Equality/Inequality of Mixed Part. Deriv. of Functs. from Rn to Rm

Equality and Inequality of Mixed Partial Derivatives of Functions from Rn to Rm

On the Higher Order Partial Derivatives of Functions from Rn to Rm we defined the notion of higher order partial derivatives for a function. We will now focus our attention on the equality / inequality of mixed partial derivatives of a function.

Let $f : \mathbb{R}^2 \to \mathbb{R}$ be defined by $f(x, y) = xy^2$. Then the partial derivatives of $f$ are:

(1)

\begin{align} \quad D_1 f(x, y) = y^2 \quad , \quad D_2 f(x, y) = 2xy \end{align}

The second order partial derivatives of $f$ are:

(2)

\begin{align} \quad D_{1, 1} f(x, y) &= 0 \quad , \quad D_{2, 1} f(x, y) &= 2y \\ \quad D_{1, 2} f(x, y) &= 2y \quad , \quad _{2, 2} &= 2x \\ \end{align}

Notice that $D_{1, 2} f(x, y) = D_{2, 1} f(x, y)$. One might ask if this is true in general. The answer unfortunately is not.

For example, consider the function $f : \mathbb{R}^2 \to \mathbb{R}$ defined by:

(3)

\begin{align} \quad f(x, y) = \left\{\begin{matrix} xy \frac{x^2 - y^2}{x^2 + y^2}& \mathrm{if} \: (x, y) \neq (0, 0)\\ 0 & \mathrm{if} \: (x, y) = (0, 0) \end{matrix}\right. \end{align}

Then the first order partial derivatives of $f$ are:

(4)

\begin{align} \quad D_1 f(x, y) = \left\{\begin{matrix} y \frac{(x^2 + y^2)(3x^2 - y^2) - (x^3 - xy^2)(2x + y^2)}{(x^2 + y^2)^2}& \mathrm{if} \: (x, y) \neq (0, 0)\\ 0 & \mathrm{if} \: (x, y) = (0, 0) \end{matrix}\right. \end{align}

(5)

\begin{align} D_2 f(x, y) = \left\{\begin{matrix} x\frac{(x^2 + y^2)(x^2 - 3y^2) - (x^2y - y^3)(x^2 + 2y)}{(x^2 + y^2)^2}& \mathrm{if} \: (x, y) \neq (0, 0)\\ 0 & \mathrm{if} \: (x, y) = (0, 0) \end{matrix}\right. \end{align}

We now compute the mixed partial derivatives of $f$ at $0$. We have that:

(6)

\begin{align} \quad D_{2, 1} f(0,0) = \lim_{h \to 0} \frac{f_1(0, h) - f_1(0, 0)}{h} = \lim_{h \to 0} \frac{\frac{-h^5}{h^4} - 0}{h} = -1 \end{align}

And also:

(7)

\begin{align} \quad D_{1, 2} f(0, 0) = \lim_{h \to 0} \frac{f_2(h, 0) - f(0, 0)}{h} = \lim_{h \to 0} \frac{\frac{h^5}{h^4}-0}{h} = 1 \end{align}

Therefore $D_{1, 2} f(0, 0) \neq D_{2, 1} f(0, 0)$!

So mixed partial derivatives of a function need not be equal. We will now state a sufficient condition for the equality of mixed partial derivatives.

Theorem 1: Let $S \subseteq \mathbb{R}^n$ be open, $\mathbf{c} \in S$, and $\mathbf{f} : S \to \mathbb{R}^m$. If: 1) $D_j \mathbf{f}$ and $D_k \mathbf{f}$ exist on an open ball centered at $\mathbf{c}$, $B(\mathbf{c}, r)$. 2) $D_j \mathbf{f}$ and $D_k \mathbf{f}$ are differentiable at $\mathbf{c}$. Then $D_{j, k} \mathbf{f}(\mathbf{c}) = D_{k, j} \mathbf{f} (\mathbf{c})$.