Estimates of the modular-type operator norm of the general geometric mean operator

Abstract

In this paper, the modular-type operator norm of the general geometric mean operator over spherical cones is investigated. We give two applications of a new limit process, introduced by the present authors, to the establishment of Pólya-Knopp-type inequalities. We not only partially generalize the sufficient parts of Persson-Stepanov’s and Wedestig’s results, but we also provide new proofs to these results.

1 Introduction

Let E be a spherical cone in $\Bbb{R}^{n}$. By this, we mean that $E=\bigcup_{s>0} sA$ for some Borel measurable subset A of the unit sphere $\Sigma^{n-1}$. Let $\|\Bbb{K}\|_{D_{\Bbb{K}}\cap L_{\Phi}^{p}(v\,dx)\mapsto L_{\Phi}^{q}(u\,dx)}$ (in brief, $\|\Bbb{K}\|_{*}$) denote the smallest constant C in (1.1):

$$ \biggl\{ \int_{E} \bigl(\Phi\circ\Bbb{K}f(x) \bigr)^{q} u(x)\,dx \biggr\} ^{1/q} \le C \biggl\{ \int _{E} \bigl(\Phi\circ f(x) \bigr)^{p} v(x)\,dx \biggr\} ^{1/p} $$

(1.1)

for all $f\in D_{\Bbb{K}}\cap L_{\Phi}^{p}(v\,dx)$, where $p, q>0$, $u(x)\ge0$, $v(x)>0$, $\Phi\in CV^{+}(I)$, $\Phi\circ f(x)=\Phi(f(x))$, and $\Bbb{K}f(x)$ is of the form

$$ \Bbb{K}f(x):=\int_{\tilde{S}_{x}} k(x,t)f(t)\,dt\quad (x\in E). $$

(1.2)

Here $CV^{+}(I)$ denotes the set of all nonnegative convex functions defined on an open interval I in $\Bbb{R}$, $D_{\Bbb{K}}$ is the space of those f such that $\Bbb{K}f(x)$ is well defined for almost all $x\in E$, and $L_{\Phi}^{p}(v\,dx)$ is the set of all real-valued Borel measurable f with

$$\|f\|_{\Phi, p,v}:= \biggl\{ \int_{E} \bigl(\Phi\circ f(x) \bigr)^{p}v(x)\,dx \biggr\} ^{1/p}< \infty. $$

Moreover, $\tilde{S}_{x}=\bigcup_{0< s\le\|x\|} sA$, $S_{x}=\tilde{S}_{x}\setminus\| x\|A$, and $k(x,t)\ge0$ is locally integrable over $\Bbb{E}\times\Bbb{E}$.

We write $L^{p}(v\,dx)$ and $\|f\|_{p,v}$ instead of $L^{p}_{\Phi}(v\,dx)$ and $\| f\|_{\Phi,p,v}$, respectively, for the case $\Phi(s)=|s|$. We also write $L^{p}(E,v\,dx)$ for $L^{p}(v\,dx)$, whenever the integral region E is emphasized.

Clearly,

$$\|\Bbb{K}\|_{*}=\sup_{f} \frac{\|\Phi\circ{\Bbb{K}}f\|_{q,u}}{\|\Phi\circ f\|_{p,v}}, $$

where the supremum is taken over all $f\in D_{\Bbb{K}}\cap L_{\Phi}^{p}(v\,dx)$ with $\|\Phi\circ f\|_{p,v}\neq0$. This number reduces to the operator norm of $\Bbb{K}$ for the case $\Phi(s)=|s|$. The investigation of the value $\|\Bbb{K}\|_{*}$ has a long history in the literature. In [1], the present authors introduced a generalized Muckenhoupt constant $A_{M}(p,q)$ and established the following Muckenhoupt-type estimate for $\|\Bbb{K}\|_{*}$:

$$ \|\Bbb{K}\|_{*}\le \biggl(\frac{q}{p^{*}}+\frac{q}{\eta} \biggr)^{1/q} \biggl(1+\frac{p^{*}}{\eta} \biggr)^{\eta^{*}/(p^{*}q^{*})}A_{M}(p,q), $$

(1.3)

where $1\le p, q\le\infty$, $\eta=\max(p,q)$, and $(\cdot)^{*}$ is the conjugate exponent of $(\cdot)$ in the sense that $1/(\cdot)+1/(\cdot)^{*}=1$. For the particular case that

$$ \Phi(s)=|s|,\qquad k(x,t)=1, $$

(1.4)

there are two other types of estimates. They are

$$ \|\Bbb{K}\|_{*}\le p^{*}A_{PS}(p,q) $$

(1.5)

and

$$ \|\Bbb{K}\|_{*}\le A_{W}(p,q):=\inf_{1< s< p} A_{W}(s,p,q) \biggl(\frac {p-1}{p-s} \biggr)^{1/p^{*}}. $$

(1.5a)

These two inequalities were proved in [2] and [3], Theorem 3.1 and Lemma 7.4, for the case $1< p\le q<\infty$ (see also [4], Theorem 2.1). We refer the readers to Section 2 for details.

In this paper, we focus on the evaluation of $\|\Bbb{K}\|_{*}$ for the following case of (1.1):

$$\Phi(s)=e^{s},\qquad k(x,t)=g(t)/G(x),\qquad f(t)\longrightarrow \log f(t), $$

where $f(t)>0$, $g(t)>0$, and

$$ G(x)=\int_{\tilde{S}_{x}} g(t)\,dt\quad (x\in E). $$

(1.6)

The corresponding inequality to (1.1) takes the form

$$ \biggl(\int_{E} \biggl\{ \exp \biggl(\frac{1}{G(x)}\int _{\tilde{S}_{x}} g(t)\log f(t)\,dt \biggr) \biggr\} ^{q} u(x)\,dx \biggr)^{1/q}\le C \biggl\{ \int_{E} \bigl(f(x) \bigr)^{p} v(x)\,dx \biggr\} ^{1/p}, $$

(1.7)

which is known as the Pólya-Knopp-type inequality.

In [4], Theorem 3.1, [2, 5], and [3], Theorem 7.3, the particular case $g(t)=1$ of (1.7) was considered. They obtained the following estimates by means of the formula $(G_{\Bbb{K}}f)(x)=\lim_{\epsilon\to0^{+}} [\Bbb{K}(f^{\epsilon})]^{1/\epsilon}(x)$:

$$ \|\Bbb{K}\|_{*}\le e^{1/p}D^{*}_{PS} \quad\mbox{and}\quad \|\Bbb{K}\|_{*}\le \inf_{s>1} e^{(s-1)/p} D^{*}_{OG}(s), $$

(1.8)

where $0< p\le q<\infty$. The definitions of $D^{*}_{PS}$ and $D^{*}_{OG}(s)$ are given in Section 3.

The purpose of this paper is two-fold. We not only extend the aforementioned sufficient parts of [2, 4, 5], and [3] from $u(x)>0$ and $g(t)=1$ to $u(x)\ge0$ and

$$ \min \biggl(\sup_{x\in E} \bigl|g(x)\bigl|, \sup_{x\in E} \biggl| \frac{g(x)}{v(x)} \biggr| \biggr)< \infty, $$

(1.9)

but we also provide a new proof of (1.8) from the viewpoint of (1.10):

$$ \|\Bbb{K}\|_{*} \le\inf_{\epsilon\in\frak{F}_{\Phi}^{+}} (A_{p/\epsilon, q/\epsilon})^{1/\epsilon} \le\liminf_{\epsilon\to0^{+}} \bigl\{ (A_{p/\epsilon, q/\epsilon })^{1/\epsilon} \bigr\} , $$

(1.10)

where $0< p,q<\infty$, $\frak{F}_{\Phi}^{+}=\{\epsilon>0: \Phi^{\epsilon}\in CV^{+}(I)\}$, and $A_{p,q}$ are absolute constants subject to the condition

$$ \biggl(\int_{E} \bigl|\Bbb{K}f(x) \bigr|^{q} u(x)\,dx \biggr)^{1/q} \le A_{p,q} \biggl(\int_{E} \bigl|f(x)\bigr|^{p} v(x)\,dx \biggr)^{1/p} \quad(f\ge0). $$

(1.11)

It is clear that (1.10) is applicable to the case $\Phi(s)=e^{s}$. In this case, $\frak{F}_{\Phi}^{+}=\{\epsilon>0\}$ and the second inequality in (1.10) holds. We remark that it may not be an equality (cf. [6]). On the other hand, we have $p/\epsilon\to\infty$ and $q/\epsilon\to\infty$ as $\epsilon\to0^{+}$. This indicates that the infimum in (1.10) can be estimated by evaluating those $A_{p,q}$ with p, q large enough.

