OPTIMAL REGULARITY FOR THE PSEUDO INFINITYLAPLACIANJULIO D. ROSSI AND MARIEL SAEZAbstract. In this paper we find the optimal regularity for viscositysolutions of the pseudo infinity Laplacian. We prove that the solutionsare locally Lipschitz and show an example that proves that this resultis optimal. We also show existence and uniqueness for the Dirichletproblem.1. Introduction.The main goal of this article is to study the optimal regularity of viscositysolutions to the pseudo infinity Laplacian. We find that the solutions areLipschitz but not necessarily C 1 .The pseudo infinity Laplacian is the second order nonlinear operator givenby(1.1) ˜∆∞ u = ∑u xi x i|u xi | 2 ,i∈I(∇u)where the sum is taken over the indexes in I(∇u) = {i : |u xi | = max j |u xj |}.This operator appears naturally as a limit of p−Laplace type problems. Infact, let u p be a sequence of solutions toN∑(1.2) ˜∆p u = (|u xi | p−2 u xi ) xi = 0.i=1Suppose that u 0 is a uniform limit of the sequence u p , then u 0 is a viscositysolution to˜∆ ∞ u = 0.A proof of this fact is contained in this paper for completeness, the mainarguments being taken from [4].Eigenvalue problems for this operator were studied in [4]. Belloni andKawohl point out that this operator arises in the problem of finding anoptimal Lipschitz extension of boundary data when the Euclidean norm issubstituted by the l 1 -norm. The Lipschitz extension problem is a relevantquestion that has been studied by several other authors, see for example[1, 2, 7, 10]. Since, the Lipschitz constant is dependent on the norm usedto measure distances in the domain, it is natural to expect that differentelliptic operators will arise from different norms. In our case, the operatorKey words and phrases. Viscosity solutions, optimal regularity, pseudo infinity Laplacian.2000 Mathematics Subject Classification. 35B65, 35J15, 35A05.1

2 J.D. ROSSI AND M. SAEZ(1.1) arises by considering the l 1 −norm (for details see [4]). When thestandard Euclidean norm is considered the corresponding equation is(1.3) ∆ ∞ u =N∑i,j=1∂u ∂ 2 u ∂u= (Du) T D 2 u(Du) = 0.∂x j ∂x j ∂x i ∂x iThis widely studied operator is known in the literature as the infinity Laplacian.A comprehensive survey in the subject of Lipschitz extensions and theinfinity Laplacian can be found in [2]. Limits of p−Laplacians are also relevantin mass transfer problems, see [6], [9].Concerning regularity results for these operators the best known resultis contained in [11] where Savin proved that a solution to the standardinfinity Laplacian (1.3) in two space dimensions is C 1 . This regularity resultcontrasts with our main result. More specifically, we show:Theorem 1. Let u : Ω → R be a viscosity solution to(1.4) ˜∆∞ u = 0,where Ω ⊂ R N . Then u is locally Lipschitz.Moreover, this result is optimal for N ≥ 2, since(1.5) u(x, y) = x + 1 2 |y|,is viscosity solution to (1.4) that has no further regularity than Lipschitz.Given that for N = 1 the infinity Laplacian and pseudo infinity Laplaciancoincide and the similar motivation for these two operators, at first one couldexpect a similar regularity for solutions to (1.3) and (1.4). Nevertheless, thecontrol over second derivatives of solutions to (1.3) is better, since for asolution to (1.4), the equation may not give any control over the secondderivatives of some of the variables (like y in our example (1.5)).Further discussion about regularity and the Proof of Theorem 1 can befound in Section 2. The main ingredient of the proof is that solutions to(1.4) verify a comparison with l 1 -cones property. This property analogousto the one satisfied by solutions to the usual infinity Laplacian with l 2 -cones,see [2] and [5].For sake of completeness, we also show existence and uniqueness to theDirichlet problem.Theorem 2. Given a bounded smooth domain Ω ⊂ R N , for any Lipschitzboundary data g(x), there exists a unique viscosity solution to˜∆ ∞ u = 0 in Ω,u = gon ∂Ω.The proof of Theorem 2 can be found in Section 3. The existence resultis proved using arguments from [4]. The strategy is to take limits (alongsubsequences) of variational solutions to (1.2) as p → ∞. Uniqueness followsby adapting results in [3].

