Prove if $u_n \in W^{1,p}$ such that $u_n\to u$ in $L^p$ and $u_n'$ is uniformly bounded in $L_p$ then $u \in W^{1,p}$

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Let $U\subset \Bbb{R}^n$ be some open set .Prove for $1<p\le \infty$ we have the following result:

$u_n \in W^{1,p}(U)$ such that $u_n\to u$ in $L^p(U)$ and $u_n'$ is uniformly bounded in $L^p(U)$ then $u \in W^{1,p}$.

My solution:

Using Banach–Alaoglu theorem we can show $u_n'$ uniformly bounded sequence has weak star convergence subsequence and we know for $1\le q<\infty$ we have $(L^q(U))' = L^p(U)$ where $q$ conjugate to $p$.

Which means for all $1<p\le \infty$ we have subsequence $u_n\in W^{1,p}(U)$ that weak star convergence to some $g\in L^p(U)$ such that :

$$\int vu_n' \to \int vg$$

For all $v\in L^q(U)$.

In particular for $\varphi\in C^\infty_c(U)$ we have :

$$\int \varphi' u_n = -\int u_n' \varphi$$

such that $\int u_n'\varphi \to \int g\varphi$ and $\int u_n \varphi' \to \int u \varphi'$

Hence we have $$\int g\varphi = -\int u\varphi'$$

which means $u\in W^{1,p}(U)$ with $u' = g\in L^p(U)$

Is my proof correct?(For simplicity we just pick $n= 1$ in the proof above but the proof is similar for rest of $n$)