This is the published version of a paper published in Annali di Matematica Pura ed Applicata.
Citation for the original published paper (version of record):
Adamowicz, T., Lundström, N L. (2016)
The boundary Harnack inequality for variable exponent p-Laplacian, Carleson estimates, barrier
functions and p(⋅)-harmonic measures.
Annali di Matematica Pura ed Applicata, 195(2): 623-658
http://dx.doi.org/10.1007/s10231-015-0481-3
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DOI 10.1007/s10231-015-0481-3
The boundary Harnack inequality for variable exponent
p-Laplacian, Carleson estimates, barrier functions and
p
(·)-harmonic measures
Tomasz Adamowicz · Niklas L. P. Lundström
Received: 13 June 2014 / Accepted: 28 January 2015 / Published online: 20 February 2015 © The Author(s) 2015. This article is published with open access at Springerlink.com
Abstract We investigate various boundary decay estimates for p(·)-harmonic functions. For domains inRn, n ≥ 2 satisfying the ball condition (C1,1-domains), we show the boundary Harnack inequality for p(·)-harmonic functions under the assumption that the variable expo-nent p is a bounded Lipschitz function. The proof involves barrier functions and chaining arguments. Moreover, we prove a Carleson-type estimate for p(·)-harmonic functions in NTA domains inRnand provide lower and upper growth estimates and a doubling property for a p(·)-harmonic measure.
Keywords Ball condition· Boundary Harnack inequality · Harmonic measure · NTA domain· Nonstandard growth equation · p-harmonic
Mathematics Subject Classification Primary 31B52; Secondary 35J92· 35B09 · 31B25
1 Introduction
The studies of boundary Harnack inequalities for solutions of differential equations have a long history. In the setting of harmonic functions on Lipschitz domains, such a result was first proposed by Kemper [41] and later studied by Ancona [11], Dahlberg [23] and Wu [60]. Subsequently, Kemper’s result was extended by Caffarelli et al. [21] to a class of elliptic equations, by Jerison and Kenig [40] to the setting of nontangentially accessible
T. Adamowicz was supported by a Grant of National Science Center, Poland (NCN), UMO-2013/09/D/ST1/03681.
T. Adamowicz (
B
)Institute of Mathematics of the Polish Academy of Sciences, 00-656 Warsaw, Poland e-mail: T.Adamowicz@impan.pl
N. L. P. Lundström
Department of Mathematics and Mathematical Statistics, Umeå University, 90187 Umeå, Sweden e-mail: niklas.lundstrom@math.umu.se
(NTA) domains, Bañuelos et al. [14] and Bass and Burdzy [15] studied the case of Hölder domains while Aikawa [6] the case of uniform domains. The extension of these results to the more general setting of p-harmonic operators turned out to be difficult, largely due to the nonlinearity of p-harmonic functions for p = 2. However, recently, there has been a substantial progress in studies of boundary Harnack inequalities for nonlinear Laplacians: Aikawa et al. [7] studied the case of p-harmonic functions in C1,1-domains, while in the same time, Lewis and Nyström [45,47,48] began to develop a theory applicable in more general geometries such as Lipschitz and Reifenberg-flat domains. Lewis–Nyström results have been partially generalized to operators with variable coefficients, Avelin et al. [12], Avelin and Nyström [13], and to p-harmonic functions in the Heisenberg group, Nyström [55]. Moreover, in [52], the second author proved a boundary Harnack inequality for p-harmonic functions with n < p ≤ ∞ vanishing on a m-dimensional hyperplane inRn for 0≤ m ≤ n − 1. We also refer to Bhattacharya [18] and Lundström and Nyström [53] for the case p= ∞, where the latter investigated A-harmonic and Aronsson-type equations in planar uniform domains. Concerning the applications of boundary Harnack inequalities, we mention free boundary problems and studies of the Martin boundary.
Another recently developing branch of nonlinear analysis is the area of differential equa-tions with nonstandard growth (variable exponent analysis) and related variational function-als. The following equation, called the p(·)-Laplace equation, serves as the model example:
div(|∇u|p(·)−2∇u) = 0, (1.1)
for a measurable function p: → [1, ∞] called a variable exponent. The variational origin of this equation naturally implies that solutions belong to the appropriate Musielak–Orlicz space W1,p(·)() (see Preliminaries). If p = const, then this equation becomes the classical
p-Laplacian.
Apart from interesting theoretical considerations, such equations arise in the applied sci-ences, for instance in fluid dynamics, see, e.g., Diening and R˚užiˇcka [25], in the study of image processing, see, for example, Chen et al. [22] and electro-rheological fluids, see, e.g., Acerbi and Mingione [1,2]; we also refer to Harjulehto et al. [35] for a recent survey and further references. In spite of the symbolic similarity to the constant exponent p-harmonic equation, various unexpected phenomena may occur when the exponent is a function, for instance the minimum of the p(·)-Dirichlet energy may not exist even in the one-dimensional case for smooth functions p; also smooth functions need not be dense in the corresponding variable exponent Sobolev spaces. Although Eq. (1.1) is the Euler-Lagrange equation of the p(·)-Dirichlet energy and thus is natural to study, it has many disadvantages comparing to the p= const case. For instance, solutions of (1.1) are, in general, not scalable, also the Harnack inequality is nonhomogeneous with constant depending on solution. In a consequence, the analysis of nonstandard growth equation is often difficult and leads to technical and non-trivial estimates (nevertheless, see Adamowicz and Hästö [4,5] for a variant of Eq. (1.1) that overcomes some of the aforementioned difficulties, the so-called strong p(·)-harmonic equation).
The main goal of this paper is to show the boundary Harnack inequality for p(·)-harmonic functions on domains satisfying the ball condition (see Theorem5.4below). Let us briefly describe the main ingredients leading to this result, as it requires number of auxiliary lemmas and observations which are interesting per se and can be applied in other studies of variable exponent PDEs.
In Sect.3, we study oscillations of p(·)-harmonic functions close to the boundary of a domain and prove, among other results, variable exponent Carleson estimates on NTA domains, cf. Theorem3.7. Similar estimates play an important role, for instance in studies
of the Laplace operator, in particular in relations between the topological boundary and the Martin boundary of the given domain, also in the p-harmonic analysis (see presentation in Sect.3for further details and references). The main tools used in the proof of Theorem3.7are Hölder continuity up to the boundary, Harnack’s inequality and an argument by Caffarelli et al. [21] which, in our situation, relies on various geometric concepts such as quasihyperbolic geodesics and related chaining arguments, also on characterizations of uniform and NTA domains.
Section4is devoted to introducing two types of barrier functions, called Wolanski-type and Bauman-type barrier functions, respectively. In the analysis of PDEs, barrier functions appear, for example, in comparison arguments and in establishing growth conditions for functions, see, e.g., Aikawa et al. [7], Lundström [52], Lundström and Vasilis [54] for the setting of p-harmonic functions. Furthermore, barriers can be applied in the solvability of the Dirichlet problem, especially in studies of regular points, see, e.g., Chapter 6 in Heinonen et al. [38] and Chapter 11 in Björn and Björn [19]. We would like to mention that our results on barriers enhance the existing results in variable exponent setting, see Remark4.2.
In Sect.5, we prove our main results, a boundary Harnack inequality and growth estimates for p(·)-harmonic functions vanishing on a portion of the boundary of a domain ⊂Rn satisfying the ball condition. We refer to Sect.2for a definition of the ball condition and point out that a domain satisfies the ball condition if and only if its boundary is C1,1-regular.
Let us now briefly sketch our results. Letw ∈ ∂, r > 0 be small and suppose that p is a bounded Lipschitz continuous variable exponent. Assume that u is a positive p (·)-harmonic function in ∩ B(w, r) vanishing continuously on ∂ ∩ B(w, r). Then, we prove that 1 C d(x, ∂) r ≤ u(x) ≤ C d(x, ∂) r whenever x∈ ∩ B(w, r/˜c), (1.2) for constants˜c and C whose values depend on the geometry of , variable exponent p and certain features of u andv, see the statement of Theorem5.4. Here, d(x, ∂) denotes the Euclidean distance from x to∂. Inequality (1.2) says that u vanishes at the same rate as the distance to the boundary when x approaches the boundary.
