Analytic and Probabilistic Perspectives on the Hardy-Littlewood …HLSIneq).pdf · 2015. 5. 28. · in “Analytic Methods in Interdisciplinary Applications”, Springer Proc. Math

Analytic and Probabilistic Perspectives on theHardy-Littlewood-Sobolev Inequality

David Applebaum

School of Mathematics and Statistics, University of Sheffield, UK

Talk at "Wales Mathematical Colloquium 2015",Gregynog.

18- 20 May 2015

Talk based on joint work with Rodrigo Bañuelos (Purdue)

in “Analytic Methods in Interdisciplinary Applications”, SpringerProc. Math. Stat. Vol. 116, pp. 17-40 (2014)

Dave Applebaum (Sheffield UK) Analytic and Probabilistic Perspectives on the Hardy-Littlewood-Sobolev InequalityMay 2015 1 / 32

Outline of Talk

Background on fractional integral operators. Work of Hardy andLittlewood.Analytic approach using ultracontractive semigroups. Simplifiedproof of Varopoulos’ generalisation of Hardy-Littlewood-Sobolev.Probabilistic approach using Brownian motion and quadraticvariation.


Historical Background on Fractional Integration

Fractional integration/differentation has origins in work of Leibnitz andEuler. The first papers were due to Liouville (1832).

Riemann (1872) introduced what is now called the Riemann-Liouvilleoperator for f ∈ L1loc(R),0 < α < 1:

(Iαf )(x) =1

Γ(α)

∫ x0

f (t)(x − t)α−1dt .

This generalised to fractional order a formula due to Cauchy:

(Inf )(x) =∫ x

0

∫ y10· · ·∫ yn−1

0f (yn)dyndyn−1 · · · dy1.


Historical Background on Fractional Integration

Fractional integration/differentation has origins in work of Leibnitz andEuler. The first papers were due to Liouville (1832).Riemann (1872) introduced what is now called the Riemann-Liouvilleoperator for f ∈ L1loc(R),0 < α < 1:

(Iαf )(x) =1

Γ(α)

∫ x0

f (t)(x − t)α−1dt .

This generalised to fractional order a formula due to Cauchy:

(Inf )(x) =∫ x

0

∫ y10· · ·∫ yn−1

0f (yn)dyndyn−1 · · · dy1.


Riemann also introduced the Riemann-Liouville fractional derivative

Dα =ddx◦ I1−α,

which enables the solution of Abel’s integral equation, i.e. for given f ,to find φ so that

f (x) =1

Γ(α)

∫ t0

φ(s)(x − s)1−α

ds.

The case α = 12 is the “tautochrome” problem which finds the timetaken for a particle to fall under gravity along a specified path f .


Hardy and Littlewood wrote two papers on this topic (on real andcomplex analytic aspects, resp). In their first paper (1928) they wrote“Our first object is to determine the Lebesgue class to which Iαfbelongs when f ∈ Lp.”

Sobolev extended this to d - dimensions, where we have the Rieszpotential operator, for 0 < α < d :

(Iαf )(x) =1cd

∫Rd

f (y)|x − y |d−α

dy = (f ∗ rd )(x),

where the Riesz kernel is rd (x) = 1cd |x |d−α , and cd =Γ( d−α2 )

Γ(α2 )2α2 π

d2

.

Theorem (Hardy-Littlewood-Sobolev)

If 0 < α < d ,1 < p < dα and1q =

1p −

αd , then there exists C ≥ 0 so that

for all f ∈ Lp,

||Iαf ||q ≤ C||f ||p. ...(HLS)


Hardy and Littlewood wrote two papers on this topic (on real andcomplex analytic aspects, resp). In their first paper (1928) they wrote“Our first object is to determine the Lebesgue class to which Iαfbelongs when f ∈ Lp.”Sobolev extended this to d - dimensions, where we have the Rieszpotential operator, for 0 < α < d :

(Iαf )(x) =1cd

∫Rd

f (y)|x − y |d−α

dy = (f ∗ rd )(x),

where the Riesz kernel is rd (x) = 1cd |x |d−α , and cd =Γ( d−α2 )

Γ(α2 )2α2 π

d2

.

