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Given $p\in\mathbb{R}$, consider the space:

$$ Lip(p) = \left\{f:[0,1] \longrightarrow \mathbb{R} : \mbox{ $f$ is $p$-Lipschitz} \right\}$$

i.e.: there is $M>0$ such that $|f(s)-f(t)|<M|s-t|^p \quad\forall s,t\in 0,1]$

We can define a norm on $Lip(p)$ by $$\Vert f\Vert = |f(0)| + \sup \left\{ \frac{|f(t)-f(s)|}{|t-s|^p}: t\neq s, \quad t,s\in [0,1] \right\}$$

It's easy to show that $\Vert\cdot\Vert$ is a norm in $Lip(p)$, but I was not able to proove that $(Lip(p),\Vert\cdot\Vert)$ is a Banach space.

Given a Cauchy sequence $(f_n) \subseteq Lip(p)$, I couldn't find a candidate to conclude the convergence proof. Any hint? (I DO NOT want an entire proof)

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    $\begingroup$ If it's a Cauchy sequence in $Lip(p)$, it's a Cauchy sequence with the uniform topology. (You need to prove that) This allows you to construct a limit function. Now prove that (a) the limit function is also in $Lip(p)$ and that (b) convergence is actually in $Lip(p)$. $\endgroup$ Commented Mar 30, 2015 at 21:47
  • $\begingroup$ I was stuck in this problem, i'll try to write it! Thanks @HansEngler $\endgroup$ Commented Mar 30, 2015 at 22:56

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Given $\varepsilon > 0$, there is $n_0 \in \mathbb{N}$ such that: $$ m,n \geq n_0 \implies \Vert f_m - f_n \Vert < \frac{\varepsilon}{4}$$

Then, we have:

  1. $|f_m(0) - f_n(0)|< \varepsilon /4$

  2. $\sup \left\{ \frac{|(f_m-f_n)(t)-(f_m-f_n)(s)|}{|t-s|^p}: t\neq s, \quad t,s\in [0,1] \right\} < \varepsilon /4$

Given $t\in]0,1]$, we have for all $m,n > n_0$: $$|f_m(t) - f_n(t)| \leq |(f_m-f_n)(t) - (f_m-f_n)(0)| + |f_m(0) - f_n(0)| < \varepsilon $$

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    $\begingroup$ OK, this proves that $(f_n)_n$ is Cauchy sequence in the uniform norm. Therefore, there is a limit function $f$. By uniform convergence, it is continuous. Next, prove that $f \in Lip(p)$. $\endgroup$ Commented Mar 31, 2015 at 1:40
  • $\begingroup$ This is the most important part of the solution, later I'll write everything in details. Thank you very much for your help! $\endgroup$ Commented Mar 31, 2015 at 3:23

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