Improved Poincar´ e inequalities

Improved Poincar´ e inequalities

Jean Dolbeault

and Bruno Volzone

aCeremade (UMR CNRS no. 7534), Universit´e Paris-Dauphine, Place de Lattre de Tassigny, 75775 Paris Cedex 16, France.

bUniversit`a degli Studi di Napoli “Parthenope”, Facolt`a di Ingegneria, Dipartimento per le Tecnologie, Centro Direzionale Isola C/4 80143 Napoli, Italy.

Abstract

Although the Hardy inequality corresponding to one quadratic singularity, with optimal constant, does not admit any extremal function, it is well known that such a potential can be improved, in the sense that a positive term can be added to the quadratic singularity without violating the inequality, and even a whole asymptotic expansion can be built, with optimal constants for each term. This phenomenon has not been much studied for other inequalities. Our purpose is to prove that it also holds for the gaussian Poincar´e inequality. The method is based on a recursion formula, which allows to identify the optimal constants in the asymptotic expansion, order by order. We also apply the same strategy to a family of Hardy-Poincar´e inequalities which interpolate between Hardy and gaussian Poincar´e inequalities.

Key words: Hardy inequality, Poincar´e inequality, Best constant, Remainder terms, Weighted norms

MSC (2010): 26D10, 35P15, 39B22, 39B62, 46E35

1 Introduction

A considerable effort has been devoted to get improvements of Hardy inequal- ities. On H₀¹(Ω), we define the Hardy functional by

H[u] :=

Z

Ω

|∇u|² dx− 1

4(d−2)²

Z

Ω

|u|²

|x|² dx

Email addresses: [email protected] (Jean Dolbeault), [email protected](Bruno Volzone).

URL: www.ceremade.dauphine.fr/∼dolbeaul(Jean Dolbeault).

where Ω = R^d or Ω is a bounded domain in R^d containing the origin, and d≥3. The standard Hardy inequality asserts that

H[u]≥0 ∀u∈H₀¹(Ω). (1) For an extension of (1) to a finite number of singularities, see [27]. Inequal- ity (1) can be improved in various directions and we can list three lines of thought:

(1) Prove thatH[u] controlskuk_L^q_(Ω)for someq ∈[2,2^∗) with 2^∗ :=2d/(d−2), ork∇uk_L^q_(Ω)for someq ∈[1,2). See [5] for a recent result in this direction, and [14,30,28,4,13] for earlier contributions.

(2) Improve on the^R_Ω ^|u|_|x|²2 dxterm by showing that, with respect to the 1/|x|² weight, not only|u|² is controlled, but also|u|²log|u|². See [15,17,16] for recent papers in this direction.

(3) Improve on the 1/|x|² weight: see [22,23,6,7,18,25,24,3].

A simple and well known method to establish (1) is based on an expansion of the square which goes as follows. Let u be a smooth function with compact support in Ω and observe that

0≤

Z

Ω

|∇u+ ^d−2_{2 |x|}^x2 u|² dx

=

Z

Ω

|∇u|² dx+^(d−2)₄ ²

Z

Ω

|u|²

|x|² dx− ^d−2₂

Z

Ω

|u|²

∇ ·_|x|^x2

dx =H[u]

where we have used an integration by parts and noted that ∇ · _|x|^x2 = ^d−2_|x|2. The Poincar´e inequality with gaussian weight, orgaussian Poincar´e inequality, reads

Z

R^d

|u−u|¯² dµ≤

Z

R^d

|∇u|² dµ ∀u∈H¹(R^d, dµ) (2) with dµ(x) := µ(x)dx, µ(x) := (2π)^−d/2e^−|x|2/2 and ¯u := ^R_Rdu dµ. Our pur- pose is to study improvements of (2) in the spirit of what has been done for (1). Let us list some known results for (2):

(1) Spectral improvements are easily achieved under appropriate orthogonal- ity conditions. See [8] for results and further references in this direction.

(2) Replacing|u|² by|u|²log|u|²amounts to consider the logarithmic Sobolev inequality instead of the Poincar´e inequality; see [26] for an historical reference. There is a huge literature on this subject, which is out of the scope of the present paper.

(3) A very standard argument based on theexpansion of the square has been repeatedly used in the literature. Let us give some details, in the gaussian case, as it is the starting point of our strategy.

For any open set Ω inR^d, let us define the functional G_Ω[u] =

Z

Ω

|∇u|² dµ+d 2

Z

Ω

|u|²dµ− 1 4

Z

Ω

|x|²|u|²dµ.

By expanding ^R

R^d|∇(u e^−|x|2/4)|² dx, we find that

G_R^d[u]≥0. (3)

If ¯u = 0, the middle term in G_R^d[u] can be estimated by (2), thus showing that the followingimproved Poincar´e inequality holds:

Z

R^d

|x|²|u|² dµ≤2 (d+ 2)

Z

R^d

|∇u|² dµ (4) (this inequality is an improvement in the sense that, as |x| → ∞, the |x|² weight diverges). A slightly more general case has been considered for instance in [19,20] (also see, e.g., [29]). The expansion of the square method raises the following question. By (3) we know that G_R^d[u]≥0 for any u ∈ H¹(R^d, dµ).

