Let f be a measurable function on a measure space (X, μ) and 0 < p < ∞, 0 < q≤∞. Define

$ \|f\|_{L^{p, q}}= \begin{cases}p^{\frac{1}{q}}\left(\int_0^{\infty}\left(d_f(\alpha)^{\frac{1}{p}} \alpha\right)^q \frac{\mathrm{d} \alpha}{\alpha}\right)^{\frac{1}{q}}, & \text { if } q <\infty , \\ \sup \limits_{\alpha>0} \alpha \; d_f(\alpha)^{\frac{1}{p}}, & \text { if } q=\infty ,\end{cases} $

(1)

where $d_f(\alpha)=\mu\{x:|f(x)|>\alpha\}$ is the distribution function of $f$. The set of all $f$ with $|f|_{L^{p, q}}<\infty$ is denoted by L^{p, q}(X, μ) and is called the Lorentz space with indices p and q.

There are many simple properties of Lorentz space. For 0 < p < ∞, we have that

$ L^{p, p}(X, \mu)=L^p(X, \mu), $

and

$ L^{p, \infty}(X, \mu)=W L^p(X, \mu) . $

For 0 < p < ∞ and 0 < q < r≤∞, there exists a constant c_{p, q, r} such that

$ \|f\|_{L^{p, r}} \leqslant c_{p, q, r}\|f\|_{L^{p, q }}. $

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

The limiting property of distribution functions of L^p functions has been proved in Ref.[5] for 1≤p < ∞. That is, when $ f \in L^{p, p}\left(\mathbb{R}^n\right)$, we have that

$ \lim\limits _{\alpha \rightarrow 0^{+}} \alpha^p d_f(\alpha)=0. $

(2)

However, for $f \in L^1\left(\mathbb{R}^n\right)$, we know that $M f \in$ $L^{1, \infty}\left(\mathbb{R}^n\right)$, where $M$ is the maximal operator. In Ref. [6], P. Janakiraman has proved that

$ \lim\limits _{\alpha \rightarrow 0^{+}} \alpha d_{M f}(\alpha)=\|f\|_{L^1\left(\mathbb{R}^n\right)}. $

This means that the limiting equality of distribution functions does not hold for L^{1, ∞} function.

It is well-known that the following inclusion relation

$ L^{p, p}\left(\mathbb{R}^n\right) \varsubsetneqq L^{p, q}\left(\mathbb{R}^n\right) \varsubsetneqq L^{p, \infty}\left(\mathbb{R}^n\right) $

holds for 0 < p < q < ∞.

It is worth investigating the limiting property of distribution functions of f∈L^{p, q}(X, μ).

1 Main results and proof

Now we formulate our main theorem.

Theorem 1.1   Let μ be a σ-finite positive measure on some σ-algebra in set X. For f∈L^{p, q}(X, μ) with 0 < p, q < ∞, denote the distribution functions of f as

$ d_f(\alpha):=\mu(\{x \in X:|f(x)|>\alpha\}). $

Then we have

$ \lim\limits _{\alpha \rightarrow 0^{+}} \alpha^p d_f(\alpha)=\lim\limits _{\alpha \rightarrow \infty} \alpha^p d_f(\alpha)=0 $

(3)

Proof  First, we prove that

$ \limsup\limits _{\alpha \rightarrow 0^{+}} \alpha^p d_f(\alpha)=0. $

If the conclusion does not hold, then we conclude that

$ \limsup\limits _{\alpha \rightarrow 0^{+}} \alpha^p d_f(\alpha)=C_0>0, $

or

$ \limsup\limits _{\alpha \rightarrow 0^{+}} \alpha^p d_f(\alpha)=\infty . $

Thus there exist a positive sequence {x_n}_n=1^∞ and a positive number C such that

$ \lim\limits _{n \rightarrow \infty} x_n=0, $

and

$ x_n^p d_f\left(x_n\right) \geqslant C $

for $ n \in \mathbb{N}$. Obviously, there exists a strictly decreasing subsequence, still denoted as {x_n}.

