I'm trying to learn Uniform Continuity in Real Analysis, so I made myself some exercises to solve and improve on $\epsilon-\delta$ proofs, but I couldn't solve one of the exercises:
"Prove that all functions of the form $f(x) = a\sqrt{x + b}$, where $a, b \in \mathbb{R}$ are uniformly continuous on $[0, \infty)$". Here's what I tried:
Let $x, y \in \mathbb{R}$. For $\delta = ...$, we have that
$|f(x) - f(y)| < \epsilon \iff$
$|a\sqrt{x + b} - a\sqrt{y + b}| < \epsilon \iff$
$|a(\sqrt{x + b} - \sqrt{y + b})| < \epsilon \iff$
$|a||\sqrt{x + b} - \sqrt{y + b}| < \epsilon \iff$
$|\sqrt{x + b} - \sqrt{y + b}| < \frac{\epsilon}{|a|} \iff$
But I don't understand how to continue from here. How should I continue this?
Thanks!
Given $\epsilon > 0$, you want to find $\delta > 0$ so if $x,y \in [-b,\infty)$ and $|x-y| < \delta$, $|a\sqrt{b+x} - a \sqrt{b+y}| < \epsilon$. Of course if $a = 0$, any $\delta$ will do, so let's suppose $a \ne 0$. Now you want $|\sqrt{b+x} - \sqrt{b+y}| < \epsilon/|a|$, so it will be convenient to write $\epsilon/|a| = \nu$. While we're at it, we may as well get rid of that $b$, so let $s = b + x$ and $t = b + y$. So now you want $\delta > 0$ so if $s, t \in [0,\infty)$ and $|s-t| < \delta$, $|\sqrt{s} - \sqrt{t}| < \nu$.
We break things up into two cases, one where $s$ is close to $0$ and one where it isn't. Let's see: if $s \le \alpha$ and $|s-t| \le \alpha$, then $t \le 2 \alpha$, and $|\sqrt{s} - \sqrt{t}| \le \max(\sqrt{s}, \sqrt{t}) \le \sqrt{2\alpha}$. So for this part, we want $\sqrt{2\alpha} < \nu$, and we can take $0 < \alpha < \nu^2/2$.
Thus if $s \in [0, \alpha]$ $t \in [0,\infty)$ and $|s-t| < \alpha$, $|\sqrt{s} - \sqrt{t}| < \nu$.
For the second case, where $s \ge \alpha$, we can use
$$ \left|\sqrt{s} - \sqrt{t}\right| = \frac{|s-t|}{\sqrt{s}+\sqrt{t}} \le \frac{|s-t|}{\sqrt{\alpha}}$$
This $< \nu$ if $|s-t| < \nu \sqrt{\alpha}$.
So we take $\delta < \min(\alpha, \nu \sqrt{\alpha})$, and we're done.