Gazzola F., Grunau H.-C., Mitidieri E. - Hardy inequalities with optimal constants and remainder terms(2000)(17).pdf

7/25/2019 Gazzola F., Grunau H.-C., Mitidieri E. - Hardy inequalities with optimal constants and remainder terms(2000)(17).pdf

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Hardy inequalities with optimal constants and remainder terms

Filippo Gazzola

Dipartimento di Scienze T.A.Via Cavour 84

I-15100 Alessandria

Hans-Christoph Grunau

Fachgruppe MathematikUniversitat

D-95440 Bayreuth

Enzo Mitidieri

Dipartimento di Scienze MatematicheVia A. Valerio 12/1

Universita degli Studi di TriesteI-34100 Trieste

May 4, 2000

Abstract

We show that in the classical Hardy inequalities with optimal constants inW1,p0

(),W2,20

(),W2,2 W1,2

0 () and also in further higher order Sobolev spaces remainder terms may be added.

Here is any bounded domain. For the Hardy inequality inW1,p0

there is a further Lp-normprovided p 2, while for 1 < p < 2 we obtain a remainder term in Lq-norms with q < p. Inhigher order Sobolev spaces besides theL2-norm further singularly weighted L2-norms arise.

1 Introduction

Hardys inequality in dimensions n > 2

u W1,20 () :

|u|2 dx (n 2)2

4

u2

|x|2dx. (1)

is one of the really classical Sobolev embedding inequalities, see [H, HLP]. Here and in what follows, Rn is a bounded domain. Although we do not explicitly assume that 0 , we always havethis particularly interesting case in mind. Closely related and proven analogously is theLp-versionwe have

u W1,p0 () :

|u|p dx

n pp

p

|u|p

|x|pdx; (2)

here we assume n > p 1. With Rellichs inequality in dimensions n >4, see[R]:

u W2,20 () :

(u)2 dx n2(n 4)2

16

u2

|x|4dx (3)

e-mail: [email protected]@[email protected]

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and its generalizations to estimates for |x|

(u)2 dx, see Lemma 2 below, it is not difficult to

obtain Hardy inequalities in higher order Sobolev spacesWk,20 (), wheren >2k. See e.g. [DH, Mi].One has, ifk = 2m even

(

m

u)2

dx 2m=1

(n+ 4m 4)2

4

u2

|x|4mdx (4)

and ifk = 2m+ 1 is odd:

|mu|2 dx

2m+1=1

(n+ 4m+ 2 4)2

4

u2

|x|4m+2dx. (5)

All the constants given here cannot be enlarged. For a more extensive survey and bibliography andalso for historical remarks we refer again to [DH].

The need to have refined versions of Hardy inequalities like (1) became first obvious in thepaper [BV] of Brezis and Vazquez, where among others the Gelfand problem

u= exp(u), u 0, u W1,20 () (6)

was studied for positive parameters and in dimensions n > 2. There exists a >0, such thatfor 0 , this problem is solvable, while for > it is not. Further, for= 2(n 2) and =B the unit ball, one has the singular solution

using= 2 log |x| W1,20 (B).

The linearization of (6) around this singular solution leads to the Hardy-type operator

Llin= 2(n 2)

|x|2 .

This operator is studied in [BV] in order to find out among others whether or not the singularsolution is also extremal in the sense that it corresponds to . For this purpose Brezis andVazquez prove by means of a reduction of dimension-technique the following Hardy inequalitywith optimal constant and a remainder term:

u W1,20 () :

|u|2 dx (n 2)2

4

u2

|x|2dx+ 2

en||

2/n

u2 dx. (7)

Here n 2, and 2 denotes the first eigenvalue of the Laplace operator (the subscript 2 refers tothe p-Laplacian with p= 2, see Definition 1 below) in the two dimensional unit disk. Further enand || denote the n-dimensional Lebesgue measure of the unit ball B Rn and the domain

respectively. For further applications and variants of (7) we refer to [VZ].Remainder terms appear also in other Sobolev inequalities, and in the context of nonlineareigenvalue problems, see e.g. [BL, GG].

The goal of the present paper is to find remainder terms for all the Hardy inequalities (2), (3),(4), (5) mentioned above. In a future paper we also want to address applications of these refinedHardy inequalities.

