Prove that if $f:[0,1] \rightarrow \mathbb{R}$ is a continuous function such that $\limsup_{y \to x} \frac{|f(x) - f(y)|}{|x-y|^{\frac{1}{2}}} = \infty$ on a dense set of $x \in [0,1]$, then $f$ does not have bounded variation.
I am having trouble approaching this problem and would really appreciate any help!
I think I found a counterexample.
(we mostly describe the construction below, and details for the proof are mostly left out. Ask in the comments if some details require additional explanations.)
In order to construct this function $f$ of bounded variation, we first define the functions $$ h:[-1,1]\to\Bbb R, \quad y\mapsto \operatorname{sgn}(y) |y|^{\frac13}, \\ h_a:[0,1]\to\Bbb R, \quad y\mapsto h(y-a), $$ where $a\in [0,1]$.
One can show that $h_a$ is monotone, continuous and satisfies $$ \sup_{y\in[0,1]}|h_a(y)|\leq 1, \qquad \limsup_{y\to a}\frac{|h_a(a)-h_a(y)|}{|a-y|^\frac12}=\infty. $$ Let $\alpha:\Bbb N\to \Bbb Q\cap [0,1]$ a bijective function, i.e. an enumeration of the rational numbers in $[0,1]$. We then define the function $$ f:[0,1]\to\Bbb R, \quad y \mapsto \sum_{n=1}^\infty 2^{-n}h_{\alpha(n)}(y). $$ The function $f$ is monotone, continuous (as the uniform limit of continuous functions), and has bounded variation. Moreover, one can show that $$ \limsup_{y\to x}\frac{|f(x)-f(y)|}{|x-y|^\frac12}=\infty $$ holds for all $x\in \Bbb Q\cap [0,1]$, which is a dense set in $[0,1]$.