OPTIMAL REGULARITY 32. Optimal regularity. Proof of Theorem 1First, let us recall the standard definition of viscosity solution, see [8].Definition 2.1. ConsiderF (x, Du, D 2 u) = 0 in Ω.(1) An upper semi-continuous function u is a subsolution if for everyφ ∈ C 2 (Ω) such that u − φ has a strict maximum at the point x 0 ∈ Ωwith u(x 0 ) = φ(x 0 ) we have:F (x 0 , Dφ(x 0 ), D 2 φ(x 0 )) ≤ 0.(2) A lower semi-continuous function u is a viscosity supersolution if forevery φ ∈ C 2 (Ω) such that u − φ has a strict minimum at the pointx 0 ∈ Ω with u(x 0 ) = φ(x 0 ) we have:F (x 0 , Dφ(x 0 ), D 2 φ(x 0 )) ≥ 0.(3) Finally, u is a viscosity solution if it is a super and a subsolution.We will use this definition withF (x, ξ, M) = − ∑where I(ξ) = {i : |ξ i | = max j |ξ j |}.i∈I(ξ)M ii |ξ i | 2 ,Now we prove that viscosity solutions enjoy comparison with l 1 -cones.These cones are defined byN∑C x0 (x) = a + b |x i − (x 0 ) i |.We denote the l 1 ball byB r (x 0 ) ={x :i=1}N∑|x i − (x 0 ) i | ≤ r .i=1Lemma 2.2. Let u be a viscosity solution to (1.4). If u(x) ≥ C x0 (x) forx ∈ ∂(B r (x 0 ) \ {x 0 }), then u(x) ≥ C x0 (x) for x ∈ B r (x 0 ).Proof. We follow [7] and argue by contradiction. Suppose that u(y) 1 close to 1). If R is large enough andε is small enough we obtain w(x) ≤ u(x) for x ∈ ∂(B r (x 0 ) \ {x 0 }) andmax(w − u) = w(z) − u(z) > 0. A direct computation shows w xi (z) ≠ 0and w xi x i(z) > 0, for ε small, which contradicts the fact that u is a viscositysolution.□Similarly we have thati=1

4 J.D. ROSSI AND M. SAEZLemma 2.3. Let u be a viscosity solution to (1.4). If u(x) ≤ C x0 (x) forx ∈ ∂(B r (x 0 ) \ {x 0 }), then u(x) ≤ C x0 (x) for x ∈ B r (x 0 ).Now we are ready to prove Theorem 1.Proof of Theorem 1. First we show, following [7], that every viscosity solutionis locally Lipschitz. Let{ }u(x) − u(x0 )S r (x 0 ) =.rmaxP Ni=1 |x i−(x 0 ) i |=rConsider the cones centered at x 0 ,and letb r = infC b x 0(x) = u(x 0 ) + bN∑|x i − (x 0 ) i |i=1{}b : u(x) ≤ Cx b 0(x) for x ∈ ∂(B r (x 0 ) \ {x 0 }) .This number b r is well defined since u is continuos and hence bounded on∂B r (x 0 ).Taking b = b r we haveu(x) − u(x 0 )r≤ Cb x 0(x) − u(x 0 )r= b r .HenceS r (x 0 ) ≤ b r .From Lemma 2.3 we get that b r is nondecreasing, if r ′ < r then b r ′ ≤ b r .Therefore(2.1) S r ′ ≤ b r or all r ′ ≤ r.A similar argument using Lemma 2.2 proves that{ }u(x) − u(x0 )(2.2) T r (x 0 ) =PminNi=1 |x i−(x 0 ) i |=r ris bounded below.From (2.1) and (2.2) we obtain that there exists a constant C such that{ }|u(x) − u(x0 )|≤ C, for all r small.rmaxP Ni=1 |x i−(x 0 ) i |≤rOr equivalently, u is locally Lipschitz.To finish the proof of the theorem we have to show that(2.3) u(x, y) = x + 1 2 |y|is a viscosity solution of (1.4). To see this fact, we need to check Definition2.1. First, let us verify (1) in 2.1. Assume that u − φ has a maximum at(x 0 , y 0 ) with y 0 ≠ 0. Then, since u − φ is smooth and satisfiesand(u − φ) x (x 0 , y 0 ) = 0, (u − φ) y (x 0 , y 0 ) = 0(u − φ) xx (x 0 , y 0 ) ≤ 0.

