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Prove that if $f: [a, \infty) \rightarrow \Bbb R$ is continuous on $[a, \infty)$ and has a finite limite $ \lim_{x\to \infty} f(x)$, then f is uniformly continuous.

I found such definition of uniform continuity: $\forall x_n,y_n \subset [a, \infty) \lim_{x\to \infty} x_n-y_n=0 \Rightarrow \lim_{x\to \infty} f(x_n)-f(y_n)=0$

and tried to do this using Heine's definitions but i got stuck at one point and can't move on. I'd really appreciate any hints!

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up vote 4 down vote accepted

I will do this with the classical definition of uniform continuity: $$\forall \epsilon>0\exists \delta>0:\ \forall x,y\ge a\ \ \left|x-y\right|<\delta\implies\left|f(x)-f(y)\right|<\epsilon$$ Since $\lim_{x\to +\infty}f(x)=L$, for $\epsilon>0$, $$\exists M>0:\forall x\ge a\ \ x\ge M\implies\left|f(x)-L\right|<\frac{\epsilon}3$$ But $f$ is continuous on $[a,M]$ and so uniformly continuous there: $$\exists \delta>0:\forall x,y\in [a,M]\ \left|x-y\right|<\delta\implies\left|f(x)-f(y)\right|<\frac{\epsilon}{3}$$ But for $x,y>M$, $$\left|f(x)-f(y)\right|\le \left|f(x)-L\right|+\left|f(y)-L\right|<\epsilon$$ The case that remains is $x\in [a,M],y>M$: $f$ is continuous at $M$ and so $$\exists \delta_1>0:\forall x\in [a,+\infty]\ \left|x-M\right|<\delta_1\implies\left|f(x)-f(M)\right|<\frac{\epsilon}{3}$$ Then, if $x\in [a,M],y>M$ and $\delta_2=\min\left\{\delta,\delta_2\right\}$, $$\left|f(x)-f(y)\right|\le \left|f(x)-f(M)\right|+\left\|f(M)-L\right|+\left|f(y)-L\right|<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon$$ whenever $\left|x-y\right|<\delta_2$ and $x\in [a,M],y>M$

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I was going to post more or less the same argument with the following simplification: Choose $M\ge a$ as above. Now select a $\delta_0$ so that $|f(x)-f(y)|<\epsilon$ whenever both $x$ and $y$ are members of $[a, \color{maroon}{M+1}]$. Finally, choose $\delta=\text{min}\{1, \delta_0\}$. Suppose $|x-y|<\delta$. Then it follows that either both $x$ and $y$ are in $[a,M+1]$, or both are in $[M,\infty$). In either case, we have $|f(x)-f(y)|<\epsilon$. – David Mitra Dec 24 '12 at 15:58
I have a couple of questions: 1. why is f uniformly continuous on $[a,M]$? 2. Why is it that $\left|f(x)-f(y)\right|\le \left|f(x)-L\right|+\left|f(y)-L\right|$ – Max Dec 24 '12 at 16:28
@Max The first is by the Heine Cantor Theorem (you should have been taught this). The second is because $\left|f(x)-f(y)\right|=\left|(f(x)-L)-(f(y)-L)\right|\le \left|f(x)-L\right|+\left|f(y)-L\right|$ – Nameless Dec 24 '12 at 16:31
I have just checked my notes and yes - I have been taught. Also why $|f(M)−L|<\frac{\epsilon}{3}$? – Max Dec 24 '12 at 17:15
yes, but $|f(x)-f(y)|< \frac{\epsilon}{3}$ was for $x,y \in [a,M]$ – Max Dec 24 '12 at 17:22

Let $\epsilon>0$. Let $\lim_{x\rightarrow\infty}f(x)=L$. Choose a real number $M>a$ such that: $$\forall x\geq M[|f(x)-L|<\frac{\epsilon}{2}]$$ Since $f$ is continuous, therefore $f$ is uniformally continionus on $[a,M]$. Thus, there exists $\delta_1>0$ such that $\forall x,y\in[a,M] [|x-y|<\delta_1\implies|f(x)-f(y)|<\epsilon]$. Since $f$ is continuous at $M$, there exists $\delta_2>0$ such that $\forall x [|x-M|<\delta_2\implies|f(x)-f(M)|<\epsilon/2$. Now let $\delta=min\{\delta_1,\delta_2\}$

*Claim:*$\forall x,y\in [a,\infty)[|x-y|<\delta\implies|f(x)-f(y)|<\epsilon]$

Proof: Consider the cases when $x,y$ belong to the interval $[a,M]$ or do not . The idea is that we have made sure that $x,y$ are close enough such that $f(x)$ is close to $f(y)$.

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