Int. J. Anal. Appl. (2022), 20:13

An Extension of a Variational Inequality in the Simader Theorem to a Variable

Exponent Sobolev Space and Applications: The Dirichlet Case

Junichi Aramaki∗

Division of Science, Tokyo Denki University, Hatoyama-machi, Saitama, 350-0394, Japan

∗Corresponding author: aramaki@hctv.ne.jp

Abstract. In this paper, we shall extend a fundamental variational inequality which is developed by

Simader in W1,p to a variable exponent Sobolev space W1,p(·). The inequality is very useful for the

existence theory to the Poisson equation with the Dirichlet boundary conditions in Lp(·)-framework,

where Lp(·) denotes a variable exponent Lebesgue space. Furthermore, we can also derive the existence

of weak solutions to the Stokes problem in a variable exponent Lebesgue space.

1. Introduction

In Simader [25], the author derived a variational inequality of a bilinear form. More precisely, let G

is a bounded domain of Rd (d ≥ 2) with a C1-boundary ∂G and 1 < p < ∞. The author proved that
there exists a positive constant C = C(p,G) > 0 such that

‖∇∇∇u‖Lp(G) ≤ C sup
06=v∈W 1,p

′
0 (G)

|〈∇∇∇u,∇∇∇v〉G|
‖∇∇∇v‖

Lp
′
(G)

for all u ∈ W 1,p0 (G), (1.1)

where 〈∇∇∇u,∇∇∇v〉G =
∫
G
∇∇∇u ·∇∇∇vdx, ∇∇∇ denotes the gradient operator and p′ is the conjugate exponent

of p, that is, 1
p

+ 1
p′ = 1. He also considered the case where G is an exterior domain and got a

variational inequality like as in (1.1).

This inequality has many applications. For example, let v ∈Lp(G), then it follows from (1.1) that
the Dirichlet problem for the Poisson equation{

∆u = divv in G,

u = 0 on ∂G
(1.2)

Received: Jan. 5, 2022.

2010 Mathematics Subject Classification. 35A15; 58E35; 35J25; 35D05; 35B30.
Key words and phrases. variational inequality; Dirichlet problem for the Poisson equation; the Helmholtz decomposi-

tion; the Stokes problem; variable exponent Sobolev spaces.

https://doi.org/10.28924/2291-8639-20-2022-13
ISSN: 2291-8639

© 2022 the author(s).

https://doi.org/10.28924/2291-8639-20-2022-13


2 Int. J. Anal. Appl. (2022), 20:13

has a unique solution in a generalized sense. The equation (1.2) plays an essential role for the existence

of a solution to the Stokes problem (cf. Fujiwara and Morimoto [15] and Kozono and Yanagisawa [20]).

It is also basic for the treatment of the Navier-Stokes equation, for example, see [15], Miyakawa [22].

In this paper, we attempt to derive an improvement of the above variational inequality (1.1) in the

Sobolev opace W 1,p0 (G) to a variable exponent Sobolev space W
1,p(·)
0 (G) (Theorem 3.1). We restrict

ourselves to the case where G is a bounded domain. Though we follow the argument of [25], we have

to proceed the analysis very carefully. The result brings about the existence theory of weak solutions

to the Dirichlet problem for the Laplacian in the variable exponent Sobolev space, that is, for given

functions f ∈ W−1,p(·)(G) and g ∈ Tr(W 1,p(·)(G)), where W 1,p(·)(G) is a variable exponent Sobolev
space, W−1,p(·)(G) is the dual space of W 1,p

′(·)
0 (G) and Tr(W

1,p(·)(G)) denotes the trace space,{
−∆u = f in G,
u = g on ∂G

has a unique weak solution. According to our best knowledge, the result for the Dirichlet problem in

a variable exponent Sobolev space is simplest. Furthermore, we show that the Stokes problem in a

variable exponent Sobolev space has a unique weak (strong) solution by a new approach which is an

application of Theorem 3.1.

The study of differential equations with p(·)-growth conditions is a very interesting topic recently.
Studying such problem stimulated its application in mathematical physics, in particular, in elastic

mechanics (Zhikov [29]), in electrorheological fluids (Diening [10], Halsey [18], Mihăilescu and Răd-

ulescu [21], Růžička [23]).

For the Neumann case of the variational inequality, we gave a result in the previous work Aramaki [5]

(cf. Simader and Sohr [24] for the case p(·) = p = const.).
The paper is organized as follows. In section 2, we give some preliminaries on variable exponent

Lebesgue-Sobolev spaces. In section 3, we give main theorems (Theorem 3.1) which is an extension

of variational inequality of type (1.1) to one in a variable exponent Sobolev space. Section 4 is a

preparation of a proof of the main theorem. In section 5, we give a proof of the main theorem. In

section 6, we consider the Dirichlet problem of the Poisson equation. Finally, section 7 is devoted to

the existence of a weak (strong) solution for the Stokes problem by a new approach.

2. Preliminaries

Throughout this paper, we only consider vector spaces of real valued functions over R. For any
space B, we denote Bd by the boldface character B. Hereafter, we use this character to denote vectors

and vector-valued functions, and we denote the standard inner product of vectors a = (a1, . . . ,ad)

and b = (b1, . . . ,bd) in Rd by a ·b =
∑d
i=1 aibi and |a| = (a ·a)

1/2. Occasionally, we also use the

same character for matrix values functions. Moreover, for the dual space B′ of B (resp. B′ of B), we

denote the duality bracket between B′ and B (resp. B′ and B) by 〈·, ·〉B′,B (resp. 〈·, ·〉B′,B).



Int. J. Anal. Appl. (2022), 20:13 3

In this section, we recall some well-known results on variable exponent Lebesgue-Sobolev spaces. See

Diening et al. [11], Fan and Zhao [14], Fan and Zhang [12], Kovác̆ik and Rácosnic [19] and references

therein for more detail. Let G be a (Lebesgue) measurable subset of Rd (d ≥ 2) with the measure
|G| > 0. Then we define a set of variable exponents by P(G) = {p; G → [1,∞); p is measurable in G}
and for p ∈P(G), define

p− = ess inf
x∈G

p(x) and p+ = ess sup
x∈G

p(x).

For any real valued measurable function u on G and p ∈P(G), a modular ρp(·),G is defined by

ρp(·),G(u) =

∫
G

|u(x)|p(x)dx.

The variable exponent Lebesgue space is defined by

Lp(·)(G) =
{
u; u is a measurable function on G satisfying ρp(·),G(u) < ∞

}
equipped with the Luxemburg norm

‖u‖Lp(·)(G) = inf
{
λ > 0; ρp(·),G

(u
λ

)
≤ 1
}
.

Then Lp(·)(G) is a Banach space and it is separable if p+ < ∞ and reflexive if 1 < p− ≤ p+ < ∞.
Define

P+(G) = {p ∈P(G); 1 < p− ≤ p+ < ∞}.

The following proposition is well known (see Fan et al. [13], Wei and Chen [26], [14], Zhao et

al. [28], Yücedağ [27]).

Proposition 2.1. Let G be a measurable set of Rd, p ∈P+(G) and let u,un ∈ Lp(·)(G) (n = 1, 2, . . .).
Then we have

(i) ‖u‖Lp(·)(G) < 1(= 1,> 1) ⇐⇒ ρp(·),G(u) < 1(= 1,> 1).
(ii) ‖u‖Lp(·)(G) > 1 =⇒‖u‖

p−

Lp(·)(G)
≤ ρp(·),G(u) ≤‖u‖

p+

Lp(·)(G)
.

(iii) ‖u‖Lp(·)(G) < 1 =⇒‖u‖
p+

Lp(·)(G)
≤ ρp(·),G(u) ≤‖u‖

p−

Lp(·)(G)
.

(iv) limn→∞‖un −u‖Lp(·)(G) = 0 ⇐⇒ limn→∞ρp(·),G(un −u) = 0.
(v) ‖un‖Lp(·)(G) →∞ as n →∞⇐⇒ ρp(·),G(un) →∞ as n →∞.

The following proposition is a generalized Hölder inequality.

Proposition 2.2. Let G be a measurable set of Rd and p ∈ P+(G). For any u ∈ Lp(·)(G) and
v ∈ Lp

′(·)(G), we have∫
G

|uv|dx ≤
(

1

p−
+

1

(p′)−

)
‖u‖Lp(·)(G)‖v‖Lp′(·)(G) ≤ 2‖u‖Lp(·)(G)‖v‖Lp′(·)(G),

where p′(·) is the conjugate exponent of p(·), that is, 1
p(x)

+ 1
p′(x)

= 1.



4 Int. J. Anal. Appl. (2022), 20:13

When G is a domain (open and connected subset) of Rd and p ∈P+(G), we can define a Sobolev
space, for any integer m ≥ 0,

Wm,p(·)(G) = {u ∈ Lp(·)(G); ∂αu ∈ Lp(·)(G) for |α| ≤ m},

where α = (α1, . . . ,αd) is a multi-index, |α| =
∑d
i=1 αi, ∂

α = ∂
α1
1 · · ·∂

αd
d

and ∂i = ∂/∂xi, endowed

with the norm

‖u‖Wm,p(·)(G) =
∑
|α|≤m

‖∂αu‖Lp(·)(G).

Of course, W 0,p(·)(G) = Lp(·)(G). The local Sobolev space is defined by

W
m,p(·)
loc

(G) = {u; for all open subset U b G,u ∈ Wm,p(·)(U)},

where U b G means that the closure U of U is compact and U ⊂ G.
For p ∈P+(G), define

p∗(x) =

{
dp(x)
d−p(x) if p(x) < d,

∞ if p(x) ≥ d

and

p∂(x) =

{
(d−1)p(x)
d−p(x) if p(x) < d,

∞ if p(x) ≥ d.

Proposition 2.3. Let p ∈P+(G) and m ≥ 0 be an integer. Then we can see the following properties.
(i) The space Wm,p(·)(G) is a separable and reflexive Banach space.

(ii) Let G be a bounded domain of Rd. If q(·) ∈ P+(G) satisfies q(x) ≤ p(x) for all x ∈ G, then
Wm,p(·)(G) ↪→ Wm,q(·)(G), where ↪→ means that the embedding is continuous.

(iii) Let G be a bounded domain of Rd. If p,q ∈P+(G) ∩C(G) satisfies that q(x) < p∗(x) for all
x ∈ G, then the embedding W 1,p(·)(G) ↪→ Lq(·)(G) is compact.

Next we consider the trace. Let G be a domain of Rd with a Lipschitz-continuous boundary ∂G
and p ∈ P+(G). Since Wm,p(·)(G) ⊂ Wm,1loc (G), the trace γ0(u) = u

∣∣
∂G

to ∂G of any function u in

Wm,p(·)(G) is well defined as a function in L1loc(∂G). We define

Tr(Wm,p(·)(G)) = {f ; γ0(u) = f for some function u ∈ Wm,p(·)(G)}

equipped with the norm

‖f‖Tr(Wm,p(·)(G)) = inf{‖u‖Wm,p(·)(G); u ∈ W
m,p(·)(G) satisfying γ0(u) = f on ∂G}

for f ∈ Tr(Wm,p(·)(G)). Then Tr(Wm,p(·)(G)) is a Banach space. More precisely, see [11, Chapter
12]. We define a space

◦
W

m,p(·)(G) = Wm,p(·)(G) ∩Wm,10 (G).



Int. J. Anal. Appl. (2022), 20:13 5

For a general measurable subset G of Rd, we say that p ∈ Plog(G) if p ∈ P+(G) and p has the
globally log-Hölder continuity and globally log-Hölder decay condition in G, that is, p : G → R satisfies
that there exist a constant Clog(p) > 0 and p∞ ∈ R such that the following inequalities hold:

|p(x) −p(y)| ≤
Clog(p)

log(e + 1/|x −y|)
for all x,y ∈ G,

and

|p(x) −p∞| ≤
Clog(p)

log(e + |x|)
for all x ∈ G,

respectively.

Proposition 2.4. If G is a domain of Rd and p ∈Plog(G), then p has an extension q ∈Plog(Rd) with
Clog(q) = Clog(p), q− = p− and q+ = p+. If G is unbounded, then additionally q∞ = p∞.

For the proof, see [11, Proposition 4.1.7].

We note that if G is bounded, the global log-Hölder decay condition always holds. For a domain

G, we write Plog+ (G) = P
log(G) ∩P+(G). Let G be a domain of Rd and p ∈P

log
+ (G), define

W
m,p(·)
0 (G) = the closure of C

∞
0 (G) in W

m,p(·)(G).

From definition, we can easily see that Wm,p(·)0 (G) ⊂
◦
W

m,p(·)(G) and
◦
W

m,p(·)(G) is a closed sub-

space of Wm,p(·)(G). When p(·) = p is a constant, Wm,p0 (G) =
◦
W

m,p(G), however, in general,

W
m,p(·)
0 (G) (

◦
W

m,p(·)(G).

Theorem 2.5. If G is a bounded domain with a Lipschitz-continuous boundary ∂G and p ∈Plog+ (G),
then

(i) C∞(G) is dense in Wm,p(·)(G).

(ii) W
m,p(·)
0 (G) =

◦
W

m,p(·)(G), In addition, when m ≥ 3, assume that G has a Cm,1-boundary. Then
we have

W
m,p(·)
0 (G) = {u ∈ W

m,p(·)(Ω); γ0(u) = · · · = γm−1(u) = 0 a.e. on ∂G},

where γj(u) =
∂ju
∂nj

=
∑
|α|=j n

α∂αu, n = (n1, . . . ,nd) is the outer unit normal vector to ∂G and

nα = n
α1
1 · · ·n

αd
d
.

For the proof, see [14, Theorem 2.6] and Galdi [16, Theorem 3.2].

Lemma 2.6. Let G be a bounded domain of Rd with a Lipschitz-continuous boundary ∂Ω and let
p ∈ Plog+ (G). Assume that q ∈ P

log
+ (∂Ω) such that q(x) < p

∂(x) for all x ∈ ∂G. Then the trace
operator Tr = γ0 : W 1,p(·)(G) → Lq(·)(∂G) is compact, In particular, Tr : W 1,p(·)(G) → Lp(·)(∂G) is
compact.

For the proof, see Deng [9, Theorem 2.1].

Frequently we use the following Poincaré inequality later.



6 Int. J. Anal. Appl. (2022), 20:13

Theorem 2.7. (i) If G is a bounded domain of Rd and p ∈ Plog+ (G), then there exists a constant c
depending only on d and Clog(p) such that

‖u‖Lp(·)(G) ≤ c diam(G)‖∇∇∇u‖Lp(·)(G) for all u ∈ W
1,p(·)
0 (G),

where diam(G) denotes the diameter of G.

