$f(n) = \mathcal{o}(g(n))$ if for any constant $c$, there exists some constant $n_0$ such that $0 \le f(n) \le cg(n), n \ge n_0 $
$f(n) = \pi(g(n))$ if for any constant $c$, there exists some constant $n_0$ such that $0 \le f(n) < cg(n), n \ge n_0 $
Are both the above definitions equivalent? In other words is the following true:
$$f(n) = \mathcal{o}(g(n)) \leftrightarrow f(n) = \pi(g(n))$$
Yes, they are equivalent, provided that $g$ is a positive function. It’s clear that if $f$ is $\pi(g)$, then $f$ is $o(g)$. Conversely, if for each $c>0$ you have an $n_c$ such that $0\le f(n)\le cg(n)$ for $n\ge n_c$, then you have $0\le f(n)\le\frac{c}2g(n)<cg(n)$ for all $n\ge n_{c/2}$, say.