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1 Introduction
Let
$\mu $
be a Borel measure on
$\mathbb {R}^n$
; that is, every Borel set is
$\mu $
-measurable and
$\mu (K)<\infty $
for any compact set
$K\subset \mathbb {R}^n$
. For a
$\mu $
-locally integrable function
$f:\mathbb {R}^n\rightarrow \left [ 0,\infty \right ] $
, the uncentered maximal function of f with respect to
$\mu $
is defined by
$$ \begin{align} M_{\mu}f(x):=\sup\limits_{B\ni x}\frac{1}{\mu(B)}\int_{B}fd\mu,\quad x\in{\mathbb{R}^n}, \end{align} $$
where the supremum is taken over all open balls B containing x. The integral average in (1.1) is equal to
$f(x)$
if
$\mu (B)=0$
.
As we know, when
$n=1$
, the operator
$M_{\mu }$
always maps
$L^1(\mathbb {R},\mu )$
into the Lorenz space
$L^{1,\infty }(\mathbb {R},\mu )$
and maps
$L^p(\mathbb {R},\mu )$
into itself for
$1<p\leq \infty $
(see [Reference Bernal1, Reference Sjögren10]). It is therefore natural to investigate the best constant
$C_1^{\mu }$
such that for any
$f\in L^1(\mathbb {R},\mu )$
and
$\lambda>0$
,
$$ \begin{align} \mu(\{x\in\mathbb{R}:M_{\mu}f(x)>\lambda\} )\leq\frac{C_1^{\mu}}{\lambda}\Vert f\Vert_{L^1(\mathbb{R},\mu)}, \end{align} $$
as well as the best constant
$C_p^{\mu }$
such that for any
$f\in L^p(\mathbb {R},\mu )$
,
$$ \begin{align} \Vert M_{\mu}f \Vert_{L^p(\mathbb{R},\mu)}\leq C_p^{\mu}\Vert f\Vert_{L^p(\mathbb{R},\mu)}. \end{align} $$
In their remarkable work [Reference Grafakos and Montgomery-Smith7], Grafakos and Montgomery-Smith showed that when
$\mu $
is the n-dimensional Lebesgue measure, the norm of the “strong” uncentered maximal operator on
$L^p(\mathbb {R}^n)$
for
$p\in (1, \infty )$
tends to
$\infty $
as
$n\rightarrow \infty $
. Moreover, when
$n=1$
, they also showed that the best constant
$C_p^{\mu }$
for
$p\in (1,\infty )$
is the unique positive solution to the equation
$$ \begin{align} (p-1)x^p-px^{p-1}-1=0. \end{align} $$
When
$\mu $
is a non-zero Borel measure satisfying the following assumption.
Assumption (A): For any point
$a\in \mathbb {R}$
,
$\mu (\{a\})=0$
,
Bernal [Reference Bernal1] proved that the best constant
$C_1^{\mu }$
in (1.2) is
$2$
. On the other hand, Grafakos and Kinnunen in [Reference Grafakos and Kinnunen6] showed that for a general Borel measure
$\mu $
on
$\mathbb {R}$
, the best constants in (1.2) and (1.3), respectively, have upper bounds
$2$
and
$\theta _p$
with
$1<p<\infty $
, where
$\theta _p$
is the unique positive solution of equation (1.4) (see Lemma 2.1 below). Recently, there are still scholars actively engaged in this field. Wu et al. [Reference Wu, Wei, Yan and Liu11] obtained the best constant of truncated Hardy–Littlewood maximal function on
$L^{1}(\mathbb {R})$
. Qin et al. [Reference Qin, Wu and Xue9] established the limiting weak type behaviors of the strong maximal operators. There are numerous papers on the best constants of maximal functions (see [Reference Bi, Li and Zhang2, Reference González-Riquelme and Madrid4, Reference Nie, Wu and Wang8]).
The purpose of this note is twofold. For
$p=1$
, by an alternative method, we reobtain the result of Bernal in [Reference Bernal1]. That is, we have the conclusion as follows.
Theorem 1.1 Let
$\mu $
be a non-zero Borel measure on
$\mathbb {R}$
that satisfies Assumption (A). Then,
$C_1^{\mu }=2$
.
The second purpose of this note is to study the best constant
$C_p^{\mu }$
of
$M_{\mu }$
with a Borel measure
$\mu $
for
$p\in (1,\infty )$
. Under some additional assumptions, we verify that the upper bound
$\theta _p$
obtained in [Reference Grafakos and Kinnunen6] is just the norm of
$M_\mu $
. More precisely, we have the following result.
