ 〈µ, L〉 for all µ ∈ C(0).


We remark that, in the above, 〈µ, L〉 can be +∞.

The proof below is almost identical to the first part of the proof of Lemma 7.

Proof. In view of the continuity and convex property of H , mollifying u, we may choose, for

each ε > 0, a function uε ∈ C1(Tn)m such that λuε + H [uε] ≤ ε 1 in Tn and ‖u − uε‖∞ < ε.

Hence, we have

λuε + ξ ·Duε(x) + η · uε(x) 1 ≤ L(x, ξ, η) + ε 1.

When λ > 0, fixing (z, k) ∈ Tn × I, recalling (11), and integrating the inequality above with

respect to µ ∈ C(z, k, λ), we obtain

uεk(z) = 〈µ, ξ ·Duε + η · uε 1+ λuε〉 ≤ 〈µ, L+ ε 1〉 .

Similarly, if λ = 0, then we get for any µ ∈ C(0),

0 = 〈µ, ξ ·Duε + η · uε〉 ≤ 〈µ, L+ ε 1〉 .

Taking the limit as ε→ 0, we finish the proof. �

In the proof below, an essential step is to construct a new Hamiltonian, say, H satisfying

(H1)–(H3) such that vλ is a solution of (Pλ), with H replaced by H, and such that, if |p|+ |u| >C, then H(x, p, u) > H(x, p, u). Notice that the function H(x, p, u) + h(p, u), with h satisfying

(23), on Tn × R

n+m does not satisfy the monotonicity (H3).

Proof of Theorem 13. Let h ∈ C1(Rn × Rm) be a function having the properties in (23). We

set Gh(x, p, u) = H(x, p, u) + h(p, u) for (x, p, u) ∈ Tn × R

n × Rm. Let Kh = (Kh

i )i∈I be the

Lagrangian of Gh, and, since Gh grows superlinearly as |p| + |u| → ∞, we see that Kh ∈C(Tn × R

n × Rm)m. Note that

Gh ≥ H and Kh ≤ L on Tn × R

n × Rm.

According to Proposition 4, Gh does not satisfy (H3), and we need to modify Gh, to remove

the drawback. We note by Proposition 4 that

L(x, ξ, η) + 0Y (η) = L(x, ξ, η) for (x, ξ, η) ∈ Tn × R

n × Rm,

where 0Y := (0Yi)i∈I. Hence, we have

Lh(x, ξ, η) := Kh(x, ξ, η) + 0Y (η) ≤ L(x, ξ, η) for (x, ξ, η) ∈ Tn × R

n × Rm.

Let Hh = (Hhi )i∈I be the Hamiltonian of Lh, and note that

H ≤ Hh ≤ Gh on Tn × R

n × Rm.


In particular, we have

H(x, p, u) = Hh(x, p, u) = Gh(x, p, u) if |p|+ |u| ≤ C,

which shows, together with (22), that vλ is a solution of λu+Hh[u] = 0 in Tn. It is clear that

Hh satisfies (H1) and (H2). Moreover, Hh satisfies (H3) due to Proposition 4.

Now, since L ≥ Lh on Tn × R

n × Rm, it follows immediately that

vλk (z) = 〈µ, L〉 ≥⟨µ, Lh

⟩.

Since λvλ +Hh[vλ] = 0 in Tn, thanks to Lemma 15, we get

vλk (z) ≤⟨µ, Lh

⟩.

Combining these yields

vλk (z) = 〈µ, L〉 =⟨µ, Lh

⟩.

Consequently, we have

(25)⟨µ, L− Lh

⟩= 0 and L ≥ Lh.

Noting that for all i ∈ I, Yi is a closed subset of Rm and Li = ∞ on Tn × R

n × (Rm \ Yi),and Li − Lhi is lower semicontinuous on T

n × Rn × Yi, we easily deduce from (25) that

supp µi ⊂ {(x, ξ, η) ∈ Tn × R

n × Yi : Li(x, ξ, η) = Lhi (x, ξ, η)}.

It remains to show that for all i ∈ I,

(26) {(x, ξ, η) ∈ Tn × R

n × Yi : Li(x, ξ, η) = Lhi (x, ξ, η)} ⊂ Tn ×Q.

