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Remarks on inequalities of Hardy-Sobolev Type

Journal of Inequalities and Applications20112011:132

https://doi.org/10.1186/1029-242X-2011-132

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

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 in n reads, for all f C 0 ( n ) and n ≥ 3,
n | f | 2 d x ( n - 2 ) 2 4 n f 2 | x | 2 d x .
(1.1)
The Sobolev inequality states that, for all f C 0 ( n ) and n ≥ 3,
n | f | 2 d x S n n | f | 2 * d x 2 2 * ,
(1.2)
where 2 * = 2 n n - 2 and S n = π n ( n - 2 ) ( Γ ( n 2 ) Γ ( n ) ) 2 n is the best constant (cf. [1, 2]). A result of Stubbe [3] states that for 0 δ < ( n - 2 ) 2 4 ,
n | f | 2 d x - δ n f 2 | x | 2 d x ( n - 2 ) 2 4 - δ n - 1 n ( n - 2 ) 2 4 n - 1 n S n n | f | 2 * d x 2 2 *
(1.3)
and the constant in (1.3) is sharp. Recently, Balinsky et al. [4] prove analogous inequalities for the operator L : = x . One of the results states that, for 0 ≤ δ < n2/4 and f C 0 ( n ) ,
n | L f | 2 d x - δ n f 2 d x C n 2 4 - δ n - 1 n S n n | r F | 2 * d x 2 2 * ,
(1.4)
where F(r) is the integral mean of f over the unit sphere S n - 1 , i.e.,
F ( r ) = 1 | S n - 1 | S n - 1 f ( r ω ) d ω ,

and | S n - 1 | = S n - 1 d ω = 2 π n 2 Γ ( n 2 ) . Here, we use the polar coordinates x = . 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 ( n ) and n ≥ 3. There holds, for 0 δ < n 2 4 ,
n | L f | 2 d x - δ n f 2 d x n 2 4 - δ n - 1 n ( n - 2 ) 2 4 n - 1 n S n n | r F ( r ) | 2 * d x 2 2 *
(1.5)

and the constant in (1.5) is sharp.

When δ = 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 A R := {x n : R-1 < |x| < R}, then
A R | L f | 2 d x - n 2 4 A R f 2 d x [ 2 ( n - 2 ) ln R ] - 2 ( n - 1 ) n S n A R | r F ( r ) | 2 * d x 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 0 s g ( t ) d t q s r - q d s p q C p , q 0 | g ( t ) | p d t ,
where
C p , q = ( q - r - 1 ) - p q r Γ ( q r ) Γ ( 1 r ) Γ ( ( q - 1 ) r ) r p q
is the sharp constant. Equality is attained for functions of the form
g ( t ) = c 1 ( c 2 s r + 1 ) - r + 1 r , c 1 > 0 , c 2 > 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 ( [ 0 , ) ) .

Lemma 2.2. Let f ( | x | ) C 0 ( n ) and n ≥ 3. There holds, for 0 δ < n 2 4 ,
n | L f | 2 d x - δ n f 2 d x n 2 4 - δ n - 1 n ( n - 2 ) 2 4 n - 1 n S n n | r F ( r ) | 2 * d x 2 2 *
(2.1)

and the constant in (2.1) is sharp.