OPTIMAL REGULARITY 5Since u x (x 0 , y 0 ) = 1, |u y (x 0 , y 0 )| = 1/2 and u xx (x 0 , y 0 ) ≤ 0 we getHence, we obtainφ x (x 0 , y 0 ) = 1 > |φ y (x 0 , y 0 )| = 1/2 and − φ xx (x 0 , y 0 ) ≤ 0.− ˜∆ ∞ φ(x 0 , y 0 ) ≤ 0.Analogously if u − φ has a minimum at (x 0 , y 0 ) with y 0 ≠ 0 we have− ˜∆ ∞ φ(x 0 , y 0 ) ≥ 0.Now, if y 0 = 0 and φ is smooth, u − φ cannot have a minimum at (x 0 , y 0 ).On the other hand, if it has a maximum at this point we obtainThereforeφ x (x 0 , y 0 ) = 1 > |φ y (x 0 , y 0 )| ≤ 1/2 and − φ xx (x 0 , y 0 ) ≤ 0.− ˜∆ ∞ φ(x 0 , y 0 ) ≤ 0.Combining the previous inequalities, we conclude that u is a viscositysolution to (1.4), finishing the proof of Theorem 1.□Remark 2.4. Note that l 1 -cones are not differentiable along the axes (theyare only Lipschitz) and that the l 2 -cones are differentiable everywhere awayfrom the vertex. Here l 1 -cones play the same role as the one played by thel 2 -cones in the theory for the usual infinity Laplacian, ∆ ∞ , see [2]. This isa good argument to explain why solutions to the pseudo infinity Laplacian,˜∆ ∞ , are Lipschitz but not C 1 .Remark 2.5. In the definition of the pseudo infinity Laplacian the sum istaken over all indexes where the l ∞ -norm of the gradient is attained. Onemay think that the lack of C 1 regularity comes from the fact that the indexesof the sum may change from one point to another. This is not always thecase, as in our example where |u x | = max{|u x |, |u y |}.Remark 2.6. As trivial examples of solutions we may consider bilinearfunctions, that is, u(x, y) = a 1 xy +a 2 x+a 3 y +a 4 . With these examples it iseasy to find solutions in which the indexes of the maximum of the derivativesdepend on the point. Also, by the above proof, u(x, y) = a(y)x + b(y) is asolution if a(y) > a ′ (y)x + b ′ (y).(3.1)(3.2)3. Existence and uniqueness. Proof of Theorem 2We will obtain a solution of (1.4) taking limit as p → ∞ of solutions to˜∆ p u = 0u(x) = g(x)in Ωfor x ∈ ∂Ω.Solutions to (3.1)-(3.2) can be obtained by variational arguments. Noticefirst that since g is Lipschitz it can be extended in Ω to a function inW 1,∞ (Ω), that we still denote as g. Hence, we can define∫ N∑(3.3) E p (g) = inf|u xi | p .v∈W 1,p (Ω) : u−g∈W 1,p0 (Ω) Ωi=1

6 J.D. ROSSI AND M. SAEZLet us denote by|||u||| =( ∫Ω) 1/pN∑|u xi | p .This is a seminorm in W 1,p (Ω) equivalent to the usual one.i=1By a standard compactness argument, we can show that there is a u p ∈W 1,p (Ω) that realizes E p (g). This function u p is a weak solution to (1.2).From definition (3.3), using g as a test function, we can easily obtain aconstant C, independent of p, such that‖u p ‖ W 1,p (Ω) ≤ C‖g‖ W 1,∞ (Ω).Therefore, fixing q < p and using Holder’s inequality, we get(3.4) ‖u p ‖ W 1,q (Ω) ≤ C.By a diagonal procedure we obtain a sequence p i → ∞ such thatu pi → uweakly in W 1,q (Ω).Again by compactness, the limit verifies u = g on the boundary. Using (3.4)we conclude u ∈ W 1,∞ (Ω).Now, we want to prove that u is a viscosity solution to (1.4). To this endwe need a lemma that shows that u p are also viscosity solutions. Then, wepass to the limit in the viscosity sense.Lemma 3.1. Every weak solution u p to (3.1) is also viscosity solution to(3.1).Proof. We will follow the ideas in [4]. Let us prove first that u p is a viscositysubsolution. Fix x 0 ∈ Ω and φ smooth such that u−φ has a strict maximumat x 0 with u(x 0 ) = φ(x 0 ). Assume, arguing by contradiction, that thereexists r > 0 such that− ˜∆ p φ(z) > 0, z ∈ B r (x 0 ).Set M = sup{(φ − u p )(y) : y ∈ ∂B r (x 0 )} and Φ = φ − M/2. Hence we haveΦ > u on ∂B r (x 0 ), Φ(x 0 ) < u(x 0 ) and− ˜∆ p Φ(z) > 0, z ∈ B r (x 0 ).We multiply by (u p − Φ) + and integrate by parts to obtain∫ N∑|Φ xi | p−2 Φ xi (u p − Φ) xi dx > 0.{u>Φ} i=1Now, we use the fact that u p is a weak solution to obtain∫ N∑|u xi | p−2 u xi (u p − Φ) xi dx = 0.Therefore∫0 ≤{u>Φ}i=1{u>Φ}i=1which is a contradiction.N∑ (|uxi | p−2 u xi − |Φ xi | p−2 )Φ xi (up − Φ) xi dx < 0,

OPTIMAL REGULARITY 7The proof of u p being a viscosity supersolution is completely analogous.This finishes the proof of the Lemma.□Now we are ready to prove our second result.Proof of Theorem 2. We will divide the proof in two steps. In the first stepwe prove that the uniform limit of u p is a viscosity solution to (1.4). Then,in the second step, we finish by proving uniqueness of viscosity solutions.Step 1:We prove that u is subsolution. As before, we skip the proof of u being asupersolution, since it is analogous.Let φ be a smooth function such that u − φ has a strict maximum atx 0 ∈ Ω. By uniform convergence of u pi , we have that there is sequence ofpoints x pi → x 0 such that u pi − φ attains a maximum at x pi . By Lemma3.1 we know thatN∑− |φ xj | pi−2 φ xj x j(x pi ) ≤ 0.j=1If the max j |φ xj |(x 0 ) = 0 the result is trivial. Otherwise, since φ is smooth,dividing by max j |φ xj | pi−4 (x pi ) and taking limits we obtain that∑− |φ xj | 2 φ xj x j(x 0 ) ≤ 0.j∈I(∇φ)(x 0 )This shows that the limit is a viscosity subsolution, finishing step 1.Step 2:We will follow closely the arguments in [3]. The main point of the proofis to obtain an equivalent result to their Lemma 3.2, that is,Lemma 3.2 (Hopf’s Lemma). Assume that w is a viscosity supersolutionto equation (1.4) with a local minimum at y 0 . Then w is constant in aneighborhood of y 0 .Proof. Let w β be the inf-convolution of w, that is)|y − z|2w β (y) = inf(w(z) +zβ 2 .It is possible to show that w β is also a viscosity supersolution. Moreover,w β is semi-convex.We will prove the result by contradiction. Since w β is a semi-convexfunction, by translating it we can assume that there is an Euclidean ball B Rsuch that w β > 0 on ∂B R and w β (z 0 ) = 0 for some z 0 ∈ B R \ B R .22We considerχ(x) = e −α|x|2 − e −αR2 .A straightforward computation shows that− ˜∆∑∞ χ(x) = −8α 3 e −3α|x|2 |x i | 2 (2α|x i | 2 − 1){i: |x i |=max j |x j |}

