Counterexample second order mixed partial derivatives

Question

I know a few examples of functions $f:\mathbb{R}^2\to\mathbb{R}$ for which $\frac{\partial^2f}{\partial x\partial y}(0,0)$ and $\frac{\partial^2f}{\partial y\partial x}(0,0)$ both exist and are different but I cannot find one where one exists and the other does not. See here for the case when they both exist. counterexample One of my students asked me in class and I couldn’t come up with one. Has anyone seen one? If possible I would like one where both first order partial derivatives exist. Thanks!

Answer 1

$$f(x,y)=|x|(y^2+1). {}{}{}{}$$

Answer 2

For $(x, s) \in \mathbb{R}^{2}$, define $$ g(x, s) = \begin{cases} x - \frac{|s|}{2} & x > |s| \\ \frac{x^{2}}{2|s|} & -|s| \leq x \leq |s| \\ -x - \frac{|s|}{2} & x < -|s|. \end{cases} $$ Let $f(x, y) = \int_{0}^{y} g(x, s) \mathrm{d} s$. For $(x, y) \in \mathbb{R}^{2}$, $$ \frac{\partial f}{\partial y}(x, y) =\lim_{h \rightarrow 0} \frac{f(x, y + h) - f(x, y)}{h} = \lim_{h \rightarrow 0} \frac{\int_{y}^{y + h} g(x, s) \mathrm{d} s}{h} = g(x, y), $$ since $g$ is continuous in $s$ for each fixed $x$. Along the line $y = 0$, $g(x, 0) = |x|$, so $\frac{\partial^{2} f}{\partial x \partial y}(0, 0)$ does not exist. Moreover, for $y \neq 0$, $$ \frac{\partial f}{\partial x}(x, y) = \lim_{h \rightarrow 0} \frac{f(x + h, y) - f(x, y)}{h} = \lim_{h \rightarrow 0} \int_{0}^{y} \frac{g(x + h, s) - g(x, s)}{h} \mathrm{d} s. $$ Since $g$ is Lipschitz in $x$ with constant at most 1 (independent of $s$), we can apply dominated convergence to get $$ \frac{\partial f}{\partial x}(x, y) = \int_{0}^{y} \frac{\partial g}{\partial x}(x, s) \mathrm{d} s. $$ In the case $y = 0$, we have $$ \frac{\partial f}{\partial x}(x, 0) = 0 $$ since $f \equiv 0$ along that line. Finally, $$ \frac{\partial^{2} f}{\partial y \partial x}(0, 0) = \lim_{h \rightarrow 0} \frac{\frac{\partial f}{\partial x}(0, h) - \frac{\partial f}{\partial x}(0, 0)}{h} = \lim_{h \rightarrow 0} \frac{\int_{0}^{h} 0 \mathrm{d} s - 0}{h} = 0, $$ because $\frac{\partial g}{\partial x}(0, s) = 0$ for any $s \neq 0$.