Failure of Clairaut's theorem where both mixed partials are defined but not equal

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 both the second-order mixed partial derivatives of $f$ exist at $(x_0,y_0)$, i.e., both the numbers $f_{xy}(x_0,y_0)$ and $f_{yx}(x_0,y_0)$ exist, but they are not equal.

For a function of two variables overall

It is possible to have a function $f$ of two variables $x,y$ such that both the second-order mixed partial derivatives $f_{xy}(x,y)$ and $f_{yx}(x,y)$ exist everywhere on $\R^2$ but they are not equal as functions, i.e., there exists a point where the values of the second-order mixed partial derivatives are not equal.

Related facts

• Failure of Clairaut's theorem where only one of the mixed partials is defined
• Separately continuous not implies continuous
• Continuous in every linear direction not implies continuous
• Existence of partial derivatives not implies differentiable
• Existence of directional derivatives in every direction not implies differentiable

Proof

Example

Consider the function:

$f(x,y) := \left\lbrace \begin{array}{rl} \frac{xy(x^2 - y^2)}{x^2 + y^2}, & (x,y) \ne (0,0) \\ 0, & (x,y) = (0,0) \\\end{array}\right.$

We do some computations:

Item	Value	Explanation
$f_x(x,y)$ for $(x,y) \ne (0,0)$	$\frac{(x^2 + y^2)(3x^2y - y^3) - 2x^2y(x^2 - y^2)}{(x^2 + y^2)^2} = \frac{x^4y + 4x^2y^3 - y^5}{(x^2 + y^2)^2}$	use the quotient rule for differentiation and simplify
$f_y(x,y)$ for $(x,y) \ne (0,0)$	$\frac{(x^2 + y^2)(x^3 - 3xy^2) - 2xy^2(x^2 - y^2)}{(x^2 + y^2)^2} = \frac{x^5 - 4x^3y^2 - xy^4}{(x^2 + y^2)^2}$	use the quotient rule for differentiation and simplify
$f_x(0,y)$ for $y \ne 0$	$-y$	plug in $x = 0$ in the general expression for $f_x(x,y)$ and simplify.
$f_y(x,0)$ for $x \ne 0$	$x$	plug in $y = 0$ in the general expression for $f_y(x,y)$ and simplify.
$f_x(0,0)$	0	start with $\lim_{x \to 0} \frac{f(x,0) - f(0,0)}{x}$ and simplify, noting that the numerator is identically zero
$f_y(0,0)$	0	start with $\lim_{y \to 0} \frac{f(0,y) - f(0,0)}{y}$ and simplify, noting that the numerator is identically zero
$f_{xy}(0,0)$	-1	start with $\lim_{y \to 0} \frac{f_x(0,y) - f_x(0,0)}{y}$ and simplify using the expressions obtained above for $f_x(0,y)$ and $f_x(0,0)$
$f_{yx}(0,0)$	1	start with $\lim_{x \to 0} \frac{f_y(x,0) - f_y(0,0)}{x}$ and simplify using the expressions obtained above for $f_y(x,0)$ and $f_y(0,0)$