H\" Older Regularity of Solutions to Generated Jacobian Equations

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( ( ,T x, ( ψ ( ( x ,T x, ,p u p, x, ,p u p, x, Introduction ) : φ , X ( ( ,p u p, x, x ( ( ,p u p, x, x ) × φ , ) ) 1 ))= )))) V , V , R ( x ) × ( ( V , )) ) ,p u p, x, ,p u p, x, V , R ψ ( ihunkown with , ,p u p, x, = ( ( ,p u p, x, ,Dφ x, ψ )= )) )= )) ¯ ( ,Dφ x, )) ))) ( x u o oegeneral more for rue ,otmltransport optimal e, u p ) qaincase. equation n ψ φ , o0]drcl to directly Loe09] ( nti ae,we paper, this In rnpr a not can transport ( x hs difficulties, these ,p u p, x, , ( ) x φ , φ )) oa Hölder local a edreflector field far d ( : , x X ) )) cba equa- acobian → oegiven some r ost gen- to ions ions R fthere If . 2 SEONGHYEON JEONG domain Y Rn, the image of X under the map T ( , Dφ( ), φ( )) is equal to Y . As ⊂ · · · a special case, if we have measures which are supported on X and Y (we call them the source and the target respectively), then this second boundary condition can be replaced by defining weak solutions using the measures. We discuss this in 2.3. The main theorem of this paper is local Hölder regularity of solutions to (GJE). In the literature, it is known that, if the source and the target are bounded away from 0 and with respect to the Lebesgue measure, and if the generating function ∞ satisfies (G3w) together with some other conditions, then the solution φ to the (GJE) has local Hölder regularity(See [GK17]). But in optimal transport, which is a special case of (GJE), the author of [Loe09] showed the local Hölder regularity using (As) condition, but weaker assumption on the source measure. The main theorem of this paper is parallel to his result in optimal transport. The idea of the proof of the main theorem of this paper will be imported from [Loe09] and the arXiv version of [KM10]. But, since there are structural differences between (GJE) and optimal transport, we need to adjust the idea in [Loe09] to import it to (GJE). One of the big differences is the dependency of the generating function on the scalar parameter v. In optimal transport case, the generating function is G(x, y, v)= c(x, y) v, so that by taking the derivative with respect to − − x or y, dependency on v vanishes. This is not necessarily true in more general (GJE) case. Hence every estimate about the cost function c(x, y) should be combined with an estimate about the scalar parameter v to import them to (GJE). Another big difference is that the conditions on optimal transport hold on the whole space X Y R, and all the derivatives have bounded norm. More general generating × × functions, however, do not necessarily satisfy the conditions on the whole space X Y R, but only on a subset g X Y R. An example can be found × × ⊂ × × in [LT16]. Moreover, because of the dependency on the scalar parameter, the derivatives of the generating function might not be bounded. Hence whenever we want to use conditions and derivatives of the generating function, we need to check that the points used are in a “nice” set g and stay inside a compact set. Since we are going to focus on the local regularity, this can be done by localizing the argument. Lemma 3.5 will allow us to choose points in a nice compact set. We introduce some related results. Optimal transport case is done in [Loe09] and the arXiv version of [KM10]. For other Hölder regularity in optimal transport, see [FKM13] and [GK15]. In [GK15], the authors prove the Hölder regularity under a condition known as (QQ-conv), which is equivalent to (A3w) when the cost function is smooth, and the same authors prove analogous result for (GJE) in [GK17]. In reflector antenna design, some cases are done by various authors. See [GH14], [AGT16], [GT19]. For other regularity theory of (GJE), see [Tru14], [Jha17], [GK17]. 3

2. Setting of the problem

2.1. Conditions for the generating function G. To have the local holder regularity result, we need some structural conditions on the domains X and Y and on the generating function G. We ﬁrst assume the regularity and monotonicity of the generating function :

(Regular) G C4(X Y R) ∈ × ×

(G-mono) DvG< 0

From (G-mono), we get a function H : X Y R R that satisﬁes × × → (1) G(x,y,H(x,y,u)) = u.

Note that the implicit function theorem ensures that H C4 by (Regular) and ∈ this, with (G-mono), implies

(H-mono) DuH < 0.

As in optimal transport case, we need conditions like (A1), (A2)(sometimes these are called (twisted) and (non-deg) respectively) and (As) in [Loe09]. But in some examples of (GJE), the function G does not satisfy these conditions on the whole domain X Y R. Instead, there is a subset g X Y R on which the generating × × ⊂ × × function G satisﬁes conditions corresponding to (A1), (A2) and (As). Therefore, we assume that there is a subset g X Y R such that the following conditions ⊂ × × hold :

(G-twist) (y, v) (D G(x, , ), G(x, , )) is injective on g 7→ x · · · · x

∗ DyG (G -twist) x ( ,y,v) is injective on gy,v 7→ −DvG ·

D G (G-nondeg) det D2 G D2 G y =0 on g xy − xv ⊗ D G 6 v where g = (y, v) (x, y, v) g and g = x (x, y, v) g . We will use E to x { | ∈ } y,v { | ∈ } denote the matrix in the condition (G-nondeg). Under these conditions, we can deﬁne the G exponential function on some subset of T ∗X and T ∗Y . − x y

G Deﬁnition 2.1. We deﬁne the expx,u and Vx by

D G(x, expG (p), V (p,u)) = p x x,u x . G ( G(x, expx,u(p), Vx(p,u)) = u 4 SEONGHYEON JEONG

G We call expx,u the G-exponential function with focus (x, u). We deﬁne another ∗ G exponential map expy,v by

∗ DyG G (expy,v(q),y,v)= q. −DvG ∗ G ∗ We call expy,v the G -exponential function with focus (y, v).

G Note that by the implicit function theorem, the functions expx,u, Vx(p,u) and ∗ G 3 expy,v are C on the domain of each function. In optimal transport case, the equations which are used to define the c-exponential maps and c∗-exponential maps are symmetric. In the above definition for the G- ∗ G G exponential function, the equations that we used to define expx,u and expy,v do not look symmetric, but they actually play symmetric roles. See Remark 9.5 in [GK17].

Remark 2.2. Let I = v R (x, y, v) g and let J = H(x, y, I ) and x,y { ∈ | ∈ } x,y x,y h = (x,y,u) u J . We will assume that h is open relative to X Y R. Let { | ∈ x,y} × × h = y Y (x,y,u) h . x,u { ∈ | ∈ } (DomOpen) h is open relative to X Y R × × ∗ We will denote the image of hx,u under the map DxG(x, ,H(x, ,u)) by hx,u. Then G ∗ · · the G-exponential map expx,u is deﬁned on hx,u. Moreover, we get the expression G Vx(p,u)= H(x, expx,u(p),u). With this expression, we can compute

G −1 G Dpexpx,u(p)= E (x, expx,u(p), Vx(p,u)). To impose geometric conditions on the sets X and Y , we deﬁne G-convexity of sets.

∗ Deﬁnition 2.3. X is said to be G-convex if gy,v, which is the image of gy,v under D G the map y ( ,y,v), is convex for any (y, v) Y R. Y is said to be G∗-convex −DvG · ∈ × if h∗ is convex for any (x, u) X R. x,u ∈ × We also add convexity conditions to the supports of source and target.

(hDomConv) X is G-convex

(vDomconv) Y is G∗-convex

Deﬁnition 2.4. For x X and u R, let y ,y h and let p = D G(x, y ,H(x, y ,u)). ∈ ∈ 0 1 ∈ x,u i x i i A G-segment that connects y0 and y1 with focus (x, u) is the image of [p0,p1] under G the map expx,u expG ((1 θ)p + θp ) θ [0, 1] . { x,u − 0 0 | ∈ } DyG ∗ For y Y and v R, let x0, x1 gy,v and let qi = (xi,y,v). The G - ∈ ∈ ∈ −DvG segment that connects x0 and x1 with focus (y, v) is the image of [q0, q1] under the 5

∗ G map expy,v : ∗ expG ((1 θ)q + θq ) θ [0, 1] . { y,v − 0 0 | ∈ }

Remark 2.5. The definition of G-convexity is different from the usual G-convex definition from the literature (See [GK17]). The two definitions serve the same purpose, however, as these convexity conditions are used to ensure that G-segments are well defined. For example, if y ,y h then the G-segment that connects y 0 1 ∈ x,u 0 and y1 with focus (x, u) is well defined by Definition 2.3. In [KM10], the authors define convexity of domains using a subset W of M M¯ (Definition 2.5). The × convexity we have defined coincides with their horizontal and vertical convexity of W when the generating function is given by a cost function.

In optimal transport, there is an important condition called (A3w). This condition is a sign condition of a 4-tensor that was first introduced in [MTW05], and [TW09] (the tensor is called the MTW tensor sometimes). In optimal transport, the tensor is defined by D2 D2 c(x, expc (p)) , and it has the coordinate pp − xx x expression MT W = (c cr,sc c ) cp,kcq,l ijkl ij,r s,pq − ij,pq (See, for example, [MTW05]). Then (A3w) condition assume that MT W [ξ,ξ,η,η] ≥ 0 for any ξ η. (A3w) is used to show many regularity results. In [FKM13], the ⊥ arXiv version of [KM10], and [GK15], (A3w) is used to show the Hölder regularity of potential functions. In [DPF13], (A3w) is used to show the Sobolev regularity of potential functions. In [Loe09] However, the author uses the strengthened condition (As) to show the Hölder regularity result.

