Best Sobolev constants in the presence of sharp Hardy terms

Best Sobolev constants in the presence of sharpHardy terms

Gerassimos BarbatisNational and Kapodistrian University of Athens

joint work with A. Tertikas

University of Crete

PDE Seminar

“Athens”, March 5th, 2021

Sobolev inequality

Assume n ≥ 3. Then∫Rn

|∇u|2dx ≥ Sn(∫

Rn

|u|2∗dx)2/2∗

, u ∈ C∞c (Rn),

where

2∗ =2n

n − 2(Sobolev exponent)

Talenti (1976): Best constant

Sn = πn(n − 2)(Γ(n2 )

Γ(n)

) 2n

Extremalu(x) =

(1 + |x |2

)− n−22

The best constant remains the same if Rn is replaced by a smallerdomain; but no extremals in this case.

Sobolev inequality


|∇u|2dx ≥ Sn(∫

Rn

|u|2∗dx)2/2∗

, u ∈ C∞c (Rn),

where

2∗ =2n



Sn = πn(n − 2)(Γ(n2 )

Γ(n)

) 2n

Extremalu(x) =

(1 + |x |2

)− n−22


Sobolev inequality


|∇u|2dx ≥ Sn(∫

Rn

|u|2∗dx)2/2∗

, u ∈ C∞c (Rn),

where

2∗ =2n



Sn = πn(n − 2)(Γ(n2 )

Γ(n)

) 2n

Extremalu(x) =

(1 + |x |2

)− n−22


Hardy inequality

∫Rn

|∇u|2dx ≥(n − 2

2

)2∫Rn

u2

|x |2dx , u ∈ C∞c (Rn)

• The power |x |2 is optimal

• The constant(n−2

2

)2is sharp

• No extremals; |x |−n−2

2 solves the Euler equation

To prove sharpness use

uε(x) =

|x |− n−2

2+ε, |x | < 1,

|x |−n−2

2−ε, |x | > 1.

Hardy inequality∫Rn

|∇u|2dx ≥(n − 2

2

)2∫Rn

u2

|x |2dx , u ∈ C∞c (Rn)



2

)2is sharp




uε(x) =

|x |− n−2

2+ε, |x | < 1,

|x |−n−2

2−ε, |x | > 1.

Hardy inequality∫Rn

|∇u|2dx ≥(n − 2

2

)2∫Rn

u2

|x |2dx , u ∈ C∞c (Rn)



2

)2is sharp




uε(x) =

|x |− n−2

2+ε, |x | < 1,

|x |−n−2

2−ε, |x | > 1.

Interpolation between Sobolev and Hardy inequalities gives that forany 2 < p < 2∗ there holds

(p−HS)

∫Rn

|∇u|2dx ≥ Sn,p(∫

Rn

|x |p(n−2)

2−n|u|pdx

)2/p, u ∈ C∞c (Rn)

Sharp constant computed by Lieb (’83)

Sn,p = 2p(n − 2

2

) p+22

[2πn/2Γ2( p

p−2 )

(p − 2)Γ(n2 )Γ( 2pp−2 )

] p−2p

Extremalu(x) =

(1 + |x |

(p−2)(n−2)2

)− 2p−2

Interpolation between Sobolev and Hardy inequalities gives that forany 2 < p < 2∗ there holds

(p−HS)

∫Rn

|∇u|2dx ≥ Sn,p(∫

Rn

|x |p(n−2)

2−n|u|pdx

)2/p, u ∈ C∞c (Rn)

Sharp constant computed by Lieb (’83)

Sn,p = 2p(n − 2

2

) p+22

[2πn/2Γ2( p

p−2 )

(p − 2)Γ(n2 )Γ( 2pp−2 )

] p−2p

Extremalu(x) =

(1 + |x |

(p−2)(n−2)2

)− 2p−2

Problem: Combine Hardy and Sobolev inequalilties

More specifically: can we improve a Hardy inequality by adding aSobolev term to the RHS ?∫

Ω|∇u|2dx ≥ c∗

∫Ω

u2

d2dx +c

(∫Ω|u|2∗W (x)dx

)2/2∗

, u ∈ C∞c (Ω)

with d(x) some distance function, W (x) some weight and c∗ thesharp Hardy constant ?Most important cases:

(i) d(x) = |x | with 0 ∈ Ω

(ii) d(x) = dist(x , ∂Ω)

(iii) d(x) = |x | with 0 ∈ ∂Ω

−→ a number of such results

We are interested in Sobolev improvements involving explicit/sharpSobolev constant c .


More specifically: can we improve a Hardy inequality by adding aSobolev term to the RHS ?

∫Ω|∇u|2dx ≥ c∗

∫Ω

u2

d2dx +c


)2/2∗

, u ∈ C∞c (Ω)


(i) d(x) = |x | with 0 ∈ Ω


(iii) d(x) = |x | with 0 ∈ ∂Ω





Ω|∇u|2dx ≥ c∗

∫Ω

u2

d2dx +c


)2/2∗

, u ∈ C∞c (Ω)

with d(x) some distance function, W (x) some weight and c∗ thesharp Hardy constant ?

Most important cases:

(i) d(x) = |x | with 0 ∈ Ω


(iii) d(x) = |x | with 0 ∈ ∂Ω





Ω|∇u|2dx ≥ c∗

∫Ω

u2

d2dx +c


)2/2∗

, u ∈ C∞c (Ω)


(i) d(x) = |x | with 0 ∈ Ω


(iii) d(x) = |x | with 0 ∈ ∂Ω





Ω|∇u|2dx ≥ c∗

∫Ω

u2

d2dx +c


)2/2∗

, u ∈ C∞c (Ω)


(i) d(x) = |x | with 0 ∈ Ω


(iii) d(x) = |x | with 0 ∈ ∂Ω



Hardy inequalities involving the distance to the boundary

Ω ⊂ Rn, d(x) = dist(x , ∂Ω)∫Ω|∇u|2dx ≥ c

∫Ω

u2

d2dx , u ∈ C∞c (Ω)

If Ω is bounded with smooth boundary then the Hardy inequality isvalid and the best constant is ≤ 1/4.

If Ω is convex then the best constant is 1/4.

B., Filippas, Tertikas (’04). If

∆d ≤ 0 , in Ω,

then the best constant is 1/4.Proof. The function d1/2 is a positive supersolution to the Eulerequation.



∫Ω

u2

d2dx , u ∈ C∞c (Ω)




∆d ≤ 0 , in Ω,




∫Ω

u2

d2dx , u ∈ C∞c (Ω)




∆d ≤ 0 , in Ω,




∫Ω

u2

d2dx , u ∈ C∞c (Ω)




∆d ≤ 0 , in Ω,




∫Ω

u2

d2dx , u ∈ C∞c (Ω)




∆d ≤ 0 , in Ω,

then the best constant is 1/4.