With no additional assumption onu, is there a nonnegative function W such that

G_Rd[u]≥

Z

R^d

W|u|² dµ

for any u ∈ H¹(R^d, dµ) and, if yes, can we give an asymptotic expansion as

|x| → ∞ of the best possible function W, order by order ?

The purpose of this paper is to systematically investigate such improvements for gaussian Poincar´e inequalities, following the same scheme as for the Hardy inequality. More precisely, using an elaborateexpansion of the square method, we derive an asymptotic expansion of the largest possible nonnegative func- tion W and, order by order, find the best possible constants for any finite truncation of the asymptotic expansion.

To clarify our purpose, we will first recall in Section 2 what can been done for the Hardy inequality and give a short proof of it based on the method used in [18]. Then we shall adapt it to the gaussian Poincar´e inequality, which provides us with our first main result: see Theorem 3 in Section 3. A striking parallel appears, which will be further investigated in Section 4, in the case of a family of inequalities interpolating between Hardy and Poincar´e inequalities.

Before giving details, let us quote a few additional references. Improvements of the Hardy inequality already have a quite long history. In [14], Brezis and V´azquez have shown that in the case of a bounded domain Ω, there exists a constant λ_Ω >0 such that

λ_Ω

Z

Ω

|u|² dx≤H[u] ∀u∈H₀¹(Ω).

The striking result of [1], by Adimurthi, Chaudhuri and Ramaswamy, is that a whole expansion in terms of iterated logarithms can be done close to the singularity (see also [2] for a generalization inW^1,p(Ω)). Filippas and Tertikas gave in [22] the expression of the best constants for all terms of the expansion;

also see [3] for a more recent result, concerning |u|^p in the remaining term, for the limit case p= 2^∗. In view of the generalization to relativistic models, Dolbeault, Esteban, Loss and Vega gave in [18] an algebraic property which simplifies the computation of the expansion, while Ghoussoub and Moradifam in [24] have established a rather simple characterization of the best constants.

Some of the results of [22] are summarized in Theorem 1 below, with a sim- plified proof inspired by the combination of all above mentioned works. This proof will be a source of inspiration for the results on the gaussian Poincar´e inequality, which are entirely new, and also for the Hardy-Poincar´e inequality of Section 4.

2 A key example: the improved Hardy inequality

Letr =|x| for any x∈Ω, and set

X₁(r) := (a−logr)⁻¹ for some a≥1 and

Xk :=X1◦Xk−1

for all k ≥2. We also define W_k := 1

4

k

Y

j=1

X_j² ∀k≥1.

We shall always assume that 0∈ Ω. With δ_Ω = max_∂Ω|x|, we choose a = a_Ω such that a≥1 is the unique solution ofδΩ = 1/(a−logδΩ),i.e. a= logδΩ+ 1/δ_Ω, so that the interval (0, δ_Ω] is stable under the action of X₁.

Theorem 1 Let Ω be a bounded domain containing the origin and assume that a=aΩ. With W =^P∞_j=1Wj, we have

Z

Ω

W |u|²

|x|² dx≤H[u] ∀u∈H₀¹(Ω). (5) Moreover, such a function W is optimal in the following sense. Assume that (5) holds for some nonnegative, bounded, radial function W. Then we have:

(i) if W converges as r→0+ to some limit ` ∈[0,+∞] then `= 0, (ii) iflimr→0₊W = 0 and if _W^W

1 converges asr →0₊ to some limit`₁ ∈[0,+∞]

then `₁ ≤1,

(iii) for any N ≥2 and with the convention W₀ := 0, if

r→0lim₊W = 0 , lim

r→0₊

W −^Pk−1_j=0W_j

Pk

j=1W_j = 1 ∀k ∈ {1, 2. . . N −1}

and if ^W−

PN−1 j=1 Wj

PN j=1Wj

converges as r → 0₊ to some limit `_N ∈ [0,+∞], then

`N ≤1.

The first part of Theorem 1 has been obtained by Filippas and Tertikas in [22, Theorem D, p. 190] and the statement on the characterization of the best constants in the asymptotic expansion for allW can be found in [22, Theorem B’, p. 192]. Here we give a detailed proof based on the approach used in [18].

Notice that if W is not radially symmetric, some results can be recovered by applying Schwarz’ symmetrization to W(x)/|x|².

Also notice that this expansion is independent of the value of a used in the definition ofX₁, as the behaviour ofW_j forrclose to 0₊does not depend ona:

what we have achieved is only an asymptotic expansion of the improvementW at the singularity.

Proof. For simplicity, we split the proof in three steps.

Step 1. Expansion of the square. Suppose that f = f(r) is a continuously differentiable function in an interval [0, R] with R > 0 such that Ω ⊂ B_R, where B_R denotes the ball of radiusR centered at the origin. Expanding the square |∇u+f(r)_r^x2 u|² with r=|x| and integrating by parts, we have

0≤

Z

Ω

∇u+f x r² u

2

dx=

Z

Ω

|∇u|²dx+

Z

Ω

f² r² − f⁰

r −d−2 r² f

!

|u|²dx ,

that is

Z

Ω

r f⁰+ (d−2)f −f²|u|²

|x|² dx≤

Z

Ω

|∇u|²dx . Settingg =f −(d−2)/2, we get

1

4(d−2)²

Z

Ω

|u|²

|x|² dx+

Z

Ω

|u|²

|x|²

r g⁰ −g² dx≤

Z

Ω

|∇u|²dx .

