AN ELEMENTARY PROOF OF THE CONTINUITY FROM _{L}2
0(Ω)

TO _{H}1

0(Ω)n OF BOGOVSKII’S RIGHT INVERSE OF THE

DIVERGENCE

RICARDO G. DUR ´AN

Abstract. The existence of right inverses of the divergence as an operator from H1 0(Ω)

n to

L2

0(Ω) is a problem that has been widely studied because of its importance in the analysis of the classic equations of fluid dynamics. When Ω is a bounded domain which is star-shaped with respect to a ballB, a right inverse given by an integral operator was introduced by Bogovskii, who also proved its continuity using the Calder´on-Zygmund theory of singular integrals.

In this paper we give an alternative elementary proof of the continuity using the Fourier transform. As a consequence, we obtain estimates for the constant in the continuity in terms of the ratio between the diameter of Ω and that ofB. Moreover, using the relation between the existence of right inverses of the divergence with the Korn and improved Poincar´e inequalities, we obtain estimates for the constants in these two inequalities. We also show that one can proceed in the opposite way, that is, the existence of a continuous right inverse of the divergence, as well as estimates for the constant in that continuity, can be obtained from the improved Poincar´e inequality. We give an interesting example of this situation in the case of convex domains.

1. Introduction

Let Ω⊂Rn _{be a bounded domain. Given a smooth vector field}_{u}_{defined in Ω}

we will denote withDu_{its differential matrix, namely,}

Du_{=}

_{∂u}

i

∂xj

and for a tensor field (aij) we define its norm by

kak2L2_{(Ω)}=

n

X

i,j=1

kaijk2L2_{(Ω)}.

The existence of solutionsu_{∈}_{H}1
0(Ω)n of

divu_{=}_{f} _{(1.1)}

satisfying

kDu_{k}

L2_{(Ω)}≤Cdiv,ΩkfkL2(Ω), (1.2)

2000 *Mathematics Subject Classification.* Primary: 76D07, 42B20.

*Key words and phrases.* Divergence operator, singular integrals, Stokes equations.

wheref ∈L2

0(Ω) :={f ∈L2(Ω) : R

Ωf = 0}and the constantCdiv,Ωdepends only on Ω, is a problem that has been widely analyzed because of its several applications and connections with other important results.

Assume that Ω ⊂ Rn _{is a domain with diameter} _{R} _{which is star-shaped with}

respect to a ball B ⊂Ω, which we assume centered at the origin and of radiusρ. For a functionω∈C∞

0 (B) such that R

Ωω= 1, a solution of (1.1) is given by

u_{(x) =}

Z

Ω

G(x, y)f(y)dy (1.3)

whereG= (G1,· · ·, Gn) is defined by

G(x, y) = Z 1

0

(x−y) t ω

y+x−y t

_{dt}

tn .

Moreover,u_{∈}_{H}1

0(Ω)n and (1.2) is satisfied.

This formula was introduced in [6] by Bogovskii who proved the estimate (1.2), as well as its generalization forLp, 1< p <∞, using the general Calder´on-Zygmund theory of singular integrals developed in [7].

More recently, several papers have considered extensions and applications of this formula. In [10], a weighted version of (1.2), which is of interest in finite element analysis, was proved. In [1], an extension of Bogovskii’s formula was introduced for the rather general class of John domains and the estimate (1.2) was proved using again the Calder´on-Zygmund theory. Also, extensions of (1.2) for fractional order positive and negative Sobolev norms have been obtained in [8,13].

The goal of this paper is twofold:

First, we want to give a simple proof of the estimate (1.2) for the solution given by (1.3) using elementary properties of the Fourier transform. In this way we avoid the use of the complicated general theory of singular integral operators. We believe that this can be interesting for teaching purposes.

Second, we are interested in obtaining some information on the constant in terms of the ratioR/ρ. As a byproduct, this result can be used to give estimates for the constants in some Korn and improved Poincar´e inequalities.

is, for example, the case of planar star-shaped domains. Consequently, we obtain new estimates for the constant in the improved Poincar´e inequality in this case.

2. Boundedness of the singular integral operator

In order to work with functions defined inRn _{we extend}_{f} _{by zero outside of Ω}

in (1.3).

Let us recall the basic properties of the Fourier transform that we will need (see
for example [24]). The Fourier transform is defined forf ∈L1_{(}_{R}n_{) by}

b f(ξ) =

Z

e−2πix·ξ_{f}_{(x)dx.}

Here and in the rest of the paper, when we do not indicate the domain of integration
it is understood that it isRn_{. The Fourier transform can be extended to}_{f} _{in the}

class of tempered distributionsS′_{, in particular, it is defined in} _{L}2_{(}_{R}n_{) and it is}

an isometry, i.e.,

kfbkL2_{(}_{R}n_{)}=kfk_{L}2_{(}_{R}n).

We will use the well known equality c ∂f ∂xj

(ξ) = 2πiξjfb(ξ).

Thek-component ofu_{is given by}

uk=uk,1−uk,2,

where

uk,1(x) =

Z 1 0

Z yk+

(xk−yk)

t

ω

y+x−y t

f(y)dydt tn.

and

uk,2(x) =

Z 1 0

Z ykω

y+x−y t

f(y)dydt tn.

These double integrals exist, if for example we assume thatf ∈ L1_{(}_{R}n_{) and has}

compact support. Indeed, if suppf ⊂B(0, M) then both integrands vanish unless

y+x−y_{t} < ρ and |y| < M, and so, assuming that ρ < M, we can restrict the
domain of integration to|x−y|<2M t. Therefore, integrating first in thetvariable,
it follows that, fori= 1,2,

|uk,i(x)| ≤C

Z

|f(y)| |x−y|n−1dy,

where the constantC depends only on ω, n, and M. Since f ∈ L1_{(}_{R}n_{) the last}

integral is finite for almost everyx.