Theorem (Hardy-Littlewood-Sobolev)

If 0 < α < d ,1 < p < dα and1q =

1p −

αd , then there exists C ≥ 0 so that

for all f ∈ Lp,

||Iαf ||q ≤ C||f ||p. ...(HLS)


Varopoulos’ Theory

For x ∈ Rd , t ≥ 0, introduce Gauss kernel: k(t ; x) = 1(2πt)

d2

e−|x|22t , and

the Gaussian semigroup

(Tt f )(x) = (f ∗ k(t ; ·))(x) =∫Rd

f (y)k(t ; x − y)dy ,

for f ∈ Lp(Rd ).

A straightforward calculation using the definition of Γ(α) shows that∫ ∞0

tα2−1k(t ; x)dt = rd (x),

i.e. the Mellin transform of the Gauss kernel is the Riesz kernel.It follows that

(Iαf )(x) =∫ ∞

0tα2−1Tt f (x)dt ,

i.e. the Mellin transform of the heat semigroup is the Rieszpotential operator.




d2

e−|x|22t , and


(Tt f )(x) = (f ∗ k(t ; ·))(x) =∫Rd

f (y)k(t ; x − y)dy ,

for f ∈ Lp(Rd ).A straightforward calculation using the definition of Γ(α) shows that∫ ∞

0tα2−1k(t ; x)dt = rd (x),

i.e. the Mellin transform of the Gauss kernel is the Riesz kernel.

It follows that

(Iαf )(x) =∫ ∞






d2

e−|x|22t , and


(Tt f )(x) = (f ∗ k(t ; ·))(x) =∫Rd

f (y)k(t ; x − y)dy ,

for f ∈ Lp(Rd ).A straightforward calculation using the definition of Γ(α) shows that∫ ∞

0tα2−1k(t ; x)dt = rd (x),

i.e. the Mellin transform of the Gauss kernel is the Riesz kernel.It follows that

(Iαf )(x) =∫ ∞




Varopoulos based his generalisation of HLS on this last identity(J.Funct. Anal. 63, 240 (1985)).

Let (S,Σ, µ) be a measure space and (Tt , t ≥ 0) be a C0-contractionsemigroup on Lp(S) for 1 ≤ p 0 |Tt f (x)|.


The Ultracontractivity Assumption

We say that the semigroup (Tt , t ≥ 0) satisfies (n,p)-ultracontractivityif there exists an n > 0 (not required to be an integer) such that for all1 ≤ p 0 so that for all t > 0, f ∈ Lp(S),

||Tt f ||∞ ≤ Cp,nt−n

2p ||f ||p (U).

The number n will be referred to as the dimension of the semigroup.e.g. By Jensen’s inequality, (n,p)-ultracontractivity holds wheneverTt f (x) =

∫S f (y)kt (x , y)µ(dy) and kt is a symmetric kernel satisfying∫

S kt (x , y)µ(dy) = 1, and an estimate of the form

kt (x , y) ≤ Ct−n2 ,

for all t ≥ 0, x , y ∈ S.


Examples of Ultracontractivity

The heat kernel on Euclidean space and some Riemannianmanifolds. In this case n = d .

If 0 < β < 1 we can subordinate the heat kernel to get 2β-stabletransition kernels in Euclidean space, and stable-like kernels onsome Riemannian manifolds. In this case n = dβ .

Some fractals such as the Sierpinski gasket. In that case n = 2Hβwhere H is the Hausdorff dimension and β is the walk dimension.Strictly elliptic operators on domains in Euclidean space, whereagain n = d .

For ultracontractive Schrödinger semigroups, see Davies and Simon,J.Funct. Anal. 59, 335 (1984)



The heat kernel on Euclidean space and some Riemannianmanifolds. In this case n = d .If 0 < β < 1 we can subordinate the heat kernel to get 2β-stabletransition kernels in Euclidean space, and stable-like kernels onsome Riemannian manifolds. In this case n = dβ .






Some fractals such as the Sierpinski gasket. In that case n = 2Hβwhere H is the Hausdorff dimension and β is the walk dimension.

Strictly elliptic operators on domains in Euclidean space, whereagain n = d .



From now on, we define for 0 < α < n, and (Tt , t ≥ 0) an(n,p)-ultracontractive semigroup:

(Iαf )(x) =1

Γ(α/2)

∫ ∞0

tα2−1Tt f (x)dt .

Its not hard to show that (Iαf )(x) is absolutely convergent.

We aim to prove (HLS). We proceed as follows:


Analytic Proof of HLS

Let δ > 0 to be chosen later. Let x ∈ S be arbitrary and choosef ∈ L1(S) ∩ Lp(S) with f 6= 0.

