Startseite A trace problem for associate Morrey potentials
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A trace problem for associate Morrey potentials

  • Jie Xiao EMAIL logo
Veröffentlicht/Copyright: 28. September 2016

Abstract

As a solution to the restriction question for associate Morrey potentials (Question 1.1), this paper characterizes a Radon measure μ on n such that the Riesz operator Iα maps the associate Morrey spaces H1p,κHp,κ to the μ-induced Morrey spaces Lμq,λLμ,q,λ continuously. The discovered restriction/trace principle (Theorem 1.2) is brand-new, and its demonstration is non-trivial.

MSC 2010: 31B15; 42B35; 46E35

1 Introduction

1.1 Description of Question 1.1

Stemming from an extensive investigation of [2, 7, 9, 10, 45], this paper attempts to answer the coming-up-next restriction question, i.e., a trace problem for associate Morrey potentials. Namely, the Riesz images IαH1p,κIαHp,κ of the spaces H1p,κHp,κ associated with the Morrey spaces Lpp-1,κLpp-1,κ that have been proven to be of independent interest in geometric potential analysis (see, e.g., [5]) but also of great value in treating the local behaviour of an appropriate solution to certain elliptic partial differential equation or the incompressible Navier-Stokes system (see, e.g., [33, 29, 21, 22, 42, 43, 16, 34, 39, 38, 32, 28, 31, 20, 35, 11, 14, 15]):

Question 1.1.

Let μ be a Radon measure on the Euclidean space n. What ball condition must μ have so that the Riesz mapping fIαf is continuous (i.e., the potential Iαf is restricted to the support of μ) in the chain

H1p,κHp,κIαLμq,λLμ,q,λ

for the associate Morrey spaces H1p,κHp,κ and the μ-based Morrey spaces Lμq,λLμ,q,λ?

A resolution of the above question relies on the following fundamentals.

First of all, from a geometric measure theory perspective, a Radon measure on n is defined as a Borel regular measure assigning finite mass to any compact subset of n (cf. [18, p. 5]). For a given Radon measure μ on n and a nonnegative number β, denote by

|||μ|||βsup(x,r)n×(0,)r-βμ(B(x,r))

the β-th order ball variation of μ. Trivially, |||μ|||0 is the total variation of μ. Obviously, this variation |||μ|||β is finite whenever β{1,,n} and μ is the β-dimensional Hausdorff measure β restricted on a smooth manifold 𝕄n with β(𝕄)<.

Next, given α(0,n) the Riesz operator Iα acting on a Lebesgue ν-measurable function f in n is determined by

Iαf(x)=n|y-x|α-nf(y)dν(y),

whose constant multiple

u=(Γ(n-α2)πn22αΓ(α2))Iαf

defines an inverse for a power of the Laplace operator on n (cf. [5, p. 9]), i.e., such an singular integral u solves the α2-th order Laplace equation (-Δ)α2u=f using the Fourier transform.

After that, for (q,λ)[1,)×(0,n], let Lμ,q,λ be the weak Morrey space of all μ-measurable functions f on n with

fLμ,q,λ=sup(x,r,t)n×(0,)×(0,)(rλ-ntqμ({yB(x,r):|f(y)|>t}))1q<,

where B(x,r) is the Euclidean open ball with centre x and radius r. Naturally, as a proper subspace of Lμ,q,λ, the symbol Lμq,λ is employed to represent the Morrey space of all μ-measurable functions f with

fLμq,λ=sup(x,r)n×(0,)(rλ-nB(x,r)|f|qdμ)1q<.

In particular, Lμ,q=Lμ,q,n and Lμq=Lμq,n. Moreover, if μ=ν (the n-dimensional Lebesgue measure on n) and (q,λ)=(p,κ)[1,)×(0,n], then Lμ,q,λ and Lμq,λ will be simply written as Lp,κ and Lp,κ respectively, and the endpoint spaces Lp,n and Lp,n coincide with the weak-Lp space Lp and Lp, respectively; but if q=, then L,κ=L,κ=L. Furthermore, for p[1,] and pp/(p-1) with 1/(1-1) and 1/(-1), set

H1p,κ={ν-measurable functions f:fH1p,κsupgLp,κ1|nfgdν|<}[Lp,κ],
Hp,κ={ν-measurable functions f:fHp,κsupgLp,κ1|nfgdν|<}[Lp,κ].

Naturally, H1p,κ and Hp,κ are called the associate spaces [Lp,κ] and [Lp,κ] of the Morrey spaces Lp,κ and Lp,κ, respectively (cf. [36]). Especially, one has H11,κ=H1,κ=L1 (for 0<κn) and H1,nH,n=L (due to L1=L1,nL1,n, cf. [17]).

Last but not least, note that x|x|-κp is an element of Lp,κ for 0κ<n (cf. [46, p. 587]) and that x(1+|x|η)-1 belongs to Hp,κ for η>κp (cf. [24, Example 2.5]). So, the next implication seems to be very natural and motivative (cf. [26, p.85] and [6, Theorem 3.3]):

fH1p,κfHp,κfLqfLqfor q=pnn+(n-κ)(p-1)p.

In the above and below, XY stands for that there exists a constant c>0 such that XcY, XYX will be simply written as XY, and 1E represents the characteristic function of a set En.

1.2 Answering Question 1.1

Working on the above question plus the last implication one discovers the following trace and embedding principle for the associate Morrey potentials IαHp,κ (regarded as the space of tempered distributions having derivatives of order α in the associate Morrey space Hp,κ).

Theorem 1.2.

