A curl-free improvement of the Rellich-Hardy inequality with weight
aa r X i v : . [ m a t h . A P ] J a n A CURL-FREE IMPROVEMENT OF THE RELLICH-HARDYINEQUALITY WITH WEIGHT
NAOKI HAMAMOTO AND FUTOSHI TAKAHASHI
Abstract.
We consider the best constant in the Rellich-Hardy inequality(with a radial power weight) for curl-free vector fields on R N , originally foundby Tertikas-Zographopoulos [12] for unconstrained fields. This inequality isconsidered as an intermediate between Hardy-Leray and Rellich-Leray inequal-ities. Under the curl-free condition, we compute the new explicit best constantin the inequality and prove the non-attainability of the constant. This paperis a sequel to [6, 7]. Introduction
Let N ∈ N be an integer with N ≥
2, let γ ∈ R and put x = ( x , · · · , x N ) ∈ R N . In the following, the notation D γ ( R N ) denotes the set of compact supported,smooth vector fields on R N u = ( u , u , · · · , u N ) : R N ∋ x u ( x ) ∈ R N with the integrability condition Z R N | u | | x | γ − dx < ∞ . Preceding results and motivation.
It is well known that the
Hardy-Lerayinequality (cid:0) γ + N − (cid:1) Z R N | u | | x | | x | γ dx ≤ Z R N |∇ u | | x | γ dx (1)holds for any vector field u ∈ D γ ( R N ), with the best constant on the left-hand side.This was first proved by J. Leray [10] for ( N, γ ) = (3 , u to be divergence-free (under the additional assumption of axisymmetryfor N ≥ N = 2, it was shown that the inequality C γ Z R | u | | x | | x | γ dx ≤ Z R |∇ u | | x | γ dx holds with the best constant C γ = ( γ − γ − γ (cid:0) | γ + 1 | ≤ √ (cid:1) γ + 1 (cid:0) | γ + 1 | > √ (cid:1) for divergence-free vector fields u ∈ D γ ( R ). Since there is an isometry on R between divergence-free fields and curl-free fields, the same conclusion also holds for the two-dimensional Mathematics Subject Classification.
Primary 26D10; Secondary 35A23.
Key words and phrases.
Rellich-Hardy inequality, curl-free vector fields, best constant. curl-free fields. As a generalization of this result, we have derived in recent papers[6, 7] the Hardy-Leray inequality H N,γ Z R N | u | | x | | x | γ dx ≤ Z R N |∇ u | | x | γ dx (2)for curl-free fields u ∈ D γ ( R N ) with the best constant H N,γ = (cid:0) γ + N − (cid:1) N − ( γ + N − ) N − ( γ + N − ) if (cid:12)(cid:12) γ + N (cid:12)(cid:12) ≤ √ N + 1 , (cid:0) γ + N − (cid:1) + N − C γ = H ,γ , this result recovers Costin-Maz’ya’s one for N = 2.The Rellich-Leray inequality is given by B N,γ Z R N | u | | x | | x | γ dx ≤ Z R N |△ u | | x | γ dx (4)for unconstrained fields u ∈ D γ − ( R N ), where the constant B N,γ is sharp when B N,γ = min ν ∈ N ∪{ } (cid:16) ( γ − − (cid:0) ν + N − (cid:1) (cid:17) . (5)This was found by Rellich [11] for γ = 0 and Caldiroli-Musina [2] for γ = 0. Inrecent papers [6, 7], we additionally considered the curl-free improvement of theRellich-Leray inequality: if u ∈ D γ − ( R N ) is assumed to be curl-free, then thesame inequality (4) holds with the best constant B N,γ : B N,γ = min (cid:26)(cid:16) ( γ − − N (cid:17) , min ν ∈ N ( γ + N − ) + α ν ( γ + N − ) + α ν (cid:16) ( γ − − (cid:0) ν + N − (cid:1) (cid:17) (cid:27) ;(6)here and hereafter we use the notation α s = s ( s + N −
2) (7)for any s ∈ R .In this paper, we are interested in another version of Rellich-Leray inequality[1, 5, 12]: Z R N |△ u | | x | γ dx ≥ A N,γ Z R N |∇ u | | x | | x | γ dx, ∀ u ∈ D γ − ( R N ) (8)holds with the best constant A N,γ = min ν ∈ N ∪{ } A N,γ,ν , where A N,γ,ν := (cid:0) γ − N (cid:1) for ν = 0 , (cid:16) ( γ − − ( ν + N − ) (cid:17) ( γ + N − ) + α ν for ν ∈ N . (9)We call (8) the Rellich-Hardy inequality . This inequality was first found for N ≥ N ∈ { , } and γ = 0, with the best constants A , = and A , = 3. See also Cazacu [3] for theunified proof of the inequality when γ = 0.Now, let us assume that γ satisfies A N,γ = min ν ∈ N ∪{ } A N,γ,ν = A N,γ, = (cid:0) γ − N (cid:1) . ELLICH-HARDY INEQUALITY FOR CURL-FREE FIELDS 3
Then we see that a successive application of Rellich-Hardy and Hardy-Leray in-equalities reproduces Rellich-Leray inequality (4): we have Z R N |△ u | | x | γ dx ≥ A N,γ, Z R N |∇ u | | x | γ − dx (from (8)) ≥ A N,γ, (cid:0) γ + N − (cid:1) Z R N | u | | x | | x | γ − dx (cid:18) from (1)with γ replaced by γ − (cid:19) = (cid:16) ( γ − − (cid:0) N − (cid:1) (cid:17) Z R N | u | | x | | x | γ dx ≥ B N,γ Z R N | u | | x | | x | γ dx with B N,γ given by (5). Hence, Rellich-Hardy inequality can be considered as astronger version of Rellich-Leray inequality and plays a role as an intermediatebetween Rellich-Leray and Hardy-Leray inequalities.In the context of the curl-free improvement, it seems also natural to ask whetherthe same phenomenon will happen in (8); in particular, the main interest of ourproblem is how the best constant in (8) will change when u is assumed to becurl-free.1.2. Results.
Motivated by the observation above, we aim to derive the best con-stant in Rellich-Hardy inequality for curl-free fields. Now, our main result reads asfollows:
Theorem 1.
Let N ≥ . Let u ∈ D γ − ( R N ) be a curl-free vector field. Then theinequality Z R N |△ u | | x | γ dx ≥ C N,γ Z R N |∇ u | | x | | x | γ dx (10) holds with the best constant C N,γ expressed as C N,γ = min ν ∈ N ∪{ } C N,γ,ν , where C N,γ, = (cid:16) ( γ − − N (cid:17) (cid:0) γ + N − (cid:1) + N − A N,γ, , (11) C N,γ, = (cid:0) γ − N − (cid:1) (cid:16)(cid:0) γ + N − (cid:1) + N − (cid:17)(cid:0) γ + N − (cid:1) + 3( N − and C N,γ,ν = (cid:16) ( γ − − ( ν + N − ) (cid:17) (cid:16) ( γ + N − ) + α ν (cid:17)(cid:16) ( γ − − ( ν + N − ) (cid:17) +2( γ − (cid:16) (2 γ + N − α ν +( N − ( γ + N − ) (cid:17) for ν ≥ . Moreover, we obtain a stronger inequality by adding a remainder term to theright-hand side of (10).
N. HAMAMOTO AND F. TAKAHASHI
Theorem 2.
Let C N,γ be the same constant as in Theorem 1. Then there existsan absolute constant c > such that the inequality Z R N |△ u | | x | γ dx − C N,γ Z R N |∇ u | | x | | x | γ dx ≥ c Z R N (cid:12)(cid:12)(cid:12) ∇ (cid:16) | x | − N − γ ( x · ∇ ) (cid:0) | x | γ + N − u (cid:1)(cid:17)(cid:12)(cid:12)(cid:12) | x | γ − dx (12) holds for all curl-free fields u ∈ D γ − ( R N ) . Remark 3.
From the proof below, we see that the constant c on the right-hand sideof (12) can be estimated by c ≥ γ ≤ ,c ≥ / N ≥ γ > ,c ≥ / N = 2 and γ > . However, the best possible (the largest) value of c is unknown. As a direct consequence of Theorem 2, we can conclude that the best constant C N,γ of the inequality (10) is never attained in D γ − ( R N ) \ { } : Corollary 4.
If the equation Z R N |△ u | | x | γ dx = C N,γ Z R N |∇ u | | x | | x | γ dx holds for a curl-free field u ∈ D γ − ( R N ) , then u ≡ .Proof. Let (10) holds true. Then the right-hand side of (12) must vanish. Thus r∂ r (cid:16) r γ + N − u ( r σ ) (cid:17) = r γ + N − x holds for some constant vector filed x , where ( r, σ ) = ( | x | , x / | x | ). Integratingboth sides on any interval [ s, r ] ⊂ R + with respect to the measure r dr , we have u ( r σ ) = (cid:0) sr (cid:1) γ + N − u ( s σ ) + ( sr ) γ + N − − γ + N − x ( γ = 2 − N ) u ( s σ ) + x log sr ( γ = 2 − N ) . In the case γ = 2 − N , take the limit s → ∞ (resp. s → +0) when γ < − N (resp. γ > − N ), then we obtain u ( r σ ) ≡ − γ + N − x and hence u ( x ) ≡ u ( ) . This fact together with the integrability condition R R N | u | | x | γ − dx < ∞ says that u ( ) must vanish, whence u ≡ . In the case γ = 2 − N , taking s → u ( r σ ) = u ( ) + x lim s → log sr and the finiteness of the right-hand side yields x = . Therefore, we see againthat u ≡ u ( ) and hence u ≡ . (cid:3) As another direct consequence of Theorem 2, we have the following fact:
ELLICH-HARDY INEQUALITY FOR CURL-FREE FIELDS 5
Corollary 5.
