Let $\alpha \in (0,1)$, $f:[0,1]\rightarrow \mathbb{R}$ be a continuous monotone function and $\varepsilon>0$. Does there exist a function $\phi_{\varepsilon} \in \mathcal{C}^{\alpha}$ such that $\lambda(\{t \in [0,1]: f(t)=\phi_{\varepsilon}(t)\})\geqslant 1-\varepsilon$?
Ideas: A monotone continuous function is a.e. differentiable and therefore by regularity of the Lebesgue measure we can find an arbitrarily small open set that includes all the points where the function is not differentiable. Then one can write this open set as a countable disjoint union of open sets, on which we could "smoothly interpolate". Then we have a function that is smooth on this open set, but I don't know how to ensure regularity on the whole interval…
The prototype of a counterexample, the Cantor-function, does not work as we can take an approximation with Lipschitz-functions agreeing up to small measure with the Cantor-function. Maybe the "fat-Cantor-function" could be an idea of a counterexample.
Best Answer
Yes, you can actually find a Lipschitz continuous function. Take a look at this post BV. The proof is not easy though but this is because it is for functions of several variables. There should be a simpler proof in the one dimensional case.