On Hardys and Caffarelli, Kohn, Nirenbergs inequalites. S eminaire - - PowerPoint PPT Presentation

On Hardy’s and Caffarelli, Kohn, Nirenberg’s inequalites.

S´ eminaire EDP-Analyse de l’Institut Camille Jordan, Lyon, April 2019

Hoai-Minh Nguyen

´ Ecole Polytechnique F´ ed´ erale de Lausanne EPFL, Switzerland joint with Marco Squassina

Outline

1 Known results related to Hardy’s and Caffarelli, Kohn,

Nirenberg’s (CKN’s) inequalities

2 CKN’s inequalities for fractional Sobolev spaces 3 New perspectives of Hardy’s and CKN’s inequalities in

Sobolev spaces.

Section 1: Known results related to Hardy’s and CKN’s inequalities

Hardy’s inequalities

1 For 1 p < d,

ˆ

Rd

|u|p |x|p dx C

ˆ

Rd |∇u|p dx

∀ u ∈ C1

c(Rd). 2 For p > d,

ˆ

Rd

|u|p |x|p dx C

ˆ

Rd |∇u|p dx

∀ u ∈ C1

c(Rd \ {0}). 3 For p = d > 1, ....

Standard proof is based on integration by parts.

Hardy’s inequalities

1 For 1 p < d,

ˆ

Rd

|u|p |x|p dx C

ˆ

Rd |∇u|p dx

∀ u ∈ C1

c(Rd). 2 For p > d,

ˆ

Rd

|u|p |x|p dx C

ˆ

Rd |∇u|p dx

∀ u ∈ C1

c(Rd \ {0}). 3 For p = d > 1, ....

Standard proof is based on integration by parts.

Hardy’s inequalities

1 For 1 p < d,

ˆ

Rd

|u|p |x|p dx C

ˆ

Rd |∇u|p dx

∀ u ∈ C1

c(Rd). 2 For p > d,

ˆ

Rd

|u|p |x|p dx C

ˆ

Rd |∇u|p dx

∀ u ∈ C1

c(Rd \ {0}). 3 For p = d > 1, ....

Standard proof is based on integration by parts.

Hardy’s inequalities

1 For 1 p < d,

ˆ

Rd

|u|p |x|p dx C

ˆ

Rd |∇u|p dx

∀ u ∈ C1

c(Rd). 2 For p > d,

ˆ

Rd

|u|p |x|p dx C

ˆ

Rd |∇u|p dx

∀ u ∈ C1

c(Rd \ {0}). 3 For p = d > 1, ....

Standard proof is based on integration by parts.

Caffarelli, Kohn, Nirenberg’s inequalities, CM 84

Let p 1, q 1, τ > 0, 0 < a 1, α, β, γ ∈ R. One has

|x|γuLτ(Rd) C|x|α∇ua

Lp(Rd)|x|βu(1−a) Lq(Rd)

∀ u ∈ C1

c(Rd)

under the following conditions 1

τ + γ d = a 1 p + α − 1 d

• + (1 − a)

1 q + β d

• ,

with γ = aσ + (1 − a)β, 0 α − σ and

• α − σ 1

if 1

τ + γ d = 1 p + α − 1 d

• ,

and 1

τ + γ d,

1

p + α d,

1

q + β d > 0.

Comments on the CKN inequality

• This inequality is related to Gagliardo-Nirenberg’s inequality

when α = β = γ = 0, Gagliardo RM 59, Nirenberg ASNSP 59.

• A full story of Gagliardo-Nirenberg’s inequality for fractional

Sobolev spaces is due to Brezis and Mironescu AIHP 18.

• The proof of CKN’s inequality is based on

Integration by parts and symmetrization in the case 0 α − σ 1. Interpolation & the application of the previous case when

α − σ > 1 and 1

τ + γ d = 1 p + α−1 d

• CKN’s inequality generalizes Hardy’s inequality when 1 < p < d.

Comments on the CKN inequality

• This inequality is related to Gagliardo-Nirenberg’s inequality

when α = β = γ = 0, Gagliardo RM 59, Nirenberg ASNSP 59.

• A full story of Gagliardo-Nirenberg’s inequality for fractional

Sobolev spaces is due to Brezis and Mironescu AIHP 18.

• The proof of CKN’s inequality is based on

Integration by parts and symmetrization in the case 0 α − σ 1. Interpolation & the application of the previous case when

α − σ > 1 and 1

τ + γ d = 1 p + α−1 d

• CKN’s inequality generalizes Hardy’s inequality when 1 < p < d.

Comments on the CKN inequality

• This inequality is related to Gagliardo-Nirenberg’s inequality

when α = β = γ = 0, Gagliardo RM 59, Nirenberg ASNSP 59.

• A full story of Gagliardo-Nirenberg’s inequality for fractional

Sobolev spaces is due to Brezis and Mironescu AIHP 18.

• The proof of CKN’s inequality is based on

Integration by parts and symmetrization in the case 0 α − σ 1. Interpolation & the application of the previous case when

α − σ > 1 and 1

τ + γ d = 1 p + α−1 d

• CKN’s inequality generalizes Hardy’s inequality when 1 < p < d.

Comments on the CKN inequality

• This inequality is related to Gagliardo-Nirenberg’s inequality

when α = β = γ = 0, Gagliardo RM 59, Nirenberg ASNSP 59.

• A full story of Gagliardo-Nirenberg’s inequality for fractional

Sobolev spaces is due to Brezis and Mironescu AIHP 18.

• The proof of CKN’s inequality is based on

Integration by parts and symmetrization in the case 0 α − σ 1. Interpolation & the application of the previous case when

α − σ > 1 and 1

τ + γ d = 1 p + α−1 d

• CKN’s inequality generalizes Hardy’s inequality when 1 < p < d.

Comments on the CKN inequality

• This inequality is related to Gagliardo-Nirenberg’s inequality

when α = β = γ = 0, Gagliardo RM 59, Nirenberg ASNSP 59.

• A full story of Gagliardo-Nirenberg’s inequality for fractional

Sobolev spaces is due to Brezis and Mironescu AIHP 18.

• The proof of CKN’s inequality is based on

Integration by parts and symmetrization in the case 0 α − σ 1. Interpolation & the application of the previous case when

α − σ > 1 and 1

τ + γ d = 1 p + α−1 d

• CKN’s inequality generalizes Hardy’s inequality when 1 < p < d.

Comments on the CKN inequality

• This inequality is related to Gagliardo-Nirenberg’s inequality

when α = β = γ = 0, Gagliardo RM 59, Nirenberg ASNSP 59.

• A full story of Gagliardo-Nirenberg’s inequality for fractional

Sobolev spaces is due to Brezis and Mironescu AIHP 18.

