Q:

$f:[0,\infty) \rightarrow \mathbb{R}$ is continuous, prove uniform continuity on $[0,\infty)$ given...

$f:[0,\infty) \rightarrow \mathbb{R}$ is continuous.
Given $f$ satisfies $\epsilon>0$ there exists $M_{\epsilon} > 0$ such that $x>M_{\epsilon} \implies |f(x)| < \epsilon$
Prove uniform continuity on $[0,\infty)$. 

Since $[0,\infty)$ is a subset of $\mathbb{R}$, $f$ is uniformly continuous if for each $\epsilon > 0$ there is a $\delta > 0$ such that $||F(x) - F(y) ||<\epsilon$ when $x,y \in [0,\infty)$ and $||x-y||<\delta$
I also know that a continuous function on a closed bounded interval is uniformly continuous but I'm not sure that this fact helps us. I am stuck on how the given information helps us.

A:

Hint: You already said you know a continuous function on a closed bounded interval is uniformly continuous. And your function tends to zero as $x \to \infty$. So just choose $M$ large enough so that the absolute value of the function is guaranteed to be less than $\epsilon/2$ for $x > M$, and then choose $\delta$ for uniform continuity on the interval $[0,M]$.

A:

Let $\epsilon > 0$. There exists $M_{\epsilon/2}$ such that $|f| < \epsilon/2$ on $[M_{\epsilon/2}, \infty)$. Since $f$ is continuous on the closed interval $[0, M_{\epsilon/2}]$, $f$ is uniformly continuous on $[0,M_{\epsilon/2}]$. Hence, there exists $\delta > 0$ such that for all $x,y \in [0,M_{\epsilon/2}]$, $|x - y| < \delta$ implies $|f(x) - f(y)| < \epsilon/2$. Therefore, given $x,y \in [0,\infty)$ and $|x - y| < \delta$, either

$x,y\in [0,M_{\epsilon/2}]$, which implies $|f(x) - f(y)| < \epsilon/2 < \epsilon$.
$x\in [0,M_{\epsilon/2}], y\in [M_{\epsilon/2},\infty)$, which implies $$|f(x) - f(y)| \le |f(x) - f(M_{\epsilon/2})| + |f(M_{\epsilon/2}) - f(y)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} < \epsilon.$$
$x\in [M_{\epsilon/2}, \infty), y\in [0, M_{\epsilon/2}]$, which gives the same result as in $2$.
$x,y\in [M_{\epsilon/2}, \infty)$, which implies $$|f(x) - f(y)| \le |f(x)| + |f(y)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$

Thus, for all $x,y\in [0,\infty)$, $|x - y| < \delta$ implies $|f(x) - f(y)| < \epsilon$. Since $\epsilon$ was arbitrary, $f$ is uniformly continuous on $[0,\infty)$.