At this point, we observe that any bounded, positive solution on a neighbor- hood of r= 0₊ of the equation

r g⁰−g² =W ≥0 (6)

is such that g(r) ≤ (a0 −logr)⁻¹ for some a0 ∈ R as r → 0+ and, as a consequence, limr→0₊g(r) = 0. This proves that the constant (d−2)²/4 in the expression ofH[u] is optimal and establishes Property (i).

Step 2. Optimal behavior at first order in the asymptotic expansion.It is worth- while to notice that the function r7→X₁(r) = (a−logr)⁻¹ solves

r g⁰−g² = 0. Also observe that g(r) = α X₁(r) solves

r g⁰−g² = (α−α²)X₁² ≤ 1 4X₁² with equality if and only if α= 1/2.

Let H be such that g(r) = X₁(r)H(s), with s = −log(X₁(r)), so that s → +∞ as r → 0+. Now we consider a solution to (6) and only assume that W ≥0. We claim that if

W(r)

W₁(r) = 4r g⁰(r)−g²(r)

X₁²(r) = 4−H⁰(s) +H(s)−H²(s)

has a limit`<sub>1</sub> asr →0<sub>+</sub>, then`₁ ≤1. Let us prove it. If we have`₁ >1, then

−H⁰ −

H−1 2

2

∼ `₁−1 4 >0

and then we know that −_(H−1/2)^H0 2 ≥1, so that for some constantC, we have 1

H(s)− ¹₂ > C+s

for any s large enough. This means that lims→∞H(s) = 1/2. Then we also know that

H⁰ ∼(1−`1)/4<0, a contradiction. This proves (ii).

If`₁ = 1, we can define h by g(r) = t

h(t) + 1 2

with t =X₁(r). Then we find that

r g⁰(r)−g²(r) t² −1

4 =t h⁰(t)−h²(t), which is a good preparation for the next step.

Step 3. Induction. Consider the sequence (h_k)k≥1 of functions defined by h₁(r) :=g(r), h_k(r) =th_k+1(t) + ¹₂, t=X₁(r)∈(0, δ_Ω) ,

whereg solves (6) andW =^P∞_j=1W_j. An elementary computation shows that r h⁰_k(r)−h²_k(r)

t² −1

4 =t h⁰_k+1(t)−h²_k+1(t). This implies that for all k ≥1 we find

r g⁰(r)−g²(r) =

k

X

j=1

W_j(r) +W_k(r)z h⁰_k+1(z)−h²_k+1(z)

with z = X_k(r) and r ∈ (0, δ_Ω). With this formula, it is clear that (5) holds with h_k+1 =h_k for any k ≥ 1, while proving the optimality of the constants in the asymptotic expansion goes at each iteration as in the computations of Step 2.

As a consequence of Theorem 1, we also have an asymptotic expansion as

|x| → ∞ of an improved Hardy inequality. By the Kelvin transformation, to any u∈H₀¹(Ω), we associate v such that

v(x) =|x|^2−du|x|⁻²x, (7) where v is defined on Ω^K := ⁿx∈R^d : x/|x|² ∈Ω^o. By standard computa- tions, we know that

Z

Ω

|∇u|² dx=

Z

Ω^K

|∇v|² dx ,

Z

Ω

|u|²

|x|² dx=

Z

Ω^K

|v|²

|x|² dx ,

and we can defineW_k^K(r) :=W_k(1/r) using the notations of Theorem 1, which can now be rewritten in the exterior domain Ω^K as follows.

Corollary 2 Let Ω be a bounded domain containing the origin and assume that a=a_Ω. With W =^P∞_j=1W_j^K, we have

Z

Ω^K

W |v|²

|x|² dx≤

Z

Ω^K

|∇v|² dx− 1

4(d−2)²

Z

Ω^K

|v|²

|x|² dx ∀v ∈H₀¹(Ω^K). Moreover, such a function W is optimal in the following sense. Assume that the above inequality holds for some nonnegative, radial function W. Then we have:

(i) if W converges as r→ ∞ to some limit `∈[0,+∞] then ` = 0, (ii) iflimr→∞W = 0 and if _W^WK

1

converges asr→ ∞ to some limit`₁ ∈[0,+∞]

then `₁ ≤1,

(iii) for any N ≥2 and with the convention W₀^K := 0, if

r→∞lim W = 0, lim

r→∞

W −^Pk−1_j=0 W_j^K

Pk

j=1W_j^K = 1 ∀k ∈ {1, 2. . . N −1}

and if ^W−

PN−1 j=1 W_j^K

PN

j=1W_j^K converges as r → 0₊ to some limit `_N ∈ [0,+∞], then

`_N ≤1.

3 Improved Poincar´e inequality: the gaussian case

Let us consider now the gaussian measure

dµ(x) = µ(x)dx , µ(x) = e^−|x|2/2 (2π)^d/2 . Suppose that d >2 and define the functions

t:= 1

1 +r^d−2 , δ :=− t

log(1−t) , X(t) := log(1−t) log(1−t)−1 .

For any r > 0 we have t ∈ (0,1) and the functions X, δ are well defined.