Assume (α,β,λ)(0,n)×(0,n]×(0,n].

  1. If

    1<q=β+λ-nn-α<,0<κn<β+λ2n,

    then Iα:H1,κLμq,λ is continuous if and only if

    |||μ|||α,β,λsupKn compact with ν(K)>0Iα1KLμq,λν(K)<,

    while Iα:H1,κLμ,q,λ is continuous if and only if |||μ|||β<.

  2. If

    1<q=p(β+λ-n)κ+p(n-α-κ)p<,0<κ<n<β+λ2n,

    then Iα:H1p,κHp,κLμq,λLμ,q,λ is continuous if and only if |||μ|||β<.

  3. If

    1<q=β+λ-nn-α-κ<,0<κ<n<β+λ2n,

    then Iα:H,κLμq,λ or Iα:H,κLμ,q,λ is continuous if and only if |||μ|||β<.

Interestingly, as an immediate application of Theorem 1.2, the next corollary completely describes the embedding of the k-th order associate Morrey–Sobolev space into Lμq,λ. More precisely, if

kf={(-Δ)k2ffor k even,(-Δ)k-12ffor k odd,

then

|f|Ik(|kf|)

for all C-functions f with compact support in n (see, e.g., [4, p. 393] or [30, p. 19]), and hence applying Theorem 1.2, together with a small modification of the argument used in the proof of [30, p. 71, Theorem] as p=1, produces the following corollary.

Corollary 1.3.

If (k,β,λ){1,,n-1}×(0,n]×(0,n], 1p,

1<q=β+λ-np-1κ+n-k-κ,0<κ<n<β+λ2n,

then

sup0<|kf|Hp,κfLμq,λ|kf|Hp,κ-1<|||μ|||β<.

Through exploiting the local nature of some indicator-like functions, introducing Lemmas 2.1, 3.1 and 4.1, and utilising the dual structure of a local Lμq-space, the non-trivial arguments for (i), (ii) and (iii) of Theorem 1.2 and their crucial Remarks 2.2, 3.2 and 4.2 will be arranged in Sections 2, 3 and 4, respectively. It is believed that the techniques developed in the forthcoming three sections can be applied to investigate a similar trace problem for the following Cordes–Nirenberg potentials (see, e.g., [26, 27, 16, 34]):

{Iαf:supxnn|f(y)|p|x-y|-γdν(y)<},1p<,0<γ<n.

Nevertheless, this investigation will be the subject of a future article.

2 Demonstration of Theorem 1.2 (i)

2.1 A lemma for Theorem 1.2(i)

To prove Theorem 1.2 (i), one needs the following result which improves [2, Theorem 5.1].

Lemma 2.1.

If 0<α<n, 0<λκn,

1p<κα,n-αp<βn,1q=p(β+λ-n)κ-αppnκ-αp,

then

|||μ|||β<sup0fLp,κIαfLμ,q,λfLp,κ-1<.

Proof.

Suppose

sup0fLp,κIαfLμ,q,λfLp,κ-1<.

For a given ball B(x,r), we compute (cf. [6])

1B(x,r)Lp,κrκp,rαinfyB(x,r)Iα1B(x,r)(y),μ(B(x,r))μ({yB(x,r):rαIα1B(x,r)(y)}).

Note that 1B(x,r) belongs to Lp,κ, so

μ(B(x,r))rαq+λ-nμ({yB(x,r):rαIα1B(x,r)(y)})rαq+λ-n1B(x,r)Lp,κqrκqp.

This inequality ensures that |||μ|||β< due to

0<β=(κp-1-α)q+n-λn.

Next, suppose |||μ|||β<. Without loss of generality, one may assume 0fLp,κ. For a positive number t and a given ball B(x,r) with B(x,r)c=nB(x,r) let 0fLp,κ, f1=f1B(x,r), f2=f1B(x,r)c,

Ft={yB(x,r):Iαf(y)>t},Ft,1={yB(x,r):Iαf1(y)>t},Ft,2={yB(x,r):Iαf2(y)>t}.

Then

μ(Ft)μ(Ft/2,1)+μ(Ft/2,2).

It remains to control the last two terms separately. Step 1: In order to prove the desired estimate

μ(Ft/2,1)|||μ|||βt-p(κ+β-n)κ-αprn-κfor all (x,r,t)n×(0,)×(0,),

one utilises the Fubini theorem to obtain

tμ(Ft,1)Ft,1Iαf1(z)𝑑μ(z)
=Ft,1(n|y-z|α-nf1(y)dν(y))dμ(z)
Ft,1(0(B(z,s)f1(y)dν(y))sα-n-1ds)dμ(z)
0(Ft,1B(z,s)f1(y)dν(y)dμ(z))sα-n-1ds
𝖩0(δ)+𝖩(δ),

where δ>0 will be determined later with

𝖩j(δ)={0δ(Ft,1B(z,s)f1(y)dν(y)dμ(z))sα-n-1dsfor j=0,δ(Ft,1B(z,s)f1(y)dν(y)dμ(z))sα-n-1dsfor j=.

First, if p>1, then an application of the Hölder inequality with p=p/(p-1) plus fLp,κ derives

Ft,1B(z,s)f1(y)𝑑ν(y)𝑑μ(z)nμ(B(y,s)Ft,1)(1B(x,r)f)(y)𝑑ν(y)
(B(x,r)(f(y))p𝑑ν(y))1p(B(x,r)(μ(B(y,s)Ft,1))p𝑑ν(y))1p
rn-κpfLp,κ(B(x,r)(μ(B(y,s)Ft,1))(μ(B(y,s)Ft,1))p-1𝑑ν(y))1p
rn-κpfLp,κsβp|||μ|||β1p(B(x,r)μ(B(y,s)Ft,1)𝑑ν(y))1p
rn-κpfLp,κsβp|||μ|||β1p(B(x,r)1B(y,s)(z)𝑑ν(z))1p(μ(Ft,1))1p
rn-κpfLp,κsβp+np|||μ|||β1p(μ(Ft,1))1p.

