# Existence of a limit of alpha-difference quotient for Hölder functions

Let $$f:\mathbb{R}\to \mathbb{R}^d,d\geq 1,$$ be an Hölder function with exponent $$\alpha\in (0,1)$$, meaning that $$$$\sup_{x, y \in \mathbb R, \,x\neq y}\frac{|f(x)-f(y)|}{|x-y|^\alpha}<\infty.$$$$ Define a set $$$$A_f:=\left\{x\in \mathbb{R}: u_f(x)>l_f(x)\right\},$$$$ where $$$$u_f(x):=\limsup_{y\to x}\frac{|f(x)-f(y)|}{|x-y|^\alpha},\quad l_f(x):=\liminf_{y\to x}\frac{|f(x)-f(y)|}{|x-y|^\alpha}.$$$$ The set $$A_f$$ is the set of all points $$x\in \mathbb{R}$$ where the limit $$\lim_{y\to x}\frac{|f(x)-f(y)|}{|x-y|^\alpha}$$ does not exist.

Does there exist an Hölder function $$f$$ with exponent $$\alpha\in (0,1)$$ as above such that $$\mathcal{L}^1(A_f)>0\; ?$$ Here $$\mathcal{L}^1$$ is the one-dimensional Lebesgue measure on $$\mathbb{R}$$.

Remark: If $$f:\mathbb{R}\to \mathbb{R}^d$$ is differentiable almost everywhere, then at each differentiable point $$x_0\in \mathbb{R}$$ it follows that $$$$\lim_{y\to x_0}\frac{|f(x_0)-f(y)|}{|x_0-y|^\alpha}=\lim_{y\to x_0}\frac{|f(x_0)-f(y)|}{|x_0-y|}|x_0-y|^{1-\alpha}=0.$$$$ Hence, $$\mathcal{L}^1(A_f)=0$$. So if there exists an Hölder function as above, then it has to be non-differentiable on a set of positive Lebesgue measure.

• Denote by $C(f)$ the first supremum in your question. If I remember correctly, the functions for which at every point $u_f=C$ and $\liminf_{y\to x} (f(x)-f(y))/|x-y|^\alpha=-C$ (without the modulus at the numerator) are residual (therefore dense) in the space of Holder functions with $C(f)\leq C$.
– Del
Commented Sep 20, 2022 at 13:22
• Hey Del, thank you for your response. I'm not familiar with the density property you have mentioned. Can you give some reference?
– Paz
Commented Sep 20, 2022 at 14:11
• Unfortunately I don't have any reference, I just remember that it should be proved in a standard way, similar to the way you prove similar Baire-type results. I will try to recall it
– Del
Commented Sep 20, 2022 at 19:38