On a new Kato class and positive solutions of Dirichlet problems for the fractional Laplacian in bounded domains

Nonlinear Analysis 74 (2011) 1555–1576

Contents lists available at ScienceDirect

Nonlinear Analysis journal homepage: www.elsevier.com/locate/na

On a new Kato class and positive solutions of Dirichlet problems for the fractional Laplacian in bounded domains Rym Chemmam, Habib Mâagli, Syrine Masmoudi ∗ Département de Mathématiques, Faculté des Sciences de Tunis, Campus universitaire, 2092 Tunis, Tunisia

article

info

Article history: Received 7 May 2010 Accepted 18 October 2010 MSC: 31B05 31C35 34B27 60J50

abstract The purpose of this paper is to extend some results of the potential theory of an elliptic operator to the fractional Laplacian (−∆)α/2 , 0 < α < 2, in a bounded C 1,1 domain D in Rn . In particular, we introduce a new Kato class Kα (D) and we exploit the properties of this class to study the existence of positive solutions of some Dirichlet problems for the fractional Laplacian. © 2010 Elsevier Ltd. All rights reserved.

Keywords: Fractional Laplacian Green function Symmetric stable process Dirichlet problem Martin representation Martin boundary Harmonic function

1. Introduction This paper deals with one of the most important subfamilies of Levy processes called symmetric stable processes. A symmetric α -stable process (Xt )t >0 on Rn is a Levy process whose transition density p(t , x − y) relative to the Lebesgue measure is uniquely determined by its Fourier transform

∫ Rn

exp(ix.ξ )p(t , x)dx = exp(−t |ξ |α ).

The constant α must be in (0, 2], and we always assume that n ≥ 2. When α = 2, we get a Brownian motion running with a time clock twice as fast as the standard one. In this paper, we only consider 0 < α < 2 and compare our results to corresponding ones in the classical case α = 2. Recently, there has been intense interest in studying stable processes, due to their applications in many branches of sciences such as physics, operations research, queuing theory, mathematical finance, and risk estimation. For these and more applications of stable processes, please see the interesting book [1], and the article [2] and the references therein. There is significant difference between Brownian motion and a symmetric α -stable process. In fact, unlike Brownian motion, the symmetric α -stable process (Xt )t >0 has discontinuous sample paths. Its Levy measure is given by

ν(dx) = cn,α |x|−(n+α) dx, ∗

Corresponding author. E-mail addresses: [email protected] (R. Chemmam), [email protected] (H. Mâagli), [email protected] (S. Masmoudi).

0362-546X/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2010.10.027

1556

R. Chemmam et al. / Nonlinear Analysis 74 (2011) 1555–1576

and it has no drift and no Gaussian part. Moreover, the infinitesimal generator of (Xt )t >0 is the fractional Laplacian

− (−∆)α/2 f (x) = lim

E x (f (Xt )) − f (x) t

t ↘0

f (y) − f (x)

∫ = cn,α lim

ε↘0 (|x−y|>ε)

|x − y|n+α

dy,

which is a prototype of non-local operators. Brownian motion and elliptic differential equations have been extensively studied in many papers (see for example [3–8,10–15] and the references therein). However, there is little detailed analysis of the symmetric α -stable process and fractional differential equations. Throughout this paper, we consider a bounded C 1,1 domain D in Rn , n ≥ 2. Let X = (Ω , F , Xt , Ft , P x ) be a symmetric α -stable process on Rn , and define

τD = inf {t > 0 : Xt ̸∈ D} , 



the first exit time of D. The killed symmetric stable process X D = XtD t >0 in D is defined by



XtD

=

if t < τD if t ≥ τD ,

Xt

∂

where ∂ is the cemetery point. The infinitesimal generator of X D is the Dirichlet fractional Laplacian − (−∆)α/2 |D (the fractional Laplacian with zero exterior condition) and the Green function GαD (x, y) of X D is a function that is continuous on D × D except along the diagonal such that, for every Borel measurable function f ≥ 0 on D, Ex

τD

[∫

]

f (Xs )ds =

0

∫

GαD (x, y)f (y)dy,

x ∈ D.

(1.1)

D

Here E x denotes the expectation with respect to P x of the process starting from x ∈ D. The following sharp estimates on the Green function GαD were obtained in [16]. Theorem 1. There exists a constant c = c (D, α) > 1 such that, for each x, y ∈ D, 1

α



1

c |x − y|n−α

(δ(x)δ(y)) 2 min 1, |x − y|α

 ≤ GD (x, y) ≤ c

α



1

α

|x − y|n−α

(δ(x)δ(y)) 2 min 1, |x − y|α

 ,

(1.2)

where δ(x) = d(x, ∂ D) is the Euclidean distance between x and ∂ D. An analogous result is known for the killed Brownian motion case (see [13] or [14]). Suppose that n ≥ 3, and let GD be the Green function of the killed Brownian motion in D; then there exists a constant c = c (D) > 1 such that, for each x, y ∈ D, 1

δ(x)δ(y) min 1, |x − y|2 

1

c |x − y|n−2



≤ GD (x, y) ≤ c

 δ(x)δ(y) min 1, . |x − y|2 

1

|x − y|n−2

(1.3)

The estimates given in (1.2) are very useful in studying other properties of symmetric stable processes. As an example of applications of these estimates is the 3G-Theorem for symmetric stable processes, which is given in [17] in the following form: GαD (x, y) GαD (y, z ) GαD (x, z )

 ≤ cα

1

|x − y|n−α

+

1

|y − z |n−α



for all x, y, z ∈ D.

This 3G-Theorem is useful for the study of functions belonging to the Kato class Jα (D), which is defined as follows. Definition 1 (See [17]). A Borel function ϕ belongs to the Kato class Jα (D) if ϕ satisfies either of the two equivalent conditions

 |ϕ(y)| lim sup dy = 0, n−α r ↓0 x∈D (|x−y|≤r )∩D |x − y|   ∫ t lim sup Ps |ϕ| (x)ds = 0. 

t ↓0

∫

x∈D

0

Here, (Pt )t >0 is the semigroup generated by X D defined by Pt f (x) = E x f XtD , for any nonnegative Borel function f on D.

 



R. Chemmam et al. / Nonlinear Analysis 74 (2011) 1555–1576

1557

For elliptic case, the class J2 (D) was introduced by Aizenman and Simon in [3]. In [18], the authors improve the 3GTheorem as follows. Theorem 2 (3G-Theorem). There exists a positive constant C0 such that, for all x, y and z in D, we have GαD (x, z ) GαD (z , y) GαD (x, y)

 ≤ C0

δ(z ) δ(x)

 α2

α

GD (x, z ) +



δ(z ) δ (y)

 α2

 α

GD (y, z ) .

This new form of the 3G-Theorem allows us to introduce a new class of functions denoted by Kα (D) (see Definition 2) and which contains the class Jα (D) (see Corollary 1). Definition 2. A Borel measurable function ϕ in D belongs to the Kato class Kα (D), if ϕ satisfies the following condition:

 lim

r →0



∫ sup D∩B(x,r )

δ(y) δ(x)

 α2

 α

GD (x, y) |ϕ(y)| dy

= 0.

(1.4)

The first purpose of this paper is to give interesting properties and examples of functions belonging to the class Kα (D). In particular, we prove that, for α > 1 and 1 ≤ λ < α , the function x → (δ(x))−λ is in Kα (D) but not in Jα (D). For the classical case α = 2, the class Kα (D) was introduced in [11] for n ≥ 3 and in [15] for n = 2, and it is widely used in the study of elliptic differential equations. Our second purpose in this paper is to investigate existence results for fractional equations of the form α

(−∆) 2 u = f (., u)

in D.

Solutions of such equation are understood as distributional solutions in D. The main body of the paper is organized as follows. In Section 2, we recall some basic definitions and we collect some known facts concerning killed symmetric α -stable processes. In Section 3, we establish some interesting properties and we give some original examples of functions in the class Kα (D). Also a careful analysis about continuity is performed in Section 3. In Section 4, we are interested in the existence of a positive continuous solution for the following nonlinear problem:

 α (−∆) 2 u = ϕ (., u) (δ(x))  lim x→ z

1− α 2

in D

u( x ) = 0 .

(1.5)

z ∈∂ D

We impose the following conditions on the nonlinear term ϕ .