In order to take the derivatives ofuk,i it is convenient to write

uk,1(x) = lim ε→0

Z 1

ε

Z yk+

(xk−yk)

t

ω

y+x−y t

and

uk,2(x) = lim ε→0

Z 1

ε

Z ykω

y+x−y t

f(y)dydt tn,

where, as we will see, the limits exist in S′_{. Consider the first integral and, to}

simplify notation, defineϕ(x) =xkω(x). Then, giveng∈ S we have to show that

Z Z 1

ε

Z ϕ

y+x−y t

f(y)dydt tn

g(x)dx

→

Z Z 1 0

Z ϕ

y+x−y t

f(y)dydt tn

g(x)dx

whenε→0. It is enough to see that

Iε:=

Z ε

0
Z Z _{}

ϕ

y+x−y t

_{}

|f(y)| |g(x)|dx dydt_{t}n →0. (2.1)

But, making the change of variablez=x−y_{t} in the interior integral we have

Iε=

Z ε

0 Z Z

|ϕ(y+z)| |f(y)| |g(y+tz)|dz dy dt≤ kgkL∞_{(}_{R}n)kϕk_{L}1_{(}_{R}n)kfk_{L}1_{(}_{R}n)ε

which proves (2.1). The integral defining uk,2 can be treated in the same way, indeed, defining now ϕ(x) =ω(x), the only difference with the case of uk,1 is the factoryk appearing in the integrand, but it can be bounded assuming again that

f has compact support.

Now, forε >0 fixed, we can take the derivative inside the integral, and therefore, ∂uk

∂xj

=Tkj,1f+Tkj,2(ykf) (2.2)

whereTkj,1 andTkj,2are of the form T f(x) = lim

ε→0 Z 1

ε

Z _{∂}

∂xj

ϕ

y+x−y t

f(y)dydt

tn (2.3)

withϕ(x) =xkω(x) forTkj,1 andϕ(x) =ω(x) forTkj,2.

We are going to prove continuity of operators of the form given in (2.3) where ϕ∈C∞

0 (B) withB =B(0, ρ). With this goal we decompose the operator as T f=T1f+T2f

where

T1f(x) = lim

ε→0 Z 1

2

ε

Z _{∂}

∂xj

ϕ

y+x−y t

f(y)dy dt

tn (2.4)

and

T2f(x) = Z 1

1 2

Z _{∂}

∂xj

ϕ

y+x−y t

f(y)dydt tn

An estimate ofkT2fkL2_{(R}n_{)} for L2-functions f vanishing outside Ω can be

Lemma 2.1. _{If}_{f} _{∈}_{L}2_{(}Rn_{)}_{vanishes outside} _{Ω}_{then}

kT2fkL2_{(Ω)}≤2n|Ω|

∂x∂ϕj

L∞_{(}_{R}n_{)}

kfkL2_{(Ω)}

Proof._{We have}

T2f(x) =

Z (Z 1

1 2

_{∂ϕ}

∂xj

y+x−y t

_{dt}

tn+1 )

f(y)dy. (2.5)

Then,

|T2f(x)| ≤2n ∂x∂ϕj

L∞_{(R}n_{)}

Z

Ω| f(y)|dy

and the result follows immediately using the Schwarz inequality.

We now proceed to bound the operator T1 in L2_{. This will be done using}
the Fourier transform. By standard density arguments it is enough to bound the
operator acting onf smooth enough. In the following lemma we give a simple form
forT1 in terms of Fourier transforms.

Lemma 2.2. _{For} _{f} _{∈}_{C}_{0}∞_{(}_{R}n_{)}_{we have}

d

T1f(ξ) = 2πiξj

Z 1 2

0 b

ϕ(tξ)fb((1−t)ξ)dt (2.6)

Proof._{From (}_{2.4}_{) we have}

T1f = lim

ε→0T1,εf, where

T1,εf(x) =

Z 1 2

ε

Z _{∂}

∂xj

ϕ

y+x−y t

f(y)dydt tn

and the limit is taken inS′_{.}

Now, we have

[

T1,εf(ξ) =

Z Z 1 2

ε

Z _{∂}

∂xj

ϕ

y+x−y t

f(y)e−2πix·ξdydt tndx,

and, since this triple integral exists, we can interchange the order of integration. Therefore, integrating by parts we obtain

[

T1,εf(ξ) = 2πiξj

Z 1 2

ε

Z Z ϕ

y+x−y t

f(y)e−2πix·ξdx dydt tn,

and making the change of variable

in the interior integral,

[

T1,εf(ξ) = 2πiξj

Z 1 2

ε

Z Z

ϕ(z)e−2πi(tz+(1−t)y)·ξ_{f}_{(y)dz dy dt}

= 2πiξj

Z 1 2

ε

Z Z b

ϕ(tξ)e−2πi(1−t)y·ξf(y)dy dt, and therefore,

[

T1,εf(ξ) = 2πiξj

Z 1 2

ε b

ϕ(tξ)fb((1−t)ξ)dt,

and takingε→0 we conclude the proof.