• The proof of CKN’s inequality is based on

Integration by parts and symmetrization in the case 0 α − σ 1. Interpolation & the application of the previous case when

α − σ > 1 and 1

τ + γ d = 1 p + α−1 d

• CKN’s inequality generalizes Hardy’s inequality when 1 < p < d.

Section 2: CKN’s inequality for fractional Sobolev spaces

CKN’s inequality for fractional Sobolev spaces

• Goals: extending CKN’s inequality for fractional Sobolev spaces

and searching for variants of Hardy’s inequality when p > d. Recall

|u|p

Ws,p :=

ˆ

Rd

ˆ

Rd

|u(x) − u(y)|p |x − y|d+sp dx dy for u ∈ Lp(Rd).

• Known results: (Hardy’s type-inequalities):

Mazya & Shaposhnikova JFA 02 (harmonic analysis, extension technique), Frank & Seiringer JFA 08 (ground state representation formula): a = 1, τ = p, α = 0 and γ = −s. Abdellaoui & Bentifour JFA 17 (Picone’s inequality): a = 1,

τ = pd/(d − sp), −(d − sp)/p < α = γ < 0, and 1 < p < d/s.

Sharp constant and the remainder are considered by Frank & Seiringer and Abdellaoui & Bentifour.

• Notation:

|u|p

Ws,p,α(Rd) =

ˆ

Rd

ˆ

Rd

|x|

αp 2 |y| αp 2 |u(x) − u(y)|p

|x − y|d+sp dx dy.

CKN’s inequality for fractional Sobolev spaces

• Goals: extending CKN’s inequality for fractional Sobolev spaces

and searching for variants of Hardy’s inequality when p > d. Recall

|u|p

Ws,p :=

ˆ

Rd

ˆ

Rd

|u(x) − u(y)|p |x − y|d+sp dx dy for u ∈ Lp(Rd).

• Known results: (Hardy’s type-inequalities):

Mazya & Shaposhnikova JFA 02 (harmonic analysis, extension technique), Frank & Seiringer JFA 08 (ground state representation formula): a = 1, τ = p, α = 0 and γ = −s. Abdellaoui & Bentifour JFA 17 (Picone’s inequality): a = 1,

τ = pd/(d − sp), −(d − sp)/p < α = γ < 0, and 1 < p < d/s.

Sharp constant and the remainder are considered by Frank & Seiringer and Abdellaoui & Bentifour.

• Notation:

|u|p

Ws,p,α(Rd) =

ˆ

Rd

ˆ

Rd

|x|

αp 2 |y| αp 2 |u(x) − u(y)|p

|x − y|d+sp dx dy.

CKN’s inequality for fractional Sobolev spaces

• Goals: extending CKN’s inequality for fractional Sobolev spaces

and searching for variants of Hardy’s inequality when p > d. Recall

|u|p

Ws,p :=

ˆ

Rd

ˆ

Rd

|u(x) − u(y)|p |x − y|d+sp dx dy for u ∈ Lp(Rd).

• Known results: (Hardy’s type-inequalities):

Mazya & Shaposhnikova JFA 02 (harmonic analysis, extension technique), Frank & Seiringer JFA 08 (ground state representation formula): a = 1, τ = p, α = 0 and γ = −s. Abdellaoui & Bentifour JFA 17 (Picone’s inequality): a = 1,

τ = pd/(d − sp), −(d − sp)/p < α = γ < 0, and 1 < p < d/s.

Sharp constant and the remainder are considered by Frank & Seiringer and Abdellaoui & Bentifour.

• Notation:

|u|p

Ws,p,α(Rd) =

ˆ

Rd

ˆ

Rd

|x|

αp 2 |y| αp 2 |u(x) − u(y)|p

|x − y|d+sp dx dy.

CKN’s inequality for fractional Sobolev spaces

• Goals: extending CKN’s inequality for fractional Sobolev spaces

and searching for variants of Hardy’s inequality when p > d. Recall

|u|p

Ws,p :=

ˆ

Rd

ˆ

Rd

|u(x) − u(y)|p |x − y|d+sp dx dy for u ∈ Lp(Rd).

• Known results: (Hardy’s type-inequalities):

Mazya & Shaposhnikova JFA 02 (harmonic analysis, extension technique), Frank & Seiringer JFA 08 (ground state representation formula): a = 1, τ = p, α = 0 and γ = −s. Abdellaoui & Bentifour JFA 17 (Picone’s inequality): a = 1,

τ = pd/(d − sp), −(d − sp)/p < α = γ < 0, and 1 < p < d/s.

Sharp constant and the remainder are considered by Frank & Seiringer and Abdellaoui & Bentifour.

• Notation:

|u|p

Ws,p,α(Rd) =

ˆ

Rd

ˆ

Rd

|x|

αp 2 |y| αp 2 |u(x) − u(y)|p

|x − y|d+sp dx dy.

CKN’s inequality for fractional Sobolev spaces

• Goals: extending CKN’s inequality for fractional Sobolev spaces

and searching for variants of Hardy’s inequality when p > d. Recall

|u|p

Ws,p :=

ˆ

Rd

ˆ

Rd

|u(x) − u(y)|p |x − y|d+sp dx dy for u ∈ Lp(Rd).

• Known results: (Hardy’s type-inequalities):

Mazya & Shaposhnikova JFA 02 (harmonic analysis, extension technique), Frank & Seiringer JFA 08 (ground state representation formula): a = 1, τ = p, α = 0 and γ = −s. Abdellaoui & Bentifour JFA 17 (Picone’s inequality): a = 1,

τ = pd/(d − sp), −(d − sp)/p < α = γ < 0, and 1 < p < d/s.

Sharp constant and the remainder are considered by Frank & Seiringer and Abdellaoui & Bentifour.

• Notation:

|u|p

Ws,p,α(Rd) =

ˆ

Rd

ˆ

Rd

|x|

αp 2 |y| αp 2 |u(x) − u(y)|p

|x − y|d+sp dx dy.

Statement of the results

Let p > 1, q 1, τ > 0, 0 < s < 1, 0 < a 1, α, β, γ ∈ R be s.t. 1

τ + γ d = a 1 p + α − s d

• + (1 − a)

1 q + β d

• ,

(2.1) and, with γ = aσ + (1 − a)β, 0 α − σ and

• α − σ s if 1

τ + γ d = 1 p + α − s d

• .

(2.2) Theorem (Ng. & Squassina JFA 17) Assume (2.1) and (2.2). If 1/τ + γ/d > 0 then

|x|γuLτ(Rd) C|u|a

Ws,p,α(Rd)|x|βu(1−a) Lq(Rd)

∀ u ∈ C1

c(Rd),

and if 1/τ + γ/d < 0 then

|x|γuLτ(Rd) C|u|a

Ws,p,α(Rd)|x|βu(1−a) Lq(Rd)

∀ u ∈ C1

c(Rd \ {0}).

