5
$\begingroup$

Assume that $\Omega\subset\mathbb{R}^N$ is a bounded Lipschitz domain and let $p\in [1,\infty)$. Suppose that $u\in W_0^{1,p}(\Omega)\cap L^\infty (\Omega)$.

Is it possible to approximate $u$ by a sequence of function $u_k\in C_0^\infty(\Omega)$ such that $\|u_k\|_\infty\leq M$ for some positive constant $M$?

I was trying to do the following. Extend $u$ by zero outside $\Omega$ and let $\eta_\delta$ be a mollifier sequence. Hence $u_\delta=\eta_\delta\star u$ is such that $u_\delta\in C_0^\infty (\mathbb{R}^N)$ and $u_\delta\to u$ in $W^{1,p}(\Omega)$.

Now for each $\delta>0$ let $\Omega_\delta=\{x\in \Omega:\ \operatorname{d}(x,\partial\Omega)\geq \delta\}$. Take $\lambda_\delta\in C_0^\infty(\Omega)$ such that $\lambda_\delta =1$ in $\Omega_\delta$, $\lambda_\delta\in [0,1]$ in $\Omega_{\delta}\setminus\Omega_{2\delta}$ and $\lambda_\delta =0$ in $\Omega\setminus \Omega_\delta$. Define $g_\delta=\lambda_\delta u_\delta$ and note that $g_\delta\in C_0^\infty(\Omega)$.

Does $g_\delta$ converges to $u$ in $W^{1,p}(\Omega)$?

Update: Note that $$\|g_\delta-u\|_p\leq \|\eta_\delta\star u-u\|_p+\|\lambda_\delta u-u\|_p\tag{1}$$

and $$\left\|\frac{\partial g_\delta}{\partial x_i}-\frac{\partial u}{\partial x_i}\right\|_p\leq \left\| \frac{\partial \lambda_\delta}{\partial x_i}(\eta_\delta\star u)\right\|_p+\left\|\eta_\delta\star \frac{\partial u}{\partial x_i}-\frac{\partial u}{\partial x_i}\right\|_p+\left\|\lambda_\delta \frac{\partial u}{\partial x_i}-\frac{\partial u}{\partial x_i}\right\|_p\tag{2}$$

By using Lebesgue theorem and the definition of $u_\delta$, we have that $(1)$ converges to $0$. By the same argument, we have that the two terms on the right of $(2)$ does converges to $0$. Hence, the question is: Can we choose $\lambda_\delta$ in such a way that $$\left\| \frac{\partial \lambda_\delta}{\partial x_i}(\eta_\delta\star u)\right\|_p\to 0$$

$\endgroup$
  • $\begingroup$ Actually, the convolution will give you a function bounded by the $L^\infty$ norm of $u$. Indeed, $\|u \ast \eta_\delta\|_{L^\infty} \leq \|\eta_\delta\|_{L^1}\|u\|_{L^\infty}$ $\endgroup$ – i like xkcd Oct 18 '13 at 15:19
  • $\begingroup$ You are right @Guillermo, however $u\star\eta_\delta$ does not need to belong to $C_0^\infty(\Omega)$. $\endgroup$ – Tomás Oct 18 '13 at 15:25
  • $\begingroup$ Oh that's right... I'll have to think more about this. $\endgroup$ – i like xkcd Oct 18 '13 at 15:31
  • $\begingroup$ For a different approach, see math.stackexchange.com/a/404467 $\endgroup$ – user98130 Oct 20 '13 at 18:59
  • $\begingroup$ @user98130, could you please undelete your answer. Although this approach is valid too, I have alread answered a question by using your approach which is also correct. $\endgroup$ – Tomás Oct 20 '13 at 19:11
2
$\begingroup$

Here is another, simpler, approach. Let $m=\|f\|_{L^\infty}$. Fix a smooth function $\psi:\mathbb R\to\mathbb R$ such that $\psi(t)=t$ when $|t|\le m$, $\psi(t)=m+1$ when $t\ge m+1 $ and $\psi(t)=-m-1$ when $t<-m-1 $. Since $\psi $ is smooth (and $\Omega$ is bounded), the composition operator $g\mapsto \psi\circ g$ is Lipschitz on $W^{1,p}(\Omega)$. (It suffices to check the Lipschitz property on smooth functions, for which it follows from the chain rule.)

Since $u\in W^{1,p}_0(\Omega)$, there is a sequence $v_k$ of $C_c^\infty(\Omega)$ functions that converges to $u$ in $W^{1,p}$. Let $u_k=\psi\circ v_k$. By the above, $u_k\to \psi\circ u=u$ in $W^{1,p}$. And $|u_k|\le m+1$ by construction.

$\endgroup$
  • $\begingroup$ I think that you want to say in the third line: "Since $\psi'$ is bounded"? $\endgroup$ – Tomás Oct 20 '13 at 14:40
  • $\begingroup$ You also should change $f\mapsto \psi\circ f$ to $g\mapsto \psi\circ g$ to avoid confusion (in the beggining you've defined $m=\|f\|_\infty$). Moreover the composition operator is not linear, hence, continuity is not the same as boundedness. $\endgroup$ – Tomás Oct 20 '13 at 15:36
  • $\begingroup$ @Tomás Thanks for the comments, I revised the post. $\endgroup$ – user98130 Oct 20 '13 at 15:51
  • $\begingroup$ When you say that the composition operator is Lipschitz on $W^{1,p}$, do you mean $\|\psi\circ g-\psi\circ h\|_{1,p}\leq c\|g-h\|_{1,p}$? $\endgroup$ – Tomás Oct 20 '13 at 16:51
  • $\begingroup$ @Tomás Never mind; I'm going to delete this answer and refer you to another one. $\endgroup$ – user98130 Oct 20 '13 at 18:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.