Using the expression given in (2.6) we will give an estimate for the operatorT1
inL2_{. First we prove an auxiliary result.}

Lemma 2.3. _{Define}_{C}_{ϕ,ρ}_{=}_{ρ}−1_{k}_{ϕ}_{k}

L1_{(R}n_{)}+ρ

∂∂x2ϕ2

j

L1_{(}_{R}n_{)}. Then,

2π|ξj|

Z ∞

0 |b

ϕ(tξ)|dt≤Cϕ,ρ

Proof._{We have}

2π|ξj|

Z ∞

0 |b

ϕ(tξ)|dt= 2π|ξj|

Z 1 2πρ|ξj|

0 |b

ϕ(tξ)|dt+ 2π|ξj|

Z ∞

1
2πρ|_{ξj}|

|ϕ(tξ)b |dt:=I+II

Now,

I≤ρ−1kϕbkL∞_{(}_{R}n_{)}≤ρ−1kϕk_{L}1_{(}_{R}n_{)}

and

II = 2π Z ∞

1 2πρ|ξj|

t2_{|}_{ξ}

j|2|ϕ(tξ)b |

t2_{|}_{ξ}

j|

dt≤2πkξj2ϕbkL∞_{(}_{R}n_{)}

Z ∞

1 2πρ|ξj|

1
t2_{|}_{ξ}

j|

dt

but

−4π2_{ξ}2

jϕb=

d
∂2_{ϕ}
∂x2

j

and therefore,

II≤ 1
2π
d
∂2_{ϕ}
∂x2
j
_{L}∞_{(}_{R}n_{)}

Z ∞

1 2πρ|ξj|

1
t2_{|}_{ξ}

j|

dt≤ρ

∂2_{ϕ}
∂x2
j
_{L}1_{(}_{R}n_{)}

and the lemma is proved.

As a consequence of this lemma we obtain the following estimate for the operator T1.

Lemma 2.4. _{If} _{C}_{ϕ,ρ} _{is the constant defined in the previous lemma, then}

kT1fkL2_{(}_{R}n_{)}≤2
n−1

2 C

Proof._{Applying the Schwarz inequality in (}_{2.6}_{) we have}

|T1fd(ξ)|2≤ Z 1

2

0

2π|ξj||ϕ(tξ)b |dt

! Z 1 2

0

2π|ξj||ϕ(tξ)b ||fb((1−t)ξ)|2dt

!

and so, from Lemma2.3,

|T1fd(ξ)|2_{≤}_{C}

ϕ,ρ

Z 1 2

0

2π|ξj||ϕ(tξ)b ||fb((1−t)ξ)|2dt

Then, integrating inξand making the change of variableη = (1−t)ξ, we obtain Z

|T1fd(ξ)|2_{dξ}_{≤}_{C}

ϕ,ρ

Z 1 2

0

Z _{2π}

(1−t)n+1|ηj| bϕ

_{tη}

1−t
_{}

|fb(η)|2dη dt and, integrating first in the variablet and making now the changes =t/(1−t), we get

Z

|T1fd(ξ)|2dξ≤2n−1Cϕ,ρ

Z Z 1 0

2π|ηj||ϕ(sη)b |ds

|fb(η)|2dη. Therefore, applying again Lemma2.3,

Z

|T1fd(ξ)|2dξ≤2n−1Cϕ,ρ2

Z

|fb(η)|2dη,

and we conclude the proof recalling that the Fourier transform is an isometry in
L2_{(}_{R}n_{).}

Summing up the lemmas we obtain the main result of this section.

Theorem 2.1. _{If}_{T} _{is the operator given in (}_{2.3}_{) and} _{f} _{vanishes outside} _{Ω, then}

kT fkL2_{(Ω)}≤C_{ϕ,ρ,}_{Ω}kfk_{L}2_{(Ω)}

with

Cϕ,ρ,Ω= 2

n−1

2 ρ−1kϕk

L1_{(}_{R}n_{)}+ 2
n−1

2 ρ

∂2_{ϕ}
∂x2

j

L1_{(R}n_{)}

+ 2n|Ω| ∂x∂ϕj

L∞_{(R}n_{)}

.

3. Dependence of the constant on Ω

An interesting question is what can be said, in terms of the geometry of the domain Ω, about the behavior of the constantCdiv,Ω in the estimate (1.2). Recall that we are assuming that the domain Ω has diameterRand that it is star-shaped with respect to a ball of radiusρwhich, to simplify notation, we assume centered at the origin.

It is known that the constant cannot be bounded independently of the ratio R/ρ. Indeed, this can be seen by the following elementary example which also shows that, in some cases,

Cdiv,Ω≥c1(R/ρ) (3.1)

Given positive numbersaandε, consider the rectangular domain Ωa,ε:= (−a, a)×

(−ε, ε) and suppose that, for any f ∈ L2

0(Ωa,ε), there exists u ∈ H01(Ωa,ε)

solv-ing (1.1) and satisfying the estimate (1.2) with a constant Cdiv,Ω = Ca,ε. Take

f(x1, x2) =x1 and the corresponding solutionu_{, then}

kx1k2L2_{(Ω}

a,ε)=

Z

Ωa,ε

x1divu_{=}_{−}

Z

Ωa,ε

u1= Z

Ωa,ε

x2∂u1 ∂x2

≤ kx2kL2_{(Ω}_{a,ε}_{)}

∂u1_{∂x2}

L2_{(Ω}

a,ε)

≤Ca,εkx2kL2_{(Ω}_{a,ε})kx1k_{L}2_{(Ω}_{a,ε}_{)}

and so,

kx1kL2_{(Ω}

a,ε)≤Ca,εkx2kL2(Ωa,ε)

but,

kx1kL2_{(Ω}

a,ε)=

2 √

3ε

1 2a

3

2 and kx2k

L2_{(Ω}

a,ε)=

2 √

3ε

3 2a

1 2

and therefore,

Ca,ε≥(a/ε).

Consequently, ifa > ε, it follows that in this example (3.1) holds.

