C^{1}-regularity of the Aronsson equation in R^{2}

Changyou Wang^{1}, Yifeng Yu^{2} 1 Deptartment of Mathematics, University of Kentucky, Lexington, KY 40506 2 Department of Mathematics, University of Texas at Austin, Austin, TX 78712

Abstract

For a nonnegative, uniformly convex H ∈ C^{2}(R^{2}) with H(0) = 0, if u ∈ C(Ω), Ω ⊂ R^{2}, is a viscosity solution of the Aronsson equation (1.7), then u ∈ C^{1}(Ω). This generalizes the C^{1}-regularity theorem on infinity harmonic functions in dimension two by Savin [S] to the Aronsson equation.

§1. Introduction

Calculus of variations in L^{∞} was initiated by Aronsson [A1,2,3,4] in 1960’s. Thanks to both the development of theory of viscosity solutions of elliptic equations by Crandall-Lions (cf. [CIL]) and several applications to applied fields (cf. Barron [B], Barron-Jensen [BJ], and Aronsson-Crandall-Juutinen [ACJ]), there have been great interests in last years to study the minimization problem of the supremal functional:

F (u, Ω) = ess sup_{x∈Ω}H(x, u(x), ∇u(x)), Ω ⊂ R^{n}, u ∈ W^{1,∞}(Ω, R^{l}).  (1.1)

Barron-Jensen-Wang [BJW1] have established both necessary and sufficient conditions for the sequentially lower semicontinuity property of the supremal functional F in W^{1,∞}, which are suitable L^{∞}-versions of Morrey’s quasiconvexity (cf. [M]) for integral functionals. In the scalar case (l = 1), Barron-Jensen-Wang [BJW2] (see also Crandall [C1]) have established, under appropriate conditions, the existence of absolute minimizers and proved that any absolute minimizer is a viscosity solution of the Aronsson equation

−∇_{x}(H(x, u(x), ∇u(x))) · H_{p}(x, u(x), ∇u(x)) = 0, x ∈ Ω.  (1.2)

Among other results in [Y], the second author has showed that the convexity of H(·, p) are sufficient for viscosity solutions of the Aronsson equation (1.2) to be absolute minimizers of F.

Partially motivated by [Y] and Crandall-Evans-Gariepy [CEG], Gariepy-Wang-Yu [GWY] have established the equivalence between absolute minimizers of F and viscosity solutions of Aronsson equation (1.2) for quasiconvex Hamiltonians H = H(p) ∈ C^{2}(R^{n}) by introducing the comparison principle of generalized cones (see Theorem 2.1 in §2 below).

In this paper, we are mainly interested in the regularity (e.g. differentiability or C^{1}) of viscosity solutions of the Aronsson equation (1.7).

Before stating the main results, we would like to review some of known theorems for H(p) = |p|^{2}, p ∈ R^{n}. It is well-known (cf. [A5,6] [J]) that equation (1.2) is the infinity-Laplace equation, of which a viscosity solution is called an infinity harmonic function,

−Δ_{∞}u := − ∑_{i,j=1}^{n} u_{i}u_{j}u_{ij} = 0, in Ω  (1.3)

and the absolute minimality is the absolute minimizing Lipschitz (or AML) property: for any open subset U ⊂⊂ Ω and v ∈ W^{1,∞}(U) with v = u on ∂U, we have

∥∇u∥_{L^{∞}(U)} ≤ ∥∇v∥_{L^{∞}(U)}.  (1.4)

Aronsson [A5] proved that any C^{2}-infinity harmonic function satisfies the AML property. Jensen [J] has not only proved the equivalence between an infinity harmonic function and the AML property, but also established the unique solvability of the Dirichlet problem of equation (1.3).

Crandall-Evans-Gariepy [CEG] have recently showed that u ∈ C^{0}(Ω) is an infinity harmonic function iff u enjoys comparison with cones in Ω:

for any open subset U ⊂⊂ Ω, a, b ∈ R, and x_{0} ∈ Ω,

u(x) ≤ (≥)a + b|x − x_{0}| on ∂(U \ {x_{0}}) ⇒ u(x) ≤ (≥)a + b|x − x_{0}| in U.  (1.5)

In a very recent important paper [S], Savin has utilized [CEG] and Crandall-Evans [CE] to prove that any C^{0}-infinity harmonic function in Ω ⊂ R^{2} is in C^{1}(Ω). The interested readers may refer to Crandall [C2] which includes comprehensive expositions on [S].

In this paper, we extend the main theorems of [S] to the Aronsson equation (1.7) for a large class of Hamiltonian functions H ∈ C^{2}(R^{2}).

First we extend [CE] and obtain the following theorem on blow-up limits of viscosity solutions of (1.7) on R^{n} for n ≥ 2, which may have its own interests.

Theorem A. Assume that H ∈ C^{2}(R^{n}) is a nonnegative, uniformly convex function, i.e. there is α_{H} > 0 such that

p^{T} · H_{pp}(p) · p ≥ α_{H}|p|^{2}, ∀p ∈ R^{n},  (1.6)

and H(0) = 0. Suppose that u ∈ C^{0}(Ω), Ω ⊂ R^{n}, is a viscosity solution of the Aronsson equation:

−H_{p}(∇u(x)) ⊗ H_{p}(∇u(x)) : ∇^{2}u(x) = 0 in Ω  (1.7)

then for any x ∈ Ω, there exists a vector e_{x,r} ∈ R^{n}, with H(e_{x,r}) = S^{+}(H, u, x) (see §2 for the definition of S^{+}(H, u, x)), such that

lim_{r→0} max_{B_{r}(x)} \frac{|u(y) − u(x) − e_{x,r} · (y − x)|}{r} = 0.  (1.8)

Based on theorem A and the main theorem of [GWY], we are able to make necessary modifications of the idea of [S] and prove

Theorem B. Under the same assumptions of theorem A. If u ∈ C^{0}(Ω), Ω ⊂ R^{2}, is a viscosity solution of the Aronsson equation (1.7), then u ∈ C^{1}(Ω).

A compactness argument from the proof of theorem B yields

Theorem C. Under the same assumptions of theorem A. There exists a monotonically nondecreasing function ρ : [0, 1] → R_{+}, with lim_{r→0} ρ(r) = 0, such that if u is a viscosity solution of the Aronsson equation (1.7) in B_{1} ⊂ R^{2}, with ∥H(∇u)∥_{L^{∞}(B_{1})} ≤ C, then

osc _{B_{r}}|∇u| ≤ ρ(r), ∀r ≤ \frac{1}{2}.  (1.9)

A direct consequence of theorem C is the following Liouville property.

Theorem D. Under the same assumptions of theorem A. If u ∈ C^{0}(R^{2}) is a viscosity solution of the Aronsson equation (1.7) in R^{2} and

|u(x)| ≤ C(1 + |x|)

then u is a linear function.

Due to the local Lipschitz continuity for viscosity solutions of (1.7), the uniform convexity condition (1.6) can be relaxed. In fact, we have the following remark.

Remark E. The conclusions of theorem B,C, and D remain to be true, provided that H ∈ C^{2}(R^{2}) satisfies:

(i) H(0) = 0 and H(p) ≥ 0 for any p ∈ R^{2}. (ii) H is coercive, i.e. lim_{|p|→+∞} H(p) = +∞.

(iii) (H_{p_{i}p_{j}})(p) is positive definite for any p ∈ R^{2}.

The paper is written as follows. In §2, we review the main theorems of [GWY]. In §3, we outline some basic properties of the Aronsson equation (1.7) and prove of Theorem A. In §4, we establish three key propositions which are needed to prove theorem B. In §5, we prove both Theorem B, C, D, and Remark E.

Acknowledgement. The authors would like to thank Professor Crandall for his many constructive suggestions that help to improve an earlier version of the paper to the current form. The first author is partially supported by NSF grant 0400718.

§2. Review of the main theorems by [GWY]

In this section, we review the main theorems of [CWY] which are needed for the paper. We assume that the Hamiltonian function H(x, u, p) = H(p) ∈ C^{2}(R^{n}) depends only on the p-variable. First, we recall a few definitions.

Definition 2.1. A function u ∈ Wloc^{1,∞}(Ω) is an absolute minimizer of the supremal functional F, if for any open subset U ⊂⊂ Ω and any v ∈ W^{1,∞}(U) with v = u on ∂U, then

F (u, U) = ∥H(∇u(x))∥_{L^{∞}(U)} ≤ F (v, U) = ∥H(∇v(x))∥_{L^{∞}(U)}.  (2.1)

Definition 2.2. For any k ≥ 0, the generalized cone centered at the origin with slope k, C_{k}^{H}(x), is defined by

C_{k}^{H}(x) = max_{{p∈R^{n}:H(p)=k}} p · x, x ∈ R^{n}  (2.2)

where · is the inner product in R^{n}.

Definition 2.3. A upper semicontinuous function u ∈ USC(Ω) enjoys comparison with generalized cones from above ( abbreviated u ∈ CGCA(Ω)) if we have, for any V ⊂⊂ Ω, x_{0} ∈ Ω, and k ≥ 0,

u(x) ≤ u(x_{0}) + C_{k}^{H}(x − x_{0}) ∀x ∈ ∂V ⇒ u(x) ≤ u(x_{0}) + C_{k}^{H}(x − x_{0}), ∀ x ∈ V. (2.3)

A lower semicontiuous function v ∈ LSC(Ω) enjoys comparison with generalized cones from below (abbreviated v ∈ CGCB(Ω)) if −v enjoys comparison with generalized cones from above, with H replaced by Ĥ (Ĥ(p) ≡ H(−p) for all p ∈ R^{n}). A continuous function w ∈ C(Ω) enjoys comparison with generalized cones (or abbreviated w ∈ CGC(Ω)) if w ∈ CGCA(Ω) ∩ CGCB(Ω).