(As) MT W [ξ,ξ,η,η] δ ξ 2 η 2 whenever ξ η, for some δ > 0. ≥ | | | | ⊥ In optimal transport, (As) is equivalent to

(As’) MT W [ξ,ξ,η,η] > 0 for ξ η. ⊥ Obviously (As) implies (As’). For the other implication, note that X Y is compact. × Then (Regular) and (G-nondeg) implies that there exists δ > 0 such that (As) is true with the δ. Since optimal transport is a special case of (GJE), we should impose a condition that corresponds to the condition (As) to import the idea from [Loe09]. Therefore, we assume the following condition.

(G3s) D2 (x,p,u)[ξ,ξ,η,η] > 0 for any ξ η ppA ⊥ where (x,p,u) = D2 G x, expG (p), V (p,u) , ξ T X and η T ∗X. Here A xx x,u x ∈ x ∈ x ξ η means that the dual pairing η(ξ) is 0. Note that p must be in h∗ to well- ⊥ x,u deﬁne (x,p,u). Here we point out that, unlike optimal transport, (G3s) is not A 6 SEONGHYEON JEONG equivalent to, but weaker than the following.

D2 (x,p,u)[ξ,ξ,η,η] >δ ξ 2 η 2 for any ξ η for some δ > 0. ppA | | | | ⊥ ∗ ∗ ∗ This is because δ depends on (x,p,u) h = R h and h is not compact ∈ (x,u)∈X× x,u in general. We use (G3s) in this paper.S Then some arguments that use δ in (As) from [Loe09] can not be applied. We resolve this problem by localizing the argument. See Remark 2.10. In this paper, we have that X and Y are subsets of Rn. Then we can identify ∗ Rn Rn TxX and Tx X with , and the dual pairing with usual inner product in . As such we will often use Rn for the tangent space and the cotangent space. (G3w) condition is the same condition (G3s), but with instead of >. ≥ Remark 2.6. The condition (A3w) is known to be equivalent to the next inequality

c(y, x¯(t)) + c(x, x¯(t)) max c(y, x¯(0)) + c(x, x¯(0)), c(y, x¯(1)) + c(x, x¯(1)) − ≤ {− − } wherex ¯(t) is the c-segment that connectsx ¯(0) andx ¯(1) with focus x (See [Loe09], [Vil03]). This is called Loeper’s property. This has the following geometric mean- ing : If two c-affine functions, which have c subdifferentialsx ¯(0) andx ¯(1) respec- − tively, meet at x, then the c-affine functions that pass though that point with c- subdifferentialx ¯(t) lies under max of the two original c-affine functions. Therefore, this property is called geometric Loeper’s property (gLp) sometimes. In [Loe09], he uses (As) to get (gLp) in some quantitative way. This can be done in the G-convex setting as well. We obtain a property corresponding to (gLp) for G-affine functions, and along with a quantitative version of (gLp) using (G3s) in the lemma 3.3.

2.2. G convex functions. The solutions of (GJE) belong to a class known as − G-convex function. We deﬁne the G-convexity of a function by generalizing the usual convexity for the c-convexity in optimal transport case :

Deﬁnition 2.7. A function φ : X R is said to be a G-convex function if for any → x X there exists y Y and v R such that (x ,y , v ) g and 0 ∈ 0 ∈ 0 ∈ 0 0 0 ∈

(2) φ(x0)= G(x0,y0, v0) φ(x) G(x, y , v ) x X. ≥ 0 0 ∀ ∈

We say that (y0, v0) is a G-focus of φ at x0 and the function G(x, y0, v0) is a G supporting function of φ at x with focus (y , v ). We deﬁne the G-subdiﬀerential − 0 0 0 of φ at x0 by

∂ φ(x )= y Y (y, v) is a G-focus of φ at x for some v R G 0 { ∈ | 0 ∈ } and we deﬁne ∂Gφ(A)= ∂Gφ(x). x[∈A 7

With the deﬁnition above and Remark 2.2, we can express v0 as follows :

v0 = H(x0,y0, φ(x0)).

Therefore, the G-supporting function of a G-convex function φ at x0 with G- sudifferential y0 can be expressed as G(x, y0,H(x0,y0, φ(x0))). Also, note that (x ,y , v ) g is equivalent to (x ,y , φ(x )) h. We will use this fact often. We 0 0 0 ∈ 0 0 0 ∈ call the function of the form G( ,y,v), G-affine function with focus (y, v), or simply · a G-affince function.

Remark 2.8. The conditions on the generating function G imply G-convexity of the G-subdiﬀerential at a point which in turn implies the following proposition :

Proposition 2.9. Ifa G-aﬃne function G( ,y , v ) supports a G-convex function φ · 0 0 at x0 locally, i.e.

φ(x0)= G(x0,y0, v0) φ(x) G(x, y , v ) on some neighborhood of x ≥ 0 0 0 and if (x ,y , v ) g, then y ∂ φ(x ). 0 0 0 ∈ 0 ∈ G 0 For example, it is proved in [GK17] (Corollary 4.24), but under an extra condition on Jx,y, namely (unif) and the authors require the solution of the (GJE) to be “nice”. (unif) u, u such that [u, u] J for any x X and y Y . ∃ ⊂ x,y ∈ ∈ (nice) The solution φ to the (GJE) lies in (u, u): φ(x) (u, u), x X. ∈ ∀ ∈ In [GK17], the authors used these conditions for two reasons. First, they use these to check that the G-segments they are using are well-defined. Under these conditions, if u (u, u), then (x,y,u) h for any x X and y Y so that (DomConv*) ∈ ∈ ∈ ∈ assures that the G-segement for any pair of points in Y with focus (x, u) can be defined. Second, they get a compact set X Y [u, u] that lies inside h with (unif). × × Then the norms of derivatives of the functions G and H will be bounded, and will be bounded away from 0 if it is signed. For the local regularity, we do not need the fixed interval [u, u] inside J for any x X and y Y . Instead, we will use the x,y ∈ ∈ following weaker conditions : (unifw) a,b : X Y R which are continuous and [a(x, y),b(x, y)] ∃ × → ⊂ J for any x X and y Y . x,y ∈ ∈ (nicew) The solution φ to the (GJE) lies in (a,b): φ(x) (a(x, y),b(x, y)) ∈ for any x X and y ∂ φ(x). ∈ ∈ G Note that the G-segments used in [GK17] to show Proposition 2.9 can be well- defined by (hDomConv) and (vDomconv). In fact, the G-segments constructed in [GK17] connect two points in ∂ φ(x) h . Hence (vDomconv) is enough G ⊂ x,φ(x) to define the G-segments. With similar reasoning, (hDomConv) ensures that the 8 SEONGHYEON JEONG

G∗-segments that are used are well-deﬁned. Moreover, the sets Φ := (x,y,u) u [a(x, y),b(x, y)] h (3) { | ∈ }⊂ Ψ := (x, y, v) v H(x, y, [a(x, y),b(x, y)]) { | ∈ } are compact. Note that (hDomConv) and (vDomconv) do not imply that (x, y ,u) θ ∈ Φ, where y is the G segment connecting y and y with focus (x, u), so we can not θ − 0 1 bound norms of G and its derivatives on a G-segment only using Φ. But we know that Φ lies in a compact set X Y [min a, max b], and Ψ lies in the corresponding × × compact set. Therefore, we use norms on these compact sets that contain Φ and Ψ and we still can use the same proof of the proposition 2.9 with assumption (unifw) and (nicew) instead of (unif) and (nice).

Remark 2.10. For a compact subset S of h, the condition (G-nondeg) implies that we have a constant Ce that depends on S such that 1 (4) E Ce Ce ≤k k≤ where E is the operator norm of E. This with Remark 2.2 implies that k k 1 G G (5) p1 p0 expx,u(p1) expx,u(p0) Ce p1 p0 Ce | − | ≤ | − |≤ | − | when (x, expG ((1 θ)p + θp ),u) S, θ [0, 1]. Also, the (G3s) condition x,u − 0 1 ∈ ∀ ∈ implies that we have a constant α that depends on S such that

(6) D2 (x,p,u)[ξ,ξ,η,η] > α ppA for any unit ξ and η such that ξ η if (x, expG (p),u) S. Note that by tensori- ⊥ x,u ∈ ality, we have

(7) D2 (x,p,u)[ξ,ξ,η,η] > α ξ 2 η 2 ppA | | | | for non unit ξ and η. Moreover, compactness of X Y [min a, max b] implies that × × we have β > 0 such that

(8) D G< β on X Y [min a, max b]. v − × × With these conditions, we show some simple propositions for the G subdiﬀerential − of φ. We will denote the r neighborhood of a set A by (A). Nr Proposition 2.11. The subdiﬀerential of φ at a point x is a closed subset of Y and it lies compactly in hx,φ(x) :

∂ φ(x) ⋐ h Y. G x,φ(x) ⊂ Note that Proposition 2.11 for optimal transport case is much easier to prove because J = R for any (x, y) X Y i.e. h = X Y R so that we only need x,y ∈ × × × 9 to check the inequality (2). But in (GJE) case, compactness is not trivial since showing the inequality (2) is not enough, and we need to check (x, y, φ(x)) h. ∈

Proof. First we show that ∂ φ(x) is closed. Suppose y ∂ φ(x). Then there exists G ∈ G a sequence y ∂ φ(x) that converges to y. Then from (unifw) and (nicew), we i ∈ G have φ(x) ∞ (a(x, y ),b(x, y )) ∈ i=1 i i [sup a(x, y ), inf b(x, y )] ⊂ T i i lim a(x, yi), lim b(x, yi) ⊂ i→∞ i→∞ =h [a(x, y),b(x, y)]. i This implies that (x, y, φ(x)) h by (unifw). Moreover, from the deﬁnition of ∈ G subdiﬀerential, we have − G(z,y ,H(x, y , φ(x))) φ(z), z X. i i ≤ ∀ ∈ Taking i , we get → ∞ G(z,y,H(x, y, φ(x))) φ(z), z X. ≤ ∀ ∈ This inequality with (x, y, φ(x)) h shows that y ∂ φ(x). Therefore ∂ φ(x) ∈ ∈ G G is closed, and hence compact as a closed subset of Y . Now note that ∂ φ(x) G ⊂ hx,φ(x). Moreover, from the openness assumption of h (Remark 2.2), hx,φ(x) is relatively open with respect to Y . Therefore compactness of ∂Gφ(x) shows that

∂Gφ(x) ⋐ hx,φ(x).

