Let $f:\mathbb{R}\to \mathbb{R}^d,d\geq 1,$ be an Hölder function with exponent $\alpha\in (0,1)$, meaning that \begin{equation} \sup_{x, y \in \mathbb R, \,x\neq y}\frac{|f(x)-f(y)|}{|x-y|^\alpha}<\infty. \end{equation} Define a set \begin{equation} A_f:=\left\{x\in \mathbb{R}: u_f(x)>l_f(x)\right\}, \end{equation} where \begin{equation} 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}. \end{equation} 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
\begin{equation}
\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.
\end{equation}
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.

everypoint $u_f=C$ 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 $C(f)\leq C$. $\endgroup$