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

by J.Dane   Last Updated August 25, 2019 19:20 PM

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?

Tags :

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 $$\mathbb R^2.$$

zhw.
August 25, 2019 19:03 PM

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