Statement of the results

Let p > 1, q 1, τ > 0, 0 < s < 1, 0 < a 1, α, β, γ ∈ R be s.t. 1

τ + γ d = a 1 p + α − s d

• + (1 − a)

1 q + β d

• ,

(2.1) and, with γ = aσ + (1 − a)β, 0 α − σ and

• α − σ s if 1

τ + γ d = 1 p + α − s d

• .

(2.2) Theorem (Ng. & Squassina JFA 17) Assume (2.1) and (2.2). If 1/τ + γ/d > 0 then

|x|γuLτ(Rd) C|u|a

Ws,p,α(Rd)|x|βu(1−a) Lq(Rd)

∀ u ∈ C1

c(Rd),

and if 1/τ + γ/d < 0 then

|x|γuLτ(Rd) C|u|a

Ws,p,α(Rd)|x|βu(1−a) Lq(Rd)

∀ u ∈ C1

c(Rd \ {0}).

Statement of the results

Let p > 1, q 1, τ > 0, 0 < s < 1, 0 < a 1, α, β, γ ∈ R be s.t. 1

τ + γ d = a 1 p + α − s d

• + (1 − a)

1 q + β d

• ,

(2.1) and, with γ = aσ + (1 − a)β, 0 α − σ and

• α − σ s if 1

τ + γ d = 1 p + α − s d

• .

(2.2) Theorem (Ng. & Squassina JFA 17) Assume (2.1) and (2.2). If 1/τ + γ/d > 0 then

|x|γuLτ(Rd) C|u|a

Ws,p,α(Rd)|x|βu(1−a) Lq(Rd)

∀ u ∈ C1

c(Rd),

and if 1/τ + γ/d < 0 then

|x|γuLτ(Rd) C|u|a

Ws,p,α(Rd)|x|βu(1−a) Lq(Rd)

∀ u ∈ C1

c(Rd \ {0}).

Statement of the results

Let p > 1, q 1, τ > 0, 0 < s < 1, 0 < a 1, α, β, γ ∈ R be s.t. 1

τ + γ d = a 1 p + α − s d

• + (1 − a)

1 q + β d

• ,

(2.1) and, with γ = aσ + (1 − a)β, 0 α − σ and

• α − σ s if 1

τ + γ d = 1 p + α − s d

• .

(2.2) Theorem (Ng. & Squassina JFA 17) Assume (2.1) and (2.2). If 1/τ + γ/d > 0 then

|x|γuLτ(Rd) C|u|a

Ws,p,α(Rd)|x|βu(1−a) Lq(Rd)

∀ u ∈ C1

c(Rd),

and if 1/τ + γ/d < 0 then

|x|γuLτ(Rd) C|u|a

Ws,p,α(Rd)|x|βu(1−a) Lq(Rd)

∀ u ∈ C1

c(Rd \ {0}).

Statement of the results

Let p > 1, q 1, τ > 0, 0 < s < 1, 0 < a 1, α, β, γ ∈ R be s.t. 1

τ + γ d = a 1 p + α − s d

• + (1 − a)

1 q + β d

• ,

(2.1) and, with γ = aσ + (1 − a)β, 0 α − σ and

• α − σ s if 1

τ + γ d = 1 p + α − s d

• .

(2.2) Theorem (Ng. & Squassina JFA 17) Assume (2.1) and (2.2). If 1/τ + γ/d > 0 then

|x|γuLτ(Rd) C|u|a

Ws,p,α(Rd)|x|βu(1−a) Lq(Rd)

∀ u ∈ C1

c(Rd),

and if 1/τ + γ/d < 0 then

|x|γuLτ(Rd) C|u|a

Ws,p,α(Rd)|x|βu(1−a) Lq(Rd)

∀ u ∈ C1

c(Rd \ {0}).

Statement of the results

Let p > 1, q 1, τ > 0, 0 < s < 1, 0 < a 1, α, β, γ ∈ R be s.t. 1

τ + γ d = a 1 p + α − s d

• + (1 − a)

1 q + β d

• ,

(2.1) and, with γ = aσ + (1 − a)β, 0 α − σ and

• α − σ s if 1

τ + γ d = 1 p + α − s d

• .

(2.2) Theorem (Ng. & Squassina JFA 17) Assume (2.1) and (2.2). If 1/τ + γ/d > 0 then

|x|γuLτ(Rd) C|u|a

Ws,p,α(Rd)|x|βu(1−a) Lq(Rd)

∀ u ∈ C1

c(Rd),

and if 1/τ + γ/d < 0 then

|x|γuLτ(Rd) C|u|a

Ws,p,α(Rd)|x|βu(1−a) Lq(Rd)

∀ u ∈ C1

c(Rd \ {0}).

Our approach

Starting point: Sobolev’s and Poincare’s inequalities; NO integration by parts. Inspiration: harmonic analysis; however, instead of localizing frequency, we localize space variables. The interpolation part is inspired from the proof of CKN’s inequality.

Our approach

Starting point: Sobolev’s and Poincare’s inequalities; NO integration by parts. Inspiration: harmonic analysis; however, instead of localizing frequency, we localize space variables. The interpolation part is inspired from the proof of CKN’s inequality.

Our approach

Starting point: Sobolev’s and Poincare’s inequalities; NO integration by parts. Inspiration: harmonic analysis; however, instead of localizing frequency, we localize space variables. The interpolation part is inspired from the proof of CKN’s inequality.

Our approach

Starting point: Sobolev’s and Poincare’s inequalities; NO integration by parts. Inspiration: harmonic analysis; however, instead of localizing frequency, we localize space variables. The interpolation part is inspired from the proof of CKN’s inequality.

A new proof of Hardy’s inequality

Recall Hardy’s inequality, for 1 p < d, ˆ

Rd

|u|p |x|p dx C

ˆ

Rd |∇u|p dx

for u ∈ C1

c(Rd).

Here is the proof. Set

Ck :=

• x ∈ Rd : 2k |x| < 2k+1.

By Poincare’s inequality, we have

Ck

• u −

Ck

u

• p

dx C2kp

Ck

|∇u|p dx,

which yields ˆ

Ck

|u|p |x|p dx ∼ 2−kp

ˆ

Ck

|u|p C

ˆ

Ck

|∇u|p dx + C2k(d−p)

• Ck

u

• p

. By Poincare’s inequality, we also have

• Ck+1

u −

Ck

u

• p

dx C2kp

Ck∪Ck+1

|∇u|p dx.

A new proof of Hardy’s inequality

Recall Hardy’s inequality, for 1 p < d, ˆ

Rd

|u|p |x|p dx C

ˆ

Rd |∇u|p dx

for u ∈ C1

c(Rd).

Here is the proof. Set

Ck :=

• x ∈ Rd : 2k |x| < 2k+1.