Theorem 1.2 Let
$\mu $
be a Borel measure on
$\mathbb {R}$
that satisfies Assumption (A) and the following condition that is
Assumption (B): There exists a non-negative locally integrable function w on
$\mathbb {R}$
, such that for any subset
$A\subset \mathbb {R}$
,
$\mu (A)=\int _{A}w(t)dt$
, and
$$ \begin{align*}\int_{-\infty}^{0}w(t)dt=\int_{0}^{\infty}w(t)dt=\infty.\end{align*} $$
Then,
$C_p^{\mu }=\theta _p$
, where
$\theta _p$
is the unique positive solution of equation (1.4) with
$1<p<\infty $
.
Remark 1.3 (i) It is not clear whether Assumption (B) is necessary.
(ii) Let
$w\in A_p$
, the class of Muckenhoupt weights (see, for example, [Reference Grafakos5]). It is not difficult to see that the Borel measure
$\mu =wdx$
satisfies Assumptions (A) and (B). This implies that for a given
$w\in A_p$
, the norm of
$M_\mu $
on
$L^p(\mathbb {R},\mu )$
or from
$L^1(\mathbb {R},\mu )$
to
$L^{1,\infty }(\mathbb {R},\mu )$
is independent of w.
We mention that when
$p=1$
, by the result in [Reference Grafakos and Kinnunen6], the best constant
$C_1^{\mu }\leq 2$
(see Lemma 2.1 below). Thus, to show Theorem 1.1, it suffices to prove the reverse inequality. A novelty of this note is that we take a new method to achieve this aim by finding some suitable “test functions.” This method is also used to prove Theorem 1.2.
At the end of this article, we also briefly discuss the “strong” maximal operator
$N_{\vec {\mu }}^n$
with the measure
$\vec {\mu }$
on
$\mathbb {R}^n$
. By an argument similar to Theorem 1.2, we show that the operator norm of
$N_{\vec {\mu }}^n$
on
$L^p(\mathbb {R}^n, \vec {\mu })$
is
$\theta _p^n$
,
$1<p<\infty $
. Moreover, we point out that
$\theta _p^n$
grows exponentially with n, as
$n\rightarrow \infty $
.
2 Proofs of Theorems 1.1 and 1.2
In this section, we firstly provide the proofs of Theorems 1.1 and 1.2, and then we consider the n-dimensional maximal operator
$N_{\vec {\mu }}^n$
in the end. Let
$M_{\mu }f$
,
$C_1^{\mu }$
, and
$C_p^{\mu }$
be as in (1.1)–(1.3), respectively, and
$\theta _p$
be the unique positive solution of (1.4). For convenience, we write
$Mf:=M_{\mu }f$
,
$C_1:=C_1^{\mu }$
, and
$C_p:=C_p^{\mu }$
. We begin with recalling the following result by Grafakos and Kinnunen in [Reference Grafakos and Kinnunen6].
Lemma 2.1 Let
$\mu $
be a Borel measure on
$\mathbb {R}$
and
$p\in (1,\infty )$
. Then,
$C_1\leq 2$
and
$C_p\leq \theta _p$
.
Theorem 1.1 is a consequence of Lemma 2.1 and the following proposition.
Proposition 2.2 Let
$\mu $
be a non-zero Borel measure on
$\mathbb {R}$
satisfying Assumption (A). Then, there exists a collection
$\{(a_i,b_i)\}_{i=1}^{\infty }$
of open intervals such that
$$ \begin{align} a_i<a_{i+1}<b_{i+1}<b_i, \quad\int_{a_i}^{b_i}d\mu>0, \quad\lim\limits_{i\rightarrow\infty}\int_{a_i}^{b_i}d\mu=0, \end{align} $$
and
$$ \begin{align*} \lim\limits_{i\rightarrow\infty}\frac{\sup\limits_{t>0}t\mu( \{x\in\mathbb{R}:M\chi_{(a_i,b_i)}(x)>t\})}{\mu((a_i,b_i))}=2. \end{align*} $$
Proof: We claim that there exists
$x_0\in \mathbb {R}$
, such that
$\mu (I)>0 $
for any open interval
${I\ni x_0}$
. Otherwise, for any
$x\in \mathbb {R}$
, there exists an open interval
$I_x\ni x$
such that
$\mu (I_x)=0$
. By the Lindelöf covering theorem, we deduce that there exists a countable collection
$\{I_j\} _{j=1}^{\infty }$
of open intervals, such that
$\mu (I_j)=0$
for any
$j\in {\mathbb {N}}$
, and
$$ \begin{align*}\mathbb{R}=\mathop{\bigcup}\limits_{j\in{\mathbb{N}}}I_j.\end{align*} $$
Therefore,
$$ \begin{align*}0<\mu(\mathbb{R})=\mu\Big(\mathop{\bigcup}\limits_{j\in{\mathbb{N}}}I_j\Big)\leq\sum\limits_{j=1}^{\infty}\mu(I_j)=0,\end{align*} $$
which yields a contradiction. From this claim, we see that there exists a collection
$\{(a_i,b_i)\}_{i=1}^{\infty }$
of open intervals satisfying (2.1).
