Let $X\subset{\mathbb{R}^d}$ be bounded and let $f:X\rightarrow{}\mathbb{R}^d$ be an $\alpha$-Hölder map, that is to say:
$$\|f(x)-f(y)\|_2\le{}C\|x-y\|_2^{\alpha},\text{ for some }\alpha>0\text{ and some }C>0.$$
I want to show $f$ is also $\beta$-Hölder continuous if $\beta<\alpha$.
My thoughts:
I understand that $X$ being bounded means there exists some $M>0$ such that $\|x-y\|_2\le{M}$ for every $x,y\in{X}$. I've tried considering cases for when $\|x-y\|_2$ is between $0$ and $1$, equal to $1$, and greater than $1$ with $\beta>\alpha$, but this seems rather excessive, is there a quicker way?
I have come up with an answer:
Since X is bounded we can define $D:=\sup\{\|x-y\|_2:x,y\in{X}\}$. Then we have $$\|x-y\|_2\le{D},$$ hence $$\|x-y\|_2^{\alpha-\beta}\le{D}^{\alpha-\beta}.$$
hence $$\|x-y\|^\alpha=\|x-y\|^{\alpha-\beta}\|x-y\|^\beta\le{D^{\alpha-\beta}\|x-y\|^\beta},$$ the result follows.