By Poincare’s inequality, we have

Ck

• u −

Ck

u

• p

dx C2kp

Ck

|∇u|p dx,

which yields ˆ

Ck

|u|p |x|p dx ∼ 2−kp

ˆ

Ck

|u|p C

ˆ

Ck

|∇u|p dx + C2k(d−p)

• Ck

u

• p

. By Poincare’s inequality, we also have

• Ck+1

u −

Ck

u

• p

dx C2kp

Ck∪Ck+1

|∇u|p dx.

A new proof of Hardy’s inequality

Recall Hardy’s inequality, for 1 p < d, ˆ

Rd

|u|p |x|p dx C

ˆ

Rd |∇u|p dx

for u ∈ C1

c(Rd).

Here is the proof. Set

Ck :=

• x ∈ Rd : 2k |x| < 2k+1.

By Poincare’s inequality, we have

Ck

• u −

Ck

u

• p

dx C2kp

Ck

|∇u|p dx,

which yields ˆ

Ck

|u|p |x|p dx ∼ 2−kp

ˆ

Ck

|u|p C

ˆ

Ck

|∇u|p dx + C2k(d−p)

• Ck

u

• p

. By Poincare’s inequality, we also have

• Ck+1

u −

Ck

u

• p

dx C2kp

Ck∪Ck+1

|∇u|p dx.

A new proof of Hardy’s inequality

Recall Hardy’s inequality, for 1 p < d, ˆ

Rd

|u|p |x|p dx C

ˆ

Rd |∇u|p dx

for u ∈ C1

c(Rd).

Here is the proof. Set

Ck :=

• x ∈ Rd : 2k |x| < 2k+1.

By Poincare’s inequality, we have

Ck

• u −

Ck

u

• p

dx C2kp

Ck

|∇u|p dx,

which yields ˆ

Ck

|u|p |x|p dx ∼ 2−kp

ˆ

Ck

|u|p C

ˆ

Ck

|∇u|p dx + C2k(d−p)

• Ck

u

• p

. By Poincare’s inequality, we also have

• Ck+1

u −

Ck

u

• p

dx C2kp

Ck∪Ck+1

|∇u|p dx.

A new proof of Hardy’s inequality

Recall Hardy’s inequality, for 1 p < d, ˆ

Rd

|u|p |x|p dx C

ˆ

Rd |∇u|p dx

for u ∈ C1

c(Rd).

Here is the proof. Set

Ck :=

• x ∈ Rd : 2k |x| < 2k+1.

By Poincare’s inequality, we have

Ck

• u −

Ck

u

• p

dx C2kp

Ck

|∇u|p dx,

which yields ˆ

Ck

|u|p |x|p dx ∼ 2−kp

ˆ

Ck

|u|p C

ˆ

Ck

|∇u|p dx + C2k(d−p)

• Ck

u

• p

. By Poincare’s inequality, we also have

• Ck+1

u −

Ck

u

• p

dx C2kp

Ck∪Ck+1

|∇u|p dx.

A new proof of Hardy’s inequality

Recall Hardy’s inequality, for 1 p < d, ˆ

Rd

|u|p |x|p dx C

ˆ

Rd |∇u|p dx

for u ∈ C1

c(Rd).

Here is the proof. Set

Ck :=

• x ∈ Rd : 2k |x| < 2k+1.

By Poincare’s inequality, we have

Ck

• u −

Ck

u

• p

dx C2kp

Ck

|∇u|p dx,

which yields ˆ

Ck

|u|p |x|p dx ∼ 2−kp

ˆ

Ck

|u|p C

ˆ

Ck

|∇u|p dx + C2k(d−p)

• Ck

u

• p

. By Poincare’s inequality, we also have

• Ck+1

u −

Ck

u

• p

dx C2kp

Ck∪Ck+1

|∇u|p dx.

It follows that 2k(d−p)

• Ck

u

• p

Cc

ˆ

Ck∪Ck+1

|∇u|p dx+c2(k+1)(d−p)

• Ck+1

u

• p

. for c > 2p−d, in particular for c = (2p−d + 1)/2 < 1. This implies

• k

2k(d−p)

• Ck

u

• p

C

• k

ˆ

Ck∪Ck+1

|∇u|p dx C

ˆ

Rd |∇u|p.

We derive that

• k

2−kp ˆ

Ck

|u|p C

• k

ˆ

Ck

|∇u|p dx + C

• k

2k(d−p)

• Ck

u

• p

C

ˆ

Rd |∇u|p.

The proof is complete.

It follows that 2k(d−p)

• Ck

u

• p

Cc

ˆ

Ck∪Ck+1

|∇u|p dx+c2(k+1)(d−p)

• Ck+1

u

• p

. for c > 2p−d, in particular for c = (2p−d + 1)/2 < 1. This implies

• k

2k(d−p)

• Ck

u

• p

C

• k

ˆ

Ck∪Ck+1

|∇u|p dx C

ˆ

Rd |∇u|p.

We derive that

• k

2−kp ˆ

Ck

|u|p C

• k

ˆ

Ck

|∇u|p dx + C

• k

2k(d−p)

• Ck

u

• p

C

ˆ

Rd |∇u|p.

The proof is complete.

It follows that 2k(d−p)

• Ck

u

• p

Cc

ˆ

Ck∪Ck+1

|∇u|p dx+c2(k+1)(d−p)

• Ck+1

u

• p

. for c > 2p−d, in particular for c = (2p−d + 1)/2 < 1. This implies

• k

2k(d−p)

• Ck

u

• p

C

• k

ˆ

Ck∪Ck+1

|∇u|p dx C

ˆ

Rd |∇u|p.

We derive that

• k

2−kp ˆ

Ck

|u|p C

• k

ˆ

Ck

|∇u|p dx + C

• k

2k(d−p)

• Ck

u

• p

C

ˆ

Rd |∇u|p.

The proof is complete.

It follows that 2k(d−p)

• Ck

u

• p

Cc

ˆ

Ck∪Ck+1

|∇u|p dx+c2(k+1)(d−p)

• Ck+1

u

• p

. for c > 2p−d, in particular for c = (2p−d + 1)/2 < 1. This implies

• k

2k(d−p)

• Ck

u

• p

C

• k

ˆ

Ck∪Ck+1

|∇u|p dx C

ˆ

Rd |∇u|p.

We derive that

• k

2−kp ˆ

Ck

|u|p C

• k

ˆ

Ck

|∇u|p dx + C

• k

2k(d−p)

• Ck

u

• p

C

ˆ

Rd |∇u|p.

The proof is complete.

Proof of CKN’s inequality for the main case

Let p > 1, q 1, τ > 0, 0 < s < 1, 0 < a 1, α, β, γ ∈ R be s.t. 1

τ + γ d = a 1 p + α − s d

• + (1 − a)

1 q + β d

• ,

(2.3) and, with γ = aσ + (1 − a)β, 0 α − σ s (2.4) Theorem Assume (2.3) and (2.4). We have, for 1/τ + γ/d > 0,

|x|γuLτ(Rd) C|u|a

Ws,p,α(Rd)|x|βu(1−a) Lq(Rd)

∀ u ∈ C1

c(Rd),

|u|p

Ws,p,α(Rd) =

ˆ

Rd

ˆ

Rd

|x|

αp 2 |y| αp 2 |u(x) − u(y)|p

|x − y|d+sp dx dy.

