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?enter image description here

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    $\begingroup$ what do you mean "end the proof". you haven't done anything at all $\endgroup$ – mathworker21 Mar 2 at 19:29
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    $\begingroup$ This is what I have tried before math.stackexchange.com/questions/3091638/… @mathworker21 $\endgroup$ – Cos Mar 2 at 19:31
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    $\begingroup$ 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 $\beta > \alpha$. $\endgroup$ – mathworker21 Mar 2 at 19:31
  • $\begingroup$ 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 $\endgroup$ – Cos Mar 2 at 19:52
  • $\begingroup$ perfect! good job $\endgroup$ – mathworker21 Mar 2 at 19:53

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