I would like some feedback on my work if possible. I was asked to work on the following question:
Suppose $(u_{n})\in X$ such that $u_{n}\to v$ pointwise a.e and $u_{n}\rightharpoonup u$ in $X$. When do we have that $u=v$ a.e?
Step 1
I began by assuming $X=L^{p}(\Omega)$, $\Omega$ bounded and $1<p<\infty$. Now assume $(u_{n})\in L^{p}(\Omega)$ such that $u_{n}\to v$ pointwise a.e and $u_{n}\rightharpoonup u$ in $L^{p}(\Omega)$. Then we can consider $(u_{n})$ as elements in $L^{p''}(\Omega)$ and so, \begin{align} \sup_{n\in\mathbb{N}}|g(u_{n})|<\infty\quad\forall g\in L^{p'}(\Omega), \end{align} since, \begin{align} \lim_{n\to\infty}\int_{\Omega}gu_{n}dx<\infty\quad\forall g\in L^{p'}(\Omega). \end{align} Then by the Uniform Boundedness Principle, \begin{align} \sup_{n\in\mathbb{N}}\|u_{n}\|_{p}=\sup_{n\in\mathbb{N}}\|u_{n}\|_{p''}<\infty. \end{align} So we have that $(u_{n})\in L^{p}(\Omega)$, $\sup_{n\in\mathbb{N}}\|u_{n}\|<\infty$ and $u_{n}\to v$ pointwise a.e, then by the Dominated Convergence Theorem we have, \begin{align} |\varphi(u_{n}-v)|\leq\int_{\Omega}|g(u_{n}-u)|dx\leq\|g\|_{p'}\|u_{n}-v\|_{p}\to 0\quad\forall \varphi\in (L^{p}(\Omega))^{*}. \end{align} So $u_{n}\rightharpoonup v$ in $L^{p}(\Omega)$, however weak limits are unique and therefore $u=v$ a.e on $\Omega$.
Step 2
The next assumption I made on $X$ was $X=W^{1,\,p}(\Omega)$ for $\Omega$ bounded and $1<p<\infty$. Since $u_{n}\rightharpoonup u$ in $W^{1,\,p}(\Omega)$ implies $u_{n}\rightharpoonup u$ in $L^{p}(\Omega)$ and $\nabla u_{n}\rightharpoonup \nabla u$ in $L^{p}(\Omega)$ then by the argument for $X=L^{p}(\Omega)$ we have the same result for $W^{1,\,p}(\Omega)$.
Questions
It seems to me that the assumption that $\Omega$ can be dropped. My only concern is the bounding function when applying DCT. However, it seems to me that from the UBP since $\sup_{n\in\mathbb{N}}\|u_{n}\|_{p}<\infty$ then regardless whether or not $\Omega$ is of finite measure the argument above will always work? Is this correct?
I have also considered the case when $p=1$ and $p=\infty$. For $p=1$ since dual is isomorphic to $L^{\infty}(\Omega)$ then it seems to me that the above argument will also hold. Regardless of the fact that $L^{1}(\Omega)\subset (L^{\infty}(\Omega))^{*}$ since we can always identify $u_{n}\in L^{1}(\Omega)$ with $<\cdot,u_{n}>\in (L^{\infty}(\Omega))^{*}$. For $p=\infty$ I am very cautious, my feeling is that the argument above does not work but I cannot figure out why.
Finally, is it possible to weaken the assumption on $X$ here and still be able to show this? I initially thought of taking $X$ to just be a general Banach space but the notion of pointwise convergence only makes sense for functions, and so the weakest assumption I could come up with was $X= L^{p}(\Omega)$.
If there are any mistakes are corrections please let me know. Thank you.