# Proposed proof for Sobolev space result

I have the following result which seems that it must be true, but I would like to prove it: This is my proposed proof.

If $U \subset \mathbb{R}^{n}$. Given $u \in W^{1,p}(U)$, where $u$ has compact support in $U$. I want to show that $u \in W^{1,p}_{o}(U)$.

Proof: We want to show that there exists $\{ u_{m} \}_{m}^{\infty} \subset C^{\infty}_{c}(U)$ such that $u_{m} \rightarrow u$ in $W^{1,p}(U)$. Consider the mollifier $u^{\epsilon}(x) = \int_{U}\eta_{\epsilon}(x-y)f(y)dy$. We consider the mollifier on supp($u$), therefore $u^{\epsilon} \in C^{\infty}(\text{supp($u$)})$. It follows then from properties of mollifiers that $u^{\epsilon} \rightarrow u$ in $L^{p}(\text{supp}($u$))$. Similarly, we can show that $(\nabla u)^{\epsilon} \rightarrow \nabla u$ in $L^{p}(\text{supp}(u))$. It follows then that $u^{\epsilon} \rightarrow u$ in $W^{1,p}(\text{supp}(u))$. We then have $u \in W^{1,p}(\text{supp}(u))$. Since $u = 0$ on $U \setminus \text{supp}(u)$, it follows that $u \in W^{1,p}_{o}(U)$. $\square$

Is the above proof fine? Also, does anyone, have a good recommended text which defines 'cut-off functions' and gives good examples?

Thanks.

• Hi John Doe. You have posted this question long time ago so maybe you already know whether the proof you gave is correct or you have found another ? – Svetoslav Jan 3 '16 at 15:48

Lemma. Let $1 \leq p<\infty$ and let $u \in W^{1,p}(\Omega) \cap C(\overline{\Omega})$. If $u=0$ on $\partial\Omega$, then $u \in W_0^{1,p}(\Omega)$.
Proof. By standard regularization, $W^{1,p}(\Omega) \cap \mathcal{K}(\Omega) \subset W_0^{1,p}(\Omega)$, where $\mathcal{K}(\Omega)$ is the set of continuous functions with compact support in $\Omega$. Assume the support of $u$ is bounded. Let $f \in C^1(\Omega)$ be such that $|f(t)| \leq |t|$ and $f(t)=0$ when $|t| \leq 1$, $f(t)=t$ when $|t| \geq 2$. Now define $u_n = f(nu)/n$. It follows that $u_n \in \mathcal{K}(\Omega)$ and $u_n \in W^{1,p}(\Omega)$ by the rule of composition in Sobolev spaces. By dominated convergence, $u_n \to u$ in $W^{1,p}(\Omega)$, and therefore $u \in W_0^{1,p}(\Omega)$. If the support of $u$ is unbounded, we define $u_n = \theta_n u$, where $\theta_n$ is a truncation function ($\theta(t)=1$ if $t \leq 1$ and $\theta(t)=0$ if $t \geq 2$). Now the support of $u_n$ is bounded, and $u_n \to u$ in $W^{1,p}(\Omega)$, so that $u \in W_0^{1,p}(\Omega)$.
• Hi, could I ask your opinion on something. Do you think that in the paper. On the second page of the paper it says "$V=W^{1,p}$, $p>1$ ($V' = W^{-1,p'})$ with $p' = \frac{p}{p-1}$". Do you think this is a typo? Shouldn't it be $V = W^{1,p}_{0}$. See the statement of the Theorem 1 at the bottom of the second page also. Thanks, sorry to bother. – user100431 Feb 5 '15 at 20:20