The obstacle and Dirichlet problems associated with p-harmonic functions in unbounded sets in Rn and metric spaces
aa r X i v : . [ m a t h . A P ] N ov THE OBSTACLE AND DIRICHLET PROBLEMSASSOCIATED WITH p -HARMONIC FUNCTIONSIN UNBOUNDED SETS IN R n AND METRIC SPACES
DANIEL HANSEVI
Abstract.
We study the obstacle problem for unbounded sets in a propermetric measure space supporting a ( p, p )-Poincar´e inequality. We prove thatthere exists a unique solution. We also prove that if the measure is doublingand the obstacle is continuous, then the solution is continuous, and moreover p -harmonic in the set where it does not touch the obstacle. This includes, asa special case, the solution of the Dirichlet problem for p -harmonic functionswith Sobolev type boundary data. Introduction
The classical Dirichlet problem is the problem of finding a harmonic function,that is, a solution of the Laplace equation that takes prescribed boundary values.According to Dirichlet’s principle, this is equivalent to minimizing the Dirichletenergy integral, Z Ω |∇ u | dx, among all functions u , in the domain Ω, that have the required boundary valuesand continuous partial derivatives up to the second order.A more general (nonlinear) Dirichlet problem considers the p -Laplace equation,∆ p u := div( |∇ u | p − ∇ u ) = 0 , < p < ∞ (which reduces to the Laplace equation when p = 2). Solving this problem isequivalent to the variational problem of minimizing the p -energy integral, Z Ω |∇ u | p dx, among all admissible functions u , and a minimizer/solution that is continuous issaid to be p -harmonic .The nonlinear potential theory of p -harmonic functions has been studied sincethe 1960s. Initially for R n , and later generalized to weighted R n , Riemannianmanifolds, and other settings. The interested reader may consult the monographHeinonen–Kilpel¨ainen–Martio [20] for a thorough treatment in weighted R n .It is not clear how to employ partial differential equations in a general metricmeasure space. However, by using the notion of minimal p -weak upper gradients,as substitutes for the modulus of the usual gradients, the variational approach Date : August 18, 2018.2010
Mathematics Subject Classification.
Primary: 31E05; Secondary: 31C45, 35D30, 35J20,35J25, 35J60, 47J20, 49J40, 49J52, 49Q20, 58J05, 58J32.
Key words and phrases.
Dirichlet problem, Dirichlet space, doubling measure, metric space,minimal p -weak upper gradient, Newtonian space, nonlinear, obstacle problem, p -harmonic,Poincar´e inequality, potential theory, upper gradient. becomes available. This has led to the more recent development of nonlinearpotential theory on complete metric spaces equipped with a doubling measuresupporting a p -Poincar´e inequality.In this paper, instead of just studying the Dirichlet problem for p -harmonicfunctions, we study the associated obstacle problem with a given obstacle andgiven boundary values. We minimize the p -energy integral among admissiblefunctions lying above the obstacle ψ . This problem reduces to the Dirichletproblem when ψ ≡ −∞ . The obstacle problem has been studied for bounded sets in (weighted) R n (see, e.g., Heinonen–Kilpel¨ainen–Martio [20] and the ref-erences therein) and later also in metric spaces (see, e.g., Bj¨orn–Bj¨orn [3], [4],[5], Bj¨orn–Bj¨orn–M¨ak¨al¨ainen–Parviainen [6], Bj¨orn–Bj¨orn–Shanmugalingam [8],Eleuteri–Farnana–Kansanen–Korte [12], Farnana [13], [14], [15], [16], Kinnunen–Martio [25], Kinnunen–Shanmugalingam [26], and Shanmugalingam [31]).Suppose that Ω is a nonempty (possibly unbounded) open subset of a propermetric measure space that supports a ( p, p )-Poincar´e inequality. Furthermore,suppose that the capacity of the complement of Ω is nonzero (this is needed forthe boundary data to make sense). Let ψ be an extended real-valued function andlet f be a function in D p (Ω) (see Section 2 for definitions). In this setting, weprove Theorem 3.4, which asserts that there exists a unique (up to sets of capacityzero) solution of the K ψ,f (Ω)-obstacle problem whenever the space of admissiblefunctions is nonempty.Moreover, by adding the assumption of the measure being doubling, we obtainTheorem 4.4, which, as a special case, implies that there is a unique solution of theDirichlet problem for p -harmonic functions with boundary values in D p (Ω) takenin Sobolev sense (i.e., that the K ψ,f (Ω)-obstacle problem has a unique continuous solution whenever ψ ≡ −∞ ).To the best of the author’s knowledge, these results are new also for R n .2. Notation and preliminaries
We assume throughout the paper that ( X, M , µ, d ) is a metric measure space(which we will refer to as X ) equipped with a metric d and a measure µ such that0 < µ ( B ) < ∞ for all balls B in X (we make the convention that balls are nonempty and open).The σ -algebra M on which µ is defined is the completion of the Borel σ -algebra.We start with the assumption that 1 ≤ p < ∞ . However, in the next section(and for the rest of the paper), we will assume that 1 < p < ∞ .The measure µ is said to be doubling if there exists a constant C µ ≥ < µ (2 B ) ≤ C µ µ ( B ) < ∞ for all balls B in X . We use the notation that if B is a ball with radius r , thenthe ball with radius λr that is concentric with B is denoted by λB .The characteristic function χ E of a set E is defined by χ E ( x ) = 1 if x ∈ E and χ E ( x ) = 0 if x / ∈ E . The set E is compactly contained in A if E (the closureof E ) is a compact subset of A . We denote this by E ⋐ A . The extended realnumber system is denoted by R := [ −∞ , ∞ ]. Recall that f + = max { f, } and f − = max {− f, } , and hence that f = f + − f − and | f | = f + + f − .By a curve in X , we mean a rectifiable nonconstant continuous mapping γ from a compact interval into X . Since our curves have finite length, they may be HE OBSTACLE PROBLEM FOR UNBOUNDED SETS IN METRIC SPACES 3 parametrized by arc length, and we will always assume that this has been done.We will abuse notation and denote both the mapping and the image by γ .Unless otherwise stated, the letter C will be used to denote various positiveconstants whose exact values are unimportant and may vary with each usage.We follow Heinonen–Koskela [21], [22] in introducing upper gradients. (Heinonenand Koskela, however, called them very weak gradients.) Definition 2.1.
A Borel function g : X → [0 , ∞ ] is said to be an upper gradient of a function f : X → R whenever | f ( x ) − f ( y ) | ≤ Z γ g ds (2.1)holds for all pairs of points x, y ∈ X and every curve γ in X joining x and y . Wemake the convention that the left-hand side is infinite when at least one of theterms is.Recall that a Borel function g : X → Y is a function such that the inverse image g − ( G ) = { x ∈ X : g ( x ) ∈ G } is a Borel set for every open subset G of Y .Observe that upper gradients are not unique (if we add a nonnegative Borelfunction to an upper gradient of f , then we obtain a new upper gradient of f )and that g ≡ ∞ is an upper gradient of all functions. Note also that if g and ˜ g are upper gradients of u and ˜ u , respectively, then g − ˜ g is not in general an uppergradient of u − ˜ u . However, upper gradients are subadditive, that is, if g and ˜ g are upper gradients of u and ˜ u , respectively, and α ∈ R , then | α | g and g + ˜ g areupper gradients of αu and u + ˜ u , respectively.A drawback of upper gradients is that they are not preserved by L p -convergence.Fortunately, it is possible to overcome this problem by relaxing the conditions.Therefore, we define the p -modulus of a curve family, and then follow Koskela–MacManus [27] in introducing p -weak upper gradients. Definition 2.2.
