How to show that $f(x)=1/x^2$ is uniformly continuous on $x\ge1$?

Question

For my homework I have to show that $f(x)=1/x^2$ is uniformly continuous on the interval $[1,∞)$, and I have absolutely no idea what to do.

Please don't give me the answer, but if you could push me in the right direction, I would really appreciate it.

Answer 1

First way

You have in general that if $f:[a,\infty [\to\mathbb R$ is continuous and such that $\lim_{x\to\infty }f(x)=\ell\in\mathbb R$, then $f$ is uniformly continuous. Let $\varepsilon>0$ and $M>a$ s.t. $$|f(x)-\ell|<\frac{\varepsilon}{2}$$ if $x\geq M$. Then $f$ is uniformly continuous on $[a,M]$ (continuous on a compact), and if $x,y>M$

$$|f(x)-f(y)|\leq |f(x)-\ell|+|f(y)-\ell|<\varepsilon$$ therefore it's also uniformly continuous on $[M,+\infty [$. Therefore if $\varepsilon>0$, there is a $\delta>0$ such that $$|x-y|<\delta\implies |f(x)-f(y)|<\varepsilon$$ for all $x,y\in [a,M]$ or $x,y\geq M$.

If $x<M<y$ are such that $|x-y|<\delta$, then

$$|f(x)-f(y)|\leq \underbrace{|f(x)-f(M)|}_{\leq \varepsilon}+\underbrace{|f(M)-f(y)|}_{\leq \varepsilon}<2\varepsilon.$$

The case $y\leq M\leq x$ is the same, what prove the claim.

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Second way

Your function is derivable on $[1,\infty [$. By mean value theorem, for all $x,y\geq 1$ there is a $|c_{x,y}-x|\leq |x-y|$ s.t. $$|f(x)-f(y)|= \underbrace{|f'(c_{x,y})|}_{\leq 2}|x-y|\leq 2|x-y|$$ Then, if $\varepsilon>0$, set $\delta=\frac{\varepsilon}{2}$ and you get$$\forall x,y\geq 1,|x-y|<\delta\implies |f(x)-f(y)|<\varepsilon,$$ what prove the claim.

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Third way

(after the deleted answer of clarinetist)

$$\left|\frac{1}{x^2}-\frac{1}{y^2}\right|=\left|\frac{x^2-y^2}{x^2y^2}\right|=\frac{x+y}{x^2y^2}|x-y|.$$

But if $x,y\geq 1$, $$\frac{x+y}{x^2y^2}=\frac{1}{xy^2}+\frac{1}{x^2y}\leq 2$$ and thus $$|\frac{1}{x^2}-\frac{1}{y^2}|\leq 2|x-y|.$$

Let $\varepsilon>0$. If you set $\delta=\frac{\varepsilon}{2}$ you get $$\forall x,y\geq 1, |x-y|<\delta\implies \left|\frac{1}{x^2}-\frac{1}{y^2}\right|<\varepsilon,$$ what prove the claim.

Answer 2

Since $f'(x)=-\frac{2}{x^3}$ is bounded on $I=[1,+\infty)$, $\left|f'(x)\right|\leq 2$ gives that $f(x)$ is Lipschitz-continuous on $I$, hence uniformly continuous.