We introduce games associated with second-order partial differential equations given by arbitrary products of eigenvalues of the Hessian. We prove that, as a parameter that controls the step length goes to zero, the value functions of the games converge uniformly to a viscosity solution of the partial differential equation. The classical Monge-Ampère equation is an important example under consideration.
Citation: Pablo Blanc, Fernando Charro, Juan J. Manfredi, Julio D. Rossi. Games associated with products of eigenvalues of the Hessian[J]. Mathematics in Engineering, 2023, 5(3): 1-26. doi: 10.3934/mine.2023066
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We introduce games associated with second-order partial differential equations given by arbitrary products of eigenvalues of the Hessian. We prove that, as a parameter that controls the step length goes to zero, the value functions of the games converge uniformly to a viscosity solution of the partial differential equation. The classical Monge-Ampère equation is an important example under consideration.
Dedicated to our friend Giuseppe Rosario Mingione, great mathematician and Master of Regularity, on the occasion of his 50th birthday.
In this article, we introduce a family of games related to second-order partial differential equations (PDEs) given by arbitrary products of eigenvalues of the Hessian. More precisely, given Ω⊂RN, and k indices i1,…,ik∈{1,…,N} (which could be repeated), we consider PDEs of the form
{Pi1,...,ik(D2u):=k∏j=1λij(D2u)=f, in Ω,u=g, on ∂Ω. | (1.1) |
Here, λ1≤⋯≤λN denote the eigenvalues of D2u and the right-hand side f is a continuous, non-negative function. Notice that operators given by Pi1,...,ik are degenerate and do not preserve order in the space of symmetric matrices, the usual condition for ellipticity. Moreover, we want to emphasize that the eigenvalues λi1,…,λik in (1.1) could be repeated. For instance, one could take i1=i2=1 and i3=3 to produce the equation λ21λ3=f. Similarly, one could also take k=N and ij=j and consider the product of all the eigenvalues, which corresponds to the classical Monge-Ampère equation, det(D2u)=∏Ni=1λi=f. We refer the reader to [11,13] for general references on the Monge-Ampère equation.
Our main goal is to design a game whose value functions approximate viscosity solutions to (1.1) as a parameter that controls the step size of the game goes to zero. The connection between games and PDEs has developed significantly over the last decade; see [10,16,19,20,21,24,25] (and we refer also to [12] in connection with mean-field games). We also refer to the recent books [6,15] and the references therein for general references on this program. The relation between games and PDEs has proven fruitful in obtaining qualitative results, see [4], and regularity estimates; see [1,18,23,27]. A game-theoretical interpretation is available in the case of Hamilton-Jacobi equations in the presence of gradient constraints (both in the convex and non-convex settings), see [28]. The case k=1 for the smallest eigenvalue λ1 gives rise to the Dirichlet problem for the convex envelop studied in [22] and it is also related to the truncated Laplacians considered in [2].
Our starting point is the case of only one eigenvalue and f=0, which was recently tackled in [5] and has connections with convexity theory. Let us describe the two-player, zero-sum game called "a random walk for λj" introduced there. Given a domain Ω⊂RN, ε>0, and a final payoff function g:RN∖Ω→R, the game is played as follows. The game starts with a token at an initial position x0∈Ω. Player Ⅰ (who wants to minimize the expected payoff) chooses a subspace S of dimension j, and then Player Ⅱ (who wants to maximize the expected payoff) chooses a unitary vector v∈S. Then, the token moves to x±εv with equal probabilities. The game continues until the token leaves the domain at a point xτ, and the first player gets −g(xτ) while the second receives g(xτ) (we can think that Player Ⅰ pays the amount g(xτ) to Player Ⅱ). This game has a value function uε (see below for the precise definition) defined in Ω, which depends on the step size ε. One of the main results in [5] is showing that, under an appropriate condition on ∂Ω, these value functions converge uniformly in ¯Ω to a continuous limit u characterized as the unique viscosity solution to
{Pj(D2u)=λj(D2u)=0, in Ω,u=g, on ∂Ω. | (1.2) |
The right-hand side f is obtained by considering a running payoff, that is, a nonnegative function f:Ω→R such that 12ε2f(xn) represents an amount paid by Player Ⅱ to Player Ⅰ when the token reaches xn. Then, the game value approximates viscosity solutions to
{Pj(D2u)=λj(D2u)=f, in Ω,u=g, on ∂Ω. | (1.3) |
Now, we introduce a new game that allows us to obtain a product of eigenvalues. Assume a list of eigenvalues (λij)j=1,…,k is given. We consider the set
Ikε={(αj)j=1,…,k∈Rk:k∏j=1αj=1 and 0<αj<ϕ2(ε)}, | (1.4) |
where ϕ(ε) is a positive function such that
limε→0ϕ(ε)=∞andlimε→0εϕ(ε)=0 | (1.5) |
(for example one can take ϕ(ε)=ε−1/2). At every round, Player Ⅰ chooses k positive real numbers (αj)j=1,…,k∈Ikε, after which, an index j∈{1,…,k} is selected uniformly at random. Then, the players play a round of the game "a random walk for λij" described above. Specifically, Player Ⅰ chooses a subspace S of dimension ij, Player Ⅱ chooses a unitary vector v∈S, and the token is moved to x±ε√αjv with equal probabilities. The running payoff at every round is given by 12ε2[f(xn)]1k (Player Ⅱ pays this ammount to Player Ⅰ) and the final payoff is given by g (if xτ denotes the first position outside Ω, the game ends and Player Ⅰ pays g(xτ) to Player Ⅱ).
Notice that in this game we adjust the length of the token jump according to the corresponding αj, and Player Ⅰ may choose to enlarge the game steps associated with λi at the expense of shortening others (since the product of the αj must equal one). Also notice that the restriction ϕ(ε)>√αj>0 implies that the maximum step size is bounded as |x−(x±ε√αjv)|<εϕ(ε)→0 as ε→0.
When both players fix their strategies, SI for the first player (a choice of (αj)j=1,…,k and ij−dimensional subspaces S at every step of the game), and SII for the second player (a choice of a unitary vector v in each possible subspace at every step of the game), then we can compute the expected outcome (the amount that Player Ⅱ receives) as
Ex0SI,SII[g(xτ)−12ε2τ−1∑n=0[f(xn)]1k]. |
Then, the value for Player Ⅰ of the game starting at any given x0∈Ω is defined as
uεI(x0)=infSIsupSIIEx0SI,SII[g(xτ)−12ε2τ−1∑n=0[f(xn)]1k], |
while the value for Player Ⅱ is
uεII(x0)=supSIIinfSIEx0SI,SII[g(xτ)−12ε2τ−1∑n=0[f(xn)]1k]. |
When these two values coincide we say that the game has a value.
