Let $\alpha \geq 0$. We say that $f \colon D \to \mathbb{R}^m$ is $\alpha$-Hölder continuous if there is a constant $c$ such that for each $x,x_0\in D$, $|f(x) - f(x_0)| \leqslant c\cdot |x - x_0|^{\alpha}.$
Prove that if $D$ is bounded, $\alpha \leqslant \beta $, and $f$ is $\beta$-Hölder continuous, then $f$ is $\alpha$-Hölder continuous.
I know how to prove if there's only $\alpha$-Hölder continuity but not sure how to prove if $\beta$ comes in.
Write for any $x$ and $y$ in $D$ that $$|x-y|^{\beta}=|x-y|^{\alpha}\cdot |x-y|^{\beta-\alpha}\leqslant M\cdot |x-y|^\alpha,$$ where $M:=2^\alpha\sup_{s\in D}|s|^\alpha$.