In Section 2 the W1,p-case is studied, corresponding to the p-Laplace operator. For p 2,carefully exploiting the convexity properties of the function p, we show that an extra Lp-norm may be added. For p (1, 2) we find it somehow unexpected that we can show the existenceonly ofLq-remainder terms with 1 q < p.

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In Sections 3 and 4 we show that remainder terms

u2 d x , . . . ,

u2

|x|2k2dx

may be added in (4) and (5), in Section 3 under Navier boundary conditions 0 = u = u= . . . on and in Section 4 under Dirichlet conditions 0 = u = u= u= . . . on .As the class of functions is larger in Section 3, the constants in front of the remainder terms are

smaller than in Section 4. The constants which appear in Section 4 seem to be more natural, whenthe order k of the Sobolev space becomes large. On the other hand, the reduction to the radiallysymmetric case is more involved, because the symmetrization argument of Section 3 fails andbecause iterative procedures will not give largest possible constants. For this reason it may be alsoof interest in its own, how the result in Wk,20 (B) for radial functions is extended to nonsymmetricfunctions.

2 In W1,p

0 , 1< p < n

We are always restricted to 1 < p < , if not stated differently.

Definition 1. LetX := {v C1([0, 1]) : v(0) =v(1) = 0} \ {0} and

p= infX

10 r

p1|v(r)|p dr10 r

p1|v(r)|p dr.

Remark. For all p > 1 we have p> 0 and ifp N, then p is the first eigenvalue ofp in Rp.

Theorem 1. Let Rn be a bounded domain, p (1, n), and leten = |B1(0)| and||denote then-dimensional Lebesgue measure of the unit ball and of the domain resp.

(a) Ifp 2, then for everyu W1,p0 () we have

|u|p dx

n p

p

p

|u|p

|x|pdx+ p

en||

p/n

|u|p dx. (8)

(b) If1< p 0 such that foreveryu W1,p0 () there holds

|u|p dx

n pp

p

|u|p|x|p

dx +C

en||

(p/n)+(p/q)1

|u|q dxp/q

. (9)

Remarks.

(i) Let n 3. If we put q := np/(n+ 4 2p) in the case p < 2 the constant C in front ofthe Lq-norm in (9) may be chosen such that C(n,p,q) 2 as p 2, cf. (17) below: thisshows the continuity between (8) and (9) for p approaching 2. For p 1 our constantsC(n,p,q) 0 for any choice of admissible q.

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(ii) Ifp = 1 and = B1(0) the Hardy constant

n 1 = inf uW1,10 (B1(0))\{0}

|u| dx

|u||x|dx

is attained on any positive smooth and radially decreasing function u with u||x|=1 = 0.This means that for p = 1 no remainder term may be added and shows that necessarilyC(n,p,q) 0 as p 1.

(iii) For related inequalities with weights being the inverse of the distance from the boundary of we refer to [BM, MS].

To prove Theorem 1 we need the following elementary inequalities, which give a quantitative esti-mate from below for the convexity behaviour of power-like functions. Similar inequalities in a moregeneral context but with smaller constants were obtained in [L].

Lemma 1. Letp 1 and, be real numbers such that 0 and 0. Then

( )p +p p1 p

||p, ifp 2,

1

2p(p 1)

2

(+ ||)2p, if1 p 2.

Proof. By means of integration by parts or Taylors formula we find

( )p +p p1 p =p(p 1)2 1

0

(1 t) ( t )p2 dt. (10)

We assume first thatp 2. If 0 one has t t || and the remainder term in (10) becomes

p(p 1)2 10

(1 t) ( t )p2 dt p(p 1)||p 10

(1 t) tp2 dt= ||p.

For 0 we estimate t (1 t) ||and obtain

p(p 1)2 10

(1 t) ( t )p2 dt p(p 1)||p 10

(1 t)p1 = (p 1) ||p ||p.

Let now 1 p < 2, that means that p 2< 0. Here we estimate ( t )p2 (+ ||)p2, andthere follows:

p(p 1)2 10

(1 t) ( t )p2 dt p(p 1) 2

(+ ||)2p

10

(1 t) dt=1

2p(p 1)

2

(+ ||)2p.

Proof of Theorem 1. After (Schwarz) symmetrization and rescaling we may assume that is theunit ball, which we simply denote with B = B1(0), and that u W

1,p0 () is nonnegative, radially

symmetric and nonincreasing. We recall that symmetrization leaves Lq-norms of functions itself

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unchanged, increases Lp-norms with the singular weight |x|p (see e.g. [AL, Theorem 2.2]) anddecreases Lp-norms of the gradient (see [AL, Theorem 2.7]). By density we may further assumethat u is as smooth as needed.