Let C N,γ be the same constant as in Theorem 1. Then there existsan absolute constant c > such that the inequality Z R N |∇△ φ | | x | γ dx − C N,γ Z R N | D φ | | x | | x | γ dx ≥ c Z R N (cid:12)(cid:12)(cid:12) ∇ (cid:16) | x | − N − γ ( x · ∇ ) (cid:0) | x | γ + N − ∇ φ (cid:1)(cid:17)(cid:12)(cid:12)(cid:12) | x | γ − dx holds for all scalar field φ such that ∇ φ ∈ D γ − ( R N ) . Here D φ denotes a Hessianmatrix of φ . Overview of the remaining content of the present paper.
The rest of thispaper is organized as follows: Section 2 provides a minimum required notationsand definitions, and reviews a representation of curl-free fields. Section 3 gives theproof of Theorem 1: we recall from [7] the scalar-potential expression of L integralsof curl-free fields; after that, we derive Lemma 7 as a key tool for evaluating theratio of the two integrals in (10), which also plays a computational part in theproof of Theorem 2. The proof of Lemma 7 is separated into two cases. Since boththe cases use similar techniques and consist of long calculations, we prove onlyone case in the same section, and postpone the other case in Section 6. Section 4proves Theorem 2 by using an operator-polynomial representation of Rellich-Hardyintegral quotient and by making full use of Lemma 7. Section 5 observes curl-freeimprovement phenomena of best constants in some cases.2. Preliminary for the proof of main theorem
Notations and definitions in vector calculus on ˙ R N . Here we summarizethe minimum required notations and definitions for the proof of our main theorems.We basically use the notation˙ R N = R N \ { } and S N − = (cid:8) x ∈ R N : | x | = 1 (cid:9) . For every vector x ∈ ˙ R N , the notation r = | x | > , σ = x / | x | ∈ S N − denotes the radius of x and its unit-vector part, which defines the smooth trans-formation ˙ R N → R + × S N − , x ( r, σ )together with its inverse R + × S N − → ˙ R N , ( r, σ ) r σ . Every vector field u = ( u , u , · · · , 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 = 0for 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 ∇ = (cid:16) ∂∂x , · · · , ∂∂x N (cid:17) can be decomposedinto the radial derivative ∂ r and the spherical gradient ∇ σ as ∇ = σ ∂ r + 1 r ∇ σ , N. HAMAMOTO AND F. TAKAHASHI in order that ∂ r f = ( ∇ f ) R = σ · ∇ f and r ∇ σ f = ( ∇ f ) S for all f ∈ C ∞ ( ˙ R N ). Thenotation ∂ := r∂ r = x · ∇ (13)denotes an alternative radial derivative, in order that the above decompositionformula of ∇ can be rewritten as r ∇ = σ ∂ + ∇ σ . The Laplace operator △ = P Nk =1 ∂ /∂x k is known to be expressed in terms of ( r, σ )by the formula △ = 1 r N − ∂ r (cid:0) r N − ∂ r (cid:1) + 1 r △ σ = 1 r (cid:0) ∂ + ( N − ∂ + △ σ (cid:1) , where △ σ denotes the Laplace-Beltrami operator on S N − . We understand thatthe action of the operator ∂ r or ∂ on a vector field u is associated with the function r u ( r σ ) for σ ∈ S N − 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 △ onthe scalar field r = | x | or its powers gives ∇ r = σ and △ r s = α s r s − for all s ∈ R ,where α s is the same as in (7).2.2. Radial-spherical-scalar representation of curl-free fields.
Every vectorfield 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 ∈ { , · · · , N } , or equivalently if there exists a scalar field φ ∈ C ∞ ( R N ) satisfying u = ∇ φ on R N . (14)In view of this equation, we say that u has a scalar potential φ . As anotherrepresentation of curl-free fields, let us recall the following fact: Proposition 6 ([7]) . Let λ ∈ R . Then a vector field u ∈ C ∞ ( R N ) N is curl-free ifand only if there exist two scalar fields f, ϕ ∈ C ∞ ( ˙ R N ) satisfying ( f is radially symmetric and R S N − ϕ ( r σ )d σ = 0 ∀ r > ,r − λ u = σ (cid:0) f + ( λ + ∂ t ) ϕ (cid:1) + ∇ σ ϕ on R N \ { } . Moreover, such f and ϕ are uniquely determined, and they are explicitly given bythe equations f = r − λ ∂φ and ϕ = r − λ (cid:0) φ − φ (cid:1) , where we set φ ( x ) = | S N − | R S N − φ ( | x | σ )d σ as the spherical mean of the scalarpotential φ given by (14) . In particular, if u has a compact support on ˙ R N , thenso do f and ϕ . Later, we will use Proposition 6 by choosing λ = 2 − N − γ . (See (20)).3. Proof of Theorem 1
In this section, we prove Theorem 1. Roughly speaking, the proof consists oftheoretical part ( § § § § L formulae of curl-free fieldsgiven in [7]. Instead, emphasis is placed on the computational part. ELLICH-HARDY INEQUALITY FOR CURL-FREE FIELDS 7
Reduction to the case of compact support on ˙ R N . Let φ be the scalarpotential of the curl-free field u satisfying φ ( ) = 0. The integrability condition u ∈ D γ − ( R N ) (cid:0) namely R R N | u | | x | γ − dx < ∞ (cid:1) together with the smoothness of u on R N implies that there exists an integer m > − γ satisfying u ( x ) = ∇ φ ( x ) = O ( | x | m ) , and hence φ ( x ) = O ( | x | m +1 ) , ∇ u ( x ) = O ( | x | m − ) , △ u ( x ) = O ( | x | m − )as x → . Then it additionally follows that the integrals Z R N φ | x | γ − dx, Z R N |∇ u | | x | γ − dx and Z R N |△ u | | x | γ dx (15)are all finite.For the purpose of deriving the best constant C N,γ in inequality (10), it is enoughto 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 asequence of curl-free fields by u n ( x ) := ∇ (cid:18) ζ (cid:18) n log | x | (cid:19) φ ( x ) (cid:19) for every n ∈ N , where ζ ∈ C ∞ ( R ) such that ζ ( t ) = (cid:26) t ≤ −
11 for 1 ≤ t . We use the abbreviationssuch as ζ n = ζ (cid:0) n log | x | (cid:1) , ζ ′ n = ζ ′ (cid:0) n log | x | (cid:1) , and ζ ′′ n = ζ ′′ (cid:0) n log | x | (cid:1) . Noticing theasymptotic formulae ∂ζ n = 1 n ζ ′ n = O ( n − ) , ∇ ζ n = σ nr ζ ′ n = σ r O ( n − ) , ∇ ζ ′ n = σ nr ζ ′′ n = σ r O ( n − ) N. HAMAMOTO AND F. TAKAHASHI as n → ∞ , we have the following calculations: u n = ∇ ( ζ n φ ) = ( ∇ ζ n ) φ + ζ n ∇ φ = σ φr O (1 /n ) + ζ n u , ∇ u n = ∇ (cid:16) σ nr ζ ′ n φ + ζ n u (cid:17) = σ nr ( ∇ ζ ′ n ) φ + ζ ′ n n ∇ σ φr + ( ∇ ζ n ) u + ζ n ∇ u = σσ φr O ( n − ) + ζ ′ n n (cid:16) σ r ∇ φ + (cid:16) ∇ σ r (cid:17) φ (cid:17) + (cid:16) σ r u (cid:17) O ( n − ) + ζ n ∇ u = σσ φr O ( n − ) + σ u r O ( n − ) + ζ n ∇ u (cid:18) where we abbreviate as vw := v ⊗ w = ( v i w j ) i,j ∈{ , ··· ,N } the tensor product of two vector fields (cid:19) , △ u n = △∇ ( ζ n φ ) = ∇ (cid:0) ( △ ζ n ) φ + 2( ∂ r ζ n ) ∂ r φ + ζ n △ φ (cid:1) = ∇ (cid:18)(cid:0) ∂ ζ n + ( N − ∂ζ n (cid:1) r − φ + 2 ∂ζ n r ∂ r φ (cid:19) + ∇ ( ζ n △ φ )= ∇ (cid:18)(cid:0) n − ζ ′′ n + ( N − n − ζ ′ n (cid:1) φr + 2 ζ ′ n n ∂ r φr (cid:19) + ∇ ( ζ n △ φ )= (cid:16) n − ∇ ζ ′′ n + ( N − n − ∇ ζ ′ n | {z } σ r O ( n − ) (cid:17) φr + (cid:16) n − ζ ′′ n + ( N − n − ζ ′ n | {z } O ( n − ) (cid:17) ∇ φr + 2 ∇ ζ ′ n n ∂ r φr + 2 ζ ′ n n ∇ ∂ r φr + ( ∇ ζ n ) △ φ + ζ n △∇ φ = σ r O ( n − ) φr + O ( n − ) ∇ φr + σ ∂ r φr O ( n − ) + O ( n − ) ∇ ∂ r φr + σ O ( n − ) △ φr + ζ n △∇ φ = σ φr O ( n − ) + u r O ( n − ) + σ u R r O ( n − ) + ∂ r u r O ( n − )+ σ div u r O ( n − ) + ζ n △ u hold as n → ∞ . Therefore, taking the L ( | x | γ dx ) integration yields Z R N | u n | | x | | x | γ dx = Z R N | ζ n u | | x | | x | γ dx + O ( n − ) → Z R N | u | | x | | x | γ dx, Z R N |∇ u n | | x | | x | γ dx = Z R N | ζ n ∇ u | | x | | x | γ dx + O ( n − ) → Z R N |∇ u | | x | | x | γ dx, Z R N |△ u n | | x | γ dx = Z R N | ζ n △ u | | x | γ dx + O ( n − ) → Z R N |△ u | | x | γ dx with the aid of the integrability conditions (15). This fact shows that the twointegrals 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 ofthe present section, we use the notation t = log | x | = log r ELLICH-HARDY INEQUALITY FOR CURL-FREE FIELDS 9 for an alternative radial coordinate obeying the differential rules ∂ t = r∂ r = ∂ = x · ∇ and dt = dr/r, (16)which reproduces the same notation ∂ given in (13). For any parameter λ ∈ R ,let f and ϕ be the scalar fields determined by the curl-free field u , as given inProposition 6, and we set v ( x ) = | x | − λ u ( x ) (17)as a new vector field in C ∞ c ( ˙ R N ) N . Then the equation v = σ (cid:0) f + ( λ + ∂ ) ϕ (cid:1) + ∇ σ ϕ (18)holds on ˙ R N . Here we keep in mind that the equations (17) and (18) are invariantunder the following replacement of the quadruple:( f, ϕ, v , u ) (cid:0) ∂f, ∂ϕ, ∂ v , r λ − ∂ ( r − λ u ) (cid:1) . (19)Now, we choose λ = 2 − N − γ, (20)and let us recall from [7, § v , f, ϕ ) as follows: Z R N |∇ u | | x | γ dx = Z Z R × S N − (cid:0) ( λ − | v | + | ∂ v | + |∇ σ v | (cid:1) dt d σ, (21) Z Z R × S N − |∇ σ v | dt d σ = Z Z R × S N − (cid:16) ( △ σ ϕ ) + (cid:0) ( λ − − N (cid:1) |∇ σ ϕ | (cid:17) dt d σ + Z Z R × S N − (cid:0) | ∂ ∇ σ ϕ | + ( N − | v | (cid:1) dt d σ, (22) Z Z R × S N − | v | dt d σ = Z Z R × S N − (cid:0) f + ( ∂ϕ ) + λ ϕ + |∇ σ ϕ | (cid:1) dt d σ = Z Z R × S N − (cid:0) f + ϕ (cid:0) λ − ∂ − △ σ (cid:1) ϕ (cid:1) dt d σ. (23)Here the last equality follows from integration by parts together with the supportcompactness of v or f, ϕ . Applying (23) to (19), we also obtain Z Z R × S N − | ∂ v | dt d σ = Z Z R × S N − (cid:16) ( ∂f ) + ( ∂ ϕ ) + λ ( ∂ϕ ) + | ∂ ∇ σ ϕ | (cid:17) dt d σ = Z Z R × S N − (cid:16) f ( − ∂ ) f + ϕ (cid:0) λ − ∂ − △ σ (cid:1) ( − ∂ ϕ ) (cid:17) dt d σ (24) by integration by parts. After plugging (22) into (21), substitute (23) and (24) intothe L terms of v and ∂ v ; then we get Z R N |∇ u | | x | γ dx = (cid:0) ( λ − + N − (cid:1) Z Z R × S N − | v | dt d σ + Z Z R × S N − | ∂ v | dt d σ + Z Z R × S N − (cid:16) ( △ σ ϕ ) + (cid:0) ( λ − − N (cid:1) |∇ σ ϕ | + | ∂ ∇ σ ϕ | (cid:17) dt d σ = (cid:0) ( λ − + N − (cid:1) Z Z R × S N − (cid:0) f + ϕ (cid:0) λ − ∂ − △ σ (cid:1) ϕ (cid:1) dt d σ + Z Z R × S N − (cid:16) f ( − ∂ ) f + ϕ (cid:0) λ − ∂ − △ σ (cid:1) ( − ∂ ϕ ) (cid:17) dt d σ + Z Z R × S N − ϕ (cid:16) △ σ − (cid:0) ( λ − − N (cid:1) △ σ + ∂ △ σ (cid:17) ϕdt d σ = Z Z R × S N − (cid:0) ϕ P ( − ∂ , −△ σ , λ ) ϕ + f P ( − ∂ , λ ) f (cid:1) dt d σ by integration by parts, where we have defined two polynomials P and P by P ( τ, a, λ ) = (cid:0) ( λ − + N − (cid:1) (cid:0) λ + τ + a (cid:1) + (cid:0) λ + τ + a (cid:1) τ + a + (cid:0) ( λ − − N (cid:1) a + aτ = a + (cid:0) λ − λ + 4 − N + 2 τ (cid:1) a + (cid:0) λ + τ (cid:1) (cid:16) ( λ − + N − τ (cid:17) , P ( τ, λ ) = ( λ − + N − τ. Now let us replace γ by γ −
1; in view of (20), this manipulation is equivalent toreplacing λ by λ + 1. Then the result of the above integral computation changesinto Z R N |∇ u | | x | | x | γ dx = Z Z R × S N − (cid:0) ϕP ( − ∂ , −△ σ ) ϕ + f P ( − ∂ ) f (cid:1) dt d σ, where and hereafter we abbreviate as P ( τ, a ) := P ( τ, a, λ + 1)= a + (cid:16) (cid:0) λ − λ + τ (cid:1) − N (cid:17) a + (cid:0) ( λ + 1) + τ (cid:1) (cid:0) λ + N − τ (cid:1) ,P ( τ ) := P ( τ, λ + 1) = λ + N − τ, (25)as the expression in terms of f, ϕ for the integral on the right-hand side of theRellich-Hardy inequality (10). To express the left-hand side, we exploit the result ELLICH-HARDY INEQUALITY FOR CURL-FREE FIELDS 11 of [7, Eq.(30),(31) with λ replaced by λ + 1]: it holds that Z R N |△ u | | x | γ dx = Z Z R × S N − (cid:16) ϕ Q ( − ∂ , −△ σ ) ϕ + f Q ( − ∂ ) f (cid:17) dt d σ, where Q and Q are the polynomials given by Q ( τ, a ) = (cid:0) τ + a + ( λ − (cid:1) (cid:0) τ + a + ( λ + 1) (cid:1) (cid:0) τ + a + ( λ + N − (cid:1) − (2 λ + N ) a ! = (cid:0) τ + a + ( λ − (cid:1)(cid:16) τ + (cid:0) a + α λ +1 ) + ( N − (cid:1) τ + ( a − α λ +1 ) (cid:17) ,Q ( τ ) = (cid:0) τ + ( λ − (cid:1) (cid:0) τ + ( λ + N − (cid:1) . (26)To proceed further, let us apply to ϕ and f the one-dimensional Fourier transfor-mation with respect to t : we set b ϕ ( τ, σ ) = 1 √ π Z R e − iτt ϕ ( e t σ ) dt, b f ( τ ) = 1 √ π Z R e − iτt f ( e t σ ) dt for ( τ, σ ) ∈ R × S N − , where i = √−
1. Also we apply to b ϕ the spherical harmonicsdecomposition: b ϕ = X ν ∈ N b ϕ ν , ( −△ σ b ϕ ν = α ν b ϕ ν ,α ν = ν ( ν + N − ∀ ν ∈ N . Now, we are in a position to evaluate the quantity R R N |△ u | | x | γ dx R R N |∇ u | | x | γ − dx , (27)which we simply call the R-H quotient . To this end, by using (25), (26) and the L ( R ) isometry of the Fourier transformation, we have R R N |△ u | | x | γ dx R R N |∇ u | | x | γ − dx = Z Z R × S N − X ν ∈ N Q ( τ , α ν ) | c ϕ ν | + Q ( τ ) | b f | ! dτ d σ Z Z R × S N − X ν ∈ N P ( τ , α ν ) | c ϕ ν | + P ( τ ) | b f | ! dτ d σ ≥ min (cid:26) inf τ ∈ R \{ } Q ( τ ) P ( τ ) , inf ν ∈ N inf τ ∈ R \{ } Q ( τ , α ν ) P ( τ , α ν ) (cid:27) = min (cid:26) inf τ> Q ( τ ) P ( τ ) , inf ν ∈ N inf τ> Q ( τ, α ν ) P ( τ, α ν ) (cid:27) . (28)Hence, our goal is reduced to evaluate the fractions Q /P and Q /P . In thefollowing subsections, we will show that the infimum values of these fractions areachieved at τ = 0.3.3. Evaluation of Q /P . A direct calculation yields Q ( τ ) P ( τ ) = (cid:0) τ + ( λ − (cid:1) (cid:0) τ + ( λ + N − (cid:1) λ + N − τ = τ + ( λ + N − + ( N − − (2 λ + N − τ + λ + N − ! for all τ ≥
0. The last expression is of the form g ( τ ) = τ + a − bτ + c for someconstants a, b ≥ c >