We splitIαf (x) = Jαf (x) + Kαf (x),

where the integrals on the right hand side range from 1 to δ and δ to∞(respectively).Using f ∗(x) := supt>0 |Tt f (x)| we have

|Jαf (x)| ≤2α

1Γ(α/2)

f ∗(x)δα2 .

Now using (U) we obtain

|Kαf (x)| ≤ Cp,n,α∫ ∞δ

tα2−

n2p−1||f ||p

≤ Cp,n,αδα2−

n2p ||f ||p,



Let δ > 0 to be chosen later. Let x ∈ S be arbitrary and choosef ∈ L1(S) ∩ Lp(S) with f 6= 0.We split

Iαf (x) = Jαf (x) + Kαf (x),

where the integrals on the right hand side range from 1 to δ and δ to∞(respectively).

Using f ∗(x) := supt>0 |Tt f (x)| we have

|Jαf (x)| ≤2α

1Γ(α/2)

f ∗(x)δα2 .



tα2−

n2p−1||f ||p

≤ Cp,n,αδα2−

n2p ||f ||p,



Let δ > 0 to be chosen later. Let x ∈ S be arbitrary and choosef ∈ L1(S) ∩ Lp(S) with f 6= 0.We split

Iαf (x) = Jαf (x) + Kαf (x),

where the integrals on the right hand side range from 1 to δ and δ to∞(respectively).Using f ∗(x) := supt>0 |Tt f (x)| we have

|Jαf (x)| ≤2α

1Γ(α/2)

f ∗(x)δα2 .



tα2−

n2p−1||f ||p

≤ Cp,n,αδα2−

n2p ||f ||p,


so that|Iαf (x)| ≤ Cp,n,α(f ∗(x)δ

α2 + δ

α2−

n2p ||f ||p).

Picking

δ =

(||f ||pf ∗(x)

)2p/nto minimize the right hand side gives

|Iαf (x)| ≤ Cp,n,α (f ∗(x))1−αp/n ||f ||αp/np = Cp,n,α (f ∗(x))p/q ||f ||αp/np .

Thus for 1 < p < nα and using (S),

||Iαf ||qq ≤ Cp,n,α||f ||αpq/np ||f ∗||

pp

≤ Cn,p,α||f ||p(1+αqn )p

= Cn,p,α||f ||qp,

and the required result follows by density.


so that|Iαf (x)| ≤ Cp,n,α(f ∗(x)δ

α2 + δ

α2−

n2p ||f ||p).

Picking

δ =

(||f ||pf ∗(x)

)2p/nto minimize the right hand side gives

|Iαf (x)| ≤ Cp,n,α (f ∗(x))1−αp/n ||f ||αp/np = Cp,n,α (f ∗(x))p/q ||f ||αp/np .

Thus for 1 < p < nα and using (S),

||Iαf ||qq ≤ Cp,n,α||f ||αpq/np ||f ∗||

pp

≤ Cn,p,α||f ||p(1+αqn )p

= Cn,p,α||f ||qp,

and the required result follows by density.Dave Applebaum (Sheffield UK) Analytic and Probabilistic Perspectives on the Hardy-Littlewood-Sobolev InequalityMay 2015 12 / 32

Consequences of HLS 1: Fractional Powers

Let −A be the (self-adjoint) infinitesimal generator of the semigroup(Tt , t ≥ 0) and assume that A is a positive operator in L2(S). For eachγ ∈ R, we can construct the self-adjoint operator Aγ in L2(S) byfunctional calculus, and we denote its domain in L2(S) by Dom(Aγ).

TheoremFor all f ∈ Dom(A−

α2 ),

Iα(f ) = A−α2 f ,

in the sense of linear operators acting on L2(S)


Proof.We use the spectral theorem to write Tt =

∫∞0 e

−tλP(dλ) for all t ≥ 0where P(·) is the projection-valued measure associated to A.

For allf ∈ Dom(A−

α2 ),g ∈ L2(S) we have, using Fubini’s theorem

〈Iα(f ),g〉 =1

Γ(α/2)

∫ ∞0

∫ ∞0

tα/2−1e−λt〈P(dλ)f ,g〉dt

=1

Γ(α/2)

(∫ ∞0

tα/2−1e−tdt)(∫ ∞

0

1λα2〈P(dλ)f ,g〉

)= 〈A−

α2 f ,g〉.