The next proposition is about continuity of the G-subdifferential of a G-convex function. Note that the G-subdifferential is a set valued function, so that the continuity here is not the ordinary continuity of a single valued function. But if the G-subdifferential is single valued, the next proposition actually coincides with the continuity of a single valued function.

Proposition 2.12. Let φ be a G-convex function with (nicew). Let x X. Then ∈ for ǫ> 0, there exists δ > 0 such that if z x δ then | − |≤ ∂ φ(z) (∂ φ(x)) . G ⊂ Nǫ G Proof. Suppose it is not true. Then we get a sequence x X and y ∂ φ(x ) k ∈ k ∈ G k such that x x as k but y / (∂ φ(x)) for any k. Since Y is compact, k → → ∞ k ∈ Nǫ G we can extract a subsequence such that y y. Note that by (nicew), we have k → (x, y, φ(x)) Φ h. From the choice of y , we have y / (∂ φ(x)). But since y ∈ ⊂ k ∈ Nǫ G k are in subdiﬀerentials of φ, we have

(9) φ(z) G(z,y ,H(x ,y , φ(x ))), z X. ≥ k k k k ∀ ∈ 10 SEONGHYEON JEONG

We can take limit on (9) because φ, G, and H are continuous. Hence we get

φ(z) G(z,y,H(x, y, φ(x))) ≥ which implies y ∂ φ(x), which contradicts to y / (∂ φ(x)). ∈ G ∈ Nǫ G 2.3. weak solutions to the (GJE). Let µ be the source measure, a probability measure supported on X, and let ν be the target measure, a probability measure supported on Y . Then we can interpret the solutions to the (GJE) with second boundary condition in the following ways, which are analogies of the Alexandrov solution and the Brenier solution in optimal transport case.

Deﬁnition 2.13. Let φ : X R be a G-convex function. Then → 1. φ is called a weak Alexandrov solution to the (GJE) if

µ(A)= ν(∂ φ(A)), A X. G ∀ ⊂ 2. φ is called a weak Brenier solution to the (GJE) if

ν(B)= µ(∂−1φ(B)), B Y. G ∀ ⊂ Note that the G-subdifferential of φ(x) is single valued µ a.e. in X as it is − semi-convex by Proposition 3.4. Hence a weak Brenier solution satisfies the push- forward condition ∂Gφ♯µ0 = µ. In optimal transport, it is well known that the Brenier solution is not necessarily an Alexandrov solution. To prevent this, we need some convexity condition on the domains. See 4.6 in [Fig17] for the Monge- Ampère case, [MTW05] for the general optimal transport case. To develop the local regularity theory in this paper, we will use a weak Alexandrov solution.

2.4. conditions for measures and main result. We will assume the target measure ν is bounded away from 0 and with respect to the Lebesgue measure on ∞ Y . For the local regularity theory, we will assume one of the following assumptions : 1. p (n, ] and Cµ such that for any x X and r 0 we have ∃ ∈ ∞ n(1− 1 ) ∈ ≥ µ(Br(x)) Cµr p . + ≤ + 2. f : R R such that limr→0 f(r) = 0 and for any x X and ∃ → 1 ∈ n(1− ) r 0 we have µ(B (x)) f(r)r n . ≥ r ≤ Then the main theorem of this paper is the following

Theorem 2.14. Suppose X and Y are compact domains in Rn and let G : X Y × × R R be the generating function. Let µ and ν be probability measures on X and Y → respectively. Assume that G satisﬁes (Regular), (G-mono), (G-twist), (G∗-twist), (G-nondeg), (G3s), and (unifw). Assume also that X and Y satisfy (hDomConv) and (vDomconv) and the target measure ν is bounded away from 0 and with ∞ respect to the Lebesgue measure on Y . Let φ be an weak Alexandrov solution to the 11 equation (GJE) that satisﬁes (nicew). Then we have the followings : n(1− 1 ) 1. If there exist p (n, ] and Cµ such that µ(Br(x)) Cµr p ∈ ∞ 1,σ ≤ for all r 0, x X, then φ Cloc (X). ≥ ∈ + ∈+ 2. If there exist f : R R such that limr→0 f(r) = 0 and 1 → n(1− ) 1 µ(B (x)) f(r)r n for all r 0, x X, then φ C (X). r ≤ ≥ ∈ ∈ loc Here, ρ =1 n and σ = ρ . − p 4n−2+ρ Note that the exponents that appears in above theorem are the same as in [Loe09].

3. Proof of local holder regularity

For the proof of local Hölder regularity, we will do most computations in the set X Y [min a, max b]. Since this set is compact we will be able to get finite × × quantities in each computation with a localizing argument. Then we will be able to apply the idea from [Loe09] for the proof of each lemma. The main difference between [Loe09] and this paper comes from the structural difference of a cost function and a generating function. In particular, a generating function has its own nice subdomain where the structural conditions hold true, whereas the corresponding conditions hold on the whole domain in optimal transport case. Therefore, we need to check that the points at which we use the conditions are in the nice subdomain. Moreover, each derivative of G still depends on the scalar parameter v, hence we need to take care of extra terms that come from the dependency on the scalar parameter v. In addition, estimates on the cost function c(x, y) should be done on G(x, y, v). Through out this paper, we will use tensor notation for derivatives many times. For 2 example, we view DxxG as a 2-tensor, and we use square bracket “[ , ]” for tensor notation. 2 t 2 DxxG[ξ, ξ]= ξ DxxGξ.

3.1. Quantitative (glp) with (G3s). We start with some estimation on G-aﬃne functions. In this subsection, x is a point in X, u R, and y ,y h . Also, m ∈ 0 1 ∈ xm,u for θ [0, 1], we denote the G-segment that connects y and y with focus (x ,u) ∈ 0 1 m by yθ, and we use vθ = H(xm,yθ,u) and pi = DxG(xm,yi, vi). Moreover, we will assume that the points (x ,y ,u) S for some compact set S ⋐ h. Then by m θ ∈ remark2.10, we get constants Ce and α that depend on S and satisfy (4),(5), and (6).

3 1 Lemma 3.1. For some constant C1 that depends on the C norm of G, C norm of H, and Ce, we have

2 2 ′ 2 (10) D G(x ,y , v ) D G(x ,y ′ , v ′ ) [ξ, ξ] C θ θ p p ξ xx m θ θ − xx m θ θ ≤ 1| − || 1 − 0|| |

12 SEONGHYEON JEONG

Proof.

D2 G(x ,y ,H(x ,y ,u)) D2 G(x ,y ′ ,H(x ,y ′ ,u)) k xx m θ m θ − xx m θ m θ k D3 G y y ′ + D3 G D H y y ′ ≤k xxy k| θ − θ | k xxv kk y k| θ − θ | ( D3 G + D3 G D H )C θ θ′ p p . ≤ k xxy k k xxv kk y k e| − || 1 − 0| We set C = ( D3 G + D3 G D H )C . 1 k xxy k k xxv kk y k e

Lemma 3.2. Let ξp = Projp1−p0 (ξ), where Projp is the orthogonal projection onto 4 p. Then for some constants ∆1 and ∆2 that depend on α, the C norm of G, we have 2 2 2 (1 θ)DxxG(xm,y0, v0)+ θDxxG(xm,y1, v1) [ξ, ξ] D G(xm,yθ, vθ)[ξ, ξ] − xx ≤ +θ(1 θ) p p 2( ∆ ξ 2 + ∆ ξ 2). − | 1 − 0| − 1| | 2| p| Proof. Let f : [0, 1] R be such that ξ → 2 fξ(θ)= DxxG(xm,yθ, vθ)[ξ, ξ]. Let ξ′ = ξ ξ so that ξ′ ξ . Then we can apply (G3s), and we obtain − p ⊥ p ′′ 2 ′ 2 f ′ α p p ξ ξ ≥ | 1 − 0| | | and from this uniform convexity, we have