Lemma (Gagliardo-Nirenberg’s inequality) Let d 1, 0 < s < 1, p > 1, q 1, τ > 0, and 0 < a 1 be s.t. 1

τ = a 1 p − s d

• + (1 − a)1

q.

We have

uLτ(Rd) C|u|a

Ws,p(Rd)u1−a Lq(Rd)

for u ∈ C1

c(Rd).

Lemma (Gagliardo-Nirenberg’s inequality) Let d 1, 0 < s < 1, p > 1, q 1, τ > 0, and 0 < a 1 be s.t. 1

τ = a 1 p − s d

• + (1 − a)1

q.

We have

uLτ(Rd) C|u|a

Ws,p(Rd)u1−a Lq(Rd)

for u ∈ C1

c(Rd).

Lemma Let d 1, p > 1, 0 < s < 1, q 1, τ > 0, and 0 < a 1 be s.t. 1

τ a 1 p − s d

• + (1 − a)1

q.

Let λ > 0, 0 < r < R, and set D :=

• x ∈ Rd : λr < |x| < λR

. Then, ∀u ∈ C1(¯ D),

• D
• u −

D

u

• τ

dx 1/τ C

• λsp−d|u|p

Ws,p(D)

a/p

D

|u|q dx (1−a)/q

.

Lemma Let d 1, p > 1, 0 < s < 1, q 1, τ > 0, and 0 < a 1 be s.t. 1

τ a 1 p − s d

• + (1 − a)1

q.

Let λ > 0, 0 < r < R, and set D :=

• x ∈ Rd : λr < |x| < λR

. Then, ∀u ∈ C1(¯ D),

• D
• u −

D

u

• τ

dx 1/τ C

• λsp−d|u|p

Ws,p(D)

a/p

D

|u|q dx (1−a)/q

.

Lemma Let d 1, p > 1, 0 < s < 1, q 1, τ > 0, and 0 < a 1 be s.t. 1

τ a 1 p − s d

• + (1 − a)1

q.

Let λ > 0, 0 < r < R, and set D :=

• x ∈ Rd : λr < |x| < λR

. Then, ∀u ∈ C1(¯ D),

• D
• u −

D

u

• τ

dx 1/τ C

• λsp−d|u|p

Ws,p(D)

a/p

D

|u|q dx (1−a)/q

.

• Proof. By scaling, one can assume that λ = 1. Let 0 < s′ s and

τ′ τ be such that

1

τ′ = a 1 p − s′ d

• + (1 − a)1

q.

From the previous lemma, we derive that

• u −

D

u

• Lτ′(D)

C |u|a

Ws′,p(D) u1−a Lq(D).

Since |u|Ws′,p(D) C |u|Ws,p(D), uLτ(D) CuLτ′(D), the conclusion follows.

• Proof. By scaling, one can assume that λ = 1. Let 0 < s′ s and

τ′ τ be such that

1

τ′ = a 1 p − s′ d

• + (1 − a)1

q.

From the previous lemma, we derive that

• u −

D

u

• Lτ′(D)

C |u|a

Ws′,p(D) u1−a Lq(D).

Since |u|Ws′,p(D) C |u|Ws,p(D), uLτ(D) CuLτ′(D), the conclusion follows.

• Proof. By scaling, one can assume that λ = 1. Let 0 < s′ s and

τ′ τ be such that

1

τ′ = a 1 p − s′ d

• + (1 − a)1

q.

From the previous lemma, we derive that

• u −

D

u

• Lτ′(D)

C |u|a

Ws′,p(D) u1−a Lq(D).

Since |u|Ws′,p(D) C |u|Ws,p(D), uLτ(D) CuLτ′(D), the conclusion follows.

• Proof. By scaling, one can assume that λ = 1. Let 0 < s′ s and

τ′ τ be such that

1

τ′ = a 1 p − s′ d

• + (1 − a)1

q.

From the previous lemma, we derive that

• u −

D

u

• Lτ′(D)

C |u|a

Ws′,p(D) u1−a Lq(D).

Since |u|Ws′,p(D) C |u|Ws,p(D), uLτ(D) CuLτ′(D), the conclusion follows.

Proof of CKN’s inequality

Recall Ck :=

• x ∈ Rd : 2k |x| < 2k+1. We have,
• α − σ 0 +

balance law

• ,

1

τ a 1 p − s d

• + (1 − a)1

q.

It follows that

• Ck
• u −

Ck

u

• τ

dx 1/τ C

• 2−(d−sp)k|u|p

Ws,p(Ck)

a/p

Ck

|u|q dx (1−a)/q

. Using the balance law, we derive that ˆ

Ck

|x|γτ|u|τ dx C|u|aτ

Ws,p,α(Ck)|x|βu(1−a)τ Lq(Ck) +C2(γτ+d)k

• Ck

u

• τ

, (2.5)

Proof of CKN’s inequality

Recall Ck :=

• x ∈ Rd : 2k |x| < 2k+1. We have,
• α − σ 0 +

balance law

• ,

1

τ a 1 p − s d

• + (1 − a)1

q.

It follows that

• Ck
• u −

Ck

u

• τ

dx 1/τ C

• 2−(d−sp)k|u|p

Ws,p(Ck)

a/p

Ck

|u|q dx (1−a)/q

. Using the balance law, we derive that ˆ

Ck

|x|γτ|u|τ dx C|u|aτ

Ws,p,α(Ck)|x|βu(1−a)τ Lq(Ck) +C2(γτ+d)k

• Ck

u

• τ

, (2.5)

Proof of CKN’s inequality

Recall Ck :=

• x ∈ Rd : 2k |x| < 2k+1. We have,
• α − σ 0 +

balance law

• ,

1

τ a 1 p − s d

• + (1 − a)1

q.

It follows that

• Ck
• u −

Ck

u

• τ

dx 1/τ C

• 2−(d−sp)k|u|p

Ws,p(Ck)

a/p

Ck

|u|q dx (1−a)/q

. Using the balance law, we derive that ˆ

Ck

|x|γτ|u|τ dx C|u|aτ

Ws,p,α(Ck)|x|βu(1−a)τ Lq(Ck) +C2(γτ+d)k

• Ck

u

• τ

, (2.5)

Proof of CKN’s inequality

Recall Ck :=

• x ∈ Rd : 2k |x| < 2k+1. We have,
• α − σ 0 +

balance law

• ,

1

τ a 1 p − s d

• + (1 − a)1

q.

