An easy proof of Jensen's theorem on the uniqueness of infinity harmonic functions
aa r X i v : . [ m a t h . A P ] J un AN EASY PROOF OF JENSEN’S THEOREM ON THEUNIQUENESS OF INFINITY HARMONIC FUNCTIONS
SCOTT N. ARMSTRONG AND CHARLES K. SMART
Abstract.
We present a new, easy, and elementary proof of Jensen’s Theoremon the uniqueness of infinity harmonic functions. The idea is to pass to a finitedifference equation by taking maximums and minimums over small balls.
In this short article, we present a new proof of the famous result of R. Jensen[8], which establishes the uniqueness of viscosity solutions of the infinity Laplaceequation(1) − ∆ ∞ u := − n X i,j =1 u x i x j u x i u x j = 0 , in a bounded domain Ω ⊆ R n , subject to a given Dirichlet boundary condition.Our argument is elementary, and other than our use of a well-known equivalence(Theorem 3, below), our presentation is self-contained. In contrast, previous proofsemploy either intricate viscosity solution methods as well as a deep result of Alek-sandrov [8, 3, 2, 7], or follow a less direct path using probabilistic arguments [11]. Definition 1.
A viscosity subsolution (supersolution, solution) of (1) is called infinity subharmonic (superharmonic, harmonic) .We refer to the survey articles [2, 6] for an introduction to the infinity Laplaceequation, as well as the definition of viscosity solution. In this article we do notapply the definition of viscosity solution directly. Instead, we use the notion of comparisons with cones . Definition 2. A cone function with vertex x ∈ R n is a function of the form ϕ ( x ) = a + b | x − x | , where a, b ∈ R . A function u ∈ C (Ω) is said to enjoycomparisons with cones from above if it possesses the following property: for everyopen V ⊆ R n for which ¯ V ⊆ Ω, and every cone function ϕ with vertex x ∈ R n \ V , u ≤ ϕ on ∂V implies u ≤ ϕ in V. We say that u enjoys comparisons with cones from below if − u enjoys compar-isons with cones from above. Finally, u enjoys comparisons with cones if it enjoyscomparisons with cones from above and below.The following result of Crandall, Evans and Gariepy [4] establishes the equiv-alence between the notions of infinity harmonic functions and functions enjoyingcomparisons with cones. Date : June 30, 2009.2000
Mathematics Subject Classification.
Primary 35J70.
Key words and phrases.
Infinity Laplace equation, comparison principle.
Theorem 3 ([4]) . A function u ∈ C (Ω) is infinity subharmonic (superharmonic)if and only if u enjoys comparisons with cones from above (below). A less general version of Theorem 3 was also obtained by Jensen [8, Lemma 3.1].In addition to [4], elementary proofs of this result can be found in [2, 6].We now introduce some notation. For ε > x ∈ R n , we write B ( x, ε ) := { y ∈ R n : | x − y | < ε } . Let Ω ε denote the set of points x ∈ Ω for which ¯ B ( x, ε ) ⊆ Ω.If u ∈ C (Ω) and x ∈ Ω ε , then we denote S + ε u ( x ) := max y ∈ ¯ B ( x,ε ) u ( y ) − u ( x ) ε and S − ε u ( x ) := max y ∈ ¯ B ( x,ε ) u ( x ) − u ( y ) ε . The first step in our proof of Jensen’s theorem is the following comparison lemmafor a finite difference equation. We adapt an argument due to Le Gruyer [9], whoproved a similar lemma for a difference equation on a finite graph.
Lemma 4.
Assume that u, v ∈ C (Ω) ∩ L ∞ (Ω) satisfy the inequalities (2) S − ε u ( x ) − S + ε u ( x ) ≤ ≤ S − ε v ( x ) − S + ε v ( x ) for every x ∈ Ω ε . Then sup Ω ( u − v ) = sup Ω \ Ω ε ( u − v ) . Proof.