Proof. We note if f is radial, then F(r) = f(r) and L f = r f ( r ) . Therefore, inequality (2.1) is equivalent to
0 | f ( r ) | 2 r n + 1 d r - δ 0 | f ( r ) | 2 r n - 1 d r n 2 4 - δ n - 1 n ( n - 2 ) 2 4 n - 1 n S n | S n - 1 | - 2 n 0 | f ( r ) | 2 * r 2 * + n - 1 d x 2 2 * .
(2.2)
Let 0 ≤ β < n/2 and set g(r) = r β f(r). Through integration by parts, we have that
0 | g ( r ) | 2 r n + 1 - 2 β d r = 0 | f ( r ) | 2 r n + 1 d r - β ( n - β ) 0 | f ( r ) | 2 r n - 1 d r .
(2.3)
Make the change of variables s = rn-2β,
0 | g ( r ) | 2 r n + 1 - 2 β d r = ( n - 2 β ) 0 s 2 g s 2 d s .
(2.4)
On the other hand, set h ( s ) = g s so that g = - s + h ( t ) d t , we have
0 s 2 g s 2 d s = 0 s 2 h 2 d s = 0 | w ( s ) | 2 d s ,
where w(s) = s-2h(s-1). By Bliss lemma,
0 | w ( s ) | 2 d s n n - 2 n - 2 n Γ ( n 2 ) Γ ( 1 + n 2 ) Γ ( n ) 2 n 0 0 s | w ( t ) | d t 2 * s 2 - 2 n n - 2 d s 2 2 * ,
i.e.,
0 s 2 g s 2 d s = 0 s 2 h 2 d s = 0 | w ( s ) | 2 d s n n - 2 n - 2 n Γ ( n 2 ) Γ ( 1 + n 2 ) Γ ( n ) 2 n 0 0 s | w ( t ) | d t 2 * s 2 - 2 n n - 2 d s 2 2 * = n n - 2 n - 2 n Γ ( n 2 ) Γ ( 1 + n 2 ) Γ ( n ) 2 n 0 s + | h ( t ) | d t 2 * s 2 n - 2 d s 2 2 * n n - 2 n - 2 n Γ ( n 2 ) Γ ( 1 + n 2 ) Γ ( n ) 2 n 0 g 2 * s 2 n - 2 d s 2 2 * .
(2.5)
Recall that s = rn-2βand g(r) = r β f(r),
0 g 2 * s 2 n - 2 d s = ( n - 2 β ) 0 ( r 1 - β g ) 2 * r n - 1 d r = ( n - 2 β ) 0 ( r f ) 2 * r n - 1 d r .
(2.6)
Therefore, by (2.3), (2.4), (2.5) and (2.6),
0 | f ( r ) | 2 r n + 1 d r - β ( n - β ) 0 | f ( r ) | 2 r n - 1 d r = ( n - 2 β ) 0 s 2 g s 2 d s ( n - 2 β ) 1 + 2 2 * n n - 2 n - 2 n Γ ( n 2 ) Γ ( 1 + n 2 ) Γ ( n ) 2 n 0 ( r f ) 2 * r n - 1 d r 2 2 * = ( n - 2 β ) 2 n - 2 n n n - 2 n - 2 n Γ ( n 2 ) Γ ( 1 + n 2 ) Γ ( n ) 2 n 0 ( r f ) 2 * r n - 1 d r 2 2 * .
Since | S n - 1 | = S n - 1 d ω = 2 π n 2 Γ ( n 2 ) and S n = π n ( n - 2 ) ( Γ ( n 2 ) Γ ( n ) ) 2 n , we have
n | L f | 2 d x - β ( n - β ) n f 2 d x = | S n - 1 | 0 | f ( r ) | 2 r n + 1 d r - β ( n - β ) | S n - 1 | 0 | f ( r ) | 2 r n - 1 d r | S n - 1 | ( n - 2 β ) 2 n - 2 n n n - 2 n - 2 n Γ ( n 2 ) Γ ( 1 + n 2 ) Γ ( n ) 2 n 0 ( r f ) 2 * r n - 1 d r 2 2 * = | S n - 1 | 2 n ( n - 2 β ) 2 n - 2 n n n - 2 n - 2 n Γ ( n 2 ) Γ ( 1 + n 2 ) Γ ( n ) 2 n n | r f ( r ) | 2 * d x 2 2 * = n - 2 β n - 2 2 n - 2 n S n n | r f ( r ) | 2 * d x 2 2 * .
Let β = n - n 2 - 4 δ 2 when 0 ≤ δ < n2/4. Then, 0 ≤ β < n/2 and δ = β (n - β). Therefore,
n | L f | 2 d x - δ n f 2 d x n 2 - 4 δ ( n - 2 ) 2 n - 1 n S n n | r f ( r ) | 2 * d x 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 ϕ k (σ) are the orthonormal eigenfunctions of the Laplace-Beltrami operator with responding eigenvalues
c k = k ( N + k - 2 ) , k 0 .
The functions g k (r) belong to C 0 ( n ) , satisfying g k (r) = O(r k ) and g k ( r ) = O ( r k - 1 ) as r → 0. By orthogonality,
F ( r ) = 1 | S n - 1 | S n - 1 f ( r ω ) d ω = g 0 ( r ) .
On the other hand,
L f ( x ) = k = 0 r ( g k ( r ) ϕ k ) r = k = 0 r g k ( r ) ϕ k ( σ ) .
Here, we use the radial derivative r = x | x | = L | x | . Therefore,
n | L f | 2 d x - δ n f 2 d x = k = 0 n r 2 | g k ( r ) | 2 d x - δ n g k 2 d x n r 2 | g 0 ( r ) | 2 d x - δ n g 0 2 d x = n r 2 | F ( r ) | 2 d x - δ n F ( r ) 2 d x
since
n | L u | 2 d x n 2 4 n u 2 d x
holds for all u C 0 ( n ) and L u = r u ( r ) if u is radial. By Lemma 2.2,
n r 2 | F ( r ) | 2 d x - δ n F ( r ) 2 d x n 2 4 - δ n - 1 n ( n - 2 ) 2 4 n - 1 n S n n | r F ( r ) | 2 * d x 2 2 *
Therefore,
n | L f | 2 d x - δ n f 2 d x n r 2 | F ( r ) | 2 d x - δ n F ( r ) 2 d x n 2 4 - δ n - 1 n ( n - 2 ) 2 4 n - 1 n S n n | r F ( r ) | 2 * d x 2 2 * .

