a hardy type inequality on fractional order sobolev spaces ... · plan of the talk 1 classical...

$: A Hardy type inequality on fractional order Sobolev spaces ... · Plan of the talk 1 Classical Hardy inequality 2 Fractional Hardy inequality on the Euclidean space 3 Overview of$
A Hardy type inequality on fractional order Sobolevspaces on the Heisenberg group

AdimurthiTIFR-CAM, Bangalore

Batsheva de Rotschild seminar on Hardy-type Inequalities and Elliptic Problems.

January 8, 2018

Joint work with Arka Mallick.To appear on Ann. Sc. Norm. Super. Pisa Cl. Sci. (5)

Adimurthi TIFR-CAM, Bangalore ( Batsheva de Rotschild seminar on Hardy-type Inequalities and Elliptic Problems.)Fractional order Sobolev Spaces January 8, 2018 1 / 52

Plan of the talk

1 Classical Hardy inequality

2 Fractional Hardy inequality on the Euclidean space

3 Overview of the Heisenberg Group

4 Sub-elliptic Sobolev and Hardy Type Inequalities on The Heisenberg Group

5 Fractional Sobolev Spaces on The Heisenberg Group

6 Fractional Sobolev and Hardy Type Inequalities on The Heisenberg Group

7 Sketch Proofs of the Sobolev and Hardy Inequality

8 Morrey Type Embedding

9 Comactness of Sobolev Type Embedding


Classical Hardy’s Inequality

The classical Hardy inequality asserts that for any domain Ω ⊂ Rn (n ≥ 3)with 0 ∈ Ω ∫

Ω

|∇u|2dx ≥ λ∗∫

Ω

u2

|x |2dx , ∀u ∈ H1

0 (Ω), (1.1)

where λ∗ := (n−2)2

4 is the optimal constant and is never achieved in H10 (Ω).

Usually the constant λ∗ plays a crucial role in analysing the behaviour ofheat equation with inverse square potential.

The inequality (1.1) is strict for any u ∈ H10 (Ω). So, there is a scope of

improving (1.1). This opportunity first exploited by Brezis and Vazquez(Rev. Mat. Univ. Complut. Madrid, 1997) followed by Adimurthi,Chaudhuri, Ramaswamy(Proc.AMS, 2002) , Barbatis, Filippas,Tertikas(Indiana Univ. Math. J.,2003) and Pinchover, Tintarev (IndianaUniv. Math. J., 2005) to derive various improvements of (1.1) by imposingdifferent conditions on Ω, whereas for Ω = Rn it has been shown byDevyver, Fraas, Pinchover (JFA,2014) and Fillipas, Tertikas (JFA 2002) thatadditional correction term cannot be added.


The Fractional order Sobolev Spaces on the Euclideanspace

To recall the fractional Hardy inequality let us define the following fractional orderSobolev space. For 1 ≤ p <∞ we define

W s,p(Ω) := u ∈ Lp(Rn) :

∫Ω

∫Ω

|u(x)− u(y)|p

|x − y |n+spdxdy <∞,

endowed with the norm

||u||W s,p(Ω) := ||u||Lp(Ω) + [u]W s,p(Ω),

where [u]W s,p(Ω) :=(∫

Ω

∫Ω|u(x)−u(y)|p|x−y |n+sp dxdy

) 1p

is the fractional seminorm and Ω is

any open set in Rn.


Fractional Hardy inequality with precise constant

Maz’ya and Shaposhnikova (JFA,2002) proved the following Hardy type inequality∫Rn

|u|p

|x |spdx ≤ C (n, p)

s(1− s)

(n − sp)p[u]pW s,p(Rn) , for all u ∈W s,p(Rn), (2.1)

where sp < n and C (n, p) > 0 is a constant depending only on n, p. Also1 ≤ p <∞ and 0 < s < 1. They used it to prove the following versions of theSobolev inequality with precise constant

(i)

||u||pLq(Rn) ≤ C (n, p)s(1− s)

(n − sp)p−1[u]pW s,p(Rn) , for all u ∈W s,p(Rn), (2.2)

where sp < n, q = npn−sp and C (n, p) > 0 is a function of n, p only and


Fractional Hardy inequality with precise constant

(ii) for any cube Q ⊂ Rn∣∣∣∣∣∣∣∣u − ∫Q

u

∣∣∣∣∣∣∣∣pLq(Q)

≤ C (n, p)(1− s)

(n − sp)p−1[u]pW s,p(Q) , for all u ∈W s,p(Q),

(2.3)

where s ∈ (0, 1), 1 ≤ p <∞, sp < n, q = npn−sp and C (n, p) > 0 is a

constant depending only on n, p.

