The minus Laplacian operator is positive definite In a textbook of functional analysis I found this equation derived from Green's first identity
$$\int _{ \Omega  }^{  }{ u{ \nabla  }^{ 2 }ud\tau  } =\int _{ \partial \Omega  }^{  }{ u\frac { \partial u }{ \partial n } ds } -\int _{ \Omega  }^{  }{ \left| \nabla u \right| ^{2}d\tau  }$$ 
Then it goes on saying that if the boundary conditions on u are such that the integral over the boundary vanishes then the operator $ -\nabla^{2}$ is positive definite.
Why ?
What I can see is that $$\int _{ \Omega  }^{  }{ u{ \nabla  }^{ 2 }u+{ \left| \nabla u \right|  }^{ 2 }d\tau  } =0$$
and what I'd need to declare that the operator is positive definite is : $$\left< -{ \nabla  }^{ 2 }u,u \right> >0\Leftrightarrow \int _{ \Omega  }^{  }{ -{ \nabla  }^{ 2 }u\bar { u } d\tau  } >0$$
So far I don't see how to prove that the operator is positive definite...
Thanks for any kind of help.
 A: Green's identity reads:
$$\int_U \left( \psi \nabla^{2} \varphi + \nabla \varphi \cdot \nabla \psi\right)\, dV  = \oint_{\partial U} \psi \left( \nabla \varphi \cdot \mathbf{n} \right)\, dS$$
Select $\psi=\bar{u}$ and $\varphi=u$ and negate:
$$-\int_U\bar{u}\nabla^2u+\nabla \bar{u}\cdot\nabla u\; dV=-\oint_{\partial U}\bar{u}(\nabla u\cdot\mathbf{n})dS.$$
Of course $\nabla u\cdot\mathbf{n}=0$ by hypothesis on the boundary conditions, so we may rearrange this to
$$\left\langle -\nabla^2 u,u\right\rangle=\int_U \overline{\nabla u}\cdot \nabla u\;dV.$$
The integrand on the right is $\sum_i|\partial u/\partial x_i|^2$, so of course it is nonnegative, and is in fact only zero when $\nabla u=0$. In fact the integral on the right displays a means to defining an inner product for complex-valued vector functions, hence $\langle\nabla u,\nabla u\rangle $ in Andrew's answer, and knowing this a priori would provide a very direct means to seeing positive definiteness. The inner product is
$$\langle \mathbf v,\mathbf w\rangle =\int_U \mathbf v\cdot \overline{\mathbf w}\; dV.$$
It is somewhat unclear to me if the statement about $-\Delta^2$'s positive definiteness is in the context of real-valued or complex-valued functions $u$. In the former situation $u=\bar{u}$ so you already had all you needed, and in the latter situation the identity it had was slightly off (didn't involve a complex conjugate) for the purpose at hand, albeit a slight modification was all that was necessary.
A: According to last but one equation 
$$\left< -{ \nabla  }^{ 2 }u,u \right>= \left< { \nabla }u,\nabla u \right>>0
$$
for $u\not\equiv 0\;$.