At the same time, if p=1, then a slight modification of the above estimation gives

Ft,1B(z,s)f1(y)𝑑ν(y)𝑑μ(z)rn-κfL1,κsβ|||μ|||β.

Second, for 𝖩0(δ) one uses the foregoing inequalities to gain that if p>1, then

𝖩0(δ)=0δ(Ft,1B(z,s)f1(y)𝑑ν(y)𝑑μ(z))sα-n-1𝑑s
rn-κpfLp,κ|||μ|||β1p(μ(Ft,1))1p0δsβp+np+α-n-1𝑑s
rn-κpδα+β-npfLp,κ|||μ|||β1p(μ(Ft,1))1p,

and if p=1, then

𝖩0(δ)rn-κδα+β-nfL1,κ|||μ|||β.

Of course, the condition β>n-αp is needed for the treatment of 𝖩0(δ). Meanwhile, for 𝖩(δ) one utilises fLp,κ to deduce

𝖩(δ)fLp,κδsα-κp-1μ(Ft,1)𝑑sδα-κpμ(Ft,1)fLp,κ.

Here the condition αp<κ has been used.

Third, putting the estimates for 𝖩0(δ) and 𝖩(δ) together gives

tμ(Ft,1){rn-κpδα+β-npfLp,κ|||μ|||β1p(μ(Ft,1))1p+δα-κpμ(Ft,1)fLp,κfor p>1,rn-κδα+β-nfL1,κ|||μ|||β+δα-κμ(Ft,1)fL1,κfor p=1.

Now, choosing

δ=(μ(Ft,1)rn-κ|||μ|||β)1β+κ-n

gives

μ(Ft,1)|||μ|||βrn-κ(t-1fLp,κ)p(n-β-κ)αp-κ.

Step 2: In order to prove the estimate

μ(Ft/2,2)|||μ|||βrn-κ(t-1fLp,κ)p(κ+β-n)κ-αpfor all (x,r,t)n×(0,)×(0,),

one can borrow the idea used in Step 1 to get

tμ(Ft,2)𝖪0(δ)+𝖪(δ),

where

𝖪j(δ)={0δ(Ft,2B(z,s)f2(y)𝑑ν(y)𝑑μ(z))sα-n-1𝑑sfor j=0,δ(Ft,2B(z,s)f2(y)𝑑ν(y)𝑑μ(z))sα-n-1𝑑sfor j=.

Notice that

yB(x,r)c and zB(y,s)Ft,2yB(x,r+s).

So, if p>1, then

Ft,2B(z,s)f2(y)dν(y)dμ(z)B(x,r+s)f(y)(n1B(y,s)(z)1Ft,2(z)dμ(z))dν(y)
(B(x,r+s)(f(y))pdν(y))1p(B(x,r+s)(n1B(y,s)(z)1Ft,2(z)dμ(z))pdν(y))1p
fLp,κ(r+s)n-κp(μ(Ft,2))1psn+β-np|||μ|||β1p,

and if p=1, then

Ft,2B(z,s)f2(y)𝑑ν(y)𝑑μ(z)fL1,κ(r+s)n-κsβ|||μ|||β,

whence

𝖪0(δ){fLp,κδα+β-np(δ+r)n-κp(μ(Ft,2))1p|||μ|||β1pfor p>1,fL1,κδα+β-n(δ+r)n-κ|||μ|||βfor p=1.

Similarly, one has

𝖪(δ)fLp,κδsα-κp-1μ(Ft,2)𝑑sδα-κpμ(Ft,2)fLp,κ.

Putting together the estimates for 𝖪0(δ) and 𝖪(δ) yields

tμ(Ft,2){fLp,κ(δα+β-np(δ+r)n-κp(μ(Ft,2))1p|||μ|||β1p+δα-κpμ(Ft,2))for p>1,fL1,κ(δα+β-n(δ+r)n-κ|||μ|||β+δα-κμ(Ft,2))for p=1.

Below is a treatment of two cases: δr and δr. Case δr. Under this situation, one has

{δα+β-np(δ+r)n-κp(μ(Ft,2))1p|||μ|||β1p+δα-κpμ(Ft,2)δα+β-κp(μ(Ft,2))1p|||μ|||β1p+δα-κpμ(Ft,2)for p>1,δα+β-n(δ+r)n-κ|||μ|||β+δα-κμ(Ft,2)δα+β-κ|||μ|||β+δα-κμ(Ft,2)for p=1,

and hence selecting

δ(μ(Ft,2)|||μ|||β)1β

yields

(μ(Ft,2)|||μ|||β)κ-αpβpt-1fLp,κ.

Taking the p(β+κ-n)/(κ-αp)-th power on this last estimate gives

μ(Ft,2)|||μ|||β(t-1fLp,κ)p(β+κ-n)κ-αp(μ(Ft,2)|||μ|||β)n-κβp(t-1fLp,κ)p(β+κ-n)κ-αp|||μ|||βrn-κ.

