I understand that the Hessian determinant (detH) for a function f (x,y) is defined as:
\begin{vmatrix} f_{xx} & f_{xy}\\ f_{yx} & f_{yy}\\\end{vmatrix}
Where the determinant is a factor for determining the nature of a critical point of the function.
My question is: Why is it that detH is evaluated as
$$ detH = f_{xx}(x,y)f_{yy}(x,y)-\left(f_{xy}(x,y)\right)^{2} $$
When the condition for a 'well behaved function':
$$ f_{xy} = f_{yx} $$
Hasn't yet been proven/determined. It seems to me like this is assumed off the bat. Is there something I'm missing?
Theorem. A twice differentiable real-valued function defined on an open subset of $\mathbf{R}^d$ has mixed partials that coincide (that is to say, if $u$ is a twice-differentiable function then $u_{x,y} = u_{y, x}$).
The proof of this stronger version is more difficult that the one usually given (where $u$ is assumed with continuous derivatives of higher order). It can be found in Foundations of Modern Analysis, by Jean Dieudonne, chapter 8.
So, "well-behaved" function simply means twice-differentiable.