# Cases for only one mixed partial derivative is continuous

During studying Clairaut's theorem(In the case of 2 variable functions), I started wondering why the condition of the theorem is written as 'both mixed partial derivative (i.e. $$f_{xy}$$ and $$f_{yx}$$) are continuous'. There can be two possibilities.

1. If $$f_{xy}$$ is continuous, then $$f_{yx}$$ has to be continuous.
2. There exist a function $$f$$ such that $$f_{xy}$$ is continuous but $$f_{yx}$$ doesn't.

However, I couldn't prove the first statement, nor found the counterexample, so if you know what is the right statement, please help us.

If $$f_{xy}$$ and $$f_{yx}$$ are not required to exist, then obviously we can have a function where $$f_{xy}$$ is continuous but $$f_{yx}$$ does not exist, such as $$f=|y|$$.
Theorem. If $$f_{xy}$$ and $$f_{yx}$$ exist everywhere, and $$f_{xy}$$ is continuous at $$(0,0)$$, then $$f_{yx}$$ is continuous at $$(0,0)$$.
Proof. First, the two mixed partial derivatives are equal at $$(0,0)$$: \begin{aligned}f_{yx}(0,0)&=\lim_{x\to0}\frac{1}{x}(f_y(x,0)-f_y(0,0))\\&=\lim_{x\to0}\lim_{y\to0}\frac{1}{xy}(f(x,y)-f(x,0)-f(0,y)+f(0,0))\\&=\lim_{x\to0}\lim_{y\to0}\frac1y(f_x(\theta x,y)-f_x(\theta x,0))\\&=\lim_{x\to0}f_{xy}(\theta x,0)\\&=f_{xy}(0,0).\end{aligned}
By continuity of $$f_{xy}$$ at $$(0,0)$$, given $$\varepsilon>0$$ there exists $$\delta>0$$ such that if $$|x|+|y|<\delta$$ then $$|f_{xy}(x,y)-f_{yx}(0,0)|=|f_{xy}(x,y)-f_{xy}(0,0)|<\frac{\varepsilon}{3}.\qquad(1)$$ Now fix this $$(x,y)$$. By definition of derivative there exists $$x'$$ with $$|x'|<(\delta-|x|-|y|)/2$$ such that $$\left|f_{yx}(x,y)-\frac{1}{x'}(f_y(x+x',y)-f_y(x,y))\right|<\frac{\varepsilon}{3}.\qquad(2)$$ Again, by definition of derivative there exists $$y'$$ with $$|y'|<(\delta-|x|-|y|)/2$$ such that $$\left|f_y(x+x',y)-f_y(x,y)-\frac{1}{y'}(f(x+x',y+y')-f(x+x',y)-f(x,y+y')+f(x,y))\right|<\frac{\varepsilon|x'|}{3}.\quad(3)$$ Applying mean value theorem twice, there exist $$\eta,\theta\in(0,1)$$ such that $$f(x+x',y+y')-f(x+x',y)-f(x,y+y')+f(x,y)=x'y'f(x+\eta x',y+\theta y').\qquad(4)$$ Combining formulas (2–4), we get $$|f_{yx}(x,y)-f_{xy}(x+\eta x',y+\theta y')|<\frac{2\varepsilon}{3}.$$ By formula (1), noticing that $$|x+\eta x'|+|y+\theta y'|\le|x|+|x'|+|y|+|y'|<\delta$$, we have $$|f_{yx}(x,y)-f_{yx}(0,0)|<\varepsilon.$$ Therefore $$f_{yx}$$ is continuous at $$(0,0)$$. QED