Differentiability of $f(x, y) = |xy|$ at $0$; is this a mistake by Munkres?

Question

I am reading Analysis on Manifolds by Munkres, and question $1$ on page $54$ says

Show that the function $f(x, y) = |xy|$ is differentiable at $0$, but it is not of class $C^1$ in any neighborhood of $0$.

I know that if a function $f$ is differentiable at a point, then all its partial derivatives exist at that point. However,

$$D_1f(x, y) = \begin{cases} |y|&\text{if}\, x>0 \\ -|y| &\text{if}\, x<0 \end{cases} $$

and $D_1f(0, 0)$ does not exist. How can $f$ be differentiable at $0$?

Answer 1

We have that

$$\frac{x^2y^2}{x^2+y^2}\leq x^2+y^2,$$

so if $||(x,y)||=r$ then

$$\frac{|xy|}{||(x,y)||}=\sqrt{\frac{x^2y^2}{x^2+y^2}}\leq r,$$

and as such

$$\lim_{x,y\to 0} \frac{|xy|}{||(x,y)||}=0$$

as $(x,y)\to(0,0)$ implies that $||(x,y)||\to 0$. The reason your statement does not imply nondifferentiability is that the fact that

$$D_1f(x, y) = \begin{cases} |y|&\text{if}\, x>0 \\ -|y| &\text{if}\, x<0 \end{cases} $$

is only problematic when $y$ is away from $0$, otherwise $|y|\approx-|y|$. The partials still very much exist at $f(0,0)$, as $|y|,-|y|\to 0$. However, you can use your result to conclude that $D_1 f(x,y)$ does not exist if $y\neq 0$.

Answer 2

How did you conclude that $D_1f(0,0)$ does not exist? Try using the definition directly. You are probably confusing differentiable and continuously differentiable.

Answer 3

Since the first part of your question has been addressed already, let me address the second part.

By any neighbourhood of $(0,0)$, I assume you mean any open neighbourhood of $(0,0)$, which should also include the whole plane $\Bbb R^2$. Let $U\subseteq\Bbb R^2$ be an arbitrary open neighbourhood of $(0,0)$. Since $U$ is open and $(0,0)\in U,$ there is some $r>0,$ s. t. $B((0,0),r)\subseteq U,$ but $B((0,0),r)$ must contain some points on the $x$ and $y$ axis. So, let's take some point $(x_0,0)\in B((0,0)r)\setminus\{(0,0)\}.$ Then the limits $$\lim_{h\to 0^+}\frac{f(x_0,0+h)-f(x_0,0)}h=\lim_{h\to 0^+}\frac{|(x_0(0+h)|-|x_0\cdot 0|}h=|x_0|$$ and $$\lim_{h\to 0^-}\frac{f(x_0,0+h)-f(x_0,0)}h=-|x_0|$$ are different, so $\frac{\partial f}{\partial y}(x_0,0)$ doesn't exist. Similarly, $\frac{\partial f}{\partial x}(0,y_0)$ doesn't exist for any $y_0\ne 0$, hence, $f$ isn't differentiable at any point on the axes, exept for the origin (that part has been proven).