1 Introduction
Define the centered Hardy-Littlewood maximal function by and the uncentered Hardy-Littlewood maximal function by The basic real-variable construct was introduced by Hardy and Littlewood [1] for \(n=1\), and by Wiener [2] for \(n\ge2\). It is well known that the Hardy-Littlewood maximal function plays an important role in many parts of analysis. It is a classical mean operator, and it is frequently used to majorize other important operators in harmonic analysis.
$$ M^{c}f(x)=\sup_{r>0 }\frac{1}{|B(x,r)|} \int_{B(x,r)} \bigl|f(y) \bigr|\,dy, $$
(1.1)
$$ M f(x)=\sup_{B \ni x }\frac{1}{|B|} \int_{B} \bigl|f(y) \bigr|\,dy. $$
(1.2)
It is clear that holds for all \(x\in\Bbb {R}^{n}\). Both M and \(M^{c}\) are sublinear operators. Although the study of the boundedness for M or \(M^{c}\) is fairly completed, it is very hard to calculate the precise norm about M or \(M^{c}\).
$$ M^{c}f(x)\le Mf(x)\le2^{n} M^{c}f(x) $$
(1.3)
Anzeige
As is well known, the truncated operator has some important properties. In fact, in most situations, \(L^{p}\) boundedness of the truncated operator and the corresponding oscillatory operator is equivalent. There are many works in this regard and the reader can refer to [3] and [4].
Now we define the truncated centered Hardy-Littlewood maximal operator and the truncated uncentered Hardy-Littlewood maximal operator.
Define and for \(x\in \Bbb {R}^{n}\) and some real positive number γ.
$$ M^{c}_{\gamma}f(x) := \sup_{0< r< \gamma} \frac{1}{|B(x,r)|} \int_{B(x,r)}\bigl|f(y)\bigr|\,dy $$
(1.4)
$$ {M}_{\gamma}f(x) := \sup_{0< r< \gamma,|y-x|< r} \frac{1}{|B(y,r)|} \int _{B(y,r)}\bigl|f(t)\bigr|\,dt, $$
(1.5)
Obviously, like the inequality (1.3), in the pointwise sense, we immediately deduce from the definition (1.4) and (1.5) that and for all \(x\in \Bbb {R}^{n}\), as long as \(\gamma\le\rho\). Consequently, referring to the two truncated operators \(M^{c}_{\gamma}\) and \(M_{\gamma}\), as the sublinear operators, we naturally obtain and if \(\gamma\le\rho\), for \(1< p\le\infty\). Clearly, when γ is fixed, for example \(\gamma=1\), \(\Vert M^{c}_{1}\Vert _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}\) and \(\|M^{c}\| _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}\) are two fixed numbers. We think that it is very significant to make certain the precise relation of the two numbers. In the paper, we will consider the question. Surprisingly, the two numbers are equal whenever \(\gamma>0\). The same is true for \(p=1\).
$$M^{c}_{\gamma}f(x)\le M^{c}_{\rho}f(x)\le M^{c}f(x) $$
$$M_{\gamma}f(x)\le M_{\rho}f(x)\le Mf(x), $$
$$\bigl\Vert M^{c}_{\gamma} \bigr\Vert _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}\le \bigl\| M^{c}_{\rho}\bigr\| _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}\le\bigl\| M^{c}\bigr\| _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}, $$
$$\Vert M_{\gamma} \Vert _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}\le \|M_{\rho} \|_{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}\le\|M\|_{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}, $$
Anzeige
Now we formulate our main theorems.
Theorem 1.1
Let
\(M^{c}_{\gamma}\)
be defined by (1.4) and
\(\gamma>0\). Then
holds for
\(1< p\le\infty\).
$$\bigl\Vert M^{c}_{\gamma} \bigr\Vert _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}= \bigl\| M^{c}\bigr\| _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})} $$
Theorem 1.2
Let
\(M^{c}_{\gamma}\)
be defined by (1.4) and
\(\gamma>0\). Then
holds.
$$\bigl\Vert M^{c}_{\gamma} \bigr\Vert _{L^{1}(\Bbb {R}^{n})\rightarrow L^{1,\infty }(\Bbb {R}^{n})}= \bigl\| M^{c}\bigr\| _{L^{1}(\Bbb {R}^{n})\rightarrow L^{1,\infty}(\Bbb {R}^{n})} $$
For the truncated uncentered Hardy-Littlewood Maximal operator, we have similar conclusions.
Theorem 1.3
Let
\(M_{\gamma}\)
be defined by (1.5) and
\(\gamma>0\). Then
holds for
\(1< p\le\infty\).
$$\Vert M_{\gamma} \Vert _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}= \| M\|_{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})} $$
Theorem 1.4
Let
\(M_{\gamma}\)
be defined by (1.5) and
\(\gamma>0\). Then
holds.
$$\Vert M_{\gamma} \Vert _{L^{1}(\Bbb {R}^{n})\rightarrow L^{1,\infty }(\Bbb {R}^{n})}= \|M\|_{L^{1}(\Bbb {R}^{n})\rightarrow L^{1,\infty}(\Bbb {R}^{n})} $$
In Section 4, we will investigate the properties of the iterated Hardy-Littlewood maximal function.