0, which leads to g ( τ ) − g (0) = τ + bτc ( τ + c ) ≥ τ . Thus wehave 1 τ (cid:18) Q ( τ ) P ( τ ) − Q (0) P (0) (cid:19) ≥ ∀ τ > , (29)whence in particular we obtain inf τ> Q ( τ ) P ( τ ) = Q (0) P (0) = ( λ − ( λ + N − λ + N − The case when P has zeros. Here we specify when P ( τ, α ν ) = 0 happens.Notice from (25) that P ( τ, a ) is strictly monotone increasing in τ ≥ a >
0, and hence it holds that P ( τ, a ) > P (0 , a )= a + (cid:0) λ − / − − N (cid:1) a + ( λ + 1) (cid:0) λ + N − (cid:1) for all τ >
0, as well as that P ( τ, α ) > P (0 , α ) = P (0 , N −
1) = λ (cid:0) ( λ + 1) + 3( N − (cid:1) for all τ >
0. Notice on the right-hand side of the (three lines) above inequalitythat the center of the graph of the quadratic function a P (0 , a ) is located at a = − (cid:0) λ − (cid:1) + + N ≤ N = α . Then we see that for all τ > ν ≥ P ( τ, α ν ) > P (0 , α ν ) ≥ P (0 , α )= λ + 2 λ + 5 N λ − N + 1) λ + ( N + 1)(2 N − λ ( λ + 1) + 2(2 N − λ + ( N + 1)( λ − + 2( N − ≥ N − > . In view of the above discussion, we see that P ( τ, α ν ) = 0 holds if and only if ( τ, ν, λ ) = (0 , , Q /P , we have to deal withthe case λ = 0 (cid:0) or equivalently γ = 2 − N (cid:1) as a special one. For this reason, inthe rest of this paper we always assume λ = 0 (cid:0) ⇔ γ = 2 − N (cid:1) unless others arespecified.3.5. Evaluation of Q /P . Let us check thatinf τ> Q ( τ, a ) P ( τ, a ) = Q (0 , a ) P (0 , a ) ∀ a ∈ { α ν } ν ∈ N in order to evaluate (28) from below. This equation is equivalent to the in-equality Q ( τ,a ) P ( τ,a ) ≥ Q (0 ,a ) P (0 ,a ) ( ∀ τ > Q ( τ, a ) P (0 , a ) − P ( τ, a ) Q (0 , a ) to be nonnegative. However, we further showthe following stronger fact, which serves as a key tool for the proof of Theorem 2: Lemma 7.
There exists a constant number c > such that the inequality τ (cid:18) Q ( τ, a ) P ( τ, a ) − Q (0 , a ) P (0 , a ) (cid:19) ≥ c holds for all τ > and a ∈ { α ν } ν ∈ N . ELLICH-HARDY INEQUALITY FOR CURL-FREE FIELDS 13
Here we give the proof of the lemma only for the case γ ≤
1. Since the proof for γ > § Maxima in thecourse of the proof. However, we need many computational techniques to simplifythe calculations and ideas to make the proof understandable, even with the use ofMaxima.
Proof of Lemma 7 for γ ≤ (cid:0) or equivalently λ ≥ − N (cid:1) . It suffices to check theinequality for c = 1:1 τ (cid:18) Q ( τ, a ) P ( τ, a ) − Q (0 , a ) P (0 , a ) (cid:19) ≥ ∀ τ > , ∀ a ∈ { α ν } ν ∈ N . To this end, we directly compute the left-hand side minus right-hand side: by using(25) and (26) we get1 τ (cid:18) Q ( τ, a ) P ( τ, a ) − Q (0 , a ) P (0 , a ) (cid:19) −
1= 1 τ (cid:0) τ + a + ( λ − (cid:1) (cid:16) τ + (cid:0) a + α λ +1 ) + ( N − (cid:1) τ + ( a − α λ +1 ) (cid:17) a + (cid:0) λ − λ + τ ) − N (cid:1) a + (( λ + 1) + τ ) ( λ + N − τ ) − (cid:0) a + ( λ − (cid:1) ( a − α λ +1 ) a + (cid:0) λ − λ ) − N (cid:1) a + ( λ + 1) ( λ + N − ! −
1= (2 λ + N − | {z } ≥ G ( a ) + G ( a ) τP (0 , a ) P ( τ, a ) , (30)where we have defined G ( a ) := (2 λ + N ) a + (cid:16) (cid:0) λ − N + 5 (cid:1) (2 λ + N ) − N − (cid:17) a + (cid:18) λ + ( N − λ − N λ − N + 2 N − λ − N − λ − N − N + 4 (cid:19) a + ( N − λ + N − λ + 1) ,G ( a ) := (2 λ + N ) a + (cid:16) (2 λ + N ) (cid:0) ( λ − − N + 1 (cid:1) − N − (cid:17) a + ( N − λ + N − λ + 1) (31)as cubic and quadratic polynomials in a . Then the necessary and sufficient conditionfor the nonnegativity of (30) ( ∀ τ ≥
0) is given by the inequalities G ( a ) ≥ G ( a ) ≥ ∀ a ∈ { α ν } ν ∈ N , whence our goal is reduced to showing them. The first inequality is easier to prove,by considering the Taylor series of G ( a ) at a = α : a straightforward calculationyields G ( α + s ) = s (2 λ + N ) + s (cid:16) ( N − N + 2 λ −
2) + ( λ − (2 λ + N ) (cid:17) + 2( N − λ (2 λ + N −
1) (32)for all s ∈ R . Since λ ≥ − N , notice here that the coefficients of the powers of s are all nonnegative, which tells us that G ( α + s ) ≥ s ≥
0. This factdirectly implies G ( a ) ≥ a ∈ { α ν } ν ∈ N , as desired. Now, all we have to do is to show G ( a ) ≥ a ∈ { α ν } ν ∈ N . To do so, let usconsider the Taylor series of G ( a ) at a = α , and we get G ( α + s ) = s (2 λ + N ) + s G ( λ ) + s G ( λ ) + 2( N − λ (2 λ + N −
1) (33)by a straightforward calculation, where G ( λ ) := (2 λ + N − (cid:16) ( λ + 1) + λ + N + 3 (cid:17) + ( N − + 9 , (34) G ( λ ) := λ (2 λ + N −
8) + 2 λ (cid:0) − λ − N + N (cid:1) + N (2 λ + N ) . (35)Noticing that G ( λ ) ≥ G ( α ) ≥
0, we aim to prove the following fact:if 1 − N ≤ λ ≤ G ( λ ) ≥ , (36)or if 1 < λ then G ( α + s ) ≥ ∀ s ≥ , (37)which implies the desired inequality G ( a ) ≥ ∀ a ∈ { α ν } ν ∈ N .For the proof of (36), let λ be parameterized as λ = 1 − N s , ≤ s ≤ . Then we directly compute G ( λ ) = (cid:18) − N s (cid:19) ( N − N s −
6) + 2 (cid:18) − N s (cid:19) (cid:0) N s − − N + N (cid:1) + N ( N − N s + 2)= N (1 − s ) s + N ( s − s + N (1 − s ) (cid:0) − s − s (cid:1) + 2 N (2 + 3 s − s ) + N (19 s − − ( N − (1 − s ) s + ( N − s (cid:16) s + (1 − s ) (cid:0) s + 4 (cid:1) (cid:17) + ( N − (cid:16) (1 − s ) s (cid:0) (1 − s ) + 4 s (cid:1) + 2 (cid:0) − s + 3 s (cid:1) (cid:17) + ( N − (cid:16) (1 − s ) (cid:0) s + 2(1 − s ) + 5(1 − s ) (cid:1) + 3 − s (cid:17) + ( N − (cid:16) (1 − s ) s (17 − s − s ) + 21 − s − s (cid:17) + 2 s (cid:16) (1 − s )(3 + s ) (cid:0) − s + s (cid:1) + 4 (cid:17) as a Taylor series of the function N
7→ G ( λ ) = G (cid:0) − Ns (cid:1) at N = 2. Notice herethat the coefficients of the powers of N − ≤ s ≤ G ( λ ) ≥
0, as desired.Now, all that is left is to show (37). To this end, notice from (33) that1 s G ( α + s ) ≥ s G ( λ ) + G ( λ ) ≥ s (2 λ + N − (cid:16) ( λ + 1) + λ (cid:17) + λ (2 λ + N −
8) + 2 λ (cid:0) − λ − N + N (cid:1) ELLICH-HARDY INEQUALITY FOR CURL-FREE FIELDS 15 holds for all s >
0. Replacing s by N + 1 + s on both sides, we then get G ( α + s ) N + 1 + s = 1 N + 1 + s G ( α + N + 1 + s ) ≥ ( N + 1 + s )(2 λ + N − (cid:0) ( λ + 1) + λ (cid:1) + λ (2 λ + N −
8) + 2 λ (cid:0) − λ − N + N (cid:1) = 2( λ − + ( λ − ( N + 2) + 4( λ − ( s + 2 N − λ − (cid:0) ( N + 6) s + 2 N + 6 N − (cid:1) + 2( λ − (cid:0) (3 N + 5) s + 5 N + 2 N − (cid:1) + 5 N s + 7 N − N − s ≥
0. Notice here that the coefficients of the powers of λ − N ≥
2. Therefore, from the assumption λ > G ( α + s ) ≥
0, as desired. (cid:3)
Since the polynomial function P ( τ, a ) is quadratic in τ , it is clear from (30) thatlim τ →∞ τ (cid:18) Q ( τ, a ) P ( τ, a ) − Q (0 , a ) P (0 , a ) (cid:19) = 1for each a ≥ α . Therefore, the constant number c of Lemma 7 is optimal when c = 1, in the sense thatinf τ> inf ν ∈ N τ (cid:18) Q ( τ, α ν ) P ( τ, α ν ) − Q (0 , α ν ) P (0 , α ν ) (cid:19) = 1holds as far as γ ≤ A lower bound for the R-H quotient.