Case δr. Under this situation, one has

{δα+β-np(δ+r)n-κp(μ(Ft,2))1p|||μ|||β1p+δα-κpμ(Ft,2)(rn-κpδn-βp-α)(μ(Ft,2))1p|||μ|||β1p+μ(Ft,2)δκp-αfor p>1,δα+β-n(δ+r)n-κ|||μ|||β+δα-κμ(Ft,2)δα+β-nrn-κ|||μ|||β+δα-κμ(Ft,2)for p=1.

By selecting

δ(rn-κ|||μ|||βμ(Ft,2))1n-κ-β,

one achieves

tfLp,κ(rn-κ|||μ|||βμ(Ft,2))α-κn-κ-β,

thereby arriving at

μ(Ft,2)|||μ|||β(t-1fLp,κ)p(κ+β-n)κ-αprn-κ.

Now, a combination of the preceding two steps gives

μ(Ft)|||μ|||β(t-1fLp,κ)p(κ+β-n)κ-αprn-κfor all (x,r,t)n×(0,)×(0,).

Since λκ, one gets

q=p(β+λ-n)κ-αpp(β+κ-n)κ-αp=q~,

thereby implying

(μ(Ft))1q~=(μ(Ft))1q(μ(Ft))1q~-1q(μ(Ft))1qrβ(1q~-1q)|||μ|||β1q~-1q.

As a result, one establishes

μ(Ft)|||μ|||βt-qrn-λfLp,κqfor all (x,r,t)n×(0,)×(0,),

as desired. ∎

2.2 Proof of Theorem 1.2 (i)

On the one hand, since

1KH1,κ=1KL1=ν(K)for all compact Kn,

one has

sup0fH1,κIαfLμq,λfH1,κ-1<|||μ|||α,β,λ<.

Conversely, assume |||μ|||α,β,λ<. Without loss of generality, one may assume 0fH1,κ=L1. Then such a function f can be written as f=limjfj (cf. [44, p. 88]), where

fj(y)=i=1j2j(j-1)2-j1Ki,j(y)+j1Kj(y),Ki,j={yn:(i-1)2-jf(y)<i2-j},Kj={yn:f(y)j}.

Thus, applying the Minkowski inequality for Lμq(B(x,r)) (all Lμq-integrable functions on a given ball B(x,r)) derives

IαfjLμq(B(x,r))i=1j2j(j-1)2-jIα1Ki,jLμq(B(x,r))+jIα1KjLμq(B(x,r))
rn-λq(i=1j2j(j-1)2-jIα1Ki,jLμq,λ+jIα1KjLμq,λ)
rn-λq|||μ|||α,β,λ(i=1j2j(j-1)2-jν(Ki,j)+jν(Kj))
rn-λq|||μ|||α,β,λfjL1.

Now, sending j to leads to

IαfLμq(B(x,r))rn-λq|||μ|||α,β,λfL1,

whence

IαfLμq,λ|||μ|||α,β,λfH1,κ|||μ|||α,β,λfH11,κ.

On the other hand, an application of Lemma 2.1 with 0λκ=n, p=1 and q=(β+λ-n)/(n-α)1 gives

|||μ|||β<sup0fL1IαfLμ,q,λfL1-1<.

So, if |||μ|||β<, then Iα:H1,κLμ,q,λ is continuous, and vice versa.

Remark 2.2.

In the first equivalence of Theorem 1.2 (i), |||μ|||α,β,λ< implies easily |||μ|||β< (cf. the beginning of the argument for the proof of Lemma 2.1). However, if λ=n, μ=ν (ensuring β=n) and q=n/(n-α), then

|||μ|||α,β,λIα1B(0,1)Ln/(n-α)(ν(B(0,1)))-1=,

and hence IαL1 is not subset of Lnn-α according to the first equivalence of Theorem 1.2 (i).

Moreover, as a key tool of verifying the second equivalence of Theorem 1.2 (i), Lemma 2.1 can be used, along with the well-known layer-cake formula, to imply the following. If 0<α<n, 0<λκn,

1p<κα,n-αp<βn,1q^<q=p(β+λ-n)κ-αp,|||μ|||β<,fLp,κ,

then

B(x,r)|Iαf(y)|q^dμ(y)=(0δ+δ)μ({yB(x,r):|Iαf(y)|>t})dtq^|||μ|||β(rβδq^+fLp,κqrn-λδq^-q),

and hence

B(x,r)|Iαf(y)|q^dμ(y)|||μ|||βfLp,κq^rβ+q^(n-λ-β)q,

through choosing δ=(rn-λ-βfLp,κq)1q. Consequently,

IαfLμq^,λ^with λ^=n-β+q^(κ-αp)p<n.

In other words,

|||μ|||β<sup0fLp,κIαfLμq^,λ^fLp,κ-1<.

Naturally, if q^=p in the last implication, then

|||μ|||β<sup0fLp,κIαfLμp,λ~=n-β+κ-αpfLp,κ-1<,

whose speciality μ=ν recovers the second imbedding of [8, Theorem 16 (i)] which in turn yields the following new k-th order Morrey–Poincaŕe inequality:

fLp,κ-kp|kf|Lp,κfor all |kf|Lp,κ>0 with kp<κn.

3 Demonstration of Theorem 1.2 (ii)

3.1 A lemma for Theorem 1.2 (ii)

To demonstrate Theorem 1.2(ii), one needs an equivalent characterisation of |||μ|||β.

Lemma 3.1.

For (β,γ)[0,n]×[0,n], let

|||μ|||β,γsup(x,y,r)n×n×(0,)r-βB(x,r)|z-y|-γdμ(z)

be the (β,γ)-order ball variation of μ. If γβ, then

|||μ|||β|||μ|||β-γ,γ.

