Show that the set of non-$\beta$-Hölder functions is dense in the space of $\alpha$-Hölder function, $0<\alpha<\beta<1$

Question

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.

Answer 1

Let $f\in C^\alpha.$ We want to find a sequence $f_n$ in $C^\alpha \setminus C^\beta$ such that $f_n \to f$ in $C^\alpha.$ If $f\notin C^\beta,$ we can of course take $f_n=f$ for every $n.$ If $f\in C^\beta,$ define $f_n(x) = f(x) + x^\alpha/n, n=1,2,\dots$ Then each $f_n\in C^\alpha,$ and $f_n\to f$ in $C^\alpha.$ However $f_n\notin C^\beta$ for every $n.$ To see this, observe that for a fixed $n$ and small $h>0,$

$$\tag 1\frac{f_n(h)-f_n(0)}{h^\beta} = \frac{f(h)-f(0)}{h^\beta} + \frac{1}{n}\frac{h^\alpha}{h^\beta}.$$

Because $f\in C^\beta,$ the first fraction on the right of $(1)$ stays bounded as $h\to 0^+,$ while the second fraction $\to \infty.$ Thus the left side of $(1) \to \infty.$ Hence every $f_n\notin C^\beta$ as claimed, and we are done.

Answer 2

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 $C^\beta$.