So, in what follows we may write u(r), u(r) = ddru(r); it holds |u(x)|= |u(r)| with r = |x|

and we have to find a lower bound for

I :=

10

rn1|u(r)|p dr

n p

p

p 10

rnp1u(r)p dr.

Similarly as in [BV] a suitable transformation reduces the dimension and replaces not only thepartial integration, which is usually performed to show Hardy inequalities, but also uncovers aremainder term. We set

v(r) :=r(n/p)1 u(r), u(r) =r1(n/p) v(r), (11)

so that

u(r) = n p

p rn/p v(r) +r1(n/p) v(r).

Since u is radially non-increasing we have for r (0, 1]:

n p

p

v(r)

r v(r), (12)

and the expression Ito be estimated becomes

I=

10

rp1

n p

p

v(r)

r v(r)

pdr

n p

p

p 10

v(r)p

r dr.

By (12) we may apply Lemma 1 with

= n pp

v(r)r

and = v (r).

For p 2 we conclude by virtue ofv(0) =v(1) = 0:

I p

n p

p

p1 10

v(r)p1 v(r) dr+

10

rp1|v(r)|p dr

p

10

rp1v(r)p dr= p

10

rn1u(r)p dr

and (8) follows.The case 1< p < 2 requires greater effort. Again by Lemma 1 and observing

10 v

p1 v dr= 0,

it follows thatI

1

2p(p 1)

10

rp1|v(r)|2npp

v(r)r + |v

(r)|2pdr. (13)

In order to estimate this term from below we introduce a further regularizing factor r, where > 0 will be chosen below in dependence on n,p, q. Consequently we will miss the Lp-norm forthe remainder term. Application of Holders inequality yields

10

r+p1|v(r)|p dr

2/p

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=

1

0

rp(p1)/2|v(r)|pnpp

v(r)r + |v

(r)|(2p)p/2

r+(p1)(2p)/2

n p

p

v(r)

r + |v(r)|

(2p)p/2dr

2p

1

0rp1

|v

(r)|2

npp

v(r)r + |v

(r)|2pdr 1

0r2/(2p)rp1

n pp

v(r)r

+ |v(r)|p

dr 2pp

(14)

For the second integral we find by means of a Hardy inequality in dimension (p+), which isproved for radial functions as in any integer dimension:

10

r2/(2p)rp1

n p

p

v(r)

r + |v(r)|

pdr

2p1

n p

p

p 10

r+p1

v(r)

r

pdr+ 2p1 10

r+p1|v(r)|p dr

2p1n p

p

+ 1 1

0r+p1|v(r)|p dr

where we have also used the inequality (a+ b)p 2p1(ap +bp) (a, b 0) and the fact thatr2/(2p) < r for all r (0, 1). Combining this estimate with (14) and inserting the result into(13) we arrive at:

I1

2p(p 1)2(p1)(p2)/p

n p

p+ 1

(p2)/p 10

r+p1|v(r)|p dr. (15)

In order to estimate

10 r

+p1|v(r)|p dr from below we need a slightly different constant than inDefinition 1. For the same set Xand for p >1 we set

,p= infX

10 r

1|v(r)|p dr10 r

1|v(r)|p dr.

With help of this first eigenvalue of the radial p-Laplacian in dimension , we obtain from (15):

I1

2p(p 1)2(p1)(p2)/p

n p

p+ 1

(p2)/pp+,p

10

r+p1v(r)p dr. (16)

For q [1, p) we choose := n(p q)/(2q), apply again Holders inequality and substitute uaccording to (11):

10

rn1u(r)q dr

p/q=

10

r(+n1)q/pu(r)q

r(n1) pq

p q

p dr

p/q

2

n

(pq)/q 10

r+n1u(r)p dr=

2

n

(pq)/q 10

r+p1v(r)p dr.

We insert this estimate into (16) and obtain finally

B

|u|p dx

n p

p

p B

|u|p

|x|pdx +C(n,p,q)

B

|u|q dx

p/q

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with

C(n,p,q) =1

2p(p1)2(p1)(p2)/p

1

2en

(pq)/q 2q(n p)n(p q)

p+ 1

(p2)/pp+n(pq)/(2q),p. (17)

With the choice q= pn/(n+ 4 2p) this constant tends to 2 as p 2 as stated in the remarkabove, because 2,p depends continuously on p.