Definition 2.4. A function u ∈ USC(Ω) is a viscosity subsolution of the Aronsson equation (1.7), if for any (x_{0}, ϕ) ∈ Ω × C^{2}(Ω)

u(x_{0}) − ϕ(x_{0}) ≥ u(x) − ϕ(x), ∀x ∈ Ω

then

−H_{p}(∇ϕ) ⊗ H_{p}(∇ϕ) : ∇^{2}ϕ|_{x=x_{0}} ≤ 0.  (2.4)

A function u ∈ LSC(Ω) is a viscosity supersolution of the Aronsson equation (1.7) , if −u is a viscosity subsolution of (1.7) with H replaced by Ĥ, where Ĥ(p) = H(−p) for p ∈ R^{n} . A function u ∈ C(Ω) is a viscosity solution of the Aronsson equation (1.7), if u is both a viscosity subsolution and supersolution of (1.7).

Now we state the main theorem proved by [GWY] on the equivalence among absolute minimizers, viscosity solutions of the Aronsson equation (1.7), and comparison with generalized cones.

Theorem 2.1. (see [GWY]) Assume that H satisfies: H1. H ∈ C^{2}(R^{n}), H(0) = 0, and H(p) ≥ 0 for any p ∈ R^{n}. H2. H is quasi-convex, i.e.

H(tp + (1 − t)q) ≤ max{H(p), H(q)}, ∀ p, q ∈ R^{n}, 0 ≤ t ≤ 1.  (2.5)

H3. H is coercive:

lim_{|p|→+∞} H(p) = +∞.  (2.6)

Then the following three statements are equivalent:

(a) u ∈ Wloc^{1,∞}(Ω) is an absolute minimizer of F (v, ·) = ∥H(∇v)∥_{L^{∞}(·)}.

(b) u ∈ CGC(Ω).

(c) u ∈ C(Ω) is a viscosity solution of the Aronsson equation (1.7).

In particular, Theorem 2.1 implies that any viscosity solution of (1.7) enjoys comparison with C_{k}^{H}(·). This plays a crucial role in the paper.

In [GWY] Proposition 3.1, the following properties of CGC(Ω) have been established.

Proposition 2.2. (see [GWY]) Under the same assumptions as in theorem 2.1. If u ∈ CGC(Ω), then u ∈ Wloc^{1,∞}(Ω), and, for any x_{0} ∈ Ω and 0 < r < d(x_{0}, ∂Ω), we have (i)

S_{r}^{+}(H, u, x_{0}) := inf{k ≥ 0 | u(x) ≤ u(x_{0}) + C_{k}^{H}(x − x_{0}), ∀x ∈ ∂B_{r}(x_{0})}  (2.7)

exists and is monotonically nondecreasing with respect to r > 0. In particular,

S^{+}(H, u, x_{0}) = lim_{r↓0^{+}} S_{r}^{+}(H, u, x_{0})  (2.8)

exists for any x_{0} ∈ Ω and is upper semicontinuous; (ii) S_{r}^{−}(H, u, x_{0}) := S_{r}^{+}(Ĥ, −u, x_{0}) exists and is monotonically nondecreasing with respect to r > 0, where Ĥ(p) = H(−p) for p ∈ R^{n}. In particular, S^{−}(H, u, x_{0}) = lim_{r↓0^{+}} S_{r}^{−}(H, u, x_{0}) exists for any x_{0} ∈ Ω and is upper semicontinuous; and (iii) for any x_{0} ∈ Ω, we have S^{+}(H, u, x_{0}) = S^{−}(H, u, x_{0}). Moreover, if ∇u(x_{0}) exists for x_{0} ∈ Ω, then we have

H(∇u(x_{0})) = S^{+}(H, u, x_{0}) = S^{−}(H, u, x_{0}).  (2.9)

Proposition 2.3. (see [GWY]) Assume u ∈ CGC(Ω). For any x_{0} ∈ Ω, 0 < r < d(x_{0}, ∂Ω), if x_{r} ∈ ∂B_{r}(x_{0}) satisfies

u(x_{r}) − u(x_{0}) = C_{S_{r}^{+}(H,u,x_{0})}^{H}(x_{r} − x_{0})  (2.10)

then for any 0 < R < d(x_{r}, ∂Ω) we have

S_{R}^{+}(H, u, x_{r}) ≥ S_{r}^{+}(H, u, x_{0}).  (2.11)

For this paper, we also need the comparison property of viscosity solutions for (1.7) with linear functions.

Proposition 2.4. Under the same assumptions as in theorem 2.1, let ψ(x) = a · x + b (a ∈ R^{n}, b ∈ R) be a linear function such that if a ≠ 0 then H_{p}(a) ≠ 0. For any open subset U ⊂⊂ Ω, if u ∈ C^{0}(Ω) is a viscosity solution of the Aronsson equation (1.7) and u ≤ ψ on ∂U, then u ≤ ψ in U.

Proof. If a = 0, then ψ = b so that ψ(x) = b + C_{0}^{H}(x) is a generalized cone with slope k = 0. Hence Proposition 2.4 follows from theorem 2.1.

If a ≠ 0, then we have |H_{p}(a)| = δ > 0 for some δ > 0. Suppose that the conclusion were false, then there exist an open subset V̂ ⊂⊂ Ω and x̂ ∈ V such that u = ψ on ∂V̂, and

max_{V̂}(u − ψ) = (u − ψ)(x̂) > 0.

Let R > 0 be such that V ⊂ B_{R}, and ϵ > 0 be sufficiently small. Then we have

u(x) ≤ ψ(x) + ϵ(R^{2} − |x|^{2}) on ∂V̂ ,

but

u(x̂) > ψ(x̂) + ϵ(R^{2} − |x̂|^{2}).

For simplicity, we may assume

{u − (ψ + ϵ(R^{2} − |x|^{2}))}(x̂) = max_{V̂}(u − (ψ + ϵ(R^{2} − |x|^{2}))).

Since ψ + ϵ(R^{2} − |x|^{2}) ∈ C^{2}(V̂) and u is a viscosity solution of (1.7), we have

0 ≥ − ∑_{i,j} H_{p_{i}}(a − 2ϵx̂)H_{p_{j}}(a − 2ϵx̂)(−2ϵδ_{ij}) = 2ϵ|H_{p}(a − 2ϵx̂)|^{2}  (2.12)

This is impossible, since

|H_{p}(a − 2ϵx̂)| ≥ \frac{1}{2}|H_{p}(a)| = \frac{δ}{2} > 0

if ϵ > 0 is sufficiently small. This completes the proof of Proposition 2.4.

§3. Some preliminary results of (1.7) and proof of Theorem A

This section is devoted to the proof of theorem A. To do it, we need several Lemmas. The first Lemma asserts that the assumption of theorem A implies that of theorem 2.1.

Lemma 3.1. Suppose that H ∈ C^{2}(R^{n}) is a nonnegative, uniformly convex function, with H(0) = 0. Then H satisfies the conditions (H1)-(H3) in theorem 2.1. Moreover, H(p) > 0 and H_{p}(p) ≠ 0 for all 0 ≠ p ∈ R^{n}.

Proof. Since a uniformly convex function is quasi-convex, H satisfies H2. Since H is uniformly convex, there exists a α_{H} > 0 such that

H_{pp}(p) ≥ α_{H}I_{n}, ∀p ∈ R^{n}.  (3.1)

This, combined with the Taylor expansion up to second order, implies

H(p) ≥ H(q) + H_{p}(q) · (p − q) + \frac{α_{H}}{2} |p − q|^{2}, ∀p, q ∈ R^{n}.  (3.2)

In particular, we have

H(p) ≥ H_{p}(0) · p + \frac{α_{H}}{2} |p|^{2} ≥ \frac{α_{H}}{4} |p|^{2} − 4α_{H}^{−1}|H_{p}(0)|^{2}, ∀p ∈ R^{n}.

Hence lim_{|p|→∞} H(p) = +∞ and H3 is satisfied.

Suppose that there exists 0 ≠ p_{0} ∈ R^{n} such that H(p_{0}) = 0. Then the convexity and nonnegativity of H imply H(tp_{0}) = 0 for all 0 ≤ t ≤ 1. This implies

0 = \frac{d^{2}}{dt^{2}} |_{t=0^{+}}H(tp_{0}) = p_{0}^{T} · H_{pp}(0) · p_{0} ≥ α_{H}|p_{0}|^{2},

this is impossible.

Suppose that there exists 0 ≠ p_{1} ∈ R^{n} such that H(p_{1}) > 0, but H_{p}(p_{1}) = 0. Then (3.2) implies

0 = H(0) ≥ H(p_{1}) − H_{p}(p_{1}) · p_{1} = H(p_{1})

This is also impossible. Hence the proof of Lemma 3.1 is complete.

Denote by S^{n−1} ⊂ R^{n} the unit sphere of R^{n}, and H^{−1}(c) = {p ∈ R^{n} : H(p) = c} (c > 0) the level set of H. Now we have

Lemma 3.2. Under the same assumptions as in Lemma 3.1. Then for any c > 0, the Gauss map ν(p) = \frac{H_{p}(p)}{|H_{p}(p)|} : H^{−1}(c) → S^{n−1} is a C^{1} diffeomorphism.

Proof. Lemma 3.1 and the implicit function theorem imply that for any c > 0, H^{−1}(c) ⊂ R^{n} is a strictly convex C^{1}-hypersurface. Hence H^{−1}(c) is diffeomorphic to S^{n−1}. Moreover, (3.1) implies ν is locally one to one, and onto so that it is a finite covering map. Hence ν is globally one to one and a C^{1}-diffeomorphism.

Lemma 3.3. Under the same assumptions as in Lemma 3.1. For any k ≥ 0, C_{k}^{H}(·) is convex and positively homogeneous of degree one. Moreover, C_{k}^{H}(x) ∈ C^{1}(R^{n} \ {0}).

Proof. It follows from the definition that C_{k}^{H}(·) is the supremum of a family of linear functions so that it convex and satisfies

C_{k}^{H}(tx) = tC_{k}^{H}(x), ∀t > 0, ∀x ∈ R^{n}.