In view of the estimate (28) for theR-H quotient (27), it follows from § § § Z R N |△ u | | x | γ dx ≥ C N,γ Z R N |∇ u | | x | γ − dx holds for curl-free fields u with the constant number C N,γ = min (cid:26) inf ν ∈ N inf τ> Q ( τ, α ν ) P ( τ, α ν ) , inf τ> Q ( τ ) P ( τ ) (cid:27) = min (cid:26) min ν ∈ N Q (0 , α ν ) P (0 , α ν ) , Q (0) P (0) (cid:27) . Notice from (25) and (26) that the last two fractions are explicitly written as Q (0 , α ν ) P (0 , α ν ) = (cid:0) α ν + ( λ − (cid:1) ( α λ +1 − α ν ) ( α λ +1 − α ν ) + (2 λ + N − (cid:0) (2 λ + 1) α ν − ( N − λ + 1) (cid:1) = (cid:16) ( γ − − ( ν + N − ) (cid:17) (cid:16) α ν + ( γ + N − ) (cid:17)(cid:16) ( γ − − ( ν + N − ) (cid:17) +2( γ − (cid:16) (2 γ + N − α ν +( N − ( γ + N − ) (cid:17) ,Q (0) P (0) = ( λ − ( λ + N − λ + N − (cid:0) ( γ − − N (cid:1) (cid:0) γ + N − (cid:1) + N − , by recalling the notation (20) together with the aid of the identity α s − α t = (cid:0) s + N − (cid:1) − (cid:0) t + N − (cid:1) ∀ s, t ∈ R ; (38) in other words, we have Q (0 , α ν ) P (0 , α ν ) = C N,γ,ν , Q (0) P (0) = C N,γ, in terms of the same notation in Theorem 1. Therefore, we have obtained C N,γ = min ν ∈ N ∪{ } C N,γ,ν as a lower bound for the R-H quotient (27), which coincides with the same constantnumber C N,γ given in Theorem 1.3.7.
Sharpness of C N,γ . We show here the optimality for the constant C N,γ in theinequality (10). To this end, we construct a sequence of curl-free fields minimizingthe value of the R-H quotient (27). First of all, choose ν ∈ N ∪ { } to be such thatmin ν ∈ N ∪{ } C N,γ,ν = C N,γ,ν . If γ = 2 − N , by the same computation as (44) below ( §
5) we have C N,γ, > C N,γ, for all N ≥
2, which implies that ν = 1. Hence it follows from § P (0 , α ν ) > { u n } n ∈ N ⊂ C ∞ c ( ˙ R N ) N by the formula u n ( x ) := ( x | x | λ − h (cid:0) n log | x | (cid:1) if ν = 0 ∇ (cid:0) | x | λ +1 ϕ n ( x ) (cid:1) otherwisetogether with ϕ n ( x ) = h (cid:0) n log | x | (cid:1) Y ( x / | x | )for any h ∈ C ∞ c ( R ) \{ } such that R R ( h ( t )) dt = 1. Here Y ∈ C ∞ ( S N − ) \ { } denotes a spherical harmonic function satisfying the eigenequation −△ σ Y = α ν Y on S N − . 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 (17). Then we have v n = ( σ f n if ν = 0 σ ( ∂ + λ + 1) ϕ n + ∇ σ ϕ n otherwise , where f n is given by f n ( x ) = h ( n log | x | ). In this setting, let us now apply theformulae (25) and (26) to the case ( u , f, ϕ ) = ( u n , f n ,
0) or ( u , f, ϕ ) = ( u n , , ϕ n ).Then we have R R N |△ u n | | x | γ dx R R N |∇ u n | | x | γ − dx = R R h ( tn ) Q ( − ∂ t ) h ( tn ) dt R R h ( tn ) P ( − ∂ t ) h ( tn ) dt if ν = 0 , R R h ( tn ) Q ( − ∂ t , α ν ) h ( tn ) dt R R h ( tn ) P ( − ∂ t , α ν ) h ( tn ) dt otherwise . = R R h ( t ) Q (cid:0) − n − ∂ t (cid:1) h ( t ) dt R R h ( t ) P ( − n − ∂ t ) h ( t ) dt if ν = 0 , R R h ( t ) Q ( − n − ∂ t , α ν ) h ( t ) dt R R h ( t ) P ( − n − ∂ t , α ν ) h ( t ) dt otherwise . ELLICH-HARDY INEQUALITY FOR CURL-FREE FIELDS 17
Notice on the right-hand side that the denominator always exceeds a fixed positivenumber, since P (0) ≥ N − > P (0 , α ν ) > n → ∞ , we get R R N |△ u n | | x | γ dx R R N |∇ u n | | x | γ − dx = O (1 /n ) + Q (0) O (1 /n ) + P (0) if ν = 0 O (1 /n ) + Q (0 , α ν ) O (1 /n ) + P (0 , α ν ) otherwise −→ C N,γ,ν = C N,γ , which gives the desired sharpness of C N,γ .Now, the proof of Theorem 1 has been completed. (cid:3) Proof of Theorem 2
Let ν denote the positive integer such that C N,γ,ν = min ν ∈ N C N,γ,ν . In order to estimate the difference between both sides of the inequality (10), recallfrom the same calculation in the first line of (28) the expression of the integrals: Z R N |△ u | | x | γ dx = Z Z R × S N − Q ( τ ) | b f | + X ν ∈ N Q ( τ , α ν ) | c ϕ ν | ! dτ d σ, Z R N |∇ u | | x | | x | γ dx = Z Z R × S N − P ( τ ) | b f | + X ν ∈ N P ( τ , α ν ) | c ϕ ν | ! dτ d σ. Then we have the following estimate: Z R N |△ u | | x | γ dx − C N,γ Z R N |∇ u | | x | | x | γ dx = Z R N |△ u | | x | γ dx − min (cid:26) Q (0 , α ν ) P (0 , α ν ) , Q (0) P (0) (cid:27) Z R N |∇ u | | x | | x | γ dx ≥ Z R N |△ u | | x | γ dx − Z Z R × S N − Q (0) P (0) P ( τ ) | b f | − Z Z R × S N − Q (0 , α ν ) P (0 , α ν ) X ν ∈ N P ( τ , α ν ) | c ϕ ν | dτ d σ = Z Z R × S N − (cid:18) Q ( τ ) − Q (0) P (0) P ( τ ) (cid:19) | b f | dτ d σ + X ν ∈ N Z Z R × S N − (cid:18) Q ( τ , α ν ) − Q (0 , α ν ) P (0 , α ν ) P ( τ , α ν ) (cid:19) | c ϕ ν | dτ d σ ≥ Z Z R × S N − P ( τ ) τ | b f | dτ d σ + c X ν ∈ N Z Z R × S N − P ( τ , α ν ) τ | c ϕ ν | dτ d σ ≥ min { , c } Z Z R × S N − P ( τ ) τ | b f | + X ν ∈ N P ( τ , α ν ) τ | c ϕ ν | ! dτ d σ = min { , c } Z Z R × S N − (cid:16) ( ∂f ) P ( − ∂ ) ∂f + ( ∂ϕ ) P ( − ∂ , −△ σ ) ∂ϕ (cid:17) dt d σ = min { , c } Z R N (cid:12)(cid:12) ∇ (cid:0) | x | λ ∂ (cid:0) | x | − λ u (cid:1)(cid:1)(cid:12)(cid:12) | x | | x | γ dx, where the last equation follows by applying the replacement (19) to the integralequation in (25), and where we the fourth inequality follows by using the inequalities(29) and 1 τ (cid:18) Q ( τ , α ν ) P ( τ , α ν ) − Q (0 , α ν ) P (0 , α ν ) (cid:19) ≥ τ (cid:18) Q ( τ , α ν ) P ( τ , α ν ) − Q (0 , α ν ) P (0 , α ν ) (cid:19) ≥ c ∀ ( τ, ν ) ∈ ( R \ { } ) × N , as verified by using the same constant c given in Lemma 7. Finally, by restoringthe notations (20) and (16), we obtain Z R N |△ u | | x | γ dx − C N,γ Z R N |∇ u | | x | | x | γ dx ≥ c Z R N (cid:12)(cid:12)(cid:12) ∇ (cid:16) | x | − N − γ ( x · ∇ ) (cid:0) | x | γ + N − u (cid:1)(cid:17)(cid:12)(cid:12)(cid:12) | x | γ − dx for some absolute constant c >
0. The proof of Theorem 2 is now complete, althoughthe optimal value of c is not known. (cid:3) An observation of the best constant C N,γ in Theorem 1
Concerning the constants in the inequalities (8) and (10), it holds that C N,γ ≥ A N,γ (39)
ELLICH-HARDY INEQUALITY FOR CURL-FREE FIELDS 19 as a matter of course. Here we wish to evaluate whether the strict inequality C N,γ > A
N,γ holds or not. However, since the expression for C N,γ is complicatedand its full picture seems difficult to reveal, we describe it for only some specificvalues of γ .5.1. Preliminary: a review of A N,γ,ν . In view of the original best constant (9)in Rellich-Hardy inequality, let us observe the increase or decrease of the function ν A N,γ,ν . In terms of the notation (20), the expression of A N,γ,ν in (9) can berewritten as A N,γ,ν = ( λ + N − for ν = 0 , ( α ν − α λ ) α ν + λ for ν ∈ N . (40)with the aid of (38). Then a direct calculation from this expression yields A N,γ, − A N,γ, N − − λ + 4( N − λ + N − N + 5 λ + N − A N,γ,ν +1 − A N,γ,ν ν + N − α ν α ν +1 + λ (cid:16) ν ( ν + N −