The limit process (1.10) differs from the scheme by means of the formula $(G_{\Bbb{K}}f)(x)=\lim_{\epsilon\to0^{+}} [\Bbb{K}(f^{\epsilon})]^{1/\epsilon}(x)$. It was introduced in [6] to get different types of Pólya-Knopp inequalities, including the n-dimensional extensions of the Levin-Cochran-Lee-type inequalities and Carleson’s result. We showed that the infimum in (1.10) can easily be evaluated by applying the following choice of $A_{p,q}$ for $1< p,q<\infty$:

$$A_{p,q}\le \biggl(\frac{q}{p^{*}}+\frac{q}{\eta} \biggr)^{1/q} \biggl(1+\frac{p^{*}}{\eta} \biggr)^{\eta^{*}/(p^{*}q^{*})}A_{M}(p,q). $$

This choice is due to (1.3). We also pointed out that for some cases, the values of $\|\Bbb{K}\|_{*}$ obtained from (1.10) are better than the known constants in the literature. In this paper, we consider two other choices of $A_{p,q}$ with $1< p\le q<\infty$, that is, $A_{p,q}\le p^{*}\tilde{A}_{PS}(p,q)$ and $A_{p,q}\le\tilde{A}_{W}(p,q)$, which are general forms of (1.5) and (1.5a). We shall derive them from (1.5) and (1.5a) and relax the conditions on $u(x)$ and $g(t)$ from $u(x)>0$ and $g(t)=1$ to $u(x)\ge0$ and $g(t)>0$ (cf. Section 2). Based on such choices, we prove that (1.8) follows from (1.10). Moreover, (1.8) can be extended from $u(x)>0$ and $g(t)=1$ to $u(x)\ge0$ and $g(t)$ of the form (1.9). This extension gives Persson-Stepanov-type and Opic-Gurka-tpye estimates of the modular-type operator norm of the general geometric mean operator corresponding to $g(t)$. We remark that the particular case $g(t)=|{\tilde{S}}_{t}|^{s-1}$ can lead us to the Levin-Cochran-Lee-type inequality (see Section 3 for details).

2 General forms of (1.5) and (1.5a)

Let $1< p\le q<\infty$, $g(t)>0$, $u(x)\ge0$, and $v(x)>0$. Consider the inequality:

$$ \biggl(\int_{E} \biggl\{ \frac{1}{G(x)}\int _{\tilde{S}_{x}} g(t)f(t)\,dt \biggr\} ^{q} u(x)\,dx \biggr)^{1/q}\le C \biggl(\int_{E} \bigl(f(x) \bigr)^{p} v(x)\,dx \biggr)^{1/p} \quad(f\ge0), $$

(2.1)

where $G(x)$ is defined by (1.6). This corresponds to the case $\Phi(s)=|s|$ and $k(x,t)=g(t)/G(x)$ of (1.1). Inequality (2.1) reduces to the form (2.2) for the case $g(t)=1$:

$$ \biggl(\int_{E} \biggl\{ \int_{\tilde{S}_{x}} f(t)\,dt \biggr\} ^{q} \tilde{u}(x)\,dx \biggr)^{1/q}\le C \biggl( \int_{E} \bigl(f(x)\bigr)^{p} v(x)\,dx \biggr)^{1/p}\quad (f\ge0), $$

(2.2)

where $\tilde{u}(x)=u(x)/G(x)^{q}$. In [4], Theorem 2.1, [2] and [3], Lemma 7.4(a), it was proved that under the conditions $u(x)>0$ and $A_{PS}(p,q)<\infty$, (1.5) holds, in other words, (2.2) with $\tilde{u}(x)$ replaced by $u(x)$ is true for $C=p^{*}A_{PS}(p,q)$, where

$$A_{PS}(p,q):=\sup_{x\in E} \biggl(\int _{\tilde{S}_{x}} v(t)^{1-p*}\,dt \biggr)^{-1/p} \biggl(\int _{\tilde{S}_{x}} \biggl\{ \int_{\tilde{S}_{t}} v(y)^{1-p*}\,dy \biggr\} ^{q} u(t)\,dt \biggr)^{1/q}. $$

This result will be extended below from $g(t)=1$ and $u(x)>0$ to $g(t)>0$ and $u(x)\ge0$. We shall see its application in the proof of Theorem 3.2.

Theorem 2.1

Let $1< p\le q<\infty$, $u(x)\ge0$, $v(x)>0$, $g(t)>0$, and $0< G(x)<\infty$, where $G(x)$ is defined by (1.6). If $\tilde{A}_{PS}(p,q)<\infty$, then (2.1) holds for $C\le p^{*}\tilde{A}_{PS}(p,q)$, where

$$\tilde{A}_{PS}(p,q)=\sup_{x\in E} \biggl(\int _{\tilde{S}_{x}} \biggl(\frac{g(t)}{v(t)} \biggr)^{p*}v(t)\,dt \biggr)^{\frac{-1}{p}} \biggl(\int_{\tilde{S}_{x}} \biggl\{ \frac{1}{G(t)}\int_{\tilde{S}_{t}} \biggl(\frac{g(y)}{v(y)} \biggr)^{p*}v(y)\,dy \biggr\} ^{q} u(t)\,dt \biggr)^{\frac{1}{q}}. $$

Proof

The case $u(x)>0$ follows from [4], Theorem 2.1, or [3], Lemma 7.4(a), under the following substitutions:

$$ f(t)\longrightarrow g(t)f(t),\qquad u(x)\longrightarrow\frac {u(x)}{(G(x))^{q}},\qquad v(x) \longrightarrow\frac {v(x)}{(g(x))^{p}}. $$

(2.3)

As for $u(x)\ge0$, let $u_{\tau}(x)=u(x)+\rho_{\tau}(x)$, where $0<\tau<1$ and $\rho_{\tau}(x)>0$ is subject to the condition

$$ \int_{\tilde{S}_{x}} \biggl\{ \frac{1}{G(t)}\int_{\tilde{S}_{t}} \biggl(\frac {g(y)}{v(y)} \biggr)^{p^{*}}v(y)\,dy \biggr\} ^{q} \rho_{\tau}(t)\,dt\le \tau \biggl\{ \int_{\tilde{S}_{x}} \biggl( \frac{g(t)}{v(t)} \biggr)^{p^{*}}v(t)\,dt \biggr\} ^{q/p}. $$

(2.4)

Such $\rho_{\tau}(x)$ exists. We have $u_{\tau}(x)>0$ on E. Moreover, the condition $1/q<1$ implies that $(a+b)^{1/q}\le a^{1/q}+b^{1/q}$ for all $a,b\ge0$. Putting this together with (2.4) yields

$$\begin{aligned} & \biggl(\int_{\tilde{S}_{x}} \biggl\{ \frac{1}{G(t)}\int _{\tilde{S}_{t}} \biggl(\frac{g(y)}{v(y)} \biggr)^{p^{*}}v(y)\,dy \biggr\} ^{q} u_{\tau}(t)\,dt \biggr)^{1/q} \\ &\quad\le \biggl(\int_{\tilde{S}_{x}} \biggl\{ \frac{1}{G(t)}\int _{\tilde{S}_{t}} \biggl(\frac{g(y)}{v(y)} \biggr)^{p^{*}} v(y)\,dy \biggr\} ^{q} u(t)\,dt \biggr)^{\frac{1}{q}}+ \tau^{\frac{1}{q}} \biggl\{ \int_{\tilde{S}_{x}} \biggl(\frac {g(t)}{v(t)} \biggr)^{p^{*}}v(t)\,dt \biggr\} ^{\frac{1}{p}}. \end{aligned}$$

This leads us to

$$ \tilde{A}_{PS}(p,q,\tau)\le\tilde{A}_{PS}(p,q)+ \tau^{1/q}< \infty, $$

(2.5)

where $\tilde{A}_{PS}(p,q,\tau)$ is the number obtained from $\tilde{A}_{PS}(p,q)$ by replacing $u(t)$ by $u_{r}(t)$. We have $u_{\tau}(x)>u(x)$ on E. By the result of the case $u(x)>0$, the following inequality holds for $f\ge0$:

$$\begin{aligned} & \biggl(\int_{E} \biggl\{ \frac{1}{G(x)} \int_{\tilde{S}_{x}} g(t)f(t)\,dt \biggr\} ^{q} u(x)\,dx \biggr)^{\frac{1}{q}} \\ &\quad\le \biggl(\int_{E} \biggl\{ \frac{1}{G(x)}\int _{\tilde{S}_{x}} g(t)f(t)\,dt \biggr\} ^{q} u_{\tau}(x)\,dx \biggr)^{\frac{1}{q}} \\ &\quad\le p^{*}\tilde{A}_{PS}(p,q,\tau) \biggl(\int_{E} \bigl(f(x)\bigr)^{p} v(x)\,dx \biggr)^{\frac{1}{p}}. \end{aligned}$$

(2.6)

It follows from (2.5) that $\liminf_{\tau\to0^{+}} \tilde{A}_{PS}(p,q,\tau)\le\tilde{A}_{PS}(p,q)$. Putting this together with (2.6) yields the desired inequality. The proof is complete. □

Next, consider (1.5a). The number $A_{W}(s,p,q)$ in (1.5a) is defined by the formula:

$$\begin{aligned} A_{W}(s,p,q)=\sup_{x\in E} \biggl(\int _{\tilde{S}_{x}} v(t)^{1-p*}\,dt \biggr)^{\frac{s-1}{p}} \biggl(\int _{E\setminus S_{x}} \biggl\{ \int_{\tilde{S}_{t}} v(y)^{1-p*}\,dy \biggr\} ^{\frac{q(p-s)}{p}} u(t)\,dt \biggr)^{\frac{1}{q}}. \end{aligned}$$

In [3], Lemma 7.4(b), $A_{W}(s,p,q)$ is replaced by another notation $A^{*}_{W}(s)$. Like (1.5), (1.5a) can be generalized in the following way, in which $g(t)=1$ and $u(x)>0$ are relaxed to $g(t)>0$ and $u(x)\ge0$. We shall see its application in the proof of Theorem 3.3.