To see C_{k}^{H} ∈ C^{1}(R^{n} \ {0}), it suffices to prove that for k > 0 C_{k}^{H} ∈ C^{1}(S^{n−1}). By the Lagrange Multiple Theorem, for any x ∈ S^{n−1} there exists a p̂ ∈ H^{−1}(k) such that

C_{k}^{H}(x) = p̂ · x, and \frac{H_{p}(p̂)}{|H_{p}(p̂)|} = x  (3.3)

so that ν(p̂) = x or p̂ = ν^{−1}(x). Therefore, we have C_{k}^{H}(x) = ν^{−1}(x) · x ∈ C^{1}(S^{n−1}).

Lemma 3.4. Under the same assumptions as in Lemma 3.1. If u ∈ C^{0}(R^{n}) is a viscosity solution of the Aronsson equation (1.7) satisfying S^{+}(H, u, 0) = ∥H(∇u)∥_{L^{∞}(R^{n})} < ∞, then there exists p_{0} ∈ R^{n}, with H(p_{0}) = S^{+}(H, u, 0), such that

u(x) = u(0) + p_{0} · x, ∀x ∈ R^{n}.  (3.4)

Proof. First we claim that

S_{r}^{+}(H, u, 0) = S^{+}(H, u, 0), ∀r > 0,  (3.5)
S_{R}^{+}(H, u, y) ≤ S^{+}(H, u, 0), 0 ≠ y ∈ R^{n}, ∀R > 0.  (3.6)

To see (3.6), observe that for any y ∈ R^{n} and x ∈ R^{n}, with |x − y| = R, we have

u(x) = u(y) + ∫_{0}^{1} ∇u(tx + (1 − t)y) · (x − y) dt

≤ u(y) + ∫_{0}^{1} max_{H(p)≤∥H(∇u)∥_{L^{∞}(R^{n}))}} p · (x − y)

= u(y) + C_{∥H(∇u)∥_{L^{∞}(R^{n})}}^{H}(x − y).

This implies

S_{R}^{+}(H, u, y) ≤ ∥H(∇u)∥_{L^{∞}(R^{n})} = S^{+}(H, u, 0).

It is easy to see that (3.5) follows from (3.6) and the fact S_{r}^{+}(H, u, 0) ≥ S^{+}(H, u, 0).

If S^{+}(H, u, 0) = 0, then H(∇u(x)) = 0 for a.e. x ∈ R^{n}. Hence ∇u(x) = 0 for a.e. x ∈ R^{n} so that u is constant and (3.4) holds with p_{0} = 0.

We may assume that S^{+}(H, u, 0) > 0. For R > 1, there exists x_{R} ∈ ∂B_{R} such that

u(x_{R}) = u(0) + C_{S^{+}(H,u,0)}^{H}(x_{R}).  (3.7)

On the other hand, for any x ∈ R^{n}, we have, by (3.6),

u(x_{R}) − u(x) ≤ C_{S_{|x_{R}−x|}^{+}(H,u,x)}^{H}(x_{R} − x) ≤ C_{S^{+}(H,u,0)}^{H}(x_{R} − x).  (3.8)

(3.7) and (3.8) imply

u(x) ≥ u(0) + |x_{R}|\{C_{S^{+}(H,u,0)}^{H}(\frac{x_{R}}{|x_{R}|}) − C_{S^{+}(H,u,0)}^{H}(\frac{x_{R} − x}{|x_{R}|})\}.  (3.9)

For any x ∈ R^{n}, let R > 1 be sufficiently large so that x_{R} − x ≠ 0. By Lemma 3.3, we have that C_{S^{+}(H,u,0)}^{H} ∈ C^{1}(R^{n} \ {0}) is convex. Hence we have

C_{S^{+}(H,u,0)}^{H}(\frac{x_{R}}{|x_{R}|}) − C_{S^{+}(H,u,0)}^{H}(\frac{x_{R} − x}{|x_{R}|}) ≥ ∇C_{S^{+}(H,u,0)}^{H}(\frac{x_{R} − x}{|x_{R}|}) · \frac{x}{|x_{R}|}.  (3.10)

Combining (3.9) with (3.10), we obtain

u(x) ≥ u(0) + ∇C_{S^{+}(H,u,0)}^{H}(\frac{x_{R} − x}{|x_{R}|}) · x, ∀x ∈ R^{n}.  (3.11)

We may assume that there exists a q ∈ S^{n−1} such that lim_{R→∞} \frac{x_{R}}{|x_{R}|} = q. It is clear that lim_{R→∞} \frac{x_{R}−x}{|x_{R}|} = q. Hence (3.11) yields

u(x) ≥ u(0) + ∇C_{S^{+}(H,u,0)}^{H}(q) · x, ∀x ∈ R^{n}.  (3.12)

Applying the same argument with u and H replaced by −u and Ĥ(p) = H(−p) for p ∈ R^{n}, we have that there is a q̂ ∈ S^{n−1} such that

u(x) ≤ u(0) + ∇C_{S^{+}(H,u,0)}^{H}(q̂) · x, ∀x ∈ R^{n}.  (3.13)

Comparing (3.12) with (3.13), we conclude that

∇C_{S^{+}(H,u,0)}^{H}(q) = ∇C_{S^{+}(H,u,0)}^{H}(q̂) = p_{0}

and

u(x) = u(0) + p_{0} · x, ∀x ∈ R^{n}.

It is easy to see that H(p_{0}) = S^{+}(H, u, 0). Hence the proof of Lemma 3.4 is complete. Now we are ready to give the proof of theorem A.

Proof of Theorem A.

For simplicity, we assume that x_{0} = 0 and u(0) = 0. It follows from theorem 2.1 and Proposition 2.2 that u ∈ Wloc^{1,∞}(Ω). For any r_{k} ↓ 0, let v_{k}(x) = \frac{u(r_{k}x)}{r_{k}} : B_{r_{k}^{−1}} → R. Then for any R > 0, we have

∥v_{k}∥_{L^{∞}(B_{R})} ≤ ∥∇u∥_{L^{∞}(B_{r_{k}R})}R ≤ CR, ∥∇v_{k}∥_{L^{∞}(B_{R})} ≤ ∥∇u∥_{L^{∞}(B_{r_{k}R})} ≤ C. (3.14)

Therefore we may assume that there exists v_{∞} ∈ Lloc^{∞}(R^{n}), with ∇v_{∞} ∈ L^{∞}(R^{n}), such that for any R > 0

∥v_{k} − v_{∞}∥_{C^{0}(B_{R})} → 0, ∇v_{k} → ∇v_{∞}weak^{∗}in L^{∞}(B_{R}).  (3.15)

Since v_{k} is a viscosity solution of (1.7) on B_{r_{k}}^{−1}, we see that v_{∞} is a viscosity solution of (1.7) on R^{n}. Moreover, by the lower semicontinuity (cf. [BJW1]), we have

∥H(∇v_{∞})∥_{L^{∞}(B_{R})} ≤ lim inf_{k→∞} ∥H(∇v_{k})∥_{L^{∞}(B_{R})} = lim inf_{k→∞} ∥H(∇u)∥_{L^{∞}(B_{r_{k}R})}
= lim inf_{k→∞} ∥S^{+}(H, u, x)∥_{L^{∞}(B_{r_{k}R})} = S^{+}(H, u, 0).  (3.16)

In particular, we have

S^{+}(H, v_{∞}, 0) ≤ ∥H(∇v_{∞})∥_{L^{∞}(R^{n})} ≤ S^{+}(H, u, 0).  (3.17)

On the other hand, for any r > 0 there exists a x_{r}^{k} ∈ ∂B_{r} such that

v_{k}(x_{r}^{k}) − v_{k}(0) = C_{S_{r}^{+}(H,v_{k},0)}^{H}(x_{r}^{k}).  (3.18)

It is readily seen that

S_{r}^{+}(H, v_{k}, 0) = S_{r_{k}r}^{+}(H, u, 0) ≥ S^{+}(H, u, 0).  (3.19)

After passing to subsequences, we may assume that there exists x_{r} ∈ ∂B_{r} such that lim_{k→∞} x_{r}^{k} = x_{r}. Then (3.18) and (3.19) imply

v_{∞}(x_{r}) − v_{∞}(0) ≥ C_{S^{+}(H,u,0)}^{H}(x_{r}).  (3.20)

This implies

S_{r}^{+}(H, v_{∞}, 0) ≥ S^{+}(H, u, 0), ∀r > 0.  (3.21)

Taking r into zero, (3.17) and (3.21) imply

S^{+}(H, v_{∞}, 0) = ∥H(∇v_{∞})∥_{L^{∞}(R^{n})} = S^{+}(H, u, 0).  (3.22)

Hence Lemma 3.4 implies that v_{∞} = a · x, for some a ∈ R^{n}, and we have

lim_{r_{k}→0} ∥\frac{u(r_{k}x) − u(0)}{r_{k}} − a · x∥_{L^{∞}(B_{1})} = 0.  (3.23)

This implies (1.8). Suppose not, then there exists r_{k} ↓ 0^{+}, x_{k} ∈ R^{n} with |x_{k}| = r_{k}, and δ > 0 such that for any e ∈ H^{−1}(S^{+}(H, u, 0)), we have

\frac{|u(x_{k}) − e · x_{k}|}{r_{k}} ≥ δ, ∀k ≥ 1.  (3.24)

But, on the other hand, there exists a ∈ H^{−1}(S^{+}(H, u, 0)) such that after passing to subsequences, v_{k}(x) ≡ \frac{u(|x_{k}|x)}{r_{k}} → a·x uniformly on B_{1}. This contradicts (3.24). Therefore the proof of theorem A is complete.

§4. Propositions

This section is to establish the differentiability for any viscosity solution of the Aronsson equation (1.7) by showing that there is a unique blow-up limit at any point for n = 2: for any x_{0} ∈ Ω, lim_{r↓0} e_{x_{0},r} exists and its limit is continuous with respect to x_{0}.

