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Find a continuous bounded function $f:(0,1]\to \mathbb{R}$ that is not uniformly continuous. Extend $f$ with continuity in such a way that $f(0)=0$ and find the oscillation $\omega _f(0)$ of $f$ at $0$.

I'm not sure if I have to extend the function with continuity, but I think so. I'm struggling to find a function, I've tried $\sin(1/x)$ but can't extend it with continuity. Any hint?

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  • $\begingroup$ the unique possibility that you have to find such function is that $\lim_{x\to 0^+}f(x)$ doesn't exists $\endgroup$
    – Masacroso
    Apr 12, 2022 at 23:28
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    $\begingroup$ The question is wrong. If $f$ extends to a continuous function on $[0,1]$ then it is necessarily uniformly continuous. $\endgroup$ Apr 12, 2022 at 23:28
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    $\begingroup$ It isn't possible. If you could extend the function to $[0, 1]$ continuously, then the extended function would be a continuous function on a compact interval, which makes it uniformly continuous. If the function is uniformly continuous on $[0, 1]$, then it will be on $(0, 1]$ too. $\endgroup$ Apr 12, 2022 at 23:30
  • $\begingroup$ (I got uniform continuity and bounded variance confused; mea culpa for my incorrect comment.) $\endgroup$ Apr 12, 2022 at 23:36
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    $\begingroup$ To repeat the advice of @Masacroso, translate this to an equivalent problem that does not mention uniform continuity: Find a continuous, bounded function $f:(0,1]\to\mathbb R$ such that the limit $\lim_{x\to 0+} f(x)$ does not exist. $\endgroup$ Apr 13, 2022 at 0:13

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How about $f(x) = sin(\frac{1}{x})$? Clearly $sin$ is bounded and continuous. As $x$ goes to $0^+$ , $sin(\frac{1}{x})$ will oscillate between -1 and +1.

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