# Alternate definitions of $C^{1,\alpha}(S^1)$ and $C^{1,\alpha}(\bar{D})$ maps

My question is about the precise definitition regarding the following:

Let $f$ be an orientation-preserving $C^1$ diffeomorphism of the unit circle $S^1$. So $f'(b)$ exists and can be thought as a positive real number for every $b$ in $S^1$. Now I want to define when $f$ is $C^{1,\alpha}, 0<\alpha<1,$ near $1\in S^1$. I know one definition can be $f'(a)-f'(1)= O|a-1|^{\alpha}, i.e. |f'(a)-f'(1)|\le K.|a-1|^{\alpha}$. But I was wondering whether we can use the following alternate definition, motivated by $C^{1,\alpha}$-maps on $\mathbb{R}^1$:

1) Can we say $f$ is $C^{1,\alpha}$ near $1$ if $|f(a)-f(1)-f'(1)(a-1)|= O(|a-1|^{1+\alpha})$ ?

2) I have also another related question. let $F$ be a $C^1$ diffeomorphism on an open set containing the closed unit disk $\bar{D}$, and let $lim_{z\to1}F_z(z)=p, lim_{z\to1}F_{\bar{z}}(z)=q$. then can we say that $F$ is $C^{1,\alpha}$ near $1$(near in the sense of the topology of $\bar{D}$) if $|F(z)-F(1)-p(z-1)-q(\bar{z}-1)|=O|z-1|^{1+\alpha}$.

-

The circle with the arclength metric is locally isometric to the line; therefore any local estimates for $f$ either hold on both spaces or fail on both.
The assumption in (1) is rather weak: it says nothing about the continuity of $f'$; in fact it does not imply the existence of $f'$ at any point other than $1$. For example, it is satisfied by the function $f(x)=(x-1)^2\chi_{\mathbb Q}$ where $\chi_{\mathbb Q}$ is the characteristic function of rationals. I don't think you would want to include this $f$ in $C^{1,\alpha}$.
As a matter of opinion, I find the condition $f'(a)=f'(1)+O(|a-1|^\alpha)$ too weak to justify saying that "$f$ is $C^{1,\alpha}$ near $1$". To me, the latter means that $f\in C^{1,\alpha}(U)$ for some open set (i.e., interval) $U$ containing $1$. This requires strictly more than $f'(a)=f'(1)+O(|a-1|^\alpha)$.
I do not like your proposal in 2) either (nothing personal!). The properties you listed say that the derivative is continuous at $1$, and that the function is squeezed at $1$ in a certain way, but they say nothing about the Hölder continuity of derivative. For example, take $f(x)=x^2\sin \frac{1}{x\log x}$ (near $x=0$). The main term in the derivative is $\frac{1}{\log x}\cos \frac{1}{x\log x}$, which makes the derivative continuous at $0$, but not Hölder continuous. Yet, $|f(x)-f(0)-xf'(0)|=O(x^2)$.