Theorem 2.2

Let $1< p\le q<\infty$, $u(x)\ge0$, $v(x)>0$, $g(t)>0$, and $0< G(x)<\infty$, where $G(x)$ is defined by (1.6). If $\tilde{A}_{W}(s,p,q)<\infty$ for some $1< s< p$, then (2.1) holds for $C\le\tilde{A}_{W}(p,q)$, where

$$ \tilde{A}_{W}(p,q):=\inf_{1< s< p} \tilde{A}_{W}(s,p,q) \biggl(\frac {p-1}{p-s} \biggr)^{1/p^{*}} $$

(2.7)

and

$$\begin{aligned} \tilde{A}_{W}(s,p,q)={}&\sup_{x\in E} \biggl(\int_{\tilde{S}_{x}} \biggl(\frac{g(t)}{v(t)} \biggr)^{p^{*}}v(t)\,dt \biggr)^{\frac {s-1}{p}} \\ &{}\times \biggl(\int_{E\setminus S_{x}} \biggl\{ \int_{\tilde{S}_{t}} \biggl(\frac{g(y)}{v(y)} \biggr)^{p^{*}}v(y)\,dy \biggr\} ^{\frac{q(p-s)}{p}} \frac{u(t)\,dt}{(G(t))^{q}} \biggr)^{\frac{1}{q}}. \end{aligned}$$

(2.8)

Proof

The case $u(x)>0$ follows from [3], Lemma 7.4(b), under the substitutions (2.3). For the case $u(x)\ge0$, we modify the proof of Theorem 2.1 in the following way. Let $1< s< p$ and $0<\tau<1$. Set $u_{\tau}(x,s)=u(x)+\rho_{\tau}(x,s)$, where $\rho_{\tau}(x,s)>0$ and satisfies the condition

$$\begin{aligned} &\int_{E\setminus S_{x}} \biggl\{ \int_{\tilde{S}_{t}} \biggl( \frac {g(y)}{v(y)} \biggr)^{p^{*}}v(y)\,dy \biggr\} ^{\frac{q(p-s)}{p}} \frac{\rho_{\tau}(t,s)}{(G(t))^{q}}\,dt \le \tau \biggl(\frac{p-1}{p-s} \biggr)^{\frac{-q}{p^{*}}} \biggl\{ \int _{\tilde{S}_{x}} \biggl(\frac{g(t)}{v(t)} \biggr)^{p^{*}}v(t)\,dt \biggr\} ^{\frac{q(1-s)}{p}}. \end{aligned}$$

Such $\rho_{\tau}(x,s)$ exists. We have $u_{\tau}(x,s)>0$ on $x\in E$. Moreover,

$$ \tilde{A}^{\tau}_{W}(s,p,q)\le\tilde{A}_{W}(s,p,q)+ \tau^{1/q} \biggl(\frac {p-1}{p-s} \biggr)^{-1/p^{*}}, $$

(2.9)

where $\tilde{A}^{\tau}_{W}(s,p,q)$ is obtained from $\tilde{A}_{W}(s,p,q)$ by making the change in (2.8): $u(t)\longrightarrow u_{\tau}(t,s)$. Obviously, $u_{\tau}(x,s)>u(x)$. Applying the preceding result of the case $u(x)>0$ to $u_{\tau}(x,s)$, we get

$$\begin{aligned} & \biggl(\int_{E} \biggl\{ \frac{1}{G(x)} \int_{\tilde{S}_{x}} g(t)f(t)\,dt \biggr\} ^{q} u(x)\,dx \biggr)^{1/q} \\ &\quad\le \biggl(\int_{E} \biggl\{ \frac{1}{G(x)}\int _{\tilde{S}_{x}} g(t)f(t)\,dt \biggr\} ^{q} u_{\tau}(x,s)\,dx \biggr)^{1/q} \\ &\quad\le \biggl\{ \inf_{1< s'< p} \tilde{A}^{\tau}_{W} \bigl(s',p,q\bigr) \biggl(\frac{p-1}{p-s'} \biggr)^{1/p^{*}} \biggr\} \biggl(\int_{E} \bigl(f(x)\bigr)^{p} v(x)\,dx \biggr)^{1/p} \\ &\quad\le\tilde{A}^{\tau}_{W}(s,p,q) \biggl(\frac{p-1}{p-s} \biggr)^{1/p^{*}} \biggl(\int_{E} \bigl(f(x) \bigr)^{p} v(x)\,dx \biggr)^{1/p}. \end{aligned}$$

(2.10)

Taking ‘$\inf_{1< s< p}$’ for both sides of (2.10), we get

$$ \biggl(\int_{E} \biggl\{ \frac{1}{G(x)}\int _{\tilde{S}_{x}} g(t)f(t)\,dt \biggr\} ^{q} u(x)\,dx \biggr)^{1/q}\le\tilde{A}^{\tau}_{W}(p,q) \biggl(\int _{E} \bigl(f(x)\bigr)^{p} v(x)\,dx \biggr)^{1/p}. $$

(2.11)

Here

$$\tilde{A}^{\tau}_{W}(p,q)=\inf_{1< s< p} \tilde{A}^{\tau}_{W}(s,p,q) \biggl(\frac {p-1}{p-s} \biggr)^{1/p^{*}}. $$

From (2.9), we obtain $\tilde{A}^{\tau}_{W}(p,q)\le \tilde{A}_{W}(p,q)+\tau^{1/q}$. Taking $\tau\to0^{+}$ for both sides of (2.11), we get the desired inequality. This completes the proof. □

3 Extensions and new proofs of (1.8)

To derive the extensions of (1.8), we need the following lemma.

Lemma 3.1

Let $0< p<\infty$, $v(x)>0$, $g(t)>0$, and $0< G(x)<\infty$, where $G(x)$ is defined by (1.6). If $\sup_{x\in E} \{g(x)/v(x)\}<\infty$, then, for all $t\in E$,

$$ \lim_{\epsilon\to0^{+}} \biggl(\frac{1}{G(t)}\int_{\tilde{S}_{t}} \biggl(\frac{g(y)}{v(y)} \biggr)^{\frac{\epsilon}{p-\epsilon}}g(y)\,dy \biggr)^{\frac{1}{\epsilon}} = \biggl\{ \exp \biggl(\frac{1}{G(t)}\int_{\tilde{S}_{t}}g(y) \biggl( \log\frac{g(y)}{v(y)} \biggr)\,dy \biggr) \biggr\} ^{\frac{1}{p}}. $$

(3.1)

Proof

Let $\alpha\ge\sup_{x\in E} \{g(x)/v(x)\}$. Without loss of generality, we may assume $\alpha>1$. We first consider the case that $\int_{\tilde{S}_{t}}g(y) |\log (\frac{g(y)}{v(y)} ) |\,dy<\infty$. Let

$$h(\epsilon)=\frac{1}{G(t)}\int_{\tilde{S}_{t}} \biggl( \frac {g(y)}{v(y)} \biggr)^{\epsilon/(p-\epsilon)}g(y)\,dy \quad(0\le \epsilon< p/2). $$

We have

$$\int_{\tilde{S}_{t}} \biggl(\frac{g(y)}{v(y)} \biggr)^{\epsilon/(p-\epsilon)}g(y)\,dy \le \alpha^{\epsilon/(p-\epsilon)}G(t)< \infty, $$

so $h(\epsilon)$ is well defined and has a finite value. For $\epsilon \in[0,p/2)$ and $0<\tau<\min(p/2-\epsilon,\epsilon)$, it follows from the mean value theorem that

$$\begin{aligned} \frac{h(\epsilon+\tau)-h(\epsilon)}{\tau}&=\frac{1}{G(t)}\int_{\tilde{S}_{t}} \frac{1}{\tau}\biggl\{ \biggl(\frac{g(y)}{v(y)} \biggr)^{\frac {\epsilon+\tau}{p-\epsilon-\tau}}- \biggl(\frac{g(y)}{v(y)} \biggr)^{\frac{\epsilon}{p-\epsilon}} \biggr\} g(y)\,dy \\ &=\frac{p}{G(t)}\int_{\tilde{S}_{t}} \frac{1}{(p-\epsilon _{0})^{2}} \biggl( \frac{g(y)}{v(y)} \biggr)^{\epsilon_{0}/(p-\epsilon _{0})}g(y) \biggl(\log\frac{g(y)}{v(y)} \biggr)\,dy, \end{aligned}$$

