I'm trying understand a specific part of the proof of the following result in a lecture notes of a course in Sobolev spaces:
$\textit{Theorem.}$ Let be $\Omega$ an open set in $\mathbb{R}^N$. If $1 < p < \infty$, then $W^{m,p}(\Omega)$ is a Banach reflexive space.
I will put the whole proof here, my doubt and my thought about what I thought to answer my doubt. I would like to know if my thought is right and, otherwise, I would like an answer for my doubt.
Notation: $D'(\Omega)$ is the space of continuous distributions defined on $\Omega$.
$\textit{Proof.}$ Let be $u_{\nu}$ a bounded sequence in $W^{m,p}(\Omega)$, then $(D^{\alpha} u_{\nu})$ is a bounded sequence in $L^p(\Omega)$ $\forall |\alpha| \leq m$. In particular, $(u_{\nu})$ is bounded in $L^p(\Omega)$, which imply that exists a subsequence $(u_{\nu}')$ of $(u_{\nu})$ such that $u_{\nu}' \rightarrow u_{(0,\cdots,0)} = u$ weakly in $L^p(\Omega)$. Also and, in particular, $\left( \frac{\partial u_{\nu}'}{\partial x_1} \right)$ is bounded in $L^p(\Omega)$, which imply that exists a subsequence $(u_{\nu}'')$ of $(u_{\nu}')$ such that $u_{\nu}'' \rightarrow u_{(1,\cdots,0)}$ weakly in $L^p(\Omega)$. By sucessive applications to this argument, we find a sequence $(v_{\nu})$ of $(u_{\nu})$ and a function $u_{\alpha}$ in $L^p(\Omega)$ such that $D^{\alpha} v_{\nu} \rightarrow u_{\alpha}$ weakly in $L^p(\Omega)$ for all $|\alpha| \leq m$. In particular, $v_{\nu} \rightarrow u_{(0,\cdots,0)} = u$ weakly in $L^p(\Omega)$ because $(v_{\nu}) \subset (u_{\nu}')$. Follow from this, $D^{\alpha} v_{\nu} \rightarrow D^{\alpha} u$ in $D'(\Omega)$. We have also that $D^{\alpha} v_{\nu} \rightarrow u_{\alpha}$ in $D'(\Omega)$, which allow us to conclude by the uniqueness of limit in $D'(\Omega)$ that $D^{\alpha} u = u_{\alpha}$ $\forall |\alpha| \leq m$ and $u \in W^{m,p}(\Omega)$ consequently.
We need to show that $(v_{\nu})$ converges weak to $u$ in $W^{m,p}(\Omega)$. Indeed, let be $T \in (W^{m,p}(\Omega))'$. According to the last observation (see the observation in the final of the proof), exist $g_{\alpha} \in L^q(\Omega)$ such that $\langle T, w \rangle = \sum\limits_{|\alpha| \leq m} \int_{\Omega} g_{\alpha}(x) D^{\alpha} w(x) dx \forall w \in W^{m,p}(\Omega)$, then
$$\langle T, v_{\nu} \rangle = \sum\limits_{|\alpha| \leq m} \int_{\Omega} g_{\alpha}(x) D^{\alpha} v_{\nu}(x) dx \rightarrow \sum\limits_{|\alpha| \leq m} \int_{\Omega} g_{\alpha}(x) D^{\alpha} u(x) dx = \langle T, u \rangle$$
because $D^{\alpha} v_{\nu} \rightarrow D^{\alpha} u$ weak in $L^p(\Omega)$. Thus, we proved that $v_{\nu} \rightarrow u$ weak in $W^{m,p}(\Omega)$. $\square$
My doubt is relative to this part:
Follow from this, $D^{\alpha} v_{\nu} \rightarrow D^{\alpha} u$ in $D'(\Omega)$.
I think this is true because
$\begin{align*} \langle D^{\alpha} v_{\nu} - D^{\alpha} u, \varphi \rangle &= \langle D^{\alpha} (v_{\nu} - u), \varphi \rangle\\ &= (-1)^{|\alpha|} \langle v_{\nu} - u, D^{\alpha} \varphi \rangle\\ &\overset{\text{Holder's inequality}}{\leq} (-1)^{|\alpha|} || v_{\nu} - u ||_{L^p(\Omega)} || D^{\alpha} \varphi ||_{L^q(\Omega)}\\ &\rightarrow 0, \forall \varphi \in C_0^{\infty}(\Omega) \end{align*}$
because $v_{\nu} \rightarrow u$ weakly in $L^p(\Omega)$.
Thanks in advance!