For the kind of domains that we are considering the following estimate for the constantCdiv,Ω is given in [12]

Cdiv,Ω≤C0(R/ρ)n+1

with a constantC0 independent of Ω. The reader can check that the result given in Theorem2.1 recovers this estimate. However, as we will show, this result can be improved.

Indeed, Theorem 2.1 does not give a good estimate of the constant in terms of the function ϕ (or equivalently on ρ). Curiously, this is due to the estimate obtained in Lemma2.1for the operatorT2which in some sense is easier to handle than T1. Then, in order to obtain a sharper bound, we will give in the following lemmas a different argument to boundT2.

Lemma 3.1. _{If} _{1}_{≤}_{p <} n
n−1 then
kT2fkLp_{(}_{R}n_{)}≤

2pn′ (1−pn′)

_{∂x}∂ϕ_{j}

L1_{(}_{R}n_{)}

kfkLp_{(}_{R}n).

Proof._{From (}_{2.5}_{) we have}

|T2f(x)| ≤ Z 1

1 2

Z

_{∂ϕ}

∂xj

y+x−y t

_{}

|f(y)|dytndt+1. Making the change of variable

in the interior integral, we obtain

|T2f(x)| ≤2 Z 1

1 2

Z _{}
_{∂x}∂ϕ_{j}(z)

f
_{tz}
−x
t−1

_{}

_{(1}_{−}1_{t)}n dz dt.

Applying now the Minkowski inequality for integrals we have

kT2fkLp_{(R}n_{)}≤2

Z 1

1 2

Z ∂x∂ϕj(z)

Z f tz−x

t−1
p
dx
1
p _{1}

(1−t)ndz dt

and, by the change of variable

x= tz−x t−1 in the interior integral, it follows that

kT2fkLp_{(}_{R}n_{)}≤2

∂x∂ϕj

_{L}1_{(}_{R}n_{)}

kfkLp_{(}_{R}n_{)}

Z 1

1 2

1 (1−t)pn′

dt

therefore, sincep′ _{> n, the integral on the right hand side of this inequality is finite}

and so we obtain the lemma.

Unfortunately the restriction for the value of pin the previous lemma excludes
the casep= 2. However, using well known interpolation theorems we can obtain
an estimate for theL2 _{case.}

Lemma 3.2. _{If}_{f} _{∈}_{L}2_{(}_{R}n_{)}_{vanishes outside} _{Ω}_{then, for} _{1}_{≤}_{p <} n
n−1,
kT2fkL2_{(Ω)}≤ 2

n

2

(1−pn′)

p

2|

Ω|1−p2

∂x∂ϕj

p 2

L1_{(R}n_{)}

∂x∂ϕj

1−p

2

L∞_{(R}n_{)}

kfkL2_{(Ω)}

Proof._{From the definition of}_{T2} _{(}_{2.5}_{) it is easy to see that}

kT2fkL∞_{(Ω)}≤2n|Ω|
∂x∂ϕj

_{L}∞_{(}_{R}n_{)}

kfkL∞(Ω).

Then, the result follows immediately from this estimate together with Lemma3.1

and the well known Riesz-Thorin interpolation theorem (see for example [14, p. 34]) which states that

kT2kL(L2_{,L}2_{)}≤ kT2k

p

2

L(Lp_{,L}p)kT2k

1−p

2

L(L∞_{,L}∞_{)}.

Summing up we obtain the following estimate in terms of the functionϕ.

Theorem 3.1. _{If} _{T} _{is the operator given in (}_{2.3}_{),} _{f} _{vanishes outside} _{Ω, and}

1≤p < _{n−}n_{1}, then

kT fkL2_{(Ω)}≤ 2

n−1 2 C

ϕ,ρ+

2n2

(1−pn′)

p

2

e

Cϕ,p|Ω|1− p

2

!

kfkL2_{(Ω)}

where

Cϕ,ρ=ρ−1kϕkL1_{(R}n_{)}+ρ

and

e Cϕ,p=

∂x∂ϕj

p 2

L1_{(}_{R}n_{)}

∂x∂ϕj

1−p

2

L∞_{(Ω)}

Proof._{The result follows immediately from Lemmas}_{2.4}_{and}_{3.2}_{.}

We want to bound k∂uk

∂xjkL2(Rn) using the expression (2.2). This is the goal of

the following theorem.

In what followsCn denotes a constant depending only onn, not necessarily the

same at each occurrence, and A ∼B means that A/B is bounded by above and below by positive constants which may depend onnandponly.

Theorem 3.2. _{Let} _{Ω} _{⊂} _{R}n _{be a bounded domain of diameter} _{R} _{which is }

star-shaped with respect to a ball B ⊂Ωof radius ρ andu_{∈}_{H}_{0}1_{(Ω)} _{be the solution of}

(1.1) given by (1.3). Then, there exists a constant Cn depending only on n such

that

kDu_{k}

L2_{(Ω)}≤Cn

R ρ

|Ω| |B|

n−2 2(n−1)

log |Ω| |B|

n

2(n−1)

kfkL2_{(Ω)}

Proof._{As we have mentioned, both operators on the right hand side of (}_{2.2}_{) are}

of the form given in (2.3). We will estimate the termTkj,2(ykf) which is the worst

part due to the presence ofyk. The reader can check that the termTkj,1f can be

bounded analogously.