(3.2)

where $\epsilon_{0}:=\epsilon_{0}(y)$ lies between ϵ and $\epsilon+\tau$. We know that

$$\frac{\chi_{\tilde{S}_{t}}(y)}{(p-\epsilon_{0})^{2}} \biggl(\frac {g(y)}{v(y)} \biggr)^{\epsilon_{0}/(p-\epsilon_{0})}g(y) \biggl|\log \biggl(\frac{g(y)}{v(y)} \biggr) \biggr|\le\frac{\alpha\chi_{\tilde{S}_{t}}(y)g(y)}{(p-\epsilon)^{2}} \biggl|\log \biggl( \frac {g(y)}{v(y)} \biggr) \biggr| \in L^{1}(E,dy). $$

By (3.2) and the Lebesgue dominated convergence theorem, h is differentiable on $[0, p/2)$. In addition,

$$h'(\epsilon)=\lim_{\tau\to0^{+}}\frac{h(\epsilon+\tau)-h(\epsilon)}{\tau}= \frac{p}{(p-\epsilon)^{2}G(t)}\int_{\tilde{S}_{t}} \biggl(\frac{g(y)}{v(y)} \biggr)^{\epsilon/(p-\epsilon)}g(y) \biggl(\log\frac{g(y)}{v(y)} \biggr)\,dy. $$

Thus,

$$\begin{aligned} &\lim_{\epsilon\to0^{+}}\log \biggl(\frac{1}{G(t)}\int _{\tilde{S}_{t}} \biggl(\frac{g(y)}{v(y)} \biggr)^{\epsilon/(p-\epsilon)}g(y)\,dy \biggr)^{1/\epsilon} \\ &\quad=\lim_{\epsilon\to0^{+}}\frac{\log h(\epsilon)-\log h(0)}{\epsilon}\\ &\quad=\frac{d}{d\epsilon} \bigl(\log h(\epsilon) \bigr) \Big|_{\epsilon=0} = \frac{h'(0)}{h(0)}=\frac{1}{pG(t)}\int_{\tilde{S}_{t}}g(y) \biggl( \log\frac{g(y)}{v(y)} \biggr)\,dy. \end{aligned}$$

We get the desired result for the case $\int_{\tilde{S}_{t}} g(y) |\log (\frac{g(y)}{v(y)} ) |\,dy<\infty$. Next, consider the case $\int_{\tilde{S}_{t}} g(y) |\log (\frac {g(y)}{v(y)} ) |\,dy=\infty$. This implies

$$ \infty=\int_{\Omega_{1}} g(y) \biggl|\log \biggl(\frac {g(y)}{v(y)} \biggr) \biggr|\,dy +\int_{\Omega_{2}} g(y) \biggl|\log \biggl(\frac{g(y)}{v(y)} \biggr) \biggr|\,dy, $$

(3.3)

where $\Omega_{1}=\{y\in\tilde{S}_{t}: g(y)/v(y)\le1\}$ and $\Omega_{2}=\{y\in \tilde{S}_{t}: g(y)/v(y)> 1\}$. We have

$$\int_{\Omega_{2}}g(y) \biggl|\log \biggl(\frac{g(y)}{v(y)} \biggr) \biggr|\,dy \le(\log\alpha) G(t)< \infty. $$

Combining this with (3.3), we find that $\int_{\Omega_{1}}g(y) |\log (\frac{g(y)}{v(y)} ) |\,dy=\infty$. This leads us to

$$\int_{\tilde{S}_{t}}g(y) \biggl(\log\frac{g(y)}{v(y)} \biggr)\,dy= - \int_{\Omega_{1}}g(y) \biggl|\log \biggl(\frac{g(y)}{v(y)} \biggr) \biggr|\,dy+ \int_{\Omega_{2}} g(y) \biggl|\log \biggl(\frac{g(y)}{v(y)} \biggr) \biggr|\,dy=- \infty. $$

We shall show

$$\lim_{\epsilon\to0^{+}} \biggl(\frac{1}{G(t)}\int_{\tilde{S}_{t}} \biggl(\frac{g(y)}{v(y)} \biggr)^{\epsilon/(p-\epsilon)}g(y)\,dy \biggr)^{1/\epsilon} =0. $$

If so, the desired equality follows. Let $0<\epsilon<p/2$ and $y\in \tilde{S}_{t}$. By the mean value theorem, we get

$$\biggl(\frac{g(y)}{v(y)} \biggr)^{\epsilon/(p-\epsilon)}-1=\frac {\epsilon p}{(p-\epsilon_{0})^{2}} \biggl( \frac{g(y)}{v(y)} \biggr)^{\epsilon _{0}/(p-\epsilon_{0})} \biggl(\log\frac{g(y)}{v(y)} \biggr) $$

for some $\epsilon_{0}\in(0, \epsilon)$. This implies

$$\begin{aligned} &\frac{1}{G(t)}\int_{\tilde{S}_{t}} \biggl( \frac {g(y)}{v(y)} \biggr)^{\epsilon/(p-\epsilon)}g(y)\,dy \\ &\quad=1+ \biggl(\frac {\epsilon p}{G(t)}\int_{\tilde{S}_{t}}\frac{1}{(p-\epsilon_{0})^{2}} \biggl(\frac {g(y)}{v(y)} \biggr)^{\epsilon_{0}/(p-\epsilon_{0})}g(y) \biggl(\log\frac{g(y)}{v(y)} \biggr)\,dy \biggr). \end{aligned}$$

(3.4)

By Fatou’s lemma, we get

$$\begin{aligned} &\liminf_{\epsilon\to0^{+}}\frac{p}{G(t)}\int_{\tilde{S}_{t}} \frac{1}{(p-\epsilon_{0})^{2}} \biggl(\frac{g(y)}{v(y)} \biggr)^{\epsilon_{0}/(p-\epsilon_{0})}g(y)\biggl\vert \log \biggl(\frac {g(y)}{v(y)} \biggr)\biggr\vert \,dy \\ &\quad\ge\frac{1}{pG(t)}\int_{\tilde{S}_{t}} \biggl\{ \liminf _{\epsilon\to0^{+}} \biggl(\frac{g(y)}{v(y)} \biggr)^{\epsilon_{0}/(p-\epsilon_{0})} \biggr\} g(y)\biggl\vert \log \biggl(\frac{g(y)}{v(y)} \biggr)\biggr\vert \,dy \\ &\quad=\frac{1}{pG(t)}\int_{\tilde{S}_{t}}g(y)\biggl\vert \log \biggl( \frac {g(y)}{v(y)} \biggr)\biggr\vert \,dy=\infty. \end{aligned}$$

Like (3.3), decompose the integral $\int_{\tilde{S}_{t}} (\cdots)$ as the sum $\int_{\Omega_{1}} (\cdots) +\int_{\Omega_{2}} (\cdots)$. For the $\Omega_{2}$ term, we have

$$\begin{aligned} &\frac{p}{G(t)}\int_{\Omega_{2}}\frac{1}{(p-\epsilon_{0})^{2}} \biggl( \frac {g(y)}{v(y)} \biggr)^{\epsilon_{0}/(p-\epsilon_{0})}g(y)\biggl\vert \log \biggl( \frac{g(y)}{v(y)} \biggr)\biggr\vert \,dy \\ &\quad\le\frac{4\alpha\log\alpha}{pG(t)}\int_{\Omega_{2}} g(y)\,dy\le \frac{4\alpha\log\alpha}{p}< \infty, \end{aligned}$$

which implies

$$\lim_{\epsilon\to0^{+}}\frac{p}{G(t)}\int_{\tilde{S}_{t}} \frac{1}{(p-\epsilon_{0})^{2}} \biggl(\frac{g(y)}{v(y)} \biggr)^{\epsilon_{0}/(p-\epsilon_{0})}g(y) \biggl( \log\frac{g(y)}{v(y)} \biggr)\,dy=-\infty. $$

