Estimations of the best constant involving the <span class="mathjax-formula">$L^2$</span> norm in Wente’s inequality and compact <span class="mathjax-formula">$H$</span>-surfaces in euclidean space

ESTIMATIONS OF THE BEST CONSTANT INVOLVING

THE

L

^{2 NORM IN WENTE'S INEQUALITY AND}

COMPACT

H

^{-SURFACES IN EUCLIDEAN SPACE}

GE YUXIN

Abstract. In the rst part of this paper, we study the best constant involving the ^L2 norm in Wente's inequality. We prove that this best constant is universal for any Riemannian surface with boundary, or re- spectively, for any Riemannian surface without boundary. The second part concerns the study of critical points of the associate energy func- tional, whose Euler equation corresponds to^H-surfaces. We will estab- lish the existence of a non-trivial critical point for a plan domain with small holes.

1. Introduction

Let be a smooth and bounded domain in ^R2. We denote

V

= ^f

a

²

H

¹()

a

⁶= constant^gand

V

⁰ =

V

^\

H

⁰¹(). Given two functions

a

,

b

²

V

, we denote by

'

the unique solution in

W

¹¹() of the Dirichlet problem

;4

'

=

a

^x

b

^y;

a

^y

b

^x

in

'

= 0

on

@

^(1.1)

where subscripts denote partial dierentiation with respect to coordinates.

By developing a previous work from H. Wente 22], H. Brezis and J.-M.

Coron 7] showed the following result:

Theorem 1.1. The solution

'

of equation (1.1) is a continuous function on and

'

²

H

¹(). Moreover there exists a constant

C

⁰() which depends only on such that

k

'

^kL1()+^kr

'

^kL2()

C

⁰()^kr

a

^kL2()kr

b

^kL2() (1.2) This result is sharp in the sense that since the right hand side of (1.1) is in

L

¹(), the classical theory of Calderon-Zygmund does provide estimates for

'

only in

L

^q() and

W

^1p() for

q <

¹and

p <

2. Note that equation (1.1) appears in many problems arising in physics and geometry, and Theorem 1.1 has many applications.

Later on, F. Bethuel and J.-M. Ghidaglia 5] proved that in fact one can nd a constant

C

⁰() which does not depend on . We are interested here in the optimal (i.e. smallest) value of this constant such that estimates analogous to (1.2) hold. To be more precise we denote by

C

¹() the best

Yuxin Ge, CMLA, CNRS URA-1611, ENS Cachan, 61 Av. du Pr esident Wilson, 94235 Cachan Cedex, France. E-mail: [email protected].

Received by the journal February 4, 1998. Revised February 13, 1998. Accepted for publication April 21, 1998.

c Soci et e de Math ematiques Appliqu ees et Industrielles. Typeset by L^ATEX.

constant involving the

L

-norm in the estimations and by

C

²() for the

L

²-norm, i.e.

C

¹() = sup

ab2V

k

'

^k1

kr

a

^k2kr

b

^k2

(1.3)

C

²() = sup

ab2V

kr

'

^k22

kr

a

^k22kr

b

^k22

:

(1.4) S. Baraket 3] obtained that

C

¹() = ²¹ for simply connected domain . This result has been recently extended to any domain by P. Topping 21].

Our aim in this paper is to study

C

²(). Thus we consider the following energy functional dened on

V V

E

(

ab

) = ^kr

'

^k22

kr

a

^k22kr

b

^k22

(1.5) where

a

,

b

²

V

, and

'

is given by (1.1).

In this paper, we will prove the following main results.

Theorem 1.2. Let be a smooth bounded domain in^R2. Then we have

C

²() = 316

:

Moreover, the best constant is achieved if and only if is simply connected.

Notice that the functional

E

(

ab

) is invariant under the action of con- formal dieomorphisms on the domain (see 15]). As a consequence we deduce that

C

²() and

C

¹() depend only on the conformal type of . Moreover it implies that the functional

E

makes sense on any Riemann sur- face (i.e. a surface equipped with a conformal structure) with or without boundary. In section 4, we prove generalizations of Theorem 1.2, namely Theorem 1.3. Let

M

be a Riemann surface with a non empty boundary,

then

C

²(

M

) = 316

and the maximum in (1.5) is achieved if and only if

M

is topologically a disc.

Theorem 1.4. Let

M

be a Riemann surface without boundary, then

C

²(

M

) = 332

and the maximum in (1.5) is achieved if and only if

M

is topologically a sphere.

