Proving that $f(x) = 6x^2$ is not uniformly continuous

Question

The question is:

Prove $f(x) = 6x^2$ is not uniformly continuous on the interval $[2, \infty)$.

I have absolutely no idea what to do. Any help is appreciated.

Answer 1

$f:[2,\infty)\to \Bbb R$ is uniformly continuous if for every $\epsilon>0$ you can find some $\delta>0$ where any time there are $x,y\in [2,\infty)$ that satisfy $|x-y|<\delta$ you have $|6x^2-6y^2|<\epsilon$.

To show that this doesn't hold, you need to find some $\epsilon>0$ where there is no such $\delta>0$.

Say that you choose $\epsilon=1$. Is this possible? Choose some $y$ based on $\delta$ and $x$.

Answer 2

VERY SIMPLE TRICK

Given $a>0$ if we take $$x = \frac{1}{a}+2 ~~~and ~~~y = \frac{1}{a}+2+\frac{a}{2} \color{red}{= x +\frac{a}{2}} $$ we have $$x,y\in [2,\infty)~~~ and~~~|x-y| =\frac{a}{2}<a $$

But since $ax = 1+ 2a$ we have

$$|f(x)-f(y)| =6 |y^2-x^2|= 6| x^2 + 2\frac{a}{2}x +\frac{a^2}{4} - x^2 |\\ = 6(ax +\frac{a^2}{4}) = 6(1+2a+\frac{a^2}{4} )>6$$

That is $$|f(x)-f(y)| >6$$

Thus $$\color{blue}{\exists \varepsilon_0 =6,\forall~a>0, \exists ~x,y\in [2,\infty): |x-y|< a~~and ~~|f(x)-f(y)|>6}$$

just take $$x = \frac{1}{a}+2 ~~~and ~~~y = \frac{1}{a}+2+\frac{a}{2} \color{red}{= x +\frac{a}{2}} $$