Let
$f_i(x):=\chi _{(a_i,b_i)}(x)$
, where
$i\in \mathbb {N}$
and
$x\in \mathbb {R}$
. We then see that for
$x<a_i$
,
$$ \begin{align*} Mf_i(x)=\sup\limits_{a_i\leq y \leq b_i}\frac{\int_{a_i}^{y}d\mu}{\int_{x}^{y}d\mu}=\frac{\int_{a_i}^{b_i}d\mu}{\int_{x}^{b_i}d\mu}. \end{align*} $$
We can treat similarly when
$x>b_i$
. Therefore, we conclude that for any i,
$$ \begin{align*} Mf_i(x)= \left\{\begin{array}{@{}ll} \frac{\int_{a_i}^{b_i}d\mu}{\int_{x}^{b_i}d\mu},\,\,\,\,\, &x<a_i;\\ \,\\[-5pt] 1,\,\,\,\,\, &a_i\leq x\leq b_i;\\ \,\\[-5pt] \frac{\int_{a_i}^{b_i}d\mu}{\int_{a_i}^{x}d\mu},\,\,\,\,\, &x>b_i. \end{array}\right. \end{align*} $$
By (2.1), we see that
$\mu ((a_i,b_i))\in (0,\infty )$
. Without loss of generality, we may further assume that
$$ \begin{align*} \gamma_i:=\max\{\frac{\int_{a_i}^{b_i}d\mu}{\int_{-\infty}^{b_i}d\mu},\,\,\frac{\int_{a_i}^{b_i}d\mu}{\int_{a_i}^{\infty}d\mu}\}<1. \end{align*} $$
Moreover, since
$Mf_i$
is continuous, increasing on
$(-\infty ,a_i)$
, and decreasing on
$(b_i,\infty )$
, we have that
$$ \begin{align*} E_i:=\{x\in \mathbb {R}:Mf_i(x)=\lambda \} \neq \varnothing \text{ for } \gamma _i<\lambda <1.\end{align*} $$
Let
$$ \begin{align*}\kern-2pt\alpha_i:=\inf\{y : y\in E_i\},\end{align*} $$
$$ \begin{align*}\beta_i:=\sup\{y : y\in E_i\}.\end{align*} $$
Then, by the continuity of
$Mf_i$
, we see that
$$ \begin{align*} \alpha_i<a_i,\quad \beta_i>b_i, \end{align*} $$
$$ \begin{align*} \alpha_i,\beta_i\in E_i, \end{align*} $$
$$ \begin{align*} \int_{a_i}^{b_i}d\mu=\lambda\int_{\alpha_i}^{b_i}d\mu, \quad \int_{a_i}^{b_i}d\mu=\lambda\int_{a_i}^{\beta_i}d\mu, \end{align*} $$
and
$$ \begin{align*} \{x\in\mathbb{R}:Mf_i(x)>\lambda\}=(\alpha_i,\beta_i). \end{align*} $$
These facts imply that
$$ \begin{align*} \lambda\mu( \{x\in\mathbb{R}:Mf_i(x)>\lambda\}) &=\lambda\int_{\alpha_i}^{\beta_i}d\mu\\ &=\lambda\int_{\alpha_i}^{b_i}d\mu+\lambda\int_{a_i}^{\beta_i}d\mu-\lambda\int_{a_i}^{b_i}d\mu\\ &=( 2-\lambda)\int_{a_i}^{b_i}d\mu. \end{align*} $$
Hence,
$$ \begin{align*} \sup\limits_{\lambda>0}\lambda\mu( \{x\in\mathbb{R}:Mf_i(x)>\lambda\}) &\geq\sup\limits_{\gamma_i<\lambda<1}\lambda\mu( \{x\in\mathbb{R}:Mf_i(x)>\lambda\})\\ &=\sup\limits_{\gamma_i<\lambda<1}(2-\lambda) \int_{a_i}^{b_i}d\mu\\ &=(2-\gamma_i)\mu((a_i,b_i)). \end{align*} $$
By letting
$i\rightarrow \infty $
, we deduce that
$$ \begin{align} \varliminf\limits_{i\rightarrow\infty}\frac{\sup\limits_{\lambda>0}\lambda\mu( \{x\in\mathbb{R}:M\chi_{( a_i,b_i) }(x)>\lambda\})}{\mu((a_i,b_i))}\geq\varliminf\limits_{i\rightarrow\infty}(2-\gamma_i)=2. \end{align} $$
On the other hand, from Lemma 2.1, we also have
$$ \begin{align*} \varlimsup\limits_{i\rightarrow\infty}\frac{\sup\limits_{\lambda>0}\lambda\mu( \{x\in\mathbb{R}:M\chi_{( a_i,b_i) }(x)>\lambda\})}{\mu((a_i,b_i))}\leq 2. \end{align*} $$
This, together with (2.2), yields that Proposition 2.2 holds and the proof is finished.
