A PDE Perspective of The Normalized Infinity Laplacian

Transcription

1 A PDE Perspective of The ormalized Infinity Laplacian Guozhen Lu & Peiyong Wang Department of Mathematics Wayne State University 656 W.Kirby, 1150 FAB Detroit, MI & Abstract The inhomogeneous normalized infinity Laplace equation was derived from the tug-of-war game in [PSSW] with the positive right-hand-side as a running payoff. The existence, uniqueness and comparison with polar quadratic functions were proved in [PSSW] by the game theory. In this paper, the normalized infinity Laplacian, formally written as u = u 2 n i,j=1 x i u xj u 2 x i x j u, is defined in a canonical way with the second derivatives in the local maximum and minimum directions, and understood analytically by a dichotomy. A comparison with polar quadratic polynomials property, the counterpart of the comparison with cones property, is proved to characterize the viscosity solutions of the inhomogeneous normalized infinity Laplace equation. We also prove that there is exactly one viscosity solution of the boundary value problem for the infinity Laplace equation u = f with positivef in a bounded open subset of R n. The stability of the inhomogeneous infinity Laplace equation u = f with strictly positive f and of the homogeneous equation u = 0 by small perturbation of the right-hand-side and the boundary data is established in the last part of the work. Our PDE method approach is quite different from those in [PSSW]. The stability result in this paper appears to be new. Introduction The homogeneous infinity Laplace equation n u := xi u xj u x 2 i x j u = 0, i,j=1 G. Lu is partially supported by an SF Grant. 1

2 a quasi-linear degenerate elliptic partial differential equation, has received extensive study since 1960 s. To help the reader to trace the development of the theory, the authors would like to just list a few references such as [A1]-[A3], [J], [LW], [Ju], [CE], [CEG], [BB], [C], [BJW], [ACJ], [CW], [EY] and [CGW]. The viscosity solutions of the infinity Laplace equation u = 0 in the sense defined by Crandall, Evans and Lions have been adopted due to the absence of classical solutions in general settings. The well-posedness, initially proposed by Hadamard for any partial differential equation, of the homogeneous infinity Laplace equation in bounded domains in R n or other settings is well developed. See, for example, the manuscripts [J] and [ACJ]. The existence (see [ACJ]) and uniqueness (see [CGW]) of a viscosity solution with prescribed growth rate of the homogeneous infinity Laplace equation in an unbounded domain are also known, at least to a large extent. Study of the singular inhomogeneous normalized infinity Laplace equation u = u 2 n i,j=1 x i u xj u x 2 i x j u = f in this work was inspired by recent works, [LW], [KS] and [PSSW], on the inhomogeneous infinity Laplace equation by the theory of partial differential equations, and on its normalized counterpart by the game theory. In [LW], the well-posedness problem of the inhomogeneous infinity Laplace equation u = f was investigated. A counterexample shows that the well-posedness fails if f changes its sign. On the other hand, under the assumption inf Ω f > 0, existence and uniqueness of a viscosity solution of the Dirichlet problem for the inhomogeneous infinity Laplace equation u = f in a bounded domain were proved. In addition, a family of special solutions of u = 1 were found to fully characterize the viscosity solutions of the equation u = f via a comparison principle with those cone-like special solutions. Stability of the inhomogeneous infinity Laplace equation u = f with inf Ω f > 0 and of the homogeneous infinity Laplace equation u = 0 has been justified with the existence, uniqueness and comparison with the cone-like functions results being employed. Moreover, the restriction inf Ω f > 0 can be relaxed to f > 0 in Ω in most results in the work except in the stability problem of u = f. In the manuscripts [KS] by Kohn and Serfaty and [PSSW] by Peres, Schramm, Scheffield and Wilson, they studied the differential games and their connection to normalized inhomogeneous infinity Laplace equation. In particular, in the tug-of-war game, the underlying equation that the value u of the game verifies is the discrete infinity Laplace equation ( sup u(y) + inf u(y)) 2u(x) = 2f(x), (x,y) E (x,y) E where E is the transition graph and f is the payoff function. Let E ε be the transition graph that admits all pairs (x, y) with d(x, y) < ε for any ε > 0. The limiting value u = lim ε 0 u ε if existing verifies the normalized infinity Laplace equation u = 2f. This continuum value u is proved to be unique as long as the payoff function f stays strictly positive. A counter-example was provided to show this assumption is necessary. We also refer the reader to the recent works of Barron, Evans and Jensen [BEJ], and of Evans [E], in which nice interpretations of relevance of infinity Laplace equation to two-person game theory with random order of play, rapid switching of states in control problems, etc. are given. Motivated by the method we used for the non-normalized inhomogeneous infinity Laplacian u in [LW], we will revisit the normalized infinity Laplacian u studied 2

3 earlier in [PSSW] using game theory. Our approach is based on the PDE method and is thus quite different from those in [PSSW]. We first give a definition of the normalized infinity Laplacian u as a possibly multi-valued function by applying the Hessian matrix to the maximum and minimum directions. This definition is consistent with the one used in [PSSW] and other places in the literature. The idea using the maximum and minimum directions in the definition appears to be new. The dichotomy that follows inevitably makes the proofs of uniqueness and existence considerably more difficult. However, the comparison with cone-like functions property which characterizes the viscosity solutions of the normalized equation is neater than that for the non-normalized equation considered in [LW] in the sense that we do not require here the assumption Du > 0 for the normalized equation. We also improve the approach given in [LW] so that we can relax the restriction inf Ω f > 0 on the right-hand-side of the equation u = f to f > 0 in Ω in most of the theory except the stability for u = f. This improvement also applies to the results in [LW]. Our current work also helps to build a further connection between the partial differential equation theory and the differential game theory about the infinity Laplacian. Such a connection has already been recently explored in [BEJ] and [E]. This paper is organized in the following order. In Section 1, we introduce the normalized infinity Laplacian u. We prove a comparison with polar quadratic functions for a viscosity solution of the inhomogeneous infinity Laplace equation in Section 2. In Section 3, we prove a strict comparison principle which is by itself a very useful tool in our approach and which immediately implies the uniqueness theorem. Section 4 is devoted to the proof of the existence theorem. In the last section, we establish the uniform convergence of the viscosity solutions of the uniformly perturbed inhomogeneous infinity Laplace equation to the viscosity solution of the homogeneous infinity Laplace equation. 1 Definition of u and The Main Results We adopt the standard notations in analysis and the set theory. For example, Ω and Ω mean the boundary and closure of a set Ω respectively, while xi u denotes the partial derivative of u with respect to x i. V Ω means V is compactly contained in Ω, i. e. V is a subset of Ω whose closure is also contained in Ω. Also, for two positive numbers λ and µ, λ << µ means λ is bounded above by a sufficiently small multiple of µ. (ε) denotes quantities whose quotients by ε approach 0 as ε does, while O(ε) denotes quantities that are comparable to ε. For two vectors x = (x 1, x 2,..., x n ) and y = (y 1, y 2,..., y n ) R n, < x, y >= n i=1 x iy i is the inner product of x and y, while x y is the tensor product yx t, or [y i x j ] n n in the matrix form, of the vectors x and y. For x R n, x denotes the Euclidean norm < x, x > 1 2 of x and ˆx = x denotes the normalized vector for x 0. B x r (x) denotes the ball with radius r and centered at x. S 1 denotes the unit sphere of the Euclidean space R n. Suppose S is a subset of R n. A function f : S R is said to be Lipschitz contin- 3

