# Can a nowhere locally Hölder function be differentiable almost everywhere?

Fix $$0 < \alpha < 1$$. Suppose $$f$$ is nowhere locally $$\alpha$$-Hölder continuous - that is, it is not $$\alpha$$-Hölder on any open subinterval of $$\mathbb R$$. Is it possible for $$f$$ to be differentiable almost everywhere?

• I managed to get it differentiable at a point. As usual it is not trivial, if even possible to extend the local construction to a.e. Dec 8, 2023 at 16:20
• Hm this is the Lipschitz case - for which I think it is easy to construct a differnetiable a.e. example. Holder though… @user479223 Dec 8, 2023 at 16:45
• Yes the weak derivative (if it even exists) must definitely be unbounded in $L^1$ on every subinterval. Definitely. I think. Dec 8, 2023 at 16:53
• Well that means the weak derivative must not even be a function.. Dec 8, 2023 at 17:06
• Weak derivative is always a function by definition. I'm cooking something up... Dec 8, 2023 at 17:06

Define $$\psi(x)=\begin{cases} 1/|\log x| &\text{if } x\in (0,1/2] \\ 0 &\text{if }x\leq 0\\ 1/\log 2& \text{ if } x>1/2.\end{cases}$$

Note that $$\psi$$ is increasing and bounded (and even continuous).

Consider an enumeration of the rationals $$q_n$$ and the function

$$f(x)=\sum_{n=1}^\infty 2^{-n} \psi(x-q_n).$$

$$f$$ is a sum of increasing positive functions hence is increasing. Hence differentiable ae.

However for any $$q_n$$, any $$x>q_n$$ and any $$\alpha\in (0,1]$$ we have that $$\frac{|f(x)-f(q_n)|}{|x-q_n|^\alpha}\geq 2^{-n} \frac{1}{|x-q_n|^\alpha|\log(x-q_n)|}.$$

As $$x\to q_n$$ this diverges. Hence nowhere locally $$\alpha$$ Hölder for any $$\alpha$$.

Edit, by replacing $$\log$$ by a continuous increasing function that diverges arbitrarily slowly you can construct a continuous, increasing and bounded function with arbitrarily low local regularity.

• @NateRiver Continuous, increasing, bounded but nowhere locally $\alpha$ Holder for any $\alpha$. What an awful function. Dec 8, 2023 at 18:06
• @NateRiver Actually this is a really nice counterexample for something else too. Everyone knows that $\alpha$ Holder implies finite $1/\alpha$ variation. However finite $p$ variation doesn't imply $1/p$ Holder - only after a reparameterization. This is bounded variation but nowhere locally $\alpha$ Holder. I wonder what the reparameterization is... Dec 8, 2023 at 18:18
• I was just thinking about that. Actually, how bad is the reparametrization allowed to be? Because since $f$ is increasing continuous, we can reparametrize by its inverse and it just becomes the identity, which is Lipschitz. Dec 8, 2023 at 18:19
• @NateRiver hah. What a stupid answer. Beautiful Dec 8, 2023 at 18:20
• The theorem about reparameterization uses the variation. The variation of an increasing function is always the function itself. So it is the inverse of the function. So stupid. I love math. Dec 8, 2023 at 18:22