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A curl-free improvement of the Rellich–Hardy inequality with weight

  • Naoki Hamamoto ORCID logo and Futoshi Takahashi EMAIL logo
Published/Copyright: July 2, 2025

Abstract

We consider the sharpness of functional inequality which we call Rellich–Hardy inequality with power weight for curl-free vector fields on R N . This inequality can be considered as an intermediate between Hardy–Leray and Rellich–Leray inequalities and its best constant was originally found by Tertikas-Zographopoulos (“Best constants in the Hardy–Rellich inequalities and related improvements,” Adv. Math., vol. 209, no. 2, pp. 407–459, 2007) for unconstrained fields. Under the curl-free condition, we compute a new best value of the constant in the same inequality and show it is unattainable. This paper is a sequel to (N. Hamamoto and F. Takahashi, “Sharp Hardy–Leray and Rellich–Leray inequalities for curl-free vector fields,” Math. Ann., vol. 379, no. 1, pp. 719–742, 2021, N. Hamamoto and F. Takahashi, “Sharp Hardy-Leray inequality for curl-free fields with a remainder term,” J. Funct. Anal., vol. 280, no. 1, 2021, Art. no. 108790).

2020 Mathematics Subject Classification: Primary 26D10; Secondary 35A23

1 Introduction

Let N N be an integer with N ≥ 2, let γ R and put x = ( x 1 , , x N ) R N . In the following, the notation D γ ( R N ) denotes the set of smooth vector fields

u = ( u 1 , u 2 , , u N ) : R N x u ( x ) R N

with compact support on R N and with the integrability condition

R N | u | 2 | x | 2 γ 2 d x < .

1.1 Preceding results and motivation

It is well known that the Hardy–Leray inequality

(1) γ + N 2 1 2 R N | u | 2 | x | 2 | x | 2 γ d x R N | u | 2 | x | 2 γ d x

holds for all vector fields u D γ ( R N ) with the best constant on the left-hand side; when the integral on the right-hand side diverges (namely R N | u | 2 | x | 2 γ d x = ), we understand that inequality (1) holds in the sense that the left-hand this is less than ∞. Historically, this inequality was first proved by Leray [1] for (N, γ) = (3, 0), as a higher-dimensional extension of Hardy’s inequality in one dimension [2], see also the book by Ladyzhenskaya [3]. Costin and Maz’ya [4] improved the best value of the constant by assuming u to be divergence-free (under the additional assumption of axisymmetry for N ≥ 3): for the case N = 2, they showed that the inequality

C γ R 2 | u | 2 | x | 2 | x | 2 γ d x R 2 | u | 2 | x | 2 γ d x

holds with the best constant C γ = 3 + ( γ 1 ) 2 1 + ( γ 1 ) 2 γ 2 | γ + 1 | 3 γ 2 + 1 | γ + 1 | > 3 for divergence-free vector fields u D γ ( R 2 ) . Since there is an isometry on R 2 between divergence-free fields and curl-free fields, the same conclusion also applies to the two-dimensional curl-free fields. As a generalization of this result, we have derived in recent papers [5], [6] the Hardy–Leray inequality

(2) H N , γ R N | u | 2 | x | 2 | x | 2 γ d x R N | u | 2 | x | 2 γ d x

for curl-free fields u D γ ( R N ) with the best constant

(3) H N , γ = γ + N 2 1 2 3 ( N 1 ) + γ + N 2 2 2 N 1 + γ + N 2 2 2 if γ + N 2 N + 1 , γ + N 2 1 2 + N 1 otherwise.

Since C γ = H2,γ, this result recovers Costin-Maz’ya’s one for N = 2.

On the other hand, the Rellich–Leray inequality is given by

(4) B N , γ R N | u | 2 | x | 4 | x | 2 γ d x R N | u | 2 | x | 2 γ d x

for unconstrained fields u D γ 1 ( R N ) , where the constant BN,γ is sharp when

(5) B N , γ = min ν N { 0 } γ 1 2 ν + N 2 1 2 2 .

This was found by Rellich [7] for γ = 0 and Caldiroli-Musina [8] for γ ≠ 0. In recent papers [5], [6], we additionally considered the curl-free improvement of the Rellich–Leray inequality: if u D γ 1 ( R N ) is assumed to be curl-free, then the same inequality (4) holds with the best constant

(6) B N , γ = min ( γ 1 ) 2 N 2 4 2 , min ν N γ + N 2 1 2 + α ν γ + N 2 3 2 + α ν ( γ 2 ) 2 ν + N 2 1 2 2 ;

here and hereafter we use the notation

(7) α s = s ( s + N 2 )

for any s R .

In this paper, we are interested in another version of Rellich–Leray inequality:

(8) R N | u | 2 | x | 2 γ d x A N , γ R N | u | 2 | x | 2 | x | 2 γ d x , u D γ 1 ( R N )

holds with the best constant A N , γ = min ν N { 0 } A N , γ , ν , where

(9) A N , γ , ν γ N 2 2 for  ν = 0 , γ 1 2 ν + N 2 1 2 2 γ + N 2 2 2 + α ν for  ν N .

We call (8) the Rellich–Hardy inequality. This inequality was first found for N ≥ 5 by Tertikas-Zographopoulos [9], Theorem 1.7]. Subsequently, Beckner [10] and Ghoussoub-Moradifam [11] established the same inequality when N ∈ {3, 4} and γ = 0, with the best constants A 3,0 = 25 36 and A4,0 = 3. See also Cazacu [12] for the unified proof of the inequality when γ = 0.

Here let us consider the special case where γ satisfies

A N , γ = min ν N { 0 } A N , γ , ν = A N , γ , 0 = γ N 2 2 .

Then we see that a successive application of Rellich–Hardy and Hardy–Leray inequalities reproduces Rellich–Leray inequality (4): we have

R N | u | 2 | x | 2 γ d x A N , γ , 0 R N | u | 2 | x | 2 ( γ 1 ) d x from ( 1 ) A N , γ , 0 γ + N 2 2 2 R N | u | 2 | x | 2 | x | 2 ( γ 1 ) d x from ( 1 ) with γ replaced by γ 1 = γ 1 2 N 2 1 2 2 R N | u | 2 | x | 4 | x | 2 γ d x B N , γ R N | u | 2 | x | 4 | x | 2 γ d x

with BN,γ given by (5). Hence, Rellich–Hardy inequality can be considered as a stronger version of Rellich–Leray inequality and plays a role as an intermediate between Rellich–Leray and Hardy–Leray inequalities.

We have mentioned inequalities (1), (5) and (8) in the context of unconstrained vector fields; this is essentially the same as in the context of scalar fields in the sense that the optimal constants remain unchanged in both contexts. For further details of important contributions and developments on the optimality of these types of inequalities in the scalar case, we refer the reader to [13], [14], [15], [16], [17]. In the context of constrained vector fields such as curl-free fields, however, there seems less developments on the optimality of Rellich–Hardy type inequalities. Then we ask how the best contant in (8) will change when u is assumed to be curl-free; this is the main interest of our present problem.

As a side note, it is also worth mentioning that in a recent paper [18] the best constant of Rellich–Hardy inequality was computed for divergence-free (instead of curl-free) fields.

1.2 Results

Motivated by the observation above, we aim to derive the best constant in Rellich–Hardy inequality for curl-free fields. Now, our main result reads as follows:

Theorem 1.

Let N ≥ 2. Let u D γ 1 ( R N ) be a curl-free vector field. Then the inequality

(10) R N | u | 2 | x | 2 γ d x C N , γ R N | u | 2 | x | 2 | x | 2 γ d x

holds with the best constant C N,γ expressed as

(11) C N , γ = min ν N { 0 } C N , γ , ν ,

where

(12) C N , γ , 0 = ( γ 1 ) 2 N 2 4 2 γ + N 2 2 2 + N 1 = A N , γ , 1 , C N , γ , 1 = γ N 2 2 2 γ + N 2 1 2 + N 1 γ + N 2 3 2 + 3 ( N 1 )

and

C N , γ , ν = ( γ 2 ) 2 ν + N 2 1 2 2 γ + N 2 1 2 + α ν ( γ 2 ) 2 ν + N 2 1 2 2 + 2 ( γ 1 ) 2 γ + N 5 α ν + ( N 1 ) γ + N 2 3 2 f o r ν 2 .

Remark 2.

When | γ 1 | N 2 , the expression (11) has a simpler form: more precisely, one can check from a numerical calculation that

C N , γ = min C N , γ , k 1 , C N , γ , k k + N 2 γ k + 1 + N 2 min C N , γ , k + 3 , C N , γ , k + 2 k N 2 γ k + 1 N 2 min C N , γ , 3 , C N , γ , 0 N 2 γ 1 N 2

holds for every integer k ≥ 1.

In addition, we obtain a stronger inequality by adding a remainder term to the right-hand side of (10).

Theorem 3.

Let C N,γ be the same constant as in Theorem 1. Then there exists a constant number c > 0 which satisfies the inequality

(13) R N | u | 2 | x | 2 γ d x C N , γ R N | u | 2 | x | 2 | x | 2 γ d x c R N | x | 2 N 2 γ ( x ) | x | γ + N 2 2 u 2 | x | 2 γ 2 d x

for all curl-free fields u D γ 1 ( R N ) . Here the operation of x ⋅ ∇ on (vector- or scalar-valued) functions f means that ( x ) f = k = 1 N x k f x k .

Remark 4.

From Section 3.5 below, we know that the constant c in this theorem has the estimate

c 1 w h e n γ 1 , c 1 / 2 w h e n N 3 a n d γ > 1 , c 1 / 3 w h e n N = 2 a n d γ > 1 .

However, the best possible (namely the largest) value of c is unknown.

Remark 5.

If we further assume that u is compactly supported on a bounded domain Ω instead of R N , the remainder term is expected to have a simpler form; for example, we expect that there exist a positive-valued function W = W( x ) on Ω satisfying the inequality

Ω | u | 2 | x | 2 γ d x C N , γ Ω | u | 2 | x | 2 | x | 2 γ d x Ω W u 2 d x

for curl-free fields u D γ 1 R N with compact support on Ω; then we may ask what is the optimal form of the function W, which is also an unsolved problem in addition to the previous remark. In the case of scalar fields, however, there has been enormous development; in particular, the generalization to a notion of Bessel pairs has been established by [11]. For further details of recent developments on this topic, we refer the reader to [19], [20], [21], [22], [23], [24].