(ii) If G is a bounded domain of Rd with a Lipschitz-continuous boundary ∂G and p ∈ Plog+ (G),
then there exists a constant c depending only on d and Clog(p) such that

‖u −〈u〉G‖Lp(·)(G) ≤ c diam(G)‖∇∇∇u‖Lp(·)(G) for all u ∈ W
1,p(·)(G),

where 〈u〉G = 1|G|
∫
G
udx.

For the proof, see [11, Theorem 8.2.4].

Corollary 2.8. Let G be a bounded domain of Rd with a Lipschitz-continuous boundary and let
p ∈Plog+ (G). Furthermore, let A ⊂ G such that |A| ≈ |G|. Then there exists a constant c depending
only on d and Clog(p) such that

‖u −〈u〉A‖Lp(·)(G) ≤ c diam(G)‖∇∇∇u‖Lp(·)(G) for all u ∈ L
1
loc(G) with ∇∇∇u ∈L

p(·)(G).

For the proof, see [11, Corollary 8.2.6].

We introduce a generalized Poincaré inequality which is found in Ciarlet and Dinca [8, Theorem

4.1].

Theorem 2.9. Let G be a bounded domain of Rd with a Lipschitz-continuous boundary Γ = ∂Ω, G
being locally on the same side of Γ. Moreover, let Γ0 be a measurable subset of Γ such that |Γ0| > 0,
and let p ∈Plog+ (G). Define

U = {v ∈ W 1,p(·)(G); v
∣∣
Γ0

= 0}.

Then there exists a constant C = C(p,d,U) such that

‖v‖Lp(·)(G) ≤ Cdiam(G)‖∇∇∇v‖Lp(·)(G) for all v ∈ U.

3. The weak Dirichlet problem for the Laplacian ∆ in a variable exponent Sobolev space in a

bounded domain

In this section, we state main theorems of this paper. Let G be a bounded domain of Rd (d ≥ 2)
and p ∈ Plog+ (G). Then taking the Poincaré inequality (Theorem 2.7) into consideration, we may
assume that the space W 1,p(·)0 (G) is equipped with the norm ‖∇∇∇v‖Lp(·)(G).

The first theorem is a variational inequality in W 1,p(·)0 (G).



Int. J. Anal. Appl. (2022), 20:13 7

Theorem 3.1. Let G be a bounded domain of Rd (d ≥ 2) with a C1-boundary and p ∈ Plog+ (G).
Then there exists a constant Cp = C(p,G) > 0 such that

‖∇∇∇u‖Lp(·)(G) ≤ Cp sup
06=v∈W 1,p

′(·)
0 (G)

|〈∇∇∇u,∇∇∇v〉G|
‖∇∇∇v‖

Lp
′(·)(G)

for all u ∈ W 1,p(·)0 (G). (3.1)

The second theorem is a functional representation in W 1,p(·)0 (G) which shows existence of weak

solution to the Dirichlet problem of the Poisson equation in a bounded domain G.

Theorem 3.2. Let G be a bounded domain of Rd (d ≥ 2) with a C1-boundary and p ∈Plog+ (G). For
every F ′ ∈ (W 1,p

′(·)
0 (G))

′ = W−1,p(·)(G), there exists a unique u ∈ W 1,p(·)0 (G) such that

F ′(v) = 〈∇∇∇u,∇∇∇v〉G for all v ∈ W
1,p′(·)
0 (G). (3.2)

Furthermore, with Cp > 0 in Theorem 3.1, the following inequality holds.

C−1p ‖∇∇∇u‖Lp(·)(G) ≤‖F
′‖

(W
1,p′(·)
0 (G))

′ ≤ 2‖∇∇∇u‖Lp(·)(G), (3.3)

where

‖F ′‖
(W

1,p′(·)
0 (G))

′ = sup{|F
′(v)|; v ∈ W 1,p

′(·)
0 (G) and ‖∇∇∇v‖Lp′(·)(G) ≤ 1}.

Before a proof of Theorem 3.1, we show that Theorem 3.1 implies Theorem 3.2. For this purpose,

we use the following proposition (cf. Amrouche and Seloula [1, Theorem 4.2]).

Proposition 3.3. Let X and M be two reflexive Banach spaces, and a be a continuous bilinear form

defined on X ×M. Assume that A ∈L(X,M′) and A′ ∈L(M,X′) are operators defined by

〈Au,v〉 = 〈u,A′v〉 = a(u,v) for u ∈ X,v ∈ M

and put V = KerA. Then the following statements are equivalent.

(i) There exists β > 0 such that

inf
0 6=w∈M

sup
0 6=v∈X

a(v,w)

‖v‖X‖w‖M
≥ β. (3.4)

(ii) A : X/V → M′ is an isomorphism and 1/β is the continuity constant of A−1.
(iii) A′ : M → V ⊥ := {f ∈ X′;〈f ,v〉 = 0 for all v ∈ V} is also an isomorphism and 1/β is the

continuity constant of (A′)−1.

Let X = (W 1,p(·)0 (G),‖∇∇∇·‖Lp(·)(G)) and M = (W
1,p′(·)
0 (G),‖∇∇∇·‖Lp′(·)(G)). Since p,p

′ ∈ Plog+ (G),
X and M are reflexive Banach spaces (Proposition 2.3). Define

a(u,v) = 〈∇∇∇u,∇∇∇v〉G =
∫
G

∇∇∇u ·∇∇∇vdx for u ∈ X,v ∈ M.

Then clearly a is a bilinear form on X × M, and it follows from the generalized Hölder inequality
(Proposition 2.2) that a is continuous. If u ∈ KerA, then a(u,v) = 〈Au,v〉 = 0 for all v ∈ W 1,p

′(·)
0 (G).

From Theorem 3.1, we have ∇∇∇u = 000 in Lp(·)(G). From the Poincaré inequality, we have u = 0 in



8 Int. J. Anal. Appl. (2022), 20:13

Lp(·)(G), so KerA = {0}. From Theorem 3.1, (i) in Proposition 3.3 holds with β = 1/Cp. Thus for
any F ′ ∈ (W 1,p

′(·)
0 (G))

′ = M′, there exists uniquely u ∈ W 1,p(·)0 (G) such that F
′ = Au, that is,

F ′(v) = 〈Au,v〉 = 〈∇∇∇u,∇∇∇v〉G for all v ∈ W 1,p
′(·)(G)

and

‖∇∇∇u‖Lp(·)(G) ≤
1

β
‖F ′‖

(W
1,p′(·)
0 (G))

′.

Therefore,

C−1p ‖∇∇∇u‖Lp(·)(G) ≤‖F
′‖

(W
1,p′(·)
0 (G))

′.

Since by the generalized Hölder inequality (Proposition 2.2),

‖F ′‖
(W

1,p′(·)
0 (G))

′ = sup{|F
′(v)|; v ∈ W 1,p

′(·)
0 (G),‖∇∇∇v‖Lp′(·)(G) ≤ 1}

= sup{|〈∇∇∇u,∇∇∇v〉G|; v ∈ W
1,p′(·)
0 (G),‖∇∇∇v‖Lp′(·)(G) ≤ 1}

≤ 2‖∇∇∇u‖Lp(·)(G),

we can see that (3.3) holds.

4. Preparation to a proof of Theorem 3.1

We use the localization method for a proof of Theorem 3.1. For any open set G ⊂ Rd (d ≥ 2)
(not necessarily bounded), we say that G satisfies (GA) if G has a C1-boundary and there exists a

non-empty open set K in Rd such that G = Rd \K.

Definition 4.1. If G satisfies (GA) and p ∈Plog+ (G), define

Ŵ
1,p(·)
0 (G) = {v : G → R; v is measurable in G,v ∈ L

p(·)(GR) for each R > 0,

∇∇∇v ∈Lp(·)(G), there exists a sequence {vi}∞i=1 ⊂ C
∞
0 (G) such that

‖v −vi‖Lp(·)(GR) → 0 for each R > 0 and ‖∇∇∇v −∇∇∇vi‖Lp(·)(G) → 0 as i →∞},

where GR = G ∩BR,BR = {x ∈ Rd; |x| < R}.

Definition 4.2. If G satisfies (GA) and p ∈Plog+ (G), define

Ŵ
1,p(·)
• (G) = {v : G → R; v is measurable in G,v ∈ Lp(·)(GR) for each R > 0,

∇∇∇v ∈Lp(·)(G) and for any η ∈ C∞0 (R
d),ηv ∈ W 1,p(·)0 (G)}.

We note that if G is bounded, then

W
1,p(·)
0 (G) = Ŵ

1,p(·)
0 (G) = Ŵ

1,p(·)
• (G).

We examine the properties of the spaces Ŵ 1,p(·)0 (G) and Ŵ
1,p(·)
• (G).



Int. J. Anal. Appl. (2022), 20:13 9

Lemma 4.3. Suppose (GA). Let p ∈ Plog+ (G) and let v ∈ Ŵ
1,p(·)
• (G). Then for every R > 0, there

exists a sequence {vi}⊂ C∞0 (G) possibly depending on R > 0 such that

‖v −vi‖Lp(·)(GR) + ‖∇∇∇v −∇∇∇vi‖Lp(·)(GR) → 0 as i →∞.

Proof. For given R > 0, choose η ∈ C∞0 (B2R) such that η = 1 on BR. Since ηv ∈ W
1,p(·)
0 (G), there

exists a sequence {vi} ⊂ C∞0 (G) such that ‖ηv − vi‖W 1,p(·)(G) → 0 . Since η = 1 on BR, we have
‖v −vi‖W 1,p(·)(GR) → 0. �

Theorem 4.4. Suppose (GA) and let p ∈Plog+ (G). Then the following properties hold.
(i) W

1,p(·)
0 (G) ⊂ Ŵ

1,p(·)
0 (G) ⊂ Ŵ

1,p(·)
• (G).

(ii) For v ∈ Ŵ 1,p(·)• (G), ‖∇∇∇v‖Lp(·)(G) is a norm on Ŵ
1,p(·)
• (G).

(iii) The space Ŵ 1,p(·)• (G) equipped with the norm ‖∇∇∇·‖Lp(·) is a reflexive Banach space.
(iv) The space Ŵ 1,p(·)0 (G) is a closed subspace of Ŵ

1,p(·)
• (G) and

Ŵ
1,p(·)
0 (G) = the closure of C

∞
0 (G) with respect to ‖∇∇∇·‖Lp(·)(G)-norm.

(v) The space W 1,p(·)0 (G) is dense in Ŵ
1,p(·)
0 (G) with respect to ‖∇∇∇·‖Lp(·)(G)-norm.

(vi) If we define E∞0 (G) = {∇∇∇φ; φ ∈ C
∞
0 (G)} and E

p(·)
0 (G) = {∇∇∇v; v ∈ Ŵ

1,p(·)
0 (G)}, then the

closure of E∞0 (G) in L
p(·)(G) is equal to Ep(·)0 (G).

Proof. (i) It is trivial that W 1,p(·)0 (G) ⊂ Ŵ
1,p(·)
0 (G). Let v ∈ Ŵ

1,p(·)
0 (G) and let η ∈ C

∞
0 (R

d). Choose

R > 0 so that supp η ⊂ BR. By definition of Ŵ
1,p(·)
0 (G), there exists a sequence {vi}⊂ C

∞
0 (G) such

that ‖v −vi‖Lp(·)(GR) → 0 and ‖∇∇∇v −∇∇∇vi‖Lp(·)(G) → 0 as i →∞. Then ηvi ∈ C
∞
0 (G) and

‖ηv −ηvi‖W 1,p(·)(G) ≤ C1(‖η‖L∞(G) + ‖∇∇∇η‖L∞(G))‖v −vi‖Lp(·)(GR)

+ C2‖η‖L∞(G)‖∇∇∇v −∇∇∇vi‖Lp(·)(G) → 0.

Thus ηv ∈ W 1,p(·)0 (G), so v ∈ Ŵ
1,p(·)
• (G).

(ii) Clearly Ŵ 1,p(·)• (G) is a linear space. If v ∈ Ŵ
1,p(·)
• (G) satisfies ‖∇∇∇v‖Lp(·)(G) = 0, then we show

that v = 0 in G. To do so, it suffices to show that v = 0 in GR for every R > 0. Choose η ∈ C∞0 (R
d)

such that η = 1 on BR and supp η ⊂ B2R. Since ηv ∈ W
1,p(·)
0 (G) by definition of Ŵ

1,p(·)
• (G), there

exists a sequence {vi} ⊂ C∞0 (G) such that ‖ηv − vi‖W 1,p(·)(G) → 0. Thus ηv ∈ W
1,p−

0 (G2R) and

‖ηv −vi‖W 1,p−(G2R) → 0. By [25, Lemma 1.2], we have

‖vi‖Lp−(G2R) ≤ CR
d/p−+1−1/p−‖∇∇∇vi‖Lp−(G2R).

By the limit process, we have

‖v‖
Lp
−

(GR)
≤ CRd/p

−+1−1/p−‖∇∇∇v‖
Lp
−

(GR)
.

Since Lp(·)(GR) ↪→Lp
−

(GR) and ∇∇∇v = 0 in Lp
−

(GR), we have v = 0 in GR. The other properties of

norm clearly hold.



10 Int. J. Anal. Appl. (2022), 20:13

(iii) We already showed that Ŵ 1,p(·)• (G) is a normed linear space. We derive the completeness of

Ŵ
1,p(·)
• (G) equipped with the norm ‖∇∇∇·‖Lp(·)(G). Let {vi}

∞
i=1 ⊂ Ŵ

1,p(·)
• (G) be a Cauchy sequence,

that is, ‖∇∇∇vi −∇∇∇vj‖Lp(·)(G) → 0 as i, j →∞. For k ∈ N, put Gk = G∩Bk. Then there exists k0 ∈ N
such that for k ≥ k0, Gk has a portion Γk of ∂G with |Γk| > 0. Since vi ∈ W 1,p(·)(Gk) and vi

∣∣
Γk

= 0,

it follows from a generalized Poincaré inequality (Theorem 2.9) that

‖vi −vj‖Lp(·)(Gk ) ≤ C(k)‖∇∇∇vi −∇∇∇vj‖Lp(·)(Gk ) ≤ C(k)‖∇∇∇vi −∇∇∇vj‖Lp(·)(G).

Thus {vi
∣∣
Gk
} is a Cauchy sequence in Lp(·)(Gk). Therefore, there exists v(k) ∈ Lp(·)(Gk) such that

vi
∣∣
Gk
→ v(k) in Lp(·)(Gk). After choosing a subsequence, we may assume that vi

∣∣
Gk
→ v(k) a.e. in

Gk. Eventually after changing v(k+1) on a subset Nk ⊂ Gk with measure zero, we may assume that
v(k+1)

∣∣
Gk

= v(k). Define a unique measurable function v : G → R so that v(x) = v(k)(x) for x ∈ Gk.
Hence for each R > 0 ‖vi −v‖Lp(·)(GR) → 0 as i →∞. Since {∇∇∇vi} is a Cauchy sequence in L

p(·)(G),

there exists f = (f1, . . . , fd) ∈ Lp(·)(G) such that ∇∇∇vi → f in Lp(·)(G). Let φ ∈ C∞0 (G). Then
supp φ ⊂ Gk for some k ∈ N. For l = 1, . . . ,d,

〈v,∂lφ〉G = lim
i→∞
〈vi,∂lφ〉Gk = − lim

i→∞
〈∂lvi,φ〉Gk = −〈fl,φ〉G.

