http://www.aimspress.com/journal/mine
Mathematics in Engineering, 5(3): 1–21.
DOI:10.3934/mine.2023052
Received: 01 July 2022
Revised: 09 August 2022
Accepted: 19 September 2022
Published: 23 September 2022
Research article
Bloch estimates in non-doubling generalized Orlicz spaces†
Petteri Harjulehto1, Peter Hästö2,∗ and Jonne Juusti2
1 Department of Mathematics and Statistics, FI-00014 University of Helsinki, Finland
2 Department of Mathematics and Statistics, FI-20014 University of Turku, Finland
† This contribution is part of the Special Issue: PDEs and Calculus of Variations–Dedicated to
Giuseppe Mingione, on the occasion of his 50th birthday
Guest Editors: Giampiero Palatucci; Paolo Baroni
Link: www.aimspress.com/mine/article/6240/special-articles
* Correspondence: Email: peter.hasto@utu.fi.
Abstract: We study minimizers of non-autonomous functionals
inf
u
ˆ
Ω
ϕ(x, |∇u|) dx
when ϕ has generalized Orlicz growth. We consider the case where the upper growth rate of ϕ is
unbounded and prove the Harnack inequality for minimizers. Our technique is based on “truncating”
the function ϕ to approximate the minimizer and Harnack estimates with uniform constants via a Bloch
estimate for the approximating minimizers.
Keywords: non-doubling; Harnack’s inequality; generalized Orlicz space; Musielak–Orlicz spaces;
nonstandard growth; variable exponent; double phase
Dedicated to Giuseppe Mingione on his 50th anniversary.
1. Introduction
Minimizers of the variable exponent energy
´ |∇u|p(x)dx have been studied in hundreds of papers.
In almost all cases, it is assumed that there exist constants c,C ∈ (1,∞) such that c 6 p(x) 6 C for all
x. However, it is possible to use limiting procedures to study the borderline cases when p(x) = 1 or
p(x) = ∞ for some points [34, 35]. In recent years, minimizers of non-autonomous functionals
inf
u
ˆ
Ω
ϕ(x, |∇u|) dx
2have been studied when ϕ has generalized Orlicz growth with tentative applications to anisotropic
materials [57] and image processing [31]. Again, the upper and lower growth rates are usually assumed
to lie in (1,∞). In this article we consider the case when the upper growth rate is allowed to equal ∞
in some points and the lower growth rate is greater than n, the dimension. We prove the Harnack
inequality for minimizers of such energies.
Let us recall some information of the context by way of motivation. PDE with generalized Orlicz
growth have been studied in many papers lately, both in the general setting and in particular special
cases, such as the double phase case (e.g., [3, 5, 16, 17, 53]), perturbed variable exponent [52], Orlicz
variable exponent [27], degenerate double phase [4], Orlicz double phase [6, 10], variable exponent
double phase [18, 49, 50], multiple-phase [7, 22], and double variable exponent [56]. Our framework
includes all these cases.
In the generalized Orlicz case it is known that solutions with given boundary values
exist [15, 28, 33], minimizers or solutions with given boundary values are locally bounded, satisfy
Harnack’s inequality and belong to C0,αloc [9, 36, 37, 55] or C
1,α
loc [38, 39], quasiminimizers satisfy a
reverse Hölder inequality [32], minimizers for the obstacle problem are continuous [41] and the
boundary Harnack inequality holds for harmonic functions [12]. Some articles deal with the
non-doubling [13] or parabolic [54] case as well as with the Gauss image problem [44]. We refer to
the surveys [11, 48] and monographs [14, 30, 42] for an overview. Advances have also been made in
the field of (p, q)-growth problems [19–21, 45–47].
In [8,9], the Harnack inequality was established in the doubling generalized Orlicz case for bounded
or general solutions. In the current paper, we consider the effect of removing the assumption that the
growth function is doubling thus allowing the upper growth rate to equal∞. The approach is based on
ideas from [35, 43] involving approximating the energy functional. This is more difficult compared to
the variable exponent case, since the form of the approximating problem is unclear as is the connection
between solutions and minimizers. Additionally, the challenge in taking limits without the doubling
assumption is to track the dependence of various constants on the parameters and to ensure that no
extraneous dependence is introduced in any step. Nevertheless, we improve even the result for the
variable exponent case.
Let us consider an example of our main result, Theorem 5.5. In the variable exponent case ϕ(x, t) :=
tp(x) we compare with our previous result [35, Theorem 6.4]. In the previous result, we assumed that
1
p is Lipschitz continuous, but now we only need the more natural log-Hölder continuity. Furthermore,
the previous result applied only to small balls in which the exponent was (locally) bounded. The next
example shows that the new result applies even to some sets where the the exponent is unbounded.
Example 1.1 (Variable exponent). Define p : B1 → (n,∞] on the unit ball B1 as p(x) := 2n log e|x| .
Hence p(0) = ∞ but p < ∞ a.e. Assume that f ∈ W1,p(·)(B1) with %p(·)(|∇ f |) < ∞. If u ∈ f + W1,p(·)0 (B1)
is a minimizer of the p(·)-energy, then the Harnack inequality
sup
Br
(u + r) 6 C inf
Br
(u + r)
holds for r 6 14 . The constant C depends only on n and %p(·)(|∇ f |). Note that Br we have, in the notation
of Theorem 5.5, p− = 2n log er and q
◦ = 2n log 2er so that
q◦
p− =
1+log 2+log 1r
log 2+log 1r
is bounded independent of r.
In the double phase case we also obtain a corollary of Theorem 4.6 which improves earlier results in
that the dependence of the constant is only on qp , not p and q. Note that the usual assumption of Hölder
Mathematics in Engineering Volume 5, Issue 3, 1–21.
3continuity of a is a special case of the inequality in the lemma, see [30, Proposition 7.2.2]. Also note
that the “+r” in the Harnack inequality is not needed in this case, since the double phase functional
satisfies (A1) in the range [0, K|B| ] rather than [1,
K
|B| ].
Corollary 1.2 (Double phase). Let Ω ⊂ Rn be a bounded domain, n < p < q and H(x, t) := tp + a(x)tq.
Assume that f ∈ W1,H(Ω) and
a(x) . a(y) + |x − y|α with qp 6 1 + αn
for every x, y ∈ Ω. Then any minimizer u of the ϕ-energy with boundary value function f satisfies the
Harnack inequality
sup
Br
u 6 C inf
Br
u.
The constant C depends only on n, qp and %H(|∇ f |).
2. Definitions and notation
We briefly introduce our definitions. More information on Lϕ-spaces can be found in [30]. We
assume that Ω ⊂ Rn is a bounded domain, n > 2. Almost increasing means that there exists a constant
L > 1 such that f (s) 6 L f (t) for all s < t. If there exists a constant C such that f (x) 6 Cg(x) for almost
every x, then we write f . g. If f . g . f , then we write f ≈ g.