2 Auxiliary and some lemmas
To prove our main theorems, we first provide some definitions and lemmas which will be used in the follows. Some lemmas can be found in the classic literature and here we omit their proofs.
Definition 2.1
Let f be a measurable function on \(\Bbb {R}^{n}\). The distribution function of f is the function \(d_{f}\) defined on \([0,+\infty)\) as follows: where \(|A|\) is the Lebesgue measure of the measurable set A.
$$ d_{f}(\alpha)= \bigl\vert \bigl\{ x\in \Bbb {R}^{n}: \bigl|f(x)\bigr|>\alpha \bigr\} \bigr\vert , $$
(2.1)
Lemma 2.1
For
\(f\in L^{p}(\Bbb {R}^{n})\)
with
\(0< p<\infty\), we have
$$ \|f\|_{L^{p}(\Bbb {R}^{n})}^{p}=p \int_{0}^{\infty}\alpha^{p-1}d_{f}( \alpha)\,d\alpha. $$
(2.2)
It is easy for us to verify the lemma by Fubini’s theorem. For more details as regards this lemma, one can refer to [5].
Lemma 2.2
Suppose that
μ
is a positive measure on a
σ-algebra
\(\mathbb{M}\). If
\(A_{1}\subset A_{2}\subset A_{3}\cdots\), \(A_{n}\in\mathbb {M}\), and
\(A=\bigcup_{n=1}^{\infty}A_{n}\), then
$$\lim_{n\rightarrow\infty}\mu(A_{n})=\mu(A). $$
Lemma 2.2 can be found in the book [6]. Using Lemma 2.2, we can formulate the following conclusions.
Lemma 2.3
Proof
For a fixed \(x\in \Bbb {R}^{n}\), by the definition of \(M^{c}\) in (1.1), associate to each ε a ball \(B(x,r_{\varepsilon})\) which satisfies Now taking \(\gamma>r_{\varepsilon}\), it follows from the definition of \(M^{c}_{\gamma}\) that Note that \(M^{c}_{\gamma}f(x)\) increases as \(\gamma\rightarrow\infty \). Thus we have
$$ \frac{1}{|B(x,r_{\varepsilon})|} \int_{B(x,r_{\varepsilon })}\bigl|f(y)\bigr|\,dy>M^{c}f(x)-\varepsilon. $$
(2.4)
$$ M^{c}_{\gamma}f(x)\geq\frac{1}{|B(x,r_{\varepsilon})|} \int _{B(x,r_{\varepsilon})}\bigl|f(y)\bigr|\,dy>M^{c}f(x)-\varepsilon. $$
(2.5)
$$ \lim_{\gamma\rightarrow\infty}M^{c}_{\gamma}f\geq Mf. $$
(2.6)
Clearly, we have Hence combining (2.6) with (2.7) yields
$$ M^{c}_{\gamma}f\leq Mf. $$
(2.7)
$$ \lim_{\gamma\rightarrow\infty}M^{c}_{\gamma}f= M^{c}f. $$
(2.8)
Obviously it implies from (2.8) that Set and We have \(A_{n}\subset A_{n+1}\) for \(n=1,2,\ldots\) , and \(A=\bigcup _{n=1}^{\infty}A_{n}\). It follows from Lemma 2.2 and the definition of the distribution function that This is our desired result. □
$$ \lim_{n\rightarrow\infty}M^{c}_{n} f= M^{c}f. $$
(2.9)
$$A_{n}= \bigl\{ x\in \Bbb {R}^{n}: M^{c}_{n} f(x)>\lambda \bigr\} , $$
$$A= \bigl\{ x\in \Bbb {R}^{n}: M^{c} f(x)>\lambda \bigr\} . $$
$$d_{M^{c}f}(\lambda)=|A|=\lim_{n\rightarrow\infty}|A_{n}|=\lim _{n\rightarrow\infty}d_{M_{n}^{c} f}(\lambda)=\lim_{\gamma\rightarrow \infty}d_{M^{c}_{\gamma}f}( \lambda). $$
Lemma 2.4
Lemma 2.5
Let
\(1< p<\infty\). For
\(\varepsilon>0\), there exists a function
\(g\in C_{c}^{\infty}(\Bbb {R}^{n})\)
such that
where
$$\begin{aligned} \frac{\|M^{c}g\|_{L^{p}(\Bbb {R}^{n})}}{\|g\|_{L^{p}(\Bbb {R}^{n})}} \geq\bigl\| M^{c}\bigr\| _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}- \varepsilon, \end{aligned}$$
(2.11)
$$\bigl\| M^{c}\bigr\| _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}=\sup_{\|f\|_{L^{p}(\Bbb {R}^{n})}\neq0} \frac{\|M^{c}f\|_{L^{p}(\Bbb {R}^{n})}}{\|f\|_{L^{p}(\Bbb {R}^{n})}}. $$
Proof
By the definition of the operator norm of \(M^{c}\), we can find a function \(f\in L^{p}(\Bbb {R}^{n})\) such that Since \(C_{c}^{\infty}(\Bbb {R}^{n})\) is dense in \(L^{p}(\Bbb {R}^{n})\), for \(\delta>0\), there exists a function \(g\in C_{c}^{\infty}(\Bbb {R}^{n})\) which satisfies Thus it implies from (2.13) that where the constant A is a bound of the operator \(M^{c}\).
