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Suppose $f$ is a continuous real valued function on $[0,+\infty)$ where $\lim_{x\to\infty}f(x)=M$ for some $M\in \mathbb{R}$. Prove that $f$ is uniform continuous.

Attempts: Suppose the contrary $f$ is not uniform continuous and hence $\exists \epsilon>0,\forall \delta>0 \exists x,y\in\mathbb{R},s.t. d(x,y)<\delta\implies d(f(x),f(y))\ge\epsilon $. Then i try to show that it is not converging or not continuous but not sure how show it.

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What is the domain? $[0,\infty)$? If so: Note that you may first choose $N$ so that $f$ is uniformly close to $M$ on $[N,\infty)$. Then you can use the fact that $f$ is uniformly continuous on $[0,N+1]$. –  David Mitra Jan 30 '13 at 13:57
    
Counterexample: Let $f:(0,\infty)\to \mathbb{R}$, $f(x)=\sin(1/x)$ –  tetori Jan 30 '13 at 13:57

2 Answers 2

Initial remark: The only little trick is that on have to use uniform continuity on a larger interval than $[0,A]$ where $A$ is obtained from the limit condition. This is to cvoer the cases $x<A<y$ and yet $|x-y|<\delta$.

Take $\epsilon>0$.

By the limit condition, there exists $A>0$ such that $M-\epsilon/2<f(x)<M+\epsilon/2$ for all $x\geq A$.

Now $f$ is continuous on the compact interval $[0,A+1]$, so it is uniformly continous and there exists $\delta>0$ such that $|f(x)-f(y)|<\epsilon$ for all $x,y\in [0,A+1]$ such that $|x-y|<\delta$.

Now replace $\delta$ by $\min(\delta,1)$ if needed.

It follows that $|f(x)-f(y)|<\epsilon$ for all $x,y\in [0,+\infty)$ such that $|x-y|<\delta$.

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$\lim_{x\to\infty}f(x)=M \implies $ given $\epsilon >0 $ $\exists A >0$

s.t.$\forall x \geq A$ $ |f(x) -M| \le \frac{\epsilon}{2}\tag {1}$

$\forall x_1$ and $x_2 \geq A$ $ |f(x_1) -f(x_2)| < \epsilon$ from $(1)$

$f$ is continuous on $[0,A] \implies f$ is uniformly continuous on $[0.A]$ and from $(1)$ $ f$ is uniformly continuous on $[A ,\infty) \implies f $ is uniformly continuous on $[0,\infty)$

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You need to use uniform continuity on a larger interval thatn $[0,A]$. Because what happens if $x<A<y$ and yet $|x-y|<\delta$? As it is stated, your proof is not complete. –  1015 Jan 30 '13 at 14:13
    
@julien $|f(y) - f(x)| \leq |f(y) - f(A) | + |f(A) - f(x) | \leq \epsilon$ –  jim Jan 30 '13 at 14:24
    
Of course! Thanks. –  1015 Jan 30 '13 at 16:49

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