# Capacities in Generalized Orlicz Spaces

Authors: Debangana Baruah; Petteri Harjulehto; Peter Hästö
Journal: Journal of Function Spaces
 (2018)

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## Abstract

In this paper basic properties of both Sobolev and relative capacities are studied in generalized Orlicz spaces. The capacities are compared with each other and the Hausdorff measure. As an application, the existence of quasicontinuous representative of generalized Orlicz functions is proved.


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## Body

## 1. Introduction

In the calculus of variations one studies existence and properties of solutions to minimization problems such as(1)minu∈W1,1∫Fx,∇udx. Classical techniques, by De Giorgi and Moser, cover both the linear case and the p-growth case, where F(t)≈tp. Marcellini [1] developed the theory of (p,q)-growth, which is based on the growth assumption tp-1≲F(x,t)≲tq+1,q>p. Zhikov [2] studied such minimizers as models of anisotropic materials and also observed that they exhibit the so-called Lavrentiev phenomenon whereby minimizers do not have improved regularity and may even be discontinuous.In the variable exponent case,F(x,t)≈tp(x), the change in the anisotropy (growth rate) is gradual owing to the continuity of p. For instance, in electrorheological fluid dynamics, where the anisotropy depends on the smooth electrical field, this is a reasonable assumption [3]. In other situations, such as composite materials, a more clear-cut transition is better. To this end, Baroni, Colombo, and Mingione [4–7] have developed a regularity theory of the double phase functional F(x,t)=tp+a(x)tq, q>p, which has the property that the growth rate changes abruptly from p to q in the sets {a=0} and {a>0} (see also [8–12]). Recently, we were able to generalize their first step, showing Hölder continuity, to the general Φ-growth case [13]. Also other results have recently been obtained for partial differential equations with generalized Orlicz growth, cf. [13–16].A different approach to differential equations is based on (nonlinear) potential theory. The foundation of nonlinear potential theory includes general notions of a Radon measure, a capacity and generalized functions. The sets of capacity zero are the exceptional sets for representatives of the function. In this paper we give basic properties of both Sobolev capacity and relative capacity in the generalized Orlicz setting. We follow the general framework of [17] and its previous adaption to the variable exponent setting [18, Chapter 10]. The results can be applied, e.g., in the study of boundary behavior of solutions to PDE.We consider generalΦ-functions which need not be convex. Many of the proofs in this paper follow a standard pattern, since they do not depend on the exact form of the integrand used in the definition of capacity. We have therefore omitted or abbreviated several proofs (e.g., Section 6). Nevertheless, it is necessary to check these basic building blocks in order to proceed with constructing the theory, since the results are not covered by earlier results, merely similar. There are also some proofs which are new, namely, Theorem 9, Example 13, and Proposition 20. Furthermore, this general setting clarifies the necessity of various assumptions for different properties (e.g., Theorem 8). In particular, we find that the relative capacity is Choquet also for nonconvex Φ-function, whereas convexity is needed for the Sobolev capacity.It should be noted that Ohno and Shimomura [19] have recently studied (Sobolev) capacity in the generalized Orlicz case. However, they consider the capacity in a metric measure space setting with Hajłasz gradients. These results therefore work in the Euclidean setting only when the maximal operator is bounded, since the Hajłasz gradient corresponds to M(∇f).The outline of the paper is as follows. We start by introducing our notation and basic definitions. Then we study Sobolev capacity and compare it with the Hausdorff measure. Next we derive existence of a quasicontinuous representative of generalized Orlicz function. Finally, we study relative capacity and compare it with the Sobolev capacity.