Definition 2.1. We say that ϕ : Ω × [0,∞)→ [0,∞] is a weak Φ-function, and write ϕ ∈ Φw(Ω), if the
following conditions hold for a.e. x ∈ Ω:
• y 7→ ϕ(y, f (y)) is measurable for every measurable function f : Ω→ R.
• t 7→ ϕ(x, t) is non-decreasing.
• ϕ(x, 0) = lim
t→0+
ϕ(x, t) = 0 and lim
t→∞ϕ(x, t) = ∞.
• t 7→ ϕ(x,t)t is L-almost increasing on (0,∞) with constant L independent of x.
If ϕ ∈ Φw(Ω) is additionally convex and left-continuous with respect to t for almost every x, then ϕ is a
convex Φ-function, and we write ϕ ∈ Φc(Ω). If ϕ does not depend on x, then we omit the set and write
ϕ ∈ Φw or ϕ ∈ Φc.
For ϕ ∈ Φw(Ω) and A ⊂ Rn we denote ϕ+A(t) := ess sup
x∈A∩Ω
ϕ(x, t) and ϕ−A(t) := ess infx∈A∩Ω
ϕ(x, t).
We next define the un-weightedness condition (A0), the almost continuity conditions (A1) and the
growth conditions (aInc) and (aDec). Note that the constants Lp and Lq are independent of x even
though p and q can be functions.
Definition 2.2. Let s > 0, p, q : Ω→ [0,∞) and let ω : Ω× [0,∞)→ [0,∞) be almost increasing with
respect to the second variable. We say that ϕ : Ω × [0,∞)→ [0,∞) satisfies
(A0) if there exists β ∈ (0, 1] such that ϕ(x, β) 6 1 6 ϕ(x, 1
β
) for a.e. x ∈ Ω;
(A1-ω) if for every K > 1 there exists β ∈ (0, 1] such that, for every ball B,
ϕ+B(βt) 6 ϕ
−
B(t) when ω
−
B(t) ∈
[
1,
K
|B|
]
;
Mathematics in Engineering Volume 5, Issue 3, 1–21.
4(A1-s) if it satisfies (A1-ω) for ω(x, t) := ts;
(A1) if it satisfies (A1-ϕ);
(aInc)p(·) if t 7→ ϕ(x,t)tp(x) is Lp-almost increasing in (0,∞) for some Lp > 1 and a.e. x ∈ Ω;
(aDec)q(·) if t 7→ ϕ(x,t)tq(x) is Lq-almost decreasing in (0,∞) for some Lq > 1 and a.e. x ∈ Ω.
We say that (aInc) holds if (aInc)p holds for some constant p > 1, and similarly for (aDec). If in the
definition of (aInc)p(·) we have Lp = 1, then we say that ϕ satisfies (Inc)p(·), similarly for (Dec)q(·).
Note that if ϕ satisfies (aInc)p with a constant Lp, then it satisfies (aInc)r for every r ∈ (0, p) with
the same constant Lp. This is seen as follows, with s < t:
ϕ(x, s)
sr
= sp−r
ϕ(x, s)
sp
6 sp−rLp
ϕ(x, t)
tp
= Lp
( s
t
)p−rϕ(x, t)
tr
6 Lp
ϕ(x, t)
tr
.
Condition (A1) with K = 1 was studied in [30] under the name (A1′). The condition (A1-ω) was
introduced in [8] to combine (A1) and (A1-n) as well as other cases. It is the appropriate assumption
if we have a priori information that the solution is in W1,ω or the corresponding Lebesgue or Hölder
space. The most important cases are ω = ϕ and ω(x, t) = ts, that is (A1) and (A1-s).
Definition 2.3. Let ϕ ∈ Φw(Ω) and define the modular %ϕ for u ∈ L0(Ω), the set of measurable functions
in Ω, by
%ϕ(u) :=
ˆ
Ω
ϕ(x, |u(x)|) dx.
The generalized Orlicz space, also called Musielak–Orlicz space, is defined as the set
Lϕ(Ω) :=
{
u ∈ L0(Ω) : lim
λ→0+
%ϕ(λu) = 0
}
equipped with the (Luxemburg) quasinorm
‖u‖Lϕ(Ω) := inf
{
λ > 0 : %ϕ
(u
λ
)
6 1
}
.
We abbreviate ‖u‖Lϕ(Ω) by ‖u‖ϕ if the set is clear from context.
Definition 2.4. A function u ∈ Lϕ(Ω) belongs to the Orlicz–Sobolev space W1,ϕ(Ω) if its weak partial
derivatives ∂1u, . . . , ∂nu exist and belong to Lϕ(Ω). For u ∈ W1,ϕ(Ω), we define the quasinorm
‖u‖W1,ϕ(Ω) := ‖u‖Lϕ(Ω) + ‖∇u‖Lϕ(Ω).
We define Orlicz–Sobolev space with zero boundary values W1,ϕ0 (Ω) as the closure of {u ∈ W1,ϕ(Ω) :
supp u ⊂ Ω} in W1,ϕ(Ω).
In the definition ‖∇u‖Lϕ(Ω) is an abbreviation of
∥∥∥|∇u|∥∥∥
Lϕ(Ω)
. Again, we abbreviate ‖u‖W1,ϕ(Ω) by ‖u‖1,ϕ
if Ω is clear from context. W1,ϕ0 (Ω) is a closed subspace of W
1,ϕ(Ω), and hence reflexive when W1,ϕ(Ω)
is reflexive. We write f + W1,ϕ0 (Ω) to denote the set { f + v : v ∈ W1,ϕ0 (Ω)}.
Mathematics in Engineering Volume 5, Issue 3, 1–21.
5Definition 2.5. We say that u ∈ W1,ϕloc (Ω) is a local minimizer ifˆ
supp h
ϕ(x, |∇u|) dx 6
ˆ
supp h
ϕ(x, |∇(u + h)|) dx
for every h ∈ W1,ϕ(Ω) with supp h b Ω. We say that u ∈ W1,ϕ(Ω) is a minimizer of the ϕ-energy with
boundary values f ∈ W1,ϕ(Ω), if u − f ∈ W1,ϕ0 (Ω), andˆ
Ω
ϕ(x, |∇u|) dx 6
ˆ
Ω
ϕ(x, |∇v|) dx
for every v ∈ f + W1,ϕ0 (Ω).
Let h ∈ W1,ϕ(Ω) have compact support in Ω, f ∈ W1,ϕ(Ω) and u ∈ f + W1,ϕ0 (Ω) is a minimizer of the
ϕ-energy. Then u + h ∈ f + W1,ϕ0 (Ω) by the definition. By the ϕ-energy minimizing property,ˆ
Ω
ϕ(x, |∇u|) dx 6
ˆ
Ω
ϕ(x, |∇(u + h)|) dx;
the integrals over the set Ω \ supp h cancel, and so u is a local minimizer. Hence every minimizer
u ∈ W1,ϕ(Ω) of the ϕ-energy with boundary values f is a local minimizer.
3. Auxiliary results
We denote by ϕ∗ the conjugate Φ-function of ϕ ∈ Φw(Ω), defined by
ϕ∗(x, t) := sup
s>0
(st − ϕ(x, s)).
