Let $\Omega$ be a bounded open subset of $\mathbb{R}^n$ with smooth ($C^\infty$) boundary. For $k > 1$, assume that $u \in C^{k-1}(\Omega)$ such that its order $k$ derivatives exist in $\Omega$ and in $\mathbb{R}^n \setminus \overline{\Omega}$ and can be extended to continuous functions on $\overline{\Omega}$ and $\mathbb{R}^n \setminus \Omega$ respectively.
Prove that the distributional derivative $\partial^\alpha u$ for $|\alpha| = k$ is given by $f_\alpha \in L^1_{\text{loc}}(\mathbb{R^n})$, where $$f_\alpha = \begin{cases} \partial^\alpha u,& \mbox{in} \quad \Omega \cup (\mathbb{R} \setminus \overline{\Omega}) \\ 0, &\mbox{on} \quad \partial \Omega. \end{cases}$$
This is the statment of Theorem 3.20 on page 44 of Grubb's Distributions and Operators (this is GTM 252).
I think I have the main idea of the proof, but I am getting bogged down in a calculation. Earlier in the text, the following lemma is proved (it is Lemma 3.6 on page 34 , and I think I understand its proof well enough):
Consider the ball $B(0, R) \subseteq \mathbb{R}^n$. For $k > 1$, assume that $u \in C^{k-1}(\overline{B(0, R)})$ such that its order $k$ derivatives exist in $B(0,R) \cap \mathbb{R}^n_+$ and $B(0,R) \cap \mathbb{R}^n_-$ and can be extended to continuous functions on $\overline{B(0,R)} \cap \{x_n \ge 0\}$ and $\overline{B(0,R)} \cap \{x_n \le 0\}$ respectively. Then, for $|\alpha|=k$, the distributional derivative $\partial^\alpha u$ on $B(0,R)$ is given by $v_{\alpha} \in L^1_{\text{loc}}(B(0,R))$, where
$$v_\alpha = \begin{cases} \partial^\alpha u,& \mbox{in} \quad (B(0,R) \cap \mathbb{R}^n_+) \cup (B(0,R) \cap \mathbb{R}^n_-) \\ 0, &\mbox{on} \quad B(0,R) \cap \{x_n = 0\}. \end{cases}$$ The idea is to use the geometry of $\Omega$ along with this lemma to prove the theorem above. Here's what I've got so far.
Because $\partial \Omega$ is $C^\infty$ and compact, we obtain finitely many points $x_i \in \partial \Omega$ ($i = 1, \dots, N$) along with open sets $U_i \ni x_i$ so that
the $U_i$ cover $\partial \Omega$, and
for each $i$ there is a $C^\infty$ diffeomorphism $\kappa_i$ from $U_i$ onto $B(0,1)$ so that $k_i(U_i \cap \Omega) = B(0,1) \cap \mathbb{R}^n_+$, $k_i(U_i \cap (\mathbb{R}^n \setminus \Omega)) = B(0,1) \cap \mathbb{R}^n_-$, $k_i(U_i \cap \partial \Omega) = B(0,1) \cap \{x_n = 0\}$. Let $J_i$ denote the determinant of the Jacobian of $k_i$.
Next, set $U'_i = \kappa_i^{-1}(B(0,\frac{1}{2}))$, and define $v_i(y) = u(k_i^{-1}(y))$ for $y \in B(0,1)$. Then each $v_i$ satisfies the hypotheses of Lemma 3.6 for $B(0,R) = B(0, \frac{1}{2})$.
So then, we make the following calculation (by change of variables). For $\phi \in C^\infty_0(U'_i)$ and a multiindex $\alpha$ with $|\alpha| = k$:
$$\langle \partial_x^\alpha u, \phi \rangle_{U'_i} = (-1)^{|\alpha|}\langle u, \partial_x^{\alpha} \phi \rangle_{U'_i} = (-1)^{|\alpha|}\int_{U'_i} u(x) \partial_x^{\alpha} \phi(x) dx =$$
$$(-1)^{|\alpha|}\int_{B(0, \frac{1}{2})} v(y) \partial_x^\alpha \phi (k^{-1}(y)) \frac{1}{J_i(k^{-1}(y))} dy. $$
And this is where I have gotten stuck. Ultimately, I would like to be able to continue this calculation to show that
$$\langle \partial_x^\alpha u, \phi \rangle_{U'_i} = \int_{U'_i} \partial^{\alpha}u(x) \phi(x).$$
To achieve this goal, I think I will need to make my last integral above look something like $$(-1)^{|\alpha|}\int_{B(0, \frac{1}{2})} v(y) \partial_x^\alpha \phi (k^{-1}(y)) \frac{1}{J_i(k^{-1}(y))} dy = \langle \partial_y^\alpha v, \psi \rangle_{B(0, \frac{1}{2})} = \int_{B(0, \frac{1}{2})} \partial^\alpha_y v(y) \psi(y) dy $$
for some appropriate $\psi \in C^\infty_0(B(0, \frac{1}{2}))$ (the second equality following from Lemma 3.6). Then, I think doing another change of variables would get me where I want to go.
The integral $(-1)^{|\alpha|}\int_{B(0, \frac{1}{2})} v(y) \partial_x^\alpha \phi (k^{-1}(y)) \frac{1}{J_i(k^{-1}(y))} dy$ is almost what I want. But the problem is that my partial derivative is in the $x$ variable rather than the $y$ variable. And the Jacobian is really mucking things up.
Any hints or suggestions about how to clean up this calculation are greatly appreciated. I think I know how to finish the proof after getting this calculation right!