I am confused how to do this question. I need to use Green's first identity and if $\nabla(f)=0$ then $f$ is constant on $\Omega$ since $\Omega$ is path connected.
I have subbed in the information into green's identity but I don't get anything useful.
$$\iiint_\Omega \nabla f\cdot \nabla g \,dV =\iint_{\partial\Omega} f\nabla g\cdot n \,dA – \iiint_\Omega f\cdot \Delta g \,dV$$
We get:
$$\iiint_\Omega \nabla f\cdot \nabla g \,dV =-\iiint_\Omega f\cdot \Delta g \,dV$$
$$\iiint_\Omega \nabla. (f \nabla g) \,dV =0$$
$$\iint_{\partial\Omega} f\nabla g\cdot n \,dA=0$$
Best Answer
Edit:More direct method, use integration by parts on the Dirichlet energy, $$ \iiint |\nabla f |^2 dV = \underbrace{\iint f \nabla f \cdot n dS}_{f=0 \text{ on } \partial \Omega} - \underbrace{\iiint f \Delta f dV}_{\Delta f = 0 \text{ on } \Omega} =0$$ Thus this means that $\nabla f =0$, so you can use your other result.