Boundedness in $L^2$ implies almost everywhere convergence?

Question

Let $\Omega\subset \mathbb{R}^N$ be a bounded domain and $\{u_n\}\subset L^2(\Omega)$ a non-negative sequence. Suppose that $$ \int_\Omega u_n^2dx=\int_\Omega u_ndx, \quad \quad \forall n=1,2,3,\cdots. \tag{1} $$ By (1) and Holder inequality, one easily gets $$ \|u_n\|_{L^2(\Omega)}\leq \text{mes} (\Omega)^{\frac{1}{2}}. \tag{2} $$

It follows from (2), up to a subsequence, we have $$u_n\rightharpoonup u\quad \text{ weakly in } L^2(\Omega) \tag{3} $$ for some $u\in L^2(\Omega)$. Now, by (3) and (1), we get $$ \int_\Omega u_n^2dx=\int_\Omega u_ndx\rightarrow \int_\Omega udx. \tag{4} $$ Now, the question is: whether we can deduce from these information, especially (4) that, up to a subsequence, $u_n$ converges almost everywhere to $u$ in $\Omega$?

Here, I would like to mention that sequences satisfying (2) and (3) may fail to have an a.e. convergent subsequence: take $u_n(x)=\sin(nx),x\in (0,2\pi)$. Then $$ u_n\rightharpoonup 0:=u \text{ weakly in } L^2(0,2\pi), \quad \int_0^{2\pi}u_n^2dx=\pi\neq0= \int_0^{2\pi}udx. $$ But $u_n$ cannot have subsequence that converges a.e. to $u=0$.

Answer 1

Thank you for your support. A careful read of the example shows that the question posed above is false. Indeed, let $$ u_n(x)=\frac{2}{3}(1+\sin(nx)). $$ Then one can easily see that $$ u_n\rightharpoonup \frac{2}{3} \text{ weakly in } L^2(0,2\pi), \quad \quad \int_0^{2\pi}u_n^2dx=\int_0^{2\pi}u_ndx=\frac{4}{3}\pi. \tag{*} $$ Now, if there is a subsequence $\{n^{'}\}$ of $\{n\}$ such that $u_{n^{'}}(x)\rightarrow 2/3$ a.e. in $(0,2\pi)$, then the Lebesgue dominated convergence theorem ($0\leq u_n^2\leq 16/9$) gives $$ \lim_{n^{'}\rightarrow \infty}\int_0^{2\pi}u_{n^{'}}^2dx=\int_0^{2\pi}(\frac{2}{3})^2dx=\frac{8}{9}\pi, $$ which is incompatible with (*). Therefore, $u_n$ has no subsequence that converges a.e. to $2/3$ in $(0,2\pi)$.

An added note: if ever there is a subsequence $\{n^{'}\}$ of $\{n\}$ such that $u_{n^{'}}(x)\rightarrow f$ a.e. in $(0,2\pi)$, then by using Egoroff' theorem we must have $f=2/3$ a.e. in $(0,2\pi)$