To present the proof of Theorem 1.2, we also need the following proposition.
Proposition 2.3 Let
$T(x):=\frac {x^{1-\frac {1}{p}}+1}{x+1}$
,
$x>0$
and
$1<p<\infty $
. Then, the maximum value of the function T is
$\frac {p-1}{p}\theta _p$
.
Proof: We firstly see that
$$ \begin{align*} T'(x)=\frac{-\frac{1}{p}x^{1-\frac{1}{p}}+(1-\frac{1}{p})x^{-\frac{1}{p}}-1}{(x+1)^2},\quad x>0. \end{align*} $$
It is not difficult to prove that there exists a unique point
$x_0>0$
such that
$T'(x_0)=0$
and
$\max \limits _{x>0} T(x)=T(x_0)$
. Therefore,
$$ \begin{align*} \max\limits_{x>0} T(x)=\frac{x_0^{1-\frac{1}{p}}+1}{x_0+1}=\frac{x_0^{1-\frac{1}{p}}+\frac{\frac{1}{p}x_0^{1-\frac{1}{p}}}{(1-\frac{1}{p})x_0^{-\frac{1}{p}}-1}}{x_0+\frac{\frac{1}{p}x_0^{1-\frac{1}{p}}}{(1-\frac{1}{p})x_0^{-\frac{1}{p}}-1}}=\frac{p-1}{p}x_0^{-\frac{1}{p}}. \end{align*} $$
Since
$x_0^{-\frac {1}{p}}>0$
satisfies (1.4), we set
$\theta _p:=x_0^{-\frac {1}{p}}$
, which implies
$$ \begin{align*} \max\limits_{x>0}T(x)=\frac{p-1}{p}\theta_p. \\[-40pt]\end{align*} $$
Next, we give the proof of Theorem 1.2.
proof Proof of Theorem 1.2:
Let
$1<p<\infty $
and
$$ \begin{align*} f(x):=\int^{x}_{0}w(t)dt,\quad F(x):=|f(x)|^{-\frac{1}{p}},\quad x\in{\mathbb{R}}. \end{align*} $$
From assumptions on
$\mu $
, we see that f is well-defined, continuous, and increasing on
$\mathbb {R}$
and absolutely continuous in any closed interval on
$\mathbb {R}$
. Also let
$$ \begin{align*} \mathscr {A}:=\left\{H:H \text{ is a maximal closed interval such that } \int _{H}w(t)dt=0\right\}, \end{align*} $$
$E:=\cup _{H\in \mathscr {A}}H$
, and
$E^c:=\mathbb {R}\setminus E$
. Obviously,
$\mathscr {A}$
is a countable set, and
$\mu (E)=0$
.