To do this, we fix i ∈ I and

(x, ξ, η) ∈ Tn × R

n × Yi

such that Li(x, ξ, η) = Lhi (x, ξ, η), set ζ = (ξ, η) and show that ζ ∈ Q. We argue by contradic-

tion and thus suppose that ζ 6∈ Q.

Note that, since ζ ∈ Rn × Yi,

(27) Khi (x, ζ) = Lhi (x, ζ) = Li(x, ζ).

In view of Lemma 14, (i) applied to Kh, we can select qζ = (pζ, uζ) ∈ ∂(ξ,η)Khi (x, ζ), which

implies by the convex duality (Lemma 14, (ii)) that ζ ∈ ∂(p,u)Ghi (x, qζ) and

(28) Khi (x, ζ) +Gh

i (x, qζ) = ζ · qζ .


We claim that h(qζ) > 0. Indeed, if, to the contrary, h(qζ) = 0, then we have |pζ|+ |uζ| ≤ C

by (23) and, by (24), (27), and (28),

∂(p,u)Hi(x, qζ) ⊂ Q and ζ · qζ = Khi (x, ζ) +Gh

i (x, qζ) = Li(x, ζ) +Hi(x, qζ),

which imply by Lemma 14, (ii) that

ζ ∈ ∂(p,u)Hi(x, qζ) ⊂ Q.

This contradicts the choice of ζ , which confirms that h(qζ) > 0.

Now, we observe that

Li(x, ζ) ≥ ζ · qζ −Hi(x, qζ) = ζ · qζ −Ghi (x, qζ) + h(qζ)

= Khi (x, ζ) + h(qζ) > Kh

i (x, ζ) = Li(x, ζ),

which is a contradiction, and we conclude that (26) is valid. The proof is complete. �

In the following proof of Theorem 12, we approximate the Hamiltonian H(x, p, u) by Hamil-

tonians which satisfy (H1)–(H3) and (H5). In the first step of the approximation of H , we

follow the argument in the proof above, with h replaced by h/r, with r ∈ N. In the proof

above, the function Gh(x, p, u) has the superlinear growth in (p, u) because of the addition of

h and its nice effect is the continuity of Kh on Tn × R

n+m. The continuity on Tn × R

n+m of

the Lagrangians of the approximating Hamiltonians, obtained in the first step, is important for

the second and final step of building the approximating Hamiltonians, which have at most the

linear growth due to (H5).

Proof of Theorem 12. We choose a constant C > 0 and a compact convex set Q ⊂ Rn+m so that

(22) and (24) hold. We may assume that Q = Q1×Q2 for some Q1 ⊂ Rn and Q2 ⊂ R

m, where,

moreover, Q1 is a neighborhood of the origin of Rn. Let h ∈ C1(Rn+m) be a function satisfying

(23). As in the proof of Theorem 13, we define sequences (Hr)r∈N, (Lr)r∈N, (G

r)r∈N, (Kr)r∈N

of functions, with h replaced by h/r. That is, Gr = (Gri )i∈I is defined by

Gri (x, p, u) = Hi(x, p, u) +

1

rh(p, u) for (x, p, u) ∈ T

n × Rn+m,

Kr is the Lagrangian of Gr, Lr is given by

Lr(x, ξ, η) = Kr(x, ξ, η) + 0Y (η) for (x, ξ, η) ∈ Tn × R

n+m,

and Hr is the Hamiltonian of Lr. We have already checked in the proof of Theorem 13 that Hr

satisfies (H1)–(H3), vλ is a solution of λvλ+Hr[vλ] = 0 in Tn, and Lr ∈ ∏

i∈I C(Tn×R

n× Yi).


Moreover, it is easily seen that for (x, p, u) ∈ Tn × R

n+m and i ∈ I, if |p|+ |u| ≤ C,

(29) H(x, p, u) = Hr(x, p, u) = Gr(x, p, u) and ∂(p,u)Hri (x, p, u) ⊂ Q.

Next we define function HrQ = (Hr

Q,i)i∈I as the Hamiltonian of the function

LrQ(x, ξ, η) := Lr(x, ξ, η) + 0Q(ξ, η).