Suppose thatv satisfies the same assumptions as u above. An immediate consequence of (1.2) is then the following boundary Harnack inequality:
1 C ≤
u(x)
v(x)≤ C whenever x ∈ ∩ B(w, r/˜c),
saying that u andv vanishes at the same rate as x approaches the boundary (see Theorem
5.4in Sect.5). Among main tools used in the proof of boundary Harnack estimates, let us mention Lemmas5.1and5.3where we show the lower and upper estimates for the rate of decay of a p(·)-harmonic function close to a boundary of the domain. It turns out that the geometry of the domain affects the number and type of parameters on which the rate of decay depends. Namely, our estimates depend on whether a domain satisfies the inte-rior ball condition or the ball condition in Lemma 5.1 , cf. parts (i) and (ii) of Lemma 5.1. Besides the ball condition, the proof of (1.2) uses the barrier functions derived in Sect.4, the comparison principle and Harnack’s inequality. Our approach extends arguments from Aikawa et al. [7] to the case of variable exponents. We point out that the constants in (1.2), and thus also in the boundary Harnack inequality, depend on u andv. Such a dependence is expected for variable exponent PDEs and difficult to avoid, as, e.g., parame-ters in the Harnack inequality Lemma3.1and the barrier functions depend on solutions as well.
Finally, in Sect.6, we define and study lower and upper estimates for a p(·)-harmonic measure. We also prove a weak doubling property for such measures. In the constant exponent setting, similar results were obtained by Eremenko and Lewis [26], Kilpeläinen and Zhong [43] and Bennewitz and Lewis [17]. For p= const, p-harmonic measures were employed to prove boundary Harnack inequalities, see, e.g., [17], Lewis and Nyström [46] and Lundström and Nyström [53]. The p-harmonic measure, defined as in the aforementioned papers, as well as boundary Harnack inequalities, have played a significant role when studying free boundary problems, see, e.g., Lewis and Nyström [48].
2 Preliminaries
We let ¯ and ∂ denote, respectively, the closure and the boundary of the set ⊂Rn, for n ≥ 2. We define d(y, ) to equal the Euclidean distance from y ∈ Rn to, while ·, · denotes the standard inner product onR2and|x| = x, x 1/2 is the Euclidean norm of x.
Furthermore, by B(x, r) = {y ∈Rn: |x − y| < r}, we denote a ball centered at point x with radius r> 0, and we let dx denote the n-dimensional Lebesgue measure onRn. If ⊂Rn is open and 1≤ q < ∞, then by W1,q(), W01,q() we denote the standard Sobolev space and the Sobolev space of functions with zero boundary values, respectively. Moreover, let (w, r) = B(w, r) ∩ ∂. By fA, we denote the integral average of f over a set A.
For background on variable exponent function spaces, we refer to the monograph by Diening et al. [24].
A measurable function p: → [1, ∞] is called a variable exponent and we denote p+A:= ess sup x∈A p(x), p−A:= ess inf x∈A p(x), p +:= p+ and p−:= p−
for A⊂ . If A = or if the underlying domain is fixed, we will often skip the index and set pA= p= p.
In this paper, we assume that our variable exponent functions are bounded, i.e., 1< p−≤ p(x) ≤ p+< ∞ for almost every x ∈ .
The set of all such exponents in will be denotedP().
The functionα defined in a bounded domain is said to be log-Hölder continuous if there is constant L> 0 such that
|α(x) − α(y)| ≤ L log(e + 1/|x − y|)
for all x, y ∈ . We denote p ∈ Plog() if 1/p is log-Hölder continuous, the smallest constant for which 1p is log-Hölder continuous is denoted by clog(p). If p ∈Plog(), then
|B|p+B ≈ |B|p−B ≈ |B|p(x)≈ |B|pB (2.1) for every ball B ⊂ and x ∈ B; here pB is the harmonic average, p1B :=
B
1
p(x)dx.
The constants in the equivalences depend on clog(p) and diam . One of the immediate
1 cr
−p(w)≤ r−p(x) ≤ cr−p(w) (2.2)
with c depending only on constants in (2.1).
In this paper, we study only log-Hölder continuous or Lipschitz continuous variable exponents. Both types of exponents can be extended to the wholeRn with their constants unchanged, see [24, Proposition 4.1.7] and McShane-type extension result in Heinonen [37, Theorem 6.2], respectively. Therefore, without loss of generality, we assume below that variable exponents are defined in the wholeRn.
We define a (semi)modular on the set of measurable functions by setting Lp(·)()(u) :=
|u(x)| p(x)dx;
here, we use the convention t∞= ∞χ(1,∞](t) in order to get a left continuous modular, see
[24, Chapter 2] for details. The variable exponent Lebesgue space Lp(·)() consists of all measurable functions u: →Rfor which the modularLp(·)()(u/μ) is finite for some
μ > 0. The Luxemburg norm on this space is defined as uLp(·)():= inf μ > 0 : Lp(·)() u μ ≤ 1.
Equipped with this norm, Lp(·)() is a Banach space. The variable exponent Lebesgue space is a special case of an Orlicz-Musielak space. For a constant function p, it coincides with the standard Lebesgue space. Often, it is assumed that p is bounded, since this condition is known to imply many desirable features for Lp(·)().
There is not functional relationship between norm and modular, but we do have the fol-lowing useful inequality:
min Lp(·)()( f ) 1 p−, Lp(·)()( f ) 1 p+ ≤ f Lp(·)() ≤ maxLp(·)()( f ) 1 p−, Lp(·)()( f ) 1 p+ . (2.3) One of the consequences of these relations is the so-called unit ball property:
Lp(·)()( f ) ≤ 1 ⇒ f Lp(·)()≤ 1 and Lp(·)()( f ) 1 p− ≤ f Lp(·)()≤ Lp(·)()( f ) 1 p+. (2.4)
If E is a measurable set of finite measure and p and q are variable exponents satisfying q ≤ p, then Lp(·)(E) embeds continuously into Lq(·)(E). In particular, every function u ∈ Lp(·)() also belongs to Lp−(). The variable exponent Hölder inequality takes the
form
f g dx≤ 2 f Lp(·)()gLp(·)(), (2.5)
where pis the point-wise conjugate exponent, 1/p(x) + 1/p(x) ≡ 1.
The variable exponent Sobolev space W1,p(·)() consists of functions u ∈ Lp(·)() whose distributional gradient∇u belongs to Lp(·)(). The variable exponent Sobolev space W1,p(·)() is a Banach space with the norm
uLp(·)()+ ∇uLp(·)().
In general, smooth functions are not dense in the variable exponent Sobolev space, see Zhikov [61] but the log-Hölder condition suffices to guarantee that they are, see Diening et al. [24,
Section 8.1]. In this case, we define the Sobolev space with zero boundary values, W01,p(·)(), as the closure of C0∞() in W1,p(·)().
The Sobolev conjugate exponent is also defined point-wise, p∗(x) := nnp(x)−p(x)for p+< n. If p is log-Hölder continuous, the Sobolev–Poincaré inequality
u − uLp∗(·)()≤ c ∇uLp(·)() (2.6)
holds when is a nice domain, for instance convex or John [24, Section 7.2]. If u ∈ W01,p(·)(), then the inequality uLp∗(·)()≤ c ∇uLp(·)()holds in any open set.
Definition 2.1 The Sobolev p(·)-capacity of a set ⊂Rnis defined as Cp(·)() := inf
u
Rn(|u|
p(x)+ |∇u|p(x)) dx,
where the infimum is taken over all u∈ W1,p(·)(Rn) such that u ≥ 1 in a neighborhood of .
The properties of p(·)-capacity are similar to those in the constant case, see Theorem 10.1.2 in [24]. In particular, Cp(·)is an outer measure, see Theorem 10.1.1 in [24].
Another type of capacity used in the paper is the so-called relative p(·)-capacity which appears for instance in the context of uniform p(·)-fatness (see next section and Chapter 10.2 in [24] for more details).
