Equivalence and self-improvement of p-fatness and Hardy's inequality, and association with uniform perfectness

0.18 MB·English

by  Riikka Korte

EQUIVALENCE AND SELF–IMPROVEMENT OF
p–FATNESS AND HARDY’S INEQUALITY, AND
ASSOCIATION WITH UNIFORM PERFECTNESS
RIIKKAKORTEANDNAGESWARISHANMUGALINGAM
8
Abstract. Wepresentaneasyproofthatp–Hardy’sinequalityimplies
0
uniformp–fatnessoftheboundarywhenp=n. Theproofworksalsoin
0
metric space setting and demonstrates the self–improving phenomenon
2
ofthep–fatness. Wealsoexploretherelationshipbetweenp–fatness,p–
n Hardyinequality,andtheuniformperfectnessforallp≥1,anddemon-
a
strate that in the Ahlfors Q–regular metric measure space setting with
J
p = Q, these three properties are equivalent. When p 6= 2, our results
4
are new even in the Euclidean setting.
1

]
A
F 1. Introduction
.
h
t The purpose of this paper is to study the relation between p–Hardy’s in-
a
m equality
|u(x)|p
[ dx ≤ C |∇u(x)|pdx
dist(x,∂Ω)p
2 ZΩ ZΩ
for all u ∈ C∞(Ω) and C independent of u, the uniform perfectness of ∂Ω,
v 0
3 and the uniform p–fatness of X \ Ω in the metric space setting. By p–
1 fatness we mean a capacitary version of the measure thickness condition.
0
Rather surprisingly, these analytic, metric and geometric conditions turn
2
. out to be equivalent in certain situations. We also consider self–improving
9
phenomena related to these conditions. Our results are new even in the
0
7 Euclidean setting, when p 6= 2.
0
The fact that when p = n, a domain satisfies p–Hardy’s inequality if and
:
v only if the complement is uniformly p–fat, was first proved by Ancona [1]
i
X in R2. Later, these results were generalized for all n = p > 1 by Lewis [14].
r Sugawa proved in [19] that for n = p = 2 these conditions are equivalent to
a
the uniform perfectness of the complement in the Euclidean plane. See also
Buckley–Koskela [3] for studies relevant to Orlicz–Sobolev spaces.
In metric spaces, for all p > 1, it has been shown that uniform p–fatness of
the complement of a domain implies that the domain supports p–Hardy’s
inequality under some conditions, see [2]. See also [9] for similar results
involving a measure thickness condition. In [13], the equivalence of the p–
fatness and a pointwise Hardy’s inequality has been studied. In this paper,
we prove that if a metric space is Ahlfors Q–regular and satisfies a weak
2000 Mathematics Subject Classification. 31B15, 46E35.
The first author is supported by the Finnish Academy of Science and Letters, Vilho,
Yrj¨o and Kalle V¨ais¨al¨a Foundation, and the second author is partially supported by the
NSFgrant DMS-0355027.
1
(1,Q)–Poincar´e inequality, then the support of a Q–Hardy inequality on a
domain implies uniform Q–fatness of the complement of the domain. Our
proof is rather transparent and it is based on estimating the Hausdorff–
content of the boundary.
We will also prove a self–improvement property for both uniform Q–fatness
andQ–Hardy’sinequalityinthesettingofAhlforsQ–regularmetricmeasure
spaces. That is, if a set satisfies Q–Hardy’s inequality or is uniformly Q–
fat, then there exists q < Q such that the set satisfies q–Hardy’s inequality
or is uniformly q–fat, respectively. The self–improving property of Hardy’s
inequality has been studied in [12] and that of uniform p–fatness in [2] and
in [16]. Our approach gives a more elementary proof of self–improvement of
uniform p–fatness when p = Q.
