Throughout this section, we fix any (positive) solution v of (P) (belonging to \(W^{1,p}(\mathbb{R}^{N})\)). Define \(v_{M}:=\max\{v,M\}\) for \(M>0\). Here, we choose \(\overline{p}^{*}\) satisfying \(p^{2}<\overline{p}^{*}\) if \(N\le p\) and set \(\overline{p}^{*}=p^{*}=Np/(N-p)\) if \(N>p\). For \(R^{\prime}>R>0\), we take a smooth function \(\eta_{R,R^{\prime}}\) such that \(0\le\eta_{R,R^{\prime}}\le1\), \(\|\eta_{R,R^{\prime}}^{\prime}\|_{\infty}\le2/(R^{\prime}-R)\), \(\eta_{R,R^{\prime}}(t)=1\) if \(t\le R\), and \(\eta_{R,R^{\prime}}=0\) if \(t\ge R^{\prime}\).
5.1 Boundedness of solutions
Lemma 4
Let
\(x_{0}\in\mathbb{R}^{N}\), \(M>0\), \(R^{\prime}>R>0\), \(\tilde{p}>1\), \(\gamma_{i}>1\), and
\(1/\gamma_{i}+1/\gamma_{i}^{\prime}=1\)
\((i=0, 1)\). Denote
\(\eta(x):=\eta_{R,R^{\prime}}(|x-x_{0}|)\). Assume that
\(\gamma_{i}^{\prime}\le\tilde{p}\)
\((i=0,1)\)
and
\(v\in L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))\)
with
\(\alpha\ge0\). Then:
$$\begin{aligned}& \int_{B(x_{0},R^{\prime})} a_{0}vv_{M}^{\alpha} \eta^{p}\,dx \le\|a_{0}\|_{L^{\gamma_{0}}(B(x_{0},R^{\prime}))} \|v \|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{1+\alpha} B_{R^{\prime}}, \end{aligned}$$
(23)
$$\begin{aligned}& \int_{B(x_{0},R^{\prime})} a_{1} v^{r_{1}+1}v_{M}^{\alpha} \eta^{p}\,dx \le\|a_{1}\|_{L^{\gamma_{1}}(B(x_{0},R^{\prime}))} \|v \|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{r_{1}+1+\alpha} B_{R^{\prime}}, \end{aligned}$$
(24)
where
\(B_{R^{\prime}} :=(1+|B(0,R^{\prime})|)\), and
\(|B(0,R^{\prime})|\)
denotes the Lebesgue measure of the ball
\(B(0,R^{\prime})\). Moreover, if
\(\gamma_{2}>p/(p-r_{2})\)
and
\(\gamma_{3}:=\frac{(p-r_{2})\gamma_{2}}{(p-r_{2})\gamma_{2}-p}\le\tilde{p}\), then
$$\begin{aligned} \int_{B(x_{0},R^{\prime})} a_{2} |\nabla v|^{r_{2}}vv_{M}^{\alpha} \eta^{p}\,dx \le&\frac{1}{4} \int_{B(x_{0},R^{\prime})} |\nabla v|^{p}v_{M}^{\alpha} \eta^{p}\,dx \\ &{}+ 4^{\frac{r_{2}}{p-r_{2}}}\|a_{2}\|_{L^{\gamma_{2}}(B(x_{0},R^{\prime}))}^{\frac {p}{p-r_{2}}} \|v \|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{\frac {p}{p-r_{2}}+\alpha} B_{R^{\prime}}. \end{aligned}$$
(25)
Proof
According to Hölder’s inequality, we easily show our assertions (23) and (24). So, we prove (25) only. By Young’s inequality and recalling that \(r_{2}< p-1\) and \(\eta^{p}\ge\eta ^{p^{2}/r_{2}}\), we have
