Finding a bounded function with unbounded derivative at zero

I am working on a linear analysis problem where we have boiled down the problem to finding a continuous function $f:\mathbb{R} \to \mathbb{R}$ that is bounded, but has infinite derivative at zero. So far, we have conjured up the example $$f_n(x) = \frac{2}{\pi}\arctan(nx)$$ This sequence of functions will have infinite derivative at $0$ when $n\to \infty$, and is bounded by $1$. I believe this will work for the sake of our problem, but I would like to find a function that doesn't depend on $n$. I can picture what this should look like, but I can't come up with an example function. Any ideas? All appreciated.

• This doesn't quite work, since it's unbounded. That was my first thought, too. – BSplitter Apr 4 '18 at 19:47
• Just take $f(x) = x^{1/3}$ for $x \in [-1,1]$, $f(x) = -1$ for $x < -1$ and $f(x) = 1$ for $x > 1$. – User8128 Apr 4 '18 at 19:49

$f(x)=\arctan(\sqrt[3]{x})$, for example.

• Fantastic, thank you! – BSplitter Apr 4 '18 at 19:52

A quarter of a unit circle (no, the other quarter) up and down: $$f(x) = \begin{cases} 0 , & x < -1, \\ 1-\sqrt{1-(x+1)^2}, & -1 \leq x < 0, \\ 1-\sqrt{1-(x-1)^2}, & 0 \leq x < 1, \\ 0 , & 1 \leq x \end{cases} \text{.}$$

$g(x)=arccotx^{1/2}$ another one.

• By this, do you mean $g(x) = \frac{1}{arccot(x^{1/2})}$ or $arccot(x^{1/2})$? – BSplitter Apr 4 '18 at 20:02
• no check the edit please – MathCosmo Apr 4 '18 at 20:03
• I think this looks mostly okay. I worry that the domain isn't $\mathbb{R}$ though. – BSplitter Apr 4 '18 at 20:14

My first thought was something like

$$f(x) = x\sin\left(\frac1x\right)$$ $$f'(x) = \sin\left(\frac1x\right) - \frac1x\cos\left(\frac1x\right)$$

This is bounded between $1$ and $-1$ (note that $\lim_{x\to\infty}f(x)=1$), and its derivative has an infinite oscillatory discontinuity.

Also, $$\lim_{x\to 0}f(x)=0$$ so $f$ itself has a removable discontinuity; it can be made continuous by defining $f(0)=0$.