Definition 2.2 The relative p(·)-capacity of a compact set K ⊂ is a number defined by capp(·)(K, ) = inf
u
|∇u| p(x)dx,
where the infimum is taken over all u∈ C0∞() ∩ W1,p(·)() such that u ≥ 1 in K. The definition extends to the setting of general sets inRnin the same way as in the case of constant p, cf. [24] for details and further properties of the relative p(·)-capacity. In what follows, we will need the following estimate, see Proposition 10.2.10 in [24]: For a bounded log-Hölder continuous variable exponent p: B(x, 2r) → (1, n), it holds that
c(n, p)rn−p(x)≤ capp(·)(B(x, r), B(x, 2r)). (2.7) The similar upper estimate holds for r≤ 1, cf. Lemma 10.2.9 in [24].
Definition 2.3 A function u∈ Wloc1,p(·)() is a (sub)solution if
|∇u|
p(x)−2∇u · ∇φ dx(≤) = 0 (2.8)
for all (nonnegative)φ ∈ C0∞().
In what follows, we will exchangeably be using terms (sub)solution and p(·)-(sub)solution. Similarly, we say that u is a supersolution ( p(·)-supersolution) if −u is a subsolution. A function which is both a subsolution and a supersolution is called a (weak) solution to the p(·)-harmonic equation. A continuous weak solution is called a p(·)-harmonic function.
Among properties of p(·)-harmonic functions, let us mention that they are locally C1,α, see, e.g., Acerbi and Mingione [1] or Fan [27, Theorem 1.1]. Another tool, crucial from our point of view, is the comparison principle.
Lemma 2.4 (cf. Lemma 3.5 in Harjulehto et al. [32]) Let u be a supersolution and v a subsolution such that u> v on ∂ in the Sobolev sense. Then, u > v a.e. in .
By the standard reasoning, the comparison principle implies the following maximum princi-ple: If u∈ W1,p(·)()∩C() is a p(·)-subsolution in , then the maximum of u is attained at the boundary of. For further discussion on comparison principles in the variable exponent setting, we refer, e.g., to Section 3 in Adamowicz et al. [3].
We close our discussion of basic definitions and results with a presentation of the geometric concepts used in the paper.
Definition 2.5 A domain ⊂ Rn is called a uniform domain if there exists a constant M ≥ 1, called a uniform constant, such that whenever x, y ∈ there is a rectifiable curveγ : [0, l(γ )] → , parameterized by arc length, connecting x to y and satisfying the following two conditions:
l(γ ) ≤ M|x − y|,
and
min{|x − z|, |y − z|} ≤ Md(z, ∂) for each point z ∈ γ.
Definition 2.6 A uniform domain ⊂ Rn with constant M
is called a nontangentially accessible (NTA) domain if and its complementRn\ satisfy, additionally, the so-called corkscrew condition:
For some r > 0 and for any w ∈ ∂ and r ∈ (0, r), there exists a point ar(w) ∈
such that r M< |ar(w) − w| < r and d ar(w), ∂ > r M.
We note that in fact the (interior) corkscrew condition is implied by a uniform domain, see Bennewitz and Lewis [17] and Gehring [30]. Among examples of NTA domains, we mention quasidisks, bounded Lipschitz domains and domains with fractal boundary such as the von Koch snowflake. A domain with the internal power-type cusp is an example of a uniform domain which fails to be NTA domain. Uniform domains are necessarily John domains, the latter one enclosing, e.g., bounded domains satisfying the interior cone condition. See Näkki and Väisälä [56] and Väisälä [58] for further information on uniform and John domains.
Recall that a quasihyperbolic distance k between points x, y in a domain Rn is defined as follows k(x, y) = inf γ γ ds(t) d(γ (t), ∂), (2.9)
where the infimum is taken over all rectifiable curvesγ joining x and y in . Any two points in a uniform domain can always be joined by at least one quasihyperbolic geodesic, i.e., a curve for which the above infimum can be achieved. See Bonk et al. [20, Section 2] and Gehring and Osgood [31] for more information.
We end this section by recalling the following geometric definition.
Definition 2.7 A domain ⊂Rn is said to satisfy the interior ball condition with radius ri > 0 if for every w ∈ ∂ there exists ηi ∈ such that B(ηi, ri) ⊂ and ∂ B(ηi, ri)∩∂ =
{w}. Similarly, a domain ⊂Rn is said to satisfy the exterior ball condition with radius
re > 0 if for every w ∈ ∂ there exists ηe ∈Rn\ such that B(ηe, re) ⊂Rn \ and
∂ B(ηe, r
e) ∩ ∂ = {w}. A domain ⊂Rn is said to satisfy the ball condition with radius
It is well known that ⊂Rnsatisfies the ball condition if and only if is a C1,1-domain. See Aikawa et al. [7, Lemma 2.2] for a proof. We also note that if ⊂Rn satisfies the ball
condition then is a NTA domain and hence also a uniform domain.
Throughout the paper, unless otherwise stated, c and C will denote constants whose values may vary at each occurrence. If c depends on the parameters a1, . . . , an, we sometimes write
c(a1, . . . , an). When constants depend on the variable exponent p(·), we write “depending
on p−, p+, clog” in place of “depending on p” whenever dependence on p easily reduces to
p−, p+, clog.
3 Oscillation and Carleson estimates for p(·)-harmonic functions
This section is devoted to discussing some important auxiliary results used throughout the rest of the paper. Namely, in Lemmas3.4,3.5and3.6, we study oscillations of p(·)-harmonic functions over the balls intersecting the boundary of the underlying domain. We also employ geometric concepts such as NTA and uniform domains, quasihyperbolic geodesics and dis-tance together with the Harnack inequality to obtain a supremum estimate for a p(·)-harmonic function over a chain of balls. Such estimates, discussed in p= const setting for instance in Aikawa and Shanmugalingam [8] or Holopainen et al. [39], require extra attention for vari-able exponent p(·) as now constant in the Harnack inequality depends on a p(·)-harmonic function and the inequality is nonhomogeneous. In Theorem3.7, we show the main result of this section, namely the variable exponent Carleson estimate. Such estimates play a crucial role in studies of positive p-harmonic functions, see, e.g., Aikawa and Shanmugalingam [8], also Garofalo [29] for an application of Carleson estimates for a class of parabolic equa-tions. According to our best knowledge, Carleson estimates in the setting of equations with nonstandard growth have not been known so far in the literature. We apply Lemma3.7in the studies of p(·)-harmonic measures in Sect.5. Moreover, the geometry of the underly-ing domain turns out to be important in our investigations, in particular properties of NTA domains and uniform p(·)-fatness of the complement come into play.
We begin with recalling the Harnack estimate for p(·)-harmonic functions.
Lemma 3.1 (Variable exponent Harnack inequality) Let p be a bounded log-Hölder contin-uous variable exponent. Assume that u is a nonnegative p(·)-harmonic function in B(w, 4r), for somew ∈Rnand 0< r < ∞. Then, there exists a constant cH, depending on n, p and
sup∩B(w,4r)u, such that
sup B(w,r)u≤ cH inf B(w,r)u+ r .
Remark 3.2 The variable Harnack inequality in the above form was proved by Alkhutov [9] (see also Alkhutov and Krasheninnikova [10]) and subsequently improved to embrace the case of unbounded solutions by Harjulehto et al. [36, Theorem 3.9]. There, cHdepends only
on n, p and the Lqs(B(w, 4r))-norm of u for 1 < q < n−1n and s> p+B(w,4r)− p−B(w,4r). In what follows we will often iterate the Harnack inequality, and therefore, we need to carefully estimate the growth of constants involved in such iterations. Let ⊂Rn be a
uniform domain with constant M(for the definition of uniform domains and related concepts see the discussion in the end of Sect.2). We follow the argument in the proof of Lemma 3.9 in Holopainen et al. [39] and note that a quasihyperbolic geodesic joining two points in is an M-uniform curve with Mdepending only on M, cf. discussion in Gehring and Osgood
[31]. Let now x and y be given points in B(w,Mr) ∩ for w ∈ ∂ and some fixed r > 0. As in [39], we find a sequence of balls Bi, i= 1, . . . , N covering quasihyperbolic geodesic
γ joining x and y in (such a geodesic always exists for points in uniform domains, see discussion preceding the proof of [39, Lemma 3.9]) and satisfying the following conditions [recall that k(x, y) stands for a quasihyperbolic distance between points x and y and is given in (2.9)]:
1. Bi∩ Bi+1= ∅ for each i,
2. 2Bi ⊂ B(w, 4r) ∩ ,
3. N ≤ 3k(x, y).
We estimate the quasihyperbolic distance k(x, y) similarly as in formula (16) in Aikawa
and Shanmugalingam [8, Section 4]. Among other facts, we employ the definition of John curve. Assume that d(x, ∂) ≤ d(y, ∂) and note that then for a John curve γ , parametrized by arc length so thatγ (0) = x and γ (l(γ )) = y, the following is true. For all z ∈ γ , we have Md(z, ∂) ≥ l(γx z), where γx zis the sub curve from x to z. Using this, we see that
k(x, y) ≤ γ ds(t) d(γ (t), ∂) ≤ 1 2d(x,∂) 0 ds 1 2d(x, ∂) + l(γ ) 1 2d(x,∂) ds d(γ (t), ∂) ≤ 1 + M l(γ ) 1 2d(x,∂) ds s = 1 + Mlog s| l(γ ) 1 2d(x,∂) ≤ 1 + Mlog s|Md(y,∂)1 2d(x,∂) = 1 + M2 + Mlog 2+ Mlog d(y, ∂) d(x, ∂) . Combining this with the estimate for the number of balls N , we get
N ≤ 9M2 + 3Mlog d(y, ∂) d(x, ∂) , (3.1)
whenever d(x, ∂) ≤ d(y, ∂). This estimate can be used in the iteration of Harnack inequality as follows.
Suppose that x, y ∈ B(w,Mr). Then, by the variable exponent Harnack inequality (Lemma3.1) and the construction of the chain of balls Biabove, we have that
u(x) ≤ sup B1(x,r1) u(x) ≤ cH inf B1 u+ r1 ≤ · · · ≤ ≤ cN Hu(y) + cNHr1+ cNH−1r2+ · · · + rN ≤ cNHu(y) + cNHN r ≤ CNu(y) + r).
By using (3.1), we find that CN ≤ C9M2+3Mlog d(y,∂) d(x,∂) ≤ C9M2 Clog d(y,∂) d(x,∂) 3M ≤ C9M2 d(y, ∂) d(x, ∂) 3Mlog C , (3.2) whenever d(x, ∂) ≤ d(y, ∂).
In some results of this section, we appeal to notion of uniform p(·)-fatness. For the sake of completeness of the presentation, we recall necessary definitions, cf. Lukkari [50, Sections 3 and 4] and Holopainen et al. [39, Section 3].
Definition 3.3 We say that has uniformly p(·)-fat complement, if there exist a radius r0> 0 and a constant c0> 0 such that
capp(·)((Rn\ ) ∩ B(x, r), B(x, 2r)) ≥ c0capp(·)(B(x, r), B(x, 2r)) (3.3)
for all x∈Rn\ and all r ≤ r0.
The next lemma provides an oscillation estimate. Similar result was proven by Lukkari in [50, Proposition 4.2]. However, here, we adapt the discussion from [50] to our case; for instance, we do not require the boundary data to be Hölder continuous.
Lemma 3.4 Let ⊂Rnbe a domain having a uniformly p(·)-fat complement with constants c0 and r0. Let further p be a bounded log-Hölder continuous variable exponent satisfying
either p+≤ n or p−> n. Suppose that w ∈ ∂, r > 0 and u is a p(·)-harmonic function in ∩ B(w, r), continuous on ∩ B(w, r) with u = 0 on ∂ ∩ B(w, r). Then, there exist β, 0 < β ≤ 1, a constant c > 0 and a radius ˆr such that
sup B(w,ρ)∩u≤ c ρ r β sup B(w,r)∩u+ r
for allρ ≤ r/2 and r ≤ ˆr. The constants β and c depend on n, p, supB(w,r)∩u and c0,
whileˆr depends on n, p−, p+, clogand r0.
Proof Denote p0 := p(w) and split the discussion into two cases: p0 > n and p0≤ n. We
start by proving the lemma for p0> n. By assumptions, u is continuous on B(w, r)∩ with
u= 0 on B(w, r) ∩ ∂. Hence, we may use Theorem 1.2 in Alkhutov and Krasheninnikova [10], with D = B(w, r) ∩ and f = u. In a consequence, f (w) = 0 and osc∂ Df ≤ supB(w,r)∩u and we obtain that there exists c= c(n, p, supB(w,r)∩u) such that
sup B(w,ρ)∩u≤ c ρ r 1−n/p0 sup B(w,r)∩u, (3.4)
for allρ ≤ r/4 and with r ≤ ˆr(n, p+, p−, clog, r0). The dependence of ˆr on the listed
parameters follows from the proof of Theorem 1.2 in [10]. Hence, we conclude the lemma for p0> n by taking β = β(p0, n) = 1 − n/p0.
Assume now that p0 ≤ n. To prove the lemma in this case, we will follow the steps and
notation of the proof of Proposition 4.2 in Lukkari [50]. In the applications of Lemma3.4, we will need to understand the exact dependence on constants, and therefore, we repeat parts of the proof from [50].
Letη ∈ C0∞(B(w, r)). Then, ηu+ ∈ W01,p(·)(B(w, r) ∩ ). Further, supB(w,r)∩u+ = supB(w,r)∩u as u attains the boundary values u ≡ 0 continuously on B(w, r) ∩ ∂. As in Lukkari’s proof, we defineφ(r) := supB(w,r)∩∂u− u(w) and λ(r) := supB(w,r)∩∂u and note that under our assumptionsφ ≡ λ ≡ 0. Then, we use [50, Formula (3.4)] and [50, Formula (4.2)] which requires r ≤ ˆr(n, p−, p+), cf. Formula (3.2) in [50]. Namely, [50, Formula (3.4)] in our case reads
sup B(w,r)∩u+ r C−1γ (r) ≤ sup B(w,r)∩u−B(w,r/2)∩sup u+ r. (3.5)
The analysis of the proof of [50, Formula (3.4)] and the proof of [50, Theorem 3.3] reveals that sup B(w,r)∩u+ r p(w)−p(x)≤ c(c log) sup B(w,r)∩u+ 1 p+−p− := C.
The p(·)-fatness of the complement of together with the capacity estimate (2.7) imply the following inequalities (cf. [50, Formula (3.5)] and [50, Formula (4.2)]):
γ (r) : = cap p(·)((Rn\ ) ∩ B(w, r/2), B(w, r)) rn−p(w) 1 p(w)−1 ≥ c0capp(·)(B(w, r/2), B(w, r)) rn−p(w) 1 p(w)−1 ≥ (c0c(n, p)) 1 p(w)−1.
Thus,γ0 := C−1γ (r) satisfies c(c0, n, p, u+L∞(B(w,r)∩)) < γ0 < 1 and (3.5) reads:
sup B(w,r/2)∩u≤ γ1 sup B(w,r)∩u+ r ,
whereγ1 := max{γ0, 1 − γ0} < 1. This inequality is a counterpart of [50, Formula (4.3)].
Note also that12 ≤ γ1< 1. We iterate the above inequality to obtain
sup B(w,r 2m)∩ u≤ γ1m sup B(w,r)∩u+ c(γ1)r , where c(γ1) < 1 if γ1= 12 and c(γ1) ≤ 2γ m+1 1
2γ1−1 for the remaining values ofγ1 ∈ (
1 2, 1). We
continue as in [50] to find that forβ = log2(γ1
1) it holds γm 1 ≤ 2β ρ r β ,
whereβ depends on c0, n, p and supB(w,r)∩u. Hence, the proof is completed.
To prove Hölder continuity up to the boundary, we will also use the following oscillation estimate which follows from Theorem 4.2, Lemma 2.8 in Fan and Zhao [28] and Lemma 4.8 in Ladyzhenskaya and Ural’tseva [44]. The careful scrutiny of the presentation in [28] reveals the dependance of c andκ on supu and structure constants (cf. Lemma3.5). A similar result is given by Theorem 2.2 in Lukkari [50], but under the assumption that p+ ≤ n.
