A GEOMETRICAL VERSION OF HARDY’S INEQUALITY
M. HOFFMANN-OSTENHOF1 , T. HOFFMANN-OSTENHOF2,3 AND A.LAPTEV4
A BSTRACT. We prove a version of Hardy’s type inequality in a domain
Ω ⊂ Rn which involves the distance to the boundary and the volume of
Ω. In partucular, we obtain a result which gives a positive answer to a
question asked by H.Brezis and M.Marcus.
0. I NTRODUCTION
Let Ω be a domain in Rn with Lipschitz boundary. It is known that the
following extension of Hardy’s inequality is valid
Z
Z
|u(x)|2
2
|∇u(x)| dx ≥ µ
(0.1)
dx,
∀u ∈ H01 (Ω),
2 (x)
δ
Ω
Ω
where µ is a positive constant and δ(x) = dist(x, ∂Ω). The best constant
µ = µ(Ω) in (0.1) depends on the domain Ω. It is also known that for convex
domains µ(Ω) = 1/4, but there are smooth domains such that µ(Ω) < 1/4
(see[6], [7]).
H.Brezis and M.Marcus [3], Theorem I, have shown that for every domain Ω of class C 2 there exists a constant λ = λ(Ω) ∈ R such that
Z
Z
Z
1
|u(x)|2
2
(0.2)
|∇u(x)| dx ≥
dx + λ |u(x)|2 dx,
2
4 Ω δ (x)
Ω
Ω
∀u ∈ H01 (Ω).
Note that there are examples ([6], [7]) which confirm that there are smooth
domains with λ ≤ 0. However, if Ω is convex then it is proved in [3] (see
Theorem II) that
(0.3)
λ(Ω) ≥
1
.
4 diam2 (Ω)
In this paper Brezis and Marcus have asked whether the diameter of Ω, in
(0.3) can be replaced by an expression depending on |Ω| := vol Ω, namely,
whether λ ≥ c |Ω|−2/n with some c = c(n) > 0.
1991 Mathematics Subject Classification. Primary 35P99; Secondary 35P20, 47A75,
47B25.
1
2
M. HOFFMANN-OSTENHOF, T. HOFFMANN-OSTENHOF AND A. LAPTEV
The aim of this short article is to prove that this is the case indeed and that
for convex domains
(0.4)
λ≥
c(n)
,
|Ω|2/n
2/n
n(n−2)/n sn−1
,
4
c(n) =
where sn−1 := |Sn−1 |. In particular, if n = 2 then c(2) = π/2.
The proof of this result is based on a one-dimensional version of Hardy’s
inequality which is obtained in Section 1. In Section 2 we extend the
one-dimensional result to the many-dimensional case using arguments of
E.B.Davies [4], Ch.5.3. In Section 3 we prove (0.4) and consider some
other generalizations of this result.
Note that various types of Hardy’s inequalities can be found in books [8]
and [9] and in the recent review article [5]. Some results from [3] were
recently extended to cases with weights in [2] and to Lp spaces in [1].
1. O NE DIMENSIONAL RESULTS
We start with a simple statement which is just a corollary of the CauchySchwarz inequality and partial integration.
Let f be a function defined on (0, b), b > 0, and whose derivative is finite
on (0, b). We say that f belongs to the class Φ(0, b) if f is real valued and
there is a constant C = C(f ) such that
(1.1)
sup0<t≤b (t|f (t)| + t2 |f 0 (t)|) ≤ C.
Lemma 1.1. Let u ∈ C 1 (0, b), b > 0, u(0) = 0 and let f ∈ Φ(0, b). Then
Rb 0
Z b 2
( 0 f (t)|u|2 dt)2
1
du (1.2)
.
dt ≥ R b
dt
4 (f (t) − f (b))2 |u|2 dt
0
0
Proof. For any constant c taking into account (1.1) we have
2
(f (b) − c)|u(b)| −
Z
b
0
Z
2 Z b
2
f (t) |u| dt =
(f (t) − c) (|u|2 )0 dt
0
2
0
b
2
=
(f (t) − c) (u ū + uū ) dt
0
Z b
Z b
0 2
≤4
|u | dt
(f (t) − c)2 |u|2 dt .
0
0
0
0
We complete the proof by substituting c = f (b).
The next result shows that (1.2) is often sharp unlikely many other
Hardy’s type inequalities.
HARDY INEQUALITIES
3
Lemma 1.2. Let us assume that f ∈ Φ(0, b) and
Rb
• t f (s)ds → +∞, as t → 0.
• There is a constant a, a > 0 such that
f (t)e−a
Rb
t
f (s) ds
∈ L2 (0, b).
Then Lemma 1.1 is sharp.
