STRONGLY NONLINEAR POTENTIAL THEORY ON METRIC SPACES

We define Orlicz-Sobolev spaces on an arbitrary metric space with a Borel regular outer measure, and we develop a capacity theory based on these spaces. We study basic properties of capacity and several convergence results. We prove that each Orlicz-Sobolev function has a quasi-continuous representative. We give estimates for the capacity of balls when the measure is doubling. Under additional regularity assumption on the measure, we establish some relations between capacity and Hausdorff measures.


Introduction
The introduction and the extensive study of Sobolev spaces on arbitrary metric spaces by Franchi et al. [9], Hajłasz [12], Hajłasz and Koskela [13], Hajłasz and Martio [14], and others, have given a great impulse to several developments in geometric analysis on metric measure spaces.Important examples are the substantial progress of various domains such as fractals, partial differential equa-functions by Lipschitz functions, both in Lusin and in norm sense.This generalizes a result by Hajłasz in [12] relative to the Sobolev case.
Then we develop a capacity theory based on the definition of Orlicz-Sobolev spaces on an arbitrary metric measure space.Basic properties of capacity and several convergence results are studied.Moreover, we prove that each Orlicz-Sobolev function has a quasi-continuous representative and we give estimates for the capacity of balls when the measure is doubling.Some relations between capacity and Hausdorff measures are established under additional regularity assumption on the measure.
This paper is organized as follows.In Section 2, we list the prerequisites from the Orlicz theory.Section 3 is dedicated to Lipschitz characterization of Orlicz-Sobolev spaces in the Euclidean case, to the study of Orlicz-Sobolev spaces on metric spaces and to establish an approximation theorem of Orlicz-Sobolev functions by Lipschitz functions.Section 4 is reserved to establish important properties of capacity on metric spaces.Section 5 deals with the comparison between capacity and measure in the metric space, and between capacity and the Hausdorff measure.
The ∆ 2 condition for Φ can be formulated in the following equivalent way: for every C > 0 there exists C > 0 such that Φ(Ct) ≤ C Φ(t) for all t ≥ 0.
The Orlicz space L Φ,µ (X) is a Banach space with the following norm, called the Luxemburg norm, Let Ω be an open set in R N , C ∞ (Ω) be the space of functions which, together with all their partial derivatives of any order, are continuous on Ω, and The space C k (Ω) stands for the space of functions having all derivatives of order ≤ k continuous on Ω, and C(Ω) is the space of continuous functions on Ω.
The (weak) partial derivative of f of order |β| is denoted by Let Φ be an N-function and m ∈ N. We say that a function f : R N → R has a distributional (weak partial) derivative of order m, denoted by Let Ω be an open set in R N and denote L Φ,L (Ω) by L Φ (Ω).The Orlicz-Sobolev space W m L Φ (Ω) is the space of real functions f , such that f and its distributional derivatives up to the order m, are in L Φ (Ω).
The space W m L Φ (Ω) is a Banach space equipped with the norm where ).For more details on the theory of Orlicz spaces, see [1,18,19,20,21].

Orlicz-Sobolev spaces on metric spaces
3.1.The Euclidean case.We begin by two lemmas which lead to a Lipschitz characterization of Orlicz-Sobolev spaces.
where f Q = 1/µ(Q) Q f dµ, and the constant C depends only on N and Q.
Proof.It is enough to establish (3.1) for C 1 (Q).But in this case the proof can be found in [11,Lemma 7.16].
We omit the proof of the following lemma (Hedberg's inequality) since it is exactly the same as the one in [15] or in [23,Lemma 2.8.3]. ) Proof.Let Q be a cube in R N and x, y ∈ Q.Then we can find a subcube Then by (3.1) and (3.2) we get where ᏹ(h)(x) = sup 0<r (1/|B(x,r)|) B(x,r) |h(y)| dy is the Hardy-Littlewood maximal function.Since Φ * verifies the ∆ 2 condition, by [10] the maximal operator is bounded in L Φ (R N ).Hence, there is a nonnegative function a.e.(3.5)

Noureddine Aïssaoui 361
For the reverse implication, it suffices to show that, due to Riesz representation and Radon-Nikodym theorem, there is a nonnegative function for all φ ∈ C ∞ 0 .Now, integrating (3.3) twice over a ball B(x,ε), we get Since (∂ψ ε /∂x i )(x)dx = 0, we get Hence Now by [10], ᏹ(g) ∈ L Φ (R N ) since Φ * satisfies the ∆ 2 condition.The proof is complete.

