## Counter example: a function has second order weak derivative but does not have first order weak derivataive

Give a counter example of a function $u(x,y)$ so that the first order weak derivative of $u(x,y)$ does not exist, but the second order weak derivative $\partial_{xy} u$ exists.
The general counter example is the following:
Suppose that $f(x)$ is a continuous but not absolutely continuous function on $[0,1]$, then we know $f(x)$ does not have first order weak derivative (otherwise $f(x)$ is absolutely continuous). Then we set $u(x,y) = f(x) + f(y)$, and it is clear that $u(x,y)$ does not have first order weak derivative. However, we can show that the second order weak derivative $\partial_{xy}u$ exists and equals $0$.
A typical continuous but not absolutely continuous function is the Cantor function.
=== References
[[http://math.stackexchange.com/questions/125345/why-is-the-cantor-function-not-absolutely-continuous|Why is the cantor function not absolutely continuous]]