Question

Show that the function f defined by f(x) =1//x^2 is uniformly continuous on [5 ,∞].

Answer 1

ANSWER

Let $\epsilon >0.$ By the definition, we need to show there exists $\delta >0$ , such that

$a\geq 5 , b\geq 5$ and $|a-b|<\delta$ imply $|f(a) - f(b)|< \epsilon$ (1)

The function $f(x)=\frac{1}{x^{2}}$ is differentiable on the set $[5,\infty)$ . $f'(x)=\frac {-2} { x^{3}}$ and $|f'(x)|\leq \frac{2}{5^{3}}<1$ for all $x\in [5,\infty).$ Let $b>a\geq 5$ , according to the mean value theorem, there exists point $c\in (a,b)$ such that $f(b)-f(a) =f'(c)\cdot (b-a)$. So, for all $a,b\in [5,\infty)$ we have

$|f(b)-f(a)|\leq |b-a|$ .

Hence, if $\epsilon >0$ and $\delta=\epsilon$ , then

$|a-b|<\delta =\epsilon \Rightarrow |f(b)-f(a)|\leq |b-a|<\delta=\epsilon$

Those, (1) is satisfied, which means that the function is uniformly continuous.