As a direct consequence of Theorem 3, we can conclude that the best constant CN,γ of the inequality (10) is never attained in D γ 1 ( R N ) \ { 0 } :

Corollary 6.

If the equation

R N | u | 2 | x | 2 γ d x = C N , γ R N | u | 2 | x | 2 | x | 2 γ d x

holds for a curl-free field u D γ 1 ( R N ) , then u 0.

Proof.

By assuming that both the equation in this corollary and inequality (13) in Theorem 3 hold true, we find that the right-hand side of (13) must vanish. Thus

r r r γ + N 2 2 u ( r σ ) = r γ + N 2 2 x 0

holds for some constant vector field x 0, where (r, σ ) = (| x |, x /| x |). Integrating both sides on any interval [ s , r ] R + with respect to the measure 1 r d r , we have

u ( r σ ) = s r γ + N 2 2 u ( s σ ) + s r γ + N 2 2 1 γ + N 2 2 x 0 ( γ 2 N 2 ) u ( s σ ) + x 0 log s r ( γ = 2 N 2 ) .

In the case γ 2 N 2 , take the limit s → ∞ (resp. s → +0) when γ < 2 N 2 (resp. γ > 2 N 2 ), then we obtain

u ( r σ ) 1 γ + N 2 2 x 0  and hence  u ( x ) u ( 0 ) .

This fact together with the integrability condition R N | u | 2 | x | 2 γ 4 d x < says that u (0) must vanish, whence u 0. In the case γ = 2 N 2 , taking s → 0 leads to

u ( r σ ) = u ( 0 ) + x 0 lim s 0 log s r

and the finiteness of the right-hand side yields x 0 = 0. Therefore, we see again that u u (0) and hence u 0.□

As another direct consequence, by expressing Theorem 3 in terms of scalar potentials of curl-free fields, we have the following fact:

Corollary 7.

Let C N,γ be the same constant as in Theorem 1. Then there exists a constant number c > 0 such that the inequality

R N | ϕ | 2 | x | 2 γ d x C N , γ R N | D 2 ϕ | 2 | x | 2 | x | 2 γ d x c R N | x | 2 N 2 γ ( x ) | x | γ + N 2 2 ϕ 2 | x | 2 γ 2 d x

holds for all scalar field ϕ such that ϕ D γ 1 ( R N ) . Here D2ϕ denotes a Hessian matrix of ϕ.

1.3 Overview of the remaining content of the present paper

The rest of this paper is organized as follows: Section 2 provides basic notations and definitions, and reviews a representation of curl-free fields. Section 3 gives the proof of Theorem 1: we recall from [6] the scalar-potential expression of L2 integrals of curl-free fields; after that, we derive Lemma 9 as a key tool for evaluating the ratio of the two integrals in (10), which also plays a computational part in the proof of Theorem 3. The proof of Lemma 9 is separated into two cases. Since both the cases use similar techniques and consist of long calculations, we prove only one case in the same section, and postpone the other case in Section 6. Section 4 proves Theorem 3 by using an operator-polynomial representation of Rellich–Hardy integral quotient and by making full use of Lemma 9. Section 5 observes curl-free improvement phenomena of best constants in some cases.

2 Preliminary for the proof of main theorem

2.1 Notations and definitions in vector calculus on R ̇ N

Here we summarize the minimum required notations and definitions for the proof of our main theorems. We basically use the notation

R ̇ N = R N \ { 0 } and S N 1 = x R N : | x | = 1 .

For every vector x R ̇ N , the notation

r = | x | > 0 , σ = x / | x | S N 1

denotes the radius of x and its unit-vector part, which defines the smooth transformation

R ̇ N R + × S N 1 , x ( r , σ )

together with its inverse

R + × S N 1 R ̇ N , ( r , σ ) r σ .

Every vector field u = ( u 1 , u 2 , , u N ) : R ̇ N R N has its radial scalar component u R = u R ( x ) and spherical vector part u S = u S ( x ) given by the formulae

u = σ u R + u S , σ u S = 0

for all x R ̇ N ; the two fields are explicitly given by

u R = σ u  and  u S = u σ u R .

In a similar way, the gradient operator = x 1 , , x N can be decomposed into the radial derivative r and the spherical gradient ∇ σ as

= σ r + 1 r σ ,

in order that r f = (∇f) R = σ ⋅ ∇f and 1 r σ f = ( f ) S hold for all f C ( R ̇ N ) . The radial derivative operator which we denote by

(14) : = r r = x

allows us to express the gradient operator by the formula

r = σ + σ .

The Laplace operator = k = 1 N 2 / x k 2 is known to be expressed in terms of (r, σ ) as

= 1 r N 1 r r N 1 r + 1 r 2 σ = r 2 + N 1 r r + 1 r 2 σ = 1 r 2 2 + ( N 2 ) + σ ,

where △ σ denotes the Laplace–Beltrami operator on S N 1 . We understand that the action of the operator r or on a vector field u is associated with the function r u (r σ ) for σ S N 1 fixed, whereas ∇ σ or △ σ is associated with the function σ u (r σ ) for r = | x | fixed. As a simple example, the operation of ∇ and △ on the scalar field r = | x | or its powers gives ∇r = σ and △r s = α s rs−2 for all s R , where α s is the same as in (7).

2.2 Radial-spherical-scalar representation of curl-free fields

Every vector field u C ( R N ) N is said to be curl-free if

u k x j = u j x k  on  R N j , k { 1 , , N } ,

or equivalently if there exists a scalar field ϕ C ( R N ) satisfying

(15) u = ϕ  on  R N .

In view of this equation, we say that u has a scalar potential ϕ. As another representation of curl-free fields, let us recall the following fact:

Proposition 8

([6]). Let λ R . Then a vector field u C ( R N ) N is curl-free if and only if there exist two scalar fields f , φ C ( R ̇ N ) satisfying

f i s r a d i a l l y s y m m e t r i c a n d S N 1 φ ( r σ ) d σ = 0 r > 0 , r 1 λ u = σ f + ( λ + ) φ + σ φ o n R N \ { 0 } .

Moreover, such f and φ are uniquely determined, and they are explicitly given by the equations

f = r λ ϕ ̄ a n d φ = r λ ϕ ϕ ̄ ,

where ϕ ̄ ( x ) = 1 | S N 1 | S N 1 ϕ ( | x | σ ) d σ denotes the spherical mean of the scalar potential ϕ in (15). In particular, if u has compact support on R ̇ N , then so do f and φ.

Later, we will use Proposition 8 by choosing λ = 2 N 2 γ . (See (21)).

3 Proof of Theorem 1

In this section, we prove Theorem 1. Roughly speaking, the proof consists of theoretical part (Sections 3.1 and 3.2) and computational part (from Sections 3.3 to 3.7). Since the theoretical part is already well established in our previous work, we will only state its minimum required content: we exploit some L2 formulae of curl-free fields given in [6]. Instead, emphasis is placed on the computational part.

3.1 Reduction to the case of compact support on R ̇ N

Let ϕ be a smooth scalar potential of a curl-free field u in D γ 1 R N satisfying ϕ(0) = 0. By considering Taylor expansion of ϕ( x ) at x = 0, one can check from the integrability condition R N | u | 2 | x | 2 γ 4 d x < that there exists an integer m > 2 N 2 γ satisfying

u ( x ) = ϕ ( x ) = O ( | x | m ) ,  and hence  ϕ ( x ) = O ( | x | m + 1 ) , u ( x ) = O ( | x | m 1 ) , u ( x ) = O ( | x | m 2 )

as x 0. Here and hereafter, for any scalar- or vector-valued functions f (x) and positive-valued functions g(x) of (vector or scalar) variable x, the notation “ f (x) = O(g(x)) as xa” means that

lim sup x a | f ( x ) | g ( x ) < .

Then it additionally follows that the integrals

(16) R N ϕ 2 | x | 2 γ 6 d x , R N | u | 2 | x | 2 γ 2 d x  and  R N | u | 2 | x | 2 γ d x

are all finite.

For the purpose of deriving the best constant CN,γ in inequality (10), it is enough to consider the case where the curl-free field u = ∇ϕ is compactly supported on R ̇ N . Here let us verify this fact. First of all let us define { u n } C c ( R ̇ N ) N as a sequence of curl-free fields by

u n ( x ) ζ 1 n log | x | ϕ ( x ) for every  n N ,

where ζ C ( R ) such that ζ ( t ) = 0  for  t 1 1  for  0 t . We use the abbreviations such as ζ n = ζ 1 n log | x | , ζ n = ζ 1 n log | x | , and ζ n = ζ 1 n log | x | . Notice that the asymptotic formulae

ζ n = 1 n ζ n = O ( n 1 ) , ζ n = σ n r ζ n = σ r O ( n 1 ) , ζ n = σ n r ζ n = σ r O ( n 1 )

hold as n → ∞. Hereafter, whenever we write such as

f = g O ( ξ ( n ) ) + h O ( η ( n ) )  as  n

for functions f , g and h of ( x , n) and positive-valued functions ξ and η of n, it means that there exist positive constant numbers C and n0 independent of ( x , n) and functions F and G of ( x , n) satisfying f = g F + h G and

