# Let $0 < \alpha < \beta \leq 1$. Prove $\operatorname{Lip}_{\beta}[a,b] \subset \operatorname{Lip}_{\alpha}[a,b]$.

Let $$0 < \alpha < \beta \leq 1\DeclareMathOperator{\Lip}{Lip}$$. Prove $$\Lip_{\beta}[a,b] \subset \Lip_{\alpha}[a,b]$$. By contention I got that having a function $$f \in \Lip_{\beta}$$ means that there is a $$M > 0$$ such for every $$x,y \in [a,b]$$ the following inequality holds:

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

So in order to proof $$f \in \Lip_{\alpha}$$ I need to find another $$M'>0$$ such

$$|f(x)-f(y)| \le M|x-y|^{\alpha}$$.

I asked this question before some weeks ago but for being honest, I cannot end the proof. I conjecture that the $$M'>0$$ im looking for is $$M'=\sup \lbrace M|x-y|^{\beta-\alpha} \rbrace$$ but another guy in college told me the $$M'>0$$ I'm looking for is something has |b-a| in its values and is not the supremum I mentioned. So I'm confused and not able to end this proof. I really need help finishing this proof. Thanks!

*A new attempt of the proof in the following photo. Is this proof right? • what do you mean "end the proof". you haven't done anything at all – mathworker21 Mar 2 at 19:29
• This is what I have tried before math.stackexchange.com/questions/3091638/… @mathworker21 – Cos Mar 2 at 19:31
• and the choice $M' = \sup\{M |x-y|^{\beta-\alpha}\}$ obviously works; it doesn't need to be optimal. Note that its value is merely $M' = M(b-a)^{\beta-\alpha}$, since $\beta > \alpha$. – mathworker21 Mar 2 at 19:31
• Already a complete attempt of the proof based on your comments as a photo of my chart. Can you check and help me by telling me if its already right? @mathworker21 – Cos Mar 2 at 19:52
• perfect! good job – mathworker21 Mar 2 at 19:53