I need to show that the integral of the derivative of a harmonic function $h$ along the surface normal $\nu$ multiplied by the function over the surface of a smooth compact set G is equal to $\int_G || \nabla h ||^2 d \lambda$. I have some progress on both ends, but I am missing a key insight on the step in the middle. Does anybody have a tip how I could show that?
$$ \begin{align*} \int_{\partial G} h \frac{\partial h}{\partial \nu} dS &= \int_{\partial G} h \left\langle \nabla h, \nu\right\rangle dS \\ &= \text{some steps and the usage of the divergence theorem} \\ &= \int_{G} \sum_i \left( \frac{\partial h}{\partial x_i} \right)^2 d \lambda \\ &= \int_{G} || \nabla h ||^2 d\lambda \end{align*} $$
By a form of the product rule, $h\nabla h = \frac{1}{2}\nabla(h^2)$, so by the divergence theorem,
\begin{align} \int_{\partial G} h\langle \nabla h,\nu\rangle\, dS = \frac{1}{2}\int_{\partial G}\left\langle \nabla (h^2), \nu\right\rangle\, dS = \frac{1}{2}\int_G \nabla \cdot [\nabla(h^2)]\, dV. \end{align} Now, you just have to calculate that carefully. $\frac{1}{2}\nabla(h^2) = \sum_{i=1}^nh\frac{\partial h}{\partial x^i}\, e_i$, where $e_i$ is the $i^{th}$ basis vector. Now, taking the divergence of this yields: \begin{align} \frac{1}{2}\nabla^2(h^2)&=\sum_{i=1}^n\frac{\partial}{\partial x^i}\left(h\frac{\partial h}{\partial x^i}\right) = \sum_{i=1}^n\left(\frac{\partial h}{\partial x^i}\right)^2 + h\frac{\partial^2h}{(\partial x^i)^2} \end{align} Since $h$ is harmonic, what can you now conclude?