Let, $f$ be a cont. func. on $\Bbb{R}$ s.t. $|f(x)| \le {A \over 1+x^2} \forall x \in \Bbb{R}$, $A \gt 0$. Prove, $f$ is unif'ly cont. on $\Bbb{R}$

Question

Firstly, I am sorry, I have not given the full question on the title part due to word limit. The actual question says-

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Let, $f:\Bbb{R} \rightarrow \Bbb{R}$ be a continuous function such that $|f(x)| \le {A \over 1+x^2} \forall x \in \Bbb{R}$, where $A \gt 0$ is a constant. Show that, $f(x) \to 0$ as $x \to \pm\infty$. Use this prove that, $f$ is uniformly continuous on $\mathbb{R}$.

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I have proved the first part of this question which is $f(x) \to 0$ as $x \to \pm\infty$ by applying Squeeze Test(or Sandwich Theorem) on $-{A \over 1+x^2} \le f(x) \le {A \over 1+x^2} \forall x \in \Bbb{R}$ because $\lim_{x \to \pm\infty}{A \over 1+x^2}=0$.
But I am stuck with the second part of the problem which asks to prove the uniform continuity of the function on the whole real line using the above result of the first part.
Can anyone help me find out a proper wayout to prove the second part using the result on first part(i.e.$f(x) \to 0$ as $x \to \pm\infty$)?
Thank you for your help in advance.

Answer 1

Let $\varepsilon>0$. We need to exhibit a $\delta_{\varepsilon}>0$ such that $|x-y|\leq \delta_{\varepsilon}$ implies $|f(x)-f(y)|\leq \varepsilon$. The idea is to partition $\mathbb{R}$ into two sets where we can control the variation of $f$ and then use the triangle inequality to obtain uniform convergence on the whole $\mathbb{R}$.

1. Since $\lim_{|x|\to+\infty}f(x)\to 0$, there is a $M>0$ such that $|f(x)|\leq \varepsilon/4$ for all $x\in \mathbb{R}\setminus (-M,M)$. Hence for all $x,y\in \mathbb{R}\setminus (-M,M)$ we have $|f(x)-f(y)|\leq \varepsilon/4+\varepsilon/4\leq \varepsilon/2$ (notice that this includes $x,y=\pm M$; this will be used later).

2. Since $f$ is uniformly continuous on the compact $[-M,M]$ (by the Heine-Cantor theorem), there is a $\delta_{\varepsilon}>0$ such that $|x-y|\leq \delta_{\varepsilon}$ implies $|f(x)-f(y)|\leq \varepsilon/2$ for all $x,y\in [-M,M]$ (again this includes $x,y=\pm M$).

3. Now let $x,y\in\mathbb{R}$, with $|x-y|\leq \delta_{\varepsilon}$. We have already obtained $|f(x)-f(y)|\leq \varepsilon$ if both $x$ and $y$ lie either in $[-M,M]$ or outside, so suppose for instance $x\in [-M,M]$ and $y>M$. Then $x,y\in [-M,M]$ and $|x-M|\leq |x-y|\leq \delta_{\varepsilon}$ and $M,y\in \mathbb{R}\setminus (-M,M)$, so $$|f(x)-f(y)|\leq |f(x)-f(M)|+|f(M)-f(y)|\leq \varepsilon/2+\varepsilon/2\leq \varepsilon $$ the same goes in the case where $x<-M$ and $y\in [-M,M]$.

Bonus facts: the proof can be easily adapted to the more general case where $\lim_{x\to +\infty}f(x)=L$ and $\lim_{x\to -\infty}f(x)=L'$ for any $L,L'\in \mathbb{R}$. On the other hand, it is not true that any continuous and bounded function on $\mathbb{R}$ is uniformly continuous, take for instance $f(x)=\sin x^2$.