Lecture 1: Overview. The subdifferential of a convex function. Thereal Monge-Ampere operator - definition

There are many ways to introduce Monge-Ampere equations, including optimal transport as in previousclasses. In the following we will introduce it by considering geometry of surfaces in R

3. The simplest typeof surfaces in R

3 are graph of a function above a domain D ⊂ R2:

gr(u)

∣∣∣∣D

= (x, y, u(x, y))|(x, y) ∈ D,

Gaussian curvature can be computed by the following recipe: the unit normal can be computed by crossingtangent vectors (

1, 0,∂u

∂x

),

(0, 1,

∂u

∂y

)and normalizing, which gives at point (x, y, u(x, y))

~n =

(∂u∂x ,

∂u∂y ,−1

)√

1 + |∇u|2.

~n can be viewed as a map to the unit sphere ~n : D → S2 and the Gauss map is its differential:

G = d~n : D → TS2

Gaussian curvature is then defined as

K := det d~n =uxxuyy − u2

xy

(1 + |∇u|2)2

where K stands for Krummung (German for curvature). Thus we have

det∇2u = K(x, y)(1 + |∇u|2)2.

A Monge-Ampere type equation is the fully nonlinear PDE of the form

det∇2u = F (∇u(~x), u(~x), ~x)

in Rn. The case where F = 0 in n = 2 corresponds to K = 0, i.e., the graph of solution will be a flat surface

above the interior of domain D.

Weak Solution

The idea to define weak solution is to replace the equation of determinant of Hessian by equation of ameasure, i.e., for C2 function u, we replace det∇2u by the measure det∇2udx where dx = dx1 ∧ . . . ∧ dxn.Observe that we have for any C2 function u in domain E,∫

E

det∇2udx =

∫∇u(E)

dy = meas(∇u(E)),

1

since Jac(∇u) = det∇2u, that is, we really only need information about gradient ∇u. This inspires us todefine

Definition For u ∈ C1, MA(u) is a Boreal measure given by

MA(u)(E) = Lebesgue measure of ∇u(E)

for Borel set E.

Remark It is easy to check that this is indeed a Boreal measure, in particular, for any Borel set E the set∇u(E) is indeed a Borel set since ∇u is continuous for u ∈ C1. On the other hand note that

MA(u)(E) =

∫E

det∇2udx

for u ∈ C2, therefore MA(u) is indeed a generalization of measure det∇2udx.

Now what if u /∈ C1? Actually we can generalize Monge-Ampere measure to convex functions, this is doneby generalizing the notion of gradient:

Definition For u a finite convex function,

∂u(x) := y ∈ Rn|u(z) ≥ u(x) + 〈y, z − x〉 ∀z

is called the subdifferential of u at x and is in fact a convex set.

Remark The subdifferential of u can be understood as set of supporting planes. It is a singleton if u isdifferentiable at x, consist of just the tangent plane at x while when u is not differentiable there can be morethan one supporting plane:

Definition For u convex,MA(u)(E) = Lebesgue measure of ∂u(E)

is called Monge-Ampere measure of E with respect to u

We will show MA(u) is a nonnegative Borel measure

Claim A := E ⊆ D : ∂u(E) is Lebesgue measurable is a σ-algebra containing the Borel σ-algebra.

Proof will be given next time.

Lecture 2: The real Monge-Ampere operator - construction as aBorel measure using Alexandrov’s theorem. Solution of the Dirich-let problem for the homogeneous real Monge-Ampere equation us-ing an upper envelope

Recall MA(u)(E) := Lebesgue measure of ∂u(E) with u ∈ Cvx(D) for any Borel set E.

2

Theorem For convex function u, the Monge–Ampere measure MA(u) is well-defined, in other words, MA(u)is a Borel measure.

Theorem is a consequence of the following Claims:

Claim 1A := E ⊆ D : ∂u(E) is Lebesgue measurable

is a σ-algebra containing the Borel σ-algebra.

Claim 2 For any x1 6= x2

B := y ∈ Rn : y ∈ ∂u(x1) ∩ ∂u(x2), x1 6= x2is Lebesgue measurable and is in fact an Lebesgue null set. (Note that only proving ∂u(x1) ∩ ∂u(x2) forfixed x1 6= x2 is null set is not enough for subsequent steps!)

Proof of Claim 1: (1) Compact sets are in A: Let K be compact, ∂u(K) =⋃x∈K ∂u(x), let yi ⊂ ∂u(K)

and yi → y, then yi ∈ ∂u(xi) for some xi, there is a convergent subsequence, thus without loss of generalityassume xi → x and thus for all z, u(z) ≥ u(xi) + 〈z, yi〉 → u(x) + 〈z, y〉 therefore y ∈ ∂u(x) ⊂ ∂u(K) hence∂u(K) is closed thus Lebesgue measurable.

(2) For E ∈ A, the complement D\E ∈ A. One has

∂u(D\E) = (∂u(D)\∂u(E)) ∪ (∂u(D\E) ∩ ∂u(E))

where by Claim 2 ∂u(D\E) ∩ ∂u(E) is a Lebesgue null set, and the other two sets (∂u(D) and ∂u(E)) arealready Lebesgue measurable, the conclusion follows.

(3) A is closed under countable unions: it suffices to show closedness under union, let E1, E2 ∈ A,

∂u(E1 ∪ E2) = ∂u(E1) ∪ ∂u(E2)

is measurable since it is union of two Lebesgue measurable sets.

(4) A is closed under countable intersection. Let E1, E2 ∈ A, it is easy to check that

∂u(E1 ∩ E2) = (∂u(E1) ∩ ∂u(E2))\(∂u(E1\E2) ∩ ∂u(E2\E1))

where the last set ∂u(E1\E2) ∩ ∂u(E2\E1) is null set by Claim 2, the conclusion follows.

Proof of Claim 2: (1) Any convex function u over D is Lipschitz over ∀Γ b D, i.e. ∃C = C(Γ, D) suchthat

‖u‖C0,1(Γ) ≤ C

Proof : By (1) of proof of Claim 1, ∂u(Γ) is compact, therefore ∃R > 0 such that ∂u(Γ) ⊂ B(0, R). Letx1, x2 ∈ Γ, we want to show that |u(x2)− u(x1)| ≤ C|x2 − x1|. By definition of subdifferentials,

u(x) ≥ u(x1) + 〈α1, x− x1〉u(x) ≥ u(x2) + 〈α2, x− x2〉

3

for all x ∈ D, and with αi ∈ ∂u(xi) ⊂ B(0, R). Taking x = x2, x1 respectively in the above two inequalities,

R|x2 − x1| ≥〈α1, x1 − x2〉 ≥ u(x1)− u(x2)

R|x2 − x1| ≥〈α2, x2 − x1〉 ≥ u(x2)− u(x1)

where we used that |αi| ≤ R. Taking R as constant C above, the conclusion follows.