In our first result, we state that this game has a value and the value verifies an equation in Ω, called a Dynamic Programming Principle (DPP) in the literature.
Theorem 1. The game has a value
uε:=uεI=uεII, |
which is characterized as the unique solution to
{uε(x)=infαj∈Ikε1kk∑j=1infdim(S)=ijsupv∈S,|v|=1{12uε(x+ε√αjv)+12uε(x−ε√αjv)}−12ε2[f(x)]1kx∈Ω,uε(x)=g(x)x∉Ω. | (1.6) |
Let us show intuitively why this holds. At each step, Player Ⅰ chooses (αj)j=1,…,k, and then j∈{1,…,k} is selected with probability 1k. Player Ⅰ chooses a subspace S of dimension ij and then Player Ⅱ chooses one unitary vector, v, in the subspace S. The token is then moved with probability 12 to x+ε√αjv or x−ε√αjv. Finally, the expected payoff at x is given by −12ε2[f(x)]1k (the running payoff) plus the expected payoff for the rest of the game. Then, the equation in (1.6) follows by considering all the possibilities (recall that Player Ⅰ seeks to minimize the expected payoff and Player Ⅱ to minimize it).
Our next goal is to look for the limit as ε→0.
Theorem 2. Assume that Ω is strictly convex. Let uε be the values of the game. We have
uε→uasε→0 |
uniformly in ¯Ω, where u is the unique viscosity solution to (1.1).
We devote Section 3 to proving the theorems. The uniqueness statement in Theorem 2 follows from [8]; see Remark 2.2 below. We need the convexity of ∂Ω to prove that the sequence converges by means of an Arzelà-Ascoli type lemma. In fact, for strictly convex domains, we show that for every point y∈∂Ω, a game that starts near y ends nearby with a high probability regardless of the players' strategies. This allows us to obtain a sort of asymptotic equicontinuity near the boundary, which leads to uniform convergence in the whole ¯Ω. Note that, in general, the value functions uε are discontinuous in Ω since we take discrete steps.
Observe that the result implies the existence of a solution to the PDE. The strict convexity of Ω is needed if a solution to the equation λ1=0 is to exist for every continuous boundary data; see [5]. Since our general setting includes this case, we require the strict convexity. However, in some cases, such as λ2=0, this hypothesis may be relaxed; see [5].
Let us see intuitively why u is a solution to Eq (1.1). By subtracting uε(x) and dividing by ε2 on both sides we get the term:
uε(x+ε√αjv)−2uε(x)+uε(x−ε√αjv)ε2 |
which in the limit approximates the second derivative of u in the direction of v multiplied by αj. Hence, by the Courant-Fischer min-max principle, the expression
infdim(S)=ijsupv∈S,|v|=1uε(x+ε√αjv)−2uε(x)+uε(x−ε√αjv)ε2 |
approximates the ij eigenvalue of D2u(x) multiplied by αj. Taking into account the running payoff, we obtain that u is a solution to
[f(x)]1k=infαj∈Ikε1kk∑j=1αjλij. |
Then, the result follows from the identity
(β1β2…βk)1k=infαj>0,k∏j=1αj=11kk∑j=1αjβjwhenever β1,β2,…,βk≥0, |
which is proved in detail in Lemma 2.4 below.
The Monge-Ampère equation. Notice that when all the eigenvalues are involved, we have a two-player game that approximates solutions to the Monge-Ampère equation det(D2u)=f. However, in the special case of the Monge-Ampère equation, we can also design a one-player game (a control problem) to approximate the solutions. This game is based on a recent asymptotic mean value formula that characterizes viscosity solutions to the Monge-Ampère equation. In fact, in [3], it is proved that u is a viscosity solution to the Monge-Ampère equation
det(D2u(x))=f(x) |
if and only if
u(x)=infV∈O infαi∈INε{1NN∑i=112u(x+ε√αivi)+12u(x−ε√αivi)}−ε22[f(x)]1n+o(ε2) |
as ε→0, holds in the viscosity sense. Here we denoted by O the set of all orthonormal bases V={v1,…,vN} of Rn, and INε is given by (1.4).
Now, let us describe a one-player game (control problem). At each play, the player (controller), who aims to minimize the expected total payoff, chooses an orthonormal basis V=(v1,...,vN) and coefficients (αi)i=1,…,N∈INε. Then, the new position of the game goes to x±ε√αivi with equal probability 1/(2N). In addition, there is a running payoff given by −ε22[f(x)]1N at every play and a final payoff g(x) (as before the game ends when the token leaves Ω). Then, the value of the game for any x0∈Ω for is given by
uε(x0)=infSIEx0SI[g(xτ)−12ε2τ−1∑i=0[f(xi)]1N]. |
Here, we take the infimum over all the possible player strategies.
For this game we have the following result.
Theorem 3. The game value uε is the unique solution to
{uε(x)=infV∈O infαi∈INε{1NN∑i=112uε(x+ε√αivi)+12uε(x−ε√αivi)}−12ε2[f(x)]1Nx∈Ω,uε(x)=g(x)x∉Ω. |
Moreover, if we assume that Ω is strictly convex. Then,
uε→u,asε→0, |
uniformly in ¯Ω, where u is the unique viscosity solution to
{det(D2u)=N∏i=1λi(D2u)=f,inΩ,u=g,on∂Ω. | (1.7) |
Notice that here the strict convexity of the domain is a natural assumption for the solvability of (1.7); see [7,26].
The paper is organized as follows: in Section 2, we collect some preliminary results and include the definition of viscosity solutions. In Section 3, we prove our main results concerning the two-player game, Theorems 1 and 2. In Section 4, we include some details for the control problem for Monge-Ampère. Finally, in Section 5, we present a variant of the game for Monge-Ampère that involves an integral average in the corresponding DPP.
We begin by stating the definition of a viscosity solution to (1.1). Recall that f:Ω→R is a non-negative function. Following [8] (see also [14]), we have that the equation is associated with the cone
F={M:k∏j=1λij(M)≥f and λij(M)≥0 for every j=1,…,k}. |
Here we do not state the definition of viscosity solutions with an explicit reference to the cone, but we prefer to present it using the usual notation; see [9].