Moreover, if γ<β, then

|||μ|||β|||μ|||β-γ,γ.

But, if 0<γ=β, then there exists a Radon measure μ0 such that

|||μ0|||β<,|||μ0|||β-γ,γ=.

Proof.

On the one hand, one utilises the above definition to discover that if γβ, then

|||μ|||β-γ,γsup(x,r)n×(0,)r-(β-γ)B(x,r)|z-x|-γdμ(z)sup(x,r)n×(0,)r-βμ(B(x,r))=|||μ|||β.

On the other hand, if γ<β, then [29, Lemma 1.27] is employed to estimate

|||μ|||β-γ,γsup(x,y,r)n×n×(0,)r-(β-γ)(0r+r)μ(B(x,r)B(y,s))s-γ-1ds
|||μ|||βsup(x,y,r)n×n×(0,)r-(β-γ)(0rsβ-γ-1𝑑s+rrβs-γ-1𝑑s)|||μ|||β.

Unfortunately, this last estimation is not valid for γ=β=j (a natural number less than n). In fact, for the origin {0}n-j of n-j, the Radon measure dμj=1j×{0}n-jdν satisfies

|||μj|||j<,|||μj|||0,j=.

3.2 Proof of Theorem 1.2 (ii)

The argument comprises two parts.

Part 1:

Verify that Iα maps Hp,κ to either Lμq,λ or Lμ,q,λ continuously if and only if |||μ|||β<.

Since Lμq,λLμ,q,λ, it is enough to check

sup0fHp,κIαfLμ,q,λfHp,κ-1<|||μ|||β<sup0fHp,κIαfLμq,λfHp,κ-1<.

To this end, recall that (cf. [6, Theorem 3.3]) if fHp,κ, then

fHp,κinf{j|cj|:f=jcjaj},

where aj is a (p,n-κ)-atom, that is, aj is supported on a ball Bjn and satisfies

ajLp(ν(Bj))κ-nnp,where p=pp-1.

So, for a given ball B(x,r), the function

fx,r=(ν(B(x,r)))κ-nnp-1p1B(x,r)

belongs to Hp,κ. Consequently,

fx,rHp,κ1,rα-n+κpinfyB(x,r)Iαfx,r(y),μ(B(x,r))μ({yB(x,r):rα-n+κpIαfx,r(y)}).

Now, if

sup0fHp,κIαfLμ,q,λfHp,κ-1<,

then Iαfx,rLμ,q,λ1, and hence

rλ-n+q(α-n+κp)μ(B(x,r))rλ-n+q(α-n+κp)μ({yB(x,r):rα-n+κpIαfx,r(y)})Iαfx,rLμq,λq1.

This in turn validates |||μ|||β<.

Next, suppose |||μ|||β<. The desired continuity follows from a consideration of the forthcoming two cases. Case λ<n. Since 0<n-λ<β2n-λ, Lemma 3.1 is utilised to ensure |||μ|||n-λ,β-(n-λ)<. Let f=jcjaj belong to Hp,κ. Given a ball B(x,r) and the conjugate index q=q/(q-1), let Lμq(B(x,r)) be the μ-measurable functions g with

gLμq(B(x,r))=(B(x,r)|g|qdμ)1q<.

Then, from duality and the Fubini theorem it follows that

(B(x,r)|Iαf|qdμ)1q=supgLμq(B(x,r))1|B(x,r)gIαfdμ|
=supgLμq(B(x,r))1|B(x,r)g(z)nf(y)|z-y|α-ndν(y)dμ(z)|
=supgLμq(B(x,r))1|n(B(x,r)g(z)|z-y|α-ndμ(z))f(y)dν(y)|
supgLμq(B(x,r))1j|cj|n(B(x,r)|g(z)||z-y|α-ndμ(z))|aj(y)|dν(y).

Next, choose the parameter pair {α1,α2} such that

n-β<α1=2n-β-λ<n,0<α2<n,αq=α1+(q-1)α2.

Then the Hölder inequality on the supporting ball Bj of aj, with qp, yields

n(B(x,r)|g(z)||z-y|α-ndμ(z))|aj(y)|dν(y)
Bj(B(x,r)|g(z)||z-y|α-ndμ(z))|aj(y)|dν(y)
(Bj(B(x,r)|g(z)||z-y|α-ndμ(z))qdν(y))1q(Bj|aj|qdν)1q
(ν(Bj))1q-1p(Bj|aj|pdν)1p(Bj(B(x,r)|g(z)||z-y|α-ndμ(z))qdν(y))1q
(ν(Bj))1q-1+κpn(Bj(B(x,r)|g(z)||z-y|α-ndμ(z))qdν(y))1q.

Another application of the Hölder inequality, along with |||μ|||n-λ,β-(n-λ)<, gives

(B(x,r)|g(z)||z-y|α-ndμ(z))q(B(x,r)|g(z)|q|z-y|n-α2dμ(z))(B(x,r)dμ(z)|z-y|n-α1)q-1
(rn-λ|||μ|||n-λ,β-(n-λ))q-1B(x,r)|g(z)|q|z-y|n-α2𝑑μ(z).