1 2 ′ 2 (11) f ′ (θ) θf ′ (1) + (1 θ)f ′ (0) α p p ξ θ(1 θ). ξ ≤ ξ − ξ − 2 | 1 − 0| | | − Let g = f f ′ . Then ξ ξ − ξ ′′ ′′ ′′ g (θ)= f (θ) f ′ (θ) ξ ξ − ξ = D2 [ξ,ξ,p p ,p p ] D2 [ξ′, ξ′,p p ,p p ] ppA 1 − 0 1 − 0 − ppA 1 − 0 1 − 0 =2D2 [ξ′, ξ ,p p ,p p ]+ D2 [ξ , ξ ,p p ,p p ]. ppA p 1 − 0 1 − 0 ppA p p 1 − 0 1 − 0 Therefore, bounding ξ′ by ξ , we obtain | | | | g′′ 3 D2 p p 2 ξ ξ | ξ |≤ k ppAk| 1 − 0| | || p| and from this bound, we have 3 (12) g (θ) θg (1) + (1 θ)g (0) + D2 p p 2 ξ ξ θ(1 θ). ξ ≤ ξ − ξ 2k ppAk| 1 − 0| | || p| − Combining (11) and (12), we obtain

2 ′ DxxG(xm,yθ, vθ)= gξ + fξ 3 θg(1) + (1 θ)g(0) + D2 p p 2 ξ ξ θ(1 θ) ≤ − 2k ppAk| 1 − 0| | || p| − 1 2 ′ 2 + θf ′ (1) + (1 θ)f ′ (0) α p p ξ θ(1 θ) ξ − ξ − 2 | 1 − 0| | | − = θD2 G(x ,y ,u)[ξ, ξ]+(1 θ)D2 G(x ,y ,u)(ξ, ξ) xx m 1 − xx m 0 13

α 3 + θ(1 θ) p p 2 ξ′ 2 + D2 ξ ξ − | 1 − 0| − 2 | | 2k ppAk| || p| θD2 G(x ,y ,u)[ξ, ξ]+(1 θ)D2 G(x ,y ,u)(ξ, ξ) ≤ xx m 1 − xx m 0 α 3 + θ(1 θ) p p 2 ξ 2 + ( D2 + α) ξ ξ . − | 1 − 0| − 2 | | 2k ppAk | || p| Here, we use weighted Young’s inequality 3 α 3 ( D2 + α) ξ ξ ξ 2 + α−1( D2 + α)2 ξ 2. 2k ppAk | || p|≤ 4 | | 2k ppAk | p| Then we obtain

D2 G(x ,y , v ) θD2 G(x ,y ,u)[ξ, ξ]+(1 θ)D2 G(x ,y ,u)[ξ, ξ] xx m θ θ ≤ xx m 1 − xx m 0 α 3 + θ(1 θ) p p 2 ξ 2 + α−1( D2 + α)2 ξ 2 . − | 1 − 0| − 4 | | 2k ppAk | p| Hence we get the inequality with ∆ = α and ∆ = α−1( 3 D2 + α)2. 1 4 2 2 k ppAk

The next lemma is the quantitative version of (gLp). We will use the (G3s) condition through Lemma 3.3 later.

Lemma 3.3. Deﬁne φ¯(x): X R by → φ¯(x) = max G(x, y , v ), G(x, y , v ) { 0 0 1 1 } Then we have the quantitative (gLp) :

(13) φ¯(x) G(x, y , v )+ δ θ(1 θ) y y 2 x x 2 γ x x 3 ≥ θ θ 0 − | 1 − 0| | − m| − | − m| for θ [ǫ, 1 ǫ] and x x Cǫ. ∈ − | − m|≤ Proof. Note that by taking the Taylor series, 1 G(x, y , v )= u+ D G(x ,y , v ), (x x ) + D2 G(x, y , v )[x x , x x ]+o( x x 2). i i h x m i i − m i 2 xx i i − m − m | − m| Therefore, we have

φ¯(x) θG(x, y , v )+(1 θ)G(x, y , v ) ≥ 0 0 − 1 1 = u + θp + (1 θ)p , x x h 1 − 0 − mi 1 + θD2 G(x, y , v )+(1 θ)D2 (x, y ,u ) [x x , x x ]+ o( x x 2). 2 xx 0 0 − xx 1 1 − m − m | − m| Applying Lemma 3.2, we obtain (14) 1 φ¯(x) u + θp + (1 θ)p , x x + D2 G(x ,y , v )[x x , x x ] ≥ h 1 − 0 − mi 2 xx m θ θ − m − m 1 θ(1 θ) p p 2( ∆ x x 2 + ∆ (x x ) 2)+ o( x x 2) − 2 − | 1 − 0| − 1| − m| 2| − m p| | − m| 14 SEONGHYEON JEONG for any θ [0, 1]. Let θ′ [0, 1], then we can write (14) with θ′. Let us call this ∈ ∈ inequality (14’). Then adding and subtracting the right hand side of (14) to the right hand side of (14’) and reordering some terms, we get 1 φ¯(x) u + θp + (1 θ)p , x x + D2 (x ,y , v )[x x , x x ] ≥ h 1 − 0 − mi 2 xx m θ θ − m − m 1 + ∆ θ(1 θ) p p 2 x x 2 2 1 − | 1 − 0| | − m| 1 + (θ′ θ) p p , x x θ(1 θ)∆ p p 2 (x x ) 2 − h 1 − 0 − mi− 2 − 2| 1 − 0| | − m p| 1 2 2 (15) + D G(x ,y ′ , v ′ ) D G(x ,y , v ) [x x , x x ] 2 xx m θ θ − xx m θ θ − m − m 1 + ∆ ((θ′(1 θ′) θ(1 θ)) p p 2 x x 2 2 1 − − − | 1 − 0| | − m| 1 + ∆ ((θ(1 θ) θ′(1 θ′)) p p 2 (x x ) 2 + o( x x 2) 2 2 − − − | 1 − 0| | − m p| | − m| Note that by deﬁnition of (x x ) , we have p p (x x ) = p p , x x . − m p | 1− 0|| − m p| |h 1− 0 − mi| Hence, we can write the third line L3 as follows 1 L = [θ′ θ θ(1 θ)∆ p p , x x ] p p , x x . 3 − − 2 − 2h 1 − 0 − mi h 1 − 0 − mi Therefore, if we choose

(16) θ′ = θ + θ(1 θ)∆ p p , x x , − 2h 1 − 0 − mi ′ we can make L3 = 0. To ensure θ is in [0, 1], we ﬁrst assume that θ is away from 0 and 1, i.e. we assume θ [ǫ, 1 ǫ] for ǫ> 0. Then we can make the second term ∈ − θ(1 θ)∆ p p , x x small. by assuming − 2h 1 − 0 − mi 4ǫ ǫ x x . | − m|≤ ∆ p p ≤ θ(1 θ)∆ p p 2| 1 − 0| − 2| 1 − 0| Under these assumptions and (16), we get θ′ [0, 1] and L = 0. We can apply ∈ 3 Lemma 3.1 to the forth line L4 of (15) to get

1 2 2 L = D G(x ,y ′ , v ′ ) D G(x ,y , v ) [x x , x x ] 4 2 xx m θ θ − xx m θ θ − m − m (17) C θ θ′ p p x x 2 ≥− 1| − || 1 − 0|| − m| C θ(1 θ)∆ p p 2 x x 3 ≥− 1 − 2| 1 − 0| | − m| 1 C ∆ p p 2 x x 3 ≥−4 1 2| 1 − 0| | − m|

For the ﬁfth and sixth line, L5 and L6, note that by (16),

θ′(1 θ′) θ(1 θ) = (θ θ′)(θ + θ′ 1) − − − − − = θ(1 θ)∆ p p , x x (θ + θ′ 1) − − 2h 1 − 0 − mi − 15 so that we can bound

L = ∆ [(θ′(1 θ′) θ(1 θ)] p p 2 x x 2 | 5| 1 − − − | 1 − 0| | − m| (18) θ (1 θ)(θ + θ′ 1)∆ ∆ p p 3 x x 3 ≤ − − 1 2| 1 − 0| | − m| 1 ∆ ∆ p p 3 x x 3 ≤ 4 1 2| 1 − 0| | − m|

L = ∆ [(θ(1 θ) θ′(1 θ′)] p p 2 (x x ) 2 | 6| 2 − − − | 1 − 0| | − m p| (19) θ(1 θ)(θ + θ′ 1)(∆ )2 p p 3 x x 3 ≤ − − 2 | 1 − 0| | − m| 1 (∆ )2 p p 3 x x 3. ≤ 4 2 | 1 − 0| | − m| Combining (17), (18), and (19), we can bound (15) from below 1 φ¯(x) u + θp + (1 θ)p , x x + D2 G(x ,y , v )[x x , x x ] ≥ h 1 − 0 − 0i 2 xx m θ θ − m − m (20) + ∆ θ(1 θ) p p 2 x x 2 1 − | 1 − 0| | − m| C ( p p 2 + p p 3) x x 3 + o( x x 2) − 2 | 1 − 0| | 1 − 0| | − m| | − m| where C2 depends on C1, ∆1, and ∆2. We apply Taylor’s Theorem to the ﬁrst line of (20) and change it to G(x, y , v ) with o( x x 2). Moreover, the little o θ θ | − m| term o( x x 2) is at least O( x x 3) because the generating function is C4. | − m| | − m| Therefore we can put it with the x x 3 term, and we get | − m| φ¯(x) G(x, y , v ) + ∆ θ(1 θ) p p 2 x x 2 ≥ θ θ 1 − | 1 − 0| | − m| C (1 + p p 2 + p p 3) x x 3 − 2 | 1 − 0| | 1 − 0| | − m| possibly taking larger C2 then before. Finally, we bound p1 p0 by Cediam(Y ) 1 | − | and y1 y0 from above and below to get Ce | − | ¯ ∆1 2 2 3 φ(x) G(x, yθ, vθ)+ 2 θ(1 θ) y1 y0 x xm γ x xm . ≥ Ce − | − | | − | − | − | 2 2 2 Hence we obtain the lemma with δ0 = ∆1/Ce and γ = C2(1 + Ce diam(Y ) + 3 3 Ce diam(Y ) ).