By Proposition 2.3 and Assumption (B), we see that for any
$x\in E^c\cap (0,\infty )$
,
$f(x)>0$
and
$$ \begin{align*} MF(x)&=\sup\limits_{a<x<b}\frac{\int_{a}^{b}F(t)d\mu(t)}{\int_{a}^{b}w(t)dt}\\&\geq\sup\limits_{a<0}\frac{\int_{a}^{x}F(t)d\mu(t)}{\int_{a}^{x}w(t)dt}\\ &=\sup\limits_{a<0}\frac{p}{p-1}\frac{|f(x)|^{1-\frac{1}{p}}+|f(a)|^{1-\frac{1}{p}}}{|f(x)|+|f(a)|}\\ &=\frac{p}{p-1}F(x)\sup\limits_{a<0}\frac{|\frac{f(a)}{f(x)}|^{1-\frac{1}{p}}+1}{|\frac{f(a)}{f(x)}|+1}\\ &=\frac{p}{p-1}F(x)\frac{p-1}{p}\theta_p\\ &=\theta_pF(x). \end{align*} $$
The above inequality also holds for
$x\in E^c\cap (-\infty ,0)$
. Hence,
$$ \begin{align*} MF(x)\geq\theta_pF(x),\quad \mu-a. e.\quad x\in\mathbb{R}. \end{align*} $$
Observe that
$F\notin L^p(\mathbb {R},\mu )$
. So instead, we consider the following function:
$$ \begin{align} F_{\varepsilon}(x):=F(x)\min\{|f(x)|^{\varepsilon},|f(x)|^{-\varepsilon}\} \in L^p(\mathbb{R},\mu),\quad \varepsilon>0. \end{align} $$
We claim that
$$ \begin{align} MF_{\varepsilon}(x)\geq\frac{1}{\varepsilon+\frac{1}{p'}}F_{\varepsilon}(x)\frac{1+\theta_p^{1-p-\varepsilon p}}{1+\theta_p^{-p}},\quad \mu-a.e.\quad x\in\mathbb{R}. \end{align} $$
In fact, assume that
$x\in E^c\cap (0,\infty )$
, and set
$$ \begin{align} \varphi(u):=\sup\{t:f(t)\leq uf(x)\},\quad u\in\mathbb{R}. \end{align} $$
We denote by m the Lebesgue measure. From the property of f,
$\varphi (u)$
is well-defined, strictly increasing, right continuous, and
$m-a. e.$
differentiable on
$\mathbb {R}$
. We can further verify the following fact:
$$ \begin{align*}\varphi(1)\geq x,\end{align*} $$
$$ \begin{align*}f(\varphi(u))=uf(x),\quad u\in\mathbb{R},\end{align*} $$
$$ \begin{align*}w(t)=0,\quad m-a. e.\quad t\in[x,\varphi(1)],\end{align*} $$
and
$$ \begin{align*}w(\varphi(u))\varphi'(u)=f(x),\quad m-a. e.\quad u\in\mathbb{R}. \end{align*} $$
From these facts, we deduce that
$$ \begin{align*} MF_{\varepsilon}(x) &\geq\sup\limits_{y\leq0}\frac{\int_{y}^{x}F_{\varepsilon}(t)d\mu(t)}{\int_{y}^{x}w(t)dt}\\ &\geq\sup\limits_{y\leq0}\frac{\int_{y}^{x}|f(t)|^{-\frac{1}{p}}\min\{|f(t)|^{\varepsilon},|f(t)|^{-\varepsilon}\} w(t)dt}{|f(x)|+|f(y)|}\\ &=\sup\limits_{y\leq0}\frac{\int_{y}^{\varphi(1)}|f(t)|^{-\frac{1}{p}}\min\{|f(t)|^{\varepsilon},|f(t)|^{-\varepsilon}\} w(t)dt}{|f(x)|+|f(y)|}\\ &\geq\frac{\int_{\varphi(-\theta_p^{-p}f(x))}^{\varphi(1)}|f(t)|^{-\frac{1}{p}}\min\{|f(t)|^{\varepsilon},|f(t)|^{-\varepsilon}\} w(t)dt}{|f(x)|+|\theta_p^{-p}f(x)|}\\ &=\frac{\int_{-\theta_p^{-p}}^{1}|u|^{-\frac{1}{p}}|f(x)|^{-\frac{1}{p}}\min\{|uf(x)|^{\varepsilon},|uf(x)|^{-\varepsilon}\} f(x)du}{|f(x)|+|\theta_p^{-p}f(x)|}\\ &=|f(x)|^{-\frac{1}{p}}\frac{\int_{-\theta_p^{-p}}^{1}|u|^{-\frac{1}{p}}\min\{|uf(x)|^{\varepsilon},|uf(x)|^{-\varepsilon}\} du}{1+\theta_p^{-p}}\\ &=|f(x)|^{-\frac{1}{p}}\min\{|f(x)|^{\varepsilon},|f(x)|^{-\varepsilon}\} \frac{\int_{-\theta_p^{-p}}^{1}|u|^{-\frac{1}{p}}\frac{\min\{|uf(x)|^{\varepsilon},|uf(x)|^{-\varepsilon}\} }{\min\{|f(x)|^{\varepsilon},|f(x)|^{-\varepsilon}\} }du}{1+\theta_p^{-p}}\\ &\geq F_{\varepsilon}(x)\frac{\int_{-\theta_p^{-p}}^{1}|u|^{\varepsilon-\frac{1}{p}}du}{1+\theta_p^{-p}}\\ &=\frac{1}{\varepsilon+\frac{1}{p'}}F_{\varepsilon}(x)\frac{1+\theta_p^{1-p-\varepsilon p}}{1+\theta_p^{-p}}, \end{align*} $$
where in the fifth equation, we use integration by substitution with
$t=\varphi (u)$
(see [Reference Folland3]). When
$x\in E^c\cap (-\infty ,0)$
, we can treat similarly. Therefore, the claim holds.
