For any $ x\in\mathbb R^n $, $ f\in L_{loc}^1(\mathbb R^n) $, the Hardy-Littlewood maximal operator over the Euclidean ball of radius $ r $ and centered at $ x $ is defined by

$ \begin{align*} Mf(x)=\sup\limits_{\epsilon>0}\frac{1}{|B(x,\epsilon)|}\displaystyle{\int}_{B(x,\epsilon)}|f(y)|dy, \end{align*} $

where $ |B(x,\epsilon)| $ is the Lebesgue measure of the ball. It is well known that the Hardy-Littlewood maximal operator is of weak type (1, 1). That is, for any $ \lambda>0 $, there exists a positive constant $ C_n $ only depends on $ n $ such that

$ \begin{align} |\{x\in\mathbb R^n: Mf(x)>\lambda\}|\leq\frac{C_n}{\lambda}\|f\|_{L_1}. \end{align} $

(1.1)

For any function $ f $ on $ \mathbb{R}^n $ and $ \epsilon>0 $, we denote $ f_\epsilon(x)=\frac{1}{\epsilon^n}f(\frac{x}{\epsilon}) $, $ k(\cdot)=\frac{1}{v_n}\chi_{B(0,1)}(\cdot) $. Then, Hardy-Littlewood maximal operator can be written as

$ \begin{align*} Mf(x) =\sup\limits_{\epsilon>0}\frac{1}{v_n\epsilon^n}\displaystyle{\int}_{\mathbb R^n}|f(x-y)|\chi_{B(0,1)}(\frac{y}{\epsilon})dy =\sup\limits_{\epsilon>0}(|f|*k_\epsilon)(x), \end{align*} $

where $ v_n $ is the volume of the unit ball. The usual way to prove (1.1) is based on Vitali type covering lemma, for details see [1]. Until 1981, Guzmán introduced the discretization technique in [2, Theorem 4.1.1].

Lemma 1.1    ([2])Let $ \{k_j\}_{j=1}^\infty\subseteq L^1(\Omega) $ be an ordinary sequence of functions, $ \{K_j\}_{j=1} $ the sequence of convolution operators associated to it, and $ K^* $ the corresponding maximal operator. Then $ K^* $ is of weak type (1, 1) if and only if $ K^* $ is of weak type (1, 1) over finite sums of Dirac deltas.

Hence, by Lemma 1.1, we only need to prove (1.1) holds for combinations of finite delta functions, then the weak type (1, 1) inequality of $ M $ holds for every $ f\in L^1(\mathbb R^n) $. Based on this fact, Carlesson [3] proved that the Hardy-Littlewood maximal operator is of weak type (1, 1) in another way. Other applications of discretization technique can be found in [4], [5], [6], [7] and [8].

For $ f\in L_{loc}^1(\mathbb R^n) $ and $ 0<\alpha<n $, the fractional maximal operator is defined as

$ \begin{align*} M_\alpha f(x)=\sup\limits_{\epsilon>0}\frac{1}{|B(x,\epsilon)|^{1-\frac{\alpha}{n}}}\displaystyle{\int}_{B(x,\epsilon)}|f(y)|dy. \end{align*} $

And for any $ \lambda>0 $,

$ \begin{align} |\{x\in\mathbb R^n: M_\alpha f(x)>\lambda\}|\leq\big(\frac{C_n\|f\|_1}{\lambda}\big)^\frac{n}{n-\alpha}, \end{align} $

(1.2)

where $ C_n $ is a constant only depending on $ n $. It means that $ M_\alpha $ is bounded from $ L^1 $ to $ L^{\frac{n}{n-\alpha},\infty} $. The usual way to prove (1.2) is by using Riesz potential $ I_\alpha $ to dominate $ M_\alpha $ in some sense, that is

$ \begin{align*} M_\alpha f(x) \leq \gamma(\alpha)\cdot I_\alpha(|f|) (x), \end{align*} $

where $ \gamma(\alpha)=\pi^{\frac{n}{2}}2^\alpha \Gamma(\frac{\alpha}{2})/ \Gamma({\frac{n-2}{2}}) $ and $ I_\alpha(f)(x)=\frac{1}{\gamma(\alpha)}\int_{\mathbb R^n}\frac{f(y)}{|x-y|^{n-\alpha}}dy $. For an in-depth understand of fractional integral operators, one can refer to [9] and [10].

Motivated by the work of Guzmán[2], Carlesson[3], we want to use the discrete method to prove the weak type (1, $ \frac{n}{n-\alpha} $) inequality for $ M_\alpha $. The fractional maximal operator also can be written as the convolution form,

$ \begin{align*} M_\alpha f(x)&=\sup\limits_{\epsilon>0}\frac{1}{|B(0,\epsilon)|^{1-\frac{\alpha}{n}}}\displaystyle{\int}_{B(0,\epsilon)}|f(x-y)|dy\\ &=\sup\limits_{\epsilon>0}\frac{1}{v_n^\frac{n-\alpha}{n}\epsilon^{n-\alpha}}\displaystyle{\int}_{\mathbb R^n}|f(x-y)|\chi_{B(0,\epsilon)}(y)dy\\ &=\sup\limits_{\epsilon>0}\frac{\epsilon^\alpha}{v_n^\frac{n-\alpha}{n}\epsilon^n}\displaystyle{\int}_{\mathbb R^n}|f(x-y)|\chi_{B(0,1)}(\frac{y}{\epsilon})dy\\ &=\sup\limits_{\epsilon>0}(|f|*k_\epsilon)(x), \end{align*} $

where $ k_\varepsilon(\cdot)=\frac{\epsilon^\alpha}{v_n^\frac{n-\alpha}{n}}\chi_{B(0,1)}(\cdot) $.

To prove our main result, one key lemma is needed.

Lemma 1.2    Let $ \{k_j\}_{j=1}^\infty\subseteq L^1(\Omega) $ be an ordinary sequence of functions, $ \{K_j\}_{j=1} $ the sequence of convolution operators associated to it, and $ K^* $ the corresponding maximal operator. Then $ K^* $ is of weak type (1, $ \frac{n}{n-\alpha} $) if and only if $ K^* $ is of weak type (1, $ \frac{n}{n-\alpha} $) over finite sums of Dirac deltas.

Proof.    By modifying the proof process of Lemma 1.1 step by step, we can prove Lemma 1.2, we omit the proof here. □

Let $ \mu=\sum\limits_{i=1}^{N}\delta_{t_i} $ and $ \|\mu\|_1=N $, where $ \delta_{t_i} $ is the Dirac delta function at $ t_i $. For any $ \lambda>0 $, (1.2) is equivalent to

$ \begin{align*} |\{x\in\mathbb R^n: M_\alpha \mu(x)>\lambda\}|\leq\big(\frac{C_nN}{\lambda}\big)^\frac{n}{n-\alpha}. \end{align*} $

The main result of the paper is the following.

Theorem 1.3    For any $ \lambda>0 $, $ |\{x\in\mathbb R^n: M_\alpha \mu(x)>\lambda\}|\leq \big(\frac{2^{n-\alpha}N}{\lambda}\big)^\frac{n}{n-\alpha}. $

Remark 1    Theorem 1.3 shows that the weak type (1, $ \frac{n}{n-\alpha} $) bound of $ M_\alpha $ is at worst $ 2^{n-\alpha} $. By using the discrete method, the weak type (1, $ \frac{n}{n-\alpha} $) bound of $ M_\alpha $ can be estimated more directly and easily, which is different from other ways, see for example [9].

Remark 2    We get the Hardy-Littlewood maximal operator $ M $ is of weak type (1, 1) and the weak (1, 1) bound is at worst $ 2^n $ if $ \alpha \rightarrow 0 $.

Proof. Denote $ E_\lambda=\{x\in\mathbb R^n:M_\alpha\mu(x)>\lambda\} $. The proof is by induction on $ N $.

When $ N=1 $, $ \mu=\delta_{t_1} $,

$ \begin{align*} M_\alpha\mu(x)&=\sup\limits_{\epsilon>0}\frac{1}{|B(x,\epsilon)|^{1-\frac{\alpha}{n}}}\displaystyle{\int}_{B(x,\epsilon)}\delta_{t_1}(y)dy =\frac{1}{(v_n^{1/n}|x-t_1|)^{n-\alpha}}, \end{align*} $

then $ E_\lambda=B\big(t_1,\frac{1}{v_n^{\frac{1}{n}}\lambda^\frac{1}{n-\alpha}}\big) $. Hence, $ |E_\lambda|=\big(\frac{1}{\lambda}\big)^{\frac{n}{n-\alpha}}\leq2^n\big(\frac{1}{\lambda}\big)^{\frac{n}{n-\alpha}} $.

When $ N>1 $, $ \forall\lambda>0 $, we can choose a ball $ B(x,\epsilon_x) $ for any $ x\in E_\lambda $ such that

$ \begin{align} |B(x,\epsilon_x)|^{1-\frac{\alpha}{n}}<\frac{1}{\lambda}\displaystyle{\int}_{B(x,\epsilon_x)}|\mu(y)|dy. \end{align} $

(2.3)

Let $ \epsilon=\sup\limits_{x\in E_\lambda}\epsilon_x $, then $ 0<\epsilon<\infty $. Morever, fix $ 0<\gamma<1 $, we can choose $ x_0\in E_\lambda $ such that $ \epsilon_{x_0}>\gamma\epsilon $.

Set

$ \begin{align*} \mu'(x)=\sum\limits_{t_i\in B(x_0,\epsilon_{x_0})}\delta_{t_i},\quad \mu''(x)=\sum\limits_{t_i\in B(x,\epsilon_x),t_i\notin B(x_0,\epsilon_{x_0})}\delta_{t_i}. \end{align*} $

Then $ \mu=\mu'+\mu'' $.

If $ x\in E_\lambda\backslash B(x_0,\frac{2}{\gamma}\epsilon_{x_0}) $, we have

$ |x-x_0|\geq \frac{2}{\gamma}\epsilon_{x_0}>2 \epsilon \geq \epsilon+\epsilon_{x_0}. $

Thus, $ B(x_0,\epsilon_{x_0})\cap B(x,\epsilon_x)=\varnothing $. By (2.3), we get

$ \begin{align*} \lambda&<\frac{1}{|B(x,\epsilon_x)|^{1-\frac{\alpha}{n}}}\displaystyle{\int}_{B(x,\epsilon_x)}|\mu(y)|dy\\ &=\frac{1}{|B(x,\epsilon_x)|^{1-\frac{\alpha}{n}}}\sum\limits_{t_i\in B(x,\epsilon_x)}1\\ &=\frac{1}{|B(x,\epsilon_x)|^{1-\frac{\alpha}{n}}}\sum\limits_{t_i\in B(x,\epsilon_x),t_i\in B(x_0,\epsilon_{x_0})}1+\frac{1}{|B(x,\epsilon_x)|^{1-\frac{\alpha}{n}}}\sum\limits_{t_i\in B(x,\epsilon_x),t_i\notin B(x_0,\epsilon_{x_0})}1\\ &=\frac{1}{|B(x,\epsilon_x)|^{1-\frac{\alpha}{n}}}\sum\limits_{t_i\in B(x,\epsilon_x),t_i\notin B(x_0,\epsilon_{x_0})}1\\ &=\frac{1}{|B(x,\epsilon_x)|^{1-\frac{\alpha}{n}}}\displaystyle{\int}_{B(x,\epsilon_x)}|\mu''(y)|dy\\ &=M\mu''(x). \end{align*} $

That is, $ M\mu''(x)>\lambda $.

Therefore, by the induction hypothesis, we have

$ \begin{align} \big|E_\lambda\backslash B(x_0,\frac{2}{\gamma}\epsilon_{x_0})\big|\leq2^n\big(\frac{\|\mu''\|_1}{\lambda}\big)^{\frac{n}{n-\alpha}}. \end{align} $

(2.4)

By using the following fact

$ \begin{align*} \big|B(x_0,\frac{2}{\gamma}\epsilon_{x_0})\big|=\big(\frac{2}{\gamma}\big)^n|B(x_0,\epsilon_{x_0})| \end{align*} $

and

$ \begin{align*} |B(x_0,\epsilon_{x_0})|^{1-\frac{\alpha}{n}} \leq\frac{1}{\lambda}\displaystyle{\int}_{B(x_0,\epsilon_{x_0})}|\mu(y)|dy =\frac{1}{\lambda}\sum\limits_{t_i\in B(x_0,\epsilon_{x_0}}\delta_{t_i}(y)dy =\frac{1}{\lambda}\displaystyle{\int}_{\mathbb R^n}|\mu'(y)|dy =\frac{\|\mu'\|_1}{\lambda}, \end{align*} $

we get the estimate

$ \begin{align} \big|B(x_0,\frac{2}{\gamma}\epsilon_{x_0})\big|\leq(\frac{2}{\gamma})^n\big(\frac{\|\mu'\|_1}{\lambda}\big)^{\frac{n}{n-\alpha}}. \end{align} $

(2.5)

Consequently, from (2.4) and (2.5), we get

$ \begin{align*} |E_\lambda|&\leq|E_\lambda\backslash B(x_0,\frac{2}{\gamma}\epsilon_{x_0})\big|+|B(x_0,\frac{2}{\gamma}\epsilon_{x_0})\big|\\ &\leq2^n\big(\frac{\|\mu''\|_1}{\lambda}\big)^{\frac{n}{n-\alpha}}+(\frac{2}{\gamma})^n\big(\frac{\|\mu'\|_1}{\lambda}\big)^{\frac{n}{n-\alpha}}\\ &\leq(\frac{2}{\gamma})^n\big(\frac{\|\mu''\|_1}{\lambda}\big)^{\frac{n}{n-\alpha}}+(\frac{2}{\gamma})^n\big(\frac{\|\mu'\|_1}{\lambda}\big)^{\frac{n}{n-\alpha}}\\ &\leq(\frac{2}{\gamma})^n\big(\frac{\|\mu\|_1}{\lambda}\big)^{\frac{n}{n-\alpha}}. \end{align*} $

Finally, let $ \gamma\rightarrow1 $. Then we conclude

$ \begin{align*} |E_\lambda|\leq2^n\big(\frac{N}{\lambda}\big)^{\frac{n}{n-\alpha}}. \end{align*} $

We complete the proof of Theorem 1.3. □