To do this, we follow the clever idea of [S] and the strategy is as follows. Consider the generic case that x_{0} ∈ Ω such that S^{+}(H, u, x_{0}) > 0 and u is nonlinear near x_{0}. Then there exists r > 0 and a linear segement [y_{0}, z_{0}] ⊂ B\frac{r}{6} (x_{0}) and a linear function l(x) = a_{0} · x + b_{0} such that for some x_{0} ∈ (y_{0}, z_{0}), one has either (i) u(x) ≥ l(x), ∀x ∈ [y_{0}, z_{0}]; u(x_{0}) = l(x_{0}); u(y_{0}) > l(y_{0}), u(z_{0}) > l(z_{0}), or (ii) u(x) ≤ l(x), ∀x ∈ [y_{0}, z_{0}]; u(x_{0}) = l(x_{0}); u(y_{0}) < l(y_{0}), u(z_{0}) < l(z_{0}). Assume the first case happens, then the first observation is that there is an affine plane P containing the graph of l such that y_{0}, z_{0} belong to two different connected components of {u > P } ∩ B_{r}(x_{0}) (see Lemma 4.2 below). Since theorem A implies that there exists e_{x_{0},r} ∈ R^{2} such that max_{x∈B_{r}(x_{0})} \frac{|u(x)−u(x_{0})−e_{x,r}·(x−x_{0})|}{r} << 1 for 0 < r << 1, we want to prove |e_{x_{0},r} − a_{0}| << 1 for 0 < r << 1 (see Lemma 4.3 and 4.4 below). The proof of Lemma 4.5 is different from that of [S] Lemma 3, here we use the discrete gradient flow which has originated from [CE] when they studied the regularity of the infinity Laplace equation. A continuous version of this type of gradient flow by [CE] was also communicated to us by Crandall [C3].

The following proposition is the crucial step to prove Theorem B.

Proposition 4.1. Under the same assumptions as in theorem A, let u ∈ C^{0,1}(B_{6}), B_{6} ⊂ R^{2}, be a viscosity solution of the Aronsson equation (1.7). For any ϵ > 0, there is δ = δ(ϵ, H, ∥∇u∥_{L^{∞}(B_{6})}) > 0 such that if

∥u − e_{6} · x∥_{L^{∞}(B_{6})} ≤ δ, 1 ≤ H(e_{6}) ≤ 2,  (4.1)

and if, for 0 < r ≤ \frac{1}{2},

∥u − e_{0,6r} · x∥_{L^{∞}(B_{6r})} ≤ δr,  (4.2)

for some e_{0,6r} ∈ R^{2} with H(e_{0,6r}) = S^{+}(H, u, 0). Then

|e_{6} − e_{0,6r}|^{2} ≤ Cϵ  (4.3)

for some 1 ≤ C = C(H) < ∞.

Proof. We divide the proof into two cases. Case A. u is non-linear in B_{r}: There are a line segement [z_{1}, z_{2}] ⊂ B_{r}, a linear function l(x) = a_{0} · x + b_{0}, x ∈ [z_{1}, z_{2}] with a_{0} = \frac{u(z_{2})−u(z_{1})}{|z_{2}−z_{1}|}, and a z_{3} ∈ (z_{1}, z_{2}) such that either (1)

u ≥ l on [z_{1}, z_{2}], u(z_{1}) > l(z_{1}), u(z_{3}) = l(z_{3}), u(z_{2}) > l(z_{2}),

or (2)

u ≤ l on [z_{1}, z_{2}], u(z_{1}) < l(z_{1}), u(z_{3}) = l(z_{3}), u(z_{2}) < l(z_{2}).

For simplicity, assume (1) holds. Then we have

Lemma 4.2. Under the same notations. Then there exists e ∈ R^{2}, with H(e) = S^{+}(H, u, z_{3}), such that z_{1} and z_{2} belong to two distinct connected components of the set {y ∈ R^{2} : u(y) > u(z_{3}) + e · (y − z_{3})} ∩ B_{6}.

Proof of Lemma 4.2. It follows from Theorem A that for s_{k} ↓ 0 there exists e ∈ R^{2}, with H(e) = S^{+}(H, u, z_{3}), such that

lim_{s_{k}↓0} \frac{∥u(y) − u(z_{3}) − e · (y − z_{3})∥_{L^{∞}(B_{s_{k}} (z_{3}))}}{s_{k}} = 0.  (4.4)

Suppose that z_{1}, z_{2} belong to the same connected component of {y ∈ R^{2} : u(y) > u(z_{3}) + e · (y − z_{3})} ∩ B_{6}. Then there exists a polygonal line Γ ⊂ {y ∈ R^{2} : u(y) > u(z_{3}) + e · (y − z_{3})} ∩ B_{6} joining z_{1} to z_{2}. Let S = Γ ∪ [z_{1}, z_{2}] be the closed curve and U ⊂ B_{6} be the open set such that S = ∂U. Without loss of generality, we assume that there exists a β > 0 such that

B_{β}^{+}(z_{3}) = B_{β}(z_{3}) ∩ {y ∈ R^{2} : 0 < ^{}(y − z_{3}, z_{2} − z_{1}) < π} ⊂ U.

Note that there exists an δ_{0} > 0 such that u(y) − u(z_{3}) − e · (y − z_{3}) ≥ δ_{0} for any y ∈ Γ. Therefore there are a small ϵ > 0 and a unit vector v ∈ R^{2}, with ^{}(v, z_{2} − z_{1}) = \frac{π}{2} and e + ϵv ≠ 0, such that

u(y) ≥ u(z_{3}) + (e + ϵv) · (y − z_{3}), ∀y ∈ S.

Note that ψ(y) = u(z_{3}) + (e + ϵv) · (y − z_{3}) is linear and ∇ψ = e + ϵv ≠ 0 so that Lemma 3.1 implies that H(∇ψ) > 0 and H_{p}(∇ψ) ≠ 0. Therefore, Proposition 2.3 implies

u(y) ≥ u(z_{3}) + (e + ϵv) · (y − z_{3}), ∀y ∈ U.

In particular, we have

lim_{s_{k}↓0} max_{y∈B_{s_{k}} (z_{3})∩B_{β}^{+}(z_{3})} \frac{|u(y) − u(z_{3}) − e · (y − z_{3})|}{s_{k}} ≥ ϵ > 0,

this contradicts (4.4). Therefore Lemma 4.2 is proven.

Now we want to prove

|e_{6} − e|^{2} ≤ Cϵ.  (4.5)

To prove (4.5), we may assume |e_{6} − e| ≥ \frac{ϵ}{2}. Denote f = e_{6} − e and form the strip S := \{y ∈ R^{2}| |f ·(y −z_{3})| ≤ 2δ\}. It is easy to see the width of S is at most \frac{4δ}{ϵ}. Moreover, (4.1) implies

|u(y) − u(z_{3}) − e_{6} · (y − z_{3})| ≤ 2δ, ∀y ∈ B_{6}.  (4.6)

One can check

{y ∈ R^{2} | f · (y − z_{3}) < −2δ} ∩ B_{6} ⊂ {y ∈ R^{2} : u(y) < u(z_{3}) + e · (y − z_{3})}
{y ∈ R^{2} | f · (y − z_{3}) > 2δ} ∩ B_{6} ⊂ {y ∈ R^{2} : u(y) > u(z_{3}) + e · (y − z_{3})}.

Therefore, by Lemma 4.2, there is a connected component of {y ∈ R^{2} : u(y) > u(z_{3}) + e · (y − z_{3})} ∩ B_{6}, called U, that intersects B_{1} and is contained in the strip S.

We claim that U ⊄ B_{6}. For, otherwise, we have u(y) = u(z_{3}) + e · (y − z_{3}) for all y ∈ ∂U, Proposition 2.3 implies u ≡ u(z_{3}) + e · (y − z_{3}) in U. This contradicts with the definition of U. Therefore there exists a polygonal line Γ inside U starting in B_{1} and exiting B_{6}.

Now we can find a point z_{4} ∈ B_{6}, with |z_{4} − z_{3}| = 3 and z_{4} − z_{3} ⊥ f, such that (A1)

∥u(y) − u(z_{3}) − e_{6} · (y − z_{3})∥_{L^{∞}(B_{2}(z_{4}))} ≤ 2δ, 1 ≤ H(e_{6}) ≤ 2.

(A2) {y ∈ R^{2} : u(y) > u(z_{3}) + e · (y − z_{3})} ∩ B_{6} has a connected component U ⊂ S that contains a polygonal line Γ connecting the two arcs S ∩ ∂B_{2}(z_{4}).

The estimate (4.5) follows from the following two Lemmas.

Lemma 4.3. Under the same notations. There is δ = δ(ϵ, H) > 0 such that

H(e) ≥ H(e_{6}) − 2ϵ.  (4.7)

Proof of Lemma 4.3. We may assume that H(e) ≤ H(e_{6}) − ϵ. The convexity of H implies

H_{p}(e_{6}) · f ≥ H(e_{6}) − H(e) ≥ ϵ.  (4.8)

Let y_{1} = z_{4} + \frac{2H_{p}(e_{6})}{f·H_{p}(e_{6})}δ be the intersection of \{z_{4} + tH_{p}(e_{6}) | t ≥ 0\} with the strip S. Now we have

Claim 4.3(a). There are δ = δ(H) > 0 and C = C(H) > 0, with Cδ ≤ 2, such that that for any x ∈ U ∩ B_{1}(z_{4}), we have

C_{H(e)+Cδ}^{H}(y − x) ≥ e · (y − x) + 4δ, ∀y ∈ ∂B_{1}(x).  (4.9)

Proof of Claim 4.3(a). As H(e) ≤ H(e_{6}) − ϵ ≤ 2 and Cδ ≤ 2, we have H(p) ≤ 4 for any p ∈ H^{−1}(H(e) + Cδ). For any y ∈ ∂B_{1}(x), let p_{0} ∈ H^{−1}(H(e) + Cδ) be such that

C_{H(e)+Cδ}^{H}(y − x) = p_{0} · (y − x).