Let Γ be a family of curves in X . The p -modulus of Γ isMod p (Γ) := inf ρ Z X ρ p dµ, where the infimum is taken over all nonnegative Borel functions ρ such that Z γ ρ ds ≥ γ ∈ Γ . Whenever a property holds for all curves except for a curve family of zero p -modulus, it is said to hold for p -almost every ( p -a.e. ) curve.The p -modulus (as the module of order p of a system of measures) was definedand studied by Fuglede [17]. Heinonen–Koskela [22] defined the p -modulus of acurve family in a metric measure space and observed that the corresponding resultsby Fuglede carried over directly.The p -modulus has the following properties (as observed in [22]): Mod p ( ∅ ) = 0,Mod p (Γ ) ≤ Mod p (Γ ) whenever Γ ⊂ Γ , and Mod p (cid:0)S ∞ j =1 Γ j (cid:1) ≤ P ∞ j =1 Mod p (Γ j ).If Γ and Γ are two curve families such that every curve γ ∈ Γ has a subcurve γ ∈ Γ , then Mod p (Γ) ≤ Mod p (Γ ). For proofs of these properties and all otherresults in this section, we refer to Bj¨orn–Bj¨orn [4]. (Some of the references thatwe mention below may not provide a proof in the generality considered here, butsuch proofs are given in [4].) DANIEL HANSEVI
Definition 2.3.
A measurable function g : X → [0 , ∞ ] is said to be a p -weak uppergradient of a function f : X → R if (2.1) holds for all pairs of points x, y ∈ X and p -a.e. curve γ in X joining x and y .Note that a p -weak upper gradient, as opposed to an upper gradient, is not required to be a Borel function. It is convenient to demand upper gradients to beBorel functions, since then the concept of upper gradients becomes independentof the measure, and all considered curve integrals will be defined. The situationis a bit different for p -weak upper gradients, as the curve integrals need only bedefined for p -a.e. curve, and therefore, it is in fact enough to require that p -weakupper gradients are measurable functions. There is no disadvantage in assumingonly measurability, since the concept of p -weak upper gradients would depend onthe measure anyway (as the p -modulus depends on the measure). The advantageis that some results become more appealing (see, e.g., Bj¨orn–Bj¨orn [4]).Since the p -modulus is subadditive, it follows that p -weak upper gradients sharethe subadditivity property with upper gradients. Definition 2.4.
The
Dirichlet space on X , denoted by D p ( X ), is the space ofall extended real-valued functions on X that are everywhere defined, measurable,and have upper gradients in L p ( X ).If E is a measurable set, then we can consider E to be a metric space in its ownright (with the restriction of d and µ to E ). Thus the Dirichlet space D p ( E ) isalso given by Definition 2.4. Note, however, that the collection of upper gradientswith respect to E can differ from those with respect to X (unless E is open).The local Dirichlet space is defined analogously to the local space L p loc ( X ).Thus we say that a function f on X belongs to D p loc ( X ) if for every x ∈ X thereis a ball B such that x ∈ B and f ∈ D p ( B ).Lemma 2.4 in Koskela–MacManus [27] asserts that if g is a p -weak upper gra-dient of a function f , then for all q such that 1 ≤ q ≤ p , there is a decreasingsequence { g j } ∞ j =1 of upper gradients of f such that k g j − g k L q ( X ) → j → ∞ .This implies that a measurable function belongs to D p ( X ) whenever it (merely)has a p -weak upper gradient in L p ( X ).If u belongs to D p ( X ), then u has a minimal p -weak upper gradient g u ∈ L p ( X ).It is minimal in the sense that g u ≤ g a.e. for all p -weak upper gradients g of u .This was proved for p > p ≥ p -weak upper gradients g u are true substitutes for |∇ u | in metric spaces.One of the important properties of minimal p -weak gradients is that they arelocal in the sense that if two functions u, v ∈ D p ( X ) coincide on a set E , then g u = g v a.e. on E . Moreover, if U = { x ∈ X : u ( x ) > v ( x ) } , then g u χ U + g v χ X \ U is a minimal p -weak upper gradient of max { u, v } , and g v χ U + g u χ X \ U is a minimal p -weak upper gradients of min { u, v } . These results are from Bj¨orn–Bj¨orn [2].It is well-known that the restriction of a minimal p -weak upper gradient toan open subset remains minimal with respect to that subset. As a consequence,the results above about minimal p -weak upper gradients extend to functions in D p loc ( X ) having minimal p -weak upper gradients in L p loc ( X ).With the help of p -weak upper gradients, it is possible to define a type ofSobolev space on the metric space X . This was done by Shanmugalingam [30].We will, however, use a slightly different (semi)norm. The reason for this is thatwhen we define the capacity in Definition 2.6, it will be subadditive. HE OBSTACLE PROBLEM FOR UNBOUNDED SETS IN METRIC SPACES 5
Definition 2.5.
The
Newtonian space on X is N ,p ( X ) := { u ∈ D p loc ( X ) : k u k N ,p ( X ) < ∞} , where k · k N ,p ( X ) is the seminorm defined by k u k N ,p ( X ) = (cid:18)Z X | u | p dµ + Z X g pu dµ (cid:19) /p . We emphasize the fact that our Newtonian functions are defined everywhere,and not just up to equivalence classes of functions that agree almost everywhere.This is essential for the notion of upper gradients to make sense.The associated normed space defined by e N ,p ( X ) = N ,p ( X ) / ∼ , where u ∼ v if and only if k u − v k N ,p ( X ) = 0, is a Banach space (see Shanmugalingam [30]).Note that some authors denote the space of the everywhere defined functions by e N ,p ( X ), and then define the Newtonian space, which they denote by N ,p ( X ),to be the corresponding space of equivalence classes.The local space N ,p loc ( X ) and the space N ,p ( E ) when E is a measurable setare defined analogously to the Dirichlet spaces.Recall that a metric space is said to be proper if all bounded closed subsets arecompact. In particular, this is true if it is complete and the measure is doubling.If X is proper and Ω is an open subset of X , then f ∈ L p loc (Ω) if and only if f ∈ L p (Ω ′ ) for all open Ω ′ ⋐ Ω. This is the case also for D p loc and N ,p loc .Various definitions of capacities for sets can be found in the literature (see,e.g., Kinnunen–Martio [24] and Shanmugalingam [30]). We will use the followingdefinition. Definition 2.6.
