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?

  • 4
    $\begingroup$ Show that $f$ is differentiable and its derivative is zero. $\endgroup$
    – Fnacool
    Mar 22, 2016 at 2:56
  • $\begingroup$ @Fnacool We know that $f(x)$ is continuous over the interval, but how do I show differentiability? $\endgroup$ Mar 22, 2016 at 3:02
  • 1
    $\begingroup$ Just divide by $|x-y|$ and let $y\to x$. $\endgroup$ Mar 22, 2016 at 3:03
  • 2
    $\begingroup$ Possible duplicate of How to show that every $\alpha$-Hölder function, with $\alpha>1$, is constant? $\endgroup$
    – user228113
    Mar 22, 2016 at 3:04
  • $\begingroup$ Once you know it's constant on $(a,b)$, you can show it's constant on $[a,b]$ by continuity. $\endgroup$
    – user228113
    Mar 22, 2016 at 3:05

2 Answers 2


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.

  • $\begingroup$ I still don't see how this shows the derivative is $0$. $\endgroup$ Mar 22, 2016 at 3:09
  • $\begingroup$ So, if you take a limit as $y \to x$, the left hand side is the definition of derivative, and the right hand side will go to zero noting that $\alpha-1>0$ $\endgroup$ Mar 22, 2016 at 3:11

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.


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