Is the following function uniformly continuous?

• March 17th 2010, 06:38 PM
Pinkk
Is the following function uniformly continuous?
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.
• March 17th 2010, 06:46 PM
Drexel28
Quote:

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.
• March 17th 2010, 07:07 PM
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.
• March 17th 2010, 07:08 PM
Drexel28
Quote:

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.