4 uous on S if there is a constant L such that f(x) f(y) L x y, for any x and y in S. The least of such constants is denoted by L f (S). If S is an open subset Ω of R n, we use the symbol Lip(Ω) to denote the set of all Lipschitz continuous functions on Ω. If instead S = Ω is the boundary of an open subset Ω of R n, we use the symbol Lip (Ω) to denote the set of all Lipschitz continuous functions on Ω. Ω always denotes an open subset of R n and is usually bounded. C(Ω) denotes the set of continuous functions defined on Ω and C( Ω) denotes the set of continuous functions on Ω. C 2 (Ω) denotes the set of functions which are continuously twice differentiable on Ω. A smooth function usually means a C 2 function in this paper. If f C(Ω), then f L (Ω) := sup x Ω f(x) denotes the L -norm of f on Ω. S n n denotes the set of all n n symmetric matrices with real entries. We use I to denote the identity matrix in S n n. For an element S S n n, S denotes its operator <Sx,x> norm, namely S = sup x R n \{0}. And we use λ x 2 1 (S), λ 2 (S),..., λ n (S) to denote the eigenvalues of an n n symmetric matrix S. u x0 ϕ means u ϕ has a local maximum at x 0. In this case, we say ϕ touches u by above at x 0. Almost always in this paper, u x0 ϕ is understood as u(x) ϕ(x) for all x Ω in interest and u(x 0 ) = ϕ(x 0 ), as subtracting a constant from ϕ does not cause any problem in the standard viscosity solution argument applied in the paper. On the other hand, if ϕ x0 u, we say ϕ touches u by below at x 0. For u C(Ω), x 0 Ω, and r > 0 with B r (x 0 ) Ω, we define g(r) = max x x0 =r u(x) and h(r) = min x x0 =r u(x). In addition, x + r denotes any point with x + r x 0 = r such that u(x + r ) = g(r), while x r denotes any point with x r x 0 = r such that u(x r ) = h(r). If x 0 Ω and u C(Ω) such that u is twice differentiable at x 0, we define the set of maximum directions of u at x 0 to be the set E + (x 0 ) = {e S 1 : e = lim k x + r k x 0 r k for some sequence r k 0} and the set of minimum directions of u at x 0 to be the set E (x 0 ) = {e S 1 : e = lim k x r k x 0 r k for some sequence r k 0}. Definition 1.1 If u C(Ω) is twice differentiable at x 0, we define the upper infinity Laplacian of u at x 0 to be the set + u(x 0 ) = {< D 2 u(x 0 )e, e >: e E + (x 0 )}. Similarly, the lower infinity Laplacian of u at x 0 is defined to be the set u(x 0 ) = {< D 2 u(x 0 )e, e >: e E (x 0 )}. Proposition 1.2 Suppose u C(Ω) is twice differentiable at x 0. (a) If u(x 0 ) 0, then + u(x 0 ) = u(x 0 ) = { u(x 0 ) 2 < D 2 u(x 0 ) u(x 0 ), u(x 0 ) >}. (b) If u(x 0 ) = 0, then both + u(x 0 ) and u(x 0 ) contain a single element. 4

5 Proof. (a) There exists a positive-valued function ρ with ρ(r) 0 as r 0, defined for all small positive numbers r, such that for all x with x x 0 = r. Take x + r = x 0 + r u(x 0 ). Then u(x) u(x 0 ) u(x 0 ) (x x 0 ) ρ(r)r (1.1) u(x 0 ) + u(x 0 ) (x + r x 0 ) ρ(r)r u(x + r ) u(x 0 ) + u(x 0 ) ( x + r x 0 ) + ρ(r)r. The second inequality is due to the choice of x + r. So u(x 0 ) (x + r x + r ) 2ρ(r)r. On the other hand, the chain of inequalities implies u(x 0 ) (x + r x + r ) 2ρ(r)r. So u(x 0 ) + u(x 0 ) ( x + r x 0 ) ρ(r)r u( x + r ) u(x + r ) u(x 0 ) + u(x 0 ) (x + r x 0 ) + ρ(r)r u(x 0 ) (x + r x + r ) 2ρ(r)r. (1.2) If we denote the angle between x + r x 0 and x + r x 0 by α(r), then the above estimate (1.2) gives u(x 0 ) (1 cos α(r)) 2ρ(r), (1.3) or equivalently sin( 1 2 α(r)) ( ρ(r) u(x 0 ) ) 1 2. (1.4) So α(r) 0 as r 0. So for any e E + (x 0 ), e = u(x 0 ) holds. Similarly, E (x 0 ) = { u(x 0 )}. Therefore + u(x 0 ) = u(x 0 ) = { u(x 0 ) 2 < D 2 u(x 0 ) u(x 0 ), u(x 0 ) >}. (b) If u(x 0 ) = 0, we denote D 2 u(x 0 ) by S. Clearly, u(x) = u(x 0 ) < S(x x 0), x x 0 > + ( x x 0 2 ). (1.5) Let λ max (S) and λ min (S) be the maximum and minimum eigenvalues of S. It is not difficult to provide a proof similar to that of (a) to show that E + (x 0 ) and E (x 0 ) are precisely the sets of normalized eigenvectors corresponding to λ max (S) and λ min (S) respectively. So + u(x 0 ) = {λ max (S)} and u(x 0 ) = {λ min (S)}. From now on, we do not distinguish + u(x 0 ) or u(x 0 ) from its single element. It is implied by the above proof that + u(x 0 ) u(x 0 ). 5

6 Definition 1.3 Suppose u C(Ω) is twice differentiable at x 0. We define the normalized infinity Laplacian of u at x 0 to be the closed interval u(x 0 ) = [ u(x 0 ), + u(x 0 )], (1.6) and if u(x 0 ) contains only one real number, we do not distinguish u(x 0 ) from its single element. Remark 1.4 If u is twice differentiable at x 0 with u(x 0 ) 0, clearly u(x 0 ) = u(x 0 ) 2 < D 2 u(x 0 ) u(x 0 ), u(x 0 ) > (1.7) which is the usual normalized infinity Laplacian. If u C 2 (Ω), then u(x) is defined at every point in Ω. We define viscosity solutions of the normalized infinity Laplacian as follows. Definition 1.5 A continuous function u defined in an open subset Ω of R n is called a viscosity sub-solution, or simply a sub-solution, of the partial differential equation u(x) = f(x) in Ω, if + ϕ(x 0 ) f(x 0 ), (1.8) whenever u x0 ϕ for any x 0 Ω and any C 2 test function ϕ. Similarly, u is called a viscosity super-solution, or simply a super-solution, of the partial differential equation u(x) = f(x) in Ω, if ϕ(x 0 ) f(x 0 ), (1.9) whenever ϕ x0 u for any x 0 Ω and any C 2 test function ϕ. A viscosity solution, or simply a solution, of the partial differential equation u(x) = f(x) in Ω is both a viscosity sub-solution and super-solution of the equation. According to this definition, for a C 2 function u, u = f simply means f(x) u(x) for every x in consideration. We will need the concepts of superjets and subjets in our approach. Definition 1.6 Suppose u C(Ω). The second-order superjet of u at x 0 is defined to be the set J 2,+ Ω u(x 0) = {(Dϕ(x 0 ), D 2 ϕ(x 0 )) : ϕ is C 2 and u x0 ϕ}, whose closure is defined to be J 2,+ Ω u(x 0) ={(p, X) R n S n n : (x n, p n, X n ) Ω R n S n n such that (p n, X n ) J 2,+ Ω u(x n) and (x n, u(x n ), p n, X n ) (x 0, u(x 0 ), p, X)}. The second-order subjet of u at x 0 is defined to be the set J 2, Ω u(x 0) = {(Dϕ(x 0 ), D 2 ϕ(x 0 )) : ϕ is C 2 and ϕ x0 u}, whose closure is defined to be J 2, Ω u(x 0) ={(p, X) R n S n n : (x n, p n, X n ) Ω R n S n n such that (p n, X n ) J 2, Ω u(x n) and (x n, u(x n ), p n, X n ) (x 0, u(x 0 ), p, X)}. 6

