If $F$ is conservative.... does $\mathrm{curl}(F)$ really equal zero? My notes say that if $F$ is conservative (i.e. $F=\nabla f$) then $\text{curl }(F)=0$.  But I feel this is not quite right.
There is a theorem that says that if $f(x,y,z)$ has continuous second order partial derivatives then $\text{curl }(\nabla f)=0$. (I'm fine with this theorem, it makes sense.) 
Obviously the 'proof' for my question is 
$$\text{curl }(F)=\text{curl }(\nabla f)=0$$
But I don't see why it is necessarily true that $f$ has continuous second order partial derivatives.  Is there something I am missing?
 A: The proof is essentially $\partial_{ij}f = \partial_{ji}f$, and the only hypotheses needed are "whenever this holds". Recall that the curl of $F$ is zero iff $\partial_i F_j = \partial_j F_i$ for all $i,j$. When $F$ is conservative, we have that $\partial_i F_j = \partial_i \partial_j f$ and $\partial_j F_i = \partial_j \partial_i f$.
Thus, if $F$ has first order partial derivatives and $F$ is conservative with potential $f$ which satisfies $\partial_{ij}f = \partial_{ji}f$ for all $i,j$, then $F$ is curl free.
Since the version of Schwartz theorem usually quoted in books requires $f$ to be $C^2$, most authors will subsequently choose to assume $F$ to be $C^1$, which guarantees that $f$ is $C^2$.
A: When you take the divergence of a function you doing $\left \langle \frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z} \right \rangle\cdot \bar{F}$ and then when you take the curl of that you're doing another derivative, the second partial derivative. If the second order derivatives are not continuous then you cannot do the arithmetic with it. Hope this clarifies. 
