Answer to Question #140416 in Calculus for Mathew

Question

The function f(x)=x is uniformly continuous on R but the function f(x)=x^2 is not.Prove

Answer 1

Definition: A function $f:Dom(f)\to \mathbb{R}$ is said to be uniformly continuous if given $\epsilon >0$ , there exists $\delta>0$ such that whenever $|x-y|<\delta, |f(x)-f(y)|<\epsilon \forall x,y\in Dom(f)$ .

We want to show that there exists $\epsilon>0$ such that for every $\delta >0 \exist x,y\in\mathbb{R}$ such that $|x-y|<\delta$ but, $|f(x)-f(y)|\geq\epsilon$ . That is, $|x^2-y^2|\geq\epsilon$

Since $f(x)=x^2$ ,

$|f(x)-f(y)|=|x^2-y^2| =|(x-y)(x+y)|\\ =|x-y||x+y|$ .

Let $\epsilon=1$ , for any $\delta>0$ , consider $y=\frac{1}{\delta}$ ,

From $|x-y|<\delta,x=\delta+y$

We have,

$|x-y||x+y|=|\delta+y-y||\delta+y+y|\\=\delta\left(\delta+\frac{2}{\delta}\right)=\delta^2+2>1=\epsilon$

$\implies |x^2-y^2|>\epsilon$

Hence, $f(x)=x^2$ is not uniformly continuous on $\mathbb{R}$ .