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?

$\begingroup$ 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. $\endgroup$– Nate RiverDec 8, 2023 at 16:20

$\begingroup$ Hm this is the Lipschitz case  for which I think it is easy to construct a differnetiable a.e. example. Holder though… @user479223 $\endgroup$– Nate RiverDec 8, 2023 at 16:45

$\begingroup$ Yes the weak derivative (if it even exists) must definitely be unbounded in $L^1$ on every subinterval. Definitely. I think. $\endgroup$– Nate RiverDec 8, 2023 at 16:53

$\begingroup$ Well that means the weak derivative must not even be a function.. $\endgroup$– Nate RiverDec 8, 2023 at 17:06

$\begingroup$ Weak derivative is always a function by definition. I'm cooking something up... $\endgroup$– user479223Dec 8, 2023 at 17:06
1 Answer
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
$$f(x)=\sum_{n=1}^\infty 2^{n} \psi(xq_n).$$
$f$ 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 $$\frac{f(x)f(q_n)}{xq_n^\alpha}\geq 2^{n} \frac{1}{xq_n^\alpha\log(xq_n)}.$$
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.

1$\begingroup$ @NateRiver Continuous, increasing, bounded but nowhere locally $\alpha$ Holder for any $\alpha$. What an awful function. $\endgroup$ Dec 8, 2023 at 18:06

2$\begingroup$ @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... $\endgroup$ Dec 8, 2023 at 18:18

1$\begingroup$ 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. $\endgroup$ Dec 8, 2023 at 18:19

1$\begingroup$ @NateRiver hah. What a stupid answer. Beautiful $\endgroup$ Dec 8, 2023 at 18:20

1$\begingroup$ 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. $\endgroup$ Dec 8, 2023 at 18:22