Corollary Non-differentiability set Nondiff(u) of a convex function u is a Lebesgue null set.

Proof : This follow from Rademacher’s theorem which states that Lipschitz function is differentiable almosteverywhere.

We will need the definition of Legendre transform:

Definition For u convex, the Legendre transform is a function u∗ given by

u∗(α) = supx

[〈α, x〉 − u(x)]

Note in particular that u∗ is also convex since it is pointwise supremum of convex (actually affine) functionsα 7→ 〈α, x〉 − u(x) thus locally Lipschitz by previous step.

By last corollary it suffices to showB ⊂ Nondiff(u∗)

The following two observations are easy to check:

• If α ∈ ∂u(x1) then the supremum in definition of u∗(α) is attained at x = x1.

• Assume α ∈ B, α ∈ ∂u(x1) ∩ ∂u(x2) with x1 6= x2, then by previous observation

u∗(α) = 〈α, x1〉 − u(x1) = 〈α, x2〉 − u(x2)

Let F (x) := 〈x, α〉 − u(x), then F is concave and achieves maximum at 2 distinct points, it followsthat F is constanton tx2 + (1− t)x1 for 0 ≤ t ≤ 1.

Claim: If α ∈ ∂u(x1) ∩ ∂u(x2), then u∗ is not differentiable at α.

Proof : α ∈ ∂u(x1) and α ∈ ∂u(x2) implies x1, x2 ∈ ∂u∗(α), therefore u∗ is not differentiable at α.

As a corollary we now have B ⊂ Nondiff(u∗) which is a null set by previous Corollary, this finishes the prooffor Claim 2. .

With Claim 1, MA(u)(E) is well-defined for any subset E ⊂ D, to prove the theorem we will also need tocheck that,

• MA(u)(E) ≥ 0 for all E. This is obvious since Lebesgue measure is nonnegative

4

• MA(u)(∅) = 0. This is also trivial.

• MA(u) has countable additivity, that is for any countable disjoint collection of subsets Ej we have

∑j

MA(u)(Ej) = MA(u)

⋃j

Ej

this also follow from Claim 2 since although the sets whose Lebesgue measure defines Monge–Amperemeasure may overlap, the overlaps are null sets. We have

MA(u)

∐j

Ej

= m

⋃j

⋃x∈Ej

∂u(x)

≤∑j

m

⋃x∈Ej

∂u(x)

=∑j

MA(u)(Ej)

MA(u)

∐j

Ej

= m

⋃j

⋃x∈Ej

∂u(x)

≥ m⋃

j

⋃Ej

∂u(x)−X

= m

⋃j

⋃x∈Ej

(∂u(x)−X)

= m

∐j

⋃x∈Ej

(∂u(x)−X)

=∑j

m

⋃x∈Ej

∂u(x)

−X

≥∑j

m ⋃x∈Ej

∂u(x)

−m(X)

=∑j

m

⋃x∈Ej

∂u(x)

=∑j

MA(u)(Ej),

whereX := α ∈ Rn|∃x1 6= x2 ∈ domu, α ∈ ∂u(x1) ∩ ∂u(x2)

is the set of subdifferentials coming from more than one points in dom(u) (i.e. domain of u), it is thenclear that ⋃

x∈Ej

(∂u(x)−X) and⋃x∈Ei

(∂u(x)−X)

are disjoint and that the set X is measure zero since it is a subset of the non-differentiable locus of afinite convex function (i.e., u∗).

This completes the proof of the Theorem.

Theorem 2.8 [1] For Ω ⊂ Rn a bounded strictly convex domain, g : ∂Ω → R continuous, the Dirichlet

problem for homogeneous Monge-Ampere equationMA(u) = 0 in Ω

u = g on ∂Ω

has unique solution u ∈ Cvx(Ω) ∩ C0(Ω).

Sketch of proof : Defineu = supa : a affine in Ω, a ≤ g on ∂Ω.

Note since u is defined as supremum of a family of convex functions, it is itself convex and we can talk aboutsubdifferentials of u. Next we will see that for this upper envelope and any point in interior of domain, thereis a point on the boundary that has the same subdifferential.

Claim Let u be the upper envelope defined above, then we have

∂u(Ω) ⊂ ∂u(∂Ω).

5

Proof : Let α ∈ ∂u(x0) for some x0 ∈ Ω. Define

a(x) := 〈α, x− x0〉,

note that this is an affine function in x. Since Ω is bounded, g(x)− a(x) attains minimum at some x1 ∈ ∂Ω.Observe that x 7→ a(x)+g(x1)−a(x1) is affine and is ≤ g on ∂Ω thus by definition of upper envelope functionu we have u(x) ≥ a(x) + g(x1)− a(x1) but two sides agree when x = x1 ∈ ∂Ω thus α ∈ ∂u(x1) ⊂ ∂u(∂Ω).

To show u ≥ g on the boundary we will need some barrier function. Geometrically u is the lower boundaryof the convex hull of the graph of boundary value function (x, g(x))|x ∈ ∂Ω

Lecture 3: The Cauchy problem for the homogeneous real Monge-Ampere equation. The Legendre transform in more detail. Convexhulls and the double Legendre dual: regularity and basic properties

We will move on to Cauchy problem of homogeneous Monge–Ampere equation which is more motivated inphysics but also more complicated and less studied. First consider boundary value problem:

MA(u) = 0 in [0, 1]× R→ R

u(0, ·) = u0 ∈ C∞ ∩ SCvx, ui : R→ R

u(1, ·) = u1 ∈ C∞ ∩ SCvx

where we specified some Dirichlet data at t = 0, t = 1. We have shown above that MA(u) = 0 iff Im(∂u)is an Lebesgue nulls set. It turns out that the solution fills in between u0 and u1 with a surface of zerocurvature and gives a foliation of the product domain [0, 1] × R with lines on which unit normal remain aconstant vector. We can give an analytic expression to the solution using partial Legendre transform definedas

u∗t = u∗(t, y) = supx

[〈x, y〉 − u(t, x)]

where ut(x) := u(t, x). To interpret Im(∂u) in terms of partial Lebesgue transform, we need some usefultricks. Below we assume derivative exists wherever derivative is taken. We have

∂u∗

∂t(t, y) =

∂

∂t

[〈(∇ut)−1(y), y〉 − u(t, (∇ft)−1(y))

]= −∂u

∂t(t, (∇ut)−1(y))

where we assume that ut ∈ C1 and strictly convex therefore ∇ut is always invertible and we used the factthat (∇ut) (∇ut)−1 = id. For u ∈ C1 this allow us to express the subdifferential of u as

∂u(t, x) =

(∂u

∂t(t, x),∇ut(x)

)=

(−∂u

∗

∂t(t, (∇ut)(x)),∇ut(x)

)therefore the subdifferential sets will lie inside graphs of functions u∗t where · stands for time derivative, i.e.

