# Example of a uniformly continuous function on R but with derivative not uniformly continuous.

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.

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

$$f(x) = \frac{\sin(e^x)}{1+x^2}$$ 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 $$f'(x_{2k}) - f'(x_{2k-1}) = \frac{ 2k \pi}{1 + (\ln(2k \pi))^2} + \frac{ (2k-1) \pi}{1 + (\ln((2k-1) \pi))^2} \to \infty \, .$$