Let $0< \alpha < \beta \leq 1$. Prove $Lip_{\beta}[a,b] \subset Lip_{\alpha}[a,b]$.

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!

• $$M|x-y|^{\beta}= M|x-y|^{\alpha}|x-y|^{\beta-\alpha}\le M'|x-y|^{\alpha}.$$ – d.k.o. Jan 29 at 2:59
• What is your definition of $Lip_\alpha[a,b]$? – d.k.o. Jan 29 at 3:00
• $f \in Lip_{\alpha}[a,b]$ if for every $x,y \in [a,b]$, $|f(x)-f(y)| \leq M|x-y|^{\alpha}$ for some $M>0$. @d.k.o. – Cos Jan 29 at 3:02
• @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? – Cos Feb 6 at 1:40
• Yes, and this supremum can be computed. – 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, $$f\notin \operatorname{Lip}_1[0,1]$$.
• Hi @d.k.o . What does $f_{n}=n^{-1} V f$ means? :/ – Cos Mar 2 at 21:52
• @Cos $a\vee b\equiv \max{a,b}$. – d.k.o. Mar 2 at 22:30