Inequality (2.3) was established previously by Bourgain, Brezis, and Mironescu(J.Anal.Math., 2002 ) under the assumption 1

2 ≤ s < 1, 1 ≤ p <∞ and sp < 1.Finally, let us remark that using ground state representation Frank andSeiringer(JFA, 2008) derived the optimal version of the Fractional Hardyinequality (2.1).


Asymptotics of The Fractional Seminorm

Given the fact that, fractional Sobolev inequalities have already been proved longago, the inequalities (2.2) and (2.3) still hold strong ground because of thefollowing two results. The first one is due to Brezis, Bourgain and Mironescu(IOS Press, 2001) and the second one is due to Maz’ya and Shaposhnikova (JFA,2002).

(i) For any smooth bounded domain Ω ⊂ Rn and for any u ∈W 1,p(Ω)

lims↑1

(1− s)

∫Ω

∫Ω

|u(x)− u(y)|p

|x − y |n+spdxdy ∼ ||∇u||pLp(Ω) . (2.4)

Here 1 ≤ p <∞.

(ii) For any u ∈ ∪0<s<1Ws,p(Rn)

lims↓0

s

∫Rn

∫Rn

|u(x)− u(y)|p

|x − y |n+spdxdy ∼ ||u||pLp(Rn) , (2.5)

where 1 ≤ p <∞.


Definition of The Heisenberg Group

The Heisenberg group Hn is defined as

Hn := ξ = (z , t) : z = (x , y) ∈ Rn × Rn, t ∈ R, x = (x1 . . . xn), y = (y1 . . . yn)

with the following group operation:

ξ ξ′ := (x + x ′, y + y ′, t + t ′+2〈y , x ′〉 − 2〈x , y ′〉) ,

where ξ′ = (z ′, t ′) = (x ′, y ′, t ′), ξ ∈ Hn and 〈., .〉 denotes the usual euclideaninner product on Rn. Clearly, 0 ∈ Hn is the identity element and for any ξ ∈ Hn,ξ−1 = −ξ. The left invariant vector fields are given by

Xj :=∂

∂xj+ 2yj

∂

∂t, 1 ≤ j ≤ n

Xn+j :=∂

∂yj− 2xj

∂

∂t, 1 ≤ j ≤ n

T :=∂

∂t.


The Subgradient on the Heisenberg Group

Let Ω ⊂ Hn be a domain. For u ∈ C 1(Ω) define the sub-gradient ∇Hn(u) by

∇Hn(u) := (X1(u), . . . ,X2n(u)) and

|∇Hn(u)|2 :=2n∑j=1

|Xj(u)|2.

For ξ = (z , t) = (x , y , t) ∈ Hn the Koranyi-Folland non isotropic gauge is definedby

d(ξ) :=((|x |2 + |y |2

)2+ t2

) 14

=(|z |4 + |t|2

) 14 .


The Sub-elliptic Sobolev Embedding Theorem

Having this bit of knowledge, let us recall the sub-elliptic Sobolev embeddingtheorem which is due to Folland and Stein. While the result is valid for anyCarnot group, we will state it in the set up of the Heisenberg group which is thefollowing.

Sub-elliptic Sobolev Embedding(Folland and Stein)

Let 1 < p < Q and set q = pQQ−p , where Q = 2n + 2 is the homogeneous

dimension of Hn. Then there exists Sp(Hn) > 0 such that

||u||Lq(Hn) ≤ Sp(Hn) ||∇Hnu||Lp(Hn) , for all u ∈ C∞c (Hn). (4.1)

In this context let us mention Vassilev(Pacific J. Math., 2006), who proved theexistence of extremal function of the above inequality.


Sub-elliptic Hardy type inequalities

On the other hand, Garofalo and Lanconelli (Ann. Inst. Fourier, 1990) proved thefollowing Hardy type inequality

Sub-elliptic Hardy Inequality(Garofalo and Lanconelli)

∫Hn

|∇Hnd(ξ)|2

d(ξ)2|u|2dξ ≤

(2

Q − 2

)2 ∫Hn

|∇Hnu|2dξ, ∀u ∈ C∞c (Hn \ 0). (4.2)

The optimality of the constant(

2Q−2

)2

is shown by Goldstein and Zhang (JFA,

2001). Over the years, more than one generalisation of (4.2) have been derivedby different people. For example

For p 6= 2 different extensions of (4.2) have been achieved by D’Ambrozio(Differ. Uravn, 2004) and Niu, Zhang and Wang (Proc. AMS, 2001).

A sharp inequality of type (4.2) has been derived in generalCarnot-Caratheodory Spaces by Danielli, Garofalo and Phuc (PotentialAnal., 2011)


Continued

In this connection, let us recall a Non linear Hardy type inequality due toAdimurthi and Sekar (Proc. Roy. Soc. Edinburgh, 2006) which is related to theoperator Lp defined as follows. For any smooth function u, and for 1 < p <∞define

Lp(u) = −n∑

j=1

[Xj

(∣∣∣∣∇Hn(u)

|z |

∣∣∣∣p−2

Xj(u)

)+ Xn+j

(∣∣∣∣∇Hn(u)

|z |

∣∣∣∣p−2

Xn+j(u)

)].