1) + (cid:0) N − λ (cid:1)(cid:0) A N,γ, − A N,γ, (cid:1)(cid:17) ( α ν + λ ) ( α ν +1 + λ ) (42)for all ν ∈ N . Notice that the numerator of the right-hand side is monotoneincreasing in ν ≥
0, as well as that the denominator is always positive. Therefore,for every k ∈ N ∪ { } the two inequalities A N,γ,k ≤ A N,γ,k +1 and A N,γ,ν ≤ A N,γ,ν +1 ( ∀ ν ≥ k ) (43)are equivalent.5.2. The case γ = 2 − N (or equivalently λ = 0 ). Let us deal with this “singular”case, in the sense of § A N,γ and C N,γ , we have A N, − N = min (cid:26) A N, − N , , min ν ∈ N A N, − N ,ν (cid:27) = min (cid:26) (2 − N ) , min ν ∈ N α ν (cid:27) = min n ( N − , N − | {z } = α o = ( N = 2) N − N ≥ ,C N, − N = min (cid:26) C N, − N , , C N, − N , , min ν ∈ N \{ } C N, − N ,ν (cid:27) (44)= min n N − , N N − , C N, − N , o = min n N − , N N − , ( N +1)(2 N +1)2 N − o = N − . Here the third equality from the last in (44) follows with the aid of computing C N, − N ,ν = ( α ν + 1) (cid:16) − N − α ν − (cid:17) which is monotone increasing in ν ≥
2. Summarizing the results above, we haveobtained C , = C , , = 1 > A , , = A , ,C N, − N = C N, − N , = N − A N, − N , = A N, − N ( N ≥ . In particular, we see that the best constant in the two-dimensional Rellich-Hardyinequality (with γ = 1) can be really improved by the curl-free condition on thetest vector fields.5.3. The case γ = 2 − N (or equivalently λ = 0 ). Recall from § § C N,γ,ν = (cid:0) α ν + ( λ − (cid:1) ( α λ +1 − α ν ) P (0 , α ν ) ,P (0 , α ν ) = ( α λ +1 − α ν ) + (2 λ + N − (cid:0) (2 λ + 1) α ν − ( N − λ + 1) (cid:1) > ν ∈ N . By using this expression together with (40), a direct computation yields C N,γ,ν − A N,γ,ν − = − ν − λ − W ( α ν )( ν + N − α ν − + λ ) P (0 , α ν ) ,C N,γ,ν − A N,γ,ν +1 = 2 νW ( α ν )(1 − ν − N − λ ) ( α ν +1 + λ ) P (0 , α ν )for all ν ∈ N , where W ( α ν ) = λ (2 λ + N − λ + N ) + ( N − − (cid:0) α ν + λ − (cid:1) . Then we have ( C N,γ,ν − A N,γ,ν − ) ( C N,γ,ν − A N,γ,ν +1 ) ≤ { A N,γ,ν − , A N,γ,ν +1 } ≤ C N,γ,ν ≤ max { A N,γ,ν − , A N,γ,ν +1 } (45)for all ν ∈ N . Based on this fact, let us consider the following two simplest cases: The case A N,γ = A N,γ, . It holds from (11) that A N,γ = A N,γ, = C N,γ, ≥ C N,γ , and hence that C N,γ = A N,γ from (39). This fact says that the curl-free restrictioncauses no effect on the improvement of the best constant. Now, we seek for whenthe equation A N,γ = A N,γ, will happen. In view of the inequalities (43) with k = 1, this equation is equivalent to that both the inequalities A N,γ, ≤ A N,γ, and A N,γ, ≤ A N,γ, hold true; in other words, A N,γ = A N,γ, holds if and only if both the numeratorsof the right-hand sides of (41) and (42) ν =1 evaluate to3 λ + 4( N − λ + N − N + 5 ≥ , and α α + λ (cid:16) N + (cid:0) N − λ (cid:1) ( A N,γ, − A N,γ, ) (cid:17) = − λ − N − λ − ( N − N + 5) λ + 2 N ( N − ≥ . For example, if γ = 0 (cid:0) or equivalently λ = 2 − N (cid:1) , then the two inequalities become N − N − ≤ , and N − N + 12 N + 64 N − ≥ . This is the case when N ≤
4, and hence we have C , = C , , = A , , = A , = 0 ,C , = C , , = A , , = A , = ,C , = C , , = A , , = A , = 3 , which says that no curl-free improvement occurs when N ∈ { , , } and γ = 0.We may understand the above result in the following way: we can choose asequence { u n } n ∈ N of vector fields by the formula w n ( x ) = x f n ( | x | ) ( n = 1 , , · · · )in order that R R N |△ u n | | x | γ dx R R N |∇ w n | | x | γ − dx → A N,γ, ( n → ∞ ) , where { f n ( r ) } n ∈ N are a smooth functions vanishing near r = 0. We may seethis fact by noticing that each coordinate function x k ( k = 1 , · · · , N ) satisfies theeigenequation −△ σ x k = α x k . On the other hand, it is easy to check that { w n } n ∈ N are always curl-free, whence C N,γ ≤ A N,γ . Therefore, we get C N,γ = A N,γ from A N,γ = A N,γ, . The case A N,γ = A N,γ, . Since A N,γ, ≥ A N,γ, = A N,γ , we see from (43) and (45)that A N,γ,ν − ≤ C N,γ,ν ≤ A N,γ,ν +1 holds for all ν ∈ N . This fact impliesmin k ∈ N ∪{ } C N,γ, k = C N,γ, and min k ∈ N C N,γ, k − = C N,γ, , whence C N,γ = min { C N,γ, , C N,γ, } . On the other hand, a direct computation yields C N,γ, − C N,γ, = ( λ − ( A N,γ, − A N,γ, ) + ( N − λ +2 N − + N + N − λ + N − ( λ + 1) + 3( N − > A N,γ, ≥ A N,γ, . Therefore, it holds from (11) that C N,γ = C N,γ, = A N,γ, , or equivalently that C N,γ − A N,γ = A N,γ, − A N,γ, . This fact together with (41) shows that the inequality C N,γ > A
N,γ (namely theeffect of the curl-free-improvement) holds as far as the right-hand side of (41) isstrictly negative, or equivalently (cid:12)(cid:12) γ − N +46 (cid:12)(cid:12) < p N − N + 1 . For example, this is the case if γ = 0 and N ≥
5, whence we have C N, = C N, , = A N, , = ( N − N − N +3 > N = A N, , = A N, for all N ≥ { w n } n ∈ N of vector fields by the formula w n ( x ) = ( f n, ( | x | ) , · · · , f n,N ( | x | )) ( n = 1 , , · · · )in order that R R N |△ w n | | x | γ dx R R N |∇ w n | | x | γ − dx → A N,γ, ( n → ∞ ) . (46)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 ∈ { , · · · , N } , where the notation ( · · · ) ′ denotes the derivative of a one-dimensional function. Thisfact implies that f ′ n,k = ( x k /x j ) f ′ n,j for all j = k , and hence that f n,j = const ∀ j ∈ { , · · · , N } from the radial symmetry of the function x f n,j ( | x | ). Consequently, we have |△ w n | = |∇ w n | ≡
0. This phenomenon indicates that there may be no curl-freesequence { w n } n ∈ N satisfying (46), which we can naturally interpret as a result ofthe inequality C N,γ > A
N,γ .Incidentally, let us further consider the case − N + √ N + 1 ≤ γ ≤ N +46 + √ N − N + 1 for 3 ≤ N ≤ (cid:12)(cid:12) γ − N +46 (cid:12)(cid:12) ≤ √ N − N + 1 for N ≥ C N,γ = C N,γ, and H N,γ − = (cid:0) γ + N − (cid:1) + N − γ = 0, this is the case for all N ≥ ELLICH-HARDY INEQUALITY FOR CURL-FREE FIELDS 23 inequality: for all curl-free fields u ∈ D γ − ( R N ), it holds that Z R N |△ u | | x | γ dx ≥ C N,γ Z R N |∇ u | | x | γ − dx (from (10)) ≥ C N,γ H N,γ − Z R N | u | | x | | x | γ − dx (cid:18) from (2)with γ replaced by γ − (cid:19) = C N,γ, (cid:16)(cid:0) γ + N − (cid:1) + N − (cid:17) Z R N | u | | x | | x | γ − dx = (cid:16) ( γ − − N (cid:17) Z R N | u | | x | | x | γ − dx ≥ B N,γ Z R N | u | | x | | x | γ dx with B N,γ given by (6). Therefore, even under the curl-free constraint, Rellich-Hardy inequality bridges between Hardy-Leray and Rellich-Leray inequalities, andso serves as a stronger version of the Rellich-Leray inequality.6.