We say that P is a paraboloid if for every x,x0∈RN we have
P(x)=P(x0)+⟨x−x0,∇P(x0)⟩+12⟨D2P(x0)(x−x0),x−x0⟩. |
Definition 4. A continuous function u verifies
k∏j=1λij(D2u)=f |
in Ω in the viscosity sense if
1) for every paraboloid ϕ that touches u from below at x0∈Ω (ϕ(x0)=u(x0) and ϕ≤u), and with eigenvalues of the Hessian that verify λij(D2ϕ(x0))≥0, we have
k∏j=1λij(D2ϕ(x))≤f(x). |
2) for every paraboloid ψ that touches u from above at x0∈Ω (ψ(x0)=u(x0) and ψ≥u), we have λij(D2ψ(x0))≥0 for j=1,…,k and
k∏j=1λij(D2ψ(x))≥f(x). |
Remark 5. The validity of the comparison principle for our equation follows from Theorem 4.9 in [8]. In fact, the map Θ:Ω→SN×N given by
Θ(x)={M:k∏j=1λij(M)≥f(x) and λij(M)≥0 for every j=1,…,k} |
is uniformly upper semicontinuous (since f is continuous) and is elliptic (since the operator is monotone on non negative matrices and f≥0).
Also observe that the concept of Θ-subharmonic/superharmonic is equivalent to the definition of subsolution/supersolution that we have given. In fact, if we consider
Φ={M:λij(M)≥0 for every j=1,…,k} |
Then, we have
Θ(x)={M∈Φ:F(M,x)≥0} |
where F is the operator that we are considering here, that is,
F(M,x)=k∏j=1λij(M)−f(x). |
Therefore, the equivalence between being Θ-subharmonic/superharmonic and being subsolution/supersolution follows from Proposition 2.11 in [8].
To obtain a convergent subsequence uε→u we will require Ω to be strictly convex. Here this means that we have that for every x,y∈¯Ω, tx+(1−t)y∈Ω for all 0<t<1. In particular, in Lemma 3.6 we will use a geometric condition over Ω equivalent to the strict convexity. We prove the equivalence between these two notions of strict convexity in the following lemma which is a variation of the classical supporting hyperplane theorem.
Lemma 6. Given an open non-empty bounded set Ω⊂RN, the following statements are equivalent:
1) Ω is strictly convex (i.e., for every x,y∈¯Ω, tx+(1−t)y∈Ω for all 0<t<1).
2) Given y∈∂Ω there exists w∈RN of norm 1 such that ⟨w,x−y⟩>0 for every x∈¯Ω∖{y}.
3) Given y∈∂Ω there exists w∈RN of norm 1 such that for every δ>0 there exists θ>0 such that
{x∈Ω:⟨w,x−y⟩<θ}⊂Bδ(y). |
Moreover, the vector w in statements (2) and (3) is the same one.
Proof. (1) ⟹ (2): We consider yk∈RN∖¯Ω such that yk→y and zk the projection of xk over ¯Ω (which exists since Ω is convex). We define the vectors
wk=zk−yk‖zk−yk‖. |
Up to a subsequence, we may assume that wk→w∈RN. We have that
⟨zk−yk,x−zk⟩≥0 |
for every x∈¯Ω. Hence, for every x∈¯Ω, we have
⟨wk,x⟩≥⟨wk,zk⟩=⟨wk,zk−yk⟩+⟨wk,yk⟩>⟨wk,yk⟩. |
Passing to the limit we get
⟨w,x−y⟩≥0 |
for every x∈¯Ω.
It remains to prove that ⟨w,x−y⟩≠0 when x≠y. Suppose, for the sake of contradiction, that ⟨w,x−y⟩=0 for some x≠y. By the strict convexity of Ω, we have x+y2∈Ω. Hence, x+y2−εw∈Ω for ε small enough and
⟨w,(x+y2−εw)−y⟩=⟨w,x−y2−εw⟩=−ε<0, |
a contradiction.
(2) ⟹ (3): Given δ, we consider f:¯Ω∖Bδ(y)→R given by
f(x)=⟨w,x−y⟩. |
Since f is continuous and is defined in a compact set it attains its minimum. We consider θ=min¯Ω∖Bδ(y)f which is positive since ⟨w,x−y⟩>0 for every x∈¯Ω∖{y}. We have that ⟨w,x−y⟩≥θ for every x∈¯Ω∖Bδ(y), and the result follows.
(3) ⟹ (2): Given x∈¯Ω∖{y} we consider δ=dist(x,y)2>0 and xk∈Ω∖Bδ(x) such that xk→x. Since xk∉Bδ(y), we have ⟨w,xk−y⟩≥θ. Hence ⟨w,x−y⟩≥θ>0.
(2) ⟹ (1): We consider the set
C=⋂y∈∂Ω{x∈Rn:⟨wy,x−y⟩≥0} |
where wy stands for the w given by statement (2) for each y∈∂Ω. Since C is the intersection of convex sets, it is also convex. We want to prove that ¯Ω is convex by proving that it is equal to C. It is clear that ¯Ω⊂C, let us show that if z∉¯Ω then z∉C. We fix x0∈Ω. Given z∉¯Ω, there exists t∈(0,1) such that y=tz+(1−t)x0∈∂Ω. We know that ⟨wy,x0−y⟩>0 since x0∈Ω. Hence ⟨wy,z−y⟩<0 and z∉C.
It remains to prove that the convexity is strict. Given x,y∈¯Ω, we know that
tx+(1−t)y∈¯Ω |
for all 0<t<1. We want to prove that tx+(1−t)y∈Ω, that is, tx+(1−t)y∉∂Ω. Suppose, arguing again by contradiction, that z=tx+(1−t)y∈∂Ω for some 0<t<1. Then, ⟨wz,x−z⟩>0 and ⟨wz,y−z⟩>0, and this implies that 0<⟨wz,(tx+(1−t)y)−z⟩=0, which is a contradiction.
The idea that allows us to obtain the product of the eigenvalues relies on the following formula.
Lemma 7. Given β1,β2,…,βk≥0, we have
(β1β2…βk)1k=infαj>0,k∏j=1αj=11kk∑j=1αjβj. |
Proof. The inequality
(β1β2…βk)1k≤infαj>0,k∏j=1αj=11kk∑j=1αjβj. |
follows from the arithmetic-geometric mean inequality.