From this definition, we have Young’s inequality st 6 ϕ(x, s) + ϕ∗(x, t). Hölder’s inequality holds in
generalized Orlicz spaces for ϕ ∈ Φw(Ω) with constant 2 [30, Lemma 3.2.13]:ˆ
Ω
|u| |v| dx 6 2‖u‖ϕ‖v‖ϕ∗ .
We next generalize the relation ϕ∗(ϕ(t)t ) 6 ϕ(t) which is well-known in the convex case, to weak Φ-
functions. The next results are written for ϕ ∈ Φw but can be applied to ϕ ∈ Φw(Ω) point-wise.
Lemma 3.1. Let ϕ ∈ Φw satisfy (aInc)1 with constant L. Then
ϕ∗
(ϕ(t)
Lt
)
6
ϕ(t)
L
.
Proof. When s 6 t we use sϕ(t)Lt − ϕ(s) 6 sϕ(t)Lt 6 ϕ(t)L to obtain
ϕ∗
(ϕ(t)
Lt
)
= sup
s>0
(
s
ϕ(t)
Lt
− ϕ(s)
)
6 max
{
ϕ(t)
L
, sup
s>t
(
s
ϕ(t)
Lt
− ϕ(s)
)}
.
On the other hand, by (aInc)1 we conclude that s
ϕ(t)
Lt 6 ϕ(s) when s > t, so the second term is non-
positive and the inequality is established. 
Mathematics in Engineering Volume 5, Issue 3, 1–21.
6If ϕ ∈ Φw is differentiable, then
d
dt
ϕ(t)
tp
=
ϕ′(t)tp − ptp−1ϕ(t)
t2p
=
ϕ(t)
tp+1
[ tϕ′(t)
ϕ(t)
− p
]
.
Thus ϕ satisfies (Inc)p if and only if
tϕ′(t)
ϕ(t) > p. Similarly, ϕ satisfies (Dec)q if and only if
tϕ′(t)
ϕ(t) 6 q. It
also follows that if ϕ satisfies (Inc)p and (Dec)q, then
1
q tϕ
′(t) 6 ϕ(t) 6 1p tϕ
′(t) (3.2)
and so ϕ′ satisfies (aInc)p−1 and (aDec)q−1. We next show that the last claim holds even if only (aInc)
or (aDec) is assumed of ϕ which is convex but not necessarily differentiable.
For ϕ ∈ Φc we denote the left and right derivative by ϕ′− and ϕ′+, respectively. We define the left
derivative to be zero at the origin, i.e., ϕ′−(0) := 0. Assume that ϕ satisfies (aInc)p with p > 1, and let
t0 > 0 be such that ϕ(t0) < ∞. Then
ϕ′+(0) = limt→0+
ϕ(t)
t
6 lim
t→0+
Lptp−1
ϕ(t0)
tp0
= 0.
Since ϕ′+ is right-continuous we also obtain that
lim
t→0+
ϕ′+(t) = ϕ
′
+(0) = 0.
Lemma 3.3. Let ϕ ∈ Φc satisfy (aInc)p and (aDec)q with constants Lp and Lq, respectively. Then
1
(Lqe − 1)qtϕ
′
+(t) 6 ϕ(t) 6
2 ln(2Lp)
p
tϕ′−(t)
for every t > 0, and ϕ′− and ϕ
′
+ satisfy (aInc)p−1 and (aDec)q−1, with constants depending only on
q
p , Lp
and Lq.
Proof. Since ϕ is convex we have
ϕ(t) =
ˆ t
0
ϕ′+(τ) dτ =
ˆ t
0
ϕ′−(τ) dτ,
for a proof see e.g., [51, Proposition 1.6.1, p. 37]. Let r ∈ [0, 1). Since the left derivative is increasing,
we obtain
ϕ(t) − ϕ(rt) =
ˆ t
rt
ϕ′−(τ) dτ 6 (t − rt)ϕ′−(t).
Thus
tϕ′−(t) >
ϕ(t) − ϕ(rt)
1 − r > ϕ(t)
1 − Lprp
1 − r
where in the second inequality we used (aInc)p of ϕ. Choosing r := (2Lp)−1/p we get
1 − Lprp
1 − r =
1/2
1 − (2Lp)−1/p =
p
2p(1 − (2Lp)−1/p) .
Mathematics in Engineering Volume 5, Issue 3, 1–21.
7Writing h := 1p and x := 2Lp, we find that
p
(
1 − (2Lp)−1/p
)
=
1 − x−h
h
6
dxs
ds
∣∣∣∣∣
s=0
= ln x,
where the inequality follows from convexity of s 7→ xs. Thus tϕ′−(t) > p2 ln(2Lp)ϕ(t).
Let R > 1. Since ϕ′+ is increasing, we obtain
ϕ(Rt) − ϕ(t) =
ˆ Rt
t
ϕ′+(τ) dτ > (Rt − t)ϕ′+(t).
Thus
tϕ′+(t) 6
ϕ(Rt) − ϕ(t)
R − 1 6 ϕ(t)
LqRq − 1
R − 1
where (aDec)q of ϕ was used in the second inequality. With R := 1 + 1q we get
tϕ′+(t) 6
Lq(1 + 1q )
q − 1
1/q
ϕ(t) 6 q(Lqe − 1)ϕ(t).
We have established the inequality of the claim.
We abbreviate cq := 1Lqe−1 and cp := 2 ln(2Lp). Since ϕ is convex, we have ϕ
′
− 6 ϕ
′
+ and so
cq
q
tϕ′−(t) 6
cq
q
tϕ′+(t) 6 ϕ(t) 6
cp
p
tϕ′−(t) 6
cp
p
tϕ′+(t)
Thus we obtain by (aDec)q of ϕ for 0 < s < t that
ϕ′+(t)
tq−1
6
q
cq
ϕ(t)
tq
6
qLq
cq
ϕ(s)
sq
6
q
p
Lqcp
cq
ϕ′+(s)
sq−1
and (aDec)q−1 of ϕ′+ follows. The proof for (aInc)p−1 is similar as are the proofs for ϕ
′
−. 
Before Lemma 3.3 we noted that limt→0+ ϕ′+(x, t) = 0, and hence
lim
|y|→0
ϕ′+(x, |y|)
|y| y · z =
(
lim
|y|→0+
ϕ′+(x, |y|)
)(
lim
|y|→0+
y
|y| · z
)
= 0
for z ∈ Rn. In light of this, we define
ϕ′+(x, |∇u|)
|∇u| ∇u · ∇h := 0 when ∇u = 0.
Theorem 3.4. Let ϕ ∈ Φc(Ω) satisfy (aInc)p and (aDec)q with 1 < p 6 q. Denote ϕ′h := ϕ′+ χ{∇u·∇h>0} +
ϕ′− χ{∇u·∇h<0}. If u ∈ W1,ϕloc (Ω), then the following are equivalent:
(i) u is a local minimizer;
(ii)
ˆ
supp h
ϕ′h(x, |∇u|)
|∇u| ∇u · ∇h dx > 0 for every h ∈ W
1,ϕ(Ω) with supp h b Ω.
