Suppose we are given the following:
Let $f:X\rightarrow Y$ be such that $|f(x)-f(y)| \leq c|x-y|^\alpha$ for all $x,y \in X$. Let $U \subset X$. Then, for all $s$, $H^{s/\alpha}(f(U)) \leq c^{s/\alpha}H^s(U).$
How does one use this inequality to show that 2-Hölder continuous functions are constant? Say we have $f:[0,1] \rightarrow \mathbb{R}$, so we take $s = 1$ and $U = [0,1]$. $H^1$ on $\mathbb{R}$ is the Lebesgue measure, so $$H^{1/2}(f([0,1]) \leq cH^1([0,1]=c$$ I'm confused about why this means that $f$ is constant? If anyone could explain this to me or give some further insight, I would greatly appreciate it!
You should take $s=2$. You get $H^{1}(f[0,1]) \leq cH^{2}([0,1]) =0$ which means the interval $f[0,1]$ has measure (i.e. length $0$). Hence $f$ is a constant.