The (
Sobolev ) capacity of a subset E of X is C p ( E ) := inf u k u k pN ,p ( X ) , where the infimum is taken over all u ∈ N ,p ( X ) such that u ≥ E .Whenever a property holds for all points except for points in a set of capacityzero, it is said to hold quasieverywhere ( q.e. ). Note that we follow the custom ofrefraining from making the dependence on p explicit here.Trivially, we have C p ( ∅ ) = 0, and C p ( E ) ≤ C p ( E ) whenever E ⊂ E . Fur-thermore, the proof in Kinnunen–Martio [24] for capacities for Haj lasz–Sobolevspaces on metric spaces can easily be modified to show that C p is countably sub-additive, that is, C p (cid:0)S ∞ j =1 E j (cid:1) ≤ P ∞ j =1 C p ( E j ). Thus C p is an outer measure.Note that C p is finer than µ in the sense that the capacity of a set may be positiveeven when the measure of the same set equals zero.Shanmugalingam [30] showed that if two Newtonian functions are equal almosteverywhere, then they are in fact equal quasieverywhere. This result extends tofunctions in D p loc ( X ).When E is a subset of X , we let Γ E denote the family of all curves in X that intersect E . Lemma 3.6 in Shanmugalingam [30] asserts that Mod p (Γ E ) = 0whenever C p ( E ) = 0. This implies that two functions have the same set of p -weakupper gradients whenever they are equal quasieverywhere.In order to be able to compare boundary values of Dirichlet functions (andNewtonian functions), we introduce the following spaces. DANIEL HANSEVI
Definition 2.7.
The
Dirichlet space with zero boundary values in A \ E , for subsets E and A of X , where A is measurable, is D p ( E ; A ) := { f | E ∩ A : f ∈ D p ( A ) and f = 0 in A \ E } . The
Newtonian space with zero boundary values in A \ E , denoted by N ,p ( E ; A ),is defined analogously.We let D p ( E ) and N ,p ( E ) denote D p ( E ; X ) and N ,p ( E ; X ), respectively.The assumption “ f = 0 in A \ E ” can in fact be replaced by “ f = 0 q.e. in A \ E ” without changing the obtained spaces.It is easy to verify that the function spaces that we have introduced are vectorspaces and lattices. This means that if u, v ∈ D p ( X ) and a, b ∈ R , then we have au + bv, max { u, v } , min { u, v } ∈ D p ( X ), and furthermore, as a direct consequence,we also have u + , u − , | u | ∈ D p ( X ).The following lemma is useful for asserting that certain functions belong to aDirichlet space with zero boundary values. Lemma 2.8.
Suppose that E is a measurable subset of X and that u ∈ D p ( E ) .If there exist two functions u and u in D p ( E ) such that u ≤ u ≤ u q.e. in E , then u ∈ D p ( E ) . This was proved for Newtonian functions in open sets in Bj¨orn–Bj¨orn [3], andwith trivial modifications, it provides a proof for our version of the lemma. Forthe reader’s convenience, we give the proof here.
Proof.
Let v and v be functions in D p ( X ) such that v | E = u , v | E = u , and v = v = 0 outside E , and let g ∈ L p ( X ) and g ∈ L p ( X ) be upper gradients of v and v , respectively. Let g ∈ L p ( E ) be an upper gradient of u and define v = ( u in E, X \ E and ˜ g = ( g + g + g in E,g + g in X \ E. To complete the proof, it suffices to show that ˜ g ∈ L p ( X ) is a p -weak uppergradient of v .Let E ′ be a subset of E with C p ( E ′ ) = 0 and such that u ≤ u ≤ u in E \ E ′ .Let γ be an arbitrary curve in X \ E ′ with endpoints x and y . Then Mod p (Γ E ′ ) = 0,so the following argument asserts that ˜ g is a p -weak upper gradient of v .If γ ⊂ E \ E ′ , then | v ( x ) − v ( y ) | = | u ( x ) − u ( y ) | ≤ Z γ g ds ≤ Z γ ˜ g ds. On the other hand, if x, y ∈ X \ E , then | v ( x ) − v ( y ) | = 0 ≤ Z γ ˜ g ds. Hence, by splitting γ into two parts, and possibly reversing the direction, we mayassume that x ∈ E \ E ′ and y ∈ X \ E . Then it follows that | v ( x ) − v ( y ) | = | u ( x ) | ≤ | v ( x ) | + | v ( x ) | = | v ( x ) − v ( y ) | + | v ( x ) − v ( y ) |≤ Z γ g ds + Z γ g ds ≤ Z γ ˜ g ds. (cid:3) Proposition 2.9.
Let Ω be an open subset of X . Then D p (Ω) = D p (Ω; Ω) . HE OBSTACLE PROBLEM FOR UNBOUNDED SETS IN METRIC SPACES 7
The proof is very similar to the proof of Lemma 2.8 (see, e.g., Proposition 2.39in Bj¨orn–Bj¨orn [4] for a corresponding proof for Newtonian functions).The next two results from Bj¨orn–Bj¨orn–Parviainen [7] (Lemma 3.2 and Corol-lary 3.3), following from Mazur’s lemma (see, e.g., Theorem 3.12 in Rudin [29]),will play a major role in the existence proof for the obstacle problem.
Lemma 2.10.
Assume that < p < ∞ . Assume further that g j is a p -weakupper gradient of u j , j = 1 , , ... , and that { u j } ∞ j =1 and { g j } ∞ j =1 are bounded in L p ( X ) . Then there exist functions u and g , both in L p ( X ), convex combinations v j = P N j i = j a j,i u i with p -weak upper gradients ˜ g j = P N j i = j a j,i g i , j = 1 , , ... , and asubsequence { u j k } ∞ k =1 , such that (a) both u j k → u and g j k → g weakly in L p ( X ) as k → ∞ ;(b) both v j → u and ˜ g j → g in L p ( X ) as j → ∞ ;(c) v j → u q.e. as j → ∞ ;(d) g is a p -weak upper gradient of u . Recall that α v + ··· + α n v n is said to be a convex combination of v , ... , v n whenever α k ≥ k = 1 , ... , n and α + ··· + α n = 1. Corollary 2.11.
Assume that < p < ∞ . Assume also that { u j } ∞ j =1 is boundedin N ,p ( X ) and that u j → u q.e. on X as j → ∞ . Then u ∈ N ,p ( X ) and Z X g pu dµ ≤ lim inf j →∞ Z X g pu j dµ. In general, the upper gradients of a function give no control over the function.This is obviously so when there are no curves. Requiring a Poincar´e inequalityto hold is one possibility of gaining such a control by making sure that there areenough curves connecting any two points.
Definition 2.12.
