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!
$\operatorname{Lip}_{\beta}[a,b]$ 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, $f\notin \operatorname{Lip}_1[0,1]$.