In the case that all the βi are strictly positive, the equality is attained for
αi=(β1β2…βk)1kβi. |
In the case that βi=0 for some i we have to show that the infimum is zero. For that, we consider αi=nk−1 and αj=1n, which gives
limn→∞1kk∑j=1αjβj=βi=0 |
as desired.
Let us recall that in our setting we have the extra restriction αi<ϕ2(ε). To handle this issue we require the following two lemmas.
Lemma 8. Given β1,β2,…,βk≥0, it holds
(β1β2…βk)1k=limε→0infαj∈Ikε1kk∑j=1αjβj, |
where Ikε is given by (1.4).
Proof. If β1,β2,…,βk>0, for ε small enough such that
ϕ2(ε)>(β1β2…βk)1kβi |
for all i, we have
infαj∈Ikε1kk∑j=1αjβj=(β1β2…βk)1k, |
and the result follows.
Now we consider the case where βi=0 for some i=1,…,k and k>1 (if k=1 the result is obvious). We take
αi=ϕ(ε)2 and αj=(2ϕ(ε))1k−1. |
for j≠i. We have
infαj∈Ikε1kk∑j=1αjβj≤1k∑j≠i(2ϕ(ε))1k−1βj. |
By taking limit as ε→0 we conclude the proof.
Lemma 9. Given β1,β2,…,βk∈R with βi<0 for some i=1,…,k, it holds
limε→0infαj∈Ikε1kk∑j=1αjβj<0. |
Proof. If k=1 we have
limε→0infαj∈Ikε1kk∑j=1αjβj=β1<0. |
If k>1, we take
αi=ϕ(ε)2 and αj=(2ϕ(ε))1k−1 |
and in the limit we get −∞.
In this section, we describe in detail the two-player zero-sum game presented in the introduction. Let Ω⊂RN be a bounded open set and fix ε>0. The values k and (ij)j=1,…,k are given along with a positive function ϕ(ε) such that
limε→0ϕ(ε)=∞andlimε→0εϕ(ε)=0. |
A token is placed at x0∈Ω and the game begins with Player Ⅰ choosing (αj)j=1,…,k∈Ikε where
Ikε={(αj)j=1,…,k∈Rk:k∏j=1αj=1 and 0<αj<ϕ2(ε)}. |
Then, j∈{1,…,k} is selected uniformly at random. Given the value of j, Player Ⅰ chooses a subspace S of dimension ij and then Player Ⅱ chooses one unitary vector v∈S. Then, the token is moved to x±ε√αjv with equal probabilities. After the first round, the game continues from x1 according to the same rules.
This procedure yields a possibly infinite sequence of game states x0,x1,… where every xk is a random variable. The game ends when the token leaves Ω. At this point the token will be in the boundary strip of width εϕ(ε) given by
Γεϕ(ε)={x∈RN∖Ω:dist(x,∂Ω)≤εϕ(ε)}. |
We denote by xτ∈Γεϕ(ε) the first point in the sequence of game states that lies in Γεϕ(ε). In other words, τ is the first time we hit Γεϕ(ε) (τ is a stopping time for this game). The payoff is determined by two given functions: g:RN∖Ω→R, the final payoff function, and f:Ω→R, the running payoff function. We require g to be continuous, f uniformly continuous and both bounded functions. When the game ends, the total payoff is given by
g(xτ)−12ε2τ−1∑l=0[f(xl)]1k. |
Player Ⅱ earns this amount and Player Ⅰ loses it (Player Ⅰ earns −g(xτ)+12ε2∑τ−1l=0[f(xl)]1k).
A strategy SI for Player Ⅰ is a function defined on the partial histories that gives the values of αj at every step of the game
SI(x0,x1,…,xn)=(αj)j=1,…,k∈Ikε |
and that for a given value of j returns a ij−dimensional subspace S
SI(x0,x1,…,xn,j)=S. |
We call SI both functions to avoid overloading the notation. A strategy SII for Player Ⅱ is a function defined on the partial histories that gives a unitary vector in a prescribed subspace S at every step of the game
SII(x0,x1,…,xn,S,αj)=v∈S. |
When the two players fix their strategies SI and SII we can compute the expected outcome as follows: Given the sequence x0,…,xn with xj∈Ω the next game position is distributed according to the probability
πSI,SII(x0,…,xn,A)=12kk∑j=1δxn+ε√αjv(A)+δxn−ε√αjv(A). |
where
(αj)j=1,…,k=SI(x0,x1,…,xn) |
and
v=SII(x0,…,xn,SI(x0,…,xn,j),αj). |
By using the Kolmogorov's extension theorem and the one-step transition probabilities, we can build a probability measure Px0SI,SII on the game sequences H∞. We denote by Ex0SI,SII the corresponding expectation. Then, when starting from x0 and using the strategies SI,SII, the expected payoff is
Ex0SI,SII[g(xτ)−12ε2τ−1∑l=0[f(xl)]1k]. | (3.1) |
The value of the game for Player I is given by
uεI(x0)=infSIsupSIIEx0SI,SII[g(xτ)−12ε2τ−1∑l=0[f(xl)]1k], |
while the value of the game for Player II is given by
uεII(x0)=supSIIinfSIEx0SI,SII[g(xτ)−12ε2τ−1∑l=0[f(xl)]1k]. |
Intuitively, the values uεI(x0) and uεII(x0) are the best expected outcomes each player can guarantee when the game starts at x0. If uεI=uεII, we say that the game has a value and we denote it by uε:=uεI=uεII.
Let us observe that the game ends almost surely, and therefore the expectation (3.1) is well defined. Let us be more precise at this point. If we consider the square of the distance to x0, at every step, this quantity increases by at least ε2 with probability 12k (a value of j such that αj≥1 is selected with probability at least 1k and given a direction v at least one of the vectors xn±ε√αjv is at least at a distance ε2 greater that the distance from xn to the initial point). As the distance to x0 is bounded (since we assumed that Ω is bounded), with a positive probability the game ends after a finite number of steps. Iterating this argument we get that the game ends almost surely. See Lemma 13 for details concerning this argument.
To see that the game has a value, we first observe that we have existence of uε, a function that satisfies the DPP. The existence of such a function can be seen by Perron's method. In fact, the operator given by the right-hand side of the DPP, that is,
u↦infαj∈Ikε1kk∑j=1infdim(S)=ijsupv∈S,|v|=1{12u(x+ε√αjv)+12u(x−ε√αjv)}−12ε2[f(x)]1k, |
is in the hypotheses of the main result of [17].