7 The following listed are the main theorems of this paper. The first is a characteristic property of the viscosity solutions of the inhomogeneous infinity Laplace equation. Theorem 1.7 Assume u C(Ω) and f C(Ω), where Ω is an open subset of R n. Then u = f(x) in the viscosity sense in Ω if and only if u enjoys comparison with polar quadratic polynomials in Ω with respect to f. The concept of the comparison with polar quadratic polynomials property will be made clear in Section 2. The second theorem is about the Dirichlet problem, or equivalently the boundary value problem, for the inhomogeneous infinity Laplace equation. Theorem 1.8 Suppose Ω is a bounded open subset of R n, f C(Ω) with f > 0 and g C( Ω). Then there exists a unique u C( Ω) such that u = g on Ω and in Ω in the viscosity sense. u(x) = f(x) The last is a connection between the homogeneous and inhomogeneous infinity Laplace equations. Theorem 1.9 Let Ω be a bounded open subset of R n. Suppose {g k } is a sequence of functions in Lip (Ω) which converges to g Lip (Ω) uniformly on Ω, and {f k } is a sequence of continuous functions on Ω which converges uniformly to 0 in Ω. If, for each k, u k C( Ω) is a viscosity solution of the Dirichlet problem { u k = f k in Ω (1.10) u k = g k on Ω and u C( Ω) is the unique viscosity solution of the Dirichlet problem { u = 0 in Ω u = g on Ω, (1.11) Then u k converges to u uniformly on Ω, i. e. as k. sup u k u 0 (1.12) Ω Because of the singularity of the normalized infinity Laplacian u when u = 0 in the viscosity sense, we need to prove the following lemma which does not automatically follow from the standard viscosity solution theory. 7

8 Lemma 1.10 Assume Ω is an open subset of R n and f C(Ω). Λ is an index set. (a)suppose u(x) = sup λ Λ u λ (x) <, x Ω, where u λ f in Ω in the viscosity sense for every λ Λ. Then u f in Ω in the viscosity sense. (b)suppose u(x) = inf λ Λ u λ (x) >, x Ω, where u λ f in Ω in the viscosity sense for every λ Λ. Then u f in Ω in the viscosity sense. Proof. Because the proof of (b) is similar to that of (a), we only present the proof of (a). Suppose u f in the viscosity sense is not true in Ω. Then there exists a function ϕ C 2 (Ω) and a point x 0 Ω such that u x0 ϕ and + ϕ(x 0 ) < f(x 0 ). If we replace ϕ by ϕ δ defined by ϕ δ (x) = ϕ(x) + δ x x 0 2, (1.13) then u x0 ϕ δ and + ϕ δ (x 0 ) = + ϕ(x 0 ) + O(δ) < f(x 0 ) (1.14) if δ is taken small enough. So we may simply assume in addition that the original test function ϕ satisfies ϕ(x) u(x) + δ x x 0 2, (1.15) for some δ > 0. We claim that + ϕ(x) < f(x) in an open neighborhood B r (x 0 ) of x 0. In fact, we prove the claim via a dichotomy. If ϕ(x 0 ) 0, then ϕ(x) 0 in a neighborhood B R (x 0 ) of x 0. The continuity of D 2 ϕ and of f implies that in a neighborhood B r (x 0 ) B R (x 0 ) of x 0, + ϕ(x) =< D 2 ϕ(x) Dϕ(x), Dϕ(x) >< f(x). (1.16) If ϕ(x 0 ) = 0, then λ max (D 2 ϕ(x 0 )) = + ϕ(x 0 ) < f(x 0 ). B r (x 0 ) of x 0, λ max (D 2 ϕ(x)) < f(x). As a result, So in a neighborhood + ϕ(x) λ max (D 2 ϕ(x)) < f(x) (1.17) in this neighborhood B r (x 0 ) of x 0. The claim is proved. For any ε with 0 < ε < δr 2, λ Λ such that u λ (x 0 ) > u(x 0 ) ε. Let ˆϕ(x) = ϕ(x) ε. Then ˆϕ(x 0 ) = u(x 0 ) ε < u λ (x 0 ) and, on B r (x 0 ), ˆϕ(x) u(x) + δr 2 ε > u(x) u λ (x). (1.18) So there exists x B r (x 0 ) such that u λ x ˆϕ. As u λ f in Ω in the viscosity sense, + ˆϕ(x ) f(x ) (1.19) holds, which is in contradiction with the claim we just have derived, in B r (x 0 ). + ˆϕ = + ϕ < f (1.20) 8

9 2 Comparison with Polar Quadratic Polynomials We define the domain D(x 0, b) of differentiability of a polar quadratic polynomial ψ(x) = a x x 0 2 +b x x 0 +d, where a, b, d R, and x 0 R n, as D(x 0, b) = { R n \{x 0 }, if b 0 R n, if b = 0. (2.1) and If b = 0, it is clear that ψ(x) = 2a, for x R n. If b 0, for any x D(x 0, b), D 2 ψ(x) = 2aI + Dψ(x) = (2a x x 0 + b) (x x0 ) (2.2) b x x 0 {I (x x 0 ) (x x 0 )}. (2.3) If 2a x x 0 + b 0, then Dψ(x) 0 and ψ(x) = 2a. If 2a x x 0 + b = 0, then D 2 ψ(x) = 2a(x x 0 ) (x x 0 ). The eigenvalues of D 2 ψ are λ 1 = 2a and λ 2 =... = λ n = 0. In this case, a 0 as b 0, and ψ(x) = { [2a, 0], if a < 0 ( i. e. b > 0) [0, 2a], if a > 0 ( i. e. b < 0). (2.4) In particular, Lemma 2.1 ψ(x) = a x x b x x 0 + d is a viscosity solution of ψ = 2a in D(x 0, b). Proof. The fact that a classical solution is a viscosity solution follows easily from the definition of a viscosity solution. One also finds that the constant function x 2a is the only continuous value of ψ in D(x 0, b). For a continuous function u defined in Ω and any open set V Ω, we use the notation u MaxP (V ) to denote the fact that u verifies the weak maximum principle sup u = max u (2.5) V V on V. Similarly, u MinP (V ) means u verifies the weak minimum principle inf V u = min V u (2.6) on V. We now prove a one-sided comparison principle for viscosity sub-solutions of the inhomogeneous normalized infinity Laplace equation u = f with continuous righthand-side. 9

10 Theorem 2.2 Assume u C(Ω) and f C(Ω), where Ω is an open subset of R n. (a) If u f in the viscosity sense in Ω, then u ψ MaxP (V ) for any polar quadratic polynomial ψ(x) = a x x b x x 0 + d and any open set V Ω with V D(x 0, b), where a, b, d R with a 1 2 inf V f and x 0 R n. (b) If for any polar quadratic polynomial ψ(x) = a x x b x x 0 + d, where x 0 R n, a 1 2 inf V f and b, d R, and for any open set V Ω with V D(x 0, b), the maximum principle u ψ MaxP (V ) holds, then u(x) f(x) in Ω in the viscosity sense. Remark 2.3 We say u enjoys comparison with polar quadratic polynomials from above in Ω with respect to f if the hypothesis in part (b) of the theorem holds. Proof. Part (a) in the case a < 1 inf 2 V f follows directly from the strict comparison principle, Theorem 3.1, the proof of which is independent of the results in this section, in the next section and the preceding lemma. If a = 1 inf 2 V f, then is the limiting result of sup V (u ψ) max(u ψ) (2.7) V sup(u ψ k ) max (u ψ k), (2.8) V V where ψ k (x) = a k x x b x x 0 + d and {a k } is a sequence that increases to a = 1 inf 2 V f. We turn to the proof of (b). Assume u enjoys comparison with polar quadratic polynomials from above in Ω. Suppose that u(x) f(x) does not hold in Ω in the viscosity sense. So there exist a C 2 function ϕ and a point x Ω such that u x ϕ and + ϕ(x ) < f(x ). We will construct a polar quadratic polynomial ψ(x) = a x x b x x 0 + d and an open neighborhood V of x satisfying the assumption specified in (b) such that x is a strict maximum point of u ψ in V. Here x 0 is taken to be different from x. Then u ψ violates the weak maximum principle in the small open neighborhood V of x. In order to make x a strict maximum point of u ψ in V, we simply construct ψ so that x is a strict maximum point of ϕ ψ in V. It suffices to make x a strict local maximum point of ϕ ψ if ψ verifies ψ(x ) = ϕ(x ) and D 2 ψ(x ) > D 2 ϕ(x ). Without loss of generality, we assume x = 0. Denote p = Dϕ(0) and S = D 2 ϕ(0). Again, we handle the problem with a dichotomy. If ϕ(0) = 0, then the largest eigenvalue of D 2 ϕ(0), λ max (D 2 ϕ(0)), verifies λ max (D 2 ϕ(0)) < f(0). (2.9) Then there exists an open neighborhood V = B r (0) of 0 such that inf x V f(x) > λ max (D 2 ϕ(0)) as f is continuous at 0. If we take ψ(x) = ϕ(0) + a x 2 where a = 1 2 x V f(x), then ψ(0) = 0 and D 2 ψ(0) = 2aI > λ max (D 2 ϕ(0))i D 2 ϕ(0). (2.10) 10

