Is $f(x) = \frac{1}{x^2}$ uniformly continuous on $(0,1]$ or $[1, \infty)$ or both?

Question

Here is the question I am trying to answer:

Are the following functions uniformly continuous? Prove it.

$(a)$ $f: (0, 1] \to \mathbb R$ with $f(x) = \frac{1}{x^2}.$

$(b)$ $f: [1, \infty) \to \mathbb R$ with $f(x) = \frac{1}{x^2}.$

My guess for the second, yes as it is continuous on a the given closed interval and hence it is uniformly continuous. For the first one I do not know, could someone help me please in answering these questions?

Answer 1

I will make my comment into an answer. First let's show that if $f$ is unifromly continuous on $(0,1]$ then it's continuous on $[0,1]$

Let $f:(0,1]\rightarrow \Bbb R$ be uniformly continuous and consider a sequence $\{x_n\}$ converging to $0$, so $\{x_n\}$ is cauchy. Now a uniformly continuous function sends Cauchy sequences to Cauchy sequences i.e. $\{f(x_n)\}$ is also Cauchy, hence $\{f(x_n)\}$ converges to some limit $l\in \Bbb R$. Again using uniform continuity you can show this limit is independent of choice of sequence converging to $0$ i.e. if both $x_n,y_n$ are sequences converging to $0$, this implies that $|x_n-y_n|$ can be made arbitrarily small for large $n$, so by uniform continuity $|f(x_n)-f(y_n)|$ can be made arbitrarily small for large $n$ and since both $\{f(x_n)\},\{f(y_n)\}$ are convergent, they converge to the same limit. So we can extend $f$ by defining $f(0)=l$. This gives a continuous extension of $f$ on $[0,1]$.

Now if the first one was uniformly continuous, we'd have that $\frac{1}{x^2}$ is continuous on $[0,1]$ which is not true.

For the second half, you can do what I wrote in the comment or notice that $f'(x)=\frac{-2}{x^3}$. Since $f':[1,\infty)\to\mathbb{R}$ and $\lim_{x\to\infty}f(x)=0$, you can show that $f'$ is bounded by some $M$. Therefore by Lagrange's theorem, for every $x,y\in[1,\infty)$ we have: $$|\frac{f(x)-f(y)}{x-y}|=|f'(c)|\leq M$$ And therefore $|f(x)-f(y)|\leq M|x-y|$. So $f$ is what we call Lipschitz in it's domain of definition. It's easy to see that if a function is Lipschitz, it's also uniformly continuous.

Answer 2

Ad a): It is $\int_0^1\frac{1}{x^2}dx = \infty$, hence not uniformly continuous. Ad b): It is $|\frac{1}{x^2}-\frac{1}{y^2}| = |\frac{-2}{\xi^3}||x-y| \leq 2|x-y|$, where $\xi \in [x,y]$, hence Lipschitz and therefore uniformly continuous.