## 2. Preliminaries

We study spaces of functions defined inRn or opens sets Ω⊂Rn.A real-valued function isL-almost increasing, L⩾1, if Lf(s)⩾f(t) for s>t. So a 1-almost increasing function is increasing. L-almost decreasing is defined analogously.Definition 1.
We say thatφ:Ω×[0,∞)→[0,∞] is aweakΦ-function, and write φ∈Φw(Ω), if the following conditions hold: (i)
For everyt∈[0,∞) the function x↦φ(x,t) is measurable and for every x∈Ω the function t↦φ(x,t) is non-decreasing and left-continuous.(ii)
φ(x,0)=limt→0+φ(x,t)=0 and limt→∞φ(x,t)=∞ for every x∈Ω.(iii)
There existsL⩾1 such that t↦φ(x,t)/t is L-almost increasing in (0,∞), for every x∈Ω.
Ifφ∈Φw(Ω) is convex with respect to the second variable, then it is called aconvexΦ-function, and we write φ∈Φc(Ω).If there existsβ∈(0,1] such that φ(x,β)⩽1 and φ(x,1/β)⩾1 for all x∈Ω, then we say that (A0) holds. Further, we say that φ∈Φw(Ω) satisfiesaIncp
if there existsL⩾1 such that t↦φ(x,t)/tp is L-almost increasing in (0,∞) for every x∈Ω,aInc
if there existp>1 such that (aInc)p holds,aDecq
if there existsL⩾1 such that t↦φ(x,t)/tq is L-almost decreasing in (0,∞) for every x∈Ω,aDec
if there existq>1 such that (aDec)q holds.The corresponding conditions withL=1 are denoted by (Inc) and (Dec). Note that the definition of weak Φ-function includes assumption (aInc)1.Definition 2.
Letφ∈Φw(Ω) and define themodularϱφ(·) for f∈L0(Ω) by (2)ϱφ·f≔∫Ωφx,fxdx. Thegeneralized Orlicz space, also called Musielak–Orlicz space, is defined as the set (3)Lφ·Ω≔f∈L0Ω:limλ→0+ϱφ·λf=0 equipped with the (Luxemburg) norm (4)fφ·≔inf⁡λ>0:ϱφ·fλ⩽1.
Forφ∈Φc(Ω), Lφ(·)(Ω) is a Banach space [18, Theorem 2.3.13.].Definition 3.
A functionf∈Lφ(·)(Ω) belongs to Sobolev space W1,φ(·)(Ω), if its weak partial derivatives ∂1f,…,∂nf exist and belong to Lφ(·)(Ω), that is, (5)W1,φ·Ω≔f∈Wloc1,1Ω:f,∇f∈Lφ·Ω. We define a semimodular on W1,φ(·)(Ω) by (6)ϱ1,φ·f≔ϱφ·f+ϱφ·∇f. which induces a quasinorm by f1,φ(·)≔inf⁡{λ>0:ϱ1,φ(·)(f/λ)⩽1}.Lemma 4.
Letφ∈Φc(Ω) satisfy (aInc) and (aDec), and let fj,gj∈W1,φ(·)(Ω) for j=1,2,… Assume further that the sequence (ϱ1,φ(·)(gj))j=1∞ is bounded. If ϱ1,φ(·)(fj-gj)→0 as j→∞, then (7)ϱ1,φ·fj-ϱ1,φ·gj→0asj→∞.Proof.
Sinceφ is convex and satisfies (aDec) it satisfies (Dec) for some possible larger exponent q by Lemma 2.6 of [13]. Let λ∈(0,1). By convexity and (Dec) we obtain(8)φx,fj=φx,fj-gj+gj⩽φx,λλfj-gj+1-λ1-λgj⩽λφx,1λfj-gj+1-λφx,11-λgj⩽λ1-qφx,fj-gj+1-λ1-qφx,gj. Next we subtract φ(x,gj) from both sides and integrate over Ω. Hence (9)ϱφ·fj-ϱφ·gj⩽λ1-qϱφ·fj-gj+1-λ1-q-1ϱφ·gj. Note that the choice λ=1/2 implies that (ϱφ(·)(fj)) is a bounded sequence.
Swappingfj and gj gives a similar inequality, and, combining the inequalities, we find that (10)ϱφ·fj-ϱφ·gj⩽λ1-qϱφ·fj-gjdx+1-λ1-q-1ϱφ·fj+ϱφ·gj. The same argument can be applied also to the weak gradient. Hence (11)ϱ1,φ·fj-ϱ1,φ·gj⩽λ1-qϱ1,φ·fj-gj+1-λ1-q-1ϱ1,φ·fj+ϱ1,φ·gj.
Letε>0 be given. Since ϱ1,φ(·)(fj)+ϱ1,φ(·)(gj)⩽2c, we can choose λ so small that (12)1-λ1-q-1ϱ1,φ·fj+ϱ1,φ·gj⩽ε2. We can then choose j0 so large that λ1-qϱ1,φ(·)(fj-gj)⩽ε/2 when j⩾j0 and it follows that ϱ1,φ·fj-ϱ1,φ·gj⩽ε.The first claim of the following lemma has been proved in [16, Lemma 4.4]. The proof for the second claim is similar.Lemma 5.
Letφ∈Φw(Rn) satisfy (A0) and let E⊂Rn be bounded. (a)
Ifφ satisfies (aInc)p, then Lφ(·)(E)↪Lp(E).(b)
Ifφ satisfies (aDec)q, then Lq(E)↪Lφ(·)(E).Lemma 6.
LetΩ⊂Rn. Let φ∈Φw(Ω) satisfy (aInc)p and (aDec)q. Then (13)fφ·⩽max⁡Lϱφ·f1/q,Lϱφ·f1/p where L is the maximum of the constants from (aInc) and (aDec).Proof.
Ifϱφ(·)(f)=0, then fφ(·)=0 and the claim holds.
Letϱφ(·)(f)>0 and assume first that Lϱφ(·)(f)⩽1.Then (aDec) gives that (14)φx,fxLϱφ·f1/q⩽LLϱφ·f-1φx,fx=ϱφ·f-1φx,fx.Integrating over Ω, we find that ϱφ(·)(f/(Lϱφ(·)(f))1/q))⩽1, which yields fφ(·)⩽(Lϱφ(·)(f))1/q. If Lϱφ(·)(f)>1 we similarly use (aInc) to conclude that fφ(·)⩽(Lϱφ(·)(f))1/p. The claim follows from these two cases.