Lemma 3.5 Let p be a bounded log-Hölder continuous variable exponent and let u be a p(·)-harmonic function in and let B(w, r) . Then, there exist c and κ, 0 < κ < 1, such that for all 0< ρ ≤ r, it holds that
oscB(w,ρ)u≤ c ρ r κ oscB(w,r)u+ r. The constants c andκ depend on n, p+, p−and supu.
We are now ready to formulate the version of Hölder continuity up to the boundary which will be needed in this paper.
Lemma 3.6 Let ⊂Rnbe a domain having a uniformly p(·)-fat complement with constants c0 and r0. Let further p be a bounded log-Hölder continuous variable exponent. Suppose
thatw ∈ ∂, r > 0 and u is a p(·)-harmonic function in ∩ B(w, 2r), continuous on ∩ B(w, 2r) with u = 0 on ∂ ∩ B(w, 2r). Let γ = min{κ, β} and r < ˆr for β and ˆr as in Lemma3.4andκ as in Lemma3.5. Then, there exists C > 0 such that
|u(x) − u(y)| ≤ C |x − y| r γ sup B(w,2r)∩u+ r whenever x, y ∈ B(w, r) ∩ .
Proof Let x, y ∈ B(w, r) ∩ and let x0 ∈ ∂ be such that d(x, ∂) = |x − x0|. We
distinguish two cases.
Case 1.|x − y| < 12d(x, ∂). Lemma3.5applied withρ = |x − y| and r = τ/2 for τ = d(x, ∂) together with Lemma3.4imply the following inequalities:
|u(x) − u(y)| ≤ oscB(x,ρ)u≤ c |x − y| τ/2 κ oscB(x,τ/2)u+τ 2 ≤ c2κ |x − y| τ κ oscB(x 0,32τ)∩u+ τ 2 ≤ c2κ |x − y| τ κ sup B(x0,32τ)∩ u+τ 2 ≤ c2κ |x − y| τ κ 2β 3 2τ 2r β sup B(w,2r)∩u+ 2r +τ 2 ≤ c3β2κ−β |x − y| τ κ τ r β sup B(w,2r)∩u+ 2r + r βτ1−β1 2 ≤ C |x − y| r κ τ r β−κ sup B(w,2r)∩u+ 2r .
Ifβ − κ > 0, thenτrβ−κ< 1 and we get the assertion for γ = κ. Otherwise, if β − κ ≤ 0, then since|x − y| < 12τ, we have that
|x − y| r κτ r β−κ < |x − y| r κ r 2|x − y| κ−β ≤ 2β−κ|x − y| r β . Thus, the estimate holds forγ = min{κ, β}.
Case 2.|x − y| ≥ 12d(x, ∂). Since u(x0) = 0, we have by Lemma3.4that
|u(x) − u(y)| ≤ |u(x) − u(x0)| + |u(y) − u(x0)|
≤ 2β |x − x0| r β sup B(x0,r)∩ u+ r+ 2β |y − x0| r β sup B(x0,r)∩ u+ r ≤ 4β |x − y| r β sup B(x0,r)∩ u+ r ≤ 4β |x − y| r β sup B(w,2r)∩u+ r .
Since|x − y| < r, the last inequality holds as well with exponent γ = min{κ, β}, giving us the assertion of the lemma in this case. The proof of Lemma3.6is, therefore, completed. Following the proof of Theorem 6.31 in Heinonen et al. [38], one can show that if the complement of satisfies the corkscrew condition at w ∈ ∂, thenRn \ is p(·)-fat at w. Indeed, using the elementary properties of the relative p(·)-capacity (see Section 10.2 in Dieninig et al. [24], in particular Lemma 10.2.9 in [24] and the discussion following it), one shows that (3.3) holds atw. Here, the log-Hölder continuity of p(·) plays an important role as one also employs property (2.2). Hence, the complement of a NTA domain is uniformly p(·)-fat, see Definition2.6.
Theorem 3.7 (Variable exponent Carleson-type estimate) Assume that ⊂Rnis an NTA domain with constants Mand r. Letw ∈ ∂, 0 < r ≤ rand p be a bounded log-Hölder continuous variable exponent satisfying either p+ ≤ n or p− > n. Suppose that u is a positive p(·)-harmonic function in ∩ B(w, r), continuous on ¯ ∩ B(w, r) with u = 0 on ∂ ∩ B(w, r). Then, there exist constants c and csuch that
sup ∩B(w,r)u≤ c u(ar(w)) + r ,
where r= r/c. The constant c depends on n, p, supB(w,r)∩u and Mwhile cdepends on n, p−, p+, clogand M, r.
Proof We proceed following the main lines of Caffarelli et al. [21]. Let k be a large number to be determined later and assume that
ku(ar(w)) + r
< sup
∩B(w,r)u= u(x1) (3.6)
where x1 ∈ ∂ B(w, r) ∩ by the maximum principle. We want to derive a contradiction if
k is chosen large enough.
Suppose first that d(x1, ∂) ≥ r/100. Since is an NTA domain, it is in particular
uniform. Hence, we may assume that ris so small that any two points in B(w, 2r) ∩ can be connected by a Harnack chain totally contained in B(w, r) ∩ . Then, r= r/cdepends only on M and r. Since the L∞-norm of u is bounded in B(w, r) ∩ , we can iterate Harnack’s inequality using the same constant for each ball contained in B(w, r) ∩ . Thus, the Harnack inequality yields the existence of a constant c0, which by (3.2) depends only on
cHand M, and such that
u(x1) ≤ c0
u(ar(w)) + r
. (3.7)
This gives us a contradiction if k> c0, and hence, the proof of Theorem3.7follows in the
case when d(x1, ∂) ≥ r/100.
Next, assume that d(x1, ∂) < r/100. It follows by the Harnack inequality and discussion
before (3.2) that there exist constants ˆc, λ ∈ [1, ∞), depending only on Mand cH, such
that u(x1) ≤ ˆc d(ar(w), ∂) d(x1, ∂) λ u(ar(w)) + r . (3.8)
From (3.6) and (3.8), we see that
d(x1, ∂) d(ar(w), ∂) < ˆc k 1/λ . (3.9)
Let x1+∈ B(w, r) ∩ ∂ be a point minimizing |x1+− x1|. By decreasing rif necessary, we
apply Lemma3.6for B(x1+, r/2) to obtain u(x1) − u(x1+) = u(x1) ≤ C
d(x 1, ∂) r/4 γ sup B(x1+,r/2)∩ u+r 4 , (3.10)
whereγ and C depend on n, p, supB(w,r)∩u and M. The constant cnow depends on n, p−, p+, c
assumption (3.6) together with (3.9) and (3.10) we obtain, for some x2∈ ∂ B(x1+, r/2) ∩ ,
the existence ofˇc = ˇc(cH, M) such that
kˇc−1 u(ar 2(x + 1)) + r ≤ k ˇc−1ˇcu(a r(w)) + r + r= ku(a r(w)) + kr+ k ˇcr < k 1+1 ˇc u(ar(w)) + r < 1+1 ˇc u(x1) ≤ 1+1 ˇc C 4d(x1, ∂) r γ u(x2) + r 4 ≤ 1+1 ˇc C ˆc k γ /λ u(x2) + r 4 . (3.11)
In the last inequality, we have also used d(ar(w), ∂) ≤ r. Define constant k1such that
ˇc 1+1 ˇc C ˆc k1 γ /λ = 1. By demanding k> max{c0, k1}, we obtain
k u(ar 2(x + 1)) + r ≤ u(x2) + r 4 and u(x1) < u(x2) + r 4. Let k> 1. Then, kr/2 ≥ r/4 and the above inequalities take the following form:
k u(ar 2(x + 1)) + r 2
≤ u(x2) and u(x1) < u(x2) +
r
4. (3.12)
We will now repeat the above argument starting from (3.6) with (3.12) replacing (3.6). As now the initial condition has an additional term on the right-hand side, we provide details of the reasoning. Once those are explained, it will become more apparent how to continue with the recurrence argument. Suppose first that d(x2, ∂) ≥ r/200. Then, similarly as for
x1we get from (3.12) and the Harnack inequality that
k u(ar 2(x + 1)) + r 2 ≤ u(x2) ≤ c0 u(ar 2(x + 1)) + r 2 ,
where c0is the constant from (3.7). Hence, we again obtain the contradiction if k> c0.