It follows that

• Ck
• u −

Ck

u

• τ

dx 1/τ C

• 2−(d−sp)k|u|p

Ws,p(Ck)

a/p

Ck

|u|q dx (1−a)/q

. Using the balance law, we derive that ˆ

Ck

|x|γτ|u|τ dx C|u|aτ

Ws,p,α(Ck)|x|βu(1−a)τ Lq(Ck) +C2(γτ+d)k

• Ck

u

• τ

, (2.5)

Since

• Ck

u −

Ck+1

u

• τ

C

• 2−(d−sp)k|u|p

Ws,p(Ck∪Ck+1)

aτ/p

Ck∪Ck+1

|u|q dx (1−a)τ/q

, we obtain, with c = 2/(1 + 2γτ+d) < 1, 2(γτ+d)k

• Ck

u

• τ

C|u|aτ

Ws,p,α(Ck∪Ck+1)|x|βu(1−a)τ Lq(Ck∪Ck+1)

+ c2(γτ+d)(k+1)

• Ck+1

u

• τ

. This yields

• k

2(γτ+d)k

• Ck

u

• τ

C

• k

|u|aτ

Ws,p,α(Ck∪Ck+1)|x|βu(1−a)τ Lq(Ck∪Ck+1).

(2.6)

Since

• Ck

u −

Ck+1

u

• τ

C

• 2−(d−sp)k|u|p

Ws,p(Ck∪Ck+1)

aτ/p

Ck∪Ck+1

|u|q dx (1−a)τ/q

, we obtain, with c = 2/(1 + 2γτ+d) < 1, 2(γτ+d)k

• Ck

u

• τ

C|u|aτ

Ws,p,α(Ck∪Ck+1)|x|βu(1−a)τ Lq(Ck∪Ck+1)

+ c2(γτ+d)(k+1)

• Ck+1

u

• τ

. This yields

• k

2(γτ+d)k

• Ck

u

• τ

C

• k

|u|aτ

Ws,p,α(Ck∪Ck+1)|x|βu(1−a)τ Lq(Ck∪Ck+1).

(2.6)

Combining (2.5) and (2.6) yields ˆ

Rd |x|γτ|u|τ dx C

• k

|u|aτ

Ws,p,α(Ck∪Ck+1)|x|βu(1−a)τ Lq(Ck∪Ck+1).

One has, for s 0, t 0 with s + t 1, and for xk 0 and

yk 0,

n

• k=m

xs

kyt k

• n
• k=m

xk s

n

• k=m

yk t

. Applying this inequality with s = aτ/p and t = (1 − a)τ/q, we

• btain that

ˆ

Rd |x|γτ|u|τ dx C|u|aτ Ws,p,α(∞

k=m Ck)|x|βu(1−a)τ

Lq(∞

k=m Ck),

since a/p + (1 − a)/q 1/τ thanks to the fact α − σ s.

Combining (2.5) and (2.6) yields ˆ

Rd |x|γτ|u|τ dx C

• k

|u|aτ

Ws,p,α(Ck∪Ck+1)|x|βu(1−a)τ Lq(Ck∪Ck+1).

One has, for s 0, t 0 with s + t 1, and for xk 0 and

yk 0,

n

• k=m

xs

kyt k

• n
• k=m

xk s

n

• k=m

yk t

. Applying this inequality with s = aτ/p and t = (1 − a)τ/q, we

• btain that

ˆ

Rd |x|γτ|u|τ dx C|u|aτ Ws,p,α(∞

k=m Ck)|x|βu(1−a)τ

Lq(∞

k=m Ck),

since a/p + (1 − a)/q 1/τ thanks to the fact α − σ s.

Combining (2.5) and (2.6) yields ˆ

Rd |x|γτ|u|τ dx C

• k

|u|aτ

Ws,p,α(Ck∪Ck+1)|x|βu(1−a)τ Lq(Ck∪Ck+1).

One has, for s 0, t 0 with s + t 1, and for xk 0 and

yk 0,

n

• k=m

xs

kyt k

• n
• k=m

xk s

n

• k=m

yk t

. Applying this inequality with s = aτ/p and t = (1 − a)τ/q, we

• btain that

ˆ

Rd |x|γτ|u|τ dx C|u|aτ Ws,p,α(∞

k=m Ck)|x|βu(1−a)τ

Lq(∞

k=m Ck),

since a/p + (1 − a)/q 1/τ thanks to the fact α − σ s.

Combining (2.5) and (2.6) yields ˆ

Rd |x|γτ|u|τ dx C

• k

|u|aτ

Ws,p,α(Ck∪Ck+1)|x|βu(1−a)τ Lq(Ck∪Ck+1).

One has, for s 0, t 0 with s + t 1, and for xk 0 and

yk 0,

n

• k=m

xs

kyt k

• n
• k=m

xk s

n

• k=m

yk t

. Applying this inequality with s = aτ/p and t = (1 − a)τ/q, we

• btain that

ˆ

Rd |x|γτ|u|τ dx C|u|aτ Ws,p,α(∞

k=m Ck)|x|βu(1−a)τ

Lq(∞

k=m Ck),

since a/p + (1 − a)/q 1/τ thanks to the fact α − σ s.

On the limiting case

Theorem (Ng. & Squassina JFA 17) Let d 1, p > 1, 0 < s < 1, q 1, τ > 1, 0 < a 1, α, β, γ ∈ R be such that (2.3) holds and 0 a − σ s. Let u ∈ C1

c(Rd), and 0 < r < R. We have

i) if 1/τ + γ/d = 0 and supp u ⊂ BR, then

ˆ

Rd

|x|γτ

lnτ(2R/|x|)|u|τ dx

1/τ C|u|a

Ws,p,α(Rd)|x|βu(1−a) Lq(Rd),

ii) if 1/τ + γ/d = 0 and supp u ∩ Br = ∅, then

ˆ

Rd

|x|γτ

lnτ(2|x|/r)|u|τ dx

1/τ C|u|a

Ws,p,α(Rd)|x|βu(1−a) Lq(Rd).

On the limiting case

Theorem (Ng. & Squassina JFA 17) Let d 1, p > 1, 0 < s < 1, q 1, τ > 1, 0 < a 1, α, β, γ ∈ R be such that (2.3) holds and 0 a − σ s. Let u ∈ C1

c(Rd), and 0 < r < R. We have

i) if 1/τ + γ/d = 0 and supp u ⊂ BR, then

ˆ

Rd

|x|γτ

lnτ(2R/|x|)|u|τ dx

1/τ C|u|a

Ws,p,α(Rd)|x|βu(1−a) Lq(Rd),

ii) if 1/τ + γ/d = 0 and supp u ∩ Br = ∅, then

ˆ

Rd

|x|γτ

lnτ(2|x|/r)|u|τ dx

1/τ C|u|a

Ws,p,α(Rd)|x|βu(1−a) Lq(Rd).