Besides, since X(t) < 1 for any t ∈ (0,1), we have that the interval (0,1) is stable under the action ofX. Moreover, for all k ≥0, we set

X₀(t) =t , X_k+1 =X◦X_k,

Y₀(t) = 1, Y_k+1 = (δ◦X_k)² = _log2(1−X^Xk2(t)_k(t)), Z₀(t) = 1 , Z_k+1(t) = X_k²(t) fork ≥0, W_k(t) =Z_k(t) if k= 0,1 andW_k(t) =

^k−1 Y

j=1

Y_j(t)

Z_k(t) if k≥2.

(8)

Theorem 3 Suppose that d≥3. With the above notations, (8), we have G_R^d[u]≥ 1

4(d−2)²

Z

R^d

u²

|x|²

∞

X

k=0

Wk(t)

!

dµ

for any u∈H¹(dµ), where t= 1/(1 +r^d−2), r=|x|. Moreover, the expansion is asymptotically optimal, in the sense that at any order N ≥0, if we consider an improved inequality of the form

G_Rd[u]≥ 1

4(d−2)²

Z

R^d

u²

|x|²





N

X

k=0

W_k(t) +

N

Y

j=0

Y_j(t)R_N(t)



 dµ

and if R_N(t) converges as t →0to some limit `<sub>N</sub> ∈[0,∞), then `_N ≤1.

We may notice that no improvement can be achieved on the terms of order|x|² and 1: if we hadG[u]≥`−2R

R^d|u|²|x|²dµ+`−1R

R^d|u|²dµfor any u∈H¹(dµ), then testing the inequality withu(x) = exp(−(1−ε)|x|²/4) shows that`−2 ≤0 and `₋₁ ≤0.

Proof. As we did for Theorem 1, we split the proof in three steps.

Step 1. Expansion of the square. Let g be any radial smooth function. Then, for any u ∈ H¹(dµ) by expanding the square ^R_Rd|∇u + g(r)u x|²dµ and integrating by parts, we get

Z

R^d

|u|²r g⁰+d g−r²(g²+g) dµ≤

Z

R^d

|∇u|²dµ . With the functionh defined by

h(r) :=r²g(r) + ¹₂ , we obtain a correction term to (3), namely

Z

R^d

|u|²

|x|²

r h⁰+ (d−2)h−h²dµ

≤

Z

R^d

|∇u|²dµ+ d 2

Z

R^d

|u|²dµ−1 4

Z

R^d

|u|²|x|²dµ=G_R^d[u]. Consider the function f such that

(d−2)²f(r) := r h⁰(r) + (d−2)h(r)−h²(r). (9) The expansion of the square now amounts to

(d−2)²

Z

R^d

|u|²

|x|² f(r)dµ≤G_R^d[u]. Our purpose is to identify the best possible function f.

Step 2. Optimal behavior at zero order in the asymptotic expansion.Our goal is to maximize f as r → +∞. Assume first that 4f(r) has a limit `₀ > 1 as r→ ∞. If we set h(r) = (d−2)H(s) with s= (d−2) logr, then H solves

H⁰+H−H² ∼ `₀

4 as s→ ∞. Fors >0, large enough, there exists ε >0 such that

H⁰ ≥H− ¹₂²+ε≥ε ,

so that lims→∞H(s) = ∞. But we can also write that _(H−1/2)^H0 2 ≥1, so that, for some constant C,

1

H(s)− ¹₂ < C−s ,

if s is taken large enough. Thus we get lims→∞H(s) = ¹₂, a contradiction.

On the other hand H(s) = ¹₂ for any s ∈ R is admissible, thus proving that limr→∞f(r) = 1/4 can be obtained.

Step 3. Induction.Observe that the nontrivial global solutions to the equation r h⁰+ (d−2)h−h² = 0

are given by h(r) = _1+C(d−2)^d−2 _rd−2 for an arbitrary constant C. This suggests to set

t =t(r) := 1 1 +r^d−2 . If

h(r) = (d−2)h₀(t), then f(r) can be rewritten in terms of t as

f(r) =−t(1−t)h⁰₀(t) +h₀(t)−h²₀(t).

Ifh₀(t) =α tfor someα∈R, thenf(r) = t²α(1−α) takes its largest possible value, namely f(r) =t²/4, for α= 1/2. Now if

h₀(t) = t

2 +H₀(t), we get

f(r)− t²

4 =−t(1−t)H₀⁰(t) + (1−t)H₀(t)−H₀²(t). If we set H₀(t) :=δ(t)h₁(s) wheres=X(t), then we have

H₀⁰(t) =δ⁰(t)h₁(s) +δ(t)h⁰₁(s)X⁰(t)

and by the definition of X and δ, it is not difficult to check that f(r)− ^t₄²

δ²(t) =−s(1−s)h⁰₁(s) +h₁(s)−h²₁(s). (10) Hence the r.h.s. in (10) exactly takes the form off(r), with t and h₀ replaced bysandh1 respectively. Since limt→0X(t) = 0, we can iterate this procedure.

Assume first that W =^P∞_k=1Wk. By (8) and (10), we find that h₁ =h₀ .

