Is this multivariable function uniformly continuous?

Question

$f(x,y)=\sin{\frac{\pi}{1-(x^{2}+y^{2})}}$

I concluded that it probably is uniformly continuous. Since: if $\sqrt{a^{2}+b^{2}}<\delta$, then $\frac{1}{1-\sqrt{x^{2}+y^{2}}}<\frac{1}{\delta}$. Then I can conclude the thesis for $\delta=\frac{2}{\epsilon}$. Are my thoughts correct?

Answer 1

It is uniformly continuous on any compact set inside the unit circle. This is true of any function that is continuous on a compact set.

However, $f$ is not uniformly continuous in the entire open unit disk $D = \{(x,y):x^2 + y^2 < 1\}$. This is shown by producing sequences of points $p_n,q_n \in D$ such that $\lim_{n \to \infty}\|p_n - \hat{p}_n\| = 0$ but $\lim_{n \to \infty}\|f(p_n) - f(\hat{p}_n)\| \neq 0$.

Switching to polar coordinates we have $f(r,\theta) = \sin \frac{\pi}{1 - r^2}$. Choose the sequences $p_n = (r_n,\theta_n)$ and $\hat{p}_n = (\hat{r}_n,\hat{\theta}_n)$ where $\theta_n = \hat{\theta}_n$ and

$$r_n = \sqrt{\frac{2n-1/2}{2n + 1/2}} \implies \frac{\pi}{1- r_n^2} = 2n\pi + \pi/2 \implies \sin \frac{\pi}{1 - r_n^2} = 1 \\ \hat{r}_n = \sqrt{\frac{2n-1}{2n }} \implies \frac{\pi}{1- \hat{r}_n^2} = 2n\pi \implies \sin \frac{\pi}{1 - \hat{r}_n^2} = 0 .$$

Thus,

$$\lim_{n \to \infty}\|p_n - \hat{p}_n\| = \lim_{n \to \infty}|r_n - \hat{r_n}| = \lim_{n \to \infty} \left|\sqrt{\frac{2n-1/2}{2n + 1/2}} - \sqrt{\frac{2n-1}{2n }}\right| = |1-1| = 0,$$

but

$\lim_{n \to \infty} |f(r_n,\theta_n) - f(\hat{r}_n,\hat{\theta}_n)| = 1$

Answer 2

Hint: If $f$ were uniformly continuous on the open unit disc, then $x\to f(x,0) = \sin(\pi/(1-x))$ would be uniformly continuous on $[0,1).$