Proof. The function d1/2 is a positive supersolution to the Eulerequation.



∫Ω

u2

d2dx , u ∈ C∞c (Ω)




∆d ≤ 0 , in Ω,


A Hardy-Sobolev inequality. Consider Ω = R3+. Then∫

R3+

|∇u|2dx ≥ S3

(∫R3

+

u6dx)1/3

, u ∈ C∞c (R3+)

and ∫R3

+

|∇u|2dx ≥ 1

4

∫R3

+

u2

x23

dx , u ∈ C∞c (R3+).

Benguria, Frank, Loss (’08). There holds∫R3

+

|∇u|2dx ≥ 1

4

∫R3

+

u2

x23

dx + S3

(∫R3

+

u6dx)1/3

for all u ∈ C∞c (R3+) !!

A Hardy-Sobolev inequality. Consider Ω = R3+. Then∫

R3+

|∇u|2dx ≥ S3

(∫R3

+

u6dx)1/3

, u ∈ C∞c (R3+)

and ∫R3

+

|∇u|2dx ≥ 1

4

∫R3

+

u2

x23

dx , u ∈ C∞c (R3+).

Benguria, Frank, Loss (’08). There holds∫R3

+

|∇u|2dx ≥ 1

4

∫R3

+

u2

x23

dx + S3

(∫R3

+

u6dx)1/3

for all u ∈ C∞c (R3+) !!

Proof. Write points in Rn+ as x = (x , y), x ∈ Rn−1, y ∈ R+. Let

H = −∆− 14y2 , a self-adjoint operator on L2(Rn

+).

Let GHpar (x, x′, t) be the corresponding parabolic Green function:ut = ∆u + 1

4y2 u

u(x, 0) = u0(x)⇒ u(x, t) =

∫Rn

+

GHpar (x, x′, t)u0(x ′)dx′

Change variables: u =√yv . Then the problem is transformed to

vt = ∆xv + vyy + 1y vy

v(x, 0) = v0(x)

This is the heat equation on

Rn+1 = (x , z) : x ∈ Rn−1 , z ∈ R2

acting on functions that are radial with respect to z ∈ R2 (soy = |z |).

Proof. Write points in Rn+ as x = (x , y), x ∈ Rn−1, y ∈ R+. Let

H = −∆− 14y2 , a self-adjoint operator on L2(Rn

+).

Let GHpar (x, x′, t) be the corresponding parabolic Green function:ut = ∆u + 1

4y2 u

u(x, 0) = u0(x)⇒ u(x, t) =

∫Rn

+

GHpar (x, x′, t)u0(x ′)dx′

Change variables: u =√yv . Then the problem is transformed to

vt = ∆xv + vyy + 1y vy

v(x, 0) = v0(x)

This is the heat equation on

Rn+1 = (x , z) : x ∈ Rn−1 , z ∈ R2

acting on functions that are radial with respect to z ∈ R2 (soy = |z |).

Write v(x , y , t) = v(x , z , t) , v radial w.r.t. z ∈ R2. Then

v(x , z , t) = (4πt)−n+1

2

∫Rn−1

∫R2

e−|x−x′|2+|z−z′|2

4t v0(x ′, z ′)dx ′dz ′

that is

v(x , y , t) =

= (4πt)−n+1

2

∫Rn−1

∫ ∞0

∫ 2π

0e−|x−x′|2+y2+y′2−yy′ cos θ

4t y ′v0(x ′, y ′)dx ′dy ′dθ

Going back to the functions u, u0 this gives

u(x , y , t) =

= (4πt)−n+1

2

∫Rn−1

∫ ∞0

∫ 2π

0

√yy ′e−

|x−x′|2+y2+y′2−yy′ cos θ4t u0(x ′, y ′)dx ′dy ′dθ

So

GHpar (x, x′, t) = (4πt)−

n+12

√yy ′e−

|x−x′|2+y2+y′24t

∫ 2π

0e

yy′ cos θ2t dθ

Hence we can compute the elliptic Green function for H,

GHell(x, x

′) =

∫ ∞0

GHpar (x, x′, t)dt (x, x′ ∈ Rn

+)

= cn√

yy ′∫ 2π

0

(|x − x ′|2 + y2 + y ′2 − 2yy ′ cos θ

)− n−12dθ

LetG−∆ell (x, x′) = c ′n|x− x′|2−n , (x, x′ ∈ Rn)

denote the Green function for the Laplacian in Rn. It may be seenthat if n = 3 then

GHell(x, x

′) ≤ G−∆ell (x, x′) , (x, x′ ∈ R3

+)

Completion of proof. The Sobolev inequality in R3 is

〈−∆u, u〉L2(R3) ≥ S3‖u‖2

L2nn−2 (R3)

or equivalently S3‖(−∆)−1/2g‖2

L2nn−2 (R3)

≤ ‖g‖2L2(R3)

By duality S3‖(−∆)−1/2g‖2L2(R3) ≤ ‖g‖

2

L2nn+2 (R3)

Let f , g ∈ C∞c (R3+). Then

|〈f , g〉L2(R3+)|

2 = |〈H1/2f ,H−1/2g〉L2(R3+)|

2

≤ 〈Hf , f 〉L2(R3+)〈H

−1g , g〉L2(R3+)

≤ 〈Hf , f 〉L2(R3+)〈(−∆)−1g , g〉L2(R3)

≤ 〈Hf , f 〉L2(R3+)

1

S3‖g‖2

L2nn+2 (R3

+)

Hence

‖f ‖2

L2nn−2 (R3

+)≤ 1

S3〈Hf , f 〉L2(R3

+) QED


〈−∆u, u〉L2(R3) ≥ S3‖u‖2

L2nn−2 (R3)


L2nn−2 (R3)

≤ ‖g‖2L2(R3)

By duality S3‖(−∆)−1/2g‖2L2(R3) ≤ ‖g‖

2

L2nn+2 (R3)


|〈f , g〉L2(R3+)|

2 = |〈H1/2f ,H−1/2g〉L2(R3+)|

2

≤ 〈Hf , f 〉L2(R3+)〈H

−1g , g〉L2(R3+)

≤ 〈Hf , f 〉L2(R3+)〈(−∆)−1g , g〉L2(R3)

≤ 〈Hf , f 〉L2(R3+)

1

S3‖g‖2

L2nn+2 (R3

+)

Hence

‖f ‖2

L2nn−2 (R3

+)≤ 1

S3〈Hf , f 〉L2(R3

+) QED


〈−∆u, u〉L2(R3) ≥ S3‖u‖2

L2nn−2 (R3)