Note by Lemma 14, (ii) that for (x, i, p, u) ∈ Tn × I× R

n+m and ζ ∈ Rn+m, if

ζ ∈ ∂(p,u)HrQ,i(x, p, u),

then

(p, u) ∈ ∂(ξ,η)LrQ,i(x, ζ),

and hence, by the definition of LrQ,i, we have ζ ∈ Q. That is, we have

∂(p,u)HrQ,i(x, p, u) ⊂ Q for (x, i, p, u) ∈ T

n × I× Rn+m.

It is now easy to see that HrQ satisfies (H1), (H2) and (H5). Note also by the inclusion in (29)

that if |p|+ |u| ≤ C,

Hri (x, p, u) = max

(ξ,η)∈Q(p · ξ + u · η − Lri (x, ξ, η))

= max(ξ,η)∈(Rn×Yi)∩Q

(p · ξ + u · η − Lri (x, ξ, η)) = HrQ,i(x, p, u).

We may now invoke Theorem 5, to conclude that there is µ ∈ C(z, k, λ) such that

(30) vλk (z) =⟨µ, LrQ

⟩= min

ν∈C(z,k,λ)

⟨ν, LrQ

⟩.

Theorem 13 and (29) ensure that for any minimizer ν = (νi)i∈I ∈ C(z, k, λ) of the optimization

in (30), we have the property

supp νi ⊂ Tn × [(Rn × Yi) ∩Q].

For each r ∈ N, we select a minimizer µr = (µri ) ∈ C(z, k, λ) of the optimization in (30).

Since supp µri ⊂ Tn ×Q, we see immediately from(30) that

(31) vλk (z) = 〈µr, Lr〉 .

In view of Lemma 11, we may assume that (µr)r∈N, after passing to a subsequence which is

denoted again by the same symbol, converges weakly in the sense of measures to a measure

µ = (µi) ∈ Pλ having the property that supp µi ⊂ T

n × [(Rn × Yi) ∩Q] for all i ∈ I.


The weak convergence of (µr) implies that

⟨µ, Sλ1

⟩= 1,

ψk(z) = 〈µ, ξ ·Dψi + η · ψ1+ λψ〉 for all ψ = (ψi) ∈ C1(Tm)m.

These ensure that µ ∈ C(z, k, λ).

It is easily checked that, as r → ∞, Kr(x, ξ, η) → L(x, ξ, η) monotonically pointwise. Since

Kr ≤ Kr+1 and Kr ≤ Lr for r ∈ N, we obtain from (31),

vλk (z) ≥ 〈µr, Kq〉 ≥ 〈µr, j ∧Kq〉 if r ≥ q, for all j, q ∈ N,

where j ∧Kq := (min{j, Kqi })i∈I ∈ Cb(T

m × Rn+m)m. Sending r → ∞ yields

vλk (z) ≥ 〈µ, j ∧Kq〉 for all j, q ∈ N.

By the monotone convergence theorem, after sending j, q → ∞, we obtain

vλk (z) ≥ 〈µ, L〉 ,

while, by Lemma 15, we have

vλk (z) ≤ infν∈C(z,k,λ)

〈ν, L〉 .

Thus, we conclude that

vλk (z) = 〈µ, L〉 = minν∈C(z,k,λ)

〈ν, L〉 . �

5. A convergence result for the vanishing discount problem

We study the asymptotic behavior of the solution vλ of (Pλ), with λ > 0, as λ→ 0.

Theorem 16. Assume (H1)–(H4). Let vλ be the solution of (Pλ) for λ > 0. Then there exists

a solution v0 of (P0) such that the functions vλ converge to v0 in C(Tn)m as λ→ 0+.

Lemma 17. Under the hypotheses of Theorem 16, there exists a constant C0 > 0 such that for

any λ > 0,

(32) |vλi (x)| ≤ C0 for (x, i) ∈ Tn × I.