Remark. We have to leave open the question, whether in the case 1 < p < 2 in Theorem 1 theLq-remainder term may be replaced with a suitable multiple of the Lp-norm. However, we havethe following proposition showing that the result forp 2 cannot hold equally for any smaller p.

Proposition 1. Let =B Rn (n 2)be the unit ball. For anyp (1, 2)there is a nonnegativeradially nonincreasing functionup W

1,p0 (B) such that

B

|up|p dx 0 such that we have on [r0, r1] uniformly in < :

n p

p v(r)

r v(r)p +p

n pp

p1

v(r)r

p1 v(r)

n pp

p

v(r)r

p |v(r)|

p .

On the whole interval [0, 1] the l.h.s. of this expression is at least nonpositive. We conclude thatwe have a positive constant c0= c0(p, r0, r1) such that for all < :

I

r1r0

rp1 dr p

n p

p

p1 10

v(r)v(r)p1 dr+

10

rp1|v(r)|p dr

c0

n p

p

p1(v(1)

p v(0)p) +

10

rp1|v(r)|p dr

+p 1

0

rp1 |v(r)|p1 + |v(r)|p1 |v(r) v(r)| dr c0+ p

10

rp1|v(r)|p dr+C

10

rp1 (1 rp) dr

+C

10

rp1

1 +p1r(1)(p1)

(1 r) +r1

dr

c0+ p

10

rn1|u(r)|p dr+ O

p1

< p

10

rn1|u(r)|p dr

for small enough; up(x) :=u(|x|) satisfies the stated reverse inequality.

Remark. LetX := {v C1([0, 1]) : v(0) =v(1) = 0} \ {0} and

p= infX

10 r|v

(r)|p dr10 r|v(r)|

p dr,

then by choosing = 2 p in (15) we obtainB

|u|p

n p

p

p B

|u|p

|x|p+

1

2p(p 1)2(p1)(p2)/p

n p

2 p

p+ 1

(p2)/pp

B

|x|2p|u|p

for all radially symmetric u W1,p0 (B). As p 2 this inequality tends to (8) with p= 2 and thisshows further continuity as p approaches 2.

3 In higher order Sobolev spaces under Navier boundary condi-

tions

Also here a similar reduction of dimension-technique as in [BV] and as in the preceding sectionwill be employed. In order to find the largest possible constants also for the remainder terms weneed a notation for first eigenvalues of the Laplacian in appropriate space dimensions.

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Definition 2. LetX= {v C1([0, 1]) : v(0) =v(1) = 0, v 0} and forn N let

(n) = infX

10 r

n1|v(r)|2 dr

10 r

n1v(r)2 dr.

This notation should not be confused with that of Section 2.

Proposition 2. Let B Rn be the unit ball. The first eigenvalue 0 of the following Navierboundary value problem

2u= u inB,u= u= 0 onB,

satisfies

(n)2 =0 = inf uW2,2W1,20 (B)\{0}

B(u)

2 dx

Bu2 dx .

Proof. The first equality is due to the fact that the biharmonic operator u 2u under Navierboundary conditions u = u = 0 on B is actually the square of the Laplacian under Dirichletconditions. For the second equality, see e.g. [V, Lemma B3].

Theorem 2. Let Rn,n 4, be a bounded domain and denote as above by || itsn-dimensionalLebesgue measure and byen = |B|, where B = B1(0) is the unit ball. Then for any u W

2,2 W1,20 () one has:

(u)2 dx n2(n 4)216

u2|x|4

dx

+n(n 4)

2 (2)

en||

2/n

u2

|x|2dx + (4)2

en||

4/n

u2 dx. (18)

Proof. Step 1. By scaling it suffices to consider || = en. Further we want to show first thatwe may restrict ourselves to = B and radial functions u(r). For this purpose we assume that(18) has already been shown in the radial setting, and let now be arbitrary with || = en andu W2,2 W1,20 (). We define f= u, let w W

2,2 W1,20 (B) be a (radial) strong solution of

w= f

in B,w= 0 on B,

where f 0 denotes the Schwarz symmetrization off. By [T, Theorem 1] we know thatw u 0, so that

(u)2 dx =

f2 dx=

B

(f)2 dx=

B

(w)2 dxB

w2

|x|2kdx

B

(u)2

|x|2k dx

u2

|x|2kdx, k= 0, 1, 2;

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for the last inequality we again refer to [AL, Theorem 2.2]. Assuming (18) in the radial setting,the same follows for any domain and any u W2,2 W1,20 ().