From (3.4) and the fact that $\lim_{\epsilon\to 0}(1+\epsilon\theta)^{1/\epsilon}=e^{\theta}$ for any $\theta\in\Bbb{R}$, we get

$$\limsup_{\epsilon\to0^{+}} \biggl(\frac{1}{G(t)}\int _{\tilde{S}_{t}} \biggl(\frac{g(y)}{v(y)} \biggr)^{\epsilon/(p-\epsilon)}g(y)\,dy \biggr)^{1/\epsilon} \le\limsup_{\epsilon\to0^{+}} (1+\epsilon \theta )^{1/\epsilon}=e^{\theta} $$

for any $\theta<0$. Letting $\theta\to-\infty$, we get the desired result. □

Lemma 3.1 may be false for the case that $\sup_{x\in E} g(x)/v(x)=\infty$. A counterexample is given as follows. Consider $n=1$, $t=1$, $g(t)=1$, and $v(x)=\sum_{m=2}^{\infty}e^{-m}\chi_{(\frac{1}{m}-\frac{1}{m^{3}},\frac{1}{m}]}(x)+\chi_{\Bbb{R}\setminus\bigcup_{m\ge2} (\frac{1}{m}-\frac{1}{m^{3}}, \frac{1}{m}]}(x)$. We have

$$\int_{0}^{1} \biggl(\frac{g(y)}{v(y)} \biggr)^{\epsilon/(p-\epsilon)}g(y)\,dy= \int_{0}^{1} v(y)^{\epsilon/(\epsilon-p)}\,dy\ge\sum_{m=2}^{\infty}\frac{1}{m^{3}}e^{\frac{m\epsilon}{p-\epsilon}}=\infty \quad(0< \epsilon< p/2) $$

and

$$\int_{0}^{1} g(y) \biggl(\log\frac{g(y)}{v(y)} \biggr)\,dy=\int_{0}^{1}\log\frac{1}{v(y)}\,dy=\sum _{m=2}^{\infty}\frac{1}{m^{2}}< \infty. $$

From these, we know that (3.1) is false for this example.

Now, we go back to the investigation of the first part of (1.8). Set

$$\tilde{D}_{PS}:=\sup_{x\in E}\frac{1}{G(x)^{\frac{1}{p}}} \biggl( \int_{\tilde{S}_{x}} \biggl\{ \exp \biggl(\frac{1}{G(t)}\int _{\tilde{S}_{t}}g(y) \biggl(\log\frac{g(y)}{v(y)} \biggr)\,dy \biggr) \biggr\} ^{\frac{q}{p}}u(t)\,dt \biggr)^{\frac{1}{q}}, $$

where $G(x)$ is defined by (1.6). The case $g(t)=1$ of $\tilde{D}_{PS}$ reduces to $D^{*}_{PS}$ mentioned in (1.8). We shall establish the following result, which extends the first inequality in (1.8) from $u(x)>0$ and $g(t)=1$ to $u(x)\ge0$ and those $g(t)$ subject to the condition (1.9). This extension gives the Persson-Stepanov-type estimate of the modular-type operator norm of the general geometric mean operator corresponding to $g(t)$. In particular, $g(t)$ can be of the form $g(t)=|{\tilde{S}}_{t}|^{s-1}$. An elementary calculation of this case will lead us to the Levin-Cochran-Lee-type inequality. We leave such a calculation to the readers. Our result partially generalizes the sufficient parts of [4], Theorem 3.1, [2], and [3], Theorem 7.3(a).

Theorem 3.2

Let $0< p\le q<\infty$, $u(x)\ge0$, $v(x)>0$, $g(t)>0$, and $0< G(x)<\infty$, where $G(x)$ is defined by (1.6). If (1.9) is true and $\tilde{D}_{PS}<\infty$, then (1.7) holds for $C\le e^{1/p}\tilde{D}_{PS}$.

Proof

Let $\Phi(s)=e^{s}$, $k(x,t)=g(t)/G(x)$, and $f(t)\longrightarrow\log f(t)$. The proof is the same as to prove that $\|\mathbb{K}\|_{*}\le e^{1/p}\tilde{D}_{PS}$. We first assume that $\sup_{x\in E} \{g(x)/v(x)\}<\infty$. Consider the case that u is bounded on $\tilde{\Omega}_{r}$ and $u(x)=0$ on $E\setminus\tilde{\Omega}_{r}$, where $r\ge1$ and $\tilde{\Omega}_{r}=\{x\in E: 1/r\le\|x\|\le r\}$. By (1.10)-(1.11) and Theorem 2.1, we know that

$$ \|\mathbb{K}\|_{*}\le\liminf_{\epsilon\to0^{+}} \bigl((p/\epsilon)^{*}\tilde{A}_{PS}(p/\epsilon,q/\epsilon) \bigr)^{1/\epsilon}, $$

(3.5)

provided that the term $(\cdots)^{1/\epsilon}$ in (3.5) is finite for all sufficiently small $\epsilon>0$. By an elementary calculation, we obtain $\lim_{\epsilon\to0^{+}} ((p/\epsilon)^{*} )^{1/\epsilon}=\lim_{\epsilon\to0^{+}} (\frac{p}{p-\epsilon } )^{1/\epsilon}=e^{1/p}$. On the other hand, let $0<\epsilon<p$. Then $p/\epsilon>1$ and $q/\epsilon>1$. Moreover, we have $(p/\epsilon)^{*}=p/(p-\epsilon)$, so

$$\biggl(\frac{g(t)}{v(t)} \biggr)^{(p/\epsilon)^{*}}v(t)= \biggl(\frac {g(t)}{v(t)} \biggr)^{p/(p-\epsilon)}v(t)= \biggl(\frac {g(t)}{v(t)} \biggr)^{\epsilon/(p-\epsilon)}g(t). $$

It follows from the definition of $\tilde{A}_{PS}(p/\epsilon,q/\epsilon)$ that

$$ \begin{aligned}[b] \bigl(\tilde{A}_{PS}(p/\epsilon,q/\epsilon) \bigr)^{1/\epsilon} ={}&\sup_{x\in E} \biggl(\int _{\tilde{S}_{x}} \biggl(\frac{g(t)}{v(t)} \biggr)^{\epsilon/(p-\epsilon)}g(t)\,dt \biggr)^{-1/p}\\ &{}\times \biggl(\int_{\tilde{S}_{x}} \biggl\{ \frac{1}{G(t)}\int _{\tilde{S}_{t}} \biggl(\frac{g(y)}{v(y)} \biggr)^{\epsilon/(p-\epsilon)}g(y)\,dy \biggr\} ^{q/\epsilon}u(t)\,dt \biggr)^{1/q}. \end{aligned} $$

(3.6)

We have assumed that $u(x)=0$ on $E\setminus\tilde{\Omega}_{r}$. Moreover, for $t\in\tilde{S}_{x}$, we have

$$\begin{aligned} \frac{1}{G(t)}\int_{\tilde{S}_{t}} \biggl(\frac{g(y)}{v(y)} \biggr)^{\epsilon/(p-\epsilon)}g(y)\,dy&\le \biggl\{ \sup_{y\in\tilde{S}_{x}} \biggl( \frac{g(y)}{v(y)} \biggr) \biggr\} ^{\epsilon/(p-\epsilon)} \biggl\{ \frac{1}{G(t)} \int _{\tilde{S}_{t}} g(y)\,dy \biggr\} \\ &= \biggl\{ \sup_{y\in\tilde{S}_{x}} \biggl(\frac{g(y)}{v(y)} \biggr) \biggr\} ^{\epsilon/(p-\epsilon)}. \end{aligned}$$

These imply

$$\begin{aligned} \bigl(\tilde{A}_{PS}(p/\epsilon,q/\epsilon) \bigr)^{1/\epsilon} \le{}&\biggl(\int_{\tilde{B}_{1/r}} \biggl( \frac{g(t)}{v(t)} \biggr)^{\epsilon/(p-\epsilon)}g(t)\,dt \biggr)^{-1/p} \\ &{}\times \biggl\{ \sup_{y\in E} \biggl(\frac {g(y)}{v(y)} \biggr) \biggr\} ^{1/(p-\epsilon)} \biggl(\int_{\tilde{\Omega}_{r}} u(t)\,dt \biggr)^{1/q}< \infty, \end{aligned}$$

(3.7)

where $\tilde{B}_{\rho}=\{x\in E: \|x\|\le\rho\}$. The above argument guarantees the validity of (3.5). Now, we try to estimate the limit infimum given in (3.5). It suffices to show that

$$ \liminf_{\epsilon\to0^{+}} \bigl(\tilde{A}_{PS}(p/\epsilon,q/ \epsilon) \bigr)^{1/\epsilon}\le\tilde{D}_{PS}. $$

(3.8)