Mathematics in Engineering Volume 5, Issue 3, 1–21.
8Proof. Let h ∈ W1,ϕ(Ω) with E := supp h b Ω be arbitrary. Define g : Ω × [0, 1] → [0,∞] by
g(x, ε) := |∇(u(x) + εh(x))|; in the rest of the proof we omit the first variable and abbreviate g(x, ε) by
g(ε).
Note that g(ε)2 = |∇u(x)|2 + ε2|h(x)|2 + 2ε∇u(x) · ∇h(x) and g > 0. Thus in [0, 1] the function g has
a local minimum at zero for x ∈ Ω with ∇u(x) · ∇h(x) > 0 and a maximum otherwise. This determines
whether we obtain the right- or left-derivative and so
lim
ε→0+
ϕ(x, g(ε)) − ϕ(x, g(0))
ε
= ϕ′h(x, g(0))g
′(0) =
ϕ′h(x, |∇u|)
|∇u| ∇u · ∇h (3.5)
for almost every x ∈ E.
Let us then find a majorant for the expression on the left-hand side of (3.5). By convexity,∣∣∣∣ϕ(x, g(ε)) − ϕ(x, g(0))
ε
∣∣∣∣ 6 ϕ′+(x,max{g(ε), g(0)}) |g(ε) − g(0)|ε
for a.e. x ∈ E. Since ε ∈ [0, 1] we have
max{g(ε), g(0)} 6 max{|∇u| + ε|∇h|, |∇u|} 6 |∇u| + |∇h|.
By the triangle inequality,∣∣∣∣g(ε) − g(0)
ε
∣∣∣∣ = ∣∣∣∣ |∇u + ε∇h| − |∇u|
ε
∣∣∣∣ 6 |∇h| 6 |∇u| + |∇h|.
Combining the estimates above, we find that∣∣∣∣ϕ(x, g(ε)) − ϕ(x, g(0))
ε
∣∣∣∣ 6 ϕ′+(x, |∇u| + |∇h|)(|∇u| + |∇h|).
By Lemma 3.3, ϕ′+(x, t)t . ϕ(x, t) for every t > 0, so that
ϕ′+(x, |∇u| + |∇h|)(∇u| + |∇h|) . ϕ(x, |∇u| + |∇h|).
By (aDec),
ϕ(x, |∇u| + |∇h|) 6 ϕ(x, 2|∇u|) + ϕ(x, 2|∇h|) 6 Lq2q(ϕ(x, |∇u|) + ϕ(x, |∇h|)) a.e.
Combining the estimates, we find that∣∣∣∣ϕ(x, g(ε)) − ϕ(x, g(0))
ε
∣∣∣∣ . ϕ(x, |∇u|) + ϕ(x, |∇h|) a.e.
The right hand side is integrable by [30, Lemma 3.1.3(b)], since |∇u|, |∇h| ∈ Lϕ(Ω) and ϕ satisfies
(aDec). Thus we have found a majorant. By dominated convergence and (3.5), we find that
ˆ
E
ϕ′h(x, |∇u|)
|∇u| ∇u · ∇h dx = limε→0+
ˆ
E
ϕ(x, g(ε)) − ϕ(x, g(0))
ε
dx. (3.6)
Let us first show that (i) implies (ii). By (i),
ˆ
E
ϕ(x, g(ε)) − ϕ(x, g(0))
ε
dx > 0
Mathematics in Engineering Volume 5, Issue 3, 1–21.
9for ε ∈ (0, 1], and hence (ii) follows by (3.6).
Let us then show that (ii) implies (i). For θ ∈ [0, 1] and s, t > 0 we have
g(θt + (1 − θ)s) = |θ∇u + θt∇h + (1 − θ)∇u + (1 − θ)s∇h|
6 |θ∇u + θt∇h| + |(1 − θ)∇u + (1 − θ)s∇h| = θg(t) + (1 − θ)g(s),
so g(ε) is convex. Since t 7→ ϕ(x, t) and g(ε) are convex for almost every x ∈ E, and t 7→ ϕ(x, t) is also
increasing, the composed function t 7→ ϕ(x, g(t)) is convex for a.e. x ∈ E. Thus
ˆ
E
ϕ(x, g(1)) − ϕ(x, g(0)) dx >
ˆ
E
ϕ(x, g(ε)) − ϕ(x, g(0))
ε
dx.
Since the above inequality holds for every ε ∈ (0, 1), (3.6) implies that
ˆ
E
ϕ(x, g(1)) − ϕ(x, g(0)) dx >
ˆ
E
ϕ′h(x, |∇u|)
|∇u| ∇u · ∇h dx > 0,
which is (i). 
We conclude the section by improving the Caccioppoli inequality from [8]; in this paper we only
need the special case ` = 1 and s = q, but we include the general formulation for possible future use.
We denote by η a cut-off function in BR, more precisely, η ∈ C∞0 (BR), χBσR 6 η 6 χBR and |∇η| 6 2(1−σ)R ,
where σ ∈ (0, 1). Note that the auxiliary function ψ is independent of x in the next lemma. Later on we
will choose ψ to be a regularized version of ϕ+B. Note also that the constant in the lemma is independent
of q1.
Lemma 3.7 (Caccioppoli inequality). Suppose ϕ ∈ Φc(Ω) satisfies (aInc)p and (aDec)q with constants
Lp and Lq, and let ψ ∈ Φw be differentiable and satisfy (A0), (Inc)p1 and (Dec)q1 , p1, q1 > 1. Let
β ∈ (0, 1] be the constant from (A0) of ψ. If u is a non-negative local minimizer and η is a cut-off
function in BR ⊂ Ω, thenˆ
BR
ϕ(x, |∇u|)ψ(u+R
βR )
−`ηs dx 6 K
ˆ
BR
ψ( u+R
βR )
−`ϕ
(
x,K u+R
βR
)
ηs−q dx
for any ` > 1p1 and s > q, where K :=
8sq(Lqe − 1)Lq ln(2Lp)
p(p1` − 1)(1 − σ) + Lp.
Proof. Let us simplify the notation by writing u˜ := u + R and v := u˜
βR . Since ∇u = ∇u˜, we see that u˜ is
still a local minimizer. By (A0) of ψ and v > 1
β
, we have 0 6 ψ(v)−` 6 1.
We would like to use Theorem 3.4 with h := ψ(v)−`ηsu˜. Let us first check that h is a valid test
function for a local minimizer, that is h ∈ W1,ϕ(BR) and has compact support in BR ⊂ Ω. As u˜ ∈ Lϕ(BR)
and |h| 6 u˜, it is immediate that h ∈ Lϕ(BR). By a direct calculation,
∇h = −`ψ(v)−`−1ηsu˜ψ′(v)∇v + sψ(v)−`ηs−1u˜∇η + ψ(v)−`ηs∇u˜.