An interesting observation, due to F. Helein 15], is that the study of

E

leads to a solution of the

H

-surface equation ^;4

u

=

u

^x

u

^y, satised by surfaces of constant mean curvature in ^R3 in conformal representation.

For this purpose, we will look for critical points of

E

. Note that direct variational approaches on that problem were developed in 7], 17] and 22].

In view of Theorem 1.2, we can not maximise the problem if is not simply connected. The major obstruction in proving the existence of a maximum comes from the fact that the norms^kr

a

^kL2 and ^kr

b

^kL2 are not continuous under weak convergence in

L

². Indeed, for any smooth bounded domain in plan, concentration phenomena occur in the maximizing sequence as shown

in section 7 of this paper. However, making use of a topological method, invented by J.-M. Coron 8], we establish the following result of existence.

Theorem 1.5. Let be the unit disc perfored with small holes. Then

E

admits a non trivial critical point.

This paper consists of two parts sections 2-5 are concerned with the estimations of the best constant involving the

L

²norm in Wente's inequality, the remainder is devoted to search of a critical point for

E

: a study of the compactness of minimizing sequences, of the Palais-Smale condition and some existence results through a topological argument.

Part ^A. Estimations of the best constant involving the

L

2 norm

2. Outline

In this part, we will study the energy functional

E

and estimate the value of

C

²(). Our approach is the following. In section 3, we will look for the Euler-Lagrange equation for critical points of the functional

E

(

ab

) on the \manifold" where ^kr

a

^k2 = ^kr

b

^k2 = 1. After a scaling which uses the Lagrange multiplier, we see that any critical point leads by a canonical way to a solution of the

H

-surface equation, that is, the equation satised by a conformal parameterization of a surface when its mean curvature is constant.

In section 4, we will calculate

C

²() in the case where is a smooth bounded domain in ^R2. With the help of the isoperimetric inequality, we will show that

C

²() = ¹⁶³ . If is a disc, it is easy to show also that this constant is achieved. The next question is to know whether the maximum of

E

is achieved for a multiply connected domain. This is an interesting problem related to surfaces of constant mean curvature. Recall that for a long time, we thought that there does not exist an immersion with constant mean curvature from torus into ^R3. In 1984, H. Wente has given a coun- terexample. In view of Euler equation, the torus of Wente gives rise to a critical point of our functional

E

on an annulus. Indeed, let = (

ab'

) be a critical point of

E

on an annulus, we construct a compact oriented Riemannian surface

M

= ^S@ ~ by sticking and a copy of , pro- vided with opposing orientation and dene a

C

¹ map ~ from

M

into ^R3 by ~ = on and ~ = (

ab

^;

'

) on ~. Would this map be conformal, then its image would be a torus of constant mean curvature. Conversely the torus of Wente corresponds to a critical point of our functional

E

on some annulus. Unfortunately, this surface can not be obtained by maximizing the energy functional

E

and Wente tori thus correspond to nonmaximizing critical points of

E

. We will prove this fact in section 5.

At end of this part, we will also generalize all these results on a compact manifold without boundary. An interesting fact is that

C

²(

M

) is also uni- versal and is just half of

C

²(). Furthermore, a maximal critical point on a domain in the plan gives rise to a maximal critical point on a compact manifold, by sticking.

ESAIM: Cocv, June 1998, Vol. 3, 263{300

3. The Euler-Lagrange Equation

Definition 3.1. A point (

ab

)²

V V

is critical for the energy functional

E

if it satises the following conditions:

(i) ^r

E

(

a

+

tb

+

s

)^j(st)=(00)= 0, for all

, ²

H

¹(),

(ii) if

^t: ^;! is a family of dieomorphisms, depending dierentiably on

t

, with

⁰=

id

, then we have

dt d

t=0

E

(

a

^t

b

^t

) = 0

:

We remark that

E

is invariant under a conformal transformation of and

E

(

ab

) =

E

(

ab

) for all

,

^2R. Hence, without loss of generality, we can assume that ^kr

a

^k2 =^kr

b

^k2 = 1.