Hence ∂lv = fl ∈ Lp(·)(G), so ∇∇∇v = f ∈ Lp(·)(G). For η ∈ C∞0 (R
d), choose R > 0 such that

supp η ⊂ BR. Choose ζ ∈ C∞0 (B2R) so that 0 ≤ ζ ≤ 1 and ζ = 1 on BR. Since ζvi ∈ W
1,p(·)
0 (G)

by definition of Ŵ 1,p(·)• (G), there exists φi ∈ C∞0 (G) such that ‖ζvi − φi‖W 1,p(·)(G) ≤ 2
−i. Since

ηv = ηζv, we have

‖ηv −ηφi‖Lp(·)(G) = ‖ηζv −ηφi‖Lp(·)(G)

≤ ‖ηζv −ηζvi‖Lp(·)(G) + ‖ηζvi −ηφi‖Lp(·)(G)

≤ ‖η‖L∞(Rd )(‖ζv −ζvi‖Lp(·)(G) + ‖ζvi −φi‖Lp(·)(G))

≤ ‖η‖L∞(Rd )(‖v −vi‖Lp(·)(G) + 2
−i ) → 0 as i →∞

and

‖∇∇∇(ηv −ηφi )‖Lp(·)(G) = ‖∇∇∇(ηζv −ηφi )‖Lp(·)(G)

≤ ‖∇∇∇(ηζv −ηζvi )‖Lp(·)(G) + ‖∇∇∇(ηζvi −ηφi )‖Lp(·)(G)

≤ ‖ηζ‖L∞(Rd )(‖∇∇∇(v −vi )‖Lp(·)(G)

+‖∇∇∇(ηζ)‖L∞(Rd )‖vi −φi‖Lp(·)(GR)

+(‖η‖L∞(Rd ) + ‖∇∇∇η‖L∞(Rd ))‖ζvi −φi‖W 1,p(·)(G)

→ 0

as i → ∞. Since ηφ ∈ C∞0 (G), we can see that ηv ∈ W
1,p(·)
0 (G), so v ∈ Ŵ

1,p(·)
• (G). Hence

Ŵ
1,p(·)
• (G) is complete.



Int. J. Anal. Appl. (2022), 20:13 11

We show the reflexivity of Ŵ 1,p(·)• (G). If we define

E
p(·)
• (G) = {∇∇∇v; v ∈ Ŵ

1,p(·)
• (G)},

then the gradient operator ∇∇∇ : Ŵ 1,p(·)• (G) → E
p(·)
• (G) is isometric isomorphism. Since Ŵ

1,p(·)
• (G) is

complete, Ep(·)• (G) is a closed subspace of a reflexive Banach space Lp(·)(G). Therefore, E
p(·)
• (G) is

reflexive, so Ŵ 1,p(·)• (G) is also reflexive.

(iv) Let {vi} ⊂ Ŵ
1,p(·)
0 (G) and v ∈ Ŵ

1,p(·)
• (G) such that ‖∇∇∇v −∇∇∇vi‖Lp(·)(G) → 0 as i → ∞. By

definition of Ŵ 1,p(·)0 (G), there exists φi ∈ C
∞
0 (G) such that ‖∇∇∇vi −∇∇∇φi‖Lp(·)(G) ≤ 2

−i. Hence

‖∇∇∇v −∇∇∇φi‖Lp(·)(G) ≤‖∇∇∇v −∇∇∇vi‖Lp(·)(G) + ‖∇∇∇vi −∇∇∇φi‖Lp(·)(G) → 0 as i →∞.

By the generalized Poincaré inequliaty (Theorem 2.9), for large R > 0,

‖v −φi‖Lp(·)(GR) ≤ C(R)‖∇∇∇v −∇∇∇φi‖Lp(·)(GR) ≤ C(R)‖∇∇∇v −∇∇∇φi‖Lp(·)(G) → 0.

From definition of Ŵ 1,p(·)0 (G), we can see that v ∈ Ŵ
1,p(·)
0 (G), so Ŵ

1,p(·)
0 (G) is a closed subspace of

Ŵ
1,p(·)
• (G).

Since C∞0 (G) ⊂ Ŵ
1,p(·)
0 (G), the closure of C

∞
0 (G) with respect to ‖∇∇∇ · ‖Lp(·)(G) is contained in

the closure of Ŵ 1,p(·)0 (G) with respect to ‖∇∇∇·‖Lp(·)(G) which is equal to Ŵ
1,p(·)
0 (G). Conversely, let

v ∈ Ŵ 1,p(·)0 (G). Then there exists {φi} ⊂ C
∞
0 (G) such that ‖∇∇∇v −∇∇∇φi‖Lp(·)(G) → 0 as i → ∞.

Thereby v is contained in the closure of C∞0 (G) with respect to ‖∇∇∇·‖Lp(·)(G)-norm.
(v) By definition of Ŵ 1,p(·)0 (G), the space C

∞
0 (G) is dense in Ŵ

1,p(·)
0 (G) with respect to ‖∇∇∇·‖Lp(·)(G)-

norm. Since C∞0 (G) ⊂ W
1,p(·)
0 (G) ⊂ Ŵ

1,p(·)
0 (G), we see that W

1,p(·)
0 (G) is dense in Ŵ

1,p(·)
0 (G) with

respect to ‖∇∇∇·‖Lp(·)(G)-norm.
(vi) From (iv), it is clear that the closure of E∞0 (G) in L

p(·)(G) is contained in Ep(·)0 (G). Let

v ∈ Ŵ 1,p(·)0 (G). Then there exists a sequence {φi} ⊂ C
∞
0 (G) such that ‖∇∇∇v −∇∇∇φi‖Lp(·)(G) → 0 as

i →∞. Therefore, ∇∇∇v is contained in the closure of E∞0 (G) in L
p(·)(G). �

We can improve Lemma 4.3.

Lemma 4.5. Suppose (GA). Let p ∈ Plog+ (G) and let v ∈ Ŵ
1,p(·)
• (G). Then there exists a sequence

{vi}⊂ C∞0 (G) such that for every R > 0,

‖v −vi‖Lp(·)(GR) + ‖∇∇∇v −∇∇∇vi‖Lp(·)(GR) → 0 as i →∞,

that is, we can choose {vi}⊂ C∞0 (G) independent of R > 0.

Proof. Choose ζ ∈ C∞0 (G) such that 0 ≤ ζ ≤ 1, and ζ(x) = 1 for |x| ≤ 1 and ζ = 0 for |x| ≥ 2.
Put ζi (x) = ζ(i−1x). By definition of Ŵ

1,p(·)
• (G), we see that ζiv ∈ W

1,p(·)
0 (G). Hence there exists

{vi}⊂ C∞0 (G) such that ‖ζiv −vi‖W 1,p(·)(G) ≤ i
−1. For each R > 0, let i ≥ R. Then

‖v −vi‖Lp(·)(GR) + ‖∇∇∇v −∇∇∇vi‖Lp(·)(GR) ≤ i
−1 → 0 as i →∞.

�



12 Int. J. Anal. Appl. (2022), 20:13

Here we characterize of Ŵ 1,p(·)• (G).

Theorem 4.6. Suppose (GA) and let p ∈Plog+ (G). Then we have Ŵ
1,p(·)
• (G) = Mp(·), where

Mp(·) = {v : G → R; v is measurable ,v ∈ L
p(·)(GR) for each R > 0,

∇∇∇v ∈Lp(·)(G) and there exists {vi}⊂ C∞0 (G) such that

‖v −vi‖Lp(·)(GR) + ‖∇∇∇v −∇∇∇vi‖Lp(·)(GR) → 0 for each R > 0 as i →∞}.

Proof. By Lemma 4.5, Ŵ 1,p(·)• (G) ⊂ Mp(·). Conversely, let v ∈ Mp(·), and let η ∈ C∞0 (R
d). Choose

R > 0 such that supp η ⊂ BR. Then

‖ηv −ηvi‖Lp(·)(G) ≤‖η‖L∞(Rd )‖v −vi‖Lp(·)(GR) → 0

and

‖∇∇∇(ηv − ηvi )‖Lp(·)(G) ≤ ‖η‖L∞(Rd )‖∇∇∇v − ∇∇∇vi‖Lp(·)(GR) + ‖∇∇∇η‖L∞(Rd )‖v − vi‖Lp(·)(GR) → 0.

] Since ηvi ∈ C∞0 (G) and ηvi → ηv in W
1,p(·)(G), we see that ηv ∈ W 1,p(·)0 (G), so v ∈ Ŵ

1,p(·)
• (G). �

Definition 4.7. Let G be a domain of Rd (d ≥ 2) such that Rd \G 6= ∅, and let s ∈Plog+ (G).
(a) We say that G has the property Pa(s) if there exists a constant Cs = C(s,G) > 0 such that

‖∇∇∇u‖Ls(·)(G) ≤ Cs sup
0 6=v∈Ŵ 1,s

′(·)
• (G)

|〈∇∇∇u,∇∇∇v〉G|
‖∇∇∇v‖

Ls
′(·)(G)

for all u ∈ Ŵ 1,s(·)• (G). (4.1)

(b) Let the bounded linear operator σs : Ŵ
1,s(·)
• (G) → (Ŵ

1,s′(·)
• (G))

′ be defined by

σs(u)(φ) = 〈∇∇∇u,∇∇∇φ〉G for u ∈ Ŵ
1,s(·)
• (G) and φ ∈ Ŵ

1,s′(·)
• (G). (4.2)

We say that G has the property Pb(s) if σs is a bijection and there exists a constant C̃s = C̃(s,G) > 0

such that

C̃s‖∇∇∇u‖Ls(·)(G) ≤‖σs(u)‖(Ŵ 1,s′(·)• (G))′
≤ 2‖∇∇∇u‖Ls(·)(G) for all u ∈ Ŵ

1,s(·)
• (G). (4.3)

Theorem 4.8. Let G be a domain of Rd (d ≥ 2) such that Rd \G 6= ∅ and let p ∈ Plog+ (G). Then
G has the property Pa(s) for s = p and s = p′ if and only if G has the property Pb(s) for s = p and

s = p′.

Proof. Assume that G has the property Pa(s) for s = p and s = p′. Let u ∈ Ŵ
1,s(·)
• (G) and define

Ss(u) = sup{〈∇∇∇u,∇∇∇φ〉G; φ ∈ Ŵ
1,s′(·)
• (G),‖∇∇∇φ‖Ls′(·)(G) ≤ 1}.

By (4.1) and the Hölder inequality (Proposition 2.2),

C−1s ‖∇∇∇u‖Ls(·)(G) ≤ Ss(u) = ‖σs(u)‖(Ŵ 1,s′(·)• (G))′

≤ 2‖∇∇∇u‖Ls(·)(G) for all u ∈ Ŵ
1,s(·)
• (G).



Int. J. Anal. Appl. (2022), 20:13 13

Hence (4.3) holds with C̃s = C−1s . From this, we see that σs(Ŵ
1,s′(·)
• (G)) is a closed subspace of

(Ŵ
1,s′(·)
• (G))

′. Suppose σs(Ŵ
1,s(·)
• (G)) ( (Ŵ

1,s′(·)
• (G))

′. By the Hahn-Banach theorem, there exists

F ′′ ∈ (Ŵ 1,s
′(·)

• (G))
′′ such that F ′′ 6= 0 and

F ′′
∣∣
σs (Ŵ

1,s(·)
• (G))

= 0.

Since Ŵ 1,s
′(·)

• (G) is reflexive, there exists uniquely φ ∈ Ŵ
1,s′(·)
• (G) such that F ′′(F ′) = F ′(φ) for

all F ′ ∈ (Ŵ 1,s
′(·)

• (G))
′ and ‖F ′′‖

(Ŵ
1,s′(·)
• (G))

′′ = ‖∇∇∇φ‖Ls′(·)(G) > 0. On the other hand, for all u ∈

Ŵ
1,s(·)
• (G),

0 = F ′′(σs(u)) = σs(u)(φ) = 〈∇∇∇u,∇∇∇φ〉G.

By the property Pa(s′), we have ‖∇∇∇φ‖Ls′(·)(G) ≤ Cs′Ss′(φ) = 0. This is a contradiction.
Conversely, assume that G has the property Pb(s) for s = p and s = p′. Let u ∈ Ŵ

1,s(·)
• (G). Since

σs′ is a bijection, for F
′ ∈ (Ŵ 1,s(·)• (G))′, there exists uniquely φ ∈ Ŵ

1,s′(·)
• (G) such that F ′ = σs′(φ).

Hence

‖∇∇∇u‖Ls(·)(G) = sup

{
|F ′(u)|

‖F ′‖
(Ŵ

1,s(·)
• (G))

′
; 0 6= F ′ ∈ (Ŵ 1,s(·)• (G))′

}

≤ sup

{
|〈∇∇∇u,∇∇∇φ〉G|
C̃s′‖∇∇∇φ‖Ls′(·)(G)

; 0 6= φ ∈ Ŵ 1,s
′(·)

• (G)

}
.

Thus (4.1) holds with Cs = C̃
−1
s′ . �

Now we consider the case G = Rd.

Lemma 4.9. If we define M := {∆v; v ∈D(Rd) := C∞0 (R
d)}, then M is dense in Lp(·)(Rd).

Proof. Suppose that M  Lp(·)(Rd), where M is the closure of M in Lp(·)(Rd). By the Hahn-
Banach theorem, there exists F ′ ∈ (Lp(·)(Rd))′ with ‖F ′‖(Lp(·)(Rd ))′ > 0 and F

′
∣∣
M

= 0. Since we can

regard (Lp(·)(Rd))′ = Lp
′(·)(Rd) isometrically, there exists v ∈ Lp

′(·)(Rd) such that ‖v‖
Lp
′(·)(Rd ) =

‖F ′‖(Lp(·)(Rd ))′ > 0 and F
′(w) = 〈v,w〉Rd :=

∫
Rd v(x)w(x)dx for all w ∈ L

p(·)(Rd). Since F ′
∣∣
M

= 0,

we can see that 〈v, ∆φ〉Rd = 0 for all φ ∈D(Rd), so ∆v = 0 in D′(Rd). By the hypoellipticity of the
Laplacian, we can regard that v ∈ C∞(Rd) (eventually after change of a set of measure zero), so v
is harmonic in Rd. For any x ∈ Rd fixed, it follows from the second mean value theorem for harmonic
functions that

v(x) =
1

|BR(x)|

∫
BR(x)

v(y)dy,

where BR(x) = {y ∈ Rd : |y−x| < R} and |BR(x)| denotes the volume of BR(x). By the generalized
Hölder inequality (Proposition 2.2),

|v(x)| ≤
2

|BR(x)|
‖v‖

Lp
′(·)(BR(x))

‖1‖Lp(·)(BR(x)).