Note that u˜∇v = v∇u˜. Since ψ is differentiable we may use (3.2) to get∣∣∣`ψ(v)−`−1ηsψ′(v)v∇u˜∣∣∣ 6 `ψ(v)−`−1q1ψ(v)|∇u˜| 6 q1`|∇u˜| ∈ Lϕ(BR).
Mathematics in Engineering Volume 5, Issue 3, 1–21.
10
For the third term in ∇h, we obtain |ψ(v)−`ηs∇u˜| 6 |∇u˜| ∈ Lϕ(BR). The term with ∇η is treated as h
itself. Thus h ∈ W1,ϕ(BR). Since s > 0 and η ∈ C∞0 (BR), h has compact support in BR ⊂ Ω and so it is a
valid test-function for a local minimizer.
We next calculate
∇u˜ · ∇h = −ψ(v)−`−1ηs[`ψ′(v)v − ψ(v)]|∇u˜|2 + sψ(v)−`ηs−1u˜∇u˜ · ∇η.
The inequality p1ψ(t) 6 ψ′(t)t from (3.2) implies that `ψ′(v)v − ψ(v) > (p1` − 1)ψ(v) > 0. Since u˜ is a
local minimizer, we can use the implication (i)⇒(ii) of Theorem 3.4 to conclude that
[p1` − 1]
ˆ
BR
ϕ′h(x, |∇u˜|)|∇u˜|ψ(v)−`ηs dx 6 s
ˆ
BR
ϕ′h(x, |∇u˜|)ψ(v)−`u˜ |∇η| ηs−1 dx.
Since ϕ′− 6 ϕ
′
h 6 ϕ
′
+, we obtain
1
q(Lqe−1)) tϕ
′
h(x, t) 6 ϕ(x, t) 6
2 ln(2Lp)
p tϕ
′
h(x, t) from Lemma 3.3. Using also
|∇η| u˜ 6 21−σv, we have
ˆ
BR
ϕ(x, |∇u˜|)ψ(v)−`ηs dx 6 4sq(Lqe − 1) ln(2Lp)
p(p1` − 1)(1 − σ)
ˆ
BR
ϕ(x, |∇u˜|)
|∇u˜| η
s−1ψ(v)−`v dx,
Note that the constant in front of the integral can be estimated from above by K2Lq .
Next we estimate the integrand on the right hand side. By Young’s inequality
ϕ(x, |∇u˜|)
|∇u˜| v 6 ϕ
(
x, ε−
1
q′ Lpv
)
+ ϕ∗
(
x, ε
1
q′ L−1p
ϕ(x,|∇u˜|)
|∇u˜|
)
,
where 1q +
1
q′ = 1. We choose ε :=
Lp
K η(x) ∈ (0, 1] and use (aInc)q′ of ϕ∗ [30, Proposition 2.4.9] (which
holds with constant Lq) and Lemma 3.1 to obtain
ϕ∗
(
x, ε
1
q′ L−1p
ϕ(x,|∇u˜|)
|∇u˜|
)
6 Lqεϕ∗
(
x, ϕ(x,|∇u˜|)Lp |∇u˜|
)
6 LqεLp ϕ(x, |∇u|) =
Lq
K η(x)ϕ(x, |∇u|).
In the other term we estimate ε−
1
q′ Lp 6 K
1− 1q′ L
1
q′
p ε
− 1q′ = η−
1
q′ K and use (aDec)q of ϕ:
ϕ
(
x, ε−
1
q′ Lpv
)
6 Lqη1−qϕ
(
x,Kv
)
.
With these estimates we obtain that
ˆ
BR
ϕ(x, |∇u˜|)ψ(v)−`ηs dx 6 1
2
ˆ
BR
ϕ(x, |∇u˜|)ψ(v)−`ηs dx + K
2
ˆ
BR
ψ(v)−`ϕ(x,Kv)ηs−q dx.
The first term on the right-hand side can be absorbed in the left-hand side. This gives the claim. 
The next observation is key to applications with truly non-doubling growth.
Remark 3.8. In the previous proof the assumption (aDec)q is only needed in the set ∇η , 0 since
we can improve the estimate on the right-hand side integral to |∇η| u˜ 6 21−σvχ{∇η,0} and only drop the
characteristic function in the final step.
Mathematics in Engineering Volume 5, Issue 3, 1–21.
11
4. Bloch-type estimate for bounded supersolutions
The following definition is like [8, Definition 3.1], except ϕ+Br has replaced ϕ
−
Br . Furthermore, we
are more precise with our estimates so as to avoid dependence on p and q.
Definition 4.1. Let ϕ ∈ Φw(Br) satisfy (aInc)p with p > 1 and constant Lp. We define ψBr : Br → [0,∞]
by setting
ψBr (t) :=
ˆ t
0
τp−1 sup
s∈(0,τ]
ϕ+Br (s)
sp
dτ for t > 0.
It is easy to see that ψBr ∈ Φw. Using that ϕ is increasing for the lower bound and (aInc)p for the
upper bound, we find that
ln(2)ϕ+Br
( t
2
)
=
ˆ t
t/2
τp−1
ϕ+Br (t/2)
τp
dτ 6 ψBr (t) 6
ˆ t
0
tp−1Lp
ϕ+Br (t)
tp
dτ = Lpϕ+Br (t). (4.2)
As in [8, Definition 3.1], we see that ψBr is convex and satisfies (Inc)p. If ϕ satisfies (A0), so does ψBr ,
since ϕ+Br ' ψBr . If ϕ satisfies (aDec)q, then ψBr is strictly increasing and satisfies (aDec)q, and, as a
convex function, also (Dec) [30, Lemma 2.2.6].
We note in both the above reasoning and in the next theorem that constants have no direct
dependence on p or q, only on Lp, Lq and
q
p .
Theorem 4.3 (Bloch-type estimate). Let ϕ ∈ Φc(Ω) satisfy (A0) and (A1). Let B2r ⊂ Ω with r 6 1 and
ϕ|B2r satisfy (aInc)p and (aDec)q with p, q ∈ [n,∞). If u is a non-negative local minimizer, then
ˆ
Br
|∇ log(u + r)|n dx 6 C,
where C depends only on n, Lp, Lq,
q
p , the constants from (A0) and (A1), and %ϕ(|∇u|).
Proof. Let us first note that ϕ satisfies (aInc)n with the constant Lp. Let β be the smaller of the
constants from (A0) and (A1). Denote v := u+2r2βr and γ :=
2K
β
, where K is from Caccioppoli inequality
(Lemma 3.7) with ` = 1, s = q and σ = 12 . Since p > n, we see that
K 6
16q2(Lqe − 1)Lq ln(2Lp)
p(p − 1) + Lp 6 16(Lqe − 1)Lq ln(2Lp)
(q
p
)2 n
n − 1 + Lp.
When |∇u| > γv, we use (aInc)n to deduce that
ϕ−B2r (γv)
vn
6 γn
ϕ(x, γv)
(γv)n
6 γnLp
ϕ(x, |∇u|)
|∇u|n
for a.e. x ∈ Br. Rearranging gives |∇u|nvn . ϕ(x,|∇u|)ϕ−B2r (γv) . Since v >
1
β
and γ > 1, we obtain by (A0) that
ϕ−B2r (γv) > 1. If also ϕ
−
B2r (γv) 6
1
|B2r | , then ϕ
+
B2r (βγv) 6 ϕ
−
B2r (γv) by (A1). Otherwise, (ϕ
−
B2r (γv))
−1 6 |B2r|.
In either case,
|∇u|n
vn
.
ϕ(x, |∇u|)
ϕ−B2r (γv)
. ϕ(x, |∇u|)
( 1
ϕ+B2r (βγv)
+ |B2r|
)
Mathematics in Engineering Volume 5, Issue 3, 1–21.
12
for a.e. x ∈ Br. When |∇u| 6 γv, we use the estimate |∇u|nvn 6 γn instead. Since u + r > 12 (u + 2r) = βrv,
we obtain thatˆ
Br
|∇ log(u + r)|n dx =
ˆ
Br
|∇u|n
(u + r)n
dx 6
1
(βr)n
ˆ
Br
|∇u|n
vn
dx .
 
Br
ϕ(x, |∇u|)
ϕ+B2r (βγv)
+ |B2r|ϕ(x, |∇u|) + 1 dx
=
 
Br
ϕ(x, |∇u|)
ϕ+B2r (βγv)
dx + 2n%ϕ(|∇u|) + 1.
It remains to bound the integral on the right-hand side.
Let ψB2r be as in Definition 4.1, let η ∈ C∞0 (B2r) be a cut-off function such that η = 1 in Br and
choose ψ(t) := ψB2r (βγt). Then 
Br
ϕ(x, |∇u|)
ϕ+B2r (βγv)
dx .
 
B2r
ϕ(x, |∇u|)
ϕ+B2r (βγv)
ηq dx 6 Lp
 
B2r
ϕ(x, |∇u|)
ψ(v)
ηq dx,
where the second inequality follows from (4.2). We note that ψ satisfies (A0), (Inc)p and (Dec). Now
we use the Caccioppoli inequality (Lemma 3.7) for ϕ and ψ with ` = 1, s = q and σ = 12 to get 
B2r
ϕ(x, |∇u|)
ψ(v)
ηq dx 6 K
 
B2r
ϕ(x,Kv)
ψ(v)
dx 6 ln(2)K;
the last inequality holds by (4.2) and γ = 2K
β
since
ϕ(x,Kv)
ψ(v)
=
ϕ(x,Kv)
ψB2r (βγv)
6 ln(2)
ϕ(x,Kv)
ϕ+B2r (
1
2βγv)
6 ln(2). 
We next show that the Bloch estimate implies a Harnack inequality for suitable monotone functions.
We say that a continuous function u is monotone in the sense of Lebesgue, if it attains its extrema on
the boundary of any compact set in its domain of definition. We say that ϕ ∈ Φw(Ω) is positive if
ϕ(x, t) > 0 for every t > 0 and a.e. x ∈ Ω. If ϕ satisfies (aDec)q(·) for q < ∞ a.e., then it is positive.
Lemma 4.4. If ϕ ∈ Φw(Ω) is positive, then every continuous local minimizer is monotone in the sense
of Lebesgue.
Proof. Let u ∈ W1,ϕloc (Ω) ∩ C(Ω) be a local minimizer and D b Ω. Fix M > max∂D u and note that
(u − M)+ is zero in some neighborhood of ∂D since u is continuous. Thus h := (u − M)+χD belongs to
W1,ϕ(Ω) ∩C(Ω) and has compact support in Ω. Using that u is a local minimizer, we obtain that
ˆ
supp h
ϕ(x, |∇u|) dx 6
ˆ
supp h
ϕ(x, |∇(u − h)|) dx = 0.
Since ϕ(x, t) > 0 for every t > 0 and a.e. x ∈ Ω, it follows that ∇u = 0 a.e. in supp h. Thus ∇h = 0 a.e.
in Ω. Since h is continuous and equals 0 in Ω \ D, we conclude that h ≡ 0. Hence u 6 M in D. Letting
M → max∂D u, we find that u 6 max∂D u. The proof that min∂D u 6 u in D is similar. 
For x ∈ Ω we write rx := 12 dist(x, ∂Ω). Let 1 6 p < ∞. In [43, Definition 3.6] a function u : Ω→ R
is called a Bloch function if
sup
x∈Ω
rpx
 
Brx
|∇u|p dx < ∞.
Mathematics in Engineering Volume 5, Issue 3, 1–21.
13
Note that if u is an analytic function in the plane and p = 2, then by the mean value property
sup
x∈Ω
rx
( 
Brx
|u′|2 dx
) 1
2 ≈ sup
x∈Ω
d(x, ∂Ω) |u′(x)| ≈ sup
x∈Ω
(1 − |x|2)|u′(x)|,
which connects this with Bloch functions in complex analysis. In the next theorem we assume for log u
a Bloch-type condition. In the case p = n the next result was stated in [35, Lemma 6.3].
Lemma 4.5. Let u : Ω → (0,∞) be continuous and monotone in the sense of Lebesgue. If B4r b Ω,
and
rp
 
B2r
|∇ log u|p dx 6 A
for p > n − 1, then
sup
Br
u 6 C inf
Br
u
for some C depending only on A, p and n.
Proof. Denote v := log u. Since the logarithm is increasing, v is monotone in the sense of Lebesgue
because u is.
As u is continuous and positive in B3r ⊂ Ω, it is bounded away from 0. Thus v ∈ W1,p(B2r) is
uniformly continuous in B3r. Mollification gives a sequence (vi)∞i=0 of functions in C
∞(B2r)∩W1,p(B2r),
such that v is the limit of vi in W1,p(B2r) and vi → v pointwise uniformly in B2r, as i→ ∞ [25, Theorem
4.1 (ii), p. 146]. By the Sobolev–Poincaré embedding W1,p(∂BR)→ C0,1− n−1p (∂BR),(
osc
∂BR
vi
)p
. Rp−n+1
ˆ
∂BR
|∇vi|p dS
for every R ∈ (0, 2r), where dS denotes the (n − 1)-dimensional Hausdorff measure and the constant
depends only on p and n (see, e.g., [26, Lemma 1], stated for the case n = p = 3). Integrating with
respect to R gives
ˆ 2r
r
(
osc
∂BR
vi
)p
dR .
ˆ 2r
r
Rp−n+1
ˆ
∂BR
|∇vi|p dS dR . rp−n+1
ˆ
B2r
|∇vi|p dx.
Since vi → v uniformly, we obtain that (osc∂BR v)p = limi→∞(osc∂BR vi)p for every R. Using this and
vi → v in W1,p(B2r), it follows by Fatou’s Lemma that
ˆ 2r
r
(
osc
∂BR
v
)p
dR 6 lim inf
i→∞
ˆ 2r
r
(
osc
∂BR
vi
)p
dR
6 lim inf
i→∞ Cr
p−n+1
ˆ
B2r
|∇vi|p dx = Crp−n+1
ˆ
B2r
|∇v|p dx.
As v is continuous and monotone in the sense of Lebesgue, we have that oscBr v 6 oscBR v = osc∂BR v
for R ∈ (r, 2r), and therefore
r
(
osc
Br
v
)p
6
ˆ 2r
r
(
osc
∂BR
v
)p
dR 6 Crp−n+1
ˆ
B2r
|∇v|p dx 6 CrA.
Mathematics in Engineering Volume 5, Issue 3, 1–21.
14
Since
osc
Br
v = sup
x,y∈Br
|v(x) − v(y)| = sup
x,y∈Br
∣∣∣∣∣ log u(x)u(y)
∣∣∣∣∣ = log supBr uinfBr u ,
it now follows that
sup
Br
u 6 exp
(
(CA)1/p
)
inf
Br
u. 
We conclude this section with the Harnack inequality for local minimizers. The novelty of the next
theorem, apart from the technique, is that the constant depends only on qp , not on p and q separately.
Theorem 4.6 (Harnack inequality). Let ϕ ∈ Φc(Ω) satisfy (A0) and (A1). We assume that B2r ⊂ Ω
with r 6 1 and ϕ|B2r satisfies (aInc)p and (aDec)q with p, q ∈ (n,∞).
Then any non-negative local minimizer u ∈ W1,ϕloc (Ω) satisfies the Harnack inequality
sup
Br
(u + r) 6 C inf
Br
(u + r)
when B4r ⊂ Ω. The constant C depends only on n, β, Lp, Lq, qp and %ϕ(|∇u|).
Proof. Let u ∈ W1,ϕloc (Ω) be a non-negative local minimizer. By Theorem 4.3,ˆ
Br
|∇ log(u + r)|n dx 6 C,
where C depends only on n, Lp, Lq,
q
p , the constants from (A0) and (A1), and %ϕ(|∇u|). Since p > n, ϕ
satisfies (aInc)p and u ∈ W1,ϕloc (Ω), u is continuous and Lemma 4.4 yields that u + r is monotone in the
sense of Lebesgue. Thus we can apply Lemma 4.5 to u + r, which gives the Harnack inequality. 
5. Minimizers with non-doubling growth
Let us study minimizers with given boundary values.
Definition 5.1. Let p ∈ [1,∞), ϕ ∈ Φc(Ω) and define, for λ > 1,
ϕλ(x, t) :=
ˆ t
0
p
λ
τp−1 + min{ϕ′−(x, τ), pλτp−1} dτ = 1λ tp +
ˆ t
0
min{ϕ′−(x, τ), pλτp−1} dτ.
Note that since t 7→ ϕλ(x, t) is convex for a.e. x ∈ Ω, the left derivative ϕ′− exists for a.e. x ∈ Ω, and
therefore the above definition makes sense.
Lemma 5.2. If ϕ ∈ Φc(Ω), then ϕλ ∈ Φc(Ω) satisfies ϕλ(·, t) ≈ tp with constants depending on λ.
Furthermore,
min{ϕ(x, t2 ), λ( t2 )p} + 1λ tp 6 ϕλ(x, t) 6 ϕ(x, t) + 1λ tp
for λ > 1, ϕλ(x, t) 6 ϕΛ(x, t) + 1λ t
p for any Λ > λ > 1, and ϕλ → ϕ as λ→ ∞.
Proof. It follows from the definition that p
λ
τp−1 6 ϕ′λ(x, τ) 6 p(
1
λ
+ λ)τp−1. Integrating over τ ∈ [0, t]
gives ϕλ(·, t) ≈ tp. Let Λ > λ > 1. Since the minimum in the integrand is increasing in λ, we see that
ϕλ(x, t) 6 ϕΛ(x, t) +
( 1
λ
− 1
Λ
)
tp 6 ϕ(x, t) + 1
λ
tp,
Mathematics in Engineering Volume 5, Issue 3, 1–21.
15
and thus lim supλ→∞ ϕλ 6 ϕ. On the other hand, Fatou’s Lemma gives
ϕ(x, t) =
ˆ t
0
lim
λ→∞(
p
λ
τp−1 + min{ϕ′−(x, τ), pλτp−1}) dτ
6 lim inf
λ→∞
ˆ t
0
p
λ
τp−1 + min{ϕ′−(x, τ), pλτp−1} dτ = lim inf
λ→∞ ϕλ(x, t).
One of the terms in the minimum min{ϕ′−(x, τ), pλτp−1} is achieved in at least a set of measure t2 . Since
both terms are increasing in τ, this implies that
ϕλ(x, t) > min
{ ˆ t/2
0
ϕ′−(x, τ) dτ,
ˆ t/2
0
pλτp−1 dτ
}
+ 1
λ
tp = min{ϕ(x, t2 ), λ( t2 )p} + 1λ tp. 
Since ϕλ(x, t) ≈ tp, it follows by [30, Proposition 3.2.4] that W1,ϕλ(Ω) = W1,p(Ω) and the norms
‖ · ‖ϕλ and ‖ · ‖p are comparable. However, the embedding constant blows up as λ → ∞ unless ϕ also
satisfies (aDec)p. This approximation approach is similar to that in [24]. Note in the next results that f
is bounded by the Sobolev embedding in W1,p(Ω).
Lemma 5.3. Let q : Ω → (n,∞), and let ϕ ∈ Φc(Ω) satisfy (A0), (aInc)p and (aDec)q(·), p > n.
Assume that f ∈ W1,ϕ(Ω) with %ϕ(∇ f ) < ∞. Then there exists a sequence (uλk) of Dirichlet ϕλk-energy
minimizers with the boundary value function f and a minimizer of the ϕ-energy u∞ ∈ f + W1,ϕ0 (Ω) such
that uλk → u∞ uniformly in Ω as λk → ∞.
Proof. Note that we use W1,ϕλ(Ω) = W1,p(Ω) and W1,ϕλ0 (Ω) = W
1,p
0 (Ω) several times in this proof. Let
λ > p. Note that f ∈ W1,p(Ω) since tp . ϕ(x, t) + 1 by (A0) and (aInc)p. By [29, Theorem 6.2] there
exists a minimizer uλ ∈ f + W1,p0 (Ω) of ˆ
Ω
ϕλ(x, |∇u|) dx.
Fix λ > 1. By Lemma 5.2 and tp . ϕ(x, t)+1, we have tp . min{ϕ(x, t2 ), λtp}+1 . ϕλ(x, t)+1. Also
by the same lemma, ϕλ . ϕ+ 1λ t
p . ϕ+ 1. Since f is a valid test-function and uλ is a ϕλ-minimizer, we
have ˆ
Ω
|∇uλ|p dx .
ˆ
Ω
ϕλ(x, |∇uλ|) + 1 dx 6
ˆ
Ω
ϕλ(x, |∇ f |) + 1 dx .
ˆ
Ω
ϕ(x, |∇ f |) + 1 dx < ∞,
and hence %p(∇uλ) is uniformly bounded. Note that the implicit constants do not depend on λ.
