How to show that $x^{1/n}$ is uniformly continuous

Question

Let $n\in\mathbb{N}$ and $n\geq 2$. For $x,y\in\mathbb{R}_{\geq 0}$, we have $$(x-y)=(x^{1/n}-y^{1/n})\sum_{i=0}^{n-1}(x^{1/n})^{n-i-1}(y^{1/n})^{i},$$ which can be equivalently stated as $$\frac{x-y}{\sum_{i=0}^{n-1}(x^{1/n})^{n-i-1}(y^{1/n})^{i}}=x^{1/n}-y^{1/n}.$$ For $\varepsilon>0$, why is it enough to take $|x-y|<\varepsilon^n$? This will only work if $$\frac{1}{\sum_{i=0}^{n-1}(x^{1/n})^{n-i-1}(y^{1/n})^{i}}\leq\frac{1}{\varepsilon^{n-1}},$$ but why would this hold?

Answer 1

Suppose $x \ge y$ and $\eta_n := x^{1/n} - y^{1/n}$. Then $\eta_n \ge 0$ and $$x = (y^{1/n} + \eta_n)^n \ge (y^{1/n})^n + \eta_n^n = y + \eta_n^n$$ Thus $0 \le \eta_n \le (x - y)^{1/n}$. By symmetry it follows that $|x^{1/n} - y^{1/n}| \le |x - y|^{1/n}$. Therefore, if $|x - y| < \epsilon^n$ then $|x^{1/n} - y^{1/n}| < (\epsilon^n)^{1/n} = \epsilon$.

Answer 2

The point is that the rate of increase is the fastest at $x_0=0$, and in that case you have $|f(x)-f(0)|=x^{1/n}<\epsilon$ if $x<\epsilon^n$. So what you want to show is that $|f(x+h)-f(x)| \leq |f(h)-f(0)|$ if $x>0$ and $h>0$. This ultimately follows from the fact that the function is concave down.