Clearly, the term $(\int_{\tilde{S}_{x}} (\cdots) )^{-1/p}$ in (3.6) becomes bigger whenever x with $\|x\|>r$ is replaced by $rx/\|x\|$. Moreover, the term $(\int_{\tilde{S}_{x}} \{\cdots\} ^{q/\epsilon}u(t)\,dt )^{1/q}$ in (3.6) is zero for $\|x\|<1/r$ and it keeps the same value for the change: x with $\|x\|>r\longrightarrow rx/\|x\|$. Hence, the term ‘$\sup_{x\in E}$’ in (3.6) can be replaced by ‘$\sup_{x\in\tilde{\Omega}_{r}}$’. By the Heine-Borel theorem, we can choose $0<\epsilon_{m}<p/2$, $\alpha_{m}>0$, and $x_{0}, x_{m}\in\tilde{\Omega}_{r}$, such that $\epsilon_{m}\to0$, $\alpha_{m}\to0$, $x_{m}\to x_{0}$, and the following inequality holds for all m:

$$\begin{aligned} &\bigl(\tilde{A}_{PS}(p/\epsilon_{m},q/ \epsilon_{m}) \bigr)^{1/\epsilon_{m}} \\ &\quad\le \biggl(\int _{\tilde{S}_{x_{m}}} \biggl(\frac{g(t)}{v(t)} \biggr)^{\epsilon_{m} /(p-\epsilon _{m})}g(t)\,dt \biggr)^{-1/p} \\ &\qquad{}\times \biggl(\int_{\tilde{S}_{x_{m}}} \biggl\{ \frac{1}{G(t)}\int _{\tilde{S}_{t}} \biggl(\frac{g(y)}{v(y)} \biggr)^{\epsilon_{m}/(p-\epsilon _{m})}g(y)\,dy \biggr\} ^{q/\epsilon_{m}}u(t)\,dt \biggr)^{1/q}+\alpha_{m}. \end{aligned}$$

(3.9)

We have

$$\begin{aligned} \biggl|\chi_{\tilde{S}_{x_{m}}}(t) \biggl(\frac{g(t)}{v(t)} \biggr)^{\epsilon_{m}/(p-\epsilon_{m})}g(t) \biggr| \le \chi_{\tilde{B}_{r}}(t) \biggl\{ \sup_{y\in E} \biggl( \frac{g(y)}{v(y)} \biggr)+1 \biggr\} g(t) \in L^{1}(E,dt) \quad(m=1,2,\ldots). \end{aligned}$$

By the Lebesgue dominated convergence theorem, we infer that

$$\begin{aligned} &\lim_{m\to\infty} \biggl(\int_{\tilde{S}_{x_{m}}} \biggl(\frac{g(t)}{v(t)} \biggr)^{\epsilon_{m}/(p-\epsilon_{m})}g(t)\,dt \biggr)^{-1/p} \\ &\quad= \biggl(\int_{\tilde{S}_{x_{0}}} \lim_{m\to\infty} \biggl\{ \biggl(\frac{g(t)}{v(t)} \biggr)^{\epsilon_{m}/(p-\epsilon_{m})} \biggr\} g(t)\,dt \biggr)^{-1/p}=\bigl(G(x_{0})\bigr)^{-1/p}. \end{aligned}$$

(3.10)

Similarly, the hypotheses on $u(t)$ and $g(t)/v(t)$ imply

$$\begin{aligned} & \biggl|\chi_{\tilde{S}_{x_{m}}}(t) \biggl\{ \frac{1}{G(t)}\int_{\tilde{S}_{t}} \biggl(\frac{g(y)}{v(y)} \biggr)^{\epsilon_{m}/(p-\epsilon_{m})}g(y)\,dy \biggr\} ^{q/\epsilon_{m}}u(t) \biggr| \\ &\quad\le\chi_{\tilde{B}_{r}}(t) \biggl\{ \sup_{y\in E} \biggl( \frac {g(y)}{v(y)} \biggr) \biggr\} ^{q/(p-\epsilon_{m})} \biggl\{ \frac{1}{G(t)}\int _{\tilde{S}_{t}}g(y)\,dy \biggr\} ^{q/\epsilon_{m}}u(t) \\ &\quad\le\chi_{\tilde{B}_{r}}(t) \biggl\{ \sup_{y\in E} \biggl( \frac {g(y)}{v(y)} \biggr)+1 \biggr\} ^{2q/p}u(t)\in L^{1}(E,dt). \end{aligned}$$

Applying the Lebesgue dominated convergence theorem again, it follows from Lemma 3.1 that

$$\begin{aligned} &\lim_{m\to\infty} \biggl(\int_{\tilde{S}_{x_{m}}} \biggl\{ \frac{1}{G(t)}\int_{\tilde{S}_{t}} \biggl(\frac{g(y)}{v(y)} \biggr)^{\epsilon_{m}/(p-\epsilon_{m})}g(y)\,dy \biggr\} ^{q/\epsilon_{m}}u(t)\,dt \biggr)^{1/q} \\ &\quad= \biggl(\int_{\tilde{S}_{x_{0}}} \lim_{m\to\infty} \biggl\{ \frac{1}{G(t)}\int_{\tilde{S}_{t}} \biggl(\frac{g(y)}{v(y)} \biggr)^{\epsilon_{m}/(p-\epsilon_{m})}g(y)\,dy \biggr\} ^{q/\epsilon_{m}}u(t)\,dt \biggr)^{1/q} \\ &\quad= \biggl(\int_{\tilde{S}_{x_{0}}} \biggl\{ \exp \biggl(\frac{1}{G(t)} \int_{\tilde{S}_{t}} g(y) \biggl(\log \frac{g(y)}{v(y)} \biggr)\,dy \biggr) \biggr\} ^{q/p}u(t)\,dt \biggr)^{1/q}. \end{aligned}$$

(3.11)

Putting (3.9)-(3.11) together yields (3.8). This finishes the proof for those u and v with the restrictions stated above. Now, we come back to the proof of the case $u\ge0$ and $\sup_{x\in E} \{ g(x)/v(x)\}<\infty$. Let $u_{r}(x)=\min\{u(x),r\}\chi_{\tilde{\Omega}_{r}}(x)$, where $r=1,2,\ldots$ . By the preceding result,

$$\begin{aligned} & \biggl(\int_{E} \biggl\{ \exp \biggl( \frac{1}{G(x)}\int_{\tilde{S}_{x}} g(t)\log f(t)\,dt \biggr) \biggr\} ^{q}u_{r}(x)\,dx \biggr)^{1/q} \\ &\quad\le e^{1/p}\tilde{D}_{PS}(r) \biggl(\int_{E} \bigl(f(x)\bigr)^{p} v(x)\,dx \biggr)^{1/p}\quad (f>0), \end{aligned}$$

(3.12)

where

$$\tilde{D}_{PS}(r)=\sup_{x\in E}\bigl(G(x) \bigr)^{-\frac{1}{p}} \biggl(\int_{\tilde{S}_{x}} \biggl\{ \exp \biggl( \frac{1}{ G(t)}\int_{\tilde{S}_{t}}g(y) \biggl(\log\frac{g(y)}{v(y)} \biggr)\,dy \biggr) \biggr\} ^{\frac{q}{p}}u_{r}(t)\,dt \biggr)^{\frac{1}{q}}. $$

We have $u_{r}(t)\le u(t)$, so $\tilde{D}_{PS}(r)\le\tilde{D}_{PS}$. Replacing $\tilde{D}_{PS}(r)$ in (3.12) by $\tilde{D}_{PS}$ first and then applying the monotone convergence theorem to (3.12), we get the desired inequality for this case.

Next, we deal with the case $\sup_{x\in E} g(x)<\infty$. Let $v_{\ell}(x)=v(x)+1/\ell$, where $\ell=1,2,\ldots$ . Then $\sup_{x\in E} \{g(x)/v_{\ell}(x)\}<\infty$ for each ℓ. By the preceding result,

$$\begin{aligned} & \biggl(\int_{E} \biggl\{ \exp \biggl( \frac{1}{G(x)}\int_{\tilde{S}_{x}} g(t)\log f(t)\,dt \biggr) \biggr\} ^{q}u(x)\,dx \biggr)^{1/q} \\ &\quad\le e^{1/p}\tilde{D}^{\ell}_{PS} \biggl(\int _{E} \bigl(f(x)\bigr)^{p} v_{\ell}(x)\,dx \biggr)^{1/p}\quad (f>0), \end{aligned}$$

(3.13)

where

$$\tilde{D}^{\ell}_{PS}=\sup_{x\in E} \frac{1}{(G(x))^{\frac{1}{p}}} \biggl(\int_{\tilde{S}_{x}} \biggl\{ \exp \biggl( \frac{1}{ G(t)}\int_{\tilde{S}_{t}}g(y)\log \biggl(\frac {g(y)}{v_{\ell}(y)} \biggr)\,dy \biggr) \biggr\} ^{\frac{q}{p}}u(t)\,dt \biggr)^{\frac{1}{q}}. $$

We have $v_{\ell}(x)\ge v(x)$, so $\tilde{D}^{\ell}_{PS}\le\tilde{D}_{PS}$. This says that (3.13) can be replaced by (3.14):

$$\begin{aligned} & \biggl(\int_{E} \biggl\{ \exp \biggl( \frac{1}{G(x)}\int_{\tilde{S}_{x}} g(t)\log f(t)\,dt \biggr) \biggr\} ^{q}u(x)\,dx \biggr)^{1/q} \\ &\quad\le e^{1/p}\tilde{D}_{PS} \biggl(\int_{E} \bigl(f(x)\bigr)^{p} v_{\ell}(x)\,dx \biggr)^{1/p}\quad (f>0). \end{aligned}$$