Since uλ − f ∈ W1,p0 (Ω), the Poincaré inequality implies that
‖uλ − f ‖p . ‖∇(uλ − f )‖p . ‖∇uλ‖p + ‖∇ f ‖p 6 c.
Therefore, ‖uλ‖p 6 ‖uλ − f ‖p + ‖ f ‖p 6 c and so ‖uλ‖1,p is uniformly bounded. Since f + W1,p0 (Ω) is a
closed subspace of W1,p(Ω), it is a reflexive Banach space. Thus there exists a sequence (λk)∞k=1 tending
to infinity and a function u∞ ∈ f + W1,p0 (Ω) such that uλk ⇀ u∞ in W1,p(Ω). Since p > n, the weak
convergence uλk − f ⇀ u∞− f in W1,p0 (Ω) and compactness of the Sobolev embedding [1, Theorem 6.3
(Part IV), p. 168] imply that uλk − f → u∞ − f in the supremum norm. Hence uλk → u uniformly in Ω.
Mathematics in Engineering Volume 5, Issue 3, 1–21.
16
We note that the modular %ϕλ satisfies the conditions of [23, Definition 2.1.1]. Hence, it is weakly
lower semicontinuous by [23, Theorem 2.2.8], and we obtain that
ˆ
Ω
ϕλ(x, |∇u∞|) dx 6 lim inf
k→∞
ˆ
Ω
ϕλ(x, |∇uλk |) dx 6 lim infk→∞
ˆ
Ω
(1 + C
λk
)ϕλk(x, |∇uλk |) + Cλk dx
6 lim inf
k→∞
ˆ
Ω
ϕλk(x, |∇uλk |) dx .
ˆ
Ω
ϕ(x, |∇ f |) + 1 dx
(5.4)
for fixed λ > 1, where in the second inequality we used Lemma 5.2 and the fact that tp . ϕλ(x, t) + 1.
It follows by monotone convergence that
ˆ
Ω
ϕ(x, |∇u∞|) dx = lim
λ→∞
ˆ
Ω
min{ϕ(x, |∇u∞|), λ|∇u∞|p} dx 6 lim sup
λ→∞
ˆ
Ω
ϕλ(x, |∇u∞|) dx,
and hence |∇u∞| ∈ Lϕ(Ω). Since p > n, Ω is bounded and u∞− f ∈ W1,p0 (Ω), we obtain by [58, Theorem
2.4.1, p. 56] that u∞ − f ∈ L∞(Ω). Moreover, L∞(Ω) ⊂ Lϕ(Ω) since Ω is bounded and ϕ satisfies (A0).
These and f ∈ Lϕ(Ω) yield that u∞ ∈ Lϕ(Ω). Hence we have u∞ ∈ W1,ϕ(Ω). Since u∞ − f ∈ W1,p0 (Ω)
and p > n, it follows that u∞ − f can be continuously extended by 0 in Ωc [2, Theorem 9.1.3]. Then
we conclude as in [40, Lemma 1.26] that u∞ − f ∈ W1,ϕ0 (Ω).
We conclude by showing that u∞ is a minimizer. Suppose to the contrary that there exists u ∈
f + W1,ϕ0 (Ω) with ˆ
Ω
ϕ(x, |∇u∞|) dx −
ˆ
Ω
ϕ(x, |∇u|) dx =: ε > 0.
By ϕλ . ϕ + 1, ϕλ → ϕ and dominated convergence, there exists λ0 such that
ˆ
Ω
ϕλ(x, |∇u∞|) dx −
ˆ
Ω
ϕλ(x, |∇u|) dx > ε2
for all λ > λ0. From the lower-semicontinuity estimate (5.4) we obtain k0 such that
ˆ
Ω
ϕλ(x, |∇u∞|) dx 6
ˆ
Ω
ϕλk(x, |∇uλk |) dx + ε4
for all k > k0. By increasing k0 if necessary, we may assume that λk > λ0 when k > k0. For such k we
choose λ = λk above and obtain that
ˆ
Ω
ϕλk(x, |∇u|) + ε2 6
ˆ
Ω
ϕλk(x, |∇uλk |) dx + ε4 .
This contradicts uλk being a ϕλk-minimizer, since u ∈ f + W1,ϕ0 (Ω) ⊂ f + W
1,ϕλk
0 (Ω). Hence the counter-
assumption was incorrect, and the minimization property of u∞ is proved. 
We conclude this paper with the Harnack inequality for ϕ-harmonic functions. Here we use
Remark 3.8 to handle the possibility that q could be unbounded and thus ϕ non-doubling, like in
Example 1.1. This is possible since q◦ is the supremum of q only in the annulus, not the whole ball.
Mathematics in Engineering Volume 5, Issue 3, 1–21.
17
Theorem 5.5 (Harnack inequality). Let p, q : Ω→ (n,∞), and ϕ ∈ Φc(Ω) be strictly convex and satisfy
(A0), (A1), (aInc)p(·) and (aDec)q(·) with inf p > n. Assume that f ∈ W1,ϕ(Ω) with %ϕ(|∇ f |) < ∞. Then
there exists a unique minimizer u of the ϕ-energy with boundary values f . Let B4r ⊂ Ω, p− := inf
B2r
p
and q◦ := sup
B2r\Br
q. If q
◦
p− < ∞, then the Harnack inequality
sup
Br
(u + r) 6 C inf
Br
(u + r)
holds for all non-negative minimizers with C depending only on n, β, Lp, Lq,
q◦
p− and %ϕ(|∇ f |).
Proof. By Lemma 5.3, there exists a sequence (uk) ⊂ f + W1,p0 (Ω) of minimizers of the ϕλk energy
which converge uniformly to a minimizer u∞ ∈ f + W1,ϕ0 (Ω) of the ϕ-energy. Since ϕ is strictly convex,
the minimizer is unique and so u = u∞.
From (A0) and (aInc)p− we conclude that tp
−
. ϕ(x, t) + 1. It follows from Lemma 5.2 that ϕλ(·, t) '
ϕ(·, t) + 1
λ
tp
−
. Thus ϕλ satisfies (A0) and (A1) with the same constants as ϕ. Since uk → u in L∞(Ω) and
u is non-negative we can choose a sequence εk → 0+ such that uk +εk is non-negative. By Theorem 4.3
with Remark 3.8, ˆ
Br
|∇ log(uk + εk + r)|n dx 6 C,
where C depends only on n, Lp, Lq,
q◦
p− , β from (A0) and (A1), and %ϕλk (|∇uk|). Since uk is a minimizer,
%ϕλk (|∇uk|) 6 %ϕλk (|∇ f |) . %ϕ(|∇ f |) + 1. Thus by Lemma 4.5, we have
sup
Br
(uk + εk + r) 6 C inf
Br
(uk + εk + r),
with C independent of k. Since uk + εk → u∞ uniformly, the claim follows. 
Acknowledgments
Peter Hästö was supported in part by the Jenny and Antti Wihuri Foundation.
Conflict of interest
The authors declare no conflict of interest.
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