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Math Help - Is the following function uniformly continuous?

  1. #1
    Senior Member Pinkk's Avatar
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    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.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Pinkk View Post
    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.
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  3. #3
    Senior Member Pinkk's Avatar
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    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.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Pinkk View Post
    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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