Let $\Omega = [0,1]$ for simplicity. Take a sequence $\{u_{n}\}$ which is bounded in $W^{1,1}(\Omega)$. Then up to a subsequence we have there exists $u$ such that $u_{n} \rightharpoonup u$ weak-* in $L^{\infty}(\Omega)$. However does it necessarily mean that:
- $u$ is continuous?
- $u_{n} \to u$ uniformly? i.e. $$\sup_{x \in \Omega}|u_{n}(x)-u(x)| \to 0 $$ as $n \to \infty$.
My thoughts:
I know that the $u_{n}$ are all uniformly continuous by Sobolev embedding. My gut feeling is that the statement shouldn't be true, i think a counterexample like a sequence which converges to $sgn(x)$ might work but to be honest I am getting a bit confused between the different notions of convergence. I would appreciate any help.
If you take $u_n(x)=nx$ for $-\frac1n\le x\le\frac1n$, $u_n=-1$ for $x\le-\frac1n$ and $u_n=1$ for $x\ge\frac1n$, you have that $u_n^\prime=n\chi_{(-1/n,1/n)}$ and $|u_n|\le 1$, so the sequence is bounded in $W^{1,1}$. The weak limit is discontinuous. $u(0)=0$, $u=-1$ for $x<0$ and $u=1$ for $x>0$. Since the sequence of continuous functions converges pointwise to a discontinuous function, you cannot have uniform convergence.