Now, the foregoing two groups of estimates are put together, and using the Fubini theorem and [29, Lemma 1.30], as well as the equalities

q=p(β+λ-n)κ-p(α+κ-n),α2=q(α-2n-β-λq),1q+α2nq=1-κnp,

gives

n(B(x,r)|g(z)||z-y|α-ndμ(z))|aj(y)|dν(y)
(ν(Bj))1q-1+κpnrn-λq|||μ|||n-λ,β-(n-λ)1q(BjB(x,r)|g(z)|q|z-y|n-α2𝑑μ(z)𝑑ν(y))1q
(ν(Bj))1q-1+κpnrn-λq|||μ|||n-λ,β-(n-λ)1q(B(x,r)(Bjdν(y)|z-y|n-α2)|g(z)|qdμ(z))1q
(ν(Bj))1q-1+κpn+α2nqrn-λq|||μ|||n-λ,β-(n-λ)1q(B(x,r)|g(z)|qdμ(z))1q
rn-λq|||μ|||n-λ,β-(n-λ)1q(B(x,r)|g(z)|qdμ(z))1q.

This in turn yields

(B(x,r)|Iαf|qdμ)1qrn-λq|||μ|||n-λ,β-(n-λ)1qj|cj|,

and so by Lemma 3.1,

IαfLμq,λ|||μ|||n-λ,β-(n-λ)1qfHp,κ|||μ|||β1qfHp,κ.

Case λ=n. Under this situation, one has Lμq,λ=Lμq. So, a particular care is needed. However, the initial issue is to take such a parameter pair (α1,α2) such that

n-β<α1<n,0<α2<n,αq=α1+(q-1)α2.

Fix a point x0 in the supporting ball B of a given (p,n-κ)-atom a and select B(x0,r0) so that ν(B(x0,r0))=ν(B) and BB(x0,2r0). One can use [29, Lemma 1.30] to get

B|y-z|α2-ndν(z)(ν(B))α2n,

and then use the Fubini theorem, [29, Lemma 1.27] and the Hölder inequality with qp to compute

B(x0,2r0)(B|a(z)|q|y-z|α1-ndν(z))dμ(y)=B(B(x0,2r0)|y-z|α1-ndμ(y))|a(z)|qdν(z)
B(02r0+2r0μ(B(x0,2r0)B(z,s))sα1-n-1ds)|a(z)|qdν(z)
|||μ|||β(B|a|qdν)r0β+α1-n
|||μ|||β(B|a|pdν)qp(ν(B))1-qpr0β+α1-n
|||μ|||β(ν(B))β+α1n+qκn(1-1p)-q.

Now, one utilises α=α1/q+(q-1)α2/q, the Hölder inequality and q=βp/(p(n-α)-(p-1)κ) to find the following estimation:

B(x0,2r0)(Iα|a|)qdμB(x0,2r0)(B|a(y)|qdν(y)|y-z|n-α1)(Bdν(y)|y-z|n-α2)q-1dμ(z)
|||μ|||β(ν(B))β+α1n+qκn(1-1p)-q+α2(q-1)n
|||μ|||β.

Meanwhile, a combination of the Minkowski inequality, [29, Lemma 1.27], the Hölder inequality and the fact that q=βp/(p(n-α)-(p-1)κ)>β/(n-α) yields

(B(x0,2r0)c(Iα|a|)qdμ)1qB(B(x0,2r0)c|y-z|(α-n)qdμ(y))1q|a(z)|dν(z)
B(2r0s(α-n)q-1μ(B(x0,2r0)cB(z,s))𝑑s)1q|a(z)|𝑑ν(z)
(B|a|pdν)1p(ν(B))1-1p(ν(B))αq+βnq-1|||μ|||β1q
(ν(B))αq+βnq-1p+κnp-1p|||μ|||β1q
|||μ|||β1q.

Putting the integrals B(x0,2r0)()𝑑μ and B(x0,2r0)c()𝑑μ together implies

Iα|a|Lμq|||μ|||β1q.

Upon expressing fHp,κ as f=jcjaj, one uses the foregoing estimate and the Minkowski inequality to establish the following inequality:

IαfLμqj|cj|Iα|aj|Lμq|||μ|||β1qj|cj|.

Consequently, one finds

IαfLμq|||μ|||β1qfHp,κ.

Part 2:

Verify that Iα maps H1p,κ to either Lμq,λ or Lμ,q,λ continuously if and only if |||μ|||β<.

Thanks to both H1p,κHp,κ and Lμq,λLμ,q,λ, using Part 1, one has that |||μ|||β< implies that Iα maps H1p,κ to either Lμq,λ or Lμ,q,λ continuously. So, it remains to check that if Iα:H1p,κLμ,q,λ is continuous, then |||μ|||β<. To this end, utilising [20, Lemma 3.1] one gets that fH1p,κ if and only if there exist a sequence of numbers {cj} and a sequence of functions {aj} such that

f=jcjajwith fH1p,κj|cj|<,

for which there exists a ball Bj containing the support of aj such that

(ν(Bj))n-κpn0(ν({xn:1Bj(x)|aj(x)|>t}))1p𝑑t1.

Upon choosing

fx,r=(ν(B(x,r)))κ-nnp-1p1B(x,r)

for a given ball B(x,r), one gets

fx,rH1p,κ1.

Now, using the induced condition

Iαfx,rLμ,q,λfx,rH1p,κ

and the calculation (with fx,r) done at the beginning of Part 1, one can readily reach |||μ|||β<.

Remark 3.2.

If Hp,0 or Hp,n is defined as the space of all ν-measurable functions f with

fHp,κinf{j|cj|:f=jcjaj}<,

and aj is a (p,n)-atom or a (p,0)-atom, i.e., aj is supported on a ball Bjn and satisfies ajLp(ν(Bj))-1p or ajLp1, then the above argument is still valid for Hp,0 and Hp,n, and hence Theorem 1.2 can be extended to Hp,0κn. Therefore, not only this extension gives a new approach to [1, Theorem 2], but also the limiting case α0 of Theorem 1.2 (ii) for λ=n and μ=ν actually gives (cf. [26, p.85])

Hp,κLqwith q=pnκ+p(n-κ).