Let now d(x2, ∂) < r/200. The discussion similar to that for (3.8) gives us
u(x2) ≤ ˆc ⎛ ⎝d(ar 2(x + 1), ∂) d(x2, ∂) ⎞ ⎠ λ u(ar 2(x + 1)) + r 2 . (3.13)
From (3.12) and (3.13), we see that d(x2, ∂) d(ar 2(x + 1), ∂) < ˆc k 1/λ .
We take point x2+ ∈ B(x1+,r2) ∩ ∂ minimizing |x2− x2+| and then apply Lemma3.6for
B(x2+,r4). In a result, we get u(x2) ≤ C d(x2, ∂) r/8 γ sup B(x+2,r 4)∩ u+r 8 .
Following the same reasoning as in (3.11), we obtain, for some x3∈ ∂ B(x2+,r 4) ∩ , that kˇc−1 u(ar 4(x + 2)) + r 2 ≤ k ˇc−1ˇcu(a r 2(x + 1)) + r 2 +r 2 = ku(ar 2(x + 1)) + kr 2 + k 2ˇcr < k 1+1 ˇc u(ar 2(x + 1)) + r 2 < 1+1 ˇc u(x2) ≤ 1+1 ˇc C 8d(x1, ∂) r γ u(x3) + r 8 ≤ 1+1 ˇc C ˆc k γ /λ u(x3) + r 8 . Since k> k1and kr/4 ≥ r/8, we arrive at
k u(ar 4(x + 2)) + r 4
≤ u(x3) and u(x2) < u(x3) +
r 8.
Having established first two steps of the iteration, we now choose points xm, x+min the similar
way as we found x1, x1+and x2, x2+and get that
k u(ar 2m(x + m)) + r 2m
≤ u(xm+1) and u(xm) < u(xm+1) +
r 2m+1.
If m→ ∞, then xm→ y ∈ ∂ ∩ B(w, 2r). Since u is assumed continuous on ∩ B(w, r)
with u= 0 on ∂ ∩ B(w, r), we obtain that u(xm) → u(y) = 0. Hence, we conclude that
ku(ar(w)) + r < u(x1) < u(x2) + r 4 < u(x3) + r 8 + r 4 < · · · < < u(xm) + r 2 → r 2 for m→ ∞. This gives ku(ar(w)) + r < r 2
which leads to k < 1/2 and results in the contradiction by demanding k > max{1, c0, k1}.
Thus, the proof of Theorem3.7is completed.
4 Constructions of p(·)-barriers
Below, we present two types of barrier functions. The first type is based on a work of Wolanski [59]; however, our Lemma4.1improves result of [59], see Remark4.2. We employ Wolanski-type barriers in the upper and lower boundary Harnack estimates, see Sect.5. The second type of barriers has been inspired by a work of Bauman [16] who uses barriers in studies of a boundary Harnack inequality for uniformly elliptic equations with bounded coefficients. Both approaches have advantages. On one hand, a radius of a ball for which a Wolanski-type barrier exists, depends on less number of parameters then a radius of a corresponding ball for a Bauman-type barrier, but on the other hand, exponents in Wolanski-type barriers depend on larger number of parameters than exponents in Bauman-type barriers, cf. Lemmas4.1and
4.1 Upper and lower p(·)-barriers of Wolanski-type
Lemma 4.1 Let y ∈ Rn and r > 0 be fixed and let p be a Lipschitz continuous variable exponent on B(y, 2r). Let further M > 0 be given and for x ∈ B(y, 2r) define functions
ˆu(x) = M e−μ− e−4μ e−μ− e−μ|x−y|2r 2 and ˇu(x) = M e−μ− e−4μ e−μ|x−y|2r 2 − e−4μ . Then, there exist r∗= r∗(p−, ||∇ p||L∞) and μ∗= μ∗(p+, p−, n, ||∇ p||L∞, M) such that
ˆu(x) is a p(·)-supersolution and ˇu(x) is a p(·)-subsolution in B(y, 2r) \ B(y, r) whenever μ ≥ μ∗and r≤ r∗. Furthermore, it holds that
ˆu(x) = M on ∂ B(y, 2r) and ˆu(x) = 0 on ∂ B(y, r), ˇu(x) = 0 on ∂ B(y, 2r) and ˇu(x) = M on ∂ B(y, r).
Remark 4.2 We would like to point out that the above theorem improves substantially some results on barrier functions in variable exponent setting, see Corollary 4.1 in Wolanski [59]. Namely in [59], the radius r depends also on M, whereas here, we manage to avoid such a dependence [see (4.7) and (4.8) for details]. This plays a role in the proof of Lemma5.1. Proof We begin the proof by noting that for any twice differentiable function u, we have
p(x)u= div
|∇u|p(x)−2∇u =∇|∇u|p(x)−2 , ∇u+ |∇u|p(x)−2u. Now, ∇|∇u|p(x)−2 , ∇u = n i=1 ∂ ∂xi |∇u|p(x)−2 ∂u ∂xi = n i=1 ∂ ∂xi
e(p(x)−2) log |∇u| ∂u ∂xi = n i=1 ∂ ∂xi (p(x) − 2) log |∇u| |∇u|p(x)−2∂u ∂xi = n i=1 ∂p ∂xi log|∇u| + (p(x) − 2) 1 |∇u| ∂ ∂xi(|∇u|) |∇u|p(x)−2∂u ∂xi = n i=1 ⎧ ⎨ ⎩ ∂p ∂xi log|∇u| + (p(x) − 2) 1 |∇u|2 n j=1 ∂u ∂xj ∂2u ∂xj∂xi ⎫ ⎬ ⎭|∇u|p(x)−2 ∂u ∂xi =
∇ p, ∇u log |∇u| + (p(x) − 2) 1 |∇u|2∞u
|∇u|p(x)−2. Moreover, assuming that|∇u| > 0, we obtain the following:
p(·)u≤ (≥) 0 ⇐⇒ ∇ p, ∇u log |∇u| + (p(x) − 2)∞u
|∇u|2 + u ≤ (≥) 0. (4.1)
From (4.1), we see that comparing to the constant p case, we have the extra term involving no second derivatives but the gradient of both u and p(·) instead.
We begin by showing that ˆu is a supersolution. We will find μ, A, B and r such that the function ˆu(x) = −Ae−μ |x−y| r 2 + B, where r < |x − y| < 2r (4.2)
has the desired properties. Differentiation of ˆu yields ˆuxi = 2 Aμ r2 e −μ|x−y|r 2 (xi− yi), |∇ ˆu| = 2 Aμ r2 e −μ|x−y|r 2 |x − y|, ˆuxixj = 2 Aμ r2 e −μ|x−y|r 2 δi j− 2μ r2(xi− yi)(xj− yj) . (4.3)
Next, we observe that ˆu = n i=1 ˆuxixi = 2 Aμ r2 e −μ|x−y|r 2n i=1 1−2μ r2(xi− yi) 2 = 2 Aμ r2 e −μ|x−y|r 2 n−2μ r2|x − y| 2 , (4.4)
and since!ni, j=1δi j(xi− yi)(xj− yj) = |x − y|2and
!n i, j=1(xi− yi)2(xj− yj)2= |x − y|4 we also have ∞ˆu = n i, j=1
ˆuxixjˆuxiˆuxj = 2 Aμ r2 3 e−3μ |x−y| r 2 n i, j=1 (xi− yi)(xj− yj)δi j− 2μ r2(xi− yi) 2(x j− yj)2 = 2 Aμ r2 3 e−3μ |x−y| r 2 |x − y|2 1−2μ r2|x − y| 2 . (4.5)
We collect expressions (4.4) and (4.5) and insert them into (4.1) to obtain the following inequality.
∇ p, ∇ ˆu log |∇ ˆu| +2 Aμ r2 e −μ|x−y|r 2 (p(x) − 2) 1−2μ r2|x − y| 2 + n−2μ r2|x − y| 2 ≤ 0.
We simplify the above condition by using ∇ p, ∇u = 2 Ar2μe
−μ|x−y|r 2 ∇ p, x − y : ∇ p, x − y log |∇ ˆu| −2μ r2|x − y| 2(p(x) − 1) + n + p(x) − 2 ≤ 0.
