Prove the following $\Omega(n\cdot f(n)) = n\cdot \Omega(f(n))$

Question

I want to prove the following by definition of asymptotic notation
$$\Omega(n\cdot f(n)) = n\cdot \Omega(f(n))$$ Any suggestions?

Answer 1

Assuming that you have a definition that involves a positive values $\epsilon$, a constant $c$, and a constant $N$, you do the following:

write out what it means for a function $g$ to be in $Omega(f(n))$; then observe that the same values of $c$ and $N$ show that $n \mapsto n \cdot g(n)$ is in $\Omega(n \cdot f(n))$. That'll show that $n \cdot \Omega(f(n)) \subset \Omega(n \cdot f(n))$. Then you do almost exactly the same thing to show the reverse inclusion.