Let $ 1 < p < 2 $ and $ u \in W_0^{1,p}( \Omega) $, where $ \Omega $ is some smooth open set of $ \mathbb{R}^N,\ N \geq 2. $ Suppose that $$ \int_{\Omega} \nabla u \nabla \phi dx = 0,\ \forall\ \phi \in C_0^{\infty}( \Omega). $$ Is it true that $ u = 0? $
The use of approximation of $ u $ by smooth functions is not useful. Any idea is welcome.
This is true for bounded domains $\Omega$ with smooth boundary ($C^1$ is sufficient also), as a consequence of the invertibility of the Laplacian as a map $$ \Delta : W^{1,p}_0(\Omega) \to W^{-1,p}(\Omega) $$ for all $1<p<\infty.$ These follow from the Calderón and Zygmund $L^p$ estimates, and I've written some further details and included relevant references in this answer.
The above does require some fairly heavy machinery however, and unfortunately I don't think this can be avoided. The necessity of requiring some regularity of the boundary is shown by Hajłasz in Theorem 1 of his paper A counterexample to the $L^p$ Hodge decomposition; there he constructs a bounded domain $\Omega \subset \Bbb R^2$ satisfying the cone condition, along with a non-trivial harmonic function $u$ which lies in $W^{1,p}_0(\Omega)$ for all $1 \leq p < \frac43.$ This suggests you do need to use the boundary regularity in a non-trivial way, which is why a direct approximation argument doesn't work.