For Tkj,2 the function ϕ is exactly ω, which is supported in B(0, ρ) and has integral equal to one. Therefore,ϕcan be taken as

ϕ(x) =ρ−nψ(ρ−1x),

whereψis a smooth function supported in the unit ball and with integral equal to one. Then,

∂ϕ ∂xj

(x) =ρ−n−1∂ψ ∂xj

(ρ−1x), ∂
2_{ϕ}
∂x2

j

(x) =ρ−n−2∂
2_{ψ}
∂x2

j

(ρ−1x) and so,

Cϕ,ρ=ρ−1kϕkL1_{(R}n_{)}+ρ

∂2_{ϕ}
∂x2
j

L1_{(R}n_{)}

∼ ρ−1 (3.2)

and

e Cϕ,p=

∂x∂ϕj

p 2

L1_{(}_{R}n_{)}

∂x∂ϕj

1−p

2

L∞_{(}_{R}n_{)}

∼ ρ−1−n(1−p2). (3.3)

Therefore, applying Theorem3.1forT =Tkj,2, using|yk| ≤Rand the relations

(3.2) and (3.3), we obtain, for 1≤p < _{n−}n_{1},
kDu_{k}

L2_{(Ω)}≤Cn

R ρ

1 (1−pn′)

p

2

|Ω| |B|

1−p

2

kfkL2_{(Ω)}

=Cn R

ρ 1 (1− n

p′)

p

2

|Ω| |B|

n−2
2(n−1)_{|}_{Ω}_{|}

|B| 1

2(nn−1−p)

Now, assuming that |_{|B|}Ω| is large enough, we can choosepsuch that
1

2
_{n}

n−1 −p

= 1

log|_{|B|}Ω|
obtaining

kDu_{k}

L2_{(Ω)}≤Cn

R ρ

1 (1−pn′)

p

2

|Ω| |B|

n−2 2(n−1)

ekfkL2(Ω),

and so, we conclude the proof using that

1−_{p}n′ =

(n−1) p

_{n}

n−1 −p

= 2(n−1)
plog|_{|B|}Ω|
andp < _{n−}n_{1}.

Remark 3.1. _{In the particular case} _{n}_{= 2} _{the theorem gives}

kDu_{k}

L2_{(Ω)}≤C

_{R}

ρ

log
_{R}

ρ

kfkL2_{(Ω)}

In view of the example given above this estimate is almost optimal (i.e., optimal up to the logarithmic factor).

4. The Korn inequality

As it is well known, Korn type inequalities are strongly connected with the
exis-tence of solutions of (1.1) satisfying (1.2). For example, in the particular case of two
dimensional simple connected domains with aC1 _{boundary, the explicit relation}
between the best constant in (1.2) and that in the so-called second case of Korn
inequality was given in [17]. More generally, for arbitrary domains inndimensions,
n≥2, the Korn inequality can be derived from the existence of solutions of the
divergence satisfying (1.2), and therefore, information on the constant in the Korn
inequality can be obtained from estimates for the constant in (1.2).

A lot of work has been done in order to obtain the behavior of the constant in the different versions of Korn inequality in terms of the domain (see [16] and its references).

We are going to show how our results in the previous section can be used to obtain estimates for the constant in the second case of Korn inequality. Let us mention that domains which are star-shaped with respect to a ball were considered by Kondratiev and Oleinik in [20,21] where the authors obtain sharp estimates for the constant in a Korn inequality in terms ofR/ρ. However, their results are for a different type of Korn inequality than the one that we are considering and it is not clear what is the relation between the constants in the two different Korn type inequalities.

For a vector field v_{∈}_{H}1_{(Ω)}n_{,} _{ε(}_{v}_{) and} _{µ(}_{v}_{) denote the symmetric and skew}

symmetric part ofDv_{respectively, i.e.,}

εij(v) =

1 2

_{∂v}

i

∂xj

+∂vj ∂xi

and

µij(v) =1

2

∂vi

∂xj −

∂vj

∂xi

Then, the so-called second case of Korn inequality states that there exists a constantCK,Ωsuch that

kDv_{k}_{L}2_{(Ω)}≤C_{K,}Ωkε(v)k_{L}2_{(Ω)}

for vector fieldsv_{∈}_{H}1_{(Ω)}n _{satisfying}

Z

Ω

µij(v) = 0, for i, j= 1, . . . , n. (4.1)

The argument used in the proof of the following theorem is known but we include
it for the sake of completeness. For an arbitrary domain Ω we will say that it admits
a right inverse of the divergence with constantCdiv,Ω if, for anyf ∈L20(Ω), there
existsu_{∈}_{H}_{0}1_{(Ω)}n _{satisfying}

divu_{=}_{f}

and

kDu_{k}

L2_{(Ω)}≤Cdiv,ΩkfkL2(Ω).

Theorem 4.1. _{If}_{Ω}_{admits a right inverse of the divergence with constant}_{C}_{div,}_{Ω,}

then the second case of Korn inequality holds in Ω with a constant CK,Ω which satisfies

CK,Ω≤(1 + 4n2)1/2Cdiv,Ω

Proof._{Let}v_{∈}_{H}1_{(Ω)}n _{such that (}_{4.1}_{) holds. By density we can assume that}v

is smooth. By orthogonality we have

kDv_{k}2

L2_{(Ω)}=kε(v)k2_{L}2_{(Ω)}+kµ(v)k2_{L}2_{(Ω)}

and so, observing thatCdiv,Ω≥1, it is enough to prove that

kµ(v_{)}_{k}2_{L}2_{(Ω)}≤4n2C_{div,}2 Ωkε(v)k2_{L}2_{(Ω)}. (4.2)

Giveniand j, since _{Z}

Ω

µij(v) = 0,

there existsuij _{∈}_{H}_{0}1_{(Ω)}n _{such that}

divuij_{=}_{µ}_{ij}_{(}v_{)}

and _{}

Duij

L2_{(Ω)}≤Cdiv,Ωkµij(v)kL2(Ω). (4.3)