14 Int. J. Anal. Appl. (2022), 20:13

Since ‖1‖Lp(·)(BR(x)) ≤ ρp(·),BR(x)(1)
1/p− = |BR(x)|1/p

−
for large R > 0 and p− > 1, we have

|v(x)| ≤ 2|BR(x)|−1+1/p
−
‖v‖

Lp
′(·)(Rd ) → 0 as R →∞.

Therefore, we have v(x) = 0. Since x ∈ Rd is arbitrary, v ≡ 0 in Rd. This is a contradiction. �

Define ∇∇∇2v = (∂i∂jv)i,j=1,...,d for v ∈D(Rd). Then there exists a constant C = C(p,d) > 0 such
that

C‖∇∇∇2v‖Lp(·)(Rd ) ≤‖∆v‖Lp(·)(Rd ) for all v ∈D(R
d). (4.4)

For the proof, see [11, Corollary 14.1.7] (cf. when p(·) = p (constant), see Gilbarg and Trudinger [17,
Corollary 9.10]).

For p ∈Plog+ (R
d), we have Ep(·)(Rd) = {∇∇∇u; u ∈ Lp(·)

loc
(Rd),∇∇∇u ∈Lp(·)(Rd)} by definition.

Lemma 4.10. Let p,q ∈Plog+ (R
d). If u ∈ Lq(·)

loc
(Rd) with ∇∇∇u ∈Lq(·)(Rd) satisfies

sup
0 6=v∈D(Rd )

|〈∇∇∇u,∇∇∇v〉Rd |
‖∇∇∇v‖

Lp
′(·)(Rd )

< ∞, (4.5)

then u ∈ Lp(·)
loc

(Rd) with ∇∇∇u ∈ Lp(·)(Rd). Furthermore, there exists a constant C1 = C1(p,d) > 0
such that

‖∇∇∇u‖Lp(·)(Rd ) ≤ C1 sup
0 6=v∈D(Rd )

|〈∇∇∇u,∇∇∇v〉Rd |
‖∇∇∇v‖

Lp
′(·)(Rd )

(4.6)

for all u ∈ Lp(·)
loc

(Rd) with ∇∇∇u ∈Lp(·)(Rd).
In particular, if u ∈ Lp(·)

loc
(Rd) with ∇∇∇u ∈Lp(·)(Rd), then

‖∇∇∇u‖Lp(·)(Rd ) ≤ C1 sup
0 6=v∈D(Rd )

|〈∇∇∇u,∇∇∇v〉Rd |
‖∇∇∇v‖

Lp
′(·)(Rd )

(4.7)

Proof. Let u ∈ Lq(·)
loc

(Rd) with ∇∇∇u ∈Lq(·)(Rd). For every i = 1, . . . ,d, using (4.4),

∞ > sup
06=v∈D(Rd )

|〈∇∇∇u,∇∇∇v〉Rd |
‖∇∇∇v‖

Lp
′(·)(Rd )

≥ sup
06=w∈D(Rd )

|〈∇∇∇u,∇∇∇(∂iw)〉Rd |
‖∇∇∇∂iw‖Lp′(·)(Rd )

≥ sup
06=w∈D(Rd )

|〈∂iu, ∆w〉Rd |
‖∇∇∇2w‖

Lp
′(·)(Rd )

≥ C sup
06=w∈D(Rd )

|〈∂iu, ∆w〉Rd |
‖∆w‖

Lp
′(·)(Rd )

, (4.8)

where C is the constant in (4.4). Define a bounded linear functional L∗ by

L∗(∆w) = 〈∂iu, ∆w〉Rd on the dense subspace M of L
p′(·)(Rd)



Int. J. Anal. Appl. (2022), 20:13 15

(cf. Lemma 4.9). Then the functional L∗ has a unique and norm-preserving extension as a continuous

linear functional on Lp
′(·)(Rd). Thus there exists g ∈ Lp(·)(Rd) such that 〈∂iu,v〉Rd = 〈g,v〉Rd for all

v ∈ M, that is,

〈∂iu −g, ∆w〉Rd = 0 for all w ∈D(R
d).

If we define W = ∂iu −g, then ∆W = 0 in D′(Rd), so W is harmonic in Rd. By the same argument
as in the proof of Lemma 4.9, we can regard W (x) ≡ 0, so ∂iu = g ∈ Lp(·)(Rd). Hence we have
∇∇∇u ∈Lp(·)(Rd). Since ∇∇∇u ∈Lp(·)(Rd) and u ∈ Lq(·)

loc
(Rd) ⊂ L1loc(R

d), for any ball B, it follows from

Theorem 2.7 (ii) that

‖u −〈u〉B‖Lp(·)(B) ≤ C(p,d,B)‖∇∇∇u‖Lp(·)(B).

This implies that u ∈ Lp(·)
loc

(Rd), that is, ∇∇∇u ∈Ep(·)(Rd). By continuity, we have

sup
06=w∈D(Rd )

|〈∂iu, ∆w〉Rd |
‖∆w‖

Lp
′(·)(rrd )

= sup
0 6=f∈Lp′(·)(Rd )

|〈∂iu,f 〉Rd |
‖f‖

Lp
′(·)(rrd )

= ‖∂iu‖Lp(·)(Rd ).

From (4.8),

‖∂iu‖Lp(·)(Rd ) ≤ C
−1
p′ sup

06=φ∈D(Rd )

|〈∇∇∇u,∇∇∇φ〉Rd |
‖∇∇∇φ‖

Lp
′(·)(Rd )

.

Therefore, we get (4.6). �

Remark 4.11. We can show that (4.6) implies (4.4). Indeed, let φ ∈D(Rd) and put u = ∂iφ. From
(4.6) replaced p with p′,

‖∇∇∇(∂iφ)‖Lp′(·)(Rd ) ≤ C1(p) sup
0 6=v∈D(Rd )

|〈∇∇∇(∂iφ),∇∇∇v〉Rd |
‖∇∇∇v‖Lp(·)(Rd )

≤ C1(p) sup
0 6=v∈D(Rd )

|〈∆φ,∂iv〉Rd |
‖∇∇∇v‖Lp(·)(Rd )

≤ 2C1(p)‖∆φ‖Lp′(·)(Rd ).

Next we consider the case where G is a half-space or a bended half-space. Let ω be a C1-function de-

fined on Rd−1, H = {x = (x′,xd); x′ = (x1, . . . ,xd−1) ∈ Rd−1,xd < 0} and Hω = {x = (x′,xd); xd <
ω(x′)}.

Lemma 4.12. Let ω be a C1-function defined on Rd−1 with ‖∇∇∇′ω‖L∞(Rd−1) < ∞, where ∇∇∇
′ =

(∂1, . . . ,∂d−1), and let p ∈P
log
+ (Hω). Then we have Ŵ

1,p(·)
• (Hω) = Ŵ

1,p(·)
0 (Hω).

Proof. Step 1. Let 0 < ρ < ∞, 0 < R < ∞ and put

Zωρ,R = {x = (x
′,xd) ∈ Rd; |x′| < ρ,−R < xd < ω(x′)}.

Assume u ∈ Ŵ 1,p(·)• (Hω). Then since u
∣∣
∂Hω

= 0, it follows from the generalized Poincaré ineequality

(Theorem 2.9) that there exists a constant C = C(d,Clog(p)) such that

‖u‖Lp(·)(Zω
ρR

) ≤ C(ρ + R)‖∇∇∇u‖Lp(·)(Zω
ρ,R

). (4.9)



16 Int. J. Anal. Appl. (2022), 20:13

Since |ω(x′) −ω(0)| ≤ ‖∇∇∇′ω‖L∞(Rd−1)|x′| ≤ ‖∇∇∇
′ω‖L∞(Rd−1)ρ for |x′| < ρ.

Hence |ω(x′)| ≤ ‖∇∇∇′ω‖L∞(Rd−1)ρ + ω0, where ω0 = |ω(0)|.
Step 2. Choose τ ∈ C∞0 (R

d) such that 0 ≤ τ ≤ 1 and τ(x) = 1 for |x| ≤ 1 and τ(x) = 0 for
|x| ≥ 2, and for k ∈ N, put τk(x) = τ(k−1x). Then |∇∇∇τk(x)| ≤ k−1‖∇∇∇τ‖L∞(Rd ), and supp(∇∇∇τk) ⊂
Ak := {x ∈ Rd; k < |x| < 2k}. Put ρk = 2k,Rk = 2k(‖∇∇∇′ω‖L∞(Rd−1) + 1) + ω0 and Zk = Zωρk,Rk. If
x ∈ Hω ∩B2k, then |x′| < 2k and −2k < xd < ω(x′). Hence

ω(x′) −Rk ≤‖∇∇∇′ω‖L∞(Rd−1)2k + ω0 −Rk = −2k < xd < ω(x
′).

Therefore, Hω ∩B2k ⊂ Zk. From (4.9), we have

‖u∇∇∇τk‖Lp(·)(Hω) ≤ k
−1‖∇∇∇′ω‖L∞(Rd−1)‖u‖Lp(·)(Hω∩B2k ) (4.10)

′ ≤ k−1‖∇∇∇′ω‖L∞(Rd−1)‖u‖Lp(·)(Zk )

≤ Ck−1‖∇∇∇′ω‖L∞(Rd−1)(ρk + Rk)‖∇∇∇u‖Lp(·)(Hω)

≤ C1‖∇∇∇u‖Lp(·)(Hω),

where C1 is a constant independent of k. By definition of Ŵ
1,p(·)
• (Hω), τku ∈ W

1,p(·)
0 (Hω) ⊂

Ŵ
1,p(·)
0 (Hω).

Step 3. Let F ′ ∈ (Ŵ 1,p(·)• (Hω))′, that is,

|F ′(φ)| ≤ ‖F ′‖
(Ŵ

1,p(·)
• (Hω))

′‖∇∇∇φ‖Lp(·)(Hω) for all φ ∈ Ŵ
1,p(·)(Hω).

Since Ŵ 1,p(·)• (Hω) is complete and E
p(·)
0 (Hω) is a closed subspace of L

p(·)(Hω), we can regard F ′ as

a continuous linear functional on Ep(·)0 (Hω). By the Hahn-Banach theorem, F
′ may be extended to

a functional F̃ ′ ∈ (Lp(·)(Hω))′ which is norm-preserving. Hence there exists f ∈Lp
′(·)(Hω) such that

‖f‖
Lp
′(·)(Hω)

= ‖F̃ ′‖(Lp(·)(Hω))′ = ‖F‖(Ŵ 1,p(·)• (Hω))′ and

F ′(φ) = F̃ ′(φ) = 〈f ,∇∇∇φ〉Hω for all φ ∈ Ŵ
1,p(·)
• (Hω).

Then

F ′(u) −F ′(τku) = 〈(1 −τk)f ,∇∇∇u〉Hω −〈f ,u∇∇∇τk〉Hω.

We have

|〈(1 −τk)f ,∇∇∇u〉Hω| ≤ 2‖(1 −τk)f‖Lp′(·)(Hω)‖∇∇∇u‖Lp(·)(Hω).

By the Lebesgue dominated convergence theorem, we have

ρp′(·),Hω ((1 −τk)f ) =
∫
Hω

|(1 −τk)f |p
′(x)dx → 0 as k →∞.

So it follows from Proposition 2.1 that ‖(1 − τk)f‖Lp′(·)(Hω) → 0 as k → ∞. By (4.10) and
supp(∇∇∇τk) ⊂ Ak,

|〈f ,u∇∇∇τk〉Hω| ≤ 2C1‖f‖Lp′(·)(Hω∩Ak )‖∇∇∇u‖Lp(·)(Hω) → 0 as k →∞.



Int. J. Anal. Appl. (2022), 20:13 17

Hence F ′(τku) → F ′(u) as k → ∞. Since F ′ ∈ (Ŵ
1,p(·)
• (Hω))

′ is arbitrary, we see that τku → u
weakly in Ŵ 1,p(·)• (Hω). By Step 2, τku ∈ Ŵ

1,p(·)
0 (Hω). Since Ŵ

1,p(·)
0 (Hω) is a closed subspace of

Ŵ
1,p(·)
• (Hω) (Theorem 4.4), it is weakly closed. Therefore u ∈ Ŵ

1,p(·)
0 (Hω). �

Lemma 4.13. Let p ∈Plog+ (H). For x ∈ R
d, define

p̃(x) =

{
p(x) for xd ≤ 0,
p(x′,−xd) for xd > 0.

Then clearly p̃ ∈Plog+ (R
d). For u ∈ Ŵ 1,p(·)• (H), define

u1(x) =



u(x) for xd < 0,

0 for xd = 0,

−u(x′,−xd) for xd > 0.

Then u1 ∈ W
1,p̃(·)
loc

(Rd), ∇∇∇u1 ∈Lp̃(·)(Rd), and furthermore,

∂iu1(x) =




(∂iu)(x) for xd < 0,

0 for xd = 0,

−(∂iu)(x′,−xd) for xd > 0

for i = 1, . . . ,d − 1 and

∂du1(x) =

{
(∂du)(x) for xd < 0,

(∂du)(x
′,−xd) for xd > 0.

In addition,

‖∇∇∇u‖Lp(·)(H) ≤‖∇∇∇u1‖Lp̃(Rd ) ≤ 2‖∇∇∇u‖Lp(·)(H).

For φ ∈ D(Rd), let (T1φ)(x) = φ(x) − φ(x′,−xd) for x ∈ H. Then T1φ ∈ Ŵ
1,p(·)
0 (H) ∩ C

1(H),

(T1φ)(x
′, 0) = 0 and there exists R = R(φ) > 0 such that (T1φ)(x) = 0 for |x| > R and

‖∇∇∇(T1φ)‖Lp(·)(H) ≤ 2‖∇∇∇φ‖Lp̃(·)(Rd ).

Furthermore, for u ∈ Ŵ 1,p(·)0 (H) and φ ∈D(R
d),

〈∇∇∇u1,∇∇∇φ〉Rd = 〈∇∇∇u,∇∇∇(T1φ)〉H.

Since this lemma follows from elementary calculations (cf. [25, Lemma 2.3]). we omit the proof.

