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. □