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Let $f$ be a smooth, bounded function such that the limits $\lim_{x\to\pm \infty}f(x)$ exist. Then is the derivative $f'$ uniformly continuous?

I am looking for a counterexample as I think such a situation not necessarily implies that $f'$ is uniformly continuous. I know that any such function $f$ is always uniformly continuous. But don't know about the derivative.

One example I am thinking is $$f(x)=\frac1{1+x^2}.$$ One can show that $f$ is uniformly continuous. Its derivative $$f'(x)=-\frac{2x}{(1+x^2)^2}.$$ Is the derivative uniformly continuous? Please help. If my example is wrong, please provide any other example.

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    $\begingroup$ In your example $f'$ is uniformly continuous because it is continuous and $f(\pm \infty)=0$. $\endgroup$ Oct 10, 2022 at 8:45
  • $\begingroup$ So, my example won't work. $\endgroup$
    – Mittal G
    Oct 10, 2022 at 8:47
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    $\begingroup$ Something like $\sin(x^4)/(1+x^2)$ should work. $\endgroup$
    – Martin R
    Oct 10, 2022 at 8:48

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$$f(x) = \frac{\sin(e^x)}{1+x^2}$$ is bounded with $\lim_{x\to\pm \infty}f(x) = 0$. The derivative $$ f'(x) = \frac{e^x \cos(e^x)}{1+x^2}- \frac{2x \sin(e^x)}{(1+x^2)^2} $$ is not uniformly continuous: For $x_k = \ln (k \pi)$ is $$ f'(x_k) = \frac{(-1)^k k \pi}{1 + (\ln(k \pi))^2} \, , $$ so that $x_{2k} - x_{2k-1} \to 0$, but $$ f'(x_{2k}) - f'(x_{2k-1}) = \frac{ 2k \pi}{1 + (\ln(2k \pi))^2} + \frac{ (2k-1) \pi}{1 + (\ln((2k-1) \pi))^2} \to \infty \, . $$

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  • $\begingroup$ Thank you very much for the answer. $\endgroup$
    – Mittal G
    Oct 10, 2022 at 11:08

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