Best Sobolev constants in the presence of sharp Hardy terms
Transcript of 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
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
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.
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)
• 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)
• 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.
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 .
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 .
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 .
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 .
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 .
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.
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.
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.
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.
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.
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.
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
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
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
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
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
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
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
1. Point singularity in Euclidean space
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 .
1. Point singularity in Euclidean space
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 .
1. Point singularity in Euclidean space
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 .
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
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
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
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.
2. Point singularity in hyperbolic space
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 |
).
2. Point singularity in hyperbolic space
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,
for all u ∈ C∞c (Hn).
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,
for all u ∈ C∞c (Hn).
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?
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?
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?
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?
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?
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?
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).
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).
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) 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 all u ∈ C∞c (Rn+).
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
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
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
3. Boundary point singularity in Euclidean space
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
3. Boundary point singularity in Euclidean space
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
3. Boundary point singularity in Euclidean space
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
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)
is a positive solution to the Euler equation.
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)
is a positive solution to the Euler equation.
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)
is a positive solution to the Euler equation.
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)
is a positive solution to the Euler equation.
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).
Theorem (B., Filippas, Tertikas ’18). There holds∫CΣ
|∇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).
Theorem (B., Filippas, Tertikas ’18). There holds∫CΣ
|∇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
,
for all u ∈ C∞c (CΣ). The inequality is sharp.
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.
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.
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.
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.
What about Sobolev improvements?
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.
What about Sobolev improvements?
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
for all u ∈ C∞c (Rn+).
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
for all u ∈ C∞c (Rn+).
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...
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...
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...
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
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
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
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 .