Let q ≥
1. We say that X supports a ( q, p )- Poincar´e inequality (or that X is a ( q, p )-Poincar´e space) if there exist constants C PI > λ ≥ B in X , all integrable functions u on X ,and all upper gradients g of u , it is true that (cid:18)Z B | u − u B | q dµ (cid:19) /q ≤ C PI diam( B ) (cid:18)Z λB g p dµ (cid:19) /p , where u B := Z B u dµ := 1 µ ( B ) Z B u dµ. For short, we say p -Poincar´e inequality instead of (1 , p )-Poincar´e inequality, andif X supports a p -Poincar´e inequality, we say that X is a p -Poincar´e space.By using H¨older’s inequality, one can show that if X supports a ( q, p )-Poincar´einequality, then X supports a (˜ q, ˜ p )-Poincar´e inequality for all ˜ q ≤ q and ˜ p ≥ p .From the next section on, we will assume X to support a ( p, p )-Poincar´e inequality.Then we have the following useful assertion that implies that a function can becontrolled by its minimal p -weak upper gradient. This was proved for Euclideanspaces by Maz ′ ya (see, e.g., [28]), and later J. Bj¨orn [9] observed that the proofgoes through also for metric spaces. The following version is from Bj¨orn–Bj¨orn [4](Theorem 5.53). DANIEL HANSEVI
Theorem 2.13 (Maz ′ ya’s inequality.) . Suppose that X supports a ( p, p ) -Poincar´einequality. Then there exists a constant C MI > such that if B is a ball in X , u ∈ N ,p loc ( X ), and S = { x ∈ X : u ( x ) = 0 } , then Z B | u | p dµ ≤ C MI (diam ( B ) p + 1) µ (2 B ) C p ( B ∩ S ) Z λB g pu dµ, where λ is the dilation constant in the ( p, p ) -Poincar´e inequality. The following result from Bj¨orn–Bj¨orn [4] (Proposition 4.14) is also a usefulconsequence of the ( p, p )-Poincar´e inequality.
Proposition 2.14.
Suppose that X supports a ( p, p ) -Poincar´e inequality. Let Ω be an open subset of X . Then D p loc (Ω) = N ,p loc (Ω) . The obstacle problem
In this section , we assume that < p < ∞ , that X is proper and supports a ( p, p ) -Poincar´e inequality with dilation constant λ , and that Ω is a nonempty opensubset of X such that C p ( X \ Ω) > . Kinnunen–Martio [25] defined an obstacle problem for Newtonian functionsin open sets in a complete p -Poincar´e space with a doubling measure. Theyproved that there exists a unique solution whenever the set is bounded and suchthat the complement has nonzero measure and the set of feasible solutions isnonempty (Theorem 3.2 in [25]). Shanmugalingam [30] had earlier solved theDirichlet problem (i.e., the obstacle problem with obstacle ψ ≡ −∞ ).Roughly, Kinnunen and Martio defined their obstacle as follows. Definition 3.1.
Suppose that V is a nonempty bounded open subset of X with C p ( X \ V ) >
0. Let ψ : V → R and let f ∈ N ,p ( V ). Define K B ψ,f ( V ) = { v ∈ N ,p ( V ) : v − f ∈ N ,p ( V ) and v ≥ ψ q.e. in V } . Then u is said to be a solution of the K B ψ,f ( V )- obstacle problem if u ∈ K B ψ,f ( V )and Z V g pu dµ ≤ Z V g pv dµ for all v ∈ K B ψ,f ( V ) . They required that µ ( X \ V ) > v ≥ ψ a.e. instead of q.e.This does not matter if the obstacle ψ is in D p loc ( V ), since then v ≥ ψ a.e. impliesthat v ≥ ψ q.e. This follows from Corollary 3.3 in Shanmugalingam [30]; see alsoCorollary 1.60 in Bj¨orn–Bj¨orn [4]. However, the distinction may be important.For example, if K is a compact subset of V such that C p ( K ) > µ ( K ) = 0, thenthe solution of the K B χ K , ( V )-obstacle problem takes the value 1 on K , whereas thesolution of the corresponding obstacle problem defined by Kinnunen–Martio [25] isthe trivial solution (because their candidate solutions do not “see” this obstacle).Moreover, it is possible to have no solution of the K B ψ,f ( V )-obstacle problem whenthere is a solution of the corresponding obstacle problem defined by [25] (see, e.g.,the discussion following Definition 3.1 in Farnana [13]).See also Farnana [13], [14], [15], [16] for the double obstacle problem, andBj¨orn–Bj¨orn [5] for obstacle problems on nonopen sets.Now we define our obstacle problem (without the boundedness requirement). HE OBSTACLE PROBLEM FOR UNBOUNDED SETS IN METRIC SPACES 9
Definition 3.2.
Suppose that V is a nonempty (possibly unbounded) open subsetof X such that C p ( X \ V ) >
0. Let ψ : V → R and let f ∈ D p ( V ). Define K ψ,f ( V ) = { v ∈ D p ( V ) : v − f ∈ D p ( V ) and v ≥ ψ q.e. in V } . We say that u is a solution of the K ψ,f ( V )- obstacle problem ( with obstacle ψ andboundary values f ) if u ∈ K ψ,f ( V ) and Z V g pu dµ ≤ Z V g pv dµ for all v ∈ K ψ,f ( V ) . When V = Ω, we denote K ψ,f (Ω) by K ψ,f for short.Observe that we only define the obstacle problem for V with C p ( X \ V ) > u − f ∈ D p ( V ) becomes empty when C p ( X \ V ) = 0,since then we have D p ( V ) = D p ( V ).Note also that we solve the obstacle problem for boundary data f ∈ D p ( V ).Since such a function is not defined on ∂V , we do not really have boundary values,and hence the definition should be understood in a weak Sobolev sense. Remark 3.3. If V is bounded, then Proposition 2.7 in Bj¨orn–Bj¨orn [5] asserts that D p ( V ) = N ,p ( V ), and hence we have K ψ,f ( V ) = K B ψ,f ( V ). Thus Definition 3.2 isa generalization of Definition 3.1 to Dirichlet functions and to unbounded sets.The main result in this paper shows that the K ψ,f -obstacle problem has aunique solution under the natural condition of K ψ,f being nonempty. Theorem 3.4.
