How to prove $x^{1/n} $ is uniformly continuos in $[0, a]$ where $a$ is a positive real.

Question

I proved that this funcion is not Lispchitz continuos making $y = 2x$ and making $x \rightarrow 0$. But I'm stuck proving the uniformly continuity.

Answer 1

The fast answer is that $f(x)=x^{1/n}$ is continuous and any continuous function on a compact set is automatically uniformly continuous. However, I suspect that you haven't learned this theorem yet so let's prove it in a more constructive way.

You are right about $f(x)=x^{1/n}$ not being a Lipschitz function. Still, it is a Holder continuous function which is almost as good, i.e. we have $$ |f(x)-f(y)| \le C|x-y|^{1/n}. $$ Can you prove that $C=1$ works?

Having shown that $f(x)$ is Holder continuous, can you conclude that it is uniformly continuous?