This is a follow up question of the question [Prove that $\dfrac{g(x,u\_{n})}{\left\Vert u\_{n}\right\Vert ^{p-1}}\rightarrow g\_{0}$ weakly in $L^{\overline{p}}$](https://mathoverflow.net/questions/202404/prove-that-dfracgx-u-n-left-vert-u-n-right-vert-p-1-rightarrow-g)


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 $p\geq N$
 .


The case $p>N$
 :


$\left\Vert g(x,u)\right\Vert \_{L^{1}}\leq\left\Vert a\right\Vert \_{L^{1}}\left\Vert u^{p-1}\right\Vert \_{L^{\infty}}+\left\Vert d\right\Vert \_{L^{1}}$


The case $p=N$
 :


$\left\Vert g(x,u)\right\Vert \_{L^{1}}\leq\left\Vert a\right\Vert \_{L^{r}}\left\Vert u^{p-1}\right\Vert \_{L^{r'}}+\left\Vert d\right\Vert \_{L^{1}}=\left\Vert a\right\Vert \_{L^{r}}\left\Vert u\right\Vert \_{L^{r'(p-1)}}^{p-1}+\left\Vert d\right\Vert \_{L^{1}}\leq C\left\Vert a\right\Vert \_{L^{r}}\left\Vert u\right\Vert \_{W\_{0}^{1,p}}^{p-1}+\left\Vert d\right\Vert \_{L^{1}}$


How should I do with the case $p>N$
 to obtain an estimate similar with the case $p=N$
 ? What should I do next after that because $\dfrac{g(x,u\_{n})}{\left\Vert u\_{n}\right\Vert ^{p-1}}$
 is bounded in $L^{1}\left(\Omega\right)
 \nRightarrow$
 we can 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 $L^{1}(\Omega)$
 as $k\to\infty$
 , for some $g\_{0}\in L^{1}(\Omega)$
 .


Thanks a lot. I appreciate your help.



Follow up question to: Prove that $\dfrac{g(x,u_{n})}{\left\Vert u_{n}\right\Vert ^{p-1}}\rightarrow g_{0}$ weakly in $L^{\overline{p}}$
Let $\displaystyle w\_n=\frac{g(x,u\_n)}{\|u\_n\|\_{W^{1,p}}^{p-1}}$ and suppose that $\|u\_n\|\_{W^{1,p}}\geq1$ for all $n$.


We treat case $p>N$ first. In this case, $W^{1,p}(\Omega)\subset L^\infty(\Omega)$. So, the sequence $\{w\_n\}$ is bounded in $L^1(\Omega)$. In order to extract a convergent subsequence for the weak-$\*$ topology, it suffices show that this sequence is uniformly integrable, i.e., 
$$
 \lim\_{M\to \infty} \sup\_n \int\_{|w\_n|\geq M} |w\_n|\,dx = 0.
$$
But
$$
 \sup\_n\int\_{|w\_n|\geq M} |w\_n|\,dx \leq \int\_{a+d\,\gtrsim\, M} (a+d)\,dx \to 0 \quad\text{as $M\to\infty$,}
$$
which finishes the argument for $p>N$.


The case $p=N$ is a just a bit more complicated. Now, one has the estimate
$$
 |w\_n| \lesssim a^r + |v\_n|^{r'} + d. 
$$
The sequence $\{|v\_n|^{r'}\}$ is uniformly integrable, since the embedding $W^{1,N}(\Omega)\subset L^{r'}(\Omega)$ is compact. In view of $a^r+d\in L^1(\Omega)$, one then proceeds as in the previous case.