Proof.We use the spectral theorem to write Tt =

∫∞0 e

−tλP(dλ) for all t ≥ 0where P(·) is the projection-valued measure associated to A. For allf ∈ Dom(A−

α2 ),g ∈ L2(S) we have, using Fubini’s theorem

〈Iα(f ),g〉 =1

Γ(α/2)

∫ ∞0

∫ ∞0

tα/2−1e−λt〈P(dλ)f ,g〉dt

=1

Γ(α/2)

(∫ ∞0

tα/2−1e−tdt)(∫ ∞

0

1λα2〈P(dλ)f ,g〉

)= 〈A−

α2 f ,g〉.


Consequences of HLS 2: Sobolev Inequality

Corollary

For all 1 < p < n, f ∈ Dom(A12 ), if A

12 f ∈ Lp(S) then f ∈ L

npn−p (S) and

||f || npn−p≤ Cn,p,1||A

12 f ||p.

Proof.Take α = 1 so that so that q = npn−p . Writing the Riesz potential

operator as a fractional power in (HLS) yields ||A−12 f ||q ≤ Cn,p,1||f ||p.

Replacing f with A12 f , we find that ||f ||q ≤ Cn,p,1||A

12 f ||p as required.


Remark.1 In most cases of interest the operator A and the space S will be

such that Dom(A)12 contains a rich set of vectors such as

Schwartz space (in Rd ) or the smooth functions of compactsupport (on a manifold). In practice, we would only apply theinequality to vectors in that set.

2 Note that in the case where n > 2 and p = 2 we have

||f ||22nn−2≤ DE(f ),

where E(f ) := 〈Af , f 〉 is a Dirichlet form. If S is a completeRiemannian manifold with bounded geometry and −A is theLaplacian ∆, then we have n = d , the dimension of the manifold,and our Sobolev inequality is more familiar to those who areknowledgeable about that topic.


Probabilistic Approach

We return to the case S = Rd . Then Tt = et∆2 is the heat semigroup.

Our first goal is to find a probabilistic interpretation of the Rieszpotential operator Iα:Let (B(t), t ≥ 0) be standard Brownian motion on Rd . It makes lifeeasier if we use a non-standard notion for “expectation”:

E(f (B(t))) :=∫Rd

E(f (B(t))|B(0) = x)dx

=

∫Rd

∫Rd

f (y)k(t , x − y)dydx

=

∫Rd

f (y)(∫

Rdk(t , x − y)dx

)dy

=

∫Rd

f (y)dy ...(E)

which is intuitively the same as taking B(0) to have Lebesgue measureas its “distribution.”




Our first goal is to find a probabilistic interpretation of the Rieszpotential operator Iα:

Let (B(t), t ≥ 0) be standard Brownian motion on Rd . It makes lifeeasier if we use a non-standard notion for “expectation”:

E(f (B(t))) :=∫Rd

E(f (B(t))|B(0) = x)dx

=

∫Rd

∫Rd


=

∫Rd

f (y)(∫

Rdk(t , x − y)dx

)dy

=

∫Rd

f (y)dy ...(E)





Our first goal is to find a probabilistic interpretation of the Rieszpotential operator Iα:Let (B(t), t ≥ 0) be standard Brownian motion on Rd . It makes lifeeasier if we use a non-standard notion for “expectation”:

E(f (B(t))) :=∫Rd

E(f (B(t))|B(0) = x)dx

=

∫Rd

∫Rd


=

∫Rd

f (y)(∫

Rdk(t , x − y)dx

)dy

=

∫Rd

f (y)dy ...(E)



Two Useful Martingales

For f ∈ S(Rd ) (Schwartz space of rapidly decreasing functions) andfixed a > 0, consider the martingales:

Maf (t) =∫ a∧t

0∇(Ta−sf )(Bs) · dBs

Ma,αf (t) =∫ a∧t

0(a− s)α/2∇(Ta−sf )(Bs) · dBs.

By Itô’s formula,

Ta−t f (Bt ) = Taf (B0) + Maf (t), 0 < t ≤ a ...(I)

Quadratic variations:

[Maf ](t) =∫ a∧t

0|∇(Ta−sf )(Bs)|2ds

[Ma,αf ](t) =∫ a∧t

0(a− s)α|∇(Ta−sf )(Bs)|2ds.