F ( x , n ) C ξ ( n ) and G ( x , n ) C η ( n )

for all nn0. Then we have the following calculations:

u n = ζ n ϕ = ( ζ n ) ϕ + ζ n ϕ = σ ϕ r O ( 1 / n ) + ζ n u , u n = σ n r ζ n ϕ + ζ n u = σ n r ζ n ϕ + ζ n n σ ϕ r + ( ζ n ) u + ζ n u = σ σ ϕ r 2 O ( n 2 ) + ζ n n σ r ϕ + σ r ϕ + σ r u O ( n 1 ) + ζ n u = σ σ ϕ r 2 O ( n 1 ) + σ u r O ( n 1 ) + ζ n u where we abbreviate as  v w v w = v i w j i , j { 1 , , N } 2 the tensor product of two vector fields , u n = ζ n ϕ = ( ζ n ) ϕ + 2 ( r ζ n ) r ϕ + ζ n ϕ = 2 ζ n + ( N 2 ) ζ n r 2 ϕ + 2 ζ n r r ϕ + ζ n ϕ = n 2 ζ n + ( N 2 ) n 1 ζ n ϕ r 2 + 2 ζ n n r ϕ r + ζ n ϕ = n 2 ζ n + ( N 2 ) n 1 ζ n σ r O ( n 1 ) ϕ r 2 + n 2 ζ n + ( N 2 ) n 1 ζ n O ( n 1 ) ϕ r 2 + 2 ζ n n r ϕ r + 2 ζ n n r ϕ r + ( ζ n ) ϕ + ζ n ϕ = σ r O ( n 1 ) ϕ r 2 + O ( n 1 ) ϕ r 2 + σ r ϕ r 2 O ( n 2 ) + O ( n 1 ) r ϕ r + σ O ( n 1 ) ϕ r + ζ n ϕ = σ ϕ r 3 O ( n 1 ) + u r 2 O ( n 1 ) + σ u R r 2 O ( n 1 ) + r u r O ( n 1 ) + σ div u r O ( n 1 ) + ζ n u

hold as n → ∞. Therefore, taking the L2(| x |2γdx) integration yields

R N | u n | 2 | x | 4 | x | 2 γ d x = R N | ζ n u | 2 | x | 4 | x | 2 γ d x + O ( n 1 ) R N | u | 2 | x | 4 | x | 2 γ d x , R N | u n | 2 | x | 2 | x | 2 γ d x = R N | ζ n u | 2 | x | 2 | x | 2 γ d x + O ( n 1 ) R N | u | 2 | x | 2 | x | 2 γ d x , R N | u n | 2 | x | 2 γ d x = R N | ζ n u | 2 | x | 2 γ d x + O ( n 1 ) R N | u | 2 | x | 2 γ d x

with the aid of the integrability conditions (16). This fact shows that the two integrals in (10) can be approximated by curl-free fields with compact support on R ̇ N , as desired.

3.2 Radial- and spherical-scalar expression of the integrals

In the rest of the present section, we use the notation

t = log | x | = log r

which serves as an alternative radial coordinate obeying the differential rules

(17) t = r r = = x  and  d t = d r / r

and hence generates the same derivative operator given in (14). For any real number λ, let f and φ be the scalar fields determined by the curl-free field u , as given in Proposition 8, and we set

(18) v ( x ) = | x | 1 λ u ( x )

as a new vector field in C c ( R ̇ N ) N . Then the equation

(19) v = σ f + ( λ + ) φ + σ φ

holds on R ̇ N . Here we keep in mind that, by taking the derivative on both sides of (18) and of (19), one easily get

v = | x | 1 λ r λ 1 r 1 λ u , and v = σ f + ( λ + ) φ + σ φ ,

which are (in this order) the same equations (18) and (19) with v , u , f and φ replaced by v , u , ∂f and ∂φ respectively; in other words, the equations (18) and (19) are invariant under the following replacement of the quadruple:

(20) ( f , φ , v , u ) f , φ , v , r λ 1 ( r 1 λ u ) .

Now, we choose

(21) λ = 2 N 2 γ ,

and let us recall from [6], Section 3.2] that the integral on the right-hand side of the Hardy–Leray inequality (2) can be expressed in terms of ( v , f, φ) as follows:

(22) R N | u | 2 | x | 2 γ d x = R × S N 1 ( λ 1 ) 2 | v | 2 + | v | 2 + | σ v | 2 d t d σ ,
(23) R × S N 1 | σ v | 2 d t d σ = R × S N 1 ( σ φ ) 2 + ( λ 2 ) 2 2 N | σ φ | 2 d t d σ + R × S N 1 | σ φ | 2 + ( N 1 ) | v | 2 d t d σ ,
(24) R × S N 1 | v | 2 d t d σ = R × S N 1 f 2 + ( φ ) 2 + λ 2 φ 2 + | σ φ | 2 d t d σ = R × S N 1 f 2 + φ λ 2 2 σ φ d t d σ .

Here the last equality follows from integration by parts together with the support compactness of v or f, φ. Applying (24) to (20), we also obtain

(25) R × S N 1 | v | 2 d t d σ = R × S N 1 ( f ) 2 + ( 2 φ ) 2 + λ 2 ( φ ) 2 + | σ φ | 2 d t d σ = R × S N 1 f ( 2 ) f + φ λ 2 2 σ ( 2 φ ) d t d σ

by integration by parts. After plugging (23) into (22), substitute (24) and (25) into the L2 terms of v and v ; then we get

R N | u | 2 | x | 2 γ d x = ( λ 1 ) 2 + N 1 R × S N 1 | v | 2 d t d σ + R × S N 1 | v | 2 d t d σ + R × S N 1 ( σ φ ) 2 + ( λ 2 ) 2 2 N | σ φ | 2 + | σ φ | 2 d t d σ = ( λ 1 ) 2 + N 1 R × S N 1 f 2 + φ λ 2 2 σ φ d t d σ + R × S N 1 f ( 2 ) f + φ λ 2 2 σ ( 2 φ ) d t d σ + R × S N 1 φ σ 2 ( λ 2 ) 2 2 N σ + 2 σ φ d t d σ = R × S N 1 φ P 1 2 , σ , λ φ + f P 0 ( 2 , λ ) f d t d σ

by integration by parts, where we have defined two polynomials P 1 and P 0 by

P 1 ( τ , a , λ ) = ( λ 1 ) 2 + N 1 λ 2 + τ + a + λ 2 + τ + a τ + a 2 + ( λ 2 ) 2 2 N a + a τ = a 2 + 2 λ 2 6 λ + 4 N + 2 τ a + λ 2 + τ λ 1 2 + N 1 + τ , P 0 ( τ , λ ) = ( λ 1 ) 2 + N 1 + τ .

Now let us replace γ by γ − 1; in view of (21), this manipulation is equivalent to replacing λ by λ + 1. Then the result of the above integral computation changes into

(26) R N | u | 2 | x | 2 | x | 2 γ d x = R × S N 1 φ P 1 2 , σ φ + f P 0 ( 2 ) f d t d σ , where and hereafter we abbreviate as P 1 ( τ , a ) P 1 ( τ , a , λ + 1 ) = a 2 + 2 λ 2 λ + τ N a + ( λ + 1 ) 2 + τ λ 2 + N 1 + τ , P 0 ( τ ) P 0 ( τ , λ + 1 ) = λ 2 + N 1 + τ ,

as the expression in terms of f, φ for the integral on the right-hand side of the Rellich–Hardy inequality (10). To express the left-hand side, we exploit the result of [6], Eqs. (30), (31) with λ replaced by λ + 1]: it holds that

(27) R N | u | 2 | x | 2 γ d x = R × S N 1 φ Q 1 2 , σ φ + f Q 0 ( 2 ) f d t d σ , where  Q 1  and  Q 0  are polynomials given by Q 1 ( τ , a ) = τ + a + ( λ 1 ) 2 τ + a + ( λ + 1 ) 2 τ + a + ( λ + N 1 ) 2 ( 2 λ + N ) 2 a = τ + a + ( λ 1 ) 2 τ 2 + 2 a + α λ + 1 + ( N 2 ) 2 τ + a α λ + 1 2 , Q 0 ( τ ) = τ + ( λ 1 ) 2 τ + ( λ + N 1 ) 2 .

To proceed further, let us apply to φ and f the one-dimensional Fourier transformation with respect to t: we set

φ ̂ ( τ , σ ) = 1 2 π R e i τ t φ ( e t σ ) d t , f ̂ ( τ ) = 1 2 π R e i τ t f ( e t σ ) d t

for ( τ , σ ) R × S N 1 , where i = 1 . Also we apply to φ ̂ the spherical harmonics decomposition:

φ ̂ = ν N φ ̂ ν , σ φ ̂ ν = α ν φ ̂ ν , α ν = ν ( ν + N 2 ) ν N .

Now, we are in a position to evaluate the quantity

(28) R N | u | 2 | x | 2 γ d x R N | u | 2 | x | 2 γ 2 d x ,

which we simply call the R–H quotient. To this end, by using (26), (27) and the L 2 ( R ) isometry of the Fourier transformation, we have

(29) R N | u | 2 | x | 2 γ d x R N | u | 2 | x | 2 γ 2 d x = R × S N 1 ν N Q 1 τ 2 , α ν | φ ν ̂ | 2 + Q 0 ( τ 2 ) | f ̂ | 2 d τ d σ R × S N 1 ν N P 1 τ 2 , α ν | φ ν ̂ | 2 + P 0 ( τ 2 ) | f ̂ | 2 d τ d σ min inf τ R \ { 0 } Q 0 ( τ 2 ) P 0 ( τ 2 ) , inf ν N inf τ R \ { 0 } Q 1 τ 2 , α ν P 1 τ 2 , α ν = min inf τ > 0 Q 0 ( τ ) P 0 ( τ ) , inf ν N inf τ > 0 Q 1 ( τ , α ν ) P 1 ( τ , α ν ) .

Then our goal is reduced to evaluate the fractions Q0/P0 and Q1/P1. In the following subsections, we will show that the infimum values of these fractions can be achieved at τ = 0.

3.3 Evaluation of Q0/P0

A direct calculation yields

Q 0 ( τ ) P 0 ( τ ) = τ + ( λ 1 ) 2 τ + ( λ + N 1 ) 2 λ 2 + N 1 + τ = τ + λ + N 2 2 + ( N 1 ) 1 2 λ + N 2 2 τ + λ 2 + N 1

for all τ ≥ 0. The last expression is of the form g ( τ ) = τ + a b τ + c for some constants a, b ≥ 0 and c > 0, which leads to g ( τ ) g ( 0 ) = τ + b τ c ( τ + c ) τ . Thus we have

(30) 1 τ Q 0 ( τ ) P 0 ( τ ) Q 0 ( 0 ) P 0 ( 0 ) 1 τ > 0 ,

whence in particular we obtain inf τ > 0 Q 0 ( τ ) P 0 ( τ ) = Q 0 ( 0 ) P 0 ( 0 ) = ( λ 1 ) 2 λ + N 1 2 λ 2 + N 1 .