By the Lagrange Multiple Theorem, we have

\frac{H_{p}(p_{0})}{|H_{p}(p_{0})|} = y − x.

By the convexity, we have

Cδ = H(p_{0}) − H(e) ≤ H_{p}(p_{0}) · (p_{0} − e).

Hence we have

C_{H(e)+Cδ}^{H}(y − x) − e · (y − x) = (p_{0} − e) · (y − x) = (p_{0} − e) · \frac{H_{p}(p_{0})}{|H_{p}(p_{0})|}

≥ \frac{C}{|H_{p}(p_{0})|}δ ≥ \frac{C}{max_{H(p)≤4} |H_{p}(p)|}δ ≥ 4δ

provided that C ≥ 4 max_{{H(p)≤4}} |H_{p}(p)|. This implies (4.9).

Claim 4.3(b). There are δ = δ(H) > 0 and C = C(H) > 0, with Cδ ≤ 2, such that that for any x ∈ U ∩ B_{1}(z_{4}), we have

S^{+}(H, u, x) ≤ H(e) + Cδ.  (4.10)

Proof of Claim 4.3(b). For any x_{0} ∈ B_{1}(z_{4}) ∩ U, we have that B_{1}(x_{0}) ⊂ B_{2}(z_{4}) and u(z_{3}) + e · (x_{0} − z_{3}) < u(x_{0}). Note that

u(y) = u(z_{3}) + e · (y − z_{3}) < u(x_{0}) + e · (y − x_{0}) ≤ u(x_{0}) + C_{H(e)}^{H}(y − x_{0}), ∀y ∈ ∂U ∩ B_{1}(x_{0}),

and

u(y) ≤ u(z_{3}) + e · (y − z_{3}) + 4δ < u(x_{0}) + e · (y − x_{0}) + 4δ, ∀y ∈ ∂B_{1}(x_{0}) ∩ U. (4.11)

It follows from (4.9) that we have

u(y) ≤ u(x_{0}) + C_{H(e)+Cδ}^{H}(y − x_{0}), on ∂(U ∩ B_{1}(x_{0})).

Hence theorem 2.1 implies

u(y) ≤ u(x_{0}) + C_{H(e)+Cδ}^{H}(y − x_{0}), in U ∩ B_{1}(x_{0}).

This, combined with Proposition 2.2, implies (4.10).

Now we have Claim 4.3(c). There exists y_{0} ∈ L = \{z_{4} + t\frac{H_{p}(e_{6})}{|H_{p}(e_{6})|} : −\frac{1}{3} ≤ t ≤ −\frac{1}{4}\} such that

S^{+}(H, u, y_{0}) ≥ H(e_{6}) − 50|H_{p}(e_{6})|δ.  (4.12)

Proof of Claim 4.3(c). Since L ⊂ B_{2}(z_{4}), A1 implies

u(z_{4} − \frac{1}{3} \frac{H_{p}(e_{6})}{|H_{p}(e_{6})}) − u(z_{4} − \frac{1}{4} \frac{H_{p}(e_{6})}{|H_{p}(e_{6})|}) ≥ \frac{1}{12}e_{6} · \frac{H_{p}(e_{6})}{|H_{p}(e_{6})|} − 4δ.  (4.13)

On the other hand, let c = sup_{x∈L} S^{+}(H, u, x), then we have

u(z_{4} − \frac{1}{3} \frac{H_{p}(e_{6})}{|H_{p}(e_{6})|}) − u(z_{4} − \frac{1}{4} \frac{H_{p}(e_{6})}{|H_{p}(e_{6})|}) ≤ \frac{1}{12}C_{c}^{H}(\frac{H_{p}(e_{6})}{|H_{p}(e_{6})|}).  (4.14)

In fact, for any large positive integer n, let v = \frac{1}{12} \frac{H_{p}(e_{6})}{|H_{p}(e_{6})|} and r_{n} = \frac{|v|}{n} and x_{0} = z_{4} − \frac{1}{3} \frac{H_{p}(e_{6})}{|H_{p}(e_{6})|}, x_{1} = x_{0} + \frac{v}{n}, ⋯, x_{n−1} = x_{0} + \frac{n−1}{n}v, x_{n} = z_{4} − \frac{1}{4} \frac{H_{p}(e_{6})}{|H_{p}(e_{6})|}. Then we have

u(x_{i}) − u(x_{i−1}) ≤ C_{S_{rn}^{+} (H,u,x_{i−1})}^{H}(x_{i} − x_{i−1}), ∀1 ≤ i ≤ n.  (4.15)

The monotonicity of S_{r}^{+}(H, u, x) and upper semicontinuity of S^{+}(H, u, x) imply

lim_{n→∞} sup_{x∈L} S_{r_{n}}^{+}(H, u, x) ≤ c ≡ sup_{x∈L} S^{+}(H, u, x).

Therefore for any ϵ > 0 there is a sufficiently large n_{0} > 0 such that

max_{0≤i≤n_{0}−1} S_{r_{n_{0}}}^{+} (H, u, x_{i}) ≤ c + ϵ.

Therefore (4.15) implies

u(z_{4} − \frac{1}{4} \frac{H_{p}(e_{6})}{|H_{p}(e_{6})|}) − u(z_{4} − \frac{1}{3} \frac{H_{p}(e_{6})}{|H_{p}(e_{6})|}) = ∑_{i=1}^{n_{0}}(u(x_{i}) − u(x_{i−1}))

≤ ∑_{i=1}^{n_{0}} C_{c+ϵ}^{H}(x_{i} − x_{i−1})

= \frac{1}{12}C_{c+ϵ}^{H}(\frac{H_{p}(e_{6})}{|H_{p}(e_{6})|}).

Since ϵ > 0 is arbitrary, this implies (4.14).

Combining (4.13) with (4.14), we obtain

e_{6} · H_{p}(e_{6}) − 48|H_{p}(e_{6})|δ ≤ C_{c}^{H}(H_{p}(e_{6})).  (4.16)

Now we have

C_{H(q)}^{H}(H_{p}(q)) = q · H_{p}(q), ∀0 ≠ q ∈ R^{2}.  (4.17)

To see (4.17), observe that for q ≠ 0, Lemma 3.1 implies H(q) > 0 and H_{p}(q) ≠ 0. If v ∈ H^{−1}(H(q)) is such that v·H_{p}(q) = C_{H(q)}^{H}(H_{p}(q)), then the Lagrange Multiple Theorem implies \frac{H_{p}(v)}{|H_{p}(v)|} = \frac{H_{p}(q)}{|H_{p}(q)|}. In particular, the Gauss map ν(v) = ν(q) and Lemma 3.2 implies v = q and (4.17) follows. (4.16), combined with (4.17), implies

C_{H(e_{6})}^{H}(H_{p}(e_{6})) − 48|H_{p}(e_{6})|δ ≤ C_{c}^{H}(H_{p}(e_{6})).  (4.18)

Now we need

C_{H(e_{6})−48|H_{p}(e_{6})|δ}^{H}(H_{p}(e_{6})) ≤ C_{H(e_{6})}^{H}(H_{p}(e_{6})) − 48|H_{p}(e_{6})|δ.  (4.19)

To see (4.19), note that for any q ∈ H^{−1}(H(e_{6}) − 48|H_{p}(e_{6})|δ), the convexity of H implies that

H(e_{6}) − 48|H_{p}(e_{6})|δ = H(q) ≥ H(e_{6}) + H_{p}(e_{6}) · (q − e_{6})

so that

q · H_{p}(e_{6}) ≤ e_{6} · H_{p}(e_{6}) − 48|H_{p}(e_{6})|δ.

Taking maximum of q ∈ H^{−1}(H(e_{6}) − 48|H_{p}(e_{6})|δ), we obtain (4.19).

From (4.18) and (4.19), we have c ≥ H(e_{6}) − 48|H_{p}(e_{6})|δ and hence (4.12) is proven. Now we proceed with the discrete gradient flow as follows. Let t = d(Γ, ∂U ∩B_{2}(z_{4})) ≤

\frac{2δ}{|f|} be the step size, let y_{i} ∈ B_{2}(z_{4}) (1 ≤ i ≤ m = m(δ)) be such that

|y_{i} − y_{i−1}| = t, u(y_{i}) = u(y_{i−1}) + C_{S_{t}^{+}(H,u,y_{i−1})}^{H}(y_{i} − y_{i−1}), 1 ≤ i ≤ m,  (4.20)

and

d(y_{m}, ∂B_{2}(z_{4})) ≤ t.  (4.21)

Then it follows from Proposition 2.3 that

S^{+}(H, u, y_{i}) ≥ S^{+}(H, u, y_{0}) ≥ H(e_{6}) − 50|H_{p}(e_{6})|δ, 0 ≤ i ≤ m.  (4.22)

To see (4.21), observe that

u(y_{m}) − u(y_{0}) = ∑_{i=1}^{m}(u(y_{i}) − u(y_{i−1})) ≥ ∑_{i=1}^{m} C_{S^{+}(H,u,y_{i−1})}^{H}(y_{i} − y_{i−1})

≥ ∑_{i=1}^{m} C_{S^{+}(H,u,y_{0})}^{H}(y_{i} − y_{i−1})

≥ ∑_{i=1}^{m} C_{H(e_{6})−50|H_{p}(e_{6})|δ}^{H}(y_{i} − y_{i−1}) ≥ Cmt  (4.23)

where we have used the fact that there exists C > 0 depending only on H such that

C_{H(e_{6})−50|H_{p}(e_{6})|δ}^{H}(x) ≥ C|x|, ∀|x| = 1.  (4.24)

Since |u(y_{m}) − u(y_{0})| ≤ ∥∇u∥_{L^{∞}(B_{6})}|y_{m} − y_{0}|, (4.23) implies (4.21) holds for some m = m(δ) ≥ 1.

