WU Di， DENG Yangkendi， YU Dandan， YAN Dunyan?

(1 School of Science, Zhejiang University of Science and Technology，Hangzhou 310023, China；2 School of Mathematical Sciences, University of Chinese Academy of Sciences，Beijing 100049, China)

Abstract In this paper, we give a novel proof for the following equality

for f∈Lp,q(X,μ) with 0

Keywords limiting behavior; distribution functions; Lorentz spaces

Letfbe a measurable function on a measure space (X,μ) and 0

(1)

wheredf(α)=μ{x:|f(x)|>α} is the distribution function off. The set of allfwith |f|Lp,q<∞ is denoted byLp,q(X,μ) and is called the Lorentz space with indicespandq.

There are many simple properties of Lorentz space. For 0

Lp,p(X,μ)=Lp(X,μ),

and

Lp,∞(X,μ)=WLp(X,μ).

For 0

‖f‖Lp,r≤cp,q,r‖f‖Lp,q.

More properties of Lorentz space can be found in Refs.[1-4].

The limiting property of distribution functions ofLpfunctions has been proved in Ref.

[5] for 1≤p<∞. That is, whenf∈Lp,p(n), we have that

(2)

However, forf∈L1(n), we know thatMf∈L1,∞(n), whereMis the maximal operator. In Ref.

[6], P. Janakiraman has proved that

This means that the limiting equality of distribution functions does not hold forL1,∞function.

It is well-known that the following inclusion relation

Lp,p(n)Lp,q(n)Lp,∞(n)

holds for 0

It is worth investigating the limiting property of distribution functions off∈Lp,q(X,μ).

1 Main results and proof

Now we formulate our main theorem.

Theorem1.1Letμbe aσ-finite positive measure on someσ-algebra in setX. Forf∈Lp,q(X,μ) with 0

df(α):=μ({x∈X:|f(x)|>α}).

Then we have

(3)

ProofFirst, we prove that

If the conclusion does not hold, then we conclude that

or

and

forn∈.Obviously, there exists a strictly decreasing subsequence, still denoted as {xn}.

Sincedfis a decreasing function, it follows that

(4)

Settingλn=(xn+1/xn)q∈(0,1), the basic principle of mathematics analysis tells us that the infinite product ∏λnconverges to a nonzero number if and only if ∑(1-λn) converges. However we deduce that

It implies from (4) that

‖f‖Lp,q=∞.

This leads to contradiction.

Next, we prove

such that

and

Therefore, we conclude that

(5)

Note that the infinite product ∏(xn/xn+1)qdiverges to 0. Forα>0,αpdf(α)≥0 implies

That leads to (3).

□

Now, we begin to prove that the functionαpin (2) can not be improved. For this purpose, we first give the following lemma which characterize the special property of some function.

Lemma1.1Letg:(0,+∞)→[0,+∞) be a right-continuous non-increasing function. Then there exists a functionf:n→[0,+∞] such thatdf(α)=g(α) for anyα∈(0,+∞).

ProofLet

(6)

wherevn=m(B(0,1)) andB(x,r) denotes a ball with the center atxand the radiusr.

Then we merely need to prove that

(7)

holds, for anyα∈(0,+∞).

Assume

Note thatgis a non-increasing function. We conclude from definition offin (6) that

≤α.

(8)

Thus the inequality (8) implies that the set of the left hand of (7) is contained in the right hand.

Assume

Note thatgis a right-continuous function. Then there existsα′>αsuch that

Then we conclude that

=α′>α.

This means that

In a word, (7) holds.

□

As an application of Lemma 1.2, we prove that the functionαpcan not be improved for some sense.

Theorem1.2Suppose 0

(9)

there exists a functionf∈Lp,q(n) such that

h(αk)≥4k

and

xk+1

fork∈andαk∈[2-1xk,xk).

Define a functiong:(0,+∞)→[0,+∞) as

(10)

By Lemma 1.1, there exists a functionf:n→[0,+∞] such thatdf(α)=g(α) for anyα∈(0,+∞).And we have from the definition ofgin (10) that

This impliesf∈Lp,q(n).

On the other hand, it follows that

Consequently, we obtain

□

RemarkIn the fact, we can also prove that, for anyhsatisfying

there exists a functionf∈Lp,q(n) such that

where 0

Theorem 1.2 tells us that the functionαpin (3) can not be improved. Forh(α)=|logα|swheres>0, we have some more explicit conclusions.

Corollary1.1Supposesis a positive constant and 0p/q, then there exists a functionf∈Lp,q(n) such that

if 0

(11)

ProofFirst, we assume thats>p/q.Letg:(0,+∞)→[0,+∞) be defined by

whereais a sufficiently small positive constant such thatg(α) is non-increasing.

By Lemma 1.1, there exists a functionf:n→[0,+∞] such thatdf(α)=g(α) for anyα∈(0,+∞).

In fact, we can choose the functionfas in (6). It implies from simple calculation that

f∈Lp,q(n)

and

Next, we assume that 0

or

Thus there exists constantsC>0 and 0

Observe that 0

=∞.

This leads to a contradiction withf∈Lp,q(n).

□

RemarkBy Corollary 1.1, we know that upper limit can not be replaced into limit in Theorem 1.2.

The following corollary tells us the limiting property does not hold forLp,∞function.

Corollary1.2Let 00, there exists a functionf∈Lp,∞(n) such that

ProofLetg:(0,+∞)→[0,+∞) be defined by

By Lemma 1.1, there exists a functionf:n→[0,+∞] such thatdf(α)=g(α) for anyα∈(0,+∞).

We can easily obtain that

f∈Lp,∞(n)

and

□