Now, concerning the value functions of our game, we know that uεI≥uεII (this is immediate from their definition). Hence, to obtain the desired result, it is enough to show that uε≥uεI and uεII≥uε.
Given η>0 we can consider the strategy S0II for Player Ⅱ that at every step almost maximize u(x+ε√αjv)+u(x−ε√αjv), that is
S0II(x0,x1,…,xn,S,αj)=w∈S |
such that
{12uε(x+ε√αjw)+12uε(x−ε√αjw)}≥supv∈S,|v|=1{12uε(x+ε√αjv)+12uε(x−ε√αjv)}−η2−(k+1). |
We have
Ex0SI,S0II[uε(xk+1)+ε2k∑n=0f(xn)−η2−(k+1)|x0,…,xk]≥infαj∈Ikε1kk∑j=1infdim(S)=ijsupv∈S,|v|=1{12u(x+ε√αjv)+12u(x−ε√αjv)}−η2−(k+1)+12ε2k∑n=0f(xn)−η2−(k+1)≥uε(xk)−12ε2f(xk)+12ε2k∑n=0f(xn)−η2−k≥uε(xk)+12ε2k−1∑n=0f(xn)−η2−k, |
where we have estimated the strategy of Player Ⅰ by inf and used the DPP. Then,
Mk=uε(xk)+12ε2k−1∑n=0f(xn)−η2−k |
is a submartingale.
Now, we have
uεII(x0)=supSIIinfSIEx0SI,SII[g(xτ)+12ε2τ−1∑n=0f(xn)]≥infSIEx0SI,S0II[g(xτ)+12ε2τ−1∑n=0f(xn)−η2−τ]≥infSIlim infk→∞Ex0SI,S0II[Mτ∧k]≥infSIEx0SI,S0II[M0]=uε(x0)−η, |
where τ∧k=min(τ,k), and we used the optional stopping theorem for Mk. Since η is arbitrary this proves that uεII≥uε. An analogous strategy can be considered for Player Ⅰ to prove that uε≥uεI.
Given a solution to the DPP we have proved that it coincides with the game value. Then, the game value satisfies the DPP and, even more, any solution coincides with it. Uniqueness follows. We have proved Theorem 1.
Remark 10. From our argument it can be deduced that
uε(x0)=supSIIinfSIEx0SI,SII[uε(x˜τ)+12ε2˜τ−1∑n=0f(xn)] |
for any stopping time ˜τ≤τ. That is, as long as the game has not ended we can separate the payoff as the cumulative amount payed during the game, 12ε2∑τ−1n=0f(xn) and the expected one for the rest of the game, uε(x˜τ).
Remark 11. We have a comparison principle for solutions to the DPP. Assume that f1≥f2 and that g1≤g2 then the corresponding solutions verify
uε1≤uε2 |
in Ω. In terms of the game this is quite intuitive since playing with g1 and f1 the palyers have a larger final payoff and a smaller running playoff than playing with g2 and f2.
Now our aim is to pass to the limit in the values of the game. We aim to prove that, along a subsequence,
uε→u,as ε→0 |
uniformly in ¯Ω and then obtain in this limit process a viscosity solution to (1.1).
To obtain a convergent subsequence uε→u we will use the following Arzela-Ascoli type lemma. For its proof see Lemma 4.2 from [21].
Lemma 12. Let {uε:¯Ω→R, ε>0} be a set of functions such that
1) there exists C>0 such that |uε(x)|<C for every ε>0 and every x∈¯Ω,
2) given η>0 there are constants r0 and ε0 such that for every ε<ε0 and any x,y∈¯Ω with |x−y|<r0 it holds
|uε(x)−uε(y)|<η. |
Then, there exists a uniformly continuous function u:¯Ω→R and a subsequence still denoted by {uε} such that
uε→uuniformlyin¯Ω, |
as ε→0.
Our task now is to show that the family uε satisfies the hypotheses of the previous lemma.
Lemma 13. There exists C=C(Ω)>0 such that
Ex0SI,SII[τ]≤Cε−2. |
for every ε>0, SI, SII and x0∈Ω.
Proof. Here we write E for Ex0SI,SII. We consider R>0 such that Ω⊂BR(0) and Mn=‖xn‖2. Given that the token lies at xn, we have that
E[Mn+1|xn]=12kk∑j=1‖xn+ε√αjvj‖2+‖xn−ε√αjvj‖2=1kk∑j=1‖xn‖2+ε2αj‖vj‖2=‖xn‖2+ε21kk∑j=1αj≥‖xn‖2+ε2k∏j=1αj=‖xn‖2+ε2=Mn+ε2, |
where vj stands for the selected unitary vector when that value of j is chosen.
We have obtained
E[Mn+1|Mn]≥Mn+ε2. |
Hence,
Mn−nε2 |
is a submartingale. According to the optional stopping theorem for submartingales
E[Mτ∧n−(τ∧n)ε2]≥M0. |
Therefore
E[τ∧n]ε2≤E[Mτ∧n]−M0≤E[Mτ∧n]≤R2. |
By taking limit in n, we obtain a bound for the expected exit time,
E[τ]≤R2ε−2, |
as desired.
Corollary 14. There exists C=C(Ω,f,g)>0 such that |uε(x)|<C for every ε>0 and every x∈¯Ω.
Proof. Since Ex0SI,SII[τ]≤Cε−2, we have
|uε(x0)|≤supSIIinfSIEx0SI,SII[|g(xτ)|+12ε2τ−1∑n=0|f(xn)|]≤maxg+Cmaxf. |
To prove that uε satisfies the second hypothesis we will make the following geometric assumption on the domain. Given y∈∂Ω we assume that for every δ>0 there exists w∈RN of norm 1 and θ>0 such that
{x∈Ω:⟨w,x−y⟩<θ}⊂Bδ(y). | (3.2) |
This condition is equivalent to Ω being strictly convex as proved in Section 2.
