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Suppose that $f : [a, ∞) → R$ is a continuous function. If $\lim\limits_{x→∞} f (x) = L$, prove that $f$ is uniformly continuous on $[a, ∞)$.

My attempt at the proof:

Well since I have to use both facts, I think my proof needs to be divided into two parts:

1- I have to prove that $f$ is uniformly continuous on $(N,∞)$ where $N>a$ (by using the limit definition somehow)

2- Using compactness, I can easily show that $f$ is uniformly continuous on $[a,N]$

I am having trouble with $(1)$ because I can't find an $\delta>0$ that works for all $\epsilon>0$

Just need a hint in the right direction, thank you!!

Martin Sleziak
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wwbb90
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  • Why do you think you need a $\delta > 0$ that works for all $\epsilon > 0$ ? You need a $\delta$ that works for all $x$ – mercio Oct 28 '13 at 08:14
  • Hmmm I don't know, I understood from the definition of uniform continuity that i need to find a unique $\delta $ that works for every $x$ and for every $\epsilon$. Not sure if it's correct though – wwbb90 Oct 28 '13 at 08:35
  • the definition of uniform continuity is $\forall \epsilon > 0 \exists \delta > 0 \forall x,y \in [a,\infty), |x-y| < \delta \implies |f(x)-f(y)| < \epsilon$. If someone gives you an $\epsilon$, you have to show them a $\delta$ such that the final implication holds for all $x,y$. – mercio Oct 28 '13 at 08:37
  • yup its a duplicate alright – wwbb90 Oct 28 '13 at 09:11

1 Answers1

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Suppose $\lim_{x\to \infty} f(x) = L$, then for any $\epsilon >0$, there is $N \in \mathbb{R}$ such that $$ x \geq N \Rightarrow |f(x) - L| < \epsilon/3 $$ Hence, for any $x,y > N$ $$ |f(x) - f(y)| < \epsilon/3 $$ Now, $f$ is uniformly continuous on $[a,N]$, so there is a $\delta > 0$ such that $$ |x-y|<\delta, \text{ and } x,y \in [a,N]\Rightarrow |f(x) - f(y)| <\epsilon/3 $$ Hence, for any $x,y \in [a,\infty)$, if $x, y \in [a,N]$ or $x,y \geq N$, then $$ |x-y|<\delta \Rightarrow |f(x) - f(y)| < \epsilon \qquad\text{(1)} $$ Furthermore, if $x < N < y$, then $|x-y|<\delta$ implies that $$ |x-N| < \delta, \text{ and } |f(y) - L| < \epsilon/3, |f(N) - L|<\epsilon/3 $$ hence $$ |f(x)-f(y)| \leq |f(x) - f(N)| + |f(N) - L| + |L - f(y)| < \epsilon $$ Hence, (1) holds for all $x,y\in [a,\infty)$

Prahlad Vaidyanathan
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