Is the following function uniformly continuous?

**Pinkk** · March 17th 2010, 06:38 PM

Let $f(x)=x^{2}sin(\frac{1}{x}), x\ne 0; f(0)=0$ . Is $f$ uniformly continuous on $\mathbb{R}$ ? Now this one I'm completely stumped on and I don't even know where to begin.

**Drexel28** · March 17th 2010, 06:46 PM

Originally Posted by Pinkk

Let $f(x)=x^{2}sin(\frac{1}{x}), x\ne 0; f(0)=0$ . Is $f$ uniformly continuous on $\mathbb{R}$ ? Now this one I'm completely stumped on and I don't even know where to begin.

Note that the function is differentiable and that $f'(x)=\begin{cases}0 & \mbox{if} \quad x=0 \\ x\sin\left(\tfrac{1}{x}\right) -\cos\left(\tfrac{1}{x}\right) & \mbox{if} \quad x\ne 0\end{cases}$ and so $|f'(x)|\leqslant |x\sin\left(\tfrac{1}{x}\right)-\cos\left(\tfrac{1}{x}\right)|\leqslant |x\sin\left(\tfrac{1}{x}\right)|+|\cos\left(\tfrac {1}{x}\right)|\leqslant 1+1=2$ . Thus, your function has a bounded derivative and is thus Lipschitz and so trivially uniformly continuous.

**Pinkk** · March 17th 2010, 07:07 PM

Ah okay, makes sense because if $f'\le M, M>0$ then if $|b-a|<\frac{\epsilon}{M}$ , then $|f(b)-f(a)| = |f'(x)||b-a|\le M|b-a| < M\frac{\epsilon}{M}=\epsilon$ , if I'm not mistaken. Thanks again, man.

**Drexel28** · March 17th 2010, 07:08 PM

Originally Posted by Pinkk

Ah okay, makes sense because if $f'\le M, M>0$ then if $|b-a|<\frac{\epsilon}{M}$ , then $|f(b)-f(a)| = |f'(x)||b-a|\le M|b-a| < M\frac{\epsilon}{M}=\epsilon$ , if I'm not mistaken. Thanks again, man.

Good call.

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