Step 2. We assume now that = B and thatu is radially symmetric: u= u(r),r = |x|. Similarlyas in (11) we introduce the transformation

v(r) :=r(n/2)2u(r), u(r) =r2(n/2)v(r),

so that

u(r) =r2(n/2)

v +3

rv

1

4n(n 4)rn/2v,

and the 4-dimensional Laplacian of v arises. In the following integration by parts only theboundary datumu(1) =v(1) = 0 is needed: 1

0rn1 (u)2 dr

n2(n 4)2

16

10

rn5u2 dr

= 1

0

r3v +3r

v2

dr+1

2

n(n 4) 1

0

rv 1r

v v dr n(n 4) 1

0

v v dr

=

10

r3 (4v)2 dr+

1

2n(n 4)

10

r (2v) v dr,

where 4v and 2v denote the radial Laplace operator in 4 and 2 dimensions respectively. Ac-cording to Proposition 2 we proceed with 1

0rn1 (u)2 dr

n2(n 4)2

16

10

rn5u2 dr

(4)2 10

r3 v2 dr+1

2n(n 4) (2)

10

r v2 dr

= (4)2 10

rn1 u2 dr+1

2 n(n 4)(2) 10

rn1u2

r2 dr,

thereby completing the proof of Theorem 2.

Combining this result with the original extended Hardy inequality of Brezis-Vazquez (7) alreadymentioned in the introduction, we have for third order Sobolev spaces:

Theorem 3. Let Rn, n 6, be a bounded domain. Then for anyu W3,2 W1,20 () with

u= 0 on, i.e. u W1,20 (), one has:

|u|2 dx (n+ 2)2(n 2)2(n 6)2

64

u2|x|6

dx

+1

16

3(n 2)4 + 8(n 2)2 + 16

(2)

en||

2/n

u2

|x|4dx

+

(n 2)2

4 (4)2 +

n(n 4)

2 (2)2

en||

4/n

u2

|x|2dx

+(2)(4)2

en||

6/n

u2 dx.

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Proof. First we may reduce to =B and the radial setting by combining the argument from theproof of Theorem 2 and [AL, Theorem 2.7].

The application of the Brezis-Vazquez inequality (7) reduces the order of terms by one:

10 r

n1

|u|

2

dr

(n 2)2

4 10 r

n3

(u)

2

dr+ (2) 10 r

n1

(u)

2

dr.

The second term will be estimated directly with help of Theorem 2, while for the singular term

10

rn3 (u)2 dr

we have to extend that proof by means of the transformation

v(r) :=r(n/2)3 u(r), u(r) =r3(n/2) v(r),

so that 10

rn3 (u)2 dr (n+ 2)2(n 6)2

16

10

rn7 u2 dr

=

10

r3

4v+2

rv2

dr (n+ 2)(n 6)

2

10

r

v +

5

rv

v dr

=

10

r3 (4v)2 dr+ 4

10

r2 (4v) v dr

+4

10

r

v2

dr+(n+ 2)(n 6)

2

10

r (2v)v dr

= 1

0

r3 (4v)2 dr+ 2 r

2v(r)21

0+

1

2 (n 2)2 + 8

1

0

r (2v) v dr

(4)2 10

r3v2 dr+1

2

(n 2)2 + 8

(2)

10

rv2 dr

= (4)2 10

rn3u2 dr+1

2

(n 2)2 + 8

(2)

10

rn5u2 dr.

Note that only u(1) =v(1) = 0 was needed. Collecting terms yields the stated inequality.

In order to iterate further we quote from [DH, Theorem 12], [Mi, Theorem 3.3]:

Lemma 2. Let Rn be a sufficiently smooth bounded domain and < n 4. Then for everyu C2() withu|= 0 we have

(u)2

|x| dx

(n 4 )2(n+)2

16

u2

|x|+4dx.

We emphasize that it is sufficient to have u|= 0.

With help of this result it is not difficult to see that there are remainder terms in Hardy inequalitiesof arbitrary order.