Since d_f is a decreasing function, it follows that

$ \begin{aligned} \|f\|_{L^{p, q}}^q & =p \int_0^{\infty}\left(\alpha d_f(\alpha)^{\frac{1}{p}}\right)^q \frac{\mathrm{d} \alpha}{\alpha} \\ & =p \int_0^{\infty} \alpha^{q-1} d_f(\alpha)^{\frac{q}{p}} \mathrm{~d} \alpha \\ & =p \int_0^{x_1} \alpha^{q-1} d_f(\alpha)^{\frac{q}{p}} \mathrm{~d} \alpha+p \int_{x_1}^{\infty} \alpha^{q-1} d_f(\alpha)^{\frac{q}{p}} \mathrm{~d} \alpha \\ & \geqslant p \sum\limits_{n=1}^{\infty} \int_{x_{n+1}}^{x_n} \alpha^{q-1} d_f(\alpha)^{\frac{q}{p}} \mathrm{~d} \alpha \\ & \geqslant p \sum\limits_{n=1}^{\infty} \int_{x_{n+1}}^{x_n} \alpha^{q-1} d_f\left(x_n\right)^{\frac{q}{p}} \mathrm{~d} \alpha \\ & =\frac{p}{q} \sum\limits_{n=1}^{\infty}\left(x_n^q-x_{n+1}^q\right) d_f\left(x_n\right)^{\frac{q}{p}} \\ & \geqslant \frac{p C^{\frac{q}{p}}}{q} \sum\limits_{n=1}^{\infty}\left(1-\left(\frac{x_{n+1}}{x_n}\right)^q\right) . \end{aligned} $

(4)

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

$ \begin{aligned} \prod\limits_{n=1}^{\infty} \lambda_n & =\lim _{N \rightarrow \infty} \prod\limits_{n=1}^N \lambda_n=\lim _{N \rightarrow \infty} \prod\limits_{n=1}^N\left(\frac{x_{n+1}}{x_n}\right)^q \\ & =\lim _{N \rightarrow \infty}\left(\frac{x_{N+1}}{x_1}\right)^q=0 . \end{aligned} $

It implies from (4) that

$ \|f\|_{L^{p, q}}=\infty . $

This leads to contradiction.

Next, we prove

$ \limsup\limits _{\alpha \rightarrow \infty} \alpha^p d_f(\alpha)=0. $

Similarly, if the upper limit is a positive number or ∞, there exist a positive number C and a strictly increasing sequence {x_n}_n=1^∞

such that

$ \lim\limits _{n \rightarrow \infty} x_n=\infty $

and

$ x_n^p d_f\left(x_n\right) \geqslant C. $

Therefore, we conclude that

$ \begin{aligned} \|f\|_{L^{p, q}}^q & =p \int_0^{\infty}\left(\alpha d_f(\alpha)^{\frac{1}{p}}\right)^q \frac{\mathrm{d} \alpha}{\alpha} \\ & =p \int_0^{\infty} \alpha^{q-1} d_f(\alpha)^{\frac{q}{p}} \mathrm{~d} \alpha \\ & =p \int_0^{x_1} \alpha^{q-1} d_f(\alpha)^{\frac{q}{p}} \mathrm{~d} \alpha+p \int_{x_1}^{\infty} \alpha^{q-1} d_f(\alpha)^{\frac{q}{p}} \mathrm{~d} \alpha \\ & \geqslant p \sum\limits_{n=1}^{\infty} \int_{x_n}^{x_{n+1}} \alpha^{q-1} d_f(\alpha)^{\frac{q}{p}} \mathrm{~d} \alpha \\ & \geqslant p \sum\limits_{n=1}^{\infty} \int_{x_n}^{x_{n+1}} \alpha^{q-1} d_f\left(x_{n+1}\right)^{\frac{q}{p}} \mathrm{~d} \alpha \\ & =\frac{p}{q} \sum\limits_{n=1}^{\infty}\left(x_{n+1}^q-x_n^q\right) d_f\left(x_{n+1}\right)^{\frac{q}{p}} \\ & \geqslant \frac{p C^{q / p}}{q} \sum\limits_{n=1}^{\infty}\left(1-\left(\frac{x_n}{x_{n+1}}\right)^q\right) .\end{aligned} $

(5)

Note that the infinite product ∏(x_n/x_n+1)^q diverges to 0. For α>0, α^pd_f(α)≥0 implies

$ \liminf\limits _{\alpha \rightarrow \infty} \alpha^p d_f(\alpha) \geqslant 0. $

That leads to (3).

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

Lemma 1.1   Let g: (0, +∞)→[0, +∞) be a right-continuous non-increasing function. Then there exists a function f: $\mathbb{R}^n $→[0, +∞] such that d_f(α)=g(α) for any α∈(0, +∞).

Proof   Let

$ f(x)=\int_0^{\infty} \chi{ }_{B\left(0, \left(\frac{g(u)}{v_n}\right)^{\frac{1}{n}}\right)}(x) \mathrm{d} u, $

(6)

where v_n=m(B(0, 1)) and B(x, r) denotes a ball with the center at x and the radius r.

Then we merely need to prove that

$ \{x: f(x)>\alpha\}=\left\{x:|x|<\left(\frac{g(\alpha)}{v_n}\right)^{\frac{1}{n}}\right\} $

(7)

holds, for any α∈(0, +∞).

