Continuity of $f(x,y) = \begin{cases} xy & \text{if} \ xy >0 \\ 0 & \text{if} \ xy \leq 0 \end{cases}$

Question

Where is the following function continuous on $\mathbb{R}^2$?

$f(x,y) = \begin{cases} xy & \text{if} \ xy >0 \\ 0 & \text{if} \ xy \leq 0 \end{cases}$

Is it continuous on positive values of $x$ and $y$?

Answer 1

Clearly, both $xy$ and $0$ are continuous functions, hence $f(x,y)$ is continuous for $xy > 0$ and $xy \leq 0.$ Our main concern here is what happens when we transition from $f(x, y) = xy$ to $f(x, y) = 0.$ Observe that this occurs precisely when the sign of $x$ or $y$ changes. Explicitly, we have that $xy > 0$ if and only if $x, y > 0$ or $x, y < 0,$ and likewise, we have that $xy \leq 0$ if and only if $x \leq 0$ and $y > 0$ or $x > 0$ and $y \leq 0,$ so when the sign of $x$ or $y$ changes, then $f(x, y)$ changes.

But in order for either $x$ or $y$ to change sign, there must exist a neighborhood in which $|x|$ or $|y|$ is arbitrarily small, hence the distance $|xy - 0|$ must be arbitrarily small, i.e., $f(x, y)$ is continuous.