The proof of Theorem 1.1 is completed.

Proof of Theorem 1.2. We denote by B R N the unit ball centered at zero.

Step 1. Assume f is radial and f C 0 ( B R ) . Then,
B R | L f | 2 d x - n 2 4 B R f 2 d x = B R | r f ( r ) | 2 d x - n 2 4 B R f 2 ( r ) d x = B R | ( r f ( r ) ) | 2 d x - ( n - 2 ) 2 4 B R ( r f ) 2 | x | 2 d x .
Therefore, by Theorem B in [7],
B R | ( r f ( r ) ) | 2 d x - ( n - 2 ) 2 4 B R ( r f ) 2 | x | 2 d x ( n - 2 ) - 2 ( n - 1 ) n S n B R X 1 2 ( n - 1 ) n - 2 a , | x | R | r f | 2 n n - 2 d x n - 2 n ,
where
X 1 ( a , s ) : = ( a - ln s ) - 1 , a > 0 , 0 < s 1 .
Thus,
B R | L f | 2 d x - n 2 4 B R f 2 d x = B R | r f ( r ) | 2 d x - n 2 4 B R f 2 ( r ) d x ( n - 2 ) - 2 ( n - 1 ) n S n B R X 1 2 ( n - 1 ) n - 2 a , | x | R | r f | 2 n n - 2 d x n - 2 n .
Step 2. Assume f is not radial and f C 0 ( B R ) . We extend f as zero outside B R . So f C 0 ( n ) . Decomposing f into spherical harmonics, we have
f = k = 0 f k : = k = 0 g k ( r ) ϕ k ( σ ) ,
where ϕ k (σ) are the orthonormal eigenfunctions of the Laplace-Beltrami operator with responding eigenvalues
c k = k ( N + k - 2 ) , k 0 .
The functions f k (r) belong to C 0 ( B R ) . By the proof of Theorem 1.1 and Step 1,
n | L f | 2 d x - n 2 4 n f 2 d x = k = 0 n r 2 | g k ( r ) | 2 d x - n 2 4 n g k 2 d x n r 2 | g 0 ( r ) | 2 d x - n 2 4 n g 0 2 d x = n r 2 | F ( r ) | 2 d x - n 2 4 n F ( r ) 2 d x ( n - 2 ) - 2 ( n - 1 ) n S n B R X 1 2 ( n - 1 ) n - 2 a , | x | R | r F | 2 n n - 2 d x n - 2 n .
Step 3. By Step 1 and Step 2, the following inequality holds for f C 0 ( B R )
n | L f | 2 d x - n 2 4 n f 2 d x ( n - 2 ) - 2 ( n - 1 ) n S n B R X 1 2 ( n - 1 ) n - 2 a , | x | R | r F | 2 n n - 2 d x n - 2 n .
We note if R-1 < |x| < R, then
X 1 2 ( N - 1 ) N - 2 a , | x | D = 1 a - ln | x | R 2 ( N - 1 ) N - 2 1 a + 2 ln R 2 ( N - 1 ) N - 2 .
Therefore, If f is supported in the annulus A R := {x n : R-1 < |x| < R}, then
A R | L f | 2 d x - n 2 4 A R f 2 d x [ ( n - 2 ) ( 2 ln R + a ) ] - 2 ( n - 1 ) n S n A R | r F ( r ) | 2 * d x 2 2 * .
Letting a → 0, we have
A R | L f | 2 d x - n 2 4 A R f 2 d x [ 2 ( n - 2 ) ln R ] - 2 ( n - 1 ) n S n A R | r F ( r ) | 2 * d x 2 2 * .

The proof of Theorem 2 is completed.

Declarations

Acknowledgements

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

Authors’ Affiliations

(1)
School of Mathematics and Statistics, Xiaogan University, Xiaogan, Hubei, 432000, People's Republic of China

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Copyright

© Xiao; licensee Springer. 2011

This article is published under license to BioMed Central Ltd. 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 the original work is properly cited.

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