3.2. Local estimates for G convex functions. If a G-convex function φ is C2, − then for any x X we get a G-supporting function at x. From the deﬁnition of ∈ a G-supporting function, the diﬀerence of φ and the G-supporting function attains global minimum at x and the regularity condition implies

D2 φ D2 G D2 G I. xx ≥ xx ≥ −k xx k Hence it is semi convex. We show the semi convexity without C2 assumption on the G- convex function in the following proposition. 16 SEONGHYEON JEONG

Proposition 3.4. Let φ be a G-convex function that satisﬁes (nicew). Then we have following inequality 1 (21) φ(x ) (1 t)φ(x )+ tφ(x )+ t(1 t) D2 G x x 2 t ≤ − 0 1 2 − k xx k| 0 − 1| where x = (1 t)x + tx . In particular, φ is semi-convex. t − 0 1 Proof. Since φ is G convex, we have y Y and v R such that (x ,y,v) Ψ and − ∈ ∈ t ∈

φ(xt)= G(xt,y,v), φ(x) G(x, y, v), x X. ≥ ∀ ∈ Moreover, we have 1 G(x, y, v) φ(x )+ p , x x D2 G (x x )2 ≥ t h t − ti− 2k xx k − t where pt = DxG(xt,y,v). Evaluate this at x = x0 and x = x1 and add them with weight (1 t) and t respectively. − (1 t)φ(x )+ tφ(x ) (1 t)G(x ,y,v)+ tG(x ,y,v) − 0 1 ≥ − 0 1 (22) φ(x )+ p, (1 t)(x x )+ t(x x ) ≥ t h − 0 − t 1 − t i 1 D2 G (1 t)(x x )2 + t(x x )2 . − 2k xx k − 0 − t 1 − t Note that by the choice of x , we have (1 t)(x x )+ t(x x ) = 0 and t − 0 − t 1 − t x x = t x x and x x = (1 t) x x . | 0 − t| | 0 − 1| | 1 − t| − | 0 − 1| Then (22) becomes 1 (1 t)φ(x )+ tφ(x ) φ(x ) t(1 t) D2 G x x 2, − 0 1 ≥ t − 2 − k xx k| 0 − 1| which is the desired inequality. This implies that φ is semi-convex because t(1 − t) x x 2 + x 2 = (1 t) x 2 + t x 2 so that | 0 − 1| | t| − | 0| | 1| 1 φ(x )+ D2 G x 2 (1 t)φ(x )+ tφ(x ) t 2k xx k| t| ≤ − 0 1 1 + D2 G (t(1 t) x x 2 + x 2) 2k xx k − | 1 − 0| | t| 1 (1 t)(φ(x )+ D2 G x 2) ≤ − 0 2k xx k| 0| 1 + t(φ(x )+ D2 G x 2). 1 2k xx k| 1| Therefore φ(x)+ 1 D2 G x 2 is convex. 2 k xx k| |

When we use a norm of some derivatives of G and H, we need to check that the points lie in h and g, and they are in a compact subset of X Y R. So far, × × we choose points on the graph of a G-convex functions. Then this choice with the 17

G-convexity of G-subdiﬀerential ensures that the points which we are using are in compact subsets of h and g. But we need to use other points later (for example,

(xt,yi,u) in the lemma 3.7). Hence we localize the argument and use the next lemma to show our points lie in a compact subset of h.

Lemma 3.5. Let φ be a G-convex function with (nicew) and let x X˚. Then 0 ∈ there exists δ(x ) > 0 and S ⋐ h such that if x x <δ(x ), then 0 | 1 − 0| 0 (x ,y , G(x ,y ,H(x ,y , φ(x ))) S (23) t θ t 0 0 0 0 ∈ (x ,y , φ(x )) S t θ t ∈ for any x = (1 t)x + tx , t [0, 1] and y , the G-segment connecting y and y t − 0 1 ∈ θ 0 1 with focus (x , φ(x )) where y ∂ φ(x ), y ∂ φ(x ). t t 0 ∈ G 0 1 ∈ G 1 Proof. Note that by (nicew), we have that (x ,y , φ(x )) ˚h for any y ∂ φ(x ). 0 0 0 ∈ 0 ∈ G 0 Therefore, we have r1, r2, r3 > 0 such that

(24) S := B (x ) ( (∂ φ(x )) Y ) (φ(x ) r , φ(x )+ r ) ⋐ ˚h r1 0 × Nr2 G 0 ∩ × 0 − 3 0 3 that is, S is compact and S is contained in ˚h. Let Ce be the constant from Remark −1 2.10, and let ∂∗ φ(x )= expG (∂ φ(x )). Then by the same remark, we G 0 x0,φ(x0) G 0 have −1 (∂∗ φ(x )) h∗ expG ( (∂ φ(x )) Y ) . Nr2/Ce G 0 ∩ x0,φ(x0) ⊂ x0,φ(x0) Nr2 G 0 ∩ −1 Note that D G(x, ,H(x, ,u)) = expG ( ). Since the function (x,y,u) x · · x,u · 7→ DxG(x,y,H(x,y,u)) is uniformly continuous on S, there exist δx,δu > 0 such that if x x <δ and u φ(x ) <δ , then | − 0| x | − 0 | u r2 DxG(x,y,H(x,y,u)) DxG(x0,y,H(x0,y,φ(x0))) < | − | 4Ce for any y r2 (∂Gφ(x0)) Y . Hence, for any y r2 (∂Gφ(x0)) Y such that ∈ N−1 ∩ ∈ N ∩ G ∗ exp (y) 4 (∂ φ(x0)), we have x0,φ(x0) ∈ Nr2/ Ce G −1 expG (y) (∂∗ φ(x )) x,u ∈ Nr2/2Ce G 0 if x x < δ and u φ(x) < δ . Note that the set (∂∗ φ(x )) is | − 0| x | − 0 | u Nr2/2Ce G 0 convex by G convexity of G subdiﬀerentials. Again from Remark 2.10, if y − − −1 ∈ G ∗ 4 2 (∂Gφ(x0)) then exp (y) 4 (∂ φ(x0)). By Proposition Nr2/ Ce x0,φ(x0) ∈ Nr2/ Ce G 2.12, there exists δ1 such that if x x0 <δ1, then | − |

∂Gφ(x) 2 (∂Gφ(x0)) . ⊂ Nr2/4Ce Moreover, by continuity of G,H, and φ, we have δ such that if x x <δ , then 2 | − 0| 2 for any y ∂ φ(x ), 0 ∈ G 0 G(x, y ,H(x ,y , φ(x ))) φ(x ) < min δ , r | 0 0 0 0 − 0 | { u 3} 18 SEONGHYEON JEONG

φ(x) φ(x ) < min δ , r . | − 0 | { u 3} We take δ(x ) small enough so that δ(x ) min δ ,δ ,δ ,δ , r , r . Suppose 0 0 ≤ { x u 1 2 1 3} x1 x0 <δ(x0). Then y1 ∂Gφ(x1) 2 (∂Gφ(x0)). Moreover, for any y0 | − | ∈ ⊂ Nr2/4Ce ∈ ∂ φ(x ), we have φ(x ), G(x ,y ,H(x ,y , φ(x ))) (φ(x ) r , φ(x )+r ). Hence G 0 t t 0 0 0 0 ∈ 0 − 3 0 3 (x ,y ,u), (x ,y ,u) S h where u is either φ(x ) or G(x ,y ,H(x ,y , φ(x ))). t 0 t 1 ∈ ⊂ t t 0 0 0 0 By our construction, the G-segment that connects y0 and y1 with focus (xt,u) is contained in (∂ φ(x )) Y . Therefore, we have Nr2/2 G 0 ∩ (x ,y ,u) B (x ) (∂ φ(x )) Y (φ(x ) r , φ(x )+ r ) S. t θ ∈ δ(x0) 0 × Nr2/2 G 0 ∩ × 0 − 3 0 3 ⊂

Remark 3.6. The constant δ(x0) depends on the modulus of continuity of φ and ∂Gφ at x0. If we have estimates on these apriori, then we can get rid of this dependency on the solution.

Note that (24) shows structure of S other than (23). We will use this structure of S too. With help of the above lemma, we can bound various norms necessary to obtain the next lemma.

Lemma 3.7. Let φ be a G-convex function and let x X. Choose x such that 0 ∈ 1 x x <δ(x ). Let G(x, y , v ) and G(x, y , v ) be supporting G-aﬃne functions | 0 − 1| 0 0 0 1 1 that supports φ at x and x respectively. Let x [x , x ] such that 0 1 t ∈ 0 1

G(xt,y0, v0)= G(xt,y1, v1) =: u.

