Define $$\langle -\Delta_p u, v \rangle_{(W^{1,p}_0)', W^{1,p}_0} = \int_{\Omega}|\nabla u |^{p-2}\nabla u \nabla v.$$ Show that $-\Delta_p$ is continous from $W^{1,p}_0$ to $(W^{1,p}_0)'$.
I showed that for every sequence $(u_n)_{n \in \mathbb{N}}$ in $W^{1,p}_0$ such that $u_n \rightarrow u$ in $W^{1,p}_0$, and for all $v \in W^{1,p}_0$ : $$\langle -\Delta_p u_n, v \rangle \rightarrow \langle -\Delta_p u, v \rangle$$ How to conclude that $-\Delta_p u_n \rightarrow -\Delta_p u $ in $ (W^{1,p}_0)'$
I tried to use Banach Steinhaus theorem, but I couldn't reach the result.
Since $u_n \to u$ in $W^{1,p}$ there is a subsequence (denoted by same index) such that $\nabla u_n(x) \to \nabla u(x)$ for almost all $x$, and there is a function $g\in L^p$ such that $|\nabla u_n(x)| \le g(x)$ for almost all $x$.
Then conclude that $|\nabla u_n|^{p-2}\nabla u_n \to |\nabla u|^{p-2}\nabla u$ in $L^{p'}$ by dominated convergence theorem, where $\frac1p+\frac1{p'}=1$.