## Proof of Identity for Laplacian

Theorem
$\mathbf{\nabla} \cdot (\mathbf{\nabla} \times g + \mathbf{\nabla} f = \nabla^2 f$
where
$g$
is an arbitrary function.
Proof
$\mathbf{\nabla} \cdot (\mathbf{\nabla} \times g + \mathbf{\nabla} f = \mathbf{\nabla} (\mathbf{\nabla} \times g + \mathbf{\nabla} f = \mathbf{\nabla} \cdot ( \mathbf{\nabla} \times g) + \mathbf{\nabla} \cdot ( \mathbf{\nabla} f) = \mathbf{\nabla} \cdot ( \mathbf{\nabla} f) = \nabla^2 f$

since
$\mathbf{\nabla} \cdot ( \mathbf{\nabla} \times g)=0$
if
$g$
is twice differentiable.