Looking for $f$ with a uniform derivative

Question

let's say that $f$ is differentiable and I use the following definition:

$f$ is uniformly differentiable if for all $\epsilon>0$ there is a $\delta>0$ so $|\frac{f(y)-f(x)}{y-x} - f'(x)|<\epsilon$ when $|x-y|<\delta$ for every $x, y \in\mathbb R$.

I'm looking for an example that has a derivative at every point in $\mathbb R$ but no uniform derivative. I tried to prove it for $x^{2}$ but i can't find a proper $\epsilon$.

Answer 1

You can take $f(x)=x^3$. Then that property becomes$$(\forall\varepsilon>0)(\exists\delta>0)(\forall x,y\in\Bbb R):|x-y|<\delta\implies\left|\frac{y^3-x^3}{y-x}-3x^2\right|<\varepsilon.$$But$$\left|\frac{y^3-x^3}{y-x}-3x^2\right|<\varepsilon\iff|y^2+xy-2x^2|<\varepsilon.$$Now, take $\varepsilon=1$. For any $\delta>0$, take some $x\in\Bbb R$ and let $y=x+\frac\delta2$. Then $|y-x|<\delta$. But you don't always have$$\left|\left(x+\frac\delta2\right)^2+x\left(x+\frac\delta2\right)-2x^2\right|<1,$$since$$\left|\left(x+\frac\delta2\right)^2+x\left(x+\frac\delta2\right)-2x^2\right|=\left|\frac{\delta ^2}{4}+\frac{3 \delta x}2\right|$$and$$\lim_{x\to\infty}\left|\frac{\delta ^2}{4}+\frac{3 \delta x}2\right|=\infty.$$

Answer 2

$x^{2}$ has a uniform derivative. Instead of $x^{2}$ take $f(x)=x^{3}$ and consider the points $x=n, y=n+\frac 1 n$.

You can use the identity $y^{3}-x^{3}=(y-x)(y^{2}+x^{2}+xy)$ and take $\epsilon =1$ for example.