TITLE: Little $\lambda$-Lipschitz functions
QUESTION [6 upvotes]: I've read in a few papers that for $\lambda \in (0,1)$ the set of functions given by
$$c^{0,\lambda}([0,1]) = \lbrace f:[0,1]\to\mathbb{R}: \lim_{x\to y} \frac{|f(x)-f(y)|}{|x-y|^{\lambda}} = 0 \rbrace $$
is a closed subspace of 
$$C^{0,\lambda}([0,1]) = \lbrace f:[0,1]\to\mathbb{R} : \sup_{x\neq y} \frac{|f(x)-f(y)|}{|x-y|^{\lambda}} < \infty \rbrace$$
equipped with the norm $\|f\|_{C^{0,\lambda}([0,1])} = \|f\|_{\infty}+\sup_{x\neq y} \frac{|f(x)-f(y)|}{|x-y|^{\lambda}}$. 
However, I can not find a proof for this. Any help would be greatly appreciated for an explanation!

REPLY [2 votes]: Introduce the modulus of continuity 
$$
\omega_f(\delta)= \sup_{|x-y|\le \delta}|f(x)-f(y)|
$$
and observe that 
$$\omega_f(\delta)\le \|f\|_{C^{0,\lambda}}\delta^\lambda\tag{1}$$
$$f\in c^{0,\lambda} \iff \lim_{\delta\to0} \delta^{-\lambda}\omega_f(\delta)=0  \tag{2}$$
Given $f\notin c^{0,\lambda}$,  let $r = \limsup_{\delta\to0} \delta^{-\lambda}\omega_f(\delta)>0$. If $g\in c^{0,\lambda}$, then by the triangle inequality 
$$
\limsup_{\delta\to0} \delta^{-\lambda}\omega_{f-g}(\delta)
= \limsup_{\delta\to0} \delta^{-\lambda}\omega_{f}(\delta) = r
$$
hence $\|f-g\|_{C^{0,\lambda}}\ge r$. This implies that the complement of $c^{0,\lambda}$ is open.