Proof. The Cauchy-Schwarz inequality which has been used in the proof
of Lemma 1.1 is sharp if there exists a real valued function u such that u0 (t)
and u(t)(f (t) − f (b)) are linearly dependent
(1.3)
u0 (t) = a u(t)(f (t) − f (b)).
Solving (1.3) we obtain
u(t) = Ce−a
Rb
t
f (s) ds−af (b) t
.
If now a > 0, then the first assumption implies u(0) = 0. The second one
provides the inclusion du/dt ∈ L2 (0, b).
Example.
If f (t) = 1/t and a > 1/2, then substituting in (1.2)
u(t) = ta e−at/b
we find that the left and the right hand sides of this inequality are the same.
Moreover, u(0) = 0, u ∈ C 1 (0, b) and therefore the inequality (1.1) has an
extremizer.
Although in many cases Lemma 1.1 gives sharp results, the right hand
side in the inequality (1.2) is not linear with respect to u. We would now
like to give the following linearized version of this inequality.
By using (1.2) we obviously have
Z
Z b
1 b 0
0 2
2
(1.4)
|u | dt ≥
2f (t) − (f (t) − f (b)) |u|2 dt.
4 0
0
If we rewrite this inequality for the interval [b, 2b] with u ∈ C 1 (b, 2b),
u(2b) = 0, then
Z 2b
Z
1 2b 0
0 2
(1.5)
2f (2b − t) − (f (2b − t) − f (b)2 |u|2 dt.
|u | dt ≥
4 b
b
Adding up (1.4) and (1.5) and by using standard density arguments we can
finally state our main one-dimensional result.
Lemma 1.3. Let u ∈ H01 (0, 2b), b > 0 and let f ∈ Φ(0, b). Then
Z 2b
Z
1 2b 0
0
2
2
|u (t)| dt ≥
(1.6)
2f (ρ(t)) − (f (ρ(t)) − f (b)) |u|2 dt,
4 0
0
where
4
M. HOFFMANN-OSTENHOF, T. HOFFMANN-OSTENHOF AND A. LAPTEV
ρ(t) = min (t, 2b − t).
2. A
RESULT FOR HIGHER DIMENSIONS
Let Ω be a domain in Rn . In order to formulate the main result of this
section we need some notations. Denote by τν (x) the distance between
x ∈ Ω and its nearest point belonging to the boundary ∂Ω in the direction
ν ∈ Sn−1 ,
(2.1)
τν (x) = min{s > 0 : x + sν 6∈ Ω}.
Let us also introduce the “distance” to the boundary ρν and the “diameter”
Dν along the line defined by ν via:
(2.2)
ρν (x) = min(τν (x), τ−ν (x))
(2.3)
Dν (x) = τν (x) + τ−ν (x).
n−1
RBy dω(ν) we denote the normalized measure on the unit sphere S ,
dω(ν) = 1.
Sn−1
Theorem 2.1. Let Ω be a domain in Rn , D ∈ (0, ∞] be its diameter and
f ∈ Φ(0, D/2). Then for any u ∈ H01 (Ω) we have
Z
Z Z
n 2
|∇u| dx ≥
(2f 0 (ρν (x)) − f 2 (ρν (x)) + 2f (ρν (x))f (Dν (x)/2)
4
n−1
Ω
Ω
S
(2.4)
− f 2 (Dν (x)/2)) dω(ν) |u(x)|2 dx,
Proof. We proceed by using E.B.Davies’ arguments (see [4]). Let ∂ν denote
partial differentiation in the direction ν ∈ Sn−1 . Then Lemma 1.3 implies
Z
Z
1 0
2
2f (ρν (x)) − f 2 (ρν (x)) + 2f (ρν (x))f (Dν (x)/2)
|∂ν u| dx ≥
4
Ω
Ω
2
− f (Dν (x)/2) |u|2 dx,
where the function ρν and Dν are defined in (2.2) and (2.3). Let us introduce
an orthonormal basis {ēj }nj=1 in Rn . Then
Z
n Z
1X
2
(2f 0 (ρēj (x)) − f 2 (ρēj (x))
|∇u| dx ≥
4
Ω
j=1 Ω
+ 2f (ρēj (x))f (Dēj /2) − f 2 (Dēj /2))|u|2 dx.
HARDY INEQUALITIES
5
Averaging both sides of the last inequality over orthonormal bases using the
group O(n) we complete the proof.
3. A PLICATIONS OF T HEOREM 2.1
3.1. On a question of Brezis and Marcus. Let
f (t) = −1/t,
(3.1)
t > 0.
Then the integral over Sn−1 in the right hand side of (2.4) becomes equal to
Z
2f 0 (ρν ) − f 2 (ρν ) + 2f (ρν )f (Dν /2) − f 2 (Dν /2) dω(ν)
Sn−1
Z
1
4
4 (3.2)
=
+
−
dω(ν).
