Showing that a function is continuous, using polar coordinates.

Question

I have to decide, if $f:\mathbb{R}^2\to\mathbb{R}$, $f(x,y)=\begin{cases} xy\frac{x^2-y^2}{x^2+y^2}\space\text{if}\, (x,y)\neq (0,0)\\ 0\space\text{else}\end{cases}$

is continuous or not.

I sketched the function with wolframalpha, and it looks continuous at $(0,0)$, hence I want to prove it. Obviously we just have to take a closer look at what happens at $(0,0)$.

I tried to involve polar coordinates, because $f$ looks like it could help. But I am not that familiar with polar coordinates, and would like to get some feedback:

$x=r\sin(\varphi), y=r\cos(\varphi)$ with $r\in\mathbb{R}$ and $-2\pi\leq\varphi\leq 2\pi$.

Then:

$r^2\sin(\varphi)\cos(\varphi)\frac{r^2(\sin(\varphi)^2-\cos(\varphi)^2)}{r^2(\sin(\varphi)^2+\cos(\varphi)^2)}=r^2\sin(\varphi)\cos(\varphi)\frac{r^2(\sin(\varphi)^2-\cos(\varphi)^2)}{r^2\cdot 1}\\=r^2\sin(\varphi)\cos(\varphi)\cdot (\sin(\varphi)^2-\cos(\varphi)^2)$

Using addition theorems such as $\sin(x)\cos(x)=\frac12\sin(2x)$ and $\sin(x)^2-\cos(x)^2=-\cos(2x)$

This simplifies to:

$=-\frac{r^2}{4}\sin(4\varphi)$

But how to proceed from here? Do I use $\epsilon$-$\delta$-criterion as always?

Talking just about the use of polar coordinates (no matter if it is useful or not), have I done it correct so far, in formal terms?

Thanks in advance for your comments/answers.

Answer 1

$|\sin 4\phi| <\le 1\\ d((x,y),(0,0)) = r$

$r<\delta \implies |f(x,y)|<\frac {r^2}{4}$

let $\delta = \sqrt {4\epsilon}$

$\forall \epsilon>0, \exists \delta >0: \ d((x,y),(0,0)) < \delta \implies d(f(x,y),(0,0)) \le \epsilon$

Answer 2

As you have shown in detail, by polar coordinates we have that

$$xy\frac{x^2-y^2}{x^2+y^2}=r^2\cdot f(\varphi) $$

and by squeeze theorem, indicating with $M$ an upper bound for $|f(\varphi)|$

$$0\le |r^2\cdot f(\varphi)|\le M\cdot r^2 \to 0$$

indeed as $(x,y)\to (0,0)$ we have $r^2=x^2+y^2 \to 0$.

Therefore $f(x,y)$ is continuous at $(0,0)$.

Answer 3

The way I would show this is that given an arbitrary $\epsilon>0$, the goal should be to find a neighborhood around $(0,0)$ such that $|f(x,y)|<\epsilon$ in this neighborhood. This will prove continuity since it is equivalent to the epsilon delta definition of continuity.

So, we know that no matter what, $|f(0,0)|<\epsilon\quad\forall\epsilon>0$, so we can check only its neighbors.

The appropriate neighborhood to pick would be $r=\sqrt{\epsilon}$. You can do the rest of the math from here relatively easily.

Answer 4

An option:

Let $(x,y)\not =0$.

$f(x,y) := |xy\dfrac{x^2-y^2}{x^2+y^2}| \le $

$(1/2)(x^2+y^2)\dfrac{(|x^2| + |y^2|)}{x^2+y^2}$

$\le (1/2)(x^2+y^2).$

Le $\epsilon >0$ be given.

Choose $\delta =√2\sqrt{\epsilon}$, then

$\sqrt{x^2+y^2} \lt \delta$ implies

$f(x,y) \le (1/2)(x^2+y^2) \lt (1/2)(\delta)^2 =\epsilon.$

Used: $x^2 +y^2 \ge 2|xy|,$ and

$|x^2-y^2| \le |x^2|+|y^2|.$