# Follow up question to: Prove that $\dfrac{g(x,u_{n})}{\left\Vert u_{n}\right\Vert ^{p-1}}\rightarrow g_{0}$ weakly in $L^{\overline{p}}$

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 $L^{\overline{p}}$

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.

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.
• What does the notation $\lesssim$ mean? – user109584 Apr 12 '15 at 0:47