2. Preliminaries
We assume that X = (X,d,µ) is a metric measure space equipped with a
metric d and a Borel regular outer measure µ such that 0 < µ(B) < ∞ for
all balls B = B(x,r) = {y ∈ X : d(x,y) < r}. The measure µ is said to
be doubling if there exists a constant c ≥ 1, called the doubling constant,
D
such that
µ(B(x,2r)) ≤ c µ(B(x,r))
D
for all x∈ X and r > 0. The measure is Q–regular if there exists a constant
c ≥1 such that
A
1
rQ ≤µ(B(x,r)) ≤ c rQ
A
c
A
for all x ∈ X and 0 < r < diam(X). The n–dimensional Lebesgue measure
on Rn is n–regular. The Hausdorff s–content of E ⊂ X is
Hs (E) = inf rs, (2.1)
∞ i
i∈I
X
where the infimum is taken over all countable covers {B(x ,r )} of E,
i i i∈I
with each B(x ,r )∩E non-empty. In addition, we may assume that x ∈E
i i i
for every i ∈ I, because that may increase the Hausdorff content at most by
a multiplicative factor 2s.
A non-negative Borel measurable function g on X is said to be a p–weak
u
upper gradient of a function u on X if there is a non-negative Borel measur-
able function ρ∈ Lp(X) such that for all rectifiable curves γ in X, denoting
the end points of γ by x and y, we have either
|u(x)−u(y)| ≤ g ds,
u
Zγ
or ρds = ∞. Let1 ≤ p < ∞. Ifu is afunction that is integrable to power
γ
p in X, let
R
1
p
kuk = |u|pdµ+inf gpdµ ,
N1,p(X) u
(cid:18)ZX gu ZX (cid:19)
where the infimum is taken over all p–weak upper gradients of u. The
Newtonian space on X is the quotient space
N1,p(X) = {u : kuk < ∞}/ ∼,
N1,p(X)
2
1,p
where u ∼ v if and only if ku−vk = 0, see [17]. We define N (Ω)
N1,p(X) 0
to be the set of functions u ∈ N1,p(Ω) that can be extended to N1,p(X) so
that the extensions are zero on X \Ω p–quasieverywhere.
We say that X supports a weak (1,p)–Poincar´e inequality if there exist
constants c > 0 and τ ≥ 1 such that for all balls B(x,r) of X, all locally
p
integrable functions u on X and for all p–weak upper gradients g of u, we
u
have
1
p
|u−u |dµ ≤ c r gpdµ , (2.2)
B(x,r) p u
ZB(x,r) ZB(x,τr) !
where
1
u = udµ = udµ.
B
µ(B)
ZB ZB
We point out here that if X is the Euclidean space Rn equipped with the
n–dimensionalLebesguemeasureandtheEuclideanmetric, thenN1,p(X) =
W1,p(Rn), theclassical Sobolevspace. Moreover, Rn supportsaweak (1,1)–
Poincar´e inequality.
Definition 2.3. Let Ω be an open set in X and E be a closed subset of Ω.
The p–capacity of E with respect to Ω is
cap (E,Ω) = inf gpdµ,
p u
ZX
wheretheinfimumis taken over allfunctionsuwithp–weak uppergradients
g such that u = 1 and u = 0. Should there be no such function u,
u |E |X\Ω
then cap (E,Ω) = ∞.
p
A metric space X is said to be linearly locally connected (LLC) if there is
a constant C ≥ 1 so that for each x ∈ X and r > 0, the following two
conditions hold:
(1) any pair of points in B(x,r) can be joined in B(x,Cr),
(2) any pair of points in X \B(x,r) can be joined in X \B(x,r/C).
By joining we mean joining by a path. Note that if a complete Q–regular
space, Q > 1, supports a weak (1,Q)–Poincar´e inequality, then it satisfies
the LLC–condition, see for example [6] or [11].
Definition 2.4. We say that a set E ⊂ X is uniformly perfect if E is not
a singleton set, and there is a constant c ≥ 1 so that for each x ∈ E
UP
and r > 0 the set E ∩B(x,c r)\B(x,r) is nonempty whenever the set
UP
E\B(x,c r) is nonempty.
UP
Formoreinformationaboutuniformperfectness,seeforexample[7]and[19].