$$\begin{aligned} \int_{B(x_{0},R^{\prime})} a_{2} |\nabla v|^{r_{2}}vv_{M}^{\alpha} \eta^{p}\,dx \le& \frac{1}{4} \int_{B(x_{0},R^{\prime})} |\nabla v|^{p}v_{M}^{\alpha} \eta^{p}\,dx \\ &{}+4^{\frac{r_{2}}{p-r_{2}}} \int_{B(x_{0},R^{\prime})} a_{2}^{\frac{p}{p-r_{2}}}v^{\frac{p}{p-r_{2}}} v_{M}^{\alpha}\,dx. \end{aligned}$$
Moreover, because of \(\gamma_{2}>p/(p-r_{2})\), \(p>p/(p-r_{2})\), and \(\tilde{p}\ge\gamma_{3}\), applying Hölder’s inequality, we obtain
$$\begin{aligned} \int_{B(x_{0},R^{\prime})} a_{2}^{\frac{p}{p-r_{2}}}v^{\frac{p}{p-r_{2}}} v_{M}^{\alpha}\,dx \le& \|a_{2}\|_{L^{\gamma_{2}}(B(x_{0},R^{\prime}))}^{\frac{p}{p-r_{2}}} \|v\|_{L^{\gamma_{3}(\frac{p}{p-r_{2}}+\alpha)}(B(x_{0},R^{\prime}))} ^{\frac{p}{p-r_{2}}+\alpha} \\ \le& \|a_{2}\|_{L^{\gamma_{2}}(B(x_{0},R^{\prime}))}^{\frac{p}{p-r_{2}}} \|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))} ^{\frac{p}{p-r_{2}}+\alpha}\bigl(1+\bigl|B\bigl(0,R^{\prime}\bigr)\bigr|\bigr). \end{aligned}$$
Hence, (25) follows. □
Lemma 5
Let
\(x_{0}\in\mathbb{R}^{N}\), \(R^{\prime}>R>0\), \(\tilde{p}>1\), \(\gamma_{i}>1\), and
\(1/\gamma_{i}+1/\gamma_{i}^{\prime}=1\)
\((i=0,1)\). Assume that
\(\gamma_{i}^{\prime}\le\tilde{p}\)
\((i=0,1)\), \(\gamma_{2}>p/(p-r_{2})\), and
\(\gamma_{3}:=\frac{(p-r_{2})\gamma_{2}}{(p-r_{2})\gamma_{2}-p}\le\tilde{p}\). If
\(v\in L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))\)
with
\(\alpha\ge 0\), then
$$\begin{aligned} &\|v\|_{L^{\frac{\overline{p}^{*}}{p}(p+\alpha )}(B(x_{0},R))}^{p+\alpha} \le2^{p}(p+\alpha)^{p}C_{*}^{p}B_{R^{\prime}}(C_{R^{\prime}}+D_{R,R^{\prime}}) \max\bigl\{ 1,\|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}\bigr\} ^{p+\alpha} \end{aligned}$$
(26)
with
$$\begin{aligned}& B_{R^{\prime}} :=\bigl(1+\bigl|B\bigl(0,R^{\prime}\bigr)\bigr|\bigr), \\& C_{R^{\prime}} := \bigl\{ \|a_{0}\|_{L^{\gamma_{0}}(B(x_{0},R^{\prime}))}+ \|a_{1}\|_{L^{\gamma _{1}}(B(x_{0},R^{\prime}))} +4^{\frac{r_{2}}{p-r_{2}}}\|a_{2} \|_{L^{\gamma_{2}}(B(x_{0},R^{\prime}))}^{\frac {p}{p-r_{2}}} \bigr\} , \\& D_{R,R^{\prime}} := \biggl\{ \frac{2^{3p-2}p^{p}+2^{p-1}}{(R^{\prime}-R)^{p}} +\frac{\mu2^{3q-2}p^{q}}{(R^{\prime}-R)^{q}} \biggr\} , \end{aligned}$$
where
\(C_{*}\)
is the positive constant from embedding from
\(W^{1,p}(\mathbb{R}^{N})\)
to
\(L^{\overline{p}^{*}}(\mathbb{R}^{N})\).