2
ρν Dν
Dν2
Sn−1 ρν
Let us consider the last two terms. It is clear that ρν (x) ≤ τν (x), x ∈ Ω,
where the functions ρ and τ are defined in (2.2) and (2.1). Since Dν =
τν + τ−ν (see (2.3)) we obtain
1
1
1
1
− 2 ≥
−
ρν Dν
Dν
τν (τν + τ−ν ) (τν + τ−ν )2
τ−ν
=
.
τν (τν + τ−ν )2
This implies
Z
Sn−1
Z
1
1 τ−ν
− 2
dω(ν) ≥
dω(ν)
2
ρν (x)Dν (x) Dν (x)
Sn−1 τν (τν + τ−ν )
Z
1
τ−ν
τν
=
+
dω(ν)
2 Sn−1 τν (τν + τ−ν )2 τ−ν (τν + τ−ν )2
Z
1
1
≥
dω(ν).
4 Sn−1 τν τ−ν
In order to estimate the latter integral we apply the Cauchy-Schwarz inequality twice and obtain
Z
Z
1
dω(ν)
1≤
τν τ−ν dω(ν)
Sn−1
Sn−1 τν τ−ν
Z
Z
1
2
dω(ν),
≤
τν dω(ν)
Sn−1 τν τν
Sn−1
R
R
2
where we have used that Sn−1 τν2 dω(ν) = Sn−1 τ−ν
dω(ν).
6
M. HOFFMANN-OSTENHOF, T. HOFFMANN-OSTENHOF AND A. LAPTEV
R
When now applying Hölder’s inequality we recall that Sn−1 dω(ν) = 1.
Therefore
Z
Z
−1
1
2
dω(ν) ≥
τν (x) dω(ν)
Sn−1 τν (x)τ−ν (x)
Sn−1
Z
−2/n
n
≥
τν (x) dω(ν)
.
Sn−1
Let us introduce the domain Ωx ⊆ Ω defined as a part of Ω which can be
“seen” from point x
(3.3)
Ωx := {y ∈ Ω : x + t(y − x) ∈ Ω, ∀t ∈ [0, 1]}.
Then
Z
Sn−1
τνn (x) dω(ν) =
n
sn−1
|Ωx |,
which finally gives us
Z
s 2/n
4
4 1
n−1
(3.4)
− 2
dω(ν) ≥
.
Dν (x)
n
|Ωx |2/n
Sn−1 ρν (x)Dν (x)
Now (2.4), (3.2) and (3.4) imply the following reformulation of Theorem
2.1 in the case when the function f is defined by (3.1).
Theorem 3.1. For any Ω ⊂ Rn and any u ∈ H01 (Ω) we have
Z
Z Z
n
1
2
|∇u| dx ≥
dω(ν) |u(x)|2 dx
2 (x)
4
ρ
n−1
Ω
Ω S
ν
(n−2)/n 2/n Z
n
sn−1
|u(x)|2
(3.5)
+
dx.
2/n
4
Ω |Ωx |
Clearly in (3.5) the value |Ωx | can always be replaced by |Ω|. If Ω is
convex then it is known (see for example [4] Exercise 5.7 and [5]) that
Z
n
1 1
(3.6)
ρ−2
,
ν (x) dω(ν) ≥
4 Sn−1
4 δ 2 (x)
Moreover, in this case Ωx = Ω, x ∈ Ω, and we obtain
Theorem 3.2. For any convex domain Ω ⊂ Rn and any u ∈ H01 (Ω)
Z
Z
2/n Z
n(n−2)/n sn−1
1
|u(x)|2
2
|∇u| dx ≥
dx +
|u(x)|2 dx.
2 (x)
2/n
4
δ
4
|Ω|
Ω
Ω
Ω
Note that the domain Ω in Theorem 3.1 can be unbounded and it is valid
for a variety of domains with fractal boundaries, for example, such as the
Koch snowflake in R2 . So often the inequality (3.6) might hold true with a
HARDY INEQUALITIES
7
constant µ < 1/4 instead of 1/4 in the right hand side, whereas the second
integral of the inequality (3.5) is very stable.
In particular, following E.B.Davies, Lemma 3 from [5] we can obtain:
Corollary 3.1. Suppose that there is a constant κ such that for each y ∈ ∂Ω
and each a > 0 there exists a disjoint from Ω ball B with centre z and radius
β ≥ aκ, where |z − y| = a. Then there exists a constant µ ≤ 1/4 such that
Z
n
1
ρ−2
ν (x) dω(ν) ≥ µ 2
4 Sn−1
δ (x)
and hence
Z
Z
2/n Z
n(n−2)/n sn−1
|u(x)|2
|u(x)|2
2
|∇u| dx ≥ µ
dx
+
dx.