Im(∂u)∣∣t×Rn = gr

(−∂u

∗t

∂t

) ∣∣∣∣Im(∇ut)

Im(∂u) =⋃t

Im ∂ut

∣∣∣∣t×Rn

which suggests that we look for functions u such that u∗t is actually constant in t in which case all thesubdifferentials will lie in graph of a single function. If instead the graphs of u∗t fill up a 2-dimensional

6

region, it is possible that Monge-Ampere measure will charge it with a positive number. To be more precise,if u ∈ C2 and strictly convex, we have the following formula of second variation of Legendre transform:

∂2u∗t∂t2

∣∣∣∣∣(t,y)

= −∂2ut∂t2

∣∣∣∣∣(t,x)

+ 〈∇x∂ut∂t

∣∣∣∣∣(t,x)

,∇y∂ut∂t

∣∣∣∣∣(t,y)

〉

= −∂2ut∂t2

∣∣∣∣∣(t,x)

− 〈∇x∂ut∂t

,(∇2xut)−1∇x

∂ut∂t〉

where y = ∇ut(x). RHS vanishes if u solves HRMA. To see this we need a bit of linear algebra, for blockmatrix with D invertible we have

det

(A BC D

)= det(D) det(A−BD−1C)

therefore assuming ∇xu 6= 0 HRMA can be reduced to

det

(ut ∇xut

(∇xut)T ∇2xut

)= ut det∇2

xut − 〈(∇2xut)

∗∇xut,∇xut〉

where ∗ means adjoint matrix. We can divide by det∇2xu 6= 0 to get

u− 〈(∇2xu)−1∇xu,∇xu〉 = 0

using the first and second variation formula above we can show that partial Legendre transformation linearizesthe equation, giving

u∗t (y) = (1− t)u∗0 + tu∗1

where we can take another partial Legendre transform to get

u(t, x) = ((1− t)u∗0 + tu∗1)∗(x)

Now let us consider initial value problem on [0, T ]× Rn:

MA(u) = 0

u(0, ·) = u0

∂u

∂t(0, ·) = u0

where u0, u0 are given. Previous computation tells us we should expect solution of the form u∗t (y) =u∗0− tu0 (∇u0)−1 but we need to be more careful: in the boundary value case, all functions are assumed tobe C2 and strictly convex so ∇2

xu is always invertible while in IVP this is no longer true. Actually we candefine a maximal existence time

Tcvx := supt > 0 : u∗0 − t · u0 (∇u0)−1 is convex

which can be shown to be positive but is not always infinity. With linearization under partial Legendretransform it is not hard to see that if u0 is concave, then Tcvx =∞, otherwise Tcvx <∞.

7

In the following we will state a theorem about solution of this IVP where we switch our notation and recallthe form of solution to be

ψ(t, x) =(ψ∗0 − tψ0 (∇ψ0)−1

)∗where we are interested in application to Kahler geometry which will give some assumptions on ψ0 and ψ0,in particular Im∇ψ0 = P will be a compact convex polytope and ψ0 is assumed to be bounded on thatpolytope.

Theorem 1 [2]

1. ψ solves HRMA on the regular locus of ψ (i.e. where ψ is C1) In particular, on [0, Tcvx]× Rn

2. If T > Tcvx, ψ fails to sovle HRMA. Futhermore, MA(ψ) whose total mass is positive has an aprioriupper bound given by ∫

[0,T ]×Rn

MAψ ≤ Vol (epi(−u0)\ epi(−(u0)∗∗))

where epi stand for epigraph

Open Question: Does there exist a solution with poorer regularity? (It is shown in [2] that there does notexist any C1 solution)

Corollary If a solution exists, it must be C0,1 but not C1.

Conjecture Such a solution does not exist.

Lecture 4: Obstruction to the solution of the Cauchy problem forthe homogeneous real Monge-Ampere equation: upper and lowerbounds on the subdifferential and strict convexity of the Legendresubsolution

We will continue the sketch of the proof that

ψ(s, x) :=(ψ∗0 − sψ0 (∇ψ0)−1

)∗does not solve HRMA but does solve the equation on regular locus

Fact 1 ψs(x) := ψ(s, x) is strictly convex on Rn ⇔ ψ∗s (y) ∈ C1(P\∂P ).

Recall thatus(y) := u0 + su0

where u0 = ψ∗0 , u0 = −ψ0 (∇ψ0)−1 and recall the underlying assumption ∇xψs ∈ P a fixed closed compactconvex polytope with Im∇xψs = P . There is a standard result in convex analysis: a function is strictlyconvex iff its dual f∗ is C1 away from boundary. (c.f. Rockafellar)

Lemma 4.1

(i) ψ∗s = u∗∗s

(ii) Let ∆n+1 be the unit simplex in Rn+1 defined by

∆n+1 = λ = (λ1, . . . , λn+1 ∈ Rn+1 : λi ≥ 0,∑

λi = 1

8

then we haveu∗∗s (y) = infλ · (us(y1), . . . , us(yn+1)) : λ ∈ ∆n+1, yi ∈ P,

∑λiyi = y

Claim:

(y, u∗∗s (y)) =

n+1∑i

λ(yi, us(yi))

that is, in particular, the infimum in representation formula of the biconjugate function is actually attained.

Proof : We know that by convexity and since epi(u∗∗s ) = co(epi(us)), that

(y, u∗∗s (y)) =

m∑i=1

λi(yi, ri)

with ri ≥ us(yi), but since∑i λi(yi, us(yi)) also belongs to epigraph of us, we have ri = us(yi) for all i ∈ I

(see below), since otherwise the point (y, u∗∗s (y)) will lie directly above another point in convex hull of epius.