On the limiting case

Theorem (Ng. & Squassina JFA 17) Let d 1, p > 1, 0 < s < 1, q 1, τ > 1, 0 < a 1, α, β, γ ∈ R be such that (2.3) holds and 0 a − σ s. Let u ∈ C1

c(Rd), and 0 < r < R. We have

i) if 1/τ + γ/d = 0 and supp u ⊂ BR, then

ˆ

Rd

|x|γτ

lnτ(2R/|x|)|u|τ dx

1/τ C|u|a

Ws,p,α(Rd)|x|βu(1−a) Lq(Rd),

ii) if 1/τ + γ/d = 0 and supp u ∩ Br = ∅, then

ˆ

Rd

|x|γτ

lnτ(2|x|/r)|u|τ dx

1/τ C|u|a

Ws,p,α(Rd)|x|βu(1−a) Lq(Rd).

On the limiting case

Theorem (Ng. & Squassina JFA 17) Let d 1, p > 1, 0 < s < 1, q 1, τ > 1, 0 < a 1, α, β, γ ∈ R be such that (2.3) holds and 0 a − σ s. Let u ∈ C1

c(Rd), and 0 < r < R. We have

i) if 1/τ + γ/d = 0 and supp u ⊂ BR, then

ˆ

Rd

|x|γτ

lnτ(2R/|x|)|u|τ dx

1/τ C|u|a

Ws,p,α(Rd)|x|βu(1−a) Lq(Rd),

ii) if 1/τ + γ/d = 0 and supp u ∩ Br = ∅, then

ˆ

Rd

|x|γτ

lnτ(2|x|/r)|u|τ dx

1/τ C|u|a

Ws,p,α(Rd)|x|βu(1−a) Lq(Rd).

Section 3: New perspectives of Hardy’s and Caffarelli, Kohn, Nirenberg’s inequalities

Motivations

Define, for d 1 and p 1,

Iδ(u) :=

ˆ

Rd

ˆ

Rd |u(x)−u(y)|>δ

δp |x − y|d+p dx dy ∀ u ∈ Lp(Rd).

1 Iδ is related to the semi-norm of Ws,q(Rd):

|u|q

Ws,q(Rd) :=

ˆ

Rd

ˆ

Rd

|u(x) − u(y)|q |x − y|d+sq dx dy.

2 Iδ appears in an estimate for the topological degree due to

Bourgain, Brezis, & Ng., CRAS 05, and Ng. JAM 07

| deg u| Cd

ˆ

Sd

ˆ

Sd |u(x)−u(y)|ℓd

1

|x − y|2d dx dy, ∀ u ∈ C(Sd, Sd),

where ℓd =

• 2 +

2 d+1.

Motivations

Define, for d 1 and p 1,

Iδ(u) :=

ˆ

Rd

ˆ

Rd |u(x)−u(y)|>δ

δp |x − y|d+p dx dy ∀ u ∈ Lp(Rd).

1 Iδ is related to the semi-norm of Ws,q(Rd):

|u|q

Ws,q(Rd) :=

ˆ

Rd

ˆ

Rd

|u(x) − u(y)|q |x − y|d+sq dx dy.

2 Iδ appears in an estimate for the topological degree due to

Bourgain, Brezis, & Ng., CRAS 05, and Ng. JAM 07

| deg u| Cd

ˆ

Sd

ˆ

Sd |u(x)−u(y)|ℓd

1

|x − y|2d dx dy, ∀ u ∈ C(Sd, Sd),

where ℓd =

• 2 +

2 d+1.

Motivations

Define, for d 1 and p 1,

Iδ(u) :=

ˆ

Rd

ˆ

Rd |u(x)−u(y)|>δ

δp |x − y|d+p dx dy ∀ u ∈ Lp(Rd).

1 Iδ is related to the semi-norm of Ws,q(Rd):

|u|q

Ws,q(Rd) :=

ˆ

Rd

ˆ

Rd

|u(x) − u(y)|q |x − y|d+sq dx dy.

2 Iδ appears in an estimate for the topological degree due to

Bourgain, Brezis, & Ng., CRAS 05, and Ng. JAM 07

| deg u| Cd

ˆ

Sd

ˆ

Sd |u(x)−u(y)|ℓd

1

|x − y|2d dx dy, ∀ u ∈ C(Sd, Sd),

where ℓd =

• 2 +

2 d+1.

Motivations

Define, for d 1 and p 1,

Iδ(u) :=

ˆ

Rd

ˆ

Rd |u(x)−u(y)|>δ

δp |x − y|d+p dx dy ∀ u ∈ Lp(Rd).

1 Iδ is related to the semi-norm of Ws,q(Rd):

|u|q

Ws,q(Rd) :=

ˆ

Rd

ˆ

Rd

|u(x) − u(y)|q |x − y|d+sq dx dy.

2 Iδ appears in an estimate for the topological degree due to

Bourgain, Brezis, & Ng., CRAS 05, and Ng. JAM 07

| deg u| Cd

ˆ

Sd

ˆ

Sd |u(x)−u(y)|ℓd

1

|x − y|2d dx dy, ∀ u ∈ C(Sd, Sd),

where ℓd =

• 2 +

2 d+1.

Theorem (Ng. JFA 06, Bourgain & Ng. CRAS 06) Let d 1, 1 < p < +∞ and u ∈ Lp(Rd). Then

1

Iδ(u) Cd,p

ˆ

Rd |∇u|p

∀ u ∈ W1,p(Rd).

2

lim

δ→0 Iδ(u) = Kd,p

ˆ

Rd |∇u|p

∀ u ∈ W1,p(Rd).

3 If

lim inf

δ→0 Iδ(u) < +∞,

then u ∈ W1,p(Rd). Related works: Bourgain, Brezis, & Mironescu 01, Davila 02.

Theorem (Ng. JFA 06, Bourgain & Ng. CRAS 06) Let d 1, 1 < p < +∞ and u ∈ Lp(Rd). Then

1

Iδ(u) Cd,p

ˆ

Rd |∇u|p

∀ u ∈ W1,p(Rd).

2

lim

δ→0 Iδ(u) = Kd,p

ˆ

Rd |∇u|p

∀ u ∈ W1,p(Rd).

3 If

lim inf

δ→0 Iδ(u) < +∞,

then u ∈ W1,p(Rd). Related works: Bourgain, Brezis, & Mironescu 01, Davila 02.

Theorem (Ng. JFA 06, Bourgain & Ng. CRAS 06) Let d 1, 1 < p < +∞ and u ∈ Lp(Rd). Then

1

Iδ(u) Cd,p

ˆ

Rd |∇u|p

∀ u ∈ W1,p(Rd).

2

lim

δ→0 Iδ(u) = Kd,p

ˆ

Rd |∇u|p

∀ u ∈ W1,p(Rd).

