Showing that $O(o(f))=o(O(f))=o(g)$ if $f=O(g)$ In Asymptotic Analysis by J.D.Murray on page 10 the following question is posed. If $f=O(g)$ show that:
$$O(o(f))=o(O(f))=o(g)$$
I am really confused by what this actually means. How can you have $O(o(f))$? Surely $O$ has to act on a function and $o(f)$ is not a function - rather a statement about the function it equals. Please can someone explain.
 A: As stated (Basin & Rusinowitch, 2004) in the context of O-notation the equals sign, $=$, means subset or equals, $\subseteq$. We then have to prove:
$$O(o(f)) \subseteq o(O(f))\subseteq o(g)$$
Showing that $O(o(f))\subseteq o(O(f))$
To show this first statement we do as Jack's advice in the comments above. We let $u\in O(o(f))$ which means that:
$$ \newcommand{\p}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\f}[2]{\frac{ #1}{ #2}} \newcommand{\l}[0]{\left(} \newcommand{\r}[0]{\right)} \newcommand{\mean}[1]{\langle #1 \rangle}\newcommand{\e}[0]{\varepsilon} \newcommand{\ket}[1]{\left|#1\right>} \newcommand{\bra}[1]{\left<#1\right|} \newcommand{\braoket}[3]{\left<#1\right|#2\left|#3\right>} |u| \le k|v| \text{ where } \lim_{x\rightarrow x_0} \l \f{v}{f}\r =0$$
But:
$$\lim_{x\rightarrow x_0} \l \f{v}{f}\r =0 \Rightarrow \lim_{x\rightarrow x_0} \l \f{kv}{kf}\r =0 \Rightarrow \lim_{x\rightarrow x_0} \l \f{u}{kf}\r =0$$
Since $kf \in O(f)$ we have that $u\in o(kf)\in o(O(f))$. Hence every element $u\in O(o(f))$ is also in $o(O(f))$ meaning that $O(o(f))=o(O(f))$ i.e.  $O(o(f))\subseteq o(O(f))$ .
Showing that $o(O(f))\subseteq o(g)$
Let $u \in o(O(f))$ then:
$$ \lim_{x \rightarrow x_0}\l \f{u}{v}\r=0 \text{ where } |v|\le k|f|$$
but since $f\in O(g)$ we have that $|v| \le k_2 |g|$. This implies that:
$$ \lim_{x \rightarrow x_0}\l \f{u}{k_2 g}\r=0$$
$$\Rightarrow  \lim_{x \rightarrow x_0}\l \f{u}{ g}\r=0$$
Hence every $u\in o(O(f))$ is also in $o(g)$ meaning that $o(O(f))=o(g)$ i.e. $o(O(f))\subseteq o(g)$.