Now by a corollary to Caratheodory’s theorem we can take m = n+ 1

Definition Let I = i : λi > 0. We say that yi, i ∈ I are called upon by y.

Lemma f∗∗∗ = f∗ for any continuous function f .

Claim yi ∈ P\∂P

Proof : The proof relies on the formula

∂u∗∗s (y) =⋂i∈I

∂us(yi)

To show this, take δ ∈ ∂u∗∗s (y), by definition of subdifferential we have

u∗∗s (z) ≥ u∗∗s (y) + 〈z − y, δ〉 ∀z

equivalentlyu∗∗∗s (δ) + u∗∗s (y) = 〈δ, y〉

(where if δ is not necessarily from subdifferential, equality would have been replaced by ≥) Using the factthat f∗∗∗ = f∗ as f∗ is already convex, this is equivalent to

u∗s(δ) +∑i

λiu∗∗s (yi) = 〈δ,

∑λiyi〉

since u∗s(δ) + us(yi) ≥ 〈δ, yi〉, we haveu∗s(δ) + us(yi) = 〈δ, yi〉

for all i ∈ I, therefore δ ∈ ∂us(yi) for all i ∈ I. This finishes the proof of the formula.

Now the LHS of the formula is non-empty since y ∈ P\∂P and each ∂us(yi) is a singleton, therefore theformula implies that all ∂us(yi) has to be the same singleton, proving that u∗∗s is indeed C1.

Claim Let y ∈ P\∂P , x := ∇u∗∗s (y), then

co(−u0(v), v) : v ∈ γxψ(s, x) ⊂ ∂ψ(s, x) ⊂ co(−u0(v), v) : v ∈ ∂xψ(s, x)

9

where coA is convex hull of A and

∂xψ(s, x) := ∂ψs(x)

γxψ(s, x) := γψs(x)

and γf(x) is the reachable subdifferential of f at x defined by

γf(x) := δ : ∃xk ⊂ regular locus of f s.t. δ = limk∇f(xk)

Fact (Lemma 2.1) ∂f(x) = co γf(x)

Proof : Second inclusion: Let v ∈ γψ(s, x) by the above ∂ψ(s, x) = co γψ(s, x) and since taking convexhull perserves inclusion, it suffices to show

Claim: γψ(s, x) ⊂ (−u0(v), v) : v ∈ ∂xψ(s, x)

Proof : By definition there is (sk, xk) ∈ regular locus of ψ, such that

v = limk∇ψ(sk, xk) = lim

k(−u0 ∇xψsk(xk),∇xψsk(xk))

Lecture 5: Obstruction to the solution of the Cauchy problem forthe homogeneous real Monge-Ampere equation: completion of theproof of the main theorem. Subequations: definition

Recall that MA(ψ) 6= 0 if ψ = (u0 + su0)∗ where u0 = ψ∗0 and u0 = −ψ0 (∇ψ0)−1 with ψ0 ∈ C∞ ∩SCvx(Rn), (where SCvx stands for strictly convex functions) Im∇ψ0 = P , a compact convex polytope,ψ0 ∈ C∞ ∩ L∞(Rn) with T > Tcvx = infs > 0 : (u+ su0)∗∗ 6= (u+ su0). We want to show∫ T

0

∫Rn

MA(ψ) > 0

Lemma 12.4 Let y ∈ P\∂P , and set x := ∇u∗∗s (y). Then

co(−u0(v), v)v ∈ γxψ(s, x) ⊂ ∂ψ(s, x) ⊂ co(−u0(v), v) : v ∈ ∂xψ(s, x)

This provides us upper and lower bounds of the subdifferential ∂ψ(s × Rn). This does not quite implythat MA measure charges positively once T > Tcvx but it does imply the a priori upper bound of MA mass:∫ T

0

∫Rn

MA(ψ) ≤ Vol(epi(−u0)\ epi(−(u0)∗∗))

which is a partial regularity statement. This also raises the question of whether ψ is an optimal solution inthe sense that for arbitrary subsolution η of Cauchy problem∫ ∞

0

∫Rn

MA(η) ≥∫ ∞

0

∫Rn

MA(ψ) > 0.

Two sets play an important role in the proof of main theoerm of [2]

Definition

As = y ∈ P : us(y) 6= u∗∗s (y),A′s = y ∈ P\∂P : ∃v 6= y with ∇u∗∗s (y) = ∇us(y).

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Note that the two sets coincide for generic function. The proof will be tremendously more complicated ifthey are not the same and we won’t go into these details, so we will assume

As\∂P = A′s

This is indeed ture for generic data u0, u0. Now on As we can generate positive mass as soon as the lefthand side in Lemma 12.4, i.e. co(−u0(v), v) : v ∈ γxψ(s, x) contains more than one point, it will containthe line segment in between, this will be where the function is not differentiable. We have∫

MA(ψ) =

∫Σsing

MA(ψ)

where Σsing is the singular set defined not by non-C1 locus of ψ but by

Σreg(T ) =⋃

s∈[0,T ]

s × ∂us(int(P\(∂P ∪A′s)))

and Σsing(T ) = [0, T ]×Rn\Σreg(T ). We partition the singular set by some equivalence classes of the polytopegiven by

Q(s, y) := v ∈ P : ∇u∗∗s (v) = ∇u∗∗s (y)

for y ∈ P\∂P . The partition of As is give by

As\∂P =⋃v∈As

Q(s, y) ∩ (As\∂P )

defineQ(s, y) := co(−u0(v), v) : v ∈ γxψ(s,∇u∗∗s (y))

which lies above Q(s, y). We will show that this set is ‘big’. Intuitively the graph of −u0 is not concaveabove the set Q. Define

U(s, y) = πp(Q(s, y)\ gr(−u0)∣∣Q(s,y)

)

which is contained in Q(s, y), we have

Lemma 13.2 U(s, y) is a set with non-empty interior relative to Q(s, y)

Final step is to put everything together. For this we need a continuity result whose proof is easy

Lemma 8.2 s 7→ As is continuous as a set valued map.

Intuitively when s increases, the set where us is non-convex grows continuously. On a given time slice

∂ψ(s × Rn) ⊃⋃

v∈As\∂P

Q(s, v)

nowU(s, y) := Q(s, y) ∩ π−1

p (U(s, y))

is affine and does not intersect the graph of −u0 over U(s, y). Now the conclusion that MA charges positivemass follows from a Fubini type argument.