11 Then 0 is a strict local maximum point of ϕ ψ in V. Taking a smaller neighborhood of 0 as V if necessary, we have shown that u ψ MaxP (V ) while still keeping a 1 2 inf V f for the new neighborhood V of 0. If p := ϕ(0) 0 and S := D 2 ϕ(0), then + ϕ(0) =< S ˆp, ˆp >< f(0). (2.11) We take a R so that < S ˆp, ˆp >< 2a < f(0) and a 0. We will take b R and x 0 R n \{0} so that a polar quadratic polynomial ψ(x) = a x x b x x 0 + ϕ(0) satisfies ψ(0) = p and D 2 ψ(0) > S. Let r = x 0. Then ψ(0) = (2ar + b)ˆx 0 and D 2 ψ(0) = (2a + b)i b ˆx r r 0 ˆx 0. So ψ(0) = p and D 2 ψ(0) > S are equivalent to { (2ar + b)ˆx 0 = p (2a + b)i b ˆx (2.12) r r 0 ˆx 0 > S In order to verify the first equality, we take ˆx 0 = ˆp, and b and r > 0 must be taken so that 2ar + b = p. In order to prove the second condition, we write x = αˆp + y 1 with < ˆp, y 1 >= 0 and show that < D 2 ψ(0)x, x >>< Sx, x > (2.13) for any x R n \{0}. Clearly, On the other hand, if x = αˆp + y 1 0, < D 2 ψ(0)x, x >= 2aα 2 + (2a + b r ) y1 2. (2.14) < Sx, x > = α 2 < S ˆp, ˆp > +2α < S ˆp, y 1 > + < Sy 1, y 1 > α 2 < S ˆp, ˆp > +α 2 ε S ˆp ε y1 2 + < Sy 1, y 1 > α 2 (< S ˆp, ˆp > +ε S ˆp 2 ) + ( 1 ε + S ) y1 2 2aα 2 + ( 1 ε + S ) y1 2 as < S ˆp, ˆp >< 2a and ε is small. < 2aα 2 + (2a + b r ) y1 2, as 2a + b = p + as r 0. So if we take r > 0 small enough and b = p 2ar, then r r < Sx, x ><< D 2 ψ(0)x, x > for all x R n \{0}. So 0 is a strict local maximum point of ϕ ψ and a < 1 f(0). Therefore, there is a 2 small open neighborhood V of 0 such that a 1 inf 2 V f and 0 is a strict maximum point of ϕ ψ due to the continuity of f. The proof is complete. The duality u = v between the viscosity sub-solutions of v = f and the viscosity super-solutions of u = f leads to a comparison principle for viscosity supersolutions of u = f. 11

12 Theorem 2.4 Assume u C(Ω). Then u f(x) in the viscosity sense in Ω if and only if the minimum principle u ψ MinP (V ) holds for any polar quadratic polynomial ψ(x) = a x x b x x 0 + d, where x 0 R n, a 1 2 sup V f, b and d R, and for any open set V Ω with V D(x 0, b). We say u enjoys comparison with polar quadratic polynomials from below in Ω with respect to f if the condition u ψ MinP (V ) holds for any ψ and V as specified in the theorem. If u enjoys comparison with polar quadratic polynomials from above and from below in Ω, we simply say u enjoys comparison with polar quadratic polynomials in Ω with respect to f. The two-sided comparison principle with polar quadratic polynomials, Theorem 1.7, follows from the above two one-sided theorems. Theorem 1.7 Assume u C(Ω) and f C(Ω), where Ω is an open subset of R n. Then u = f(x) in the viscosity sense in Ω if and only if u enjoys comparison with polar quadratic polynomials in Ω with respect to f. 3 A Strict Comparison Principle In this section, Ω always denotes a bounded open subset of R n. The main theorem in this section is the following strict comparison principle. Theorem 3.1 For j = 1, 2, suppose u j C( Ω) and in Ω, where f 1 < f 2, and f j C(Ω). Then sup Ω (u 2 u 1 ) max Ω (u 2 u 1 ). u 1 f 1 and u 2 f 2 Proof. Without the loss of generality, we may assume u 2 u 1 on Ω and intend to prove u 2 u 1 in Ω. Furthermore, for any small δ > 0, let u δ = u 2 δ. Then u δ < u 1 on Ω and u δ f 2 in Ω. If we can show that u δ u 1 in Ω for every small δ > 0, then it follows that u 2 u 1 in Ω. So we may additionally assume u 2 < u 1 on Ω in the following proof. We apply the sup- and inf-convolution technique here. Take any A max{ u 1 L (Ω), u 2 L (Ω)}. For any sufficiently small real number ε > 0, we take δ = 3 Aε and Ω δ = {x Ω : dist(x, Ω) > δ}. We define, on R n, u 1,ε (x) = inf y Ω (u 1(y) + 1 2ε x y 2 ) (3.1) and u ε 2(x) = sup(u 2 (y) 1 y Ω 2ε x y 2 ). (3.2) 12