Then,

kµij(v)k2L2_{(Ω)}=

Z

Ω

µij(v) divuij =−

Z

Ω∇

µij(v)·uij

but,

∂µij(v)

∂xk

=
_{∂ε}

ik(v)

∂xj −

∂εjk(v)

∂xi

and so,

kµij(v)k2L2_{(Ω)}=−

n

X

k=1 Z

Ω
_{∂ε}

ik(v)

∂xj −

∂εjk(v)

∂xi

uij_{k}

=

n

X

k=1 Z

Ω

εik(v)

∂uij_{k}
∂xj −εjk(

v_{)}∂u
ij
k

∂xi

! ,

and using now (4.3) we obtain,

kµij(v)k2L2_{(Ω)}≤Cdiv,Ωkµij(v)kL2_{(Ω)}

n

X

k=1

kεik(v)kL2_{(Ω)}+kε_{jk}(v)k_{L}2_{(Ω)}.

Therefore,

kµij(v)kL2_{(Ω)}≤C_{div,}_{Ω}n1/2

(_{X}n

k=1

kεik(v)kL2_{(Ω)}+kε_{jk}(v)k_{L}2_{(Ω)}2

)1/2

and then

kµij(v)kL22_{(Ω)}≤2C_{div,}2 _{Ω}n

n

X

k=1

kεik(v)k2L2_{(Ω)}+kεjk(v)k2L2_{(Ω)}

.

Finally, summing now iniandj we obtain (4.2).

Consequently, using the results of the previous section we obtain an estimate for the Korn inequality in star-shaped domains.

Theorem 4.2. _{Let} _{Ω} _{⊂} _{R}n _{be a bounded domain of diameter} _{R} _{which is }

star-shaped with respect to a ballB ⊂Ω of radius ρ. Then, there exists a constant Cn

depending only on n such that, for all v _{∈} _{H}1_{(Ω)}n _{satisfying} R

Ωµij(v) = 0, for i, j= 1, . . . , n,

kDv_{k}_{L}2_{(Ω)}≤C_{n}

R ρ

|Ω| |B|

n−2 2(n−1)

log|Ω| |B|

n

2(n−1)

kε(v_{)}_{k}_{L}2_{(Ω)}

Proof._{The result follows immediately from Theorems}_{3.2}_{and}_{4.1}_{.}

5. The improved Poincar´e inequality

In this section we consider another well known result usually called improved
Poincar´e inequality. To recall this inequality we need to introduce some notation.
For a bounded domain Ω⊂Rn _{and any}_{x}_{∈}_{Ω we denote with}_{d(x) the distance}

fromxto the boundary of Ω. Then, the improved Poincar´e inequality states that there exists a constantCiP,Ωsuch that, for any f ∈H1(Ω)∩L20(Ω),

kfkL2_{(Ω)}≤C_{iP,}_{Ω}kd∇fk_{L}2(Ω). (5.1)

For the star-shaped domains that we are considering in this paper, the argument given in [9], applied in this particular case, can be used to show that

CiP,Ω≤Cn(R/ρ)n+1; (5.2)

indeed, this was done in [4, Prop. 5.2] for the analogous inequality in L1_{, but it}
is easy to see that the arguments extend straightforward to the L2 _{case. We are}
going to show that the dependence onR/ρcan be improved using our estimates of
Section3, at least in the two dimensional case.

In [11] the relation between Poincar´e type inequalities and solutions of the di-vergence was analyzed in a very general context. A particular case of the results in that paper says that the improved Poincar´e inequality (5.1) implies the existence of a right inverse of the divergence as an operator from H1

0(Ω)n to L20(Ω). The interest of this result is that it allows us to obtain information on the constant in (1.2) from that for the constant in the improved Poincar´e inequality. We will give an example of this situation for the case of convex domains.

Let us reproduce the argument given in [11] for the sake of completeness. With this purpose we need to use a Whitney decomposition of Ω, i.e., a sequence of cubes {Qj} with pairwise disjoint interiors and such that, ifdj and ℓj are the distance

ofQj to the boundary of Ω and the length of its edges respectively, thendj/ℓj is

bounded by above and below by positive constants depending only onn. Associated with this decomposition there is a partition of unity{φj}, namely,Pjφj = 1 in Ω,

withφj ∈C0∞(Qej) whereQejis an expansion ofQj still with diameter proportional

to its distance to the boundary of Ω. A Whitney decomposition exists for any domain (see for example [24] for a proof).

Lemma 5.1. _{If the improved Poincar´e inequality (}_{5.1}_{) is satisfied in}_{Ω}_{then, given}

f ∈L2

0(Ω) and a Whitney decomposition of Ω, there exists a sequence {fj} such

that fj∈L20(Qej),f =P_{j}fj, and

kfk2L2_{(Ω)}≤Cn

X

j

kfjk2_{L}2_{(}Qej) (5.3)

and _{X}

j

kfjk2_{L}2_{(}_{Q}_{e}

j)≤Cn(1 +CiP,Ω)kfk

2

L2_{(Ω)}. (5.4)

Proof._{First we observe that, by duality, (}_{5.1}_{) implies that, for all} _{f} _{∈} _{L}2_{0}_{(Ω),}

there existsv_{∈}_{L}2_{(Ω)}n _{such that}

divv_{=}_{f} _{in Ω} _{,} v_{·}n_{= 0 on}_{∂Ω} _{(5.5)}

and _{}

v_{d}

L2_{(Ω)}≤CiP,ΩkfkL

2(Ω), (5.6)

where both equations in (5.5) have to be understood in a distributional sense. Indeed,

L(∇g) = Z

defines a linear form on the subspace of L2_{(Ω)}n _{formed by the gradient vector}

fields. Lis well defined becauseR_{Ω}f = 0. Moreover, it follows from (5.1) that
|L(∇g)|=

Z

Ω

f(g−g)

≤CiP,ΩkfkL2(Ω)kd∇gk_{L}2_{(Ω)}

wheregis the average of gin Ω.

