Báo cáo hóa học: " Remarks on inequalities of Hardy-Sobolev Type Ying-Xiong Xiao" ppt

math@gmail.com School of Mathematics and Statistics, Xiaogan University, Xiaogan 432000 Hubei, People ’s Republic of China Abstract We obtain the sharp constants of some Hardy-Sobolev-ty

R E S E A R C H Open Access

Remarks on inequalities of Hardy-Sobolev Type

Ying-Xiong Xiao

Correspondence: yxxiao.

math@gmail.com

School of Mathematics and

Statistics, Xiaogan University,

Xiaogan 432000 Hubei, People ’s

Republic of China

Abstract

We obtain the sharp constants of some Hardy-Sobolev-type inequalities proved by Balinsky et al (Banach J Math Anal 2(2):94-106)

2000 Mathematics Subject Classification: Primary 26D10; 46E35

Keywords: Hardy inequality, Sobolev Inequality

1 Introduction Hardy inequality inℝn

reads, for all f ∈ C∞

0 (Rn

)and n≥ 3,



Rn

|∇f |2dx≥ (n− 2)2

4



Rn

f2

The Sobolev inequality states that, for all f ∈ C∞

0(Rn)and n≥ 3,



Rn

|∇f |2dx ≥ S n

⎛

⎝

Rn

|f |2∗dx

⎞

⎠

2

2∗

where2∗= 2n

n− 2andS n=πn(n − 2)(( n

2)/(n))

2

n is the best constant (cf [1,2]) A

result of Stubbe [3] states that for0≤ δ < (n− 2)2



Rn

|∇f |2dx − δ



Rn

f2

|x|2dx≥



(n− 2)2

n − 1 n



(n− 2)2 4

n − 1 n

S n

⎛

⎝

Rn

|f |2∗dx

⎞

⎠

2

2∗

(1:3)

and the constant in (1.3) is sharp Recently, Balinsky et al [4] prove analogous inequalities for the operatorL := x · ∇ One of the results states that, for 0 ≤ δ <n2

/4 and f ∈ C∞

0(Rn),



Rn

|Lf |2dx − δ



Rn

f2dx ≥ C

n2

4 − δ

n− 1

n

S n

⎛

⎝

Rn

|rF|2∗dx

⎞

⎠

2

2∗

© 2011 Xiao; licensee Springer This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided

where F(r) is the integral mean of f over the unit sphereSn−1, i.e.,

|Sn−1|



Sn−1

f (r ω)dω,

and|Sn−1| =Sn−1d ω = (n/2)2π n/2 Here, we use the polar coordinates x = rω The aim

of this note is to look for the sharp constant of inequality (1.4) To this end, we have:

Theorem 1.1 Let f ∈ C∞

0(Rn)and n≥ 3 There holds, for0≤ δ < n2

4,



Rn

|Lf |2dx − δ



Rn

f2dx≥

n2

4 − δ

n− 1

n



(n− 2)2 4

n − 1 n

S n

⎛

⎝

Rn

|rF(r)|2∗dx

⎞

⎠

2

2∗

(1:5)

and the constant in(1.5) is sharp

Whenδ = n2

/4, we have the following Theorem, which generalize the results of [4], Corollary 4.6

Theorem 1.2 If f is supported in the annulus AR:= {xÎ ℝn

: R-1< |x| <R}, then



A R

|Lf |2dx−n2

4



A R

f2dx ≥ [2(n − 2) ln R]−

2(n− 1)

⎛

⎝

A R

|rF(r)|2∗dx

⎞

⎠

2

2∗

2 The proofs

We first recall the Bliss lemma [5]:

Lemma 2.1 For s ≥ 0, q >p > 1 and r = q/p - 1,

⎛

⎝

∞

 0

s

 0

g(t)dt

q

s r −q ds

⎞

⎠

p/q

≤ C p,q

∞

 0

|g(t)| p dt,

where

C p,q = (q − r − 1) −p/q

(1/r)((q − 1)/r)

rp/q

is the sharp constant Equality is attained for functions of the form

g(t) = c1 (c2sr+ 1)−

r + 1

r , c1 > 0, c2 > 0.