A set E ⊂ X is said to be uniformly p–fat if there exists a constant c > 0
0
so that for every point x∈ E and for all 0 < r <∞,
cap (B(x,r)∩E,B(x,2r))
p
≥ c . (2.5)
0
cap (B(x,r),B(x,2r))
p
This condition is stronger than the Wiener criterion. Uniform p–fatness
is a capacitary version of the uniform measure thickness condition, see for
example [9].
3
Definition 2.6. Let 1 < p < ∞. The set Ω ⊂ X satisfies p–Hardy’s
1,p
inequality if there exists 0< c < ∞ such that for all u∈ N (Ω),
H 0
p
|u(x)|
dµ(x) ≤ c g (x)pdµ(x). (2.7)
H u
dist(x,X \Ω)
ZΩ(cid:18) (cid:19) ZΩ
Here g is a p–weak upper gradient of u. Here we use dist(x,X\Ω) instead
u
of dist(x,∂Ω) since in general the latter quantity can be larger than the
former one.
Hardy’s inequality has been studied for example in [4], [5], [14], and [20].
Hardy’sinequality hasbeenusedalsoto characterize Sobolevfunctionswith
zero boundary values, see [10] and [14].
3. Main results
In this section, we show that Q–Hardy’s inequality on Ω implies uniform
Q–fatness of the complement. Our method also shows that Q–fatness is
a self–improving property. To simplify notation, we will assume X to be
unbounded throughout this section. However, for our arguments, it is im-
material what happens outside Ω, and therefore our arguments work also if
X is bounded,provided we adjust the conditions of uniform perfectness and
uniform fatness to the bounded setting. Notice also that if Ω is a domain,
then X \Ω can be replaced by ∂Ω in our arguments.
Theorem 3.1. Let (X,d,µ) be a complete Q–regular metric measure space
supporting a weak (1,Q)–Poincar´e inequality, and Ω ⊂ X be an open subset.
If Ω satisfies Q–Hardy’s inequality, then X \Ω is uniformly (Q−ε)–fat for
some ε > 0.
We splittheproofinto two parts. FirstinLemma3.2, weshowthatHardy’s
inequality implies uniform perfectness of the complement. Then in Theo-
rem3.6, weshow thatuniformperfectnessimplies (Q−ε)–fatness withsome
ε > 0. Recall that we assume X to be unbounded.
Lemma 3.2. Let X be as in Theorem 3.1. If Ω ⊂ X satisfies Q–Hardy’s
inequality, then X \Ω is uniformly perfect and unbounded.
Proof. Fix m > 4 and suppose that Ω satisfies Hardy’s inequality (2.7) and
that X \Ω is not uniformly perfect with respect to the constant m or that
X \Ω is bounded. In both cases, there exists x ∈ X \Ω and r > 0 such
0 0
that B(x ,mr )\B(x ,r ) ⊂ Ω. We will deduce an upper bound for such
0 0 0 0
m independent of x and r , and hence conclude that X \Ω is uniformly
0 0
perfectfor any constant larger than this upperboundandthat X\Ω cannot
be bounded.
Define u :X → [0,∞) so that
d(x0,x) −1 , d(x ,x) ≤ 2r ,
r0 + 0 0
u(x) = 1(cid:16), (cid:17) 2r < d(x ,x) < mr0,
 0 0 2

 2− 2d(x0,x) , mr0 ≤ d(x ,x).
mr0 + 2 0
(cid:16) (cid:17) 4



Now the minimal upper gradient of u satisfies
Q Q
1 2
gQdµ ≤ µ(B(x ,2r ))+ µ(B(x ,mr )) ≤ c 2Q+1.
u r 0 0 mr 0 0 A
ZΩ (cid:18) 0(cid:19) (cid:18) 0(cid:19)
(3.3)
Next, we show that
u(x)Q
dµ(x) ≥ clog(m/4), (3.4)
dist(x,X \Ω)Q
ZΩ
where c > 0 is a constant that depends only on c and Q. For x ∈ X and
A
0 < r < R, we denote the annulus A(x,r,R) = B(x,R)\B(x,r). Let n ∈ N
be the unique number such that 2n ≤ m < 2n+1. Since m > 4, we have
n ≥ 2. Then
n−1
A(x ,2r ,mr /2) ⊃ A(x ,2kr ,2k+1r ).