By the Hahn-Banach theorem L can be extended as a linear continuous func-tional to the space L2d(Ω)n, where L2d(Ω) denotes the Hilbert space with norm

kfkL2

d := kdfkL2, and therefore, there exists

v _{∈} _{L}2

d−1(Ω)n satisfying (5.6) and

such that

L(w_{) =}

Z

Ω

v_{·}w _{∀}w_{∈}_{L}2
d(Ω)n;

in particular, _{Z}

Ω

v_{· ∇}_{g}_{=}

Z

f g ∀g∈H1(Ω) which is equivalent to (5.5).

Given nowf ∈L2

0(Ω), letv∈L2(Ω)n satisfying (5.5) and (5.6), and define fj = div (φjv).

Then, we have

f = divv_{= div}vX
j

φj

=X

j

div (φjv) =

X

j

fj.

Since suppφj⊂Qej we have suppfj ⊂Qej andR fj = 0.

Moreover, using the finite superposition (with constant depending only onn) of the expanded cubesQej, we obtain immediately (5.3). On the other hand, using

again the finite superposition and thatkφjkL∞ ≤1 andk∇φ_{j}k_{L}∞ ≤C/d_{j}, we have

kfjk2_{L}2_{(}Qej)≤Cn

kfk2_{L}2_{(}Qej)+

v_{d}2

L2_{(}_{Q}_{e}

j)

,

and therefore, (5.4) follows from (5.6).

Theorem 5.1. _{If the improved Poincar´e inequality (}_{5.1}_{) is satisfied in}_{Ω, then}_{Ω}

admits a right inverse of the divergence with constantCdiv,Ωwhich satisfies
Cdiv,Ω≤Cn(1 +CiP,Ω). (5.7)
Proof._{Given}_{f} _{∈}_{L}2_{0}_{(Ω) let}_{f}_{j}_{be the functions given in the previous lemma. Since}

fj∈L20(Qej), there existsuj ∈H01(Qej)n such that

divu_{j}_{=}_{f}_{j} _{and} _{k}_{D}u_{j}_{k}
L2_{(}_{Q}_{e}

j)≤CnkfjkL2(Qej),

indeed, a scaling argument shows that the constant in this inequality is independent
of the size of the cube. Then, u _{=}P

juj ∈ H01(Ω)n is a solution of divu =f. Moreover, it follows from (5.4) that

kDu_{k}_{L}2_{(Ω)}≤C_{n}(1 +C_{iP,}Ω)kfk_{L}2_{(Ω)}

Let us now consider the case of convex domains. For these domains it was proved
in [4, Cor. 5.3] that, forf ∈W1,1_{(Ω),}

kf−fSekL1_{(Ω)}≤C_{n} R

ρ kd∇fkL1(Ω), (5.8) where Cn is a constant depending only on n and fSe is the average of f over a

particular subset of Ω. From this inequality it is easy to see that, forf ∈W1,1_{(Ω)}
with vanishing mean value,

kfkL1_{(Ω)}≤2C_{n}R

ρ kd∇fkL1(Ω). (5.9)
Indeed, ifR_{Ω}f = 0, we have

kfSekL1_{(Ω)}=

Z

Ω

(fSe−f)

≤ kf−fSekL1(Ω),

and therefore,

kfkL1_{(Ω)}≤ kf −f_{e}

SkL1(Ω)+kfSekL1(Ω)≤2kf−fSekL1(Ω)

which together with (5.8) implies (5.9).

It is known that (5.9) implies the improved Poincar´e in L2. In fact, this was
proved in a more general context in [11, Prop. 3.3]. In the following lemma we
reproduce, for a particular case, the argument given in that paper but tracing
constants in the proof in order to obtain an explicit dependence of the constant in
theL2 _{estimate in terms of the constant in (}_{5.9}_{).}

Lemma 5.2. _{Let}_{Ω}_{be an arbitrary bounded domain. If there exists a constant}_{C1}

such that, for any f ∈W1,1_{(Ω)} _{with} R

Ωf = 0,

kfkL1_{(Ω)}≤C1kd∇fk_{L}1(Ω), (5.10)

then, for anyf ∈H1_{(Ω)}_{∩}_{L}2
0(Ω),
kfkL2_{(Ω)}≤6

√

2C1kd∇fkL2(Ω). (5.11)

Proof._{The first step in the proof given in [}_{11}_{] is to show that for any measurable}

E⊂Ω such that|E| ≥ |Ω|/2 and anyf ∈W1,1_{(Ω) which vanishes on}_{E, it follows}
from (5.10) that

kfkL1_{(Ω)}≤3C1kd∇fk_{L}1(Ω). (5.12)

Indeed, callingfΩthe average off over Ω and using thatf vanishes onE, we have

|fΩ|= 1 |E|

Z

E

(fΩ−f)
≤_{|}E1|

Z

E|

fΩ−f|,

and therefore,

kfΩkL1_{(Ω)}≤ |

Ω|

|E|kfΩ−fkL1(Ω). Then,

kfkL1_{(Ω)}≤ kf −fΩk_{L}1_{(Ω)}+kfΩk_{L}1_{(Ω)}≤ kf−fΩk_{L}1_{(Ω)}+ |Ω|

and using that|E| ≥ |Ω|/2 we obtain

kfkL1_{(Ω)}≤3kf−fΩk_{L}1(Ω),

which together with (5.10) applied tof−fΩgives (5.12).

Now, for E ⊂Ω as above and f ∈H1_{(Ω) vanishing on}_{E} _{we can apply (}_{5.12}_{)}
forf2 _{to obtain,}

Z

Ω

f2_{≤}_{3}_{C1}Z
Ω

d|∇f2_{| ≤}_{6}_{C1}Z
Ω

d|f||∇f| ≤6C1kfkL2(Ω)kd∇fk_{L}2(Ω),

and therefore,

kfkL2_{(Ω)}≤6C1kd∇fk_{L}2(Ω). (5.13)

For a functionf we definef+as the function which agrees withf wheref ≥0 and is equal to zero otherwise, andf−:=f+−f. Now, givenf ∈H1(Ω), it is easy to

see by continuity arguments, that there existsλ∈Rsuch that Z

Ω

(f−λ)2+= Z

Ω

(f−λ)2−. (5.14)

On the other hand one of the two functions (f−λ)+ or (f−λ)− vanishes in a set

E such that |E| ≥ |Ω|/2, suppose that it is (f −λ)+ (if not we apply the same argument to the other function). Then, using (5.13) applied to (f−λ)+we obtain

k(f−λ)+kL2_{(Ω)}≤6C1kd∇fk_{L}2(Ω),

but, in view of (5.14) we have Z

Ω

(f−λ)2= Z

Ω

(f−λ)2++ Z

Ω

(f−λ)2−= 2

Z

Ω

(f−λ)2+≤72C1k2 d∇fk2L2_{(Ω)},

and the proof concludes by recalling that, forf ∈L2
0(Ω),
kfkL2_{(Ω)}≤ kf −λk_{L}2(Ω).

Theorem 5.2. _{Let} _{Ω} _{⊂} _{R}n _{be a bounded convex domain of diameter} _{R} _{which}

contains a ball B of radius ρ. Then, given f ∈ L2

0(Ω), there exists a solution

u_{∈}_{H}_{0}1_{(Ω)}n _{of (}_{1.1}_{) such that}

kDu_{k}

L2_{(Ω)}≤Cn R

ρ kfkL2(Ω), whereCn is a constant which depends only onn.

Proof._{The result follows immediately from Theorem} _{5.1}_{, inequality (}_{5.9}_{), and}

Lemma5.2.

Now, a natural question is whether the converse of Theorem5.1is true, namely, if the improved Poincar´e inequality can be proved assuming the existence of contin-uous right inverses of the divergence. To the author’s knowledge this is not known. However, a weaker result will allow us to obtain an estimate for the constant in the improved Poincar´e inequality for planar star-shaped domains using the results of Section3. In fact, we will see that the converse can be proved if we assume that the following inequality is satisfied in Ω,

g_{d}_{L}2_{(Ω)}≤CH,Ωk∇gkL

This is one of the many results called “Hardy inequality” although, at least to the author’s knowledge, Hardy proved only the one dimensional case. It is known that this inequality is valid for a very large class of domains (see for example [15,19,22,23]).

Theorem 5.3. _{If} _{Ω}_{admits a right inverse of the divergence with constant} _{C}_{div,}_{Ω}

and the Hardy inequality (5.15) is satisfied in Ω, then the improved Poincar´e in-equality (5.1) is valid in Ω with a constantCiP,Ωsuch that

CiP,Ω≤CH,ΩCdiv,Ω. (5.16)

Proof._{Given}_{f} _{∈}_{H}1_{(Ω)}_{∩}_{L}_{0}2_{(Ω) let}u_{∈}_{H}_{0}1_{(Ω)}n _{be such that}

divu_{=}_{f} _{and} _{k}_{D}u_{k}_{L}2_{(Ω)}≤C_{div,}Ωkfk_{L}2(Ω). (5.17)

Then,

kfk2L2_{(Ω)}=

Z

Ω

fdivu_{=}_{−}

Z

Ω∇

f ·u_{≤ k}_{d}_{∇}_{f}_{k}_{L}2_{(Ω)}

u_{d}

L2_{(Ω)}

≤CH,Ωkd∇fkL2(Ω)kDuk_{L}2_{(Ω)}

and using (5.17) we conclude the proof.

In order to apply this theorem together with our results of Section 3 we need to know estimates for CH,Ω. For example, for simply connected (in particular for

star-shaped) planar domains it has been proved that

CH,Ω≤4; (5.18)

see [2,3].

Therefore, using this estimate and the results of Section3, we obtain an estimate for the constantCiP which improves (5.2).

Theorem 5.4. _{Let} _{Ω} _{⊂} _{R}2 _{be a bounded domain of diameter} _{R} _{which is }

star-shaped with respect to a ball B ⊂ Ω of radius ρ. Then, there exists a positive
constant C such that, for allf ∈H1_{(Ω)}_{∩}_{L}2

0(Ω), we have
kfkL2_{(Ω)}≤C

_{R}

ρ

log
_{R}

ρ

kd∇fkL2(Ω).

Proof._{The result follows immediately from Theorems}_{3.2}_{and} _{5.3}_{and inequality}

(5.18).

To finish the paper let us mention that the bound given in the previous theorem is almost optimal. Indeed, in view of Theorem 5.1, the same example given in Section3 shows that in some casesCiP ≥c1(R/ρ), wherec1 is a constant.

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*Ricardo G. Dur´an*

Departamento de Matem´atica

Facultad de Ciencias Exactas y Naturales Universidad de Buenos Aires

and IMAS, CONICET, Buenos Aires, Argentina. rduran@dm.uba.ar