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Let $0< \alpha < \beta \leq 1$. Prove $Lip_{\beta}[a,b] \subset Lip_{\alpha}[a,b]$. Also, I want to know if $Lip_\beta[a,b]$ is a closed subset for $Lip_{\alpha}[a,b]$.

My attemp of proof goes as follow, let $f \in Lip_{\beta}[a,b]$, then for every $x,y \in [a,b]$ I got that there is a $M>0$ such $|f(x)-f(y)| \leq M|x-y|^{\beta}$. As someone point me below in the comments, I have that

$$|f(x)-f(y)| \leq M|x-y|^{\beta}=M|x-y|^{\beta-\alpha}|x-y|^{\alpha}.$$

So I think the $M' > 0$ im looking for is $M'=\sup \lbrace M|x-y|^{\beta-\alpha} \rbrace$, this way for every $x,y \in [a,b]$ there is an $M>0$ such that

$$|f(x)-f(y)|\leq M|x-y|^{\beta}=M|x-y|^{\beta- \alpha }|x-y|^{\alpha} \leq \sup \lbrace M|x-y|^{\beta- \alpha} \rbrace=M'|x-y|^{\alpha}.$$

Is my proof right?

For $Lip_\beta[a,b]$ is a closed subset for $Lip_{\alpha}[a,b]$ I was thinking in using the equivalence of a closed subset as a subset which contains all its limit points. Then how do I proof this subset contains all its limit point, Im working here with the supremum norm of the space of continuous functions. Thank you!

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  • $\begingroup$ $$ M|x-y|^{\beta}= M|x-y|^{\alpha}|x-y|^{\beta-\alpha}\le M'|x-y|^{\alpha}. $$ $\endgroup$ – d.k.o. Jan 29 at 2:59
  • $\begingroup$ What is your definition of $Lip_\alpha[a,b]$? $\endgroup$ – d.k.o. Jan 29 at 3:00
  • $\begingroup$ $f \in Lip_{\alpha}[a,b]$ if for every $x,y \in [a,b]$, $|f(x)-f(y)| \leq M|x-y|^{\alpha}$ for some $M>0$. @d.k.o. $\endgroup$ – Cos Jan 29 at 3:02
  • $\begingroup$ @d.k.o. I like how your idea seems, but I think the $M'$ you give me at your hint doesnt work. I think I should take $M'=sup \lbrace M|x-y|^{\beta-\alpha} \rbrace$ isnt? $\endgroup$ – Cos Feb 6 at 1:40
  • $\begingroup$ Yes, and this supremum can be computed. $\endgroup$ – d.k.o. Feb 6 at 1:58
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$\operatorname{Lip}_{\beta}[a,b]$ is not closed in $\operatorname{Lip}_{\alpha}[a,b]$ under the sup norm. Take $a=0$, $b=1$, $\alpha=1/2$, and $\beta=1$. Consider $f(x)=\sqrt{x}$, which is 1/2-Hölder continuous and let $f_n=n^{-1}\vee f$. Then each $f_n$ is Lipschitz and $\|f_n-f\|_{\infty}=n^{-1}\to 0$ as $n\to\infty$. However, $f\notin \operatorname{Lip}_1[0,1]$.

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  • $\begingroup$ Hi @d.k.o . What does $f_{n}=n^{-1} V f $ means? :/ $\endgroup$ – Cos Mar 2 at 21:52
  • $\begingroup$ @Cos $a\vee b\equiv \max{a,b}$. $\endgroup$ – d.k.o. Mar 2 at 22:30

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