Local Approximation $W^{1,p}$ functions by smooth functions.

Question

I am working through "Measure Theory and Fine Properties of Functions" by Evans and Gariepy, and Chapter 4 section 2 focuses on approximating $W^{1,p}$ functions. In particular, I am confused by Theorem 2 on page 125 that reads:

Assume $f \in W^{1,p}(U)$ for some $1 \le p < \infty$, where $U \subset \mathbb{R}^{n}$ is open. Then there exists a sequence $\{f_k\}_{k=1}^{\infty} \in C^{\infty}(U) \cap W^{1,p}(U)$ such that $$f_k \rightarrow f \quad \text{ in } W^{1,p}(U).$$

In the proof of this theorem we choose a sequence of functions $\{\zeta_k\}_{k=1}^{\infty}$ such that $\zeta_k \in C_c^{\infty}(V_k)$ for some open $V_k \subset U$. The proof requires (and this is what I don't believe) that every such functions $\zeta_k$ has the property that if $f \in W^{1,p}(U),$ then $f\zeta_k \in W^{1,p}(U)$.

I expressed my concern to a professor, specifically that I think this is only true if $V_k$ is bounded. The professor did not immediately know if it was true for unbounded $V_k$, but said it was easy to extend to the unbounded case from knowing the bounded case using the extension theorem in section 4.4 which states:

Assume $U$ is bounded $\partial U$ is Lipschitz, $1 \le p < \infty$. Let $U$ be compactly contained in $V$. Then, there exists a bounded linear operator $$ E : W^{1,p}(U) \rightarrow W^{1,p}(\mathbb{R}^n)$$ such that $$ Ef = f \text{ on } U$$ and $$\text{spt}(Ef) \subset V \quad (\forall f \in W^{1,p}(U)).$$

My two questions are:

1) Can that $\zeta_k$ be a partition of unity chosen such that $f \zeta_k \in W^{1,p}(U)$ if $V_k$ is unbounded.

2) Regardless to the answer of $1$, how does this extension theorem "trivially" generalize the scenario with bounded $V_k$ to the unbounded $V_k$ case?

Answer 1

Let $f\in W^{1,p}(U)$, $V_k\subset U$, open and $\zeta_k\in\ C_c^\infty(V_k)$, then $\zeta_k\in\ C_c^\infty(U)$, $$ \|\zeta_k f\|_p\le \max_{x\in \mathrm{supp}\,\zeta_k} |\zeta_k(x)| \cdot\|f\|_p, $$ and $$ \frac{\partial }{\partial x_j} \big(\zeta_k f\big)= f\frac{\partial }{\partial x_j}\zeta_k+\zeta_k\frac{\partial }{\partial x_j}f, $$ and hence $$ \Big\|\frac{\partial }{\partial x_j} \big(\zeta_k f\big)\Big\|_p^p\le \Big\|f\frac{\partial }{\partial x_j}\zeta_k\Big\|_p+\Big\|\zeta_k\frac{\partial }{\partial x_j}f\Big\|_p\le \max_{x\in \mathrm{supp}\,\zeta_k} |\partial_j\zeta_k(x)| \cdot \|f\|_p+ \max_{x\in \mathrm{supp}\,\zeta_k} |\zeta_k(x)| \cdot \|\partial_j f\|_p. $$ Thus $\zeta_k f\in W^{1,p}(U)$.

Answer 2

In their revised edition, the $V_k$ are intersected with $B(0,k)$; so by definition they are bounded. Notice that as $k \rightarrow \infty$ we again cover all of $U$. Maybe they have been more careful in the new edition?!