Proof
Taking \(vv_{M}^{\alpha}\eta^{p}\in W^{1,p}_{0}(B(x_{0},R^{\prime}))\) (for \(M>0\)) as a test function, where \(\eta(x)=\eta_{R,R^{\prime}}(|x-x_{0}|)\), by Lemma 4 and (F1) we obtain
$$\begin{aligned} &\|a_{0}\|_{L^{\gamma_{0}}(B(x_{0},R^{\prime}))} \|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{1+\alpha }B_{R^{\prime}} \\ &\qquad{}+\|a_{1}\|_{L^{\gamma_{1}}(B(x_{0},R^{\prime}))} \|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{r_{1}+1+\alpha }B_{R^{\prime}} \\ &\qquad{} +4^{\frac{r_{2}}{p-r_{2}}}\|a_{2}\|_{L^{\gamma_{2}}(B(x_{0},R^{\prime}))}\|v\| _{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{\frac{p}{p-r_{2}}+\alpha} B_{R^{\prime}} +\frac{1}{4} \int_{B(x_{0},R^{\prime})} |\nabla v|^{p}v_{M}^{\alpha}\eta^{p}\,dx \\ &\quad\ge \int_{B(x_{0},R^{\prime})} |\nabla v|^{p}v_{M}^{\alpha}\eta^{p}\,dx +\lambda_{1} \int_{B(x_{0},R^{\prime})}v_{M}^{p+\alpha}\eta^{p} \,dx -\frac{2p}{R^{\prime}-R} \int_{B(x_{0},R^{\prime})} |\nabla v|^{p-1} v_{M}^{\alpha}v\eta^{p-1}\,dx \\ &\qquad{}+\mu \biggl\{ \int_{B(x_{0},R^{\prime})} |\nabla v|^{q}v_{M}^{\alpha}\eta^{p}\,dx -\frac{2p}{R^{\prime}-R} \int_{B(x_{0},R^{\prime})} |\nabla v|^{q-1} v_{M}^{\alpha}v\eta^{p-1}\,dx \biggr\} , \end{aligned}$$
(27)
where we use \(|\nabla\eta|\le2/(R^{\prime}-R)\). According to Young’s and Hölder’s inequalities, for \(j=p,q\), we see that
$$\begin{aligned} &\frac{2p}{R^{\prime}-R} \int_{B(x_{0},R^{\prime})} |\nabla v|^{j-1} v_{M}^{\alpha}v\eta^{p-1}\,dx \\ &\quad\le \frac{1}{4} \int_{B(x_{0},R^{\prime})} |\nabla v|^{j} v_{M}^{\alpha}\eta^{p}\,dx +\frac{2^{j}p^{j}4^{j-1}}{(R^{\prime}-R)^{j}} \int_{B(x_{0},R^{\prime})}v^{j+\alpha}\eta^{p-j}\,dx \\ &\quad\le \frac{1}{4} \int_{B(x_{0},R^{\prime})} |\nabla v|^{j} v_{M}^{\alpha}\eta^{p}\,dx +\frac{2^{3j-2}p^{j}}{(R^{\prime}-R)^{j}} \|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{j+\alpha} B_{R^{\prime}}. \end{aligned}$$
(28)
Consequently, because of \(\mu\ge0\) and \(p+\alpha>r_{1}+1+\alpha,p/(p-r_{2})+\alpha\), it follows from (27) and (28) that
$$\begin{aligned} &B_{R^{\prime}} \biggl(C_{R^{\prime}} +\frac{2^{3p-2}p^{p}}{(R^{\prime}-R)^{p}} + \frac{\mu2^{3q-2}p^{q}}{(R^{\prime}-R)^{q}} \biggr) \max\bigl\{ 1,\|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}\bigr\} ^{p+\alpha} \\ &\quad\ge \frac{1}{2} \int_{B(x_{0},R^{\prime})} |\nabla v|^{p}v_{M}^{\alpha}\eta^{p}\,dx +\lambda_{1} \int_{B(x_{0},R^{\prime})}v_{M}^{p+\alpha}\eta^{p} \,dx. \end{aligned}$$
(29)
Moreover, by using
