Uniform Continuity of a Continuous Function on $[0, \infty)$

Question

If a function, $f:[0,\infty) \rightarrow \mathbb{R}$ is continuous, and converges to $L \in \mathbb{R}$ as $x \rightarrow \infty$, is the function uniformly continuous on $[0, \infty)$? If so, what is the proof?

I think so, and I was thinking something along the following... Let $\epsilon > 0$ be given. Then, because the function converges, we can find $k \in \mathbb{R}$ such that if $x \geq k$: $| f(x) - L | < \frac{\epsilon}{2}$. Then, $f$ is continuous on $[0, k]$, therefore it is uniformly continuous because $[0, k]$ is a closed, bounded interval. That means, there exists $\delta$ such that if $x, a \in [0,k]$ and $| x - a | < \delta$, we have that $|f(x) - f(a)| < \epsilon$. Notice that this $\delta$ also works for $[k, \infty)$, because all points in $[k, \infty)$ are within $\epsilon$ of each other!

Thus, we've shown $f$ is uniformly continuous on $[0,k]$, and $[k, \infty)$. How can I finish the proof? Is this approach correct?

Answer 1

Almost there! You have to inspect what happens when you have two points $x,y$ such that $\lvert x-y\rvert<\delta$ and $x<k<y$. In this case $$\lvert f(x)-f(y)\rvert\le \lvert f(x)-f(k)\rvert+\lvert f(k)-f(y)\rvert\le 2\varepsilon$$

So, technically a $\delta$ that works for $\varepsilon$ is $$\widehat \delta:=\min\left\{\delta_{\varepsilon /2}^{[0,k]},\delta_{\varepsilon /4}^{L,\,[k,\infty)}\right\}$$

Answer 2

Your basic idea is correct but I don't understand what you mean by "all points in $[k,\infty)$ are within $\varepsilon$ of each other".

Note that to show that $f$ is uniformly continuous, you need to say something about $|f(x) - f(y)|$ whenever $x,y$ are close enough. If you divide $[0,\infty)$ as $[0,k] \cup [k,\infty)$ then you'll be able to use uniform continuity to estimate $|f(x) - f(y)|$ when $x,y$ are both in $[0,k]$ and use the existence of a limit and the triangle inequality (you didn't write this part out) to estimate $|f(x) - f(y)|$ in where $x,y$ are both in $[k,\infty)$ but you are still left with the problem of estimating $|f(x) - f(y)|$ when one of the points is in $[0,k)$ and the other is in $[k,\infty)$. Here, the triangle inequality comes again to your rescue and you have

$$ |f(x) - f(y)| \leq |f(x) - f(k)| + |f(k) - f(y)| \leq |f(x) - f(k)| + |f(k) - L| + |L - f(y)| \\ < \varepsilon + \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = 2\varepsilon $$

(where here we assume that $x \leq k \leq y$).