Now we have

Claim 4.3(d). There exist δ = δ(ϵ) > 0 and 1 ≤ i ≤ m such that y_{i} ∈ B_{1}(z_{4}) ∩ U.

Proof of Claim 4.3(d). First observe that (4.23) also yields

u(y_{m}) − u(y_{0}) ≥ ∑_{i=1}^{m} C_{H(e_{6})−50|H_{p}(e_{6})|δ}^{H}(y_{i} − y_{i−1})

≥ C_{H(e_{6})−50|H_{p}(e_{6})|δ}^{H}(∑_{i=1}^{m}(y_{i} − y_{i−1}))

= C_{H(e_{6})−50|H_{p}(e_{6})|δ}^{H}(y_{m} − y_{0}),  (4.25)

where we have used the triangle inequality C_{k}^{H}(x + y) ≤ C_{k}^{H}(x) + C_{k}^{H}(y).

On the other hand, A1 implies

u(y_{m}) − u(y_{0}) ≤ e_{6} · (y_{m} − y_{0}) + 4δ.  (4.26)

Therefore we have

C_{H(e_{6})−50|H_{p}(e_{6})|δ}^{H}(y_{m} − y_{0}) ≤ e_{6} · (y_{m} − y_{0}) + 4δ.  (4.27)

Note that (4.21) implies |y_{m} − z_{4}| ≥ 2 − \frac{2δ}{|f|} ≥ \frac{4}{3} and hence |y_{m} − y_{0}| ≥ 1. Denote e_{m} = \frac{y_{m}−y_{0}}{|y_{m}−y_{0}|}, then (4.27) implies

C_{H(e_{6})−50|H_{p}(e_{6})|δ}^{H}(e_{m}) ≤ e_{6} · e_{m} + 4δ.  (4.28)

Now we claim that there exists η(δ) > 0, with lim_{δ→0} η(δ) = 0, such that

cos^{}(e_{m}, H_{p}(e_{6})) ≥ 1 − η(δ).  (4.29)

In fact, we may assume that there exists ê ∈ R^{2}, with |ê| = 1, such that lim_{δ→0} e_{m} = ê. Taking δ into zero, (4.28) implies

C_{H(e_{6})}^{H}(ê) = e_{6} · ê.

Hence (4.17) implies ê = H_{p}(e_{6}) and (4.29) follows.

Since H_{p}(e_{6}) · e_{6} ≥ H(e_{6}) − H(0) ≥ 1, we have |H_{p}(e_{6})| ≥ \frac{1}{|e_{6}|}. Moreover, since

H(e_{6}) ≤ 2, H(e) ≤ max_{|p|≤∥∇u∥_{L^{∞}(B_{6})}} H(p),

we conclude that there is C = C(H) > 0 such that |e_{6}| + |e| + |H_{p}(e_{6})| ≤ C, and hence

cos^{}(H_{p}(e_{6}), f) ≥ \frac{ϵ}{|H_{p}(e_{6})||f|} ≥ \frac{ϵ}{C} .  (4.30)

(4.29) and (4.30) imply that if δ = δ(ϵ, H) > 0 is sufficiently small, then y_{m} ∈ {x ∈ R^{2}|f · (y − z_{4}) ≥ 2δ}. From the choice of the step size t, we can conclude that there exists 1 ≤ i ≤ m such that y_{i} ∈ U ∩ B_{2}(z_{4}).

Now we claim that |y_{i} − z_{4}| ≤ 1. For otherwise, we have |y_{i} − y_{0}| ≥ \frac{2}{3} and the above argument again yields cos(^{}(y_{i} − y_{0}, H_{p}(e_{6}))) ≥ 1 − η(δ). But this is also impossible if δ = δ(ϵ) > 0 is chosen to be sufficiently small. Hence the claim 4.3(d) is proven.

Combining (4.10), (4.12), (4.22), and Claim 4.3(d), we obtain

H(e_{6}) − 50|H_{p}(e_{6})|δ ≤ S^{+}(H, u, y_{i}) ≤ H(e) + Cδ

or

H(e) ≥ H(e_{6}) − (C + 50|H_{p}(e_{6})|)δ.

This implies (4.7), provided that δ = δ(ϵ, H) > 0 is chosen to be sufficiently small. Therefore the proof of Lemma 4.4 is complete.

Let α = ^{}(H_{p}(e), f) ∈ [0, π] be the angle between H_{p}(e) and f. Now we have

Lemma 4.4. Under the same notations. For any ϵ > 0, there is a δ = δ(ϵ, H) > 0 such that

|α − \frac{π}{2} | ≤ 2ϵ.  (4.31)

Proof. Without loss of generality, we may assume α ∈ [0, \frac{π}{2} ]. We may assume \frac{π}{2} − α ≥ ϵ.

Let x_{δ} = z_{4} − \frac{2H_{p}(e)}{H_{p}(e)·f} δ be the intersection of L = \{z_{4} + tH_{p}(e) : t ∈ R\} and \{y ∈ R^{2} : (y − z_{4}) · f = −2δ}. Observe that

|x_{δ} − z_{4}| = \frac{2δ}{|f| cos^{}(H_{p}(e), f)} ≤ \frac{2δ}{ϵ sin(ϵ)} ≤ 1

provided that δ = δ(ϵ) > 0 is chosen to be sufficiently small. This implies B_{1}(x_{δ}) ⊂ B_{2}(z_{4}). By A1, we have

u(y)−u(z_{3})−e·(y−z_{3}) ≤ u(y)−u(z_{3})−e_{1}·(y−z_{3})+f·(y−z_{3}) ≤ 4δ, ∀y ∈ U∩B_{1}(x_{δ}), (4.32)

and

u(y) = u(z_{3}) + e · (y − z_{3}), ∀y ∈ ∂U ∩ B_{1}(x_{δ}).  (4.33)

We need Claim 4.4(a).

u(x) ≤ u(z_{3}) + e · (x_{δ} − z_{3}) + C_{H(e)}^{H}(x − x_{δ}), ∀x ∈ ∂(U ∩ B_{1}(x_{δ})).  (4.34)

Proof of Claim 4.4(a): First observe that (4.33) implies

u(x) = u(z_{3}) + e · (x_{δ} − z_{3}) + e · (x − x_{δ}) ≤ u(z_{3}) + e · (x_{δ} − z_{3}) + max_{H(p)=H(e)} p · (x − x_{δ})
= u(z_{3}) + e · (x_{δ} − z_{3}) + C_{H(e)}^{H}(x − x_{δ}), ∀x ∈ ∂U ∩ B_{1}(x_{δ}).  (4.35)

On U ∩ ∂B_{1}(x_{δ}), we have

u(x) ≤ u(z_{3}) + e · (x_{δ} − z_{3}) + e · (x − x_{δ}) + 4δ.  (4.36)

Now we need Claim 4.4(b). For any ϵ > 0, there exists δ = δ(ϵ) > 0 such that

c_{δ} := min_{S∩∂B_{1}(x_{δ})}(C_{H(e)}^{H}(x − x_{δ}) − e · (x − x_{δ})) > 4δ.  (4.37)

Proof of Claim 4.4(b): Suppose (4.37) were false. Then there exist ϵ_{0} > 0 and δ_{i} → 0 such that c_{δ_{i}} ≤ 4δ_{i}. Since e_{6}, e, α, z_{3}, z_{4}, x_{δ} all depend on δ_{i}, we write e_{6}^{i} = e_{6}, e^{i} = e, f^{i} = e_{6}^{i} − e^{i}, α_{i} = ^{}(H_{p}(e^{i}), f^{i}), z_{3}^{i} = z_{3}, z_{4}^{i} = z_{4}, S_{i} = S, x_{i} = x_{δ}. Then, by the above assumption and Lemma 4.3, we have

α_{i} ≤ \frac{π}{2} − ϵ_{0}, \frac{1}{2} ≤ H(e_{6}^{i}) − 2ϵ_{0} ≤ H(e^{i}) ≤ max_{|p|≤∥∇u∥_{L^{∞}(B_{6})}} H(p).

Let y_{i} ∈ S_{i} ∩ ∂B_{1}(x_{i}) such that
c_{δ_{i}} = C_{H(e^{i})}^{H}(y_{i} − x_{i}) − e^{i} · (y_{i} − x_{i}) ≤ 4δ_{i}.

By the Lagrange Multiple Theorem, there exists p_{i} such that

H(p_{i}) = H(e^{i}), \frac{H_{p}(p_{i})}{|H_{p}(p_{i})|} = y_{i} − x_{i}, C_{H(e^{i})}^{H}(y_{i} − x_{i}) = p_{i} · (y_{i} − x_{i}).  (4.38)

After passing to subsequences, we may assume that e_{6}^{i} → e_{6}, e^{i} → e, f^{i} → f, z_{3}^{i} → 0, z_{4}^{i} → z_{4}, S_{i} → {x ∈ R^{2} : x·f = 0}, x_{i} → z_{4}, p_{i} → p, y_{i} → y. It is easy to see that |f| ≥ ϵ_{0}, (H_{p}(e), f) ≤ \frac{π}{2} − ϵ_{0}, |z_{4}| = 3, z_{4} ⊥ f, |y − z_{4}| = 1, y − z_{4} ⊥ f,

\frac{H_{p}(p)}{|H_{p}(p)|} = y − z_{4}, H(p) = H(e), C_{H(e)}^{H}(y − z_{4}) = p · (y − z_{4}) = e · (y − z_{4}).

This implies p = e and y − z_{4} = \frac{H_{p}(e)}{|H_{p}(e)|}. Therefore H_{p}(e) ⊥ f, this gives the desired contradiction. Hence (4.37) is true.