Completion of the proof of Lemma 7
In this section, we prove Lemma 7 for the case γ > λ < − N )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 (cid:18) Q ( τ, a ) P ( τ, a ) − Q (0 , a ) P (0 , a ) (cid:19) ≥
12 when N ≥ , τ (cid:18) Q ( τ, a ) P ( τ, a ) − Q (0 , a ) P (0 , a ) (cid:19) ≥
13 when N = 2for all a ∈ { α ν } ν ∈ N , which clearly satisfies the conclusion of the lemma.6.1. The case N ≥ . By adding 1 / τ (cid:18) Q ( τ, a ) P ( τ, a ) − Q (0 , a ) P (0 , a ) (cid:19) −
12= (2 λ + N − G ( a ) + G ( a ) τP (0 , a ) P ( τ, a ) + 12= (2 λ + N −
2) ( G ( a ) + G ( a ) τ ) P (0 , a ) (cid:16) P (0 , a ) + τ + (cid:0) a + λ + λ ) + N (cid:1) τ (cid:17) + 12= τ P (0 , a ) + τ E ( a ) + E ( a )2 P ( τ, a ) P (0 , a ) , (47)where E ( a ) := 2(2 λ + N − G ( a ) + (cid:0) a + λ + λ ) + N (cid:1) P (0 , a ) , (48) E ( a ) := 2 (2 λ + N − G ( a ) + ( P (0 , a )) (49)Then it suffices to check that E ( a ) ≥ E ( a ) ≥ ∀ a ∈ { α ν } ν ∈ N or that E ( α + s ) ≥ E ( α + s ) ≥ ∀ s ≥ . (50)In the following, let us check the two inequalities in (50). Proof of E ( α + s ) ≥ . A direct computation from (25) yields P (0 , α + s ) = ( α + s ) + (cid:0) λ − λ ) − N (cid:1) ( α + s )+ ( λ + 1) (cid:0) λ + N − (cid:1) = s + s (cid:0) λ − λ + N − (cid:1) + λ (cid:0) ( λ + 1) + 3 N − (cid:1) . (51)Substitute (32) and (51) into (48) with a = α + s ; then a straightforward compu-tation yields E ( α + s ) = 2(2 λ + N − G ( α + s ) + (cid:0) α + s + λ + λ ) + N (cid:1) P (0 , α + s )= 2(2 λ + N − s (2 λ + N )+ s (cid:0) ( N − λ + N −
2) + ( λ − (2 λ + N ) (cid:1) + 2( N − λ (2 λ + N − + (cid:16) s + 2 (cid:0) λ + λ (cid:1) + 3 N − (cid:17) s + s (cid:0) λ − λ + N − (cid:1) + λ (cid:0) ( λ + 1) + 3 N − (cid:1) ! = 2 s + s E ( λ ) + sE ( λ ) + λ E ( λ ) ∀ s ≥ E ( a ) at a = α , where we set E ( λ ) := 2(2 λ + N − λ + N ) + 2 (cid:0) λ + λ (cid:1) + 3 N −
2+ 2 (cid:0) λ − λ + N − (cid:1) = 14 λ + 2(4 N − λ + ( N + 2)(2 N − (cid:0) λ + N − (cid:1) + 6 λ + 6( N −
2) + ,E ( λ ) := 2(2 λ + N − (cid:16) ( N − λ + N −
2) + ( λ − (2 λ + N ) (cid:17) + 2 λ (( λ + 1) + 3 N −
3) + (cid:16) (cid:0) λ + λ (cid:1) + 3 N − (cid:17) (cid:0) λ − λ + N − (cid:1) = 14 λ + 4(2 N − λ + 2 N ( N + 1) λ + 4( N − N − λ + (cid:0) N − N + 2 (cid:1) ( N − λ + λ (cid:16) (4 λ + 2 N − + 16 N − (cid:17) + ( N − (cid:16) (2 λ + N − + N + N + 1 (cid:17) ,E ( λ ) := 4( N − λ + N − λ + N − (cid:0) λ + λ ) + 3 N − (cid:1) (cid:0) ( λ + 1) + 3 N − (cid:1) = ( λ + 1) + ( λ + 1) λ + ( N − (cid:16)(cid:0) λ + N − (cid:1) + N +21 N +8925 (cid:17) . Notice that E ( λ ) , E ( λ ) and E ( λ ) are all nonnegative; then E ( α + s ) ≥ s ≥
0, which proves the desired inequality.
ELLICH-HARDY INEQUALITY FOR CURL-FREE FIELDS 25
Proof of E ( α + s ) ≥ . Substitute (33) and (51) into (49) with a = α + s ; thena straightforward computation yields E ( α + s ) = 2 (2 λ + N − G ( α + s ) + ( P (0 , α + s )) = 2(2 λ + N − (cid:18) s (2 λ + N ) + s G ( λ ) + s G ( λ )+ 2( N − λ (2 λ + N − (cid:19) + (cid:16) s + s (cid:0) λ − λ + N − (cid:1) + λ (cid:0) ( λ + 1) + 3 N − (cid:1) (cid:17) = s + s E ( λ ) + s E ( λ ) + 2 sE ( λ ) + λ E ( λ ) ∀ s ≥ E ( a ) at a = α , where we set E ( λ ) := 2(2 λ + N − λ + N ) + 2 (cid:0) λ − λ + N − (cid:1) = (4 λ + 2 N − + 4 λ + 4( N −
3) + ,E ( λ ) := 2(2 λ + N − G ( λ ) + (cid:0) λ − λ + N − (cid:1) + 2 λ (cid:0) ( λ + 1) + 3 N − (cid:1) , (53) E ( λ ) := (2 λ + N − G ( λ ) + λ (cid:0) ( λ + 1) + 3 N − (cid:1) (cid:0) λ − λ + N − (cid:1) , (54) E ( λ ) := 4( N − λ + N − λ + N −
1) + (cid:0) ( λ + 1) + 3 N − (cid:1) = ( λ + 1) + ( N − (cid:16) (4 λ + 2 N − + 6( λ + 1) + 9 N − (cid:17) . Hence, in order to show E ( α + s ) ≥ ∀ s ≥ { E k ( λ ) } k =0 , , , . Since it is clear that E ( λ ) ≥ E ( λ ) ≥
0, allthat is left is to show the two inequalities E ( λ ) ≥ E ( λ ) ≥ . (55)To this end, substitute (34) into (53), then we get E ( λ ) = 2(2 λ + N − (cid:16) (2 λ + N − (cid:0) ( λ + 1) + λ + N + 3 (cid:1) + ( N − + 9 (cid:17) + (cid:0) λ − λ + N − (cid:1) + 2 λ (cid:0) ( λ + 1) + 3 N − (cid:1) = 22 λ + 4(4 N − λ + 2( N + 4)(2 N + 1) λ + 4 N (4 N − λ + (cid:0) N + N + 2 (cid:1) ( N − λ + 5 λ (cid:0) λ + N − (cid:1) + N (cid:0) λ + N − (cid:1) + λ (cid:0) N + N + 3 (cid:1) + ( N − (cid:0) N + 29 N + 51 (cid:1) + . Since N ≥
3, this result implies E ( λ ) ≥
0, whence we have proved the firstinequality of (55). In the same way, the proof of the second could also be carriedout by the method of completing the square (see Appendix); however, we workhere in a different way. From the assumption λ < − N , we see that λ can beparameterized as λ = 1 − N − s, s > . Then we have G ( λ ) = G (cid:0) − N − s (cid:1) = ( − − s ) (cid:0) − N − s (cid:1) + 2 (cid:0) − N − s (cid:1) (cid:0) s − − N + N (cid:1) + 2 N (1 − s ) from (35). Substituting this expression into (54), we compute E ( λ ) = E (cid:0) − N − s (cid:1) = − s ( − − s ) (cid:0) − N − s (cid:1) + 2 N (1 − s )+ 2 (cid:0) − N − s (cid:1) (cid:0) s − − N + N (cid:1) ! + (cid:0) − N − s (cid:1) (cid:16) s + ( N − s + ( N + 2) (cid:17)(cid:16) s + 2( N − s + (cid:0) N − (cid:1) (cid:17) = 6 s + 2(7 N − s + (cid:0) ( N − N + 25) + 55 (cid:1) s + (cid:16) ( N − (cid:0) N + 2 N − (cid:1) + 34 (cid:17) s + (cid:16) ( N − (cid:0) N + 46 N + 107 (cid:1) + 740( N −
3) + 601 (cid:17) s + (cid:16) ( N − (cid:0) N ( N + N + 4) + 11 N + 14 (cid:1) + 86 (cid:17) s + (cid:0) N − (cid:1) as a Taylor series of E ( λ ) at λ = 1 − N . When N ≥
4, the coefficients of thepowers of s are all positive, and hence E ( λ ) ≥ N = 3, we directly see that E ( λ ) = E (cid:0) − − s (cid:1) = 6 s + 24 s + s + 34 s + s − s + | {z } > ≥ . Therefore, we have obtained the second inequality of (55) for all N ≥
3, as desired.6.2.