(H1 ) ϕ is a nonnegative, nontrivial measurable function defined in D × (0, ∞) such that the map t → ϕ (x, t ) is continuous and nonincreasing in (0, ∞), for every x ∈ D. (H2 ) For all c > 0, the function   α α x → (δ(x))1− 2 ϕ x, c (δ(x)) 2 −1 belongs to Kα (D). Then, using a fixed point argument, we establish our first existence result. Theorem 3. Let ϕ : D ×(0, ∞) → [0, ∞) be a measurable function satisfying (H1 ) and (H2 ). Then, problem (1.5) has a positive continuous solution u in D. For the elliptic case, i.e., α = 2, problem (1.5) was studied in [10]. In fact, the authors showed that, if ϕ satisfies (H1 ) such that, for all c > 0, the function ϕ (x, c ) belongs to J2 (D), then the equation 1u + ϕ (., u) = 0 in D has a unique positive continuous solution which vanishes continuously in ∂ D. In Section 5, we shall study the fractional nonlinear problem

 α (−∆) 2 u + uϕ (., u) = 0 in D u( x ) (1.6) = f (z ),  lim x→z M α 1(x) D z ∈∂ D  α where MD 1(x) := ∂ D MDα (x, z ) σ (dz ). Here, σ is an appropriate measure in ∂ D which will be specified later and MDα (x, z ) is the Martin kernel of the killed process X D ; we recall its definition in Section 2. In fact, as a consequence of

1558

R. Chemmam et al. / Nonlinear Analysis 74 (2011) 1555–1576

Theorem 3.18  in [19], we deduce that, if f is a nonnegative, nontrivial continuous function on ∂ D, then the function MDα f (x) := ∂ D MDα (x, z ) f (z )σ (dz ) is the unique solution of the homogeneous fractional equation

 α (−∆) 2 u = 0 in D u(x) = f (z ).  lim x→z M α 1(x) D z ∈∂ D The nonlinear term ϕ which appears in problem (1.6) is required to satisfy the following assumptions.

(H3 ) ϕ is a nonnegative measurable function defined in D × (0, ∞).    α (H4 ) For all c > 0, there exists a nonnegative function qc ∈ Kα (D) such that the map s → s qc (x) − ϕ x, s(δ(x)) 2 −1 is continuous and nondecreasing on [0, c], for every x ∈ D. To prove the following second existence result, we use a technical method that requires a potential theory approach. Theorem 4. Let f be a nonnegative, nontrivial continuous function on ∂ D and let ϕ be a measurable function on D × (0, ∞) satisfying (H3 ) and (H4 ). Then there exists a positive continuous solution u in D for problem (1.6) satisfying, for each x ∈ D, cMDα f (x) ≤ u(x) ≤ MDα f (x) ,

(1.7)

where c ∈ (0, 1). 2. Notations and preliminaries We lay out in this section a number of notations and well-known definitions. We also recall some basic properties concerning the killed α -stable process. Throughout this paper, d denotes the diameter of D and c denotes a positive constant which may vary from line to line and which depends only on D and the index α . Moreover, if f and g are nonnegative functions on a set S, we write f (x) ≈ g (x) for x ∈ S, and we say that f and g are comparable in S if there exists c > 0 such that 1c f (x) ≤ g (x) ≤ cf (x), for all x ∈ S. We denote by L1 (D) the class of all integrable functions in D and by L1loc (D) the class of functions that are locally integrable     in D. Analogously, we denote by C D the class of continuous functions in D and by C0 (D) the subclass of C D consisting of the functions which vanish continuously on ∂ D. Let Gα be the Green function of the α -symmetric stable process (Xt )t >0 . We recall that Gα is given by Gα (x, y) = 2−α π −n/2 Γ



n−α 2



Γ

 α −1 2

|x − y|α−n ,

for x, y ∈ Rn .

Moreover, the Green function GαD of the killed process X D given by (1.1) satisfies, for x, y ∈ D,

   GαD (x, y) = Gα (x, y) − E x Gα XτD , y . It is well known that GαD (x, y) > 0 on D, GαD (x, x) = ∞, if x ∈ D and GαD (x, y) = 0 provided that x ∈ Dc or y ∈ Dc . Moreover, GαD (x, y) = GαD (y, x), for x, y ∈ D. Let us recall that, if pD (t , x, y) denotes the transition density of the killed process X D with respect to the Lebesgue measure, then for x, y ∈ D we have α

GD (x, y) =

∞

∫

pD (t , x, y)dt .

(2.1)

0

In addition, it is shown in [20, Theorem 1.1] that, for every T > 0, we have on (0, T ] × D × D



pD (t , x, y) ≈ min 1,

(δ(x)δ(y))α/2 t





min t −n/α ,

t

|x − y|n+α



.

(2.2)

We fix an arbitrary ‘‘reference point’’ x0 ∈ D, and we define the Martin kernel of the killed symmetric stable process X D as follows: MDα (x, z ) = lim

D∋y→z

GαD (x, y)

GαD (x0 , y)

,

for x ∈ D and z ∈ ∂ D.

(2.3)

Then, by [21] or [22], the limit in (2.3) exists, and the function MDα (., .) is continuous in D × ∂ D and satisfies, for x ∈ D and z ∈ ∂ D, α

MDα (x, z ) ≈

(δ(x)) 2 . |x − z |n

(2.4)

R. Chemmam et al. / Nonlinear Analysis 74 (2011) 1555–1576

1559

When α = 2, the sharp estimates (2.4) is well known for the Martin kernel of the killed Brownian motion MD (., .), which is defined by G D ( x, y )

MD (x, z ) = lim

D∋y→z

GD (x0 , y)

,

for x ∈ D and z ∈ ∂ D.

Here, and throughout, GD denotes the Green function of the killed Brownian motion in D. Moreover, using the Hergoltz theorem, there exists a positive measure σ in ∂ D such that

∫ 1= ∂D

MD (x, z )σ (dz ),

for x ∈ D.

For a nonnegative measurable function f on ∂ D, we put MDα (f )(x) :=

∫ ∂D

MDα (x, z ) f (z )σ (dz ),

for x ∈ D.

Then, using (2.4), we obtain the following estimates: α

MDα 1(x) ≈ (δ(x)) 2 −1 ,

for x ∈ D.

(2.5)

For symmetric stable processes, there are two kinds of harmonic function in D: functions which are harmonic in D with respect to the process X (simply called α -harmonic in D) and functions which are harmonic in D with respect to the killed process X D . The precise definitions of these two kinds of harmonic functions are as follows. Definition 3. A locally integrable function f defined on Rn taking values in (−∞, ∞] is said to be (i) α -harmonic in D if, for each open set B with B ⊂ D, Ex f XτB  < ∞

 



and f (x) = Ex f XτB

 



,

for x ∈ B;

(ii) α -superharmonic in D if f is lower semicontinuous in D and, for each open set B with B ⊂ D, Ex f − XτB







   < ∞ and f (x) ≥ Ex f XτB ,

for x ∈ B;

(iii) singular α -harmonic in D if it is α -harmonic in D and it vanishes outside D. Definition 4. A locally integrable function f defined on D taking values in (−∞, ∞] is said to be (i) α -harmonic with respect to X D if, for each open set B with B ⊂ D, Ex f XτDB  < ∞ and

 



f (x) = Ex f XτDB

 



,

for x ∈ B;

(ii) α -superharmonic with respect to X if f is lower semicontinuous in D and, for each open set B with B ⊂ D, D

Ex f − XτDB







   < ∞ and f (x) ≥ Ex f XτDB ,

for x ∈ B.

Remark 1. (1) If a function f is α -harmonic in D or harmonic with respect to X D , then f is continuous in D. (2) A singular α -harmonic function in D is α -harmonic with respect to X D . Also, an α -superharmonic function in D is α superharmonic with respect to X D . (3) Conversely, if a function f is α -superharmonic (respectively α -harmonic) with respect to X D and we extend it to be zero off the domain D, then the resulting function is α -superharmonic (respectively singular α -harmonic) in D. (4) Unlike the case α = 2, the only bounded singular α -harmonic function in D is the constant zero (see [22]). As example of an α -harmonic and α -superharmonic function is the function x → GαD (x, y), which is α -harmonic in D \{y} and α -superharmonic in D, for each y ∈ D. Since GαD is symmetric, the same is true when the roles of x and y are interchanged. Moreover, the function x → MDα (x, z ) is singular α -harmonic in D, for each z ∈ ∂ D. Hence, the function MDα (f ) is singular α -harmonic in D, for every continuous function f on ∂ D. As in the classical case, the notion of α -harmonicity can also be formulated in terms of the generator. Namely, as stated in [23], a nonnegative Borel measurable function f defined on all Rn is α -harmonic in D if and only if it is continuous in D α and (−∆) 2 f = 0 in D, in the distributional sense. Furthermore, as is shown in [22,21], we have the following Martin representation of a nonnegative α -harmonic and α -superharmonic function with respect to X D . Theorem 5. If f is a nonnegative α -superharmonic function with respect to X D , then f can be written uniquely as f ( x) =

∫

α

GD (x, y) ν(dy) + D

∫ ∂D

MDα (x, z ) µ(dz ),

where ν and µ are nonnegative measures in D and ∂ D, respectively. If f is a nonnegative α -harmonic function with respect to X D (or singular α -harmonic in D), then the measure ν = 0.

1560

R. Chemmam et al. / Nonlinear Analysis 74 (2011) 1555–1576

The following result is due to [19, Theorem 3.18]. Theorem 6. Let f be a nonnegative continuous function in ∂ D and let h be a singular α -harmonic in D such that lim

x→ z z ∈∂ D

h (x) MDα 1(x)

= f (z );

then h(x) = MDα f (x), for x ∈ D. In what follows, for a nonnegative measurable function f on D, we denote α

GD (f )(x) :=

∫

GαD (x, y) f (y)dy,

for x ∈ D

GD (x, y) f (y)dy,

for x ∈ D.

D

and GD (f )(x) :=

∫ D

It is easy to see that the function GαD (f ) is α -superharmonic with respect to X D , for every nonnegative measurable function f on D. As in the classical case, the potential kernel GαD satisfies the following properties.

