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 $\mathbb {R} ^{2}$ 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:

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

We do some computations:

Item	Value	Explanation
$f_{x}(x,y)$ for $(x,y)\neq (0,0)$	${\frac {(x^{2}+y^{2})(3x^{2}y-y^{3})-2x^{2}y(x^{2}-y^{2})}{(x^{2}+y^{2})^{2}}}={\frac {x^{4}y+4x^{2}y^{3}-y^{5}}{(x^{2}+y^{2})^{2}}}$	use the quotient rule for differentiation and simplify
$f_{y}(x,y)$ for $(x,y)\neq (0,0)$	${\frac {(x^{2}+y^{2})(x^{3}-3xy^{2})-2xy^{2}(x^{2}-y^{2})}{(x^{2}+y^{2})^{2}}}={\frac {x^{5}-4x^{3}y^{2}-xy^{4}}{(x^{2}+y^{2})^{2}}}$	use the quotient rule for differentiation and simplify
$f_{x}(0,y)$ for $y\neq 0$	$-y$	plug in $x=0$ in the general expression for $f_{x}(x,y)$ and simplify.
$f_{y}(x,0)$ for $x\neq 0$	$x$	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 $f_{y}(0,0)$

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