Proving a function with a certain property on a dense set does not have bounded variation 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!
 A: 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 $[0,1]$.
