Timeline for Alternate definitions of $C^{1,\alpha}$ and $C^{1,\alpha}(\bar{D})$ maps
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Nov 13, 2012 at 4:11 | comment | added | Connor Mooney | @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 $Cr^{\alpha}$. | |
| Nov 13, 2012 at 3:59 | comment | added | Analysis Now | (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. | |
| Nov 13, 2012 at 3:54 | comment | added | Analysis Now | (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 | |
| Nov 13, 2012 at 3:43 | comment | added | Analysis Now | 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): | |
| Nov 13, 2012 at 2:16 | history | answered | Connor Mooney | CC BY-SA 3.0 |