3.4 The case when P1 has zeros

Here we specify when P1(τ, α ν ) = 0 happens. Notice from (26) that P1(τ, a) is strictly monotone increasing in τ ≥ 0 for any a > 0, and hence it holds that

P 1 ( τ , a ) > P 1 ( 0 , a ) = a 2 + 2 ( λ 1 / 2 ) 2 1 2 N a + ( λ + 1 ) 2 λ 2 + N 1

for all τ > 0, as well as that

P 1 ( τ , α 1 ) > P 1 ( 0 , α 1 ) = P 1 ( 0 , N 1 ) = λ 2 ( λ + 1 ) 2 + 3 ( N 1 )

for all τ > 0. Notice on the right-hand side of the (three lines) above inequality that the center of the graph of the quadratic function aP1(0, a) is located at a = λ 1 2 2 + 1 4 + 1 2 N 2 N = α 2 . Then we see that for all τ > 0 and ν ≥ 2,

P 1 ( τ , α ν ) > P 1 ( 0 , α ν ) P 1 ( 0 , α 2 ) = λ 4 + 2 λ 3 + 5 N λ 2 2 ( N + 1 ) λ + ( N + 1 ) ( 2 N 1 ) = λ 2 ( λ + 1 ) 2 + 2 ( 2 N 1 ) λ 2 + ( N + 1 ) ( λ 1 ) 2 + 2 ( N 2 1 ) 2 ( N 2 1 ) > 0 .

In view of the above discussion, we see that

P 1 ( τ , α ν ) = 0  holds if and only if  ( τ , ν , λ ) = ( 0,1,0 )

Hence, every time we treat the rational polynomial Q1/P1, we have to deal with the case λ = 0 or equivalently γ = 2 N 2 as a special one. For this reason, in the rest of this paper we always assume λ ≠ 0 γ 2 N 2 unless otherwise specified.

3.5 Evaluation of Q1/P1

Let us check that

inf τ > 0 Q 1 ( τ , a ) P 1 ( τ , a ) = Q 1 ( 0 , a ) P 1 ( 0 , a ) a { α ν } ν N

in order to evaluate (29) from below. This equation is equivalent to the inequality Q 1 ( τ , a ) P 1 ( τ , a ) Q 1 ( 0 , a ) P 1 ( 0 , a ) ( τ > 0 ) , which can be verified by directly evaluating Q1(τ, a)P1(0, a) − P1(τ, a)Q1(0, a) to be nonnegative. However, we further show the following stronger fact, which serves as a key tool for the proof of Theorem 3:

Lemma 9.

There exists a constant number c0 > 0 such that the inequality

1 τ Q 1 ( τ , a ) P 1 ( τ , a ) Q 1 ( 0 , a ) P 1 ( 0 , a ) c 0

holds for all τ > 0 and a { α ν } ν N .

Here we give a proof of this lemma only for the case γ ≤ 1. Since the proof for γ > 1 follows by a similar technique, we postpone it in later section (see Section 6). The proof of Lemma 9 includes a large amount of numerical computations, and we used Maxima in the course of it. However, we also made a lot of ingenuity to simplify the calculations and ideas to make it understandable, even with the use of Maxima.

Proof of Lemma 9

for γ ≤ 1 or equivalently λ 1 N 2 . It suffices to check the inequality for c0 = 1:

1 τ Q 1 ( τ , a ) P 1 ( τ , a ) Q 1 ( 0 , a ) P 1 ( 0 , a ) 1 τ > 0 , a α ν ν N .

To this end, we directly compute the left-hand side minus right-hand side: by using (26) and (27) we get

(31) 1 τ Q 1 ( τ , a ) P 1 ( τ , a ) Q 1 ( 0 , a ) P 1 ( 0 , a ) 1 = 1 τ τ + a + ( λ 1 ) 2 τ 2 + 2 a + α λ + 1 + ( N 2 ) 2 τ + a α λ + 1 2 a 2 + 2 λ 2 λ + τ N a + ( λ + 1 ) 2 + τ λ 2 + N 1 + τ a + ( λ 1 ) 2 a α λ + 1 2 a 2 + 2 ( λ 2 λ ) N a + ( λ + 1 ) 2 λ 2 + N 1 1 = ( 2 λ + N 2 ) 0 G 0 ( a ) + G 1 ( a ) τ P 1 ( 0 , a ) P 1 ( τ , a ) ,

where we have defined

(32) G 0 ( a ) ( 2 λ + N ) a 3 + 2 λ 2 N + 5 ( 2 λ + N ) 2 ( N 1 ) a 2 + 2 λ 5 + ( N 8 ) λ 4 8 N λ 3 2 ( N 2 + 2 N 2 ) λ 2 2 ( 6 N 7 ) λ 2 N 2 N + 4 a + ( N 1 ) ( 2 λ + N 2 ) ( λ + 1 ) 4 , G 1 ( a ) ( 2 λ + N ) a 2 + ( 2 λ + N ) ( λ 1 ) 2 N + 1 2 ( N 1 ) a + ( N 1 ) ( 2 λ + N 2 ) ( λ + 1 ) 2

as cubic and quadratic polynomials in a. Recall from the discussion in subsection 3.4 that P1(0, a) and P1(τ, a) are positive; then, in order to obtain the nonnegativity of (31) for all τ ≥ 0, it is enough to show the two inequalities

G 1 ( a ) 0  and  G 0 ( a ) 0 a { α ν } ν N ,

The first inequality is easier to prove, by considering the Taylor series of G1(a) at a = α1: a straightforward calculation yields

(33) G 1 ( α 1 + s ) = s 2 ( 2 λ + N ) + s ( N 1 ) ( N + 2 λ 2 ) + ( λ 1 ) 2 ( 2 λ + N ) + 2 ( N 1 ) λ 2 ( 2 λ + N 1 )

for all s R . Since λ 1 N 2 , notice here that the coefficients of the powers of s are all nonnegative, which tells us that G1(α1 + s) ≥ 0 for all s ≥ 0; this fact directly implies G1(a) ≥ 0 for all a { α ν } ν N , as desired. To show the second inequality, that is, G0(a) ≥ 0 for a { α ν } ν N , let us compute the Taylor series of G0(a) at a = α1 as follows:

(34) G 0 ( α 1 + s ) = s 3 ( 2 λ + N ) + s 2 G 2 ( λ ) + s G 1 ( λ ) + 2 ( N 1 ) λ 4 ( 2 λ + N 1 ) ,

where

(35) G 2 ( λ ) ( 2 λ + N 2 ) ( λ + 1 ) 2 + λ 2 + N + 3 + ( N 1 ) 2 + 9 ,
(36) G 1 ( λ ) λ 4 ( 2 λ + N 8 ) + 2 λ 2 2 4 λ 4 N + N 2 + N 2 ( 2 λ + N ) .

Noticing that G 2 ( λ ) 0 and G0(α1) ≥ 0, we aim to prove the following fact:

(37) if 1 N 2 λ 1  then G 1 ( λ ) 0 ,
(38) or  if 1 < λ  then G 0 ( α 2 + s ) 0 s 0 ,

which implies the desired inequality G0(a) ≥ 0 a { α ν } ν N .

For the proof of (37), let λ be parameterized as

λ = 1 N s 2 , 0 s 1 .

Then we directly compute

G 1 ( λ ) = 1 N s 2 4 N N s 6 + 2 1 N s 2 2 2 N s 2 4 N + N 2 + N 2 N N s + 2 = 1 16 N 5 ( 1 s ) s 4 + 1 8 N 4 ( s 2 ) 2 s 2 + 1 2 N 3 ( 1 s ) 2 4 s 5 s 2 + 2 N 2 ( 2 + 3 s 6 s 2 ) + N ( 19 s 7 ) 10 = 1 16 ( N 2 ) 5 ( 1 s ) s 4 + 1 8 ( N 2 ) 4 s 2 s 2 + ( 1 s ) 5 s 2 + 4 + 1 2 ( N 2 ) 3 ( 1 s ) s 2 ( 1 s ) 2 + 4 s 2 + 2 1 3 s + 3 s 2 + ( N 2 ) 2 ( 1 s ) 2 4 s + 2 ( 1 s 2 ) + 5 ( 1 s 3 ) + 3 2 s + ( N 2 ) ( 1 s ) 2 s ( 17 s 5 s 2 ) + 21 10 s 3 s 2 + 2 s ( 1 s ) ( 3 + s ) 5 4 s + s 2 + 4

which is a Taylor series of the function N G 1 ( λ ) = G 1 1 N s 2 at N = 2. Notice here that the coefficients of the powers of N − 2 are all nonnegative since 0 ≤ s ≤ 1. Therefore, we get G 1 ( λ ) 0 , as desired.

Now, all that is left is to show (38). To this end, notice from (34) that

1 s G 0 ( α 1 + s ) s G 2 ( λ ) + G 1 ( λ ) s ( 2 λ + N 2 ) ( λ + 1 ) 2 + λ 2 + λ 4 ( 2 λ + N 8 ) + 2 λ 2 2 4 λ 4 N + N 2

holds for all s > 0. Replacing s by N + 1 + s on both sides, we then get

G 0 ( α 2 + s ) N + 1 + s = 1 N + 1 + s G 0 ( α 1 + N + 1 + s ) ( N + 1 + s ) ( 2 λ + N 2 ) ( λ + 1 ) 2 + λ 2 + λ 4 ( 2 λ + N 8 ) + 2 λ 2 2 4 λ 4 N + N 2 = 2 ( λ 1 ) 5 + ( λ 1 ) 4 ( N + 2 ) + 4 ( λ 1 ) 3 ( s + 2 N 4 ) + 2 ( λ 1 ) 2 ( N + 6 ) s + 2 N 2 + 6 N 18 + 2 ( λ 1 ) ( 3 N + 5 ) s + 5 N 2 + 2 N 14 + 5 N s + 7 N 2 2 N 10

for all s ≥ 0. Notice here that the coefficients of the powers of λ − 1 are all nonnegative since N ≥ 2. Therefore, from the assumption λ > 1 we have obtained G0(α2 + s) ≥ 0, as desired.□

Since the polynomial function P1(τ, a) is quadratic in τ, it is clear from (31) that

lim τ 1 τ Q 1 ( τ , a ) P 1 ( τ , a ) Q 1 ( 0 , a ) P 1 ( 0 , a ) = 1

for each aα1. Therefore, the constant number c0 of Lemma 9 is optimal when c0 = 1, in the sense that

inf τ > 0 inf ν N 1 τ Q 1 ( τ , α ν ) P 1 ( τ , α ν ) Q 1 ( 0 , α ν ) P 1 ( 0 , α ν ) = 1

holds as far as γ ≤ 1.

