Check if the mappings
$\mathbb{R}\to\mathbb{R},x\mapsto x^2$ and $[0,\infty[:\mathbb{R},x\mapsto \sqrt{x}$
are uniformly continuous.
I was going through some old exams our teacher gave us and this was one of the problems in there.
I have troubles appraoching this. I mean, I know the definition of uniform continuity but I don't know how to apply it to find solutions to explicit problems.
Can anyone show me how to solve this type of things?
For $f:\Bbb R \to \Bbb R$ with $f(x)=x^2$, we proceed as follows:
Suppose $f$ is uniformly continuous on $\Bbb R$. Then for every $\epsilon \gt 0$, $\exists$ $\delta \gt 0$ such that $|f(x)-f(y)| \lt \epsilon$ whenever $|x-y| \lt \delta$ for any $x,y \in \Bbb R$.
Let $x=a$ and $y=a+\frac \delta2$. Then we have $|x-y|=|a-(a+\frac \delta2)| \lt \delta$. Hence $|f(x)-f(y)|=|x^2-y^2|=|a^2-(a+ \frac \delta2)^2| \lt \epsilon$. $\Rightarrow$ $|a\delta+\frac {\delta^2}{4}| \lt \epsilon$. Hence $|a\delta| \lt \epsilon$ too. $\Rightarrow$ $|a| \lt \frac {\epsilon}{\delta}$ $\forall a \in \Bbb R$. This is impossible! Hence $f$ is not uniformly continuous on $\Bbb R$.
For $g:[0,\infty) \to \Bbb R$ with $g(x)=\sqrt x$, we proceed as follows:
Consider the interval $[0,2]$. Then this interval is closed and bounded and hence from Heine-Borel theorem it is compact in $\Bbb R$. Also since $g$ is continuous on $[0,\infty)$ it is continuous on $[0,2]$ too. So we get that $g$ is a continuous function on a compact set. Hence $g$ is uniformly continuous on the interval $[0,2]$.
Now consider the interval $[1,\infty)$. Then for $x,y \in [1,\infty)$ we have $|g(x)-g(y)|=|\sqrt x-\sqrt y|=\frac {|x-y|}{\sqrt x+\sqrt y} \lt \frac {|x-y|}{2}$. We can see that $g$ is a Lipschitz function on $[1,\infty)$ with the Lipschitz constant $\frac 12$. Hence $g$ is uniformly continuos on $[1,\infty)$ too!
Combining the two arguments(about the interval $[0,2]$ and the interval $[1,\infty)$), we see that $g$ is uniformly continuous on $[0,\infty)$.