Then, from (2.4), it follows that
$$ \begin{align*}\lim\limits_{\varepsilon\rightarrow0}\frac{\Vert MF_{\varepsilon}\Vert_{L^p(\mathbb{R},\mu)}}{\Vert F_{\varepsilon}\Vert_{L^p(\mathbb{R},\mu)}}\geq p'\frac{1+\theta_p^{1-p}}{1+\theta_p^{-p}}=\theta_p,\end{align*} $$
which means
$C_p\geq \theta _p$
. Hence, we complete the proof of Theorem 1.2.
Finally, we briefly discuss the n-dimensional case. Following the method of Grafakos and Montgomery-Smith [Reference Grafakos and Montgomery-Smith7], we present a complete adaptation of their argument here for completeness. Denote by
$\mathbf {x}:=(x_1,\ldots ,x_n)$
in
$\mathbb {R}^n$
. For a non-negative locally integrable function f on
$\mathbb {R}^n$
and
$w_j(t)$
on
$\mathbb {R}$
,
$j=1,\ldots ,n$
, satisfying the conditions in Theorem 1.2, we define the “strong” maximal function on
$\mathbb {R}^n$
, by
$$ \begin{align*}N_{\vec{\mu}}^{n}(f)(\mathbf{x}):=\sup\limits_{a_1<x_1<b_1}\cdots\sup\limits_{a_n<x_n<b_n}\frac{\int_{a_n}^{b_n}\ldots\int_{a_1}^{b_1}f(\mathbf{y})d\vec{\mu}(\mathbf{y})}{\int_{a_n}^{b_n}\ldots\int_{a_1}^{b_1}d\vec{\mu}(\mathbf{y})},\end{align*} $$
where
$d\vec {\mu }(\mathbf {y}):=w_1(y_1)\ldots w_n(y_n)dy_1\ldots dy_n$
. Clearly,
$N_{\vec {\mu }}^{1}=M_\mu $
. Moreover, observe that
$$ \begin{align*}N_{\vec{\mu}}^{n}\leq M_\mu^{(1)}\circ\cdots\circ M_\mu^{(n)},\end{align*} $$
where
$M_\mu ^{(j)}$
denotes the maximal operator
$M_\mu $
on
$\mathbb {R}$
applied to the
$x_j$
coordinate. This implies that the operator norm of
$N_{\vec {\mu }}^{n}$
on
$L^p(\mathbb {R}^n, \vec {\mu })$
is less than or equal to
$\theta _p^n$
,
$1<p<\infty $
. By considering the function
$$ \begin{align*}G_{\varepsilon}(\mathbf{x}):=\prod_{j=1}^{n}F_{\varepsilon}(x_j),\quad \mathbf{x}\in{\mathbb{R}^n},\end{align*} $$
where
$F_{\varepsilon }$
is as in (2.3), and using an argument similar to Theorem 1.2, we can obtain the reverse inequality. We state this result as follows.
Corollary 2.4 For
$1<p<\infty $
, the operator norm of
$N_{\vec {\mu }}^n$
on
$L^p(\mathbb {R}^n, \vec {\mu })$
is
$\theta _p^n$
.
Furthermore, it is easy to verify
$\frac {p}{p-1}<\theta _p<\frac {2p}{p-1}$
, which implies that the operator norm of
$N_{\vec {\mu }}^n$
on
$L^p(\mathbb {R}^n, \vec {\mu })$
grows exponentially with n, as
$n\rightarrow \infty $
.
Acknowledgements
The authors would like to thank Professor Dongyong Yang for helpful discussions and also express their deep thanks to the referee for his/her very careful reading and useful comments which do improve the presentation of this article.
