I want to show that $x^2\sin \frac {1}{x}$ is not uniformly continuous on $(0,\infty)$. I tried to find sequences $(x_n),(y_n)$ in $(0,\infty)$ such that $|x_n-y_n|\to 0$ but $|f(x_n)-f(y_n)|$ does not go to $0$. Please help.

  • $\begingroup$ Why do you think it's not uniformly continuous? $\endgroup$ – RRL Jan 21 '17 at 9:10
  • $\begingroup$ @RRL yes you are correct. It is uniformly continuous. $\endgroup$ – Anupam Jan 21 '17 at 11:21

For $f(x) = x^2 \sin(1/x)$ we have

$$f'(x) = 2x \sin(1/x) - \cos(1/x).$$

For $ x \in (0,\infty)$ we have

$$|f'(x)| \leqslant 2|x\sin(1/x)| + |\cos(1/x)| \leqslant 3.$$

Since the derivative is bounded the function is uniformly continuous on $(0,\infty).$

Note that $|x\sin(1/x)| \leqslant 1$ for $x \in (0,\infty)$. On $[1,\infty)$ we have $0 < 1/x \leqslant 1$ and $|x \sin(1/x)| = \frac{\sin(1/x)}{1/x} \leqslant 1$ since $\frac{sin y}{y} \uparrow 1$ as $y \to 0+$. On $(0,1)$ we have $|x \sin(1/x)| \leqslant 1$ since $|x \sin(1/x)| \to 0$ as $x \to 0$.


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