(1) GαD satisfies the complete maximum principle (see [5, Proposition 7.1 p. 76]); i.e., for each nonnegative measurable function f on D and nonnegative α -superharmonic function v with respect to X D such that GαD f ≤ v in [f > 0], we have GαD f ≤ v in D. (2) If f is nonnegative measurable function in D then the following assertions are equivalent: α α (i) G  D f is aαpotential that is GD f ̸= ∞; (ii) D δ(y) 2 f (y)dy < ∞. In particular, for any nonnegative measurable function f in D such that L1loc (D), and we have in the distributional sense α

(−∆) 2 GαD f = f

α

 D

δ(y) 2 f (y)dy < ∞, the functions f and GαD f are in

in D.

α

We recall that (−∆) u = f in the distributional sense in D if 2

∫

f (x)ϕ(x)dx =

∫

D

α

u(x) (−∆) 2 ϕ(x)dx, D

for all functions ϕ of class C ∞ with compact support in D. In the special case of balls, the explicit formulas for the Green function and Martin kernel for a symmetric stable process are known (see [24,22,25]). In fact, let B = {x ∈ Rn : |x| < r } , r > 0; then α

α−n

GB (x, y) = cn,α |x − y|

∫ (r 2 −|x|2 )(r 2 −|y|2 )

sα/2

|x−y|2

(1 + s)n/2

0

ds,

for x, y ∈ B,

and, for x0 = 0, we have MBα (x, z ) = r n−α

r 2 − |x|2



α/2

|x − z |n

,

for x ∈ B and z ∈ ∂ B.

3. The Kato class Kα (D) We start this section by proving some inequalities for the Green function GαD , which we will use later. Proposition 1. For each x, y ∈ D, we have GαD (x, y) ≈

α

(δ(x)δ(y)) 2

1

α

|x − y|n−α (|x − y|2 + δ(x)δ(y)) 2

(3.1)

and α

(δ(x)δ(y)) 2 ≤ cGαD (x, y).

(3.2)

Moreover, if |x − y| ⩾ r, then α

α

GD (x, y) ≤ c

(δ(x)δ(y)) 2 rn

.

(3.3)

ts Proof. Using (1.2) and the fact that min (t , s) ≈ t + , for t , s > 0, we obviously obtain (3.1). Now, assertions (3.2) and (3.3) s hold immediately from (3.1). 

R. Chemmam et al. / Nonlinear Analysis 74 (2011) 1555–1576

1561

Corollary 1. For each x, y ∈ D, we have



δ(y) δ(x)

 α2

GαD (x, y) ≤ c

1

|x − y|n−α

.

In particular, the class Kα (D) contains the classical Kato class Jα (D). Proof. Using (3.1), we have



δ(y) δ(x)

 α2

GαD (x, y) ≤ c

δ(y)α

1

|x − y|

n−α

α

(|x − y| + δ(x)δ(y)) 2 2

.

Since 1

|x − y|2 + δ(x)δ(y) ≥

4

 |δ(x) − δ(y)|2 + 4δ(x)δ(y) ,

we obtain

|x − y|2 + δ(x)δ(y) ≥ c δ (y)2 , which proves the result.

(3.4)



Proposition 2. Let ϕ be a function in Kα (D). Then the function x → δ(x)α ϕ(x) is in L1 (D). In particular, Kα (D) ⊂ L1loc (D). Proof. Let ϕ ∈ Kα (D); then, by (1.4), there exists r > 0 such that, for each x ∈ D,



∫ D∩B(x,r )

δ(y) δ(x)

 α2

GαD (x, y) |ϕ(y)| dy ≤ 1.

Let x1, x2 , x3 , . . . , xm in D such that D ⊂ and y ∈ B (xi , r ) ∩ D, we have

(δ(y))α ≤ c



δ(y) δ ( xi )

 α2

m

i =1

B (xi , r ). Then, by (3.2), there exists c > 0 such that, for all i ∈ {1, 2, . . . , m}

GαD (xi , y).

Hence, we have

∫

α

δ(y) |ϕ(y)| dy ≤ c D

m ∫ − i=1

 D∩B(xi ,r )

δ(y) δ ( xi )

 α2

GαD (xi , y) |ϕ(y)| dy

≤ cm < ∞. This completes the proof.



In what follows, we use the notation

α δ(y) 2 α ‖ϕ‖D := sup GD (x, y) |ϕ (y)| dy. δ(x) x∈D D ∫ α GD (x, z ) GαD (z , y) |ϕ (z )| dz . aα (ϕ) := sup GαD (x, y) x,y∈D D ∫ 

Proposition 3. Let ϕ be a function in Kα (D); then aα (ϕ) ≤ 2C0 ‖ϕ‖D < ∞, where C0 is the positive constant given in Theorem 2. Proof. Let ϕ ∈ Kα (D). Then, by Theorem 2, it follows immediately that aα (ϕ) ≤ 2C0 ‖ϕ‖D . Next, we will prove that ‖ϕ‖D < ∞. Using (1.4), there exists r > 0 such that, for x ∈ D, we have



∫ D∩B(x,r )

δ(y) δ(x)

 α2

GαD (x, y) |ϕ(y)| dy ≤ 1.

1562

R. Chemmam et al. / Nonlinear Analysis 74 (2011) 1555–1576

Then

∫  D

 α2

δ(y) δ(x)

α

GD (x, y) |ϕ (y)| dy ≤ 1 +



∫ D∩Bc (x,r )

δ(y) δ(x)

 α2

GαD (x, y) |ϕ(y)| dy.

Now, by (3.3), we have

δ(y) δ(x)



∫ D∩Bc (x,r )

 α2

∫

C

α

GD (x, y) |ϕ(y)| dy ≤

rn

D∩Bc (x,r )

(δ(y))α |ϕ(y)| dy,

which is finite from Proposition 2. This completes the proof.



Lemma 1. For each r > 0 and x, y in D, we have r

∫

pD (t , x, y)dt ≤ GαD (x, y) .

(3.5)

0

Moreover, if |x − y| ≤ r 1/α , then GαD (x, y) ≤ c

r

∫

pD (t , x, y)dt .

(3.6)

0

Proof. Let r > 0 and let x, y in D. Then the inequality (3.5) is obviously obtained from (2.1). Let us prove (3.6). Using (2.2) and the fact that min(1, ab) ≥ min (1, a) min (1, b), for a, b > 0, we have, for |x − y| ≤ r 1/α , r

∫

pD (t , x, y)dt ≥ c

r

∫

0



min 1,

(δ(x)δ(y))α/2



t

0 r



t −n/α min 1,

t (n+α)/α



|x − y|n+α

dt

   (δ(x)δ(y))α/2 −n/α s min 1, s(n+α)/α ds α | | s x − y 0 ∫ 1      1 (δ(x)δ(y))α/2 α−n min 1 , s−n/α min 1, s(n+α)/α ds ≥ c |x − y| min 1, |x − y|α s 0   (δ(x)δ(y))α/2 α−n . ≥ c |x − y| min 1, |x − y|α ≥ c |x − y|α−n

So the result holds from (1.2).

∫

|x−y|α



min 1,



Proposition 4. A measurable function ϕ in D belongs to the class Kα (D) if and only if

 lim

t →0

∫ ∫ t sup x∈D

D

0

δ(y) δ ( x)

 α2

 pD (s, x, y) |ϕ (y)| dsdy

 Proof. Suppose that limt →0



∫ D∩B(x,r )

δ(y) δ(x)

 α2

supx∈D

  t  δ(y)  α2 D

α

0

δ(x)

GD (x, y) |ϕ(y)| dy ≤ c



pD (s, x, y) |ϕ (y)| dsdy rα

∫ ∫ D

= 0.

0



δ(y) δ(x)

 α2

= 0. Then, using (3.6), we have, for r > 0,

pD (t , x, y) |ϕ(y)| dtdy,

which proves that ϕ satisfies (1.4), and so ϕ ∈ Kα (D). Conversely, suppose that ϕ ∈ Kα (D). First, we prove that, for r > 0,

 sup

t ∈]0,1]



∫ sup x∈D

(|x−y|≥r )∩D

δ(y) δ(x)

 α2

 pD (t , x, y) |ϕ(y)| dy

= C (r ) < ∞.

Let 0 < t ≤ 1 and 0 < r ≤ |x − y| < d; then, using (2.2) and (1.2), we have

   (δ(x)δ(y))α/2 1 1 pD (t , x, y) ≤ c min t , min , |x − y|α t (n+α)/α |x − y|n+α   c (δ(x)δ(y))α/2 1 ≤ 2α min 1, |x − y|α r |x − y|n−α 

≤

c r 2α

GαD (x, y).

(3.7)

R. Chemmam et al. / Nonlinear Analysis 74 (2011) 1555–1576

1563

It follows, for 0 < t ≤ 1, that



∫ sup (|x−y|≥r )∩D

x∈D

δ(y) δ(x)

 α2

∫ 

c

pD (t , x, y) |ϕ(y)| dy ≤

sup r 2α x∈D

D

δ(y) δ(x)

 α2

GαD (x, y) |ϕ(y)| dy =

c r 2α

‖ϕ‖D ,

which is finite from Proposition 3, and so (3.7) holds. Now, since ϕ satisfies (1.4), for each ε > 0 there exists r > 0 such that, for x ∈ D, we have



∫ D∩B(x,r )

δ(y) δ(x)

 α2

GαD (x, y) |ϕ(y)| dy ≤ ε.