L2nn−2 (R3)

≤ ‖g‖2L2(R3)

By duality S3‖(−∆)−1/2g‖2L2(R3) ≤ ‖g‖

2

L2nn+2 (R3)


|〈f , g〉L2(R3+)|

2 = |〈H1/2f ,H−1/2g〉L2(R3+)|

2

≤ 〈Hf , f 〉L2(R3+)〈H

−1g , g〉L2(R3+)

≤ 〈Hf , f 〉L2(R3+)〈(−∆)−1g , g〉L2(R3)

≤ 〈Hf , f 〉L2(R3+)

1

S3‖g‖2

L2nn+2 (R3

+)

Hence

‖f ‖2

L2nn−2 (R3

+)≤ 1

S3〈Hf , f 〉L2(R3

+) QED


〈−∆u, u〉L2(R3) ≥ S3‖u‖2

L2nn−2 (R3)


L2nn−2 (R3)

≤ ‖g‖2L2(R3)

By duality S3‖(−∆)−1/2g‖2L2(R3) ≤ ‖g‖

2

L2nn+2 (R3)


|〈f , g〉L2(R3+)|

2 = |〈H1/2f ,H−1/2g〉L2(R3+)|

2

≤ 〈Hf , f 〉L2(R3+)〈H

−1g , g〉L2(R3+)

≤ 〈Hf , f 〉L2(R3+)〈(−∆)−1g , g〉L2(R3)

≤ 〈Hf , f 〉L2(R3+)

1

S3‖g‖2

L2nn+2 (R3

+)

Hence

‖f ‖2

L2nn−2 (R3

+)≤ 1

S3〈Hf , f 〉L2(R3

+) QED

Brezis and Vazquez (’97). Let Ω be a bounded domain in Rn,n ≥ 3, and 2 ≤ p < 2∗. There exists c > 0 such that∫

Ω|∇u|2dx ≥

(n − 2

2

)2∫

Ω

u2

|x |2dx + c

(∫Ω|u|pdx

)2/p

Filippas and Tertikas (’02). Let

X (t) :=1

1− ln t, t ∈ (0, 1).

Let Ω be a bounded domain and D = supΩ |x |. Then there existsc > 0 such that∫

Ω|∇u|2dx ≥

(n − 2

2

)2∫

Ω

u2

|x |2dx+c

(∫ΩX

2(n−1)n−2 (|x |/D)|u|2∗dx

)2/2∗

for all u ∈ C∞c (Ω). The exponent of X is sharp.

Brezis and Vazquez (’97). Let Ω be a bounded domain in Rn,n ≥ 3, and 2 ≤ p < 2∗. There exists c > 0 such that∫

Ω|∇u|2dx ≥

(n − 2

2

)2∫

Ω

u2

|x |2dx + c

(∫Ω|u|pdx

)2/p

Filippas and Tertikas (’02). Let

X (t) :=1

1− ln t, t ∈ (0, 1).

Let Ω be a bounded domain and D = supΩ |x |. Then there existsc > 0 such that∫

Ω|∇u|2dx ≥

(n − 2

2

)2∫

Ω

u2

|x |2dx+c

(∫ΩX

2(n−1)n−2 (|x |/D)|u|2∗dx

)2/2∗

for all u ∈ C∞c (Ω). The exponent of X is sharp.

Adimurthi, Filippas and Tertikas (’09). Let Ω be a boundeddomain in Rn, n ≥ 3, and D = supΩ |x |. There holds∫

Ω|∇u|2dx ≥

(n − 2

2

)2∫

Ω

u2

|x |2dx

+(n − 2)−2(n−1)

n Sn(∫

ΩX

2(n−1)n−2 (

|x |D

)|u|2∗dx)2/2∗

for all u ∈ C∞c (Ω). The constant is the best possible.

Our aim: Find an explicit Sobolev improvement for a sharp Hardyinequality

We obtain such improvements in three different contexts:

1. Point singularity in Euclidean space

2. Point singularity in hyperbolic space

3. Boundary point singularity in Euclidean space


Ω|∇u|2dx ≥

(n − 2

2

)2∫

Ω

u2

|x |2dx

+(n − 2)−2(n−1)

n Sn(∫

ΩX

2(n−1)n−2 (

|x |D

)|u|2∗dx)2/2∗








Ω|∇u|2dx ≥

(n − 2

2

)2∫

Ω

u2

|x |2dx

+(n − 2)−2(n−1)

n Sn(∫

ΩX

2(n−1)n−2 (

|x |D

)|u|2∗dx)2/2∗








Theorem. Let n ≥ 3 and 2 < p ≤ 2∗. There exists αn > 0 suchthat for all 0 < α ≤ αn there holds∫

B1

|∇u|2dx −(n − 2

2

)2∫B1

u2

|x |2dx

≥ (n − 2)−p+2p Sn,p

(∫B1

|x |p(n−2)

2−nX (α|x |)

p+22 |u|pdx

)2/p,

for all u ∈ C∞c (Ω). Moreover the inequality is sharp for anyα ≤ αn.

Note: For any α, α′ > 0 there holds

limx→0

X (α|x |)X (α′|x |)

= 1 .



B1

|∇u|2dx −(n − 2

2

)2∫B1

u2

|x |2dx

≥ (n − 2)−p+2p Sn,p

(∫B1

|x |p(n−2)

2−nX (α|x |)

p+22 |u|pdx

)2/p,



limx→0

X (α|x |)X (α′|x |)

= 1 .



B1

|∇u|2dx −(n − 2

2

)2∫B1

u2

|x |2dx

≥ (n − 2)−p+2p Sn,p

(∫B1

|x |p(n−2)

2−nX (α|x |)

p+22 |u|pdx

)2/p,



limx→0

X (α|x |)X (α′|x |)

= 1 .

Two improved versions:

Theorem. Let n ≥ 3, 2 < p ≤ 2∗, θ ∈ (0, 2). There exist R > 1and αn,θ < 1 such that for any 0 < α ≤ αn,θ there holds∫

B1

|∇u|2dx −(n − 2

2

)2∫B1

u2

|x |2dx − θ2

∫B1

u2

|x |2−θ(Rθ − |x |θ)dx

≥ (n − 2)−p+2p Sn,p

(∫B1

|x |p(n−2)

2−nX

p+22 (α|x |)|u|pdx

)2/p

.

for all u ∈ C∞c (B1). Moreover the inequality is sharp for allα ≤ αn,θ.

Rθ = 1+1√n − 2

, − lnαn,θ = R2θ−1+

∫ 1

0

sθ−1(2Rθ − sθ)

(Rθ − sθ)2ds



B1

|∇u|2dx −(n − 2

2

)2∫B1

u2

|x |2dx − θ2

∫B1

u2

|x |2−θ(Rθ − |x |θ)dx

≥ (n − 2)−p+2p Sn,p

(∫B1

|x |p(n−2)

2−nX

p+22 (α|x |)|u|pdx

)2/p

.