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Corollary 1. Let Rn be a sufficiently smooth domain and let k N, 2k n. Then thereexist constants c1, . . . , ck, depending only onk, n and ||, such that for every u W

2,k() withju|= 0 forj N0 and2j < k there holds:

(a) ifk even, k= 2m, m N:

(mu)2 dx 1

42m

2mj=1

(n+ 4(m j))2

u2

|x|4mdx +

2m=1

c

u2

|x|4m2dx;

(b) ifk odd, k= 2m+ 1, m N0:

|mu|2 dx 1

42m+1

2m+1

j=1(n+ 4m+ 2 4j)2

u2

|x|4m+2dx

+2m+1=1

c

u2

|x|4m+22dx.

The existence of some constants c1, . . . , ck as above is easily shown, of particular interest would betheir respective largest possible value. But already the proof of Theorem 3 indicates that iterativemethods under Navier boundary conditions 0 = u = u= . . .on yield constants which cannotbe easily given in closed form and which presumably will not be optimal.

One may hope that things will improve under the more restrictive Dirichlet conditions, i.e.

u W

k,2

0 (). This will be discussed in the following section.

4 In higher order Sobolev spaces under Dirichlet boundary con-

ditions

Our first goal will be to adapt Theorem 2 to Dirichlet conditions, i.e. to the space W2,20 () instead

ofW2,2 W1,20 (). At least when is a ball or close to a ball the estimation constants will beconsiderably larger. On the other hand the symmetrization argument in the proof of Theorem 2fails: the function w used in that proof is not in the class W2,20 (B) to be considered here. Weovercome this difficulty by a relatively involved argument, which bases on positivity properties of

Green functions and works only if is replaced with the circumscribedball.

As the technical difficulties to obtain best possible constants increase rapidly with the orderk of the space Wk,20 (), we give a complete discussion only for the improved Hardy inequality in

W2,20 (). For the general case we only give a conjecture to show the expected behaviour of theremainder terms, when the order of spaces become arbitrarily large.

In this section also eigenvalues of biharmonic and more general polyharmonic operators willplay an important role.

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Definition 3. LetB Rn denote again the unit ball and let

()k , n =

infWk,20 (B)\{0}

B(

mu)2 dxBu

2 dx , k= 2m, m N;

infWk,2

0 (B)\{0}

B |

m

u|

2

dxBu

2 dx , k= 2m+ 1, m N0.

Remark. The notation (n) of Section 3 is a special case by (n) = (() , n).

Taking into account Proposition 2 we immediately see

()2 , n

(, n)2. With help

of qualitative properties of first eigenfunctions in particular under Navier boundary conditions,also strict inequality can be shown. Moreover the ratio of these eigenvalues is rather large, by

elementary calculations one finds e.g.

()2 , 1

= 31.285243 . . .while (() , 1)2 =4/16 =

6.088068 . . .. Forn = 4, the case which is needed below, a very roughestimate according to [PS, p.

57] gives

()

2

, 4

j2

1j2

2 = 387.23 . . . while (() , 4)

2

=j4

1 = 215.56 . . ..

Theorem 4. Let Rn, n 4, be a bounded domain, BR(0). Then for everyu W2,20 ()

one has:

(u)2 dx n2(n 4)2

16

u2

|x|4dx

+n(n 4)

2 (, 2) R2

u2

|x|2dx +

()2, 4

R4

u2 dx. (19)

Proof. By trivial extension we have W2,2

0 () W2,2

0 (BR(0)), after suitable scaling we may assumethat =B .

Ifu is additionally radially symmetric, we proceed as in Step 2 of the proof of Theorem 2. In itslast conclusion we may instead exploit that u satisfies homogeneous Dirichlet boundary conditions

and replace (4)2 = (, 4)2 with

()2 , 4

, thereby proving (19) for radial u.

It remains to extend (19) to arbitrary functionsu W2,20 (B). The idea is that the formal Euler-Lagrange equation for a minimum problem associated with (19) and with the eigenvalue parameterin front of theBu

2 dx-term is linear and has only radial coefficients. Hence radialization of anysolution would give again a solution. That the latter solution is not the trivial one has to be avoidede.g. with starting with a positive solution. Hence a minimizer could be chosen radial. However,the mentioned eigenvalue problem is critical in so far, as the infimum may not be attained.