3.6 A lower bound for the R–H quotient

In view of the estimate (29) for the R–H quotient (28), it follows from Sections 3.3, 3.4 and 3.5 that the inequality

R N u 2 | x | 2 γ d x C N , γ R N | u | 2 | x | 2 γ 2 d x

holds for curl-free fields u with the constant number

C N , γ = min inf ν N inf τ > 0 Q 1 ( τ , α ν ) P 1 ( τ , α ν ) , inf τ > 0 Q 0 ( τ ) P 0 ( τ ) = min min ν N Q 1 ( 0 , α ν ) P 1 ( 0 , α ν ) , Q 0 ( 0 ) P 0 ( 0 ) .

Notice from (26) and (27) that the last two fractions are explicitly written as

Q 1 ( 0 , α ν ) P 1 ( 0 , α ν ) = α ν + ( λ 1 ) 2 ( α λ + 1 α ν ) 2 α λ + 1 α ν 2 + 2 λ + N 2 ( 2 λ + 1 ) α ν ( N 1 ) ( λ + 1 ) 2 = ( γ 2 ) 2 ν + N 2 1 2 2 α ν + γ + N 2 1 2 ( γ 2 ) 2 ν + N 2 1 2 2 + 2 ( γ 1 ) 2 γ + N 5 α ν + ( N 1 ) γ + N 2 3 2 , Q 0 ( 0 ) P 0 ( 0 ) = ( λ 1 ) 2 ( λ + N 1 ) 2 λ 2 + N 1 = ( γ 1 ) 2 1 4 N 2 2 γ + N 2 2 2 + N 1 ,

by recalling the notation (21) together with the aid of the identity

(39) α s α t = s + N 2 1 2 t + N 2 1 2 s , t R ;

in other words, we have

Q 1 ( 0 , α ν ) P 1 ( 0 , α ν ) = C N , γ , ν , Q 0 ( 0 ) P 0 ( 0 ) = C N , γ , 0

in terms of the same notation in Theorem 1. Therefore, we have obtained

C N , γ = min ν N { 0 } C N , γ , ν

as a lower bound for the R–H quotient (28), which coincides with the same constant number CN,γ given in Theorem 1.

3.7 Sharpness of CN,γ

We show here the optimality of the constant CN,γ in the inequality (10). To this end, we construct a sequence of curl-free fields minimizing the value of the R–H quotient (28). First of all, choose ν 0 N { 0 } to be such that

min ν N { 0 } C N , γ , ν = C N , γ , ν 0 .

If γ = 2 N 2 , by the same computation as (45) below (Section 5) we have CN,γ,1 > CN,γ,0 for all N ≥ 2, which implies that ν0 ≠ 1. Hence it follows from Section 3.4 that we can always assume that P 1 ( 0 , α ν 0 ) > 0 .

Let h be any one-variable function in C c ( R ) satisfying R ( h ( t ) ) 2 d t = 1 . Define a sequence of curl-free fields { u n } n N C c ( R ̇ N ) N by the formula

u n ( x ) | x | ρ λ h 1 n log ρ d ρ = x | x | λ 1 h 1 n log | x | if  ν 0 = 0 , | x | λ + 1 φ n ( x ) otherwise ,

where { φ n } n N are functions in C c ( R ̇ N ) given by

φ n ( x ) = h 1 n log | x | Y ( x / | x | ) ,

where Y C ( S N 1 ) \ { 0 } denotes a spherical harmonic function satisfying the eigenequation

σ Y = α ν 0 Y on  S N 1 .

Also define { v n } C c ( R ̇ N ) N as a sequence of vector fields by the formula

u n ( x ) = | x | λ v n ( x )

in the same way as (18). Then we have

v n = σ f n  if  ν 0 = 0 σ ( + λ + 1 ) φ n + σ φ n  otherwise ,

where f n is given by f n ( x ) = h ( 1 n log | x | ) . In this setting, let us now apply the formulae (26) and (27) to the case ( u , f, φ) = ( u n , f n , 0) or ( u , f, φ) = ( u n , 0, φ n ). Then we have

R N | u n | 2 | x | 2 γ d x R N | u n | 2 | x | 2 γ 2 d x = R h t n Q 0 t 2 h t n d t R h t n P 0 t 2 h t n d t  if  ν 0 = 0 , R h t n Q 1 t 2 , α ν 0 h t n d t R h t n P 1 t 2 , α ν 0 h t n d t  otherwise . = R h ( t ) Q 0 n 2 t 2 h ( t ) d t R h ( t ) P 0 n 2 t 2 h ( t ) d t  if  ν 0 = 0 , R h ( t ) Q 1 n 2 t 2 , α ν 0 h t d t R h ( t ) P 1 n 2 t 2 , α ν 0 h t d t  otherwise .

Notice on the right-hand side that the denominator always exceeds a fixed positive number, since P0(0) ≥ N − 1 > 0 and P 1 ( 0 , α ν 0 ) > 0 as mentioned above. Therefore, passing to n → ∞, we get

R N | u n | 2 | x | 2 γ d x R N | u n | 2 | x | 2 γ 2 d x = O ( 1 / n 2 ) + Q 0 ( 0 ) O ( 1 / n 2 ) + P 0 ( 0 ) if  ν 0 = 0 O ( 1 / n 2 ) + Q 1 ( 0 , α ν 0 ) O ( 1 / n 2 ) + P 1 ( 0 , α ν 0 ) otherwise C N , γ , ν 0 = C N , γ ,

which gives the desired sharpness of CN,γ.

Now, the proof of Theorem 1 has been completed.□

4 Proof of Theorem 3

Let ν1 denote the positive integer such that

C N , γ , ν 1 = min ν N C N , γ , ν .

In order to estimate the difference between both sides of the inequality (10), recall from the same calculation in the first line of (29) the expression of the integrals:

R N | u | 2 | x | 2 γ d x = R × S N 1 Q 0 ( τ 2 ) | f ̂ | 2 + ν N Q 1 τ 2 , α ν | φ ν ̂ | 2 d τ d σ , R N | u | 2 | x | 2 | x | 2 γ d x = R × S N 1 P 0 ( τ 2 ) | f ̂ | 2 + ν N P 1 τ 2 , α ν | φ ν ̂ | 2 d τ d σ .

Then we have the following estimate:

R N | u | 2 | x | 2 γ d x C N , γ R N | u | 2 | x | 2 | x | 2 γ d x = R N | u | 2 | x | 2 γ d x min Q 1 ( 0 , α ν 1 ) P 1 ( 0 , α ν 1 ) , Q 0 ( 0 ) P 0 ( 0 ) R N | u | 2 | x | 2 | x | 2 γ d x R N | u | 2 | x | 2 γ d x R × S N 1 Q 0 ( 0 ) P 0 ( 0 ) P 0 ( τ 2 ) | f ̂ | 2 R × S N 1 Q 1 ( 0 , α ν 1 ) P 1 ( 0 , α ν 1 ) ν N P 1 τ 2 , α ν | φ ν ̂ | 2 d τ d σ = R × S N 1 Q 0 ( τ 2 ) Q 0 ( 0 ) P 0 ( 0 ) P 0 ( τ 2 ) | f ̂ | 2 d τ d σ + ν N R × S N 1 Q 1 τ 2 , α ν Q 1 ( 0 , α ν 1 ) P 1 ( 0 , α ν 1 ) P 1 τ 2 , α ν | φ ν ̂ | 2 d τ d σ R × S N 1 P 0 ( τ 2 ) τ 2 | f ̂ | 2 d τ d σ + c 0 ν N R × S N 1 P 1 τ 2 , α ν τ 2 | φ ν ̂ | 2 d τ d σ min 1 , c 0 R × S N 1 P 0 ( τ 2 ) τ 2 | f ̂ | 2 + ν N P 1 τ 2 , α ν τ 2 | φ ν ̂ | 2 d τ d σ = min 1 , c 0 R × S N 1 ( f ) P 0 ( 2 ) f + ( φ ) P 1 2 , σ φ d t d σ = min 1 , c 0 R N | x | λ | x | λ u 2 | x | 2 | x | 2 γ d x ,

where the last equation follows by applying (26) to the replacement (20), and where we the fourth inequality follows by using the inequalities (30) and

1 τ 2 Q 1 τ 2 , α ν P 1 τ 2 , α ν Q 1 ( 0 , α ν 1 ) P 1 ( 0 , α ν 1 ) 1 τ 2 Q 1 τ 2 , α ν P 1 τ 2 , α ν Q 1 ( 0 , α ν ) P 1 ( 0 , α ν ) c 0 ( τ , ν ) R \ { 0 } × N ,

as verified by using the same constant c0 given in Lemma 9. Finally, by restoring the notations (21) and (17), we obtain

R N | u | 2 | x | 2 γ d x C N , γ R N | u | 2 | x | 2 | x | 2 γ d x c R N | x | 2 N 2 γ ( x ) | x | γ + N 2 2 u 2 | x | 2 γ 2 d x

for some constant c > 0. The proof of Theorem 3 is now complete, although the optimal value of c is not known.□

5 An observation of the best constant CN,γ in Theorem 1

Concerning the constants in the inequalities (8) and (10), it holds that

(40) C N , γ A N , γ

as a matter of course. Here we wish to evaluate whether the strict inequality CN,γ > AN,γ holds or not. However, since the expression for CN,γ is complicated and its full picture seems difficult to reveal, we describe it for only some specific values of γ.