Then, using (3.5) and (3.7), we obtain, for 0 < t < 1,

∫ ∫ t 0

D

δ(y) δ (x)

 α2

α δ(y) 2 pD (s, x, y) |ϕ (y)| dsdy ≤ pD (s, x, y) |ϕ(y)| dsdy δ(x) D∩(|x−y|
∫

Then we deduce that

 lim

∫ ∫ t sup

t →0

x∈D

D

0

δ(y) δ (x)

 α2

 pD (s, x, y) |ϕ (y)| dsdy

= 0. 

Proposition 5. Let ϕ ∈ Kα (D) and let h be a positive α -superharmonic function with respect to X D . Then, for all x ∈ D, we have

∫

GαD (x, y) h(y) |ϕ(y)| dy ≤ aα (ϕ) h(x).

(3.8)

D

Moreover, we have, for x0 ∈ D,

 lim

sup

r →0

x∈D



∫

1 h( x )

α

(

D∩B x0 ,r

)

GD (x, y) h(y) |ϕ(y)| dy

= 0.

(3.9)

Proof. Let h be a positive α -superharmonic function with respect to X D . Then, by [26], h is excessive with respect to X D . So, using [5, Proposition 3.11 p. 51], there exists a sequence (fk )k of nonnegative measurable functions in D such that h(y) = sup k

∫

GαD (y, z ) fk (z )dz . D

Hence, it is enough to prove (3.8) and (3.9) for h(y) = GαD (y, z ), uniformly in z ∈ D. Let ϕ ∈ Kα (D). Then, by the definition of aα (ϕ), the assertion (3.8) is obviously satisfied for h(y) = GαD (y, z ), uniformly in z ∈ D. Now, we shall prove (3.9) for h(y) = GαD (y, z ), uniformly in z ∈ D. Let ε > 0; then, by (1.4), there exists r1 > 0 such that



∫ sup D∩B(x,r1 )

x∈D

δ(y) δ(x)

 α2

GαD (x, y) |ϕ(y)| dy ≤ ε.

Let r > 0. Then, using Theorem 2, we have

∫

1 GαD (x, z )

D∩B(x0 ,r )

GαD (x, y) GαD (y, z ) |ϕ(y)| dy

 α  α δ(y) 2 α δ(y) 2 α ≤ C0 GD (x, y) + GD (z , y) |ϕ(y)| dy δ(x) δ (z ) D∩B(x0 ,r )  α ∫ δ(y) 2 α ≤ 2C0 sup GD (x, y) |ϕ(y)| dy. δ(x) x∈D D∩B(x0 ,r ) ∫



(3.10)

1564

R. Chemmam et al. / Nonlinear Analysis 74 (2011) 1555–1576

On the other hand, it follows from (3.3) and (3.10) that



∫ D∩B(x0 ,r )

∫

δ(y) δ(x) 

≤ D∩B(x,r1 )

≤ε+

c

 α2

GαD (x, y) |ϕ(y)| dy

δ (y) δ(x)

 α2

∫

r1n

D∩B(x0 ,r )

GαD (x, y) |ϕ(y)| dy +



∫ D∩B(x0 ,r )∩Bc (x,r1 )

δ (y) δ(x)

 α2

GαD (x, y) |ϕ(y)| dy

δ(y)α |ϕ(y)| dy.

This together with Proposition 2 ends the proof by letting r → 0.



Corollary 2. Let ϕ be a nonnegative function in Kα (D); then we have

 α  G (ϕ) D

∫ ∞

GαD (x, y) ϕ(y)dy < ∞.

:= sup x∈D

D

Proof. Putting h = 1 in (3.8) and using Proposition 3, we obtain the result.

 α

Corollary 3. Let ϕ be a measurable function in D such that GαD (|ϕ|) < ∞; then the function x → δ(x) 2 ϕ (x) is in L1 (D). α

In particular, if ϕ ∈ Kα (D), then the function x → δ(x) 2 ϕ(x) is in L1 (D). Proof. Let x0 ∈ D such that GαD (|ϕ|) (x0 ) < ∞. Then, by (3.2), we get

∫

α

α

δ(y) 2 |ϕ (y)| dy ≤ c δ (x0 )− 2 D

∫

GαD x0 , y |ϕ (y)| dy < ∞.





D

So the result holds.



Remark 2. Let q be the function defined in D by q(x) =

1

δ(x)λ

. α

If GαD (q) < ∞, then, by Corollary 3, the function x → δ(x) 2 −λ ∈ L1 (D). Hence it follows by [7, Lemma p. 726] that λ < 1 + α2 . For the case α = 2, the condition λ < 2 is equivalent to GD (q) < ∞. In fact, in [4], the authors give the following estimates on GD (q). Proposition 6. Let q be the function defined in D by q(x) =

1

δ(x)λ

.

Then, for all x ∈ D, we have

  δ(x)2−λ     2d GD (q)(x) ≈ δ(x) log  δ(x)   δ(x)

if 1 < λ < 2, if λ = 1, if λ < 1.

In the following proposition, we prove an analogous resultfor α ∈]0, 2[. In particular, we show that the condition α is sufficient to have GαD (q) < ∞. Moreover, we have GαD (q)∞ < ∞ if and only if λ ≤ α . 2

λ<1+

Proposition 7. Let λ < 1 + α2 and let q be the function defined in D by q(x) =

1

δ(x)λ

.

R. Chemmam et al. / Nonlinear Analysis 74 (2011) 1555–1576

1565

Then, for x in D, we have

α <λ<1+ , 2 2 α if λ = , 2 α if λ < .

 δ(x)α−λ     2d α GαD (q)(x) ≈ δ(x) 2 Log  δ(x)   δ(x) α2

α

if

2

To prove Proposition 7, we need the following two lemmas. The first one is a technical lemma given in [9]. The second one is due to Proposition 6. In the following, for x ∈ D we denote D1 = y ∈ D, |x − y|2 ≤ δ (x) δ(y)





and D2 = y ∈ D, |x − y|2 ≥ δ (x) δ(y) .





Lemma 2. Let x ∈ D; then we have the following. (i) If y ∈ D2 , then

√ max (δ(x), δ(y)) ≤

5+1 2

|x − y| .

(3.11)

(ii) If y ∈ D1 , then

√ 3−

√ 5

δ(x) ≤ δ(y) ≤

2

3+

√

5

2

δ(x) and

|x − y| ≤

√



1+

5

2

min (δ(x), δ(y)) .

(3.12)

In particular,

√



5−1

B x,

2





δ(x) ⊂ D1 ⊂ B x,

5+1 2

δ(x) .

Lemma 3. There exists a constant c > 0 such that, for each x ∈ D, we have

 log

2d



∫

dy

≈1+

δ(x)

D2

|x − y|n

.

Proof of Proposition 7. The upper bound estimate of GαD (q) is given in [9]. Let us prove the lower ones. If λ < α2 , the result follows immediately from (3.2). To prove the result for α2 ≤ λ < 1 + α2 , we need to show that

∫

GαD (x, y) D1

1

δ(y)λ

dy ≈ c δ(x)α−λ ,

for x ∈ D.

(3.13)

Using Lemma 2 and (1.2), we have, for x ∈ D,

∫

GαD (x, y) D1

1

δ(y)

dy ≈ λ

1

δ(x)λ

∫

dy

|x − y|n−α

D1

.

Now, using again Lemma 2, it follows that

∫

α

GD (x, y) D1

1

δ(y)λ

dy ≥

c

∫

dy  √  B x, 52−1 δ(x)

δ(x)λ

|x − y|n−α

√

≥

c

∫

δ(x)λ 0 ≥ c δ(x)α−λ .

5−1 δ(x) 2

r α−1 dr

The reverse inequality is obtained by a similar argument. This proves (3.13).

1566

R. Chemmam et al. / Nonlinear Analysis 74 (2011) 1555–1576

For the rest of the proof, we distinguish two cases.

< λ < 1 + α2 , then, going back to (3.13), we have, for x ∈ D, ∫ 1 dy GαD (q)(x) = GαD (x, y) δ(y)λ ∫D 1 GαD (x, y) ≥ dy δ (y)λ D1

• Case 1: If

α 2

≥ c δ(x)α−λ . α

• Case 2: If λ = 2 , then, using (1.2), we have, for x ∈ D, ∫ ∫ 1 dy α/2 GαD (x, y) dy ≥ c δ( x ) n. α/2 δ (y) D2 D2 |x − y| This, together with (3.13), implies that, for x ∈ D,

∫

∫

1

1

dy + GαD (x, y) dy δ(y)α/2 δ (y)α/2 D2 D1   ∫ dy ≥ c δ(x)α/2 1 + . | x − y|n D2

α

GD (q)(x) =

α

GD (x, y)

Then, using Lemma 3, we deduce that, for x ∈ D, α

α/2

GD (q)(x) ≥ c δ(x)

 log



2d

δ(x)

.

This completes the proof of Proposition 7.



