I'm trying to prove that if
$$\dfrac{\partial^2 f}{\partial x \partial y} \quad \text{and} \quad \dfrac{\partial^2 f}{\partial y \partial x}$$
are continuous in an open set containing $a \in \mathbb{R}^2$, then they are equal using the definition of derivative:
First, I say
$$ \dfrac{\partial^2f}{\partial x \partial y} = \dfrac{\partial}{\partial x} \left( \dfrac{\partial f}{\partial y} \right) = \dfrac{\partial}{\partial x} \left( \lim_{h \to 0} \dfrac{f(x,y+h)-f(x,y)}{h} \right) = \lim_{k \to 0} \dfrac{1}{k} \left( \lim_{h \to 0} \dfrac{f(x+k,y+h)-f(x+k,y)}{h} - \lim_{h \to 0} \dfrac{f(x,y+h)-f(x,y)}{h} \right) = \\ \lim_{k \to 0} \left( \lim_{h \to 0} \dfrac{f(x+k,y+h)-f(x+k,y)-f(x,y+h)+f(x,y)}{kh} \right) $$
Then I do the same thing for $ \dfrac{\partial^2 f}{\partial y \partial x} $, and I get
$$ \lim_{h \to 0} \left( \lim_{k \to 0} \dfrac{f(x+k,y+h)-f(x+k,y)-f(x,y+h)+f(x,y)}{kh} \right)$$
My question is: am I allow to say that those limits are equal because both
$$\dfrac{\partial^2 f}{\partial x \partial y} \quad \text{and} \quad \dfrac{\partial^2 f}{\partial y \partial x}$$ are continuous on $a$, or do I need something extra? Please let me know if the question is clear enough, or if I made some silly mistakes :)
Thanks in advance!
Under these hypotheses the standard way to prove that is to consider some new auxiliary functions and use MVT theorem (refer to Schwarz's theorem).
The same result can be proved also under the hypotheses of the existence of $f_x(x,y)$ and $f_y(x,y)$ in the open domain and that $f_x(x,y)$ and $f_y(x,y)$ are differentiable at the point $a$.