Failure of Clairaut's theorem where only one of the mixed partials is defined

Statement

For a function of two variables at a point

It is possible to have a function $f$ of two variables $x,y$ and a point $(x_0,y_0)$ in the domain of $f$ such that the second-order mixed partial derivative $f_{xy}(x_0,y_0)$ exists but the second-order mixed partial derivative $f_{yx}(x_0,y_0)$ does not exist.

Proof

Example

Consider a function defined on all of $\R^2$ as:

$f(x,y) := \left\lbrace \begin{array}{rl} 0, & y = 0 \\ 1, & y \ne 0 \\\end{array} \right.$

We note that:

• Since $f$ depends only on $y$, $f_x(x,y)$ is identically the zero function.
• Thus, the second-order mixed partial derivative $f_{xy}(x,y)$ is identically the zero function.
• On the other hand, $f_y(x,y) = 0$ for $y \ne 0$ and $f_y(x,y)$ is undefined for $y = 0$.
• In particular, this means that $f_{yx}(x,y)$ is not defined on the line $y = 0$ (i.e., the $x$-axis).