Directional Derivatives of Functions from Rn to Rm and Continuity

Recall that if $f : S \to \mathbb{R}$ where $S \subseteq \mathbb{R}$ is open, then $f$ being differentiable at a point $x_0 \in S$ implies that $f$ is continuous at $x_0$ (the converse being false).

We would like to develop a derivative definition for functions $\mathbf{f} : S \to \mathbb{R}^m$ where $S \subseteq \mathbb{R}^n$ is open that preserves this nice property. Recall from the Directional Derivatives of Functions from Rn to Rm page that the directional derivative of $\mathbf{f}$ at $\mathbf{c}$ in the direction of the vector $\mathbf{u} \in \mathbb{R}^n$ is defined as the following limit provided that it exists:

(1)

\begin{align} \quad \mathbf{f}'(\mathbf{c}, \mathbf{u}) = \lim_{h \to 0} \frac{\mathbf{f}(\mathbf{c}+ h\mathbf{u}) - \mathbf{f}(\mathbf{c})}{h} \end{align}

One might ask, does the existence of all directional derivatives of a function $\mathbf{f}$ at $\mathbf{c}$ imply that $\mathbf{f}$ is continuous at $\mathbf{c}$? The answer is sadly NO. For example, consider the following function $f : \mathbb{R}^2 \to \mathbb{R}$ defined by:

(2)

\begin{align} \quad f(x, y) = \left\{\begin{matrix} \frac{x^2y}{(x^4 + y^2)} & \mathrm{if} \: (x, y) \neq (0, 0) \\ 0 & \mathrm{if} \: (x, y) = (0, 0) \end{matrix}\right. \end{align}

Let's compute the directional derivatives of this function at the origin $\mathbf{c} = (0, 0)$. Let $\mathbf{u} = (u_1, u_2) \in \mathbb{R}^2$. Then:

(3)

\begin{align} \quad f' (\mathbf{c}, \mathbf{u}) &= \lim_{h \to 0} \frac{f(\mathbf{c} + h\mathbf{u}) - f(\mathbf{c})}{h} \\ &= \lim_{h \to 0} \frac{f(hu_1, hu_2) - f(0, 0)}{h} \\ &= \lim_{h \to 0} \frac{\frac{(hu_1)^2(hu_2)}{(hu_1)^4 + (hu_2)^2} - 0}{h} \\ &= \lim_{h \to 0} \frac{\frac{h^3u_1^2u_2}{h^2(h^2u_1^4 + u_2^2)}}{h} \\ &= \lim_{h \to 0} \frac{h^3u_1^2u_2}{h^3(h^2u_1^4 + u_2^2)} \\ &= \lim_{h \to 0} \frac{u_1^2u_2}{h^2u_1^4 + u_2^2} \\ &= \frac{u_1^2}{u_2} \end{align}

Notice that $f'(\mathbf{c}, \mathbf{u})$ exists whenever $u_2 \neq 0$. Suppose now that $u_2 = 0$. I.e., let's compute the directional derivative of $f$ at $\mathbf{c}$ in the direction of $\mathbf{u} = (1, 0)$ (the partial derivative of $f$ with respect to the first variable). Then:

(4)

\begin{align} \quad f'(\mathbf{c}, \mathbf{u}) &= \lim_{h \to 0} \frac{f(\mathbf{c} + h\mathbf{u}) - f(\mathbf{c})}{h} \\ &= \lim_{h \to 0} \frac{f(hu_1, 0) - 0}{h} \\ &= \lim_{h \to 0} \frac{\frac{(hu_1)^2(0)}{(hu_1)^4 + (0)^2}}{h} \\ &= 0 \end{align}

So indeed all directional derivatives of $f$ at $\mathbf{c}$ exist. However, we claim that $f$ is discontinuous at $\mathbf{c}$. We first give a graph of $f$ below to visualize the discontinuity at $\mathbf{c} = (0, 0)$:

Now, along the line $y = x^2$ we have that:

(5)

\begin{align} \quad \lim_{x \to 0, y=x^2} f(x, y) &= \lim_{x \to 0} f(x, x^2) \\ &= \lim_{x \to 0} \frac{x^4}{x^4 + x^4} \\ &= \lim_{x \to 0} \frac{x^4}{2x^4} \\ &= \frac{1}{2} \\ \end{align}

But $\displaystyle{f(0, 0) = 0 \neq \frac{1}{2}}$ so $f$ is discontinuous at $\mathbf{c} = (0, 0)$ despite having all of its directional derivatives existing at $\mathbf{c}$.