Lemma 4.14. Let p,q ∈Plog+ (H). If u ∈ Ŵ
1,q(·)
0 (H) satisfies

Sp(u) := sup
0 6=φ∈C∞0 (H)

|〈∇∇∇u,∇∇∇φ〉H|
‖∇∇∇φ‖

Lp
′(·)(H)

< ∞,

then u ∈ Ŵ 1,p(·)0 (H) and
‖∇∇∇u‖Lp(·)(H) ≤ C2(p)Sp(u),

where C2(p) = 2C1 > 0, C1 is a constant as in (4.6).



18 Int. J. Anal. Appl. (2022), 20:13

Proof. For any function φ ∈ D(Rd), we consider T1φ. If supp φ ⊂ BR, then supp T1φ ⊂ BR,
(T1φ)(x

′, 0) = 0 and ∇∇∇(T1φ) ∈ L∞(H). Choose η ∈ C∞(Rd) such that 0 ≤ η ≤ 1 and η(x) = 1
for |x| ≥ 1 and η(x) = 0 for |x| ≤ 1/2. For k ∈ N, put ηk(x) = η(kx) and φk(x) = ηk(x)(T1φ)(x).
Then for s = q′ and s = p′,

‖∇∇∇(T1φ) −∇∇∇φk‖Ls(·)(H) ≤‖(1 −ηk(x))∇∇∇(T1φ)‖Ls(·)(H) + ‖(∇∇∇ηk(x))T1φ‖Ls(·)(H).

Here from the Lebesgue dominated convergence theorem,

‖(1 −ηk(x))∇∇∇(T1φ)‖Ls(·)(H) → 0 as k →∞.

Since supp ηk ⊂ {x ∈ Rd; 1/(2k) < |x| < 1/k} =: Ak, It follows from the Poincaré inequality
(Theorem 2.9) that

‖(∇∇∇ηk(x))T1φ‖Ls(·)(H) ≤ k‖∇∇∇η‖L∞(Rd )‖T1φ‖Ls(·)(H∩Ak )

≤ k‖∇∇∇η‖L∞(Rd )
1

k
‖∇∇∇(T1φ)‖Ls(·)(H∩Ak ) = ‖∇∇∇η‖L∞(Rd )‖∇∇∇(T1φ)‖Ls(·)(H∩Ak ) → 0

as k →∞. Therefore, since u ∈ Ŵ 1,q(·)0 (H), we have

Sp(u) ≥
|〈∇∇∇u,∇∇∇φk〉H|
‖∇∇∇φk‖Lp′(·)(H)

→
|〈∇∇∇u,∇∇∇(T1φ)〉H|
‖∇∇∇(T1φ)‖Lp′(·)(H)

.

Hence for 0 6= φ ∈ C∞0 (R
d) such that T1φ 6= 0, by Lemma 4.13

|〈∇∇∇u1,∇∇∇φk〉Rd |
‖∇∇∇φ‖

Lp̃
′(·)(Rd )

≤ 2
|〈∇∇∇u,∇∇∇(T1φ)〉H|
‖∇∇∇T1φ‖Lp′(·)(H)

≤ 2Sp(u).

By Lemma 4.10, we see that ∇∇∇u1 ∈ Lp̃(Rd), so ∇∇∇u ∈ Lp(·)(H). Since u ∈ Ŵ
1,q(·)
0 (H), we can see

that u ∈ Ŵ 1,p(·)0 (H) as in the proof of Lemma 4.12, and

‖∇∇∇u‖Lp(·)(H) ≤‖∇∇∇u1‖Lp̃(·)(Rd ) ≤ 2C1Sp(u).

�

Lemma 4.15. Let ω be a C1-function on Rd−1 such that there exists R = R(ω) > 0 such that
ω(x′) = 0 for |x′| > R and let p ∈Plog+ (Hω). Assume that there exists a constant Kp = K(p,d) > 0
such that

‖∇∇∇′ω‖L∞(Rd−1) ≤ Kp.

Then there exists a constant C3(s) = C3(s,d,Kp) > 0 such that for all u ∈ Ŵ
1,s(·)
0 (Hω),

‖∇∇∇u‖Ls(·)(Hω) ≤ C3(s) sup
0 6=φ∈C∞0 (Hω)

|〈∇∇∇u,∇∇∇φ〉Hω|
‖∇∇∇φ‖

Ls
′(·)(Hω)

(4.11)

for s = p,p′.



Int. J. Anal. Appl. (2022), 20:13 19

Proof. Let y : Rd → Rd be a map defined by{
yi (x) = xi for i = 1, . . . ,d − 1,
yd(x) = xd −ω(x′) .

Then y is a C1-map and bijective, y(Hω) = H,y(∂Hω) = ∂H and the Jacobian J(y(x)) = 1 for

x ∈ Rd. The inverse map x : Rd → Rd is given by{
xi (y) = yi for i = 1, . . . ,d − 1,
xd(y) = yd + ω(y

′).

For s ∈ Plog+ (Hω), define s̃(y) = s(x(y)). Then s̃ ∈ P
log
+ (H). If u ∈ Ŵ

1,s(·)
0 (Hω) and define

ũ(y) = u(x(y)) for y ∈ H, then ũ ∈ Ŵ 1,s̃(·)0 (H). Conversely, if ũ ∈ Ŵ
1,s̃(·)
0 (H), then u(x) = ũ(y(x))

for x ∈ Hω belongs to Ŵ
1,s(·)
0 (Hω). Since{

∂iu(x) = (∂iũ)(y(x)) − (∂dũ)(y(x))∂iω(x′) for i = 1, . . . ,d − 1,
∂du(x) = (∂dũ)(y(x)),

there exists a constant d1(s) = d1(s,d) > 0 such that

‖∇∇∇u‖Ls(·)(Hω) ≤ d1(s)(1 + ‖∇∇∇
′ω‖L∞(Rd−1))‖∇∇∇ũ‖Ls̃(·)(H),

‖∇∇∇ũ‖Ls̃(·)(H) ≤ d1(s)(1 + ‖∇∇∇
′ω‖L∞(Rd−1))‖∇∇∇u‖Ls(·)(Hω).

Thus the map Ŵ 1,s(·)0 (Hω) 3 u 7→ ũ ∈ Ŵ
1,s̃(·)
0 (H) is continuous, linear and bijective.

Let u ∈ Ŵ 1,s(·)0 (Hω) and φ ∈ Ŵ
1,s′(·)
0 (Hω). By elementary calculations, we have

〈∇∇∇u,∇∇∇φ〉Hω =
∫
Hω

∇∇∇u(x) ·∇∇∇φ(x)dx =
∫
H

(∇∇∇ũ)(y) · (∇∇∇φ̃)(y)dy −Bω[∇∇∇ũ,∇∇∇φ̃],

where

Bω[∇∇∇ũ,∇∇∇φ̃] = −
d−1∑
i=1

∫
H

(
(∂dũ)(∂iφ̃) + (∂iũ)(∂dφ̃)∂iω(y

′)
)
dy +

∫
H

(∂dũ)(∂dφ̃)|∇∇∇ω|2dy.

By the generalized Hölder inequality, there exists a constant d2(s) > 0 such that

|Bω[∇∇∇ũ,∇∇∇φ̃]| ≤ d2(s)‖∇∇∇′ω‖L∞(Rd−1)(1 + ‖∇∇∇
′ω‖L∞(Rd−1))‖∇∇∇ũ‖Ls̃(·)(H)‖∇∇∇φ̃‖Ls̃′(·)(H).

Therefore, for 0 6= φ ∈ Ŵ 1,s
′(·)

0 (Hω),

|〈∇∇∇u,∇∇∇φ〉Hω|
‖∇∇∇φ‖

Ls
′(·)(Hω)

≥
(
d1(s

′)(1 + ‖∇∇∇′ω‖L∞(Rd−1)
))−1{|〈∇∇∇ũ,∇∇∇φ̃〉H|

‖∇∇∇φ̃‖
Ls̃
′(·)(H)

−d2(s)‖∇∇∇′ω‖L∞(Rd−1)(1 + ‖∇∇∇
′ω‖L∞(Rd−1))‖∇∇∇ũ‖Ls̃(·)(H)

}
.

Define

Kp = min

{
1

2
, min

{
(4C2(s)d2(s))

−1; s = p,p′
}}

,



20 Int. J. Anal. Appl. (2022), 20:13

where C2(s) > 0 is a constant defined in Lemma 4.14. If ‖∇∇∇′ω‖L∞(Rd−1) ≤ Kp, then

sup
06=φ∈Ŵ 1,s

′(·)
0 (Hω)

|〈∇∇∇u,∇∇∇φ〉Hω|
‖∇∇∇φ‖

Ls
′(·)(Hω)

≥ (2d1(s′))−1
{

sup
06=φ̃∈Ŵ 1,s̃′(·)(H)

|〈∇∇∇ũ,∇∇∇φ̃〉H|
‖∇∇∇φ̃‖

Ls̃
′(·)(H)

− 2d2(s)((4C2(s)d2(s))−1‖∇∇∇ũ‖Ls̃(·)(H)
}

≥ (4ds(s′)Cs(s))−1‖∇∇∇ũ‖Ls̃(·)(H) ≥ C3(s)‖∇∇∇u‖Ls(·)(Hω),

where C3(s) = (8ds(s)d1(s′)C2(s))−1. �

Lemma 4.16. Suppose (GA). Let x0 ∈ G and BR(x0) b G, and let p ∈ P
log
+ (R

d). Then for

0 < R′ < R, there exists a constant C3(p,R,R′) > 0 such that

‖∇∇∇(ηu)‖Lp(·)(G) ≤ C3(p,R,R
′) sup

06=v∈C∞0 (BR(x0))

|〈∇∇∇(ηu),∇∇∇v〉G|
‖∇∇∇v‖

Lp
′(·)(BR(x0))

for all u ∈ Ŵ 1,p(·)• (G) and η ∈ C∞0 (BR′(x0)).

Proof. Let ρ ∈ D(BR(x0)) such that 0 ≤ ρ ≤ 1 and ρ(x) = 1 for x ∈ BR′(x0). If φ ∈ D(Rd), put
cφ =

1
|BR(x0)|

∫
BR(x0)

φdx and v = ρ(φ−cφ). By the Poincaré inequlity (Theorem 2.7),

‖φ−cφ‖Lp′(·)BR′(x0)) ≤ CR‖∇∇∇v‖Lp′(·)(BR(x0)).

Here we have

‖∇∇∇v‖
Lp
′(·)(BR(x0))

≤ (1 + CR‖∇∇∇ρ‖L∞(BR(x0)))‖∇∇∇φ‖Lp′(·)(Rd ).

Since ∇∇∇ρ = 000 on BR′(x0), ρ = 1 on BR′(x0) and ∇∇∇v = (∇∇∇ρ)(φ−cφ) + ρ∇∇∇φ, we see that ∇∇∇v = ∇∇∇φ
on BR′(x0). If φ 6= 0 and v 6= 0, then we have

|〈∇∇∇(ηu),∇∇∇φ〉Rd |
‖∇∇∇φ‖

Lp
′(·)(Rd )

≤ (1 + CR‖ρ‖L∞(BR(x0)))
|〈∇∇∇(ηu),∇∇∇φ〉G|
‖∇∇∇v‖

Lp
′(·)(BR(x0))

.

By Lemma 4.10, we can see that

‖∇∇∇(ηu)‖Lp(·)(G) ≤ C1(p) sup
06=φ∈C∞0 (Rd )

|〈∇∇∇(ηu),∇∇∇φ〉Rd |
‖∇∇∇φ‖

Lp
′(·)(Rd )

≤ C1(p)(1 + CR‖∇∇∇ρ‖L∞(Rd )) sup
0 6=v∈C∞0 (BR(x0))

|〈∇∇∇(ηu),∇∇∇v〉G|
‖∇∇∇v‖

Lp
′(·)(BR(x0))

.

�

Lemma 4.17. Suppose (GA) and p ∈Plog+ (G). For each x0 ∈ ∂G, there exist R = R(p,x0,∂G) > 0
and a constant C5 = C5(R) > 0 such that

‖∇∇∇(ηu)‖Lp(·)(G) ≤ C5 sup
0 6=v∈Ŵ 1,p′(·)(GR(x0))

|〈∇∇∇(ηu),∇∇∇v〉G|
‖∇∇∇v‖

Lp
′(·)(GR(x0))

(4.12)

for all u ∈ Ŵ 1,p(·)• (G) and η ∈ C∞0 (BR/2(x0)).



Int. J. Anal. Appl. (2022), 20:13 21

Proof. There exist ρ > 0 and a C1-function σ on Bρ(x0) with (∇∇∇σ)(x0) 6= 000 such that G ∩Bρ(x0) =
{x ∈ Bρ(x0); σ(x) < 0} and ∂G ∩ BR(x0) = {x ∈ Bρ(x0); σ(x) = 0}. Then |∇∇∇σ(x0)|−1∇∇∇σ(x0)
is the unit outer normal vector at x0. Hence there exists an orthogonal matrix S such that

S(|∇∇∇σ(x0)|−1∇∇∇σ(x0)) = ed = (0, . . . , 0, 1)t. Define a transformation y = y(x) = S(x − x0).
Then y : Bρ(x0) → B̂ρ(0) = {y ∈ Rd; |y| < ρ} is a C1-bijective mapping and define σ̂(y) =
σ(x0 + S

−1y) = σ(x). Hence (∇∇∇yσ̂)(0) = S(∇∇∇xσ)(x0) = |∇∇∇σ(x0)|ed, so (∇∇∇y ′σ̂)(0) 6= 000 and
(∂ydσ̂)(0) = |∇∇∇σ(x0)| > 0. By the implicit function theorem, there exist 0 < ρ

′ < ρ, h > 0 and

ψ ∈ C1(B′
ρ′), where B

′
ρ′ = {y

′ ∈ Rd−1; |y ′| < ρ′}, such that

Z = Zρ′,h = {y = (y ′,yd) ∈ Rd; |y ′| < ρ′, |yd| < h}⊂ B̂ρ(0),

(y ′,ψ(y ′)) ∈ Z if y ′ ∈ B′ρ′ and σ̂(y
′,ψ(y ′)) = 0 for y ′ ∈ B′ρ′, Conversely, if (y

′,yd) ∈ Z and
σ(y ′,yd) = 0, then yd = ψ(y ′), ψ(0) = 0 and ∇∇∇′y ′ψ(0) = 000. Then clearly, Ĝ ∩Z = {y = (y ′,yd) ∈
Z; yd < ψ(y

′)} and ∂Ĝ ∩ Z = {y = (y ′,yd) ∈ Z; yd = ψ(y ′)}. ψ(0) = 0 and (∇∇∇′y ′ψ)(0) = 000.
Put Gρ = G ∩ Bρ(x0), Ĝ = SG and Ĝρ = Ĝ ∩ B̂ρ(0). For p ∈ P

log
+ (Gρ), u ∈ Ŵ

1,p(·)
• (Gρ) and

v ∈ Ŵ 1,p
′(·)

0 (Gρ), define p̂(y) = p(x0 + S
−1y), û(y) = u(x0 + S−1y) and v̂(y) = v(x0 + S−1y). Then

by the elementary calculations, we have

〈∇∇∇û,∇∇∇v̂〉
Ĝρ

= 〈∇∇∇u,∇∇∇v〉Gρ

and ‖∇∇∇û‖
Lp̂(·)(Ĝρ)

= ‖∇∇∇u‖Lp(·)(Gρ).
Let η ∈ D(Rd−1) such that η(y ′) = 1 for |y ′| ≤ 1 and η(y ′) = 0 for |y ′| ≥ 2. For 0 < λ < ρ′/2,

put ηλ(y ′) = η(λ−1y ′), and define

ωλ(y
′) =

{
ηλ(y

′)ψ(y ′) for |y ′| ≤ ρ′,
0 otherwise .