Corollary 4. Let α2 − 1 ≤ β ≤ α2 ; then the function x → δ(x)β is comparable to an α -superharmonic function with respect to X D. Proof. If β = α2 − 1, the result follows immediately from (2.5).   If α2 − 1 < β < α2 , we put λ = α − β . Then λ ∈ α2 , 1 + α2 and so, by Proposition 7, we have



δ(x)β ≈ GαD Since GαD



1 δ(.)α−β



1

δ (.)α−β



(x),

for x ∈ D.

is an α -superharmonic function with respect to X D , we obtain the result.

For the case β = α2 , again using Proposition 7, we obtain, for x ∈ D, α

δ(x) 2 ≈ GαD (1)(x), which is α -superharmonic with respect to X D . This completes the proof.



Proposition 8. If α2 − 1 ≤ β ≤ α2 , then there exists c > 0 such that for each φ ∈ Kα (D)

∫  sup x∈D

D

δ(y) δ(x)

β

GαD (x, y) |ϕ(y)| dy ≤ caα (ϕ) .

(3.14)

Moreover, if x0 ∈ D, we have

 lim

r →0



∫ sup x∈D

B(x0 ,r )∩D

δ(y) δ ( x)

β

 α

GD (x, y) |ϕ(y)| dy

= 0.

Proof. Using Corollary 4, (3.8), and (3.9), we deduce (3.14) and (3.15).

(3.15)



Remark 3. For β = α2 , the inequality (3.14) proves that there exists c > 0 such that, for all ϕ ∈ Kα (D), we have ‖ϕ‖D ≤ caα (ϕ). This together with Proposition 3 implies that

‖ϕ‖D ≈ aα (ϕ) ,

for all ϕ ∈ Kα (D).

Corollary 5. Let ϕ ∈ Kα (D); then the function x → δ(x)α−1 ϕ(x) ∈ L1 (D).

R. Chemmam et al. / Nonlinear Analysis 74 (2011) 1555–1576

1567

Proof. Let x0 ∈ D; then, using (3.2) and (3.14), we obtain

∫

(δ(y))α−1 |ϕ(y)| dy ≤ c D

≤

∫ 

1

δ ( x0 ) c

δ (x0 )

D

δ(y) δ (x0 )

 α2 −1

GαD (x0 , y) |ϕ(y)| dy

aα (ϕ) < ∞. 

Proposition 9. Let q be the function defined in D by q(x) =

1

δ(x)λ

.

Then q belongs to Kα (D) if and only if λ < α . Proof. Suppose that the function q belongs to Kα (D). Then, from Corollary 5, the function x → δ(x)α−λ−1 ∈ L1 (D). Now, using [7, Lemma p. 726], we deduce that λ < α . Conversely, let λ < α . Using (3.1), we have that



λ 

1

δ(y)

δ(y) δ(x)

 α2

GαD (x, y) ≤ c

δ(y)α−λ α/2 .  |x − y|n−α |x − y|2 + δ(y)δ(x)

Using (3.4), we have



α/2 |x − y|2 + δ(y)δ (x) ≥ c (|x − y|α + (δ(y))α ) ≥ c |x − y|λ (δ (y))α−λ , +

+

where λ+ = max(λ, 0). Thus we obtain that



λ 

1

δ(y)

δ(y) δ(x)

 α2

GαD (x, y) ≤ c

≤

δ(y)α−λ |x − y|n−α |x − y|λ (δ(y))α−λ +

c +

|x − y|n−α+λ

+

.

Then it follows that



∫ sup x∈D

D∩B(x,r )

δ(y) δ(x)

 α2

GαD (x, y)



λ

1

δ(y)

r

∫ dy ≤ c

+ t α−λ −1 dt

0

≤ cr α−λ , +

which tends to zero as r → 0.



In the following, we give a large class of functions belonging to Kα (D). Proposition 10. Let p > αn and q ≥ 1 such that for some η > 0, the following conditions.

1 p

α− n

+

1 q

= 1. Let θ be a nonnegative continuous function in (0, 2d) satisfying,

(i) The function t → t p θ (t ) is nondecreasing on (0, η) and limt →0+ t (ii) The function t → max (θ (t ), 1) is nonincreasing on (0, η). (iii) The function t → t

1 α−1− n− p

α− np

θ (t ) = 0.

θ (t ) ∈ Lq ((0, η)).

Then, we have

θ (δ(.)) Lp (D) ⊂ Kα (D). Proof. Let p > αn and q ≥ 1 such that ϕ ∈ Lp (D). Let r > 0 and x ∈ D; then

∫

1 p

+

1 q

= 1. Let θ : (0, 2d) → [0, ∞) be a continuous function satisfying (i)–(iii) and

α δ(y) 2 α GD (x, y) |ϕ(y)| θ (δ(y)) dy δ(x) B(x,r )∩D  α  α ∫ ∫ δ(y) 2 α δ(y) 2 α = GD (x, y) |ϕ(y)| θ (δ(y)) dy + GD (x, y) |ϕ(y)| θ (δ(y)) dy δ(x) δ(x) B(x,r )∩D1 B(x,r )∩D2 = I1 (x) + I2 (x). 

We aim to show that I1 (x) and I2 (x) tend to zero as r → 0, uniformly in x.

1568

R. Chemmam et al. / Nonlinear Analysis 74 (2011) 1555–1576

Let us estimate I1 (x). Using (3.1) and (3.12), we have, for y ∈ D1 ,



δ(y) δ(x)

 α2

c

GαD (x, y) ≤

|x − y|n−α

.

Hence, by (ii) and the Hölder inequality, we get I1 ( x ) ≤ c

∫

1

|ϕ(y)| max (θ (δ(y)) , 1) dy n−α B(x,r )∩D1 |x − y|   ∫   √  1q 3− 5 (α−n)q |x − y| dy δ(x) , 1 ≤ c ‖ϕ‖p max θ . 2

B(x,r )∩D1

√

For simplicity, we put β = 3−2

√

, σ = 1+2 5 and ρ(x) = min (r , βδ (x)). So, by (3.12), we have, for y ∈ B(x, r ) ∩ D1 , σ σ |x − y| ≤ min (r , σ δ(x)) ≤ min (r , βδ (x)) = ρ(x). β β 5

So condition (ii) implies that I1 (x) ≤ c ‖ϕ‖p max (θ (βδ(x)) , 1)

σ ρ(x) β

∫

 1q t

n−1+(α−n)q

dt

0 n

≤ c ‖ϕ‖p max (θ (ρ(x)) , 1) (ρ(x))α− p   n n ≤ c ‖ϕ‖p max (ρ (x))α− p θ (ρ(x)) , (ρ(x))α− p . Since ρ(x) ≤ r, then by (i) we reach



I1 (x) ≤ c ‖ϕ‖p max r

α− np

 n θ (r ), r α− p ,

which implies by (i) that I1 (x) tends to zero as r → 0, uniformly in x. To control I2 (x), using (1.2), we have

(δ(y))α |ϕ(y)| dy. n θ (δ(y)) B(x,r )∩D2 |x − y|  n  n α− Writing t α θ (t ) = t p θ (t ) t p , we deduce from (i) that the function t → t α θ (t ) is nondecreasing. So, using (3.11), we I2 ( x ) ≤ c

∫

have I2 ( x ) ≤ c

∫



1 B(x,r )∩D2

|x − y|n−α

θ

√ 1+

5

2

 |x − y| |ϕ(y)| dy.

By the Hölder inequality, we obtain I2 (x) ≤ c ‖ϕ‖p

∫

  |x − y|

(α−n)q

B(x,r )∩D2

5

2

q |x − y|

 1q dy

 1q

√

 ∫ ≤ c ‖ϕ‖p 

θ

√ 1+

1+ 5 r 2



t



1 α−1− n− q p

(θ (t ))q dt  .

0

Now, using condition (iii), we deduce that I2 (x) tends to zero as r → 0, uniformly in x. This ends the proof.



To illustrate Proposition 10, we give the following example. Example 1. Let m ∈ N∗ and let β be a sufficiently large positive real number such that the function

θ (t ) = t

−λ

m  ∏

logk

k=1

β

−µk

t

is defined and positive on (0, 2d), where logk x = log ◦ log ◦ · · · ◦ log x (k times). Let p > αn ; then, if one of the following conditions is satisfied,

• λ < α − np and µk ∈ R for k ∈ N∗ , • λ = α − np , µ1 = µ2 = · · · = µk−1 = 1 − 1p , µk > 1 −

1 p

and µj ∈ R for j > k,

R. Chemmam et al. / Nonlinear Analysis 74 (2011) 1555–1576

1569

we have

θ (δ(.)) Lp (D) ⊂ Kα (D). 



−2

Remark 4. Let α ≥ 1; then the function q(x) = (δ(x))−α log δ(2dx)  Indeed, we have Jα (D) ⊂ L1 (D) but, by [7, Lemma p. 726], we have D

belongs to Kα (D). On the other hand, q ̸∈ Jα (D). 1

  2 dy (δ(y))α log δ(2dy)

= ∞.

Proposition 11. Let ϕ be a nonnegative function in Kα (D) and α2 − 1 ≤ β ≤ α2 . Then the function v defined in D by

v(x) :=

∫  D

δ(y) δ(x)

β

GαD (x, y) ϕ(y)dy,

is continuous in D. If, further, α2 − 1 ≤ β < α2 , then v is in C0 (D). Proof. Let ϕ be a nonnegative function in Kα (D) and x0 ∈ D; then, by (3.15), for each ε > 0, there exists r > 0 such that



∫ sup z ∈D

B(x0 ,2r )∩D

δ(y) δ(z )

β

GαD (z , y) ϕ(y)dy ≤ ε.