2
2/n
4
Ω
Ω δ (x)
Ω |Ωx |
3.2. Some refined inequalities. The next application of Theorem 2.1 concerns the function
1
1
(3.7)
f (t) = − +
,
0 < t < D/2,
t t(1 − ln(αt/D))
where D = diam Ω and 0 < α ≤ 2. In this case the expression appearing
in the right hand side of (2.4) is equal to
2f 0 (ρν ) − f 2 (ρν ) + 2f (ρν )f (Dν /2) − f 2 (Dν /2)
1
1
= 2 + 2
ρν ρν (1 − ln(αρν /D))2
ln(αρν /D) ln(αDν /2D)
ln2 (αDν /2D)
−
+4
ρν Dν (1 − ln(αρν /D))(1 − ln(αDν /2D)) Dν2 (1 − ln(αDν /2D))2
1
1
1
1 ln2 (α/2)
+
4
−
.
≥ 2 + 2
ρν ρν (1 − ln(αρν /D))2
ρν Dν
Dν2 (1 − ln(α/2))2
Theorem 2.1 therefore gives
Z
|∇u|2 dx
Ω
Z Z
1
n
1
+
dω(ν) |u(x)|2 dx
≥
4 Ω Sn−1 ρ2ν (x) ρ2ν (1 − ln(αρν /D))2
Z Z
1
n ln2 (α/2)
1 +
− 2 dω(ν) |u(x)|2 dx.
(1 − ln(α/2))2 Ω Sn−1 ρν Dν
Dν
Application of (3.4) leads to a more refined version of Theorem 3.1.
8
M. HOFFMANN-OSTENHOF, T. HOFFMANN-OSTENHOF AND A. LAPTEV
Theorem 3.3. Let 0 < α ≤ 2. Then for any Ω ⊂ Rn and any u ∈ H01 (Ω)
we have
Z
|∇u|2 dx
Ω
Z Z
1
n
1
≥
+
dω(ν) |u(x)|2 dx
4 Ω Sn−1 ρ2ν (x) ρ2ν (1 − ln(αρν /D))2
Z
2/n
n(n−2)/n sn−1 ln2 (α/2)
|u(x)|2
+
dx.
2/n
4 (1 − ln(α/2))2
Ω |Ωx |
Remark. Theorem 3.3 is a stronger result than Theorem 3.1. Indeed, we
obtain Theorem 3.1 from Theorem 3.3 if we let α → 0.
For Ω convex we obtain via (3.6) that
Z
n
1 1
1
+
dω(ν)
4 Sn−1 ρ2ν (x)
(1 − ln(αρν (x)/D))2
1 1 1
≥
1
+
.
4 δ 2 (x)
(1 − ln(αδ(x)/D))2
The latter inequality and Theorem 3.3 implies a version of Theorem 3.2:
Theorem 3.4. Let 0 < α ≤ 2. For any convex domain Ω ⊂ Rn and any
u ∈ H01 (Ω)
Z
Z
1
|u(x)|2 1
2
|∇u| dx ≥
1
+
dx
4 Ω δ 2 (x)
(1 − ln(αδ(x)/D))2
Ω
Z
2/n
n(n−2)/n sn−1 ln2 (α/2) 1
(3.8)
|u(x)|2 dx.
+
4 (1 − ln(α/2))2 |Ω|2/n Ω
Remark. The last statement is an improvment of Theorem 5.1 from [3],
where for convex domains Ω and u ∈ H01 (Ω) the authors obtain the inequality
Z
Z
1
|u(x)|2 1
2
(3.9)
1
+
dx.
|∇u(x)| dx ≥
4 Ω δ 2 (x)
(1 − ln(δ(x)/D))2
Ω
Indeed, if we choose α = 1 in Theorem 3.4, then the first integral in the
right hand side of (3.8) coincides with the right hand side of (3.9). However,
α = 1 still allows to have an additional non-zero term in (3.8).
HARDY INEQUALITIES
9
3.3. Acknowledgements. The first and the second authors have been supported by BBWK der Republik Österreich and by the Austrian Science
Foundation, grantnumber P12864-MAT. The third author has been supported by the Swedish Natural Sciences Research Council, Grant MAA/MA 09364-320. T.Hoffmann-Ostenhof thanks the Göran Gustafssons
Stiftelse in Sweden. A.Laptev is grateful to the Erwin Schödinger Institute
for its warm hospitality. Partial financial support from the European Union
through the TMR network FMRX-CT 96-0001 is gratefully acknowledged.
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Institut für Mathematik, Universität
Wien1
Institut für Theoretische Chemie,
Universität Wien2
International Erwin Schrödinger
Institute for Mathematical Physics3
Department of Mathmetics, KTH, 100
44 Stockholm, Sweden4