The metric case.
Let (X,d) be a metric space and let µ be a nonnegative Borel regular outer measure on X.The triplet (X,d,µ) will be fixed in the sequel and will be denoted by X.
Let u : X → [−∞,+∞] be a µ-measurable function defined on X.We denote by D(u) the set of all µ-measurable functions g : for every x, y ∈ X \ F, x = y, with µ(F) = 0.The set F is called the exceptional set for g.
Note that the right-hand side of (3.12) is always defined for x = y.For the points x, y ∈ X, x = y such that the left-hand side of (3.12) is undefined, we may assume that the left-hand side is +∞.
Let Φ be an N-function.The Dirichlet-Orlicz space L 1 Φ,µ (X) is the space of all µ-measurable functions u such that D(u) ∩ L Φ,µ (X) = ∅.This space is equipped with the seminorm Proof.The proof is a simple verification.
Proof.There are two subsequences of (u i ) i and (g i ) i , which we denote again by (u i ) i and (g i ) i , such that u i → u and g i → g, µ-a.e.Let, for i = 1,2,..., F i be the exceptional set for g i and let G be a set of measure zero such that u i → u and g i → g on c G. We set ), for all x, y ∈ X \ F. This implies that g ∈ D(u).The proof is complete.
X) is complete for the norm defined by (3.14).Let (u i ) i be an arbitrary Cauchy sequence in M 1 Φ,µ (X).Taking if necessary a subsequence, we may assume that From the identity u k = u 1 − k−1 j=1 v j and Lemma 3.4, it follows that g k ∈ D(u), for all k ∈ N. Now (u k ) k and (g k ) k are Cauchy sequences in L Φ,µ (X).Thus there are limit functions u = limu k in L Φ,µ (X) and g = limg k in L Φ,µ (X).Lemma 3.5 implies that g ∈ D(u) and therefore the sequence (u k ) k has a limit in M 1 Φ,µ (X).The proof is complete.
The previous results lead to the following characterization of M 1 Φ,µ (X).
Proof.The proof is an immediate consequence of Lemma 3.5 and Theorem 3.6.
X). Proof.We prove the case (i) only, the proof of (ii) is similar.Let g = max(g 1 ,g 2 ) and suppose that F 1 and F 2 are the exceptional sets for u 1 and u 2 in (3.12), respectively.It is evident that u,g ∈ L Φ,µ (X).It remains to show that g ∈ D(u).
By the same manner we obtain, for x, y ∈ X U, u(x) − u(y) ≤ d(x, y) g 2 (x) + g 2 (y) . (3.16) For the remaining cases, let x ∈ U and y ∈ X U. If u 1 (x) ≥ u 2 (y), then (3.18)The case x ∈ X U and y ∈ U follows by symmetry.Hence, for all x, y ∈ Next, we prove the following important Poincaré inequality for Orlicz-Sobolev functions.Proposition 3.9.Let Φ be an N-function. where By the Jensen inequality (3.21) On the other hand, using the definition of D(u) and the convexity of Φ, we get The proof is complete.Now, we prove an approximation theorem of Orlicz-Sobolev functions by Lipschitz functions, both in Lusin sense and in norm.This generalizes a result in [12] relative to the Sobolev case.

