Let $f : \mathbb{R}^2 \setminus \{0\} \to \mathbb{R}$ be defined by $$f(x_1, x_2) = \frac{x_1x_2}{x_1^2 + x_2^2}. $$

I want to show that f is uniformly continuous on $\{x \in \mathbb{R}^n : \left\|x\right\| \geq r\}$ where $r$ is some positive constant.

I can see that $f(\alpha x) = f(x)$ for any $\alpha \neq 0$, so $f$ is constant on all lines through origin.

So my thought process was this:

I wanted to use the compactness of the circle of radius $r$ centered ta origin to show that $f$ was uniformly continuous on that circle and then somehow show that if two arbitrary points $x$ and $y$ in$\{x \in \mathbb{R}^n : \left\|x\right\| \geq r\}$ are close enough ($\left\|x - y\right\| < \delta$) and not on the circle, then "bringing back" (taking $x \rightarrow rx/\left\|x\right\|$ and $y \rightarrow ry/\left\|y\right\|$) the points along a line through origin to points $p$ and $q$ on the circle of radius $r$ centered at origin will ensure that $p$ and $q$ are as close as I need (I mean $\left\|p - q\right\| < \delta$, where $\delta > 0$ such that if $p$ and $q$ are on the circle and $\left\|p - q\right\| < \delta$ then $\lvert f(p) - f(q)\rvert < \varepsilon$) to show that $\lvert f(x) - f(y)\rvert = \lvert f(p) - f(q)\rvert < \varepsilon$).

But I'm not able to show that "bringing back" $x$ and $y$ to $p$ and $q$ in the manner I mentioned above will ensure $\left\|p - q\right\| \leq \left\|x - y\right\|$.

Am I going about this the wrong way? Or am I missing something?


2 Answers 2


An easy way forward here is to change to polar coordinates and apply the prosthaphaeresis formula.

Consider two points $x = (x_1,x_2)$ and $y = (y_1,y_2)$ with $x_1 = r_x \cos \theta_x$, $x_2 = r_x \sin \theta_x$, $y_1 = r_y \cos \theta_y$, and $y_2 = r_y \sin \theta_y$.

We have

$$\begin{align}\left|\frac{x_1x_2}{x_1^2 + x_2^2} - \frac{y_1y_2}{y_1^2 + y_2^2}\right| &= \left|\cos \theta_x \sin \theta_x - \cos \theta_y \sin \theta_y \right| \\ &= \frac{1}{2}\left|\sin 2\theta_x - \sin 2\theta_y \right| \\ &= \frac{1}{2}2\left|\sin \frac{(2\theta_x - 2\theta_y)}{2} \right|\left|\cos \frac{(2\theta_x + 2\theta_y)}{2} \right| \\ &\leqslant \left|\sin \frac{(2\theta_x - 2\theta_y)}{2} \right| \\ &\leqslant |\theta_x - \theta_y|\end{align}$$

Suppose $\|x - y\| \leqslant \delta.$ Then $y$ is inside a disk with center $x$ and radius $\delta$. The largest deviation in $\theta_x - \theta_y$ occurs at two points on the bounding circle where the ray with angle $\theta_y$ emanating from the origin is tangent to the circle. In each case, we have a right triangle with hypoteneuse $r_x$ and side $\delta$. Hence, the included angle has sine equal to $\delta / r_x$, and when $\|x\| \geqslant r$,

$$|\theta_x - \theta_y| \leqslant \arcsin \frac{\delta}{r_x} = \arcsin \frac{\delta}{\|x\|}\leqslant \arcsin \frac{\delta}{r}. $$

Choosing $\delta < r \sin \epsilon$, we have

$$|f(x) - f(y)| \leqslant |\theta_x - \theta_y| \leqslant \arcsin \frac{\delta}{r} < \epsilon.$$

for all $x,y \in \{x \in \mathbb{R}^2 : \left\|x\right\| \geq r\} $ with $\|x-y\| < \delta$.

  • $\begingroup$ I initially thought I could easily work it out from here, but after numerous failed attempts I realized that in my initial attempt, I got stuck on this step exactly: showing that if $\left\|x - y\right\| < \delta$ then $\left\|\theta_x - \theta_y\right\| < \varepsilon$ for some $\delta > 0$ depending only on $\varepsilon$ and $r$. Could you please elaborate this? $\endgroup$
    – Anamaki
    Jul 20, 2016 at 11:24
  • $\begingroup$ @theakholic: Sure -- I will add to the answer. $\endgroup$
    – RRL
    Jul 20, 2016 at 15:31

Here's another approach: Let $U= \{|x| >r\}, f(x_1,x_2) = x_1x_2/(x_1^2+x_2^2).$ Verify that the partial derivatives of $f$ are bounded in $U.$ Thus $\nabla f$ is bounded in $U.$

Recall the following: If $g$ is differentiable in an open convex set $V,$ and $\nabla g$ is bounded in $V,$ then $g$ is Lipschitz in $V,$ hence $g$ is uniformly continuous in $V.$

If $f$ is not uniformly continuous in $U,$ then there exists $\epsilon>0$ and points $x_n,y_n \in U$ such that $|y_n-x_n| \to 0$ and $|f(y_n)-f(x_n)| \ge \epsilon.$ If $x_n$ is bounded, then some subsequence $x_{n_k}$ converges to a point $x \in \overline U.$ The subsequence $y_{n_k}$ then also converges to $x.$ But $f$ is continuous on $\overline U,$ contradiction. If $x_n$ is not bounded, then some subsequence $x_{n_k}$ lies in one of the four closed quadrants with $|x_{n_k}| \to \infty.$ It follows that both $x_{n_k},y_{n_k}$ both lie in a fixed open half plane for large $k,$ violating the second paragraph, contradiction. These contradictions prove $f$ is uniformly continuous in $U.$


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