8 J.D. ROSSI AND M. SAEZGiven that for x ∈ B R \ B R2enough we obtainit holds |x i | = max j |x j | ≥(3.5) − ˜∆ ∞ χ(x) < 0.R2 √ , for α largeNNow we show, that for a large enough α, it holds that w β ≥ χ. Infact, 0 = χ ≤ w β on ∂B R and for α large also χ ≤ w β on ∂B R . If the2max BR \ B ¯ R2(χ − w β ) > 0, we have that the maximum is attained for some zthat satisfies χ(z) − w β (z) > 0 and, since w β is a viscosity supersolution, itmust also hold− ˜∆ ∞ χ(z) ≥ 0,which contradicts (3.5). Hence, for every x ∈ B R \ B R2w β ≥ χ,which contradicts that w β (z 0 ) = 0.it holdsThis implies that w β is locally constant in a neighborhood of z 0 . Hence,following the proof in [3], we obtain that w is constant in some neighborhoodof every minimum.□By using the sup-convolution we can prove a similar statement to Lemma3.2 when a viscosity subsolution has a local maximum.Once these Lemmas are established, the rest of the proof of uniqueness iscontained in the comparison principle proved in [3]. We briefly sketch theproof here.Let u and v be sub and supersolution to (1.4) respectively, such that u ≤ von ∂Ω. By regularizing u and v by sup and inf convolution and taking u − ηinstead of u we can assume u is semi-convex, v is semi-concave and u < von ∂Ω. We need to show u ≤ v in Ω.We establish the comparison principle by contradiction. Suppose that(3.6) max(u − v) > 0.ΩDefine for h smallwhere Ω h = {x ∈ Ω : d(x, ∂Ω) > h}.M(h) = maxx∈Ω h(u(x + h) − v(x)),Notice that (3.6) implies that M(0) > 0. Hence, if (3.6) holds, we musthave for h small enough that M(h) > 0.Before finding the contradiction we will show that necessarily one of thefollowing holds:(1) There is a sequence h n → 0 such that at any maximum point x hn ofu(· + h n ) − v(·) holds Du(x hn + h n ) = Dv(x hn ) ≠ 0 for every n. or(2) There is a neighborhood of 0 such that for every h in this neighborhoodM(h) = M(0).If (1) holds we are going to reach the contradiction by proving that necessarilyfor n large enough M(h n ) ≤ 0, which contradicts (3.6). On the other

OPTIMAL REGULARITY 9hand, if (2) holds, we show that Lemma 3.2 implies that the set where M(0)is achieved is open and closed, hence equal to Ω, contradicting that u < von ∂Ω (when M(0) > 0).Let us start by showing that either (1) or (2) holds. Suppose that (1)does not occur. Notice that M is a maximum of semi-convex functionswhich implies that M is semi-convex in a neighborhood of 0. Let x h ∈ Ωbe a maximum point of u(· + h) − v(·). By general properties of semiconvexfunctions (see (DMP) in [3]) u(· + h) and v(·) are differentiable at x h andDu(x h + h) = Dv(x h ).The semiconvexity of u implies that, if Du(x h +h) = Dv(x h ) = 0 for someh, then there is a constant r, small enough, such that for every h ′ ∈ B r (h)It follows thatWhich implies 0 ∈ ∂M(h).M(h ′ ) ≥u(x h + h ′ ) − v(x h )≥u(x h + h) − v(x h ) − C|h − h ′ | 2 .M(h ′ ) ≥ M(h) − C|h − h ′ | 2 .Since (1) does not hold, necessarily for any h in a neighborhood of 0must hold that 0 ∈ ∂M(h), or equivalently M(h) = M(0) for h in someneighborhood of 0. That is (2) holds.Now we are left to show that both of these alternatives lead us to contradiction.