5.1 Preliminary: a review of AN,γ,ν

In view of the original best constant (9) in the Rellich–Hardy inequality, let us observe the increase or decrease of the function νAN,γ,ν. In terms of the notation (21), the expression of AN,γ,ν in (9) can be rewritten as

(41) A N , γ , ν = ( λ + N 2 ) 2 for  ν = 0 , ( α ν α λ ) 2 α ν + λ 2 for  ν N .

with the aid of (39). Then a direct calculation yields

(42) A N , γ , 1 A N , γ , 0 N 1 = 3 λ 2 + 4 ( N 2 ) λ + N 2 5 N + 5 λ 2 + N 1

and

(43) A N , γ , ν + 1 A N , γ , ν 2 ν + N 1 = α ν α ν + 1 + λ 2 2 ν ( ν + N 1 ) + 1 + 1 N 1 λ 2 A N , γ , 1 A N , γ , 0 α ν + λ 2 α ν + 1 + λ 2

for all ν N . Notice that the numerator of the right-hand side of (43) is monotone increasing in ν ≥ 0, as well as that the denominator is always positive; then it is clear, for every k N { 0 } , that the whole fraction (43) is non-negative for all νk only if the same applies to ν = k. In other words, for every k the two inequalities

(44) A N , γ , k A N , γ , k + 1  and  A N , γ , ν A N , γ , ν + 1 ( ν k )

are equivalent in the sense that the former holds if and only if so does the latter.

5.2 The case γ = 2 N 2 (or equivalently λ = 0)

Let us deal with this “singular” case, in the sense of Section 3.4. Following from the definitions of AN,γ and CN,γ, we have

(45) A N , 2 N 2 = min A N , 2 N 2 , 0 , min ν N A N , 2 N 2 , ν = min 2 N 2 , min ν N α ν = min ( N 2 ) 2 , N 1 = α 1 = 0 ( N = 2 ) N 1 ( N 3 ) , C N , 2 N 2 = min C N , 2 N 2 , 0 , C N , 2 N 2 , 1 , min ν N \ { 1 } C N , 2 N 2 , ν = min N 1 , N 3 3 N 2 , C N , 2 N 2 , 2 = min N 1 , N 3 3 N 2 , ( N + 1 ) ( 2 N + 1 ) 2 N 1 = N 1 .

Here the third equality from the last in (45) follows with the aid of computing

C N , 2 N 2 , ν = ( α ν + 1 ) 1 N 2 α ν 1

which is monotone increasing in ν ≥ 2. Summarizing the results above, we have obtained

C 2,1 = C 2,1,0 = 1 > 0 = A 2,1,0 = A 2,1 , C N , 2 N 2 = C N , 2 N 2 , 0 = N 1 = A N , 2 N 2 , 1 = A N , 2 N 2 ( N 3 ) .

In particular, we see that the best constant in the two-dimensional Rellich–Hardy inequality (with γ = 1) can be really improved by the curl-free condition on the test vector fields.

5.3 The case γ 2 N 2 (or equivalently λ ≠ 0)

Recall from Sections 3.6 and 3.4 that

C N , γ , ν = α ν + ( λ 1 ) 2 ( α λ + 1 α ν ) 2 P 1 ( 0 , α ν ) , P 1 ( 0 , α ν ) = α λ + 1 α ν 2 + 2 λ + N 2 ( 2 λ + 1 ) α ν ( N 1 ) ( λ + 1 ) 2 > 0

for ν N . By using this expression together with (41), a direct computation yields

C N , γ , ν A N , γ , ν 1 = 2 ( ν λ 1 ) 2 W ( α ν ) ( ν + N 2 ) α ν 1 + λ 2 P 1 ( 0 , α ν ) , C N , γ , ν A N , γ , ν + 1 = 2 ν W ( α ν ) ( 1 ν N λ ) 2 α ν + 1 + λ 2 P 1 ( 0 , α ν )

for all ν N , where

W ( α ν ) = λ 2 ( 2 λ + N 4 ) ( 2 λ + N ) + ( N 2 ) 2 α ν + λ 2 1 2 .

Then we have

C N , γ , ν A N , γ , ν 1 C N , γ , ν A N , γ , ν + 1 0

or equivalently

(46) min { A N , γ , ν 1 , A N , γ , ν + 1 } C N , γ , ν max { A N , γ , ν 1 , A N , γ , ν + 1 }

for all ν N . Based on this fact, let us consider the following two simplest cases:

The case AN,γ = AN,γ,1. It holds from (12) that

A N , γ = A N , γ , 1 = C N , γ , 0 C N , γ ,

and hence that CN,γ = AN,γ from (40). This fact says that the curl-free restriction takes no effect on the improvement of the best constant. Now, we specify the condition on the values of γ (or λ) when the equation AN,γ = AN,γ,1 will happen. In view of the inequalities (44) with k = 1, this equation is equivalent to that both the inequalities

A N , γ , 1 A N , γ , 0  and  A N , γ , 1 A N , γ , 2

hold true. Here the first inequality is equivalent to that the numerator of (42),

3 λ 2 + 4 ( N 2 ) λ + N 2 5 N + 5

is nonnegative; to specify the second, a direct computation of the numerator of (43)ν=1 using (42) yields

α 2 α 1 + λ 2 2 N + 1 + 1 N 1 λ 2 A N , γ , 1 A N , γ , 0 = 2 N ( N 1 ) + λ 2 2 N 1 + λ 2 N 1 N 1 λ 2 + N 1 3 λ 2 + 4 ( N 2 ) λ + N 2 5 N + 5 = 3 λ 4 4 ( N 2 ) λ 3 ( N 2 7 N + 5 ) λ 2 + 2 N ( N 1 ) .

Therefore, the equation AN,γ = AN,γ,1 holds if and only if λ satisfies both the two inequalities

3 λ 2 + 4 ( N 2 ) λ + N 2 5 N + 5 0 and 3 λ 4 4 ( N 2 ) λ 3 ( N 2 7 N + 5 ) λ 2 + 2 N ( N 1 ) 0 .

For example, if γ = 0 or equivalently λ = 2 N 2 , then the two inequalities become

N 2 4 N 4 0 ,  and N 4 4 N 3 + 12 N 2 + 64 N 64 0 ( which is always true ) .

This is the case when N ≤ 4, and hence we have

C 2,0 = C 2,0,0 = A 2,0,1 = A 2,0 = 0 , C 3,0 = C 3,0,0 = A 3,0,1 = A 3,0 = 25 36 , C 4,0 = C 4,0,0 = A 4,0,1 = A 4,0 = 3 ,

which says that no curl-free improvement arises when N ∈ {2, 3, 4} and γ = 0.

We may understand the above result in the following way: we can choose a sequence { u n } n N of vector fields by the formula

w n ( x ) = x f n ( | x | ) n = 1,2 ,

in order that

R N | u n | 2 | x | 2 γ d x R N | w n | 2 | x | 2 γ 2 d x A N , γ , 1 ( n ) ,

where { f n ( r ) } n N are a smooth functions vanishing near r = 0. We may see this fact by noticing that each coordinate function x k  (k = 1, …, N) satisfies the eigenequation −△ σ x k = α1x k . On the other hand, it is easy to check that { w n } n N are always curl-free, whence CN,γAN,γ. Therefore, we get CN,γ = AN,γ from AN,γ = AN,γ,1.

The case AN,γ = AN,γ,0.

Since AN,γ,1AN,γ,0 = AN,γ, we see from (44) and (46) that

A N , γ , ν 1 C N , γ , ν A N , γ , ν + 1

holds for all ν N . This fact implies

min k N { 0 } C N , γ , 2 k = C N , γ , 0  and  min k N C N , γ , 2 k 1 = C N , γ , 1 ,

whence

C N , γ = min C N , γ , 0 , C N , γ , 1 .

On the other hand, a direct computation yields

C N , γ , 1 C N , γ , 0 = ( λ 1 ) 2 A N , γ , 1 A N , γ , 0 + ( N 2 1 ) 5 ( 5 λ + 2 N 4 ) 2 + N 2 + N 1 λ 2 + N 1 ( λ + 1 ) 2 + 3 ( N 1 ) > 0

from AN,γ,1AN,γ,0. Therefore, it holds from (12) that

C N , γ = C N , γ , 0 = A N , γ , 1 ,

or equivalently that

C N , γ A N , γ = A N , γ , 1 A N , γ , 0 .

This fact together with (42) shows that the inequality CN,γ > AN,γ (namely the effect of the curl-free-improvement) holds as far as the right-hand side of (42) is strictly negative, or equivalently

γ N + 4 6 < 1 3 N 2 N + 1 .

For example, this is the case if γ = 0 and N ≥ 5, whence we have

C N , 0 = C N , 0,0 = A N , 0,1 = N 2 4 1 2 N 2 4 N + 3 > N 2 4 = A N , 0,0 = A N , 0

for all N ≥ 5.

We may understand the above phenomenon in the following way: we can choose a sequence { w n } n N of vector fields by the formula

w n ( x ) = f n , 1 ( | x | ) , , f n , N ( | x | ) ( n = 1,2 , )

in order that

(47) R N | w n | 2 | x | 2 γ d x R N | w n | 2 | x | 2 γ 2 d x A N , γ , 0 ( n ) .

In order for w n to be curl-free, it must hold that

f n , j x k f n , k x j = r x k f n , j r x j f n , k = x k r f n , j x j r f n , k = 0 , j , k { 1 , , N } ,

where the notation (…)′ denotes the derivative of a one-dimensional function. This fact implies that f n , k = ( x k / x j ) f n , j for all jk, and hence that

f n , j = c o n s t j { 1 , , N }

from the radial symmetry of the function x fn,j(| x |). Consequently, we have |△ w n | = |∇ w n | ≡ 0. This phenomenon indicates that there may be no curl-free sequence { w n } n N satisfying (47), which we can naturally interpret as a result of the inequality CN,γ > AN,γ.

