I would just like to know if my working here is sound.
I want to show that given a sequence of function $(u_{n})\in L^{p}(\Omega)$ (with $|\Omega|<\infty$) such that $u_{n}(x)\to u(x)$ for every $x\in\Omega$ and there exists some $M>0$ such that $\|u_{n}\|_{p}\leq M$ for all $n$ then $u_{n}\rightharpoonup u$ in $L^{p'}(\Omega)$.
Working
Choose any $\varphi\in L^{p^{*}}(\Omega)$. Then we can write, \begin{align*} |\varphi(u_{n})-\varphi(u)|=|\varphi(u_{n}-u)|=\bigg|\int_{\Omega}v(u_{n}-u)dx\bigg|\leq\int_{\Omega}|v(u_{n}-u)|dx,\quad v\in L^{p'}(\Omega). \end{align*} By definition there exists some $K>0$ such that $\|v\|_{p'}=\|\varphi\|_{p^{*}}\leq K$. So $\|v\|_{\infty}<\infty$. Then, \begin{align} \int_{\Omega}|v(u_{n}-u)|dx\leq\|v\|_{\infty}\int_{\Omega}|u_{n}-u|dx. (*) \end{align} We have that $|u_{n}|\leq M$ for every $n$ since $\|u_{n}\|_{1}\leq\|u_{n}\|_{p}\leq M$ , $p>1$. Moreover, $M\in L^{1}(\Omega)$ and hence by the Dominated Convergence Theorem, \begin{align} \lim_{n\to\infty}|\varphi(u_{n})-\varphi(u)|\leq\|v\|_{\infty}\lim_{n\to\infty}\int_{\Omega}|u_{n}-u|dx=0. \end{align} Hence, since the choice of $\varphi$ is arbitrary, $\varphi(u_{n})\to\varphi(u)$ for all $\varphi\in L^{p^{*}}(\Omega)$. That is, $u_{n}\rightharpoonup u$ in $L^{p'}(\Omega)$.
If there are any errors in my working can you please leave a comment
Extra Question
In the line indicated by $(*)$ I was wondering why you cannot apply Holder's and conclude strong convergence. Clearly that would be way too easy and not true but I wanted to understand why this isn't possible. My reasoning was that we do not know that $u_{n}-u$ is in $L^{p}(\Omega)$ and so the expression, \begin{align} \int_{\Omega}|u_{n}-u|^{p}dx, \end{align} wouldn't make sense. However, this also got me thinking about what I have done in $(*)$. I don't really know whether or not $u_{n}-u$ is in $L^{1}(\Omega)$, so does it make sense to write, \begin{align} \int_{\Omega}|u_{n}-u|dx. \end{align} I have gotten myself confused with this so could someone please clarify this?
Lastly, the above assumes that $|\Omega|<\infty$. However, what if $|\Omega|=\infty$? Is it still possible to prove the desired results given the other assumptions? If so, could you please leave an indication of how one would attempt this proof?
Thank you.