$$\begin{aligned} &\bigl\| \nabla\bigl(v_{M}^{1+\alpha/p}\eta\bigr)\bigr\| _{L^{p}(\mathbb{R}^{N})}^{p}\\ &\quad\le2^{p-1} \bigl\{ \bigl\| \eta\nabla\bigl(v_{M}^{1+\alpha/p} \bigr)\bigr\| _{L^{p}(\mathbb{R}^{N})}^{p} +\bigl\| v_{M}^{1+\alpha/p}\nabla \eta\bigr\| _{L^{p}(\mathbb{R}^{N})}^{p} \bigr\} \\ &\quad\le 2^{p-1} \biggl(1+\frac{\alpha}{p} \biggr)^{p} \int_{B(x_{0},R^{\prime})} |\nabla v|^{p}v_{M}^{\alpha}\eta^{p}\,dx +\frac{2^{2p-1}}{(R^{\prime}-R)^{p}} \int_{B(x_{0},R^{\prime})}v_{M}^{p+\alpha}\,dx \end{aligned}$$
and Hölder’s inequality, due to the embedding from \(W^{1,p}(\mathbb{R}^{N})\) to \(L^{\overline{p}^{*}}(\mathbb{R}^{N})\), we have
$$\begin{aligned} &\frac{1}{2} \int_{B(x_{0},R^{\prime})} |\nabla v|^{p}v_{M}^{\alpha}\eta^{p}\,dx +\lambda_{1} \int_{B(x_{0},R^{\prime})}v_{M}^{p+\alpha}\eta^{p} \,dx \\ &\quad\ge2^{-p}p^{p}(p+\alpha)^{-p} \bigl\{ \bigl\| \nabla\bigl(v_{M}^{1+\alpha/p}\eta\bigr)\bigr\| _{L^{p}(\mathbb{R}^{N})}^{p} +\lambda_{1}\bigl\| v_{M}^{1+\alpha/p}\eta\bigr\| _{L^{p}(\mathbb{R}^{N})}^{p} \bigr\} \\ &\qquad{} -\frac{2^{p-1}p^{p}}{(p+\alpha)^{p}(R^{\prime}-R)^{p}} \int_{B(x_{0},R^{\prime})}v_{M}^{p+\alpha}\,dx \\ &\quad\ge2^{-p}p^{p}(p+\alpha)^{-p} \bigl\| v_{M}^{1+\alpha/p}\eta\bigr\| _{W^{1,p}(\mathbb{R}^{N})}^{p} \\ &\qquad{} -\frac{2^{p-1}}{(R^{\prime}-R)^{p}} \|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{p+\alpha }\bigl(1+\bigl|B \bigl(0,R^{\prime}\bigr)\bigr|\bigr) \\ &\quad\ge 2^{-p}p^{p}(p+\alpha)^{-p}C_{*}^{-p} \bigl\| v_{M}^{1+\alpha/p}\eta\bigr\| _{L^{\overline{p}^{*}}(\mathbb{R}^{N})}^{p} - \frac{2^{p-1}}{(R^{\prime}-R)^{p}} \|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{p+\alpha }B_{R^{\prime}} \\ &\quad\ge2^{-p}p^{p}(p+\alpha)^{-p}C_{*}^{-p} \|v_{M}\|_{L^{\overline{p}^{*}(p+\alpha)/p}(B(x_{0},R))}^{p+\alpha} \\ &\qquad{} -\frac{2^{p-1}}{(R^{\prime}-R)^{p}} \|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{p+\alpha} B_{R^{\prime}}. \end{aligned}$$
(30)
Therefore, (29) and (30) lead to
$$\begin{aligned} &2^{-p}p^{p}(p+\alpha)^{-p}C_{*}^{-p} \|v_{M}\|_{L^{\overline{p}^{*}(p+\alpha)/p}(B(x_{0},R))}^{p+\alpha} \\ &\quad\le B_{R^{\prime}}(C_{R^{\prime}}+D_{R,R^{\prime}}) \max\bigl\{ 1,\|v \|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}\bigr\} ^{p+\alpha}. \end{aligned}$$
(31)
Applying Fatou’s lemma and letting \(M\to\infty\) in (31), our conclusion follows. □
Proposition 1
Under the assumptions in Theorem
2, we have that
\(v\in L^{\infty}(\mathbb{R}^{N})\).
