Let $\Omega
 \subset
 \mathbb{R}^{N}$
 be a smooth bounded domain , $g:\Omega\times\mathbb{R}\rightarrow\mathbb{R}$
 is a Caratheodory function such that $g(x,t)=0$
 for $t\leq0$
 . Suppose that there exist function $a\in L^{r}$
 and $d\in L^{p'}$
 such that


$\left|g(x,t)\right|\leq a(x)t^{p-1}+d(x)$


with $r>N/p$
 if $1<p\leq N$
 and $r=1$
 if $p>N$
 ; $p'$
 is Holder conjugate of $p$


Let $\left\{ u\_{n}\right\} \subset W\_{0}^{1,p}$
 be a sequence such that $\left\Vert u\_{n}\right\Vert \rightarrow\infty$
 as $n\rightarrow\infty$
 . Let us define $v\_{n}=u\_{n}/\left\Vert u\_{n}\right\Vert $
 . Hence $\left\Vert v\_{n}\right\Vert =1$
 and we may assume that $v\_{n}\rightarrow v$
 weakly in $W\_{0}^{1,p}$
 . Prove that $\dfrac{g(x,u\_{n})}{\left\Vert u\_{n}\right\Vert ^{p-1}}\rightarrow g\_{0}$
 weakly in $L^{\overline{p}}$
 for some $\overline{p}>p\*'$
 if $p<N$
 and $\overline{p}=1$
 if $p\geq N$
 .


Here are my efforts: 


Firstly consider $p<N$
 , my intension is: prove that $\dfrac{g(x,u\_{n})}{\left\Vert u\_{n}\right\Vert ^{p-1}}$
 is bounded in $L^{\overline{p}}$
 for some $\overline{p}>p\*'$


$\dfrac{g(x,u\_{n})}{\left\Vert u\_{n}\right\Vert ^{p-1}}\leq a(x)\left|v\_{n}\right|^{p-1}+\dfrac{d(x)}{\left\Vert u\_{n}\right\Vert ^{p-1}}$


${\displaystyle \int\_{\Omega}}\left|a(x)\left|v\_{n}\right|^{p-1}\right|^{\delta}dx\leq{\displaystyle \int\_{\Omega}}\left|a(x)\right|^{\delta}\left|v\_{n}\right|^{(p-1)\delta}dx$


$\leq\left\Vert a(x)^{\delta}\right\Vert \_{L^{\frac{N}{p\delta}}}\left\Vert \left|v\_{n}\right|^{(p-1)\delta}\right\Vert \_{L^{\frac{N}{N-p\delta}}}$


$\leq\left\Vert a(x)^{\delta}\right\Vert \_{L^{\frac{N}{p\delta}}}\left\Vert v\_{n}\right\Vert \_{L^{\frac{N(p-1)\delta}{N-p\delta}}}^{(p-1)\delta}$


I expect that by Sobolev embedding, $\left\Vert a(x)^{\delta}\right\Vert \_{L^{\frac{N}{p\delta}}}\left\Vert v\_{n}\right\Vert \_{L^{\frac{N(p-1)\delta}{N-p\delta}}}^{(p-1)\delta}\leq C\left\Vert a(x)^{\delta}\right\Vert \_{L^{\frac{N}{p\delta}}}\left\Vert v\_{n}\right\Vert ^{(p-1)\delta}=C\left\Vert a(x)^{\delta}\right\Vert \_{L^{\frac{N}{p\delta}}}$
 , so we are done. 


Thus we have to pick $\delta$
 such that ${\displaystyle \frac{N(p-1)\delta}{N-p\delta}}<\dfrac{Np}{N-p}
 \Longleftrightarrow\delta<\dfrac{pN}{Np-N+p}=p\*'$


My trouble is when choose $\overline{p}=\delta$
 , we are done that $\dfrac{g(x,u\_{n})}{\left\Vert u\_{n}\right\Vert ^{p-1}}$
 is bounded in $L^{\overline{p}}$
 with $\overline{p}<p\*'$
 (not $\overline{p}>p\*'$
 ). On the other hand if choose $\overline{p}'=\delta$
 , we are done that $\overline{p}>p\*'$
 but $\dfrac{g(x,u\_{n})}{\left\Vert u\_{n}\right\Vert ^{p-1}}$
 is bounded in $L^{\overline{p}'}$
 , not $L^{\overline{p}}$
 .


Please help me to take out that trouble. I appreciate your help.



Prove that $\dfrac{g(x,u_{n})}{\left\Vert u_{n}\right\Vert ^{p-1}}\rightarrow g_{0}$ weakly in $L^{\overline{p}}$
First notice that the assumption $g(x,t)=0$ for $t\leq0$ is not essential, you can work with the estimate $|g(x,t)|\leq a(x)\,|t|^{p-1}+d(x)$ instead.


Then I take it that you have trouble with the case $1<p<N$. In this
case, you pick $(\bar p,s)$ such that $(p^\*)'<\bar{p}\leq p'$, $1\leq
s\leq p^\*$, and $\displaystyle \frac1{\bar p}=\frac1r+\frac{p-1}s$
(where $\displaystyle p^\*=\frac{N p}{N-p}$). It follows that
$$
 \|g(x,u)\|\_{L^{\bar p}}\leq \|a\|\_{L^r}\|u\|\_{L^s}^{p-1} + \|d\|\_{L^{\bar p}}
 \lesssim 1 + \|u\|\_{W^{1,p}}^{p-1}
$$
for $u\in W^{1,p}(\Omega)$.


Using this estimate, you find a *subsequence* $\{u\_{n\_k}\}$ of $\{u\_n\}$ such that $\displaystyle \frac{g(x,u\_{n\_k})}{\|u\_{n\_k}\|\_{W^{1,p}}^{p-1}} \to g\_0$ weakly in $L^{\bar p}(\Omega)$ as $k\to\infty$, for some $g\_0\in L^{\bar p}(\Omega)$.


