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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?

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  • $\begingroup$ This function is not defined at $(0,0).$ $\endgroup$ – zhw. Aug 25 at 18:57
  • $\begingroup$ 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 $(x,y) = (0,0)$. $\endgroup$ – Rick Aug 25 at 19:01
  • $\begingroup$ @Rick you're right I'll edit it $\endgroup$ – J.Dane Aug 25 at 19:02
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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.

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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 $\mathbb R^2.$

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

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