## 3. Sobolev Capacity

We define a set of test-functions for the capacity of the setE by (15)S1,φ·E≔f∈W1,φ·Rn:f⩾1inanopensetcontainingEandf⩾0. The generalized Orlicz φ(·)-capacity of E is defined by (16)Cφ·E≔inff∈S1,φ·E∫Rnφx,f+φx,∇fdx.We now prove the following properties for the set functionE↦Cφ(·)(E). We emphasize that these do not require the convexity of φ.Proposition 7.
Letφ∈Φw(Rn). (S1)
Cφ(·)(∅)=0.(S2)
IfE1⊂E2⊂Rn, then Cφ(·)(E1)⩽Cφ(·)(E2).(S3)
IfE⊂Rn, then Cφ(·)(E)=infU⊃EopenCφ(·)(U).(S4)
IfE1,E2⊂Rn, then Cφ(·)(E1∪E2)+Cφ(·)(E1∩E2)⩽Cφ(·)(E1)+Cφ(·)(E2).(S5)
IfK1⊃K2⊃… are compact, then limi→∞Cφ(·)(Ki)=Cφ(·)(⋂i=1∞Ki).Proof.
(S1) follows from testing withf=0. Since a test-function for E2 is also a test-function for E1, (S2) follows from the infimum in the definition of capacity.
IfU⊃E, then Cφ(·)(E)⩽Cφ(·)(U) by (S2). Thus, (17)Cφ·E⩽infU⊃EopenCφ·U.For the opposite inequality, let ε>0 and choose f∈S1,φ(·)(E) such that (18)ϱ1,φ·f⩽Cφ·E+ε. Denote V≔int{f⩾1}. Then f∈S1,φ(·)(V), and so (19)infU⊃EopenCφ·U⩽Cφ·V⩽ϱ1,φ·f⩽Cφ·E+ε. This implies (S3) as ε→0+.
Letε>0. Choose fj∈S1,φ(·)(Ej) such that (20)ϱ1,φ·fj⩽Cφ·Ej+εforj∈1,2.Since max{f1,f2}⩾1 in an open set containing E1∪E2, it follows that max{f1,f2}∈S1,φ(·)(E1∪E2). Similarly, min{f1,f2}∈S1,φ(·)(E1∩E2).
The lattice property of Sobolev functions [18, Proposition 8.1.9] implies that ∇max⁡{f1,f2}=∇f1 and ∇min⁡{f1,f2}=∇f2 almost everywhere in A={f1⩾f2}. Therefore (21)∫Aφx,∇max⁡f1,f2dx+∫Aφx,∇min⁡f1,f2dx=∫Aφx,∇f1dx+∫Aφx,∇f2dx. The same argument also gives the equality in the complement of A, i.e. {f1<f2}. Hence(22)ϱφ·∇max⁡f1,f2+ϱφ·∇min⁡f1,f2=ϱφ·∇f1+ϱφ·∇f2almost everywhere. An analogous results holds without the derivative ∇.
Sincemax⁡{f1,f2}∈S1,φ(·)(E1∪E2), it follows from the definition that Cφ(·)(E1∪E2)⩽ϱ1,φ(·)(max{f1,f2}), and similarly Cφ(·)(E1∩E2)⩽ϱ1,φ(·)(min{f1,f2}). Combining this with the conclusion of the previous paragraph, we find that (23)Cφ·E1∪E2+Cφ·E1∩E2⩽ϱ1,φ·max⁡f1,f2+ϱ1,φ·min⁡f1,f2=ϱ1,φ·f1+ϱ1,φ·f2⩽Cφ·E1+Cφ·E2+2ε. (S4) follows from this as ε→0+.
It remains to prove (S5). Since⋂i=1∞Ki⊂Kj for any j, the “⩾”-inequality follows from (S2). To prove the opposite inequality, we choose an open U⊃⋂i=1∞Ki. Since ⋂i=1∞Ki is compact and (Kj) is decreasing, there is a positive integer k such that, Ki⊂U for all i⩾k. Then, again by (S2), we have (24)limi→∞Cφ·Ki⩽Cφ·U. Taking the infimum of this inequality over such sets U, we obtain the claim by (S3).Notice that we need convexity for the next property.Theorem 8.
Letφ∈Φc(Rn) satisfy (aInc) and (aDec) and E1⊂E2⊂⋯⊂Rn. Then (S6)
limi→∞Cφ(·)(Ei)=Cφ(·)(⋃i=1∞Ei).Proof.
Let us denoteE≔⋃i=1∞Ei. By (S2), limi→∞Cφ(·)(Ei)⩽Cφ(·)(E).
Now to prove the opposite inequality, we may assume thatlimi→∞Cφ(·)(Ei)<∞. Let fi∈S1,φ(·)(Ei) and ϱ1,φ(·)(fi)⩽Cφ(·)(Ei)+2-i for every i∈N.