13 For any y Ω such that y x 2 Aε, u 1 (y) + 1 2ε x y 2 u 1 (x) holds. So, in Ω δ, u 1,ε (x) = inf y Ω, x y 2 (u 1 (y) + 1 Aε 2ε x y 2 ) = inf z 2 (u 1 (x + z) + 1 Aε 2ε z 2 ), (3.3) as x + z Ω for any x Ω δ and z 2 Aε. Similarly, for x Ω δ, and Let u ε 2(x) = sup (u y Ω, x y 2 2 (y) 1 Aε 2ε x y 2 ) = sup (u z 2 2 (x + z) 1 Aε 2ε z 2 ). (3.4) f ε 1(x) = f 2,ε (x) = sup f x+z Ω, z 2 1 (x + z) = sup f Aε z 2 1 (x + z) (3.5) Aε inf x+z Ω, z 2 f 2 (x + z) = inf Aε z 2 f 2 (x + z), (3.6) Aε for x Ω δ. Clearly, f ε 1 is upper-semicontinuous. It is continuous due to the equicontinuity of the one parameter family of the functions x f 1 (x + z) in any compact subset of Ω. f 2,ε is continuous for a similar reason. We notice that, for every z with z 2 Aε and x Ω δ, (u 1 (x + z) + 1 2ε z 2 ) f 1 (x + z) f ε 1(x) (3.7) and (u 2 (x + z) 1 2ε z 2 ) f 2 (x + z) f 2,ε (x). (3.8) Lemma 1.10 implies that u 1,ε f ε 1 and u ε 2 f 2,ε in Ω δ in the viscosity sense. The rest properties of u 1,ε and u ε 2 are summarized in the following proposition, the proof of which is well known (see, for example, [ACJ] Proposition 6.4.). Proposition 3.2 u 1,ε and u ε 2 are semi-convex in R n. u 1,ε u 1 and u ε 2 u 2 in Ω. u 1,ε and u ε 2 converge locally uniformly to u 1 and u 2 in Ω, as ε 0. u 1,ε and u ε 2 are both differentiable at the maximum points of u ε 2 u 1,ε. As a result, if we take the value of ε smaller if necessary, then u 1,ε > u ε 2 on Ω δ, u 1,ε f ε 1 and u ε 2 f 2,ε in Ω δ, and f ε 1 < f 2,ε in Ω δ. If we can prove u ε 2 u 1,ε in Ω δ for any small ε > 0 and δ = 3 Aε, then u 2 u 1 in Ω holds. So we may without loss of generality assume that u 1 and u 2 are semi-convex in R n. Suppose u 1 (x 0 ) < u 2 (x 0 ) for some x 0 Ω. Without the loss of generality, we may assume that u 2 (x 0 ) u 1 (x 0 ) = max Ω(u 2 u 1 ). Then δ > 0 such that for any h R n with h < δ, we have u 1 (x 0 ) < u 2 (x 0 + h), while u 2 ( + h) < u 1 ( ) in Ω\Ω δ, and f 2 (x + h) > f 1 (x), x Ω δ. For any small positive number ε and h R n with h < δ, we define w ε,h (x, y) = u 2 (x + h) u 1 (y) 1 2ε x y 2, (3.9) 13

14 (x, y) Ω δ Ω δ. Let and M 0 = max(u 2 u 1 ), (3.10) Ω M h = max Ω δ (u 2 ( + h) u 1 ( )) (3.11) M ε,h = max Ω δ Ω δ w ε,h = u 2 (x ε,h + h) u 1 (y ε,h ) 1 2ε x ε,h y ε,h 2 (3.12) for some (x ε,h, y ε,h ) Ω δ Ω δ. Our assumption implies M h > 0 for all h with 0 h < δ, and clearly lim h 0 M h = M 0. As the semi-convex functions u 2 ( + h) and u 1 are locally Lipschitz continuous, the function M h is Lipschitz continuous in h R n with h < δ, if δ is taken smaller. By Lemma 3.1 of [CIL], we know and lim M ε,h = M h, (3.13) ε 0 1 lim ε 0 2ε x ε,h y ε,h 2 = 0 (3.14) lim(u 2 (x ε,h + h) u 1 (y ε,h )) = M h. (3.15) ε 0 As a result of the second equality, lim ε 0 x ε,h y ε,h = 0. As M h > 0 max Ωδ (u 2 ( + h) u 1 ( )), we know x ε,h, y ε,h Ω 1 for some Ω 1 Ω δ and all small ε > 0. Theorem 3.2 of [CIL] implies that there exist X = X ε,h, Y = Y ε,h S n n such that ( x ε,h y ε,h 2,+, X) J ε Ω u 2(x ε + h), ( x ε,h y ε,h 2,, Y ) J ε Ω u 1(y ε ) and 3 ε ( I 0 0 I ) ( X 0 0 Y ) 3 ( I I ε I I ). (3.16) In particular, X Y. Again, we solve the problem via a dichotomy. Case 1. Suppose that h with h < δ, and ε k 0 such that x εk,h y εk,h. Then it is easy to see that f 2 (x εk,h) < X( ε k,h y εk,h ), ( x ε k,h y εk,h ) > ε k ε k (3.17) < Y ( x ε k,h y εk,h ), ( x ε k,h y εk,h ) > ε k ε k (3.18) f 1 (y εk,h). (3.19) For a subsequence of {ε k }, x εk,h x h and y εk,h y h. As lim ε 0 x εk,h y εk,h = 0, we know that x h = y h, which leads to a contradiction with the assumption f 1 (x h ) < f 2 (x h ). Case 2. For every h R n with h < δ, x ε,h = y ε,h holds for every small ε > 0. 14

15 Then M ε,h = u 2 (x ε,h + h) u 1 (y ε,h ) = M h. We simply write x ε,h = y ε,h = x h. The semi-convexity of u 2 ( + h) and u 1 ( ) implies that the two functions are differentiable at the maximum point x h of their sum. The definition of x h shows that which in turn implies u 2 (x h + h) u 1 (x h ) u 2 (y + h) u 1 (x h ) 1 2ε x h y 2, (3.20) u 2 (x h + h) u 2 (y + h) 1 2ε x h y 2, (3.21) for small ε > 0. So u 2 (x h + h) = u 1 (x h ) = 0. For small h, k R n, M h = u 2 (x h + h) u 1 (x h ) u 2 (x k + h) u 1 (x k ) = M k + u 2 (x k + h) u 2 (x k + k) M k ( h k ), as u 2 (x k + k) = 0. So DM h = 0 a. e. as M h is Lipschitz continuous, which implies M h = M 0 for all small h R n. At x 0, either f 1 (x 0 ) < 0 or f 2 (x 0 ) > 0 holds due to the fact f 1 < f 2. Without loss of generality, we assume that f 2 (x 0 ) > 0. The proof for the case f 1 (x 0 ) < 0 is parallel. So u 2 is -subharmonic in a neighborhood of x 0. For any h with h < δ, u 2 (x 0 + h) u 1 (x 0 ) u 2 (x h + h) u 1 (x h ) = u 2 (x 0 ) u 1 (x 0 ). (3.22) So u 2 (x 0 ) is a local maximum of u 2. As u 2 0, the maximum principle for infinity harmonic functions implies that u 2 is constant near x 0. So, if we denote the largest eigenvalue of a symmetric matrix M by λ max (M), we have which is a contradiction. u 2 (x 0 ) = λ max (D 2 u 2 (x 0 )) = 0 < f 2 (x 0 ), (3.23) To prove the uniqueness of viscosity solutions to the Dirichlet problem, we need to prove the following comparison principle which follows fairly easily from the strict comparison principle. Theorem 3.3 Suppose u, v C(Ω) satisfy and u f(x) (3.24) v f(x) (3.25) in the viscosity sense in the domain Ω, where f is a continuous positive function defined on Ω. Then sup(u v) max(u v). (3.26) Ω Ω 15

16 Proof. Without loss of generality, we may assume that u v on Ω and intend to prove u v in Ω. For every small δ > 0, we take u δ (x) = (1 + δ)u(x) δ u L ( Ω). (3.27) Then u δ u v on Ω, and it is easily checked by the standard viscosity solution theory that u δ (x) = (1 + δ) u(x) (1 + δ)f(x) > f(x) v(x) (3.28) in Ω in the viscosity sense. Applying the preceding strict comparison theorem to v and u δ, we have u δ v in Ω for any small δ > 0. Sending δ to 0, we have u v in Ω as desired. It is obvious that the theorem is true if the condition f > 0 in Ω is replaced by the condition f < 0 in Ω. The uniqueness theorem follows as a direct corollary of the preceding comparison principle. Theorem 3.4 Suppose Ω is a bounded open subset of R n, and u and v C( Ω) are both viscosity solutions of the inhomogeneous normalized infinity Laplace equation w = f(x) in Ω, where f is a continuous function defined on Ω such that either f > 0 in Ω or f < 0 in Ω holds. If, in addition, u = v on Ω, then u = v in Ω. The condition that f does not change sign in Ω is indispensable, as a counter-example provided in [PSSW] shows uniqueness fails without such a condition. 4 Proof of Existence Theorem We prove the existence of a viscosity solution of the normalized infinity Laplace equation by constructing a solution as the infimum of a family of admissible super-solutions. Theorem 4.1 Suppose Ω is a bounded open subset of R n, f C(Ω) with f > 0 and g C( Ω). Then there exists u C( Ω) such that u = g on Ω and in Ω in the viscosity sense. Proof. We define the admissible set to be u(x) = f(x) A f,g = {v C( Ω) : v f(x) in Ω, and v g on Ω}. (4.1) Here the differential inequality v(x) f(x) is verified in the viscosity sense. Take u(x) = inf v(x), x Ω. (4.2) v A f,g 16