Then ∇∇∇′ωλ(y ′) = (∇∇∇′ηλ(y ′))ψ(y ′) + ηλ∇∇∇′ψ(y ′). Since ψ(0) = 0 and ψ ∈ C1(B′ρ′), using the mean
value theorem,

|ψ(y ′)| = |ψ(y ′) −ψ(0)| ≤ |(∇∇∇′ψ)(θy ′)||y ′| ≤ 2λ|(∇∇∇′ψ)(θy ′)| for some 0 < θ < 1.

Hence

|(∇∇∇′ηλ(y ′))ψ(y ′)| ≤ 2λ−1|(∇∇∇′η)(λ−1y ′)|λ|∇∇∇′ψ(θy ′)| = 2|(∇∇∇′η)(λ−1y ′)||∇∇∇′ψ(θy ′)|.

Therefore, we have

sup
|y ′|≤ρ

|(∇∇∇ηλ(y ′))ψ(y ′)| ≤ ‖∇∇∇′η‖L∞(Rd−1) sup
|y ′|≤2λ

|(∇∇∇′ψ)(y ′)|→ 0

as λ → 0 because ∇∇∇′ψ is continuous function and ∇∇∇′ψ(0) = 000. Moreover,

sup
|y ′|≤ρ

|ηλ(y ′)∇∇∇′ψ(y ′)| ≤ sup
|y ′|≤2λ

‖η‖L∞(Rd−1)‖(∇∇∇
′ψ)(y ′)|→ 0 as λ → 0.



22 Int. J. Anal. Appl. (2022), 20:13

Thereby, if we choose λ > 0 small enough, then ‖∇∇∇′ωλ‖L∞(Rd−1) ≤ Kp, where Kp is as in Lemma
4.15. Let R = R(p̂, 0,∂Ĝ) = λ. Then Hωλ ∩BR = Ĝ∩BR. If η ∈D(BR/2) and u ∈ Ŵ

1,p(·)
• (G), then

ηû ∈ W 1,p̂0 (Ĝ) by definition of Ŵ
1,p̂
• (Ĝ) and ηû vanishes outside Ĝ ∩BR = Hωλ ∩BR. We extend ηû

by zero to Hωλ. Then ηû ∈ W
1,p̂
0 (Hω) ⊂ Ŵ

1,p̂
0 (Hω). By Lemma 4.15, we have

‖∇∇∇(ηû)‖Lp̂(·)(Hω) ≤ C3(p) sup
0 6=v̂∈C∞0 (Hω)

|〈∇∇∇(ηû),∇∇∇v̂〉Hω|
‖∇∇∇v̂‖Lp̂(·)(Hω)

. (4.13)

We show (4.12). Let ρ ∈ C∞0 (BR) such that 0 ≤ ρ ≤ 1 and ρ(x) = 1 on BR/2. If v̂ ∈ Ŵ
1,p̂(·)
• (Hω),

then ρv̂ ∈ W 1,p̂(·)0 (Hω) and by Poincaré inequality,

‖∇∇∇(ρv̂)‖
Lp̂
′(·)(Hωλ )

≤‖∇∇∇ρ‖L∞(BR)‖v̂‖Lp̂′(·)(Zωλ
R,R

)
+ ‖∇∇∇v̂‖

Lp̂
′(·)(Hωλ )

≤ (‖∇∇∇ρ‖L∞(BR)cR + 1)‖∇∇∇v̂‖Lp̂′(·)(Hωλ ).

If v̂ 6= 0 and ρv̂ 6= 0, then we have

|〈∇∇∇(ηû),∇∇∇v̂〉Hωλ|
‖∇∇∇v̂‖

Lp̂
′(·)(Hωλ

)
≤ (‖∇∇∇ρ‖L∞(BR)cR + 1)

|〈∇∇∇(ηû),∇∇∇(ρv̂)〉Hωλ|
‖∇∇∇(ρv̂)‖

Lp̂
′(·)(Hωλ )

.

Thus (4.12) follows from Lemma 4.15 with C5 = C2(p)(‖∇∇∇ρ‖L∞(BR)cR + 1). �

5. Proof of Theorem 3.1

First we derive the uniqueness.

Theorem 5.1. Let G be a bounded domain of Rd (d ≥ 2) with a C1-boundary ∂G, and let p ∈Plog+ (G).
If u ∈ W 1,p(·)0 (G) satisfies

〈∇∇∇u,∇∇∇φ〉G = 0 for all φ ∈ W
1,p′(·)
0 (G),

then we have u = 0 a.e. in G.

Proof. Since p(x) ≥ p− for all x ∈ G and G is a bounded domain, we see that W 1,p(·)0 (G) ⊂ W
1,p−

0 (G).

Since D(G) ⊂ W 1,p
′(·)

0 (G), we have 〈∇∇∇u,∇∇∇φ〉G = 0 for all φ ∈D(G). Hence it follows from the fact
that D(G) is dense in W 1,(p

−)′(G) that we can see that

〈∇∇∇u,∇∇∇φ〉G = 0 for all φ ∈ W
1,(p−)′(·)
0 (G)

by continuity. Therefore by [25, Theorem 3.1], we have u = 0 a.e. in G. �

We give a proof of Theorem 3.1. Suppose that (3.1) does not hold. Then there exists {uk}∞k=1 ⊂
W

1,p(·)
0 (G) such that

‖∇∇∇uk‖Lp(·)(G) = 1 (5.1)

and

εk = sup
06=φ∈W 1,p

′(·)
0 (G)

|〈∇∇∇uk,∇∇∇φ〉G|
‖∇∇∇φ‖

Lp
′(·)(G)

→ 0 as k →∞. (5.2)



Int. J. Anal. Appl. (2022), 20:13 23

By the Poincaré inequality and (5.1),

‖uk‖Lp(·)(G) ≤ c diam(G)‖∇∇∇uk‖Lp(·)(G) = c diam(G).

Hence {uk}∞k=1 is bounded in a reflexive Banach space W
1,p(·)
0 (G), so passing to a subsequence (still

denoted by {uk}), we may assume that there exists u ∈ W
1,p(·)
0 (G) such that uk → u weakly in

W
1,p(·)
0 (G). For each φ ∈ W

1,p′(·)
0 (G), from (5.2), we have

〈∇∇∇u,∇∇∇φ〉G = lim
k→∞
〈∇∇∇uk,∇∇∇φ〉G = 0.

Therefore it follows from Theorem 5.1 that u = 0. Since G is bounded, the embedding W 1,p(·)0 (G) ↪→
Lp(·)(G) is compact (cf. [11, Theorem 8.4.2], so uk → 0 strongly in Lp(·)(G).

By Lemma 4.17, for each x0 ∈ ∂G, there exist R0 = R0(p,x0,∂G) > 0 and C5 = C5(R0) > 0 such
that

‖∇∇∇(ηu)‖Lp(·)(G) ≤ C5 sup
06=v∈W 1,p

′(·)
0 (G∩BR0 (x0))

|〈∇∇∇(ηu),∇∇∇v〉G|
‖∇∇∇v‖

Lp
′(·)(G∩BR0 (x0))

(5.3)

for all u ∈ W 1,p(·)0 (G) and η ∈ C
∞
0 (BR0/2)(x0)). Since ∂G is compact, there exist finitely many

xi ∈ ∂G (i = 1, . . . ,M), Ri > 0 and Ci > 0 such that ∂G ⊂ ∪Mi=1Bi, where Bi = BRi/4(xi ), and
(5.3) holds with R0 = Ri and C5 = Ci. We note that G1 := G \ (∪Mi=1Bi ) is compact and G1 ⊂ G.
According to Lemma 4.16, for each x0 ∈ G1, thete exist R0 > 0 such that BR0 (x0) ⊂ G and C3 > 0
such that

‖∇∇∇(ηu)‖Lp(·)(G) ≤ C3 sup
0 6=v∈W 1,p

′(·)
0 (G∩BR0 (x0))

|〈∇∇∇(ηu),∇∇∇v〉G|
‖∇∇∇v‖

Lp
′(·)(G∩BR0 (x0))

(5.4)

for all u ∈ W 1,p(·)0 (G) and η ∈ C
∞
0 (BR0/2)(x0)). Since G1 is compact, there exist finitely many xi ∈ G,

Ri > 0 and Ci > 0 (i = M + 1, . . . ,N) such that G1 ⊂ ∪Ni=M+1Bi, where Bi = BRi/4(xi ) and (5.4)
holds with R0 = Ri and C3 = Ci.

For each i = 1, . . . ,N, choose ψi ∈ C∞0 (B
′
i ), where B

′
i = RRi/2(xi ) such that 0 ≤ ψi ≤ 1, ψi = 1

on Bi and denote Gi = G ∩BRi/4(xi ). Then from (5.3) and (5.4), we have

‖∇∇∇uk‖Lp(·)(Gi ) ≤‖∇∇∇(ψiuk)‖Lp(·)(G) ≤ C
i sup

06=v∈W 1,p
′(·)

0 (Gi )

|〈∇∇∇(ψiuk),∇∇∇v〉G|
‖∇∇∇v‖

Lp
′(·)(Gi )

=: d ik (5.5)

Fix i = 1, . . . ,N. For each k ∈ N, there exists vk ∈ W
1,p′(·)
0 (Gi ) satisfying ‖∇∇∇vk‖Lp′(·)(Gi ) = 1 and

0 ≤ d ik −|〈∇∇∇(ψiuk),∇∇∇vk〉G| ≤ 1/k. Therefore,

0 ≤ d ik ≤
1

k
+ |〈∇∇∇uk,∇∇∇(ψivk)〉G| + |〈∇∇∇uk,vk∇∇∇ψi〉G| + |〈uk∇∇∇ψi,∇∇∇vk〉G| (5.6)

≤
1

k
+ εk‖∇∇∇(ψkvk)‖Lp′(·)(G) + |〈∇∇∇uk,vk∇∇∇ψi〉G| + |〈uk∇∇∇ψi,∇∇∇vk〉G|.

Using again the Poincaré inequality, we can see that the sequence {vk}∞k=1 is bounded in W
1,p′(·)
0 (Gi ).

Passing to a subsequence (still denoted by {vk}), there exists v ∈ W
1,p′(·)
0 (Gi ) such that vk → v



24 Int. J. Anal. Appl. (2022), 20:13

weakly in W 1,p
′(·)

0 (Gi ), so vk → v strongly in L
p′(·)(Gi ). We estimate the right-hand side of (5.6). By

the Hölder inequality,

|〈∇∇∇uk,vk∇∇∇ψi〉G| ≤ |〈∇∇∇uk, (vk −v)∇∇∇ψi〉G| + |〈∇∇∇uk,v∇∇∇ψi〉G|

≤ 2‖∇∇∇uk‖Lp(·)(G)‖∇∇∇ψi‖L∞(Gi )‖vk −v‖Lp(·)(Gi ) + |〈∇∇∇uk,v∇∇∇ψi〉G|→ 0 as k →∞

and

|〈uk∇∇∇ψi,∇∇∇vk〉G| ≤ ‖uk∇∇∇ψi‖Lp(·)(Gi ) ≤‖∇∇∇ψi‖L∞(B′i )‖uk‖Lp(·)(G) → 0 as k →∞.

By the Poincaré inequality,

‖vk‖Lp′(·)(Gi ) ≤ c diam(Gi )‖∇∇∇vk‖Lp′(·)(Gi ) = c diam(Gi ).

Hence

‖∇∇∇(ψivk)‖Lp′(·)(G) ≤‖(∇∇∇ψi )vk‖Lp′(·)(G) + ‖ψi∇∇∇vk‖Lp′(·)(Gi ) ≤‖∇∇∇ψi‖L∞(B′i )c diam(Gi ) + 1.

Summing up the above, we see that d ik → 0 as k →∞ for every i = 1, . . . ,N. Since G ⊂∪
N
i=1Bi,

‖∇∇∇uk‖Lp(·)(G) ≤
N∑
i=1

‖∇∇∇uk‖Lp(·)(Gi ) ≤
N∑
i=1

d ik → 0 as k →∞.

This contradicts ‖∇∇∇uk‖Lp(·)(G) = 1. This completes the proof of Theorem 3.1.
We can derive the Lp(·)-regularity.

Theorem 5.2. Let G be a bounded domain of Rd (d ≥ 2) with a C1-boundary. Assume that
p,q ∈Plog+ (G) satisfies q(x) ≤ p(x) for all x ∈ G. If u ∈ W

1,q(·)
0 (G) satisfies

Sp(u) := sup
0 6=φ∈C∞0 (G)

|〈∇∇∇u,∇∇∇φ〉G|
‖∇∇∇φ‖

Lp
′(·)(G)

< ∞, (5.7)

then u ∈ W 1,p(·)0 (G) and

‖∇∇∇u‖Lp(·)(G) ≤ CpSp(u), (5.8)

where Cp is the constant in Theorem 3.1.

Proof. Define a functional F ′ such that F ′(φ) = 〈∇∇∇u,∇∇∇φ〉G for φ ∈D(G). From (5.7),

|F ′(φ)| ≤ ‖F ′‖
(W

1,p′(·)
0 (G))

′‖∇∇∇φ‖Lp′(·)(G) for φ ∈D(G),

where ‖F ′‖
(W

1,p′(·)
0 (G))

′ = Sp(u). Since D(G) is dense in W
1,p′(·)
0 (G) with respect to the norm ‖∇∇∇·

‖
Lp
′(·)(G), F

′ has an extension F̃ ′ ∈ (W 1,p
′(·)

0 (G))
′ which is unique and norm-preserving, by continuity.

