# If is true f(n) = Θ(g(n)) and if f(n) = o(h(n)) then g(n) = o(h(n))?

In asymptotic notation the transivity holds, however what happens when we have small o such as if f(n)= o(h(n)) does that means that also g(n)=o(h(n)) holds?

i take as granted that both of f(n)=o(h(n)) is true and f(n)= O(g(n)) (from THETA) then i need to show g(n) < c h(n) given that i have f(n) <= d g(n) and f(n) < c h(n) how one would proceed from there?

From $$f(n) = \Theta(g(n))$$ you know that, for some positive constants $$c \in \mathbb{R}^+$$ and $$\eta \in \mathbb{N}$$, and for all $$n \ge \eta$$, it holds that $$f(n) \ge c g(n)$$, i.e., $$g(n) \le \frac{1}{c} f(n)$$.$$^1$$
Moreover, from $$f(n) = o(h(n))$$, you know that for any constant $$c' > 0$$ there is some $$\eta'_{c'}$$ such that $$f(n) \le c' h(n)$$ for any $$n \ge \eta'_{c'}$$.
Pick any constant $$c'' > 0$$ and let $$\eta'' = \max\{\eta, \eta'_{c'}\}$$ where $$c' = c \cdot c'' > 0$$. From the previous inequalities we have that, for any $$n \ge \eta''$$: $$g(n) \le \frac{1}{c} f(n) \le \frac{1}{c} \cdot c' h(n) = c'' h(n),$$ thus proving that $$g(n) = o(h(n))$$.
$$^1$$: In fact, the weaker condition $$f(n) = \Omega(g(n))$$ suffices.