Proof
First, in the case of \(N>p\), we note that
$$\begin{aligned}& \gamma_{j}^{\prime}< \frac{p^{*}}{p} \quad\Longleftrightarrow \quad \gamma_{j}>\frac{p^{*}}{p^{*}-p}\quad (j=0,1), \\& \gamma_{2}>\frac{p}{p-r_{2}}\quad \mbox{and} \quad \gamma_{3}:=\frac{(p-r_{2})\gamma_{2}}{(p-r_{2})\gamma_{2}-p}< \frac{p^{*}}{p} \quad \Longleftrightarrow\quad \gamma_{2}>\frac{pp^{*}}{(p-r_{2})(p^{*}-p)}. \end{aligned}$$
In the cases of (i) and (ii) (case \(p<\overline{p}^{*}/p\)), we take \(\gamma_{0}=p^{\prime}\) and \(\gamma_{j}=\tilde{r}_{j}\) (\(j=1,2\)). Then, we have \(\gamma_{0}^{\prime}=p\), \(\gamma_{1}^{\prime}=\tilde{r}_{1}^{\prime}=p/(r_{1}+1)\le p\), \(\gamma_{2}=\tilde{r}_{2}=p/(p-r_{2}-1)>p/(p-r_{2})\), and \(\gamma_{3}=(p-r_{2})\tilde{r_{2}}/((p-r_{2})\tilde{r}_{2}-p)=p-r_{2}\le p\). Choose p̃ such that
$$\begin{aligned}& \bigl(\max\bigl\{ \gamma_{0}^{\prime},\gamma_{1}^{\prime}, \gamma_{3}\bigr\} = \bigr) \tilde{p}=p \biggl(< \frac{\overline{p}^{*}}{p}\biggr) \mbox{ in the cases of (i) and (ii)}, \\& \max\bigl\{ \gamma_{0}^{\prime},\gamma_{1}^{\prime}, \gamma_{3}\bigr\} \le \tilde{p}< \frac{p^{*}}{p} \mbox{ in the case of (iii)}. \end{aligned}$$
Let \(R_{*}\) be the positive constant satisfying (4) in the case of (iii) and any positive constant in the cases of (i) and (ii). Put
$$ A_{i}:=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \|a_{i}\|_{L^{\gamma_{i}}(\mathbb{R}^{N})} & \mbox{if (i) and (ii)}, \\ \sup_{x\in\mathbb{R}^{N}} \|a_{i}\|_{L^{\gamma_{i}}(B(x,2R_{*}))} &\mbox{if (iii)} \end{array}\displaystyle \right . $$
for \(i=0,1,2\). Define the sequences \(\{\alpha_{n}\}\), \(\{R_{n}^{\prime}\}\), and \(\{R_{n}\}\) by
$$\begin{aligned}& \alpha_{0}:=\frac{\overline{p}^{*}}{\tilde{p}}-p>0,\qquad \tilde{p}(p+ \alpha_{n+1})=\frac{\overline{p}^{*}}{p} (p+\alpha_{n}), \\& R_{n}^{\prime}:=\bigl(1+2^{-n}\bigr)R_{*}, \qquad R_{n}:=R_{n+1}^{\prime}. \end{aligned}$$
Recall that \(v\in W^{1,p}(\mathbb{R}^{N})\), and using the embedding of \(W^{1,p}(\mathbb{R}^{N})\) to \(L^{\overline{p}^{*}}(\mathbb{R}^{N})\), we see that \(v\in L^{\overline{p}^{*}}(\mathbb{R}^{N})=L^{\tilde{p}(p+\alpha _{0})}(\mathbb{R}^{N})\).
Fix any \(x_{0}\in\mathbb{R}^{N}\). Then Lemma 5 guarantees that if \(v\in L^{\tilde{p}(p+\alpha_{n})}(B(x_{0},R_{n}^{\prime}))\), then \(v\in L^{\frac{\overline{p}^{*}}{p}(p+\alpha_{n})}(B(x_{0},R_{n})) =L^{\tilde{p}(p+\alpha_{n+1})}(B(x_{0},R_{n+1}^{\prime}))\). Noting that