The case N = 2 . Applying the same calculation in (30) to the case N = 2,we have the identity1 τ (cid:18) Q ( τ, a ) P ( τ, a ) − Q (0 , a ) P (0 , a ) (cid:19) − λ ( G ( a ) + G ( a ) τ ) P (0 , a ) P ( τ, a ) , where P ( τ, a ) = τ + 2 (cid:0) a + λ + λ + 1 (cid:1) τ + (cid:0) a + λ − λ − (cid:1) + (4 λ + 3) λ is the same as in (25) with N = 2, and where G ( a ) and G ( a ) are the same as in(31) with N = 2. Adding 2 / τ (cid:18) Q ( τ, a ) P ( τ, a ) − Q (0 , a ) P (0 , a ) (cid:19) −
13= 23 λ ( G ( a ) + G ( a ) τ ) (cid:16) τ + 2 (cid:0) a + λ + λ + 1 (cid:1) τ (cid:17) P (0 , a ) + ( P (0 , a )) = 23 · τ P (0 , a ) + 2 τ F ( a ) + F ( a ) P (0 , a ) P ( τ, a ) , where F ( a ) := (cid:0) a + λ + λ + 1 (cid:1) P (0 , a ) + λG ( a ) , (56) F ( a ) := ( P (0 , a )) + 3 λG ( a ) . (57)Thus, it is enough to show that F ( a ) ≥ F ( a ) ≥ ∀ a ∈ { α ν = ν ( ν + 1) } ν ∈ N ELLICH-HARDY INEQUALITY FOR CURL-FREE FIELDS 27 under the assumption λ < − N = 0. Proof of F ( a ) ≥ for a ∈ { α ν } ν ∈ N . It suffices to check that F ( α + s ) ≥
0, i.e., F (1 + s ) ≥ ∀ s ≥ . In order to compute the left-hand side, apply N = 2 to (51) and (32); then we get P (0 , s ) = s + 2 sλ ( λ −
1) + λ (cid:0) ( λ + 1) + 3 (cid:1) , (58) G (1 + s ) = ( λ + 1) s + s (cid:0) λ + ( λ − ( λ + 1) (cid:1) + λ (2 λ + 1) . Substitute them into (56) and (57) with a = 1 + s , then we have F (1 + s ) = (cid:0) s + λ + λ + 2 (cid:1) P (0 , s ) + λG (1 + s )= (cid:0) s + λ + λ + 2 (cid:1) (cid:16) s + 2 sλ ( λ −
1) + λ (cid:0) ( λ + 1) + 3 (cid:1) (cid:17) + 3 λ (cid:16) ( λ + 1) s + s (cid:0) λ + ( λ − ( λ + 1) (cid:1) + λ (2 λ + 1) (cid:17) = s + 2 (cid:0) λ + λ + 1 (cid:1) s + λ (6 λ − (cid:0) λ + 1 (cid:1) s + λ (cid:0) λ + 3 λ + 14 λ + 11 λ + 8 (cid:1) = s + (cid:16) (cid:0) λ + (cid:1) + (cid:17) s − λ (1 − λ ) (cid:0) λ + 1 (cid:1) s + λ (cid:16) ( λ + 1) (cid:0) λ + (cid:1) + ( λ + 1) + (2 λ − (cid:17) In view of the right-hand side, the coefficients of the powers of s are all positivesince λ <
0, which implies that F ( a ) ≥ a ≥ α , as desired. Proof of F ( a ) ≥ for a ∈ { α ν } ν ∈ N . First of all, let us compute F (1 + s ). To thisend, apply N = 2 to (33), (34) and (35); then we get G (1 + s ) = ( λ + 1) s + s G ( λ ) + s G ( λ ) + λ (2 λ + 1)= ( λ + 1) s + s (cid:0) λ + 2 λ + 6 λ + 5 (cid:1) + s ( λ − (cid:0) λ − λ + 6 λ − λ − (cid:1) + λ (2 λ + 1) . Substitute this expression and (58) into (57) with a = 1 + s , then we have F (1 + s ) = ( P (0 , s )) + 3 λG (1 + s )= (cid:16) s + 2 sλ ( λ −
1) + λ (cid:0) ( λ + 1) + 3 (cid:1) (cid:17) + 3 λ (cid:16) λ + 1) s + s G ( λ ) + s G ( λ ) + 2 λ (2 λ + 1) (cid:17) = (cid:16) s + 2 sλ ( λ −
1) + λ (cid:0) ( λ + 1) + 3 (cid:1) (cid:17) + 6 λ ( λ + 1) s + s (cid:0) λ + 2 λ + 6 λ + 5 (cid:1) + λ (2 λ + 1)+ s ( λ − (cid:0) λ − λ − λ − λ − (cid:1) ! = s + 2 λ (5 λ + 1) s + 2 λ (9 λ + 4 λ + 24 λ + 15) s + 2 λ ( λ − (cid:0) λ − λ − λ − λ − (cid:1) s + λ (cid:0) ( λ + 1) + 9( λ + 1) + 9 λ + 6 (cid:1) (59) as a Taylor series of F ( a ) at a = α = 1. Then it is clear that F ( α ) = F (1) ≥ F ( a ) ≥ a ≥ α . For this purpose,it suffices to check that F ( α + s ) = F (4 + s ) ≥ ∀ s ≥ . To this end, notice from (59) that F (1 + s ) s ≥ F (1 + s ) − F (1) s = s + 2 λ (5 λ + 1) s + (cid:0) λ + 4 λ + 24 λ + 15 (cid:1) s + ( λ − (cid:0) λ − λ − λ − λ − (cid:1) ! holds for all s >
0. Replacing s by 3 + s on both sides, we get F (4 + s )3 + s ≥ ( s + 3) + 2 λ (5 λ + 1)( s + 3) + (cid:0) λ + 4 λ + 24 λ + 15 (cid:1) ( s + 3)+ ( λ − (cid:0) λ − λ − λ − λ − (cid:1) = s + (cid:0) ( λ + 1) + 9 λ + 8 (cid:1) s + (cid:16) λ + λ (cid:0) λ + 1) + 83 (cid:1) + 21( λ + 1) + 6 (cid:17) s + 3 λ + λ (cid:0) λ − + 27 (cid:1) + λ (2 λ − + 16(4 λ + 1) + ( λ + 2) + 7 ≥ , and therefore F (4 + s ) ≥ s ≥
0, as desired.
Appendix
Here we give another proof of the second inequality E ( λ ) ≥ E ( λ ) = (2 λ + N − (cid:18) λ (2 λ + N −
8) + N (2 λ + N )+ 2 λ (cid:0) − λ − N + N (cid:1) (cid:19) + λ (cid:0) ( λ + 1) + 3 N − (cid:1) (cid:0) λ − λ + N − (cid:1) = 2 λ + (cid:16) (4 λ + 2 N − + 8 N − (cid:17) λ + (cid:16) N (4 N −
1) + 2 (cid:17) λ + N ( N − λ + 2) + 4 N + ( N − (cid:16) (2 λ + N − λ + N (cid:17) by a straightforward computation, whence we get E ( λ ) ≥ N ≥ . ELLICH-HARDY INEQUALITY FOR CURL-FREE FIELDS 29
Moreover, the same also applies to the case N ∈ { , , } , as can be directly verifiedby the following calculation:when N = 3 , E ( λ ) = 6 λ − λ + 10 λ − λ + 41 λ + 72 λ + 27= (cid:0) λ + (cid:1) (cid:16)
40 + (cid:0) λ − (cid:1) (cid:17) + (cid:0) λ + (cid:1) (cid:0) + (6 λ − (cid:1) + (cid:0) λ + (cid:1) + ;when N = 4 , E ( λ ) = 6 λ − λ − λ − λ + 92 λ + 192 λ + 128= (cid:0) λ − λ − λ − (cid:1) + 2 λ + λ + 4 λ + 16( λ + 2) ;when N = 5 , E ( λ ) = 6 λ + 2 λ − λ + 181 λ + 400 λ + 375= 5 (cid:0) λ − (cid:1) + λ (cid:0) λ + λ − (cid:1) + ( λ + 16) + λ + . Therefore, we have obtained E ( λ ) ≥ N ≥
3, as desired.
Acknowledgments.
The second author (F.T.) was supported by JSPS Grant-in-Aid for Scientific Re-search (B), No.19H01800. This work was partly supported by Osaka City UniversityAdvanced Mathematical Institute (MEXT Joint Usage/Research Center on Math-ematics and Theoretical Physics JPMXP0619217849).
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