Are singular functions dense in the space of Hölder continuous functions?

Question

We say a non-constant function $f$ on $[0, 1]$ is singular if it is continuous, and in addition differentiable almost everywhere with $f' = 0$ a.e.

For every positive $\alpha < 1$, is the set of singular functions dense in the space of Holder continuous functions of order $\alpha$? Where we equip the space with the norm

$$\|f\|_{C^\alpha} := \sup|f| + \sup_{x, y \in [0, 1]} \frac{|f(y) - f(x)|}{|y - x|^\alpha}.$$

Comments: The prototypical example of a singular function is the Cantor function, which is Hölder continuous of order $\frac{\log 2}{\log 3}$.

Answer 1

No. Take $f(t) = \sum_{n=0}^\infty 2^{-\alpha 2^n}\cos(2^{2^n}t)$. This is $\alpha$-Hölder and there exists $c>0$ such that, for every point $t$, one has $\limsup_{s \to t} |t-s|^{-\alpha}|f(t)-f(s)| > c$. Take any "singular" functions $g$. Since there is some point $t$ such that $g'(t) = 0$, it follows from the property of $f$ that $\|f-g\|_\alpha \ge c$.