My Understanding of the curl operator is the following: it maps continuously differentiable functions $f$: $\mathbb{R}^3$ $\to$ $\mathbb{R}^3$ to continuous functions $g$: $\mathbb{R}^3$$\to$ $\mathbb{R}^3$, in particular it maps $C^k$ functions in $\mathbb{R}^3$ to $C$$^k$$^-$$^1$ functions in $\mathbb{R}^3$. however, my question is this:
why does it map functions with $k$ continuous derivatives to functions with $k-1$ continous derivatives?
Implicitly, the curl of a vector field F is defined as (where $p$ is any point in the field):
($\nabla$ $\times$ F)($p$) $\bullet$ n $\stackrel{\mathrm{def}}{=}$ $\displaystyle{\lim_{A \to \infty}}$($\frac{1}{|A|}$$\oint\limits_{C}$F $\bullet$ dr).
My only thought as to why this is true is because in the defintion shown above, ($\nabla$ $\times$ F) before it acts on the point $p$ is simply the determinant of the matrix of the component functions of the vector field where the second row of the matrix is the partial derivatives that act on the component functions of the vector field in standard cartesian coordinates.
The curl is also defined using derivatives. Check the Wikipedia definition. The components of the curl of a vector field $\mathbf{F}$ are linear combinations of the partial derivatives of $\mathbf{F}$. If $\mathbf{F}$ is $k$-times continuously differentiable, then its partial derivatives are $k-1$-times continuously differentiable. So $\nabla \times \mathbf{F}$ is in $C^{k-1}$.
You can easily construct a vector field in $C^k$ whose curl is NOT in $C^k$; for example $\mathbf{F}(x,y,z) = \min(x^{k+1}, 0) \hat{x}$.