$$\begin{aligned} \frac{\|M^{c}f\|_{L^{p}(\Bbb {R}^{n})}}{\|f\|_{L^{p}(\Bbb {R}^{n})}}\geq\bigl\| M^{c}\bigr\| _{L^{p}(\Bbb {R}^{n}) \rightarrow L^{p}(\Bbb {R}^{n})}- \frac{\varepsilon}{2}. \end{aligned}$$
(2.12)
$$\begin{aligned} \|f-g\|_{L^{p}(\Bbb {R}^{n})}< \delta. \end{aligned}$$
(2.13)
$$\begin{aligned} \bigl\| M^{c}(f-g)\bigr\| _{L^{p}(\Bbb {R}^{n})}\leq A\|f-g \|_{L^{p}(\Bbb {R}^{n})}< A\delta, \end{aligned}$$
(2.14)
Combining (2.13) with (2.14) yields If the number δ is small enough, we can immediately deduce that It implies from (2.15) and (2.16) that the inequality (2.11) holds. □
$$\begin{aligned} \frac{\|M^{c}g\|_{L^{p}(\Bbb {R}^{n})}}{\|g\|_{L^{p}(\Bbb {R}^{n})}}\geq\frac{\|M^{c}f\| _{L^{p}(\Bbb {R}^{n})}-\|M^{c}(f-g)\|_{L^{p}(\Bbb {R}^{n})}}{\|f\|_{L^{p}(\Bbb {R}^{n})}+\|f-g\| _{L^{p}(\Bbb {R}^{n})}} \geq\frac{\|M^{c}f\|_{L^{p}(\Bbb {R}^{n})}-A\delta}{\|f\|_{L^{p}(\Bbb {R}^{n})}+\delta}. \end{aligned}$$
(2.15)
$$ \frac{\|M^{c}f\|_{L^{p}(\Bbb {R}^{n})}-A\delta}{\|f\|_{L^{p}(\Bbb {R}^{n})}+\delta} \ge \frac{\|M^{c}f\|_{L^{p}(\Bbb {R}^{n})}}{\|f\|_{L^{p}(\Bbb {R}^{n})}}-\frac{\varepsilon }{2}\geq \bigl\| M^{c}\bigr\| _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}-\varepsilon. $$
(2.16)
3 Proof of main theorems
Now we shall prove our main theorems. We first consider the case \(1< p<\infty\).
Proof of Theorem 1.1
For convenience, we first prove for all \(0<\gamma<\infty\).
$$\bigl\| M^{c}_{\gamma}\bigr\| _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}= \bigl\| M^{c}_{1} \bigr\| _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})} $$
From the definition of the operator \(M^{c}_{\gamma}\) in (1.4), we have for \(x\in \Bbb {R}^{n}\) and \(0<\gamma<\infty\), where \(v_{n}\) is the volume of the unit ball in \(\Bbb {R}^{n}\).
$$\begin{aligned} M^{c}_{\gamma}f(x)=\sup_{0< r< \gamma} \frac{1}{|B(x,r)|} \int _{B(x,r)}\bigl|f(y)\bigr|\,dy=\sup_{0< r< \gamma} \frac{1}{v_{n}r^{n}} \int_{|y|\leq r}\bigl|f(x-y)\bigr|\,dy, \end{aligned}$$
(3.1)
A simple computation implies that where the dilation operator \(\tau_{\gamma}\) is defined as follows: for \(\gamma>0\) and \(x\in \Bbb {R}^{n}\).