Hence, if we define (R_k)_k≥0 by

R₀(t) := 4f(r) and R_k+1 :=R_k◦X_k+1 for any k≥0, then for any N ≥1 we obtain

R₀(t) = t²+ 4δ²(t)^h−s(1−s)h⁰₁(s) +h₁(s)−h²₁(s)ⁱ

=Z1(t) +Y1(t) (R0◦X) (t)

=Z₁(t) +Y₁(t)R₁(t)

=Z₁(t) +Y₁(t) (Z₂(t) +Y₂(t)R₂(t))

=Z1(t) +Y1(t)Z2(t) +Y1(t)Y2(t)R2(t)

=. . .=

N

X

k=1

Wk(t) +

N

Y

j=1

Yj(t)RN(t). Otherwise, we already know from Step 2 that

R₀(t) := 4^h−t(1−t)h⁰₀(t) +h₀(t)−h²₀(t)ⁱ is such that, if limt→0R₀(t) =`<sub>0</sub>, then `₀ ≤1, and if`₀ = 1, then

R₀(t) =Z₁(t) +Y₁(t)R₁(t) with R₁(t) := 4^h− t(1−t)h⁰₁(t)h₁(t)−h²₁(t)ⁱ and the conclusion follows by a straightforward iteration.

Now we study the case d = 2. First, set t = 1/logr. Then we let a > 1 and define R^? =R^?(a) = e^1/t? where t^? is given as the largest positive solution of t=X(t) with

X(t) := 1 a−logt .

Notice that t =X(t) has a unique solution such that t > 1. We also observe that [0, t^?] is stable under the action ofX (also see [18] for further properties of X). Also notice thatt^? > e^a−1. With this new definition of X andδ(t) =t, t= 1/logr, we can now construct Xk, Yk =Zk and Wk as in (8):

X₀(t) =t , X_k+1 =X◦X_k, Y₀ = 1 , Y_k+1 =X_k² for k≥0, W₀(t) = 0, and W_k(t) =

k

Y

j=1

Y_j(t) if k ≥1.

(11)

Theorem 4 Suppose that d = 2. For any function u ∈ H¹(dµ) with support outside the ball of radius R^?, we have

G_R²[u]≥ 1 4

Z

R²

u²

|x|²

∞

X

k=1

W_k(t)

!

dµ

with t = 1/logr, r =|x| and W_k defined by (11). Moreover, the expansion is asymptotically optimal, in the sense that at any order N ≥ 1, if we consider an improved inequality of the form

G_R²[u]≥ 1 4

Z

R²

u²

|x|²

N−1

X

k=0

W_k(t) +W_N(t)R_N(t)

!

dµ

and if RN(t) converges as t →0 to some limit `N ∈[0,∞), then `N ≤1.

Proof. The expansion of the square method reduces the problem to find the best possible functionf(r) = r h⁰(r)−h²(r) such that

Z

R²

|u|²

|x|² f(r)dµ≤

Z

R²

|∇u|²dµ− 1 4

Z

R²

|u|²|x|²dµ+

Z

R²

|u|²dµ .

If f(r) ∼ ` ≥ 0 as r → +∞, then it follows that H(t) =h(r) with t = logr solves

H⁰−H² ∼`

as t → ∞. On the one hand, if ` > 0, then H(t)≥ <sup>`</sup>₂t as t → ∞, i.e. h(r) ≥

`

2 logr →+∞, and on the other hand, h⁰ h² ≥ 1

r

means that for some constantC and for r large enough, we have C− 1

h(r) ≥logr

which implies that limr→∞h(r) = 0, a contradiction. As a consequence,`= 0.

In other words, we have shown that the first term in the expansion (that is, the term of order 1/|x|²) is W₀ = 0.

The nontrivial solutions to the equation r h⁰−h² = 0 are

h= 1

C−logr ,

withC being an arbitrary constant. If we sett= 1/logr and considerh₀ such that h(r) = h₀(t), we easily infer that

f(r) =−t²h⁰₀(t)−h²₀(t).

Ifh₀(t) = α tfor someα∈R, the largest possible value off(r) =−t²(α²+α) is t²/4. It is achieved forα =−1/2. Now if

h0(t) =−t

2 +H0(t),

we get

f(r)−t²

4 =−t²H₀⁰(t) +t H₀(t)−H₀²(t). As in the proof of Theorem 3, if we set

H₀(t) = t h₁(s) and s =X(t), using the definition of X, it is easy to verify that

f(r)− t²/4

t² =−s²h⁰₁(s)−h²₁(s).

Now it is enough to argue as in the proof of Theorem 3 to conclude.

As a further remark, we notice that we can combine the results of Theorems 3 and 4 with the method used for proving (4) to get, for any u ∈ H¹(R^d, dµ) such that ¯u=^R_Rdu dµ= 0,

2 (d+ 2)

Z

R^d

|∇u|²dµ≥

Z

R^d

u²|x|²dµ+ (d−2)²

∞

X

k=0

Z

R^d

u²

|x|² Wk(t)dµ if d ≥ 3 and W_k is defined as in (8), and, under the same assumptions as in Theorem 4,

4

Z

R²

|∇u|²dµ≥

Z

R²

u²|x|²dµ+

∞

X

k=1

Z

R²

u²

|x|²W_k(t)dµ

if d= 2 and W_k is defined as in (11), thus improving also (4) for any d≥2.

Exactly as for the Hardy inequality, we can use the Kelvin transform and provide an inequality for the measure

d ν(x) := 1 (2π)^d/2 e⁻

1 2|x|2

dx .

