Let ƒ be a complex-valued measurable function that is locally integrable, denoted by $f \in L_{\mathrm{loc}}^1\left(\mathbb{R}^n\right)$. The non-centered Hardy-Littlewood maximal function Mƒ and the centered Hardy-Littlewood maximal function M^cƒ are defined by

$ M f(x)=\sup _{B_r \ni x} \frac{1}{\left|B_r\right|} \int_{B_r}|f(y)| \mathrm{d} y, $

and

$ M^{\mathrm{c}} f(x)=\sup _{r>0} \frac{1}{\left|B_r(x)\right|} \int_{B_r(x)}|f(y)| \mathrm{d} y . $

Here B_r is a ball with radius r, and B_r(x) is a ball with radius r and centered at the point x. For every subset $A \subset \mathbb{R}^n$, we use |A| to denote its Lebesgue measure of the set A.

The Hardy-Littlewood maximal function was introduced by Hardy and Littlewood^[1] for n=1, and by Wiener^[2] for n≥2. It is well-known that the Hardy-Littlewood maximal function plays an important role in various fields of analysis.

The operator M: $f \mapsto M f$ generated by the Hardy-Littlewood maximal function is named as Hardy-Littlewood maximal operator. It maps L^p($\mathbb{R}^n$) to itself for 1 < p ≤ ∞, and L¹($\mathbb{R}^n$) to L^{1, ∞}($\mathbb{R}^n$).M is a classical mean operater, and usually used to control some other important operators in harmonic analysis. More properties of Hardy-Littlewood maximal functions can be found in Refs.[3-4].

For some real number γ∈ $\mathbb{R}$ satisfying 0 < γ < ∞, the truncated non-centered Hardy-Littlewood maximal function M_γƒ and the truncated centered Hardy-Littlewood maximal function $M_γ^{\mathrm{c}} f$ can be defined by

$ M_\gamma f(x)=\sup _{\substack{B_r \ni x \\ 0<r<\gamma}} \frac{1}{\left|B_r\right|} \int_{B_r}|f(y)| \mathrm{d} y, $

(1)

and

$ M_\gamma^{\mathrm{c}} f(x)=\sup _{0<r<\gamma} \frac{1}{\left|B_r(x)\right|} \int_{B_r(x)}|f(y)| \mathrm{d} y . $

Although both operators M_γ and $M_\gamma^c$ are obviously controlled by M and M^c, they have the same norms. Wei et al.^[5] proved the (p, p) norm of M_γ equals to that of M for 1 < p≤∞ and 0 < γ < ∞. For the case of p=1, they also proved that the weak (1, 1) norm of M_γ equals to that of M for 0 < γ < ∞. As for $M_\gamma^{\mathrm{c}}$, the result is true as well.

The regularity properties of Mƒ and M^cƒ are of significant interest due to their successful applications in the study of Sobolev functions and partial differential equations. A deal of work has been done since Kinnunen^[6] proved that M^c is bounded in the Sobolev space W^{1, p}($\mathbb{R}^n$) for 1 < p≤∞. Aldaz and Pérez^[7] proved that for a nondegenerate interval I if ƒ: I→$\mathbb{R}$ is of bounded variation, then Mƒ is absolutely continuous.

In 2020, Wu and Yan^[8] obtained a characterization of continuity of Mƒ and M^cƒ, and investigated the relationships between the continuity of Mƒ and the continuity or bounded variation of ƒ.

Theorem A   Let ƒ∈$L_{\mathrm{loc}}^1\left(\mathbb{R}^n\right) $ and Mƒ(a) < ∞. The non-centered Hardy-Littlewood maximal function Mƒ is continuous at the point a∈$\mathbb{R}^n$ if and only if there exists $\tilde{f}$ such that

$ \tilde{f}(x)=f(x) $

for almost every x∈$\mathbb{R}^n$, and

$ \limsup _{x \rightarrow a}|\tilde{f}(x)| \leqslant M^{\mathrm{c}} f(a) . $

Theorem B   Let ƒ∈$L_{\mathrm{loc}}^1\left(\mathbb{R}^n\right) $ and M^cƒ(a) < ∞. The centered Hardy-Littlewood maximal function Mƒ is continuous at the point a∈$\mathbb{R}^n$ if and only if there exists $\tilde{f}$ such that

$ \tilde{f}(x)=f(x) $

for almost every x∈$\mathbb{R}^n$, and

$ \limsup _{x \rightarrow a}|\tilde{f}(x)| \leqslant M^{\mathrm{c}} f(a) . $

1 Main results

Noticing the same properties on norms of M_γ and M, it is natural to consider whether M_γƒ and $M_\gamma^{\mathrm{c}} f$ have similar continuity characterizations to that in Theorem A and B. Finally, we obtain the following results.

Theorem 1.1   Ifƒ∈$L_{\mathrm{loc}}^1\left(\mathbb{R}^n\right)$ and M_γƒ(x₀) < ∞, then M_γƒ is continuous at the point x₀ if and only if there exists a function $\tilde{f}$ such that

$ \tilde{f}(x)=f(x) $

(2)

for almost every x∈$\mathbb{R}^n$, and

$ \limsup _{x \rightarrow x_0}|\tilde{f}(x)| \leqslant M_\gamma f\left(x_0\right) . $

(3)

Theorem 1.1 characterizes an equivalent condition of continuity of M_γƒ. For $M_\gamma^{\mathrm{c}} f$, we have a similar characterization in Theorem 1.2.

Theorem 1.2   If ƒ ∈ $L_{\mathrm{loc}}^1\left(\mathbb{R}^n\right)$ and $M_\gamma^{\mathrm{c}} f\left(x_0\right)$ < ∞, then $M_\gamma^{\mathrm{c}} f$ is continuous at the point x₀ if and only if there exists a function $\tilde{f}$ such that

$ \tilde{f}(x)=f(x), $

for almost every x∈$\mathbb{R}^n$, and

$ \limsup _{x \rightarrow x_0}|\tilde{f}(x)| \leqslant M_γ^{\mathrm{c}} f\left(x_0\right) . $

2 Some facts and lemmas

In this section, we give two useful lemmas. In Lemma 2.1, we prove the lower semi-continuity of M_γƒ and $M_\gamma^{\mathrm{c}} f$. In Lemma 2.2 and 2.3, we investigate the upper semi-continuity with respect to γ of the value M_γƒ(x₀) and $M_\gamma^{\mathrm{c}} f\left(x_0\right)$ for some fixed x₀∈$\mathbb{R}^n$.

It is well-known that Mƒ and M^cƒ are lower semi-continuous. So are M_γƒ and $M_\gamma^{\mathrm{c}} f$. Now we provide the proof of the lower semi-continuity of M_γƒ and $M_\gamma^{\mathrm{c}} f$.

Lemma 2.1   If ƒ∈ $L_{\mathrm{loc}}^1\left(\mathbb{R}^n\right)$, 0 < γ < ∞ and M_γƒ(x₀) < ∞, then M_γƒ and $M_\gamma^{\mathrm{c}} f$ are lower semi-continuous at the point x₀.

Proof   For any ε>0, according to the definition of M_γƒ, there exists a ball $B_r \ni x_0$ with radius r∈(0, γ) such that

$ \frac{1}{\left|B_r\right|} \int_{B_r}|f(y)| \mathrm{d} y>M_\gamma f\left(x_0\right)-\frac{\varepsilon}{2} . $

Take δ∈(0, γ-r) such that

$ \frac{\left|B_{r+\delta}\right|}{\left|B_r\right|}<\frac{2 M_\gamma f\left(x_0\right)-\varepsilon}{2 M_\gamma f\left(x_0\right)-2 \varepsilon} . $

Let $\hat{x}_0$ denote the center of B_r. For any x∈B_δ(x₀), we have

$ \left|x-\hat{x}_0\right| \leqslant\left|x-x_0\right|+\left|x_0-\hat{x}_0\right|<r+\delta, $

thus $B_\delta\left(x_0\right) \subset B_{r+\delta}\left(\hat{x}_0\right)$. For any x∈B_δ(x₀), we conclude that

$ \begin{aligned} M_\gamma f(x) & \geqslant \frac{1}{\left|B_{r+\delta}\left(\hat{x}_0\right)\right|} \int_{B_{r+\delta\left(\hat{x}_0\right)}}|f(y)| \mathrm{d} y \\ & \geqslant \frac{\left|B_r\right|}{\left|B_{r+\delta}\right|} \cdot \frac{1}{\left|B_r\right|} \int_{B_{r\left(\hat{x}_0\right)}}|f(y)| \mathrm{d} y \\ & >\frac{2 M_\gamma f\left(x_0\right)-2 \varepsilon}{2 M_\gamma f\left(x_0\right)-\varepsilon} \cdot\left(M_\gamma f\left(x_0\right)-\frac{\varepsilon}{2}\right) \\ & =M_\gamma f\left(x_0\right)-\varepsilon . \end{aligned} $

(4)

The calculation (4) implies that M_γƒ is lower semi-continuous at the point x₀.

The proof of $M_\gamma^{\mathrm{c}} f$ is quite similar. We just need to require $\hat{x}_0=x_0$ as an extra condition.□

In order to prove Theorem 1.1, we also need the following crucial lemma that proves the upper semi-continuity with respect to γ of the value M_γƒ(x₀) for fixed x₀.

Lemma 2.2   Let ƒ∈$L_{\mathrm{loc}}^1\left(\mathbb{R}^n\right)$. For fixed x₀∈ $\mathbb{R}^n$, define

$ \varphi(\gamma)=M_\chi f\left(x_0\right)=\sup _{\substack{B_r \ni x_0 \\ 0<r<\gamma}} \frac{1}{\left|B_r\right|} \int_{B_r}|f(y)| \mathrm{d} y . $

(5)

Then one of the following two cases is true:

(i) φ(γ)=∞ for any γ∈(0, ∞);

(ii) φ is upper semi-continuous on (0, ∞).

Proof    Suppose that there exists γ₀>0 such that φ(γ₀)=∞. Now we prove that (i) is true.

Notice the observation that γ>γ₀ implies

$ \begin{aligned} \varphi(\gamma) & =\sup _{\substack{B_r \ni x_0 \\ 0<r<\gamma}} \frac{1}{\left|B_r\right|} \int_{B_r}|f(y)| \mathrm{d} y \\ & \geqslant \sup _{\substack{B_r \ni x_0 \\ 0<r<\gamma}} \frac{1}{\left|B_k\right|} \int_{B_k}|f(y)| \mathrm{d} y \\ & =\varphi\left(\gamma_0\right) . \end{aligned} $

(6)

It immediately follows from (6) that φ(γ)=∞ for any γ>γ₀.

Next we just need to prove that φ(γ)=∞ for any γ∈(0, γ₀). By the definition of Mƒ, we can choose a sequence of balls $\left\{B_k\right\}_{k=1}^{\infty}$ such that each of them contains x₀, the radius r_k of B_k is less than γ₀, and

$ \frac{1}{\left|B_k\right|} \int_{B_k}|f(y)| \mathrm{d} y=\frac{1}{V_n r_{k B_k}^n}|f(y)| \mathrm{d} y \geqslant k, $

(7)

where V_n is the Lebesgue measure of the ndimensional unit ball.

Then we claim that

$ \lim _{k \rightarrow \infty} r_k=0 . $

(8)

Assume that the assertion (8) is true. Then for any γ∈(0, γ₀), there exists an integer K>0 such that r_k < γ for any k>K. It follows from (7) that

$ \begin{aligned} \varphi(\gamma) & =\sup _{\substack{B_r \ni x_0 \\ 0<r<\gamma}} \frac{1}{\left|B_r\right|} \int_{B_r}|f(y)| \mathrm{d} y \\ & \geqslant \sup _{k>K} \frac{1}{\left|B_r\right|} \int_{B_r}|f(y)| \mathrm{d} y \\ & =\infty . \end{aligned} $

Now let us prove the assertion (8) is true. Recalling that ƒ∈$L_{\mathrm{loc}}^1\left(\mathbb{R}^n\right)$ and noticing that $B_k \subset B_{2 \gamma_0}\left(x_0\right)$

$ \int\limits_{{B_k}} {f(y)|{\rm{d}}y} \int\limits_{{B_{2{\gamma _0}}}\left( {{x_0}} \right)} {|f(y)|{\rm{d}}y: = C < \infty } . $

(9)

It follows from (7) and (9) that

$ r_k \leqslant \frac{C}{V_n k} . $

(10)

The inequality (10) implies the assertion (8).

Next we investigate the other case. If φ(γ) < ∞ for any γ∈(0, ∞), we can prove case (ii) by contradiction. Assume that case (ii) is false. Then there exists a point γ₀ such that φ is not upper semi-continuous at the point. Furthermore, there exists α>0 such that

$ \limsup _{\gamma \rightarrow\gamma_0} \varphi(\gamma)>\varphi\left(\gamma_0\right)+\alpha $

(11)

Recalling the observation (6), the statement (11) implies that there exists a sequence $\left\{\boldsymbol{\gamma}_k\right\}_{k=1}^{\infty}$ such that γ₀ < γ_k < 2γ₀, monotonically decreasing,

$ \lim _{k \rightarrow \infty} \gamma_k=\gamma_0, $

(12)

and

$ \varphi\left(\gamma_k\right)>\varphi\left(\gamma_0\right)+\frac{\alpha}{2} $

holds for any k≥1. By the definition (5) of φ, there exists a sequence of balls $\left\{B_k\right\}_{k=1}^{\infty}$ such that each of them contains x₀,

$ \frac{1}{{\left| {{B_k}} \right|}}\int\limits_{{B_k}} {f(y)|{\rm{d}}y > \varphi \left( {{\gamma _0}} \right) + \frac{\alpha }{4}, } $

(13)

and 0 < r_k < γ_k, where r_k denotes the radius of B_k. The statements (12) and (13) imply that γ₀ < r_k < γ_k and r_k→γ₀ as k→∞. We denote the center of B_k by $\hat{x}_k$. Let B′_k be the ball with radius r′_k=2γ₀-r_k centered at the point

$ \begin{aligned} \hat{x}_k^{\prime} & =\left(1-\frac{r_k^{\prime}}{r_k}\right) x_0+\frac{r_k^{\prime}}{r_k} \hat{x}_k \\ & =2\left(1-\frac{\gamma_0}{r_k}\right) x_0+\left(\frac{2 \gamma_0}{r_k}-1\right) \hat{x}_k \end{aligned} $

Then we can deduce that x₀∈ $B_k^{\prime} \subset B_k$, 0 < r′_k < γ₀ and r′_k→γ₀ as k→∞. By the definition (5) of φ, the statement (13) implies that

$ \begin{aligned} \frac{1}{\left|B_k\right|} \int_{B_k}|f(y)| \mathrm{d} y & >\varphi\left(\gamma_0\right)+\frac{\alpha}{4} \\ & >\frac{1}{\left|B_k^{\prime}\right|} \int_{B_k^{\prime}}|f(y)| \mathrm{d} y+\frac{\alpha}{4} . \end{aligned} $

Then we have that

$ \begin{aligned} \limsup _{k \rightarrow \infty} & \left(\frac{1}{\left|B_k\right|} \int_{B_k}|f(y)| \mathrm{d} y-\right. \\ & \left.\frac{1}{\left|B_k^{\prime}\right|} \int_{B_k^{\prime}}|f(y)| \mathrm{d} y\right) \geqslant \frac{\alpha}{4} . \end{aligned} $

(14)

On the other hand, denote the characteristic function of $A \subset \mathbb{R}^n \text { by } \chi_A$. For any γ₀∈(0, ∞), let

$ \begin{aligned} g(x) & =|f(x)| \chi_{B_{4 \gamma_0}\left(x_0\right)}(x) \\ & =\left\{\begin{array}{cc} |f(x)|, & x \in B_{B_{4 \gamma_0}}\left(x_0\right), \\ 0, & x \notin B_{4 \gamma_0}\left(x_0\right) . \end{array}\right. \end{aligned} $

Then we obtain that g∈L¹($\mathbb{R}^n$) is implied just by $f \in L_{\mathrm{loc}}^1\left(\mathbb{R}^n\right)$. For k≥1, set

$ f_k(x)=|f(x)| \chi_{B_k \backslash B_k^{\prime}}(x), \quad x \in \mathbb{R}^n . $

Due to 0 < r′_k < γ₀ < r_k < 2γ₀ and the choice of B′_k, we have $B_k^{\prime} \subset B_k \subset B_{4 \gamma_0}\left(x_0\right)$. It follows that $B_k \backslash B^{\prime}{ }_k \subset B_{4 \gamma_0}\left(x_0\right)$, which implies that ƒ_k is controlled by g. Using the Lebesgue dominated convergence theorem, we obtain that

$ \begin{aligned} & \lim _{k \rightarrow \infty}\left(\frac{1}{\left|B_k\right|} \int_{B_k}|f(y)| \mathrm{d} y-\frac{1}{\left|B_k^{\prime}\right|} \int_{B_k^{\prime}}|f(y)| \mathrm{d} y\right) \\ & \leqslant \lim _{k \rightarrow \infty} \frac{1}{\left|B_k\right|}\left(\int_{B_k}|f(y)| \mathrm{d} y-\int_{B_k^{\prime}}|f(y)| \mathrm{d} y\right) \\ &= \lim _{k \rightarrow \infty} \frac{1}{\left|B_k\right|} \int_{B_k \backslash B_k^{\prime}}|f(y)| \mathrm{d} y \\ &= \lim _{k \rightarrow \infty} \frac{1}{\left|B_k\right|} \int_{\mathbf{R}^n}|f(y)| \chi_{B_k \backslash B_k^{\prime}}(y) \mathrm{d} y \\ &= \lim _{k \rightarrow \infty} \frac{1}{\left|B_k\right|} \int_{\mathbf{R}^n} f_k(y) \mathrm{d} y \\ &= \frac{1}{\left|B_k\right|} \int_{\mathbf{R}^n } \lim _{k\rightarrow \infty} f_k(y) \mathrm{d} y=0, \text { thus } \\ & \limsup _{k \rightarrow \infty}\left(\frac{1}{\left|B_k\right|} \int_{B_k}|f(y)| \mathrm{d} y-\frac{1}{\left|B_k^{\prime}\right|} \int_{B_k^{\prime}}|f(y)| \mathrm{d} y\right) \leqslant 0, \end{aligned} $

which contradicts with (14).

Therefore case (ii) is true.□

For the centered version, we have a similar result, and the proof can be repeated almost word for word. We just need an additional condition that the center of the ball is at the point x₀ every time we take balls.

Lemma 2.3   Let $f \in L_{\mathrm{loc}}^1\left(\mathbb{R}^n\right)$. For fixed x₀∈$\mathbb{R}^n$, define

$ \varphi_{\mathrm{c}}(\gamma)=M_{\gamma}^{\mathrm{c}} f\left(x_0\right)=\sup _{0<r<\gamma} \frac{1}{\left|B_r\left(x_0\right)\right|} \int_{B_r\left(x_0\right)}|f(y)| \mathrm{d} y . $

Then one of the following two cases is true:

(i) φ_c(γ)=∞ for any γ∈(0, ∞);

(ii) φ_c is upper semi-continuous on (0, ∞).

3 Proof of main theorems and discussions

Now we present the proof of Theorem 1.1.

Proof of Theorem 1.1   First we consider the necessity. Define

$ \tilde{f}(x)= \begin{cases}M_γ f(x), & |f(x)|>M_\gamma f(x), \\ f(x), & |f(x)| \leqslant M_\gamma f(x) .\end{cases} $

(15)

If x is one of the Lebesgue points of ƒ, then we have that

$ |f(x)| \leqslant M_\gamma^{\mathrm{c}} f(x) \leqslant M_\gamma f(x) . $

(16)

Due to ƒ∈$L_{\mathrm{loc}}^1\left(\mathbb{R}^n\right)$, almost every x∈$\mathbb{R}^n$ is a Lebesgue point of ƒ. It follows that (16) holds almost everywhere. Furthermore, $\tilde{f}$=f holds almost everywhere on $\mathbb{R}^n$.

On the other hand, (15) implies that $|\tilde{f}(x)| \leqslant M_\gamma f(x)$. Therefore the continuity of M_γƒ yields that

$ \limsup _{x \rightarrow x_0}|\tilde{f}(x)| \leqslant \limsup _{x \rightarrow x_0} M_\gamma f(x)=M_\gamma f\left(x_0\right) . $

Next we prove the sufficiency by contradiction. Suppose that M_γƒ is not continuous at the point x₀. Then M_γƒ is not upper semi-continuous at x₀, since M_γƒ is lower semi-continuous at the point x₀. Thus there exists a positive real number α>0 such that

$ \limsup _{x \rightarrow x_0} M_γ f(x)>M_γ f\left(x_0\right)+\alpha . $

(17)

The statement (17) implies that there exists a sequence of balls $\left\{\boldsymbol{B}_k\right\}_{k=1}^{\infty}$ such that x_k→x₀ as k→∞ and

$ M_γ f\left(x_k\right)>M_\gamma f\left(x_0\right)+\alpha $

holds for any k≥1. By the definition (1) of M_γƒ, for any x_k, there exists a ball $B_k \ni x_k$ such that the radius r_k of B_k is less than γ and

$ \begin{aligned} \frac{1}{\left|B_k\right|} \int_{B_k}|f(y)| \mathrm{d} y & >M_\gamma f\left(x_k\right)-\frac{\alpha}{2} \\ & >M_\gamma f\left(x_0\right)+\frac{\alpha}{2} . \end{aligned} $

(18)

We claim that

$ \lim _{k \rightarrow \infty} r_k=0 . $

(19)

Assume that the assertion (19) is true. It follows from (2) and (18) that

$ \begin{aligned} M_\gamma f\left(x_0\right)+\frac{\alpha}{2} & <\frac{1}{\left|B_k\right|} \int_{B_k}|f(y)| \mathrm{d} y \\ & =\frac{1}{\left|B_k\right|} \int_{B_k}|\tilde{f}(y)| \mathrm{d} y . \end{aligned} $

(20)

For any B_k, there exists x_k∈B_k such that

$ \left|\tilde{f}\left(\bar{x}_k\right)\right| \geqslant \frac{1}{\left|B_k\right|} \int_{B_k}|\tilde{f}(y)| \mathrm{d} y>M_\gamma f\left(x_0\right)+\frac{\alpha}{2} . $

We can deduce that x_k→x₀ as k→∞ from (19) since x_k→x₀ as k→∞. It follows that

$ \limsup _{x \rightarrow x_0}|\tilde{f}(x)| \geqslant \limsup _{k \rightarrow \infty}\left|\tilde{f}\left(\bar{x}_k\right)\right| \geqslant M_\gamma f\left(x_0\right)+\frac{\alpha}{2}, $

which contradicts with (3). Therefore M_γƒ is continuous at the point x₀.

Now we prove the assertion (19) by contradiction again. Suppose that the assertion (19) is false, then we have

$ 0<c:=\limsup _{k \rightarrow \infty} r_k \leqslant \gamma . $

Take a subsequence of $\left\{r_k\right\}_{k=1}^{\infty}$, still denoted by $\left\{r_k\right\}_{k=1}^{\infty}$ without loss of clarity, such that

$ \lim _{k \rightarrow \infty} r_k=c>0 . $

Lemma 2.2 implies that M_γƒ(x₀) is upper semi-continuous with respect to γ. Thus there exists γ′> γ such that

$ M_{\gamma^{\prime}} f\left(x_0\right) \leqslant M_\gamma f\left(x_0\right)+\frac{\alpha}{4} . $

(21)

Take ε=γ′-γ. Then there exists K>0 such that |x_k-a| < ε for any k>K. Recalling that 0 < r_k < γ, we have

$ \rho_k:=r_k+\left|x_k-x_0\right|<\gamma^{\prime} . $

(22)

Denote the center of B_k by $\hat{x}_k $. Then we have

$ \begin{aligned} \left|\hat{x}_k-x_0\right| & \leqslant\left|\hat{x}_k-x_k\right|+\left|x_k-x_0\right| \\ & <r_k+\left|x_k-x_0\right|=\rho_k \end{aligned} $

which implies that x₀∈B_ρk($\hat{x}_k $). For any k>K, it follows from (20), (21), and (22) that

$ \begin{aligned} M_\gamma f\left(x_0\right)+\frac{\alpha}{4} & \geqslant M_\gamma f\left(x_0\right) \\ & \geqslant \frac{1}{\left|B_{\rho_k}\left(\hat{x}_k\right)\right|} \int_{B_{\rho_k}\left(\hat{x}_k\right)}|f(y)| \mathrm{d} y \\ & \geqslant \frac{\left|B_k\right|}{\left|B_{\rho_k}\left(\hat{x}_k\right)\right|} \cdot \frac{1}{\left|B_k\right|} \int_{R_k}|f(y)| \mathrm{d} y \\ & >\frac{r_k^n}{\left(r_k+\left|x_k-x_0\right|\right)^n} \cdot\left(M_\gamma f\left(x_0\right)+\frac{\alpha}{2}\right) . \end{aligned} $

Let k→∞, then we deduce that

$ M_\gamma f\left(x_0\right)+\frac{\alpha}{4} \geqslant M_\gamma f\left(x_0\right)+\frac{\alpha}{2}, $

which is a contradiction. Therefore the assertion (19) is true.□

The proof of Theorem 1.2 is quite similar to the proof of Theorem 1.1. In fact, we just need an additional condition that the center of the ball is at the point x₀ every time we take balls. So we omit the proof of Theorem 1.2.

The applications and discussions of the classical Hardy-Littlewood maximal functions presented by Wu and Yan^[8] make sense for the truncated version as well, and the proof just need to be adjusted as we did in Theorem 1.1. Now we give the results without proof.

Theorem 3.1   Let ƒ∈$L_{\mathrm{loc}}^1\left(\mathbb{R}^n\right)$. If ƒ is continuous at the point x₀, then M_γƒ and $M_\gamma^{\mathrm{c}} f$ is continuous at x₀.

Theorem 3.2   Let ƒ∈ $L_{\mathrm{loc}}^1\left(\mathbb{R}^n\right)$. If ƒ is of locally bounded variation at the point x₀, then M_γƒ is continuous at x₀.

Remark 3.1   The same as the classical version, Theorem 3.2 is not true on $\mathbb{R}^n$ for n≥2. We can use the function ƒ showed by Wu and Yan^[8] to explain it as well. For the convenience of reading, we repeat the definition of ƒ. We just consider the case of n=2.

Define a set $E \subset \mathbb{R}^2$ by

E: ={(x, y)∈ $\mathbb{R}^2$ |x < 1, x+y>0, x-y>0}.

For k=1, 2, …, define B_k, $C_k \subset \mathbb{R}$ by

$ \begin{aligned} & B_k:=\left\{x \in \mathbb{R} \left\lvert\, \frac{1}{2^k} \leqslant x<\frac{1}{2^k}+\frac{1}{2^{k+2}}\right.\right\}, \\ & C_k:=\left\{x \in \mathbb{R} \left\lvert\, -\frac{1}{2^{k+1}} \leqslant <\frac{1}{2^{k+1}}\right.\right\}, \end{aligned} $

and let

A_k=B_k×C_k.

Then $A_k \subset E$ for k≥1.

Furthermore, define ƒ: $\mathbb{R}^2 \rightarrow \mathbb{R}$ by

$ f(x, y):=\sum\limits_{k=1}^{\infty} \chi_{A_k}(x, y), $

where χ_Ak is the characteristic function of A_k. Then ƒ is of locally bounded variation at the coordinate origin (0, 0). Take γ=2, and consider the point sequence $\left\{p_k=\left(\frac{1}{2^k}, 0\right)\right\}_{k=1}^{\infty} $. It is obvious that p_k→(0, 0) as k→∞. By the definition (1) of M_γƒ, we have

M_γƒ(p_k)=1,

and

$ M_\gamma f(0, 0) \geqslant \frac{1}{2} . $

Therefore, M_γƒ is not continuous at (0, 0).

Remark 3.2   The converse of the Theorem 3.2 is false as well. It is equivalent to that there exists a function ƒ∈$L_{\mathrm{loc}}^1\left(\mathbb{R}^n\right)$, a real number γ>0 and a point x₀∈ $\mathbb{R}$ such that M_γƒ is continuous at x₀ but ƒ does not equal to a function g almost everywhere, where g is of locally bounded variation at x₀. Here we give a counterexample. Consier the function

$ f(x)= \begin{cases}x \sin \left(\frac{1}{x}\right), & 0<x<1 \\ 0, & x \leqslant 0 \text { or } x \geqslant 1 .\end{cases} $

Notice that ƒ is continuous. By the Theorem 3.1, M_γƒ is continuous at the point 0. On the other hand, the continuity of ƒ implies that if g=ƒ almost everywhere, then g=ƒ everywhere. Thus we just need to show that ƒ is not of locally bounded variation at 0. For x≠0 and x≠1, calculate the derivative of ƒ, and we have

$ f^{\prime}(x)= \begin{cases}\sin \left(\frac{1}{x}\right)-\frac{1}{x} \cos \left(\frac{1}{x}\right), & 0<x<1, \\ 0, & x<0 \text { or } x>1 .\end{cases} $

Then for any δ>0, there exists an integer K≥1 such that

$ \frac{1}{2 K \pi+2 \pi / 3}<\delta . $

Then we have

$ \begin{aligned} & \int_{-\delta}^\delta\left|f^{\prime}(x)\right| \mathrm{d} x \\ \geqslant & \sum\limits_{k=K}^{\infty} \int_{\frac{1}{2 k\mathsf{π}++\mathsf{π}}}^{ \frac{1}{2 k\mathsf{π}+2 \mathsf{π} / 3}}\left|\sin \left(\frac{1}{x}\right)-\frac{1}{x} \cos \left(\frac{1}{x}\right)\right| \mathrm{d} x \\ \geqslant & \sum\limits_{k=K}^{\infty} \int_{\frac{1}{2 k \mathsf{π}+\mathsf{π}}}^{\frac{1}{2 k \mathsf{π}+\frac{2 \mathsf{π}}{3}}} k \mathsf{π} \mathrm{~d} x=\sum\limits_{k=K}^{\infty} \frac{k}{2(2 k+1)(3 k+1)} \\ \geqslant & \sum\limits_{k=K}^{\infty} \frac{1}{12(k+1)}=\infty . \end{aligned} $

Thus ƒ is not of locally bounded variation at 0.

Remark 3.3   Observe that the codomain of M_γƒ is [0, ∞]. We can consider taking ∞ as a valid value when investigating the pointwise continuity of M_γƒ. If we generalize the definition of pointwise continuity in the following way: M_γƒ is continuous at the point x₀∈$\mathbb{R}^n$, if

$ \lim _{x \rightarrow x_0} M_γ f(x)=M_γ f\left(x_0\right) . $

This would not cause confusions. Then M_γƒ is continuous at the point x₀ if M_γƒ(x₀)=∞, since M_γƒ is lower continuous. Thus in this sense, the condition M_γƒ(x₀) < ∞ can be removed from theorems or lemmas.