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My question is about the precise definition 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, using only information on $f$, but not on $f'$, 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 $F\in C^2({\mathbb{D}})$, 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 $|F(z)-F(1)-p(z-1)-q(\bar{z}-1)|=O|z-1|^{1+\alpha}$.

If the above are not correct, could you please give me or refer to me an alternate definition for each of the above ? Thank you.

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It is a good exercise to show that for a function $u: \mathbb{R}^n \rightarrow \mathbb{R}$, $C^{1,\alpha}$ regularity is equivalent to the following: there exists a linear function $l_x$ such that $$\|u-l_x\|_{L^{\infty}(B_r(x))} \leq Cr^{1+\alpha}$$ where $C$ is uniform. The forward direction is clear; for the other direction, observe that for points distance $r$ apart, the linear approximations differ by $Cr^{1+\alpha}$ nearby these points, so (up to changing $C$ by constants depending only on $n$ and $\alpha$) their slopes can differ by at most $Cr^{\alpha}$.

Actually, in the proof above I only needed that I can choose $C$ uniform on neighborhoods. This is crucial; as Pietro notes, we can get this property at a point (say 0) by taking an arbitrary bounded function and multiplying by $|x|^{1+\alpha}$, which clearly doesn't give $C^{1,\alpha}$ regularity away from $0$.

The equivalence of "pointwise $C^{1,\alpha}$ regularity" with uniform constant and $C^{1,\alpha}$ is the basis of most approaches I know to proving $C^{1,\alpha}$ estimates in PDE. This is usually done by estimating a solution with a linear function, rescaling, and improving the estimate geometrically, i.e. finding $r_k \rightarrow 0$ and $l_k$ so that $$\|u-l_k\|_{L^{\infty}(B_r(x))} \leq r_k^{1+\alpha}.$$

One can analogously define "pointwise $C^{k,\alpha}$" by approximation with $k^{th}$ order polynomials to get Holder estimates on higher derivatives.

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To Conor: first, thanks for your answer. I think in your answer, $l_x= u(0) + Du_x ?$ I am still a little doubtful about how I can pass from the difference between the difference quotient and the linear approximation to the difference quotient between the derivatives between these two points. If I use my notation, then: $r=|a-1|$, and here I am also working with a domain with boundary(which would not probably make any difference in this case): – Analysis Now Nov 13 '12 at 3:43
(To Conor, continued): $|f(a)-f(1)-f'(1)(a-1)|=O(|a-1|^{1+\alpha})$, which gives $\frac{f(a)-f(1)}{a-1}-f'(1)=O(|a-1|^{\alpha})$. From here, how do you get $|f'(a)-f'(1)|=O(|a-1|^{\alpha})$? I am tempted to use mean-value theorem, but that is not giving me the answer! About the locally $C^{1,\alpha}$ versus globally $C^1{1,\alpha}$, I think circle $S^1$ being compact, they are the same. Anyway, for my question 2, there was a mistake in my question; I know that my $F$ is $C^2(\mathbb{D})$, so if I try to prove $F\in C^{1,\alpha}(\mathbb{D})$, then isn't it enough that I prove: locally near – Analysis Now Nov 13 '12 at 3:54
(To Conor, continued): every point $\zeta \in S^1,$ I have $F(z)-F(\zeta)-l_{\zeta}=O(|z-\zeta|^{1+\alpha})$, where the constant of Holder continuity is locally uniform ? My second comment, passing from difference quotient to the derivative expresses my main concern. You can ignore the rest. – Analysis Now Nov 13 '12 at 3:59
@Analysis Now: Yes, $l_x(z) = u(x) + \langle Du(x), z-x \rangle$. Let $L = l_x-l_y$ with $|x-y| = r$. Then $L$ is linear, and $|L| \leq |l_x-u| + |u-l_y| \leq Cr^{1+\alpha}$ on say $B_{2r}(x)$. The oscillation of a linear function on a ball of radius $r$ is $r$ times the slope, hence $L$ must have slope bounded by $Cr^{\alpha}$, giving $Du(x)-Du(y)$ differing by at most $Cr^{\alpha}$. – Connor Mooney Nov 13 '12 at 4:11

The latter property does not implies the former: any $C^1$ diffeo of the form $f(a)=a+O(|a-1|^{\alpha+1})$, satisfies the latter property, but can have a derivative oscillating so strongly, that the former property does not hold.

However, the former property implies the latter, since

$|f(a)-f(1)-f'(1)(a-1)|\le \int_1^a |f'(t)-f'(1)| dt\le K \int_1^a |t-1|^\alpha dt \le K_1 |a-1|^{\alpha+1}\, .$

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To Pietro, thanks. If for the first question, I JUST need $C^{1,\alpha}-regularity$ near $1$, i.e. if I want to get $f'(a)-f'(1)=O(|a-1|)^{\alpha},$ can we get it from the alternate definition. I am unable to fill out the details. – Analysis Now Nov 13 '12 at 4:04
No: even for a $C^1$ diffeo, the $|f(a)−f(1)−f'(1)(a−1)|=O(|a−1|^{\alpha+1})$ does not imply $f'(a)−f'(1)=O(|a−1|^\alpha)$ – Pietro Majer Nov 13 '12 at 5:12

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