Little Hölder spaces

Question

In this question, we have a definition of the above spaces:

What is little Holder space?

How can we prove that the closure of the “smooth” functions, i.e. the little Hölder space, is not the whole space?

How can we construct a $C^{k,\alpha}$ function that cannot be approximated by smooth functions?

Questions involving the construction of a counterexample often demand significant creativity, and I don’t happen to be able to know how to start to answer this question

Answer 1

This is entirely covered by the answers in this MO question. The cliff notes (WLOG $k=0$):

1. Every $f$ in $c^\alpha$ satisfies $$\lim_{t\downarrow 0} \sup_{x\neq y:|x-y|\le t} \frac{|f(x) - f(y)|}{|x-y|^\alpha} = 0$$
2. But for say $\sqrt{x}$ on $[0,1]$, $\sup_{x\neq y:|x-y|\le t} \frac{|f(x) - f(y)|}{|x-y|^{1/2}} \ge \frac{|\sqrt t - 0|}{\sqrt{t-0}}=1$ for all $t\ge0$. Alternatively you can check directly that there is no smooth function close by in $C^{1/2}$. Other $\alpha\in (0,1)$ can be treated similarly.

For the proof of 1: Let $\epsilon>0$ and $f_n$ be smooth functions with $f_n\to f$ in $C^\alpha$. Then $$ \frac{|f(x) - f(y)|}{|x-y|^\alpha} \le \frac{|f_n(x) - f_n(y)|}{|x-y|^\alpha} + \frac{|(f-f_n)(x) - (f-f_n)(y)|}{|x-y|^\alpha} \le \|\nabla f_n\|_{C^0}|x-y|^{1-\alpha} +[f-f_n]_{C^\alpha}.$$ Pick $n=n(\epsilon)\gg 1$ so that $[f-f_n]_{C^\alpha} < \epsilon/2$, and then pick $t_0=t_0(n,\epsilon)$ so that $\|\nabla f_n\|_{C^0}t_0^{1-\alpha} < \epsilon/2$. It follows that for all $0<t<t_0$, $\sup_{x\neq y:|x-y|<t} \frac{|f(x) - f(y)|}{|x-y|^\alpha} < \epsilon$, QED.