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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.

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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 $g_0\in L^{\bar p}(\Omega)$.

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