In fact, like in Section 2, let Ω be an open set ofR^d containing the origin and for any function u ∈ C₀¹(Ω) we consider v(y) = |y|^2−du(x) with y = x/|x|², i.e. the Kelvin transformation (7). It follows that v ∈ C₀¹Ω^K. By standard calculations, we find

|∇v|² =|y|^−2d|∇u(x)|²+ (d−2)²|y|^2−2du²(x) + (d−2)|y|^−2dy· ∇u²(x) (12) with y = x/|x|². Integrating and performing an integration by parts to the last term in the expression of|∇v|², we get

Z

Ω^K

|∇v|²dµ=

Z

Ω

|∇u|²dν−(d−2)

Z

Ω

|u|²

|x|⁴ dν .

If^R_Ω|x|^−d−2u dν = 0, by applying (2) to v we obtain the weighted inequality

Z

Ω

|∇u|²dν ≥(d−1)

Z

Ω

|u|²

|x|⁴ dν .

With the change of variables y =x/|x|² and u given in terms ofv by (7), we find that G_Ω^K[v] =K_Ω[u], where

K_Ω[u] :=

Z

Ω

|∇u|²dν− d−4 2

Z

Ω

|u|²

|x|⁴ dν− 1 4

Z

Ω

|u|²

|x|⁶ dν

is nonnegative because of (3). This implies the existence of a constant Cd

depending only on the dimensiond, such that C_d

Z

Ω

|∇u|²dν ≥

Z

Ω

|u|²

|x|⁶ dν if ^R_Ω|x|^−d−2u dν = 0, which is the analogue of (4).

As in Theorems 3 and 4, we may add a whole asymptotic expansion as|x| →0 to the right hand side of the inequality KΩ[u]≥0.

Corollary 5 Assume that d ≥ 2. If d = 2, let u ∈ H₀¹(B_1/R^?, dµ), where R^? is defined as in Theorem 4 and B_1/R^? denotes the ball of radius 1/R^?. Then we have

K_B1/R?[u]≥ 1 4

Z

B_1/R?

u²

|x|²

∞

X

k=0

W_k(−1/log|x|)dν .

If d ≥ 3, let Ω be any open set of R^d containing the origin. Then for any u∈H₀¹(Ω, dµ) we have

K_Ω[u]≥ 1

4(d−2)²

Z

Ω

u²

|x|²

∞

X

k=0

W_k |x|^d−2 1 +|x|^d−2

!

dν .

Moreover, for alld ≥2 the asymptotic expansion as |x| →0 at the right hand side of these inequalities is optimal, in the same sense as in Theorems 3 and 4.

Proof. Assume thatd >2, as the arguments in the case d= 2 are analogous.

Ifu ∈H₀¹(Ω, dµ), let us consider the function v ∈H₀¹(Ω^K, dµ) defined by (7).

Therefore we can use the same method in the proof of Theorem 3, up to the replacement of R^d by Ω^K. Thus

G_Ω^K[v]≥(d−2)²

Z

Ω^K

|v|²

|x|²f(r)dµ

with the same f defined in (9), and it is clear that the optimality arguments

described in the proof of Theorem 3 still hold. Besides we find G_Ω^K[u]≥ 1

4(d−2)²

Z

Ω^K

v²

|x|²

∞

X

k=0

W_k(t)

!

dµ

wheret = 1/(1 +r^d−2),r=|x|. Then using the Kelvin transformation (7) we

can finally conclude.

4 Improved Hardy-Poincar´e inequalities

4.1 Hardy-Poincar´e inequalities

In this section, we shall consider improvements of a family of Hardy-Poincar´e inequalities which has been investigated in [10,9,11]. Let hα(x) := (1 +|x|²)^α and define dµ_α(x) = h_α(x)dx, for any α≤0. From [11], we know that

Λ_α,d

Z

R^d

|u−µα−1(u)|² dµα−1 ≤

Z

R^d

|∇u|² dµ_α ∀u∈H¹(R^d, dµ_α) (13) with the convention

µα−1(u) :=^R_Rdu dµα−1 if α∈(−∞,−(d−2)/2), µα−1(u) := 0 if α∈(−(d−2)/2,0).

The inequality holds not only in H¹(R^d, dµα) but also in the larger space {u∈L²(R^d, dµα−1) : ∇u∈L²(R^d, dµ_α)}. This is easy to establish by density of smooth functions with support in R^d\ {0}. The optimal value of Λ_α,d has been determined in [11]. If d≥2, we have

Λ_α,d =−2α if α∈(−∞,−d),

Λ_α,d =−2 (d+ 2α) if α ∈(−d,−(d+ 2)/2), Λ_α,d = ¹₄(d−2 + 2α)² if α ∈(−(d+ 2)/2,0).

Notice that for α=−(d−2)/2, we find Λ_α,d = 0 and the inequality becomes trivial. See [12] for more details in such a case. In the limit case α = 0, if we apply (13) to u_λ(x) = λ^d/2u(λ x) and take the limit λ → 0₊, we recover the Hardy inequality (1). If we apply (13) tou_λ(x) =u(λ x) with λ =^q2|α|

and take the limit α→ −∞, we recover the gaussian Poincar´e inequality (2).