This holds true if
2r∇ pL∞| log |∇ ˆu|| − 2μ(p−− 1) + n + p+− 2 ≤ 0. (4.6)
Next, we demand that our function ˆu satisfies ˆu(x) = M whenever x ∈ ∂ B(y, 2r) and ˆu(x) = 0 whenever x ∈ ∂ B(y, r). These assumptions imply that A = M/(e−μ− e−4μ) and
B= Me−μ/(e−μ− e−4μ).
We now bound log|∇ ˆu|. By (4.3), we have |∇ ˆu| = 2Mμ e−μ− e−4μ |x − y| r2 e −μ|x−y|r 2 ,
and hence, upon using r< |x − y| < 2r, we obtain the following estimate 2Mμe−3μ
r(1 − e−3μ) ≤ |∇ ˆu| ≤
4Mμ r(1 − e−3μ).
Thus, −3μ + log 2Mμ r(1 − e−3μ)
≤ log |∇ ˆu| ≤ log 4Mμ r(1 − e−3μ) . Therefore, we conclude
| log |∇ ˆu|| ≤ log
4 1− e−3μ
+ | log M| + | log r| + log μ + 3μ. (4.7) Assume thatμ is large and combine (4.6) together with (4.7) to obtain thatˆu is a supersolution provided that the following condition is satisfied.
2r∇ pL∞ log 4 1− e−3μ + | log M| + | log r| + 4μ − 2μ(p−− 1) + n + p+− 2 ≤ 0. (4.8)
Upon rearranging terms in (4.8), we obtain the following inequality: μ8r∇ pL∞− 2(p−− 1) + 2r∇ pL∞ log 4 1− e−3μ + | log M| + | log r| + n + p+− 2 ≤ 0.
Pick r0= (p−− 1)/(4∇ pL∞). Then, for r ≤ r0, the above inequality can be satisfied by
a large enoughμ (upon including the term | log M| into the first log-term). Moreover, taking r∗= min{r0, 1/4} ensures that r| log r| is an increasing function of r for r ≤ r∗. Thus, we
conclude that if r ≤ r∗, then there existsμ∗= μ∗(p+, p−, n, ||∇ p||L∞, M) such that ˆu is
a supersolution forμ ≥ μ∗. This completes the proof for the supersolution.
Next, we want to show thatˇu is a subsolution. We will find μ, C, D and r in the function ˇu(x) = Ce−μ |x−y| r 2 + D, where r < |x − y| < 2r. In this case, ˇuxi = − 2Cμ r2 e −μ|x−y|r 2 (xi− yi), |∇ ˇu| = 2Cμ r2 e −μ|x−y|r 2 |x − y|, ˇuxixj = − 2Cμ r2 e −μ|x−y|r 2 δi j− 2μ r2(xi− yi)(xj− yj) . (4.9)
Similarly to computations in (4.4) and (4.5), we observe that ˇu = −2Cμ r2 e −μ|x−y|r 2 n−2μ r2|x − y| 2 , ∞ˇu = − 2Cμ r2 3 e−3μ |x−y| r 2 |x − y|2 1−2μ r2|x − y| 2 , and ∇ p, ∇u = −2Cμ r2 e −μ|x−y|r 2 ∇ p, x − y .
Collecting the terms, we obtain from (4.1) that the condition forˇu to be a subsolution becomes − 2r∇ pL∞| log |∇ ˇu|| + 2μ(p−− 1) − n − p++ 2 ≥ 0. (4.10)
This is equivalent to (4.6). Finally, we check that assumptions ˇu(x) = 0 whenever x ∈ ∂ B(y, 2r) and ˇu(x) = M whenever x ∈ ∂ B(y, r) imply C = M/(e−μ− e−4μ) and D =
Me−4μ/(e−μ− e−4μ). Let A be as in the definition of supersolution ˆu, see (4.2) and cf. the discussion following (4.6). Since C= A, we obtain that bounds for log(|∇ ˇu|) are identical to the case of supersolution. Therefore, the proof of the lemma is completed. 4.2 Upper and lower p(·)-barriers of Bauman-type
Lemma 4.3 Let y ∈Rn and r > 0 be fixed, and let p be a Lipschitz continuous variable exponent on B(y, 2r). Let further M > 0 be given and for x ∈ B(y, 2r) define functions
ˆu(x) = M 1− 2−μ 1− r |x − y| μ and ˇu(x) = − 2 −μM 1− 2−μ 1− 2r |x − y| μ .
Then, there existμ∗= μ∗(p−, n) > 0 and r∗= r∗(p−, n, ∇ pL∞, M) such that ˆu(x) is
a p(·)-supersolution and ˇu(x) is a p(·)-subsolution in B(y, 2r) \ B(y, r) whenever μ ≥ μ∗
and r≤ r∗. Furthermore, it holds that
ˆu(x) = M on ∂ B(y, 2r) and ˆu(x) = 0 on ∂ B(y, r), ˇu(x) = 0 on ∂ B(y, 2r) and ˇu(x) = M on ∂ B(y, r).
Proof Let us show first that ˆu is a supersolution. This will be done by choosing parameters μ, A, B and r in the function
ˆu(x) = −A r |x − y| μ + B, where r < |x − y| < 2r. Differentiation of ˆu yields
ˆuxi = Aμrμ|x − y|−(μ+2)(xi− yi), ˆuxixj = Aμrμ|x − y|−(μ+4) |x − y|2δ i j− (μ + 2)(xi− yi)(xj− yj) .
Next, we calculate the following expressions: |∇ ˆu| = Aμrμ|x − y|−(μ+1), ˆu = n i=1 ˆuxixi = Aμrμ|x − y|−(μ+4) n i=1 |x − y|2δ ii− (μ + 2)(xi− yi)2 = Aμrμ|x − y|−(μ+4)n|x − y|2− (μ + 2)|x − y|2 = Aμrμ|x − y|−(μ+2)n− μ − 2. (4.11)
As!ni, j=1δi j(xi− yi)(xj− yj) = |x − y|2and
!n
i, j=1(xi− yi)2(xj− yj)2 = |x − y|4we
also get that ∞ˆu =
n
i, j=1
ˆuxixjˆuxiˆuxj = A3μ3r3μ|x − y|−(3μ+8) n i, j=1 |x − y|2δ i j− (μ + 2)(xi− yi)(xj− yj) ×(xi− yi)(xj− yj) = A3μ3r3μ|x − y|−(3μ+8) |x − y|4− (μ + 2)|x − y|4 = A3μ3r3μ|x − y|−(3μ+4) 1− μ − 2 .
Clearly, ˆu ∈ C2(B(y, 2r) \ B(y, r)) and (4.11) gives us that in the given annulus|∇ ˆu| > 0. Recall that by the formal computations div|∇ ˆu|p(x)−2∇ ˆu≤ 0 is equivalent to
∇ p, ∇ ˆu log |∇ ˆu| + (p(x) − 2)∞ˆu
|∇ ˆu|2 + ˆu ≤ 0. (4.12)
By collecting the above expressions and substituting them into (4.12), we obtain the following inequality:
∇ p, ∇ ˆu log |∇ ˆu| + (p(x) − 2)Aμrμ|x − y|−(μ+2)"1− μ − 2#
+Aμrμ|x − y|−(μ+2)"n− μ − 2#≤ 0. (4.13)
Use ∇ p, ∇ ˆu = Aμrμ|x − y|−(μ+2) ∇ p, x − y in order to simplify (4.13): ∇ p, x − y log |∇ ˆu| − μ(p(x) − 1) + n − p(x) ≤ 0. This holds true if
∇ pL∞|x − y|| log |∇ ˆu|| − μ(p−− 1) + n − p−≤ 0. (4.14)
We now choseμ∗= μ∗(p−, n) so that if μ ≥ μ∗, then we have
− μ(p−− 1) + n − p−≤ −1. (4.15)
Next, we demand that our function ˆu satisfies ˆu(x) = M whenever x ∈ ∂ B(y, 2r) and ˆu(x) = 0 whenever x ∈ ∂ B(y, r). This implies that A = B = M/(1 − 2−μ). Our next step
is to find conditions for r so that the first term on the left-hand side of (4.14) does not exceed value 1. Since|x − y| ≤ 2r it is enough to ensure that
∇ pL∞| log |∇ ˆu||2r ≤ 1. (4.16)
Then, the proof will be completed by collecting (4.14), (4.15) and (4.16). Hence, it only remains to satisfy (4.16). We have
|∇ ˆu| = M 1− 2−μμr μ|x − y|−(μ+1)= M 1− 2−μμr −1 r |x − y| μ+1 . Since r< |x − y| < 2r and μ > 0, it holds:
M 1− 2−μ μ r 1 2μ+1 ≤ |∇ ˆu| ≤ M 1− 2−μ μ r.
