Let \(f:[0,\infty)\to\mathbb R\) be continuous and suppose \[ \lim_{x\to\infty}f(x) \] exists and is finite. Must \(f\) be uniformly continuous? Give a proof or a counterexample.
Solution
Let \[ L\coloneqq\lim_{x\to\infty}f(x). \]
1. For every \(\varepsilon>0\), there is \(M\ge0\) such that\[ x\ge M \quad\Longrightarrow\quad \abs{f(x)-L}<\frac{\varepsilon}{3}. \]
This is the definition of the finite limit \(L\) at infinity.
2. For the \(M\) from step 1, there is\(\delta\in(0,1]\) such that whenever\(x,y\in[0,M+1]\) and \(\abs{x-y}<\delta\), one has\[ \abs{f(x)-f(y)}<\varepsilon. \]
The function \(f\) is continuous on the compact interval \([0,M+1]\). Hence it is uniformly continuous there. Choose a corresponding positive modulus and decrease it if necessary so that \(\delta\le1\).
3. If \(x,y\ge0\) and \(\abs{x-y}<\delta\), then\[ \abs{f(x)-f(y)}<\varepsilon. \]
Interchange \(x\) and \(y\) if necessary so that \(x\le y\).
If \(x\ge M\), then also \(y\ge M\), and step 1 gives \[ \begin{aligned} \abs{f(x)-f(y)} &\le \abs{f(x)-L}+\abs{L-f(y)}\\ &<\frac{2\varepsilon}{3} <\varepsilon. \end{aligned} \]
If \(x<M\), then \[ y<x+\delta\le M+1. \] Thus \(x,y\in[0,M+1]\), and step 2 gives \(\abs{f(x)-f(y)}<\varepsilon\).
4. The answer is \(\boxed{\text{yes}}\): the function \(f\) isuniformly continuous on \([0,\infty)\).
Given any \(\varepsilon>0\), steps <1>1–<1>3 produce a single \(\delta>0\) that works for every \(x,y\ge0\). This is uniform continuity.
5. Q.E.D.
Step 4 gives the required answer and proof.