Let ψ : Ω → R and let f ∈ D p (Ω) . Then there exists a unique ( upto sets of capacity zero ) solution of the K ψ,f -obstacle problem whenever K ψ,f isnonempty. The assumption that X is proper is needed only in the end of the existence partof the proof.In the uniqueness part of the proof, we use the fact that L p (Ω) is strictly convex.Clarkson [11] introduced the notions of strict convexity and uniform convexity (thelatter being a stronger condition), and proved that all L p -spaces, 1 < p < ∞ , areuniformly convex. A Banach space Y (with norm k · k ) is strictly convex if x = cy for some constant c > x and y are nonzero and k x + y k = k x k + k y k .In particular, x = y whenever k x k = k y k = (cid:13)(cid:13) ( x + y ) (cid:13)(cid:13) = 1.The idea used in the uniqueness part of the proof comes from Cheeger [10]. Proof. (Existence.) We start by choosing a ball B ⊂ X such that C p ( B \ Ω) > B ∩ Ω is nonempty. Clearly, we have B ⊂ B ⊂ B ⊂ ··· ⊂ X = S ∞ t =1 tB .Let I = inf v Z Ω g pv dµ, with the infimum taken over all v ∈ K ψ,f . Then 0 ≤ I < ∞ as K ψ,f is nonempty.Let { u j } ∞ j =1 ⊂ K ψ,f be a minimizing sequence such that I j := Z Ω g pu j dµ ց I as j → ∞ . Let w j ∈ D p ( X ) be such that w j = u j − f in Ω and w j = 0 outside Ω, j = 1 , , ... .We claim that both { w j } ∞ j =1 and { g w j } ∞ j =1 are bounded in L p ( tB ) for all t ≥ g w j ≤ ( g u j + g f ) χ Ω a.e., and hence k g w j k L p ( X ) ≤ k g u j k L p (Ω) + k g f k L p (Ω) ≤ k g u k L p (Ω) + k g f k L p (Ω) =: C ′ < ∞ . Let t ≥ S = T ∞ j =1 { x ∈ X : w j ( x ) = 0 } . Then C p ( tB ∩ S ) ≥ C p ( tB \ Ω) ≥ C p ( B \ Ω) > . Maz ′ ya’s inequality (Theorem 2.13) asserts the existence of a constant C tB > Z tB | w j | p dµ ≤ C ptB Z λtB g pw j dµ. This implies that we also have k w j k L p ( tB ) ≤ C tB k g w j k L p ( X ) ≤ C tB C ′ =: C ′ tB < ∞ , (3.1)and the claim follows.Consider the ball B . Lemma 2.10 asserts that we can find a function ϕ ∈ L p ( B )and convex combinations ϕ ,j = N ,j X k = j a ,j,k w k in D p ( X ) , j = 1 , , ... , (3.2)such that ϕ ,j → ϕ q.e. in B as j → ∞ . Because ϕ ,j = 0 outside Ω, we musthave ϕ = 0 q.e. in B \ Ω, and hence we may choose ϕ so that ϕ = 0 in B \ Ω.Let v ,j = f + ϕ ,j | Ω . Then v ,j = f + N ,j X k = j a ,j,k w k | Ω = N ,j X k = j a ,j,k ( f + w k | Ω ) = N ,j X k = j a ,j,k u k ≥ ψ q.e. in Ω . We also have g v ,j ≤ N ,j X k = j a ,j,k g u k a.e. in Ω and g ϕ ,j ≤ N ,j X k = j a ,j,k g w k a.e.A sequence of convex combinations of functions taken from a bounded sequencemust also be bounded, and therefore we can apply Lemma 2.10 repeatedly here.Hence, for every n = 2 , , , ... , we can find a function ϕ n ∈ L p ( nB ) such that ϕ n = 0 in nB \ Ω and convex combinations ϕ n,j = N n,j X k = j a n,j,k ϕ n − ,k in D p ( X ) , j = 1 , , ... , (3.3)such that ϕ n,j → ϕ n q.e. in nB as j → ∞ . Let v n,j = f + ϕ n,j | Ω . Then v n,j = N n,j X k = j a n,j,k ( f + ϕ n − ,k | Ω ) = N n,j X k = j a n,j,k v n − ,k ≥ ψ q.e. in Ω , and also g v n,j ≤ N n,j X k = j a n,j,k g v n − ,k a.e. in Ω and g ϕ n,j ≤ N n,j X k = j a n,j,k g ϕ n − ,k a.e.Let u = f + ϕ | Ω , where ϕ is the function on X defined by ϕ ( x ) = ∞ X n =1 ϕ n ( x ) χ nB \ ( n − B ( x ) , x ∈ X. We shall now show that u is indeed a solution of the K ψ,f -obstacle problem. Todo that, we first establish that u ∈ K ψ,f , and then show that u is a minimizer. HE OBSTACLE PROBLEM FOR UNBOUNDED SETS IN METRIC SPACES 11
Because ϕ = u − f in Ω and ϕ = 0 outside Ω, it suffices to show that ϕ ∈ D p ( X )in order to establish that u − f ∈ D p (Ω) and u ∈ D p (Ω).Consider the diagonal sequences { v n,n } ∞ n =1 and { ϕ n,n } ∞ n =1 . Observe that thelatter is bounded in L p ( tB ) for t ≥
1, since k ϕ n,j k L p ( tB ) ≤ C ′ tB for all n and j , by(3.1), (3.2), and (3.3).We claim that ϕ n,n → ϕ q.e. as n → ∞ . To prove that, we start by fixing aninteger n ≥ nB . Then | ϕ n +1 − ϕ n | ≤ | ϕ n +1 − ϕ n +1 ,j | + | ϕ n +1 ,j − ϕ n |≤ | ϕ n +1 − ϕ n +1 ,j | + N n +1 ,j X k = j a n +1 ,j,k | ϕ n,k − ϕ n | → nB as j → ∞ . Thus ϕ n +1 = ϕ n q.e. in nB for n = 1 , , ... .By definition, we have ϕ = ϕ in B . Now assume that ϕ = ϕ n q.e. in nB forsome positive integer n . By definition also, we have ϕ = ϕ n +1 in ( n + 1) B \ nB ,and because ϕ n +1 = ϕ n q.e. in nB , it follows that ϕ = ϕ n +1 q.e. in ( n + 1) B .Hence, by induction, we have ϕ = ϕ n q.e. in nB for n = 1 , , ... .For n = 1 , , ... , let E n be the subset of nB where ϕ n,j → ϕ n = ϕ as j → ∞ and let E = S ∞ n =1 ( nB \ E n ). Then we have C p ( E ) ≤ P ∞ n =1 C p ( nB \ E n ) = 0.Let x ∈ X \ E . Clearly, x ∈ mB and ϕ ( x ) = ϕ m ( x ) for some positive integer m .Given ε >