Theorem 3.2. Assume that(

ab

)²

V V

is a critical point of

E

such that

'

⁶= 0. Then

(i)

Z

r

a

^r

b

= 0, (ii)

@a

@n

⁼

@b

@n

^{= 0 on}

@

where

n

= (

n

¹

n

²) is the normal vector on

@

, (iii) there exists

^2Rsuch that = (

a

¹

b

¹

'

¹) = (

ab

²

'

) satises:

8

<

:

;4

'

¹ =^f

a

¹

b

^1g

;4

a

¹ =^f

b

¹

'

^1g

;4

b

¹ =^f

'

¹

a

^1g

^(3.1)

where ^f

^g=

^x

^y;

^y

^x, (iv) is

C

¹ on,

(v) the Hopf dierential

!

=^h

@

^z

@

^zi is holomorphic, i.e.

@

^zh

@

^z

@

^zi= 0

:

Moreover, if we denote

t

=^;

n

²+

in

¹ the unit complex number tangent to

@

, we have

Im(

!t

²) = 0 on

@ :

(vi) If is simply connected, then the Hopf dierential vanishes:

h

@

^z

@

^zi= 0

where

@

^z _{= 12(}

@

^x;

i@

^y), which implies that is conformal, (vii) if is an annulus, then there exists

c

^2Rsuch that

h

@

^z

@

^zi=

c z

²

:

First we prove some technical lemma.

Lemma 3.3. (see 7] and also 22]). If

'

²

H

⁰¹()^\

L

¹() (resp.

'

²

H

⁰¹()),

a

²

H

¹()^\

L

¹() (resp.

a

²

H

¹()) and

b

²

H

¹() (resp.

b

²

W

¹¹()), then we have

Z

'

^f

ab

^g=

Z

a

^f

b'

^g

:

Proof. Assuming rst that

'

,

a

,

b

²

C

(), we have

Z

'

^f

ab

^g =

Z

'

(

a

^x

b

^y;

a

^y

b

^x)

=

Z

'

(

ab

^y)^x;(

ab

^x)^y]

:

Integrating by parts and using the fact

'

= 0 on

@

, we obtain

Z

'

^f

ab

^g=

Z

a

(

b

^x

'

^y;

b

^y

'

^x) =

Z

a

^f

b'

^g

:

Now, we consider

'

²

L

¹()^\

H

⁰¹(),

b

²

H

¹() and

a

²

L

¹()^\

H

¹().

We choose three suitable sequences of smooth functions ^f

'

^ngn2N,^f

a

^ngn2N and ^f

b

^ngn2Nsatisfying the following conditions:

'

^n;!

'

in

H

¹() and

'

^n;!

'

weakly

?

in

L

¹()

b

^n;!

b

in

H

¹()

a

^n;!

a

in

H

¹() and

a

^n;!

a

weakly

?

in

L

¹()

:

We state that

j Z

'

^f

ab

^{gj k}

'

^kL1kr

a

^k2kr

b

^k2

j Z

a

^f

b'

^{gj k}

a

^kL1kr

'

^k2kr

b

^k2

:

Passing to the limit in the inequality for

a

ⁿ,

b

ⁿ and

'

ⁿ, this completes the proof.

Lemma 3.4. (see5] and see also 22]). Let ²

H

¹(^R3) be a solution of equation (3

:

1) in the sense of distributions. Then ²

C

¹(^R3).

Proof. (of Theorem 3.2). Let

a

^t =

a

+

tb

,

b

^t =

b

. We denote by

'

^t the unique solution in

H

⁰¹() of equation (1.1). Obviously, we have

'

^t =

'

for all

t

^2Rand ^kr

a

^tk22=^kr

a

^k22+ 2

t

^Rr

a

^r

b

+

O

(

t

²). Then (i) follows from the denition of a critical point.

Given

a

^t=

a

+

t

,

b

^t=

b

with

²

C

¹(). We denote

the unique solution in

H

⁰¹() of equation (1.1) with

a

=

, that is,

;4

=^f

b

^g

in

= 0

on

@ :

It is clear thatZ

jr

'

^tj2=

Z

jr

'

^j2+ 2

t

^Z

r

'

^r

+

O

(

t

²)

:

By Lemma 3.3, ^Z

'

^f

b

^g=

Z

^f

b'

^g

:

Hence, we obtain

Z

jr

'

^tj2 =

Z

jr

'

^j2+ 2

t

^Z

'

(^;4

)+

O

(

t

²)

=

Z

jr

'

^j2+ 2

t

^Z

'

^f

b

^g+

O

(

t

²)

=

Z

jr

'

^j2+ 2

t

^Z

^f

b'

^g+

O

(

t

²)

:

ESAIM: Cocv, June 1998, Vol. 3, 263{300

On the other hand,

Z

jr

a

^tj2 =

Z

jr

a

^j2+ 2

t

^Z

r

a

^r

+

O

(

t

²)

:

Thus, we have

E

(

a

^t

b

^t

) = ^kr

'

^k22+ 2

t

^Z

^f

b'

^g+

O

(

t

²)

kr

a

^k22+ 2

t

^Z

r

a

^r

^kr

b

^k22+

O

(

t

²)

:

With the denition of critical point, we conclude that

Z

^f

b'

^g=^kr

'

^k22Z

r

a

^r

⁸

²

C

¹()

:

Performing analogous deformations for

b

, we obtain

Z

f

'a

^g=^kr

'

^k22Z

r

b

^r for any ²

C

¹()

:

In particular, if we set

, ²

C

⁰¹(), we deduce that

8

>

>

>

<

>

>

>

:

;4

a

= 1

kr

'

^k22f

b'

^g

;4

b

= 1

kr

'

^k22f

'a

^g

:

^(3.2)

In order to establish the property (ii), we put

, ²

C

¹(). Setting

= 1

=

^kr

'

^k2, the property (iii) is demonstrated.

In view of Lemma 3.4, is

C

¹on . To prove the regularity of

u

up to the boundary, x

x

²

@

. So there exists a conformal map

I

from

B

(

xr

)^\ onto

B

⁺ =

B

^\f

x >

0^g, where

B

is a unit disc. Without loss of generality, we can assume that is dened on

B

⁺. We dene the extensions of on

B

as follows:

f

'

¹(

xy

) =

'

¹(

xy

)

if

x

0

;

'

¹(^;

xy

)

if

x

0

a

e¹(

xy

) =

a

¹(

xy

)

if

x

0

a

¹(^;

xy

)

if

x

0

and

b

e¹(

xy

) =

b

¹(

xy

)

if

x

0

b

¹(^;

xy

)

if

x

0

:

Clearly, ~ is in

H

¹(

B

^R3). We will prove that ~ is also a solution of equation (3.1). Thus, by Lemma 3.4, we conclude that is

C

¹ on . Set

²

C

⁰ (

B

). From the properties (ii) and (iii), we have

Z

B

re

a

^1r

= ^Z

B+

r

a

^1r

+^Z

B;

re

a

^1r

= ^Z

B+

r

a

^1r

+^Z

B+

r

a

¹(

xy

)^r(

(^;

xy

))

=

Z

B+

f

b

¹

'

^1g

+

Z

B+

f

e

b

¹

'

^e1g(

xy

)

(^;

xy

)

=

Z

B+

f

b

¹

'

^1g

+

Z

B;

f

e

b

¹

'

^e1g(^;

xy

)

(

xy

)

=

Z

B f

e

b

¹

'

^e1g

:

i.e. ^;4e

a

¹=^fe

b

¹

'

^e1g.

With the same arguments, we deduce that

;4

e

b

¹=^f

'

^e1

^e

a

^1g

:

On the other hand, we have

Z

B

r

'

e^1r

=

Z

B+

r

'

^1r

+

Z

B;

r

'

e^1r

=

Z

B+

r

'

^1r

^;Z

B+

r

'

¹(

xy

)^r(

(^;

xy

))

= ^;

Z

B+

4

'

¹

+

Z

B+

4

'

¹(

xy

)

(^;

xy

)

=

Z

B+

f

a

¹

b

^1g

^;Z

B+

f

a

¹

b

^1g(

xy

)

(^;

xy

)

=

Z

B

fe

a

¹

^e

b

^1g

that is, ^;4

'

^e1 =^fe

a

¹

^e

b

^1g.

To prove the property (v), set

a

^t=

a

^t,

b

^t=

b

^t where

^t is a family of smooth dieomorphisms of . Suppose that

d

^t

dt

t=0

= (

X

¹

X

²). Clearly, (

X

¹

X

²)

n

= 0 on

@

where

n

is the normal vector on

@

. Moreover, we have

Z

(^;4

'

^t)

'

=

Z

(^f

ab

^g

^t)det(^r

^t)

'

=

Z

f

ab

^g(

'

^;t)

=^;

Z

4

'

(

'

^;t)

=

Z

r

'

^r(

'

^;t)

ESAIM: Cocv, June 1998, Vol. 3, 263{300

i.e.

r

'

^tr

'

=

r

'

^r(

'

^;t).

