Characterization of weak convergence in $W^{1, p}(\Omega)$

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 $L^{p}(\Omega;\mathbb{R}^n)$.
• By definition I have that $\int_{\Omega}u_ng\rightarrow\int_{\Omega}ug$ for all $g\in L^{p'}(\Omega)$ and $\int_{\Omega}Du_ng\rightarrow\int_{\Omega}Dug$ for all $g\in L^{p'}(\Omega, \mathbb{R}^n)$. Can I conclude that $\int u_ng_o+\sum g_i(x)(u_n(x))_i\rightarrow\int ug_o+\sum g_i(x)u_i(x)$? – Jeji Aug 22 '17 at 9:11
• You have $\int_{\Omega}u_ng_0\rightarrow\int_{\Omega}ug_0$ for all $g_0$ and $\int_{\Omega}Du_ng\rightarrow\int_{\Omega}Dug$ for all $g$. So if you sum them you get $\int_{\Omega}u_ng_0+Du_ng\rightarrow \int_{\Omega}ug_0+Dug$. The hint is to tell you what the dual of $W^{1,p}$ is. – Gio67 Aug 22 '17 at 10:50
• No. No need to use Riesz. For every $g\in L^{p'}(\Omega)$ the functional $L_g$ is linear and continuous. Just use Holder's inequality. – Gio67 Aug 22 '17 at 14:37
• If $X$ is a Banach space, a linear functional $L:X\to \mathbb{R}$ is continuous iff $|L(x)|\le c\Vert x\Vert$ for all$x\in X$ and for some constant $c>0$. Now take $X=W^{1,p}$ and $\Vert f\Vert_{W^{1,p}}=\Vert f\Vert_{L^{p}}+\Vert Df\Vert_{L^{p}}$ and apply Holder's inequality to $L_g$. – Gio67 Aug 22 '17 at 15:40