(3.14)

We shall claim that $v_{\ell}(x)$ in (3.14) can be replaced by $v(x)$. Without loss of generality, we may assume $\int_{E} (f(x))^{p}v(x)\,dx <\infty$. Set

$$f_{r}(x)=\chi_{\tilde{B}_{r}}(x)\min\bigl(f(x),r\bigr)+ \chi_{E\setminus\tilde{B}_{r}}(x)h(x)\quad (r=1,2,\ldots), $$

where $\tilde{B}_{\rho}$ is defined before and $h:E\mapsto(0,\infty)$ is chosen so that

$$h(x)\le\min\bigl(f(x),1\bigr) \quad\mbox{and} \quad\int_{E} \bigl(h(x)\bigr)^{p}v_{1}(x)\,dx< \infty. $$

Replacing f in (3.14) by $f_{r}$, we get

$$\begin{aligned} & \biggl(\int_{E} \biggl\{ \exp \biggl( \frac{1}{G(x)}\int_{\tilde{S}_{x}}g(t)\log f_{r}(t)\,dt \biggr) \biggr\} ^{q}u(x)\,dx \biggr)^{1/q} \\ &\quad\le e^{1/p}\tilde{D}_{PS} \biggl(\int_{E} \bigl(f_{r}(x)\bigr)^{p}v_{\ell}(x)\,dx \biggr)^{1/p}. \end{aligned}$$

(3.15)

For each r, we have

$$\begin{aligned} \int_{E} \bigl(f_{r}(x)\bigr)^{p}v_{1}(x)\,dx&= \int_{\tilde{B}_{r}} \bigl(\min\bigl(f(x),r\bigr) \bigr)^{p}v_{1}(x)\,dx+ \int_{E\setminus\tilde{B}_{r}} \bigl(h(x)\bigr)^{p}v_{1}(x)\,dx \\ &\le\int_{E} \bigl(f(x)\bigr)^{p}v(x)\,dx+\int _{\tilde{B}_{r}} r^{p}\,dx+\int_{E} \bigl(h(x)\bigr)^{p}v_{1}(x)\,dx< \infty \end{aligned}$$

and $|f_{r}(x)|^{p}v_{\ell}(x)\le(f_{r}(x))^{p}v_{1}(x)$ for $\ell=1, 2,\ldots$ . Applying the Lebesgue dominated convergence theorem to the right hand side of (3.15), we get

$$ \begin{aligned}[b] &\biggl(\int_{E} \biggl\{ \exp \biggl( \frac{1}{G(x)}\int_{\tilde{S}_{x}}g(t)\log f_{r}(t)\,dt \biggr) \biggr\} ^{q}u(x)\,dx \biggr)^{1/q}\\ &\quad\le e^{1/p}\tilde{D}_{PS} \biggl(\int_{E} \bigl(f_{r}(x)\bigr)^{p}v(x)\,dx \biggr)^{1/p}. \end{aligned} $$

(3.16)

By definition, $f_{r}(x)\uparrow f(x)$ as $r\to\infty$. Applying the monotone convergence theorem to both sides of (3.16), the right hand side tends to

$$e^{1/p}\tilde{D}_{PS} \biggl(\int_{E} \bigl(f(x)\bigr)^{p}v(x)\,dx \biggr)^{1/p} \quad(\mbox{as } r\to\infty) $$

and the left hand side has the limit

$$ \biggl(\int_{E} \biggl\{ \exp \biggl(\frac{1}{G(x)}\lim _{r\to\infty} \int_{\tilde{S}_{x}}g(t)\log f_{r}(t)\,dt \biggr) \biggr\} ^{q} u(x)\,dx \biggr)^{1/q}. $$

(3.17)

Let $x\in E$. Since $\int_{\tilde{S}_{x}}g(t)\log f(t)\,dt$ is well defined, the following equality makes sense:

$$\int_{\tilde{S}_{x}}g(t)\log f(t)\,dt=\int_{\tilde{S}_{x}}g(t) \bigl(\log f(t) \bigr)^{+}\,dt-\int_{\tilde{S}_{x}}g(t) \bigl(\log f(t) \bigr)^{-}\,dt, $$

where $\xi^{+} =\max(\xi,0)$ and $\xi^{-}=\min(-\xi,0)$. Consider $r\ge \max(\|x\|,1)$. By the monotone convergence theorem,

$$\begin{aligned} \int_{\tilde{S}_{x}}g(t)\log f_{r}(t)\,dt&=\int _{\tilde{S}_{x}}g(t)\log \bigl\{ \min\bigl(f(t),r\bigr) \bigr\} \,dt \\ &=\int_{\tilde{S}_{x}}g(t)\min \bigl(\bigl(\log f(t)\bigr)^{+},\log r \bigr)\,dt- \int_{\tilde{S}_{x}}g(t) \bigl(\log f(t) \bigr)^{-}\,dt \\ &\longrightarrow\int_{\tilde{S}_{x}}g(t) \bigl(\log f(t) \bigr)^{+}\,dt- \int_{\tilde{S}_{x}}g(t) \bigl(\log f(t) \bigr)^{-}\,dt=\int _{\tilde{S}_{x}}g(t)\log f(t)\,dt. \end{aligned}$$

Inserting this limit in (3.17) yields the desired inequality. This finishes the proof. □

Theorem 3.2 gives a new proof of [3], Theorem 7.3(a). In the following, we shall display another example to show how (1.10) works well for the estimate of Opic-Gurka type. Set

$$\begin{aligned} \tilde{D}_{OG}(s):={}&\sup_{x\in E}\bigl(G(x) \bigr)^{\frac{s-1}{p}} \\ &{}\times \biggl(\int_{E\setminus S_{x}}\bigl(G(t)\bigr)^{\frac {-sq}{p}} \biggl\{ \exp \biggl(\frac{1}{G(t)}\int_{\tilde{S}_{t}}g(y) \biggl(\log \frac{g(y)}{v(y)} \biggr)\,dy \biggr) \biggr\} ^{\frac{q}{p}}u(t)\,dt \biggr)^{\frac{1}{q}}, \end{aligned}$$

where $G(x)$ is defined by (1.6). The number $D^{*}_{OG}(s)$ in (1.8) is just the case $g(t)=1$ of $\tilde{D}_{OG}(s)$. In the following, we shall extend the second inequality in (1.8) from $u(x)>0$ and $g(t)=1$ to $u(x)\ge0$ and those $g(t)$ subject to the condition (1.9). This extension gives the Opic-Gurka-type estimate of the modular-type operator norm of the general geometric mean operator corresponding to $g(t)$. In particular, $g(t)$ can be of the form $g(t)=|{\tilde{S}}_{t}|^{s-1}$, which leads us to the Levin-Cochran-Lee-type inequality. Our result partially generalizes the sufficient parts of [5] and [3], Theorem 7.3(b).

Theorem 3.3

Let $0< p\le q<\infty$, $u(x)\ge0$, $v(x)>0$, $g(t)>0$, and $0< G(x)<\infty$, where $G(x)$ is defined by (1.6). If (1.9) is true and $\tilde{D}_{OG}(s)<\infty$ for some $s>1$, then (1.7) holds for $C\le\inf_{s>1}e^{(s-1)/p}\tilde{D}_{OG}(s)$.

Proof

Let $\Phi(s)=e^{s}$, $k(x,t)=g(t)/G(x)$, and $f(t)\longrightarrow\log f(t)$. The proof is similar to Theorem 3.2. We shall show that $\|\mathbb{K}\|_{*}\le \inf_{s>1}e^{(s-1)/p}\tilde{D}_{OG}(s)$. To observe the proof of Theorem 3.2, we find that it suffices to prove this inequality for the case: u is bounded on $\tilde{\Omega}_{r}$, $u(x)=0$ on $E\setminus\tilde{\Omega}_{r}$, and $\sup_{x\in E}\{g(x)/v(x)\}<\infty$, where $\tilde{\Omega}_{r}$ is defined in the proof of Theorem 3.2. It follows from (1.10)-(1.11) and Theorem 2.2 that

$$\begin{aligned} \|\mathbb{K}\|_{*} &\le\inf_{0< \epsilon< p} \bigl(\tilde{A}_{W}(p/\epsilon,q/\epsilon) \bigr)^{1/\epsilon} \\ &= \inf_{0< \epsilon< p} \biggl\{ \inf_{1< s< p/\epsilon} \biggl( \frac{p-\epsilon}{p-\epsilon s} \biggr)^{1/\epsilon-1/p} \bigl(\tilde{A}_{W}(s,p/ \epsilon,q/\epsilon ) \bigr)^{1/\epsilon} \biggr\} \\ &\le\inf_{s>1} \biggl\{ \liminf_{\epsilon\to 0^{+}} \biggl(\frac{p-\epsilon}{p-\epsilon s} \biggr)^{1/\epsilon-1/p} \bigl(\tilde{A}_{W}(s,p/ \epsilon,q/\epsilon) \bigr)^{1/\epsilon} \biggr\} . \end{aligned}$$