However, from Theorem 1.1, we get

;

12

Z

jr

'

^j2+

Z

r

'

^tr

'

_{= 12}^Z

jr

'

^tj2+

O

(

t

²)

;

12

Z

jr

'

^j2+

Z

r

'

^r(

'

^;t_{) = 12}^Z

jr(

'

^;t)^j2+

O

(

t

²)

:

Thus, we get ^Z

jr(

'

^;t)^j2 =^Z

jr

'

^tj2+

O

(

t

²)

:

This means that

E

(

a

^t

b

^t

) = ^kr(

'

^;t)^k22

kr

a

^tk22kr

b

^{tk22 +}

O

(

t

²)

:

On the other hand, it is easy to get the following relations:

d

(^kr

a

^tk22)

dt

t=0

= 2

Z

((

@

^x

a

)^2;(

@

^y

a

)²)(

@

^x

X

^1;

@

^y

X

²) + 2

@

^x

a@

^y

a

(

@

^y

X

¹+

@

^x

X

²)]

d

(^kr

b

^tk22)

dt

t=0

= 2

Z

((

@

^x

b

)^2;(

@

^y

b

)²)(

@

^x

X

^1;

@

^y

X

²) + 2

@

^x

b@

^y

b

(

@

^y

X

¹+

@

^x

X

²)]

d

(^kr(

'

^;t)^k22)

dt

t=0

= ^;2

Z

((

@

^x

'

)^2;(

@

^y

'

)²)(

@

^x

X

^1;

@

^y

X

²) + 2

@

^x

'@

^y

'

(

@

^y

X

¹+

@

^x

X

²)]

:

Thus, we get the equality

Z

h(

@

^x

'

)^2;(

@

^y

'

)²+^kr

'

^k22((

@

^x

a

)^2;(

@

^y

a

)²+ (

@

^x

b

)^2;(

@

^y

b

)²)ⁱ (

@

^x

X

^1;

@

^y

X

²)

+ 2

@

^x

'@

^y

'

+^kr

'

^k22(

@

^x

a@

^y

a

+

@

^x

b@

^y

b

)](

@

^y

X

¹+

@

^x

X

²) = 0 i.e.

Z

(^j

@

^xj2;j

@

^yj2)(

@

^x

X

^1;

@

^y

X

²) + 2

< @

^x

@

^y

>

(

@

^y

X

¹+

@

^x

X

²)

= 0

:

(3.3)

A convenient way to rewrite this equation is to set

!

=^j

@

^xj2;j

@

^yj2;2

i <

@

^x

@

^y

>

, and we obtain Re^Z

!@

^z(

X

¹+

iX

²)

dxdy

= 0

where

@

^z = ¹²(

@

^x +

i@

^y). In particular, if we put

X

¹+

iX

^{2 2}

C

⁰¹(), we deduce that

@

^z

!

= 0

i.e.

!

is holomorphic.

Now, if we use (

X

¹

X

²) such that (

X

¹

X

²) =

f

(^;

n

²

n

¹) on

@

, where

f

is an arbitrary continuous real-valued function on

@

, we obtain

0 = Re

Z

@

^z(

!

(

X

¹+

iX

²))

dxdy

= -Im

Z

@

!

2

ft

²

ds

thus Im(

!t

²) = 0 on

@

. The property (v) is proved.

If is a disc or an annulus, from (v), we obtain Im(

!z

²) = 0 on

@

. From the principle of maximum, we have Im(

!z

²) = 0 on since Im(

!z

²) is harmonic. So we deduce that there exists

c

^2Rsuch that

!z

² =

c

. In the case where is a disc, we have moreover

lim

z !0

!z

² = 0

:

So we conclude the properties (vi) and (vii).

Remark 3.5. If (

ab

)²

V

⁰

V

⁰ is a critical point of

E

in

V

⁰

V

⁰, then all the conclusions of Theorem except (ii) are also right.

Remark 3.6. We know that every plane domain of one connectivity can be mapped conformally onto some annulus (see Ahlfors 1]). Thus, we obtain a characterization of Hopf's dierential

!

. But for a multiply connected domain , the characterization of

!

is less simple.