Furthermore, one has the following cases:

  1. If λ=n and 0<κ<n, then Theorem 1.2 (ii) recovers [10, Theorem 2.4 (ii)].

  2. If μ=ν (forcing β=n) and λ=κ in Theorem 1.2 (ii), then

    IαfLq~=pκ/(κ+p(n-α-κ)),κfHp,κfor all fHp,κ,

    which may be regarded as a sort of the associate representation of the first embedding of [8, Theorem 16 (ii)] (cf. [2, Theorem 3.1])

    IαfLq¯=pκ/(κ-αp),κfLp,κfor all fLp,κ,

    due to 1/q~+1/q¯+κ/n=1.

  3. If p=q=(β+λ-n-κ)/(n-α-κ), then Theorem 1.2 (ii) is utilised to deduce

    |||μ|||β<IαfLμp,λ¯=n+κ-β+p(n-α-κ)fHp,κfor all fHp,κ,

    which may be treated as a kind of the associate form of the last equivalence in Remark 2.2 thanks to (λ¯-λ~)/(pn)+κ/n=1.

4 Demonstration of Theorem 1.2 (iii)

4.1 A lemma for Theorem 1.2 (iii)

In accordance with [6, Theorem 2.3], [36, Theorem 4.1] and [12, Theorem 2.7], if p(1,) and γ[0,n), then

Hp,γ={fLlocp:fHp,γinfwB1(n-γ)(n|f|pw1-pdν)1p<},

where B1(n-γ) comprises all nonnegative functions w with

nw𝑑Λn-γ()0Λn-β()({yn:w(y)>t})𝑑t1,

and, for En, the symbol Λn-γ()(E) expresses the Hausdorff capacity of E with order n-β, i.e.,

Λn-γ()(E)=infjrjn-γ,

where the infimum ranges over all countable coverings of E by open balls of radius rj.

To validate Theorem 1.2 (iii), it is required to push the above description of Hp,γ to the limiting case p=, that is, to characterise H,γ by means of Λn-γ() as seen below.

Lemma 4.1.

For γ(0,n), let L1,γ be the Morrey space of all signed Radon (locally finite regular signed Borel) measures μ whose total variation measures |μ|μ~ obey |||μ~|||β<, and set LΛγ()1 be the class of all Λγ()-quasi continuous functions f on Rn for which

fLΛγ()1n|f|dΛγ()<.

Then

𝖫1,γ={μ signed Radon measure:μ[LΛγ()1]supgLΛγ()11|ngdμ|<}[LΛγ()1].

Moreover, if L0 is the class of L-functions with compact support in Rn, then

H,γ={fL0:f𝖧,γinfwB1(n-γ)fw-1L<}𝖧,γ.

Consequently, LΛn-γ()1 exists as a subspace of H,γ.

Proof.

The first identity immediately follows from [3, Proposition 1]. The second needs a demonstration.

Suppose f𝖧,γ. Then fL0 and f𝖧,γ<. Now for any gL1,κ with gL1,κ1, one has (cf. [6, Theorem 2.2])

gL1,γ=supwB1(n-γ)n|f|wdν1.

This in turn implies

|nfgdν|n|g|w|f|w-1dνgL1,γfw-1Lfor all wB1(n-γ),

and so fH,γ. As a consequence, 𝖧,γH,γ.

Since L1,γ is just the associate space of H,γ, in short, L1,γ=[H,γ], in order to prove H,γ𝖧,γ it is enough to check

[𝖧,γ]{f ν-measurable function:f[𝖧,γ]supg𝖧,γ1|nfgdν|<}L1,γ.

Now, if f lies in [𝖧,γ], then for a given ball B(x0,r0) and any function g supported in B(x0,r0) one has

g𝖧,γ=infwB1(n-γ)gw-1L(B(x0,r0)).

Since g vanishes in nB(x0,r0), one can choose

w0=r0β-n1B(x0,r0)B1(n-β),

and then get

g𝖧,γgL(B(x0,r0))r0n-γ.

Upon taking

g0(x)={1B(x0,r0)|f(x)|f(x)-1for f(x)0,0for f(x)=0,

one achieves

B(x0,r0)|f|dν=|nfg0dν|f[𝖧,γ]g0𝖧,γf[𝖧,γ]r0n-γ,

thereby reaching fL1,γ, and thus [𝖧,γ]L1,γ, as desired.

To see that LΛn-γ()1 is contained in H,γ, suppose fLΛn-γ()1. Then w=|f|fLΛn-γ()1-1 satisfies

nw𝑑Λn-γ()=0Λn-γ()({yn:|f(y)|fLΛn-γ()1-1>t})𝑑t1,

i.e., wB1(n-γ), and hence

fH,γfw-1LfLΛn-γ()1,

that is, fH,γ, as desired. ∎

4.2 Proof of Theorem 1.2 (iii)

For a given nonnegative Radon measure μ, according to Lemma 4.1 it is better to begin with proving that

Iα:LΛn-κ()1Lμq,λ

is continuous if and only if

Iα:LΛn-κ()1Lμ,q,λ

is continuous if and only if μ𝖫1,β. As a matter of fact, it is enough to check

|||μ|||β<IαfLμq,λfLΛn-κ()1for all fLΛn-κ()1.

