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Let $$H=\{x\in C[0,1]: \forall s,t\in [0,1]: |x(s)-x(t)|\leq b|s-t|^\alpha\}$$ where $C[0,1]$ are the continuous functions on $[0,1]$ with $x(0)=0$ and $\alpha\in (0,1/2)$. Take a sequence $(x_n)\in H$. Choosing $\delta<(\frac{\varepsilon}{b})^{1/\alpha}$, shows $\{x_n\}$ shows the family is equicontinuous. Furthermore $|x(t)|\leq bt^\alpha$ so the family is bounded and by Arzela Ascoli its precompact. So there exists a subsequence, also denoted $x_n$ converging to some $x_0$ uniformly. For all $s,t\in [0,1]$ we have \begin{align*} |x_0(s)-x_0(t)|&\leq |x_0(s)-x_n(s)|+|x_0(t)-x_n(t)|+|x_n(s)-x_n(t)| \\ &\leq |x_0(s)-x_n(s)|+|x_0(t)-x_n(t)|+b|s-t|^\alpha \\ &\to b|s-t|^\alpha. \end{align*} so $x\in H$. So $H$ is compact subset of $C[0,1]$. This has to be wrong right?

Davide Giraudo
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    Why do you think it has to be wrong? – Jürgen Sukumaran Sep 22 '19 at 13:14
  • I doubt its compact... –  Sep 22 '19 at 13:30
  • The space you defined is not compact. You have to assume also that one point in the domain has image in a fixed interval. – Moishe Kohan Sep 22 '19 at 13:39
  • Explain what you wrote? –  Sep 22 '19 at 13:45
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    What he means is it's nit true that $|x(t)|\le bt^\alpha$. The functions $x_n=n$ are in $H$ but have no convergent subsequence. – David C. Ullrich Sep 22 '19 at 16:21
  • Alpha strictly between 0 and 1/2 –  Sep 22 '19 at 17:49
  • You are mis-using the Arzela Ascoli theorem: This theorem has two conditions, the first of which is that the sequence of functions is uniformly bounded, say, there exists $C\ge 0$ and $x_0\in [0,1]$ such that $|f(x_0)|\le C$ for all functions in $H$. Such a space will be indeed compact since Arzela Ascoli theorem now applies and your proof of the Holder property of the limit is correct. – Moishe Kohan Sep 22 '19 at 18:51
  • I added an assumption that should fix that problem –  Sep 22 '19 at 19:22
  • Now it is all correct. – Moishe Kohan Sep 22 '19 at 19:28
  • @empty wrong. First of all; hölder continuous functions of exponent > 1 are constant –  Sep 23 '19 at 06:26
  • @Empty What???? – David C. Ullrich Sep 24 '19 at 14:08
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    It's correct now, but the way it's stated is very curious. Instead of making $x(0)=0$ part of the definition of $C[0,1]$ it would make more sense to define $C[0,1]$ as usual and add the condition $x(0)=0$ to the definition of $H$... (Also why $\alpha<1/2$? It's true for any $\alpha>0$, with the same proof.) – David C. Ullrich Sep 24 '19 at 14:11
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    @DavidC.Ullrich sorry for not replying. I'm only using this for Wiener process which is Hölder continuous for alpha less than 1/2 so it's all I needed –  Nov 03 '19 at 07:59

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