Rθ = 1+1√n − 2

, − lnαn,θ = R2θ−1+

∫ 1

0


(Rθ − sθ)2ds



B1

|∇u|2dx −(n − 2

2

)2∫B1

u2

|x |2dx − θ2

∫B1

u2

|x |2−θ(Rθ − |x |θ)dx

≥ (n − 2)−p+2p Sn,p

(∫B1

|x |p(n−2)

2−nX

p+22 (α|x |)|u|pdx

)2/p

.


Rθ = 1+1√n − 2

, − lnαn,θ = R2θ−1+

∫ 1

0


(Rθ − sθ)2ds

Theorem. Let n ≥ 3, 2 < p ≤ 2∗ and 0 ≤ θ < 1/2. There existsαn > 0 such that for any 0 < α ≤ αn there holds∫

B1

|∇u|2dx −(n − 2

2

)2∫B1

u2

|x |2dx − θ(1− θ)

∫B1

u2

|x |2X 2(α|x |)dx

≥(1− 2θ

n − 2

) p+2pSn,p

(∫B1

|x |p(n−2)

2−nX (α|x |)

p+22 |u|pdx

)2/p.

for all u ∈ C∞c (B1). Moreover the constant ( 1−2θn−2 )

p+2p Sn,p is sharp

for any α ≤ αn.


Hyperbolic space Hn: use the Poincare ball model

Equip the unit ball with the Riemannian metric

ds2 =(1− |x |2

2

)−2dx2.

Under this model we have

|∇Hnv |2 =(1− |x |2

2

)2|∇Rnv |2 , dV =

(1− |x |2

2

)−ndx

and Riemannian distance to the origin is

ρ(x) = ln(1 + |x |

1− |x |

).


Hyperbolic space Hn: use the Poincare ball model

Equip the unit ball with the Riemannian metric

ds2 =(1− |x |2

2

)−2dx2.

Under this model we have

|∇Hnv |2 =(1− |x |2

2

)2|∇Rnv |2 , dV =

(1− |x |2

2

)−ndx

and Riemannian distance to the origin is

ρ(x) = ln(1 + |x |

1− |x |

).

In Hn both Hardy and Sobolev inequalities are valid:

E. Hebey. If n ≥ 3 then

∫Hn

|∇Hnu|2dV − n(n − 2)

4

∫Hn

u2dV ≥ Sn(∫

Hn

|u|2∗dV)2/2∗

for all u ∈ C∞c (Hn).

More generally, for any 2 ≤ p ≤ 2∗,∫Hn

|∇Hnu|2dV−n(n − 2)

4

∫Hn

u2dV ≥ Sn,p(∫

Hn

(sinh ρ)p(n−2)

2−n|u|pdV

)2/p,


In Hn both Hardy and Sobolev inequalities are valid:

E. Hebey. If n ≥ 3 then

∫Hn

|∇Hnu|2dV − n(n − 2)

4

∫Hn

u2dV ≥ Sn(∫

Hn

|u|2∗dV)2/2∗


More generally, for any 2 ≤ p ≤ 2∗,∫Hn

|∇Hnu|2dV−n(n − 2)

4

∫Hn

u2dV ≥ Sn,p(∫

Hn

(sinh ρ)p(n−2)

2−n|u|pdV

)2/p,


Hardy inequality is also valid:

∫Hn

|∇u|2dV ≥(n − 2

2

)2∫Hn

u2

ρ2dV , u ∈ C∞c (Hn).

Several related results and improvements: Carron; Yang and Kong;Kombe; Berchio, Ganguly, Grillo, Pinchover; Kristaly, . . .

Q: What about Sobolev improvement?


∫Hn

|∇u|2dV ≥(n − 2

2

)2∫Hn

u2





∫Hn

|∇u|2dV ≥(n − 2

2

)2∫Hn

u2




Consider the inequality∫B1

|∇u|2dx −(n − 2

2

)2∫B1

u2

|x |2dx

≥ (n − 2)−p+2p Sn,p

(∫B1

|x |p(n−2)

2−nX (α|x |)

p+22 |u|pdx

)2/p

(valid for α ≤ αn and all u ∈ C∞c (B1)) and “translate” it to thehyperbolic context.

Obtain

Theorem. Let n ≥ 3 and 2 < p ≤ 2∗. For any 0 < α ≤ αn thereholds∫

Hn

|∇Hnu|2dV −(n − 2

2

)2∫Hn

u2

ρ2dV

≥ (n − 2)−p+2p Sn,p

(∫Hn

(sinh ρ)p(n−2)

2−nX

p+22 (α tanh(

ρ

2))|u|pdV

)2/p,

for all u ∈ C∞c (Hn). Moreover the inequality is sharp for all0 < α ≤ αn.

Can we have more?


|∇u|2dx −(n − 2

2

)2∫B1

u2

|x |2dx

≥ (n − 2)−p+2p Sn,p

(∫B1

|x |p(n−2)

2−nX (α|x |)

p+22 |u|pdx

)2/p

(valid for α ≤ αn and all u ∈ C∞c (B1)) and “translate” it to thehyperbolic context. Obtain


Hn

|∇Hnu|2dV −(n − 2

2

)2∫Hn

u2

ρ2dV

≥ (n − 2)−p+2p Sn,p

(∫Hn

(sinh ρ)p(n−2)

2−nX

p+22 (α tanh(

ρ

2))|u|pdV

)2/p,


Can we have more?


|∇u|2dx −(n − 2

2

)2∫B1

u2

|x |2dx

≥ (n − 2)−p+2p Sn,p

(∫B1

|x |p(n−2)

2−nX (α|x |)

p+22 |u|pdx

)2/p

(valid for α ≤ αn and all u ∈ C∞c (B1)) and “translate” it to thehyperbolic context. Obtain


Hn

|∇Hnu|2dV −(n − 2

2

)2∫Hn

u2

ρ2dV

≥ (n − 2)−p+2p Sn,p

(∫Hn

(sinh ρ)p(n−2)

2−nX

p+22 (α tanh(

ρ

2))|u|pdV

)2/p,


Can we have more?

Indeed we may also obtain

Theorem. Let n ≥ 3 and 2 < p ≤ 2∗. For any 0 < α ≤ αn we have∫Hn

|∇Hnu|2dV − n(n − 2)

4

∫Hn

u2dV −(n − 2

2

)2∫Hn

u2

sinh2 ρdV

≥ (n − 2)−p+2p Sn,p

(∫Hn

(sinh ρ)p(n−2)

2−nX

p+22 (α tanh(ρ/2))|u|pdV

)2/p,

for all u ∈ C∞c (Hn). Moreover the constant (n − 2)−p+2p Sn,p is

sharp for all 0 < α ≤ αn.