$$\begin{aligned} M^{c}_{\gamma}f(\gamma x) =&\sup _{0< r< \gamma}\frac{1}{v_{n}r^{n}} \int_{|y|\leq r}\bigl|f(\gamma x-y)\bigr|\,dy \\ =&\sup_{0< r< \gamma}\frac{\gamma^{n} }{v_{n}r^{n}} \int_{|y|\leq\frac {r}{\gamma}}\bigl|f(\gamma x-\gamma y)\bigr|\,dy \\ =&\sup_{0< \frac{r}{\gamma}< 1}\frac{1}{v_{n}({\frac{r}{\gamma }})^{n}} \int_{|y|\leq\frac{r}{\gamma}}\bigl|(\tau_{\gamma}f) ( x-y)\bigr|\,dy \\ =&\sup_{0< r< 1}\frac{1}{v_{n}r^{n}} \int_{|y|\leq r}\bigl|(\tau_{\gamma}f) ( x-y)\bigr|\,dy \\ =&M^{c}_{1}(\tau_{\gamma}f) (x), \end{aligned}$$
(3.2)
$$\begin{aligned} (\tau_{\gamma}f) (x)=f(\gamma x), \end{aligned}$$
(3.3)
It follows from (3.2) that
$$\begin{aligned} \frac{\|M^{c}_{\gamma}f\|_{L^{p}(\Bbb {R}^{n})}}{\|f\|_{L^{p}(\Bbb {R}^{n})}}=\frac{\| M^{c}_{\gamma}f(\gamma\cdot)\|_{L^{p}(\Bbb {R}^{n})}}{\|f(\gamma\cdot)\| _{L^{p}(\Bbb {R}^{n})}}= \frac{\|M^{c}_{1}(\tau_{\gamma}f)\|_{L^{p}(\Bbb {R}^{n})}}{\|\tau_{\gamma}f\| _{L^{p}(\Bbb {R}^{n})}}. \end{aligned}$$
(3.4)
Taking the supremum over all \(f\in L^{p}(\Bbb {R}^{n})\) with \(\|f\|_{L^{p}(\Bbb {R}^{n})}\neq0\) for the two sides of equation (3.4), we have
$$\begin{aligned} \bigl\| M^{c}_{\gamma}\bigr\| _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})} = \bigl\| M^{c}_{1}\bigr\| _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}. \end{aligned}$$
(3.5)
Next, we will use equation (3.5) to prove for all \(\gamma>0\).
$$\bigl\| M^{c}_{\gamma}\bigr\| _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}=\bigl\| M^{c} \bigr\| _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})} $$
Since we merely need to prove By Lemma 2.5, for \(\varepsilon>0\), there exists a function \(g\in C_{c}^{\infty}(\Bbb {R}^{n})\) such that
$$\bigl\| M^{c}_{\gamma}\bigr\| _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}\leq\bigl\| M^{c}\bigr\| _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}, $$
$$\bigl\| M^{c}_{\gamma}\bigr\| _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}\geq\bigl\| M^{c}\bigr\| _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}. $$
$$\begin{aligned} \frac{\|M^{c}g\|_{L^{p}(\Bbb {R}^{n})}}{\|g\|_{L^{p}(\Bbb {R}^{n})}} \geq\bigl\| M^{c}\bigr\| _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}- \varepsilon. \end{aligned}$$
(3.6)
We may assume that the support of g is contained in the ball \(B(0,R)\), where R is a positive number. Since \(g\in C_{c}^{\infty}(\Bbb {R}^{n})\) implies \(g\in L^{p}(\Bbb {R}^{n})\), naturally we have \(M^{c}g\in L^{p}(\Bbb {R}^{n})\) by the \(L^{p}\) boundedness of the operator \(M^{c}\). It is not hard to find a positive number S such that Now we set \(\gamma_{0}=R+S\). Then it can be deduced from the definition of \(M^{c}_{\gamma}\) that holds for \(|x|< S\).
$$\begin{aligned} \bigl\Vert \bigl(M^{c}g \bigr)\chi_{\{|\cdot|\geq S\}} \bigr\Vert _{L^{p}(\Bbb {R}^{n})}\leq \varepsilon\|g\|_{L^{p}(\Bbb {R}^{n})}. \end{aligned}$$
(3.7)
$$\begin{aligned} M^{c}g(x)=M^{c}_{\gamma_{0}}g(x) \end{aligned}$$
(3.8)
It follows from (3.6), (3.7), and (3.8) that Obviously, (3.9) implies that Consequently, the inequality (3.10) yields By (3.5) and (3.11), we can derive from the arbitrariness property of ε that for all \(\gamma>0\).