To this end, assume |||μ|||β<. In accordance with [6, Remark 3.4], one has

fLΛn-κ()1inf{j|cj|:f=jcjaj}<,

where aj is a (,n-κ)-atom, i.e., aj is supported on a ball Bjn and satisfies ajL(ν(Bj))κ-nn.

Next, for the given parameter q=(β+λ-n)/(n-α-κ)>1 and a function f=jcjajLΛn-κ()1, one can slightly modify the second part of the argument used for the proof of Theorem 1.2 (ii) (with not only putting p=1 when p= but also noticing that ajL(ν(Bj))κ-nn) to obtain

IαajLμq,λ|||μ|||β1q,

which, by using the Minkowski inequality, implies

IαfLμq,λj|cjIαajLμq,λ|||μ|||β1qj|cj|.

Consequently, one finds

IαfLμq,λ|||μ|||β1qfLΛn-κ()1.

Finally, if Iα:H,κLμq,λ is continuous, then Iα:H,κLμ,q,λ is continuous, and hence |||μ|||β< follows from choosing 1B(x,r) as the test function in H,κ. Now, if |||μ|||β<, then the previous analysis ensures that Iα:LΛn-κ()1Lμq,λ is continuous. Hence, for a given ball B(x,r), any wB1(n-κ) and any gLμq(B(x,r)), one applies the Fubini theorem to compute

(B(x,r)|Iαf|qdμ)1q=supgLμq(B(x,r))1|B(x,r)(Iαf)gdμ|
supgLμq(B(x,r))1B(x,r)(n|f(y)||y-z|α-ndν(y))|g(z)|dμ(z)
=supgLμq(B(x,r))1n(B(x,r)|g(z)||y-z|α-ndμ(z))|f(y)|dν(y)
fw-1LsupgLμq(B(x,r))1n(B(x,r)|g(z)||y-z|α-ndμ(z))w(y)dν(y)
=fw-1LsupgLμq(B(x,r))1B(x,r)(Iαw)|g|𝑑μ
=fw-1LIαwLμq(B(x,r))
fw-1Lrn-λqIαwLμq,λ
fw-1Lrn-λq|||μ|||β1qwLΛn-κ()1
rn-λqfw-1L|||μ|||β1q.

Therefore, one utilises Lemma 4.1 to achieve

IαfLμq,λ|||μ|||β1qf𝖧,κ|||μ|||β1qfH,κ.

That is, Iα:H,κLμq,λ is continuous.

Remark 4.2.

Perhaps, it is worth pointing out that κ=0 is suitably allowable in Theorem 1.2 (iii). To see this nature more clearly, see [3, Proposition 5] which says, as Fefferman–Stein’s predual of BMO (John Nirenberg’s space of functions of bounded mean oscillation on n; see, e.g., [25, 19, 40]), that if H1 stands for the real 1-Hardy space comprising all L1-functions f for which the {1,,n}j-Riesz transform

yRjf(y)=limϵ0Γ(n+12)π-n+12B(0,ϵ)czj|z|-n-1f(y-z)𝑑ν(z)

belongs to L1, then

IαfLΛn-α()1fH1=fL1+j=1nRjfL1for all fH1,

and hence, by [3, Corollary],

|nf𝑑μ||||μ|||n-αfH1for all (μ,f)𝖫+1,n-α×H1,

where 𝖫+1,n-α consists of all nonnegative elements in 𝖫1,n-α. Also, recall that (cf., e.g., [23, Theorem 6.6.10]) fH1 if and only if there exist a sequence of (,n)-atoms {aj} with na𝑑ν=0 and a sequence of numbers {cj} such that

f=jcjaj,fH1j|cj|<.

So, this leads to an investigation of the endpoint κ=0 of Theorem 1.2 (iii) for q=(β+λ-n)/(n-α)>1.

On the one hand, if Iα:H1Lμq,λ is continuous, then

IαaLμq,λaH1for all aH1.

For B(x-,r)=B(x,r), B(x+,r) (whose distance to B(x,r) is 3r, i.e., |x-x+|=5r) and

a=1B(x-,r)-1B(x+,r)ν(B(x,6r))H1with aH11,

via a simple geometric computation, one gets

Iαa(y)=(ν(B(x,6r)))-1(B(x-,r)dν(z)|z-y|n-α-B(x+,r)dν(z)|z-y|n-α)rα-nfor all yB(x,r),

whence finding

1IαaLμq,λqrq(α-n)+λ-nμ(B(x,r))r-βμ(B(x,r))|||μ|||β<.

On the other hand, assume |||μ|||β<. Upon writing f=j=1cjajH1 and observing that the argument for the proof of Theorem 1.2 (iii) actually reveals

IαajLμq,λ|||μ|||β1q,

one utilises the Minkowski inequality to achieve

IαfLμq,λj|cj|IαajLμq,λ|||μ|||β1qj|cj||||μ|||β1qfH1.

Surprisingly, this last estimation for μ=ν, i.e., IαH1Lλn-α,λ, is an associate form of [26, Theorem 2], which implies IαH1Hp,pα, since λ=pα amounts to p=λ/(λ-α). When λ=n, both go back to the well-known Stein–Weiss embedding IαH1Lnn-α, see, e.g., [40, 13, 37, 41].

Award Identifier / Grant number: 202979463102000

Funding statement: The author is in part supported by NSERC of Canada (202979463102000).

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Received: 2016-03-27
Accepted: 2016-08-18
Published Online: 2016-09-28
Published in Print: 2018-08-01

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