Proof. Let v0 = (v0,i)i∈I ∈ Lip(Tn)m be a solution of (P0). Choose a constant C1 > 0 so that

|v0,i(x)| ≤ C1 for (x, i) ∈ Tn × I,

and observe by Lemma 1 that the functions v0 + C11 and v0 − C11 are a supersolution and a

subsolution of (P0), respectively. Noting that v0 + C11 ≥ 0 and v0 − C11 ≤ 0, we deduce that

v0 + C11 ≥ 0 and v0 −C11 ≤ 0 are a supersolution and a subsolution of (Pλ), respectively, for


any λ > 0. By comparison (Theorem 2), we see that, for any λ > 0, v0 −C11 ≤ vλ ≤ v0 +C11

on Tn and, moreover, −2C11 ≤ vλ ≤ 2C11 on T

n. Thus, (32) holds with C0 = 2C1. �

Lemma 18. Under the hypotheses of Theorem 16, the family (vλ)λ∈(0, 1) is equi-Lipschitz con-

tinuous on Tn.

Proof. According to Lemma 17, we may choose a constant C0 > 0 so that

|vλi (x)| ≤ C0 for (x, i, λ) ∈ Tn × I× (0, ∞).

Hence, as vλ is a solution of (Pλ), we deduce by (H1) that there exists a constant C1 > 0 such

that the functions vλi , with λ ∈ (0, 1), are subsolutions of |Du| ≤ C1 in Tn. As is well-known,

this implies that the vλi are Lipschitz continuous on Tn with C1 as their Lipschitz bound. �

We remark that one can show, with a slightly more elaboration, the equi-Lipschitz property

of (vλ)λ>0 in the above lemma.

Theorem 19. Let (z, k) ∈ Tn × I. Assume (H1)–(H4). For any λ > 0, let vλ be the solution

of (Pλ) and µλ ∈ C(z, k, λ) a minimizer in (21). Then, for any sequence (λj)j∈N of positive

numbers converging to zero, there exists a subsequence of (λj), which is denoted again by the

same symbol, such that, as j → ∞,

λjµλj → ν0

weakly in the sense of measures for some ν0 = (ν0i )i∈I ∈ C(0), and ν0 satisfies

(33) 0 =⟨ν0, L

⟩= min

ν∈C(0)〈ν, L〉 .

We call any minimizing measure ν0 ∈ C(0) in (33) a Mather measure. The set of all Mather

measures ν0 ∈ C(0) is denoted by M(L). See, for example, [11,16,35,36] for some work related

to Mather measures. Notice that the limit measure ν0 in Theorem 19 is a Mather measure.

It should be noted that, in our formulation, the existence of a Mather measure is trivial since

0 ∈ C(0).

Proof. We fix (z, k) ∈ Tn × I. By Theorem 12, for each λ > 0 there exists µλ = (µλi )i∈I ∈

C(z, k, λ) such that

(34) λvλk (z) =⟨λµλ, L

⟩.

By Lemmas 17 and 18, there is a constant C > 0 such that for any λ ∈ (0, 1),

|vλ(z)| + |Dvλ(x)| ≤ C a.e. x ∈ Tn.


We choose a closed ball Q ⊂ Rn+m so that (24) holds with C given above. Thanks to Theo-

rem 13, we find that

supp µλi ⊂ Tn × [Q ∩ (Rn × Yi)] for (i, λ) ∈ I× (0, 1).

Noting that Sλ(η) ≥ λ for η ∈ Yi and suppµλi ⊂ Tn × R

n × Yi for all i ∈ I, we observe that

⟨λµλ, 1

⟩=

⟨µλ, λ1

⟩≤

⟨µλ, Sλ1

⟩= 1.

Hence, by applying Lemma 10 and passing to a subsequence, we may assume that the sequence

(λjµλj )j∈N ⊂ M+(T

n ×Rn+m)m converges weakly in the sense of measures to some ν0 = (ν0i ) ∈

M+(Tn×R

n+m)m having properties 〈ν0, 1〉 ≤ 1 and supp ν0i ⊂ Tn× [Q∩ (Rn×Yi)] for all i ∈ I.

It is an immediate consequence that ν0 ∈ P0.

Since µλ ∈ C(z, k, λ), we have

λψk(z) =⟨λµλ, ξ ·Dψ + η · ψ1+ λψ

⟩for all ψ = (ψi) ∈ C1(Tn)m.