$$\begin{aligned}& B_{R^{\prime}_{n}} \le\bigl(1+\bigl|B(0,2R_{*})\bigr|\bigr)=:B_{0}, \\& C_{R_{n}^{\prime}} \le A_{0}+A_{1}+4^{\frac{r_{2}}{p-r_{2}}}A_{2}^{\frac {p}{p-r_{2}}}+1=:C_{0}, \\& D_{R_{n},R_{n}^{\prime}} \le\frac{(1+p^{p})2^{p(n+3)}}{R_{*}^{p}} +\frac{\mu q^{q}2^{q(n+3)}}{R_{*}^{q}}=:D_{n} \le C^{\prime}2^{p(n+3)} \end{aligned}$$
for any \(n\ge0\) with sufficiently large \(C^{\prime}\) independent of n and setting
$$b_{n}:=\max\bigl\{ 1,\|v\|_{L^{\tilde{p}(p+\alpha_{n})(B(x_{0},R_{n}^{\prime}))}}\bigr\} , $$
by Lemma 5 we obtain
$$ b_{n+1}\le C^{\frac{1}{p+\alpha_{n}}}(p+\alpha_{n})^{\frac{p}{p+\alpha_{n}}} (C_{0}+D_{n})^{\frac{1}{p+\alpha_{n}}}b_{n} $$
(32)
for every \(n\ge0\) with \(C:=2^{p}(C_{*}+1)^{p}B_{0}\). Put \(P:=\tilde{p}p/\overline{p}^{*}<1\). Then, because of \(p+\alpha_{n+1}=(p+\alpha_{n})/P\), \(\alpha_{n+1}>\alpha_{n}/P>\alpha_{0}(1/P)^{n+1}\to\infty\) as \(n\to \infty\). Moreover, we see that
$$\begin{aligned} &S_{1} :=\sum_{n=0}^{\infty}\frac{1}{p+\alpha_{n}}= \frac{1}{p+\alpha_{0}}\sum_{n=0}^{\infty}P^{n} =\frac{1}{(p+\alpha_{0})(1-P)}< \infty, \\ &S_{2} :=\ln \prod_{n=0}^{\infty}(p+ \alpha_{n})^{\frac{p}{p+\alpha_{n}}} =\frac{p}{p+\alpha_{0}}\sum _{n=0}^{\infty}P^{n} \bigl(\ln (p+ \alpha_{0})+n\ln P^{-1} \bigr)< \infty, \end{aligned}$$
and
$$\begin{aligned} S_{3}&:=\ln \prod_{n=0}^{\infty}(C_{0}+D_{n})^{\frac{1}{p+\alpha_{n}}} =\sum_{n=0}^{\infty}\frac{P^{n}}{p+\alpha_{0}} \ln (C_{0}+D_{n}) \\ &\le\sum_{n=0}^{\infty}\frac{P^{n}}{p+\alpha_{0}} p(n+3)\ln \bigl(C_{0}+C^{\prime}\bigr)2< \infty. \end{aligned}$$
As a result, by iteration in (32) and the equality \(\tilde{p}(p+\alpha_{0})=\overline{p}^{*}\) we obtain
$$\|v\|_{L^{\frac{\overline{p}^{*}}{p}(p+\alpha_{n})}(B(x_{0},R_{*}))} \le b_{n} \le C^{S_{1}}e^{S_{2}}e^{S_{3}} \max\bigl\{ 1,\|v\|_{L^{\overline{p}^{*}}(B(x_{0},2R_{*}))}\bigr\} $$
for every \(n\ge1\). Letting \(n\to\infty\), this ensures that
$$ \|v\|_{L^{\infty}(B(x_{0},R_{*}))} \le C^{S_{1}}e^{S_{2}}e^{S_{3}} \max\bigl\{ 1,\|v\|_{L^{\overline{p}^{*}}(B(x_{0},2R_{*}))}\bigr\} . $$
(33)
Recalling that \(v\in W^{1,p}(\mathbb{R}^{N})\) and using the embedding of \(W^{1,p}(\mathbb{R}^{N})\) to \(L^{\overline{p}^{*}}(\mathbb{R}^{N})\), (33) yields that
$$\begin{aligned} \|v\|_{L^{\infty}(B(x_{0},R_{*}))} &\le C^{S_{1}}e^{S_{2}}e^{S_{3}} \max\bigl\{ 1,\|v\|_{L^{\overline{p}^{*}}(\mathbb{R}^{N})}\bigr\} \\ &\le C^{S_{1}}e^{S_{2}}e^{S_{3}} \max\bigl\{ 1,C_{*}\|v \|_{W^{1,p}(\mathbb{R}^{N})}\bigr\} , \end{aligned}$$
whence v is bounded in \(\mathbb{R}^{N}\) because \(x_{0}\in\mathbb{R}^{N}\) is arbitrary and the constant \(C^{S_{1}}e^{pS_{2}}e^{S_{3}}\) is independent of \(x_{0}\). □
5.2 Proof of Theorem 2
Proof of Theorem 2
