$f:I\subset \Bbb R^n \rightarrow \Bbb R^m$ is said to be Hölder continuous if $\exists$ $\alpha>0$ and $M>0$ such that $\|{f(x)-f(y)}\| \leq M\|x-y\|^\alpha$, $ \forall x,y \in I$, $0<\alpha\leq 1$. Prove that $f$ is Hölder $\Rightarrow f$ is continuous.

To prove that $f$ Hölder $\Rightarrow f$ 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 continuous.

But how can I prove continuity for the case in which $\alpha >1$? If we were on $\Bbb R$ it is clear that, by the definition of derivative, the function is constant and therefore continuous, not sure if this is the case in $\Bbb R^n$.

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    $\begingroup$ You have it backwards. If $\alpha > 1$ then $F$ is actually differentiable with identically zero derivative. If $\alpha < 1$ then $F$ is not Lipschitz. $\endgroup$ – Ian Sep 16 '14 at 0:57

It is all the same, you don't need to break it in cases for $\alpha$. Let $\epsilon > 0$ and ${\bf x},{\bf y} \in \Bbb R^n$. Choose $\delta < \sqrt[\alpha]{\frac{\epsilon}{M}}$. Then: $$\|{ \bf x - y}\| < \delta \implies \| f{\bf x} - f{\bf y}\| < M\|{\bf x - y}\|^\alpha < M\left(\sqrt[\alpha]{\frac{\epsilon}{M}}\right)^\alpha = \epsilon$$ In fact, this proves that $f$ is uniformly continuous, which is stronger than simply being continuous.


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