Sending λ→ 0 along the sequence (λj) yields

0 =⟨ν0, ξ ·Dψ + η · ψ1

⟩for all ψ = (ψi) ∈ C1(Tn)m,

which concludes that ν0 ∈ C(0).

For any function φ ∈ Cb(Tn × R

n+m)m such that φ ≤ L, we see from (34) that

0 ≥⟨ν0, φ

⟩.

Moreover, by approximating L monotonically from below by bounded continuous functions and

applying the monotone convergence theorem, we deduce that

0 ≥⟨ν0, L

⟩.

By Lemma 15, we have 0 ≤ 〈ν, L〉 for all ν ∈ C(0). It is now clear that (33) holds. �

Let V denote the set of accumulation points v = (vi)i∈I ∈ C(Tn)m of (vλ)λ>0 in the space

C(Tn)m as λ → 0. Note by the stability of the viscosity property under uniform convergence

that any v ∈ V is a solution of (P0). Let W denote the set of those solutions w ∈ C(Tn)m of

(P0) which satisfy

(35) 〈ν, w〉 ≤ 0 for all ν ∈ M(L).

Proof of Theorem 16. In view of the Ascoli-Arzela theorem, Lemmas 17 and 18 assure that the

family (vλ)λ∈(0, 1) is relatively compact in C(Tn)m. In particular, the set V is nonempty.


If V is a singleton, then it is obvious that the whole family (vλ)λ>0 converges to the unique

element of V in C(Tn)m as λ→ 0.

We need only to show that V is a singleton. For this, we first show that

(36) V ⊂ W.

To see this, let v ∈ V and ν ∈ M(L). Choose a sequence (λj)j∈N of positive numbers converging

to zero such that (vλj)j∈N converges to v in C(Tn)m. Since (L−λvλ, vλ) ∈ F(0), ν ∈ C(0), and

〈ν, L〉 = 0, using Lemma 15, we get

0 ≤⟨ν, L− λvλ

⟩= 〈ν, L〉 −

⟨ν, λvλ

⟩= −λ

⟨ν, vλ

⟩,

which yields, after dividing by λ > 0 and then sending λ→ 0 along λ = λj ,

〈ν, v〉 ≤ 0.

This proves (35), which ensures the inclusion (36).

Next, we show that

(37) w ≤ v for all w ∈ W, v ∈ V.

To check this, it is enough to show that for any v ∈ V, w ∈ W and (z, k) ∈ Tn × I, the

inequality wk(z) ≤ vk(z) holds.

Fix any v ∈ V and w ∈ W and (z, k) ∈ Tn×I. Select a sequence (λj)j∈N ⊂ (0, ∞) converging

to zero so that

vλj → v in C(Tn)m as j → ∞.

By Theorem 12, there exists a sequence (µj)j∈N such that for j ∈ N,

(38) µj ∈ C(z, k, λj) and vλjk (z) =

⟨µj, L

⟩.

In view of Theorem 19, we may assume by passing to a subsequence if necessary that, as j → ∞,

λjµj → ν weakly in the sense of measures

for some ν = (νi)i∈I ∈ M(L).

Now, note that (L+ λjw,w) ∈ F(λj) and infer by Lemma 15 and (38) that

wk(z) ≤⟨µj, L+ λjw

⟩= v

λjk (z) + λj

⟨µj, w

⟩.

Sending j → ∞ now yields

wk(z) ≤ vk(z) + 〈ν, w〉 .


This together with (35) shows that wk(z) ≤ vk(z), which ensures that (37) holds. Noting that

(37) combined with (36) shows that w ≤ v for all v, w ∈ V, that is, V is a singleton. The proof

is complete. �

Reviewing the proof above, we conclude easily the following proposition, which is a general-

ization of [11, Theorem 3.8] (see also [16, Proof of Theorem 1]).

Corollary 20. Under the assumptions and notation of Theorem 16, the limit function v0 =

(v0i )i∈I can be represented as

v0i (x) = max{wi(x) : w = (wi) ∈ W} for x ∈ Tn.