We assume y y x x , then we have | 1 − 0| ≥ | 0 − 1| (25) φ(x ) u C x x y y t − ≤ 3| 1 − 0|| 1 − 0| 2 where C3 depends on the C norm of G, hence on constant Ce from Remark 2.10, which in turn depends on S in Lemma 3.5

Proof. First of all, the deﬁnition of supporting function implies that

G(x ,y , v ) G(x ,y , v )= φ(x ) G(x ,y , v ) 0 0 0 0 − 0 1 1 0 − 0 1 1 ≥ G(x ,y , v ) G(x ,y , v )= G(x ,y , v ) φ(x ) 0 1 0 0 − 1 1 1 1 0 0 − 1 ≤ so that existence of xt is implied by the intermediate value theorem. If t was either 1 or 0, then the left hand side of (25) is 0 so that the lemma is trivial. Otherwise, by our choice of xt and u, we have the equalities

u = G(xt,y0, v0)= G(xt,y0,H(x0,y0, φ(x0))). 19

Note that by (24) of Lemma 3.5, we have (x ,y ,u) S. We use Taylor expansion t i ∈ on G(x, yi, vi) at xt to get 1 (26) φ(x ) u D G(x ,y , v ), x x + D2 G x x 2. i − ≤ h x t i i i − ti 2k xx k| i − t| From Lemma 3.4, 1 (27) φ(x ) u (1 t)(φ(x ) u)+ t(φ(x ) u)+ t(1 t) D2 G x x 2 t − ≤ − 0 − 1 − 2 − k xx k| 1 − 0| 1 (1 t)(φ(x ) u)+ t(φ(x ) u)+ D2 G x x 2. ≤ − 0 − 1 − 8k xx k| 1 − 0| If (1 t) D G(x ,y , v ), x x + t(φ(x ) u) 0, then from (26) and (27), − h x t 0 0 0 − ti 1 − ≤ 1 φ(x ) u (1 t) D G(x ,y , v ), x x + D2 G x x 2 t − ≤ − h x t 0 0 0 − ti 2k xx k| 0 − t| 1 + t(φ(x ) u)+ D2 G x x 2 1 − 8k xx k| 1 − 0| 1 1 (1 t) D2 G x x 2 + D2 G x x 2 ≤ − 2k xx k| 0 − t| 8k xx k| 1 − 0| 5 D2 G x x 2 ≤ 8k xx k| 1 − 0| 5 D2 G y y x x . ≤ 8k xx k| 1 − 0|| 1 − 0| Otherwise, since t(1 t) 1, we have − ≤ 0 (1 t) D G(x ,y , v ), x x + t(φ(x ) u) ≤ − h x t 0 0 0 − ti 1 − 1 1 D G(x ,y , v ), x x + (φ(x ) u) ≤ t h x t 0 0 0 − ti 1 t 1 − − so that

φ(x ) u (1 t) D G(x ,y , v ), x x + t(φ(x ) u) t − ≤ − h x t 0 0 0 − ti 1 − 1 1 + + (1 t) D2 G x x 2 8 2 − k xx k| 1 − 0| 1 1 D G(x ,y , v ), x x + (φ(x ) u) ≤ t h x t 0 0 0 − ti 1 t 1 − − 1 1 + + (1 t) D2 G x x 2 8 2 − k xx k| 1 − 0| 1 1 D G(x ,y , v ), x x + D G(x ,y , v ), x x ≤ t h x t 0 0 0 − ti 1 th x t 1 1 1 − ti 1 5− + D2 G x x 2 + D2 G x x 2 2(1 t)k xx k| 1 − t| 8k xx k| 1 − 0| − = D G(x ,y , v ), x x + D G(x ,y , v ), x x h x t 0 0 0 − 1i h x t 1 1 1 − 0i 1 5 + D2 G x x x x + D2 G x x 2. 2k xx k| 1 − t|| 1 − 0| 8k xx k| 1 − 0| 20 SEONGHYEON JEONG

In the last equality, we used t = |x0−xt| and 1 t = |x1−xt| . We use the fundamental |x1−x0| − |x1−x0| theorem of calculus to obtain 1 d φ(x ) u D G(x ,y ,H(x ,y ,u)), x x ds t − ≤ dθ h x t θ t θ 1 − 0i Z0 9 + D2 G x x 2 8k xx k| 1 − 0| 9 E C y y x x + D2 G x x 2 ≤k k e| 1 − 0|| 0 − t| 8k xx k| 1 − 0| 9 C2 + D2 G y y x x ≤ e 8k xx k | 1 − 0|| 1 − 0| where yθ is the G-segment connecting y0 and y1 with focus (xt,u).

Lemma 3.8. Let x be as in Lemma 3.7. There exist l, r that depend on x x t | 0 − 1| and y y and κ such that if ([x , x ]) X and | 0 − 1| Nr 0 1 ⊂ (28) y y max x x ,κ x x 1/5 | 0 − 1|≥ {| 1 − 0| | 1 − 0| } then, choosing x1 close to x0 if necessary, we have 1 3 y θ , Y ∂ φ(B (x )) Nl θ| ∈ 4 4 ∩ ⊂ G r t where yθ is the G-segment connecting y0 and y1 with focus (xt,u) as in the proof of Lemma 3.7.

Proof. Let u be as in the proof of Lemma 3.7. By the deﬁnition of G-convexity and Lemma 3.3, we have

(29) φ(x) max G(x, y , v ), G(x, y , v ) ≥ { 1 0 1 1 } 3 G(x, y , v )+ δ y y 2 x x 2 γ x x 3 ≥ θ θ 16 0| 1 − 0| | − t| − | − t| for θ [ 1 , 3 ], x x 1 C where v = H(x ,y ,u). Next, we look at a G- ∈ 4 4 | − t| ≤ 4 θ t θ aﬃne function that pass through (xt, φ(xt)) with focus (y,H(xt,y,φ(xt))) where y h , i.e. G(x,y,H(x ,y,φ(x ))). we estimate this function on the boundary ∈ xt,φ(xt) t t of Br(xt) to compare with φ. Let v(y,u)= H(xt,y,u).

G(x, y, v(y, φ(x ))) = G(x, y, v(y, φ(x ))) G(x, y , v(y , φ(x ))) t t − θ θ t (30) + G(x, y , v(y , φ(x ))) G(x, y , v(y ,u)) θ θ t − θ θ + G(x, yθ, v(yθ,u)) 21

Note that by Lemma 3.5, (x ,y ,u) S ⋐ h. For the ﬁrst line, noting that t θ ∈ φ(xt)= G(xt,y,v(y, φ(xt))), we have

G(x, y, v(y, φ(x ))) φ(x ) t − t 1 d (31) = (G(x + s(x x ),y,v(y, φ(x )))) ds ds t − t t Z0 1 = D G(x + s(x x ),y,v(y, φ(x ))), x x ds h x t − t t − ti Z0 and similar equation holds for G(x, y , v(y , φ(x ))) φ(x ). Therefore, we have θ θ t − t G(x, y, v(y, φ(x ))) G(x, y , v(y , φ(x ))) t − θ θ t 1 DxG(xt + s(x xt),y,v(y, φ(xt))) = − , x xt ds 0 h DxG(xt + s(x xt),yθ, v(yθ, φ(xt))) ! − i Z − − 1 1 d x + s(x x ),y + s′(y y ), (32) = D G t t θ θ , x x ds′ds ′ x ′− − t 0 0 ds h v(yθ + s (y yθ), φ(xt)) !! − i Z Z − 1 1 = (D G + D GD H)[y y , x x ]ds′ds xy xv y − θ − t Z0 Z0 1 1 D G = D G + D G y [y y , x x ]ds′ds xy xv D G − θ − t Z0 Z0 − v C x x y y , ≤ 4| − t|| − θ| 2 where C4 depends on the C norm of G and β (note that the functions in the last integral are evaluated at diﬀerent points so that C4 might not be equal to Ce). For the second line, we use Lemma 3.7.