Dirichlet Duality

In this section we introduce the concept of subequations which is useful in the study of a large family ofequations from geometry depending only on Hessian of the unknown function. We consider equation theform

F(∇2u) = 0

11

where F : Sym2(Rn) ⊂ Rn2 → R is a continuous function on the space of symmetric n × n matrices. We

wish to study weak solution to this equation. For some choices of F there will be a natural notion of weaksolution. For this we associate to the function F a subset F ⊂ Sym2(Rn) such that for A ∈ Sym2(Rn) onehas F(A) = 0 if A ∈ ∂F , i.e., ∂F ⊂ F−1(0). Note there is in general no formula to derive the subset F fromgiven equation F , since the later is in general not unique.

Example Let F = tr, the corresponding equation of hessian is just the Laplacian equation:

∆u = 0

and for the associated subset we can take

F = A ∈ Sym2(Rn) : trA ≥ 0

We consider the Dirichlet problem of equationF(∇2u) = 0 in Ω

u = f on ∂Ω

To get guarantee both uniqueness and existence of solution for this equation, we need some assumption forF or equivalently for the subset F .

Definition (Definition 3.1 of [3]) A proper nonempty closed subset F ⊂ Sym2(Rn) is a Dirichlet set(or a subequation if it satisfies the positivity condition

F + P ⊂ F

whereP ≡ A ∈ Sym2(Rn) : A ≥ 0

denotes the set of nonnegative quadratic forms on Rn

Definition Suppose F ⊂ Sym2(Rn) is a Dirichlet set. The Dirichlet dual of F is the set

F =∼ (− IntF ) = −(∼ IntF )

where ∼ stands for complement and − just takes the negative matrices

Fact: The subequation condition can be characterized with dual condition

A ∈ F ⇔ A+B ∈ P ∀B ∈ F

Lecture 6: Subequations: basic properties. Subaffine functions.Functions of type F. Comparison with viscosity solutions

Recall again we are studying the Dirichlet problemF(∇2u) = 0 on Ω

u = g on ∂Ω

The general idea of studying such problem is find a set F ⊂ Sym2(Rn) such that A ∈ ∂F and F(A) = 0and to construct a set of functions that will be thought of as subsolutions/competitors. Once we haveconstructed this class, one can define u := sup(subsolutions) and prove u solves the Dirichlet problem, inother words F(∇2u) = 0 and it realizes boundary condition. For such a construction to work we will need

12

w1, w2 subsolutions ⇒ maxw1, w2 also a subsolution, that is the class of subsolutions to be closed undertaking maximum.

Definition 4.4 An u.s.c function u on X (some bounded domain in Rn) is of type F (we write u ∈ F (X)) if

u+ v ∈ SA(X), ∀v ∈ C2(X) ∩ F (X)

Definition u ∈ C2(X) is of type F if ∇2u(x) ∈ F for all x ∈ X.

Definition u ∈ SA(X) called subaffine functions if ∀K ⊂ X compact and a affine in K, we have maximumprinciple of the form

u ≤ a on ∂K ⇒ u ≤ a on K

When one generalize a class of funtion in the study of fully nonlinear PDE, it is wise to first look at convexfunctions, in other words one should first check if the technique works for Monge-Ampere equation. Theconcept of subaffine functions and dual Dirichlet set is thus motivated by:

Proposition 2.5-2.6

u ∈ Cvx(X)⇔ u+ v ∈ SA(X), ∀v ∈ SA(X)

u ∈ SA(X)⇔ u+ v ∈ SA(X), ∀v ∈ Cvx(X)

Fact (Lemma 2.2) u /∈ SA(X) iff ∃x0 ∈ X, a affine, ε > 0 such that

(u− a)(x) ≤ −ε|x− x0|2 near x0

u(x0) = a(x0)

Consider SA(X), the class of subaffine functions on X and P (X) = Cvx(X) the class of convex functionson X. These two classes of functions each characterize the other one, there is clearly an underlying dualitybetween them. Thus we wish to define a notion of duality such that P (X) = SA(X) and SA(X) = P (X).It is easier if we start on the level of matrices, what allows one to do that is the following theorem due toSlodkowski ([4] Prop 2.4). Note that for F ⊂ Sym2(Rn) we defined

F = (− intF )c

it follows from definition that ˜F = F . (Note Slodkowski’s theory is more complicated where double dualneed not equal to original class while triple dual is always the same with single dual) In order to ensuresolution of Dirichlet problem some kind of ellipticity is called form, it comes in the form of our positivitycondition

F + P ⊂ F

(Note if one take arbitrary, for instance some hyperbolic equation, Dirichlet problem may very well beill-posed)

Lemma 4.3 u ∈ C2(X) ∩ F (X) iff u+ v ∈ SA(X) for all v ∈ C2(X) ∩ F (X)

Lecture 7: Subequations: rays sets and boundary defining functions

For Dirichlet problem given by F(∇2u) = 0 in Ω

u = f on ∂Ω

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Similar to Perron solution to Dirichlet problem of Laplacian equation, we will eventually construct solutionof the type u = supg∈F (Ω) g and we will need to make some geometric assumption on the boundary toguarantee that the Perron solution is continuous up to the boundary. Note that there are examples wherethe Perron solution fails to be continuous up to the boudary, e.g. for HRMA when boundary is not strictlyconvex.

Theorem 5.12 [3] F is a subequation, then ∂Ω is strictly ~F -convex iff ∃ global defining function ρ ∈ C∞(Ω)

for ∂Ω which is strictly of type ~F on Ω.

First we need to define some concepts in this theorem

Definition A smooth function ρ ∈ C∞(Ω) is called a global defining function for ∂Ω if Ω = ρ < 0 and∇ρ 6= 0 on ∂Ω.

Definition 5.8 [3]~F := A ∈ Sym2(Rn) : tA ∈ F for t 1

is the Dirichlet ray associated to F a given subequation.

