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$$f:(0,1)\rightarrow [0,1]$$ is a differentiable function . Is it uniformly continuous then $?$

Now $f$ being differentiable on $(0,1)$ is continuous , that is easy. Now I could say it is uniformly continuous if the differentiability on $(0,1)$ implied continuity on $[0,1]$. What the range being the closure of the domain contributes here $?$

Please give some lead on how to proceed .

Thanks for any help.

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    $\begingroup$ What about $\sin\frac1x$? $\endgroup$ – 5xum Sep 4 '15 at 18:43
  • $\begingroup$ the derivative should be bounded $\endgroup$ – Dr. Mohammad Alomari Sep 4 '15 at 18:46
  • $\begingroup$ It is continuous , differentiable on $(0,1)$ but not uniformly continuous . Right $?$ $\endgroup$ – user118494 Sep 4 '15 at 18:47
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    $\begingroup$ The derivative should be bounded, i.e. $|f'( \zeta)| \leq M$ for all $\zeta \in (0,1)$. Then, according to the mean value theorem, $|f(x) - f(x)| = |f(\zeta)| |x-y| \leq M |x-y|$. Which yields uniform continuity. I am not sure if this is the weakest possible assumption. $\endgroup$ – Nigel Overmars Sep 4 '15 at 18:53
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    $\begingroup$ @NigelOvermars If you consider $f(x)=\sqrt x $ you get a uniformly continuous function with an unbounded derivative. $\endgroup$ – Kitegi Sep 4 '15 at 19:02
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If we consider $$f(x)=\frac{1+\sin(1/x)}{2},$$ we get a differentiable function that is not uniformly continuous.

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