$f:I \rightarrow \mathbb R$ is said to be Hölder continuous if $\exists \alpha>0$ such that $|f(x)-f(y)| \leq M|x-y|^\alpha$, $ \forall x,y \in I$, $0<\alpha\leq1$. Prove that $f$ Hölder continuous $\Rightarrow$ $f$ uniformly continuous and if $\alpha>1$, then f is constant.

In order to prove that $f$ Hölder continuous $\Rightarrow$ $f$ uniformly continuous, it is enough to note that $|f(x)-f(y)| \leq M |x-y|^\alpha \leq M|x-y|$, since $\alpha \leq 1$. This implies that f is Lipschitz $\Rightarrow$ f is uniformly continuous.

But how can I prove that if $\alpha >1$, then $f$ is constant?


marked as duplicate by user99914, hardmath, Nosrati, Daniel Fischer real-analysis Sep 27 '17 at 11:43

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  • 1
    $\begingroup$ It does not follow that $M|x-y|^\alpha\leq M|x-y|$ if $\alpha\in (0,1)$. This is not true if $|x-y|<1$ ! $\endgroup$ – Svetoslav Oct 19 '16 at 17:35


For some $\epsilon>0$ and all $x\ne y$, you have $\Bigl|{f(x)-f(y)\over x-y}\Bigr|\le M|x-y|^\epsilon$ for some $\epsilon>0$.

Why must $f'(x)$ exist? What is the value of $f'(x)$?

  • $\begingroup$ David, if I divide $|f(x)-f(y)| \leq M|x-y|^\alpha$ by $|x-y|$, I obtain $\Bigl|{f(x)-f(y)\over x-y}\Bigr|\le M|x-y|^\epsilon$, with $\varepsilon <0$. Isn't it true? $\endgroup$ – Walter r Jun 24 '13 at 18:46
  • $\begingroup$ @Walterr With $\alpha>1$, I was assuming. So $\alpha=1+\epsilon$ for some $\epsilon>0$. $\endgroup$ – David Mitra Jun 24 '13 at 18:48
  • $\begingroup$ So, $f'(x)= 0$ when $x \rightarrow y$ and $f(x)= C$, correct? $\endgroup$ – Walter r Jun 24 '13 at 19:20
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    $\begingroup$ @Walterr Yes. Be a bit careful: Fix $x$. It first follows that $\lim_{y\rightarrow x}|{f(x)-f(y)\over x-y}|=0$. Then $\lim_{y\rightarrow x} {f(x)-f(y)\over x-y}=0$. So $f'(x)$ exists and has value $0$. Since this is true for all $x$, $f$ is a constant. $\endgroup$ – David Mitra Jun 24 '13 at 19:23

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