Complexity (prove or disprove): If $f(g(n))=O(n)$ and $f(n)=Ω(n)$ then $g(n)=O(n)$

Question

I was wondering if I could get a hint on the following question (prove or disprove):

If $f(g(n))=O(n) $ and $f(n)=Ω(n)$ then $g(n)=O(n)$.

Answer 1

Using the definition:$$ f(m) = {\mit Ω}(m) \Longleftrightarrow f(m) \geqslant cm, \quad \forall m \geqslant m_0 $$ where $c > 0$ is a constant and $m_0 \in \mathbb{N}_+$, and$$ f(g(n)) = O(n) \Longleftrightarrow f(g(n)) \leqslant Cn, \quad \forall n \geqslant n_0 $$ where $C > 0$ is a constant and $n_0 \in \mathbb{N}_+$.

In fact, assume that $f:\mathbb{N}_+ \to \mathbb{N}_+$ and$$ f(m) = {\mit Ω}(m) \Longleftrightarrow f(m) \geqslant cm. \quad \forall m \geqslant m_0 $$ Define$$ a = \min_{1 \leqslant m \leqslant m_0} f(n) > 0, \quad c' = \min(c, a), $$ then$$ f(m) \geqslant c'm. \quad \forall m \geqslant 1 $$ Also assume that $g:\mathbb{N}_+ \to \mathbb{N}_+$, then$$ Cn \geqslant f(g(n)) \geqslant c'g(n). $$

Answer 2

The question seems too fishy to be true.

Take $f(n) = \frac{1}{n} = \Omega(n)$ and $g(n) = n^2$, then $f(g(n)) = n = O(n)$, but the conclusion does not hold.