The spaceLφ(·)(Rn) is uniformly convex and reflexive [20, Theorem 1.3]. The same holds for W1,φ(·), which is a closed subspace of (Lφ(·))n+1. By reflexivity, the bounded sequence (fi) has a subsequence which converges weakly to a function f∈W1,φ(·)(Rn). It follows from the Banach-Saks theorem that (25)1m∑i=1mfi→finW1,φ·Rnasm→∞. Let gj≔1/j(j-1)∑i=j+1j2‍fi. Then (26)1j2∑i=1j2fi-gj=1j2-1jj-1∑i=1j2fi+1jj-1∑i=1jfi from which we get, by the triangle inequality, that (27)1j2∑i=1j2fi-gj1,φ·⩽1j-11j2∑i=1j2fi1,φ·+1j-11j∑i=1jfi1,φ·→0 as j→∞, so that gj→f in W1,φ(·)(Rn).
By the convexity of the modular and the choice ofgj, we obtain that (28)ϱ1,φ·gj⩽1jj-1∑i=j+1j2ϱ1,φ·fi⩽supi⩾j+1ϱ1,φ·fi⩽supi⩾j+1Cφ·Ei+2-i. Now Cφ(·)(Ei) increases in i whereas 2-i decreases. Hence(29)ϱ1,φ·gj⩽limi→∞Cφ·Ei+2-j.
By considering a subsequence if necessary, we may assume thatgj+1-gj1,φ(·)⩽2-j. Then hj≔gj+∑i=j∞‍gi+1-gi∈W1,φ(·)(Rn). By definition of a test-function, there exists an open set Ui⊃Ei such that fi⩾1 in Ui. As Ej is an increasing sequence, it follows that Ej⊂Ui for every j⩽i. In ⋂i=j+1j2Uj, fi⩾1 for all relevant i, so that gj⩾1 in the same (open) set. Since hj⩾supi⩾jgj, we have (30)Ei⊂intgi⩾1⊂inthj⩾1. Taking the union over i of the previous inclusion, we find that hj⩾1 in the open set ⋃i=j∞int{gj⩾1}⊃E, so that hj∈S1,φ(·)(E). Hence(31)Cφ·E⩽ϱ1,φ·hjforj=1,2,…
Furthermore,(32)hj-gj1,φ·⩽∑i=j∞gi+1-gi1,φ·⩽∑i=j∞2-i=2-j+1 and, hence, by [18, Corollary 2.1.15], ϱ1,φ(·)(hj-gj)⩽hj-gj1,φ(·)→0 as j→∞. Then, using Lemma 4, we also have ϱ1,φ·hj-ϱ1,φ·gj→0 as j→∞, which we apply in (31) to obtain (33)Cφ·E⩽limj→∞ϱ1,φ·hj=limj→∞ϱ1,φ·gj and further (29) implies (34)Cφ·E⩽limj→∞ϱ1,φ·gj⩽limj→∞limi→∞Cφ·Ei+2-j=limi→∞Cφ·Ei.In the next result we use a trick to get back to weakΦ-functions despite an application of (S6).Theorem 9.
Letφ∈Φw(Rn) satisfy (aInc) and (aDec) and E1,E2,…⊂Rn. Then (S7)
Cφ(·)(⋃i=1∞Ei)⩽∑i=1∞‍Cφ(·)(Ei).Proof.
Letψ∈Φc(Rn) with φ≃ψ [21, Lemma 3.1]. Since φ satisfies (aDec) φ≃ψ yields φ≈ψ. Denote Fkm≔∪i=mkEi for k=m,m+1,… By induction on (S4), we obtain that (35)Cφ·Fkm=Cφ·⋃i=mkEi⩽∑i=mkCφ·Ei⩽∑i=m∞Cφ·Ei. The same inequality holds also for ψ. Now, using (S6) for (Fk), we have (36)Cψ·⋃k=m∞Ek=Cψ·⋃k=m∞Fkm=limk→∞Cψ·Fkm⩽∑i=m∞Cψ·Ei
Furthermore, fork=1,2,…, by (S4) and φ≈ψ,(37)Cφ·⋃j=1∞Ej⩽Cφ·Fk1+Cφ·⋃j=k+1∞Ej⩽Cφ·Fk1+cCψ·⋃j=k+1∞Ej. By the two estimates in the previous paragraph, we obtain that(38)Cφ·⋃j=1∞Ej⩽∑i=1∞Cφ·Ei+c∑i=k∞Cψ·Ei. Since the sum is finite (otherwise, there is nothing to prove), the second term on the right hand side tends to zero as k→∞. This gives the claim.Remark 10.
A set function satisfying the properties (S1), (S2), and (S7) is called anouter measure. This holds if φ∈Φw(Ω) is satisfies (aInc) and (aDec). If φ is convex and satisfies (aInc) and (aDec), then it is aChoquet capacity, [22], i.e., it satisfies (S1), (S2), (S5), and (S6). Then, for every Borel set E⊂Ω,(39)Cφ·E=sup⁡Cφ·K:KiscompactandK⊂E.