Using the Bliss lemma, we can prove the Theorem 1.1 for the radial function f, i.e.,

f (x) = ˜f( |x|)for some ˜f ∈ C∞([0,∞))

Lemma 2.2 Let f ( |x|) ∈ C∞

0(Rn)and n≥ 3 There holds, for0≤ δ < n2

4,



Rn

|Lf |2dx − δ



Rn

f2dx≥

n2

4 − δ

n− 1

n



(n− 2)2 4

n − 1 n

S n

⎛

⎝

Rn

|rF(r)|2∗dx

⎞

⎠

2

2∗

(2:1)

and the constant in(2.1) is sharp

Proof We note if f is radial, then F(r) = f(r) andLf = rf(r) Therefore, inequality (2.1) is equivalent to

∞

 0

|f(r)|2r n+1 dr − δ

∞

 0

|f (r)|2r n−1dr

≥

n2

4 − δ

n− 1

n



(n− 2)2 4

n − 1 n

S n· |Sn−1|−

2

n

⎛

⎝

∞

 0

|f (r)|2∗r2∗+n−1dx

⎞

⎠

2

2∗

(2:2)

Let 0≤ b <n/2 and set g(r) = rbf(r) Through integration by parts, we have that

∞

 0

|g(r)|2r n+1 −2β dr =

∞

 0

|f(r)|2r n+1 dr − β(n − β)

∞

 0

|f (r)|2r n−1dr. (2:3)

Make the change of variables s = rn-2b,

∞

 0

|g(r)|2r n+1 −2β dr = (n − 2β)

∞

 0

s2

On the other hand, seth(s) = ∂g

∂s so thatg =−

+ ∞

s h(t)dt, we have

∞

 0

s2 ∂g ∂s 2ds =

∞

 0

s2h2ds =

∞

 0

|w(s)|2ds,

where w(s) = s-2h(s-1) By Bliss lemma,

∞



|w(s)|2ds≥

n

n− 2 n

(n/2)(1 + n/2)

⎛

⎜∞

s



|w(t)|dt

2∗

s

2− 2n

n − 2 ds

⎞

⎟

2

2∗ ,

∞



0

s2 ∂g ∂s 2ds =

∞



0

s2h2ds =

∞



0

|w(s)|2ds

≥

n

n− 2 n

(n/2)(1 + n/2)

⎛

⎜∞

0

s



0

|w(t)|dt

2 ∗

s

2− 2n

n − 2 ds

⎞

⎟

2

2∗

=

n

n− 2 n

(n/2)(1 + n/2)

⎛

⎜∞

0

+∞



s

|h(t)|dt

2∗

s

2

n − 2 ds

⎞

⎟

2

2∗

≥

n

n− 2 n

(n/2)(1 + n/2)

⎛

⎜∞

0

g 2∗

s

2

n − 2 ds

⎞

⎟

2

2∗ .

(2:5)

Recall that s = rn-2band g(r) = rbf(r),

∞

 0

g2∗s

2

n − 2 ds = (n − 2β)

∞

 0

(r1−β g)2∗r n−1dr = (n − 2β)

∞

 0

(rf )2∗r n−1dr. (2:6)

Therefore, by (2.3), (2.4), (2.5) and (2.6),

∞



0

|f(r)|2r n+1 dr − β(n − β)

∞



0

|f (r)|2r n−1dr

= (n − 2β)

∞



0

s2 ∂g ∂s 2ds

≥ (n − 2β)1+

2

2∗

n

n− 2 n

(n/2)(1 + n/2)

⎛

⎝

∞



0

(rf )2∗r n−1dr

⎞

⎠

2

2∗

= (n − 2β)

2n− 2

n

n

n− 2 n

(n/2)(1 + n/2)

⎛

⎝

∞



0

(rf )2∗r n−1dr

⎞

⎠

2

2∗ .

Since|Sn−1| =Sn−1dω = 2π n/2

(n/2)andS n=πn(n − 2)(( n

2)/(n))

2

n, we have



Rn

|Lf |2

dx − β(n − β)



Rn

f2dx

= |Sn−1 |

∞



0

|f(r)|2r n+1 dr − β(n − β)|Sn−1 |

∞



0

|f (r)|2r n−1dr

≥ |Sn−1| · (n − 2β)

2n− 2

n

n

(n/2)(1 + n/2)

⎛

⎝

∞



0

(rf )2∗r n−1dr

⎞

⎠

2

2∗

= |Sn−1 |

2

n (n − 2β)

2n− 2

n

n

(n/2)(1 + n/2)

⎛

⎝

Rn

|rf (r)|2 ∗

dx

⎞

⎠

2

2 ∗

=

n − 2β

⎛

⎝ |rf (r)|2 ∗

dx

⎞

⎠

2

2 ∗

.