A Useful Identity

Set t = a in (I) to obtain

f (Ba) = Taf (B0) +∫ a

0∇(Ta−sf )(Bs) · dBs.

If g ∈ S(Rd ), we have

g(Ba)∫ a

0(a− s)α/2∇(Ta−sf )(Bs) · dBs

= Tag(B0)∫ a


+

(∫ a0∇(Ta−sg)(Bs) · dBs

)(∫ a0

(a− s)α/2∇(Ta−sf )(Bs) · dBs).


Observe further that the expectation of the first term is zero. That is,

E(

Tag(B0)∫ a


)=

∫Rd

Ex(

Tag(B0)∫ a


)dx

=

∫Rd

Tag(x)Ex(∫ a


)dx

= 0

Thus by Itô’s isometry,

E(

g(Ba)∫ a


)= E

(∫ a0∇(Ta−sg)(Bs) · dBs

)(∫ a0

(a− s)α/2∇(Ta−sf )(Bs) · dBs)

= E(∫ a

0(a− s)α/2∇(Ta−sf )(Bs) · ∇(Ta−sg)(Bs)ds

)...(ID)


Probabilistic Interpretation of Riesz Potential Operator

From now on, let t = a and define Mαf (a) = Ma,αf (a).

Definition: Probabilistic Riesz PotentialDefine for all x ∈ Rd :

(Sa,αf )(x) = E(Mαf (a) | Ba = x).

TheoremFor all f ∈ S(Rd ), x ∈ Rd ,

1

Sa,αf (x) = −∫ a

0sα/2Ts(∆Tsf )(x)ds.

2

lima→∞

(Sa,αf )(x) = cαIα(f )(x),

(almost everywhere) where cα > 0 depends only on α.


Proof. (1) Let g ∈ S(Rd ). Using (E) and (ID), we have∫RdSa,αf (x)g(x)dx

=

∫Rd

E(∫ a

0(a− s)α/2∇(Ta−sf )(Bs) · dBs | Ba = x

)g(x)dx

= E(E(∫ a

0(a− s)α/2∇(Ta−sf )(Bs) · dBs | Ba

)g(Ba)

)= E

(E(

g(Ba)∫ a


)|Ba)

= E(

g(Ba)∫ a


)= E

(∫ a0

(a− s)α/2∇(Ta−sf )(Bs) · ∇(Ta−sg)(Bs)ds)

=

∫ a0

{sα/2

∫Rd∇(Tsf )(x) · ∇(Tsg)(x)dx

}ds


Now by integration by parts, and self-adjointness of the semigroup,∫RdSa,αf (x)g(x)dx

= −∫ a

0

{sα/2

∫Rd

Ts (∆(Tsf ))(x)g(x)dx}

ds

= −∫Rd

{∫ a0

sα/2 Ts (∆(Tsf ))(x)ds}

g(x)dx .

(2) Recall that ddt Tt f = ∆Tt f . Write u(t , ·) = Tt f , then∂∂t u(t , ·) = ∆u(t , ·) and so

∂

∂tu(2t , ·) = 2u′(2t , ·) = 2∆u(2t , ·).

This gives that

∆T2sf =12

dds

T2sf


Proof.(2)Hence

Sa,αf (x) = −∫ a

0sα/2 ∆(T2s)f (x)ds

= −12

∫ a0

sα/2dT2sf

ds(x)ds

= −12

aα/2T2af (x) +α

4

∫ a0

sα/2−1T2sf (x)ds.

Since by (U), |T2af (x)| ≤ Cad/2 ‖f‖1 and 0 < α < d , as a→∞, the righthand side of the previous equality goes to

α

4

∫ ∞0

sα/2−1T2sf (x)ds = 2−α+4

2 αΓ (α/2) Iαf (x).


Some Heat Kernel Estimates

Our next goal is to find a probabilistic proof of (HLS). We’ll sketch themain ideas rather than giving fully detailed proofs:

We begin by deriving some cheap estimates for heat kernelderivatives.

Here we write, for x ∈ Rd t ≥ 0, kt (x) := kt (x ,0) so that kt is the law ofB(t).

TheoremFor all x ∈ Rd , t > 0,

|∇xkt (x)| ≤ 2d+4

21√tk2t (x). ...(G)


Proof. Observe that

∇xkt (x) = −(x1

t, · · · xd

t

)kt (x)

so that

|∇xkt (x)| ≤1√t

√|x |2

tkt (x)

=1√t

√|x |2

t1

(2πt)d/2e−|x|22t


We now claim that the right hand side is dominated by 2d+4

2 1√tk2t (x).

