Prove that if $f$ is locally lipschitz of order $\alpha >0$ at $x_0$, then $ f$ is continuous at $x_0$.

Question

We say a function is locally Lipstchitz of order $\alpha$ at $x_0$ if there exists $\epsilon, M>0$ such that

$$ |x-x_0|<\epsilon \rightarrow |f(x)-f(x_0)|<M(x-x_0)^\alpha $$

Prove that $f$ is continuous at $x_0$.

I think that I just need to prove that $x^\alpha$ is continuous for all $\alpha>0$ and then the result follows by squeezing, but this turned out to be a bit complicated (the result is pretty obvious for $\alpha\in \Bbb N$). Is there an easier way?

I've read two similar questions here, but they used a different definition (without the $\epsilon$ part).

Also, what's an easy way to prove that if $\alpha>1$, then $f'(x_0)$ exists and it's equal to $0$?

E: This question is trivial if we assume various properties of the function $t\mapsto t^\alpha$, however, I want to prove this without assuming extra stuff (as that it is continous, or increasing, or differentiable, etc), if you use any of these things, I'd like you to post a proof.

Answer 1

Suppose that there is some $\eta>0$ such that if $|x-x_0|<\eta$ then $|f(x)-f(x_0)|\leq M|x-x_0|^{\alpha}$

Fix $\varepsilon>0$, and choose $0<\delta<\min\{(\frac{\varepsilon}{M})^{\frac{1}{\alpha}},\eta\}$. If $|x-x_0|<\delta$, then $$ |f(x)-f(x_0)|\leq M|x-x_0|^\alpha<M\delta^{\alpha}<\varepsilon$$ which shows that $f$ is continuous at $x_0$.

Answer 2

You can just use upper limit to do it. Since $$ \varlimsup_{n\to\infty}|f(x)-f(x_0)|\leqslant \varlimsup_{n\to\infty}M(x-x_0)^\alpha=0 $$ We have $$ \varlimsup_{n\to\infty}|f(x)-f(x_0)|=0 $$ And so $$ \lim_{n\to\infty}|f(x)-f(x_0)|=0\quad\text{}\quad $$ For derivative, it is similar $$ \varlimsup_{n\to\infty}\frac{|f(x)-f(x_0)|}{|x-x_0|}\leqslant \varlimsup_{n\to\infty}M(x-x_0)^{\alpha-1}=0 $$