It is easy to give examples of continuous functions $f:[0,1]\to \mathbb R\_+\cup\{0\}$ non-zero but vanishing on a Cantor set (ex: [Can Cantor set be the zero set of a continuous function?](https://mathoverflow.net/questions/24034/can-cantor-set-be-the-zero-set-of-a-continuous-function)). It is clearly non-true for analytic functions. My question is: 


1. Are there uniformly continuous non-zero functions vanishing on a Cantor set?
2. Are there α-Hölder continuous non-zero functions vanishing on a Cantor set?
3. Are there continuously differentiable non-zero functions vanishing on a Cantor set?



Non-zero smooth functions vanishing on a Cantor set
A Cantor set $C\subset[0,1]$ is closed, and that is all we need.
Therefore $f(x)=d(x,C)$ (distance from $x$ to $C$) vanishes on and only on $C$.
It is also Lipschitz-continuous.


For your third question, you can take $g(x)=f(x)^2$.
This is $C^1$ apart from peaks in the middle points of the intervals of the complement of $C$.
You can easily smooth the function near these points, since you are well away from $C$.


Choosing a higher power gives you any $C^k$ smoothness you want.
But you can also get $C^\infty$.
Let $\phi:\mathbb R\to[0,\infty)$ be a smooth function vanishing on $(-\infty,0]$ and set $g(x)=\phi(f(x))$ and mollify the tips.


Conclusion: The answer to all three questions is yes.


**Edit:**
It is not entirely clear what you mean by constructing by hand.
This is as close as I can come to making an explicit function if the Cantor set is not specified.
(I also noted that the distance was also used in [this answer](https://mathoverflow.net/a/24041/55893) to the OP's linked question. But it was not concerned with regularity.)