To see this, observe that if√|x |2

t ≤ 1, then the right hand side is

dominated by 1√t

1(2πt)d/2 e

− |x|2

2t .

If a =√|x |2

t > 1, then a < a2 = 4(a/2)2 ≤ 4e

a24 and the right hand side

is dominated by

41√t

1(2πt)d/2

e(−|x|22t +

|x|24t ) = 4

1√t

1(2πt)d/2

e−|x|24t .

Since e−|x|22t ≤ e−

|x|24t , we see that in either case, the right hand side is

dominated by

41√t

1(2πt)d/2

e−|x|24t = 2

d+42

1√tk2t (x)

and this completes the proof.


We are in the HLS framework where 0 < α < d ,1 < p < dα and1q =

1p −

αd .

The derivation of the next few results are rather lengthy and technical.We will want to use the Burkholder-Davis-Gundy inequality, and for thiswe need control of the quadratic variation of the martingale Mαf (t) attime t = a.Recall that

[Mαf ](a) =∫ a∧t


LemmaLet δ > 0 be arbitrary. Then there exists C1,C2 ≥ 0 so that

[Mαf ](a) ≤ C1(

sup0


1p −

αd .

The derivation of the next few results are rather lengthy and technical.

We will want to use the Burkholder-Davis-Gundy inequality, and for thiswe need control of the quadratic variation of the martingale Mαf (t) attime t = a.Recall that

[Mαf ](a) =∫ a∧t



[Mαf ](a) ≤ C1(

sup0


1p −

αd .

The derivation of the next few results are rather lengthy and technical.We will want to use the Burkholder-Davis-Gundy inequality, and for thiswe need control of the quadratic variation of the martingale Mαf (t) attime t = a.

Recall that

[Mαf ](a) =∫ a∧t



[Mαf ](a) ≤ C1(

sup0


1p −

αd .

The derivation of the next few results are rather lengthy and technical.We will want to use the Burkholder-Davis-Gundy inequality, and for thiswe need control of the quadratic variation of the martingale Mαf (t) attime t = a.Recall that

[Mαf ](a) =∫ a∧t



[Mαf ](a) ≤ C1(

sup0

The proof uses (G) and (U), and requires separate arguments for thetwo cases δ ≥ a and δ < a.

If we minimise this inequality with respect to δ, just as in the case ofthe analytic proof, we find that there exists Cp,α,d > 0 so that

[Mαf ](a)12 ≤ Cp,α,d

(sup

0

The proof uses (G) and (U), and requires separate arguments for thetwo cases δ ≥ a and δ < a.If we minimise this inequality with respect to δ, just as in the case ofthe analytic proof, we find that there exists Cp,α,d > 0 so that

[Mαf ](a)12 ≤ Cp,α,d

(sup

0

Probabilistic Proof of HLS

Now we proceed to derive the classical HLS. We need theBurkholder-Davis-Gundy inequality for the martingale (Ma,αf (t), t ≥ 0)at t = a, namely there exists Cq ≥ 0 so that

||Mαf (a)||q ≤ Cq||[Mαf ](a)

12 ||q

Using the definition (Sa,αf )(x) = E(Mαf (a) | Ba = x) , the fact thatconditional expectation is an Lq-contraction and (Q) we deduce that:

||Sa,αf ||q ≤ ||Mαf (a)||q≤ Cq||[Mαf ](a)

12 ||q

≤ Cp,α,d∥∥∥∥ sup

0

Now apply (S) to find that

||Sa,αf ||q ≤ Cp,α,d‖f‖pqp ‖f‖

αpd

p .

≤ Cp,α,d‖f‖p,

since1q

+α

d=

1p.

Since this bound does not depend on a, letting a→∞ and applyingFatou’s lemma gives the result for f ∈ S(Rd ).The result extends to all f ∈ Lp by density of Schwartz space therein.


Thank You For Listening.Diolch Yn Fawr Am Wrando.


Documents

Analytic and Probabilistic Perspectives on the Hardy-Littlewood …HLSIneq).pdf · 2015. 5. 28. · in “Analytic Methods in Interdisciplinary Applications”, Springer Proc. Math