G(x, y , v(y , φ(x ))) G(x, y , v(y ,u)) θ θ t − θ θ 1 d = (G(x, y , v(y ,u + s(φ(x ) u))) ds ds θ θ t − Z0 1 (33) = D GD H(φ(x ) u)ds v u t − Z0 C′ φ(x ) u ≤ 5| t − | C x x y y ≤ 5| 1 − 0|| 1 − 0| 1 where C5 depends on the C norm of G, β, and C3. applying (32) and (33) to (30),

(34) G(x, y, v(y, φ(x ))) G(x, y , v(y ,u))+C x x y y +C x x y y . t ≤ θ θ 4| − t|| − θ| 5| 1− 0|| 1− 0| Comparing (29) and (34), if we have 3 (35) C x x y y + C x x y y δ y y 2 x x 2 γ x x 3, 4| − t|| − θ| 5| 1 − 0|| 1 − 0|≤ 16 0| 1 − 0| | − t| − | − t| 22 SEONGHYEON JEONG then we can obtain G(x, y, v(y, φ(x ))) φ(x). (35) is satisﬁed if we have t ≤ 1 C x x y y δ y y 2 x x 2 5| 1 − 0|| 1 − 0|≤16 0| 1 − 0| | − t| 1 C x x y y δ y y 2 x x 2 4| − t|| − θ|≤16 0| 1 − 0| | − t| 1 γ x x 3 δ y y 2 x x 2. | − t| ≤16 0| 1 − 0| | − t| Therefore we choose 1 16C x x δ 163γ2C 5 (36) r2 = 5 | 1 − 0|, l = 0 r y y 2, κ = 5 δ y y 16C | 1 − 0| δ3 0 | 1 − 0| 4 0 so that we obtain 1 3 G(x, y, v(y, φ(x ))) φ(x) for y y θ [ , ] and x ∂B (x ). t ≤ ∈ Nl θ| ∈ 4 4 ∈ r t Note that κ does not depend on x0 and x1. From the condition (28), we know that

2 16C5 4 √δ0C5 1 3 r x1 x0 5 , l x1 x0 2 diam(Y ) 2 . ≤ κδ0 | − | ≤ 4C4 | − | 2 2 4C4 r2 Therefore, choosing x1 close enough to x0 so that x1 x0 3 , we can | − |≤ diam(Y ) δ0C5 assume that r (37) l 2 ≤ 2 where r2 is from the proof of Lemma 3.5. Since G(xt,y,v(y, φ(xt))) = φ(xt), we get a local maximum of G(x, y, v(y, φ(x ))) φ(x) at some point x B (x ) with non- t − y ∈ r t negative value. Then from the proof of Lemma 3.5, we get that ( y θ [0, 1] ) Nl { θ| ∈ } ∩ Y h . If G(x ,y,v(y, φ(x ))) = φ(x ), then G(x, y, v(y, φ(x ))) is a local ⊂ xy ,φ(xt) y t y t support of φ at xy. Since φ is G-convex, the G-aﬃne function G(x, y, v(y, φ(xt))) is a global support of φ and hence y ∂ φ(x ) ∂ φ(B (x )) by Proposition 2.9 (Re- ∈ G y ∈ G r t call by Remark 2.8, we can use this proposition). Suppose G(xy,y,v(y, φ(xt))) > φ(x ). Note that we have φ(x) G(x, y, v(y,u)). We deﬁne y ≥

fy(h)= max G(x, y, v(y,h)) φ(x) x∈Br(xt) { − }

= max G(x,y,H(xt,y,h)) φ(x) x∈Br(xt) { − } Then f (φ(x )) > 0 and f (u) 0. Moreover, since D G and D H are y t y ≤ k v k k u k bounded on Ψ and Φ, the functions in the max are equicontinuous so that fy is a continuous function. Therefore, there exists h [u, φ(x )] at which we have y ∈ t ′ fy(hy) = 0. In other words, G(x, y, v(y,hy)) supports φ at some point x in Br(xt). From (37) and the proof of Lemma 3.5, we have (x′,y,h ) h. Hence we get y ∈ y ∂ φ(B (x )). ∈ G r t 23

3.3. Some convex geometry. In this subsection, we prove a useful convex geometry lemma to estimate the volume of y θ 1 , 3 . This subsection Nl θ| ∈ 4 4 corresponds to Lemma 5.10 in [Loe09]. In this paper we will show the same type of lemma but we do not use the generating function G outside of its domain X Y R. × × Note that in [Loe09], it is used that the cost function c can be extended to and diﬀerentiated outside of Ω′.

Remark 3.9. Suppose we have a compact convex set A. Then for each point p ∂A, ∈ there is rp > 0 such that the boundary ∂A can be written as a graph of a convex function up to an isometry in Brp (p). But since A is compact, we can have r > 0 that does not depend on p ∂A but can replace r . Moreover, since the convex ∈ p functions are locally Lipschitz, by taking smaller r if needed, we can assume that each convex function that describes ∂A is a Lipschitz function in Br(p). Then using compactness again, we can assume that we have a uniform Lipschitz constant. In diamA fact, we can bound the Lipschitz constant by L = r . See corollary A.23 of [Fig17].

Lemma 3.10. Let A be a compact convex set. Then there exist rA > 0 and CA that depend on the set A such that for any r′ < r and x A, we have A ∈ (38) Vol (B ′ (x) A) C Vol (B ′ (x)) . r ∩ ≥ A r Proof. Let r be as in the Remark 3.9, and let L be the Lipschitz constant in the

′ r r ′ same remark. Let r < 2 . If x q A dist(q,∂A) 2 , Then Br (x) A so that ∈{ ∈ | ≥ } r ⊂ Vol (B ′ (x) A) = Vol (B ′ (x)). Now suppose dist(x, ∂A) < , then p ∂A such r ∩ r 2 ∃ ∈ that B ′ (x) B (p). In B (p), we have that ∂A is the graph of a convex function r ⊂ r r f up to an isometry, and f is Lipschitz with the Lipschitz constant L. Then, since x A, we know that xn f(x′) where x = (x′, xn) Rn−1 R and we have the ∈ ≥ ∈ × inclusions

′ n n−1 n ′ B ′ (x) A B ′ (x) q = (q , q ) R R q f(q ) ) r ⊃ r ∩{ ∈ × | ≥ } n ′ ′ ′ (39) \ B ′ (x) q q f(x )+ L x q ⊃ r ∩{ | ≥ | − |} n n ′ ′ B ′ (x) q q x + L x q . ⊃ r ∩{ | ≥ | − |} The last set above is the intersection of a ball centered at x and a cone with vertex at x. Hence the volume of this set is a multiple of the volume of the ball where the constant multiplied depends on L. Therefore the above inclusion implies (38) with r CA depending on L, hence on A. Therefore the lemma holds with rA = 2 .

In the next lemma, we use the term “length of a curve”. The curve that we are using is not necessarily diﬀerentiable, so we use next deﬁnition for the length of a curve 24 SEONGHYEON JEONG

Definition 3.11. Let γ :[a,b] Rn be a continuous curve. We define its length → by n Length (γ) = sup γ(t ) γ(t ) a = t t t = b | i − i−1 | 0 ≤ 1 ≤···≤ n (i=1 ) X It is well known that this definition preserves a lot of properties of arclength of C1 curves. Some continuous curves may have Length (γ) = , but Length (γ) is ∞ finite if γ is a Lipschitz curve with finite domain.

Lemma 3.12. Let A be a compact convex set and let γ : [0, 1] A be a bi-Lipschitz → curve, that is

(40) C′ s t γ(s) γ(t) C s t γ | − | ≤ | − |≤ γ | − | ′ for some constants Cγ and Cγ . Then there exists KA and lA > 0 that depend on A and C′ such that for any l l , we have γ ≤ A (41) Vol ( (γ) A) K C′ ln−1. Nl ∩ ≥ A γ Proof. We assume l< rA where r is from the previous lemma 3.10. Let m N be 2 A ∈ the smallest number such that C r m. Then by taking r smaller if necessary, γ ≤ A A we have m 2Cγ . Note that we have Length (γ) C from (40). We deﬁne rA γ ≤ ≤ ′ i i+1 Cγ γi(t) = γ((1 t) 2m + t 2m ). Then γi is bi-Lipschitz with constants 2m for lower C−γ bound and 2m for upper bound i.e. C′ C γ s t γ (s) γ (t) γ s t . 2m| − | ≤ | i − i |≤ 2m| − | In addition, we have Length (γ ) rA by our choice of m and l. Suppose (γ ) i ≤ 2 Nl i ∩ ∂A = . Then we can write ∂A as a graph of a Lipschitz convex function f 6 ∅ with Lipschitz constant L around some point p (γ ) ∂A. Moreover, for any ∈ Nl i ∩ x (γ ), there are some t,s [0, 1] such that ∈ Nl i ∈ x p < x γ (t) + γ (t) γ (s) + γ (s) p | − | | − i | | i − i | | i − | r r r 3 A + A + A = r, ≤ 2 2 2 4 where r is the radius of the ball around a point on ∂A in which we can write ∂A as a graph of a convex function (Remark 3.9). Therefore (γ ) lies in the epigraph Nl i of f in Br(p). Then the proof of Lemma 3.10 shows that at each point on the curve γ , there exists a conical sector Sec in B (γ (t)) A which is a translation of i γi(t) l i ∩ the conical sector Sec0 :

Sec = B (0) q qn L q′ . 0 l ∩{ | ≥ | |} Note that the inscribed ball in this conical sector has radius L′l where L′ is a constant that depends on L so that the inscribed ball is B ′ (v) for some v Sec . L l ∈ 0 25

Therefore, for each 0 i 2m 1, we get vi such that L′l (γi + vi) l (γ) A. ′ C≤ ≤ − N ⊂ N ∩ Now, if we have l< γ , then for any x (γ ) and y (γ ), we get that for 4m ∈ Nl i ∈ Nl i+2 some s,t [0, 1], ∈ x y γ (t) γ (s) ( x γ (t) + y γ (s) ) | − | ≥ | i − i+2 |− | − i | | − i+2 | C′ γ 2l> 0. ≥ 2m −

Therefore, L′l (γi) L′l (γi+2)= . Note that each L′l (γi) has volume bounded ′ − ′ N ∩N (∅L )n 1C N below by (L′l)n−1 γ (0) γ (1) γ ln−1 so that | i − i |≥ 2m 2m−1 Vol ( (γ) A) Vol ′ (γ + v ) Nl ∩ ≥ NL l i i i=0 ! [ m−1 Vol ′ (γ + v ) ≥ NL l 2i i i=0 ! [ (L′)n−1C′ 1 γ ln−1 m = (L′)n−1C′ ln−1. ≥ 2m × 2 γ ′ ′ ′ 1 Cγ rA Cγ Cγ Therefore we get the lemma with lA = rA min , = and KA = 8 Cγ ≤ { 2 4m } 4m 1 ′ n−1 2 (L ) .