Incidentally, let us further consider the case

1 N 2 + N + 1 γ N + 4 6 + 1 3 N 2 N + 1 for  3 N 11 γ N + 4 6 1 3 N 2 N + 1 for  N 12

which ensures both CN,γ = CN,γ,0 and

H N , γ 1 = γ + N 2 2 2 + N 1

from (3). (For γ = 0, this is the case for all N ≥ 8.) Then a successive application of Rellich–Hardy inequality and Hardy–Leray inequality reproduces Rellich–Leray inequality: for all curl-free fields u D γ 1 ( R N ) , it holds that

R N | u | 2 | x | 2 γ d x C N , γ R N | u | 2 | x | 2 γ 2 d x ( from  ( 10 ) ) C N , γ H N , γ 1 R N | u | 2 | x | 2 | x | 2 γ 2 d x from  ( 2 ) with  γ  replaced by  γ 1 = C N , γ , 0 γ + N 2 2 2 + N 1 R N | u | 2 | x | 2 | x | 2 γ 2 d x = ( γ 1 ) 2 N 2 4 2 R N | u | 2 | x | 2 | x | 2 γ 2 d x B N , γ R N | u | 2 | x | 4 | x | 2 γ d x

with BN,γ given by (6). Therefore, even under the curl-free constraint, Rellich–Hardy inequality bridges between Hardy–Leray and Rellich–Leray inequalities, and so serves as a stronger version of the Rellich–Leray inequality.

6 Completion of the proof of Lemma 9

In this section, we prove Lemma 9 for the case γ > 1 (or equivalently λ < 1 N 2 ) with the aid of Maxima, by a similar technique to the former case. More precisely, our goal is to show the two inequalities:

1 τ Q 1 ( τ , a ) P 1 ( τ , a ) Q 1 ( 0 , a ) P 1 ( 0 , a ) 1 2 when  N 3 , 1 τ Q 1 ( τ , a ) P 1 ( τ , a ) Q 1 ( 0 , a ) P 1 ( 0 , a ) 1 3 when  N = 2

for all a { α ν } ν N , which clearly satisfies the conclusion of the lemma.

6.1 The case N ≥ 3

By adding 1/2 to both sides of (31), we have

(48) 1 τ Q 1 ( τ , a ) P 1 ( τ , a ) Q 1 ( 0 , a ) P 1 ( 0 , a ) 1 2 = ( 2 λ + N 2 ) G 0 ( a ) + G 1 ( a ) τ P 1 ( 0 , a ) P 1 ( τ , a ) + 1 2 = ( 2 λ + N 2 ) G 0 ( a ) + G 1 ( a ) τ P 1 ( 0 , a ) P 1 ( 0 , a ) + τ 2 + 2 a + λ 2 + λ + N τ + 1 2 = τ 2 P 1 ( 0 , a ) + τ E 1 ( a ) + E 0 ( a ) 2 P 1 ( τ , a ) P 1 ( 0 , a ) ,

where

(49) E 1 ( a ) 2 ( 2 λ + N 2 ) G 1 ( a ) + 2 ( a + λ 2 + λ ) + N P 1 ( 0 , a ) ,
(50) E 0 ( a ) 2 2 λ + N 2 G 0 ( a ) + P 1 ( 0 , a ) 2 .

Then it suffices to check that

E 1 ( a ) 0  and  E 0 ( a ) 0 a { α ν } ν N

or that

(51) E 1 ( α 1 + s ) 0  and  E 0 ( α 1 + s ) 0 s 0 .

In the following, let us check the two inequalities in (51).

Proof

of E1(α1 + s) ≥ 0. A direct computation from (26) yields

(52) P 1 ( 0 , α 1 + s ) = ( α 1 + s ) 2 + 2 ( λ 2 λ ) N ( α 1 + s ) + ( λ + 1 ) 2 λ 2 + N 1 = s 2 + s 2 λ 2 2 λ + N 2 + λ 2 ( λ + 1 ) 2 + 3 N 3 .

Substitute (33) and (52) into (49) with a = α1 + s; then a straightforward computation yields

E 1 ( α 1 + s ) = 2 ( 2 λ + N 2 ) G 1 ( α 1 + s ) + 2 α 1 + s + λ 2 + λ + N P 1 ( 0 , α 1 + s ) = 2 ( 2 λ + N 2 ) s 2 ( 2 λ + N ) + s ( N 1 ) ( 2 λ + N 2 ) + ( λ 1 ) 2 ( 2 λ + N ) + 2 ( N 1 ) λ 2 ( 2 λ + N 1 ) + 2 s + 2 λ 2 + λ + 3 N 2 s 2 + s 2 λ 2 2 λ + N 2 + λ 2 ( λ + 1 ) 2 + 3 N 3 = 2 s 3 + s 2 E 12 ( λ ) + s E 11 ( λ ) + λ 2 E 10 ( λ ) s 0

which is a Taylor series of E1(a) at a = α1, where we set

E 12 ( λ ) 2 ( 2 λ + N 2 ) ( 2 λ + N ) + 2 λ 2 + λ + 3 N 2 + 2 2 λ 2 2 λ + N 2 = 14 λ 2 + 2 ( 4 N 5 ) λ + ( N + 2 ) ( 2 N 3 ) = 2 2 λ + N 5 4 2 + 6 λ 2 + 6 ( N 2 ) + 23 8 , E 11 ( λ ) 2 ( 2 λ + N 2 ) ( N 1 ) ( 2 λ + N 2 ) + ( λ 1 ) 2 ( 2 λ + N ) + 2 λ 2 ( ( λ + 1 ) 2 + 3 N 3 ) + 2 λ 2 + λ + 3 N 2 2 λ 2 2 λ + N 2 = 14 λ 4 + 4 ( 2 N 5 ) λ 3 + 2 N ( N + 1 ) λ 2 + 4 ( N 1 ) ( N 2 ) λ + 2 N 2 N + 2 ( N 2 ) = 6 λ 4 + 1 2 λ 2 ( 4 λ + 2 N 5 ) 2 + 16 N 9 + ( N 2 ) ( 2 λ + N 1 ) 2 + N 2 + N + 1 , E 10 ( λ ) 4 ( N 1 ) ( 2 λ + N 2 ) ( 2 λ + N 1 ) + 2 ( λ 2 + λ ) + 3 N 2 ( λ + 1 ) 2 + 3 N 3 = ( λ + 1 ) 4 + ( λ + 1 ) 2 λ 2 + ( N 1 ) 5 λ + 8 N 6 5 2 + 36 N 2 + 21 N + 89 25 .

Notice that E12(λ), E11(λ) and E10(λ) are all nonnegative; then E1(α1 + s) ≥ 0 holds from s ≥ 0, which proves the desired inequality.

Proof

of E0(α1 + s) ≥ 0. Substitute (34) and (52) into (50) with a = α1 + s; then a straightforward computation yields

(53) E 0 ( α 1 + s ) = 2 2 λ + N 2 G 0 ( α 1 + s ) + P 1 ( 0 , α 1 + s ) 2 = 2 ( 2 λ + N 2 ) s 3 ( 2 λ + N ) + s 2 G 2 ( λ ) + s G 1 ( λ ) + 2 ( N 1 ) λ 4 ( 2 λ + N 1 ) + s 2 + s 2 λ 2 2 λ + N 2 + λ 2 ( λ + 1 ) 2 + 3 N 3 2 = s 4 + s 3 E 03 ( λ ) + s 2 E 02 ( λ ) + 2 s E 01 ( λ ) + λ 4 E 00 ( λ ) s 0

which is a Taylor series of E0(a) at a = α1, where we set

(54) E 03 ( λ ) 2 ( 2 λ + N 2 ) ( 2 λ + N ) + 2 2 λ 2 2 λ + N 2 = 1 2 ( 4 λ + 2 N 3 ) 2 + 4 λ 2 + 4 ( N 3 ) + 7 2 , E 02 ( λ ) 2 ( 2 λ + N 2 ) G 2 ( λ ) + 2 λ 2 2 λ + N 2 2 + 2 λ 2 ( λ + 1 ) 2 + 3 N 3 ,
(55) E 01 ( λ ) ( 2 λ + N 2 ) G 1 ( λ ) + λ 2 ( λ + 1 ) 2 + 3 N 3 2 λ 2 2 λ + N 2 ,
E 00 ( λ ) 4 ( N 1 ) ( 2 λ + N 2 ) ( 2 λ + N 1 ) + ( λ + 1 ) 2 + 3 N 3 2 = ( λ + 1 ) 4 + ( N 1 ) ( 4 λ + 2 N 3 ) 2 + 6 ( λ + 1 ) 2 + 9 N 10 .

Hence, in order to show that E0(α1 + s) ≥ 0 (∀s ≥ 0), it suffices to check the nonnegativity of { E 0 k ( λ ) } k = 0,1,2,3 . Since it is clear that E03(λ) ≥ 0 and E00(λ) ≥ 0, all that is left is to show the two inequalities

(56) E 02 ( λ ) 0  and  E 01 ( λ ) 0 .

To this end, substitute (35) into (54), then we get

E 02 ( λ ) = 2 ( 2 λ + N 2 ) ( 2 λ + N 2 ) ( λ + 1 ) 2 + λ 2 + N + 3 + ( N 1 ) 2 + 9 + 2 λ 2 2 λ + N 2 2 + 2 λ 2 ( λ + 1 ) 2 + 3 N 3 = 22 λ 4 + 4 ( 4 N 5 ) λ 3 + 2 ( N + 4 ) ( 2 N + 1 ) λ 2 + 4 N 4 N 3 λ + 4 N 2 + N + 2 ( N 2 ) = 2 λ 4 + 5 λ 2 2 λ + 4 N 5 1 2 + N 5 λ + 8 N 6 5 2 + λ 2 4 5 N 2 + N + 3 + 1 25 ( N 3 ) 36 N 2 + 29 N + 51 + 53 25 .