4. Isoperimetric inequality

In the following denotes a smooth simply connected domain. For sim- plicity, we suppose that is a disc, that is, =

B

= ^f(

xy

)

=r <

1^g. We check easily that a stereographic representation of the upper hemi-sphere

(

ab'

) =

4

x

1 +

r

²

⁴

y

1 +

r

²

^2(1;

r

²) 1 +

r

²

veries all the properties of Theorem 3.2, i.e. is a critical point of

E

. It is just a maximum of

E

. More precisely, we have the following result.

Theorem 4.1. Let =

B

, then (i) sup

ab2V

E

(

ab

) = 316

and the map

x

1 +

r

²

y

_{1 +}

r

²

achieves the best constant,

(ii) sup

ab2V

0

E

(

ab

) = 332

and the best constant is not achieved in

V

⁰

V

⁰.

ESAIM: Cocv, June 1998, Vol. 3, 263{300

First, we will introduce the following notations. Given , !²

H

(^R ), we dene

h

!^iD =

Z

h^x

!^xi+^hy

!^yi=

Z

hr

^r!ⁱ

j^j2D = ^h

^iD

V

_{() = 13}^Z

(^{x y})

if ²

C

⁰(^R3)

L

() =

Z

q

f

'

¹

'

^2g2+^f

'

²

'

^3g2+^f

'

³

'

^1g2

where = (

'

¹

'

²

'

³)

(

ab'

)^V =

L

(!)

where ! = (

ab'

)

:

In the proof, we will make use of the following lemmas.

Lemma 4.2. (see22]) Let , !²

C

⁰(^R3)^\

H

¹(^R3) be two mappings such that ^j@ and!^j@ describe the same oriented Jordan curve

then

j

V

()^;

V

(!)^j2

L

() +

L

(!)]³

36

:

(4.1)

In fact, this Lemma is equivalent to the isoperimetric inequality.

Lemma 4.3. (see 23]). Let ²

C

⁰(^R3)^\

C

²(^R3)^\

H

⁰¹(^R3) be a solution of equation (3

:

1) then 0.

Proof. (of Theorem 4.1). Let

a

,

b

²

C

¹(^R3) and

'

be the corresponding solution of (1.1). By Lemma 3.3, we get

(

ab'

)^V =

Z

a

^f

b'

^g=

Z

b

^f

'a

^g=

Z

'

^f

ab

^g

:

Now the two vector functions

=

a

j

a

^jD

b

j

b

^jD

'

j

'

^jD

and ! =

a

j

a

^jD

b

j

b

^jD

^;

'

j

'

^jD

:

have the same boundary values. Noting that

V

() =^;

V

(!) and

L

() =

L

(!) 1

2^jj2D = 32

from Lemma 4.2, we obtain that

j

V

()^j2 3 16

:

Consequently,

kr

'

^k22=

Z

(^;4

'

)

'

=

Z

'

^f

ab

^g= (

ab'

)^V

r 3

16

^j

a

^jDj

b

^jDj

'

^jD

that is,

E

(

ab

) 3

16

. Then the density of

C

¹() into

H

¹() implies

that sup

ab2V

E

(

ab

) 3 16

:

On the other hand, it is easy to check that

E

x

1 +

r

²

y

_{1 +}

r

²

= 316

:

Hence, we deduce the property (i). Similarly, putting ! = 0, we get

E

(

ab

) 3

32

^{for all}

ab

²

H

⁰¹()

:

We set

a

^"1 =

"x

"

²+

r

^2,

b

^"1=

"y

"

²+

r

^2,

a

^"2 =

"x

1 +

"

^{2 and}

b

^"2 =

"y

1 +

"

^2. We claim that

kr

a

^"1k22 =^kr

b

^"1k22 =

^{Z "12}

0

(1 +

r

²)

dr

(1 +

r

)⁴

kr

a

^"2k22 =^kr

b

^"2k22 =

"

²

(1 +

"

²)²

where

r

=^p

x

²+

y

². Set

a

^"=

a

^"1;

a

^"2 and

b

^"=

b

^"1;

b

^"2. We denote by

'

^" the unique solution of equation (1

:

1). Then

'

^"can be written as follows:

'

^"=

"

^2;

r

²

8(

"

²+

r

²) ^;

"

^2;1

8(

"

²+ 1) +

^"

where

^" is the unique solution of the following equation

;4

^" =^;f

a

^"1

b

^"2g;f

a

^"2

b

^"1g+^f

a

^"2

b

^"2g

in

^" = 0

on

@ :

^(4.2)

Using Theorem 1.1, we have^kr

^"k22=

O

(

"

²). Hence,

kr

'

