= Solution
For <continuity> at any $x$, choose $\delta=\eta^{1/\alpha}$ for a desired error $\eta>0$. If $|y-x|<\delta$, the stated bound gives $|f(y)-f(x)|<\eta$. Thus \b[$f$ is <continuous>], in fact <uniformly continuous>.
When $\alpha>1$, the difference quotient satisfies
$$
\left|\frac{f(x+h)-f(x)}h\right|\leq|h|^{\alpha-1}\longrightarrow0.
$$
Therefore the <derivative> exists at every point and is zero. Part (i) proves \b[$f$ is constant]. This is the lemma that <Hölder exponent greater than one forces constancy>; no differentiability was assumed in advance.
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