Let β = n−

√

n2− 4δ

2 when 0≤ δ <n2

/4 Then, 0≤ b <n/2 and δ = b (n - b) There-fore,



Rn

|Lf |2dx − δ



Rn

f2dx≥

n2− 4δ (n− 2)2

n− 1

n

S n

⎛

⎝

Rn

|rf (r)|2∗dx

⎞

⎠

2

2∗

Inequality (2.1) follows

Now we can prove Theorem 1.1

Proof of Theorem 1.1 Decomposing f into spherical harmonics, we get (see e.g [6])

f =

∞



k=0

f k:=

∞



k=0

g k (r) φ k(σ ),

where jk(s) are the orthonormal eigenfunctions of the Laplace-Beltrami operator with responding eigenvalues

c k = k(N + k − 2), k ≥ 0.

The functions gk(r) belong toC∞0(Rn), satisfying gk(r) = O(rk) andgk (r) = O(r k−1)as r

® 0 By orthogonality,

|Sn−1|



Sn−1

f (rω)dω = g0 (r).

On the other hand,

Lf (x) =∞

k=0

r ∂(g k (r) φ k)

∞



k=0

rg k(r) φ k(σ ).

Here, we use the radial derivative ∂

∂r =

x· ∇

|x| =

L

|x| Therefore,



Rn

|Lf |2dx − δ



Rn

f2dx =

∞



k=0

⎛

⎝

Rn

r2|g

k (r)|2dx − δ



Rn

g k2dx

⎞

⎠

≥



Rn

r2|g0(r)|2dx − δ



Rn

g20dx =



Rn

r2|F(r)|2dx − δ



Rn F(r)2dx

since



Rn

|Lu|2dx≥n2

4



Rn

u2dx

holds for allu ∈ C∞

0(Rn)andLu = ru(r)if u is radial By Lemma 2.2,



Rn

r2|F(r)|2dx − δ



Rn F(r)2dx≥

n2

4 − δ

n− 1

n



(n− 2)2 4

n − 1 n

S n

⎛

⎝

Rn

|rF(r)|2∗dx

⎞

⎠

2

2∗

Therefore,



Rn

|Lf |2dx − δ



Rn

f2dx

≥

Rn

r2|F(r)|2dx − δ

Rn F(r)2dx

≥

n2

4 − δ

n− 1

n



(n− 2)2 4

n − 1 n

S n

⎛

⎝

Rn

|rF(r)|2∗dx

⎞

⎠

2

2∗

The proof of Theorem 1.1 is completed

Proof of Theorem 1.2 We denote by BR⊂ ℝN

the unit ball centered at zero

Step 1 Assume f is radial and f ∈ C∞

0 (B R) Then,



B R

|Lf |2dx−n2

4



B R

f2dx =



B R

|rf(r)|2dx−n2

4



B R

f2(r)dx

=



B R

|(rf (r))|2dx− (n− 2)2

4



B R

(rf )2

|x|2 dx.

Therefore, by Theorem B in [7],



B R

|(rf (r))|2dx−(n− 2)2

4



B R

(rf )2

|x|2 dx

≥ (n − 2)−

2(n− 1)

⎛

⎜

⎝



B R X

2(n− 1)

n− 2 1

a, |x|

R |rf |

2n

n − 2 dx

⎞

⎟

⎠

n− 2

n

,

where

X1 (a, s) := (a − ln s)−1, a > 0, 0 < s ≤ 1.



B R

|Lf |2dx−n2

4



B R

f2dx =



B R

|rf(r)|2dx−n2

4



B R

f2(r)dx

≥ (n − 2)−

2(n− 1)

⎛

⎜

⎝



B R X

2(n− 1)

n− 2 1

a, |x|

R |rf |

2n

n − 2 dx

⎞

⎟

⎠

n− 2

n

Step 2 Assume f is not radial and f ∈ C∞

0(B R) We extend f as zero outside BR So

f ∈ C∞

0 (Rn) Decomposing f into spherical harmonics, we have

f =

∞



k=0

f k:=

∞



k=0

g k (r) φ k(σ ),

where jk(s) are the orthonormal eigenfunctions of the Laplace-Beltrami operator with responding eigenvalues

c k = k(N + k − 2), k ≥ 0.

