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

From Calculus
Jump to: navigation, search

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).