Continued

Non Linear Sub-elliptic Hardy Inequality(A*and Sekar)

Let 1 < p < n + 2. Then, for all u ∈ FS1,p0 (Hn),∫

Hn

|∇Hnu|p

|z |p−2−(

2(n + 2− p)

p

)p ∫Hn

|z |2|u|p

(|z |2 + t2)p2

≥ 0 (4.3)

and(

2(n+2−p)p

)pis the best constant which is never achieved. Here FS1,p

0 (Hn) is

defined as the completion of C∞c (Hn) under the following norm.

|u|p1,p :=

∫Hn

|∇Hnu|p

|z |p−2dxdydt.


Fractional Sobolev Spaces on The Heisenberg Group

Loosely speaking, our version of the Hardy-Sobolev inequality is a fractionalanalog of (4.3). Before going to our results let us define the following fractionalorder Sobolev space on Hn.

W s,p,α0 (Hn) := Clf ∈ C∞c (Hn) :

∫Hn×Hn

|f (ξ′)− f (ξ)|p

d(ξ−1 ξ′)Q+ps |z ′ − z |(p−2)αdξ′dξ <∞.

Here 0 < s < 1, 1 ≤ p <∞, α ∈ R and as usual ξ = (z , t), ξ′ = (z ′, t) ∈ Hn. Forf ∈ C∞c (Hn) define

[f ]s,p,α :=

(∫Hn×Hn

|f (ξ′)− f (ξ)|p

d(ξ−1 ξ′)Q+ps |z ′ − z |(p−2)αdξ′dξ

) 1p

and

||f ||s,p,α := ||f ||Lp(Hn) + [f ]s,p,α.

The above closure is taken under the norm ||.||s,p,α.


Non triviality of the defined spaces

From the definition it is not clear whether W s,p,α0 (Hn) is non trivial or not. In

fact, if (p − 2)α ≥ Q − 2 then W s,p,α0 (Hn) is trivial. We will show that if

(p − 2)α < min Q − 2, p(1− s) then C∞c (Hn) ⊂W s,p,α0 (Hn). To prove this we

need to use the sub-Riemannian structure of the group. More precisely we willuse the following theorem which is due to Caratheodory, Rashevsky and Chow.Although the Theorem is true for any general Homogeneous Carnot Group, let usstate it, in the set up of the Heisenberg group.


Caratheodory, Rashevsky and Chow Theorm

TheoremLet g denotes the Lie algebra of vector fields of Hn and Exp : g → Hn be theusual exponential map. Then there exist a constant M ∈ N (depending only onHn) such that for any h ∈ Hn,

h = h1 . . . hMd(hj) ≤ c0d(h), j = 1 . . .M,

where hj = Exp(tjXij ), for some tj ∈ R, ij ∈ 1, . . . , 2n and c0 > 0 is a constantindependent of h and hj .

Remark

Note that, here Exp(tjXij ) = tjeij , where ei2n+1i=1 is the standard basis on R2n+1

and hence d(hj) = |tj |.


Proof of Caratheodory, Rashevsky and Chow TheormWe will prove the theorem for n = 1. The same proof will work for higher valuesof n. Note that, it is enough to find a decomposition of type mentioned in thetheorem for the elements (0, 0, t) = Exp(tT ) ∈ H1. With out loss generality, wecan take t > 0. Clearly, [X1,X2] = −4T . Easy to see that

Exp

(X1 + X2 +

1

2[X1,X2]

)= Exp(X1) Exp(X2). (5.1)

Consequently,

Exp ([X1,X2]) = Exp(X1) Exp(X2) Exp(−X1) Exp(−X2). (5.2)

This implies

Exp(tT ) = Exp

(√t

2X2

) Exp

(√t

2X1

) Exp

(−√t

2X2

) Exp

(−√t

2X1

).

Note that

d

(Exp

(±√t

2Xi

))=

√t

2≤√t = d (Exp (tT )) , for any i = 1, 2.

This proves the theorem.Adimurthi TIFR-CAM, Bangalore ( Batsheva de Rotschild seminar on Hardy-type Inequalities and Elliptic Problems.)Fractional order Sobolev Spaces January 8, 2018 17 / 52

Non triviality of the defined spaces contd..

Proposition (A*,Mallick,’17)

Let s ∈ (0, 1), 1 ≤ p <∞ and α ∈ R satisfies the following condition

(p − 2)α < min Q − 2, p(1− s). (5.3)

Then for any u ∈ C∞c (Hn), [u]s,p,α <∞.



