The purpose of this paper is to establish the continuity of L^p(Rⁿ)→L^p(Rⁿ)-norm of the truncated Hardy-Littlewood maximal operator.

1 Introduction

We begin with the definition of the truncated maximal operator studied here. Given two truncation parameters a and b which satisfy 0 < a ≤ b ≤ ∞, for a locally integrable function f on Rⁿ we define the truncated Hardy-Littlewood maximal operator M_a^b as follows:

$ M_a^b f(x)=\sup\limits _{a<r<b} \frac{1}{|B(x, r)|} \int_{B(x, r)}|f(y)| \mathrm{d} y, $

(1)

where B(x, r) denotes the ball centered at x∈Rⁿ with the radius r. Note that by the absolute continuity of the Lebesgue integral, for b < ∞, we have

$ M_a^b f(x)=\sup\limits _{a \leqslant r \leqslant b} \frac{1}{|B(x, r)|} \int_{B(x, r)}|f(y)| \mathrm{d} y, $

which means, in the case b < ∞, the definition of M_a^b has nothing to do with whether the radius r is taken as the truncation parameters a and b.

Recall that for a locally integrable function f, the classical Hardy-Littlewood maximal operator M is defined by

$ M f(x)=\sup\limits _{r>0} \frac{1}{|B(x, r)|} \int_{B(x, r)}|f(y)| \mathrm{d} y . $

The classical Hardy-Littlewood maximal operator is a fundamental tool to study harmonic analysis, potential theory, and the theory of partial differential equations^[1-2].

Meanwhile, how to compute the continuity of the operator norm of the classical Hardy-Littlewood maximal operator is an important problem.

As is known to all, the truncated operators have many significant properties and are closely related to their corresponding operators. For example, as shown in Ref.[3], the L^p-boundedness of the truncated operator is equivalent to that of its corresponding oscillatory operator. Therefore, in order to facilitate and simplify our study of the classical maximal operators, we can first study truncated maximal operators and try to figure out their relationships with the classical maximal operator.

The operator M_a^b was first introduced by Wei et al.^[4] and Zhang et al.^[5] (in their papers, they study the case a=0).

As long as 0 < a′≤a and b≤b′ < ∞, in the pointwise sense, it can be deduced immediately from the definition (1) that

$ M_a^b f(x) \leqslant M_{a^{\prime}}^{b^{\prime}} f(x) \leqslant M f(x), $

holds for any x∈Rⁿ. It then follows that

$ \begin{aligned} & \left\|M_a^b\right\|_{L^p\left(\mathbf{R}^n\right) \rightarrow L^p\left(\mathbf{R}^n\right)} \\ & \leqslant\left\|M_{a^{\prime}}^{b^{\prime}}\right\|_{L^p\left(\mathbf{R}^n\right) \rightarrow L^p\left(\mathbf{R}^n\right)} \leqslant\|M\|_{L^p\left(\mathbf{R}^n\right) \rightarrow L^p\left(\mathbf{R}^n\right)} . \end{aligned} $

(2)

It was proved in Refs.[4-5] that for 1 < p < ∞, ‖M_a^b‖_{L^p(Rⁿ)→L^p(Rⁿ)}≤‖M_c^d‖_{L^p(Rⁿ)→L^p(Rⁿ)}. if b/a=d/c by using the dilation. Therefore, with respect to the parameter θ=b/a in [1, ∞], we can set

$ g(\theta)=\left\|M_a^b\right\|_{L^p\left(\mathbf{R}^n\right) \rightarrow L^p\left(\mathbf{R}^n\right)} . $

(3)

Furthermore, applying inequality (2) to the function g in formula (3), one can show that g is precisely an increasing function. Moreover, in the case p=1, Wu et al.^[6] give the explicit expression of g (and extend their result to the case of general strong truncated Hardy-Littlewood maximal operator), and hence show the continuity of g in this setting.

Motivated by these works, in this paper, we place our emphasis on exploring the continuity of g for the case p>1.

Now we formulate our main theorems as follows.

Theorem 1.1 Let p>1 be a real number. Then the function g in formula (3) is a continuous function for θ∈[1, ∞].

Remark 1.1 It is noteworthy that the continuity of case p=1 discussed in Ref.[6] can be viewed as a complement to our theorem to yield that for any p≥1, g is a continuous function.

The rest of the present paper will be organized as follows: In section 2, we will give some lemmas.

The continuity of the L^p(Rⁿ)→L^p(Rⁿ)-norm of M_a^b will be proved in section 3. A tacit understanding in the present paper is that ‖·‖_L^p is the L^p(Rⁿ)→L^p(Rⁿ)-norm of an operator, and ‖·‖_L^p represents the normal L^p(Rⁿ)-norm of a function. For simplicity, we will abbreviate the L^p(Rⁿ)→L^p(Rⁿ)-norm of an operator as the L^p-norm. We denote by |E| the Lebesgue measure of the measurable set E in Rⁿ.

2 Some facts and lemmas

Before we proceed to prove our main results, several basic lemmas are needed. The first lemma is used to reduce the space we consider, while the second one is a technical lemma which is useful in the proof of Theorem 1.1.

Lemma 2.1 For 0 < a≤b≤∞ and 1≤p < ∞, we have

$ \left\|M_a^b\right\|_{L^p}=\sup\limits _{\phi \in C_c\left(\mathbf{R}^n\right):\|\phi\|_p=1}\left\|M_a^b \phi\right\|_p . $

The proof of Lemma 2.1 is just a repetition of the proof of Lemma 2.5 in Ref. [4], and hence we omit it.

Lemma 2.2 Suppose α, β are two positive real numbers. For 1≤p < ∞, we have

$ (\alpha+\beta)^p \leqslant \alpha^p+p 2^{p-1} \alpha^{p-1} \beta+p 2^{p-1} \beta^p . $

Proof: Applying the Taylor formula yields that

$ (\alpha+\beta)^p \leqslant \alpha^p+p(\alpha+\beta)^{p-1} \beta . $

then the desired result follows from the fact that

$ (\alpha+\beta)^{p-1} \leqslant 2^{p-1}\left(\alpha^{p-1}+\beta^{p-1}\right), $

and some simple calculations.

3 Continuity of the L^p-norm of the truncated maximal operator

In this section, we present the proof of Theorem 1.1. Recall that ɡ(θ)=‖M_a^b‖_{L^p(Rⁿ)→L^p(Rⁿ)}.

Now, we are in the position to give the proof of the continuity of g with respect to θ.

Proof of Theorem 1.1

Due to the dilation property of maximal operator, we only need to show ‖M_a^b‖_L^p is continuous with respect to b for any fixed a>0.

Fix a>0. For b∈[a, ∞], let us define the function $\hat{g}(b) $=‖M_a^b‖_L^p, so that we merely need to prove the continuity of $ \hat{g}$ on [a, ∞]. We denote $\hat{g} $ by g, which makes sense.

For clarity, we will give the proof in three steps.

(ⅰ) g is left semi-continuous.

(ⅱ) g is right semi-continuous.

(ⅲ) g is continuous at the infinity.

Step 1. We first prove that g is left semicontinuous at the point bsuch that a < b < ∞.

For any ε>0, we apply lemma 2.1 to yield that there exists some ϕ∈C_c(Rⁿ) with ‖ϕ‖_p=1, such that

$ g(b)-\left\|M_a^b \phi\right\|_p<\varepsilon . $

Suppose that ϕ is supported in the ball B(0, γ) where γ>0. Set $ \rho=\max\limits _{x \in B(0, \gamma)} \phi(x)$.

Then we further take

$ s:=p 2^{p-1} \rho^p\left(\sum\nolimits_{j=1}^n \frac{C_n^j}{a^j}+\left(\sum\nolimits_{j=1}^n \frac{C_n^j}{a^j}\right)^p\right) V_n(\gamma+b)^n, $

where C_n^j (j=1, 2, 3…n) are binomial coefficients, and V_n is the volume of the unit ball in Rⁿ. Finally, we choose

$ \delta:=\min \left\{1, \varepsilon^p / s, b-a\right\} . $

When 0 < b-c < δ, since δ≤b-a, the operator M_a^c is well defined and so does g(c) which satisfies g(c)≤g(b), and for all x∈Rⁿ, we also have M_a^bϕ(x)≥M_a^cϕ(x).

Consequently, we can obtain that

$ \begin{aligned} \left\|M_a^b \varphi\right\|_p= & \left(\int_{\left\{M_a^b \varphi>M_a^c \varphi\right\}}\left|M_a^b \varphi(x)\right|^p \mathrm{~d} x+\right. \\ & \left.\int_{\left\{M_a^b \varphi=M_a^c \varphi\right\}}\left|M_a^b \varphi(x)\right|^p \mathrm{~d} x\right)^{\frac{1}{p}} . \end{aligned} $

(4)

For all x∈{M_a^bφ>M_a^cφ}, we have

$ M_a^b \varphi(x)=\sup\limits _{c<r<b} \frac{1}{|B(x, r)|} \int_{B(x, r)}|\varphi(y)| \mathrm{d} y, $

which yields

$ \begin{aligned} & M_a^b \varphi(x) \\ & \leqslant \frac{1}{|B(x, c)|}\left(\int\limits_{B(x, c)}|\varphi(y)| \mathrm{d} y+\rho V_n\left(b^n-c^n\right)\right) \\ & \leqslant M_a^c \varphi(x)+\rho\left(\left(\frac{b}{c}\right)^n-1\right) . \end{aligned} $

(5)

Since b-c < δ, there holds b/c < 1+δ/c.

Observe that δ < min{1, b-a}, using the binomial theorem and inequality (5), one gets

$ \begin{aligned} & M_a^b \varphi(x) \\ & \leqslant M_a^c \varphi(x)+\rho\left(\left(1+\frac{\delta}{c}\right)^n-1\right) \\ & \leqslant M_a^c \varphi(x)+\rho \delta\left(\sum\nolimits_{j=1}^n \frac{C_n^j}{a^j}\right), \end{aligned} $

(6)

for all x∈{M_a^bϕ>M_a^cϕ}.

Substitute inequality (6) into formula (4), we can deduce that

$ \begin{aligned} & \left\|M_a^b \varphi\right\|_p \\ & =\left(\int_{\left\{M_a^b \varphi=M_a^c \varphi\right\}}\left|M_a^b \varphi(x)\right|^p \mathrm{~d} x+\int_{\left\{M_a^b \varphi>M_a^c \varphi\right\}}\left|M_a^b \varphi(x)\right|^p \mathrm{~d} x\right)^{\frac{1}{p}}\\ & \leqslant\left(\int_{\left\{M_d^b \varphi=M_a^c \varphi\right\}}\left|M_a^c \varphi(x)\right|^p \mathrm{~d} x+\int_{\left\{M_a^b \varphi>M_a^c \varphi\right\}}\left|M_a^c \varphi(x)+\rho \delta\left(\sum\nolimits_{j=1}^n \frac{C_n^j}{a^j}\right)\right|^p \mathrm{~d} x\right)^{\frac{1}{p}} \\ & \leqslant\left(\int_{\left\{M_a^b \varphi=M_a^c \varphi\right\}}\left|M_a^c \varphi(x)\right|^p \mathrm{~d} x+\int_{\left\{M_d^b \varphi>M_a^c \varphi\right\}}\left|M_a^c \varphi(x)\right|^p \mathrm{~d} x+\right. \\ & \left.\int_{\left\{M_a^b \varphi>M_a^c \varphi\right\}} p 2^{p-1}\left|M_a^c \varphi(x)\right|^{p-1}\left|\rho \delta\left(\sum\nolimits_{j=1}^n \frac{C_n^j}{a^j}\right)\right|+p 2^{p-1}\left|\rho \delta\left(\sum\nolimits_{j=1}^n \frac{C_n^j}{a^j}\right)\right|^p \mathrm{~d} x\right)^{\frac{1}{p}} \\ & \leqslant\left(\left\|M_a^c \varphi\right\|_p^p+\int_{\left\{M_a^b \varphi>M_a^c \varphi\right\}} p 2^{p-1}\left(\left|M_a^c \varphi(x)\right|^{p-1}\left|\rho \delta\left(\sum\nolimits_{j=1}^n \frac{C_n^j}{a^j}\right)\right|+\left|\rho \delta\left(\sum\nolimits_{j=1}^n \frac{C_n^j}{a^j}\right)\right|^p\right) \mathrm{d} x\right)^{\frac{1}{p}} \\ & \leqslant\left(\left\|M_a^c \varphi\right\|_p^p+p 2^{p-1} \delta \rho^p\left(\sum\nolimits_{j=1}^n \frac{C_n^j}{a^j}+\left(\sum\nolimits_{j=1}^n \frac{C_n^j}{a^j}\right)^p\right) \int_{\left\{M_a^b \varphi>M_a^c \varphi\right\}} 1 \mathrm{~d} x\right)^{\frac{1}{p}} \\ & \leqslant\left(\left\|M_a^c \varphi\right\|_p^p+p 2^{p-1} \delta \rho^p\left(\sum\nolimits_{j=1}^n \frac{C_n^j}{a^j}+\left(\sum\nolimits_{j=1}^n \frac{C_n^j}{a^j}\right)^p\right) V_n(\gamma+b)^n\right)^{\frac{1}{p}} \\ & \leqslant\left(\left\|M_a^c \varphi\right\|_p^p+\delta s\right)^{\frac{1}{p}} \\ & \leqslant\left(\left\|M_a^c \varphi\right\|_p^p+\varepsilon^p\right)^{\frac{1}{p}} \\ & \leqslant\left\|M_a^c \varphi\right\|_p+\varepsilon, \end{aligned} $

where the second inequality holds due to Lemma 2.2, and we use the fact that δ≤ε^p/s in the penultimate inequality.

To sum up,

$ g(b)<\left\|M_a^b \varphi\right\|_p+\varepsilon \leqslant\left\|M_a^c \varphi\right\|_p+2 \varepsilon \leqslant g(c)+2 \varepsilon . $

Therefore, for any ε>0, there exists a δ=min{1, ε^p/s, b-a} where s is defined as above, such that for all 0 < b-c < δ, there holds

$ g(c) \leqslant g(b) \leqslant g(c)+2 \varepsilon . $

As a result, g is left semicontinuous at the point b>a.

Step 2. Furthermore, we proceed to prove that the function g is right semicontinuous at the point b such that a≤b < ∞. Let us start by giving some notations that will be used in the proof of this part.

First, we set

$ C=\|M\|_{L^p}, $

that is, C is the L^p-norm of the classical Hardy-Littlewood maximal operator.

Then we define

$ t:=\left(p 2^{p-1}\left(\frac{C^{p-1}}{b^n}+\frac{1}{b^{n p}}\right)\right)^{\frac{1}{p}}. $

Now, for any ε>0, take

$ \delta=\min \left\{\left(b^n+1\right)^{\frac{1}{n}}-b, \left[(\varepsilon / t)^p+b^n\right]^{\frac{1}{n}}-b\right\} . $

Using Lemma 2.1, we can reduce our discussion to functions ϕ in C_c(Rⁿ) with ‖ϕ‖_p=1. Suppose c < b-1 < δ, then for any ϕ∈C_c(Rⁿ) satisfying ‖ϕ‖_p=1, there holds

$ \begin{aligned} & \left\|M_a^c \varphi\right\|_p \\ & =\left(\int_{\left\{M_a^c \varphi>M_a^b \varphi\right\}}\left|M_a^c \varphi(x)\right|^p \mathrm{~d} x+\int_{\left\{M_a^c \varphi=M_a^b \varphi\right\}}\left|M_a^c \varphi(x)\right|^p \mathrm{~d} x\right)^{\frac{1}{p}} \\ & =\left(\int_{\left\{M_a^c \varphi>M_a^b \varphi\right\}}\left|M_a^c \varphi(x)\right|^p \mathrm{~d} x+\right. \\ & \left.\int_{\left\{M_a^c \varphi=M_a^b \varphi\right\}}\left|M_a^b \varphi(x)\right|^p \mathrm{~d} x\right)^{\frac{1}{p}} . \end{aligned} $

(7)

For all x∈{M_a^cϕ>M_a^bϕ}, we have

$ M_a^c \varphi(x)=\sup\limits _{b<r<c} \frac{1}{|B(x, r)|} \int_{B(x, r)}|\varphi(y)| \mathrm{d} y . $

which yields

$ \begin{aligned} & M_a^c \varphi(x) \\ & \leqslant \frac{1}{|B(x, b)|}\left(\int_{B(x, c) \backslash B(x, b)}|\varphi(y)| \mathrm{d} y+\int_{B(x, b)}|\varphi(y)| \mathrm{d} y\right) \\ & \leqslant \frac{1}{V_n b^n} \int_{B(x, c) \backslash B(x, b)}|\varphi(y)| \mathrm{d} y+M_a^b \varphi(x) . \end{aligned} $

(8)

Using Lemma 2.2, it then follows from formula (7) and formula (8) that

$ \begin{aligned} & \left\|M_a^c \varphi\right\|_p \\ & =\left(\int_{\left\{M_a^c \varphi>M_a^b \varphi\right\}}\left|M_a^c \varphi(x)\right|^p \mathrm{~d} x+\int_{\left\{M_a^c \varphi=M_a^b \varphi\right\}}\left|M_a^b \varphi(x)\right|^p \mathrm{~d} x\right)^{\frac{1}{p}}\\ & \leqslant\left(\int_{\left\{M_a^c \varphi>M_a^b \varphi\right\}}\left|M_a^b \varphi(x)+\frac{1}{V_n b^n} \int_{B(x, c) \backslash B(x, b)}\right| \varphi(y)|\mathrm{d} y|^p \mathrm{~d} x+\int_{\left\{M_a^c \varphi=M_a^b \varphi\right\}}\left|M_a^b \varphi(x)\right|^p \mathrm{~d} x\right)^{\frac{1}{p}} \\ & \leqslant\left(\int_{\left\{M_a^c \varphi>M_a^b \varphi\right\}}\left|M_a^b \varphi(x)\right|^p \mathrm{~d} x+\int_{\left\{M_a^c \varphi>M_a^b \varphi\right\}} p 2^{p-1}\left|M_a^b \varphi(x)\right|^{p-1} \frac{1}{V_n b^n} \int_{B(x, c) \backslash B(x, b)}|\varphi(y)| \mathrm{d} y \mathrm{~d} x+\right. \\ & \left.\int_{\left\{M_a^c \varphi>M_a^b \varphi\right\}} p 2^{p-1}\left(\frac{1}{V_n b^n} \int_{B(x, c) \backslash B(x, b)}|\varphi(y)| \mathrm{d} y\right)^p \mathrm{~d} x+\int_{\left\{M_a^c \varphi=M_a^b \varphi\right\}}\left|M_a^b \varphi(x)\right|^p \mathrm{~d} x\right)^{\frac{1}{p}} \\ & \leqslant\left(\left\|M_a^b \varphi\right\|_p^p+\int_{R^n} p 2^{p-1}\left|M_a^b \varphi(x)\right|^{p-1} \frac{1}{V_n b^n} \int_{B(x, c) \backslash B(x, b)}|\varphi(y)| \mathrm{d} y \mathrm{~d} x+\right. \\ & \left.\int_{\mathbf{R}^n} p 2^{p-1}\left(\frac{1}{V_n b^n} \int_{B(x, c) \backslash B(x, b)}|\varphi(y)| \mathrm{d} y\right)^p \mathrm{~d} x\right)^{\frac{1}{p}} \\ & =\left(\left\|M_a^b \varphi\right\|_p^p+\sum\limits_{i=1, p} \int_{R^n} p 2^{p-1}\left|M_a^b \varphi(x)\right|^{p-i}\left(\frac{1}{V_n b^n} \int_{B(x, c) \backslash B(x, b)}|\varphi(y)| \mathrm{d} y\right)^i \mathrm{~d} x\right)^{\frac{1}{p}} . \end{aligned} $

(9)

Now we give the estimate of the quantities

$ \int\limits_{\mathbf{R}^n}\left|M_a^b \varphi(x)\right|^{p-i}\left(\int_{B(x, c) \backslash B(x, b)}|\varphi(y)| \mathrm{d} y\right)^i \mathrm{~d} x . $

(10)

in formula (9), where i=1, p. Applying Holder's inequality to quantity (10), we can calculate that

$ \begin{aligned} & \int_{\mathbf{R}^n}\left|M_a^b \varphi(x)\right|^{p-i}\left(\int_{B(x, c) \backslash B(x, b)}|\varphi(y)| \mathrm{d} y\right)^i \mathrm{~d} x \\ & \leqslant\left(\int_{\mathbf{R}^n}\left(\int_{B(x, c) \backslash B(x, b)}|\varphi(y)| \mathrm{d} y\right)^p \mathrm{~d} x\right)^{\frac{i}{p}} \times \\ & \quad\left(\int_{\mathbf{R}^n}\left|M_a^b \varphi(x)\right|{ }^p \mathrm{~d} x\right)^{\frac{p-i}{p}} \\ & \leqslant\left(\int_{\mathbf{R}^n}\left(\int_{B(x, c) \backslash B(x, b)}|\varphi(y)| \mathrm{d} y\right)^p \mathrm{~d} x\right)^{\frac{i}{p}} M_a^b \varphi_p^{p-i} \\ & \leqslant\left(\int_{\mathbf{R}^n}\left(\int_{B(x, c) \backslash B(x, b)}|\varphi(y)| \mathrm{d} y\right)^p \mathrm{~d} x\right)^{\frac{i}{p}}\|M \varphi\|_p^{p-i} \\ & \leqslant\left(\int_{\mathbf{R}^n}\left(\int_{B(x, c) \backslash B(x, b)}|\varphi(y)| \mathrm{d} y\right)^p \mathrm{~d} x\right)^{\frac{i}{p}} C^{p-i}\|\varphi\|_p^{p-i} \\ & =\left(\int_{\mathbf{R}^n}\left(\int_{B(x, c) \backslash B(x, b)}|\varphi(y)| \mathrm{d} y\right)^p \mathrm{~d} x\right)^{\frac{i}{p}} C^{p-i} . \end{aligned} $

(11)

where the penultimate inequality and the last equality hold because of our choices of C and φ.

However, our current estimate formula (11) is not sufficient to deduce our desired result. Therefore, to get a much more delicate estimate of formula (10), we should give a precise evaluation of quantities

$ \left(\int_{\mathbf{R}^n}\left(\int_{B(x, c) \backslash B(x, b)}|\varphi(y)| \mathrm{d} y\right)^p \mathrm{~d} x\right)^{\frac{i}{p}}, $

in formula (11). Let q be the conjugate exponent of p, which is, 1/p+1/q=1. By once again using Holder's inequality, we have

$ \begin{aligned} & \left(\int_{\mathbf{R}^n}\left(\int_{B(x, c) \backslash B(x, b)}|\varphi(y)| \mathrm{d} y\right)^p \mathrm{~d} x\right)^{\frac{i}{p}} \\ & =\left(\int_{\mathbf{R}^n}\left(\int_{\mathbf{R}^n} \chi_{B(x, c) \backslash B(x, b)}(y)|\varphi(y)| \mathrm{d} y\right)^p \mathrm{~d} x\right)^{\frac{i}{p}} \\ & =\left(\int_{\mathbf{R}^n}\left(\int_{\mathbf{R}^n} \chi_{B(x, c) \backslash B(x, b)}(y) \chi_{B(x, c) \backslash B(x, b)}^{\frac{1}{p}}(y)|\varphi(y)| \mathrm{d} y\right)^p \mathrm{~d} x\right)^{\frac{i}{p}} \\ & \leqslant\left(\int_{\mathbf{R}^n}\left[V_n\left(c^n-b^n\right)\right]^{\frac{p}{q}} \int_{\mathbf{R}^n} \chi_{B(x, c) \backslash B(x, b)}(y)|\varphi(y)|^p \mathrm{~d} y \mathrm{~d} x\right)^{\frac{i}{p}} \\ & \leqslant\left[V_n\left(c^n-b^n\right)\right]^{\frac{i}{q}}\left(\int_{\mathbf{R}^n} \int_{\mathbf{R}^n} \chi_{B(x, c) \backslash B(x, b)}(y)|\varphi(y)|^p \mathrm{~d} y \mathrm{~d} x\right)^{\frac{i}{p}} \\ & \leqslant\left[V_n\left(c^n-b^n\right)\right]^{\frac{i}{q}}\left(\int_{\mathbf{R}^n}|\varphi(y)|^p \mathrm{~d} y \int_{\mathbf{R}^n} \chi_{B(x, c) \backslash B(x, b)}(y) \mathrm{d} x\right)^{\frac{i}{p}} \\ & =\left[V_n\left(c^n-b^n\right)\right]^i\|\varphi\|_p^i \\ & =\left[V_n\left(c^n-b^n\right)\right]^i \\ & \leqslant V_n^i\left(c^n-b^n\right), \end{aligned} $

(12)

where we have used the fact that $ c - b < \delta \le {\left( {{b^n} + 1} \right)^{\frac{1}{n}}}$ in the last inequality.

Combining the estimates of formula (11) and inequality (12) together, it follows that

$ \begin{gathered} \int_{\mathbf{R}^n}\left|M_a^b \varphi(x)\right|^{p-i}\left(\int_{B(x, c) \backslash B(x, b)}|\varphi(y)| \mathrm{d} y\right)^i \mathrm{~d} x \\ \leqslant C^{p-i} V_n^i\left(c^n-b^n\right) . \end{gathered} $

(13)

Then, by substituting inequality (13) into formula (9), we can show that where we have used the fact that $c-b<\delta \leqslant\left(b^n+1\right)^{\frac{1}{n}} $ in the last inequality.

By the arbitrariness of ϕ, we can reach the conclusion that g(c)≤g(b)+ε.

$ \begin{aligned} & \left\|M_a^c \varphi\right\|_p \leqslant\left(M_a^b \varphi_p^p+\int_{\mathbf{R}^n} p 2^{p-1}\left|M_a^b \varphi(x)\right|^{p-1}\right.\\ & \frac{1}{V_n b^n} \int_{B(x, c) \backslash B(x, b)}|\varphi(y)| \mathrm{d} y \mathrm{~d} x+ \\ & \left.\int_{\mathbf{R}^n} p 2^{p-1}\left(\frac{1}{V_n b^n} \int_{B(x, c) \backslash B(x, b)}|\varphi(y)| \mathrm{d} y\right)^p \mathrm{~d} x\right)^{\frac{1}{p}} \\ & \leqslant\left(\left\|M_a^b \varphi\right\|_p^p+p 2^{p-1}\left(c^n-b^n\right)\left(\frac{C^{p-1}}{b^n}+\frac{1}{b^{n p}}\right)\right)^{\frac{1}{p}} \\ & \leqslant\left(\left\|M_a^b \varphi\right\|_p^p+\left(c^n-b^n\right) t^p\right)^{\frac{1}{p}} \\ & \leqslant\left\|M_a^b \varphi\right\|_p+\left(c^n-b^n\right)^{\frac{1}{p}} t \\ & \leqslant g(b)+\varepsilon . \end{aligned} $

To sum up, for any ε>0, there is a δ : = $\min \left\{\left(b^n+1\right)^{\frac{1}{n}}-b, \left[(\varepsilon / t)^p+b^n\right]^{\frac{1}{n}}-b\right\} $, where t is defined as above, such that for all 0 < c-b < δ, there holds

$ g(b) \leqslant g(c) \leqslant g(b)+\varepsilon . $

Consequently, g is right semicontinuous at the point b≥a.

Step 3. At last, we prove g is continuous at the infinity.

Following from Lemma 2.1 again, for any ε>0, there exists some ϕ∈C_c(Rⁿ) with ‖ϕ‖_p=1, such that

$ g(\infty)-\left\|M_a^{\infty} \varphi\right\|_p=\left\|M_a^{\infty}\right\|_{L^p}-\left\|M_a^{\infty} \varphi\right\|_p<\varepsilon . $

Suppose that ϕ is supported in the ball B(0, γ) with γ>0.

Takes $ s: = \max \left\{ {{{\left( {{w_1}/{\varepsilon ^p}} \right)}^{\frac{1}{{np - 1}}}} + \gamma , {{\left( {{w_2}/{\varepsilon ^p}} \right)}^{\frac{1}{{np - n}}}} + \gamma } \right\}$.

Where $w_1=\frac{2^n n\|\varphi\|_1^p \gamma^{n-1}}{(n p-1) V_n^{p-1}}, w_2=\frac{2^n\|\varphi\|_1^p}{(p-1) V_n^{p-1}} $.

We first calculate that

$ \begin{aligned} & \left\|M_a^{\infty} \varphi\right\|_p \\ & =\left(\int_{|x|>s}\left|M_a^{\infty} \varphi(x)\right|^p \mathrm{~d} x+\int_{|x| \leqslant s}\left|M_a^{\infty} \varphi(x)\right|^p \mathrm{~d} x\right)^{\frac{1}{p}} \\ & \leqslant\left(\int_{|x|>s}\left|\frac{\|\varphi\|_1}{V_n(|x|-\gamma)^n}\right|^p \mathrm{~d} x+\int_{|x| \leqslant s}\left|M_a^{\infty} \varphi\right|^p \mathrm{~d} x\right)^{\frac{1}{p}} \\ & \leqslant\left(\frac{\|\varphi\|_1^p}{V_n^p} \int_{|x|>s} \frac{1}{(|x|-\gamma)^{n p}} \mathrm{~d} x+\int_{|x| \leqslant s}\left|M_a^{\infty} \varphi(x)\right|^p \mathrm{~d} x\right)^{\frac{1}{p}} . \end{aligned} $

(14)

Moreover, one can compute that

$ \begin{aligned} & \frac{\|\varphi\|_1^p}{V_n^p} \int_{|x|>s} \frac{1}{(|x|-\gamma)^{n p}} \mathrm{~d} x=\frac{\|\varphi\|_1^p}{V_n^{p-1}} \frac{\omega_{n-1}}{V_n} \int_{r>s} \frac{r^{n-1}}{(r-\gamma)^{n p}} \mathrm{~d} r \\ & \leqslant \frac{2^n n\|\varphi\|_1^p}{V_n^{p-1}} \int_{r>s-\gamma} r^{n-1-n p}+\gamma^{n-1} r^{-n p} \mathrm{~d} r \\ & \leqslant \frac{2^n n\|\varphi\|_1^p}{V_n^{p-1}}\left(\frac{1}{n p-n} \frac{1}{(s-\gamma)^{n p-n}}+\frac{\gamma^{n-1}}{n p-1} \frac{1}{(s-\gamma)^{n p-1}}\right) \\ & \leqslant \varepsilon^p, \end{aligned} $

(15)

where ω_n-1 is the surface area of the unit ball in Rⁿ, and the last inequality holds because of our priori choice of s.

Substitute formula (15) into formula (14), we can further obtain that

$ \left\|M_a^{\infty} \varphi\right\|_p \leqslant\left(\varepsilon^p+\int\limits_{|x| \leqslant s}\left|M_a^{\infty} \varphi(x)\right|^p \mathrm{~d} x\right)^{\frac{1}{p}} . $

Let N=s+γ. Then for any c>N and any x∈{|x|≤s}, we have

$ M_a^{\infty} \varphi(x)=M_a^c \varphi(x)=M_a^N \varphi(x) . $

As a result, for any c>N, it follows that

$ \begin{aligned} & \left\|M_a^{\infty} \varphi\right\|_p \\ \leqslant & \left(\varepsilon^p+\int_{|x| \leqslant s}\left|M_a^c \varphi(x)\right|^p \mathrm{~d} x\right)^{\frac{1}{p}} \\ \leqslant & \left(\varepsilon^p+\int_{\mathbf{R}^n}\left|M_a^c \varphi(x)\right|^p \mathrm{~d} x\right)^{\frac{1}{p}} \\ = & \left(\varepsilon^p+\left\|M_a^c \varphi\right\|_p^p\right)^{\frac{1}{p}} \\ \leqslant & \varepsilon+\left\|M_a^c \varphi\right\|_p \\ \leqslant & \varepsilon+\left\|M_a^c\right\|_{L^p} . \end{aligned} $

(16)

In terms of formula (16), we can deduce that

$ \begin{aligned} & g(\infty) \\ & =\left\|M_a^{\infty}\right\|_{L^p}<\varepsilon+\left\|M_a^{\infty} \varphi\right\|_p \leqslant 2 \varepsilon+\left\|M_a^c\right\|_{L^p} \\ & =2 \varepsilon+g(c) . \end{aligned} $

To sum up, for all ε>0, there exists N=s+γ where s and γ are defined as above, such that for any c>N, there holds

$ g(c) \leqslant g(\infty) \leqslant g(c)+2 \varepsilon . $

Consequently, g is continuous at the infinity.

Now, as far as step 1, step 2, and step 3 are completed, our proof is then finished.

We sincerely appreciate Shao Liu for his insightful suggestions on the proof of our main theorem.