Consider the following proof in Lawrence Evans book 'Partial Differential Equations':
How does it follows that $v^{\epsilon} \in C^{\infty}(\bar{V})$? I could see how $v^{\epsilon} \in C^{\infty}(V)$ by using the translations, but I'm having difficulty seeing how it extends to $\bar{V}$, since it says that $u_{\epsilon}(x) := u(x^{\epsilon}) \text{ for } x \text{ in } V$, we are mollifying on $V$. Do you have any idea of how this follows? Thanks for any assistance.
How do we show that the convergent sequence $(v_m)_m$ which is stated in the Theorem statement converges uniformly to $u$?(Its actually called $(u_{m})_{m}$ in the proof stament but I will use $(v_{m})_{m}$). We use the fact that the convergence is uniform in a proof of a Theorem later in the book, so it is established as being uniformly convergent.
The idea that I am proposing: In the proof we have that $v := \sum_{i=0}^{N}\zeta_{i}v_{i}$ is a member of $(v_{m})_{m}$. I think the idea is that we form a sequence $(v_m)_m$ by letting say $\epsilon = (\frac{1}{m})$ and then by choosing a small enough $\epsilon$ such that we have the same $\epsilon$ for all the translations on each $V_{i}$ for $i=1,...N$ and $V_{0}$ and therefore we have $v_{m} := \sum_{i=0}^{N}\zeta_{i}v_{i}$ which allows us to write $||D^{\alpha}v_{m} - D^{\alpha}u||_{L^{P}(U)} \leq CN\delta$ for small enough $\epsilon$ and all $\epsilon$ smaller, so for some $N \in \mathbb{N}$ and for all $m \geq N$ we have $||v_{m} - u||_{L^{P}(U)} \leq CN\delta$. From this we have the uniform convergence from the usual definition.
Does this make sense? Is it necessary to do this or is the uniform continuity more apparent?
Let me know if anything is unclear, thanks.
For fixed $\epsilon>0$, let $A=\{x+\lambda\epsilon e_n:\ x\in V\}$. Define $f:A\to \mathbb{R}$ by $$f(y)=(\eta_{\epsilon}\star u)(y)$$
Because $\overline{A}\subset U$, we can extend $f$ and its derivatives continuouly up to the boundary. Now, for each $y\in A$, there exist $x\in V$ such that $y=x+\lambda\epsilon e_n$, hence $$v^\epsilon(x)=(\eta_\epsilon\star u_\epsilon)(x)=(\eta_\epsilon\star u)(y) $$
The last equality shows that you can also extend $v^\epsilon$ and its derivatives continuously up to the boundary of $V$.