Definition 5.1 [3] ∂Ω is strictly ~F -convex if

∇2ρ(x)

∣∣∣∣Tx∂Ω

= B

∣∣∣∣Tx∂Ω

(**)

for some B ∈ int ~F . ∀x ∈ ∂Ω and for some ρ ∈ C∞ defined on a neighborhood Ux of x ∈ Rn such that

Ux ∩ Ω = ρ < 0 ∩ Ux, ∇ρ∣∣∣∣U∩∂Ω

6= 0. Such ρ is called a local defining function

Examples Consider the Laplacian equation∆u = 0 in Ω

u = f on ∂Ω

let F = A ∈ Sym2(Rn) : trA ≥ 0 thus ~F = F and int~F = A : trA > 0. Any smooth doain is strictly~F -convex (*)

Exercise 1 For any smooth Ω (i.e. with smooth ∂Ω) there is a global defining function (hint: use partitionof unity argument) such that

∇2ρ(x)

∣∣∣∣Tx∂Ω

= B

∣∣∣∣Tx∂Ω

Exercise 2 Show (*), i.e. that any smooth domain is strictly−→P -convex as above

We know that when ∂Ω is smooth Dirichlet problem is solvable and solution is unique. We will see thatsimilar conclusion holds for general subequation.

Lemma 5.2 Definition 5.1 is independent of choice of ρ.

Corollary 5.4 ∂Ω is strictly ~F -convex at x ∈ ∂Ω iff II(x) = B

∣∣∣∣Tx∂Ω

for some B ∈ int ~F , where II is the

second fundamental form of ∂Ω as a smooth hypersurface.

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Idea of proof : Choose the signed distance function

ρ = δ(x) =

−dist(x, ∂Ω) x ∈ Ω

dist(x, ∂Ω) x /∈ Ω

Exercise 3 Orthogonally decompose Rn = R∇δ(x)⊕ (∇δ(x))⊥, then

∇2δ(x) =

(0 00 II(x)

)

Corollary 5.10 A ∈ int ~F iff ∃ε > 0, R > 0 such that C(A− εI) ∈ F for all c ≥ R.

Moreover, ∃ε, R > 0 such that

C(ρ− ε

2|x|2) ∈ F (Ω), ∀c ≥ R

Note the left hand side produces a subsolution (function of type F ) whose behaviour on ∂Ω is very simple,just like barrier function in Perron method for Laplacian equation.

Sketch of Proof of Theorem 5.12: ⇐ is trivial. For ⇒ we proceed by two steps

1st Step : ∃ global defining function ρ ∈ C∞(Ω) (See exercise below)

2nd Step : Let ρ = ρ+Cρ2. (Note: all local defining functions are locally equivalent and all are equivalentto signed distance function introduced earlier, adding a ρ2 term is to force it to be strictly convex, seebelow). We have

∇2ρ = (1 + 2Cρ)∇2ρ+ 2C∇ρ ∇ρ

where the 2nd term is a rank 1 matrix. If x ∈ ∂Ω,

∇2ρ(x) = ∇2ρ(x) + C∇ρ(x) ∇ρ(x) ∈ int ~F (***)

Claim (Lemma 5.3(ii)[3]) ∀c 1, ∇2ρ(x) ∈ int ~F

Lemma 5.3 Let ρ be local defining function, n be unit normal vector field along ∂Ω then

(∗∗)⇔ ∇2ρ(x) + tn n ∈ int ~F

for t 1. Note n n is orthogonal projection onto Rn and Rn = Rn⊕ (Rn)⊥ = Rn⊕ (Tx∂Ω).

Proof : ⇒: Given B, by (**) we can decompose according to the direct sum Rn⊕ Tx∂Ω

∇2ρ(x)−B =

(a ααT 0

)hence

∇2ρ(x) + tn n = ∇2ρ(x) + tn n+ εI −B +B − εI

. . . =

(t+ a+ ε ααT εI

)and B − εI ∈ int ~F since B is in the ray set and ε is small. We can take t large enough so thatthe underlined part is positive definite. Now the conclusion follows by the property of Dirichlet setsintF + P ⊂ intF .

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Exercise 4 Show that for t large enough the underlined matrix is indeed positive definite

Consider again (***). The constant C = C(x) depends on x but since ∂Ω is compact we can alwaystake a constant C 1 such that (***) holds for all x ∈ ∂Ω. We also take C such that this holds forx ∈ U ⊃ ∂Ω where U is a neighborhood of the boundary ∂Ω.

3rd Step : Define ρ(x) = maxρ(x),δ|x|2

2− r, δ, r > 0. This is not a smooth function, only Lipschitz.

We choose r such that U ⊃ x ∈ Rn : −r < ρ(x) < r. Note that ρ = ρ on U and where maximum

function takes value equals to second function, Hessian is δI ∈ int ~F .

4th Step : Smooth the maximum function M(u1, u2) = maxu1, u2 by defining a two variable function

Mε(t1, t2). This will finish the proof and provide a strictly ~F -plurisubharmonic global defining function.

Lecture 8: Subequations: Proof of Main Theorem in Harvey-Lawson

Recall the F -Dirichlet problem: F(∇2u) = 0 on X (∗)u = g on ∂X

and we have defined subequation F ⊂ Sym2(Rn) such that for u ∈ C2 ∩ ∂F (X) (*) holds. We call elementsof F (X) subsolutions and that of −F (X) supersolutions.

Definition 6.1 u ∈ F (X) ∩ (−F (X)) and u

∣∣∣∣∂X

= g, we say that u solves (*) in the sense of Harvey-

Lawson

Main Theorem (Theorem 6.2 [3]) Ω ⊂ Rn a bounded domain with smooth boundary, F a subequation,

suppose

(i) ∂Ω is strictly-~F convex

(ii) ∂Ω is strictly-−→F convex

then (*) admits a unique C0(Ω) solution (i.e. continuous up to boundary and attain boundary value g)

Proof of Existence: Set

u(x) := supv(x) : v ∈ F (Ω) ∩ USC(Ω), v

∣∣∣∣∂Ω

≤ g

(denote the set where supremum is over as F (Ω)) where USC(Ω) is important for making sense of thecondition that v be controlled by g on the boundary.

Facts:

(1) u ∈ USC(Ω)

(2) u ∈ F (Ω)

(3) u

∣∣∣∣∂Ω

≤ g

(4) u = usc(u) where usc(f)(x) := limsupy→x f(y) is upper semicontinuous regularization of u. This willbe used to show (2)

(1’) usc(u) ∈ F (Ω)

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(5) usc(u)

∣∣∣∣∂Ω

≤ g + δ for all δ > 0 (This ⇒(3).)