We would like (n − 1)2/4 instead of n(n − 2)/4 in front of the L2

term.

Indeed we may also obtain

Theorem. Let n ≥ 3 and 2 < p ≤ 2∗. For any 0 < α ≤ αn we have∫Hn

|∇Hnu|2dV − n(n − 2)

4

∫Hn

u2dV −(n − 2

2

)2∫Hn

u2

sinh2 ρdV

≥ (n − 2)−p+2p Sn,p

(∫Hn

(sinh ρ)p(n−2)

2−nX

p+22 (α tanh(ρ/2))|u|pdV

)2/p,

for all u ∈ C∞c (Hn). Moreover the constant (n − 2)−p+2p Sn,p is

sharp for all 0 < α ≤ αn.

We would like (n − 1)2/4 instead of n(n − 2)/4 in front of the L2

term.

Berchio, Ganguly, Grillo (’17). There holds∫Hn

|∇Hnu|2dV −(n − 1

2

)2∫Hn

u2dV

≥ 1

4

∫Hn

u2

ρ2dV +

(n − 1)(n − 3)

4

∫Hn

u2

sinh2 ρdV

for all u ∈ C∞c (Hn). The inequality is optimal, non-improvable.

Also recent article by Berchio, Ganguly, Grillo, Pinchover (’19).

Berchio, Ganguly, Grillo (’17). There holds∫Hn

|∇Hnu|2dV −(n − 1

2

)2∫Hn

u2dV

≥ 1

4

∫Hn

u2

ρ2dV +

(n − 1)(n − 3)

4

∫Hn

u2

sinh2 ρdV

for all u ∈ C∞c (Hn). The inequality is optimal, non-improvable.

Also recent article by Berchio, Ganguly, Grillo, Pinchover (’19).

We are interested in Sobolev improvements of the Poincare-Hardyinequality∫Hn

|∇Hnv |2dV ≥(n − 1

2

)2∫Hn

v2dV +(n − 2

2

)2∫Hn

v2

sinh2 ρdV ,

for u ∈ C∞c (Hn).

To state the result we need to introduce a constant Sn,p and afunction Y (t), t > 0.

We are interested in Sobolev improvements of the Poincare-Hardyinequality∫Hn

|∇Hnv |2dV ≥(n − 1

2

)2∫Hn

v2dV +(n − 2

2

)2∫Hn

v2

sinh2 ρdV ,

for u ∈ C∞c (Hn).

To state the result we need to introduce a constant Sn,p and afunction Y (t), t > 0.

The constant Sn,p, 2 < p ≤ 2∗.

Given 2 < p ≤ 2∗ we define Sn,p to be the best constant for thePoincare-Sobolev inequality∫

Hn

|∇Hnv |2dV ≥(n − 1

2

)2∫Hn

v2dV

+Sn,p

(∫Hn

(sinh ρ)p(n−2)

2−n|v |pdV

)2/p, v ∈ C∞c (Hn)

Note. The positivity of Sn,p follows from the positivity of Sn,2∗

(Mancini and Sandeep ’08)

The constant Sn,p, 2 < p ≤ 2∗.

Given 2 < p ≤ 2∗ we define Sn,p to be the best constant for thePoincare-Sobolev inequality∫

Hn

|∇Hnv |2dV ≥(n − 1

2

)2∫Hn

v2dV

+Sn,p

(∫Hn

(sinh ρ)p(n−2)

2−n|v |pdV

)2/p, v ∈ C∞c (Hn)

Note. The positivity of Sn,p follows from the positivity of Sn,2∗

(Mancini and Sandeep ’08)

The function Y (t).

Consider the following two auxiliary problems:g ′′(t) + 1

4 sinh2 tg(t) = 0, t > 0,

limt→+∞ g(t) = 1,

and, for n ≥ 3,h′′(t)− (n−1)(n−3)

4 sinh2 th(t) = 0, t > 0,

limt→+∞ h(t) = 1 .

Prove existence and uniqueness and study asymptotics near zero.Then define a function ρ = ρ(t) by∫ ρ(t)

0

dr

h(r)2=

∫ t

0

ds

g(s)2, t > 0.

The function Y (t).


4 sinh2 tg(t) = 0, t > 0,

limt→+∞ g(t) = 1,

and, for n ≥ 3,h′′(t)− (n−1)(n−3)

4 sinh2 th(t) = 0, t > 0,

limt→+∞ h(t) = 1 .


0

dr

h(r)2=

∫ t

0

ds

g(s)2, t > 0.

The function Y (t).


4 sinh2 tg(t) = 0, t > 0,

limt→+∞ g(t) = 1,

and, for n ≥ 3,h′′(t)− (n−1)(n−3)

4 sinh2 th(t) = 0, t > 0,

limt→+∞ h(t) = 1 .


0

dr

h(r)2=

∫ t

0

ds

g(s)2, t > 0.

The function Y (t) is defined by

Y (t) = (n − 2)h(ρ(t)

)2sinh t

g(t)2 sinh ρ(t), t > 0.

Theorem (Poicare-Hardy-Sobolev inequality). Let n ≥ 3 and2 < p ≤ 2∗. There holds∫Hn

|∇Hnv |2dV ≥(n − 1

2

)2∫Hn

v2dV +(n − 2

2

)2∫Hn

v2

sinh2 ρdV

+(n − 2)−p+2p Sn,p

(∫Hn

(sinh ρ)p(n−2)

2−nY

p+22 (ρ)|v |pdV

)2/p.

for all v ∈ C∞c (Hn). Moreover the inequality is sharp.

The function Y (t) is defined by

Y (t) = (n − 2)h(ρ(t)

)2sinh t

g(t)2 sinh ρ(t), t > 0.

Theorem (Poicare-Hardy-Sobolev inequality). Let n ≥ 3 and2 < p ≤ 2∗. There holds∫Hn

|∇Hnv |2dV ≥(n − 1

2

)2∫Hn

v2dV +(n − 2

2

)2∫Hn

v2

sinh2 ρdV

+(n − 2)−p+2p Sn,p

(∫Hn

(sinh ρ)p(n−2)

2−nY

p+22 (ρ)|v |pdV

)2/p.

for all v ∈ C∞c (Hn). Moreover the inequality is sharp.

Concerning the function Y (t) we have:

Proposition.

(i) There exists αn > 0 such that

Y (t) ≥ X(αn tanh

t

2

), t > 0

(ii) There holds

limt→0

Y (t)

X (t)= 1

What about Sn,p ?

