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?