Prove $\sqrt x$ is uniformly continuous on $[1,+\infty[$

Question

I have done the following:

For any given $\;\varepsilon>0\;,\;$ let $\;\delta=2\varepsilon>0\;,$

$\forall\;x,y\in[1,+\infty[\;$ with $\;|x−y|<\delta\;,$

since $\;\sqrt{x}+\sqrt{y}\ge2\;,\;$ we get that

$\left|\sqrt{x}-\sqrt{y}\right|=\dfrac{|x-y|}{\sqrt{x}+\sqrt{y}}\le\dfrac{|x-y|}{2}<\dfrac{\delta}{2}=\varepsilon\;.$

Hence, $\;\sqrt{x}\;$ is uniformly continuous on $\;[1,+\infty[\;.$

Am I missing something? Is this proof correct?

Answer 1

The function $\sqrt{x}$ is uniformly continuous on $[0,\infty)$. It is uniformly continuous on $[0,1]$ because it is continuous, and because $[0,1]$ is compact. To prove uniform continuity on $[1,\infty)$, you can do this directly, as you suggest $$ \sqrt{x}-\sqrt{y} = \frac{x-y}{\sqrt{x}+\sqrt{y}}, \\ |\sqrt{x}-\sqrt{y}| \le \frac{1}{2}|x-y|. $$ Combining the results on $[0,1]$ and $[1,\infty)$ gives you what you want.

Answer 2

The simplest way I see is to note that $\sqrt{x}$ is Lipschitz on $[1, +\infty[$. (To show that it is Lipschitz, compute the upper bound on the absolute value of the derivative of $\sqrt{x}$.) Lipschitz continuity implies uniform continuity.