# Sobolev spaces - about smooth aproximation

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 ..

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 $\limsup\|\phi\circ u_m\|_{H^1}\le \limsup\| u_m\|_{H^1}=\|u\|_{H^1}$.

• can you give a example of a $\phi$ with the properties that you writed ? please ^^ – math student May 28 '13 at 4:16
• Excellent answer. – Euler....IS_ALIVE May 28 '13 at 4:30
• @LeandroTavares Mollify $\min(\max(x,M+1),-M-1)$. – ˈjuː.zɚ79365 May 28 '13 at 4:34
• thankyou! beatiful answer! – math student May 28 '13 at 4:54
• where I can find the fact the mollification satisfies the properties that you writed ? – math student May 28 '13 at 5:28

You are looking for Meyers-Serrin Theorem. However, It may be added that it is not always possible to get an approximation from $C^{\infty}(\overline{\Omega})$. A proof for the above may be found here. I think I might have misunderstood your question. I am assuming that you mean that $u\in L^{\infty}(\Omega)\cap H^1(\Omega)$. In that case, it is well known that if $u_m\to u$ in $L^p(\Omega)$, then there is a subsequence $u_{m_k}$ which converges pointwise almost everywhere to $u$. Therefore, for almost every $x$, $\{u_{m_k}(x)\}$ is bounded. I cannot think of a counterexample.

• If a sequence converges a.e. to a bounded function, it does not immediately follow that the sequence is uniformly bounded. – ˈjuː.zɚ79365 May 28 '13 at 4:04
• yes, of course. – user38404 May 28 '13 at 4:13