Since v is bounded in \(\mathbb{R}^{N}\) by Proposition 1, we put \(M_{0}:=\|v\|_{L^{\infty}(\mathbb{R}^{N})}\). Then, as in Lemma 4, we see that
$$\begin{aligned} & \int_{B(x_{0},R^{\prime})} a_{0}vv_{M}^{\alpha} \eta^{p}\,dx \le\|a_{0}\|_{L^{\gamma_{0}}(B(x_{0},R^{\prime}))} M_{0} \|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{\alpha} B_{R^{\prime}}, \end{aligned}$$
(34)
$$\begin{aligned} & \int_{B(x_{0},R^{\prime})} a_{1} v^{r_{1}+1}v_{M}^{\alpha} \eta^{p}\,dx \le\|a_{1}\|_{L^{\gamma_{1}}(B(x_{0},R^{\prime}))} M_{0}^{1+r_{1}}\|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{\alpha} B_{R^{\prime}}, \end{aligned}$$
(35)
$$\begin{aligned} &\begin{aligned}[b] & \int_{B(x_{0},R^{\prime})} a_{2} |\nabla v|^{r_{2}}vv_{M}^{\alpha} \eta^{p}\,dx \\ &\quad \le\frac{1}{4} \int_{B(x_{0},R^{\prime})} |\nabla v|^{p}v_{M}^{\alpha} \eta^{p}\,dx + 4^{\frac{r_{2}}{p-r_{2}}}\|a_{2}\|_{L^{\gamma_{2}}(B(x_{0},R^{\prime}))}^{\frac {p}{p-r_{2}}} M_{0}^{\frac{p}{p-r_{2}}}\|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{\alpha} B_{R^{\prime}}, \end{aligned} \end{aligned}$$
(36)
and
$$ \|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{j+\alpha}\le M_{0}^{j}\|v \|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{\alpha}\quad (j=p,q) . $$
(37)
Fix any \(x_{0}\in\mathbb{R}^{N}\). It follows from the argument as in the proof of Lemma 5 with (34), (35), (36), and (37) that
$$\begin{aligned} &\|v\|_{L^{\frac{\overline{p}^{*}}{p}(p+\alpha )}(B(x_{0},R))}^{p+\alpha} \le2^{p}(p+\alpha)^{p}C_{*}^{p}B_{R^{\prime}}(C_{R^{\prime}}+D_{R,R^{\prime}}) (M_{0}+1)^{p} \|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{\alpha}, \end{aligned}$$
(38)
provided that \(v\in L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))\). Choose \(\gamma_{i}\)
\((i=0,1,2)\) and p̃ and define the sequences \(\{\alpha_{n}\}\), \(\{R_{n}^{\prime}\}\), and \(\{R_{n}\}\) as in the proof of Proposition 1. Set
$$V_{n}:=\|v\|_{L^{\tilde{p}(p+\alpha_{n})}(B(x_{0},R^{\prime}_{n}))}^{\alpha_{n}}. $$
Then, by the same argument as in the proof of Proposition 1 with (38) we obtain
$$ V_{n}^{\frac{p+\alpha_{n-1}}{\alpha_{n}}} \le C (p+\alpha_{n-1})^{p}(C_{0}+D_{n-1}) V_{n-1} $$
(39)
with \(C:=2^{p}C_{*}^{p}B_{0}(M_{0}+1)^{p}\). Recall that
$$\alpha_{n}+p=P^{-1}(p+\alpha_{n-1}) \quad \mbox{and}\quad \frac{p}{p+\alpha_{0}}=P. $$
Define
$$Q_{n}:=\prod_{k=2}^{n+1} \biggl(1+\frac{P^{k}}{1-P^{k}} \biggr) = \prod_{k=2}^{n+1} \bigl(1-P^{k} \bigr)^{-1} \quad \mbox{and}\quad W_{n}:=(C_{0}+D_{n}). $$
Then, inequality (39) leads to