$$\begin{aligned} \bigl\Vert M^{c}_{\gamma_{0}}g \bigr\Vert _{L^{p}(\Bbb {R}^{n})} \geq& \bigl\Vert \bigl(M^{c}_{\gamma_{0}}g \bigr) \chi_{\{|\cdot|< S\}} \bigr\Vert _{L^{p}(\Bbb {R}^{n})} \\ =& \bigl\Vert \bigl(M^{c}g \bigr)\chi_{\{|\cdot|< S\}} \bigr\Vert _{L^{p}(\Bbb {R}^{n})} \\ \geq& \bigl\Vert M^{c}g \bigr\Vert _{L^{p}(\Bbb {R}^{n})}- \bigl\Vert \bigl(M^{c}g \bigr)\chi_{\{ |\cdot|\geq S\}} \bigr\Vert _{L^{p}(\Bbb {R}^{n})} \\ \geq&\bigl\| M^{c}\bigr\| _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}\|g\|_{L^{p}(\Bbb {R}^{n})}-2\varepsilon\|g \|_{L^{p}(\Bbb {R}^{n})}. \end{aligned}$$
(3.9)
$$\begin{aligned} \frac{\|M^{c}_{\gamma_{0}}g\|_{L^{p}(\Bbb {R}^{n})}}{\|g\|_{L^{p}(\Bbb {R}^{n})}} \geq\bigl\| M^{c}\bigr\| _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}-2 \varepsilon. \end{aligned}$$
(3.10)
$$\begin{aligned} \bigl\| M^{c}_{\gamma_{0}}\bigr\| _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})} \geq \bigl\| M^{c}\bigr\| _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}-2\varepsilon. \end{aligned}$$
(3.11)
$$\begin{aligned} \bigl\| M^{c}_{\gamma}\bigr\| _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}= \bigl\| M^{c}\bigr\| _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})} \end{aligned}$$
(3.12)
This finishes the proof of Theorem 1.1. □
Next we will pay attention to proving the weak \((1,1)\) boundedness for the truncated centered Hardy-Littlewood maximal operator.
Proof of Theorem 1.2
First, we prove that holds for all \(0<\gamma<\infty\).
$$\bigl\| M^{c}_{\gamma}\bigr\| _{L^{1}(\Bbb {R}^{n})\rightarrow L^{1,\infty}(\Bbb {R}^{n})}=\bigl\| M^{c}_{1} \bigr\| _{L^{1}(\Bbb {R}^{n})\rightarrow L^{1,\infty}(\Bbb {R}^{n})} $$
From the identity (3.2), we have
$$\begin{aligned} M^{c}_{\gamma}f(\gamma x)=M^{c}_{1}( \tau_{\gamma}f) (x). \end{aligned}$$
(3.13)
For any \(\lambda>0\), we derive from (3.13) that Thus (3.14) implies that If \(\|f\|_{ L^{1}(\Bbb {R}^{n})}\neq0\), then it follows from (3.15) that Now taking the supremum over all \(f\in L^{1}(\Bbb {R}^{n})\) with \(\|f\|_{ L^{1}(\Bbb {R}^{n})}\neq0\) for the two sides of (3.16), we have
$$\begin{aligned} \bigl\vert \bigl\{ x: M^{c}_{1}( \tau_{\gamma}f) (x)>\lambda \bigr\} \bigr\vert =&\bigl| \bigl\{ x: M^{c}_{\gamma}f(\gamma x)>\lambda \bigr\} \bigr| \\ =& \biggl\vert \biggl\{ \frac{x}{\gamma}: M^{c}_{\gamma}f(x)>\lambda \biggr\} \biggr\vert \\ =&\frac{1}{\gamma^{n}} \bigl\vert \bigl\{ x: M^{c}_{\gamma}f(x)> \lambda \bigr\} \bigr\vert . \end{aligned}$$
(3.14)
$$\begin{aligned} \sup_{\lambda>0}\lambda\bigl| \bigl\{ x: M^{c}_{1}( \tau_{\gamma}f) (x)>\lambda \bigr\} \bigr| = \frac{1}{\gamma^{n}}\sup_{\lambda>0}\lambda \bigl\vert \bigl\{ x: M^{c}_{\gamma}f(x)>\lambda \bigr\} \bigr\vert . \end{aligned}$$
(3.15)
$$\begin{aligned} \frac{1}{\gamma^{n}}\frac{\sup_{\lambda>0}\lambda|\{x: M^{c}_{\gamma}f(x)>\lambda\}|}{\|f\|_{L^{1}(\Bbb {R}^{n})}} =&\frac{\sup_{\lambda>0}\lambda|\{x: M^{c}_{1}(\tau_{\gamma}f)(x)>\lambda\}|}{\|f\|_{L^{1}(\Bbb {R}^{n})}} \\ =&\frac{1}{\gamma^{n}}\frac{\sup_{\lambda>0}\lambda|\{x: M^{c}_{1}(\tau_{\gamma}f)(x)>\lambda\}|}{\|\tau_{\gamma}f\|_{L^{1}(\Bbb {R}^{n})}}. \end{aligned}$$
(3.16)
$$\begin{aligned} \bigl\| M^{c}_{\gamma}\bigr\| _{L^{1}(\Bbb {R}^{n})\rightarrow L^{1,\infty}(\Bbb {R}^{n})}=\bigl\| M^{c}_{1}\bigr\| _{L^{1}(\Bbb {R}^{n})\rightarrow L^{1,\infty}(\Bbb {R}^{n})}. \end{aligned}$$
(3.17)
Next we will use (3.17) to prove that holds for all \(\gamma>0\).
