Symmetry of Mixed Partial Derivatives (Clairaut’s Theorem)

Symmetry of Mixed Partial Derivatives (Clairaut’s Theorem)#

Theorem (Clairaut Schwarz)

Let f:R2R be a function such that both mixed partial derivatives 2fxy and 2fyx exist and are continuous on an open set containing a point (x0,y0)

Then:

2fxy(x0,y0)=2fyx(x0,y0)

That is, the order of differentiation can be interchanged.

Intuition#

If a function is smooth enough (specifically, if the second-order partial derivatives exist and are continuous), then the “curvature” in the x direction after differentiating in the y direction is the same as the curvature in the y direction after differentiating in the x direction.


Proof Sketch#

We will sketch a proof using the mean value theorem and the definition of partial derivatives. Let’s assume that f has continuous second partial derivatives in an open rectangle around the point (x0,y0).

Define:

F(h,k)=f(x0+h,y0+k)f(x0+h,y0)f(x0,y0+k)+f(x0,y0)hk

Then, as h,k0, F(h,k)2fyx(x0,y0) and also F(h,k)2fxy(x0,y0), provided the second partial derivatives are continuous.

Step-by-step:#

  1. By the Mean Value Theorem, the numerator of F(h,k) can be interpreted as a finite difference approximation to a mixed partial derivative.

  2. Using Taylor’s Theorem with remainder, or via integral representations of derivatives, one can show that:

    lim(h,k)(0,0)F(h,k)=2fxy(x0,y0)

    and also

    lim(h,k)(0,0)F(h,k)=2fyx(x0,y0)

    due to continuity of the second derivatives.

  3. Hence, the limits agree and the mixed partials are equal.

Therefore:

2fxy(x0,y0)=2fyx(x0,y0)

When Clairaut’s Theorem Does Not Apply#

If the second-order mixed partial derivatives exist but are not continuous, the symmetry may fail. A classic counterexample is:

f(x,y)={xy(x2y2)x2+y2,if (x,y)(0,0)0,if (x,y)=(0,0)

This function has both mixed partial derivatives at the origin, but they are not equal because they are not continuous there.