Note that (1)-(3) will imply that u ∈ F (Ω) and (5)+(1’)⇒(4). Observe that u is bounded from above: ifF ⊂ P = SA then F (Ω) satisfies the maximum principle since the subaffine functions satisfy the maximumprinciple. However even if that does not hold, we always have for some λ such that F + λI ⊂ P (where λ isa fixed number depending only on F ). (To show this, it suffices to show the dualized version F − λI ⊃ P ⇔F ⊃ P + λI but this follows from F ⊃ P + F , i.e. the positivity condition for F and the fact that for largeenough λ, λI ∈ F .) Therefore for v ∈ F (Ω) we have

∃λ v +1

2λ|x|2 ∈ SA(Ω)

Proof of (5): Assumption (ii) and theorem from last lecture guarantees existence of boundary defining

function for Ω, ρ ∈ int−→F (Ω). Let x0 ∈ ∂Ω, there exists ε > 0, R > 0 such that

C(ρ− ε|x− x0|2) ∈ F (Ω) ∀c ≥ R

on ∂Ω we haveg + C(ρ− ε|x− x0|2) = g − Cε|x− x0|2 ≤ g(x0) + δ

where the last inequality holds if we take C 1. Now let v ∈ F (Ω) from the family and define

w := v + C(ρ− ε|x− x0|2) ∈ F (Ω) + F (Ω) ⊂ SA(Ω)

which will play similar role as barrier function in Perron’s method. By maximum principal supΩ w = sup∂Ω wand since

sup∂Ω

w ≤ g(x0) + δ

so for all v in the familyv + C(ρ− ε|x− x0|2) ≤ g(x0) + δ

it follows that usc(u)(x0) ≤ g(x0) + δ

We have already showed u ≤ g on ∂Ω (Prop 6.7 [3]) To show that u

∣∣∣∣∂Ω

= g it suffices to show

Lemma (Lemma 6.9 [3]) If Ω has strictly ~F -convex boundary, then

lim infx→x0

u(x) ≥ g(x0) ∀x0 ∈ ∂Ω

Proof : Fix δ > 0, let ρ ∈ int ~F (Ω) be boundary defining function for Ω. ∃ε > 0 such that for C 0

C(ρ− ε|x− x0|2) ∈ F (Ω)

For C 0 we have on ∂Ωg(x0) + C(ρ− ε|x− x0|2) ≤ g(x) + δ

Letv(x) := g(x0)− δ + C(ρ− ε|x− x0|2)

we have v ∈ F thus v ≤ sup F (Ω) = u ∈ F (Ω),

u ≥ g(x0)− δ + C(ρ− ε|x− x0|2)

solim infx→x0

u(x) ≥ lim infx→x0

v(x) = g(x0) + δ ∀δ > 0

Next we will show interior continuity and uniqueness in a more general context [5].

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Lecture 9: Solving the Dirichlet problem for domains with non-smooth boundary: the main theorem

We prove a more general version of the result in [3] for Dirichlet Problem with weak boundary assumptions

Theorem 7.8 [5] F a subequation, X ⊂ Rn bounded domain, X a manifold with embeded corners. ϕ is a

consistent continuous function on ∂X with

(i) ∂X is strictly (F,ϕ)-convex

(ii) ∂X is strictly (F ,−ϕ)-convex

Then ∃!u ∈ C0(X) s.t. u ∈ F (X), −u ∈ F (X), u

∣∣∣∣∂X

= ϕ.

Definitions Given X manifold with corners, ∂X is also a manifold with coners equipped with a mapιX : ∂X → X which could be non-injective. We say that X has embeded corners if ∂X is a disjoint union offinitely many open and closed subsets on which ιX is injective. ϕ ∈ C0(∂X) is consistent if it is constanton fibres of ιX .

Examples [0, 1] × D with D a closed manifold is a manifold with corners. Another example is takingD = [0, 1] then X = [0, 1] × [0, 1]. Boundary ∂X has four disjoint components. An example of a manifoldwith corners whose corner is not embeded is provided by the teardrop.

Now we need to define boundary convexity (which as we can see above in our context also depends onboundary data!)

Definition 7.1 Write ∂X =∐i ∂Xi. ∂Xi is strictly ~F -convex if

II|Tx∂Xi= B|Tx∂Xi

for some B ∈ int ~F ∀x ∈ ∂Xi

Definition 7.5 Let ϕ ∈ C0(∂X) be consistent u ∈ F (X) is called a subsolution for (F,ϕ)-Dirichletproblem if u|∂X ≤ ϕ

Definition 7.6 ∂X is strictly (F,ϕ) if ∂X can be decomposed to A∐B with A, B unions of boundary

components such that

(i) ∀p ∈ A,∀δ > 0, ∃C0(X) subsolution of the (F,ϕ)-Dirichlet problem that is δ-maximal at p

(ii) B is strictly ~F .

For a subsolution of the F -Dirichlet problem for (X, g), i.e. a function u ∈ F (X)∩USC(X) such that u ≤ gon ∂X is called δ-maximal for some δ > 0 at x0 if u(x0) ≥ g(x0)− δ. The assumption (i) is natural in viewof the following lemma:

Lemma 7.9 Suppose ∂Xi ⊂ B, then there is a w satisfying (i) above

Proof : Choose boundary defining function ρ for ∂Xi, we have showed before that there is ε, R > 0 suchthat C(ρ− ε|x|2) ∈ F (X). Since we can add any affine function and still stay in F (X) we have

C(ρ− ε|x− x0|2) ∈ F (X)

given δ > 0 there is C 0 such that

−g(x) + C(ρ(x)− ε|x− x0|2) ≤ −g(x)− Cε|x− x0|2 ≤ −g(x0) + δ

18

letw = C(ρ(x)− ε|x− x0|2) + g(x0) + δ

It is easy to verify that this w has the property we wanted.

