Let $\Omega$ be a bounded domain in $\mathbb{R}^{n}$, $n\geq2$. An integrable function $u$ is said to be weakly harmonic in $\Omega$ if for all $\varphi\in C_{c}^{2}\left( \Omega\right) $, $$ \int_{\Omega}u\Delta\varphi=0. $$ Show that a continuous weakly harmonic function is harmonic. (Remark: The continuous assumption makes the proof easier but is not needed.)
Let's first assume $u\in\mathscr{C}^{2}(\bar{U}).$ Then integration by parts (or Green's identity) shows \begin{align*} 0 &=\int\limits_{U}u\Delta\phi\;dx\\ &=\int\limits_{\partial U}u\frac{\partial\phi}{\partial\nu}\;dS-\int\limits_{U}\nabla u\cdot\nabla\phi\;dx\\ &=-\int\limits_{\partial U}\phi\frac{\partial u}{\partial\nu}\;dS+\int\limits_{U}\phi\Delta u\;dx\\ &=\int\limits_{U}\phi\Delta u\;dx. \end{align*}
I am unsure about this line that $$\int\limits_{\partial U}u\frac{\partial\phi}{\partial\nu}\;dS-\int\limits_{U}\nabla u\cdot\nabla\phi\;dx\\ =-\int\limits_{\partial U}\phi\frac{\partial u}{\partial\nu}\;dS+\int\limits_{U}\phi\Delta u\;dx.$$ How do I get this and also after that why do we have $\int\limits_{\partial U}\phi\frac{\partial u}{\partial\nu}\;dS=0$?