0, choose a J such that j ≥ J implies that | ϕ m,j ( x ) − ϕ m ( x ) | < ε. Assume that for some n ≥ m , we have | ϕ n,j ( x ) − ϕ m ( x ) | < ε for j ≥ J . Then | ϕ n +1 ,j ( x ) − ϕ m ( x ) | ≤ N n +1 ,j X k = j a n +1 ,j,k | ϕ n,k ( x ) − ϕ m ( x ) | < ε for j ≥ J . By induction, it follows that | ϕ n,j ( x ) − ϕ m ( x ) | < ε for n ≥ m and j ≥ J , and hence, for n ≥ max { m, J } , we have | ϕ n,n ( x ) − ϕ ( x ) | = | ϕ n,n ( x ) − ϕ m ( x ) | < ε. We conclude that ϕ n,n → ϕ q.e., and also that v n,n → u q.e. in Ω, as n → ∞ .By using Jensen’s inequality, we can see that Z Ω g pv ,j dµ ≤ Z Ω (cid:18) N ,j X k = j a ,j,k g u k (cid:19) p dµ ≤ N ,j X k = j a ,j,k Z Ω g pu k dµ ≤ Z Ω g pu j dµ and Z X g pϕ ,j dµ ≤ Z X (cid:18) N ,j X k = j a ,j,k g w k (cid:19) p dµ ≤ N ,j X k = j a ,j,k Z Ω ( g u k + g f ) p dµ ≤ p N ,j X k = j a ,j,k Z Ω ( g pu k + g pf ) dµ ≤ p Z Ω ( g pu j + g pf ) dµ. Assume that for some positive integer n , it is true that Z Ω g pv n,j dµ ≤ Z Ω g pu j dµ and Z X g pϕ n,j dµ ≤ p Z Ω ( g pf + g pu j ) dµ. Then Z Ω g pv n +1 ,j dµ ≤ Z Ω (cid:18) N n +1 ,j X k = j a n +1 ,j,k g v n,k (cid:19) p dµ ≤ N n +1 ,j X k = j a n +1 ,j,k Z Ω g pv n,k dµ ≤ N n +1 ,j X k = j a n +1 ,j,k Z Ω g pu k dµ ≤ Z Ω g pu j dµ and Z X g pϕ n +1 ,j dµ ≤ Z X (cid:18) N n +1 ,j X k = j a n +1 ,j,k g ϕ n,k (cid:19) p dµ ≤ N n +1 ,j X k = j a n +1 ,j,k Z X g pϕ n,k dµ ≤ p N n +1 ,j X k = j a n +1 ,j,k Z Ω ( g pf + g pu k ) dµ ≤ p Z Ω ( g pf + g pu j ) dµ. By induction, and letting j = n , it follows that Z Ω g pv n,n dµ ≤ Z Ω g pu n dµ and Z X g pϕ n,n dµ ≤ p Z Ω ( g pf + g pu n ) dµ, n = 1 , , ... . Fix an integer m ≥
1. Since { ϕ n,n } ∞ n =1 and { g ϕ n,n } ∞ n =1 are bounded in L p ( mB )and ϕ n,n → ϕ q.e. in mB as n → ∞ , Corollary 2.11 asserts that ϕ ∈ N ,p ( mB ).This implies that ϕ ∈ D p loc ( X ). Note that g ϕ and g ϕ n,n are minimal p -weak uppergradients of ϕ and ϕ n,n , respectively, with respect to mB . Hence, by Corollary 2.11again, it follows that Z mB g pϕ dµ ≤ lim inf n →∞ Z mB g pϕ n,n dµ ≤ lim inf n →∞ Z X g pϕ n,n dµ ≤ p lim inf n →∞ Z Ω ( g pf + g pu n ) dµ = 2 p Z Ω g pf dµ + 2 p I. Letting m → ∞ yields Z X g pϕ dµ = lim m →∞ Z mB g pϕ dµ ≤ p Z Ω g pf dµ + 2 p I < ∞ , and hence ϕ ∈ D p ( X ). We conclude that u − f ∈ D p (Ω) and u ∈ D p (Ω).Let A n = { x ∈ Ω : v n,n ( x ) < ψ ( x ) } for n = 1 , , ... , and let A = S ∞ n =1 A n .Then, since v n,n → u q.e. in Ω as n → ∞ , it follows that u ≥ ψ q.e. in Ω \ A .Because v n,n ≥ ψ q.e. in Ω, we have C p ( A n ) = 0, and hence C p ( A ) = 0 bythe subadditivity of the capacity. Thus u ≥ ψ q.e. in Ω, and we conclude that u ∈ K ψ,f .Proposition 2.14 asserts that f ∈ N ,p loc (Ω), and hence f ∈ L p (Ω ′ ) for all openΩ ′ ⋐ Ω. Let Ω t = n x ∈ tB ∩ Ω : inf y ∈ ∂ Ω d ( x, y ) > δ/t o , ≤ t < ∞ , where δ > is nonempty. Then we haveΩ ⋐ Ω ⋐ ··· ⋐ Ω = S ∞ t =1 Ω t . Moreover, { v n,n } ∞ n =1 is bounded in L p (Ω t ), since k v n,n k L p (Ω t ) ≤ k ϕ n,n k L p (Ω t ) + k f k L p (Ω t ) ≤ C ′ tB + k f k L p (Ω t ) < ∞ . Fix an integer m ≥
1. Since { v n,n } ∞ n =1 and { g v n,n } ∞ n =1 are bounded in L p (Ω m ), v n,n → u q.e. in Ω m as n → ∞ , and g u and g v n,n are minimal p -weak upper HE OBSTACLE PROBLEM FOR UNBOUNDED SETS IN METRIC SPACES 13 gradients of u and v n,n , respectively, with respect to Ω m , by Corollary 2.11, itfollows that Z Ω m g pu dµ ≤ lim inf n →∞ Z Ω m g pv n,n dµ ≤ lim inf n →∞ Z Ω g pv n,n dµ ≤ lim inf n →∞ Z Ω g pu n dµ = I. Letting m → ∞ completes the existence part of the proof by showing that I ≤ Z Ω g pu dµ = lim m →∞ Z Ω m g pu dµ ≤ I. (Uniqueness.) Suppose that u ′ and u ′′ are solutions to the K ψ,f -obstacle problem.We begin this part by showing that g u ′ = g u ′′ a.e. in Ω.Clearly, ( u ′ + u ′′ ) ∈ K ψ,f , and hence k g u ′ k L p (Ω) ≤ k g ( u ′ + u ′′ ) k L p (Ω) ≤ (cid:13)(cid:13) ( g u ′ + g u ′′ ) (cid:13)(cid:13) L p (Ω) ≤ k g u ′ k L p (Ω) + k g u ′′ k L p (Ω) = k g u ′′ k L p (Ω) = k g u ′ k L p (Ω) . Thus g u ′ = g u ′′ a.e. in Ω by the strict convexity of L p (Ω).Now we show that g u ′ − u ′′ = 0 a.e. in Ω. Fix a real number c and let u = max { u ′ , min { u ′′ , c }} . The following shows that u ∈ K ψ,f . Clearly, u ∈ D p (Ω). Furthermore, we have u ≥ u ′ ≥ ψ q.e. in Ω, and u − f ∈ D p (Ω) by Lemma 2.8, since u − f ≤ max { u ′ , u ′′ } − f = max { u ′ − f, u ′′ − f } ∈ D p (Ω)and u − f ≥ u ′ − f ∈ D p (Ω).Let U c = { x ∈ Ω : u ′ ( x ) < c < u ′′ ( x ) } . Then we have g u = 0 a.e in U c , since U c ⊂ { x ∈ Ω : u ( x ) = c } . The minimizing property of g u ′ then implies that Z Ω g pu ′ dµ ≤ Z Ω g pu dµ = Z Ω \ U c g pu dµ = Z Ω \ U c g pu ′ dµ, since g u = g u ′ = g u ′′ a.e. in Ω \ U c . Hence g u ′ = g u ′′ = 0 a.e. in U c for all c ∈ R ,and because { x ∈ Ω : u ′ ( x ) < u ′′ ( x ) } ⊂ [ c ∈ Q U c , we have g u ′ = g u ′′ = 0 a.e. in { x ∈ Ω : u ′ ( x ) < u ′′ ( x ) } . Analogously, the same istrue for { x ∈ Ω : u ′ ( x ) > u ′′ ( x ) } , and hence g u ′ − u ′′ ≤ ( g u ′ + g u ′′ ) χ { x ∈ Ω: u ′ ( x ) = u ′′ ( x ) } = 0 a.e. in Ω . Since u ′ − u ′′ = u ′ − f − ( u ′′ − f ) ∈ D p (Ω), there exists w ∈ D p ( X ) such that w = u ′ − u ′′ in Ω and w = 0 outside Ω. We have g w = g u ′ − u ′′ χ Ω = 0 a.e.Let e S = { x ∈ X : w ( x ) = 0 } and let t ≥ C p ( tB ∩ e S ) ≥ C p ( tB \ Ω) ≥ C p ( B \ Ω) > . Maz ′ ya’s inequality (Theorem 2.13) applies to w , and hence there exists a constant e C tB > Z tB ∩ Ω | u ′ − u ′′ | p dµ ≤ Z tB | w | p dµ ≤ e C tB Z λtB g pw dµ = 0 . This implies that u ′ = u ′′ q.e. in tB ∩ Ω.Let V m = { x ∈ mB ∩ Ω : u ′ ( x ) = u ′′ ( x ) } , m = 1 , , ... , and let V = S ∞ m =1 V m .Then u ′ = u ′′ in Ω \ V . Since C p ( V m ) = 0 for all m , the subadditivity of thecapacity implies that C p ( V ) = 0, hence u ′ = u ′′ q.e. in Ω. We conclude that thesolution of the K ψ,f -obstacle problem is unique (up to sets of capacity zero). (cid:3) If v = u q.e. in Ω and u is a solution of the K ψ,f -obstacle problem, then so is v .Indeed, v = u q.e. implies that g u is a p -weak upper gradient of v . Thus v ∈ D p (Ω)and R Ω g pv dµ ≤ R Ω g pu dµ . Clearly, we have v ≥ ψ q.e., and since Lemma 2.8 assertsthat v − f ∈ D p (Ω), it follows that v ∈ K ψ,f .The following criterion for the existence of a unique solution is easy to prove. Proposition 3.5.
