Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $|F(z)-F(1)-p(z-1)-q(\bar{z}-1)|=O|z-1|^{1+\alpha}$.

share|cite|improve this question

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, $|f(x)-f(0)-xf'(0)|=O(x^2)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.