Concerning the function Y (t) we have:

Proposition.

(i) There exists αn > 0 such that

Y (t) ≥ X(αn tanh

t

2

), t > 0

(ii) There holds

limt→0

Y (t)

X (t)= 1

What about Sn,p ?

The precise value of Sn,p is not known.∫Hn

|∇Hnv |2dV ≥(n − 1

2

)2∫Hn

v2dV

+Sn,p

(∫Hn

(sinh ρ)p(n−2)

2−n|v |pdV

)2/p

Using the half-space model for Hn we find that Sn,p is the bestconstant for the Hardy-Sobolev inequality∫

Rn+

|∇u|2dx ≥ 1

4

∫Rn

+

u2

x2n

dx

+Sn,p

(∫Rn

+

( 2

|x − en| |x + en|

)n− n−22

p|u|pdx

)2/p

for all u ∈ C∞c (Rn+).

The precise value of Sn,p is not known.∫Hn

|∇Hnv |2dV ≥(n − 1

2

)2∫Hn

v2dV

+Sn,p

(∫Hn

(sinh ρ)p(n−2)

2−n|v |pdV

)2/p

Using the half-space model for Hn we find that Sn,p is the bestconstant for the Hardy-Sobolev inequality∫

Rn+

|∇u|2dx ≥ 1

4

∫Rn

+

u2

x2n

dx

+Sn,p

(∫Rn

+

( 2

|x − en| |x + en|

)n− n−22

p|u|pdx

)2/p


For p = 2∗ this becomes∫Rn

+

|∇u|2dx ≥ 1

4

∫Rn

+

u2

x2n

dx+Sn,2∗

(∫Rn

+

|u|2∗dx)2/2∗

, u ∈ C∞c (Rn+)

The result of Benguria-Frank-Loss states that

S3,2∗ = S3.

We have

Theorem. Let n = 3. For any 2 ≤ p < 2∗ there holds

S3,p = S3,p =p

22p

[4πΓ2( p

p−2 )

(p − 2)Γ( 2pp−2 )

] p−2p

Open problem: compute Sn,p for n ≥ 4

For p = 2∗ this becomes∫Rn

+

|∇u|2dx ≥ 1

4

∫Rn

+

u2

x2n

dx+Sn,2∗

(∫Rn

+

|u|2∗dx)2/2∗

, u ∈ C∞c (Rn+)

The result of Benguria-Frank-Loss states that

S3,2∗ = S3.

We have

Theorem. Let n = 3. For any 2 ≤ p < 2∗ there holds

S3,p = S3,p =p

22p

[4πΓ2( p

p−2 )

(p − 2)Γ( 2pp−2 )

] p−2p

Open problem: compute Sn,p for n ≥ 4

In case n = 3 the result is sharp also with the logarithmic functionX .

Theorem. There exists an α3 > 0 such that for all 0 < α ≤ α3, all2 < p ≤ 2∗ and all v ∈ C∞c (H3) there holds∫

H3

|∇H3v |2dV ≥∫H3

v2dV +1

4

∫H3

v2

sinh2 ρdV

+S3,p

(∫Hn

(sinh ρ)p−6

2 Xp+2

2(α tanh(ρ/2)

)|v |pdV

)2/p

The constant S3,p is sharp for all 0 < α ≤ α3.

Note. To show optimality use the sharpness of the inequality∫B1

|∇u|2dx −(n − 2

2

)2∫B1

u2

|x |2dx − θ(1− θ)

∫B1

u2

|x |2X 2(α|x |)dx

≥(1− 2θ

n − 2

) p+2pSn,p

(∫B1

|x |p(n−2)

2−nX (α|x |)

p+22 |u|pdx

)2/p



H3

|∇H3v |2dV ≥∫H3

v2dV +1

4

∫H3

v2

sinh2 ρdV

+S3,p

(∫Hn

(sinh ρ)p−6

2 Xp+2

2(α tanh(ρ/2)

)|v |pdV

)2/p



|∇u|2dx −(n − 2

2

)2∫B1

u2

|x |2dx − θ(1− θ)

∫B1

u2

|x |2X 2(α|x |)dx

≥(1− 2θ

n − 2

) p+2pSn,p

(∫B1

|x |p(n−2)

2−nX (α|x |)

p+22 |u|pdx

)2/p



H3

|∇H3v |2dV ≥∫H3

v2dV +1

4

∫H3

v2

sinh2 ρdV

+S3,p

(∫Hn

(sinh ρ)p−6

2 Xp+2

2(α tanh(ρ/2)

)|v |pdV

)2/p



|∇u|2dx −(n − 2

2

)2∫B1

u2

|x |2dx − θ(1− θ)

∫B1

u2

|x |2X 2(α|x |)dx

≥(1− 2θ

n − 2

) p+2pSn,p

(∫B1

|x |p(n−2)

2−nX (α|x |)

p+22 |u|pdx

)2/p


Let Ω be a domain in Rn and assume that 0 ∈ ∂Ω. We areinterested in the Hardy inequality∫

Ω|∇u|2dx ≥ c

∫Ω

u2

|x |2dx , u ∈ C∞c (Ω)

and related Sobolev improvements.

What about the best constant c? We know that

c =(n − 2

2

)2

works, but can we do better ? The geometry plays a role.

I. ConesII. Bounded domains with nice boundary



Ω|∇u|2dx ≥ c

∫Ω

u2

|x |2dx , u ∈ C∞c (Ω)



c =(n − 2

2

)2





Ω|∇u|2dx ≥ c

∫Ω

u2

|x |2dx , u ∈ C∞c (Ω)



c =(n − 2

2

)2



Part I. Cones.

A. Nazarov (’06). Let CΣ be a finite (or infinite) cone with vertexat zero:

CΣ = rω : ω ∈ Σ , 0 < r < 1 (or r > 0)

where Σ an open subset of Sn−1.Let µ1(Σ) be the first eigenvalue of the Dirichlet Laplacian on Σ.Then ∫

CΣ

|∇u|2dx ≥[(n − 2

2

)2+ µ1(Σ)

] ∫CΣ

u2

|x |2dx ,

for all u ∈ C∞c (CΣ). Moreover the constant is the best possible.

Proof. The function

φ(x) = r−n−2

2 ψ1(ω)

is a positive solution to the Euler equation.

Part I. Cones.

A. Nazarov (’06). Let CΣ be a finite (or infinite) cone with vertexat zero:

CΣ = rω : ω ∈ Σ , 0 < r < 1 (or r > 0)

where Σ an open subset of Sn−1.Let µ1(Σ) be the first eigenvalue of the Dirichlet Laplacian on Σ.Then ∫

CΣ

|∇u|2dx ≥[(n − 2

2

)2+ µ1(Σ)

] ∫CΣ

u2

|x |2dx ,

for all u ∈ C∞c (CΣ). Moreover the constant is the best possible.