## 4. Sobolev Capacity and Hausdorff Measure

In this section, we discuss simple relations between the generalized Orlicz capacity and the Lebesgue and Hausdorff measures.Lemma 11.
Letφ∈Φw(Rn) satisfy (aDec) and (A0). Then every measurable set E⊂Rn satisfies E⩽cCφ(·)(E), where c>1 depends on (aDec) and (A0).Proof.
Letf∈S1,φ(·)(E). By (aDec), with exponent q and constant L, we conclude that φ(x,f(x)/β)⩽Lβ-qφ(x,f(x)). We have φ(x,f(x)/β)⩾1 for f(x)⩾1 by (A0). Therefore (40)E⩽∫Rnφx,fβdx⩽Lβ-q∫Rnφx,fdx⩽Lβ-qϱ1,φ·f. Taking infimum over f∈S1,φ(·)(E), we obtain the claim.Proposition 12.
Letφ∈Φw(Rn) satisfy (A0), (aInc)p, p>1, and (aDec). If Cφ(·)(E)=0, then Cp(E)=0.Proof.
LetB=B(0,R+1) and η∈C0∞(B) be a cut-off function with η=1 in B(0,R),0⩽η⩽1 and ∇η⩽2. Let f∈Lφ(·)(Rn). By Lemma 5, we obtain (41)fηLpRn=fηLpB⩽cηfLφ·B⩽cfLφ·Rn where c depends on (aInc) and B. The same inequality holds for ∇f. Now ∇(fη)=η∇f+f∇η, and, thus ∇fη⩽∇f+2f by the assumptions on η. Since fW1,φ(·)(Rn)≈fLp(Rn)+∇fLp(Rn) we obtain that (42)fηW1,pRn≲fηLpRn+∇fηLpRn≲fW1,φ·Rn.
SinceCφ(·)(E)=0, we can choose fi∈S1,φ(·)(E∩B(0,R)) with ϱ1,φ(·)(fi)→0. Since φ satisfies (aDec), fiW1,φ(·)(Rn)→0 [18, Lemma 2.1.11]. Thus the above inequality implies that fiηW1,p(Rn)→0. Since fiη is a test-function for the p-capacity of E∩B(0,R), we get Cp(E∩B(0,R))=0, for every R>0. Since E=⋃i=1∞(E∩B(0,i)), the claim follows by the subadditivity of the p-capacity.In the previous result the assumption(aInc)p is natural, since p gives the capacity to compare with. However, the assumption (aDec) is surprising. The next example shows that it is nevertheless needed.Example 13.
Letφ(t)=0 when t∈[0,11/10] and φ(t)=t-11/10 when t>11/10. Then φ satisfies (aInc) with p=1 and (A0) with β=10/21. Let B be an open ball with a radius one. Let f be a Lipschitz-continuous function that is one in B, zero in Rn∖2B and linear in 2B∖B. Then ∇f=1 in 2B∖B and zero elsewhere. We obtain ϱ1,φ(f)=0 and thus Cφ(·)(B)=0. On the other hand Cp(B)>0.Thes-dimensional Hausdorff measure of a set E⊂Rn is denoted by Hs(E).Corollary 14.
Letφ∈Φw(Rn) satisfy (A0), (aInc)p with p>1 and (aDec). (1)
Ifp⩽n and E⊂Rn with Cφ(·)(E)=0, then Hs(E)=0 for all s>n-p.(2)
Ifp>n, then Cφ(·)(E)=0 if and only if E=∅, where E⊂Rn.Proof.
IfCφ(·)(E)=0, we then obtain Cp(E)=0 by Proposition 12. This leads to the first claim by [23, Theorem 4, p. 156].
Ifp>n, then we may choose s=0 in (1) and so H0(E)=0. Since H0 is a counting measure, E must be an empty set. On the other hand, Cφ(·)(∅)=0, by (S1).Corollary 15.
Letψ∈Φw(Rn) satisfy (A0), (aInc), and (aDec)q. Let E⊂Rn be bounded. If Cq(E)=0 or Hn-q(E)<∞, then Cψ(·)(E)=0.Proof.
LetCq(E)=0. By Lemma 5, Lq(E)↪Lψ(·)(E). The remaining proof follows the same procedure as in the proof of Proposition 12. If Hn-q(E)<∞, it follows from [23, Theorem 3, p. 154] that Cq(E)=0 and thus the claim follows from the first part.


