On real functions with the property that for each $\varepsilon>0,$ every interval contains two points whose secant line is $>\varepsilon.$

Question

Let $f:\mathbb{R}\to\mathbb{R}$ have the following property:

$$\forall\varepsilon>0,\text{ no matter how large },\forall a<b,\ \exists a<x_1<x_2<b\text{ such that }\left\lvert\frac{f(x_2)-f(x_1)}{x_2-x_1}\right\rvert>\varepsilon.$$

In other words, for every $\varepsilon>0,$ every interval contains two points whose (absolute value of) gradient of secant line is $>\varepsilon.$

Obviously $f$ can be discontinuous - in fact, every "everywhere discontinuous" function satisfies the above property. I am not sure if $f$ can be continuous, but perhaps the Weierstrass function is one example (but I would like to confirm or deny this). But I think the function cannot be differentiable. So my questions are:

Can such a function be continuous? If yes, can it be differentiable?

Answer 1

At least $f$ can't be everywhere differentiable if I'm not wrong, here's an uncomplete though long answer. Using recursively the property, for any $a<b$ there are adjacent sequences $(x_{1,n})$ and $(x_{2,n})$ in $[a;b]$ such that for any integer $n\geqslant 1$, $|f(x_{1,n})-f(x_{1,n})|>n|x_{1,n}-x_{2,n}|$ (1). Details follow.

-once you have $x_{1,n}$ and $x_{2,n}$ use the property with $\varepsilon=n+1$, $a=x_{1,n}$ and $b=x_{2,n}$ to get $x_{1,n+1}$ and $x_{2,n+1}$.

-By compacity, any subsequences of $(x_{i,n})$ have convergent subsequences : without relabeling we suppose $\lim x_{i,n}=x_i$ and $x_1\neq x_2$. Then you have a contradiction with (1) when trying to let $n$ go to $\infty$. Hence necessarily $x_1=x_2$, the sequences (in fact subsequences of the previous ones) are adjacent. This entails that if $f$ is differentiable at $x_0$ we should have $\displaystyle \lim\frac{f(x_{1,n})-f(x_{2,n})}{x_{1,n}-x_{2,n}}=f'(x_0)$ which is a contradiction with (1).

Edit As $a$ and $b$ are arbitrarily chosen you have that non differentiable points are dense in $\mathbb R$.

You certainly wonder as I do whether $f$ could be differentiable somewhere, and if so, on what kind of set. I have no answer. But concerning an example of continuous $f$ satisfying the property I think that the Bolzano example of a continuous nowhere differentiable function works. I can't give all the details now (and it would be too long I think, but maybe you can find about it elsewhere), here is just how the Bolzano function is built on $[0;1]$ : it is a limit of piecewise continuous affine functions $f_n$ where $f_0(x)=x$, $f_1$ is then such that $f_1(0)=f_0(0)$, $f_1(1)=f_0(1)$, $f_1$ affine with slope $2$ on $[0;1/3]$, affine with same slope $2$ on $[2/3;1]$ and affine on $[1/3;2/3]$ (hence with slope $-1$. For each interval $I=[u;u+1/3^n]$ on which $f_n$ is affine with slope $p_n$ you get $f_{n+1}$ likewise : $f_{n+1}(u)=f_n(u)$, $f_{n+1}(u+1/3^n)=f_n(u+1/3^n)$, $f_{n+1}$ affine on $[u,u+1/3^{n+1}]$ with slope $p_{n+1}=2p_n$, affine on $[u+1/3^{n+1},u+2/3^{n+1}]$ with slope $-p_n$ and affine on $[u+2/3^{n+1},u+1/3^n]$ with slope $2p_n$.

Answer 2

What you're asking for is a continuous function that is not Lipschitz on any interval. There has been a similar question asked and answered here.

The Weierstras function indeed satisfies your condition - it is nowhere Lipschitz.