Proof. The function

φ(x) = r−n−2

2 ψ1(ω)


Q: Is it possible to improve the above inequality by adding moreterms ?

Theorem (B., Filippas, Tertikas ’18). There holds∫CΣ

|∇u|2dx ≥[(n − 2

2

)2+ µ1(Σ)

] ∫CΣ

u2

|x |2dx

+1

4

∫CΣ

u2

|x |2X 2(|x |)dx

for all u ∈ C∞c (CΣ). The inequality is sharp.

Proof. The function

φ(x) = r−n−2

2 X−12 (r)ψ1(ω)




|∇u|2dx ≥[(n − 2

2

)2+ µ1(Σ)

] ∫CΣ

u2

|x |2dx

+1

4

∫CΣ

u2

|x |2X 2(|x |)dx


Proof. The function

φ(x) = r−n−2

2 X−12 (r)ψ1(ω)




|∇u|2dx ≥[(n − 2

2

)2+ µ1(Σ)

] ∫CΣ

u2

|x |2dx

+1

4

∫CΣ

u2

|x |2X 2(|x |)dx


Proof. The function

φ(x) = r−n−2

2 X−12 (r)ψ1(ω)


What about Sobolev improvements?

Theorem (B., Filippas, Tertikas ’18). Let n ≥ 3. There exists apositive constant C = C (Σ) such that∫

CΣ

|∇u|2dx ≥[(n − 2

2

)2+ µ1(Σ)

] ∫CΣ

u2

|x |2dx

+C (Σ)

(∫CΣ

X2n−2n−2 (|x |)|u|2∗dx

)2/2∗

,

for all u ∈ C∞c (CΣ). Moreover

(i) The exponent (2n − 2)/(n − 2) of X (|x |) is the best possible.(ii) For the best constant C (Σ) we have

C (Σ) ≤ Cn|Σ|2n ;

in particular it cannot be taken to be independent of Σ.

What about additional improvements ?Define X1(t) = X (t) and for k ≥ 2

Xk(t) = X1(Xk−1(t)), t ∈ (0, 1).


|∇u|2dx ≥[(n − 2

2

)2+ µ1(Σ)

] ∫CΣ

u2

|x |2dx

+1

4

∞∑i=1

∫CΣ

u2

|x |2X 2

1 X22 . . .X

2i dx ,

for all u ∈ C∞c (CΣ).The inequality is sharp at each step.

Proof. For each fixed m, the function

φ(x) = r−n−2

2 X1(r)−1/2 . . .Xm(r)−1/2ψ1(ω)

is a positive solution to the mth-improved Hardy inequality.

What about additional improvements ?Define X1(t) = X (t) and for k ≥ 2

Xk(t) = X1(Xk−1(t)), t ∈ (0, 1).


|∇u|2dx ≥[(n − 2

2

)2+ µ1(Σ)

] ∫CΣ

u2

|x |2dx

+1

4

∞∑i=1

∫CΣ

u2

|x |2X 2

1 X22 . . .X

2i dx ,

for all u ∈ C∞c (CΣ).The inequality is sharp at each step.

Proof. For each fixed m, the function

φ(x) = r−n−2

2 X1(r)−1/2 . . .Xm(r)−1/2ψ1(ω)

is a positive solution to the mth-improved Hardy inequality.

What about Sobolev improvements for the mth improved Hardyinequality?

Theorem Let n ≥ 3. There exists a constant C that depends onlyon Σ such that for any m ∈ N∫

CΣ

|∇u|2dx ≥[(n − 2

2

)2+ µ1(Σ)

] ∫CΣ

u2

|x |2dx

+1

4

m∑i=1

∫CΣ

u2

|x |2X 2

1 . . .X2i dx

+C

(∫CΣ

(X1 . . .Xm+1)2n−2n−2 |u|

2nn−2 dx

) n−2n

,


Part II. General bounded domains

Let Ω be a domain with nice boundary and with 0 ∈ ∂Ω. We areinterested in the Hardy inequality∫

Ω|∇u|2dx ≥ c

∫Ω

u2

|x |2dx , u ∈ C∞c (Ω)

Consider the half-space Rn+ = xn > 0.

This is a special case of a cone, hence the Hardy constant is(n − 2

2

)2+ µ1(Sn−1

+ ) =(n − 2

2

)2+ (n − 1) =

n2

4

This is the ‘right’ constant locally for a domain with smoothboundary.

Part II. General bounded domains

Let Ω be a domain with nice boundary and with 0 ∈ ∂Ω. We areinterested in the Hardy inequality∫

Ω|∇u|2dx ≥ c

∫Ω

u2

|x |2dx , u ∈ C∞c (Ω)

Consider the half-space Rn+ = xn > 0.

This is a special case of a cone, hence the Hardy constant is(n − 2

2

)2+ µ1(Sn−1

+ ) =(n − 2

2

)2+ (n − 1) =

n2

4

This is the ‘right’ constant locally for a domain with smoothboundary.

M.M. Fall (’12) Let Ω be bounded Lipschitz boundary and assumethat ∂Ω is C 2 near the origin. There exists an r = r(Ω) such thatfor all u ∈ C∞c (Ω ∩ Br ) there holds∫

Ω∩Br

|∇u|2dx ≥ n2

4

∫Ω∩Br

u2

|x |2dx

+C

∫Ω∩Br

u2

|x |2X 2

1 dx

Theorem (B., Filippas, Tertikas ’18). Let Ω ⊂ Rn, n ≥ 2, be abounded domain with 0 ∈ ∂Ω admiting an exterior ball of radius ρat 0. Let D = supΩ |x |. There exist σn > 0 such that if ρ ≥ D/σnthen ∫

Ω|∇u|2dx ≥ n2

4

∫Ω

u2

|x |2dx

for all u ∈ C∞c (Ω).

If in addition Ω satisfies an interior ball condition at 0 then theconstant is sharp.


Ω∩Br

|∇u|2dx ≥ n2

4

∫Ω∩Br

u2

|x |2dx +C

∫Ω∩Br

u2

|x |2X 2

1 dx


Ω|∇u|2dx ≥ n2

4

∫Ω

u2

|x |2dx




Ω∩Br

|∇u|2dx ≥ n2

4

∫Ω∩Br

u2

|x |2dx +C

∫Ω∩Br

u2

|x |2X 2

1 dx


Ω|∇u|2dx ≥ n2

4

∫Ω

u2

|x |2dx



Theorem. Let Ω ⊂ Rn, n ≥ 2, be a bounded domain with 0 ∈ ∂Ωhaving an exterior ball of radius ρ at 0. Let D = supΩ |x |. Thereexist σn > 0 and κ > 0 such that if ρ ≥ D/σn then∫

Ω|∇u|2dx ≥ n2

4

∫Ω

u2

|x |2dx +

1

4

∞∑i=1

∫Ω

u2

|x |2X 2

1 . . .X2i dx ,

for all u ∈ C∞c (Ω); here Xi = Xi (σn|x |/(3κD)).