Definition 7.2 ρ ∈ C∞(X) global defining function for ∂Xi if

ρ|X\∂Xi< 0 ρ|∂Xi

= 0 ∇ρ|∂Xi6= 0

Note we only require ρ to vanish on the component ∂Xi

The following proposition will be proven next time:

Proposition (Prop 7.3 [5]) For ∂Xi strictly ~F -convex, there exists global defining function ρ for ∂Xi that

is strictly ~F -convex, in particular ∃ε, R > 0 such that

C(ρ− ε|x|2) ∈ F (X), ∀C ≥ R

Proof of Theorem 7.8: Set

F (ϕ) = v subsolution for (F,ϕ)-Dirichlet Problem

It is easy to see that this set is nonempty since we can always take functions of the form C1|x|2 − C2 withC1, C2 appropriately chosen. Let

u := supv : v ∈ F (ϕ)This is well-defined since we can show that the family F is uniformly bounded from above, v ≤ C =C(X,ϕ, F ). By [3] usc(u) ∈ F (X). We now need to show usc(u) ≤ ϕ on ∂X. Either using condition (ii) ofthe theorem or by Lemma 7.9, ∃w ∈ C0(X) subsolution for (F ,−ϕ)-Dirichlet Problem that is δ-maximal atp ∈ ∂Xi for all i, p, δ, that is w(p) ≥ −ϕ(p)− δ and w ∈ F (X), therefore w ≤ −ϕ on ∂X. Let v ∈ F (ϕ) wehave v + w ∈ SA(X) and v + w ≤ 0 on ∂X. By Maximum Principle of subaffine functions v + w ≤ 0 on X,thus v ≤ −w and u ≤ −w since u is the upper envelope of v’s. Since w is continuous, taking upper-semi-continuous regularization on both sides we have usc(u) ≤ −w. Therefore usc(u)(p) ≤ −w(p) ≤ ϕ(p) + δ forall δ > 0, it follows that usc(u) ≤ ϕ.

The interior continuity of upper envelope will be proven next time.

Lecture 10: Solving the Dirichlet problem for domains with non-smooth boundary: boundary defining functions for boundary com-ponents

Recall from last lecture we have sketched the proof of Theorem 7.8 [5] for domains with non-smooth boundarywith a more general condition of boundary convexity. The crucial step is existence of a δ maximal subsolutionw which is constructed by Lemma 7.9 using Proposition 7.3. We now prove this proposition:

Proposition 7.3 Boundary component ∂Xi is strictly ~F -convex, then there exists global defining functionρ ∈ C∞(X) for ∂Xi that is strictly of type ~FRemark Note we treat boundary components separately since conclusion of the above proposition is ingeneral false! For example let X = [0, 1] × [0, 1] and g(s, t) = s(s − 1)t(t − 1) and F = P thus F -convexityis ordinary convexity. It is not hard to see that each corner is a saddle point of g thus it is not convex. Anyother global defining function must also has this property since they must vanish on four sides, be negativeinside the square and have nonvanishing normal derivative on interior of each side.

Proof : X can be covered with atlas (see [6]), i.e. a collection of maps ψi : Ui → X where Ui are open setsin R

n+ = [0,∞)n

19

By Whitney extension theorem there exists smooth extension of ψi (also denoted ψ) to Ui open in the wholespace Rn which is still diffeomorphism to its image Wi := ψi(Ui). By definition of embeded corners, there isan l such that ψ−1

i (∂Xi ∩Wi) = x ∈ Ui, xl = 0, it is easy to see −xl is a nice local defining function, set

fi = −xl ψ−1i on Wi

Now let U be an open set whose intersection with X is X with the boundary piece ∂Xi removed.

Wi for all i, U, V = Rn − X

form an open cover of Rn, let αi, αU , αV be a smooth partition of unity subordinary to it, i.e. they aresupported on each open set and

∑i αi + αU + αV ≡ 1 and set

ρ :=∑

αifi + αV − αU ∈ C∞(Rn)

it is straightforward to check that (i) ρ = 0 on ∂Xi, (ii) ρ < 0 on X\∂Xi, (iii) ∇ρ 6= 0 on ∂Xi

Now we can construct a global defining function using similar steps as in [3], first note ρ = ρ + Cρ2 is

strictly ~F local defining function on a neighborhood of ∂Xi for C sufficiently large, then use partition ofunity argument to modify it so it is negative on X\∂Xi that is strictly ~F in a neighborhood W of ∂Xi.Choose r > 0 small such that ρ > −r ∩ X is contained in W , and δ > 0 small such that δ|x|2 − r < 0

on ∂Xi, ρ = maxρ, δ|x|2 − r agrees with ρ in a neighborhood of ∂Xi where ρ is strict ~F and equals thequadratic function elsewhere, this will give the desired global defining function after a smoothing processsimilar to [3].

The last step is to show interior continuity of the upper envelope function u of the family F (ϕ). For thiswe use an argument due to Walsh [7].

20

Proposition (Prop 6.11 [3]) Let u be the upper envelope defined above, then u ∈ C(Ω)

Proof : Let Ωδ = dist(x, ∂Ω) > δ and Cδ = dist(x, ∂Ω) < δ be the δ-collar of the boundary and letuy = u(x + y) be y-translate of u where we set the undefined part equal to −∞. By continuity of u, thereis δ > 0 such that uy < u+ ε on C2δ for |y| ≤ δ. We claim

uy ≤ u+ ε on Ω

as well, for |y| ≤ δ. It is easy to see that conclusion follow from this claim. Note first that by translationproperty and affine property of Dirichlet sets uy − ε ∈ F (Ωδ). Since uy < u+ ε on collar C2δ we have

gy := maxuy − ε, u ∈ F (Ω)

by maximum property. But since gy takes value u on collar, we have in fact gy is in the family F (ϕ) hencegy ≤ u on Ω and the claim is proved.

References

[1] J. Rauch and B. A. Taylor, The Dirichlet Problem for the Multidimensional Monge-Ampere Equation,Rocky Mountain Journal of Mathematics, Volume 7, Number 2, 1977

[2] Y. A. Rubinstein and S. Zelditch, The Cauchy problem for the homogeneous Monge-Ampere equation, II.Legendre transform, Advanced in Mathematics 228 (2011) 2989-3025

[3] F. R. Harvey and H. B. Lawson Jr., Dirichlet Duality and the Nonlinear Dirichlet Problem, Communi-cations on Pure and Applied Mathematics, Vol. LXII, (2009) 0396-0443

[4] Z. Slodkowski, Pseudoconvex Classes of Functions I. Pesudoconcave and Pseudoconvex Sets, Pacific Jour-nal of Mathematics, Vol 134, No.2, 1988

[5] Y. A. Rubinstein, J. P. Solomon, The Degenerate Special Lagrangian Equation, arXiv:1506.08077

[6] D. Joyce, On manifolds with corners, in: Advances in geometric analysis, Int. Press, 2012, pp. 225-258

[7] J. B. Walsh, Continuity of envelopes of plurisubharmonic functions, J. Math. Mech. 18 (1968/1969),143-148

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