(3.16)

We distinguish two cases. Case 1: Suppose that x0 ∈ D, and let x, x′ ∈ B (x0 , r ) ∩ D. Then, by (3.16), we have

 β  β ∫     δ( y )  δ(y)  α ′ GαD (x, y) − G x , y   ϕ (y) dy D ′)   δ x δ x ( ) ( D β  ∫ δ(y) GαD (z , y) ϕ(y)dy ≤ 2 sup δ(z ) z ∈D B(x0 ,2r )∩D     ∫  δ(y) β δ(y) β α  ′   α + GD (x, y) − GD x , y  ϕ(y)dy   δ ( x′ ) Bc (x0 ,2r )∩D  δ(x)      ∫  δ(y) β δ (y) β α  ′   α ≤ 2ε + GD (x, y) − GD x , y  ϕ(y)dy   δ (x′ ) Bc (x0 ,2r )∩D  δ(x)

   v(x) − v x′  ≤

:= 2ε + I (x).     If |x0 − y| ≥ 2r , |x − x0 | ≤ r and x′ − x0  ≤ r, then |x − y| ≥ r and x′ − y ≥ r. Hence, it follows from (3.3) that      δ(y) β δ(y) β α  ′  c α  α GD (x, y) − GD x , y  ≤ n (δ(y)) 2 +β  ′  δ(x)  δ (x ) r ≤ c (δ(y))α−1 . Gα (x,y)

D Using the continuity of x → δ( outside the diagonal and particularly in D ∩ (|x0 − y| ≥ 2r ), we deduce by Corollary 5 x)β   and the dominated convergence theorem that I (x) tends to zero as x − x′  → 0. So the function v is continuous in D. Case 2: Suppose that x0 ∈ ∂ D, and let x ∈ B (x0 , r ) ∩ D. We aim to show that, if α2 − 1 ≤ β < α2 , we have limx→x0 v(x) = 0. Using (3.16), we obtain

   ∫ δ (y) β α δ (y) β α GD (x, y) ϕ(y)dy + GD (x, y) ϕ(y)dy δ(x) δ(x) D∩B(x0 ,2r ) D∩Bc (x0 ,2r )   ∫ δ(y) β α ≤ ε+ GD (x, y) ϕ(y)dy δ(x) D∩Bc (x0 ,2r ) := ε + J (x).

v(x) =

∫



If y ∈ D ∩ Bc (x0 , 2r ) then |x − y| ≥ r. So, using (3.3), we have, for all y ∈ D ∩ Bc (x, r ),



δ(y) δ(x)

β

α

GαD (x, y) ≤ c δ(y)α−1 (δ(x)) 2 −β .

Then, using the same argument as for I (x), we prove that J (x) tends to 0 as |x − x0 | → 0. So limx→x0 v(x) = 0.

1570

R. Chemmam et al. / Nonlinear Analysis 74 (2011) 1555–1576

α Now, let β = α2 . It is shown in [16, Lemma 6.5] that the function x → δ(x)− 2 GαD (x, y) is continuous in D outside the

diagonal, for each y ∈ D. Then, using a similar argument to that above, we show that v is continuous in D. This completes the proof.  Corollary 6. Let ϕ be a nonnegative function in Kα (D) and α2 − 1 ≤ β < α2 . Then the family of functions defined in D by

 Λ=

x → S (f )(x) :=

∫  D

δ(y) δ(x)

β

 α

GD (x, y) f (y)dy, |f | ≤ ϕ

is equicontinuous in D. Moreover, for each x0 ∈ ∂ D, we have limx→x0 S (f )(x) = 0, uniformly in f . Proof. Let ϕ be a nonnegative function in Kα (D) and f be a measurable function such that |f | ≤ ϕ . Then, for each x, x′ ∈ D, we have

 β  β ∫     δ( y )  δ(y)  α ′ GαD (x, y) − G x , y   ϕ (y) dy, D ′)   δ x δ x ( ) ( D   which, by the proof of Proposition 11, tends to zero as x − x′  → 0. Then the family Λ is equicontinuous in D. Now, if x0 ∈ ∂ D, then, for x ∈ D, we have  ∫  δ(y) β α |S (f )(x)| ≤ GD (x, y) ϕ(y)dy. δ(x) D   S (f )(x) − S (f )(x′ ) ≤

So, again using Proposition 11, we deduce that limx→x0 S (f )(x) = 0, uniformly in f .



4. First existence result First, we shall investigate an existence and a uniqueness result of the following nonlinear boundary problem:

(Pλ )

 α (−∆) 2 u = ϕ (., u)     u(x) =λ lim α x → z  z ∈∂ D MD 1(x)    u > 0,

in D

where λ > 0. To this aim, we need the following lemmas. Lemma 4. Let h be a nonnegative measurable function in D and v be a nonnegative α -superharmonic function with respect to X D . Let w be a measurable function in D such that GαD (h |w|) < ∞ and v = w + GαD (hw). Then 0 ≤ w ≤ v. Proof. Since GαD (h |w|) < ∞, we have

    w + GαD hw + = v + GαD hw − .         Hence, GαD hw + ≤ v + GαD hw − in [w > 0] = w + > 0 . Since v + GαD hw − is a nonnegative α -superharmonic function with respect to X D and GαD satisfies the complete maximum principle (see [5]), we have GαD hw + ≤ v + GαD hw −









in D.

That is, GαD (hw) ≤ v = w + GαD (hw). So w ≥ 0, and we deduce that 0 ≤ w ≤ w + GαD (hw) = v.



Lemma 5. Let ϕ : D × (0, ∞) → [0, ∞[ be a measurable function satisfying (H1 ) and (H2 ) and u be a positive continuous function in D such that lim

x→ z z ∈∂ D

u(x) MDα 1(x)

Then the function

= λ > 0.

Gα (ϕ(.,u)) D Mα 1 D

is in C0 (D).

(4.1)

α

In particular, the function x → δ(x) 2 ϕ (x, u(x)) ∈ L1 (D).

R. Chemmam et al. / Nonlinear Analysis 74 (2011) 1555–1576

Proof. Since the function x →

u(x) α 1(x) MD

1571

is continuous and positive in D, then it follows by (4.1) that u(x) ≈ MDα 1(x), for x ∈ D.

This implies by (2.5) that there exists c > 1 such that 1 c

α

α

(δ(x)) 2 −1 ≤ u(x) ≤ c (δ(x)) 2 −1 ,

for x ∈ D.

Now, using the monotonicity of ϕ , we have α

ψ(x) := (δ(x))1− 2 ϕ (x, u(x)) ≤ θ (x),   α α where θ (x) = (δ(x))1− 2 ϕ x, 1c (δ(x)) 2 −1 . By hypothesis (H2 ), the function θ is in Kα (D).

(4.2)

Now, we write that α

GαD (ϕ (., u)) (x)

(δ(x)) 2 −1 = MDα 1(x)

MDα 1(x)

∫  D

δ(y) δ(x)

 α2 −1

GαD (x, y) ψ(y)dy.

Then, using (2.5) and (4.2), the result holds from Corollary 6.



Theorem 7. Let ϕ : D ×(0, ∞) → [0, ∞[ be a measurable function satisfying (H1 ) and (H2 ). Then, for each λ > 0, the problem (Pλ ) has a unique continuous solution uλ in D satisfying

λMDα 1(x) ≤ uλ (x) ≤ γ MDα 1(x),

for x ∈ D,

where γ > 0. Proof. Let λ > 0. We shall convert the problem (Pλ ) into a suitable integral equation. So we aim to show an existence result for the following integral equation: u(x) = λMDα 1(x) +

∫

GαD (x, y) ϕ (y, u(y)) dy.

(4.3)

D

By (2.5), there exists c1 , c2 > 0 such that α

α

c1 (δ(x)) 2 −1 ≤ MDα 1(x) ≤ c2 (δ(x)) 2 −1 .

(4.4)

From hypothesis (H2 ), the nonnegative function θ defined on D by

  α α θ (x) = δ(x)1− 2 ϕ x, λc1 δ(x) 2 −1 , is in Kα (D). Put

γ =λ+

∫ 

1

sup c1 x∈D

D

δ(y) δ(x)

 α2 −1

GαD (x, y) θ (y)dy.

Then (3.14) implies that γ < ∞. Let Λ be the convex set given by

    Λ = v ∈ C D : λ ≤ v (x) ≤ γ . We consider the integral operator T on Λ defined by T v(x) = λ +

∫

1 α

MD 1(x)

GαD (x, y) ϕ y, MDα 1(y)v(y) dy,





x ∈ D.

D

We shall apply Schauder’s fixed theorem to prove the existence of a fixed point of T in Λ. First, we prove that

λ ≤ Tv ≤ γ ,

for all v ∈ Λ.

(4.5)

Let v ∈ Λ; it is clear that T v ≥ λ. So, using the monotonicity of ϕ and (4.4), we obtain, for x, y ∈ D, 1 MDα 1(x)

  1 ϕ y, MDα 1(y)v(y) ≤

c1



δ (y) δ(x)

 α2 −1

θ (y).