17 We may take a constant function which is bigger than the supremum of g on Ω. This constant function is clearly an element of A f,g. So the admissible set A f,g is nonempty. In addition, we take any x 0 Ω and b < 0 and d R, and define a polar quadratic polynomial ψ x0,bd(x) = 1 x x b x x 0 +d. For any v A f,g, v (inf Ω f)ψ x0,bd MinP (Ω). In particular, v A f,g is locally uniformly bounded below in Ω. So inf v Af,g v(x) > for any x Ω. Clearly, u g on Ω. As the infimum of a family of continuous functions, u is upper-semicontinuous on Ω. Lemma 1.10 implies u is a viscosity super-solution of u(x) = f(x) in Ω. We next prove u(x) f(x) in Ω in the viscosity sense. Suppose not, there exists a C 2 function ϕ and a point x 0 Ω such that and + ϕ(x 0 ) < f(x 0 ). For any small ε > 0, we define u x0 ϕ, ϕ ε (x) = ϕ(x 0 ) + ϕ(x 0 ) (x x 0 ) < D2 ϕ(x 0 )(x x 0 ), x x 0 > +ε x x 0 2. (4.3) Clearly, x 0 is a strict local maximum point of u ϕ ε in the sense that ϕ ε (x 0 ) = u(x 0 ) and ϕ ε (x) > u(x) for x near x 0 but x x 0. We claim that + ϕ ε (x) < f(x) for all x close to x 0 including x 0 and for ε sufficiently small. As before, we prove the claim by way of a dichotomy. If ϕ(x 0 ) 0, then ϕ(x) 0 in a neighborhood of x 0. So for all x close to x 0, + ϕ ε (x) =< D 2 ϕ ε (x) ϕ ε (x), ϕ ε (x) > (4.4) = + ϕ(x 0 ) + O(ε). (4.5) The continuity of + ϕ and f implies that for ε > 0 sufficiently small and x sufficiently close to x 0, + ϕ ε (x) < f(x) as a result of + ϕ(x 0 ) < f(x 0 ). On the other hand, if ϕ(x 0 ) = 0, then λ max (D 2 ϕ(x 0 )) = + ϕ(x 0 ) < f(x 0 ). As λ max (D 2 ϕ ε (x)) λ max (D 2 ϕ(x)) + Cε, λ max (D 2 ϕ ε (x)) < f(x) for sufficiently small ε and all x near x 0 due to the continuity of f and the C 2 regularity of ϕ. Therefore + ϕ ε (x) λ max (D 2 ϕ ε (x)) < f(x) (4.6) as + ϕ ε (x) =< D 2 ϕ ε (x) Dϕ ε (x), Dϕ ε (x) > λ max (D 2 ϕ ε (x)) (4.7) if ϕ ε (x) 0. The claim is proved. Fix the value of ε so that the claim holds. We take δ > 0 so small that a newly defined function ˆϕ(x) = ϕ ε (x) δ satisfies ˆϕ < u in a neighborhood of x 0 which is an open subset of the set {x Ω : + ϕ ε (x) < f(x)}, and ˆϕ u outside this neighborhood of x 0. Take ˆv = min{ ˆϕ, u}. Then ˆv = ˆϕ in a neighborhood of x 0 and ˆv = u otherwise. So ˆv g on Ω and ˆv C( Ω). 17

18 Suppose a C 2 function ψ satisfies ψ z ˆv for some z Ω. Then either ψ z ˆϕ or ψ z u depending on the location of z. In either case, + ψ(z) f(z). So ˆv A f,g. But ˆv = ˆϕ < u in a neighborhood of x 0. This is a contradiction to the definition of u. So u(x) f(x) in Ω in the viscosity sense. We now show u = g on Ω. For any point z Ω, and any ε > 0, there is a neighborhood B r (z) of z such that g(x) g(z) < ε for all x B r (z). Take a large number C > 0 such that Cr > 2 g L ( Ω). We define v(x) = g(z) + ε + C x z, (4.8) for x Ω. For x z < r and x Ω, v(x) g(z) + ε g(x); while for x z r and x Ω, v(x) g(z) + ε + Cr g L ( Ω) g(x). In addition, for any x Ω, v(x) = C(x z) 0 and v = 0. So, x Ω, v(x) = 0 f(x). So v A f,g and v(z) = g(z) + ε. Therefore g(z) u(z) g(z) + ε, (4.9) ε > 0. So u(z) = g(z), z Ω. Our proof is complete if we can show u C( Ω). We know that u 0 as u f > 0. So u is locally Lipschitz continuous in Ω. So it suffices to show that, z Ω, lim u(x) = g(z). (4.10) x Ω z Let us construct another set of admissible functions by defining S f,g = {w C( Ω) : w f(x) in Ω, and w g on Ω}. (4.11) Again w f(x) is satisfied in the viscosity sense. S f,g is nonempty with a particular element ψ(x) := C x z 2 D, where z Ω and C and D are so large that 2C > f L (Ω) and ψ g on Ω, since ψ = 2C. We take ū(x) = sup w(x) (4.12) w S f,g for every x Ω. Clearly, ū is lower-semicontinuous in Ω and ū(z) g(z) for any z Ω. We claim that lim inf ū(x) g(z), z Ω. (4.13) x Ω z z Ω and ε > 0, r > 0 such that g(x) g(z) < ε for all x Ω B r (z). We take a large number A such that and a large number C f L (Ω) such that A > sup x z (4.14) x Ω C{A 2 (A r) 2 } 2 g L ( Ω). (4.15) 18

19 One may define For x Ω, and w(x) = g(z) ε C{A 2 (A x z ) 2 }, x Ω. (4.16) w(x) = 2C( x z A) (x z) 0 (4.17) w(x) =< D 2 w(x) w(x), w(x) >= 2C f L (Ω) f(x). (4.18) On Ω B r (z), w(x) g(z) ε < g(x), while on Ω\B r (z), w(x) g(z) ε C{A 2 (A r) 2 } g(z) 2 g L ( Ω) g(x). So w S f,g and ū(z) w(z) = g(z) ε, (4.19) ε > 0, which implies ū(z) g(z). As the supremum of a family of continuous functions on Ω, ū is lower semi-continuous on Ω. Therefore lim inf ū(x) ū(z) g(z), z Ω. (4.20) x Ω z The claim is proved. The comparison principle, Theorem 3.3, implies w v on Ω for any w S f,g and v A f,g. In particular, ū(x) u(x), x Ω. So lim inf u(x) lim inf ū(x) g(z), z Ω. (4.21) x Ω z x Ω z On the other hand, the upper semi-continuity of u on Ω implies that lim sup u(x) u(z) = g(z), z Ω. (4.22) x Ω z So lim x Ω z u(x) = g(z), z Ω. This completes the proof. The following theorem is obtained from the above one by considering v = u and the proof is clear. Theorem 4.2 Suppose Ω is a bounded open subset of R n, f C(Ω) with f < 0 and g C( Ω). Then there exists u C( Ω) such that u = g on Ω and in Ω in the viscosity sense. u(x) = f(x) 19

