Consider some $f: [0,1)\times [0,1)\to \mathbb{R}$. I'm interested in conditions that guarantee that the following one-sided second partial derivatives at $(x,y)=(0,0)$ are symmetric:
$$
\partial\_x^+ \partial\_y^+ f(x,y)= \partial\_y^+ \partial\_x^+ f(x,y).
$$
(where as usual $\partial\_x^+$ and $\partial\_y^+$ indicate partial derivatives defined via one-sided limits).


Everything on the topic that I've found (such as Rudin Theorem 9.41, etc.) considers $f$ defined on an open set, where regular (not one-sided) derivatives can be defined.


Is there a set of simple condition for the above equality to hold? Is there a reference for this?


(My thought right now is to "extend" $f$ to an open set $E \supset [0,1]\times[0,1]$ in a way that preserves the derivatives.)


[Cross-posted from math.stackexchange [here]](https://math.stackexchange.com/questions/3885613/symmetry-of-one-sided-partial-derivatives)



Symmetry of one-sided partial derivatives
One such condition is that $f$ be absolutely continuous in $[0,h)^2$ for some $h\in(0,1)$ -- so that
$$f(x,y)+f(0,0)-f(x,0)-f(0,y)=\int\_0^x du\,\int\_0^y dv\,g(u,v)$$
for some function $g$ integrable on $[0,h)^2$ and for all $(x,y)\in[0,h)^2$ -- with $g$ continuous on the set $([0,h)\times\{0\})\cup(\{0\}\times[0,h))\subset[0,h)^2$. (In particular, it suffices that $\partial\_x\partial\_y f$ exist on $(0,h)^2$ and admit a continuous extension to $[0,h)^2$.)


Indeed, then, by dominated convergence, for all $x\in[0,h)$
$$(\partial^+\_y f)(x,0)=(\partial^+\_y f)(0,0)+\int\_0^x du\,g(u,0)$$
and hence
$$(\partial^+\_x \partial^+\_y f)(0,0)=g(0,0).$$
Similarly, $(\partial^+\_y \partial^+\_x f)(0,0)=g(0,0)$. So, $(\partial^+\_x \partial^+\_y f)(0,0)=(\partial^+\_y \partial^+\_x f)(0,0)$, as desired.