For that reason we consider for Na sequence of relaxed minimum problems:

:= inf W2,20 (B)\{0}

F(v)Bv

2 dx,

where

F(v) =

B

(v)2 dx n2(n 4)2

16

1

1

B

v2

|x|4dx

n(n 4)

2 (, 2)

1

1

B

v2

|x|2dx.

13


14/17

As we may already use Theorem 2 we have F(v) 1

B(v)

2 dx for every v W2,20 (B). That

means that for every minimizing sequencevk W2,20 (B) withBv

2kdx= 1 and limkF(vk) =,

we have boundedness of

ukW2,20 (B)

k

. After selecting a subsequence we may assume that

uk u W2,20 (B) and uk ustrongly in L

2(B). The bilinear form

F(v, w) :=

B

v w dx n2(n 4)2

16

1

1

B

v w

|x|4 dx

n(n 4)

2 (, 2)

1

1

B

v w

|x|2 dx

associated withF(v) defines a scalar product on W2,20 (B), which by

1

B

(v)2 dx F(v, v)

B

(v)2 dx= vW2,20 (B)

gives an equivalent norm. If we consider just for the following argument W2,2

0

(B) with F( . , . ) asscalar product, then lower semicontinuity of the corresponding norm in the weak topology gives

= lim infk

F(vk) F(v).

We have further

1 = limk

B

v2kdx=

B

v2 dx.

HenceF(v) =, andv W2,20 (B) is an optimal (nontrivial) function for . Consequently v is a

weak solution of the Euler-Lagrange equation

2v= 1 1

n

2(n 4)2

16

v

|x|4

+1 1 n(n 4)

2

(, 2) v

|x|2

+v in B ,

v= v= 0 on B. (20)

Next we show thatv is of fixed sign. Assume by contradiction that there are subsets B+, B Bboth with positive measure such that v > 0 on B+ and v < 0 on B. We may now apply adecomposition method explained in detail in [GG, Section 3], cf. also [Mo]. Let

K =

u W2,20 (B) : u 0

be the closed convex cone of nonnegative functions and

K = u W2,20 (B) : (u

, u) 0 for allu Kthe dual cone. Here (u, u) =

B(u

) (u) dx is the standard scalar product in W2,20 (B). K is

the cone of all weak subsolutions of the biharmonic equation in B under Dirichlet conditions. Bya comparison result of Boggio [Bo, p. 126] the elements ofK are nonpositive. We decompose

v= v1+v2, v1 K, v2 K, v1 v2.

As 0 v1 0, 0 v2 0, replacingv = v1+v2 with v1 v2 0 yields

F(v1+v2)> F(v1 v2),

B

(v1+v2)2 dx

B(w)

2 dx n2(n4)2

16

B

w2

|x|4dx n(n4)2 (, 2)B

w2

|x|2dx

Bw2 dx

()2, 4

.

To sum up, we have shown that for any N one has for every u W2,20 (B):

B

(u)2 dx 1 1 n2(n 4)2

16B

u2

|x|4dx

+

1

1

n(n 4)

2 (, 2)

B

u2

|x|2dx+

()2, 4

Bu2 dx.

Letting , the stated Hardy inequality (19) follows.

We conclude with some remarks concerning general higher order Sobolev spaces. We recall thatWk,20 () is equipped with the norm

u2Wk,20 ()

=

(mu)2 dx, ifk = 2m, m N;

|

m

u|

2

dx, ifk = 2m+ 1, m N0.

We expect the following result:

Conjecture 1. Let Rn be a bounded domain, BR(0). Letk N , n 2k. Then for all

u Wk,20 () there holds:

u2Wk,20 ()

k

j=0

1

4j

k

j

j=1

((n+ 2k 4)(n 2k 4 + 4))

()kj, 2k 2j

R2j2k

u2

|x|2jdx (21)

with the convention that(()0, 0) = 1.

In the radial setting the terms

()kj , 2k 2j

B

u2

|x|2jdx

originate from

nen

10

r2k2j1

(kj)/22k2j v2

dr

15


16/17

ifk j is even, and from

nen

10

r2k2j1

(kj1)/22k2j v2

dr

ifk j is odd. Here 2k2j is the radial Laplacian in dimension 2k 2j. The radial functionsuand v are related by means of the transformation

v(r) =r(n/2)ku(r), u(r) =rk(n/2)v(r).