0 0 0 0 0 0
k=1
[
As X is quasiconvex (which follows from the Poincar´e inequality, see for
example [11]) and hence path-connected, and as X \B(x ,2k+1r ) is non-
0 0
3
empty, there is a point y ∈ A(x ,2kr ,2k+1r ) such that d(x ,y ) = 2k;
k 0 0 0 0 k
2
hence the ball B(y ,2k−1r ) ⊂ A(x ,2kr ,2k+1r ). Thus
k 0 0 0 0
u(x)Q 1
dµ ≥ dµ
dist(x,X \Ω)Q d(x ,x)Q
ZΩ ZA(x0,2r0,mr0) 0
n−1
1
≥ dµ
d(x ,x)Q
k=1ZA(x0,2kr0,2k+1r0) 0
X
n−1
1
≥ dµ
k=1ZB(yk,2k−1r0) d(x0,x)Q
X
n−1
1
≥ µ(B(y ,2k−1r ))
(2k+1r )Q k 0
0
k=1
X
n−1
≥ .
4Qc
A
log(m/2) log(m/4)
Since n > , we see that n−1 > . Thus,
log(2) log(2)
u(x)Q log(m/4)
dµ > = c log(m/4).
dist(x,X \Ω)Q 4Qc log(2)
ZΩ A
By combining (3.3) and (3.4), and the fact that u satisfies Hardy’s inequal-
ity (2.7), it follows that
clog(m/4) < 2Q+1c c .
H A
Hence m < 4exp(2Q+1c c /c), and therefore X \ Ω is uniformly perfect
H A
with constant c = 4exp(2Q+1c c /c) and X \Ω is unbounded.
UP H A
The following example shows that p–Hardy’s inequality with p 6= Q does
not imply uniform perfectness.
5
Example 3.5. If X = Rn, 1 < p < n, and Ω = B(0,1) \ {0}, then Ω
supportsp–Hardy’sinequalityeventhoughX\Ωisneitheruniformlyperfect
nor uniformly p–fat, see [14, p. 179]. When p > n, even single points have
positive p–capacity and hence X \ Ω is uniformly p–fat and supports p–
Hardy’s inequality but X \Ω is not uniformly perfect.
The uniform perfectness of the boundary implies uniform q–fatness of the
complement for all q > Q−ε. We get a quantitative estimate for ε> 0 that
depends only on c .
UP
Theorem 3.6. Let (X,d,µ) be a complete Q–regular metric measure space.
Suppose that X supports a weak (1,Q)–Poincar´e inequality. Let Ω ⊂ X be
an open subset. If X \ Ω is uniformly perfect and unbounded, then there
exists a constant ε > 0 such that X \Ω is uniformly (Q−ε)–fat.
We begin the proof with an elementary inequality.
Lemma 3.7. For every C > 0 there exists 0 < ε < 1 such that for all
C
0 < ε< ε and a,b >0,
C
aε+bε ≥ (a+b+Cmin{a,b})ε.
Proof. We may assumethata ≥ b = 1. Therefore, itis enough to prove that
(a+C +1)ε −aε ≤1
when 0< ε < 1 is sufficiently small and a≥ 1. As f(a) = (a+C+1)ε−aε
is a decreasing function, and hence for every a≥ 1
(a+C +1)ε −aε ≤ (1+C +1)ε −1ε,
it is enough to choose ε so that
log2
ε ≤ ε = .
C
log(C +2)
In the proof of Theorem 3.6, we first obtain an estimate for the Hausdorff–
content of the boundary. Then the following result is needed to get capaci-
tary estimates. For a proof, see Theorem 5.9 in [6].