$$\bigl\| M^{c}_{\gamma}\bigr\| _{L^{1}(\Bbb {R}^{n})\rightarrow L^{1,\infty}(\Bbb {R}^{n})}=\bigl\| M^{c}\bigr\| _{L^{1}(\Bbb {R}^{n})\rightarrow L^{1,\infty}(\Bbb {R}^{n})} $$
We assert the following equation: holds for any \(f\in L^{1}(\Bbb {R}^{n})\) with \(\|f\|_{L^{1}}\neq0\).
$$ \sup_{\lambda>0}\lambda d_{M^{c}f}(\lambda) = \lim_{\gamma\rightarrow\infty}\sup_{\lambda>0}\lambda d_{M^{c}_{\gamma}f}( \lambda) $$
(3.18)
Clearly the left side of (3.18) is not smaller than the right side, so it suffices to prove the opposite inequality.
It follows from Lemma 2.3 that Set For \(\varepsilon>0\), there must be a positive number \(\lambda_{0}\) such that We conclude that This is equivalent to Consequently, (3.18) holds.
$$\sup_{\lambda>0}\lambda d_{M^{c}f}(\lambda)=\sup _{\lambda>0}\lambda \Bigl(\lim_{\gamma\rightarrow\infty}d_{M^{c}_{\gamma}f}( \lambda ) \Bigr). $$
$$A=\sup_{\lambda>0}\lambda d_{M^{c}f}(\lambda). $$
$$A-\varepsilon\le\lambda_{0} d_{M^{c}f}(\lambda_{0}) \le A. $$
$$\sup_{\lambda>0}\lambda \Bigl(\lim_{\gamma\rightarrow\infty }d_{M^{c}_{\gamma}f}( \lambda) \Bigr)\ge\lim_{\gamma\rightarrow \infty}\lambda_{0}d_{M^{c}_{\gamma}f}( \lambda_{0}) =\lambda_{0} d_{M^{c}f}( \lambda_{0})\ge A-\varepsilon. $$
$$\sup_{\lambda>0}\lambda \Bigl(\lim_{\gamma\rightarrow\infty }d_{M^{c}_{\gamma}f}( \lambda) \Bigr)\ge A. $$
Using equation (3.18), we deduce that
$$\begin{aligned} \bigl\| M^{c}\bigr\| _{L^{1}(\Bbb {R}^{n})\rightarrow L^{1,\infty}(\Bbb {R}^{n})} =&\sup_{\|f\| _{L^{1}(\Bbb {R}^{n})}\neq0} \frac{\sup_{\lambda>0}\lambda d_{M^{c}f}(\lambda )}{\|f\|_{L^{1}(\Bbb {R}^{n})}} \\ =&\sup_{\|f\|_{L^{1}(\Bbb {R}^{n})}\neq0} \lim_{\gamma\rightarrow\infty} \frac{\sup_{\lambda>0}\lambda d_{M^{c}_{\gamma}f}(\lambda)}{\|f\|_{L^{1}(\Bbb {R}^{n})}} \\ =&\lim_{\gamma\rightarrow\infty} \sup_{\|f\|_{L^{1}(\Bbb {R}^{n})}\neq0} \frac{\sup_{\lambda>0}\lambda d_{M^{c}_{\gamma}f}(\lambda)}{\|f\|_{L^{1}(\Bbb {R}^{n})}} \\ =&\lim_{\gamma\rightarrow\infty}\bigl\| M^{c}_{\gamma} \bigr\| _{L^{1}(\Bbb {R}^{n})\rightarrow L^{1,\infty}(\Bbb {R}^{n})}. \end{aligned}$$
(3.19)
Proof of Theorem 1.3
We conclude from the definition of the operator \(M_{\gamma}\) in (1.5) that Thus we have for all \(\gamma>0\) and \(1< p<\infty\).
