Symmetry of Mixed Partial Derivatives (Clairaut’s Theorem)

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

Theorem (Clairaut Schwarz)

Let \(f: \mathbb{R}^2 \to \mathbb{R}\) be a function such that both mixed partial derivatives \(\frac{\partial^2 f}{\partial x \partial y}\) and \(\frac{\partial^2 f}{\partial y \partial x}\) exist and are continuous on an open set containing a point \((x_0, y_0)\)

Then:

\[ \boxed{ \frac{\partial^2 f}{\partial x \partial y}(x_0, y_0) = \frac{\partial^2 f}{\partial y \partial x}(x_0, y_0) } \]

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 \((x_0, y_0)\).

Define:

\[ F(h,k) = \frac{f(x_0 + h, y_0 + k) - f(x_0 + h, y_0) - f(x_0, y_0 + k) + f(x_0, y_0)}{hk} \]

Then, as \(h, k \to 0\), \(F(h,k) \to \frac{\partial^2 f}{\partial y \partial x}(x_0, y_0)\) and also \(F(h,k) \to \frac{\partial^2 f}{\partial x \partial y}(x_0, y_0)\), 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:

    \[ \lim_{(h,k) \to (0,0)} F(h,k) = \frac{\partial^2 f}{\partial x \partial y}(x_0, y_0) \]

    and also

    \[ \lim_{(h,k) \to (0,0)} F(h,k) = \frac{\partial^2 f}{\partial y \partial x}(x_0, y_0) \]

    due to continuity of the second derivatives.

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

Therefore:

\[ \frac{\partial^2 f}{\partial x \partial y}(x_0, y_0) = \frac{\partial^2 f}{\partial y \partial x}(x_0, y_0) \]

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:

\[\begin{split} f(x, y) = \begin{cases} \frac{xy(x^2 - y^2)}{x^2 + y^2}, & \text{if } (x, y) \neq (0, 0) \\ 0, & \text{if } (x, y) = (0, 0) \end{cases} \end{split}\]

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

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