Say we have a function $f:(0,1)\to\mathbb{R}$ that is known to be $\alpha$-Hölder continuous, for $\alpha<1$.
For $x,y$ in $f$'s domain, $|x-y|<1$ and hence, for $\beta<\alpha$, $|x-y|^{\alpha}<|x-y|^{\beta}.$
So in this case, it looks to me that $f$ is also $\beta$-Hölder continuous. Can something similar (or more general) be said, if the domain of $f$ isn't $(0,1)$?
If $A$ is a bounded subset of a metric space, then $C^\alpha(A)\subset C^\beta(A)$ whenever $\alpha>\beta$. Indeed, let $D=\operatorname{diam}A$; then $d(x,y)^{\alpha-\beta} \le D^{\alpha-\beta}$ for all $x,y\in A$, hence $$ \frac{|f(x)-f(y)|}{d(x,y)^{\beta}} \le D^{\alpha-\beta} \frac{|f(x)-f(y)|}{d(x,y)^{\alpha}} $$ On an unbounded set, the Hölder spaces are not nested: for example, $f(x)=x^p$ belongs to $C^\alpha((0,\infty))$ if and only if $p=\alpha$.
On the other hand, one rarely needs a global bound $|f(x)-f(y)|\le Cd(x,y)^\alpha$ on an unbounded set; it's too restrictive, ruling out even linear functions. So the definition is often adjusted on unbounded sets so that, for example, it requires $$|f(x)-f(y)|\le C\max(d(x,y)^\alpha, d(x,y))$$ which again results in nested spaces.