My professor did this problem in our Q&A session last night and his method was long and very involved. I said I would email him my proof and ask for feedback, but when I did I got an out of office reply. Our assignment is due before he returns. So I'm asking here.
Does this proof work? It's much simpler than what he did online and that left me feeling unsure about this. If it doesn't work, can you tell me why not? Thank you!
My new proof:
On
Another approach in case you have the relevant theorems available (or for future reference):
For $x > 0$ we have $$f'(x) = \frac{1}{2\sqrt{x}}$$ so, $f$ is differentiable on $(0,\infty)$ and has a bounded derivative on, say, $[1,\infty)$. Therefore $f$ is Lipschitz, hence uniformly continuous, on $[1, \infty)$.
Furthermore, $f$ is continuous on the compact interval $[0,1]$, hence uniformly continuous on that interval.
It follows that $f$ is uniformly continuous on the union $[0,1] \cup [1,\infty) = [0,\infty)$.
You are correct. Here is the proof written in a different way.
Let $x,y \in [0,+\infty)$. Notice that $\left|\sqrt{x}-\sqrt{y} \right|\leq \left|\sqrt{x}+\sqrt{y}\right|$. Choosing $\delta:=\varepsilon^2$ in the definition, we get that if $|x-y|<\delta$ then $$\left|\sqrt{x}-\sqrt{y} \right|^2\leq \left|\sqrt{x}-\sqrt{y} \right|\cdot \left|\sqrt{x}+\sqrt{y} \right|=|x-y|<\varepsilon^2 \implies \left|\sqrt{x}-\sqrt{y}\right|<\varepsilon,$$ hence $x \mapsto \sqrt{x}$ is uniformely continious in $[0,+\infty)$.