I have the following question:
Let $\Omega\subseteq \mathbb{R}^n$ be open and connected, but not fit for the divergence theorem, only use the formula for partial integraion. Let $u\in C(\Omega),\,V\in C^1(\Omega,\mathbb{R^n})$, where one of these has compact support in $\Omega$. Show that:
$\int_\Omega V\cdot\nabla u \, dx=-\int_\Omega u \;div(V)\,dx$.
What I thought of was using the product rule of the divergence:
$div(V\cdot u)=div(V)\cdot u+ V\cdot \nabla u$ and integrating this over $\Omega$ to get:
$\int_\Omega div(V\cdot u)\, dx=\int_\Omega div(V)\cdot u \, dx+ \int_\Omega V\cdot \nabla u \, dx$
And the left hand side equals $0$ since one of those functions has compact support in $\Omega$? I know it is grasping for straws but I would appreciate any help!
Using what I explained in the commentary, you have :
\begin{aligned} \int_{\Omega} V \cdot \nabla u & = \sum_{i=1}^n \int_{\Omega} V_i \partial_i u \\ & = \sum_{i=1}^n (-\int_{\Omega} \partial_i V_i u + \underbrace{\int_{\partial \Omega} V_i \ u \ n_i}_{=0} ) \text{ using integration by parts}\\ & = - \int_{\Omega} \sum_{i=1}^n \partial_i V_i u\\ & = \int_{\Omega} div(V) u \end{aligned}
I only used the first integration by parts formule but I assumed that is what you are reffering to when you mention "the formula for partial integration".