Higher boundary regularity theorem in Evans for elliptic equation I have difficulty with the proof of this theorem:

Let $m$ be a nonnegative integer, and assume $a^{ij},b^i,c\in C^{m+1}(\overline{U})$, for $(i,j=1,...,n)$ and $f\in H^m(U)$. Suppose that $u\in H^1_0(U)$ is a weak solution of the boundary problem $$\left\{ 
 \begin{array}{c}
 Lu=f\hspace{0.2 cm} in \hspace{0.1 cm} U\\ 
 u=0\hspace{0.3 cm} su \hspace{0.1 cm} \partial U,
 \end{array}\right.  \hspace{0.8 cm} (1)$$ Assume  $\partial U$ is $C^{m+2}$.  Then $u\in H^{m+2}(U)$, and we have the estimate $$\| u\|_{H^{m+2}(U)}\le C (\| f\|_{H^m(U)}+\| u\|_{L^2(U)}),$$ with $C$ constant.

Consider $U:= B(0,s)\cap \{x_n>0\}\subset \Bbb R^n$, for $s>0$.  We prove by induction. The case $m=0$ is proved from a preceding theorem. Let $\alpha$ be any multi index with $|\alpha|=m+1$ and $\alpha_n=0$. Suppose by induction that $u\in H^{m+2}(U)$.
My question: Why $D^{\alpha} u$ vanishes along the plane $\{x_n=0\}$ in the trace sense?
My attempt: Exist $(u_m)_{m\in \Bbb N}\subset C^{\infty}(\overline{U})$ such that $$u_m \rightarrow u\hspace{0.5 cm} in \hspace{0.2 cm} H^{m+2}(U)$$ This implies that $$D^{\alpha} u_m \rightarrow D^{\alpha} u \hspace{0.5 cm} in \hspace{0.2 cm} H^1(U).$$ Then if $T$ is the trace operator  on $U$ $$ T: H^1(U)\rightarrow L^2(\partial U)$$ i have $T(D^{\alpha} u)=\lim _m T(D^{\alpha} u_m)$. How can I continue? Hints?
 A: I believe this works. Add to the induction statement on $m$ that for all multi indices $\alpha$ with $|\alpha|=m$ and $\alpha_n=0$, we have that $D^\alpha u\in H^1$ vanishes in the sense of trace. This when $m=0$ is just saying that it satisfies the Dirichlet boundary conditions.
To prove the statement for $|\alpha|=m+1$,
We write $\alpha=a+\alpha’$ for some $|\alpha’|=m$ and $v=D^{\alpha’}u$. By assumption $D^av\in H^1$. By the added induction hypothesis, we have that for sufficiently small $h\neq 0$,
$$ \mathbb D_hv:=\frac{v(x+ah)-v(x)}h \equiv 0$$
in the sense of trace, since $x,x+ah\in\partial U$.
In particular, for all $\epsilon>0$, $\|\mathbb D_h v\|_{L^2(\partial U)}\le \epsilon$. By the section on finite differences in chapter 5, Theorem 3 (ii), we know that taking a weak limit along a subsequence $h_k\to 0$, we have $\mathbb D_{h_k} v \underset{k\to\infty}\rightharpoonup  D^a v$ in $L^2(\partial U)$, with $\|D^a v\|_{L^2(\partial U)}\le \epsilon$. Since $\epsilon$ was arbitrary,
$$ D^\alpha u|_{x_n=0}=D^a v|_{x_n=0}=0,$$
proving this subclaim of the induction argument.
Would love for readers to comment if they have checked this argument!
A: Thank to @CalvinKhor i think that i have an answer. By approximation theorem exist $(u_m)_{m\in \Bbb N}\subset C^{\infty}(\overline{U})$ such that $$u_m \rightarrow u\hspace{0.5 cm} in \hspace{0.2 cm} H^{m+2}(U)$$ This implies that $$D^{\alpha} u_m \rightarrow D^{\alpha} u \hspace{0.5 cm} in \hspace{0.2 cm} H^1(U).$$ Now because $T(u)=0$, we obtain that $$\lim _m u_m (x)=0,$$ for every $x\in \{x_n=0\}$, but we can easily verify that $\lim _m D^{\alpha} u_m(x)=0$, for every $x\in \{x_n=0\}$. For the definition of the trace operator we done.
