Fractional Maximal Operator on Musielak–Orlicz Spaces Over Unbounded Quasi-Metric Measure Spaces

Abstract

The main target of this article is the boundedness of the fractional maximal operator , on Musielak–Orlicz spaces $L^{\Phi }(X)$ over unbounded quasi-metric measure spaces as an extension of recent results by Cruz-Uribe and Shukla (Studia Math 242(2):109–139, 2018) and the authors (2019), where $\eta$ is the order of the fractional maximal operator and $\lambda$ is its modification rate. Our results are new even for the Hardy–Littlewood maximal operator $M_{\lambda }$ or for the Orlicz spaces $L^{p(\cdot )}(\log L)^{q(\cdot )}(X)$. Usually, for the proof of the boundedness, the three-line theorem is employed. This new technique of using the three-line theorem enables us to extend the function spaces with ease. An example explains why we can not remove the modification parameter $\lambda$.

Appendix–An Estimate of Welland Type

The following equivalence is well known. But for the sake of self-containedness, we supply its proof. The symbol $g\sim h$ means that $C^{-1}h\le g\le Ch$ for some constant $C>0$.

Lemma 6.1

Let $\varepsilon ,A,B>0$. Then

$$\begin{aligned} \sum _{j=-\infty }^\infty \min (2^{-j\varepsilon }A,2^{j\varepsilon }B) \sim \sqrt{A B}. \end{aligned}$$

This estimate is standard and somewhat well known. Here for the sake of convenience, we supply the whole proof.

Proof

Let $j_0\equiv \left[ \frac{1}{2 \varepsilon } \log _2 \frac{A}{B}\right]$. Then $2^{-j_0\varepsilon }A \sim 2^{j_0\varepsilon }B$. Thus,

$$\begin{aligned} \sum _{j=-\infty }^\infty \min (2^{-j\varepsilon }A,2^{j\varepsilon }B)&\sim \sum _{j=-\infty }^{j_0} 2^{j\varepsilon }B + \sum _{j=j_0}^\infty 2^{-j\varepsilon }A\\&\sim 2^{j_0\varepsilon }B+2^{-j_0\varepsilon }A\\&\sim \sqrt{A B}. \end{aligned}$$

$\square$

The following lemma extends [9, Proposition 5.1]. We employ the idea of Welland [43].

Lemma 6.2

Let $0<\eta <1$, $0<\varepsilon <\min (1-\eta ,\eta )$ and $\lambda \ge 1$. Then

$$\begin{aligned} I_{\eta ,\lambda }f(x) \le C \sqrt{ M_{\eta -\varepsilon ,\lambda }f(x) M_{\eta +\varepsilon ,\lambda }f(x) } \quad (x \in X) \end{aligned}$$

for any non-negative $\mu$-measurable function f.

Noteworthy is the fact that we do not have to postulate any condition on $\mu$ such as the reverse doubling condition. In [9, Proposition 5.1], they used the assumption that $\mu$ is a reverse doubling measure, that is,

$$\begin{aligned} \mu (B(x,r/2)) \le \gamma \mu (B(x,r)) \end{aligned}$$

for every $x \in X$, $r>0$ such that $B(x,r) \subset X$ and some constant $0<\gamma <1$.

Proof

Let us set

$$\begin{aligned} r_j \equiv \max \{r>0\,:\,\mu (B(x,\lambda r)) \le 2^j\} = \sup \{r>0\,:\,\mu (B(x,\lambda r)) \le 2^j\}. \end{aligned}$$

Then we have $r_j \le r_{j+1}$ for all $j \in {{\mathbb {Z}}}$. We decompose

$$\begin{aligned} I_{\eta ,\lambda }f(x)&\equiv \int _{X {\setminus } \{x\}} \frac{f(y)}{\mu (B(x,2\lambda d(x,y)))^{1-\eta }}d\mu (y)\\&= \sum _{j=-\infty }^\infty \int _{B(x,r_j) {\setminus } B(x,r_{j-1})} \frac{f(y)}{\mu (B(x,2\lambda d(x,y)))^{1-\eta }}d\mu (y)\\&\le \sum _{j=-\infty }^\infty \int _{B(x,r_j) {\setminus } B(x,r_{j-1})}\frac{f(y)}{\mu (B(x,2\lambda r_{j-1}))^{1-\eta }}d\mu (y). \end{aligned}$$

Since $\mu (B(x,2\lambda r_{j-1}))>2^{j-1}$ thanks to the definition of $r_{j-1}$, we obtain

$$\begin{aligned} I_{\eta ,\lambda }f(x)&\le \sum _{j=-\infty }^\infty 2^{-(j-1)(1-\eta )} \int _{B(x,r_j) {\setminus } B(x,r_{j-1})}f(y)d\mu (y)\\&\le \sum _{j=-\infty }^\infty 2^{-(j-1)(1-\eta )} \int _{B(x,r_j)}f(y)d\mu (y). \end{aligned}$$

Since $\mu (B(x,\lambda r_j)) \le 2^j$, we obtain

$$\begin{aligned} I_{\eta ,\lambda }f(x)&\le \sum _{j=-\infty }^\infty 2^{-(j-1)(1-\eta )} \min ( 2^{j(1-\eta +\varepsilon )}M_{\eta -\varepsilon ,\lambda }f(x), 2^{j(1-\eta -\varepsilon )}M_{\eta +\varepsilon ,\lambda }f(x))\\&\le C \sqrt{ M_{\eta -\varepsilon ,\lambda }f(x) M_{\eta +\varepsilon ,\lambda }f(x) } \end{aligned}$$

thanks to Lemma 6.1, as required. $\square$