In the above integral, changing the variable ξ′ by ξ ξ, whereξ = (z , t) ∈ R2n × R we obtain

I1 ≤ 2p ||u||pLp(Hn)

∫Bc(0,1)

dzd t

(|z |4 + t2)Q+ps

4 |z |(p−2)α. (5.4)

One can easily show I3 :=∫Bc(0,1)

dzdt

(|z|4+t2)Q+ps

4 |z|(p−2)α

<∞ which in turn proves

I1 <∞. It remains to show that I2 <∞. Before that let us point out thefollowing fact, which will be used without mentioning. If ξ′, ξ, ξ ∈ Hn be such that

ξ = ξ′ ξ then for any u ∈ C 1(Hn) we have ∇ξ′

Hnu(ξ′ ξ) = ∇ξHnu(ξ), where ∇ξ′

Hn

and ∇ξHn denotes the sub-gradient with respect to ξ′ and ξ variable respectively.Now let us consider

I2 : =

∫Hn

∫B(ξ,1)

|u(ξ′)− u(ξ)|p

d(ξ−1 ξ′)Q+ps |z ′ − z |(p−2)αdξ′dξ.


Proof of the above PropositionChanging the variable ξ′ by ξ ξ, ξ = (z , t) ∈ Hn we have

I2 =

∫Hn

|u(ξ ξ)− u(ξ)|p∫B(0,1)

d ξ

d(ξ)Q+ps |z |(p−2)αdξ. (5.5)

By the Theorem of Caratheodory, Chow, and Rashevsky for ξ ∈ B(0, 1) thereexists M ∈ N(depending only on Hn) such that

ξ = h1 . . . hMd(hj) ≤ c0d(ξ), j = 1, . . . ,M,

where hj = Exp(tjXij ) ∈ Hn for some tj ∈ R and ij ∈ 1, . . . , 2n and c0 > 0 is a

constant independent of ξ and hj . Also d(hj) = |tj |. Using this decomposition of

ξ we will estimate |u(ξ ξ)− u(ξ)| in terms of |∇Hnu|. For this, we can taketj ≥ 0 without loosing the generality. Now,

|u(ξ ξ)− u(ξ)| ≤M∑j=1

|u(ξ h1 . . . hj)− u(ξ h1 . . . hj−1)|. (5.6)



Here we follow the convention that at j = 1, ξ h1 . . . hj−1 = ξ. Now letconsider the integral curve γ(t,Xij ) of Xij starting from ξ h1 . . . hj−1. Then

γ(t,Xij ) = ξ h1 . . . hj−1 Exp(tXij )

and

γ(0,Xij ) = ξ h1 . . . hj−1, γ(tj ,Xij ) = ξ h1 . . . hj .

So,

u(γ(tj ,Xij ))− u(γ(0,Xij )) =

∫ tj

0

d

dru(γ(r ,Xij ))dr

=

∫ tj

0

∇u(γ(r ,Xij ))γ(r ,Xij )dr

=

∫ tj

0

X ξiju(γ(r ,Xij ))dr

=

∫ 1

0

tjXξiju(ξ h1 . . . hj−1 Exp(tj rXij )

)dr .



Hence using Jensen’s inequality and the above in (5.6) we get

|u(ξ ξ)− u(ξ)|p ≤ c0d(ξ)M∑j=1

∫ 1

0

∣∣∣∇ξHnu(ξ h1 . . . hj−1 Exp(tj rXij )

)∣∣∣p dr .So, from (5.5) we have

I2 ≤M∑j=1

∫ 1

0

∫B(0,1)

1

d(ξ)Q−p(1−s)|z |(p−2)α

∫Hn

∣∣∣∇ξHnu(ξ h1 . . . hj−1 Exp(rtjXij )

)∣∣∣p dξd ξdr= c0M ||∇Hnu||pLP (Hn)

∫B(0,1)

d ξ

d(ξ)Q−p(1−s)|z |(p−2)α

≤ c0M ||∇Hnu||pLp(Hn)

∫|z|<1

∫|t|<1

dtdz

(|z |4 + t2)Q−p(1−s)

4 |z |(p−2)α(5.7)



One can easily show that

J :=

∫|z|<1

∫|t|<1

dtdz

(|z |4 + t2)Q−p(1−s)

4 |z |(p−2)α<∞.

Then (5.7) suggests that I2 <∞. This proves the Proposition.


Fractional Sobolev Embedding On the Heisenberg Group

Our first result is the Sobolev embedding which is the following.

Theorem (Fractional Sobolov Embedding (A*,Mallick) ’17 )

Let s ∈ (0, 1), 1 ≤ p <∞ and α ∈ R satisfies the following two conditions:

(a) (p − 2)α ≥ 0

(b) ps + (p − 2)α < Q.