## 5. Quasicontinuity

In this section, we prove the existence ofφ(·)-quasicontinuous representatives of generalized Orlicz functions. A function f:Rn→[-∞,∞] is φ(·)-quasicontinuous if for every ε>0 there exists an open set U with Cφ(·)(U)<ε such that f restricted to Rn∖U is continuous. We say that a claim holds φ(·)-quasieverywhere if it holds except in a set of Sobolev φ(·)-capacity zero.In this section we assume the density of continuous functions. A sufficient condition for this can be found in Theorem 6.5 of [16].Lemma 16.
Letφ∈Φw(Rn) satisfy (aInc) and (aDec). Then, for each Cauchy sequence C(Rn)∩W1,φ(·)(Rn), there is a subsequence which converges pointwise φ(·)-quasieverywhere in Rn. Moreover, the convergence is uniform outside a set of arbitrarily small Sobolev φ(·)-capacity.Proof.
Let(fi) be a Cauchy sequence in C(Rn)∩W1,φ(·)(Rn). We assume without loss of generality, by considering a subsequence if necessary, that fi-fi+11,φ(·)⩽4-i for every i=1,2,… For i∈N we denote gi≔2ifi-fi+1∈W1,φ(·)(Rn), Ei≔{gi>1} and Fj≔⋃i=j∞Ei. Note that gi1,φ(·)⩽2-i. We obtain by Corollary 2.1.15(a) of [18] that (43)Cφ·Ei⩽ϱ1,φ·gi⩽gi1,φ·⩽2-i.The subadditivity (S7) further implies (44)Cφ·Fj⩽∑i=j∞Cφ·Ei⩽∑i=j∞2-i=21-j.Since (Fj)j=1∞ is decreasing and Cφ(·)(Fj)⩾0, the limit limj→∞Cφ(·)(Fj) exists. Since ⋂j=1∞Fj⊂Fk for all k=1,2,…, we obtain by (S2) that (45)Cφ·⋂j=1∞Fj⩽limj→∞Cφ·Fj⩽limj→∞21-j=0. Now (fj) converges pointwise in Rn∖⋂j=1∞Fj and Cφ(·)(⋂j=1∞Fj)=0, so the first claim of the lemma is proved. Moreover, for x∈Rn∖Fj and k>l>j, (46)flx-fkx⩽∑i=lk-1fix-fi+1x⩽∑i=lk-12-i<21-l. Therefore, the convergence is uniform in Rn∖Fj, and the second claim follows.The existence of the so-calledφ(·)-quasicontinuous representative follows from Lemma 16 by standard arguments (e.g., [18, Theorem 11.1.3]):Theorem 17.
Letφ∈Φw(Rn) satisfy (aInc) and (aDec). Assume that C(Rn)∩W1,φ(·)(Rn) is dense in W1,φ(·)(Rn). Then for each f∈W1,φ(·)(Rn), there exists a φ(·)-quasicontinuous function g∈W1,φ(·)(Rn) such that f=g almost everywhere in Rn.Lemma 18.
Letφ∈Φw(Rn) satisfy (aDec). Suppose that f∈W1,φ(·)(Rn) is a nonnegative φ(·)-quasicontinuous function with f⩾1 in E⊂Rn. Then for every ε>0 there exists g∈S1,φ(·)(E) such that ϱ1,φ(·)(f-g)<ε.Proof.
Letε∈(0,1) and δ≔ε/2L2q(1+Lϱ1,φ(·)(f)), and let U⊂Rn be an open set such that f restricted to Rn∖U is continuous and Cφ(·)(U)<δ. Moreover, let us take h∈S1,φ(·)(U) such that ϱ1,φ(·)(h)<δ, and denote g≔(1+δ)f+h. Then g∈W1,φ(·)(Rn). The set V≔{x∈Rn∖U:(1+δ)f(x)>1}∪U is open and contains E. Since g⩾1 in V, g∈S1,φ(·)(E). It remains to estimate ϱ1,φ(·)(f-g).
Now,φ(x,h+δf)⩽φ(x,2h)+φ(x,2δf). By (aDec)p and (aInc)1 we have φ(x,2δf)⩽L2qφ(x,δf)⩽L22qδφ(x,f), and so φ(x,h+δf)⩽L2q(φ(x,h)+Lδφ(x,f)). Therefore (47)ϱφ·f-g=ϱφ·h+δf⩽L2qϱφ·h+Lδϱφ·f<L2q1+Lϱφ·fδ=12ε by the assumption on δ. An analogous inequality holds for the gradient, and so the claim follows.


