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Prove that the function $g:A=[0,\infty)\to \Bbb R$ such that $g(x)=\sqrt x$ is uniformly continuous on $A.$

My attempt so far:

Let $I=[0,2]$. If we restrict $g$ to $I$ then by uniform continuity theorem we may conclude that $g$ is uniformly continuous on $I.$

Let $J=[2,\infty).$ We note that if we restrict the domain of $g$ to $J$ then $|g(x)-g(u)|=\frac{x-u}{\sqrt x+\sqrt u}\leq \frac 12(x-u).$ This means that $g$ is a Lipschitz function in $[2,\infty).$ So, $g$ is again uniformly continuous on $J.$

Noww, $A=I\cup J.$ Given, any $\epsilon \gt 0$ we have, a $\delta_I\gt 0$ that works for all $u\in [0,2]=I$ i.e $\forall x,u\in [0,2]$ such that if $|x-u|\lt\delta_I$ then, $|f(x)-f(u)|\lt\epsilon.$ Similarly, we find a $\delta_J\gt 0$ for all points $u\in J.$

We fix $\delta=\{\delta_I,\delta_J\}.$ If $x,u\in A$ and $|x-u|\lt\delta$ then $ |f(x)-f(u)|\lt \epsilon$ whenenver $x,u\in I$ or $x,u\in J.$

However, I don't get what happens if $x\in I, u\in J$ and vice-versa. This makes my solution incomplete. Is there a way to fix this?

Thomas Finley
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1 Answers1

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You want to, for each $\epsilon>0,$ find a $\delta$ such that $|x-u|<\delta$ implies $|g(x)-g(u)|<\epsilon.$ Restrict yourself by only searching for those $\delta$ that are smaller than $1$. This is perfectly OK, because you simply need to find one such $\delta$. Let $x,u\in \mathbb{R}$ be two real numbers. Since we are going to assume $|x-u|<\delta$, you may as well assume that $|x-u|<1.$ Now $g$ is uniformly continuous on any compact interval. So if $x,u\in I$ you are done. If $x,u\in J$ then you are again done, by the Lipschitz estimate you proved above; it suffices to use the minimum function to ensure that $\delta_J$ is smaller than 1. If $x\in I$ and $u\in J$ then notice that $[x,u] \subset [1,3]$ because of $|x-u|<1.$ Since $g$ is uniformly continuous on $[1,3]$ you are again done.

Ssay
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