20 5 A Connection between The Homogeneous and Inhomogeneous Equations This section deals with the stability of the inhomogeneous and homogeneous normalized infinity Laplace equations subject to the boundary value condition. The general approach in this section is similar in spirit to a scheme that we developed in our earlier work [LW] for the non-normalized infinity Laplace equations. evertheless, we have to take extra care of the difficulties arising from the normalized infinity Laplacian. In this section, Ω again denotes a bounded open subset of R n. Lemma 5.1 Assume f C(Ω) such that either f > 0 in Ω or f < 0 in Ω. For j = 1, 2, suppose c j > 0, g j C( Ω) and u j C( Ω) is the viscosity solution of the Dirichlet problem { u j = c j f in Ω (5.1) u j = g j on Ω. Then If, in particular g 1 = g 2 = g C( Ω), then u 1 c 1 u 2 c 2 L (Ω) g 1 c 1 g 2 c 2 L ( Ω). (5.2) Proof. Let u 1 c 1 u 2 c 2 L (Ω) 1 c 1 1 c 2 g L ( Ω). (5.3) v j = 1 c j u j. Then v j is the viscosity solution of the Dirichlet problem { v j = f in Ω v j = 1 c j g j on Ω, (5.4) j = 1, 2. Applying the maximum principle, Theorem 3.3, and the remark following it, one obtains v 1 v 2 L (Ω) g 1 g 2 L c 1 c ( Ω), (5.5) 2 which implies the desired inequality. Lemma 5.2 Assume f C(Ω) such that either f > 0 in Ω or f < 0 in Ω. Suppose c k 0, g k, g C( Ω) such that g k g L ( Ω) 0, and u k and u in C( Ω) are the viscosity solutions of the following Dirichlet problems respectively { u k = (1 + c k )f in Ω (5.6) u k = g k on Ω and { u = f in Ω u = g on Ω. (5.7) 20

21 Then sup(u k u) 0 as k. (5.8) Ω Proof. The preceding lemma implies which in turn implies Therefore 1 u k u L 1 + c (Ω) g k g L k 1 + c ( Ω), k 1 u k u L 1 + c (Ω) c k u L k 1 + c (Ω) k c k g k g L + c k 1 + c k g L ( Ω). u k u L (Ω) c k ( u L (Ω) + g L (Ω)) + g k g L ( Ω). (5.9) So lim k u k u L (Ω) = 0. The following perturbation theorem secures the stability of the inhomogeneous normalized infinity Laplace equation with positive or negative right-hand-side. Theorem 5.3 Suppose {f k } is a sequence of continuous functions in C(Ω) which converges uniformly in Ω to f C(Ω) and either inf Ω f > 0 or sup Ω f < 0. Furthermore, {g k } is a sequence of functions in C( Ω) which converges uniformly on Ω to g C( Ω). Assume u k C( Ω) is a viscosity solution of the Dirichlet problem { u k = f k in Ω (5.10) u k = g k on Ω while u C( Ω) is the unique viscosity solution of the Dirichlet problem { u = f in Ω u = g on Ω. (5.11) Then sup Ω u k u 0 as k. Proof. Without loss of generality, we assume inf Ω f > 0. Let ε k = sup Ω f k f. Then ε k 0 as k and f(x) ε k f k (x) f(x) + ε k for all x Ω. (5.12) To forbid ε k = 0, we replace ε k by ε k + 1 k and still denote the new quantity by ε k, as the new ε k 0. And now f(x) ε k < f k (x) < f(x) + ε k for all x Ω. (5.13) 21

22 Since inf Ω f > 0, the sequence {c k } defined by c k = equals 0. So, for all x Ω, ε k inf Ω f converges to 0 but never (1 c k )f(x) < f k (x) < (1 + c k )f(x), (5.14) as a result of ε k c k f(x). We define u 1 k and u2 k to be the viscosity solutions of the following Dirichlet problems respectively { u 1 k = (1 c k)f in Ω u 1 k = g (5.15) k on Ω and { u 2 k = (1 + c k)f in Ω u 2 k = g k on Ω. (5.16) By Theorem 3.1, we know that u 2 k u k u 1 k on Ω, since (1 c k )f(x) < f k (x) < (1+c k )f(x) for all x Ω. In addition, the preceding Lemma 5.2 implies sup Ω u j k u 0 for j = 1, 2. Consequently, sup u k u 0 Ω as k. ow we are at a position to prove the last main theorem of this paper, Theorem 1.9. For simplicity, we fix the boundary data for the time being. Theorem 5.4 Suppose Ω is a bounded open subset of R n. Suppose g Lip (Ω) and {f k } is a sequence of continuous functions on Ω which converges uniformly to 0 in Ω. If u k C( Ω) is a viscosity solution of the Dirichlet problem { u k = f k in Ω (5.17) u k = g on Ω and u C( Ω) is the unique viscosity solution of the Dirichlet problem { u = 0 in Ω u = g on Ω, (5.18) Then u k converges to u uniformly on Ω, i. e. as k. sup Ω u k u 0 Proof. Let c k = f k L (Ω) and {ε k } denote a sequence of positive numbers that converges to 0. Let u 1 k and u2 k C( Ω) be the respective viscosity solutions of the following Dirichlet problems { u 1 k = c k ε k in Ω u 1 k = g on Ω (5.19) 22

23 and { u 2 k = c k + ε k in Ω u 2 k = g on Ω. (5.20) By Theorem 3.1, we know that u 2 k u k u 1 k on Ω. (5.21) So it suffices to show that sup Ω u j k u 0 as k, for both j = 1 and 2. As the proof of either case of the above convergence implies that of the other, we will only prove sup Ω (u u 2 k ) 0 as k. The proof of sup Ω(u 1 k u) 0 follows when one considers u 1 k and u. In other words, we reduce the problem to the case in which u k is a viscosity solution of the Dirichlet problem { u k = δ k in Ω (5.22) u k = g on Ω where δ k > 0 and δ k 0 as k, and our goal is to prove sup(u u k ) 0 as k, (5.23) Ω since u k u is clear. For simplicity, we omitted the superscript 2 in the above and will do the same in the following. We use argument by contradiction. Suppose there is an ε 0 > 0 and a subsequence {u kj } such that sup Ω (u u kj ) ε 0, for all j = 1, 2, 3,... In addition, we may assume {δ kj } is a strictly decreasing sequence that converges to 0. Without further confusion, we will abuse our notation by using {u k } for the subsequence {u kj } and δ k for δ kj. So we will derive a contradiction from the fact sup(u u k ) ε 0 > 0, k, (5.24) Ω where u k C( Ω) is a viscosity solution of the Dirichlet problem { u k = δ k in Ω u k = g on Ω (5.25) and {δ k } decreases to 0. By Theorem 3.1, one obtains u k u k+1 u in Ω, k. (5.26) So {u k } converges pointwise on Ω, as u k = g on Ω. Moreover, {u k } is equicontinuous on any compact subset of Ω. In fact, let K be any compact subset of Ω. Then the distance from K to Ω defined by dist(k, Ω) = inf{dist(x, Ω) : x K}, 23

