# Can $u\in W_0^{1,p}\cap L^\infty$ be approximated by a sequence $u_k\in C_0^\infty$ with $\|u_k\|_\infty$ bounded?

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 $$\left\| \frac{\partial \lambda_\delta}{\partial x_i}(\eta_\delta\star u)\right\|_p\to 0$$

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

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.
• I think that you want to say in the third line: "Since $\psi'$ is bounded"? – Tomás Oct 20 '13 at 14:40
• 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. – Tomás Oct 20 '13 at 15:36
• When you say that the composition operator is Lipschitz on $W^{1,p}$, do you mean $\|\psi\circ g-\psi\circ h\|_{1,p}\leq c\|g-h\|_{1,p}$? – Tomás Oct 20 '13 at 16:51