Since N ≥ 3, this result implies E02(λ) ≥ 0, whence we have proved the first inequality of (56). The proof of the second can also be carried out in the same way as follows. Substitute (36) into (55), then a straightforward computation yields

E 01 ( λ ) = ( 2 λ + N 2 ) λ 4 ( 2 λ + N 8 ) + N 2 ( 2 λ + N ) + 2 λ 2 2 4 λ 4 N + N 2 + λ 2 ( λ + 1 ) 2 + 3 N 3 2 λ 2 2 λ + N 2 = 2 λ 6 + 1 4 ( 4 λ + 2 N 9 ) 2 + 8 N 25 λ 4 + N ( 4 N 1 ) + 2 λ 2 + N 2 ( N 1 ) ( λ + 2 ) 2 + 4 N 2 + ( N 6 ) ( 2 λ + N 1 ) 2 λ 2 + N 3

whence we get

E 01 ( λ ) 0  for all  N 6 .

The same also applies to the case N ∈ {3, 4, 5}, as directly verified by the following calculation:

when  N = 3 , E 01 ( λ ) = 6 λ 6 6 λ 5 + 10 λ 4 24 λ 3 + 41 λ 2 + 72 λ + 27 = 1 63 λ + 2 3 2 40 + 28 λ 145 3 2 + λ + 2 3 4 1 18 + 1 6 ( 6 λ 11 ) 2 + 26 9 3 λ + 11 9 2 + 1405 729 ; when  N = 4 , E 01 ( λ ) = 6 λ 6 2 λ 5 6 λ 4 24 λ 3 + 92 λ 2 + 192 λ + 128 = 2 λ 3 1 2 λ 2 8 λ 8 2 + 2 λ 6 + 103 4 λ 4 + 4 λ 2 + 16 ( λ + 2 ) 2 ; when  N = 5 , E 01 ( λ ) = 6 λ 6 + 2 λ 5 16 λ 3 + 181 λ 2 + 400 λ + 375 = 5 λ 3 3 2 2 + λ 2 λ 2 + λ 1 2 2 + ( 25 2 λ + 16 ) 2 + 49 2 λ 2 + 431 4 .

Therefore, we have obtained E01(λ) ≥ 0 for all N ≥ 3, as desired.

6.2 The case N = 2

Applying the same calculation in (31) to the case N = 2, we have the identity

1 τ Q 1 ( τ , a ) P 1 ( τ , a ) Q 1 ( 0 , a ) P 1 ( 0 , a ) 1 = 2 λ G 0 ( a ) + G 1 ( a ) τ P 1 ( 0 , a ) P 1 ( τ , a ) ,

where

P 1 ( τ , a ) = τ 2 + 2 a + λ 2 + λ + 1 τ + a + λ 2 λ 1 2 + ( 4 λ + 3 ) λ 2

is the same as in (26) with N = 2, and where G0(a) and G0(a) are the same as in (32) with N = 2. Adding 2/3 to both sides of the above identity, we then get

1 τ Q 1 ( τ , a ) P 1 ( τ , a ) Q 1 ( 0 , a ) P 1 ( 0 , a ) 1 3 = 2 3 1 + 3 λ G 0 ( a ) + G 1 ( a ) τ τ 2 + 2 a + λ 2 + λ + 1 τ P 1 ( 0 , a ) + P 1 0 , a 2 = 2 3 τ 2 P 1 ( 0 , a ) + 2 τ F 1 ( a ) + F 0 ( a ) P 1 ( 0 , a ) P 1 ( τ , a ) ,

where

(57) F 1 ( a ) a + λ 2 + λ + 1 P 1 ( 0 , a ) + 3 2 λ G 1 ( a ) ,
(58) F 0 ( a ) P 1 ( 0 , a ) 2 + 3 λ G 0 ( a ) .

Hence it is enough to show that

F 1 ( a ) 0  and  F 0 ( a ) 0 a { α ν = ν ( ν + 1 ) } ν N

under the assumption λ < 1 N 2 = 0 .

Proof

of F1(a) ≥ 0 for a { α ν } ν N . It suffices to check that F1(α1 + s) ≥ 0, i.e.,

F 1 ( 1 + s ) 0 s 0 .

In order to compute the left-hand side, apply (52) and (33) to N = 2; then we get

(59) P 1 ( 0,1 + s ) = s 2 + 2 s λ ( λ 1 ) + λ 2 ( λ + 1 ) 2 + 3 , 1 2 G 1 ( 1 + s ) = ( λ + 1 ) s 2 + s λ + ( λ 1 ) 2 ( λ + 1 ) + λ 2 ( 2 λ + 1 ) .

Substitute these equations into (57) and (58) with a = 1 + s, then we have

F 1 ( 1 + s ) = s + λ 2 + λ + 2 P 1 ( 0,1 + s ) + 3 2 λ G 1 ( 1 + s ) = s + λ 2 + λ + 2 s 2 + 2 s λ ( λ 1 ) + λ 2 ( λ + 1 ) 2 + 3 + 3 λ ( λ + 1 ) s 2 + s λ + ( λ 1 ) 2 ( λ + 1 ) + λ 2 ( 2 λ + 1 ) = s 3 + 2 3 λ 2 + λ + 1 s 2 + λ ( 6 λ 1 ) λ 2 + 1 s + λ 2 λ 4 + 3 λ 3 + 14 λ 2 + 11 λ + 8 = s 3 + 6 λ + 1 6 2 + 11 6 s 2 λ ( 1 6 λ ) λ 2 + 1 s + λ 2 ( λ + 1 ) 2 λ + 1 2 2 + 27 4 ( λ + 1 ) 2 + ( 2 λ 1 ) 2 .

Here the coefficients of the powers of s are all positive since λ < 0, which implies that F1(a) ≥ 0 for all aα1, as desired.

Proof

of F0(a) ≥ 0 for a { α ν } ν N . First of all, we compute F0(1 + s); to this end, apply (34), (35) and (36) to the case N = 2, then we get

1 2 G 0 ( 1 + s ) = ( λ + 1 ) s 3 + 1 2 s 2 G 2 ( λ ) + 1 2 s G 1 ( λ ) + λ 4 ( 2 λ + 1 ) = ( λ + 1 ) s 3 + s 2 2 λ 3 + 2 λ 2 + 6 λ + 5 + s ( λ 1 ) λ 4 2 λ 3 + 6 λ 2 8 λ 4 + λ 4 ( 2 λ + 1 ) .

Substitute this expression and (59) into (58) with a = 1 + s, then we have

(60) F 0 ( 1 + s ) = P 1 ( 0,1 + s ) 2 + 3 λ G 0 ( 1 + s ) = s 2 + 2 s λ ( λ 1 ) + λ 2 ( λ + 1 ) 2 + 3 2 + 3 λ 2 ( λ + 1 ) s 3 + s 2 G 2 ( λ ) + s G 1 ( λ ) + 2 λ 4 ( 2 λ + 1 ) = s 2 + 2 s λ ( λ 1 ) + λ 2 ( λ + 1 ) 2 + 3 2 + 6 λ ( λ + 1 ) s 3 + s 2 2 λ 3 + 2 λ 2 + 6 λ + 5 + λ 4 ( 2 λ + 1 ) + s ( λ 1 ) λ 4 2 λ 3 6 λ 2 8 λ 4 = s 4 + 2 λ ( 5 λ + 1 ) s 3 + 2 λ ( 9 λ 3 + 4 λ 2 + 24 λ + 15 ) s 2 + 2 λ ( λ 1 ) 5 λ 4 2 λ 3 10 λ 2 24 λ 12 s + λ 4 ( λ + 1 ) 4 + 9 ( λ + 1 ) 2 + 9 λ 2 + 6

which is a Taylor series of F0(a) at a = α1 = 1. Then it is clear that F0(α1) = F0(1) ≥ 0. Hence all that is left is to show that F0(α ν ) ≥ 0 for all ν ≥ 2. For this purpose, it suffices to check that

F 0 ( α 2 + s ) = F 0 ( 4 + s ) 0 s 0 .

To this end, notice from (60) that

F 0 ( 1 + s ) s F 0 ( 1 + s ) F 0 ( 1 ) s = s 3 + 2 λ ( 5 λ + 1 ) s 2 + 9 λ 3 + 4 λ 2 + 24 λ + 15 s + ( λ 1 ) 5 λ 4 2 λ 3 10 λ 2 24 λ 12

holds for all s > 0. Replacing “s” by 3 + s on both sides, we get

F 0 ( 4 + s ) 3 + s ( s + 3 ) 3 + 2 λ ( 5 λ + 1 ) ( s + 3 ) 2 + 9 λ 3 + 4 λ 2 + 24 λ + 15 ( s + 3 ) + ( λ 1 ) 5 λ 4 2 λ 3 10 λ 2 24 λ 12 = s 3 + ( λ + 1 ) 2 + 9 λ 2 + 8 s 2 + 14 λ 4 + λ 2 4 ( λ + 1 ) 2 + 83 + 21 ( λ + 1 ) 2 + 6 s + 3 λ 6 + λ 4 7 ( λ 1 ) 2 + 27 + λ 2 ( 2 λ 1 ) 2 + 16 ( 4 λ + 1 ) 2 + ( λ + 2 ) 2 + 7 0

and therefore F0(4 + s) ≥ 0 for all s ≥ 0, as desired.


Corresponding author: Futoshi Takahashi, Osaka Central Advanced Mathematical Institute, 3-3-138 Sugimoto, Sumiyoshi-ku, Osaka 558-8585, Japan, E-mail:

Acknowledgments

The first author (N.H.) was supported by JSPS KAKENHI Grant number JP22KJ2604 and is supported by JSPS KAKENHI Grant number JP22K13943. He thanks Prof. Y. Kabeya (Osaka Metropolitan University) for his great support and encouragement. The second author (F.T.) was supported by JSPS Grant-in-Aid for Scientific Research (B), No.19H01800. This work was also partly supported by MEXT Promotion of Distinctive Joint Research Center Program JPMXP0723833165.

  1. Research ethics: Not applicable.

  2. Informed consent: Not applicable.

  3. Author contributions: Both authors have accepted responsibility for the entire content of this manuscript and approved its submission.

  4. Use of Large Language Models, AI and Machine Learning Tools: Not applicable.

  5. Conflict of interest: The authors have no possible conflict of interest.

  6. Research funding: None declared.

  7. Data availability: Not applicable.

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Received: 2024-03-28
Accepted: 2025-04-27
Published Online: 2025-07-02

© 2025 the author(s), published by De Gruyter, Berlin/Boston

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