By Theorem 3.2, there exists uniquely up ∈ W
1,p(·)
0 (G) such that 〈∇∇∇up,∇∇∇φ〉G = F̃ ′(φ) for all φ ∈

W
1,p′(·)
0 (G). Hence

〈∇∇∇up,∇∇∇φ〉G = F̃ ′(φ) = F ′(φ) = 〈∇∇∇u,∇∇∇φ〉G



Int. J. Anal. Appl. (2022), 20:13 25

for all φ ∈D(G). Since D(G) is dense in W 1,q
′(·)

0 (G) with respect to ‖∇∇∇·‖Lq′(·)(G)-norm and q(x) ≤
p(x) for all x ∈ G, so u −up ∈ W

1,q(·)
0 (G), we have

〈∇∇∇(u −up),∇∇∇φ〉G = 0 for all φ ∈ W
1,q′(·)
0 (G).

By Theorem 5.1, u −up = 0, so u = up ∈ W
1,p(·)
0 (G) and (3.1) holds, so (5.8) follows. �

Corollary 5.3. Let G be a bounded domain of Rd (d ≥ 2) with a C1-boundary. Assume that p,q ∈
Plog+ (G) satisfies q(x) ≤ p(x) for all x ∈ G. Suppose that u ∈ W

1,q(·)
0 (G) and there exists f ∈L

p(·)(G)

such that

〈∇∇∇u,∇∇∇φ〉G = 〈f ,∇∇∇φ〉G for all φ ∈D(G). (5.9)

Then u ∈ W 1,p(·)0 (G) and satisfies (5.7). Moreover, we have

‖∇∇∇u‖Lp(·)(G) ≤ 2Cp‖f‖Lp(·)(G),

where Cpis the constant in Theorem 3.1.

Proof. By the generalized Hölder inequality,

|〈∇∇∇u,∇∇∇φ〉G| = |〈f ,∇∇∇φ〉G| ≤ 2‖f‖Lp(·)(G)‖∇∇∇φ‖Lp′(·)(G) for all φ ∈D(G).

Hence (5.7) holds and Sp(u) ≤ 2‖f‖Lp(·)(G). Hence we have

‖∇∇∇u‖Lp(·)(G) ≤ CpSp(u) ≤ 2Cp‖f‖Lp(·)(G).

�

6. Dirichlet problem for the Poisson equation

Let G be a bounded domain of Rd (d ≥ 2) with a C1-boundary ∂G. We consider the following
Dirichlet problem for the Poisson equation.{

−∆u = f in G,
u = g on ∂G.

(6.1)

We are in a position to state the main theorem of this section.

Theorem 6.1. Let G be a bounded domain of Rd (d ≥ 2) with a C1-boundary ∂G and let p ∈Plog+ (G).
Assume that f ∈ W−1,p(·)(G) and g ∈ Tr(W 1,p(·)(G)). Then the system (6.1) has a unique weak
solution u ∈ W 1,p(·)(G) in the sense that u

∣∣
∂G

= g and

〈∇∇∇u,∇∇∇v〉G = 〈f ,v〉
W−1,p(·)(G),W

1,p′(·)
0 (G)

for all v ∈ W 1,p
′(·)

0 (G). (6.2)

Furthermore, there exists a constant C = C(p,d,G) > 0 such that

‖u‖W 1,p(·)(G) ≤ C(‖f‖W−1,p(·)(G) + ‖g‖Tr(W 1,p(·)(G))). (6.3)



26 Int. J. Anal. Appl. (2022), 20:13

Proof. First we reduce the problem (6.1) to the homogeneous Dirichlet problem. Since g ∈
Tr(W 1,p(·)(G)), there exists w ∈ W 1,p(·)(G) such that w

∣∣
∂G

= g and

‖w‖W 1,p(·)(G) = ‖g‖Tr(W 1,p(·)(G)). (6.4)

Indeed, by definition of ‖g‖Tr(W 1,p(·)(G)), there exists {wj}⊂ W
1,p(·)(G) with wj

∣∣
∂G

= g and

‖g‖Tr(W 1,p(·)(G)) = lim
j→∞
‖wj‖W 1,p(·)(G)).

Hence {wj} is bounded in a reflexive Banach space W 1,p(·)(G), so passing to a subsequence of {wj}
(still denoted by {wj}) we may assume that wj → w weakly in W 1,p(·)(G). By Lemma 2.6, wj

∣∣
∂G
→

w
∣∣
∂G

in Lp(·)(∂G), so w
∣∣
∂G

= g. Therefore,

‖g‖Tr(W 1,p(·)(G)) ≤‖w‖W 1,p(·)(G) ≤ lim inf
j→∞

‖wj‖W 1,p(·)(G) = ‖g‖Tr(W 1,p(·)(G)).

Since ∆w ∈ W−1,p(·)(G), if we replace an unknown function u with v = u −w and a known function
f with F = f + ∆w ∈ W−1,p(·)(G), the problem (6.1) is reduced the following problem.{

−∆v = F in G,
v = 0 on ∂G.

(6.5)

Therefore, we consider (6.5), that is, find v ∈ W 1,p(·)0 (G) such that

〈∇∇∇v,∇∇∇φ〉G = 〈F,φ〉
W−1,p(·)(G),W

1,p′(·)
0 (G)

for all φ ∈ W 1,p
′(·)

0 (G). (6.6)

According to Theorem 3.2, there exists a unique v ∈ W 1,p(·)0 (G) such that (6.6) holds and

C−1p ‖∇∇∇v‖Lp(·)(G) ≤ ‖F‖W−1,p(·)(G)

≤ ‖f‖W−1,p(·)(G) + ‖∆w‖W−1,p(·)(G)

≤ ‖f‖W−1,p(·)(G) + C(p,G)‖w‖W 1,p(·)(G)

≤ ‖f‖W−1,p(·)(G) + C(p,G)‖g‖Tr(W 1,p(·)(G)).

By the Poincaré inequality, ‖v‖W 1,p(·)(G) ≤ C1(p,G)‖∇∇∇v‖Lp(·)(G). If we put u = v + w, (6.2) and the
estimate (6.3) follows. �

Remark 6.2. The authors in [11] showed that if G is a bounded domain with a C1,1-boundary and

f ∈ Lp(·)(G),g ∈ Tr(W 2,p(·)(G)), the system (6.1) has a unique strong solution u ∈ W 2,p(·)(G) and
there exists a constant C depending only on p and G such that

‖u‖W 2,p(·)(G) ≤ C(‖f‖Lp(·)(G) + ‖g‖Tr(W 2,p(·)(G))). (6.7)

They used the Newton potential, and only announced the existence of a weak solution as in Theorem

6.1. However, we can easily show that Theorem 6.1 holds using Theorem 3.1 and 3.2 under the

weaker assumption of the regularity of boundary.



Int. J. Anal. Appl. (2022), 20:13 27

7. An approach to the Stokes problem

In this section, let G be a bounded domain of Rd (d ≥ 2) with a C1-boundary Γ = ∂G. We consider
the following homogeneous Stokes problem.


−∆u + ∇∇∇π = f in G,
divu = 0 in Ω,

u = 000 on Γ,

π = 0 on Γ.

(7.1)

We have the following theorem.

Theorem 7.1. Let G be a bounded domain of Rd (d ≥ 2) with a C1-boundary Γ = ∂G and let
p ∈ Plog+ (G). Assume that f ∈ L

p(·)(G). Then the problem (7.1) has a unique weak solution

(u,π) ∈W1,p(·)0 (G) ×W
1,p(·)
0 (G), in the sense of

〈∇∇∇u,∇∇∇v〉G + 〈∇∇∇π,v〉G = 〈f ,v〉G for all v ∈W
1,p′(·)
0 (G), (7.2)

and there exists a constant C = C(p,d,G) > 0 such that

‖u‖W1,p(·)(G) + ‖π‖W 1,p(·)(G) ≤ C‖f‖Lp(·)(G). (7.3)

Furthermore, if G is of class C1,1, then u ∈ W2,p(·)(G) and (u,π) is a strong solution of (7.1).
Moreover, we have

‖u‖W2,p(·)(G) + ‖π‖W 1,p(·)(G) ≤ C
′‖f‖Lp(·)(G)

where C′ is a constant depending only on p,d and G.

Proof. First we consider the following Dirichlet problem for the Laplacian ∆.{
∆π = div f in G,

π = 0 on Γ.
(7.4)

Suppose that f ∈ Lp(·)(G). Since div f ∈ W−1,p(·)(G), if follows from Theorem 6.1 that (7.4)
has a unique weak solution π ∈ W 1,p(·)0 (G) and there exist positive constants C = C(p,d,G) and
C1 = C1(p,d,G) such that

‖π‖W 1,p(·)(G) ≤ C‖div f‖W−1,p(·)(G) ≤ C1‖f‖Lp(·)(G). (7.5)

We note that

f −∇∇∇π ∈Lp(·)(G) and div (f −∇∇∇π) = 0 in the distribution sense. (7.6)

We apply Proposition 3.3. For this purpose, define

Xp(·)(G) = {u ∈W1,p(·)0 (G); divu = 0 in G}.



28 Int. J. Anal. Appl. (2022), 20:13

Then clearly Xp(·)(G) is a closed subspace of a reflexive Banach space W1,p(·)0 (G), so X
p(·)(G) is also

reflexive Banach space. Since Xp(·)(G) ⊂W1,p(·)0 (G), if follows from Theorem 3.1

0 < C−1p ≤ inf
0006=u∈W1,p(·)0 (G)

sup
000 6=v∈W1,p

′(·)
0 (G)

|〈∇∇∇v,∇∇∇u〉G|
‖∇∇∇u‖Lp(·)(G)‖∇∇∇v‖Lp′(·)(G)

(7.7)

≤ inf
0006=u∈Xp(·)(G)

sup
0006=v∈W1,p

′(·)
0 (G)

|〈∇∇∇v,∇∇∇u〉G|
‖∇∇∇u‖Lp(·)(G)‖∇∇∇v‖Lp′(·)(G)

.

Taking the Poincaré inequality into consideration, let X = W1,p
′(·)

0 (G) equipped with the norm ‖v‖X =
‖∇∇∇v‖

Lp
′(·)(G) and M = X

p(·)(G) equipped with the norm ‖u‖M = ‖∇∇∇u‖Lp(·)(G). Define a(v,u) =
〈∇∇∇v,∇∇∇u〉G for v ∈ X,u ∈ M. By the generalized Hölder inequality, we have

|a(v,u)| ≤ 2‖∇∇∇v‖
Lp
′(·)(G)‖∇∇∇u‖Lp(·)(G) = 2‖v‖X‖u‖M.

Thus a(v,u) is a continuous bilinear form on X×M. Define bounded linear operators A : X → M′ and
A′ : M → X′ by 〈Av,u〉 = 〈v,A′u〉 = a(v,u). Then |〈v,A′u〉| ≤ 2‖v‖X‖u‖M for all v ∈ X,u ∈ M.
From (7.7),

C−1p ≤ inf
000 6=u∈M

sup
0006=v∈X

|a(v,u)|
‖v‖X‖u‖M

.

Put V = KerA.

We characterize V = KerA.

Lemma 7.2. It follows that V = {v = (−∆)−1∇∇∇ϕ; ϕ ∈ Lp
′(·)(G)}. Here v = (−∆)−1g, g ∈

W−1,p
′(·)(G) means that v ∈W1,p

′(·)
0 (G) is a unique weak solution of the following problem.{

−∆v = g in G,
v = 000 on Γ.

(7.8)

Proof. Let v ∈ V . Then 〈Av,u〉M′,M = 0 for all u ∈ M, that is,

〈−∆v,u〉
W−1,p

′(·)(G),W
1,p(·)
0 (G)

= 〈∇∇∇v,∇∇∇u〉G = 〈Av,u〉M′,M = 0

for all u ∈ Xp(·)(G). By the de Rham theorem (cf. Aramaki [3,4]), there exists ϕ ∈ Lp
′(·)(G) such

that −∆v = ∇∇∇ϕ in W−1,p
′(·)(G). Since v = 000 on Γ, we have v = (−∆)−1∇∇∇ϕ. Thus we have

V ⊂{v = (−∆)−1∇∇∇ϕ; ϕ ∈ Lp
′(·)(G)}.

Conversely, let v = (−∆)−1∇∇∇ϕ for some ϕ ∈ Lp
′(·)(G). Then v is a unique weak solution of (7.8)

with g = ∇∇∇ϕ. For any u ∈Xp(·)(G), we have

〈Av,u〉M′,M = 〈∇∇∇v,∇∇∇u〉G = 〈−∆v,u〉W−1,p′(·)(G),W1,p(·)0 (G)

= 〈∇∇∇ϕ,u〉
W−1,p

′(·)(G),W
1,p(·)
0 (G)

= −〈ϕ, divu〉G = 0.

Hence Av = 000 in M′, that is, v ∈ KerA = V . �

Denote that V ⊥ = {f ∈ X′ = W−1,p(·)(G);〈f ,v〉X′,X = 0 for all v ∈ V}.



Int. J. Anal. Appl. (2022), 20:13 29

Lemma 7.3. If g ∈Lp(·)(G) satisfies divg = 0, then g ∈ V ⊥.

Proof. According to Lemma 7.2, for any v ∈ V , there exists ϕ ∈ Lp
′(·)(G) such that v = (−∆)−1∇∇∇ϕ.

Thereby, we have

〈g,v〉X′,X = 〈g, (−∆)−1∇∇∇ϕ〉
W−1,p(·)(G),W

1,p′(·)
0 (G)

= 〈(−∆)−1g,∇∇∇ϕ〉
W

1,p(·)
0 (G),W

−1,p′(·)(G)

= −〈div (−∆)−1g,ϕ〉G.

If we put w = (−∆)−1g, we have −∆w = g in Ω and w = 000 on Γ. Hence

(−∆)divw = −div ∆w = divg = 0.

Therefore, (−∆)−1(−∆)divw = divw = 0, that is, div (−∆)−1g = 0. This implies that 〈g,v〉X′,X = 0
for all v ∈ V . Thus g ∈ V ⊥. �

We continue the proof of Theorem 7.1. From (7.6), we know that f − ∇∇∇π ∈ Lp(·)(G) and
div (f −∇∇∇π) = 0. By Proposition 3.3, A′ : M → V ⊥ is an isomorphism and Cp is the continuity
constant of (A′)−1. By Lemma 7.3, we see that f −∇∇∇π ∈ V ⊥, so there exists a unique u ∈ M such
that A′u = f −∇∇∇π, that is, divu = 0 in G, u = 000 on Γ and

〈∇∇∇v,∇∇∇u〉G = 〈v, f −∇∇∇π〉X,X′ for all v ∈W
1,p′(·)
0 (G),

so (7.2) holds. Furthermore, we have

‖u‖W1,p(·)(G) ≤ Cp‖f −∇∇∇π‖W−1,p(·)(G)

≤ C′p‖f −∇∇∇π‖Lp(·)(G)

≤ C′p(‖f‖Lp(·)(G) + ‖∇∇∇π‖Lp(·)(G))

≤ C′′p‖f‖Lp(·)(G).

Summing up this inequality and (7.5), we get the estimate (7.3).