Assume

$ |x| \geqslant\left(\frac{g(\alpha)}{v_n}\right)^{\frac{1}{n}}. $

Note that g is a non-increasing function. We conclude from definition of f in (6) that

$ \begin{aligned} & f(x)=\int_0^\alpha \chi{ }_{B\left(0, \left(\frac{g(u)}{v_n}\right)^{\frac{1}{n}}\right)}(x) \mathrm{d} u+\int_\alpha^{\infty} \chi{ }_{B\left(0, \left(\frac{g(u)}{v_n}\right)^{\frac{1}{n}}\right)}(x) \mathrm{d} u \\ & \;\;\;\;=\int_0^\alpha \chi{ }_{B\left(0, \left(\frac{g(u)}{v_n}\right)^{\frac{1}{n}}\right)}(x) \mathrm{d} u \\ & \;\;\;\;\leqslant \alpha .\\ \end{aligned} $

(8)

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

Assume

$ |x|<\left(\frac{g(\alpha)}{v_n}\right)^{\frac{1}{n}}. $

Note that g is a right-continuous function. Then there exists α′>α such that

$ |x|<\left(\frac{g\left(\alpha^{\prime}\right)}{v_n}\right)^{\frac{1}{n}}. $

Then we conclude that

$ \begin{aligned} f(x) & =\int_0^{\alpha^{\prime}} \chi{ }_{B\left(0, \left(\frac{g(u)}{v_n}\right)^{\frac{1}{n}}\right)}(x) \mathrm{d} u+\int_{\alpha^{\prime}}^{\infty} \chi{ }_{B\left(0, \left(\frac{g(u)}{v_n}\right)^{\frac{1}{n}}\right)}(x) \mathrm{d} u \\ & \geqslant \int_0^{\alpha^{\prime}} \chi{ }_{B\left(0, \left(\frac{g(u)}{v_n}\right)^{\frac{1}{n}}\right)}(x) \mathrm{d} u \\ & =\alpha^{\prime}>\alpha . \end{aligned} $

This means that

$ \{x: f(x)>\alpha\} \supseteqq\left\{x:|x|<\left(\frac{g(\alpha)}{v_n}\right)^{\frac{1}{n}}\right\} . $

In a word, (7) holds.

As an application of Lemma 1.2, we prove that the function α^p can not be improved for some sense.

Theorem 1.2   Suppose 0 < p, q < ∞. For any nonnegative function h satisfying

$ \lim\limits _{\alpha \rightarrow 0^{+}} h(\alpha)=\infty, $

(9)

there exists a function f∈L^{p, q}($ \mathbb{R}^n$) such that

$ \limsup\limits _{\alpha \rightarrow 0^{+}} \alpha^p h(\alpha) d_f(\alpha)=\infty. $

Proof   If x_k>0 and x_k→0, by (9), then there exists a strictly decreasing positive subsequence, still denoted by {x_k}_k=1^∞, such that

$ h\left(\alpha_k\right) \geqslant 4^k $

and

$ x_{k+1}<\min \left\{2^{-1 / p}, 2^{-1}\right\} x_k, $

for $ k \in \mathbb{N}$ and α_k∈[2^-1x_k, x_k).

Define a function g: (0, +∞)→[0, +∞) as

$ g(\alpha)= \begin{cases}\frac{1}{2^k x_k^p}, & \text { if } x_{k+1} \leqslant \alpha <x_k, \\ 0, & \text { if } \alpha \geqslant x_1.\end{cases} $

(10)

By Lemma 1.1, there exists a function f: $ \mathbb{R}^n$→[0, +∞] such that d_f(α)=g(α) for any α∈(0, +∞). And we have from the definition of g in (10) that

$ \begin{aligned} \|f\|_{L^{p, q}}^q & =p \int_0^{\infty} \alpha^{q-1} d_f(\alpha)^{q / p} \mathrm{~d} \alpha \\ & =p \sum\limits_{k=1}^{\infty} \int_{x_{k+1}}^{x_k} \alpha^{q-1} d_f(\alpha)^{q / p} \mathrm{~d} \alpha \\ & =\frac{p}{q} \sum\limits_{k=1}^{\infty}\left(1-\left(\frac{x_{k+1}}{x_k}\right)^q\right) 2^{-\frac{q k}{p}} \\ & \leqslant \frac{p}{q} \sum\limits_{k=1}^{\infty} 2^{-\frac{q k}{p}}<\infty . \end{aligned} $

This implies f∈L^{p, q}($ \mathbb{R}^n$).