## 6. Relative Capacity

In this section, we introduce relativeφ(·)-capacity, analogous to the relative p(·)-capacity of variable exponent spaces in [18], taken with respect to an open set, Ω, in Rn.Definition 19.
LetΩ⊂Rn, K⊂Ω be compact, and φ∈Φw(Ω). Denote (48)Rφ·K,Ω≔f∈W1,φ·Ω∩C0Ω:f>1inKandf⩾0.We define capφ·∗K,Ω≔inff∈Rφ·K,Ωϱφ·(∇f). Further, for U⊂Ω open, we set (49)capφ·U,Ω≔supK⊂Ucompactcapφ·∗K,Ω, and, for an arbitrary set E⊂Ω, we define (50)capφ·E,Ω≔infU⊃Eopencapφ·U,Ω. The number capφ(·)(E,Ω) is called therelativeφ(·)-capacity of E with respect to Ω.In the next result we offer two alternate set of assumptions. It seems that (aDec) alone is not sufficient, although we have not been able to prove this.Proposition 20.
Ifφ∈Φw(Ω) satisfies either (Dec) or (A0), then (51)capφ·∗K,Ω=inff∈R~φ·K,Ω∫Ωφx,∇fdx,where R~φ·K,Ω≔{f∈W1,φ(·)(Ω)∩C0(Ω):f⩾1inKandf⩾0}.Proof.
SinceRφ(·)(K,Ω)⊂R~φ(·)(K,Ω), the inequality “⩾” is clear. Now, to prove the opposite inequality, we fix ε>0 and let f∈R~φ(·)(K,Ω), be such that (52)ϱφ·∇f⩽infg∈R~φ·K,Ωϱφ·∇g+ε.
If (Dec) holds, we setg≔(1+ε)f. Then g>1 in K, so that (53)capφ·∗K,Ω⩽ϱφ·∇g=∫Ωφx,1+ε∇fdx⩽1+εq∫Ωφx,∇fdx, where we used (Dec) for the last inequality. The claim follows as ε→0+.
If, on the other hand, (A0) holds, then we fixη∈C0∞(Ω) with η=1 in suppf. Let λ=β∇η∞-1. We set g≔f+ε′λη∈Rφ(·)(K,Ω). Since the supports of ∇f and ∇η are disjoint, we observe that(54)ϱφ·∇g=ϱφ·∇f+ϱφ·ε′λ∇η⩽ϱφ·∇f+Lε′ϱφ·λ∇η, where (aInc)1 has been used in the last inequality. The claim follows as ε,ε′→0+.Next, we show thatcapφ(·)∗(K,Ω) and capφ(·)(K,Ω) are the same; that is, the relative capacity is well defined on compact sets.Proposition 21.
Letφ∈Φw(Ω). Then capφ(·)∗(K,Ω)=capφ(·)(K,Ω) for every compact K⊂Ω.Proof.
The inequalitycapφ(·)∗(K,Ω)⩽capφ(·)(K,Ω) follows directly from the definition.
Now, to prove the opposite inequality, fixε>0 and let f∈Rφ(·)(K,Ω) be such that ϱφ(·)(∇f)⩽capφ(·)∗(K,Ω)+ε. Then, f is greater than one in U≔f-1(1,∞), which is open since f∈C0(Ω), and contains K. Thus, f is also a valid test-function for every compact set K′⊂U, so that (55)capφ·U,Ω=supK′⊂Ucapφ·∗K′,Ω⩽ϱφ·∇f⩽capφ·∗K,Ω+ε. The result follows from this as ε→0+.Next, we show that the relative capacity has the same basic properties as the Sobolev capacity.Proposition 22.
LetΩ⊂Rn be open and φ∈Φw(Rn). (R1)
capφ(·)(∅,Ω)=0.(R2)
IfE⊂E′⊂Ω′⊂Ω, then capφ·(E,Ω)⩽capφ·(E′,Ω′).(R3)
IfE⊂Ω, then capφ(·)(E,Ω)=infU⊃Eopencapφ(·)(U,Ω).(R4)
IfE,F⊂Ω, then(56)capφ·E∪F,Ω+capφ·E∩F,Ω⩽capφ·E,Ω+capφ·F,Ω.(R5)
IfK1⊃K2⊃⋯ are compact, then limi→∞capφ(·)(Ki,Ω)=capφ(·)(⋂i=1∞Ki,Ω).Proof.
Properties (R1) and (R3) follow immediately from the definition. For proving property (R2), we observe that ifE′⊂U⊂Ω′ then also E⊂U⊂Ω. Thus (57)capφ·E′,Ω′=infE′⊂U⊂Ω′capφ·U,Ω′⩾capφ·E,Ω. Properties (R4) and (R5) are proved exactly like (S4) and (S5).The proof of the following results follows from (R4) by standard arguments, see, e.g., [18, Lemma 10.2.5].Lemma 23.
Suppose thatφ∈Φw(Ω) and E1,E2,…,Ek⊂Ω. Then, (58)capφ·⋃i=1kEi,Ω-capφ·⋃i=1kAi,Ω⩽∑i=1kcapφ·Ei,Ω-capφ·Ai,Ω, whenever Ai⊂Ei,i=1,2,…,k and capφ(·)⋃i=1kEi,Ω<∞.Furthermore, the previous lemma implies directly properties (R6) and (R7), see, e.g., [18, Theorem 10.2.6]. Note that these properties hold for the relative capacity without any extra assumptions on φ.Theorem 24.
Letφ∈Φw(Ω). (R6)
IfE1⊂E2⊂⋯⊂Ω, then limi→∞capφ(·)(Ei,Ω)=capφ(·)(⋃i=1∞Ei,Ω).(R7)
IfEi⊂Ω,i=1,2,…, then capφ(·)(⋃i=1∞Ei,Ω)⩽∑i=1∞‍capφ(·)(Ei,Ω).Remark 25.
A set function satisfying the properties (R1), (R2), (R5), and (R6) is called aChoquet capacity [22]. They imply for every Borel set E⊂Ω that(59)capφ·E,Ω=sup⁡capφ·K,Ω:KiscompactandK⊂E.