Then there exists a positive constant Cn,p,s,α depending only n, p, s and α suchthat

||f ||Lq(Hn) ≤ Cn,p,s,α[f ]s,p,α, for all f ∈W s,p,α0 (Hn), (6.1)

where q := QpQ−ps−(p−2)α .


Fractional Hardy Inequality On the Heisenberg Group

Next we have the following Hardy type inequality.

Theorem (Fractional Hardy Inequality (A*,Mallick) ’17)

Let s ∈ (0, 1), 1 ≤ p <∞ and α ∈ R satisfies the following three conditions:

(a) (p − 2)α ≥ 0,

(b) ps > 2,

(c) ps + (p − 2)α < Q.

Then there exists a positive constant Cn,p,s,α depending only n, p, s and α suchthat for any f ∈W s,p,α

0 (Hn), the following holds true∫Hn

|f (ξ)|pdξd(ξ)ps |z |(p−2)α

≤ Cn,p,s,α[f ]ps,p,α, (6.2)

where ξ = (z , t) ∈ Hn.


Fractional Hardy Inequality On the Heisenberg GroupContd...

Remark

(i) The condition ps > 2 along with s ∈ (0, 1) forces us to choose p > 2. Butthe main ingredient needed holds true for p = 2. Hence, the Hardy-Sobolevtype inequality (6.2) still holds for p = 2.

(ii) Since we are dealing with the case p ≥ 2, the condition (p − 2)α ≥ 0 isnatural.

(iii) Condition (b) and (c) together implies (p − 2)α < Q − 2 which is notsurprising since as proved in Section 2, if (p− 2)α ≥ Q − 2 then W s,p,α

0 (Hn)becomes trivial.

(iv) One can easily observe that (6.2) is an analogous version of (2.1) but witha rough constant. Because of this, our proof is more elementary than theproof of (2.1).


Brief Review of Available Literature Regarding FractionalHardy Inequality

Ciatti, Cowling and Ricci(Adv.Math, 2015) derived different Hardy typeinequalities for different fractional powers of the sublaplacian for generalstratified lie groups.

Roncal and Thangavelu(Adv. Math, 2016) used a completely differentmethod to derive a Hardy type inequality with a non homogeneous weightfor the conformally equivallant fractional power of the sublaplacian Ls onthe Heisenberg group. They derived and used the following integralrepresentation of Ls to proved the inequality.

〈Ls f , f 〉 =2n−2+3sΓ

(n+1+s

2

)2

πn+1|Γ(−s)|

∫Hn

∫Hn

|f (ξ′)− f (ξ)|2

d(ξ−1.ξ′)Q+2sdξ′dξ.

The above representation makes our definition of fractional Sobolev space morerelevant. Also we remark in the case of p = 2 our inequality is a homogeneousversion of their inequality.


Sketch of the Proofs

Our proof of Sobolev inequality inspired by the idea introduced by Nezza,Palatucci and Valdinoci (Bull. Sci. Math, 2012). Although the proof of our Hardyinequality falls under the same framework we need some new estimate to derive it.We prove the inequalities via the following lemmas. The first of them estimatesthe measure of finite measurable set in terms of the kernel which is not radial.

Lemma

Let E ⊂ Hn be any measurable set with finite Lebesgue measure. Fixξ = (x , y , t) = (z , t) ∈ Hn, 1 ≤ p <∞. Then there exists a positive constantCn,p,s,α depending only on n, p, α, s such that∫

E c

dξ′

d(ξ′−1 ξ)Q+ps |z − z ′|(p−2)α≥ Cn,p,s,α|E |−

ps+(p−2)αQ , (7.1)

where ξ′ = (x ′, y ′, t ′) = (z ′, t ′) ∈ E c and |E | denotes the Lebesgue measure of E .


Proof of the above Lemma

The integral of (7.1) is of the form

I :=

∫E c

dz ′dt ′(|z − z ′|4 + (t − t ′ + 2 < x ′, y > −2 < y ′, x >)2

) Q+ps4 |z − z ′|(p−2)α

.

Put ζ = z − z ′, where ζ = (ζ1, ζ2) ∈ R2n andµ = t − t ′ + 2 < x ′, y > −2 < y ′, x > . Then

I =

∫F c

dζdµ

(|ζ|4 + µ2)Q+ps

4 |ζ|(p−2)α, (7.2)

where F is a measurable set depending on ξ with |F | = |E |. Now let R > 0 besuch that |E | = 2w2nR

Q . Denote BR = B2n(0,R)× (−R2,R2), where B2n(0,R)



is a ball in R2n with radius R and centre at the origin. Then clearly|E | = |F | = |BR |. Now (7.2) can be written as

I =

∫F c∩BR

dζdµ

(|ζ|4 + µ2)Q+ps

4 |ζ|(p−2)α+

∫F c∩Bc

R

dζdµ

(|ζ|4 + µ2)Q+ps

4 |ζ|(p−2)α

≥∫F c∩BR

dζdµ

2Q+ps

4 RQ+ps+(p−2)α+

∫F c∩Bc

R

dζdµ

(|ζ|4 + µ2)Q+ps

4 |ζ|(p−2)α. (7.3)

The last inequality came from the fact that, for (p − 2)α ≥ 0 and (ζ, µ) ∈ BR(|ζ|4 + µ2

) Q+ps4 < 2

Q+ps4 RQ+ps+(p−2)α and |ζ|(p−2)α ≤ R(p−2)α.

