Uniformly continuous at infinity

Question

Let $a \in \mathbb{R}$ and $f:[a,\infty) \rightarrow \mathbb{R}$ a continuous function. Suppose that there exists $L \in \mathbb{R}$ s.t. $\displaystyle\lim_{x\to\infty}f(x)=L$. Prove that $f$ is uniformly continuous on $[a,\infty)$.

I usually do fine however I am stuck at implementing my idea. I thought to assume the opposite*, then choose $\frac\varepsilon2$ at the limit definition where the epsilon is given by the assumption*. What do I miss?

Thanks

Please write down explicitely what you mean by "assume the opposite, then choose ε/2 at the limit definition where the epsilon is given by the assumption." — Anne Bauval, May 23 '23 at 15:39
This is pretty obvious if you try to understand what I ask instead of looking after what I did not provide. ' — MXO120, May 23 '23 at 16:23
My question was aimed to help you. I did try to understand but your writing was too allusive. If you don't make the effort to present your question and attempt clearly, we cannot help you efficiently, despite all our goodwill and expertise. Quick beginner guide for asking a well-received question + please avoid "no clue" questions. — Anne Bauval, May 23 '23 at 20:45

score 1 · Answer 1 · answered May 23 '23 at 15:57

1

You can prove this by 'splitting' the function. Because there is a limit, we see that:

There exists an $N$ such that $x\geq N\Rightarrow|f(x)-L|<\frac{\epsilon}{2}$.

This means that $|f(x_{1})-f(x_{2})|<\epsilon$ for any $x_{1},x_{2}\geq N$.

Thus on [N,$\infty$) it follows directly that the function is uniformly continuous: whichever $\delta$ you choose, the difference between two function values is always less than $\epsilon$. Thus the 'second half' of the function is uniformly continuous.

The 'first half' of the function is then defined on the closed interval $[0,N]$. It follows from a theorem that a continuous function on a closed interval is uniformly continuous.

answered May 23 '23 at 15:57

UnrulyTank

161

This is what I did however I dont see why this corresponds the definition of uniform continuity. E.g, given a smaller epsilon, the interval $[N,\infty)$ might not satisfy that condition anymore. Am I wrong? – MXO120 May 23 '23 at 16:21
It is important to look at the definition of the limit. For every $\epsilon$ there exists an $N$ such that $x>N\Rightarrow|f(x)-L|<\epsilon$. If you take a smaller epsilon, then there is a different $N$ such that the implication still holds. Intuitively: for a smaller epsilon you need a bigger $N$, but that is allowed. – UnrulyTank May 23 '23 at 18:10
But if you give me some epsilon and I give you N, how does that imply $f$ is uniformly continuous at $[N, \infty)$ – MXO120 May 23 '23 at 18:32
@UnrulyTank This is roughly the idea but there is a small hole in your racket. To locate it, and for a complete and rigorous proof, look at the careful question and answer in the duplicate I provided. – Anne Bauval May 23 '23 at 20:51
@AnneBauval the idea isnt similar to what he suggested – MXO120 May 23 '23 at 21:31
Read more carefully the whole post. – Anne Bauval May 23 '23 at 21:32

Uniformly continuous at infinity

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