3 If

lim inf

δ→0 Iδ(u) < +∞,

then u ∈ W1,p(Rd). Related works: Bourgain, Brezis, & Mironescu 01, Davila 02.

Theorem (Ng. JFA 06, Bourgain & Ng. CRAS 06) Let d 1, 1 < p < +∞ and u ∈ Lp(Rd). Then

1

Iδ(u) Cd,p

ˆ

Rd |∇u|p

∀ u ∈ W1,p(Rd).

2

lim

δ→0 Iδ(u) = Kd,p

ˆ

Rd |∇u|p

∀ u ∈ W1,p(Rd).

3 If

lim inf

δ→0 Iδ(u) < +∞,

then u ∈ W1,p(Rd). Related works: Bourgain, Brezis, & Mironescu 01, Davila 02.

Theorem (Ng. CVPDE, 11) Let p 1, Q be a cube or a ball of Rd. Then ∃C > 0 s.t. for all

δ > 0:

¨

Q2 |u(x) − u(y)|p dx dy

C

• |Q|

d+p d

¨

Q2 |u(x)−u(y)|>δ

δp |x − y|d+p dx dy + δp|Q|2

. A variant of Sobolev’s inequality also holds for Iδ for 1 < p < d.

Question: How’s about Hardy’s and Caffarelli, Kohn, Nirenberg’s inequalities?

Variants of Hardy’s inequalities

Theorem (Ng. & Squassina JAM, to appear) Let d 1, p 1, 0 < r < R, and u ∈ Lp(Rd). We have i) if 1 p < d and supp u ⊂ BR, then ˆ

Rd

|u(x)|p |x|p dx C

• Iδ(u) + Rd−pδp,

ii) if p > d and supp u ⊂ Rd \ Br, then ˆ

Rd

|u(x)|p |x|p dx C

• Iδ(u) + rd−pδp,

Similar results hold for the case p = d.

Variants of Hardy’s inequalities

Theorem (Ng. & Squassina JAM, to appear) Let d 1, p 1, 0 < r < R, and u ∈ Lp(Rd). We have i) if 1 p < d and supp u ⊂ BR, then ˆ

Rd

|u(x)|p |x|p dx C

• Iδ(u) + Rd−pδp,

ii) if p > d and supp u ⊂ Rd \ Br, then ˆ

Rd

|u(x)|p |x|p dx C

• Iδ(u) + rd−pδp,

Similar results hold for the case p = d.

Variants of Hardy’s inequalities

Theorem (Ng. & Squassina JAM, to appear) Let d 1, p 1, 0 < r < R, and u ∈ Lp(Rd). We have i) if 1 p < d and supp u ⊂ BR, then ˆ

Rd

|u(x)|p |x|p dx C

• Iδ(u) + Rd−pδp,

ii) if p > d and supp u ⊂ Rd \ Br, then ˆ

Rd

|u(x)|p |x|p dx C

• Iδ(u) + rd−pδp,

Similar results hold for the case p = d.

Theorem (Ng. & Squassina JAM, to appear) Let d 2, 1 < p < d, τ > 0, 0 < r < R, and u ∈ Lp

loc(Rd). Assume

that 1

τ + γ d = 1 p + α − 1 d

and 0 α − γ 1. We have i) if d − p + pα > 0 and supp u ⊂ BR, then

ˆ

Rd |x|γτ|u(x)|τ dx

p/τ C

• Iδ(u, α) + Rd−p+pαδp,

ii) if d − p + pα < 0 and supp u ⊂ Rd \ Br, then

ˆ

Rd |x|γτ|u(x)|τ dx

p/τ C

• Iδ(u, α) + rd−p+pαδp.
• Variants for p = d 2 and also for p = d = 1 hold.
• Variants for 0 < a < 1 hold.

Theorem (Ng. & Squassina JAM, to appear) Let d 2, 1 < p < d, τ > 0, 0 < r < R, and u ∈ Lp

loc(Rd). Assume

that 1

τ + γ d = 1 p + α − 1 d

and 0 α − γ 1. We have i) if d − p + pα > 0 and supp u ⊂ BR, then

ˆ

Rd |x|γτ|u(x)|τ dx

p/τ C

• Iδ(u, α) + Rd−p+pαδp,

ii) if d − p + pα < 0 and supp u ⊂ Rd \ Br, then

ˆ

Rd |x|γτ|u(x)|τ dx

p/τ C

• Iδ(u, α) + rd−p+pαδp.
• Variants for p = d 2 and also for p = d = 1 hold.
• Variants for 0 < a < 1 hold.

Theorem (Ng. & Squassina JAM, to appear) Let d 2, 1 < p < d, τ > 0, 0 < r < R, and u ∈ Lp

loc(Rd). Assume

that 1

τ + γ d = 1 p + α − 1 d

and 0 α − γ 1. We have i) if d − p + pα > 0 and supp u ⊂ BR, then

ˆ

Rd |x|γτ|u(x)|τ dx

p/τ C

• Iδ(u, α) + Rd−p+pαδp,

ii) if d − p + pα < 0 and supp u ⊂ Rd \ Br, then

ˆ

Rd |x|γτ|u(x)|τ dx

p/τ C

• Iδ(u, α) + rd−p+pαδp.
• Variants for p = d 2 and also for p = d = 1 hold.
• Variants for 0 < a < 1 hold.

Theorem (Ng. & Squassina JAM, to appear) Let d 2, 1 < p < d, τ > 0, 0 < r < R, and u ∈ Lp

loc(Rd). Assume

that 1

τ + γ d = 1 p + α − 1 d

and 0 α − γ 1. We have i) if d − p + pα > 0 and supp u ⊂ BR, then

ˆ

Rd |x|γτ|u(x)|τ dx

p/τ C

• Iδ(u, α) + Rd−p+pαδp,

ii) if d − p + pα < 0 and supp u ⊂ Rd \ Br, then

ˆ

Rd |x|γτ|u(x)|τ dx

p/τ C

• Iδ(u, α) + rd−p+pαδp.
• Variants for p = d 2 and also for p = d = 1 hold.
• Variants for 0 < a < 1 hold.

Theorem (Ng. & Squassina JAM, to appear) Let d 2, 1 < p < d, τ > 0, 0 < r < R, and u ∈ Lp

loc(Rd). Assume

that 1

τ + γ d = 1 p + α − 1 d

and 0 α − γ 1. We have i) if d − p + pα > 0 and supp u ⊂ BR, then

ˆ

Rd |x|γτ|u(x)|τ dx

p/τ C

• Iδ(u, α) + Rd−p+pαδp,

ii) if d − p + pα < 0 and supp u ⊂ Rd \ Br, then

ˆ

Rd |x|γτ|u(x)|τ dx

p/τ C

• Iδ(u, α) + rd−p+pαδp.
• Variants for p = d 2 and also for p = d = 1 hold.
• Variants for 0 < a < 1 hold.

Thank you for your attention!