# Show f is unif. cont. on $\Bbb{R}$ if it is cont. on $\Bbb{R}$ and $\lim\limits_{x\to +\infty}f(x)$ and $\lim\limits_{x\to -\infty}f(x)$ exist

Sequel to the question I asked Prove that if f is continuous on $[a,\infty)$ and if $\lim\limits{x\to \infty}f(x)$ exists then $f$ is uniformly continuous and the reply I got, I want to show that $f$ is uniformly continuous on $\Bbb{R}$ if it is continuous on $\Bbb{R}$ and $\lim\limits_{x\to +\infty}f(x)$ and $\lim\limits_{x\to -\infty}f(x)$ exist

Let $\epsilon>0$ be given.

$\lim\limits_{x\to -\infty}f(x):=k$ exists at $N_\epsilon \in \Bbb{R} \implies\;\exists\,N_\epsilon<a,$ s.t. $\forall \;x< N_\epsilon,\;\;|f(x)-k|<\epsilon/2.$ The set $[N_\epsilon,a]\subset(-\infty,a]$ is a compact set and so, $f$ is uniformly continuous on $[N_\epsilon,a]$, i.e. $\exists\;\delta_1=\delta_1(\epsilon)>0\;$ s.t. $\forall\;x,y\in [N_\epsilon,a]$ with $|x-y|<\delta_1$, we have $|f(x)-f(y)|<\epsilon$.

$\lim\limits_{x\to \infty}f(x):=l$ exists at $M_\epsilon \in \Bbb{R} \implies\;\exists\,M_\epsilon>a,$ s.t. $\forall \;x> M_\epsilon,\;\;|f(x)-l|<\epsilon/2.$ The set $[a,M_\epsilon]\subset[a,\infty)$ is a compact set and so, $f$ is uniformly continuous on $[a,M_\epsilon]$, i.e. $\exists\;\delta_2=\delta_2(\epsilon)>0\;$ s.t. $\forall\;x,y\in [a,M_\epsilon]$ with $|x-y|<\delta_2$, we have $|f(x)-f(y)|<\epsilon$.

Since $f$ is continuous at $N_\epsilon \in \Bbb{R} \implies\;\exists\,\delta_3=\delta_3(N_\epsilon,\epsilon)>0$ s.t. $\forall \;x\in [N_\epsilon,a]$ with $|x-N_\epsilon|<\delta_3$ we have $|f(x)-f(N_\epsilon)|<\epsilon/2.$

Also $f$, continuous at $M_\epsilon \in \Bbb{R} \implies\;\exists\,\delta_4=\delta_4(M_\epsilon,\epsilon)>0$ s.t. $\forall \;x\in [a,M_\epsilon]$ with $|x-M_\epsilon|<\delta_4$ we have $|f(x)-f(M_\epsilon)|<\epsilon/2.$

Take $\delta=\min\{\delta_1,\delta_2,\delta_3,\delta_4\}$ s.t. $\forall\;x,y\in (-\infty,\infty)$ with $|x-y|<\delta$. WLOG, we take $x<y$.

Case 1: $x<N_\epsilon<a<y$

$|f(x)-f(y)|=|f(x)-k+k-f(y)|\leq |f(x)-k|+|f(y)-k|<\epsilon/2+\epsilon/2=\epsilon$

Case 2: $a<M_\epsilon<x<y$

$|f(x)-f(y)|=|f(x)-l+l-f(y)|\leq |f(x)-l|+|f(y)-l|<\epsilon/2+\epsilon/2=\epsilon$

Case 3: $N_\epsilon<x<y<a$

$|x-y|<\delta$, we have $|f(x)-f(y)|<\epsilon$

Case 4: $a<x<y<M_\epsilon$

$|x-y|<\delta$, we have $|f(x)-f(y)|<\epsilon$

Case 5: $N_\epsilon<x<a<y$

$|f(x)-f(y)|=|f(x)-f(N_\epsilon)+f(N_\epsilon)-f(y)|\leq |f(x)-f(N_\epsilon)|+|f(y)-f(N_\epsilon)|<\epsilon/2+\epsilon/2=\epsilon$, since $|x-N_\epsilon|<|x-y|<\delta$ as well as $|y-N_\epsilon|<|x-y|<\delta$

Case 6: $N_\epsilon<x<a<y$

$|f(x)-f(y)|=|f(x)-f(M_\epsilon)+f(M_\epsilon)-f(y)|\leq |f(x)-f(M_\epsilon)|+|f(y)-f(M_\epsilon)|<\epsilon/2+\epsilon/2=\epsilon,$ since $|x-M_\epsilon|<|x-y|<\delta$ as well as $|y-M_\epsilon|<|x-y|<\delta$

Hence, $f$ is uniformly continuous on $\Bbb{R}$.

Please, I'm I right? If no, can anyone please provide a detailed proof?

## 1 Answer

Your proof looks fine, but it is a little hard to read and consequently hard to find bugs in the proof.

It helps to give the reader an idea of what you are doing. A 'proof plan' as such. This is a subjective issue, and really depends on your audience. On advantage of a narrative is that even if you make a typographical mistake, it is clear where you are going, so it is 'self repairing' to some extent.

Here is how I would write it:

Let $\epsilon>0$. We need to find a $\delta>0$ so that if $|x-y| < \delta$ then $|f(x)-f(y)| < \epsilon$. Without any loss of generality, we can decide up front that $\delta \le 1$. This simplifies dealing with infinities.

On a (compact) set of the form $[-M,M]$, $f$ is uniformly continuous, so the idea is to choose $M$ so that $f$ varies by less than $\epsilon$ on $(M,\infty)$ and $(-\infty, -M)$. The only catch is when $x,y$ straddle $\pm M$, so some finessing is required.

Choose $M$ so that if $x,y \in [M,\infty)$ or $x,y \in (-\infty,-M]$, then $|f(x)-f(y)| < \epsilon$. This is possible because of the limit assumption on $f$.

Now partition $\mathbb{R}$ into $(-\infty, -(M+1)), [-(M+1),M+1], (M+1,\infty)$.

Note that if $|x-y| < 1$ then both $x,y$ must lie in at least one of the intervals $(-\infty, -M)$, $[-(M+1),M+1]$, or $(M,\infty)$. If $x,y$ lie in the first or last interval, then we have $|f(x)-f(y)| < \epsilon$.

Since $[-(M+1),M+1]$ is compact, we can find a $\delta_1>0$ so that $|f(x)-f(y)| < \epsilon$ if $x,y \in [-(M+1),M+1]$ and $|x-y| < \delta$.

Now if we choose $\delta = \min(1,\delta_1)$ we are finished.

• I should add that one always makes mishtakes. – copper.hat May 10 '18 at 17:58