Curl of Gradient is Zero

Definition
Let $\map {\R^3} {x, y, z}$ denote the real Cartesian space of $3$ dimensions..

Let $\map U {x, y, z}$ be a scalar field on $\R^3$.

Then:
 * $\map \curl {\grad U} = \mathbf 0$

where:
 * $\curl$ denotes the curl operator
 * $\grad$ denotes the gradient operator.

Proof
From Curl Operator on Vector Space is Cross Product of Del Operator and definition of the gradient operator:

where $\nabla$ denotes the del operator.

Hence we are to demonstrate that:
 * $\nabla \times \paren {\nabla U} = \mathbf 0$

Let $\tuple {\mathbf i, \mathbf j, \mathbf k}$ be the standard ordered basis on $\R^3$.

Then:

From Partial Differentiation Operator is Commutative for Continuous Functions:
 * $\dfrac {\partial^2 U} {\partial x \partial y} = \dfrac {\partial^2 U} {\partial y \partial x}$

and the same mutatis mutandis for the other partial derivatives.

The result follows.