∗ rA Lemma 3.13. Let yθ be as in Lemma 3.8 and let A = h ( ). If l 5 , then x0,φ x0 ≤ 8Ce we have 1 3 (42) Vol y θ , Y C ln−1 y y Nl θ| ∈ 4 4 ∩ ≥ V | 0 − 1| where CV depends on x0, h, and Ce.

Proof. Note that θ y is a bi-Lipschitz curve with 7→ θ 1 ′ ′ 2 ′ 2 y1 y0 θ θ yθ yθ Ce y1 y0 θ θ . Ce | − || − | ≤ | − |≤ | − || − |

Then the reparametrized curve θ y(1 ) 1 + 3 is bi-Lipschitz with Lipschitz con- 7→ −θ 4 θ 4 2 2 G −1 stants 2 y1 y0 and 2C y1 y0 . Then the curve θ exp (y 1 3 ) Ce e xt,φ(xt) (1−θ) 4 +θ 4 | − | | − | 2 7→ is bi-Lipschitz with Lipschitz constants L = 3 y1 y0 for lower bound and Ce | − | 3 G −1 L = 2C y1 y0 for upper bound. Moreover, since the map exp is e | − | x0,φ(x0) bi-Lipschitz with Lipschitz constants 1 and C , we have Ce e

G −1 1 3 ∗ l expx0,φ(x0) yθ θ , hx0,φ(x0) N Ce | ∈ 4 4 ∩ −1 1 3 expG y θ , Y . ⊂ x0,φ(x0) Nl θ| ∈ 4 4 ∩ Note that by (vDomConv), h∗ is convex. Moreover, by our choice of l, we x0,φ(x0) l 1 have L/LrA. Then by Lemma 3.12, we get a constant Kx that depends on Ce ≤ 8 0 26 SEONGHYEON JEONG h∗ , hence on x such that x0,φ(x0) 0

G −1 1 3 ∗ l Vol expx0,φ(x0) yθ θ , hx0,φ(x0) N Ce | ∈ 4 4 ∩ 2 n−1 Kx0 3 y1 y0 (l/Ce) . ≥ Ce | − | Using bi-Lipschitzness once more, we obtain 1 3 Vol y θ , Y C ln−1 y y , Nl θ| ∈ 4 4 ∩ ≥ V | 1 − 0| 2n+2 with CV =2Kx0 /Ce .

Remark 3.14. The constant C depends on the Lipschitz constant of ∂h∗ so V x0,φ(x0) ∗ that it depends on x and the value of φ at x . If we assume that h R 0 0 { x,u}(x,u)∈X× is uniformly Lipschitz, we can get rid of this dependency.

3.4. Proof of the main theorem. In the proof of the main theorem, we will use the lemmas in previous subsections. So, we review the conditions to use those lemmas. We choose x , x X˚ and y ,y ∂ φ(x ). First of all, to localize the 0 1 ∈ 0 1 ∈ G 0 argument, we choose x1 close enough to x0. Explicitly, we choose x0 x1 smaller 2 2 | − | 4C4 r2 than δ(x ) to use Lemma 3.5, smaller than 3 to use Lemma 3.8, and smaller 0 diam δ0C5 2 2 C4 rA 1 ∗ than 12 3 (with A = h ) to get the condition in Lemma 3.13. For 4Ce δ0C5 diam(Y ) x0,φ(x0) y and y , we need y y max x x ,κ x x 1/5 to use Lemma 3.8. 0 1 | 0 − 1| ≥ {| 0 − 1| | 0 − 1| } Note that if there is no such y0 and y1, that means we have Hölder regularity with 1 exponent 5 . Proof of the main theorem 2.14. We deal with the first part of the theorem. 1. In the first case, we deal with the case p = first. If p = , then we have ∞ ∞ µ (B (x )) CVol (B (x )) C′rn r t ≤ r t ≤ for some C and C′. Moreover, since φ is an Alexandrov solution, we have µ (B (x )) = ν (∂ φ(B (x ))) ν y θ 1 , 3 (43) r t G r t ≥ Nl θ| ∈ 4 4 λC ln−1 y y . ≥ V | 0 − 1| Combining these, we get C′rn C ln−1 y y . We plug (36) into this inequality ≥ V | 0 − 1| to obtain 1 y y C x x 4n−1 | 0 − 1|≤ | 0 − 1| for some constant C. Note that this implies single valuedness and Hölder continuity of ∂Gφ. Next we deal with the case p< . If the condition on µ holds with p< , define ∞ ∞ F by

(44) F (V ) = sup µ(B) B X a ball of volume V . { | ⊂ } 27

Then we have F (Vol (B (x ))) µ(B (x )) = ν(∂ φ(B (x ))). This with (43) r t ≥ r t G r t implies x x n/2 (45) F C | 0 − 1| C′ x x (n−1)/2 y y (3n−1)/2 y y n/2 ≥ | 0 − 1| | 0 − 1| | 0 − 1| for some constants C and C′. From the assumption on µ, we have F (V ) ≤ C′′V 1−1/p for some C′′. This with above inequality (45), we have

2n−1+ 1 (1− n ) 1 (1− n ) y y 2 p C x x 2 p . | 0 − 1| ≤ | 1 − 0| Therefore, with the condition p>n, we get

ρ y y C x x 4n−2+ρ | 0 − 1|≤ | 0 − 1| where ρ = 1 n . Therefore, we have the following : for any x X˚, there exists − p 0 ∈ some constants rx0 and Cx0 that depends on x0, φ(x0), continuity of φ at x0 such that if x x < r , we have | 0 − 1| x0 ρ (46) y y C x x 4n−2+ρ . | 0 − 1|≤ x0 | 0 − 1| Then for a compact set X′ ⋐ X˚, we can cover it with a finite number of balls on which we have (46) with respect to the center of the ball. Then we can get the ′ ′ Hölder regularity of ∂Gφ on X by connecting any two points in X with a piecewise segment where each segment lies in one of the balls. Then the Hölder constant will be bounded by the sum of the Hölder constant on each ball that contains a segment times the number of the balls. To get the Hölder regularity of the potential φ, we G note that ∂Gφ(x)= expx,φ(x)(Dxφ(x)), and use remark 2.10. Now we prove the second part of the theorem. + + 2. Suppose we have f : R R such that limr→0 f(r) = 0 and for any x X → n(1− 1 ) ∈ and r 0 we have µ(B (x)) f(r)r n . Note that we can choose f strictly ≥ r ≤ increasing. Then by (44), we have

1 1 n 1− n 1 1 1 1− 1 (47) F (V ) f V n V n ≤ ω × ω n ! n n where ωn is the volume of the unit ball in R . Deﬁne f˜ by

1 1 1− n n 2 1 1 1 1 f˜(V ) n− = f V 2 . ω ω n n ! Then (47) becomes 2n−1 2 1− 1 (48) F (V ) f˜ V n V n . ≤ h i 28 SEONGHYEON JEONG

We combine (48) with (45), and we have x x (49) f˜ C′ | 0 − 1| C′′ y y y y ≥ | 0 − 1| | 0 − 1| for some constants C′, C′′ > 0. Note that we can assume that |x0−x1| 0 as |y0−y1| → x x 0 because otherwise, we get a Lipschitz estimate so that we still can | 0 − 1| → get H¨older regularity. Then (49) implies that ∂Gφ is a single valued map. Let g be the modulus of continuity of the G subdiﬀerential map ∂Gφ. We divide into two 1 − cases. If g(u) max u,κu 5 , then we get g(u) 0 as u 0. In the other case, ≤ → → from (49) we get n o u f˜ C′ C′′g(u). g(u) ≥ Since f was strictly increasing, so is f˜, so that f˜ is invertible. Therefore the above equation is equivalent to g(u) u f˜−1 (C′′g(u)) . ≥ C′ −1 ′′ z Let ω be the inverse of z f˜ (C z) ′ . Note that ω is strictly increasing. 7→ C Therefore, composing ω on above inequality shows that

g(u) ω(u). ≤ −1 ′′ z Since the function z f˜ (C z) ′ is strictly increasing and has limit 0 as z 0, 7→ C → ω(u) also has limit 0 as u 0. Therefore the above inequality implies that g(u) 0 → → as u 0. Hence the modulus of continuity of ∂ φ has limit 0 as the variable tends → G to 0 so that ∂Gφ is continuous at x0.

Remark 3.15. Note that from Lemma 3.5 and 3.13, the constants that we get depend on the value of φ and continuity of φ. Because of these dependencies, the H¨older regularity that we get in this paper might not be uniform for solutions to the (GJE). To get a bounds on the H¨older norm that do not depend on the solution φ, we need to add some conditions on the set h so that we can get a uniform Lipschitz constant for ∂h∗ for any (x, u) X R as mentioned in Remark 3.14. Moreover, x,u ∈ × we need an apiori estimate on the modulus of continuity of φ and ∂Gφ as mentioned in Remark 3.6.

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