# Prove that Hölder condition in $\Bbb R^n$ implies continuity

$$f:I\subset \Bbb R^n \rightarrow \Bbb R^m$$ 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 $$\Bbb R^n$$.

• You have it backwards. If $\alpha > 1$ then $F$ is actually differentiable with identically zero derivative. If $\alpha < 1$ then $F$ is not Lipschitz.
– Ian
Commented Sep 16, 2014 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.