(4.6)

This proves (4.5). Now by Corollary 6, the family T Λ is equicontinuous in D, and we have lim T v (x) = λ,

x→z z ∈∂ D

uniformly for all v ∈ Λ.

In particular, for all v ∈ Λ, T v ∈ C D . This, together with (4.5), implies that T Λ ⊂ Λ.

 

(4.7)

1572

R. Chemmam et al. / Nonlinear Analysis 74 (2011) 1555–1576

On the other hand, from (4.5),   the family {T v(x), v ∈ Λ} is uniformly bounded in D. It follows by Ascoli’s theorem that T Λ is relatively compact in C D . Next, let us prove the continuity of T in Λ. We consider a sequence (vk ) in Λ which converges uniformly to a function v in Λ; then we have

|T vk (x) − T v(x)| ≤

∫

1 α

MD 1(x)

GαD (x, y) ϕ y, MDα 1(y)vk (y) − ϕ y, MDα 1(y)v (y)  dy.

 







D

By (4.6), we have

     ϕ y, M α 1(y)vk (y) − ϕ y, M α 1(y)v (y)  ≤ 2 (δ(y)) α2 −1 θ (y). D D Since ϕ is continuous with respect to the second variable, using (3.14), we deduce by the dominated convergence theorem that

∀x ∈ D ,

T vk (x) → T v(x) as k → ∞.

Since T Λ is a relatively compact family in C D , we have the uniform convergence, namely

 

‖T vk − T v‖∞ → 0 as k → ∞. Thus we have proved that T is a compact mapping from Λ to itself. Hence, by Shauder’s fixed theorem, there exists vλ ∈ Λ such that

vλ (x) = λ +

∫

1 α

MD 1 (x)

GαD (x, y) ϕ y, MDα 1(y)vλ (y) dy.





(4.8)

D

Put uλ (x) = MDα 1(x)vλ (x),

x ∈ D.

Then uλ is a continuous function in D that satisfies (4.3); that is, uλ (x) = λMDα 1(x) +

∫

GαD (x, y) ϕ (y, uλ (y)) dy,

(4.9)

D

and from (4.7) we have lim

x→ z z ∈∂ D

uλ ( x )

MDα 1(x)

= lim vλ (x) = λ. x→z z ∈∂ D

α

α

α

Now, since GD (ϕ (., uλ )) is continuous in D, it follows that y → δ(y) 2 ϕ (y, uλ (y)) ∈ L1 (D). So, applying (−∆) 2 on both sides of equality (4.9), we conclude that uλ satisfies (in the distributional sense) α

(−∆) 2 uλ = ϕ (., uλ )

in D.

This proves that uλ is a positive continuous solution of the problem (Pλ ). Finally, let us prove the uniqueness of such a solution. Let u be a continuous solution of (Pλ ). Then, using Lemma 5, we have

  α  α  (−∆) 2 u − GD (ϕ (., u)) = 0 in D, u − GαD (ϕ (., u)) (x)  = λ.  lim x→ z MDα 1(x) z ∈∂ D So Theorem 6 implies that u − GαD (ϕ (., u)) = λMDα 1

in D.

It follows that, if u and v are two continuous solutions of (Pλ ), then the function w = u − v satisfies

w + GαD (hw) = 0 in D, where h is the nonnegative measurable function defined in D by

  ϕ (x, u(x)) − ϕ (x, v (x)) , h(x) = u(x) − v(x)  0,

if u(x) ̸= v(x), if u(x) = v(x).

Using Lemma 4, we deduce that w = 0, and so u = v .



R. Chemmam et al. / Nonlinear Analysis 74 (2011) 1555–1576

1573

Proposition 12. Let ϕ : D × (0, ∞) → [0, ∞[ be a measurable function satisfying (H1 ) and (H2 ). Let 0 < µ ≤ λ; then we have 0 ≤ uλ − uµ ≤ (λ − µ) MDα 1,

in D,

where uλ , uµ are respectively the solutions of problems (Pλ ) and Pµ . Proof. Let h be the function defined on D by



     ϕ (x, uλ (x)) − ϕ x, uµ (x) , h(x) = uµ (x) − uλ (x)  0,



if uµ (x) ̸= uλ (x), if uµ (x) = uλ (x).

Then h is a nonnegative measurable function in D, and we have



uλ − uµ (x) + GαD h uλ − uµ



 



(x) = (λ − µ) MDα 1(x).

Moreover, by Lemma 5, we have

     GαD h uλ − uµ  ≤ GαD (ϕ (., uλ )) + GαD ϕ ., uµ < ∞. 



So, using Lemma 4 with v = (λ − µ) MDα 1 and w = uλ − uµ , we deduce the result.  Proof of Theorem 3. Let (λk ) be a sequence of positive real numbers, nonincreasing to zero. For each k ∈ N, put

µk = λk + sup x∈D

∫

1 α

MD 1(x)

GαD (x, y)ϕ y, λk MDα 1(y) dy,





D





and denote by uk the unique solution of the problem Pλk ; then uk (x) = λk MDα 1(x) +

∫

GαD (x, y)ϕ (y, uk (y)) dy,

x ∈ D.

D

From Proposition 12, the sequence (uk ) decreases to a function u. So, using the monotonicity of ϕ , the sequence uk − λk MDα 1 increases to u, and we have, for each x ∈ D,





u(x) ≥ uk (x) − λk MDα 1(x)

∫

GαD (x, y)ϕ (y, uk (y)) dy

= D

∫

GαD (x, y)ϕ y, µk MDα 1(y) dy > 0.



≥



D

Hence, applying the monotone convergence theorem and the continuity of ϕ with respect to the second variable, we get u(x) =

∫

GαD (x, y)ϕ (y, u(y)) dy,

x ∈ D.

D

Moreover, since u = sup uk − λk MDα 1 = inf uk ,





k

k

we deduce, from the continuity of uk and MDα 1 in D, that u is continuous in D. α Now, applying (−∆) 2 on both sides of equality (4.10), we deduce that u satisfies (in the distributional sense) α

(−∆) 2 u = ϕ (., u)

in D.

Finally, since, for each x ∈ D and k ∈ N, 0<

u(x)

≤

MDα 1(x)

uk ( x ) MDα 1(x)

it follows that lim

x→z z ∈∂ D

u(x) MDα 1(x)

= 0.

Then, from (2.5), we have α

lim (δ(x))1− 2 u(x) = 0.

x→z z ∈∂ D

This completes the proof.



,

(4.10)

1574

R. Chemmam et al. / Nonlinear Analysis 74 (2011) 1555–1576

Corollary 7. Let ϕ : D × (0, ∞) → [0, ∞) be a measurable function satisfying (H1 ) and (H2 ) and f : ∂ D → [0, ∞) be a continuous function. Then the problem

 α  (−∆) 2 u = ϕ (., u) in D u (x)  = f (z ),  lim x→z M α 1(x) D z ∈∂ D has a positive continuous solution u in D. Proof. Let ψ be the function defined on D × (0, ∞) by

  ψ(x, t ) = ϕ x, t + MDα f (x) . Then it is obvious to see that ψ satisfies the same hypotheses as ϕ . So using Theorem 3, the problem

 α  (−∆) 2 v = ψ (., v) v (x)  = 0,  lim x→z M α 1(x) D z ∈∂ D

in D

has a positive continuous solution v in D. Then the function u = v + MDα f is the desired solution.



As an application of Corollary 7, we give the following example. Example 2. Let γ > 0 and let k be a nonnegative measurable function in D such that x → (δ(x))(1− 2 )(1+γ ) k(x) belongs to Kα (D). Let f : ∂ D → [0, ∞) be a continuous function; then the problem α

 α −γ  (−∆) 2 u = k(x)u u (x)  = f (z )  lim x→z M α 1(x) D z ∈∂ D

in D

has a continuous positive solution in D. 5. Second existence result First, we need to recall some potential theory tools. For more details, we refer the reader to [6] or [8]. For a nonnegative measurable function q in D, we denote by Vq the kernel defined by Vq f (x) =

∞

∫



E x e−

t

 ( ) f X D  dt . t

D 0 q Xs ds

0

One can see, by (1.1), that the kernel V := V0 is the potential kernel GαD . Using the Markov property, we have that, for each nonnegative measurable function q such that Vq < ∞, the kernel Vq satisfies the following resolvent equation: V = Vq + Vq (qV ) = Vq + V qVq .





(5.1)

It follows that, for each measurable function u in D such that V (q|u|) < ∞, we have I − Vq (q.) (I + V (q.)) u = (I + V (q.)) I − Vq (q.) u = u.









(5.2)

The following lemma are essential for the proof of Theorem 4. Lemma 6. Let ϕ be a nonnegative function in Kα (D) and h be an α -superharmonic function with respect to X D . Then, for all x ∈ D such that 0 < h(x) < ∞, we have exp (−aα (q)) h(x) ≤ h(x) − Vq (qh)(x) ≤ h (x) . Proof. The upper inequality is obviously satisfied. Let us prove the lower one. Since, by [26], h is excessive with respect to X D , using [5, Proposition 3.11 p. 51], there exists a sequence (fk )k of nonnegative measurable function in D such that h = supk Vfk . Let x ∈ D such that 0 < h(x) < ∞; then there exists k0 ∈ N such that 0 < Vfk (x) < ∞, for each k ≥ k0 . Let k ≥ k0 and consider θ (t ) = Vt .q fk (x), for t ≥ 0. Then the function θ is completely monotone on [0, ∞), and so Log θ is convex on [0, ∞). It follows that

 ′  θ (0) θ (0) ≤ θ (1) exp − , θ (0)

R. Chemmam et al. / Nonlinear Analysis 74 (2011) 1555–1576

1575

which means that Vfk (x) ≤ Vq fk (x) exp



V (qVfk ) (x)



Vfk (x)

.