$$\begin{aligned} M_{\gamma}f(\gamma x) =&\sup_{0< r< \gamma,|y-\gamma x|< r} \frac {1}{|B(y,r)|} \int_{B(y,r)}\bigl|f(t)\bigr|\,dt \\ =& \sup_{0< r< \gamma,|\gamma y-\gamma x|< r}\frac{1}{v_{n}r^{n}} \int _{|t|< r}\bigl|f(\gamma y-t)\bigr|\,dt \\ =& \sup_{0< r< \gamma,|y-x|< \frac{r}{\gamma}}\frac{\gamma ^{n}}{v_{n}r^{n}} \int_{|t|< \frac{r}{\gamma}}\bigl|f(\gamma y-\gamma t)\bigr|\,dt \\ =& \sup_{0< \frac{r}{\gamma}< 1,|y-x|< \frac{r}{\gamma}}\frac {1}{v_{n}({\frac{r}{\gamma}})^{n}} \int_{|t|< \frac{r}{\gamma}}\bigl|(\tau _{\gamma}f) (y-t)\bigr|\,dt \\ =& \sup_{0< r< 1,|y-x|< r}\frac{1}{|B(y,r)|} \int_{|t|< r}\bigl|(\tau_{\gamma}f) (x-t)\bigr|\,dt \\ =& M_{1}(\tau_{\gamma}f) (x). \end{aligned}$$
(3.20)
$$\begin{aligned} \|M_{\gamma}\|_{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}= \|M_{1} \|_{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})} \end{aligned}$$
(3.21)
Next we will prove that
$$\Vert M_{\gamma} \Vert _{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}= \| M\|_{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}. $$
If \(f\in L^{p}(\Bbb {R}^{n})\), then we have \(Mf\in L^{p}(\Bbb {R}^{n})\). It follows from Lemma 2.1, Lemma 2.4, and equation (3.21) that Since we have the obvious inequality we derive from (3.22) that This is our desired result. □
$$\begin{aligned} \|Mf\|_{L^{p}(\Bbb {R}^{n})}^{p} =&p \int_{0}^{\infty}\lambda^{p-1}d_{Mf}( \lambda)\,d\lambda \\ =&p \int_{0}^{\infty}\lambda^{p-1}\lim _{\gamma\rightarrow\infty }d_{M_{\gamma}f}(\lambda)\,d\lambda \\ =&\lim_{\gamma\rightarrow\infty}p \int_{0}^{\infty}\lambda ^{p-1}d_{M_{\gamma}f}( \lambda)\,d\lambda \\ =&\lim_{\gamma\rightarrow\infty}\|M_{\gamma}f\|_{L^{p}(\Bbb {R}^{n})}^{p} \\ \le&\lim_{\gamma\rightarrow\infty} \Vert M_{\gamma} \Vert ^{p}_{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}\|f\|_{L^{p}(\Bbb {R}^{n})}^{p} \\ =&\Vert M_{1}\Vert ^{p}_{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}\|f\| _{L^{p}(\Bbb {R}^{n})}^{p}. \end{aligned}$$
(3.22)
$$\begin{aligned} \|M\|_{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}\ge\|M_{1}\|_{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}, \end{aligned}$$
(3.23)
$$\|M\|_{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}=\|M_{1}\|_{L^{p}(\Bbb {R}^{n})\rightarrow L^{p}(\Bbb {R}^{n})}. $$
4 Iterated Hardy-Littlewood maximal function
In this section, we will consider the iterated Hardy-Littlewood maximal function.
Let M be the uncentered Hardy-Littlewood maximal function defined by (1.2). Define the iterated Hardy-Littlewood maximal function denoted by \(M^{k+1}\) as follows: for \(k=1,2,\ldots\) , and \(x\in \Bbb {R}^{n}\). Set \(M^{1}f(x):=Mf(x)\).
$$ M^{k+1}f(x):=M \bigl(M^{k}f \bigr) (x), $$
(4.1)
In order to study the properties of the iterated Hardy-Littlewood maximal function, we first introduce the following lemma.
Lemma 4.1
Suppose that a sequence
\(\{c_{i}\}_{i=1}^{\infty}\)
satisfies the following two conditions simultaneously:
Then
\(\{c_{i}\}_{i=1}^{\infty}\)
is strictly monotone increasing and we have
(i)
\(c_{1}=r\in(0,1)\);
(ii)
for any
\(k\ge1\), \(c_{k+1}=(1-r)c_{k}+r\).
$$\lim_{k\rightarrow\infty}c_{k}=1. $$
Proof
By the mathematical induction and the two conditions (i) and (ii), we can easily obtain \(0< c_{k}<1\) for each \(k\in\Bbb {N}\). Moreover, the condition (ii) implies This shows that \(\{c_{i}\}_{i=1}^{\infty}\) is strictly monotone increasing. Since \(\{c_{i}\}_{i=1}^{\infty}\) is monotone increasing and has the upper bound, the limit of \(\{c_{i}\}_{i=1}^{\infty}\) exists, and we can easily get □
$$c_{k+1}-c_{k}=(1-r)c_{k}+r-c_{k}=r(1-c_{k})> 0. $$
$$\lim_{k\rightarrow\infty}c_{k}=1. $$
By Lemma 4.1, we have the following theorem.