(3.18)

For $s>1$, we have $\lim_{\epsilon\to0^{+}} (\frac{p-\epsilon}{p-\epsilon s} )^{1/\epsilon-1/p}=e^{(s-1)/p}$. We shall prove

$$\liminf_{\epsilon\to0^{+}} \bigl(\tilde{A}_{W}(s,p/\epsilon,q/ \epsilon ) \bigr)^{1/\epsilon}\le \tilde{D}_{OG}(s). $$

If so, the desired inequality follows from (3.18). Let $0<\epsilon<p/s$. We have

$$\begin{aligned} \bigl(\tilde{A}_{W}(s,p/\epsilon,q/\epsilon) \bigr)^{1/\epsilon}={}&\sup_{x\in E} \biggl(\int _{\tilde{S}_{x}} \biggl(\frac{g(t)}{v(t)} \biggr)^{\frac {\epsilon}{p-\epsilon}}g(t)\,dt \biggr)^{\frac{s-1}{p}} \\ &{}\times \biggl(\int_{E\setminus S_{x}} \biggl\{ \int_{\tilde{S}_{t}} \biggl(\frac{g(y)}{v(y)} \biggr)^{\frac{\epsilon}{p-\epsilon}}g(y)\,dy \biggr\} ^{\frac{q(p-\epsilon s)}{\epsilon p}} \frac {u(t)\,dt}{(G(t))^{q/\epsilon}} \biggr)^{1/q}. \end{aligned}$$

(3.19)

The term ‘$(\int_{\tilde{S}_{x}} (\cdots) )^{\frac{s-1}{p}}$’ in (3.19) increases in $\|x\|$. On the other hand, the term ‘$(\int_{E\setminus S_{x}} \{\cdots \}^{q(p-\epsilon s)/(\epsilon p)}\frac{u(t)\,dt}{(G(t))^{q/\epsilon}} )^{1/q}$’ in (3.19) is zero for $\|x\|>r$ and it keeps the same value for the change: x with $\|x\|<1/r\longrightarrow(1/r)x/\|x\|$. These imply that the term ‘$\sup_{x\in E}$’ in (3.19) can be replaced by ‘$\sup_{x\in\tilde{\Omega}_{r}}$’. By the Heine-Borel theorem, we can choose $0<\epsilon_{m}<p/s$, $\alpha_{m}>0$, and $x_{0}, x_{m}\in\tilde{\Omega}_{r}$ such that $\epsilon_{m}\to0$, $\alpha_{m}\to0$, $x_{m}\to x_{0}$, and the following inequality holds for all m:

$$\begin{aligned} &\bigl(\tilde{A}_{W}(s,p/\epsilon_{m},q/ \epsilon_{m}) \bigr)^{1/\epsilon_{m}} \\ &\quad\le \biggl(\int_{\tilde{S}_{x_{m}}} \biggl(\frac{g(t)}{v(t)} \biggr)^{\frac{\epsilon _{m}}{p-\epsilon_{m}}}g(t)\,dt \biggr)^{\frac{s-1}{p}} \\ &\qquad{}\times \biggl(\int_{E\setminus S_{x_{m}}} \biggl\{ \int_{\tilde{S}_{t}} \biggl(\frac{g(y)}{v(y)} \biggr)^{\frac{\epsilon_{m}}{p-\epsilon_{m}}}g(y)\,dy \biggr\} ^{\frac{q(p-\epsilon _{m} s)}{\epsilon_{m} p}} \frac {u(t)\,dt}{(G(t))^{q/\epsilon_{m}}} \biggr)^{1/q}+\alpha_{m}. \end{aligned}$$

(3.20)

For the first integral in (3.20), we have

$$\begin{aligned} &\lim_{m\to\infty} \biggl(\int_{\tilde{S}_{x_{m}}} \biggl(\frac {g(t)}{v(t)} \biggr)^{\frac{\epsilon_{m}}{p-\epsilon_{m}}}g(t)\,dt \biggr)^{\frac{s-1}{p}} \\ &\quad= \biggl(\int_{\tilde{S}_{x_{0}}} \lim_{m\to\infty} \biggl\{ \biggl(\frac{g(t)}{v(t)} \biggr)^{\frac{\epsilon_{m}}{p-\epsilon_{m}}} \biggr\} g(t)\,dt \biggr)^{\frac{s-1}{p}}=\bigl(G(x_{0})\bigr)^{\frac{s-1}{p}}. \end{aligned}$$

(3.21)

As for the second integral, it follows from Lemma 3.1 that

$$\begin{aligned} & \biggl(\int_{\tilde{S}_{t}} \biggl(\frac{g(y)}{v(y)} \biggr)^{\frac{\epsilon_{m}}{p-\epsilon_{m}}}g(y)\,dy \biggr)^{\frac{q(p-\epsilon_{m} s)}{\epsilon_{m} p}}\frac{1}{(G(t))^{q/\epsilon_{m}}} \\ &\quad= \biggl(\int_{\tilde{S}_{t}} \biggl(\frac{g(y)}{v(y)} \biggr)^{\frac{\epsilon_{m}}{p-\epsilon_{m}}}g(y)\,dy \biggr)^{-qs/p} \biggl(\frac{1}{G(t)}\int _{\tilde{S}_{t}} \biggl(\frac{g(y)}{v(y)} \biggr)^{\frac{\epsilon_{m}}{p-\epsilon _{m}}}g(y)\,dy \biggr)^{q/\epsilon_{m}} \\ &\quad\longrightarrow\bigl(G(t)\bigr)^{\frac{-qs}{p}} \biggl\{ \exp \biggl( \frac{1}{G(t)}\int_{\tilde{S}_{t}} g(y) \biggl(\log\frac{g(y)}{v(y)} \biggr)\,dy \biggr) \biggr\} ^{\frac{q}{p}} \quad\mbox{as } m\to\infty. \end{aligned}$$

(3.22)

Moreover, for m large enough,

$$\begin{aligned} & \biggl|\chi_{E\setminus S_{x_{m}}(t)} \biggl(\int_{\tilde{S}_{t}} \biggl( \frac{g(y)}{v(y)} \biggr)^{\frac{\epsilon_{m}}{p-\epsilon_{m}}}g(y)\,dy \biggr)^{\frac{q(p-\epsilon_{m} s)}{\epsilon_{m} p}} \frac{u(t)}{(G(t))^{q/\epsilon_{m}}} \biggr| \\ &\quad\le \biggl\{ \sup_{x\in E} \biggl(\frac{g(y)}{v(y)} \biggr)+1 \biggr\} ^{q/p}\chi_{\tilde{\Omega}_{r}}(t)G(t)^{-qs/p}u(t)\in L^{1}(E, dt). \end{aligned}$$

Integrating the left hand side of (3.22) with respect to $u(t)\,dt$ first and then applying the Lebesgue dominated convergence theorem, we obtain

$$\begin{aligned} &\lim_{m\to\infty} \biggl(\int_{E\setminus S_{x_{m}}} \biggl(\int_{\tilde{S}_{t}} \biggl(\frac{g(y)}{v(y)} \biggr)^{\frac{\epsilon_{m}}{p-\epsilon_{m}}}g(y)\,dy \biggr)^{\frac{q(p-\epsilon_{m} s)}{\epsilon_{m} p}}\frac{u(t)\,dt}{(G(t))^{q/\epsilon_{m}}} \biggr)^{1/q} \\ &\quad= \biggl(\int_{E\setminus S_{x_{0}}}\lim_{m\to\infty} \biggl\{ \frac{1}{(G(t))^{q/\epsilon_{m}}} \biggl(\int_{\tilde{S}_{t}} \biggl(\frac{g(y)}{v(y)} \biggr)^{\frac{\epsilon_{m}}{p-\epsilon_{m}}}g(y)\,dy \biggr)^{\frac{q(p-\epsilon_{m} s)}{\epsilon_{m} p}} \biggr\} u(t)\,dt \biggr)^{1/q} \\ &\quad= \biggl(\int_{E\setminus S_{x_{0}}} \frac{1}{(G(t))^{qs/p}} \biggl\{ \exp \biggl( \frac{1}{G(t)}\int_{\tilde{S}_{t}} g(y) \biggl(\log\frac{g(y)}{v(y)} \biggr)\,dy \biggr) \biggr\} ^{\frac{q}{p}}u(t)\,dt \biggr)^{\frac{1}{q}}. \end{aligned}$$

(3.23)

Putting (3.20), (3.21), and (3.23) together yields the desired inequality. This finishes the proof. □

For other estimates of Hardy-type inequalities, we may use a similar limit process to Theorems 3.2 and 3.3 to get the corresponding Pólya-Knopp inequalities.