We choose r so small that the left-hand side is larger than one. Such a requirement leads to condition that r < r∗∗ := 2(2Mμμ−1)and thus r∗∗depends on M andμ∗and therefore on M, p−and n. Now,| log |∇ ˆu|| ≤ | logMμ/(1 − 2−μ)− log r| ≤ | logMμ/(1 − 2−μ)| + | log r|. As limr→0+r| log r| = 0 we have
∇ pL∞| log |∇ ˆu||2r ≤ ∇ pL∞2r
| logMμ/(1 − 2−μ)| + | log r|
≤ 1, (4.17) provided that r ≤ r∗ is small enough. Indeed, if r| log r| < 1/(4∇ pL∞) and r <
1/(4∇ pL∞| log(2μ+1r∗∗)|), then (4.17) holds. In a consequence r∗depends only on M,
∇ pL∞, p−and n. The last inequality completes the proof of (4.16), and hence, we have
shown thatˆu is a supersolution.
In order to show that ˇu is a p(·)-subsolution, we proceed in the analogous way as in the second part of Lemma4.1, cf. discussion between formulas (4.9) and (4.10). We define
ˇu(x) = C 2r |x − y| μ − D where r < |x − y| < 2r.
Similarly to computations forˆu, we obtain that with A as in the definition of the supersolution ˆu
C = D = 2−μA, ∇ ˇu = −∇ ˆu, ˇu = − ˆu, and ∞ˇu = −∞ˆu. (4.18) Upon collecting these expressions, we use them in (4.1) together with div|∇ ˇu|p(x)−2∇ ˇu≥ 0. In a consequence, we arrive at the following inequality:
∇ p, ∇ ˇu log |∇ ˇu| − (p(x) − 2)Cμ(2r)μ|x − y|−(μ+2)"1− μ − 2# −Cμ(2r)μ|x − y|−(μ+2)"n− μ − 2#≥ 0.
Using (4.18), we see that the above inequality is the same as (4.13) and also that the bounds for log|∇ ˇu| are the same as in the case of supersolution. Thus, the proof for p(·)-subsolutions,
and for Lemma4.3, is completed.
5 Upper and lower boundary growth estimates: The boundary Harnack inequality This section contains main result of the paper, namely the proof of the boundary Harnack inequality for positive p(·)-harmonic functions on domains satisfying the ball condition, see Theorem5.4. The proof relies on Lemmas5.1and5.3, where we show the lower and, respectively, the upper estimates for a rate of decay of a p(·)-harmonic function close to a boundary of the underlying domain. In particular, Lemmas5.1and5.3imply stronger result than the usual boundary Harnack inequality, namely that a p(·)-harmonic function vanishes at the same rate as the distance function. Moreover, Lemma5.1illustrates the following phenomenon: the geometry of the domain effects the sets of parameters on which the rate of decay depends. Indeed, it turns out that constants in our lower estimate depend whether domain satisfies the interior ball condition or the ball condition, cf. parts (i) and (ii) of Lemma5.1. As a corollary, we also obtain a decay estimate for supersolutions (a counterpart of Proposition 6.1 in Aikawa et al. [7]).
Forw ∈ ∂, we denote by Ar(w) a point satisfying d(Ar(w), ∂) = r and |Ar(w)−w| =
r . Existence of such a point is guaranteed by the interior ball condition (with radius ri) for
r≤ ri/2. Recall also that by cH, we denote the constant from the Harnack inequality, Lemma
Lemma 5.1 (Lower estimates) Let ⊂Rnbe a domain satisfying the interior ball condition with radius ri,w ∈ ∂ and 0 < r < ri. Let p be a bounded Lipschitz continuous variable
exponent. Assume that u is a positive p(·)-harmonic function in ∩ B(w, r) satisfying u = 0 on∂ ∩ B(w, r). Then, the following is true.
(i) There exist constants c and˜c such that if ˜r := r/˜c then c u(x) ≥ d(x, ∂)
r for x∈ ∩ B(w, ˜r).
The constant˜c depends only on riand p−, ∇ pL∞, while c depends on infw,˜ru, riand
p+, p−, n, ∇ pL∞, wherew,˜r = {x ∈ |˜r < d(x, ∂) < 3˜r} ∩ B(w, r). Moreover,
c is decreasing in infw,˜ru.
Assume in addition that satisfies the ball condition with radius rband that 0< r < rb.
(ii) Then, there exist constants cLand˜cLsuch that if˜r := r/˜cL then
cLu(x) ≥
d(x, ∂)
r for x∈ ∩ B(w, ˜r).
The constant ˜cL depends only on rb and p−, ∇ pL∞, while cL depends on
sup∩B(w,r)u, u(A2˜r(w)), rband p+, p−, n, ∇ pL∞. Moreover, cL is decreasing in
u(A2˜r(w)) and increasing in sup∩B(w,r)u.
Proof To prove (i), we start by applying Lemma 4.1 to obtain r∗, depending only on ∇ pL∞, p−, such that we can construct barriers in an annulus with radius less than r∗.
Assume˜c to be so large that ˜r ≤ min{r∗, r/6} and note that so far ˜c ≥ 6 depends only on ∇ pL∞, p−and ri.
Let x ∈ ∩ B(w, ˜r) be arbitrary. Then, there exists η ∈ ∂ such that d(x, ∂) = |x − η|. By the interior ball condition at η, we find a point ηi such that B(ηi, r
i) ⊂
andη ∈ ∂ B(ηi, r
i). Take ηi2˜r ∈ [η, ηi] with d(ηi2˜r, ∂) = 2˜r. Since ˜r ≤ r/6 we have
B(ηi
2˜r, 2˜r) ⊂ ∩ B(w, r). Thus, u is a positive p(·)-harmonic function in B(ηi2˜r, 2˜r). Next
we note that B(ηi2˜r, ˜r) ⊂ w,˜rand since u is continuous u≥ inf
w,˜r
u> 0 in w,˜r. (5.1)
Using (5.1) and Lemma 4.1, we construct a subsolution ˇu in B(ηi2˜r, 2˜r) \ B(ηi2˜r, ˜r) with boundary values ˇu ≡ infw,˜ru≡ M on ∂ B(ηi2˜r, ˜r) and ˇu ≡ 0 on ∂ B(ηi2˜r, 2˜r). Since ˇu ≤ u on∂ B(ηi2˜r, ˜r) and 0 = ˇu ≤ u on ∂ B(ηi2˜r, 2˜r), we obtain that ˇu ≤ u in B(ηi2˜r, 2˜r) \ B(ηi2˜r, ˜r) by the comparison principle (Lemma2.4). By the above discussion x∈ B(ηi2˜r, 2˜r)\B(ηi2˜r, ˜r) and the result will follow by showing thatˇu does not vanish faster than d(x, ∂) as x → ∂. In order to show this, we observe that the derivative ofˇu in a direction normal to ∂ B(ηi2˜r, 2˜r) does not vanish. Indeed, usingμ = μ∗in Lemma4.1,˜r ≤ |x − ηi2˜r| ≤ 2˜r together with computations for∇ ˇu in (4.9) results in the following estimate:
$$ $$ $ % ∇ ˇu(x), x− η i 2˜r |x − ηi 2˜r| &$$ $$ $= 2μ∗infw,˜ru ˜r2 e−μ∗ |x−ηi 2˜r | ˜r 2 e−μ∗− e−4μ∗|x − η i 2˜r| ≥ 2˜cμ∗infw,˜ru r e−3μ∗ 1− e−3μ∗ ≥ 1 cr. (5.2)