Lemma 15. Given η>0 there are constants r0 and ε0 such that for every ε<ε0 and any x,y∈¯Ω with |x−y|<r0 it holds
|uε(x)−uε(y)|<η. |
Proof. The case x,y∈Γεϕ(ε) follows from the uniform continuity of g in Γεϕ(ε). Since the rules of the game do not depend on the point, the case x,y∈Ω follows from the case x∈Ω and y∈Γεϕ(ε). The argument is as follows. Suppose that we want to prove that uε(x)−uε(y)<η, that is
supSIIinfSIExSI,SII[g(xτ)+12ε2τ−1∑n=0f(xn)]−sup˜SIIinf˜SIEy˜SI,˜SII[g(yτ)+12ε2τ−1∑n=0f(yn)]<η. |
Then, it is enough to show that given S0II and ˜S0I (strategies for Player Ⅱ in the game starting at x and for Player Ⅰ in the game starting at y, respectively) there exists ˜S0II and S0I such that
ExS0I,S0II[g(xτ)+12ε2τ−1∑n=0f(xn)]−Ey˜S0I,˜S0II[g(yτ)+12ε2τ−1∑n=0f(yn)]<η. |
We consider the strategies ˜S0II and S0I that mimic S0II and ˜S0I, that is
˜S0II(y,y1,…,yn)=S0II(x,y1−y+x,…,yn−y+x) |
and
S0I(x,x1,…,xn)=˜S0II(y,x1−x+y,…,xn−x+y). |
Even more, we couple the random steps, then we have that when the token lies at xn, in the other games it lies at yn=xn−x+y. We call E the common expectation of the coupled processes. We proceed in this way until one of the games ends, at time ˜τ, that is for the first time xn∈Γεϕ(ε) or yn∈Γεϕ(ε). By Remark 10 it is enough to show that
E[uε(x˜τ)−uε(y˜τ)+12ε2˜τ−1∑n=0(f(xn)−f(yn))]<η. |
At every point we have |xl−yl|=|x−y|, and the desired estimate follow from the one for xn∈Ω, yn∈Γεϕ(ε) or for xn,yn∈Γεϕ(ε) together with the uniform continuity of f and the bound for the exit time obtained in Lemma 13.
Now, we can concentrate on the case x∈Ω and y∈Γεϕ(ε). Due to the uniform continuity of g in Γεϕ(ε), we can assume that y∈∂Ω. In fact, if we have the bound valid for points on the boundary we can obtain a bound for a generic point y∈Γεϕ(ε) just by considering z∈∂Ω in the line segment between x and y and using the triangular inequality.
In this case we have
uε(y)=g(y), |
and we need to obtain a bound for uε(x). We observe that, for any possible strategy of the players (for any possible choice of the direction v at every point) we have that the projection of xn in the direction of the a fixed vector w of norm 1,
⟨xn−y,w⟩ |
is a martingale. We fix an arbitrary pair of strategies SI and SII, we denote by P=PxSI,SII the corresponding probability playing with these strategies and E=ExSI,SII the corresponding expectation. We take δ>0 and consider xτ, the first time x leaves Ω or Bδ(y). Hence, we have
E⟨xτ−y,w⟩≤⟨x−y,w⟩≤d(x,y)<r0. |
From the geometric assumption on Ω, by choosing w as in Lemma 6, we have that
⟨xn−y,w⟩≥−ϕ(ε)ε |
because at every step the token moves at most √αiε≤ϕ(ε)ε. Therefore, we obtain
P(⟨xτ−y,w⟩>r1/20)r1/20−(1−P(⟨xτ−y,w⟩>r1/20))ϕ(ε)ε<r0. |
Then, we take ε0>0 such that ϕ(ε)ε≤r0 for every ε<ε0 and we get
P(⟨xτ−y,w⟩>r1/20)<2r1/20. |
We consider the corresponding θ such that (3.2) holds, this implies
{dist(xτ,y)>δ}⊂{⟨xτ−y,w⟩>θ} |
and hence, for r1/20<θ, we can conclude that
P(d(xτ,y)>δ)<2r1/20. |
Repeating the argument in Lemma 13 for Mn=‖xn−y‖2 we can show that
E[τ]≤δ2ε−2. |
Therefore,
|uε(x)−g(y)|≤E[|g(xτ)−g(y)||d(xτ,y)≤δ]P(d(xτ,y)≤δ)+P(d(xτ,y)>δ)(2maxg+δ2maxf)≤supx∈Bδ(y)|g(x)−g(y)|+2r1/20(2maxg+δ2maxf)<η |
if r0 and δ are small enough.
From Corollary 14 and Lemma 15 we have that the hypotheses of the Arzela-Ascoli type lemma, Lemma 12, are satisfied. Hence we have obtained uniform convergence of uε along a subsequence.
Corollary 16. Let uε be the values of the game. Then, along a subsequence,
uε→u,asε→0, |
uniformly in ¯Ω.
Now, we prove that the uniform limit of uε is a viscosity solution to the limit PDE problem.
Theorem 17. The uniform limit of the game values uε, denoted by u, is a viscosity solution to
{k∏j=1λij(D2u)=f,inΩ,u=g,on∂Ω. | (3.3) |
Proof. First, we observe that since uε=g on ∂Ω it is immediate, form the uniform convergence, that u=g on ∂Ω. Also, notice that Lemma 12 gives that a uniform limit of uε is a continuous function.
