If $f(n)=\Omega(g(n))$ and $h(n)=\theta (g(n))$ then does this implies $f(n).g(n)=\Omega(g(n).h(n))$?
I have got the answer. It's easy.

Since $f(n)= \Omega g(n)$

$cg(n)\le f(n)$

Multiplying both sides by h(n)

$cg(n).h(n)&lt;=f(n).h(n)$

$f(n).g(n)=\Omega(g(n).h(n))$