# Analysis - Uniform Continuity [duplicate]

I need to find an example of a function $f:I$ to $R$ such that $f$ is uniformly continuous $f'$ exists but $f'$ is not bounded? I'm fairly stuck with this - it's part of a review and I guess I never handled any of these types of problems very well. Any direction is appreciated.

• $I$ is just some interval, in case that wasn't clear. – user116219 Dec 16 '13 at 20:57
• Is it a bounded interval, or do you allow the type $[a, \infty)$ – user99680 Dec 16 '13 at 20:58
• $I$ does not have to be bounded – user116219 Dec 16 '13 at 21:03
• See the accepted answer at math.stackexchange.com/questions/352321/… : with $I=\mathbb{R}$, the countexample given is $f(x)=\sin(e^x)/(1+x^2)$. You didn't ask for it, but some nice things about this counterexample are that $f'$ is continuous and $f$ is bounded. – Stefan Smith Dec 16 '13 at 22:08

Hint: Take $I=(0,1)$ and $f(x)=\sqrt{x}$.

• MSE users/workers, unite! We'll make MSE a Mathematicians' paradise! – user99680 Dec 16 '13 at 21:15
• @user99680 Let the moderators tremble at a communist revolution. The stackexchangers have nothing to lose but their chains. They have a world to win. Workingmen of all countries, unite! – Kortlek Dec 16 '13 at 21:59
• I would love to join your revolution. I am a revolutionary myself. Google me! – ILoveMath Dec 20 '13 at 11:45

Try to think to a continuous and differentiable function on a compact interval $[a,b]$, except for an infinite derivative on one of $a$ or $b$.
Then consider the same function on $(a,b)$.

(This is a standard example, it has the advantage that the issue is not an "infinite derivative".)

Take $f(x)=x^2\sin(1/x^2)$ for $x\ne0$ and $f(0)=0$. This function is continuous and therefore uniformly continuous on any bounded interval, for example $I=[0,1]$. On the other hand, $f'(0)=0$ and $f'(x)=2x\sin(1/x^2)-(2/x)\cos(1/x^2)$ for $x\ne0$, which is unbounded on any neighborhood of $0$.

If you want an example on unbounded intervals, pick $x_0>0$ with $f'(x_0)=0$, start with the $f$ from the previous paragraph, but only on $[-x_0,x_0]$ and extend it to $\mathbb R$ by setting $f(t)=f(x_0)$ for $|t|>x_0$.