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?

Answers 1

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 $\mathbb R^2.$

August 25, 2019 19:03 PM

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