(a) If (2) holds: Then for every h in a neighborhood from 0u(x 0 ) − v(x 0 ) = M(0) = M(h) ≥ u(x 0 + h) − v(x 0 ).That is x 0 is a local maximum for u. Lemma 3.2 implies that u is constantin a neighborhood of x 0 . Since u − v attains a local maximum at x 0 and uis constant, v must attain a local minimum at x 0 . Using once more Lemma3.2, we conclude that v must also be constant in a neighborhood of x 0 . Itfollows that the set where M(0) is attained is open. By continuity of u andv it must be also closed, hence it must equal Ω, which contradicts that u < von ∂Ω.(b) If (1) holds: Let ϕ ε such that(∫ t)(3.7) ϕ ′ ɛ (t) = exp exp (−ε −1 (s + ε −1 ))dsand ψ ε (t) its inverse. DefineG ε (w, q, N) = − ∑0i∈I(q)where I(q) = {i : |q i | = max j |q j |}.We denote as|q i | 2 ( ϕ ′ ɛ (w)N ii + ϕ ′′ε (w)|q i| 2) ,U ε (x) =ψ ε (u),V ε (x) =ψ ε (v).

10 J.D. ROSSI AND M. SAEZThen, given thatG ε(U ε , ∂U ε∂x i, D 2 U ε)= −(ϕ ′ ε (u))2 ˜∆∞ u,we have that U ε is a subsolution toG ε(U ε , ∂U ε∂x i, D 2 U ε)= 0.Similarly, V ε is a supersolution.By definition of ϕ ε we have that U ε and V ε are, respectively, semi-convexand semi-concave. It holds that U ε (· + h n ) − V ε (·) attains its maximum atsome interior point x ε ∈ Ω. Since U ε and V ε are semiconvex and semiconcave(property (DMP) in [3]), both of them are differentiable at x ε and |DU ε | =|DV ε |. Notice that U ε → u and V ε → v as ε → 0 . Hence, we can finda sequence ε k → 0 such that x εk → x, where x is a maximum of u ε (· +h n ) − v ε (·). Since (1) holds, we have |Du| = |Dv| ≥ δ(n) > 0. By generalproperties of semi-convex and semi-concave functions (property (PGC) in[3]) and the definition of ϕ ε , we have for ε small enough |DU ε | = |DV ε | ≥δ(n)2 .Note that one can construct a sequence of points p m and a sequencefunctionsf m (x) = U ε (x + h n ) − V ε (x) − 〈p m , x〉,such that f m has a strict maxima at x m ε and xm ε → x ε as m → ∞. LemmaA.3 in [8] shows that if r > 0 is small enough, there is a ρ > 0 such that theset of maximum points in B r (x m ε ) ofg m (x) = U ε (x + h n ) − V ε (x) − 〈p m , x〉 − 〈q, x〉with q ∈ B ρ (0) (ρ ≤ ρ), contains a set of positive Lebesgue measure. ByAlexansandrov’s result, U ε (· + h n ) and V ε (·) are twice differentiable a.e.Therefore for r small and ρ ≤ ρ, there is a z ∈ B r (x m ε ) and q ∈ B ρ(0) suchthat z is a maximum of g m and U ε and V ε are twice differentiable at z. Sincez is a maximum it holds DU ε = DV ε + p m + q. As before, for q, ρ small andm large|DU ε | = |DV ε + p m + q| ≥ δ(n)4 .Moreover, since U ε is semi-convex and V ε semi-concave, it holds−C · Id ≤ D 2 U ε (z) ≤ D 2 V ε (z) ≤ C · Id,for some constant C > 0 independent of ρ, r and m. Evaluating at z wehave by the definition of G(3.8)G ε (U ε (z + h n ), DV ε (z) + p m + q, D 2 V ε (z)) ≤ 0≤ G ε (V ε (z), DV ε (z), D 2 V ε (z)).Since DV ε (z) and D 2 V ε (z) are bounded, by taking a subsequence whenρ, r → 0 and m → ∞ we can find P ≥ δ(n)4and X such that DV ε (z) → Pand D 2 V ε (z) → X.