Now, since |F | = |BR |, we have |BcR ∩ F | = |BR ∩ F c|. Hence continuing from

(7.3) we have

I ≥∫F∩Bc

R

dζdµ

2Q+ps

4 RQ+ps+(p−2)α+

∫F c∩Bc

R

dζdµ

(|ζ|4 + µ2)Q+ps

4 |ζ|(p−2)α. (7.4)



Now let BR := Bc2n(0,R)× (−R2,R2)c. Then BR ⊂ Bc

R and for (ζ, µ) ∈ BR wehave (

|ζ|4 + µ2) Q+ps

4 |ζ|(p−2)α ≥ 2Q+ps

4 RQ+ps+(p−2)α.

Consequently, from (7.4) we obtain

I ≥∫BR

dζdµ

(|ζ|4 + µ2)Q+ps

4 |ζ|(p−2)α

=

∫|ζ|>R

∫|µ|>R2

dµdζ

(|ζ|4 + µ2)Q+ps

4 |ζ|(p−2)α. (7.5)

Let

I1 : =

∫|µ|>R2

dµ

(|ζ|4 + µ2)Q+ps

4

= 2

∫ ∞R2

dµ

(|ζ|4 + µ2)Q+ps

4

.



Put µ = |ζ|2ν, then I1 becomes

I1 =2

|ζ|Q+ps−2

∫ ∞R2

|ζ|2

dν

(1 + ν2)Q+ps

4

. (7.6)

Consider

I2 : =

∫ ∞R2

|ζ|2

dν

(1 + ν2)Q+ps

4

(Integrating by parts ) =Q + ps

2

∫ ∞R2

|ζ|2

ν2dν

(1 + ν2)Q+ps

4 +1− R2

|ζ|2(

1 + R4

|ζ|4

) Q+ps4

=Q + ps

2I2 −

Q + ps

2

∫ ∞R2

|ζ|2

dν

(1 + ν2)Q+ps

4 +1− R2

|ζ|2(

1 + R4

|ζ|4

) Q+ps4

.



Hence

I2 ≥2R2

Q − 2 + ps

|ζ|Q−2+ps

(|ζ|4 + R4)Q+ps

4

.

So, from (7.5) and (7.6) we have

I ≥ 4R2

Q + ps − 2

∫|ζ|>R

dζ

(|ζ|4 + R4)Q+ps

4 |ζ|(p−2)α

=8nw2nR

2

Q + ps − 2

∫ ∞R

rQ−3dr

(r4 + R4)Q+ps

4 r (p−2)α.



Now, in the above integral, since r > R, so 1r4+R4 ≥ 1

2r4 . Hence form above wehave

I ≥ 8nw2nR2

(Q + ps − 2)2Q+ps

4

∫ ∞R

rQ−3−(p−2)αdr

rQ+ps

=8nw2n

(2 + ps + (p − 2)α)(Q + ps − 2)2Q+ps

4

1

Rps+(p−2)α

= Cn,p,s,α|E |−ps+(p−2)α

Q .

This proves the lemma. Next we need the following algebraic lemma.


Lemma

Let s ∈ (0, 1) , 1 ≤ p <∞ and α be such that it satisfies (p − 2)α ≥ 0. Letps + (p − 2)α < Q. Let ak dk be two sequences of nonnegative real numberssatisfying the following properties:

(a) ak is decreasing and

(b) ak =∞∑j=k

dj .

Then we have

(i) for any fixed T > 0∑k∈Z

aQ−ps−(p−2)α

Q

k T k ≤ TQ

Q−ps−(p−2)α

∑k∈Z,ak 6=0

ak+1a− ps+(p−2)α

Q

k T k . (7.7)

(ii) Moreover, if T > 1 then∑i∈Z,ai−1 6=0

∑j≥i+1

T pia− ps+(p−2)α

Q

i−1 dj ≤1

T p − 1

∑i∈Z,ai−1 6=0

T pia− ps+(p−2)α

Q

i−1 di (7.8)


Proof of Sobolev inequality

It is enough to prove the theorem for 0 ≤ f ∈W s,p,α0 (Hn) with compact support.

For such an f let us define the following sets:

Ak := x ∈ Hn : f (x) ≥ 2k,Dk := Ak \ Ak+1 = x ∈ Hn : 2k ≤ f (x) < 2k+1.

