# Find a continuous bounded function $f:(0,1]\to \mathbb{R}$ that is not uniformly continuous.

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?

• the unique possibility that you have to find such function is that $\lim_{x\to 0^+}f(x)$ doesn't exists Apr 12, 2022 at 23:28
• The question is wrong. If $f$ extends to a continuous function on $[0,1]$ then it is necessarily uniformly continuous. Apr 12, 2022 at 23:28
• 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. Apr 12, 2022 at 23:30
• (I got uniform continuity and bounded variance confused; mea culpa for my incorrect comment.) Apr 12, 2022 at 23:36
• 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. Apr 13, 2022 at 0:13

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.