$$\begin{aligned} \ln V_{n} \le{}&\frac{\alpha_{n}}{p+\alpha_{n-1}} \bigl(\ln V_{n-1}+\ln C(p+\alpha_{n-1})^{p}+\ln W_{n-1} \bigr) \\ ={}&P^{-1} \bigl(1-P^{n+1} \bigr) \bigl(\ln V_{n-1}+ p\ln C P^{-n+1}(p+\alpha_{0})+\ln W_{n-1} \bigr) \\ \le{}& P^{-1} \bigl(1-P^{n+1} \bigr)\ln V_{n-1} +pP^{-1}\ln (C+1) P^{-n+1}(p+\alpha_{0})+P^{-1} \ln W_{n-1} \\ \le{}& P^{-n} \Biggl( \prod _{k=1}^{n} \bigl(1-P^{k+1} \bigr) \Biggr) \ln V_{0} +p\sum_{k=1}^{n} P^{-k}\ln (C+1) P^{-n+k}(p+\alpha_{0}) \\ &{} +\sum_{k=1}^{n} P^{-k}\ln W_{n-k} \\ ={}& P^{-n}Q_{n}^{-1} \ln V_{0} +p \sum_{k=1}^{n} P^{-k}\ln (C+1) P^{-n+k}(p+\alpha_{0}) +\sum_{k=1}^{n} P^{-k}\ln W_{n-k} \end{aligned}$$
for every n because of \(\ln (C+1) P^{-n+1}(p+\alpha_{0})> 0\) and \(\ln W_{n}>0\) for all n. Therefore, we have
$$\begin{aligned} &\ln \|v\|_{L^{\tilde{p}(p+\alpha_{n})}(B(x_{0},R^{\prime}_{n}))} \\ &\quad=\frac{\ln V_{n}}{\alpha_{n}}=\frac{P^{n}\ln V_{n}}{p+\alpha_{0}-pP^{n}} \\ &\quad\le\frac{Q_{n}^{-1}\ln V_{0}}{p+\alpha_{0}-pP^{n}} +\frac{\sum_{l=0}^{n-1} P^{l} \ln (C+1) P^{-l}(p+\alpha _{0})}{p+\alpha_{0}-pP^{n}} +\frac{\sum_{l=0}^{n-1} P^{l}\ln W_{l}}{p+\alpha_{0}-pP^{n}}. \end{aligned}$$
(40)
Here, taking a sufficiently large positive constant \(C^{\prime}\) independent of n, we see that
$$\sum_{l=0}^{n-1} P^{l} \ln (C+1) P^{-l}(p+\alpha_{0}) \le C^{\prime}\sum _{l=0}^{\infty}P^{l} (l+1)=:S_{1}< \infty $$
and
$$\sum_{l=0}^{n-1} P^{l}\ln W_{l}\le C^{\prime}\sum_{l=0}^{n-1} P^{l} (l+3)\le C^{\prime}\sum_{l=0}^{\infty}P^{l} (l+3) =:S_{2}< \infty. $$
Next, we shall show that \(\{Q_{n}\}\) is a convergent sequence. It is easy see that \(\{Q_{n}\}\) is increasing. Moreover, setting \(d_{k}:=\ln (1+\frac{P^{k}}{1-P^{k}} )\), we see that
$$\lim_{k\to\infty}\frac{d_{k+1}}{d_{k}} = \lim_{k\to\infty} \frac{\ln (1-P^{k+1})}{\ln (1-P^{k})} =\lim_{k\to\infty}\frac{1-P^{k}}{1-P^{k+1}} P=P< 1 $$
by L’Hospital’s rule. This implies that
$$\ln Q_{n}=\sum_{k=2}^{n+1} \ln \biggl(1+\frac {P^{k}}{1-P^{k}} \biggr) \le\sum_{k=1}^{\infty}\ln \biggl(1+\frac{P^{k}}{1-P^{k}} \biggr) < \infty. $$
Therefore, \(\{Q_{n}\}\) is bounded from above, whence \(\{Q_{n}\}\) converges, and
$$1< \frac{1}{1-P^{2}}=Q_{1}\le Q_{\infty}:=\lim _{n\to\infty}Q_{n}< \infty. $$
Consequently, letting \(n\to\infty\) in (40), we have
$$\|v\|_{L^{\infty}(B(x_{0},R_{*}))} \le(pS_{1}S_{2})^{\frac{1}{p+\alpha_{0}}} \|v \|_{L^{\overline{p}^{*}}(B(x_{0},2R_{*}))}^{\frac{\alpha_{0}}{(p+\alpha _{0})Q_{\infty}}}. $$
This yields our conclusion since \(\|v\|_{L^{\overline{p}^{*}}(B(x_{0},2R_{*}))}\to0\) as \(|x_{0}|\to \infty\), \(\alpha_{0}>0\), and the constant \(pS_{1}S_{2}\) is independent of \(x_{0}\). □