Let ak = |Ak | and dk = |Dk |. Then clearly, ak and dk satisfies the conditions ofthe above Lemma. Now

||f ||qLq(Hn) =∑k∈Z

∫Dk

f q(x)dx

≤∑k∈Z

2q(k+1)dk

≤ 2q∑k∈Z

2qkak (Since dk ≤ ak).



So,

||f ||pLq(Hn) ≤ 2p

(∑k∈Z

2qkak

) pq

≤ 2p∑k∈Z

2pkapq

k (Since p < q)

= 2p∑k∈Z

2pkaQ−ps−(p−2)α

Q

k

Using inequality (7.7) of Lemma 7 we obtain

||f ||pLq(Hn) ≤ 2p+q∑

k∈Z,ak 6=0

2pkak+1a− ps+(p−2)α

Q

k (7.9)



Note that if j ≤ i − 2, then for any ξ′ == (z ′, t ′) ∈ Dj and ξ = (z , t) ∈ Di onecan estimate [f ]s,p,α to get

[f ]ps,p,α ≥ Cn,s,p,α

∑i∈Z,ai−1 6=0

2p(i−1)a− ps+(p−2)α

Q

i−1 di (7.10)

= Cn,s,p,α

∑ai−1 6=0

2piaia− ps+(p−2)α

Q

i−1 −∑

ai−1 6=0

∑j≥i+1

2pia− ps+(p−2)α

Q

i−1 dj

In the last equality we have used di = ai −∑j≥i+1

dj . Hence

[f ]ps,p,α + Cn,s,p,α

∑ai−1 6=0

∑j≥i+1


Q

i−1 dj ≥ Cn,s,p,α

∑ai−1 6=0


Q

i−1

(7.11)



So, by using inequality (7.8) and (7.10) we have∑ai−1 6=0

∑j≥i+1


Q

i−1 dj ≤1

2p − 1

∑i∈Z,ai−1 6=0


Q

i−1 di ≤ Cn,s,p,α[f ]ps,p,α.

Hence, from (7.11) and the above inequality we get

[f ]ps,p,α ≥ Cn,s,p,α

∑i∈Z,ai−1 6=0


Q

i−1 (7.12)

Finally, combining the above inequality with (7.9) we get the required Sobolevembedding (6.1).


Next we need the following upper estimate of the kernel in terms of the measureof a set.

Lemma (A*, Mallick)

Let s ∈ (0, 1), 1 ≤ p <∞ and α ∈ R satisfies the following three conditions:

(a) (p − 2)α ≥ 0,

(b) ps > 2 and

(c) ps + (p − 2)α < Q.

Then there exists a positive constant Cn,s,p,α depending only on n, s, p, α suchthat for any measurable set D ⊂ R2n+1 with |D| <∞ we have∫

D

dzdt

(|z |4 + t2)ps4 |z |(p−2)α

≤ Cn,s,p,α|D|1−ps+(p−2)α

Q . (7.13)

Here, z ∈ R2n, t ∈ R and |D| denotes the Lebesgue measure of D.



Let |D| = 2w2nRQ . Then, if we denote the set B2n(0,R)× (−R2,R2) by BR ,

where B2n(0,R) is the ball in R2n centred at origin with radius R, one can easilynotice that |BR | = |D|. Let

I : =

∫D

dzdt

(|z |4 + t2)ps4 |z |(p−2)α

=

∫D∩BR

dzdt

(|z |4 + t2)ps4 |z |(p−2)α

+

∫D∩Bc

R

dzdt

(|z |4 + t2)ps4 |z |(p−2)α

Now let us define the following quantities:

BR,1 : = z ∈ R2n : |z | > R × t ∈ R : |t| > R2,BR,2 : = z ∈ R2n : |z | > R × t ∈ R : |t| < R2,BR,3 : = z ∈ R2n : |z | < R × t ∈ R : |t| > R2



and

I1 : =

∫D∩BR

dzdt

(|z |4 + t2)ps4 |z |(p−2)α

,

I2 : =

∫D∩BR,1

dzdt

(|z |4 + t2)ps4 |z |(p−2)α

,

I3 : =

∫D∩BR,2

dzdt

(|z |4 + t2)ps4 |z |(p−2)α

,

I4 : =

∫D∩BR,3

dzdt

(|z |4 + t2)ps4 |z |(p−2)α

.

Then I = I1 + I2 + I3 + I4. We were able to estimate each of I1, I2, I3, I4 separatelyto arrive at (7.13). This proves the lemma.


Some Remarks

Remark

(i) The condition ps > 2 forces us to choose p > 2. However, in the case ofp = 2, the approach of estimating I3, in the above Lemma could be followedto deduce the inequality (7.13).