If, in particular, G is of class C1,1, since −∆u = f −∇∇∇π ∈Lp(·)(G) in G and u = 000 on Γ, it follows
from [11, Theorem 14.1.2] that u ∈W2,p(·)(G) and

‖u‖W2,p(·)(G) ≤ C‖f −∇∇∇π‖Lp(·)(G) ≤ C1‖f‖Lp(·)(G).

�

Now we consider the inhomogeneous Stokes problem.

−∆u + ∇∇∇π = f in G,
divu = ϕ in G,

u = g on Γ,

π = π0 on Γ.

(7.9)



30 Int. J. Anal. Appl. (2022), 20:13

Theorem 7.4. Let G be a bounded domain of Rd with a C1,1-boundary Γ and let p ∈ Plog+ (G).
Assume that f ∈ Lp(·)(G), π0 ∈ Tr(W 1,p(·)(G)), ϕ ∈ W 1,p(·)(G) and g ∈ Tr(W2,p(·)(G)) satisfy the
compatibility condition ∫

G

ϕdx =

∫
Γ

g ·ndσ, (7.10)

where dσ is the surface measure on Γ. Then there exists a unique solution (u,π) ∈ W2,p(·)(G) ×
W 1,p(·)(G) of (7.9) and there exists a constant C = C(p,d,G) > 0 such that

‖u‖W2,p(·)(G) + ‖π‖W 1,p(·)(G)

≤ C(‖f‖Lp(·)(G) + ‖π0‖Tr(W 1,p(·)(G)) + ‖ϕ‖W 1,p(·)(G) + ‖g‖Tr(W1,p(·)(G))).

Before the proof, it is necessary to prepare some arguments.

Proposition 7.5. Let G be a bounded domain of Rd with a C1,1-boundary Γ and p ∈Plog+ (G). If we
assume that g ∈ Tr(W 1,p(·)(G)), then there exists u ∈ W 2,p(·)(Ω) such that γ1(u) = g and γ0(u) = 0.

Proof. We use the argument of Boyer and Fabrie [7, Proof of Theorem III.2.23].

Let δ(x) be the signed distance from x to Γ, that is,

δ(x) =

{
d(x, Γ) if x ∈ G,
−d(x, Γ) if x /∈ G.

Then δ is Lipschitz-continuous in Rd with the Lipschitz constant Lip(δ) ≤ 1. Let η be a standard
mollifier, that is, η ∈ C∞0 (R

d), supp η ⊂ B (the unit sphere of Rd), η ≥ 0,
∫
Rd ηdx =

∫
B
ηdx = 1

and η(x) only depends on |x|.
For x ∈ Rd and τ ∈ R, define a function

G(x,τ) =

∫
B

δ
(
x +

τ

2
z
)
η(z)dx.

Then we can clearly see that G ∈ C∞(Rd × (R\{0}) and

|G(x,τ1) −G(x,τ2)| ≤
1

2
|τ1 −τ2|.

Therefore, by Banach fixed-point theorem, for any x ∈ Rd, there exists uniquely ρ(x) ∈ R such that
ρ(x) = G(x,ρ(x)). We call ρ a regularized distance function of G. The regularized distance function

ρ has the following properties.

(i) ρ(x) = 0 ⇐⇒ δ(x) = 0 ⇐⇒ x ∈ Γ, and there exists constants C1,C2 > 0 such that C1 ≤
δ(x)/ρ(x) ≤ C2 for x ∈ Rd \ Γ.

(ii) ρ ∈ C1,1(Rd) ∩C∞(Rd \ Γ).
(iii) ∇∇∇ρ(x) = ∇∇∇δ(x) = −n(x) for all x ∈ Γ, and there exists an open neighborhood U of Γ such

that infU |∇∇∇ρ| > 0.



Int. J. Anal. Appl. (2022), 20:13 31

For g ∈ Tr(W 1,p(·)(G)), there exists v ∈ W 1,p(·)(G) such that γ0(v) = g and there exists a constant
C > 0 such that ‖v‖W 1,p(·)(G) ≤ C‖g‖Tr(W 1,p(·)(G)). We define

Rng(x) = −ρ(x)
∫
B

v(x + αρ(x)z)η(z)dz, (7.11)

where α > 0 so that α ≤ C1 and αLip(ρ) < 1, C1 is the constant of (i). For x ∈ G and z ∈ B, we
have x + αρ(x)z ∈ G, so Rng is well defined and Rng ∈ C∞(G). We show that Rng ∈ W 2,p(·)(G).
By the calculations, for i, j = 1, . . . ,d, we have

∂Rng

∂xi
(x) = −

∂ρ

∂xi
(x)

∫
B

v(x + αρ(x)z)η(z)dz

−ρ(x)
∫
B

∇∇∇v(x + αρ(x)z) ·
(
ei + α

∂ρ

∂xi
(x)z

)
η(z)dz, (7.12)

∂2Rng

∂xi∂xj
= −

∂2ρ

∂xi∂xj
(x)

∫
B

v(x + αρ(x)z)ψ(z)dz

−
∂ρ

∂xi
(x)

∫
B

∂v

∂xj
(x + αρ(x)z)ψ(z)dz

−α
∂ρ

∂xi
(x)

∂ρ

∂xj
(x)

∫
B

∇∇∇v(x + αρ(x)z) ·ψ(z)zdz

+(d − 1)
∂ρ

∂xj
(x)

∫
B

∂v

∂xi
(x + αρ(x)z)η(z)dz

+
1

α

∫
B

∂v

∂xi
(x + αρ(x)z)

∂η

∂zj
(z)dz

+
∂ρ

∂xj
(x)

∫
B

∂v

∂xi
(x + αρ(x)z)∇∇∇η(z) ·zdz,

where ψ(z) = η(z) − div z (η(z)z). Since all the derivatives of ρ up to second-order are bounded, it
suffices to show that the terms of the form

F (x) =

∫
B

f (x + αρ(x)z)ψ̃(z)dz for x ∈ G,

where f ∈ Lp(·)(G) and ψ̃ ∈ C∞0 (B), belong to L
p(·)(G).

We note that we can not use the Jensen inequality in the case of variable exponent. However,

applying a variant of the Jensen inequality (cf. [11, Theorem 4.2.4 and Corollary 4.2.5]), there exists

a constant C > 0 such that

ρp(·),G(F ) ≤ C‖f‖
p+−p−
Lp(·)(G)

ρp(·),G(f ) + C‖f‖
p+

Lp(·)(G)
.

Therefore, we see that F ∈ Lp(·)(G), so u := Rng ∈ W 2,p(·)(G). From (7.11) and property (i), we
see that γ0(u) = 0. From (7.12) and property (iii), we can see that γ1(u) = γ0(∇∇∇Rng) ·n = γ0(v) =
g. �



32 Int. J. Anal. Appl. (2022), 20:13

Lemma 7.6. Let G be a bounded domain of Rd with a C1,1-boundary Γ and let p ∈ Plog+ (G). For
(g0,g1) ∈ Tr(W 2,p(·)(G)) × Tr(W 1,p(·)(G)), there exists u ∈ W 2,p(·)(G) such that γ0(u) = g0 and
γ1(u) = g1, moreover there exists a constant C > 0 such that

‖u‖W 2,p(·)(G) ≤ C(‖g0‖Tr(W 2,p(·)(G)) + ‖g1‖Tr(W 1,p(·)(G))). (7.13)

Proof. From Theorem 2.5, we see that

W
2,p(·)
0 (G) = {v ∈ W

2,p(·)(G); γ0(v) = γ1(v) = 0}.

We consider the mapping

γ : W 2,p(·)(G)/W
2,p(·)
0 (G) 3 [u] 7→ (γ0(u),γ1(u)) ∈ Tr(W

2,p(·)(G)) × Tr(W 1,p(·)(G)).

Since n is a Lipschitz function on Γ, we can extend n to a Lipschitz function on G, so γ1(u) =

γ0(∇∇∇u · n). Thus γ is a linear continuous injection. We show that γ is surjective. Let (g0,g1) ∈
Tr(W 2,p(·)(G)) × Tr(W 1,p(·)(G)). Choose v0 ∈ W 2,p(·)(G) such that γ0(v0) = g0 and define v =
v0 + Rn(g1 −γ1(v0)) ∈ W 2,p(·)(G). Then by Proposition 7.5, γ0(v) = γ0(v0) = g0 and γ1(v) = g1.
Thereby γ is surjective. By the Banach open mapping theorem, γ−1 is also linear and continuous.

Moreover there exists a constant C > 0 such that

‖[v]‖
W 2,p(·)(G)/W

2,p(·)
0 (G)

≤ C(‖g0‖Tr(W 2,p(·)(G)) + ‖g1‖Tr(W 1,p(·)(G))).

We show that ‖[v]‖
W 2,p(·)(G)/W

2,p(·)
0 (G)

= inf{‖v + w‖W 2,p(·)(G); w ∈ W
2,p(·)
0 (G)} is achieved. Indeed,

choose wj ∈ W
2,p(·)
0 (G) such that

lim
j→∞
‖v + wj‖W 2,p(·)(G) = ‖[v]‖W 2,p(·)(G)/W 2,p(·)0 (G)

.

Then {wj} is bounded in a reflexive Banach space W
2,p(·)
0 (G). Passing to a subsequence, we may

assume that wj → w weakly in W 2,p(·)(G). Hence

‖v + w‖W 2,p(·)(G) ≤ lim inf
j→∞

‖u + wj‖W 2,p(·)(G) = ‖[v]‖W 2,p(·)(G)/W 2,p(·)0 (G)
.

If we put u = v + w ∈ W 2,p(·)(G), then we have γ0(u) = g0, γ1(u) = g1 and the estimate (7.13)
holds. �

The following Lemma is the celebrated Héron formula (cf. Amrouche and Girault [2, Lemma 3.5]).

Lemma 7.7. Let G be a bounded domain of Rd with a C1,1-boundary Γ and let p ∈ Plog+ (G). Then
for u ∈W2,p(·)(G), the following Héron formula holds.

γ0(divu) = div Γ(γ0(u)t) + γ1(u) ·n− 2Kγ0(u) ·n,

where K denotes the mean curvature of Γ, div Γ is the surface divergence and vt = v−(v ·n)n is the
tangent component of v.



Int. J. Anal. Appl. (2022), 20:13 33

Proposition 7.8. Let G be a bounded domain of Rd with a C1,1-boundary Γ and let p ∈ Plog+ (G).
Then for every g ∈ Tr(W2,p(·)(G)) and ϕ ∈ Tr(W 1,p(·)(G)), there exists u ∈ W2,p(·)(G) such that
γ0(divu) = ϕ and γ0(u) = g, and there exists a constant C > 0 depending only on p and G such

that

‖u‖W2,p(·)(G) ≤ C(‖g‖Tr(W2,p(·)(G)) + ‖ϕ‖Tr(W 1,p(·)(G))). (7.14)

Proof. Put g0 = g ∈ Tr(W2,p(·)(G)),g1 = 2Kg − ndiv Γ(gt) + ϕn. By Lemma 7.6, there exists
u ∈ W2,p(·)(G) such that γ0(u) = g,γ1(u) = 2Kg −ndiv Γ(gt) + ϕn, and (7.14) holds. Then by
Lemma 7.7, we have γ0(divu) = ϕ. �

Proposition 7.9. Let G be a bounded domain of Rd with a C1,1-boundary Γ and let p ∈Plog+ (G). For
any g ∈ Tr(W2,p(·)(G)) and any ϕ ∈ W 1,p(·)(G) satisfying the compatibility condition (7.10), there
exists u0 ∈ W2,p(·)(G) such that divu0 = ϕ in G and γ0(u0) = g, moreover, there exists a constant
C > 0 depending only on p,d and G such that

‖u0‖W2,p(·)(G) ≤ C(‖ϕ‖W 1,p(·)(G) + ‖g‖Tr(W2,p(·)(G))). (7.15)

Proof. By Proposition 7.8, there exists u ∈ W2,p(·)(G) such that γ0(divu) = γ0(ϕ), γ0(u) = g and
(7.14) holds. Then divu−ϕ ∈ W 1,p(·)(G). Since γ0(divu−ϕ) = 0, we see that divu−ϕ ∈ W

1,p(·)
0 (G).

Since it follows from the compatibility condition (7.10) and the Green theorem that∫
G

(divu−ϕ)dx =
∫

Γ

g ·ndσ −
∫
G

ϕdx = 0.

By [3, 4, Theorem 3.1] (e) (cf. Aramaki [6] for the case p(·) = p = const.), there exists w ∈
W

2,p(·)
0 (G), unique up to an additive function of Kerdiv := {v ∈ W

2,p(·)
0 (G); divv = 0 in G}, such

that divw = divu−ϕ in G, and there exists a constant C > 0 such that

‖[w]‖W2,p(·)(G)/Kerdiv ≤ C‖divu−ϕ‖W 1,p(·)(G) ≤ C1(‖g‖Tr(W2,p(·)(G)) + ‖ϕ‖W 1,p(·)(G)).

Since we can easily see that

‖[w]‖W2,p(·)(G)/Kerdiv = inf{‖w + v‖W2,p(·)0 (G)
with divv = 0 in G}

is achieved, there exists u1 ∈W
2,p(·)
0 (G) such that

‖w + u1‖W2,p(·)(G) = ‖[w]‖W2,p(·)(G)/Kerdiv ≤ C1(‖g‖Tr(W2,p(·)(G)) + ‖ϕ‖W 1,p(·)(G)).

It suffices to put u0 = w + u1. �

Proof of Theorem 7.4

By Proposition 7.9, there exists u0 ∈W2,p(·)(G) such that

divu0 = ϕ in G and γ0(u0) = g



34 Int. J. Anal. Appl. (2022), 20:13

and the estimate (7.15) holds. Moreover, there exists π̃ ∈ W 1,p(·)(G) such that γ0(π̃) = π0 and

‖π̃‖W 1,p(·)(G) ≤ C‖π0‖Tr(W 1,p(·)(G)). (7.16)

If we put v = u−u0 and π̂ = π−π̃ in problem (7.9), then the system (7.9) is reduced to the following
problem 


−∆v + ∇∇∇π̂ = f + ∆u0 + ∇∇∇π̃ in G,
divv = 0 in G,

v = 000 on Γ,

π̂ = 0 on Γ.

(7.17)

From Theorem 7.1, the estimates (7.15) and (7.16), the conclusion is clear.

Conflicts of Interest: The author(s) declare that there are no conflicts of interest regarding the

publication of this paper.

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	1. Introduction
	2. Preliminaries
	3. The weak Dirichlet problem for the Laplacian  in a variable exponent Sobolev space in a bounded domain
	4. Preparation to a proof of Theorem 3.1
	5. Proof of Theorem 3.1
	6. Dirichlet problem for the Poisson equation
	7. An approach to the Stokes problem
	References