Show $f(x)=\sqrt{x}$ is uniformly continuous on $[0,\infty)$

Question

Show $f(x)=\sqrt{x}$ is uniformly continuous on $[0,\infty)$

Let $\epsilon>0$, then there exists a $\delta=\epsilon^2$ such that $|x_1-x_2|<\delta$ for all $x_1,x_2\in[0,\infty)$.

$\begin{align}|f(x_1)-f(x_2)|&=|\sqrt{x_1}-\sqrt{x_2}|\\&<\sqrt{|x_1-x_2|}\\&<\sqrt{\delta}\\&=\epsilon\end{align}$

Can someone tell me where I did wrong? I don't see where the error is. Thanks

EDIT

From comment, people said the argument I have is right, I want to know how to show $|\sqrt{x_1}-\sqrt {x_2} |\leq\sqrt {|x_1-x_2|} $ is true.

score 3 · Answer 1 · answered Dec 10 '15 at 02:21

3

For $x,y\geq 0$, $$|\sqrt x - \sqrt y|^2 \leq |\sqrt x - \sqrt y||\sqrt x + \sqrt y|.$$

answered Dec 10 '15 at 02:21

score 2 · Answer 2 · answered Dec 10 '15 at 01:44

2

The only thing I see wrong is using $<$ instead of $\leq$ for $|\sqrt{x_1}-\sqrt{x_2}|\leq \sqrt{|x_1-x_2|}$.

answered Dec 10 '15 at 01:44

parsiad

2

I am wondering how to show $|\sqrt {x_1}-\sqrt {x_2}|\leq \sqrt {|x_1-x_2|}$ is true – Simple Dec 10 '15 at 01:49
Try squaring that. – Umang Dec 10 '15 at 02:55

2 Answers2