If in addition Ω satisfies an interior ball condition at 0 then theestimate is sharp at each step.


Theorem (Hardy-Sobolev inequality) Let Ω ⊂ Rn, n ≥ 2, be abounded domain with 0 ∈ ∂Ω having an exterior ball of radius ρ at0. There exist σn,Cn > 0 such that if ρ ≥ D/σn then∫

Ω|∇u|2dx ≥ n2

4

∫Ω

u2

|x |2dx + Cn

(∫ΩX

2n−2n−2 |u|

2nn−2 dx

) n−2n

,

for all u ∈ C∞c (Ω); here X = X (|x |/3D).

If in addition Ω satisfies an interior ball condition at 0 then theestimate is sharp.


Theorem (Hardy-Sobolev inequality) Let Ω ⊂ Rn, n ≥ 2, be abounded domain with 0 ∈ ∂Ω having an exterior ball of radius ρ at0. There exist σn,Cn > 0 such that if ρ ≥ D/σn then∫

Ω|∇u|2dx ≥ n2

4

∫Ω

u2

|x |2dx + Cn

(∫ΩX

2n−2n−2 |u|

2nn−2 dx

) n−2n

,

for all u ∈ C∞c (Ω); here X = X (|x |/3D).

If in addition Ω satisfies an interior ball condition at 0 then theestimate is sharp.

A natural question:

In order to have the Hardy inequality with constant n2/4 is itnecessary that the exterior ball is large compared to D = supΩ |x | ?

Example. For ρ ∈ (0, 1/2) and θ ∈ (0, π/2) define

Aρ,θ = x = (x ′, xn) ∈ B1 : xn < cot θ|x ′| and |x − ρen| > ρ.

Let Ω ⊃ Aρ,θ having B(ρen, ρ) as largest exterior ball at 0. Letλ1(n, θ) be the first Dirichlet eigenvalue of the Laplace operator onthe spherical cap

Σθ = (x ′, xn) ∈ Sn−1 : xn < cot θ |x ′|.

If

ρ <1

2 cos θe− π√

n−1−λ1(n,θ) ,

then the best Hardy constant of Ω is strictly smaller than n2/4.

Back to Hardy-Sobolev inequalities with explicit constants.

Let n ≥ 3, 0 ≤ γ < n/2 and 2 < p ≤ 2∗. We define S∗n,p,γ to bethe best constant for the inequality∫

Rn+

|∇u|2dx − γ(n − γ)

∫Rn

+

u2

|x |2dx

≥ S∗n,p,γ

(∫Rn

+

|x |p(n−2)

2−n|u|pdx

)2/p


Back to Hardy-Sobolev inequalities with explicit constants.

Let n ≥ 3, 0 ≤ γ < n/2 and 2 < p ≤ 2∗. We define S∗n,p,γ to bethe best constant for the inequality∫

Rn+

|∇u|2dx − γ(n − γ)

∫Rn

+

u2

|x |2dx

≥ S∗n,p,γ

(∫Rn

+

|x |p(n−2)

2−n|u|pdx

)2/p


Theorem. Let Ω ⊂ Rn, n ≥ 3, be a bounded domain with 0 ∈ ∂Ωand let D = supΩ |x |. Assume that Ω satisfies an exterior ballcondition at zero with exterior ball Bρ(−ρen). Then for any2 < p ≤ 2∗ and any γ ∈ [0, n/2) there exist an rn,γ and α∗n,γ in(0, 1) such that, if the radius ρ of the exterior ball satisfiesρ ≥ D/rn,γ then for all 0 < α ≤ α∗n,γ there holds∫

Ω|∇u|2dx ≥ n2

4

∫Ω

u2

|x |2dx

+(n − 2γ)−p+2p S∗n,p,γ

(∫Ω|x |

p(n−2)2−n( |x + 2ρen|

2ρ

) p(n−2)2−n

Xp+2

2 |u|pdx) 2

p

,

for all u ∈ C∞c (Ω); here X = X (α|x |/D).

But no sharpness...


Ω|∇u|2dx ≥ n2

4

∫Ω

u2

|x |2dx

+(n − 2γ)−p+2p S∗n,p,γ

(∫Ω|x |

p(n−2)2−n( |x + 2ρen|

2ρ

) p(n−2)2−n

Xp+2

2 |u|pdx) 2

p

,


But no sharpness...


Ω|∇u|2dx ≥ n2

4

∫Ω

u2

|x |2dx

+(n − 2γ)−p+2p S∗n,p,γ

(∫Ω|x |

p(n−2)2−n( |x + 2ρen|

2ρ

) p(n−2)2−n

Xp+2

2 |u|pdx) 2

p

,


But no sharpness...

About the proof.

For interior point singularity: Change variables so that the twoinfima are directly comparable. For example the best constant Sn,p

we have

Sn,p = inf

∫ ∞0

∫Sn−1

h(ρ)2(w2ρ +

1

sinh2ρ|∇ωw |2

)dS dρ(∫ ∞

0

∫Sn−1

(sinh ρ)−p+2

2 h(ρ)p|w |pdS dρ)2/p

.

The term involving ∇ω can be ignored by symmetrization.

For boundary point singularity: use in addition a conformal mapfrom Rn

+ onto B(ρ)c .

About the proof.

For interior point singularity: Change variables so that the twoinfima are directly comparable.

For example the best constant Sn,p

we have

Sn,p = inf

∫ ∞0

∫Sn−1

h(ρ)2(w2ρ +

1

sinh2ρ|∇ωw |2

)dS dρ(∫ ∞

0

∫Sn−1

(sinh ρ)−p+2


.



+ onto B(ρ)c .

About the proof.


we have

Sn,p = inf

∫ ∞0

∫Sn−1

h(ρ)2(w2ρ +

1

sinh2ρ|∇ωw |2

)dS dρ(∫ ∞

0

∫Sn−1

(sinh ρ)−p+2


.



+ onto B(ρ)c .

About the proof.


we have

Sn,p = inf

∫ ∞0

∫Sn−1

h(ρ)2(w2ρ +

1

sinh2ρ|∇ωw |2

)dS dρ(∫ ∞

0

∫Sn−1

(sinh ρ)−p+2


.



+ onto B(ρ)c .

The end

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Best Sobolev constants in the presence of sharp Hardy terms