It follows from (4.35), (4.36), (4.37) that for sufficiently small δ > 0, we have

u(y) ≤ u(z_{3}) + e · (x_{δ} − z_{3}) + C_{H(e)}^{H}(y − x_{δ}), ∀y ∈ ∂(U ∩ B_{1}(x_{δ})).  (4.39)

Therefore, theorem 2.1 implies

u(y) ≤ u(z_{3}) + e · (x_{δ} − z_{3}) + C_{H(e)}^{H}(y − x_{δ}), ∀y ∈ U ∩ B_{1}(x_{δ}).  (4.40)

By A2, we have that {x_{δ} + tH_{p}(e)|t ≥ 0} ∩ (U ∩ B_{1}(x_{δ})) ≠ ∅. But on the other hand, (4.17) and (4.40) imply that if y_{0} = x_{δ} + t_{0}H_{p}(e) ∈ U ∩ B_{1}(x_{δ}), then

u(y_{0}) ≤ u(z_{3}) + e · (x_{δ} − z_{3}) + C_{H(e)}^{H}(y_{0} − x_{δ})
≤ u(z_{3}) + e · (x_{δ} − z_{3}) + t_{0}C_{H(e)}^{H}(H_{p}(e))
= u(z_{3}) + e · (x_{δ} − z_{3}) + t_{0}e · H_{p}(e)
= u(z_{3}) + e · (y_{0} − z_{3}).

This is impossible, since y_{0} ∈ U. Therefore (4.31) holds and the proof of Lemma 4.5 is complete.

Now we return to the proof of (4.5) as follows. From (3.2) of Lemma 3.1, Lemma 4.3, and Lemma 4.4, we have

2ϵ ≥ H(e_{6}) − H(e) ≥ H_{p}(e) · f + \frac{1}{2}α_{H}|f|^{2}

≥ −|H_{p}(e)||f| cos^{}(H_{p}(e), f) + \frac{1}{2}α_{H}|f|^{2}

≥ −C sin(2ϵ) + \frac{1}{2}α_{H}|f|^{2}.  (4.41)

Hence we have

|f|^{2} ≤ α_{H}^{−1}C(ϵ + sin(2ϵ)).

provided that δ = δ(ϵ, H) > 0 is chosen to be sufficiently small. This yields (4.5).

Next we want to prove

|e_{0,6r} − e|^{2} ≤ Cϵ.  (4.42)

To prove (4.42), define v_{r}(x) = \frac{u(rx)}{r} : B_{6} → R. Then we have

∥v_{r} − e_{0,6r} · x∥_{L^{∞}(B_{6})} ≤ δ.

Moreover, we can check that Lemma 4.2 also implies that {y ∈ R^{2} : u(y) > u(z_{3}) + e · (y − z_{3})\} ∩ B_{6r} has two connected components that intersect B_{r}. This implies that for z_{3}^{r} = \frac{z_{3}}{r}, {y ∈ R^{2} : v_{r}(y) > v_{r}(z_{3}^{r}) + e · (y − z_{3}^{r})} ∩ B_{6} has two connected components that intersect B_{1}. Therefore the same argument as above yields (4.42), if we can verify

\frac{1}{2} ≤ H(e_{0,6r}) ≤ 4.  (4.43)

To see (4.43), first observe that Lemma 4.3 implies that for δ > 0 sufficiently small, we have

H(e) ≥ H(e_{6}) − C_{H}δ ≥ 1 − C_{H}δ ≥ \frac{3}{4}.  (4.44)
For any x ∈ ∂B_{2r}(z_{3}) ⊂ B_{6r}, we have
u(x) ≤ u(z_{3}) + e_{0,6r} · (x − z_{3}) + 2δr
≤ u(z_{3}) + C_{H(e_{0,6r})}^{H}(x − z_{3}) + 2δr
≤ u(z_{3}) + C_{H(e_{0,6r})+Cδ}^{H}(x − z_{3}).  (4.45)

Here we have used the following claim, whose proof is similar to that of Claim 4.4(a), Claim 4.1(a). There exists a C > 0 such that for any α > 0, we have

C_{H(e_{0,6r})}^{H}(x) + α|x| ≤ C_{H(e_{0,6r})+Cα}^{H}(x), ∀x ∈ B_{6r}.  (4.46)

It follows from (4.45) that

S_{2r}^{+}(H, u, z_{3}) ≤ H(e_{0,6r}) + Cδ.

This implies

H(e) = S^{+}(H, u, z_{3}) ≤ S_{2r}^{+}(H, u, z_{3}) ≤ H(e_{0,6r}) + Cδ.  (4.47)

(4.47) and (4.44) imply that for δ > 0 sufficiently small, we have

H(e_{0,6r}) ≥ H(e) − Cδ ≥ \frac{1}{2}.  (4.48)

On the other hand, (4.1) implies

u(x) ≤ e_{6} · x + δ ≤ C_{H(e_{6})}^{H}(x) + δ ≤ C_{H(e_{6})+Cδ}^{H}(x), ∀x ∈ ∂B_{3}.

This and (4.46) imply that for δ > 0 sufficiently small, we have

H(e_{0,6r}) = S^{+}(H, u, 0) ≤ H(e_{6}) + Cδ = 2 + Cδ ≤ 4.  (4.49)

It is clear that (4.48) and (4.49) imply (4.43). Therefore (4.42) holds. Note that (4.3) follows from (4.5) and (4.42). This proves Proposition 4.1 in case A.

Next we consider the second case.

Case B. u(x) ≡ e·x near 0 for some e ∈ R^{2}: Denote U as the interior of {x ∈ R^{2} | u(x) = e · x\}. If d(0, ∂U) > \frac{1}{2}, then we have e_{6} is arbitrarily close to e_{0,6r} and hence (4.3) holds trivially. If d(0, ∂U) ≤ \frac{1}{2}, then we let x_{0} ∈ ∂U be such that |x_{0}| = d(0, ∂U). Note that

∥u(y) − u(x_{0}) − e_{6} · (y − x_{0})∥_{L^{∞}(B_{5}(x_{0}))} ≤ 2δ, 1 ≤ H(e_{6}) ≤ 2,

and

lim_{r↓0} \frac{∥u(y) − u(x_{0}) − e_{x_{0},6r}(y − x_{0})∥_{L^{∞}(B_{6r}(x_{0}))}}{r} = 0.

Hence, by applying Case A in B_{5}(x_{0}), we get

|e_{x_{0},6r} − e_{6}|^{2} ≤ Cϵ.

On the other hand, it is easy to see that for r > 0 sufficiently small, e = e_{0,6r} = e_{x_{0},6r}. Hence we have

|e_{0,6r} − e_{6}|^{2} ≤ Cϵ

provided that δ = δ(ϵ, H) > 0 is chosen to be sufficiently small. This implies (4.3). Therefore the proof of Proposition 4.1 is complete.

It follows from Proposition 4.1 that we have

Corollary 4.5. Suppose that u ∈ C^{0,1}(B_{1}), B_{1} ⊂ R^{2}, is a viscosity solution of the Aronsson equation (1.7). Then for any ϵ > 0, there exists δ = δ(ϵ, H) > 0 such that if

∥u(x) − u(0) − e_{1} · x∥_{L^{∞}(B_{1})} ≤ δ, 1 ≤ H(e_{1}) ≤ 2.  (4.50)

Then u is differentiable at 0 and

|∇u(0) − e_{1}|^{2} ≤ Cϵ,  (4.51)

for some C = C(H) > 0.

Proof. By theorem A, we have that there exists r_{0} = r_{0}(δ) > 0 such that for any 0 < r ≤ r_{0} there exists e_{0,r} ∈ R^{2} such that

∥u(x) − u(0) − e_{0,r} · x∥_{L^{∞}(B_{r})} ≤ δr.  (4.52)

Therefore we can apply Proposition 4.1 to conclude that there exist δ = δ(ϵ, H) > 0 and C = C(H) > 0 such that

|e_{1} − e_{0,r}|^{2} ≤ Cϵ, ∀0 < r ≤ r_{0}.  (4.53)

This implies

|e_{0,r} − e_{0,s}|^{2} ≤ 2Cϵ, ∀0 < r, s ≤ r_{0}.

Since ϵ > 0 is arbitrary, this implies {e_{0,r}}_{r>0} is a Cauchy sequence. Hence lim_{r↓0} e_{0,r} exists, u differentiable at 0, ∇u(0) = lim_{r↓0} e_{0,r}, and (4.51) holds. The proof of Corollary 4.5 is complete.

§5. Proof of Theorem B, C, D, and Remark E

In this section, we apply Proposition 4.1 to give proofs of theorem B, C and D. We will see that the proof for the second part of theorem B follows from Proposition 4.1 and theorem C.

Proof of Theorem B.

First Theorem A implies that for any x ∈ Ω, there exists e_{x,r} ∈ R^{2}, with H(e_{x,r}) = S^{+}(H, u, x), such that for any small r > 0

max_{y∈B_{r}(x)} |u(y) − u(x) − e_{x,r} · (y − x)| ≤ σ(r)r, lim_{r→0} σ(r) = 0.

If e_{x,r} = 0, then by theorem 2.1, we have

|u(y) − u(x)| ≤ σ(r)|y − x|, ∀y ∈ B_{r}(x).

This implies that u is differentiable at x and ∇u(x) = 0. If e_{x,r} ≠ 0, then we have, by Lemma 3.1, H(e_{x,r}) > 0. Let v_{x,r}(y) = \frac{u(x+ry)−u(x)}{r} and H_{x,r}(p) = \frac{H(p)}{|H(e_{x,r})|}, then we can check that H_{x,r} satisfies the conditions of Proposition 4.1, v_{x,r} ∈ C^{0,1}(B_{1}) satisfies the Aronsson equation, with H replaced by H_{x,r}. Moreover, we have

max_{y∈B_{1}} |v_{x,r} − e_{x,r} · y| ≤ σ(r), H_{x,r}(e_{x,r}) = 1.

Therefore, by Corollary 4.5, there exists r_{0} > 0 such that v_{x,r_{0}} is differentiable at 0. This implies that u is differentiable at any x ∈ Ω.