In Conjecture 1 as well as in the proof of Theorem 4 we simply used for these terms eigenvalueestimates. One may also wish to use Hardy-like inequalities in order to have largest possibleconstants in front of the most singular remainder terms. The following Lemma may be applied,because we assume every function to satisfy homogeneous Dirichlet boundary conditions.

Lemma 3.

(a) Letk 2, j 1, v C2([0, 1]) withv(1) = 0. Then: 10

rk (jv)2 dr

(k+ 1 2j)2

4

10

rk2

v2

dr.

(b) Letk 2, v C1([0, 1]) withv(1) = 0. Then:

10

rk

v2

dr (k 1)2

4

10

rk2v2 dr.

To illustrate how the application of this elementary and well-known Lemma shifts less singularremainder terms to more singular ones, we modify the proof of Theorem 4 and obtain:

Corollary 2. Let Rn, n 4, be a bounded domain, BR(0). Then for everyu W2,20 ()

one has:

(u)2 dx n2(n 4)2

16

u2

|x|4dx

+1

2(n(n 4) + 8) (, 2) R2

u2

|x|2dx.

References

[AL] F. Almgren, E. Lieb, Symmetric decreasing rearrangement is sometimes continuous, J.Amer. Math. Soc. 2 (1989), 683-773

[Bo] T. Boggio, Sulle funzioni di Green dordinem,Rend. Circ. Mat. Palermo 20 (1905), 97-135.

[BL] H. Brezis, E. Lieb, Sobolev inequalities with remainder terms,J. Funct. Anal. 62 (1985),73-86.

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[BM] H. Brezis, M. Marcus, Hardys inequalities revisited, Ann. Sc. Norm. Sup. Pisa25 (1997),217-237

[BV] H. Brezis, J. L. Vazquez, Blow-up solutions of some nonlinear elliptic problems, Rev. Mat.Univ. Complutense Madrid10(1997), 443-469.

[DH] E. B. Davies, A. M. Hinz, Explicit constants for Rellich inequalities inLp(),Math. Z. 227(1998), 522-523.

[GG] F. Gazzola, H.-Ch. Grunau, Critical dimensions and higher order Sobolev inequalities withremainder terms, Nonlin. Diff. Eq. Appl. NoDEA, to appear.

[H] G. H. Hardy, Notes on some points in the integral calculus, Messenger Math. 48 (1919),107-112.

[HLP] G. H. Hardy, J. E. Littlewood, G. Polya,Inequalities, Cambridge: University Press, 1934.

[L] P. Lindqvist, On the equation div(|u|p2u) + |u|p2u= 0, Proc. Am. Math. Soc. 109

(1990), 157-164.

[MS] M. Marcus, I. Shafrir, An eigenvalue problem related to HardysLp inequality, preprint

[Mi] E. Mitidieri, A simple approach to Hardys inequalities, Mat. Zametki (2000), to appear,downloadable from: http://mathsun1.univ.trieste.it/mitidier/pubblicazioni.html.

[Mo] J. J. Moreau, Decomposition orthogonale dun espace hilbertien selon deux cones mutuelle-ment polaires, C. R. Acad. Sci. Paris255(1962), 238-240.

[PS] P. Pucci, J. Serrin, Critical exponents and critical dimensions for polyharmonic operators,J. Math. Pures Appl. 69(1990), 55-83.

[R] F. Rellich, Halbbeschrankte Differentialoperatoren hoherer Ordnung, in: J. C. H. Gerretsenet al. (eds.), Proceedings of the International Congress of Mathematicians Amsterdam 1954,Vol. III, pp. 243-250, Groningen: Nordhoff, 1956.

[T] G. Talenti, Elliptic equations and rearrangements,Ann. Scuola Norm. Sup. Pisa, Cl. Sci.,Ser. IV, 3 (1976), 697-718.

[VZ] J. L. Vazquez, E. Zuazua, The Hardy constant and the asymptotic behaviour of the heatequation with an inverse-square potential, J. Funct. Anal. 173(2000), 103-153.

[V] R. van der Vorst, Best constant for the embedding of the space H2 H10 () intoL2N/(N4)(), Diff. Int. Eq. 6(1993), 259-276.

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Gazzola F., Grunau H.-C., Mitidieri E. - Hardy inequalities with optimal constants and remainder terms(2000)(17).pdf