^"k22 =

16

Z 1

"

2

0

(1 +

rdr r

)⁴ +

O

(

"

)

:

It is easy to see that

E

(

a

^"

b

^"

)^;! 3

32

^as

"

^;!0

:

Finally, we obtain

sup

ab2V

0

E

(

ab

) = 332

:

Now we suppose that the best constant is achieved in the point (

ab

) ²

V

⁰

V

⁰. By Theorem 3.2, there exists

^2R such that (

ab

²

'

) sat- ises equation (3

:

1). From Lemma 4.2 and Theorem 1.1, (

ab

²

'

) ²

C

⁰(^R3)^\

C

¹(^R3). And, applying Lemma 4.3, we obtain (

ab

²

'

) = 0. Thus, this contradiction completes the proof.

Remark 4.4. Because of the isoperimetric inequality, we always have sup

ab2V

E

(

ab

) 3

16

^{and asupb2V0}

E

(

ab

) 3 32

for any multiply connected domain in^R2. Moreover in the light of 9], this theorem implies that the embedding of Hardy space^H1(^R2) into

H

^;1(^R2) is not compact. Indeed, let (

a

ⁿ

b

ⁿ) ²

V

⁰

V

⁰ be a maximizing sequence of

E

in

V

⁰

V

⁰. Clearly,^f

a

ⁿ

b

^ngis bounded in^H1(^R2), but it does not converge strongly in

H

^;1(^R2).

In the following, we consider a multiply connected domain . We set

m

() = sup

ab2V

0

E

(

ab

). The analogue of Theorem 4.1 is following result.

ESAIM: Cocv, June 1998, Vol. 3, 263{300

Theorem 4.5. Let, ¹ be two smooth bounded domains such that¹. Then

m

()

m

(¹). Moreover, we have

m

() = 332

:

^(4.3)

Furthermore, the best constant is not achieved in

V

⁰

V

⁰.

Proof. Let

a

,

b

be two functions in

H

⁰¹(). We dene a embedding of

H

⁰¹() into

H

⁰¹(¹) as follows, to any

²

H

⁰¹(), we associated

²

H

⁰¹(¹) such that

(

xy

) =

(

xy

)

if (

xy

)²

(

xy

) = 0

if (

xy

)²

= :

We dene a energy functional

E

¹ on

H

⁰¹(¹) by following:

E

¹( ) = 12

Z

1 jr j

2

; Z

1 f

a

b

^g

where ²

H

⁰¹(¹). We denote

'

¹ the unique solution of equation (1

:

1) in

H

⁰¹(¹), i.e.

;4

'

¹ =^f

a

b

^g

in ¹

'

¹ = 0

on

@

¹

:

^(4.4)

Recall that

'

¹ is the unique minimal point of functional

E

¹. Thus, we get

E

¹(

'

)

E

¹(

'

¹) where

'

is the unique solution of equation (1.1) in

H

⁰¹().

Therefore, we obtain that

E

¹(

'

_{) = 12}^Z

jr

'

^j2;Z

f

ab

^g

'

= 12

Z

jr

'

^j2;Z

(^;4

'

)

'

=^;1 2

Z

jr

'

^j2

:

Similarly,

E

¹(

'

¹) =^;1

2

Z

1

jr

'

^1j2. Consequently, we deduce that

kr

'

^k2L2()=^kr

'

^k2L2(1)kr

'

^1k2L2(1)

:

But, stating that^kr

a

^k2L2(1

)=^kr

a

^k2L2() and ^kr

b

^k2L2(1

)=^kr

b

^k2L2(), we conclude that

E

(

ab

)

E

(

a

b

¹)

that is,

m

()

m

(¹). Now we choose

B

(

z

⁰

r

⁰) =^f

z

^{2C jj}

z

^;

z

^0j

< r

^0g and

B

(

z

¹

r

¹) =^f

z

^{2C jj}

z

^;

z

^1j

< r

^1g such that

B

(

z

⁰

r

⁰)

B

(

z

¹

r

¹).

Thus, we obtain

323

⁼

m

(

B

(

z

⁰

r

⁰))

m

()

m

(

B

(

z

¹

r

¹)) = 332

:

Hence, (4.3) follows.

We suppose that the best constant is achieved in the point (

ab

)²

V

⁰

V

⁰. It is clear that

323

⁼

E

(

ab

)

E

(

a

bB

(

z

¹

r

¹)) 3 32

: