王泽群, 燕敦验
中国科学院大学数学科学学院, 北京 100049
摘要: 研究迭代的非中心型哈代-李特伍德极大函数和迭代的中心型哈代-李特伍德极大函数。证明迭代极大函数的极限是极大算子的一个不动点。作为不动点理论的一个应用，最终得到，对于非中心型哈代-李特伍德极大算子，这个不动点处处为‖f‖_∞。对于中心型哈代-李特伍德极大算子，仅在n=1，2时有相同的结果。
关键词: 哈代-李特伍德极大函数不动点迭代的哈代-李特伍德极大函数
Define the centered Hardy-Littlewood maximal function by

${M_c}f\left( x \right) = \mathop {\sup }\limits_{r > 0} \frac{1}{{\left| {B\left( {x,r} \right)} \right|}}\int_{B\left( {x,r} \right)} {\left| {f\left( y \right)} \right|{\rm{d}}y} ,$

(1)

and the non-centered Hardy-Littlewood maximal function by

$Mf\left( x \right) = \mathop {\sup }\limits_{B \ni x} \frac{1}{{\left| B \right|}}\int_B {\left| {f\left( y \right)} \right|{\rm{d}}y} ,$

(2)

where B is a ball and B(x, r) is a ball with the center at the point x and the radius r. The basic real-variable construct was introduced for n=1 by Hardy and Littlewood^[1] and for n≥2 by Wiener^[2]. 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 frequently used to majorize other important operators in harmonic analysis.
It is clear that

${M_c}f\left( x \right) \le Mf\left( x \right) \le {2^n}{M_c}f\left( x \right)$

(3)

holds for all x∈$ \mathbb{R} $ⁿ. Both M and M^c are sublinear operators and the two functions Mf and M_c f never vanish unless f=0 almost everywhere^[3]. The study of the boundedness for M or M_c is fairly complete^[4]. The primary purpose of this paper is to study the properties of the iterated Hardy-Littlewood maximal function.
Let M be the non-centered Hardy-Littlewood maximal function defined by (2). Define the iterated non-centered Hardy-Littlewood maximal function denoted by M^k+1 as follows:

${M^{k + 1}}f\left( x \right): = M\left( {{M^k}f} \right)\left( x \right),$

(4)

for k=1, 2, …, and x∈$ \mathbb{R} $ⁿ. Set M¹f(x):=Mf(x).
In the same way, we can set

$M_c^{k + 1}f\left( x \right): = {M_c}\left( {M_c^kf} \right)\left( x \right).$

(5)

We all know that both operators M_c and M have the L^p-boundedness and the two maximal functions M_c f and Mf have a little difference in the pointwise sense from inequalities (3). We want to investigate the limit of the iterated Hardy-Littlewood maximal function.
Wei et al^[5] studied the limit of M^kf and obtained Theorem A as follows.
Theorem A??For any f∈L^∞($ \mathbb{R} $ⁿ), the equation

$\mathop {\lim }\limits_{k \to \infty } {M^k}f\left( x \right) = {\left\| f \right\|_\infty }$

(6)

holds for any x∈$ \mathbb{R} $ⁿ.
For M_c, we want to know whether it has the same properties as M. Unexpectedly, the limit of M_c^kf is essentially different from the limit of M^kf.
Now we formulate our main results as follows.
Theorem B??If f∈L_loc¹($ \mathbb{R} $ⁿ) and x∈$ \mathbb{R} $ⁿ, then

$\mathop {\lim }\limits_{k \to \infty } M_c^kf\left( x \right) = {\left\| f \right\|_\infty }$

(7)

holds for every x∈$ \mathbb{R} $ⁿ if and only if n=1, 2.
Theorem C??Let f∈L_loc¹($ \mathbb{R} $ⁿ). We have

$\mathop {\lim }\limits_{k \to \infty } {M^k}f\left( x \right) = {\left\| f \right\|_\infty },$

for every x∈$ \mathbb{R} $ⁿ and any n∈N.
We remark that the range of function f in Theorem C is wider than that in Theorem A. Furthermore in this paper we will use some novel ideas to prove Theorem C.
1 Fixed point of Hardy-Littlewood maximal operatorTo prove our main theorems, we first provide some definitions and lemmas which will be used in the follows. Some lemmas can be found in classic literatures and here we omit their proofs.
Definition 1.1??A function F is called a fixed point of a operator T, if

$TF\left( x \right) = F\left( x \right)$

(8)

holds for all x∈$ \mathbb{R} $ⁿ.
Obviously if F is a fixed point of the operator T, then we have

$\mathop {\lim }\limits_{k \to \infty } {T^k}F\left( x \right) = F\left( x \right).$

By the Lesbegue differentiation theorem, for almost all x∈$ \mathbb{R} $ⁿ, we have

${M_c}f\left( x \right) \ge \left| {f\left( x \right)} \right|$

and

$Mf\left( x \right) \ge \left| {f\left( x \right)} \right|.$

For the iterated Hardy-Littlewood maximal operator, we have the following lemma.
Lemma 1.1??For x∈$ \mathbb{R} $ⁿ, and k≥1, the two inequalities

${M^{k + 1}}f\left( x \right) \ge {M^k}f\left( x \right)$

(9)

and

$M_c^{k + 1}f\left( x \right) \ge M_c^kf\left( x \right)$

(10)

hold for all f∈L_loc¹($ \mathbb{R} $ⁿ).
Proof??Set E={x:x is not the Lesbegue point of |f|}. It follows from the Lesbegue differentiation theorem that m(E)=0. Actually we merely need to prove

${M^2}f\left( x \right) \ge Mf\left( x \right)$

for all x∈$ \mathbb{R} $ⁿ.
We conclude that

$\begin{array}{l}Mf\left( x \right) = \mathop {\sup }\limits_{B \ni x} \frac{1}{{\left| B \right|}}\int_B {\left| {f\left( y \right)} \right|{\rm{d}}y} \\ = \mathop {\sup }\limits_{B \ni x} \frac{1}{{\left| B \right|}}\int_{B\backslash E} {\left| {f\left( y \right)} \right|{\rm{d}}y} \\ = \mathop {\sup }\limits_{B \ni x} \frac{1}{{\left| B \right|}}\int_{B\backslash E} {\left\{ {\mathop {\lim }\limits_{k \to 0} \frac{1}{{\left| {B\left( {y,r} \right)} \right|}}\int_{B\left( {y,r} \right)} {\left| {f\left( u \right)} \right|{\rm{d}}u} } \right\}{\rm{d}}y} \\ \le \mathop {\sup }\limits_{B \ni x} \frac{1}{{\left| B \right|}}\int_{B\backslash E} {Mf\left( y \right){\rm{d}}y} = \mathop {\sup }\limits_{B \ni x} \frac{1}{{\left| B \right|}}\int_B {Mf\left( y \right){\rm{d}}y} \\ = {M^2}f\left( x \right).\end{array}$

(11)

Using the completely same method, we can obtain that M_c² f(x)≥M_c f(x).
By Lemma 1.1, since M^kf monotonously increases, the limit of M^kf(x) exists for all x∈$ \mathbb{R} $ⁿ.
Lemma 1.2??If f∈L_loc¹($ \mathbb{R} $ⁿ) is a fixed point of M_c, then there exists another function f_t∈C^∞($ \mathbb{R} $ⁿ)∩L_loc¹($ \mathbb{R} $ⁿ) such that f_t is a fixed point of M_c as well.
Proof??Set ?∈C_c^∞($ \mathbb{R} $ⁿ) such that

$\int_{{\mathbb{R}^n}} {\phi \left( x \right){\text{d}}x} = 1.$

For t > 0, set

${f_t} = f * {\phi _t},$

where ?_t(x)=t^-n?(x/t), for all x∈$ \mathbb{R} $ⁿ. Obviously we have

${f_t} \in {C^\infty }\left( {{\mathbb{R}^n}} \right) \cap L_{{\text{loc}}}^1\left( {{\mathbb{R}^n}} \right).$

Put

${{\chi }_{r}}=\frac{1}{\left| {{B}_{r}} \right|}{{\chi }_{{{B}_{r}}}}.$

The centered Hardy-Littlewood maximal function is written by

${M_c}\left( f \right)\left( x \right) = \mathop {\sup }\limits_{r > 0} {\chi _r} * \left| f \right|\left( x \right).$

(12)

Note f is a fixed point of M_c. This implies f≥0. We have that

$\begin{array}{*{20}{c}} {{\chi _r} * \left| {{\phi _t} * f} \right|\left( x \right) = \int_{{\mathbb{R}^n}} {{\phi _t}\left( y \right)\left( {{\chi _r} * {\tau _y}f} \right)\left( x \right){\text{d}}y} } \\ { \leqslant \int_{{\mathbb{R}^n}} {{\phi _t}\left( y \right){M_c}\left( {{\tau _y}f} \right)\left( x \right){\text{d}}y} = {\phi _t} * {M_c}\left( f \right)\left( x \right)} \\ { = {f_t}\left( x \right).} \end{array}$

(13)

On the other hand, it follows from (12) and (13) that M_c(f_t)(x)=M_c(?_t*f)(x)=$ \underset{r>0}{\mathop{\text{sup}}}\, {{\chi }_{r}} $*|?_t*f|(x)≤f_t(x).Since f_t is smooth function, it follows from the definition of M_cf that

${M_c}\left( {{f_t}} \right)\left( x \right) \ge {f_t}\left( x \right).$

Thus we obtain that f_t is a fixed point of M_c.
Using the similar method, we can easily prove that f_t is a fixed point of M if f is a fixed point of M.
Lemma 1.3??There is a non-constant fixed point of M_c in L_loc¹($ \mathbb{R} $ⁿ) if and only if there exists a non-negative upper-harmonic function.
There is a non-constant fixed point of M in L_loc¹($ \mathbb{R} $ⁿ) if and only if there exists a non-negative function f∈$ \mathbb{C} $^∞($ \mathbb{R} $ⁿ)∩L_loc¹($ \mathbb{R} $ⁿ) such that f(x)≥Mf(x) for all x∈$ \mathbb{R} $ⁿ.
Lemma 1.3 is due to that for a smooth function, every point in $ \mathbb{R} $ⁿ is its Lesbegue point. We only need that f(x)≥Mf(x) to guarantee that the function is a fixed point.
Lemma 1.4??If f is a non-constant and smooth function, and f is a fixed point of M, then, in any closed ball, the minimum value of f is gotten only in the sphere.
Lemma 1.4 has the same proof as the proof of extremism principle of harmonic function. For the details please see Ref.[6].
Lemma 1.5??Suppose that f is a fixed point of M. If f∈L_loc¹($ \mathbb{R} $ⁿ), then we have f=C≤∞; if f?L_loc¹($ \mathbb{R} $ⁿ), then we have f(x)=∞.
Proof. Since f is a fixed point of M, it follows from Lemma 1.2 that there exists f_t∈$ \mathbb{C} $^∞($ \mathbb{R} $ⁿ)∩L_loc¹($ \mathbb{R} $ⁿ) such that f_t is a fixed point of M as well.
Suppose that B is a ball in $ \mathbb{R} $ⁿ and f_t∈$ \mathbb{C} $^∞($ \mathbb{R} $ⁿ)∩L_loc¹($ \mathbb{R} $ⁿ) such that Mf_t(x)=f_t(x) for all x∈$ \mathbb{R} $ⁿ. We use the proof by contradiction.
If f_t is not a constant, then, by Lemma 1.4, there is at least one point x∈?B such that f_t(y) > f_t(x) holds for all y∈B°.
Note that f_t is a fixed point of M. Thus we have that

$\begin{array}{*{20}{c}}{{f_t}\left( x \right) < \frac{1}{{m\left( {{B^ \circ }} \right)}}\int_{{B^ \circ }} {\left| {{f_t}\left( y \right)} \right|{\rm{d}}y} }\\={\frac{1}{{m\left( B \right)}}\int_B {\left| {{f_t}\left( y \right)} \right|{\rm{d}}y} \le M{f_t}\left( x \right) = {f_t}\left( x \right).}\end{array}$

(14)

This is impossible. Consequently it implies that f_t(x)=C for all x∈$ \mathbb{R} $ⁿ.
Next we will prove that f(x)=C for all x∈$ \mathbb{R} $ⁿ.
Choose a radial nonnegative function ?∈C_c^∞($ \mathbb{R} $ⁿ) such that
∫_{$ \mathbb{R} $ⁿ}?(x)dx=1, supp ??{x∈$ \mathbb{R} $ⁿ:|x|≤1} and ?(x)≥?(x′) for 0≤|x|≤|x′|.
For each t > 0, we have f_t(x)=C.
Set B_R={x∈$ \mathbb{R} $ⁿ:|x| < R}, for R > 0. We conclude from Lemma 1.2 that

$\begin{gathered} C={{f}_{t}}\left( x \right)=\int_{{{\mathbb{R}}^{n}}}{\left( f{{\chi }_{{{B}_{R}}}}+f{{\chi }_{{{\mathbb{R}}^{n}}\backslash {{B}_{R}}}} \right)\left( x-y \right){{\phi }_{t}}\left( y \right)\text{d}y} \hfill \\ = f{\chi _{{B_R}}} * {\phi _t}\left( x \right) + f{\chi _{{\mathbb{R}^n}\backslash {B_R}}} * {\phi _t}\left( x \right). \hfill \\ \end{gathered} $

(15)

It follows from (15) that

$\mathop {\lim }\limits_{t \to 0} f{\chi _{{B_R}}} * {\phi _t}\left( x \right) + \mathop {\lim }\limits_{t \to 0} f{\chi _{{\mathbb{R}^n}\backslash {B_R}}} * {\phi _t}\left( x \right) = C.$

(16)

Note that f∈L_loc¹($ \mathbb{R} $ⁿ). Thus we have fχ_BR∈L¹($ \mathbb{R} $ⁿ). This implies that

$\mathop {\lim }\limits_{t \to 0} f{\chi _{{B_R}}} * {\phi _t}\left( x \right) = f{\chi _{{B_R}}}\left( x \right)$

(17)

for almost every x∈$ \mathbb{R} $ⁿ. By the property of convolution, we get that

${\text{supp}}f{\chi _{{\mathbb{R}^n}\backslash {B_R}}} * {\phi _t} \subset \left\{ {x \in {\mathbb{R}^n}:\left| x \right| \geqslant R - t} \right\}.$

(18)

Combing (16), (17) with (18) yields that

$f{\chi _{{B_R}}}\left( x \right) + \mathop {\lim }\limits_{t \to 0} f{\chi _{{\mathbb{R}^n}\backslash {B_R}}} * {\phi _t}\left( y \right) = C$

holds for almost every x∈$ \mathbb{R} $ⁿ. This is equivalent to that

$f{\chi _{{B_R}}}\left( x \right) = C$

holds for almost every x∈$ \mathbb{R} $ⁿ. Let R→∞, then we have

$f\left( x \right) = C$

for almost every x∈$ \mathbb{R} $ⁿ. This implies that Mf(x)=C for every x∈$ \mathbb{R} $ⁿ. Note that f is a fixed point of M, that is,

$Mf\left( x \right) = f\left( x \right).$

Thus we must obtain that

$f\left( x \right) = C$

for every x∈$ \mathbb{R} $ⁿ.
If f?L_loc¹($ \mathbb{R} $ⁿ), then there is a ball B such that

$\int_B {\left| {f\left( x \right)} \right|{\rm{d}}x} = \infty .$

We have Mf(x)=∞. So we have f(x)=∞ for every x∈$ \mathbb{R} $ⁿ.
We remark that M_c has essential difference with M with respect to the fixed point. We all know that when n≥3, the function f(x)=|x|^2-n is a harmonic function in $ \mathbb{R} $ⁿ\{0}. In fact, we can easily check that f(x)=|x|^2-n is a fixed point of M_c.
Korry^[7] obtained the following lemma 1.6.
Lemma 1.6??For the M_c, if f∈L_loc¹($ \mathbb{R} $ⁿ) with n=1, 2 and M_cf=f, then we have f=C≥0.
2 Main theoremsIn the section, for any local integral function, the limit of the iterated Hardy-Littlewood maximal function is a fixed point of Hardy-Littlewood maximal operator.
Theorem 2.1??Write

$\mathop {\lim }\limits_{k \to \infty } {M^k}f\left( x \right) = F\left( x \right).$

We have

$MF\left( x \right) = F\left( x \right).$

That is, F a fixed point of M.
In the same way, write

$\mathop {\lim }\limits_{k \to \infty } M_c^kf\left( x \right) = {F_c}\left( x \right),$

then M_cF_c(x)=F_c(x). That is, F_c a fixed point of M_c.
Proof??We only prove the first part of Theorem 2.1. It follows that

$\begin{array}{*{20}{c}}{MF\left( x \right) = M\mathop {\lim }\limits_{k \to \infty } {M^k}f\left( x \right) = }\\{\mathop {\sup }\limits_{B \ni x} \frac{1}{{m\left( B \right)}}\int_B {\mathop {\lim }\limits_{k \to \infty } {M^k}f\left( y \right){\rm{d}}y} .}\end{array}$

(19)

Associate to an arbitrary ε > 0 B_ε?x such that

$\begin{array}{l}MF\left( x \right) - \varepsilon \le \frac{1}{{m\left( {{B_\varepsilon }} \right)}}\int_{{B_\varepsilon }} {\mathop {\lim }\limits_{k \to \infty } {M^k}f\left( y \right){\rm{d}}y} \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \le \mathop {\lim }\limits_{k \to \infty } \frac{1}{{m\left( {{B_\varepsilon }} \right)}}\int_{{B_\varepsilon }} {{M^k}f\left( y \right){\rm{d}}y} \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \le \mathop {\lim }\limits_{k \to \infty } {M^{k + 1}}f\left( x \right) = F\left( x \right)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \le MF\left( x \right).\end{array}$

(20)

That is

$MF\left( x \right) = F\left( x \right).$

(21)

Lemma 2.1??Write

$\mathop {\lim }\limits_{k \to \infty } {M^k}f\left( x \right) = F\left( x \right).$

If F(x)=C for all x∈$ \mathbb{R} $ⁿ, then we have C=‖f‖_∞.
Proof??Since F(x)=C, it implies from the definition of Hardy-Littlewood maximal function that C≤‖f‖_∞. By the definition of essential supremum of function, associate to an arbitrary ε > 0, a set E?$ \mathbb{R} $ⁿ with m(E) > 0, such that

$\left| {f\left( x \right)} \right| > {\left\| f \right\|_\infty } - \varepsilon ,$

for x∈E. When x∈E is the lebesgue point of f, we have that

$C = F\left( x \right) \ge Mf\left( x \right) \ge \left| {f\left( x \right)} \right| > {\left\| f \right\|_\infty } - \varepsilon .$

By the arbitrary property of ε, we immediately have

$C \ge {\left\| f \right\|_\infty }.$

Consequently we have C=‖f‖_∞
Next we will prove our main theorems.
The proof of Theorem B
Proof??It follows from Theorem 2.1 that

${F_c}\left( x \right) = \mathop {\lim }\limits_{k \to \infty } M_c^kf\left( x \right)$

(22)

is a fixed point of M_c.
By Lemm 1.6, if F_c∈L_loc¹($ \mathbb{R} $ⁿ) with n=1, 2, then F_c=C.
Since F_c(x)=C, it implies from the definition of center Hardy-Littlewood maximal function that C≤‖f‖_∞.
By the definition of essential supremum of function, associate to an arbitrary ε > 0 a set E?$ \mathbb{R} $ⁿ with m(E) > 0, such that

$\left| {f\left( x \right)} \right| > {\left\| f \right\|_\infty } - \varepsilon ,$

for x∈E. Since f∈L_loc¹($ \mathbb{R} $ⁿ), almost every points in $ \mathbb{R} $ⁿ is the lebesgue point of f. Choose x∈E is the lebesgue point of f. We have that

$C = F\left( x \right) \ge {M_c}f\left( x \right) \ge \left| {f\left( x \right)} \right| > {\left\| f \right\|_\infty } - \varepsilon .$

By the arbitrary property of ε, we immediately have

$C \ge {\left\| f \right\|_\infty }.$

Consequently we have F_c=‖f‖_∞.
If F_c?L_loc¹($ \mathbb{R} $ⁿ), we can easily have F=∞=‖f‖_∞.
The proof of Theorem C
Proof??It follows from Theorem 2.1 that

$F\left( x \right) = \mathop {\lim }\limits_{k \to \infty } {M^k}f\left( x \right)$

(23)

is a fixed point of M.
By Lemma 1.5, if F∈L_loc¹($ \mathbb{R} $ⁿ), then we have F=C.
It follows from Lemma 2.1 that F=‖f‖_∞.
If F?L_loc¹($ \mathbb{R} $ⁿ), then we have ‖f‖_∞=∞, and there exists a ball B(0, R) such that

$\int_{B\left( {0,R} \right)} {\left| {F\left( x \right)} \right|{\rm{d}}x} = \infty .$

Note that F a fixed point of M. We have

$F\left( x \right) = MF\left( x \right) = \infty .$

Consequently, we have F=‖f‖_∞.
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