Capacity on metric spaces
where Functions belonging to B(E) are called admissible functions for E. Remark 4.2.In the definition of C Φ,µ (E), we can restrict ourselves to those admissible functions u such that 0 ≤ u ≤ 1.
We define a capacity as an increasing positive set function C given on a σadditive class of sets Γ, which contains compact sets and such that C(∅) = 0 and Proof.It is obvious that C Φ,µ (∅) = 0 and that C Φ,µ is increasing.For countable subadditivity, let E i , i = 1,2,... be subsets of X and let ε > 0. We may assume that We show that v = sup i u i is admissible for ∞ i=1 E i and g = sup i g ui ∈ D(v) ∩ L Φ,µ (X).
Observe that v,g ∈ L Φ,µ (X).Define v k = max 1≤i≤k u i .By Lemma 3.8 the function It remains to prove that C Φ,µ is outer, that is, Since ε is arbitrary, we obtain the claim, and the proof is complete.
Corollary 4.4.Let Φ be an N-function.Let (K i ) i≥1 be a decreasing sequence of compact sets in X and let Proof.This is a direct consequence of the fact that C Φ,µ is an outer capacity.
Theorem 4.5.Let Φ be a uniformly convex N-function such that Φ satisfies the Proof.The hypothesis implies that L Φ,µ (X) is uniformly convex.By monotonicity, To prove the reverse inequality, we may assume that lim i→∞ C Φ,µ (O i ) < ∞.Let ε > 0 and for i = 1,2,..., u i ∈ B(O i ), and g ui ∈ D(u i ) ∩ L Φ,µ (X) be such that The sequence (u i ) i is bounded in L Φ,µ (X) and, hence, it possesses a weakly convergent subsequence, which we denote again by (u i ) i .The sequence (g ui ) i is also bounded in L Φ,µ (X) and, by passing to a subsequence, we may assume that u i → u weakly in L Φ,µ (X) and g ui → g weakly in L Φ,µ (X).We use the Banach-Saks theorem to deduce that the sequence defined by v j = j −1 j i=1 u i converges to u in L Φ,µ (X) and g vj = j −1 j i=1 g ui converges to g in L Φ,µ (X).Now there is a subsequence of the sequence (v j ) j so that v j → u µ-a.e. and g vj → g µ-a.e.By Lemma 3.7, u ∈ M 1 Φ,µ (X).On the other hand, v j → 1 µ-a.e. in O and hence u ≥ 1 µ-a.e. in O. Thus u ∈ B(O).By the weak lower semicontinuity of norms, we get By letting ε → 0, we obtain the result.
The set function C Φ,µ is called the Φ-capacity.If a statement holds except on a set E where C Φ,µ (E) = 0, then we say that the statement holds Φ-quasieverywhere (abbreviated Φ-q.e.).Lemma 4.6.Let Φ be any N-function and let u be a function in Proof.If n is any positive integer, the function u n = n −1 inf(u,n) is an admissible function for the Φ-capacity of the set PL u .It is easily seen that Theorem 4.7.Let Φ be an N-function and let E be any subset of X.Then C Φ,µ (E) = 0 if and only if for any ε > 0, there exists a nonnegative function and let ε > 0.Then, by the definition of the Φ-capacity, there is a sequence of nonnegative functions, (u n ) n , such that u n = 1 in some neighborhood of E and The converse implication is an immediate consequence of Lemma 4.6.
Theorem 4.8.Let Φ be an N-function and (u i ) i be a Cauchy sequence of functions in M 1 Φ,µ (X) ∩ C(X).Then, there is a subsequence (u i ) i of (u i ) i which converges pointwise Φ-q.e. in X.Moreover, the convergence is uniform outside a set of arbitrary small Φ-capacity.
We set .., and Y j = ∞ i= j X i .Since the functions u i are continuous by hypothesis, X i and Y j are open.Then, for all i, By subadditivity of C Φ,µ , we obtain From the convergence of the sum (4.12) we conclude that Noureddine Aïssaoui 369 This implies that (u i ) i converges uniformly in X Y j .This completes the proof.
and the restriction of u to X E is continuous.Since C Φ,µ is an outer capacity, we may assume that E is open.
Theorem 4.10.Let Φ be an N-function satisfying the ∆ 2 condition and u ∈ M 1 Φ,µ (X).Then there is a function v ∈ M 1 Φ,µ (X) such that u = v µ-a.e. and v is Φ-quasi-continuous in X.The function v is called a Φ-quasi-continuous representative of u.
Proof.We know that M 1 Φ,µ (X) is a Banach space and by Theorem 3.10, C(X) ∩ M 1 Φ,µ (X) is a dense subspace of M 1 Φ,µ (X).Hence, completeness implies that M 1 Φ,µ (X) can be characterized as the completion of C(X) ∩ M 1 Φ,µ (X) in the norm of M 1 Φ,µ (X).Thus there are sequences of functions (u i ) i ⊂ C(X) ∩ L Φ,µ (X) and (g i ) i ⊂ D(u i − u) such that u i → u and g i → 0 in L Φ,µ (X).Hence for a subsequence, which we denote again by (u i ) i , u i → u µ-a.e.From Theorem 4.8, we deduce that the limit function v of the sequence (u i ) i is Φ-quasi-continuous in X.This completes the proof.

Comparison between capacity and measures
5.1.Comparison between capacity and the measure µ.The sets of zero capacity are exceptional sets in the strongly nonlinear potential theory.We show in the next lemma that sets of vanishing capacity are also of measure zero.
Proof.(1) If n is a strictly positive integer, we can find u n ∈ B(E) such that (5.1) By the inequality nΦ(u n ) ≤ Φ(nu n ) and by the fact that u n ≥ 1 in E, we get Φ(1)µ(E) ≤ 1/n.This implies the claim.
(2) Let u ∈ B(E), then there is an open set O such that We obtain the result by taking the infimum over all u ∈ B(E).

Comparison between capacity and Hausdorff measures.
We recall the definition of Hausdorff measures.Let h : [0,+∞[→ [0,+∞[ be an increasing function such that lim r→0 h(r) = 0.For 0 < δ ≤ ∞ and E ⊂ X, we define B x i ,r i , r i ≤ δ . (5.21) The h-Hausdorff measure of E is given by A measure µ is regular with dimension s > 0, if there is c ≥ 1 such that, for each x ∈ X and 0 < r ≤ diam(X), c −1 r s ≤ µ B(x,r) ≤ cr s . (5.23) Note that if µ is regular with dimension s, then µ is doubling.Moreover, X has Hausdorff dimension s and there is a constant c > 0 such that, for every E ⊂ X, c −1 H s (E) ≤ µ(E) ≤ cH s (E).where the constant c depends only on Φ and the doubling constant.In particular, if H h (E) = 0, then C Φ,µ (E) = 0.

22 )
If h(t) = t s for 0 ≤ s < ∞, we obtain the s-dimensional Hausdorff measure which we denote by H s .For more details on Hausdorff measures, one can consult [8, Chapter 2, Section 10].
Theorem 5.3.Let Φ be any N-function, µ a doubling measure with the doubling constant C d , x 0 ∈ X, and 0