(Context: I'm self-studying Murray (1984). I learned (and have forgotten quite a lot of) real and complex analysis. I'm willing to relearn and to look up references.)
Problem: if $f=O(g)$, show that
- $O(o(f))=o(O(f))=o(g)$;
- $O(f)O(g)=O(fg)$;
- $O(f)o(g)=o(f)o(g)=o(fg)$.
Question: could you please verify my attempt below? Please feel free to suggest improvements / alternative solutions. I appreciate your time.
My attempt
The problem doesn't make explicit the domain so neither will I.
Part 1. Let $\varphi=o(f)$ and $\theta=O(\varphi)$. Then, $\theta/g=(\theta/\varphi)(\varphi/f)(f/g)\to 0$ because $\varphi/f\to 0$ and $(\theta/\varphi)$ and $(f/g)$ are bounded.
Simiarly, let $\alpha=O(f)$ and $\beta=o(\alpha)$. Then $\beta/g=(\beta/\alpha)(\alpha/f)(f/g)\to 0$ because $\beta/\alpha\to 0$ and $\alpha/f$ and $f/g$ are bounded.
The results here use $f=O(g)$.
Part 2. Let $\varphi=O(f)$ and $\theta=O(g)$. Then, because $|\varphi/f|$ and $|\theta/g|$ are both bounded, so is $|\varphi\theta/(fg)|$. We therefore have $\varphi\theta=O(fg)$. This result does not use $f=O(g)$.
Part 3. Let $\varphi=O(f)$, $\theta=o(g)$, and $\alpha=o(f)$. Then, $\varphi\theta/fg=(\varphi/f)(\theta/g)\to 0$ because $\varphi/f$ is bounded and $\theta/g\to 0$. Analogously, $\alpha\theta/fg=(\alpha/f)(\theta/g)\to 0$ because both $\alpha/f$ and $\theta/g$ converge $0$. These results do not use $f=O(g)$.