I have to prove that, given $\Omega\subset\mathbb{R}^n$ an open subset, $1\leq p<\infty$ and $\{u_n\}\subset W^{1, p}(\Omega)$, we have $u_n\rightharpoonup u$ in $W^{1, p}(\Omega)$ if and only if $u_n\rightharpoonup u$ in $L^p(\Omega)$ and $Du_n\rightharpoonup Du$ in $L^p(\Omega, \mathbb{R}^n)$.
The hint of the exercise is to consider the map $$ T:W^{1, p}(\Omega)\longrightarrow L^p(\Omega)\times L^p(\Omega, \mathbb{R}^n) $$ and prove that it is an isometry. So, my attempt is: by giving $W^{1,p}(\Omega)$ the norm $$ \|u\|_{W^{1, p}(\Omega)}=\left(\|u\|_{L^p(\Omega)}^p+\sum_{i=1}^n\left\|\frac{\partial u}{\partial x_i}\right\|_{L^p(\Omega)}^p\right)^{\frac{1}{p}} $$ and by giving $L^p(\Omega)\times L^p(\Omega, \mathbb{R}^n)$ the norm $$ (u, v) = \left(\|u\|^p_{L^p(\Omega)}+\sum_{i=1}^n\|v_i\|^p_{L^p(\Omega)}\right)^{\frac{1}{p}}, $$ the application $T$ is an isometry (right?). But now, how can I conclude that the equivalence of the assertions follows? The definition of weak convergence in $W^{1, p}(\Omega)$ is that $u_n\rightharpoonup u$ in $W^{1, p}(\Omega)$ if and only if $L(u_n)\longrightarrow L(u)$ for all $L\in (W^{1, p}(\Omega))'$. By the Riesz Representation Theorem on $W^{1, p}(\Omega)$ I know that there exist $f_0,\ldots,f_n\in L^{p'}(\Omega)$ such that, for every $L\in (W^{1, p}(\Omega))'$, $$ L(u)=\int_{\Omega}\left(f_0(x)u(x)+\sum_{i=1}^nf_i(x)\frac{\partial u}{\partial x_i}(x)\right)\ dx $$ for all $u\in W^{1, p}(\Omega)$ and $$ \|L\|_{(W^{1, p}(\Omega))'}=\left(\sum_{i=0}^n\|f_i\|_{L^{p'}(\Omega)}^{p'}\right)^{\frac{1}{p'}}. $$ Thank you
Which implication are you trying to prove? If $u_n\rightharpoonup u$ in $W^{1, p}(\Omega)$ and you take $g\in L^{p'}(\Omega)$, then by Holder's inequality the map $L_g(v)=\int_\Omega gv\,dx$ is linear and continuous in $W^{1, p}(\Omega)$ and so $L_g(u_n)\to L_g(u)$, which shows that $u_n\rightharpoonup u$ in $L^{p}(\Omega)$. Similarly, if $h_1,\ldots, h_n\in L^{p'}(\Omega)$ by Holder's inequality the map $L(v)=\int_\Omega \sum_{i=1}^nh_i(x)\frac{\partial u}{\partial x_i}(x)\,dx$ is linear and continuous in $W^{1, p}(\Omega)$ and so $L(u_n)\to L(u)$, which shows that $Du_n\rightharpoonup Du$ in $L^{p}(\Omega;\mathbb{R}^n)$.