If $f$ is Lipschitz of order $\alpha > 1$ over $[a,b]$, prove $f$ is constant

Question

If $f$ is Lipschitz of order $\alpha > 1$ over $[a,b]$, prove $f$ is constant.

We are given that for all $x,y \in [a,b]$ there exists $C>0$ such that $|f(x)-f(y)| \leq C|x-y|^{\alpha}$. We would like to show that $f(x)-f(y) = 0$ in this domain, but I don't see how to get to that conclusion since $C>0$.

Also, does the question mean $f$ is constant over $[a,b]$ or everywhere?

Answer 1

It may help if you notice that $|f(x)-f(y)| \leq C|x-y|^{\alpha}$ is equivalent to $\frac{|f(x)-f(y)|}{|x-y|} \leq C|x-y|^{\alpha-1}$, and $\alpha -1 > 0$ by assumption.

Answer 2

Without derivatives (just for fun): It suffices to show $f(b)=f(a).$ Let $a=x_0 < x_1 < \cdots < x_n$ be the uniform partition of $[a,b]$ into $n$ subintervals of length $(b-a)/n.$ Then

$$|f(b)-f(a)| = |\sum_{k=1}^{n}(f(x_k) - f(x_{k-1}))| \le \sum_{k=1}^{n}|f(x_k) - f(x_{k-1})|$$ $$\le \sum_{k=1}^{n}C(x_k-x_{k-1})^\alpha = nC[(b-a)/n]^\alpha = C(b-a)^\alpha n^{1-\alpha}.$$

Because $1-\alpha < 0,$ $n^{1-\alpha}\to 0$ as $n\to \infty.$ Since the above holds for any $n,$ we have $f(b)-f(a) = 0$ as desired.