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!

  • $\begingroup$ $$ M|x-y|^{\beta}= M|x-y|^{\alpha}|x-y|^{\beta-\alpha}\le M'|x-y|^{\alpha}. $$ $\endgroup$ – 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

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.