Lemma 3.8. Suppose that (X,d,µ) is a Q–regular space. Suppose further
that X admits a weak (1,p)–Poincar´e inequality for some 1 ≤ p ≤ Q. Let
E ⊂B(x,r) be a compact set. If
Hs (E) ≥ λrs
∞
for some s > Q−p and λ> 0, then
cap (E,B(x,2r)) ≥ cλcap (B(x,r),B(x,2r)).
p p
The constant c depends only on s and on the data associated with X.
6
Proof of Theorem 3.6. Let Ω ⊂ X be open, X \Ω uniformly perfect with
constant c > 1, and α > 1. Fix x ∈ X \ Ω and r > 0. Let A =
UP 0 0
B(x ,r )\Ω, and 0 < ε< ε , where ε is as in Lemma 3.7. First we
0 0 αcUP αcUP
estimate the Hausdorff ε–content of A. Let F be a family of balls covering
A. Because Ais compact, we may assumethat F consists of a finitenumber
of balls. We may also assume that all the balls in F are centered at A.
If there exists balls B(x ,r ) and B(x ,r ) in F such that
i i j j
r ≤ αr (3.9)
i j
and
B(x ,c r )∩B(x ,r ) 6= ∅, (3.10)
i UP i j j
then (when r ≤r )
j i
(B(x ,r )∪B(x ,r )) ⊂ B(x ,r +r +αc min{r ,r })
i i j j i i j UP i j
or (when r ≤ r )
i j
(B(x ,r )∪B(x ,r )) ⊂ B(x ,r +r +αc min{r ,r }),
i i j j j i j UP i j
and by Lemma 3.7,
rε+rε ≥ (r +r +αc min{r ,r })ε.
i j i j UP i j
Thus, we may replace balls B(x ,r ) and B(x ,r ) with
i i j j
B(x ,r +r +αc min{r ,r }) or B(x ,r +r +αc min{r ,r })
i i j UP i j j i j UP i j
in the covering F so that the sum
rε
B(x,r)∈F
X
does not increase. We continue this process until there is no pair of balls
satisfying (3.10) and (3.9). Because the number of balls in F decreases in
each step and F consists of a finite number of balls, the process ends after
a finite number of replacements.
Let B(x ,r ) ∈ F be the ball containing x . Because X \Ω is uniformly
1 1 0
perfect and unbounded, the set
(B(x ,c r )\B(x ,r ))∩(X \Ω)
1 UP 1 1 1
is nonempty. Now there are two possibilities: either
(B(x ,c r )\B(x ,r ))∩ X \B(x ,r ) 6= ∅
1 UP 1 1 1 0 0
or (cid:0) (cid:1)
(B(x ,c r )\B(x ,r ))∩A6= ∅,
1 UP 1 1 1
because (X \ Ω) ⊂ A ∪ (X \ B(x ,r )). In the latter case, there exists
0 0
B(x ,r ) ∈ F such that B(x ,r ) 6= B(x ,r ) and
2 2 1 1 2 2
B(x ,r )∩B(x ,c r ) 6= ∅,
2 2 1 UP 1
because F covers A. Now the balls B(x ,r ) and B(x ,r ) satisfy condi-
1 1 2 2
tion (3.10). Hence (3.9) fails, that is, r < r /α.
2 1
We continue inductively in the same way: For a ball B(x ,r )∈ F, either
i i
B(x ,c r )∩(X \B(x ,r )) 6= ∅
i UP i 0 0
7
or there exists a ball B(x ,r ) ∈ F such that r ≤ r /α and B(x ,r )
i+1 i+1 i+1 i i i
and B(x ,r ) satisfy the condition (3.10).
i+1 i+1
Thus we obtain a chain of distinct balls {B(x ,r )}n ⊂ F such that r ≤
i i i=1 i
α1−ir , (since r ≤ r /α, we have B(x ,r ) 6= B(x ,r ) if i 6= j).
1 i i−1 i i j j
B(x ,c r )∩B(x ,r ) 6= ∅
i UP i i+1 i+1
for every i = 1,...,n−1, and
B(x ,c r )∩ X \B(x ,r ) 6= ∅.
n UP n 0 0
It follows that
(cid:0) (cid:1)
n
α
r ≤ (c +1)r ≤ (c +1) r
0 UP i UP 1
α−1
i=1
X
and we have a lower bound for r :
1
α−1
r ≥ r .
1 0
α(c +1)
UP
We may choose α = 2 and thus
1
rε ≥ rε ≥ rε.
1 (2c +2)ε 0
UP
B(x,r)∈F
X
By [8], there exists ε > 0 such that X satisfies a weak (1,Q−ε)–Poincar´e
inequality. Fix such an ε< ε . Now by Lemma 3.8,
αcUP
cap (B(x ,r )\Ω,B(x ,2r ))≥ ccap (B(x ,r ),B(x ,2r ))
p 0 0 0 0 p 0 0 0 0
for every Q−ε< p ≤ Q, where c depends on ε and c , but is independent
UP
of x and r .
0 0
It is known that uniform p–fatness is a self–improving phenomenon, see [2].
Theorem 3.11. Let X be a proper linearly locally convex metric space
endowed with a doubling Borel regular measure supporting a weak (1,q )–
0
Poincar´e inequality for some q with 1 ≤ q < ∞. Let p > q and suppose
0 0 0
that E ⊂ X is uniformly p–fat. Then there exists q < p so that E is uni-
formly q–fat.
Remark 3.12. Theproof of Theorem 3.6 gives a new and easier proof for the
self–improvement when p = Q.
Remark 3.13. To complete the picture, note that uniform p–fatness for any
p ≤ Q implies uniform perfectness. To see this, suppose that X supports a
(1,p)–Poincar´e inequality for some 1 ≤ p < Q, and that X \Ω is uniformly
p–fat. We will show that X \ Ω is uniformly perfect. Fix x ∈ ∂Ω and
0
0 < r < 1. Suppose that B(x ,r)\B(x ,r/m) ⊂ Ω for some m > 1. We
0 0
will demonstrate that m has an upper bound that is independent of x and
0
r. Indeed,
1
cap (B(x ,r),B(x ,2r)) ≥ rQ−p.
p 0 0 C
Also, as the function
1 1
g(x) = χ
log(r/ρ) d(x ,x) B(x0,r)\B(x0,ρ)
0
8
is an upper gradient of the function
log(d(x ,x)/ρ)
0
u(x) = min 1,max 0, ,
log(r/ρ)
(cid:26) (cid:26) (cid:27)(cid:27)
with u= 0 on B(x ,ρ) and u = 1 on X \B(x ,r); hence
0 0
cap (B(x ,r)\Ω,B(x ,2r)) ≤ cap (B(x ,r/m),B(x ,2r))
p 0 0 p 0 0
C
≤ rQ−p.
log(m)p
ThelastestimatecanbeprovedinthesamewayasintheproofofLemma3.8.
We have by uniform p–fatness of X \Ω that
C
rQ−p ≥ cap (B(x ,r)\Ω,B(x ,2r))
log(m)p p 0 0
1 1
≥ cap (B(x ,r),B(x ,2r)) ≥ rQ−p,
c p 0 0 c C
0 0
where c is the uniform fatness constant, and therefore m ≤ eC. Thus X\Ω
0
is uniformly perfect.
Remark 3.14. IntheproofofTheorem3.6,weneedtoassumethatthespace
supports a weak (1,Q−ε)–Poincar´e inequality. This follows by [8] for some
positive ε if the space supports a (1,Q)–Poincar´e inequality. However, if
we assume a priori the stronger Poincar´e inequality, then our proof gives a
quantitative estimate for ε. More precisely, if
log(2)
max Q− ,1 < p < Q
log(3)
(cid:26) (cid:27)
and X supports a (1,p)–Poincar´e inequality, then there exists c > 1 such
p
that whenever X\Ω is uniformly perfect for some uniform perfectness con-
stant 1 ≤ c < c , then X \Ω is uniformly p–fat and hence Ω supports a
UP p
p–Hardy inequality by Theorem 3.17. The proof of Theorem 3.6 implies the
claim if
log(2)
p > Q−
log(αc +2)
UP
1
with some α > 1. So it is enough to have cUP < cp = 2Q−p −2. By the
assumption on p, it is clear that c > 1.
p
The following examples illustrate the sharpness of Remark 3.14.
Example 3.15. If 1 < p < Q, there is a Cantor set E ⊂ Rn such that
p
cap (E ) = 0, see [6, p. 40]. Thus the domain Rn\E has uniformly perfect
p p p
complement, which is not uniformly p–fat.
Example 3.16. If1 ≤ p < Q−1, thenanyrectifiablecurveγ inX isofzero
p–capacity. In this case, with Ω = X \γ, we have that X \Ω is uniformly
perfect with constant c = 1, but it is not uniformly p–fat.
UP
The following theorem shows that Hardy’s inequality follows from uniform
fatness for all 1 < p ≤ Q, see Corollary 6.1 in [2]. Note that the LLC–
condition is not a serious restriction in our case, since it follows from the
(1,Q)–Poincar´e inequality, see for example [11].
9
Theorem 3.17. Let X be a proper LLC metric space endowed with a dou-
bling Borel regular measure supporting a weak (1,p)–Poincar´e inequality,
and suppose that Ω is a bounded open set in X with X \Ω uniformly p–fat.
Then Ω satisfies p–Hardy’s inequality.
The converse of Theorem 3.17 is not true in general, see Example 3.5. As a
corollary of Lemma 3.2 and Theorems 3.6 and 3.17, we obtain the following
result. Note that uniform p–fatness implies uniform q–fatness for all q > p.
Theorem 3.18. Let (X,d,µ) be a complete Q–regular metric measure space
with Q > 1. Suppose that X supports a weak (1,Q)–Poincar´e inequality.
Let Ω ⊂ X be a bounded open subset. Then the following conditions are
quantitatively equivalent.
(1) Ω satisfies Q–Hardy’s inequality.
(2) X \Ω is uniformly perfect.
(3) X \Ω is uniformly Q–fat.
(4) X \Ω is uniformly (Q−ε)–fat for some ε> 0.
Theorem 3.17 is stated only for boundedsets buttheproofworks also inthe
unbounded setting. Hence Theorem 3.18 holds also when Ω is unbounded if
we require additionally that X \Ω is unbounded in conditions (2) and (3).
4. Maz′ya type characterization
In this section, we present one more characterization of open sets that is
equivalent with the Hardy’s inequality. For more information about this
kind of characterizations, see Chapter 2.3 in [15].
Theorem 4.1. Let X be a complete metric space endowed with a doubling
measure and supporting a weak (1,p)–Poincar´e inequality. Let 1 < p ≤ Q.
Then Ω⊂ X satisfies p–Hardy’s inequality if and only if for every K ⊂⊂Ω,
we have
dist(x,X \Ω)−pdµ(x) ≤ ccap (K,Ω). (4.2)
p
ZK
Proof. First assume that Ω satisfies p–Hardy’s inequality. Let u ∈ N1,p(Ω)
0
such that u= 1 in K. Then
|u(x)|p
dist(x,X \Ω)−pdµ(x) ≤ dµ(x) ≤ c gpdµ.
dist(x,X \Ω)p H u
ZK ZΩ ZΩ
By taking infimum over all such functions u, we obtain (4.2).
Now assume that equation (4.2) is satisfied. We will first prove the claim
for Lipschitz–functions that have compact support in Ω. By Theorems 2.12
1,p
and 4.8 in [18], such functions form a dense subclass of N (Ω), and thus
0
1,p
we get the result for all functions in N (Ω).
0
1,p
Let u∈ N (Ω) be compactly supported Lipschitz function, and denote
0
E = {x ∈ Ω : |u(x)| > 2k}, k = 1,2,....
k
10