Since Vfk is α -superharmonic with respect to X D , we deduce from Proposition 5 that Vfk (x) ≤ Vq fk (x) exp (aα (q)) . This, together with (5.1), yields exp (−aα (q)) Vfk (x) ≤ Vfk (x) − Vq (qVfk ) (x). So, the result holds by letting k → ∞.



Proof of Theorem 4. Let c2 be the constant given in (4.4), and denote by ‖f ‖ = supx∈∂ D |f (x)|. To simplify our statement, we put q := qc as the function in Kα (D) due to hypothesis (H4 ) for c := c2 ‖f ‖. We shall state the existence of a solution for the fractional integral equation u + V (uϕ (., u)) = MDα f

in D,

(5.3)

belonging to the convex set

  Λ = u measurable function in D : exp (−aα (q)) MDα f ≤ u ≤ MDα f . Since, by (4.4), we have α

0 < MDα f (x) ≤ ‖f ‖MDα 1(x) ≤ c2 ‖f ‖(δ(x)) 2 −1 ,

x ∈ D,

it follows from (H4 ) that 0 ≤ ϕ (., u) ≤ q,

for all u ∈ Λ.

(5.4)

We define the operator T on Λ by Tu(x) = MDα f (x) − Vq qMDα f (x) + Vq [(q − ϕ (., u)) u] (x),





x ∈ D.

We aim to show that T has a fixed point u in Λ. To this end, we first prove that T Λ ⊂ Λ. Indeed, for u ∈ Λ, we have Tu ≤ MDα f − Vq qMDα f + Vq (qu) ≤ MDα f ,





and from (5.4) and Lemma 6, we have Tu ≥ MDα f − Vq qMDα f ≥ exp (−aα (q)) MDα f .





Next, we prove that T is nondecreasing on Λ. Let u, v ∈ Λ such that u ≤ v ; then, using (H4 ), we deduce that T v − Tu = Vq [(q − ϕ (., v)) v − (q − ϕ (., u)) u] ≥ 0. Now, we consider the sequence (uk ) defined by u0 = exp (−aα (q)) MDα f

and uk+1 = Tuk ,

for k ∈ N.

Since T Λ ⊂ Λ, then, from the monotonicity of T , we immediately obtain that u0 ≤ u1 ≤ · · · ≤ uk ≤ uk+1 ≤ MDα f . Hence, by (H4 ) and the dominated convergence theorem, the sequence (uk ) converges to a function u ∈ Λ which satisfies u = MDα f − Vq qMDα f + Vq [(q − ϕ (., u)) u] ,





which means that



I − Vq (q.) u + Vq (ϕ (., u) u) = I − Vq (q.) MDα f .







So, applying the operator (I + V (q.)) on both sides of the last equality, it follows by (5.1) and (5.2) that u is a solution of (5.3). To achieve the proof, we need to show that u is a solution of (1.6). Going back to (5.4), we have 0 ≤ uϕ (., u) ≤ qMDα f . Hence, from (4.4), we get α

0 ≤ u(x)ϕ (x, u(x)) ≤ c2 (δ(x)) 2 −1 q(x).

(5.5)

1576

R. Chemmam et al. / Nonlinear Analysis 74 (2011) 1555–1576 α

Since q ∈ Kα (D), it follows by Corollary 6 that x → (δ(x))1− 2 V (uϕ (., u)) (x) is in C0 (D), and so V (uϕ (., u)) is continuous α α in D. Then, by (5.3), u is continuous in D, and the map x → δ(x) 2 u(x)ϕ (x, u(x)) ∈ L1 (D). Hence, applying (−∆) 2 on both sides of (5.3), it follows that u is a continuous solution of α

(−∆) 2 u + uϕ (., u) = 0 in D (in the distributional sense) satisfying (1.7). Now, by (5.5) and (4.4), we have 0≤

V (uϕ (., u)) (x) MDα 1(x)

≤

c2 c1

∫  D

δ(y) δ(x)

 α2 −1

GαD (x, y) q(y)dy.

It follows, from Proposition 11, that lim

V (uϕ (., u)) (x) MDα 1(x)

x→z

= 0,

for z ∈ ∂ D.

So, again using (5.3), we deduce that lim

x→z

u(x) MDα 1(x)

= f (z ),

This completes the proof.

for z ∈ ∂ D. 

As an application of Theorem 4, we give the following example. Example 3. Let p > 0 and let k be a nonnegative measurable function in D such that the map x → (δ(x))( 2 −1)p k(x) belongs to Kα (D). Let f be a nonnegative, nontrivial continuous function on ∂ D; then the problem α

 α p+1  = 0 in D (−∆) 2 u + k(x)u u (x)  = f (z ),  lim x→z M α 1(x) D z ∈∂ D has a continuous positive solution in D satisfying (1.7). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26]

A. Janicki, A. Weron, Simulation and Chaotic Behavior of α -Stable Processes, Dekker, 1994. J. Klater, M.F. Shlesinger, G. Zumofen, Beyond Brownian motion, Phys. Today 49 (2) (1996) 33–39. M. Aizenman, B. Simon, Brownian motion and Harnack inequality for Schrödinger operators, Comm. Pure Appl. Math. 35 (1982) 209–271. S. Ben Othman, H. Mâagli, S. Masmoudi, M. Zribi, Exact asymptotic behavior near the boundary to the solution for singular nonlinear Dirichlet problems, Nonlinear Anal. 71 (2009) 4137–4150. J. Bliedtner, W. Hansen, Potential Theory. An Analytic and Probabilistic Approach to Balayage, Springer-Verlag, Berlin, New York, 1986. K.L. Chung, Z. Zhao, From Brownian Motion to Schrödinger’s Equation, Springer Verlag, 1995. A.C. Lazer, P.J. Mckenna, On a singular nonlinear elliptic boundary value problem, Proc. Amer. Math. Soc. 111 (1991) 721–730. H. Mâagli, Perturbation semi-linéaire des résolvantes et des semi-groupes, Potential Anal. 3 (1994) 61–87. H. Mâagli, Inequalities for the Riesz potentials, Arch. Inequal. Appl. 1 (2003) 285–294. H. Mâagli, M. Zribi, Existence and estimates of solutions for singular nonlinear elliptic problems, J. Math. Anal. Appl. 263 (2001) 522–542. H. Mâagli, M. Zribi, On a new Kato class and singular solutions of a nonlinear elliptic equation in bounded domain of Rn , Positivity 9 (2005) 667–686. S. Port, C. Stone, Brownian Motion and Classical Potential Theory, in: Probab. Math. Statist., Academic Press, New York, 1978. K.O. Widman, Inequalities for the Green function and boundary continuity of the gradient of solutions of elliptic differential equations, Math. Scand. 21 (1967) 17–37. Z. Zhao, Green function for Schrödinger operator and conditional Feynman–Kac gauge, J. Math. Anal. Appl. 116 (1986) 309–334. N. Zeddini, Positive solutions for a singular nonlinear problem on a bounded domain in R2 , Potential Anal. 18 (2003) 97–118. Z.Q. Chen, R. Song, Estimates on Green functions and Poisson kernels for symmetric stable processes, Math. Ann. 312 (1998) 465–501. K. Bogdan, T. Byczkowski, Potential theory for the α -stable Schrödinger operator on bounded Lipschitz domains, Studia Math. 133 (1999) 53–92. Z.Q. Chen, R. Song, General gauge and conditional gauge theorems, Ann. Probab. 30 (2002) 1313–1339. P. Kim, Relative Fatou’s theorem for (−∆)α/2 -harmonic functions in bounded κ -fat open sets, J. Funct. Anal. 234 (2006) 70–105. Z.Q. Chen, P. Kim, R. Song, Heat kernel estimates for Dirichlet fractional Laplacian, J. Eur. Math. Soc. 12 (2010) 1307–1329. K. Bogdan, Representation of α -harmonic functions in Lipschitz domains, Hiroshima Math. J. 29 (1999) 227–243. Z.Q. Chen, R. Song, Martin boundary and integral representation for harmonic function of symmetric processes, J. Funct. Anal. 159 (1998) 267–294. K. Bogdan, The boundary Harnack principle for the fractional Laplacian, Studia Math. 123 (1) (1997) 43–80. R.M. Blumenthal, R.K. Getoor, D.B. Ray, On the distribution of first hits for the symmetric stable processes, Trans. Amer. Math. Soc. 99 (1961) 540–554. F. Hmissi, Fonctions harmoniques pour les potentiels de Riesz sur la boule unité, Expo. Math. 12 (1994) 281–288. Z.Q. Chen, P. Kim, Stability of Martin boundary under nonlocal Feynman–Kac perturbations, Probab. Theory Related Fields 128 (2004) 525–564.