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A bounded derivative is a sufficient condition for uniform continuity, but not necessary.

I know the counterexample $f(x) = \sqrt{x}$ on the interval $[0,\infty)$ where the derivative is unbounded at $0$, but the function is uniformly continuous.

Is there an example where $f$ is uniformly continuous and $f'(x)$ is unbounded as $x \to \infty$ but bounded on any compact interval?

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Taking $$f(x)=\frac{\cos(x^3)}{x}$$ on $[1,\infty)$ suffices since it is uniformly continuous and has derivative

$$f’(x)=-3x\sin x-\frac{\cos(x^3)}{x^2}.$$

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  • $\begingroup$ Suppose if we put more one condition that function should be unbounded as $x$ goes to infinity. Can such uniformly continuous function exist? ($f^\prime(x)$ should be unbounded as $x$ goes to infinity.) $\endgroup$ – ramanujan Nov 28 '18 at 23:51
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$x^a logx$ on $[0, \infty)$ for $\alpha\in (0,1)$.

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