TITLE: Sobolev spaces - about smooth aproximation
QUESTION [6 upvotes]: Consider $\Omega $ a open and bounded set of $\mathbb R^n$ . Let $u \in H^{1}(\Omega)$ a bounded function. I know that there exists a sequence $u_m \in C^{\infty} (\Omega) $   where $u_m \rightarrow u$ in $H^{1}(\Omega)$ . I can afirm that exists $C>0$ where $|u_m (x)| \leq C , \forall \ x , \forall m  ?$ 
I dont have idea how to prove this or how to show a counter example ... 
my english is terrible, sorry ..

REPLY [7 votes]: Yes, it is possible to find a uniformly bounded approximating sequence. Let  $\phi:\mathbb R\to\mathbb R$ be a bounded $C^\infty $ function such that 

$\phi(x)=x$ for $x\in [-M,M]$ where $M=\operatorname{ess\,sup}|u|$
$ 0\le \phi'(x)\le 1$ for all $x$. (As a consequence, $|\phi(x)|\le |x|$ for all $x$.)

For every $m$, the function $\phi\circ u_m$ is smooth and satisfies $\|\phi\circ u_m\|_{H^1}\le \|u_m \|_{H^1}$. Since the sequence $(u_m)$ converges in the $H^1$ norm, it is bounded in the $H^1$ norm. Therefore, the sequence $(\phi\circ u_m)$ is bounded in the $H^1$ norm. It follows that $(\phi\circ u_m)$ has a subsequence that converges in the weak topology of $H^1(\Omega)$. On the other hand, $\phi\circ u_m\to \phi\circ u=u$ in $L^2(\Omega)$. Therefore, the aforementioned weak limit is indeed $u$. Furthermore, it is the strong limit because $\limsup\|\phi\circ u_m\|_{H^1}\le \limsup\| u_m\|_{H^1}=\|u\|_{H^1}$.