Arguing by contradiction, we suppose thatsup Ω ( u − v ) > sup Ω \ Ω ε ( u − v ) . The set E := { x ∈ Ω : ( u − v )( x ) = sup Ω ( u − v ) } is nonempty, closed and containedin Ω ε . Define F := { x ∈ E : u ( x ) = max E u } , and select a point x ∈ ∂F . Since u − v attains its maximum at x , we see that(3) S − ε v ( x ) ≤ S − ε u ( x ) . Consider the case S + ε u ( x ) = 0. From the first inequality in (2) we also have S − ε u ( x ) = 0. From (3), we deduce that S − ε v ( x ) = 0. Now the second inequal-ity of (2) implies S + ε v ( x ) = 0. In particular, u and v are constant in ¯ B ( x , ε ),contradicting our assumption that x ∈ ∂F .It remains to examine the case S + ε u ( x ) >
0. Select z ∈ ¯ B ( x , ε ) such that εS + ε u ( x ) = u ( z ) − u ( x ). Since u ( z ) > u ( x ) and x ∈ F , we see that z E . Fromthis we deduce that(4) εS + ε v ( x ) ≥ v ( z ) − v ( x ) > u ( z ) − u ( x ) = εS + ε u ( x ) . Combining (3) and (4), we obtain S − ε v ( x ) − S + ε v ( x ) < S − ε u ( x ) − S + ε u ( x ) . This contradicts (2), and the lemma follows. (cid:3)
The next lemma allows us to modify solutions of the PDE (1) to obtain solutionsof the finite difference inequalities (2). We use the notation u ε ( x ) := max ¯ B ( x,ε ) u and u ε ( x ) := min ¯ B ( x,ε ) u, x ∈ Ω ε , which allows us to write εS + ε u ( x ) = u ε ( x ) − u ( x ) and εS − ε u ( x ) = u ( x ) − u ε ( x ). N EASY PROOF OF THE UNIQUENESS OF INFINITY HARMONIC FUNCTIONS 3
Lemma 5. If u is infinity subharmonic in Ω , then (5) S − ε u ε ( x ) − S + ε u ε ( x ) ≤ for every x ∈ Ω ε , and if v is infinity superharmonic in Ω , then (6) S − ε v ε ( x ) − S + ε v ε ( x ) ≥ for every x ∈ Ω ε . Proof.
We first prove (5). Fix a point x ∈ Ω ε . Select y ∈ ¯ B ( x , ε ) and z ∈ ¯ B ( x , ε ) such that u ( y ) = u ε ( x ) and u ( z ) = u ε ( x ). We have ε ( S − ε u ε ( x ) − S + ε u ε ( x )) = 2 u ε ( x ) − ( u ε ) ε ( x ) − ( u ε ) ε ( x ) ≤ u ε ( x ) − u ε ( x ) − u ( x )= 2 u ( y ) − u ( z ) − u ( x ) . A simple calculation verifies that the inequality u ( w ) ≤ u ( x ) + u ( z ) − u ( x )2 ε | w − x | holds for all w ∈ ∂ ( B ( x , ε ) \ { x } ). Since u enjoys comparisons with cones fromabove, the inequality therefore holds for every w ∈ B ( x , ε ) \ { x } , and thus forevery w ∈ ¯ B ( x , ε ). Substituting w = y and using the fact that | y − x | ≤ ε , weobtain 2 u ( y ) − u ( x ) − u ( z ) ≤
0. This completes the proof of (5).To obtain (6), apply (5) to − v and simplify, using the fact that ( − v ) ε = − v ε . (cid:3) Jensen’s Theorem ([8]) . Assume that u, v ∈ C ( ¯Ω) are infinity subharmonic andsuperharmonic, respectively. Then (7) max ¯Ω ( u − v ) = max ∂ Ω ( u − v ) . Proof.
According to Lemmas 4 and 5,sup Ω ε ( u ε − v ε ) = sup Ω ε \ Ω ε ( u ε − v ε )for every ε >
0. We pass to the limit ε → (cid:3) Further remarks.
M. Crandall [5] has pointed out that our arguments apply,nearly verbatim, if we replace the Euclidean norm in the definition of cone func-tion with a general norm. In this case, the PDE (1) is replaced by the notion ofan absolutely minimizing Lipschitz extension (we refer to [2] for the definition andbackground). We thereby obtain a new proof of [11, Theorem 1.4] for boundeddomains, which establishes the uniqueness of absolutely minimizing Lipschitz ex-tensions in this general setting. The hypotheses of Lemma 4 can be further relaxedto metric spaces with mild structural conditions.Our use of the functions u ε and u ε appears more natural when we recall thata function u ∈ C (Ω) is infinity subharmonic if and only if the map ε → u ε ( x ) isconvex for every x ∈ Ω (see [6, Lemma 4.1]). Crandall [5] has shown that our proofof Jensen’s theorem can be efficiently presented by using this equivalence in lieu ofTheorem 3.Generalized versions of Lemmas 4 and 5 appear in [1], wherein the idea of passingfrom a PDE to a finite difference inequality by maxing over ε -balls is used toobtain new results for the (1-homogeneous) infinity Laplace equation with a nonzerofunction f on the right-hand side. The finite difference equation S + ε u = S − ε u hasbeen previously considered on finite graphs by Oberman [10] and Le Gruyer [9], SCOTT N. ARMSTRONG AND CHARLES K. SMART both of whom proved existence and uniqueness of solutions to the discretized finitedifference equation (with given boundary data).The authors warmly thank Michael Crandall for his many valuable suggestionsand remarks. This short article was also improved by the helpful comments ofStephanie Somersille, Kelli Talaska, and Yifeng Yu.
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