Theorem 4.2
For any
\(f\in L^{\infty}(\Bbb {R}^{n})\), the equation
holds for any
\(x\in \Bbb {R}^{n}\).
$$ \lim_{k\rightarrow\infty}M^{k}f(x)=\|f\|_{\infty}$$
(4.2)
Proof
If \(\|f\|_{\infty}=0\), the proof is trivial. If \(\|f\|_{\infty}>0\), for any \(\varepsilon\in(0,\|f\|_{\infty})\), define a set Then we have \(|E_{\varepsilon}|>0\), where \(|E_{\varepsilon}|\) denotes the Lebesgue measure of \(E_{\varepsilon}\). For any fixed point \(a\in \Bbb {R}^{n}\), there exists a number \(R>0\) such that
$$ E_{\varepsilon}:= \bigl\{ x\in \Bbb {R}^{n}:\bigl|f(x)\bigr|\ge\|f \|_{\infty}-\varepsilon \bigr\} . $$
(4.3)
$$ \bigl|E_{\varepsilon}\cap B(a,R)\bigr|\ge\frac{1}{2}|E_{\varepsilon}|. $$
(4.4)
Denote \(\tilde{E}_{\varepsilon}=E_{\varepsilon}\cap B(a,R)\).
Define a set as Actually if \(f\in L^{p}(\Bbb {R}^{n})\) with \(1\le p\le\infty\), then \(|(S_{L}(f))^{c}|=0\), where \((S_{L}(f))^{c}\) denotes the complement set of \(S_{L}(f)\). When \(x\in\tilde{E}_{\varepsilon}\cap S_{L}(f)\), we derive from (4.3) that and for all \(k=1,2,\ldots\) .
$$S_{L}(f):= \bigl\{ x\in \Bbb {R}^{n}: x \mbox{ is the Lebesgue point of } f \mbox{ and } M_{k}f,k=1,2,\ldots \bigr\} . $$
$$\bigl|f(x)\bigr|\ge\|f\|_{\infty}-\varepsilon, $$
$$M_{k}f(x)\ge\|f\|_{\infty}-\varepsilon, $$
When \(x\in B(a,R)\), we consider the uncentered Hardy-Littlewood maximal function of f at the point x. It follows that Set It implies from (4.5) that A straightforward computation implies from (4.6) that
$$ Mf(x)\ge\frac{1}{|B(a,R)|} \int_{B(a,R)}\bigl|f(y)\bigr|\,dy\ge\frac{\vert \tilde{E}_{\varepsilon} \vert }{|B(a,R)|}\bigl(\|f\|_{\infty}- \varepsilon\bigr). $$
(4.5)
$$r=\frac{\vert \tilde{E}_{\varepsilon} \vert }{|B(a,R)|}>0. $$
$$ Mf(x)\ge r\bigl(\|f\|_{\infty}-\varepsilon\bigr). $$
(4.6)
$$\begin{aligned} M^{2}f(x) \ge&\frac{1}{|B(a,R)|} \int_{B(a,R)}\bigl|Mf(y)\bigr|\,dy \\ =&\frac{1}{|B(a,R)|} \int_{\tilde{E}_{\varepsilon}}\bigl|Mf(y)\bigr|\,dy+\frac {1}{|B(a,R)|} \int_{B(a,R)\setminus\tilde{E}_{\varepsilon }}\bigl|Mf(y)\bigr|\,dy \\ \ge&\frac{\vert \tilde{E}_{\varepsilon} \vert }{|B(a,R)|}\bigl(\|f\| _{\infty}-\varepsilon\bigr)+\frac{|B(a,R)|-\vert \tilde{E}_{\varepsilon }\vert }{|B(a,R)|}r\bigl(\|f\|_{\infty}-\varepsilon\bigr) \\ =& \bigl(r+(1-r)r \bigr) \bigl(\|f\|_{\infty}-\varepsilon\bigr). \end{aligned}$$
(4.7)
Denote and for \(k=1,2,\ldots\) .
$$c_{1}=r $$
$$c_{k+1}=c_{1}+(1-c_{1})c_{k}, $$
It implies from (4.7) that Using the inductive method, we can easily obtain Thus Lemma 4.1 implies that When \(\varepsilon\rightarrow0\), we have By the definition of the Hardy-Littlewood function, we obviously deduce Combining (4.8) with (4.9) yields
$$M^{2}f(x)\ge c_{2}\bigl(\|f\|_{\infty}-\varepsilon\bigr). $$
$$M^{k+1}f(x)\ge c_{k+1}\bigl(\|f\|_{\infty}-\varepsilon\bigr). $$
$$\liminf_{k\rightarrow\infty} M^{k}f(x)\ge\|f\|_{\infty}- \varepsilon. $$
$$ \liminf_{k\rightarrow\infty} M^{k}f(x)\ge\|f \|_{\infty}. $$
(4.8)
$$ \limsup_{k\rightarrow\infty} M^{k}f(x)\le\|f \|_{\infty}. $$
(4.9)
$$\lim_{k\rightarrow\infty} M^{k}f(x)=\|f\|_{\infty}. $$
By the arbitrary choice of a, we obtain for all \(x\in \Bbb {R}^{n}\). □
$$\lim_{k\rightarrow\infty} M^{k}f(x)=\|f\|_{\infty}$$
Acknowledgements
The research was supported by the National Natural Foundation of China (Grant Nos. 11471039, 11271162, and 11561062).
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.