## 7. Relationship between Capacities

Lemma 26.
Assume thatφ∈Φw(Rn) satisfies (A0) and (aDec)q. If Ω is bounded and K⊂Ω is compact, then, (60)Cφ·K⩽Cmax⁡capφ·K,Ω1/q,capφ·K,Ω,where the constant C depends on the dimension, Ω and the constants in (A0) and (aDec).Proof.
We may assume thatcapφ(·)(K,Ω)<∞. Let f∈Rφ(·)(K,Ω) with ϱφ·(∇f)<2capφ(·)(K,Ω). Extend f by zero outside of Ω and set g≔min⁡{1,f}. Since f∈C0(Ω) and f>1 in the compact set K, U={f>1}⊃K is open. Hence g∈S1,φ(·)(K) and so(61)Cφ·K⩽∫Rnφx,g+φx,∇gdx⩽∫Rnφx,g+φx,∇fdx. Using 0⩽g⩽1, (aInc)1, (aDec)q and (A0), we obtain (62)φx,gx⩽Lgxφx,1⩽L2gxβ-qφx,β⩽L2gxβ-q. Integrating over the bounded set Ω and using the classical Poincaré inequality, we find that (63)∫Ωφx,gdx⩽C∫Ωgdx⩽C∫Ωfdx⩽C∫Ω∇fdx. Then the embedding Lφ(·)(Ω)↪L1(Ω) (Lemma 5) and Lemma 6 for the function ∇f give that (64)∫Ωφx,gdx⩽Cmax⁡ϱφ·∇f1/q,ϱφ·∇f. Combining this with (61) and taking the infimum over f, we obtain the claim.In the next proof the Choquet property for Sobolev capacity is needed and hence we assume thatφ is convex.Theorem 27.
Assume thatφ∈Φc(Rn) satisfies (A0) and (aDec)q. If Ω is bounded and E⊂Ω, then (65)Cφ·E⩽Cmax⁡capφ·E,Ω1/q,capφ·E,Ω, where the constant C depends on the dimension, Ω, and the constants in (A0) and (aDec).Proof.
We may assume thatcapφ(·)(E,Ω)<∞. By the definition of capφ(·)(E,Ω), there exist open sets Ui⊃E with capφ(·)(Ui,Ω)→capφ(·)(E,Ω), as i→∞. Let U≔⋂i=1∞Ui. Then U is a Borel set, and hence, by the Choquet property (Remark 10), (66)Cφ·E⩽Cφ·U=supK⊂UCφ·K, where the supremum is taken over all compact sets K⊂U. By Lemma 26, we obtain (67)Cφ·E⩽supKCφ·K⩽CsupK⊂Umax⁡capφ·K,Ω1/q,capφ·K,Ω⩽Cmax⁡capφ·Ui,Ω1/q,capφ·Ui,Ω.The claim follows from this as i→∞.From the above result, we can conclude thatCφ(·)(E)=0 if capφ(·)(E,Ω)=0. To get the converse implication, we first prove the following results.Lemma 28.
Letφ∈Φw(Ω) satisfy (aDec). Assume that C(Rn)∩W1,φ(·)(Rn) is dense in W1,φ(·)(Rn). If K⊂Rn is compact, then (68)Cφ·K=inff∈S1,φ·cK∫Rnφx,f+φx,∇fdx, where S1,φ(·)c(K)=S1,φ(·)(K)∩C(Rn).Proof.
SinceCφ(·)(K) is defined as the infimum over the larger set S1,φ(·)(K), the inequality “⩽” is clear.
Supposef∈S1,φ(·)(K), with 0⩽f⩽1 and choose functions fj∈C(Rn)∩W1,φ(·)(Rn) converging to f in W1,φ(·)(Rn), with 0⩽fj⩽1. Choose a bounded neighborhood U of K such that f=1 in U. Let η∈C1(Rn), 0⩽η⩽1 with η=1 in Rn∖U and η=0 in a neighborhood of K.
Letgj≔1-(1-fj)η and note that gj⩾0 since 0⩽fj,η⩽1. We find that f-gj=f-1+η-ηfj=(f-fj)η+(1-η)(f-1)=(f-fj)η, as f=1 in U and η=1 in Rn∖U. Since η∈W1,∞(Rn) and fj→f in W1,φ(·)(Rn), we find that gj→f in W1,φ(·)(Rn). Further, η=0 in a neighborhood of K, so gj=1 in a neighborhood of K. Thus gj∈S1,φ(·)c(K). This and gj→f imply the “⩾” inequality.Proposition 29.
LetΩ⊂Rn be bounded and φ∈Φw(Rn) satisfy (aDec). Assume that C(Rn) is dense in W1,φ(·)(Rn). If E⊂Ω with Cφ(·)(E)=0, then capφ(·)(E,Ω)=0.Proof.
LetK⊂Ω be compact with Cφ(·)(K)=0. By Lemma 28 we may choose a sequence of functions (fi) from S1,φ(·)c(K) such that fiW1,φ(·)(Rn)→0. Let η∈C0∞(Ω) be a cut-off function that equals two in K. Since fi∈S1,φ(·)c(K), it is easy to conclude that ηfi>1 in K and ηfi⩾0, hence ηfi∈Rφ(·)(K,Ω). Since modular convergence and norm convergence are equivalent [18, Theorem 2.1.11], we obtain (69)capφ·K,Ω⩽ϱφ·∇ηfi⩽ϱφ·cfi+ϱφ·c∇fi→0.Hence the claim holds for compact sets.
By (S3) there exists a sequence of open setsUi⊃E with Cφ(·)(Ui)→0 as i→∞. Let U≔⋂i=1∞Ui∩Ω. Then, U is a Borel set containing E which satisfies Cφ(·)(U)=0. By the Choquet property of the relative capacity, we obtain (70)capφ·E,Ω⩽capφ·U,Ω=supKcapφ·K,Ω, where the supremum is taken over all compact sets K⊂U. By the first part of the proof, capφ(·)(K,Ω)=0, and the claim follows.IfE⊂B and Ω=2B for a ball B, then the Lipschitz constant of the cut-off function η in the previous proof can be chosen to be c(diamB)-1. Then an similar proof gives the following quantitative version of the previous result, cf. [18, Theorem 10.3.5].Theorem 30.
LetB⊂Rn be a ball and φ∈Φw(Rn) satisfy (aDec)q. Assume that C(Rn) is dense in W1,φ(·)(Rn). For E⊂B, (71)capφ·E,2B⩽C1+max⁡diamB,diamB-qCφ·Ewhere the constant C depends on q and L.



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