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Let $f:\Bbb{R^2}\rightarrow\Bbb{R}$ and two real parameters $a,b$ such that $$f (\mathbf{x,y})= \begin{cases}a(x^2+y^2),&(x,y)\in B_{d_{2}}(0;2)\\ \frac{b}{\sqrt{x^2+y^2}},&(x,y)\in \Bbb{R^2}- B_{d_{2}}(0;2)\end{cases}$$ $B_{d_{2}}(0;2) $ represents the ball centered at $(0,0)$ with radius 2, with respect to the Euclidean metric $d_2$ on $\Bbb{R^2}$. We have to check the uniform continuity of $f$ on $B_{d_{2}}(0;3)$.

My attempt: I start by observing the fact that $B_{d_{2}}(0;3)$ is a compact set (Heine-Borel theorem), hence proving just continuity is enough. Any ideas on how to prove continuity?

EDIT Using polar coordinates $$f(r)=\begin{cases}ar^2,&r\in(0,2)\\ \frac{1}{r}b,&r\in\Bbb{R^+}-(0,2) \end{cases} $$ I need to check continuity in $ 0,2$ respectively. Correct? But it's not continuous.

EDIT2 After some thought we need to check $$\lim_{r\to 2 }f(r)$$ I think it all comes down to $4a=\frac{1}{2}b$. Hence f is continuous iff $b=8a$ Is it now correct?

prometheus21
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  • you have to check what happens in those points where the two different definitions apply in a neighbourhood. – Thomas Jan 04 '15 at 14:18
  • I think the question is actually asking (in a rather oblique way) which values of $a$ and $b$ make $f$ continuous. It isn't continuous for every pair of values. – Rob Arthan Jan 04 '15 at 14:24
  • It isn't continuous for all $a$ and $b$, but what if they are chosen so that $ar^2 = \frac{1}{r}b$ when $r = 2$? (Note that $r \ge 0$.) – Rob Arthan Jan 04 '15 at 14:38
  • Your latest edit correctly gives $b = 8a$ as the condition for the function to be continuous on $\mathbb{R}^2$ and hence uniformly continuous on any bounded subset of $\mathbb{R}^2$. – Rob Arthan Jan 04 '15 at 14:52
  • Thank you Rob Arthan & Ivo Terek for the help in solving this. – prometheus21 Jan 04 '15 at 14:54

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Hints: First: $B_{d_2}(0,3)$ is not compact, it is open. But $B_{d_2}[0,3]$ is compact. You need to check continuity only on the boundary of $B_{d_2}(0,2)$: use polar coordinates.

Ivo Terek
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