Inequality (13) is therefore an interesting family of inequalities which inter- polates between the Hardy inequality (1) and the gaussian Poincar´e inequal- ity (2). Our purpose is to show that the results of Sections 2 and 3 can be adapted to this more general family of inequalities.

Let us take α < 0. If we expand the square ∇(u h_α/2)², an integration by parts gives

0≤

Z

R^d

∇(u h_α/2)²dx

=

Z

R^d

|∇u|²dµ_α+α(2− α)

Z

R^d

|x|²

(1 +|x|²)² u²dµ_α− α d

Z

R^d

u²dµα−1 ,

that is

Z

R^d

|∇u|²dµ_α− α d

Z

R^d

u²dµα−1 ≥α(α−2)

Z

R^d

|x|²

(1 +|x|²)² u²dµ_α . Exactly as in the Gaussian case, we can get the analogue of (4). By the Hardy- Poincar´e inequality (13), if µα−1(u) = 0, we can estimate the second term of the left-hand side by

−α d

Z

R^d

u²dµα−1 ≤ −α dΛ⁻¹_α,d

Z

R^d

|∇u|²dµ_α .

Hence for allu∈H¹(R^d, dµ_α) such that µα−1(u) = 0 we find

Z

R^d

|x|²

(1 +|x|²)² u²dµ_α≤ 1− α dΛ⁻¹_α,d α(α−2)

Z

R^d

|∇u|²dµ_α , (14) which is an improved Hardy-Poincar´e inequality. Of course all these inequal- ities are valid if R^d is replaced by any open set Ω and the space H¹(R^d, dµ_α) by H₀¹(Ω, dµ_α). Next, for any open set Ω and any u∈ H₀¹(Ω, dµ_α), we define the functional

I_Ω[u] :=

Z

Ω

|∇u|²dµ_α+α(2− α)

Z

Ω

|x|²

(1 +|x|²)² u²dµ_α− α d

Z

Ω

u²dµ_α−1

and we know that I_Ω[u]≥0 for any u∈H₀¹(Ω, dµ_α).

4.2 A scheme for improving Hardy-Poincar´e inequalities

To get a full asymptotic expansion, the strategy is similar to the one used in Theorems 1, 3 and 4, but various cases have to be distinguished depending on the dimension.

Step 1. Expansion of the square.Letgbe any smooth radial function onR^d. For any u∈ H¹(R^d, dµ_α), if we expand the square |∇u+g(r)u x|² and integrate

by parts with respect to the measuredµ_α, we find 0≤

Z

R^d

|∇u+g(r)u x|² dµ_α

=

Z

R^d

|∇u|²dµ_α+

Z

R^d

−r g⁰ −(2α+d)r²+d

1 +r² g+r²g²

u²dµ_α . Define now a functionh(r) by

h(r) = (1 +r²)g(r)− α . We find that

I_R^d[u]≥

Z

R^d

f(r)u²dµα−2 (15)

where

f(r) := (1 +r²)r h⁰+^h(d−2)r² +dⁱh−r²h². (16) The nontrivial positive global solutions to the equation

(1 +r²)r h⁰+^h(d−2)r²+dⁱh−r²h² = 0 (17) are given for d≥3 by

h(r) = (d−2) 1 +r²

r²+C(d−2)r^d ∼ 1

C r^d−2 as r →+∞

where C is an arbitrary positive constant, while the positive solutions when d= 2 are given in a neighborhood of r= 0₊ by

h(r) = 1 +r² r²(C−logr) for some C ∈R.

Step 2. Optimal behavior at zeroth order in the asymptotic expansion.Our aim is now to maximize f(r)/r² asr →+∞. To do that, assume that

f(r) r² −d h

r² = 1 +r²

r h⁰+ (d−2)h−h² has a limit <sup>`</sup><sub>4</sub>(d−2)<sup>2</sup> if d≥3 and ` if d= 2.

Assume first that d≥3. Withh(r) = (d−2)H(s) and s= ^d−2₂ log(1 +r²)→ +∞, we find that

H⁰(s) +H(s)−H²(s)∼ ` 4

and get a contradiction if ` >1, by the same arguments as in Theorems 3. As a consequence, lim sup<sub>r→+∞</sub>h(r) ≤ (d−2)/2 and f(r) ∼ <sup>1</sup><sub>4</sub>(d−2)<sup>2</sup>` r² with

` ≤ 1 as r → +∞. Finally, it is straightforward to check that if f(r)/r² has

a limit larger than (d−2)²/4 as r →+∞, then h(r)∼ C r² up to a positive constant C and we also get a contradiction.

If d = 2, we can work as in Theorem 4. If ` >0 and also get a contradiction using h(r) = H(s) and s = ¹₂log(1 +r²) → +∞. After some elementary considerations, this shows that asr→+∞, the limit off(r)/r²is non-positive if it exists.

Step 3. Induction. Assume temporarily that for some functions γ_d, δ_d and X to be determined, we have

X₀(t) =t , X_k+1 =X◦X_k,

Y₀(t) = 1, Y_k+1 = (δ_d◦X_k)², (18)

Z₀(t) = 1 for d≥3 and Z₀(t) = 0 for d= 2 , Z_k+1 =γ_d◦X_k for k ≥0, W_k(t) =Z_k(t) if k= 0,1 andW_k(t) =

k−1

Y

j=1

Y_j(t)Z_k(t) if k ≥2.