The proof of Corollary 20, with W replaced by

W− = {w ∈ C(Tn)m : w is a subsolution and satisfies (35)},

shows also that, under the hypotheses and notation of Corollary 20,

v0i (x) = max{wi(x) : w = (wi) ∈ W−} for x ∈ Tn.

6. Ergodic problem

Remark that, given a Hamiltonian H , condition (H4) is not satisfied in general. We consider

the problem of finding an m-vector c = (ci)i∈I ∈ Rm and a function u = (ui)i∈I ∈ C(Tn)m such

that u is a solution of the m-system

(39) H [u] = c in Tn,

which is stated componentwise as

Hi(x,Dui(x), u(x)) = ci in Tn for i ∈ I.

We call this problem the ergodic problem for H .

If the ergodic problem has a solution c ∈ Rm and u ∈ C(Tn)m, then we may apply the

main convergence result (Theorem 16) to (Pλ), with H replaced by Hc := H − c. As noted in

the introduction, this change of Hamiltonians, in general, does not help analyze the vanishing

discount problem for the original system (Pλ).

However, if H satisfies a certain additional condition, then the argument of switching from

the HamiltonianH to Hc makes sense for the vanishing discount problem for (Pλ). For instance,

given a solution (c, u) ∈ Rm × C(Tn)m of (39), assume that the equality

(40) H(x, p, v + tc) = H(x, p, v)


holds for all t ∈ R and (x, p, v) ∈ Tn×R

n+m. It is easily seen that if vλ ∈ C(Tn)m is a solution

of (Pλ), then wλ := vλ + λ−1c is a solution of (Pλ), with Hc in place of H . This is a situation

where one can apply Theorem 16, to observe the convergence of vλ + λ−1c as λ→ 0.

In the next result, we do not need the convexity or monotonicity of H , and we assume only

(H1).

For R > 0 and r > 0, we set

αR(r) = inf{Hi(x, p, u) : (x, i) ∈ Tn × I, u ∈ Bm

R , p ∈ Rn \Bn

r },

βR = sup{Hi(x, 0, u) : (x, i) ∈ Tn × I, u ∈ Bm

R }.

The constants αR(r) and βR are finite by the continuity of Hi and (H1). It is clear that for any

R > 0, the function r 7→ αR(r) is nondecreasing in (0, ∞) and diverges to infinity as r → ∞.

Theorem 21. Assume (H1) and that there exists a constant R > 0 such that

(41) βR < αR

(2R√n

).

Then problem (39) has a solution (c, u) ∈ Rm × C(Tn)m.

We remark that a result similar to the above has been established in [32, Theorem 1.2] in

the case of a scalar Hamilton-Jacobi equation.

Roughly speaking, the condition (41) in the theorem above is satisfied for a large R > 0 if

the growth of Hi(x, p, u) in p is higher in a certain sense than that in u as |(p, u)| → ∞. In the

case of linear coupling (and hence, Hi have the form of (1)), it is obvious that if for all i ∈ I,

the functions Gi(x, p) have the superlinear growth, i.e., satisfy

limR→∞

inf(x,p)∈Tn×(Rn\Bn

R)

Gi(x, p)

|p| = ∞,

then condition (41) is valid. We refer to [15, Theorem 2.12], [26, Theorem 17] for results, in

the linear coupling case, similar to but more subtle than the theorem above.

Condition (41) is also valid when, as a direct generalization of the linear coupling case,

Hi, i ∈ I have superlinear growth in p, i.e., for any r > 0,

limR→∞

inf(x,p,u)∈Tn×(Rn\Bn

R)×Bm

r

Hi(x, p, u)

|p| = ∞,

and uniformly Lipschitz dependence in u, i.e., there exists Θ > 0 such that for any u, v ∈ Rm

|Hi(x, p, u)−Hi(x, p, v)| ≤ Θ|u− v|.


Proof. We choose R > 0 so that (41) holds and select λ > 0 so that

(42) βR + λR < αR

(2R√n

).

Let u ∈ C(Tn)m and consider the uncoupled m-system for v = (vi)i∈I:

(43) λ(vi(x)− ui(x)) +Hi(x,Dvi(x), u(x)) = 0 in Tn for i ∈ I.

