TITLE: Show that the set of non-$\beta$-Hölder functions is dense in the space of $\alpha$-Hölder function, $0<\alpha<\beta<1$
QUESTION [6 upvotes]: I want to show that $C^\alpha([0,1]) \setminus C^\beta([0,1])$ is dense in $C^\alpha([0,1])$ where $0<\alpha<\beta<1$. 
I know that I can show that $C^\beta$ is not dense here, using the function $x^\alpha$. To show that it's complement is dense, given some function $f \in C^\alpha$ and $\epsilon > 0$, I think I can use the function $f+\iota W$, where $\iota < \epsilon$ is chosen so that $f+\iota W \in C^\alpha$ but $f+\iota W \notin C^\beta$, where $W$ is Weierstrass' “sawtooth” function.
I'm having a hard time making this idea rigorous however, and I was wondering if there are any easy ways to proceed.

REPLY [2 votes]: Your idea is roughly right, however the problem is not in choosing a small enough $\iota$, but the right (Weierstrass) function $W$. Write the Hölder-seminorm as
$$[f]_\alpha := \sup_{x,y\in [0,1]} \frac{|f(x)-f(y)|}{|x-y|^\alpha}.$$
Then $f \in C^\alpha$ iff $[f]_\alpha < \infty$.
So if we find a function $W$, which is in $C^\alpha$ but not in $C^\beta$, we can use this in the following way: Let $f\in C^\alpha$. Then either $f\notin C^\beta$ (in which case we do not need to approximate), or $f\in C^\beta$. In the latter case, we have for any $\iota \in \mathbb{R}$ that $f+\iota W \in C^\alpha$ with
$$[(f+\iota W) - f]_\alpha = |\iota| [W]_\alpha \stackrel{\iota \to 0}{\to} 0$$
as well as (using the inverse triangle inequality and splitting the supremum)
$$[f+\iota W]_\beta \geq |\iota|\sup_{x,y\in [0,1]} \frac{|W(x)- W(y)|}{|x-y|^\beta}-\sup_{x,y\in [0,1]} \frac{|f(x)-f(y)|}{|x-y|^\beta} = |\iota| [W]_\beta - [f]_\beta = \infty.$$
for $\iota \neq 0$. This already is enough to see that $f+\iota W \in C^\alpha \setminus C^\beta$ approximates $f$.
The point here is that the Hölder-continuity of $f+\iota W$ is independent of $\iota$ (apart from the pathological choice of $\iota = 0$), as sums of $C^\alpha$ functions are themselves $C^\alpha$. In fact the from above can even show that if $f\in C^\alpha$ and $g \in C^\beta$, but not higher, then $f \in C^{\min(\alpha,\beta)}$, but not higher (for $\alpha \neq \beta$, interestingly, if both are the same, they might cancel out).
Now to finish your idea, the Weierstrass function has some tuning parameters, which luckily can be chosen such that $W$ is exactly $C^\alpha$, but not $C^\beta$.