Continuity and uniform continuity on $[0,\infty)$

Question

I have a problem proving the following statement:

Let $f$ be a function which is continuous on $[0,\infty)$ and there exists $c$ for some $c>0$ such that $f$ is uniformly continuous on $[c,\infty)$. Prove that $f$ is uniformly continuous on $[0,\infty)$

Maybe I can claim that $[0,c+1]$ has an overlap of length 1 with $[c,\infty)$. Then I can argue that two points $x,y$ such that $|x-y|<1$ is contained in one of $[0,c+1]$ and $[c,\infty)$. But how would I argue $f$ is uniformly continuous on the whole $[0,\infty)$?

And, especially how would I argue that $f(x)=\sqrt{x}$ is uniformly continuous on $[0,\infty)$?

Answer 1

Suppose $f$ is uniformly continuous on $[c, \infty)$, $c > 0$, and is continuous on $[0, \infty)$. Then $f$ is uniformly continuous on $[0, c + 1]$ since $f$ is continous and $[0, c + 1]$ is compact.

Consider that for all $[0, c + 1] \cup [c, \infty) = [0, \infty)$. Consider further that for all $x \in [0, \infty)$, the open ball $B_{1/3}(x) \cap [0, \infty)$ is a subset either of $[0, c + 1]$ or of $[c, \infty)$.

Now consider some $\epsilon > 0$. Take $\delta_1$ such that for all $x, y \in [0, c + 1]$, if $|x - y| < \delta_1$ then $|f(x) - f(y)| < \epsilon$. Take $\delta_2$ similarly for $[c, \infty)$.

Define $\delta = \min(\delta_1, \delta_2, 1/3)$. Consider $x, y \in [0, \infty)$ such that $|x - y| < \delta$. Suppose WLOG that $B_{1/3}(x) \cap [0, \infty) \subseteq [0, c + 1]$. And $y \in B_{1/3}(x) \cap [0, \infty)$. Then $x, y \in [0, c + 1]$. And $|x - y| < \delta \leq \delta_1$, so $|f(x) - f(y)| < \epsilon$. The case where $B_{1/3}(x) \cap [0, \infty) \subseteq [c, \infty)$ is analogous.

So $f$ is uniformly continuous on $[0, \infty)$.

To prove this for the function $f(x) = \sqrt{x}$, note that $f'$ is bounded on the interval $[1, \infty)$ and therefore $f$ is uniformly continous on $[1, \infty)$. Since $f$ is continuous on $[0, \infty)$, the uniform continuity of $f$ follows.