To check that u is a viscosity supersolution to ∏kj=1λij(D2u)=f in Ω, let φ be a paraboloid that touches u from below at x∈Ω, and with eigenvalues of the Hessian that verify λij(D2φ(x))>0. We need to check that
k∏j=1λij(D2φ(x))−f(x)≤0. |
As uε→u uniformly in ¯Ω we have the existence of a sequence xε such that xε→x as ε→0 and
uε(z)−φ(z)≥uε(xε)−φ(xε)−ε3 |
(notice that uε is not continuous in general). As uε is a solution to (1.6),
uε(x)=infαj∈Ikε1kk∑j=1infdim(S)=ijsupv∈S,|v|=1{12uε(x+ε√αjv)+12uε(x−ε√αjv)}−ε22[f(x)]1k |
we obtain
0≥infαj∈Ikε1kk∑j=1infdim(S)=ijsupv∈S,|v|=1{12φ(xε+ε√αjv)+12φ(xε−ε√αjv)−φ(xε)}−ε22[f(xε)]1k−ε3. |
Now, using the second-order Taylor expansion of φ,
φ(y)=φ(x)+∇φ(x)⋅(y−x)+12⟨D2φ(x)(y−x),(y−x)⟩ |
we get
φ(xε+ε√αjv)=φ(xε)+ε√αj∇φ(xε)⋅v+ε2αj12⟨D2φ(xε)v,v⟩ | (3.4) |
and
φ(xε−ε√αjv)=φ(xε)−ε√αj∇φ(xε)⋅v+ε2αj12⟨D2φ(xε)v,v⟩. | (3.5) |
Therefore,
0≥infαj∈Ikε1kk∑j=1infdim(S)=ijsupv∈S,|v|=1ε22αj⟨D2φ(xε)v,v⟩−12ε2[f(xε)]1k−ε3. |
Dividing by ε22 we get
[f(xε)]1k≥infαj∈Ikε1kk∑j=1αjinfdim(S)=ijsupv∈S,|v|=1⟨D2φ(xε)v,v⟩−2ε. |
We have
infdim(S)=ijsupv∈S,|v|=1⟨D2φ(xε)v,v⟩=λj(D2φ(xε)) |
by the Courant-Fischer min-max principle. Even more, since D2φ(xε)=D2φ(x), we have λj(D2φ(xε))=λj(D2φ(x)). Hence, we conclude that
[f(xε)]1k≥infαj∈Ikε1kk∑j=1αjλj(D2φ(x))+o(1). |
Using Lemma 8 to pass to the limit as ε→0 we get
[f(x)]1k≥(k∏j=1λij(D2φ(x)))1k, |
that is,
k∏j=1λij(D2φ(x))≤f(x). |
Now, to check that u is a viscosity subsolution to ∏kj=1λij(D2u)=f in Ω, let ψ be a paraboloid that touches u from above at x∈Ω. We want to see that λij(D2ψ(x))≥0 for every j=1,…,k and
k∏j=1λij(D2ψ(x))−f(x)≥0. |
As uε→u uniformly in ¯Ω we have the existence of a sequence xε such that xε→x as ε→0 and
uε(z)−ψ(z)≥uε(xε)−ψ(xε)−ε3 |
(notice that uε is not continuous in general). Arguing as before we obtain
0≤infαj∈Ikε1kk∑j=1infdim(S)=ijsupv∈S,|v|=1{12ψ(xε+ε√αjv)+12ψ(xε−ε√αjv)−ψ(xε)}−12ε2[f(xε)]1k−ε3. |
Using a Taylor expansions we arrive to
12ε2[f(xε)]1k≤infαj∈Ikε1kk∑j=1infdim(S)=ijsupv∈S,|v|=1ε22αj⟨D2ψ(xε)v,v⟩−ε3=ε22infαj∈Ikε1kk∑j=1αjλij(D2ψ(x))−ε3. |
Since f≥0 by Lemma 9 we get that λij(D2ψ(x))≥0. Dividing by ε22 and using Lemma 8 as before to pass to the limit as ε→0 we get
[f(x)]1k≤(k∏j=1λij(D2ψ(x)))1k, |
that is,
k∏j=1λij(D2ψ(x))≥f(x). |
This concludes the proof.
Finally, since problem (1.1) has a unique solution, see Remark 5, Theorem 2 follows.
In this section we describe a one-player/control problem presented in the introduction in order to approximate solutions to the Monge-Ampère equation. As before, let Ω⊂RN be a bounded open set and fix ε>0. Also take a positive function ϕ(ε) such that
limε→0ϕ(ε)=∞andlimε→0εϕ(ε)=0. |
A token is placed at x0∈Ω and Player Ⅰ (the controller) chooses coefficients (αi)i=1,…,N∈INε where
INε={(αi)i=1,…,N∈RN:N∏i=1αi=1 and 0<αi<ϕ2(ε)}. |
The player/controller also chooses an orthonormal basis of RN, V=(v1,...,vN). Then, the token is moved to x±ε√αivi with equal probabilities. After the first round, the game continues from the new position x1 according to the same rules.
As before, the game ends when the token leaves Ω. We denote by xτ the first point in the sequence of game states that lies outside Ω, so that τ is a stopping time for this game. The payoff is determined by two given functions: g:RN∖Ω→R, the final payoff function, and f:Ω→R, the running payoff function. We require g to be continuous, f uniformly continuous and both bounded functions. When the game ends, the total payoff is given by
g(xτ)−12ε2τ−1∑l=0[f(xl)]1N. |
When the player/controller fixes a strategy SI the expected payoff is given by
Ex0SI[g(xτ)−12ε2τ−1∑l=0[f(xl)]1N]. | (4.1) |
The player/controller aims to minimize the expected payoff, hence the value of the game is defined as
uεI(x0)=infSIEx0SI[g(xτ)−12ε2τ−1∑l=0[f(xl)]1N]. |
Let us observe that the game ends almost surely, and therefore the expectation (4.1) is well defined. We can proceed as before to prove that the value of the game coincides with the solution to
{uε(x)=infV∈O infαi∈INε{1NN∑i=112uε(x+ε√αivi)+12uε(x−ε√αivi)}−12ε2[f(x)]1Nx∈Ω,uε(x)=g(x)x∉Ω. | (4.2) |
We first observe that we have existence of solutions to (4.2) and we can apply the main result of [17]. Now, given η>0 we can consider the strategy S∗I for the player that at every step of the game xk almost realizes the infimum, that is, the player chooses coefficients ^αi and an orthonormal basis (ˆvi) such that
infV∈O infαi∈INε{1NN∑i=112uε(xk+ε√αivi)+12uε(xk−ε√αivi)}≥1NN∑i=112uε(xk+ε√^αiˆvi)+12uε(xk−ε√^αiˆvi)−η2k+1. |
Playing with this strategy we have
Ex0S∗I[uε(xk+1)+12ε2k∑n=0f(xn)−η2k+1|x0,…,xk]≤infV∈O infαi∈INε{1NN∑i=112uε(xk+ε√αivi)+12uε(xk−ε√αivi)}−η2k+1+12ε2k∑n=0f(xn)−η2k+1=uε(xk)−12ε2f(xk)+12ε2k∑n=0f(xn)−η2k≤uε(xk)+12ε2k−1∑n=0f(xn)−η2k. |
Then,
Mk=uε(xk)+12ε2k−1∑n=0f(xn)−η2k |
is a supermartingale.
From this fact, arguing as before, we get
uεI(x0)=infSIEx0SI[g(xτ)+12ε2τ−1∑n=0f(xn)]≤Ex0S∗I[g(xτ)+12ε2τ−1∑n=0f(xn)]≤uε(x0)−η. | (4.3) |
Since η is arbitrary we obtain that
uεI(x)≤uε(x). |
To obtain the reverse inequality, we fix an arbitrary strategy SI for the player and we observe that
Mk=uε(xk)+12ε2k−1∑n=0f(xn) |
is a submartingale. Indeed,
Ex0SI[uε(xk+1)+12ε2k∑n=0f(xn)|x0,…,xk]≥infV∈O infαi∈INε{1NN∑i=112uε(xk+ε√αivi)+12uε(xk−ε√αivi)}+12ε2k∑n=0f(xn)=uε(xk)−12ε2f(xk)+12ε2k∑n=0f(xn)=uε(xk)+12ε2k−1∑n=0f(xn). |
Taking infimum over all the strategies SI we get
uεI(x)≥uε(x). |
Given a solution to the DPP we have proved that it coincides with the game value. Then, the game value is characterized as the unique solution to the DPP.