The functions fk(r) belong toC∞0(B R) By the proof of Theorem 1.1 and Step 1,



Rn

|Lf |2dx−n2

4



Rn

f2dx =

∞



k=0

⎛

⎝

Rn

r2|g

k (r)|2dx−n2

4



Rn

g2k dx

⎞

⎠

≥



Rn

r2|g

0(r)|2

dx−n2 4



Rn

g02dx =



Rn

r2|F(r)|2

dx− n2 4



Rn F(r)2dx

≥ (n − 2)−

2(n− 1)

⎛

⎜

⎝



B R X

2(n− 1)

n− 2 1

a, |x|

2n

n − 2 dx

⎞

⎟

⎠

n− 2

n

Step 3 By Step 1 and Step 2, the following inequality holds for f ∈ C∞

0(B R)



Rn

|Lf |2dx−n2

4



Rn

f2dx ≥ (n − 2)−

2(n− 1)

n S n

⎛

⎜

⎝



B R

X

2(n− 1)

n− 2

1

a, |x|

R |rF|

2n

n − 2 dx

⎞

⎟

⎠

n− 2

n

.

We note if R-1 < |x| <R, then

X

2(N− 1)

1

a, |x|

⎛

a− ln|x|

R

⎞

⎟

2(N− 1)

≥

1

a + 2 ln R

Therefore, If f is supported in the annulus AR:= {xÎ ℝn

: R-1< |x| <R}, then



A R

|L f|2dx−n2

4



A R

f2dx ≥ [(n − 2)(2 ln R + a)]−

2(n− 1)

⎛

⎝

A R

|rF(r)|2∗dx

⎞

⎠

2

2∗

Letting a® 0, we have



A R

|Lf |2dx−n2

4



A R

f2dx ≥ [2(n − 2) ln R]−

2(n− 1)

⎛

⎝

A R

|rF(r)|2∗dx

⎞

⎠

2

2∗

The proof of Theorem 2 is completed

Acknowledgements

The author thanks the referee for his/her careful reading and very useful comments that improved the final version of

this paper.

Authors ’ contributions

YX designed and performed all the steps of proof in this research and also wrote the paper All authors read and

approved the final manuscript.

Competing interests

The authors declare that they have no competing interests.

Received: 15 April 2011 Accepted: 5 December 2011 Published: 5 December 2011

References

1 Aubin, T: Probléme isopérimétric et espace de Sobolev J Differ Geom 11, 573 –598 (1976)

2 Talenti, G: Best constant in Sobolev inequality Ann Matem Pura Appl 110(4), 353 –372 (1976)

3 Stubbe, J: Bounds on the number of bound states for potentials with critical decay at infinity J Math Phys 31(5),

1177 –1180 (1990) doi:10.1063/1.528750

4 Balinsky, A, Evans, WD, Hundertmark, D, Lewis, RT: On inequalities of Hardy-Sobolev type Banach J Math Anal 2(2),

94 –106 (2008)

5 Bliss, G: An integral inequality J Lond Math Soc 5, 40 –46 (1930) doi:10.1112/jlms/s1-5.1.40

6 Tertikas, A, Zographopoulos, NB: Best constants in the Hardy-Rellich inequalities and related improvements Adv Math.

209, 407 –459 (2007) doi:10.1016/j.aim.2006.05.011

7 Adimurthi, , Filippas, S, Tertikas, A: On the best constant of Hardy-Sobolev inequalities Nonlinear Anal 70, 2826 –2833

(2009) doi:10.1016/j.na.2008.12.019

doi:10.1186/1029-242X-2011-132 Cite this article as: Xiao: Remarks on inequalities of Hardy-Sobolev Type Journal of Inequalities and Applications

2011 2011:132.

Submit your manuscript to a journal and benefi t from:

7 Convenient online submission

7 Rigorous peer review

7 Immediate publication on acceptance

7 Open access: articles freely available online

7 High visibility within the fi eld

7 Retaining the copyright to your article

Submit your next manuscript at 7 springeropen.com

2∗ ,

∞



0... function f, i.e.,

f (x) = ˜f( |x|)for some ˜f ∈ C∞([0,∞))

Lemma...

(1/r)((q − 1)/r)

rp/q

is the sharp constant Equality is attained for functions of the form

g(t) = c1 (c2sr+