Necessary and Sufficient Conditions for $f_{xy} = f_{yx}$

Question

When beginning to study multivariate calculus, you almost immediately come across the Schwarz-Clairaut Theorem which gives sufficient conditions to guarantee that mixed partials are equal.

For the sake of clarity and posterity, I'll explicity state the $\mathbb{R}^2$ case:

Given $f \colon \mathbb{R}^2 \to \mathbb{R}$; if the the second partials $f_{xy}$ and $f_{yx}$ both exist and are continuous at some $(x,y) \in \mathbb{R}^2$, then

$$ f_{xy}(x,y) = f_{yx}(x,y) $$

Anyway, I've always wondered about more general conditions. Are there well known theorem which imply Schwarz-Clairaut?

Better yet, is there a theorem that gives both necessary as well as sufficient conditions to guarantee mixed partial equality?

note: Naturally, to talk about equality, the second partials have to exist, so this question really comes down to the continuity condition.

In that vein, I'd also be interested in some examples of functions with second partials that exist and are discontinuous at some point but are still equal.

I don't seem to have the chops yet to figure this out myself and consulting Google et al came up with nothing as far as I could tell.

Answer 1

It's unlikely that you could get a useful necessary and sufficient condition. For example, let's suppose $g$ and $h$ are two horrible functions that have $g_{xy}(x,y) \ne g_{yx}(x,y)$ and $h_{x,y}(x,y) \ne h_{y,x}(x,y)$: those quantities have to exist at $(x,y)$ in order to talk about them, but there's nothing else nice you can say about $g$ and $h$. Let $f = a g + h$ where $a = - \dfrac{h_{x,y}(x,y) - h_{y,x}(x,y)}{g_{xy}(x,y) - g_{yx}(x,y)}$. Then $f$ is likely to be equally nasty, but $f_{x,y}(x,y) = f_{y,x}(x,y)$.

Answer 2

Partial derivatives are defined by limits in the same manner as univariate derivatives.

$\begin{align} f_{xy}(x,y) & = \lim_{k\to 0} \frac{f_y(x+k,y)-f_y(x,y)}{k} & \text{by definition} \\[1ex] & =\lim_{k\to 0} \frac{\lim_{h\to 0}\frac{f(x+k,y+h)-f(x+k, y)}{h}-\lim_{h\to 0}\frac{f(x,y+h)-f(x,y)}{h}}{k} & \text{by definition} \\[1ex] & = \lim_{k\to 0}\lim_{h\to 0}\frac{f(x+k,y+h)-f(x+k,y)-f(x,y+h)+f(x,y)}{hk} & \text{IFF both limits exist} \\[1ex] & = \lim_{h\to 0}\lim_{k\to 0}\frac{f(x+k,y+h)-f(x,y+h)-f(x+k,y)+f(x,y)}{hk} & \substack{\text{change of order of limits}\\\text{IFF we have continuity}} \\[1ex] & =\lim_{h\to 0} \frac{\lim_{k\to 0}\frac{f(x+k,y+h)-f(x, y+h)}{k}-\lim_{k\to 0}\frac{f(x+k,y)-f(x,y)}{k}}{h} & \text{IFF both limits exist} \\[1ex] & =\lim_{h\to 0} \frac{f_x(x,y+h)-f_x(x,y)}{k} & \text{by definition} \\[2ex] & = f_{yx}(x,y) & \text{by definition} \\[3ex] \therefore f_{xy}(x,y) & = f_{yx}(x,y) \end{align}$

Hence the sufficient requirement for this result is that all the limits exist at the point and are continuous through it.

However, there is no requirement that if the partials are equal that it is necessary for them to be continuous.