Now our aim is to pass to the limit in the values of the game and obtain that, along a subsequence,
uε→u,as ε→0 |
uniformly in ¯Ω.
To obtain a convergent subsequence uε→u we will use again the Arzela-Ascoli type lemma, Lemma 12. So our task now is to show that the family uε satisfies the hypotheses of the previous lemma.
First, we observe that Lemma 13 still holds here. Hence, we have that there exists a constant C=C(Ω)>0 such that
Ex0SI[τ]≤Cε−2. |
for every ε>0, any strategy SI and x0∈Ω. As an immediate consequence we get that there exists C=C(Ω,f,g)>0 such that
|uε(x)|<C |
for every ε>0 and every x∈¯Ω. In fact, using that Ex0SI,SII[τ]≤Cε−2, we have
|uε(x0)|≤infSIEx0SI[|g(xτ)|+12ε2τ−1∑n=0|f(xn)|]≤maxg+Cmaxf. |
Now we observe that uε satisfies the second hypothesis of the Arzela-Ascoli type lemma. We want to see that, given η>0 there are constants r0 and ε0 such that for every ε<ε0 and any x,y∈¯Ω with |x−y|<r0 it holds
|uε(x)−uε(y)|<η. |
The proof of this estimate is analogous to the proof of Lemma 15. In fact, the case x,y∈Γεϕ(ε) follows from the uniform continuity of g in Γεϕ(ε). As before, since the rules of the game do not depend on the point, the case x,y∈Ω follows from the case x∈Ω and y∈Γεϕ(ε); see the proof of Lemma 15. For the case x∈Ω and y∈Γεϕ(ε) we can argue as before considering the projection of xk in the direction of the a fixed vector w of norm 1, choosed as in Lemma 6,
⟨xk−y,w⟩. |
This projection is a martingale. Then, with the same computations used before (see the proof of Lemma 15) se obtain
|uε(x)−g(y)|<η |
if |x−y| is small enough.
From these computations we have obtained uniform convergence of uε along a subsequence,
uε→u, as ε→0, |
uniformly in ¯Ω. Finally, we observe that the uniform limit of uε is a viscosity solution to the limit PDE problem. This follows from the same computation done in the proof of Theorem 3.3. We observe that Lemmas 7–9 include the case where each eigenvalue is selected once, that is the Monge-Ampère equation. This concludes the proof of Theorem 3.
One can also study a game for Monge-Ampère in which the possible movements are not discrete.
In fact, for a small ε and a fixed matrix A∈SN consider the following random walk in a bounded domain Ω, from x the next position is given by x+Ay with y∈Bε(0) being chosen with uniform distribution in Bε(0). If we fix a final payoff function g in RN∖Ω and we set
vε(x)=Ex(g(xτ)) |
with τ the first time when this random walk leaves the domain (τ is a stopping time for this process), then, it follows that vε verifies
vε(x)=−∫Bε(0)vε(x+Ay)dy |
for x∈Ω and vε(x)=g(x) for x∈RN∖Ω. These value functions converge to a solution to
{trace(AtD2uA)=0 in Ω,u=g on ∂Ω. |
Now, the game/control problem for Monge-Ampère runs as follows: at each turn the player/controller choose a matrix A in the set
A={A∈SN×N:detA=1 and 0<A≤ϕ(ε)I}. |
Then the new position of the game goes to x+Ay with y∈Bε(0) chosen with uniform probability. We add a running payoff −N2(N+2)(f(x))1/Nε2 and a final payoff, g(x). In this case the DPP for the value of the game reads as
uε(x)=infdetA=1A≤ϕ(ε)I−∫Bε(0)uε(x+Ay)dy−N2(N+2)(f(x))1/Nε2. | (5.1) |
This DPP is related to an asymptotic mean value formula for Monge-Ampère; see [3].
With similar ideas as the ones used before (assuming that Ω is strictly convex) one can show that
uε→u, as ε→0, |
where the limit u is characterized as the unique convex viscosity solution to
{det(D2u)=infdet(A)=1trace(AtD2uA)=f in Ω,u=g on ∂Ω. |
Remark 18. Observe that for uε to be a solution to (5.1) it must be measurable so that the integral in the right-hand side is well defined. Therefore, in the construction of a solution to the DPP by Perron's method (as in [17]) one has to take into account this measurability issue. The set of sub solutions should be restricted to bounded measurable functions, and we have to check that if u and f are bounded measurable functions, then T(x) given by
T(x)=infdetA=1A≤ϕ(ε)I−∫Bε(0)u(x+Ay)dy−N2(N+2)(f(x))1/Nε2 |
is also measurable. This fact holds since the only problematic term is the uncountable infimum and we have
infdetA=1A≤ϕ(ε)I−∫Bε(0)u(x+Ay)dy=infA∈QN×NdetA=1A≤ϕ(ε)I−∫Bε(0)u(x+Ay)dy. | (5.2) |
The right-hand side is a countable infimum and the equality (5.2) follows from the absolute continuity of the mapping E↦∫Eu(y)dy.
P. Blanc partially supported by the Academy of Finland project no. 298641 and PICT-2018-03183 (Argentina). F. Charro partially supported by a Wayne State University University Research Grant, and grants MTM2017-84214-C2-1-P (Spain) and PID2019-110712GB-I100 (Spain), funded by MCIN/AEI/10.13039/501100011033 and by "ERDF A way of making Europe." J. J. Manfredi supported by Simons Collaboration Grants for Mathematicians Award 962828. J. D. Rossi partially supported by CONICET grant PIP GI No 11220150100036CO (Argentina), PICT-2018-03183 (Argentina) and UBACyT grant 20020160100155BA (Argentina).
The authors declare no conflict of interest.
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1. | Pablo Blanc, Mikko Parviainen, Julio D. Rossi, A bridge between convexity and quasiconvexity, 2025, 0933-7741, 10.1515/forum-2024-0190 |