If $D$ is bounded and $f$ is $\alpha$-Hölder continuous, then $f$ is also $\beta$-Hölder continuous for $\beta \in (0,\alpha]$

Question

If $D$ is bounded and $f$ is $\alpha$-Hölder continuous in $D$, then $f$ is also $\beta$-Hölder continuous for $\beta \in (0,\alpha]$

Couldn't I just use the fact that I have already proven that every $f$ that is $\alpha$-Hölder continuous, is uniformly continuous. Uniform Continuity encompasses the entire interval.

Answer 1

Let $D$ be bounded, then there exists $M>0$ s.t. $|x|\leq M$ for all $x\in D$. Therefore, \begin{align} |f(x)-f(y)|\leq K|x-y|^\alpha &=K|x-y|^{\alpha -\beta }|x-y|^\beta\\ & \leq K(|x|+|y|)^{\alpha -\beta }|x-y|^\beta \tag{$*$}\\ &\leq K(2M)^{\alpha -\beta }|x-y|^\beta , \end{align} where $(*)$ holds because $\alpha -\beta >0$.