The functions γd and δd are determined as follows. With t= 1/logr if d= 2, t=r^2−d if d≥3 and h(r) =h₀(t), with c₂ = 1 and c_d= (d−2)² if d≥3, we may write

f(r)

c_dr² =Ft, h0(t), h⁰₀(t) (19) for some function F. The above choice of t is justified by the behavior for r → ∞ of the solutions h to equation (17). Then we identify a function γ_d such that

(i) for some constantβ ∈R,Ft, h₀(t), h⁰₀(t)=β γ_d(t) +oγ_d(t)ast→0, (ii) if β takes its largest possible value, then we look for some function h₁

and s=X(t) such that

Ft, h₀(t), h⁰₀(t)−β γ_d(t) = δ_d(t)²Fs, h₁(s), h⁰₁(s). (20) The functions X , δ_d will be chosen in order to satisfy a relation of the type

At, X(t), X⁰(t), δd(t)=Bt, X(t), δd(t), δ⁰_d(t)=δ_d²(t) (21) whereA,B are suitable functions. Then the analogue of Theorems 1, 3 and 4 holds. The main difficulty is to build the functions γ_d and X. A restriction comes from the requirement that some interval is stable under the action of X. This program can be completed in dimension d = 2, 3 and 4, and also in dimension higher than 4, in exterior domains.

4.3 The case d= 2

Ifh(r) =h₀(t) witht = 1/logr, we get that f(r)

r² =−(1 +e^−2t)t²h⁰₀(t) + 2e^−2t h₀(t)−h²₀(t). (22) Implicitly define the function h₁ such that

h₀(t) :=−β t+δ₂(t)h₁(s)

where β ∈ (0,+∞) and δ₂(t), s = X(t) are two functions to be determined, and

γ₂(t) := (1 +e^−2t)t²−2e^−2t t−β t² . (23) Replacing h₀(t) in (22) we find

f(r)

r² −β γ₂(t)

=−(1 +e^−2t)t²δ₂(t)X⁰(t)h⁰₁(s)

+^h2e^−2t −β tδ₂(t)− (1 +e^−2t)t²δ₂⁰(t)ⁱh₁(s)−δ₂²(t)h²₁(s). We can then write

f(r)

r² −β γ₂(t) = −A(t) (1 +e^−2s)s²h⁰₁(s) + 2e^−2sB(t)h₁(s)−C(t)h²₁(s) where we have set

A:= (1 +e^−2t) (1 +e^−X2)

t²δ₂X⁰

X² , B:= 2 (e^−2t −β t)δ₂−(1 +e^−2t)t²δ⁰₂ 2e^−X2

and C := δ²₂. We look for functions X and δ₂ such that A = B = C, i.e.

satisfying equations (21). This amounts to δ₂ = (1 +e⁻²_t)

(1 +e^−X2) t²X⁰

X² (24)

and

X⁰

(1 +e^X2)X² =− δ₂⁰

2δ₂ + 1−β t e^2t t²(1 +e^2t) .

By taking the logarithmic derivative of equation (24), we obtain δ₂⁰

δ₂ = 2

t²(1 +e^2t) − 2X⁰

(1 +e^X2)X² +2

t −2X⁰ X +X⁰⁰

X⁰

so thatX solves the ordinary differential equation X⁰⁰

X⁰ −2X⁰

X =−2 (β+ 1)

t + 2β

t(1 +e^2t) which leads to

log X⁰

X² =−2 (β+ 1) logt+ 2β

Z t 1

ds

s(1 +e^2s)+C1

for some constant C₁ ∈ R. The solution with initial condition X(0) = 0 can be written as

X(t) =

"

C₀−e^C1

Z t 1

s^{−2 (β+1)} exp 2β

Z s 1

dσ σ(1 +e^σ2)

!

ds

#−1

for some positive constant C₀. We also notice that for t > 0 small enough and β ∈(0,1−1/e²], the function γ₂ defined by (23) is positive. Notice that t = 1/logr ranges in (0,+∞) as r ranges in (1,+∞), and so we can take anyusupported outside the unit ball without further precautions. We remark that if β > 1/2, then X(t) ∼t^2β−1 as t → 0₊, so that X satisfies the initial condition X(0) = 0. Besides, for a fixed t^? > 1, the constants C₀ and C₁ are chosen such that X(t^?) = t^?, in order that the interval [0, t^?] is stable under the action of X.

Proposition 6 Assume that d = 2 and choose β ∈ (1/2,1 −1/e²]. With the above notations and {W_k}_k defined by (18), for any u ∈ H¹(R², dµ_α), compactly supported outside the ball of radius R = e^1/t? with t^? > 1, if t = 1/log|x|, then we have

I_R²[u]≥β

Z

R²

∞

X

k=0

W_k(t)

!

|x|²u²dµα−2 .

At this point, optimality is clearly an open question because of the β factor.

4.4 The case d≥3

Assume that

γd(t) :=t^d−2d _d−2² −¹₄ t^d−4d−2 if d≥4.

Since the function γ_d for d = 3 is positive only for r ≤ 8 and we want an asymptotic expansion for r → ∞, we need a different choice of γ₃. More precisely, we set

γ3(t) := 1 4

√

t 10t²−√

t+ 2.