The continuity of ∇u follows from Theorem C.

Proof of Theorem C.

First note that ∥H(∇u)∥_{L^{∞}(B_{1})} ≤ 1 implies that there is C = C(H) > 0 such that

∥∇u∥_{L^{∞}(B_{1})} ≤ C.  (5.1)

Suppose that the conclusion of theorem C were false. Then there exist ϵ_{0} > 0, a family {u_{k}} ⊂ C^{0}(B_{1}) of viscosity solutions of the Aronsson equation (1.7) and x_{k} → 0 such that

∥∇u_{k}∥_{L^{∞}(B_{1})} ≤ C, |∇u_{k}(x_{k}) − ∇u_{k}(0)| ≥ ϵ_{0}.  (5.2)

We may assume that there is a Lipschitz continuous u_{∞} on B_{1} such that u_{k} → u_{∞} uniformly in B_{1}. Hence u_{∞} is a viscosity solution of (1.7), and

∥H(∇u_{∞})∥_{L^{∞}(B_{1})} ≤ 1, ∥∇u_{∞}∥_{L^{∞}(B_{1})} ≤ C.  (5.3)

By theorem A, we have that for any δ > 0, there exists r_{0} > 0 and e_{0} ∈ R^{2} such that we have

∥u_{∞}(x) − u_{∞}(0) − e_{0} · x∥_{L^{∞}(B_{r_{0}} )} ≤ \frac{δ}{2}r_{0}.  (5.4)

Therefore there exists a sufficiently large k_{0} > 0 such that

∥u_{k}(x) − u_{k}(0) − e_{0} · x∥_{L^{∞}(B_{r_{0}} )} ≤ δr_{0}, ∀k ≥ k_{0}.  (5.5)

This implies that for k ≥ k_{0}, we have

|u_{k}(x) − u_{k}(y) − e_{0} · (y − x)| ≤ 2δr_{0}, ∀x, y ∈ B_{r_{0}}.  (5.6)

In particular, for k ≥ k_{0}, we have

∥u_{k}(x + x_{k}) − u_{k}(x_{k}) − e_{0} · x∥_{L^{∞}(B_{\frac{r_{0}}{2}} )} ≤ 2δr_{0}.  (5.7)

If e_{0} = 0, then (5.5), (5.7), and (2.9) imply

H(∇u_{k}(0) = S^{+}(H, u_{k}, 0) ≤ max_{|p|=δ} H(p), H(∇u_{k}(x_{k})) = S^{+}(H, u_{k}, x_{k}) ≤ max_{|p|=4δ} H(p).

(5.8)

Therefore we know that there exists η(δ) > 0, with lim_{δ→0} η(δ) = 0, such that for k sufficiently large

|∇u_{k}|(x_{k}) + |∇u_{k}|(0) ≤ η(δ).  (5.9)

This contradicts with (5.2), if we choose δ > 0 to be sufficiently small.

If e_{0} ≠ 0, then (5,5), (5.7), and Corollary 4.5 imply that for any 0 < ϵ ≤ \frac{ϵ_{0}}{4} , we can find a sufficiently small δ_{0} = δ_{0}(ϵ, H(e_{0})) > 0 so that for k sufficiently large, we have

|∇u_{k}(x_{k}) − e_{0}| ≤ ϵ, |∇u_{k}(0) − e_{0}| ≤ ϵ.

In particular, |∇u_{k}(x_{k}) − ∇u(0)| ≤ 2ϵ < ϵ_{0}. This again contradicts with (5.2). Hence the proof of theorem C is complete.

Proof of Theorem D.

First we claim that the linear growth condition at infinity implies

∥H(∇u)∥_{L^{∞}(B_{R})} = sup_{x∈B_{R}} S^{+}(H, u, x) ≤ C_{1}.  (5.10)

In fact, for any x ∈ B_{R} and y ∈ R^{2} with |y − x| = R, we have

u(y) − u(x) ≤ 2C(1 + R) ≤ 4CR ≤ C_{C_{1}}^{H}(y − x)

for some C_{1} > 0 depends only on H, where we have used the coercivity condition H3. Hence, S^{+}(H, u, x) ≤ S_{R}^{+}(H, u, x) ≤ C_{1}.

For R ↑ ∞, let w_{R}(x) = \frac{u(Rx)}{R} : B_{1} → R. Then w_{R} satisfies the condition of Theorem C. Hence we have, for any fixed x_{0} ∈ R^{2}

|∇u(x_{0}) − ∇u(0)| = |∇w_{R}(R^{−1}x_{0})) − ∇w_{R}(0)| ≤ Cρ(|x_{0}|R^{−1}) → 0,  (5.11)

as R → ∞. This implies ∇u(x) ≡ ∇u(0) for all x ∈ R^{2} and u is linear. The proof is now complete.

Proof of Remark E.

We sketch the argument that theorem B and C are true under the conditions (i), (ii), and (iii). First note that H satisfies all the conditions of theorem 2.1, hence we have that u is locally Lipschitz continuous. For simplicity, we may assume that u is Lipschitz continuous on Ω so that

m = ∥∇u∥_{L^{∞}(Ω)} < ∞.

Let ϕ ∈ C^{2}(R^{2}) be a nonnegative convex function such that

ϕ(p) = 0 for |p| ≤ m, (ϕ_{p_{i}p_{j}}) ≥ I_{2} for |p| ≥ m + 1.  (5.12)

Let η ∈ C^{2}(R^{2}) be a nonnegative function such that

η(p) = 1 for |p| ≤ m + 1, η(p) = 0 for |p| ≥ m + 2.  (5.13)

For k ≥ 1, define H_{k}(p) = H(p)η(p) + kϕ(p) for p ∈ R^{2}. Since

H_{k}(p) = H(p) for |p| ≤ m,

u is also a viscosity solution of (1.7) with H replaced by H_{k}. Moreover, one can check that if k is sufficiently large, then H_{k} ∈ C^{2}(R^{2}) is uniformly convex. Therefore theorem B and C hold.

REFERENCES

[A1] G. Aronsson, Minimization problems for the functional sup xF (x, f(x), f(x)). Ark. Mat. 6 (1965), 33–53.

[A2] G. Aronsson, Minimization problems for the functional sup xF (x, f(x), f(x)). II. Ark. Mat. 6 (1966), 409–431.

[A3] G. Aronsson, Minimization problems for the functional sup xF (x, f(x), f(x)). III. Ark. Mat. 7 (1969), 509–512.

[A4] G. Aronsson, Extension of functions satisfying Lipschitz conditions. Arkiv für Mate. 6, 551–561 (1967)

[A5] G. Aronsson, On the partial differential equation u_{x}^{2}u_{xx} +2u_{x}u_{y}u_{xy} +u_{y}^{2}u_{yy} = 0. Ark. Mat. 7 (1968), 395–425.

[A6] G. Aronsson, On certain singular solutions of the partial differential equation u_{x}^{2}u_{xx} + 2u_{x}u_{y}u_{xy} + u_{y}^{2}u_{yy} = 0. Manuscripta Math. 47 (1984), no. 1-3, 133–151.

[ACJ] G. Aronsson, M. Crandall, P. Juutinen, A tour of the theory of absolutely minimizing functions. Bull. Amer. Math. Soc. (N.S.) 41 (2004), no. 4, 439–505 (electronic).

[B] N. Barron, Viscosity solutions and analysis in L^{∞}. Nonlinear analysis, differential equations and control (Montreal, QC, 1998), 1–60, NATO Sci. Ser. C Math. Phys. Sci., 528, Kluwer Acad. Publ., Dordrecht, 1999.

[BJ] N. Barron, R. Jensen, Minimizing the L^{∞}-norm of the gradient with an energy constraint. To appear.

[BJW1] N. Barron, R. Jensen, C. Y. Wang, Lower semicontinuity of L^{∞} functionals. Ann. Inst. H. Poincar Anal. Non Linaire 18 (2001), no. 4, 495–517.

[BJW2] N. Barron, R. Jensen, C. Y. Wang, The Euler equation and absolute minimizers of L^{∞} functionals. Arch. Ration. Mech. Anal. 157 (2001), no. 4, 255–283.

[C1] M. Crandall, An efficient derivation of the Aronsson equation. Arch. Ration. Mech. Anal. 167 (2003), no. 4, 271–279.

[C2] M. Crandall, The infinity-Laplace equation and the elements of calculus of variations in L-infinity. CIME lecture notes (to appear).

[C3] M. Crandall, The continuous gradient flow for the infinity-Laplace equation. Private communication.

[CE] M. Crandall, L. C. Evans, A remark on infinity harmonic functions. Proceedings of the USA-Chile Workshop on Nonlinear Analysis (Via del Mar-Valparaiso, 2000), 123–129 (electronic), Electron. J. Differ. Equ. Conf., 6.

[CEG] M. Crandall, C. Evans, R. Gariepy, Optimal Lipschitz extensions and the infinity Laplacian. Calc. Var. Partial Differential Equations 13 (2001), no. 2, 123–139.

[CIL] M. Crandall, H. Ishii, P. L. Lions, User’s guide to viscosity solutions of second order partial differential equations. Bull. Amer. Math. Soc. (N.S.) 27 (1992), no. 1, 1–67.

[GWY] R. Gariepy, C. Y. Wang, Y. Yu, Generalized cone comparison principle for viscosity solutions of the Aronsson equation and absolute minimizers. Preprint (2005).

[J] R. Jensen, Uniqueness of Lipschitz extensions: minimizing the sup norm of the gradient. Arch. Rational Mech. Anal. 123 (1993), no. 1, 51–74.

[M] C. B. Morrey, Quasi-convexity and the lower semicontinuity of multiple integrals. Pacific J. Math. 2, (1952). 25–53.

[S] O. Savin, C^{1}-regularity for infinity harmonic functions in dimensions two. Arch. Ration. Mech. Anal (to appear).

[Y] Y. Yu, L^{∞}-variational problems and the Aronsson equations. Preprint, 2004.