If f: $[0,\infty)\to \mathbb{R}$ is continuous and uniformly continuous in $[k,\infty]$, then it's uniformly continuous in $[0,\infty]$. Let f: $[0,\infty]\to \mathbb{R}$ continuous. Prove that, if f is uniformly continuous in $[k,\infty]$ for some k, then f is uniformly continuous in $[0,\infty]$ I don´t know how to do this, I would appreciate your help

(S.A. Understanding Analysis. pp 119 question 4.4.8)
Hint: It's very relevant that $[0, k]$ is compact, and every continuous function on a compact set is uniformly continuous. Can you combine uniform continuity on $[0, k]$ with uniform continuity on $[k, \infty)$?
(http://answers.yahoo.com/question/index?qid=20090401115020AACCrH3)

S.A. Theorem 4.4.8 pp 118 :
A continuous function on a closed and bounded interval is uniformly continuous.

Let $e &gt; 0$ be given. Since f is uniformly continuous on $[b, \infty)$, this means
there exists $d &gt; 0 : $ $\forall x, y \in [b, \infty)$, $|x-y| &lt; d \implies |f(x) - f(y)| &lt; e$.
We can assume that $\color{red}{d &lt; 1}$. If you aren't solid on this, think about why.

$\color{red}{ \text{If anyone is solid on this, expatiate please. } }$

The question posits f is continuous on $[0,\infty).$ Ergo it's continuous on any subset of it.
Ergo consider $[0, b+1]$, a closed and bounded interval.   $\color{red}{ \text{I think this step is the nub. I don't know how to presage it feyly. Demystify please... } }$
By the agency of Theorem 4.4.8, ergo, $f$ is uniformly continuous there.
We conclude that there is $D &gt; 0$ with the property that
whenever $x,y \in [0, b+1]$ satisfy $|x-y| &lt; D$, then $|f(x) - f(y)| &lt; e.$

Let $m = \min(d, D)$ (or their common value if d = D).
Suppose $x, y \in [0, \infty)$ satisfy $|x-y| &lt; m \le \color{red}{d &lt; 1}.$ Then x and y are less than one unit apart.
It follows that either they both lie in $[0, b+1]$ or they both lie in $[b,\infty).$ 

Why? See picture beneath. If one of them, posit $s$, were $\in [0,b+1]$ but $\not \in [b,\infty)$, then $s&lt; b$.
If the other, posit $t$, were $\not \in [0,b+1]$ but $\in [b,\infty)$, then  $t &gt; b+1$.
And then $|x-y| = |s-t| &gt; 1$.

If x and y are both in $[0, b+1]$; then we know, because $|x - y| &lt; m \le D$, that $|f(x) - f(y)| &lt; e$.

If x and y are both in $[b,\infty)$, then we know because $|x - y| &lt; m \le d$ that $|f(x) - f(y)| &lt; e.$

In both cases, we have deduced that whenever $|x-y| &lt; m$ we have $|f(x) - f(y)| &lt; e$.
So f is uniformly continuous.