24 must equal to some positive number ε. Take R > 0 such that 4R < ε. Then B 4R (z) Ω for any z K. Since u k is infinity sub-harmonic in Ω, i. e. u k 0 in the viscosity sense, it is well-known, e. g. by [ACJ] Lemma 2.9, that x y u k (x) u k (y) ( sup u k sup u k ) B 4R (z) B R (z) R, (5.27) for any x, y B 4R (z). As u 1 u k u in Ω, we have u k (x) u k (y) ( sup u sup u 1 ) x y B 4R (z) B R (z) R L R x y R, (5.28) where L R = sup Ω u inf Ω u 1 0, which is independent of k. As K can be covered by finitely many balls B 4R (z), where z K, {u k } must be equicontinuous on K. Therefore a subsequence of {u k } converges locally uniformly in Ω to some function ū C(Ω). We once again abuse our notation by denoting the convergent subsequence by {u k }. We claim that ū verifies (i) ū = 0 in the viscosity sense in Ω, (ii) x 0 Ω, lim x Ω x0 ū(x) = g(x 0 ), and (iii) ū C( Ω) if we extend the definition of ū to Ω by defining ū Ω = g. One side of (i), u 0, is implied by Lemma 1.10 as ū(x) = sup k u k (x) and u k 0. In order to prove u 0, we suppose ϕ C 2 (Ω) touches u from below at x 0 Ω. Then, for any small ε > 0, the function x u(x) (ϕ(x) ε 2 x x 0 2 ) has a strict minimum at x 0. In particular, u(x 0 ) ϕ(x 0 ) < min (u(y) (ϕ(y) ε y B r(x 0 ) 2 y x 0 2 )) (5.29) for all small r > 0 and B r (x 0 ) Ω. As {u k } converges to u uniformly on B r (x 0 ), for all large k, inf (u k(x) (ϕ(x) ε x B r (x 0 ) 2 x x 0 2 )) u k (x 0 ) ϕ(x 0 ) < min y B r (x 0 ) (u k(y) (ϕ(y) ϕ 2 y x 0 2 )). So the function x u k (x) (ϕ(x) ε 2 x x 0 2 ) assumes its minimum over B r (x 0 ) at some point x k B r (x 0 ). By the definition of viscosity solutions, (ϕ(x) ε 2 x x 0 2 ) δ k (5.30) at x = x k, i. e. ϕ(x k ) + O(ε) δ k, (5.31) 24

25 ε > 0 and k k(r), where k(r) as r 0. If we send r to 0, we obtain ϕ(x 0 ) O(ε) for any ε > 0, which implies ϕ(x 0 ) 0, i. e. u(x 0 ) 0 in the viscosity sense. As the local uniform limit of {u k } in Ω, ū is clearly in C(Ω). In order to prove (ii) and (iii), we will apply the comparison with polar quadratic polynomials properties of the viscosity sub- and super-solutions of the equation v = 1. In fact, u k δ k = 1 in the viscosity sense in Ω. Fix x 0 Ω. The comparison with polar quadratic polynomials from above property states that u k (x) δ k (a x x b x x 0 ) max y Ω (u k(y) δ k (a y x b y x 0 ) (5.32) for all x Ω, a 1 2 and b R, or equivalently u k (x) aδ k x x 0 2 bδ k x x 0 max y Ω (g(y) aδ k y x 0 2 bδ k y x 0 ). (5.33) Fix the value of a, say a = 1 2. Take b = b k > 0 large enough so that b 2 a max y Ω y x 0 and bδ k = CL g ( Ω) for some universal constant C >> 1. So, for y Ω, g(y) aδ k y x 0 2 bδ k y x 0 g(y) 1 2 bδ k y x 0 g(x 0 ), (5.34) i. e. max y Ω (g(y) aδ k y x 0 2 bδ k y x 0 ) = g(x 0 ). (5.35) As a result, for x Ω near x 0, u k (x) g(x 0 ) + aδ k x x bδ k x x 0 (5.36) g(x 0 ) + CL g ( Ω) x x 0. (5.37) On the other hand, the comparison with polar quadratic polynomials from below property implies that, for any a 1, b R and all x in Ω, 2 u k (x) δ k or equivalently a x x 0 2 b x x 0 min y Ω (u k(y) δ k a x x 0 2 b x x 0 ), (5.38) u k (x) aδ k x x 0 2 bδ k x x 0 min y Ω (g(y) aδ k y x 0 2 bδ k y x 0 ). (5.39) Fix the value of a. Take b = b k < 0 so that b > 2a max x Ω x x 0 and bδ k = CL g ( Ω) for some universal constant C >> 1. As a result, for y Ω, g(y) aδ k y x 0 2 bδ k y x 0 g(y) 1 2 bδ k y x 0 g(x 0 ), (5.40) which means min Ω (g(x) aδ k x x 0 2 b x x 0 ) = g(x 0 ). (5.41) 25

26 So, for x Ω near x 0, u k (x) g(x 0 ) + aδ k x x bδ k x x 0 (5.42) Therefore, for some C >> 1 independent of k, g(x 0 ) CL g ( Ω) x x 0. (5.43) g(x 0 ) CL g ( Ω) x x 0 u k (x) g(x 0 ) + CL g ( Ω) x x 0, (5.44) for all k and all x Ω near x 0. Sending k to, we have g(x 0 ) CL g ( Ω) x x 0 ū(x) g(x 0 ) + CL g ( Ω) x x 0, (5.45) for all k and all x Ω near x 0. ow it is clear that (ii) and (iii) hold. The uniqueness of a viscosity solution in C( Ω) of the Dirichlet problem for homogeneous equation u = 0 in Ω under u Ω = g implies that ū = u on Ω. As a result, {u k } converges to u locally uniformly in Ω. Recall that sup Ω (u u k ) > ε 0. There exists, for each k, an x k Ω such that u(x k ) > u k (x k ) + ε 0 and x k approaches the boundary Ω, since {u k } converges to u locally uniformly in Ω. Without loss of generality, we assume x k x 0 Ω. Then we will have the following contradiction by previous estimate on u k (x) for x Ω near x 0. This completes the proof. g(x 0 ) = lim u(x k ) lim sup u k (x k ) + ε 0 k k lim sup(g(x 0 ) CL g ( Ω) x k x 0 +ε 0 ) k = g(x 0 ) + ε 0. We may also perturb the boundary data and still have the uniform convergence desired. This is the content of Theorem 1.9. ow we prove Theorem 1.9. Proof of Theorem 1.9. Let c k = f k L (Ω) and {ε k } denotes a sequence of positive numbers that converges to 0. Proceeding as in the proof of the preceding theorem, we let u 1 k and u2 k C( Ω) be the respective viscosity solutions of the following Dirichlet problems { u 1 k = c k ε k in Ω u 1 k = g (5.46) k on Ω and { u 2 k = c k + ε k in Ω u 2 k = g k on Ω. By Lemma 1.10, we know that (5.47) u 2 k u k u 1 k on Ω. (5.48) 26

27 So it suffices to show that sup Ω (u j k u) 0 as k, for both j = 1 and 2. We introduce vk 1 and v2 k C( Ω) as the viscosity solutions of the following Dirichlet problems respectively { vk 1 = c k ε k in Ω vk 1 = g on Ω (5.49) and { v 2 k = c k + ε k in Ω v 2 k = g on Ω. (5.50) The comparison principle, Theorem 3.3, implies that sup Ω u j k vj k max Ω g k g 0, as k, (5.51) for j = 1, 2. The preceding Theorem 5.4 implies that sup Ω v j k u 0, as k, for j = 1, 2. Therefore we have for j = 1, 2, as expected. sup Ω u j k u 0, as k, Acknowledgement The authors would like to thank the anonymous referee for some valuable suggestions. References [A1] Aronsson,G., Minimization problems for the functional sup x F (x, f(x), f (x)), Ark. Mat., 6, 1965, (1965). [A2] Aronsson,G., Minimization problems for the functional sup x F (x, f(x), f (x)). II. Ark. Mat., 6, 1966, (1966). [A3] Aronsson,G., Extension of functions satisfying Lipschitz conditions, Ark. Mat., 6, 1967, (1967). [ACJ] Aronsson, G., Crandall, M. and Juutinen, P., A tour of the theory of absolute minimizing functions, Bull. A.M.S., 41(2004), no.4, [BB] Barles, G. and Busca, J., Existence and comparison results for fully nonlinear degenerate elliptic equations without zeroth-order term, Comm. Partial Diff. Equations, 26(2001), [BJW] Barron, E..,Jensen, R. R. and Wang, C. Y., The Euler equation and absolute minimizers of L funcitonals, Arch. Ration. Mech. Anal. 157(2001), no.4, [BEJ] Barron, E.., Evans, L. C. and Jensen, R. R., The infinity Laplacian, Aronsson s equation and their generalizations, Trans. Amer. Math. Soc. 360 (2008), no. 1,