$u_n\rightharpoonup u$ in $W^{k,p}(\Omega)$, then $D^\alpha u_n\rightharpoonup D^\alpha u$ in $L^p(\Omega)$

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Let $u_n\rightharpoonup u$ in $H_0^1(\Omega)$ where $\Omega$ is a open bounded in $\mathbb{R}^n$, then $u_n\to u$ in $L^2(\Omega)$ by Compact Embedding Theorem. For any $f\in L^2(\Omega)$, define a linear operator on $H_0^1(\Omega)$ by

$$ T_f(u):= \int_\Omega fu_{x_i}\,dx \leq \|f\|_{L_2}\|u_{x_i}\|_{L^2} \leq \|f\|_{L^2}\|u\|_{H_0^1}. $$

So $T_f\in H^{-1}(\Omega)$, the dual space of $H_0^1(\Omega)$. Since $u_n\rightharpoonup u$ in $H_0^1(\Omega)$, we have $$ T_f(u_n)\to T_f(u)\quad \text{as}\ n\to\infty, $$ i.e., $$ \int_\Omega f(u_n)_{x_i}\,dx \to \int_\Omega fu_{x_i}\,dx\quad\text{as}\ n\to\infty\quad\forall f\in L^2(\Omega). $$ Therefore $(u_n)_{x_i}\rightharpoonup u_{x_i}$ in $L^2(\Omega)$ and $Du_n\rightharpoonup Du$ in $(L^2(\Omega))^n$.

Generally, by using the same procedure as above we can say that if $u_n\rightharpoonup u$ in $W^{k,p}(\Omega)$, then $D^\alpha u_n\rightharpoonup D^\alpha u$ in $L^p(\Omega)$ for all $1\leq|\alpha|\leq k$.

Am I correct?