Suppose that ψ and f are in D p (Ω) . Then K ψ,f is nonemptyif and only if ( ψ − f ) + ∈ D p (Ω) .Proof. Suppose that K ψ,f is nonempty and let v ∈ K ψ,f . Since ( v − f ) + ∈ D p (Ω)and 0 ≤ ( ψ − f ) + ≤ ( v − f ) + q.e. in Ω , Lemma 2.8 asserts that ( ψ − f ) + ∈ D p (Ω).Conversely, suppose that ( ψ − f ) + ∈ D p (Ω). Let v = max { ψ, f } . Then we have v ∈ D p (Ω), v − f = ( ψ − f ) + , and v ≥ ψ in Ω. Thus v ∈ K ψ,f . (cid:3) The following comparison principle (for the version of the obstacle problemdefined in Kinnunen–Martio [25]) was obtained in Bj¨orn–Bj¨orn [3]. Their proof(with trivial modifications) is valid also for our obstacle problem.
Lemma 3.6.
Let ψ j : Ω → R and f j ∈ D p (Ω) be such that K ψ j ,f j is nonempty , and let u j be a solution of the K ψ j ,f j -obstacle problem for j = 1 , . If ψ ≤ ψ q.ein Ω and ( f − f ) + ∈ D p (Ω), then u ≤ u q.e. in Ω .Proof. Let h = u − f − u + f . Then h ∈ D p (Ω) and − ( f − f ) + − h − = − max {− ( f − f ) , } − max {− h, } = min { f − f , } + min { h, } ≤ min { f − f , h } ≤ h. Since − ( f − f ) + − h − ∈ D p (Ω), Lemma 2.8 asserts that min { f − f , h } ∈ D p (Ω).Let u = min { u , u } . Then u ∈ D p (Ω), and since u ≥ ψ ≥ ψ q.e. in Ω, weclearly have u ≥ ψ q.e. in Ω. Moreover, since u − f = u − f + h , we have u − f = min { u , u } − f = min { u − f , u − f } = min { u − f + h, u − f } = u − f + min { h, f − f } . Hence u − f ∈ D p (Ω), and we conclude that u ∈ K ψ ,f .Let v = max { u , u } . Then v ∈ D p (Ω) and v ≥ ψ q.e. in Ω. As v − f = max { u − f , u − f } = max { u − f , u − f − h } = u − f + max { f − f , − h } = u − f − min { f − f , h } , we see that v − f ∈ D p (Ω), and hence v ∈ K ψ ,f .Let E = { x ∈ Ω : u ( x ) ≤ u ( x ) } . Since u is a solution of the K ψ ,f -obstacleproblem, we have Z Ω g pu dµ ≤ Z Ω g pv dµ = Z E g pu dµ + Z Ω \ E g pu dµ, which implies that Z E g pu dµ ≤ Z E g pu dµ. By using the last inequality, we see that Z Ω g pu dµ = Z E g pu dµ + Z Ω \ E g pu dµ ≤ Z E g pu dµ + Z Ω \ E g pu dµ = Z Ω g pu dµ. HE OBSTACLE PROBLEM FOR UNBOUNDED SETS IN METRIC SPACES 15
Since u ∈ K ψ ,f and u is a solution of the K ψ ,f -obstacle problem, this inequalityimplies that also u is a solution. Theorem 3.4 asserts that u = u q.e. in Ω, andwe conclude that u ≤ u q.e. in Ω. (cid:3) The following local property of solutions of the obstacle problem can be useful.In some cases it may enable the use of results from the theory for bounded sets.In this paper, we will use it in the proof of Theorems 4.4 and 4.5.
Proposition 3.7.
Let ψ : Ω → R and f ∈ D p (Ω) be such that K ψ,f is nonempty , and let u be a solution of the K ψ,f -obstacle problem. Suppose that Ω ′ is an opensubset of Ω . Then u is a solution of the K ψ,u (Ω ′ ) -obstacle problem.Moreover , if Ω ′ ⋐ Ω, then u is a solution also of the K B ψ,u (Ω ′ ) -obstacle problem.Proof. Let Ω ′ be an open subset of Ω. Clearly, u ∈ K ψ,u (Ω ′ ). Let v ∈ K ψ,u (Ω ′ )be arbitrary. To complete the first part of the proof, it is sufficient to show that Z Ω ′ g pu dµ ≤ Z Ω ′ g pv dµ. (3.4)Let E = Ω \ Ω ′ and extend v to Ω by letting v = u in E . Since v − u ∈ D p (Ω ′ ),we have v = ( v − u ) + u ∈ D p (Ω) and v − f = ( v − u ) + ( u − f ) ∈ D p (Ω), andsince v ≥ ψ q.e. in Ω ′ and v = u ≥ ψ q.e. in E , we conclude that v ∈ K ψ,f .Because u is a solution to the K ψ,f -obstacle problem, we have Z Ω ′ g pu dµ + Z E g pu dµ = Z Ω g pu dµ ≤ Z Ω g pv dµ = Z Ω ′ g pv dµ + Z E g pv dµ. (3.5)Since u = v in E implies that g u = g v a.e. in E , we have Z E g pv dµ = Z E g pu dµ ≤ Z Ω g pu dµ < ∞ . Subtracting the integrals over E in (3.5) yields (3.4).For the second part, assume that Ω ′ ⋐ Ω and let v ∈ K B ψ,u (Ω ′ ) be arbitrary.Clearly, v ∈ K ψ,u (Ω ′ ). The first part of the proof asserts that u is a solution ofthe K ψ,u (Ω ′ )-obstacle problem and hence (3.4) holds. By Proposition 2.14, wehave u ∈ N ,p loc (Ω), and hence u ∈ N ,p (Ω ′ ). Thus u ∈ K B ψ,u (Ω ′ ) and the proof iscomplete. (cid:3) There are many equivalent definitions of (super)minimizers in the literature (seeProposition 3.2 in A. Bj¨orn [1]). The first definition for metric spaces was given byKinnunen–Martio [25]. Here we follow Bj¨orn–Bj¨orn–M¨ak¨al¨ainen–Parviainen [6].We also follow the custom of not making the dependence on p explicit in thenotation. Definition 3.8.
