$f(x,y)=\frac{(x,y)}{\|(x,y)\|^2}$ - Show that $f$ is not uniformly continuous

Question

Let $f: \mathbb{R^2}-\{(0,0)\} \to \mathbb{R^2}$ difined as $f(x,y)=\frac{(x,y)}{\|(x,y)\|^2}$. Show that $f$ is not uniformly continuous.

I tried to solve it in using the Cauchy-Schwarz inequality, but something blocked. I think at the end, we can use Geometric-Mean inequality, but it's unclear.

Is anyone could give me a hint to solve this problem?

Answer 1

Consider the restriction of $f$ to the positive real line, say, $g(x) = 1/x$. Show that this function is not uniformly continuous, by writing \begin{equation} g(x_1) - g(x_2) = \frac{x_2 - x_1}{x_1 x_2} \end{equation} and letting $x_1, x_2$ tend to zero.

Answer 2

Hint: Consider $f(1/4^n,0) - f(1/2^n,0)$ as $n\to \infty.$

Answer 3

Let $\varepsilon=1$ then there is no $\delta>0$ such that any two points within $\delta$ (and neither being $(0,0)$) have their $f$ images within $1$, since one can consider diametrically opposite point $(\pm a,0)$ which for small enough nonzero $a$ are within $\delta$ of each other, yet their $f$ images are $(1,0),(-1,0)$ and are $2$ units apart.

Edit: I didn't notice the denominator is squared, to that the points $(\pm a,0)$ I used actually map to $(\pm(1/a),0).$ For small $a$ this distance is still greater than 1...