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.
Does there exist a singular function that is Hölder continuous of order $\alpha$ for all $\alpha < 1$?
Here is a way to implement Christian Remling's comment. The construction is similar to the Cantor step function. Define a function $f:[0,1]\to[0,1]$ as follows.
For $x\in[0,1]$ let $x=\sum_{n=2}^\infty\frac{a_n}{n!}$ (with $a_n\in\{0,\dots,n-1\}$) be the factorial representation of $x$.
Now, from the sequence $(a_n)_n$ we obtain another sequence $(b_n)_n$ in the following way: If $a_n\neq0$ for all $n$, let $(b_n)_n=(a_n)_n$. If $a_k=0$ for some (minimal) value of $k$, let $b_n=a_n$ for all $n\leq k$ and $b_n=0$ for all $n>k$. Finally, define a sequence $(c_n)_n$ by $c_n=b_n$ if $b_n=0$ and $c_n=b_n-1$ if not.
Then we let $f(x)=\sum_n\frac{c_n}{(n-1)!}$. You can check that $f$ is well defined when $x$ is rational (that is, when $x$ has two factorial representations).
By induction on $n=1,2,\dots$ you can check that the set of numbers $x$ with $a_i=0$ for some $i=2,\dots,n$ is a union of intervals with total measure $1-\frac{1}{n}$. The function $f$ is by definition constant in each interval where $a_i=0$ for some $i$, so it follows that $f'=0$ a.e.
But $f$ is Hölder continuous of order $\alpha$ for any $\alpha<1$; indeed, we will prove that for all $x,y\in[0,1]$ we have $|f(x)-f(y)|\leq C_\alpha|x-y|^\alpha$, where $C_\alpha$ is some constant such that $\frac{1}{k!}\leq C_\alpha\left(\frac{1}{(k+2)!}\right)^\alpha$ for all $k\in\mathbb{N}$.
To see why, let $(a_n)_n$, $(a_n')_n$ be the factorial representations of $x,y$ respectively (define $(b_n)_n,(b_n')_n,(c_n)_n,(c_n')_n$ as above). We assume $x<y$. If $a_i=0$ for some $i$, change $(a_n)$ by the sequence $a_2,a_3,\dots,a_{i-1},1,1,1,\dots$; that increases $x$ but not $f(x)$. So we may assume $a_n>0$ for all $n$ (thus, $c_n=a_n-1$ for all $n$).
Now let $k$ be the smallest natural such that $a_k<a_k'$. If $a_k'>a_k+1$, then $|x-y|\geq\frac{1}{k!}$. But also $|f(x)-f(y)|<\frac{1}{(k-2)!}$ (because $c_n=c_n'$ for all $n\leq k-1$). So $|f(x)-f(y)|\leq C_\alpha|x-y|^\alpha$. So we can suppose $a_k'=a_k+1$.
If $a_{k+1}'\neq0$, then $|x-y|\geq\frac{1}{(k+1)!}$, but $|f(x)-f(y)|\leq\frac{1}{(k-1)!}$, so again $|f(x)-f(y)|\leq C_\alpha|x-y|^\alpha$. So suppose $a_{k+1}'=0$. Thus, $c_i'=0$ for all $i>k$.
Letting $l$ be the smallest index $>k$ such that $a_l<l-1$, we then have $|x-y|\geq\frac{1}{l!}$. But we also have $|f(x)-f(y)|\leq\frac{1}{(l-2)!}$ (because $c_k=c_k'-1$ and $c_i=i-2$ for all $i=k+1,\dots,l-1$). So again, $|f(x)-f(y)|\leq C_\alpha|x-y|^\alpha$.