Let V be a nonempty open subset of X . We say that a function u ∈ N ,p loc ( V ) is a superminimizer in V if Z ϕ =0 g pu dµ ≤ Z ϕ =0 g pu + ϕ dµ (3.6)holds for all nonnegative ϕ ∈ N ,p ( V ).Furthermore, u is said to be a minimizer in V if (3.6) holds for all ϕ ∈ N ,p ( V ).According to Proposition 3.2 in A. Bj¨orn [1], it is in fact only necessary to test(3.6) with (nonnegative and all, respectively) ϕ ∈ Lip c ( V ).As a direct consequence of Proposition 3.7 together with Proposition 9.25 inBj¨orn–Bj¨orn [4], we have the following result. Proposition 3.9.
Suppose that u is a solution of the K ψ,f -obstacle problem. Then u is a superminimizer in Ω . Lsc-regularized solutions and p -harmonic solutions In this section , we make the rather standard assumptions that < p < ∞ , that X is a complete p -Poincar´e space , that µ is doubling , and that Ω is a nonemptyopen subset of X such that C p ( X \ Ω) > . When µ is doubling, it is true that X is proper if and only if X is complete,and also that X supports a ( p, p )-Poincar´e inequality if and only if X supportsa p -Poincar´e inequality (the necessity follows from H¨older’s inequality, and thesufficiency was proved in Haj lasz–Koskela [19]; see also Corollary 4.24 in Bj¨orn–Bj¨orn [4]). Thus, the difference between this section and the previous is that herewe make the assumption that µ is doubling.Note that under these assumptions, Poincar´e inequalities are self-improving inthe sense that X supports a q -Poincar´e inequality for some q < p (this was provedby Keith–Zhong [23]). Hence, in this section, we make the same assumptions asKinnunen–Martio [25], and we can therefore use Theorems 5.1 and 5.5 in [25]. Theorem 4.1.
Let ψ : Ω → R and let f ∈ D p (Ω) . Then there exists a uniquelsc-regularized solution of the K ψ,f -obstacle problem whenever K ψ,f is nonempty. The lsc-regularization of a function u is the (lower semicontinuous) function u ∗ defined by u ∗ ( x ) := ess lim inf y → x u ( y ) := lim r → ess inf B ( x,r ) u. Proof.
Suppose that K ψ,f is nonempty. Theorem 3.4 asserts that there existsa solution u of the K ψ,f -obstacle problem and that all solutions are equal to u q.e. in Ω. Proposition 3.9 asserts that u is a superminimizer in Ω, and hence byTheorem 5.1 in Kinnunen–Martio [25], we have u ∗ = u q.e. in Ω. Thus u ∗ is theunique lsc-regularized solution of the K ψ,f -obstacle problem. (cid:3) The following comparison principle improves upon Lemma 3.6.
Lemma 4.2.
Let ψ j : Ω → R and f j ∈ D p (Ω) be such that K ψ j ,f j is nonempty , and let u j be the lsc-regularized solution of the K ψ j ,f j -obstacle problem for j = 1 , .Then u ≤ u in Ω whenever ψ ≤ ψ q.e in Ω and ( f − f ) + ∈ D p (Ω) .Proof. By Lemma 3.6, we have u ≤ u q.e. in Ω, and since both u and u arelsc-regularized, it follows that u ( x ) = ess lim inf y → x u ( y ) ≤ ess lim inf y → x u ( y ) = u ( x ) for all x ∈ Ω . (cid:3) Definition 4.3.
Let V be a nonempty open subset of X . We say that a function u ∈ N ,p loc ( V ) is p -harmonic in V whenever it is a continuous minimizer in V .Kinnunen and Martio proved that the solution u (if it exists) of their obstacleproblem for bounded sets is continuous in Ω and is a minimizer in the open set { x ∈ Ω : u ( x ) > ψ ( x ) } whenever the obstacle ψ is continuous in Ω (Theorem 5.5 in[25]). This is true also for the K B ψ,f (Ω)-obstacle problem (see, e.g., Theorem 8.28in Bj¨orn–Bj¨orn [4]), and also for our obstacle problem (that allows for unboundedsets). HE OBSTACLE PROBLEM FOR UNBOUNDED SETS IN METRIC SPACES 17
Theorem 4.4.
Let ψ : Ω → [ −∞ , ∞ ) be continuous and f ∈ D p (Ω) be such that K ψ,f is nonempty. Then the lsc-regularized solution u of the K ψ,f -obstacle problemis continuous in Ω and p -harmonic in the open set A = { x ∈ Ω : u ( x ) > ψ ( x ) } . We also have the following corresponding pointwise result.
Theorem 4.5.
Let ψ : Ω → [ −∞ , ∞ ) and f ∈ D p (Ω) be such that K ψ,f isnonempty. Let x ∈ Ω . Then the lsc-regularized solution u of the K ψ,f -obstacleproblem is continuous at x if ψ is continuous at x .Proof. Let x ∈ Ω and let Ω ′ be an open set such that x ∈ Ω ′ ⋐ Ω. Let u bethe lsc-regularized solution of the K ψ,f -obstacle problem. Proposition 3.7 assertsthat u is a solution of the K B ψ,u (Ω ′ )-obstacle problem. By Theorem 8.29 in Bj¨orn–Bj¨orn [4] (which is a special case of Corollary 3.4 in Farnana [16]), it follows that u is continuous at x . (cid:3) Proof of Theorem 4.4.
The first part follows directly from Theorem 4.5.Now we prove that u is a minimizer in A . The set A is open since ψ and u arecontinuous. Choose a ball B ⊂ A and δ > A n := n x ∈ nB ∩ A : inf y ∈ ∂A d ( x, y ) > δ/n o , n = 1 , , ... , are nonempty. Then A ⋐ A ⋐ ··· ⋐ A = S ∞ n =1 A n . Fix a positive integer n .Since u is a solution of the K B ψ,u ( A n )-obstacle problem, Theorem 5.5 in Kinnunen–Martio [25] asserts that u is p -harmonic in A n . From this, it follows that u is p -harmonic in A (see, e.g., Theorem 9.36 in Bj¨orn–Bj¨orn [4]). (cid:3) Due to Theorem 4.4, the following definition makes sense.
Definition 4.6.
The p -harmonic extension H Ω f of a function f ∈ D p (Ω) to Ω isthe continuous solution of the K −∞ ,f (Ω)-obstacle problem.Then H Ω f is the unique p -harmonic function in Ω such that f − H Ω f ∈ D p (Ω).Note that Definition 4.6 is a generalization of Definition 8.31 in Bj¨orn–Bj¨orn [4]to Dirichlet functions and to unbounded sets (see Remark 3.3).We conclude that we have solved the Dirichlet problem for p -harmonic functionsin open sets with boundary values in D p (Ω) taken in Sobolev sense, and we finishthe paper by giving a short proof of the following comparision principle. Lemma 4.7.
Suppose that f and f are in D p (Ω) and that ( f − f ) + ∈ D p (Ω) .Then H Ω f ≤ H Ω f in Ω .The conclusion holds also under the assumption that f and f belong to D p (Ω) and that f ≤ f q.e. on ∂ Ω . The first part is just a special case of Lemma 4.2.
Proof.
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