The functions (x, p) 7→ Hi(x, p, u(x)) are continuous and coercive and, hence, the standard

theory of viscosity solutions (also, Theorem 2 applied to each single equations) guarantees that

(43) has a unique solution v = (vi)i∈I and the functions vi are Lipschitz continuous on Tn.

For any u ∈ C(Tn)m, let v = (vi)i∈I ∈ C(Tn)m be the solution of (43). We set

Tu := v −minTn

v,

where

minTn

v := (minx∈Tn

vi(x))i∈I ∈ Rm,

which gives a mapping T from C(Tn)m to C(Tn)m. Because of the stability of viscosity solutions

under the uniform convergence and the uniqueness of solution of (43), we easily deduce that T

is a continuous mapping on the Banach space C(Tn)m, with norm ‖u‖∞ := maxx∈Tn |u(x)|.Now, fix u so that

‖u‖∞ ≤ R and u(x) ≥ 0 for all x ∈ Tn,

and observe that the function w(x) := −λ−1βR1 is a subsolution of (43). Indeed, we have

λ(wi(x)− ui(x)) +Hi(x,Dwi(x), u(x)) ≤ −βR +Hi(x, 0, u(x)) ≤ 0 for (x, i) ∈ Tn × I.

By the standard comparison theorem, we have

−βRλ1 ≤ v.

Noting that vi is Lipschitz continuous and hence it is almost everywhere differentiable, we

compute at any point x of differentiability of vi that, if Dvi(x) 6= 0,

0 ≥ λ

(−βRλ

− ui(x)

)+ αR(|Dvi(x)|) ≥ −βR − λR + αR(|Dvi(x)|),

and observe by the choice of λ that, if Dvi(x) 6= 0,

αR(|Dvi(x)|) ≤ βR + λR < αR

(2R√n

),

which yields

|Dvi(x)| <2R√n

a.e. in Tn for i ∈ I,


and moreover

0 ≤ vi(x)−minTn

vi ≤ R for all (x, i) ∈ Tn × I.

Thus, we conclude that

‖D(Tu)i‖L∞(Tn) := ess supTn

|D(Tu)i| ≤ R for i ∈ I,

(Tu)(x) ≥ 0 for x ∈ Tn and ‖Tu‖∞ ≤ R.

We set

K = {u ∈ C(Tn)m : u ≥ 0, ‖u‖∞ ≤ R, ‖Dui‖L∞(Tn) ≤ R for all i ∈ I},

and note that K is a compact convex subset of C(Tn)m. The above observations show that

T maps K into K. The Schauder fixed point theorem guarantees that there is a fixed point

u ∈ K of T . Let v be the a solution of (43), with the fixed point u. By the definition of T , we

have

u = Tu = v −minTn

v,

and u solves

λminTn

vi +Hi(x,Dui, u) = 0 in Tn for i ∈ I.

That is, the pair (−λminTn v, u) is a solution of (41). �

Acknowledgement

The authors would like to thank Wenjia Jing of Tsinghua University, who kindly shared the

ideas for the proof of [32, Theorem 1.2] before its publication, which was a great help for them

to establish Theorem 16.

The authors would like to thank the anonymous referees for their careful reading of and crit-

ical and useful comments on the original version of this paper, which have helped significantly

to improve the presentation.

HI was supported in part by the JSPS Grants KAKENHI No. 16H03948 and No. 18H00833

and by the NSF Grant No. 1440140 while in residence at the Mathematical Sciences Research

Institute in Berkeley, California, in October 2018. HI thanks the Department of Mathematics

at the Sapienza University of Rome for financial support and its hospitality while his visit there

in May 6–June 5 2019. LJ was supported by the National Natural Science Foundation of China

(Grant No. 11901293 and No. 11571166) and Start-up Foundation of Nanjing University of

Science and Technology (No. AE89991/114).


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(Hitoshi Ishii) Institute for Mathematics and Computer Science

Tsuda University

2-1-1 Tsuda, Kodaira, Tokyo, 187-8577 Japan.

E-mail address : [email protected]

(Liang Jin)

Department of Mathematics, Nanjing University of Science and Technology

Nanjing 210094, China.

E-mail address : [email protected]

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