(ii) In general, the inequality (7.13) is not true. For example if

0 < (p − 2)α < Q − 2 and ps < 2(p−2)αQ−2 , then one can choose

D = B2n(0,R)× (−1, 1) and make R → 0 to show that (7.13) is false. Note

that in this case 2(p−2)αQ−2 < 2 and so ps < 2.


Proof of our Hardy Inequality

Note that it is enough to prove the theorem for nonnegative compactly supportedsmooth function f . Let the quantities Ak ,Dk , ak , dk be the same as defined in theproof of Sobolev inequality. We define

J : =

∫Hn

|f (ξ)|p

d(ξ)ps |z |(p−2)αdξ

=∑i∈Z

∫Di

f p

(|z |4 + t2)ps4 |z |(p−2)α

dzdt

≤∑i∈Z

2(i+1)p

∫Di

dzdt

(|z |4 + t2)ps4 |z |(p−2)α

.

Now we use the previous Lemma to obtain

J ≤ Cn,s,p,α

∑i∈Z

2(i+1)pd1− ps+(p−2)α

Q

i

≤ Cn,s,p,α

∑i∈Z

2ipa1− ps+(p−2)α

Q

i (Since di ≤ ai )


Proof of our Hardy Inequality

Using inequality (7.7) with T = 2p and combining it with (7.12) we get therequired inequality (6.2). This proves the theorem.


Morrey Type Embedding

Next we have the following embedding result.

Theorem (A*,Mallick, 2017)

Let p ∈ [1,∞), s ∈ (0, 1) and α ∈ R satisfies the following two conditions:

(i) 0 ≤ (p − 2)α < Q − 2 and

(ii) ps + (p − 2)α > Q.

Then there exists a constant C > 0 depending only on n, s, p, α such that for anyu ∈ Lp(Hn)

||u||C 0,β(Hn) ≤ C(||u||pLp(Hn) + [u]ps,p,α

) 1p

, (8.1)

where β := ps+(p−2)α−Qp .


Fractional Sobolev Spaces on arbitrary domain

In analogy with the definition of W s,p,α0 (Hn) one can define W s,p,α

0 (Ω), whereΩ ⊂ Hn is any open set. More precisely, we define

W s,p,α0 (Ω) := Clu ∈ C∞c (Ω) :

∫Ω×Ω

|u(ξ′)− u(ξ)|p

d(ξ−1.ξ′)Q+ps |z ′ − z |(p−2)αdξ′dξ <∞.

Here the closure is taken under the norm ||.||s,p,α,Ω := ||.||LP (Ω) + [.]s,p,α,Ω, where

for any u ∈ C∞c (Ω),

[u]s,p,α,Ω :=

∫Ω×Ω

|u(ξ)− u(ξ′)|p

d(ξ−1.ξ′)Q+ps |z ′ − z |(p−2)αdξ′dξ.

Clearly, for (p − 2)α < min Q − 2, p(1− s), W s,p,α0 (Ω) is non trivial.


Fractional Sobolev inequality on arbitrary domain.

DefinitionLet Ω be any domain in Rn. We say that Ω is an extension domain if it satisfiesthe following property. For any f ∈W s,p,α

0 (Ω) there exists f ∈W s,p,α0 (Hn) such

that f |Ω = f and satisfies the following inequality

||f ||s,p,α ≤ Cn,s,p,α(Ω)||f ||s,p,α,Ω,

where Cn,s,p,α(Ω) > 0 is a constant depending only on n, s, p, α and Ω.

Now if Ω is an extension domain and s ∈ (0, 1), 1 ≤ p <∞, α ∈ R satisfies thefollowing conditions:

(a) (p − 2)α ≥ 0

(b) ps + (p − 2)α < Q,


Fractional Sobolev inequality on arbitrary domain.

then from Sobolev inequlaity it follows that there exists a constant Cn,s,p,α(Ω) > 0depending only on n, s, p, α and Ω such that for any u ∈W s,p,α

0 (Ω)

||u||Lq(Ω) ≤ Cn,s,p,α(Ω)||u||s,p,α,Ω. (9.1)

Here q = QpQ−ps−(p−2)α . The above inequality shows that W s,p,α

0 (Ω) is

continuously embedded in Lr (Ω) for any 1 ≤ r < q, if Ω is a bounded extensiondomain.


Compactness

Theorem

Let s ∈ (0, 1), 32 ≤ p <∞ and α ∈ R satisfies the following conditions:

(a) (p − 2)α ≥ 0

(b) ps + (p − 2)α < Q.

Also let Ω be a bounded extension domain for W s,p,α0 and F be a bounded set of

Lp(Ω). Suppose that

supu∈F

[u]s,p,α,Ω <∞.

Then for any 1 ≤ r < q, F is relatively compact in Lr (Ω).


Happy Birthday for Prof Marcus.


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