# Prove that the function $\sqrt{x^2+y^2}\sin \frac{1}{\sqrt{x^2+y^2}}$ is uniformly continuous in $\mathbb R^2$

Prove that the function $$f(x,y) = \begin{cases}\sqrt{x^2+y^2}\sin \frac{1}{\sqrt{x^2+y^2}} & \text{if}\ (x,y) \ne (0,0) \\ 0 & \text{if}\ (x,y) = (0,0) \end{cases}$$

is uniformly continuous over $$\mathbb{R}^2$$.

I know that for the function to be uniformly continuous it should hold the property

For ever $$\epsilon > 0$$ there exist $$\delta(\epsilon) > 0$$ such that for every $$x, y$$ such that $$d(x,y) < \delta$$ it exist that: $$|f(x) - f(y)| < \epsilon$$

where $$x=(x_1,y_1)$$ ,$$y=(x_2,y_2)$$ $$\in \mathbb R^2$$

From $$|f(x) - f(y)| = \left|\sqrt{x_1^2+y_1^2}\sin \frac{1}{\sqrt{x_1^2+y_1^2}}-\sqrt{x_2^2+y_2^2}\sin \frac{1}{\sqrt{x_2^2+y_2^2}}\right|$$

but I don't where to go from here

Can somebody help me with this problem, because I don't really know how to prove it?

• This function is not defined at $(0,0).$ – zhw. Aug 25 at 18:57
• I guess the O.P. has the definition as above for all $(x,y) \in \mathbb{R}^2, (x,y) \ne (0,0)$ and $0$ for $(x,y) = (0,0)$. – Rick Aug 25 at 19:01
• @Rick you're right I'll edit it – J.Dane Aug 25 at 19:02

An idea that can make things way easier: substitute $$\;t:=\sqrt{x^2+y^2}\;$$ , and now check that $$\;t\to 0\iff (x,y)\to (0,0)\;$$ , so you can take your function as

$$f(t)=\begin{cases}t\sin\frac1t\,&t\neq0\\{}\\0,&t=0\end{cases}$$

so what you need to check is simply

$$\text{If}\;\;\lim_{n\to\infty}|t_n-s_n|=0\;,\;\;\text{then}\;\lim_{n\to\infty}|f(t_n)-f(s_n)|=0$$

and we have here

$$|f(t_n)-f(s_n)|=\left|t_n\sin\frac1{t_n}-s_n\sin\frac1{s_n}\right|=\left|t_n\sin\frac1{t_n}-s_n\sin\frac1{t_n}+s_n\sin\frac1{t_n}-s_n\sin\frac1{s_n}\right|\le$$

$$\le\left|t_n-s_n\right|\left|\sin\frac1{t_n}\right|+\left|\sin\frac1{t_n}-\sin\frac1{s_n}\right||s_n|\;(**)$$

But we have that

$$\sin\frac1{t_n}-\sin\frac1{s_n}=2\,\sin\frac{\frac1{t_n}-\frac1{s_n}}2\;\cos\frac{\frac1{t_n}+\frac1{s_n}}2=2\,\sin\frac{s_n-t_n}{2t_ns_n}\;\cos\frac{t_n+s_n}{2t_ns_n}\xrightarrow[n\to\infty]{}0$$

as the last one is the limit of a sequence converging to zero times a bounded one, and thus

$$(**)\xrightarrow[n\to\infty]{}0\cdot\left|\sin\frac1{t_n}\right|+0\cdot|s_n|=0$$

and we're done.

Hint: Once you define the function properly at $$(0,0),$$ it will be continuous on $$\mathbb R^2.$$ It's better to think a little more abstractly here rather than by possibly tedious calculations. I'm thinking of this result: Any function continuous on $$\mathbb R^2$$ that has a finite limit as $$\sqrt {x^2+y^2} \to \infty$$ is uniformly continuous on $$\mathbb R^2.$$

• Can you tell me why does that theory hold – J.Dane Aug 25 at 19:08

Switching to polar coordinates will show that $$f$$ is continuous at $$x=0.$$ Looking at the Maclaurin series for sine will show it has limit $$1$$ as $$(x,y)\to \infty.$$ Now, split the problem in two parts, noting that there is an $$r>0$$ such that $$|f-1|<\epsilon$$ if $$\|(x,y)\|>r$$, and that the closed ball of radius $$r$$ is compact.