Proving that $x^{\frac{1}{n}}$ is uniformly continuous over $[0, \infty)$ with the usual metric

Solution 1:

Your proof is not correct because for $n>2$, $$|x^{1/n} - y^{1/n}| \left(x^\frac{n-1}{n} + y^\frac{n-1}{n}\right)\not=|x-y|.$$ However you can show the required property by noting that $x^{1/n}$ is u. c. in $[0,1]$ (continuous function on a compact set) and for $1\leq x<y$, by the Mean Value Theorem, there is $t\in (x,y)$ such that $$|x^{1/n} - y^{1/n}|=\frac{t^{1/n-1}}{n}|x-y|\leq \frac{1}{n}|x-y|.$$

Solution 2:

For $a>0$ and $X\ge 0$ the function $X^a$ is $\nearrow$

So for $\quad 0\le x\le y\quad$ we have $\quad 0\le x^a\le y^a$.

Let set $$\phi(y)=(y^a-x^a)-(y-x)^a$$

For $a-1\le 0$ and $X\ge 0$ the function $X^{a-1}$ is $\searrow$

So $\phi'(y)=ay^{a-1}-a(y-x)^{a-1}=a\left[(y-x)^{a-1}-y^{a-1}\right]\le 0\quad$ since $0\le y-x\le y$.

Finally for $a\in]0,1]$ the function $\phi$ is $\searrow$, this means $0\le y^a-x^a\le (y-x)^a$

It means that for $a\in]0,1]$ the function $X^a$ is $a$-Hölderian thus uniformly continuous.

Apply for $a=\frac 1n$ and $n\ge 1$.

Note:

The uniform continuity is easily deduced from Hölder property: $|f(x)-f(y)|\le C|x-y|^a$

$\forall \varepsilon>0$ let $\delta=(\frac{\varepsilon}C)^{\frac 1a}$ then $|x-y|<\delta\implies|f(x)-f(y)|\le C|x-y|^a<C{(\frac{\varepsilon}C)^{\frac aa}}\le\varepsilon$