Taking limits in (3.8) we obtainOPTIMAL REGULARITY 11(3.9) G ε (U ε (z + h n ), P , X) ≤ 0 ≤ G ε (V ε (z), P , X).On the other hand, it is easy to see by the definition of G ε and ϕ ε that∂G ε (w, q, N)∂w= − ∑i∈I(q)= − ∑i∈I(q)(∫ w× exp|q i | 2 ( ϕ ′′ɛ (w)N ii + ϕ ′′′ε (w)|q i | 2)|q i | 2 exp(−ε −1 (w + ε −1 ))0)exp(−ε −1 (s + ε −1 ))ds)×[( 1ε − exp(−ε−1 (s + ε −1 ))|q i | 2 − N ii]Hence, for any δ > 0, ε small enough, |w| ≤ δ −1 , δ ≤ |q| ≤ δ −1 and |N| ≤ δ −1it holds(3.10)∂G ε (w, q, N)∂w> 0,which contradicts (3.9), finishing the proof of uniqueness.□Acknowledgements. JDR partially supported by Fundación Antorchas,CONICET and ANPCyT PICT No. 05009.References[1] G. Aronsson, Extensions of functions satisfiying Lipschitz conditions. Ark. Math., 6(1967), 551-561.[2] G. Aronsson, M.G. Crandall and P. Juutinen, A tour of the theory of absolutelyminimizing functions. Bull. Amer. Math. Soc., 41 (2004), 439-505.[3] G. Barles and J. Busca, Existence and comparison results for fully nonlinear degenerateelliptic equations without zeroth-order term. Comm. Part. Diff. Eq. 26, No.11&12 (2001), 2323-2337.[4] M. Belloni and B. Kawohl, The pseudo-p-Laplace eigenvalue problem and viscositysolutions as p → ∞. ESAIM COCV, 10 (2004), 28-52.[5] M. Belloni, B. Kawohl and P. Juutinen, The p-Laplace eigenvalue problem as p → ∞in a Finsler metric. J. Europ. Math. Soc., to appear.[6] G. Bouchitte, G. Buttazzo and L. De Pasquale. A p−laplacian approximation forsome mass optimization problems. J. Optim. Theory Appl. 118(1) (2003), 125.[7] M. G. Crandall, L. C. Evans and R. F. Gariepy. Optimal Lipschitz extensions andthe infinity Laplacian. Calc. Var. and PDE 13 (2001), 123-139.[8] M.G. Crandall, H. Ishii and P.L. Lions. User’s guide to viscosity solutions of secondorder partial differential equations. Bull. Amer. Math. Soc., 27 (1992), 1-67.[9] L.C. Evans and W. Gangbo, Differential equations methods for the Monge-Kantorovich mass transfer problem. Mem. Amer. Math. Soc., 137 (1999), no. 653.[10] R. Jensen, Uniqueness of Lipschitz extensions: minimizing the sup norm of thegradient. Arch. Rational Mech. Anal. 123 (1993), 51-74.[11] O. Savin, C 1 regularity for infinity harmonic functions in two dimensions. Arch.Rational Mech. Anal. 176 (2005), 351361.

12 J.D. ROSSI AND M. SAEZJ. D. RossiInstituto de Matemáticas y Física FundamentalConsejo Superior de Investigaciones CientíficasSerrano 123, Madrid, Spain,on leave from Departamento de Matemática, FCEyN UBA (1428)Buenos Aires, Argentina.E-mail address: jrossi@dm.uba.arWeb page: http://mate.dm.uba.ar/∼jrossiM. SaezMax Planck Institute for Gravitational PhysicsAlbert Einstein InstituteAm Mühlenberg 1, D-14476 Golm, Germany,Germany .E-mail address: mariel.saez@aei.mpg.de

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