On the other hand, it follows that

$ \alpha_k^p h\left(\alpha_k\right) d_f\left(\alpha_k\right) \geqslant\left(\frac{x_k}{2}\right)^p 4^k \frac{1}{2^k x_k^p}=2^{k-p} . $

Consequently, we obtain

$ \limsup\limits _{\alpha \rightarrow 0^{+}} \alpha^p h(\alpha) d_f(\alpha)=\infty. $

Remark    In the fact, we can also prove that, for any h satisfying

$ \lim\limits _{\alpha \rightarrow \infty} h(\alpha)=\infty, $

there exists a function f∈L^{p, q}($\mathbb{R}^n $) such that

$ \limsup\limits _{\alpha \rightarrow \infty} \alpha^p h(\alpha) d_f(\alpha)=\infty \text {, } $

where 0 < p, q < ∞. The case can be proved similarly so we omit the details.

Theorem 1.2 tells us that the function α^p in (3) can not be improved. For h(α)=| logα|^s where s>0, we have some more explicit conclusions.

Corollary 1.1    Suppose s is a positive constant and 0 < p, q < ∞. If s>p/q, then there exists a function f∈L^{p, q}($\mathbb{R}^n $) such that

$ \lim\limits _{\alpha \rightarrow 0^{+}} \alpha^p|\log \alpha|^s d_f(\alpha)=\infty ; $

if 0 < s≤p/q, then for any f∈L^{p, q}($ \mathbb{R}^n$), we have

$ \liminf\limits _{\alpha \rightarrow 0^{+}} \alpha^p|\log \alpha|^s d_f(\alpha)=0 $

(11)

Proof   First, we assume that s>p/q. Let g: (0, +∞)→[0, +∞) be defined by

$ g(\alpha)=\chi_{(0, a)}(\alpha) \frac{1}{\alpha^p|\log \alpha|^{\frac{1}{2}(s+p / q)}}, $

where a is a sufficiently small positive constant such that g(α) is non-increasing.

By Lemma 1.1, there exists a function f: $ \mathbb{R}^n$→[0, +∞] such that d_f(α)=g(α) for any α∈(0, +∞).

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

$ f \in L^{p, q}\left(\mathbb{R}^n\right) $

and

$ \lim\limits _{\alpha \rightarrow 0^{+}} \alpha^p|\log \alpha|^s d_f(\alpha)=\infty . $

Next, we assume that 0 < s≤p/q. If the limit in (11) does not hold, then there exists a function f∈L^{p, q}($\mathbb{R}^n $) satisfying

$ \liminf\limits _{\alpha \rightarrow 0^{+}} \alpha^p|\log \alpha|^s d_f(\alpha)=C_0>0, $

or

$ \liminf\limits _{\alpha \rightarrow 0^{+}} \alpha^p|\log \alpha|^s d_f(\alpha)=\infty. $

Thus there exists constants C>0 and 0 < a < 1, such that for α < a,

$ \alpha^p|\log \alpha|^s d_f(\alpha) \geqslant C \text {. } $

Observe that 0 < s≤p/q. We have that

$ \begin{aligned} \|f\|_{L^{p, q}}^q & =p \int_0^{\infty}\left(\alpha d_f(\alpha)^{\frac{1}{p}}\right)^q \frac{\mathrm{d} \alpha}{\alpha} \\ & \geqslant p \int_0^a\left(\alpha d_f(\alpha)^{\frac{1}{p}}\right)^q \frac{\mathrm{d} \alpha}{\alpha} \\ & \geqslant p \int_0^a\left(\alpha\left(\frac{C}{\alpha^p|\log \alpha|^s}\right)^{\frac{1}{p}}\right)^q \frac{\mathrm{d} \alpha}{\alpha} \\ & =p C^{q / p} \int_0^a \frac{1}{\alpha|\log \alpha|^{s q / p}} \mathrm{~d} \alpha \\ & =\infty . \end{aligned} $

This leads to a contradiction with f∈L^{p, q}($ \mathbb{R}^n$).

Remark    By 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 for L^{p, ∞} function.

Corollary 1.2   Let 0 < p < ∞. For a fixed constant C>0, there exists a function f∈L^{p, ∞}($ \mathbb{R}^n$) such that

$ \lim\limits _{\alpha \rightarrow 0^{+}} \alpha^p d_f(\alpha)=\lim\limits _{\alpha \rightarrow \infty} \alpha^p d_f(\alpha)=C . $

Proof   Let g: (0, +∞)→[0, +∞) be defined by

$ g(\alpha)=\frac{C}{\alpha^p}. $

By Lemma 1.1, there exists a function f: $ \mathbb{R}^n$→ [0, +∞] such that d_f(α)=g(α) for any α∈(0, +∞).

We can easily obtain that

$ f \in L^{p, \infty}\left(\mathbb{R}^n\right) $

and

$ \lim\limits _{\alpha \rightarrow 0^{+}} \alpha^p d_f(\alpha)=\lim\limits _{\alpha \rightarrow \infty} \alpha^p d_f(\alpha)=C . $