Does the Laplacian and gradient commute? If I have function $u: \mathbb{R}^n \longrightarrow \mathbb{R}$ smooth, does it always hold that:
$$\nabla^2(\nabla u)= \nabla(\nabla^2 u)$$
this is true in general?
 A: This is true for a $C^3$ function defined on $\mathbb{R}^n$. To see this, note that 
$$\tag{1}\Delta (\nabla u)=\Delta\Big(\frac{\partial u}{\partial x_1},\frac{\partial u}{\partial x_2},...,\frac{\partial u}{\partial x_n}\Big)
=\Big(\Delta(\frac{\partial u}{\partial x_1}),\Delta(\frac{\partial u}{\partial x_2}),...,\Delta(\frac{\partial u}{\partial x_n})\Big).$$
On the other hand, we have 
$$\tag{2}\nabla(\Delta u)=\Big(\frac{\partial }{\partial x_1}(\Delta u),\frac{\partial}{\partial x_2}(\Delta u),...,\frac{\partial}{\partial x_n}(\Delta u)\Big).$$
Since $u$ is $C^3$, we have for $1\leq i\leq n$
$$\Delta(\frac{\partial u}{\partial x_i})=
\sum_{j=1}^n\frac{\partial^2}{\partial x_j^2}(\frac{\partial u}{\partial x_i})=\sum_{j=1}^n\frac{\partial^3u}{\partial x_j^2\partial x_i}\\
=\sum_{j=1}^n\frac{\partial^3u}{\partial x_i\partial x_j^2}
=\frac{\partial}{\partial x_i}\Big(\sum_{j=1}^n\frac{\partial^2u}{\partial x_j^2}\Big)
=\frac{\partial}{\partial x_i}(\Delta u),
$$
which implies that 
$$\Delta (\nabla u)=\nabla(\Delta u)$$
by $(1)$ and $(2)$. 
